DP (Grade 11 & 12) · Maths AA HL
Number & Algebra
505 questions across 11 sub-topics
Use the Sub-Topic filter above to focus on one.
Sequences & Series
Financial and Growth Models
Exponents & Logarithms
The Binomial Theorem
Counting Principles
Complex Numbers: Algebraic Operations
Complex Numbers: Geometric Representations
Proof by Deduction, Contradiction, and Counterexample
Proof by Induction: Series, Divisibility, and Inequalities
Proof by Induction: Differentiation & Integration
Systems of Linear Equations
Sequences & Series 50 questions
QUESTION 1
4 marks
Easy
An arithmetic sequence has first term $u_1=7$ and common difference $d=4$.
(a) Find $u_{10}$.
(b) Find $S_{10}$, the sum of the first 10 terms.
Show complete worked solution
(a) Using the formula for the $n$th term of an arithmetic sequence, $u_n=u_1+(n-1)d$:
$$u_{10}=7+(10-1)(4)$$
$$u_{10}=7+36=43$$
$$\boxed{u_{10}=43}$$
(b) Using the formula for the sum of an arithmetic series, $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$:
$$S_{10}=\dfrac{10}{2}\big(2(7)+9(4)\big)$$
$$S_{10}=5(14+36)=5(50)=250$$
$$\boxed{S_{10}=250}$$
QUESTION 2
6 marks
Medium
In an arithmetic sequence, the 3rd term is 11 and the 8th term is 31.
(a) Find the first term and the common difference.
(b) Find the sum of the first 20 terms.
Show complete worked solution
(a) Let the first term be $u_1=a$ and common difference $d$. Using $u_n=a+(n-1)d$:
$$u_3=a+2d=11 \quad (1)$$
$$u_8=a+7d=31 \quad (2)$$
Subtracting (1) from (2) to eliminate $a$:
$$(a+7d)-(a+2d)=31-11$$
$$5d=20 \implies d=4$$
Substituting into (1):
$$a+2(4)=11 \implies a=11-8=3$$
$$\boxed{a=3,\ d=4}$$
(b) Using $S_n=\dfrac{n}{2}\big(2a+(n-1)d\big)$ with $n=20$:
$$S_{20}=\dfrac{20}{2}\big(2(3)+19(4)\big)$$
$$S_{20}=10(6+76)=10(82)=820$$
$$\boxed{S_{20}=820}$$
QUESTION 3
5 marks
Easy
An arithmetic series has first term 5 and common difference 3. Given that the sum of the first $n$ terms is 390, find the value of $n$.
Show complete worked solution
Using the sum formula for an arithmetic series, $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$, with $u_1=5$, $d=3$, $S_n=390$:
$$390=\dfrac{n}{2}\big(2(5)+(n-1)(3)\big)$$
Simplify inside the brackets:
$$390=\dfrac{n}{2}(10+3n-3)=\dfrac{n}{2}(3n+7)$$
Multiply both sides by 2:
$$780=n(3n+7)$$
$$3n^{2}+7n-780=0$$
Using the quadratic formula, $n=\dfrac{-7\pm\sqrt{7^{2}-4(3)(-780)}}{2(3)}$:
$$n=\dfrac{-7\pm\sqrt{49+9360}}{6}=\dfrac{-7\pm\sqrt{9409}}{6}=\dfrac{-7\pm97}{6}$$
Taking the positive root, since $n$ must be a positive integer:
$$n=\dfrac{-7+97}{6}=\dfrac{90}{6}=15$$
$$\boxed{n=15}$$
QUESTION 4
7 marks
Medium
The first three terms of an arithmetic sequence are $k-2$, $2k+1$, $5k-4$, where $k$ is a constant.
(a) Find the value of $k$.
(b) Hence find the sum of the first 12 terms of the sequence.
Show complete worked solution
(a) For an arithmetic sequence, consecutive differences are equal, so:
$$(2k+1)-(k-2)=(5k-4)-(2k+1)$$
Simplify each side:
$$k+3=3k-5$$
$$3+5=3k-k$$
$$8=2k \implies k=4$$
$$\boxed{k=4}$$
(b) Substituting $k=4$ into the terms:
$$u_1=k-2=2,\quad u_2=2k+1=9,\quad u_3=5k-4=16$$
Check: the common difference is $d=9-2=7$ and $16-9=7$, consistent.
Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $u_1=2$, $d=7$, $n=12$:
$$S_{12}=\dfrac{12}{2}\big(2(2)+11(7)\big)$$
$$S_{12}=6(4+77)=6(81)=486$$
$$\boxed{S_{12}=486}$$
QUESTION 5
16 marks
Hard
The seats in a theatre are arranged in rows that form an arithmetic sequence. Row 1 has 18 seats, and each subsequent row has 3 more seats than the previous row.
(a) Find the number of seats in row 25. [3]
(b) Find the total number of seats if the theatre has 30 rows. [3]
(c) The theatre wants to increase its capacity to at least 3000 seats by adding further rows, continuing the same arithmetic pattern beyond row 30. Find the minimum total number of rows required for the total capacity to be at least 3000. [5]
(d) Using the 40-row configuration found in part (c), the theatre charges \$12 per seat for the first 20 rows and \$8 per seat for the remaining rows. Find the total revenue if every seat in the 40-row theatre is sold. [5]
Show complete worked solution
(a) Using $u_n=u_1+(n-1)d$ with $u_1=18$, $d=3$:
$$u_{25}=18+(25-1)(3)$$
$$u_{25}=18+72=90$$
$$\boxed{90\text{ seats in row }25}$$
(b) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $n=30$:
$$S_{30}=\dfrac{30}{2}\big(2(18)+29(3)\big)$$
$$S_{30}=15(36+87)=15(123)=1845$$
$$\boxed{1845\text{ seats}}$$
(c) Require $S_n\ge3000$:
$$\dfrac{n}{2}\big(2(18)+(n-1)(3)\big)\ge3000$$
Simplify inside the brackets:
$$\dfrac{n}{2}(36+3n-3)\ge3000$$
$$\dfrac{n(3n+33)}{2}\ge3000$$
$$3n^{2}+33n-6000\ge0$$
$$n^{2}+11n-2000\ge0$$
Solving the corresponding equation with the quadratic formula:
$$n=\dfrac{-11\pm\sqrt{121+8000}}{2}=\dfrac{-11\pm\sqrt{8121}}{2}$$
$$\sqrt{8121}\approx90.12,\qquad n\approx\dfrac{-11+90.12}{2}\approx39.56$$
Since $n$ must be a positive integer and $S_n$ is increasing, test $n=39$ and $n=40$:
$$S_{39}=\dfrac{39}{2}\big(36+38(3)\big)=19.5(150)=2925<3000$$
$$S_{40}=\dfrac{40}{2}\big(36+39(3)\big)=20(153)=3060\ge3000$$
$$\boxed{\text{minimum of }40\text{ rows}}$$
(d) With $n=40$ rows, split the seats into rows 1-20 and rows 21-40.
Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ for rows 1-20:
$$S_{20}=\dfrac{20}{2}\big(2(18)+19(3)\big)=10(93)=930$$
Seats in rows 21-40 are the remaining seats out of the 40-row total found in (c):
$$3060-930=2130$$
Revenue is $\$12$ per seat for the first 930 seats and $\$8$ per seat for the remaining 2130 seats:
$$\text{Revenue}=12\times930+8\times2130$$
$$=11\,160+17\,040=28\,200$$
$$\boxed{\text{Total revenue}=\$28\,200}$$
QUESTION 6
4 marks
Easy
A geometric sequence has first term $u_1=5$ and common ratio $r=2$.
(a) Find $u_7$.
(b) Find $S_7$, the sum of the first 7 terms.
Show complete worked solution
(a) Using $u_n=u_1r^{n-1}$:
$$u_7=5(2)^{6}$$
$$u_7=5(64)=320$$
$$\boxed{u_7=320}$$
(b) Using $S_n=\dfrac{u_1(r^{n}-1)}{r-1}$ with $r=2$ (since $r>1$):
$$S_7=\dfrac{5(2^{7}-1)}{2-1}$$
$$S_7=5(128-1)=5(127)=635$$
$$\boxed{S_7=635}$$
QUESTION 7
6 marks
Medium
In a geometric sequence, $u_2=6$ and $u_5=162$.
(a) Find the common ratio $r$ and the first term $u_1$.
(b) Find the sum of the first 6 terms.
Show complete worked solution
(a) Using $u_n=u_1r^{n-1}$:
$$u_2=u_1r=6 \quad (1)$$
$$u_5=u_1r^{4}=162 \quad (2)$$
Dividing (2) by (1) to eliminate $u_1$:
$$\dfrac{u_1r^4}{u_1r}=\dfrac{162}{6}$$
$$r^{3}=27 \implies r=3$$
Substituting into (1):
$$u_1(3)=6 \implies u_1=2$$
$$\boxed{r=3,\ u_1=2}$$
(b) Using $S_n=\dfrac{u_1(r^{n}-1)}{r-1}$ with $n=6$:
$$S_6=\dfrac{2(3^{6}-1)}{3-1}$$
$$S_6=\dfrac{2(729-1)}{2}=728$$
$$\boxed{S_6=728}$$
QUESTION 8
6 marks
Easy
A geometric series has first term 8 and common ratio 0.6.
(a) Find the sum to infinity of the series.
(b) Find the least value of $n$ for which the sum of the first $n$ terms exceeds 19.9.
Show complete worked solution
(a) Since $|r|=0.6<1$, the series converges, so use $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty=\dfrac{8}{1-0.6}$$
$$S_\infty=\dfrac{8}{0.4}=20$$
$$\boxed{S_\infty=20}$$
(b) Using $S_n=\dfrac{u_1(1-r^{n})}{1-r}$:
$$S_n=\dfrac{8(1-0.6^{n})}{0.4}=20(1-0.6^{n})$$
Require $S_n>19.9$:
$$20(1-0.6^{n})>19.9$$
$$1-0.6^{n}>0.995$$
$$0.6^{n}<0.005$$
Taking logarithms of both sides (note $\ln0.6<0$, so the inequality direction reverses):
$$n\ln0.6<\ln0.005$$
$$n>\dfrac{\ln0.005}{\ln0.6}\approx\dfrac{-5.298}{-0.511}\approx10.37$$
Since $n$ is a positive integer, test $n=10$ and $n=11$:
$$S_{10}=20(1-0.6^{10})=20(1-0.006047)=19.879<19.9$$
$$S_{11}=20(1-0.6^{11})=20(1-0.003628)=19.927>19.9$$
$$\boxed{n=11}$$
QUESTION 9
7 marks
Medium
In a geometric sequence with all terms positive, $u_3=18$ and $u_6=486$.
(a) Find the common ratio and the first term.
(b) Find the smallest value of $n$ for which $S_n>2000$.
Show complete worked solution
(a) Using $u_n=u_1r^{n-1}$:
$$u_3=u_1r^{2}=18 \quad (1)$$
$$u_6=u_1r^{5}=486 \quad (2)$$
Dividing (2) by (1) to eliminate $u_1$:
$$r^{3}=\dfrac{486}{18}=27 \implies r=3$$
(taking the real cube root, consistent with all terms positive.)
Substituting into (1):
$$u_1(3)^{2}=18 \implies 9u_1=18 \implies u_1=2$$
$$\boxed{r=3,\ u_1=2}$$
(b) Using $S_n=\dfrac{u_1(r^{n}-1)}{r-1}$:
$$S_n=\dfrac{2(3^{n}-1)}{3-1}=3^{n}-1$$
Require $3^{n}-1>2000$:
$$3^{n}>2001$$
Taking logarithms of both sides:
$$n>\dfrac{\ln2001}{\ln3}\approx\dfrac{7.601}{1.099}\approx6.92$$
Since $n$ is a positive integer, test $n=6$ and $n=7$:
$$S_6=3^{6}-1=728<2000$$
$$S_7=3^{7}-1=2186>2000$$
$$\boxed{n=7}$$
QUESTION 10
16 marks
Hard
A ball is dropped from a height of 4 m onto a hard floor. After each bounce it rebounds to $75\%$ of the height from which it fell, so the successive rebound heights form a geometric sequence with $h_n=4(0.75)^{n}$, where $h_n$ is the height reached after the $n$th bounce.
(a) Find the height reached after the 5th bounce, correct to 3 significant figures. [3]
(b) Find the total vertical distance travelled by the ball, from the moment it is dropped until it (theoretically) comes to rest. [5]
(c) Find the number of the first bounce after which the rebound height is less than 1 cm. [4]
(d) Find the total distance travelled by the ball by the time it strikes the ground for the 6th time. [4]
Show complete worked solution
(a) Using $h_n=4(0.75)^{n}$ with $n=5$:
$$h_5=4(0.75)^{5}$$
$$h_5=4(0.2373046875)=0.949218\ldots$$
$$\boxed{h_5\approx0.949\text{ m (3 s.f.)}}$$
(b) The ball falls 4 m initially. After that, for each bounce $n=1,2,3,\ldots$ it travels up a distance $h_n$ and back down the same distance $h_n$, so
$$\text{Total distance}=4+2\sum_{n=1}^{\infty}h_n$$
The sum $\sum_{n=1}^{\infty}h_n$ is a geometric series with first term $h_1=4(0.75)=3$ and common ratio $0.75$. Using $S_\infty=\dfrac{u_1}{1-r}$:
$$\sum_{n=1}^{\infty}h_n=\dfrac{3}{1-0.75}=\dfrac{3}{0.25}=12$$
Substituting:
$$\text{Total distance}=4+2(12)=4+24=28$$
$$\boxed{28\text{ m}}$$
(c) Require $h_n<0.01$ (since 1 cm $=0.01$ m):
$$4(0.75)^{n}<0.01$$
$$(0.75)^{n}<0.0025$$
Taking logarithms of both sides (note $\ln0.75<0$, so the inequality reverses):
$$n\ln0.75<\ln0.0025$$
$$n>\dfrac{\ln0.0025}{\ln0.75}\approx\dfrac{-5.991}{-0.288}\approx20.83$$
Since $n$ is a positive integer, the least candidate is $n=21$. Check the boundary bounces:
$$h_{20}=4(0.75)^{20}\approx0.0127\text{ m}=1.27\text{ cm}>1\text{ cm}$$
$$h_{21}=4(0.75)^{21}\approx0.00951\text{ m}=0.951\text{ cm}<1\text{ cm}$$
$$\boxed{\text{the 21st bounce}}$$
(d) The ball hits the ground for the 1st time after the initial drop of 4 m. It then travels up and down through heights $h_1,h_2,h_3,h_4,h_5$ before hitting the ground for the 6th time, so
$$\text{Total distance to the 6th impact}=4+2(h_1+h_2+h_3+h_4+h_5)$$
Using $S_n=\dfrac{u_1(1-r^{n})}{1-r}$ with first term 3, ratio 0.75, and $n=5$ terms:
$$S_5=\dfrac{3(1-0.75^{5})}{1-0.75}=\dfrac{3(1-0.2373046875)}{0.25}=\dfrac{3(0.7626953125)}{0.25}=9.15234375$$
Substituting:
$$\text{Total distance}=4+2(9.15234375)=4+18.3046875=22.3046875$$
$$\boxed{\approx22.3\text{ m (3 s.f.)}}$$
QUESTION 11
4 marks
Easy
An infinite geometric sequence has first term $u_1 = 12$ and common ratio $r = \dfrac{1}{3}$. Find the sum to infinity of the sequence.
Show complete worked solution
Since $|r|=\dfrac{1}{3}<1$, the sum to infinity exists, so use $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty=\dfrac{12}{1-\tfrac{1}{3}}$$
$$S_\infty=\dfrac{12}{\tfrac{2}{3}}=12\times\dfrac{3}{2}$$
$$\boxed{S_\infty=18}$$
QUESTION 12
5 marks
Medium
The sum to infinity of a geometric sequence is 45, and its first term is 27. Find the common ratio $r$, and explain why this series converges.
Show complete worked solution
Using the sum to infinity formula $S_\infty=\dfrac{u_1}{1-r}$:
$$45=\dfrac{27}{1-r}$$
Rearrange:
$$1-r=\dfrac{27}{45}=\dfrac{3}{5}$$
$$r=1-\dfrac{3}{5}=\dfrac{2}{5}$$
$$\boxed{r=\dfrac{2}{5}}$$
Since $\left|\dfrac{2}{5}\right|=\dfrac{2}{5}<1$, the condition for convergence of an infinite geometric series ($|r|<1$) is satisfied, so the series converges to a finite sum.
QUESTION 13
6 marks
Medium
The sum to infinity of a geometric series is 50. The second term of the series is 8. Find the possible value(s) of the common ratio $r$.
Show complete worked solution
Let the first term be $u_1$. From the sum-to-infinity condition, $S_\infty=\dfrac{u_1}{1-r}=50$:
$$u_1=50(1-r)$$
From the second term condition, $u_2=u_1r=8$. Substituting the expression for $u_1$:
$$50r(1-r)=8$$
$$50r-50r^2=8$$
$$50r^2-50r+8=0$$
Divide through by 2:
$$25r^2-25r+4=0$$
Using the quadratic formula, $r=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $a=25$, $b=-25$, $c=4$:
$$\text{discriminant}=(-25)^2-4(25)(4)=625-400=225$$
$$r=\dfrac{25\pm\sqrt{225}}{50}=\dfrac{25\pm15}{50}$$
So $r=\dfrac{40}{50}=0.8$ or $r=\dfrac{10}{50}=0.2$.
Both values satisfy $|r|<1$, so both are valid for a convergent infinite series.
$$\boxed{r=0.8 \text{ or } r=0.2}$$
QUESTION 14
5 marks
Easy
An infinite geometric sequence has first term $5.4$ and common ratio $-0.3$. Find the sum to infinity, giving your answer correct to 3 significant figures.
Show complete worked solution
Since $|r|=0.3<1$, the sum to infinity exists, so use $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty=\dfrac{5.4}{1-(-0.3)}$$
$$S_\infty=\dfrac{5.4}{1.3}$$
$$S_\infty=4.15384\ldots$$
$$\boxed{S_\infty=4.15\text{ (3 s.f.)}}$$
QUESTION 15
15 marks
Hard
An infinite geometric sequence has first term $u_1 = a$ and common ratio $r$, where $-1 < r < 1$ and $r \neq 0$.
(a) [3 marks] Write down an expression for $S_\infty$, the sum to infinity of the sequence, in terms of $a$ and $r$.
(b) [6 marks] The sum to infinity of the original sequence is $24$. The sequence formed by squaring every term of the original sequence, that is $u_1^2, u_2^2, u_3^2, \ldots$, is itself an infinite geometric sequence, and its sum to infinity is $172.8$. Show that $r = \dfrac{7}{13}$, and find the exact value of $a$.
(c) [6 marks] Consider the sequence formed by taking only the even-indexed terms of the original sequence, that is $u_2, u_4, u_6, \ldots$. Show that this is also an infinite geometric sequence, and find its sum to infinity as an exact fraction.
Show complete worked solution
(a) Using the sum to infinity formula for a geometric series:
$$\boxed{S_\infty=\dfrac{a}{1-r}}$$
(b) From the given sum to infinity:
$$\dfrac{a}{1-r}=24 \quad \Rightarrow \quad a=24(1-r) \quad (*)$$
The squared sequence $u_1^2,u_2^2,u_3^2,\ldots$ has first term $a^2$ and common ratio $r^2$, since $u_n=ar^{n-1} \Rightarrow u_n^2=a^2(r^2)^{n-1}$. Its sum to infinity is
$$\dfrac{a^2}{1-r^2}=172.8$$
Factorise the denominator using difference of squares, $1-r^2=(1-r)(1+r)$, and substitute $a=24(1-r)$ from $(*)$:
$$\dfrac{\big[24(1-r)\big]^2}{(1-r)(1+r)}=172.8$$
$$\dfrac{576(1-r)^2}{(1-r)(1+r)}=172.8$$
$$\dfrac{576(1-r)}{1+r}=172.8$$
Cross-multiply:
$$576(1-r)=172.8(1+r)$$
$$576-576r=172.8+172.8r$$
$$576-172.8=576r+172.8r$$
$$403.2=748.8r$$
$$r=\dfrac{403.2}{748.8}=\dfrac{7}{13}$$
as required. Substituting back into $(*)$:
$$a=24\left(1-\dfrac{7}{13}\right)=24\times\dfrac{6}{13}=\dfrac{144}{13}$$
$$\boxed{r=\dfrac{7}{13}, \quad a=\dfrac{144}{13}}$$
(c) The even-indexed terms are $u_2=ar$, $u_4=ar^3$, $u_6=ar^5,\ldots$. Consecutive terms have ratio
$$\dfrac{u_4}{u_2}=\dfrac{ar^3}{ar}=r^2$$
which is constant, so this is a geometric sequence with first term $ar$ and common ratio $r^2$. Since $-1
QUESTION 16
5 marks
Easy
The table below shows the first four terms of an arithmetic sequence.
(a) State the common difference $d$. [1]
(b) Find $u_{12}$. [2]
(c) Find $S_{12}$, the sum of the first 12 terms. [2]
| $n$ | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| $u_n$ | 9 | 15 | 21 | 27 |
Show complete worked solution
(a) Consecutive terms in the table increase by the same amount each time, $15-9=6$, $21-15=6$, $27-21=6$.
$$\boxed{d=6}$$
(b) Using $u_n=u_1+(n-1)d$ with $u_1=9$, $d=6$:
$$u_{12}=9+(12-1)(6)$$
$$u_{12}=9+66=75$$
$$\boxed{u_{12}=75}$$
(c) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $n=12$:
$$S_{12}=\dfrac{12}{2}\big(2(9)+11(6)\big)$$
$$S_{12}=6(18+66)=6(84)=504$$
$$\boxed{S_{12}=504}$$
QUESTION 17
5 marks
Easy
The first three terms of an arithmetic sequence are $2k-1$, $k+5$, $3k-4$, where $k$ is a constant.
(a) Find the value of $k$. [3]
(b) Hence find $S_{10}$, the sum of the first 10 terms. [2]
Show complete worked solution
(a) For an arithmetic sequence, consecutive differences are equal, so:
$$(k+5)-(2k-1)=(3k-4)-(k+5)$$
Simplify each side:
$$-k+6=2k-9$$
$$6+9=2k+k$$
$$15=3k \implies k=5$$
$$\boxed{k=5}$$
(b) Substituting $k=5$ into the terms:
$$u_1=2(5)-1=9,\quad u_2=5+5=10,\quad u_3=3(5)-4=11$$
Check: the common difference is $d=10-9=1$ and $11-10=1$, consistent.
Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $u_1=9$, $d=1$, $n=10$:
$$S_{10}=\dfrac{10}{2}\big(2(9)+9(1)\big)$$
$$S_{10}=5(18+9)=5(27)=135$$
$$\boxed{S_{10}=135}$$
QUESTION 18
4 marks
Easy
A geometric sequence has first term $u_1=3$ and common ratio $r=-2$.
(a) Find $u_6$. [2]
(b) Find $S_6$, the sum of the first 6 terms. [2]
Show complete worked solution
(a) Using $u_n=u_1r^{n-1}$:
$$u_6=3(-2)^{5}$$
$$u_6=3(-32)=-96$$
$$\boxed{u_6=-96}$$
(b) Using $S_n=\dfrac{u_1(r^{n}-1)}{r-1}$ with $r=-2$:
$$S_6=\dfrac{3\big((-2)^{6}-1\big)}{-2-1}$$
$$S_6=\dfrac{3(64-1)}{-3}=\dfrac{3(63)}{-3}=-63$$
$$\boxed{S_6=-63}$$
QUESTION 19
5 marks
Easy
The recurring decimal $0.454545\ldots$ can be written as an infinite geometric series:
$$0.454545\ldots=0.45+0.0045+0.000045+\cdots$$
(a) State the first term $u_1$ and the common ratio $r$ of this series. [2]
(b) Hence express $0.454545\ldots$ exactly as a fraction in simplest form. [3]
Show complete worked solution
(a) Each term is obtained from the previous one by multiplying by $0.01$, since $0.0045=0.45\times0.01$ and $0.000045=0.0045\times0.01$.
$$\boxed{u_1=0.45,\ r=0.01}$$
(b) Since $|r|=0.01<1$, the series converges, so use $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty=\dfrac{0.45}{1-0.01}=\dfrac{0.45}{0.99}$$
Multiply numerator and denominator by 100 to clear the decimals:
$$S_\infty=\dfrac{45}{99}$$
Divide numerator and denominator by their highest common factor, 9:
$$S_\infty=\dfrac{5}{11}$$
$$\boxed{0.454545\ldots=\dfrac{5}{11}}$$
QUESTION 20
4 marks
Easy
Find the value of $\displaystyle\sum_{r=1}^{20}(4r-3)$.
Show complete worked solution
The terms $4r-3$ for $r=1,2,3,\ldots$ form an arithmetic sequence, since each increase of $r$ by 1 increases $4r-3$ by exactly 4.
The first term (at $r=1$) is $u_1=4(1)-3=1$, the common difference is $d=4$, and there are $n=20$ terms (for $r=1$ to $r=20$).
Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$:
$$S_{20}=\dfrac{20}{2}\big(2(1)+19(4)\big)$$
$$S_{20}=10(2+76)=10(78)=780$$
$$\boxed{\displaystyle\sum_{r=1}^{20}(4r-3)=780}$$
QUESTION 21
4 marks
Easy
Find the value of $\displaystyle\sum_{k=1}^{7}2(3)^{k-1}$.
Show complete worked solution
The terms $2(3)^{k-1}$ form a geometric sequence with first term $u_1=2(3)^{0}=2$ and common ratio $r=3$, and there are $n=7$ terms.
Using $S_n=\dfrac{u_1(r^{n}-1)}{r-1}$:
$$S_7=\dfrac{2(3^{7}-1)}{3-1}$$
$$S_7=\dfrac{2(2187-1)}{2}=2186$$
$$\boxed{\displaystyle\sum_{k=1}^{7}2(3)^{k-1}=2186}$$
QUESTION 22
5 marks
Easy
An arithmetic sequence has first term $u_1=8$ and common difference $d=3$. The last term of the sequence is 137.
(a) Find the number of terms, $n$, in the sequence. [3]
(b) Find the sum of all the terms of the sequence. [2]
Show complete worked solution
(a) Using $u_n=u_1+(n-1)d$ with $u_n=137$, $u_1=8$, $d=3$:
$$137=8+(n-1)(3)$$
$$137-8=3(n-1)$$
$$129=3(n-1)$$
$$n-1=43 \implies n=44$$
$$\boxed{n=44}$$
(b) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $n=44$:
$$S_{44}=\dfrac{44}{2}\big(2(8)+43(3)\big)$$
$$S_{44}=22(16+129)=22(145)=3190$$
$$\boxed{S_{44}=3190}$$
QUESTION 23
5 marks
Easy
In a geometric sequence, the common ratio is $r=\dfrac{1}{2}$ and the 4th term is $u_4=10$.
(a) Find the first term, $u_1$. [3]
(b) Find $u_8$, giving your answer as an exact fraction. [2]
Show complete worked solution
(a) Using $u_n=u_1r^{n-1}$ with $n=4$:
$$u_4=u_1\left(\dfrac{1}{2}\right)^{3}=10$$
$$u_1\left(\dfrac{1}{8}\right)=10$$
$$u_1=10\times8=80$$
$$\boxed{u_1=80}$$
(b) Using $u_n=u_1r^{n-1}$ with $n=8$:
$$u_8=80\left(\dfrac{1}{2}\right)^{7}$$
$$u_8=\dfrac{80}{128}$$
Simplify by dividing numerator and denominator by 16:
$$u_8=\dfrac{5}{8}$$
$$\boxed{u_8=\dfrac{5}{8}}$$
QUESTION 24
6 marks
Easy
The diagram below shows the first three rows of a triangular arrangement of dots. Row 1 has 3 dots, row 2 has 7 dots, row 3 has 11 dots, and this pattern continues so that the number of dots in each row forms an arithmetic sequence.
(a) Write down the common difference of the sequence. [1]
(b) Find the number of dots in row 8. [2]
(c) Find the total number of dots in the first 8 rows. [3]
Show complete worked solution
(a) The number of dots increases from row to row by $7-3=4$ and $11-7=4$.
$$\boxed{d=4}$$
(b) Using $u_n=u_1+(n-1)d$ with $u_1=3$, $d=4$:
$$u_8=3+(8-1)(4)$$
$$u_8=3+28=31$$
$$\boxed{31\text{ dots in row }8}$$
(c) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $n=8$:
$$S_8=\dfrac{8}{2}\big(2(3)+7(4)\big)$$
$$S_8=4(6+28)=4(34)=136$$
$$\boxed{136\text{ dots in total}}$$
QUESTION 25
4 marks
Easy
An infinite geometric series has first term $u_1=25$ and common ratio $r=-0.4$.
(a) Explain why the series converges. [1]
(b) Find the sum to infinity, giving your answer as an exact fraction. [3]
Show complete worked solution
(a) The series converges because $|r|=|-0.4|=0.4<1$, which is the condition for convergence of an infinite geometric series.
(b) Using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty=\dfrac{25}{1-(-0.4)}=\dfrac{25}{1.4}$$
Multiply numerator and denominator by 5 to clear the decimal:
$$S_\infty=\dfrac{125}{7}$$
$$\boxed{S_\infty=\dfrac{125}{7}}$$
QUESTION 26
5 marks
Easy
An arithmetic sequence has first term $u_1=4$ and common difference $d=7$.
(a) Show that 100 is not a term of this sequence. [3]
(b) Find the two terms of the sequence that are closest to 100. [2]
Show complete worked solution
(a) Suppose 100 is a term, so $u_n=100$ for some positive integer $n$. Using $u_n=u_1+(n-1)d$:
$$100=4+(n-1)(7)$$
$$96=7(n-1)$$
$$n-1=\dfrac{96}{7}$$
$$n=1+\dfrac{96}{7}=\dfrac{103}{7}\approx14.71$$
Since $n$ is not a positive integer, 100 is not a term of the sequence.
(b) Since $n\approx14.71$ lies between 14 and 15, check $u_{14}$ and $u_{15}$:
$$u_{14}=4+13(7)=4+91=95$$
$$u_{15}=4+14(7)=4+98=102$$
$$\boxed{u_{14}=95\text{ and }u_{15}=102\text{ are the closest terms}}$$
QUESTION 27
5 marks
Easy
The $n$th term of a sequence is given by $u_n=5n-2$.
(a) Show that the sequence is arithmetic, and state the first term and common difference. [3]
(b) Find $S_{20}$, the sum of the first 20 terms. [2]
Show complete worked solution
(a) Consider consecutive terms:
$$u_{n+1}-u_n=\big(5(n+1)-2\big)-(5n-2)=5n+5-2-5n+2=5$$
Since this difference is constant (independent of $n$), the sequence is arithmetic with common difference $d=5$.
The first term is $u_1=5(1)-2=3$.
$$\boxed{u_1=3,\ d=5}$$
(b) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $n=20$:
$$S_{20}=\dfrac{20}{2}\big(2(3)+19(5)\big)$$
$$S_{20}=10(6+95)=10(101)=1010$$
$$\boxed{S_{20}=1010}$$
QUESTION 28
5 marks
Easy
The first three terms of a geometric sequence are $4$, $4r$, $4r^2$, where $r$ is a real number. The third term is 36.
(a) Find the possible value(s) of $r$. [3]
(b) Write down both possible sequences for the first three terms. [2]
Show complete worked solution
(a) Since the third term is 36:
$$4r^2=36$$
$$r^2=9$$
$$r=3 \text{ or } r=-3$$
$$\boxed{r=3\text{ or }r=-3}$$
(b) For $r=3$: the terms are $4$, $4(3)=12$, $4(3)^2=36$, giving the sequence $4, 12, 36$.
For $r=-3$: the terms are $4$, $4(-3)=-12$, $4(-3)^2=36$, giving the sequence $4, -12, 36$.
$$\boxed{4,\ 12,\ 36\quad\text{or}\quad 4,\ -12,\ 36}$$
QUESTION 29
5 marks
Easy
An arithmetic series has first term 6 and last term 130. The common difference is $d=4$.
(a) Find the number of terms, $n$. [3]
(b) Hence find the sum of the series, using the formula $S_n=\dfrac{n}{2}(u_1+u_n)$. [2]
Show complete worked solution
(a) Using $u_n=u_1+(n-1)d$ with $u_1=6$, $u_n=130$, $d=4$:
$$130=6+(n-1)(4)$$
$$124=4(n-1)$$
$$n-1=31 \implies n=32$$
$$\boxed{n=32}$$
(b) Using $S_n=\dfrac{n}{2}(u_1+u_n)$ with $n=32$, $u_1=6$, $u_n=130$:
$$S_{32}=\dfrac{32}{2}(6+130)$$
$$S_{32}=16(136)=2176$$
$$\boxed{S_{32}=2176}$$
QUESTION 30
7 marks
Medium
The numbers $a-3$, $a+1$, and $3a+13$ are consecutive terms of a geometric sequence, where $a$ is a real constant.
(a) Show that $a^2+a-20=0$, and hence find the possible value(s) of $a$. [4]
(b) For each value of $a$ found in part (a), write down the first three terms of the geometric sequence and state the common ratio. [3]
Show complete worked solution
(a) For a geometric sequence, the ratio between consecutive terms is constant, so:
$$\dfrac{a+1}{a-3}=\dfrac{3a+13}{a+1}$$
Cross-multiplying:
$$(a+1)^2=(a-3)(3a+13)$$
Expanding the left side:
$$a^2+2a+1$$
Expanding the right side:
$$3a^2+13a-9a-39=3a^2+4a-39$$
So:
$$a^2+2a+1=3a^2+4a-39$$
Rearranging:
$$0=3a^2+4a-39-a^2-2a-1$$
$$0=2a^2+2a-40$$
Dividing through by 2:
$$a^2+a-20=0$$
as required. Factorising:
$$(a+5)(a-4)=0$$
$$\boxed{a=4\text{ or }a=-5}$$
(b) For $a=4$: the terms are $a-3=1$, $a+1=5$, $3a+13=25$, giving the sequence $1, 5, 25$ with common ratio $r=5$.
For $a=-5$: the terms are $a-3=-8$, $a+1=-4$, $3a+13=-2$, giving the sequence $-8, -4, -2$ with common ratio $r=\dfrac{1}{2}$.
$$\boxed{1,\ 5,\ 25\ (r=5)\quad\text{or}\quad -8,\ -4,\ -2\ \left(r=\tfrac{1}{2}\right)}$$
QUESTION 31
7 marks
Medium
Three consecutive terms of an arithmetic sequence can be written as $a-d$, $a$, $a+d$. The sum of these three terms is 21, and their product is 280.
(a) Find the value of $a$. [2]
(b) Find the possible value(s) of $d$, and hence write down the two possible sets of three terms. [5]
Show complete worked solution
(a) The sum of the three terms is:
$$(a-d)+a+(a+d)=21$$
$$3a=21 \implies a=7$$
$$\boxed{a=7}$$
(b) The product of the three terms is 280:
$$(a-d)(a)(a+d)=280$$
Using $(a-d)(a+d)=a^2-d^2$ (difference of squares):
$$a(a^2-d^2)=280$$
Substituting $a=7$:
$$7(49-d^2)=280$$
$$49-d^2=40$$
$$d^2=9$$
$$d=3\text{ or }d=-3$$
For $d=3$: the terms are $7-3=4$, $7$, $7+3=10$, giving $4, 7, 10$.
For $d=-3$: the terms are $7-(-3)=10$, $7$, $7+(-3)=4$, giving $10, 7, 4$ (the same three numbers in reverse order).
$$\boxed{d=\pm3,\text{ terms }4,\ 7,\ 10\text{ (in some order)}}$$
QUESTION 32
8 marks
Medium
(a) By writing the sum $S=1+2+3+\cdots+n$ both forwards and backwards and adding the two versions term by term, show that
$$\displaystyle\sum_{r=1}^{n}r=\dfrac{n(n+1)}{2}$$ [4]
(b) Hence evaluate $\displaystyle\sum_{r=1}^{150}r$. [2]
(c) Hence evaluate $\displaystyle\sum_{r=51}^{150}r$. [2]
Show complete worked solution
(a) Write $S$ forwards and backwards:
$$S=1+2+3+\cdots+(n-1)+n$$
$$S=n+(n-1)+(n-2)+\cdots+2+1$$
Adding these two equations term by term, each pair of corresponding terms sums to $n+1$ (for example $1+n=n+1$, $2+(n-1)=n+1$, and so on), and there are $n$ such pairs:
$$2S=\underbrace{(n+1)+(n+1)+\cdots+(n+1)}_{n\text{ terms}}=n(n+1)$$
Dividing both sides by 2:
$$S=\dfrac{n(n+1)}{2}$$
so $\displaystyle\sum_{r=1}^{n}r=\dfrac{n(n+1)}{2}$, as required.
(b) Using the formula with $n=150$:
$$\displaystyle\sum_{r=1}^{150}r=\dfrac{150(151)}{2}=\dfrac{22\,650}{2}=11\,325$$
$$\boxed{\displaystyle\sum_{r=1}^{150}r=11\,325}$$
(c) The sum from $r=51$ to $r=150$ is the sum from $r=1$ to $150$ with the sum from $r=1$ to $50$ removed:
$$\displaystyle\sum_{r=51}^{150}r=\displaystyle\sum_{r=1}^{150}r-\displaystyle\sum_{r=1}^{50}r$$
Using the formula for $n=50$:
$$\displaystyle\sum_{r=1}^{50}r=\dfrac{50(51)}{2}=1275$$
So:
$$\displaystyle\sum_{r=51}^{150}r=11\,325-1275=10\,050$$
$$\boxed{\displaystyle\sum_{r=51}^{150}r=10\,050}$$
QUESTION 33
8 marks
Medium
An arithmetic sequence has first term $u_1=85$ and common difference $d=-6$.
(a) Find the first term of the sequence that is negative, stating its position $n$ in the sequence. [4]
(b) Hence find the value of $n$ for which $S_n$ is greatest, and find this maximum value of $S_n$. [4]
Show complete worked solution
(a) Require $u_n<0$. Using $u_n=u_1+(n-1)d$:
$$85+(n-1)(-6)<0$$
$$85-6n+6<0$$
$$91<6n$$
$$n>\dfrac{91}{6}\approx15.17$$
Since $n$ must be a positive integer, the smallest such $n$ is $n=16$. Check:
$$u_{16}=85+15(-6)=85-90=-5$$
$$\boxed{u_{16}=-5\text{ is the first negative term}}$$
(b) Since the sequence is decreasing (as $d<0$), $S_n$ is greatest at the last value of $n$ for which $u_n$ is still positive or zero, that is $n=15$ (since $u_{16}$ is the first negative term). Check $u_{15}$:
$$u_{15}=85+14(-6)=85-84=1>0$$
so terms 1 to 15 are all positive, confirming $n=15$ gives the maximum sum.
Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $n=15$:
$$S_{15}=\dfrac{15}{2}\big(2(85)+14(-6)\big)$$
$$S_{15}=7.5(170-84)=7.5(86)=645$$
$$\boxed{n=15,\ S_{15}=645\text{ is the maximum sum}}$$
QUESTION 34
7 marks
Medium
The sum of the first $n$ terms of a sequence is given by $S_n=3n^2+5n$.
(a) Find $u_1$. [1]
(b) Show that $u_n=6n+2$, and hence show that the sequence is arithmetic, stating the common difference. [4]
(c) Find $u_{25}$. [2]
Show complete worked solution
(a) Since $S_1=u_1$:
$$u_1=3(1)^2+5(1)=3+5=8$$
$$\boxed{u_1=8}$$
(b) For $n\ge2$, $u_n=S_n-S_{n-1}$. First find $S_{n-1}$ by substituting $n-1$ for $n$:
$$S_{n-1}=3(n-1)^2+5(n-1)$$
Expanding $(n-1)^2=n^2-2n+1$:
$$S_{n-1}=3(n^2-2n+1)+5n-5=3n^2-6n+3+5n-5=3n^2-n-2$$
So:
$$u_n=S_n-S_{n-1}=(3n^2+5n)-(3n^2-n-2)=6n+2$$
Checking $n=1$: $u_1=6(1)+2=8$, which agrees with part (a), so the formula $u_n=6n+2$ holds for all $n\ge1$.
Since $u_n=6n+2$ is linear in $n$, consecutive terms differ by a constant amount:
$$u_{n+1}-u_n=\big(6(n+1)+2\big)-(6n+2)=6$$
so the sequence is arithmetic with common difference $d=6$.
$$\boxed{u_n=6n+2,\ d=6}$$
(c) Using $u_n=6n+2$ with $n=25$:
$$u_{25}=6(25)+2=150+2=152$$
$$\boxed{u_{25}=152}$$
QUESTION 35
7 marks
Medium
A mosaic pattern is built from tiles arranged in concentric rings around a central point. Ring 1 contains 6 tiles, and each successive ring contains twice as many tiles as the previous ring, so the number of tiles in each ring forms a geometric sequence.
(a) Find the number of tiles in ring 8. [3]
(b) Find the total number of tiles used in the first 8 rings. [4]
Show complete worked solution
(a) Using $u_n=u_1r^{n-1}$ with $u_1=6$, $r=2$:
$$u_8=6(2)^{7}$$
$$u_8=6(128)=768$$
$$\boxed{768\text{ tiles in ring }8}$$
(b) Using $S_n=\dfrac{u_1(r^{n}-1)}{r-1}$ with $n=8$:
$$S_8=\dfrac{6(2^{8}-1)}{2-1}$$
$$S_8=6(256-1)=6(255)=1530$$
$$\boxed{1530\text{ tiles in total}}$$
QUESTION 36
6 marks
Medium
Find the value of $\displaystyle\sum_{r=6}^{25}(3r-2)$.
Show complete worked solution
The terms $3r-2$ form an arithmetic sequence as $r$ increases by 1 each time (common difference 3).
At $r=6$: $u_1=3(6)-2=16$. At $r=25$: $u_n=3(25)-2=73$. The number of terms is $n=25-6+1=20$.
Using $S_n=\dfrac{n}{2}(u_1+u_n)$:
$$S_{20}=\dfrac{20}{2}(16+73)$$
$$S_{20}=10(89)=890$$
$$\boxed{\displaystyle\sum_{r=6}^{25}(3r-2)=890}$$
QUESTION 37
8 marks
Medium
In a geometric series, the sum to infinity is equal to 4 times the first term $u_1$, where $u_1\neq0$.
(a) Show that the common ratio is $r=\dfrac{3}{4}$. [3]
(b) Given that the third term of the series is 18, find the exact value of $u_1$. [3]
(c) Find $S_5$, the sum of the first 5 terms, as an exact fraction. [2]
Show complete worked solution
(a) Using $S_\infty=\dfrac{u_1}{1-r}$ and the given condition $S_\infty=4u_1$:
$$\dfrac{u_1}{1-r}=4u_1$$
Since $u_1\neq0$, divide both sides by $u_1$:
$$\dfrac{1}{1-r}=4$$
$$1-r=\dfrac{1}{4}$$
$$r=1-\dfrac{1}{4}=\dfrac{3}{4}$$
as required.
(b) Using $u_n=u_1r^{n-1}$ with $n=3$, $r=\dfrac{3}{4}$:
$$u_3=u_1\left(\dfrac{3}{4}\right)^2=18$$
$$u_1\left(\dfrac{9}{16}\right)=18$$
$$u_1=18\times\dfrac{16}{9}=32$$
$$\boxed{u_1=32}$$
(c) Using $S_n=\dfrac{u_1(1-r^{n})}{1-r}$ with $u_1=32$, $r=\dfrac{3}{4}$, $n=5$:
$$S_5=\dfrac{32\left(1-\left(\tfrac{3}{4}\right)^5\right)}{1-\tfrac{3}{4}}=\dfrac{32\left(1-\tfrac{243}{1024}\right)}{\tfrac{1}{4}}$$
$$S_5=32\times4\times\left(1-\dfrac{243}{1024}\right)=128\times\dfrac{781}{1024}$$
$$S_5=\dfrac{128\times781}{1024}=\dfrac{781}{8}$$
$$\boxed{S_5=\dfrac{781}{8}}$$
QUESTION 38
7 marks
Medium
A stack of firewood logs is arranged so that the bottom row has 40 logs, and each row above has 2 fewer logs than the row directly below it.
(a) Find the maximum number of complete rows that can be stacked before a row would contain zero or a negative number of logs. [4]
(b) Find the total number of logs in the stack described in part (a). [3]
Show complete worked solution
(a) Let the bottom row be row 1, so $u_1=40$ and $d=-2$ (counting rows upward). Require $u_n\le0$:
$$40+(n-1)(-2)\le0$$
$$40-2n+2\le0$$
$$42\le2n$$
$$n\ge21$$
So row 21 would contain $u_{21}=40+20(-2)=40-40=0$ logs, which is not a physical row. The maximum number of complete rows with a positive number of logs is therefore $n=20$. Check:
$$u_{20}=40+19(-2)=40-38=2>0$$
$$\boxed{20\text{ complete rows}}$$
(b) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $n=20$:
$$S_{20}=\dfrac{20}{2}\big(2(40)+19(-2)\big)$$
$$S_{20}=10(80-38)=10(42)=420$$
$$\boxed{420\text{ logs in total}}$$
QUESTION 39
8 marks
Medium
(a) Show that, for any sequence $u_1, u_2, u_3,\ldots$, $$\displaystyle\sum_{r=1}^{n}\big(u_{r+1}-u_r\big)=u_{n+1}-u_1$$ [4]
(b) Let $u_r=r^2$. Show that $u_{r+1}-u_r=2r+1$, and hence use the result of part (a) to evaluate $\displaystyle\sum_{r=1}^{20}(2r+1)$. [4]
Show complete worked solution
(a) Writing out the sum term by term:
$$\displaystyle\sum_{r=1}^{n}\big(u_{r+1}-u_r\big)=(u_2-u_1)+(u_3-u_2)+(u_4-u_3)+\cdots+(u_{n+1}-u_n)$$
Every term except the first $-u_1$ and the last $u_{n+1}$ cancels with an adjacent term: $-u_2$ cancels with $+u_2$, $-u_3$ cancels with $+u_3$, and so on. What remains is:
$$\displaystyle\sum_{r=1}^{n}\big(u_{r+1}-u_r\big)=u_{n+1}-u_1$$
as required.
(b) With $u_r=r^2$:
$$u_{r+1}-u_r=(r+1)^2-r^2=r^2+2r+1-r^2=2r+1$$
as required. Using the result of part (a) with $u_1=1^2=1$ and $u_{21}=21^2=441$:
$$\displaystyle\sum_{r=1}^{20}(2r+1)=u_{21}-u_1=441-1=440$$
$$\boxed{\displaystyle\sum_{r=1}^{20}(2r+1)=440}$$
(This can be checked directly: $2r+1$ forms an arithmetic sequence with first term 3, common difference 2, and 20 terms, giving $S_{20}=\dfrac{20}{2}(2(3)+19(2))=10(6+38)=440$, confirming the answer.)
QUESTION 40
8 marks
Medium
Each time an image is copied on a certain photocopier, its area is reduced to 80% of the area of the previous copy. The original image has an area of $600\text{ cm}^2$.
(a) Find the area of the image after it has been copied 6 times, correct to 1 decimal place. [3]
(b) Find the least number of copies required for the area to first fall below $50\text{ cm}^2$. [5]
Show complete worked solution
(a) The area after $n$ copies forms a geometric sequence, $A_n=600(0.8)^n$. Using $n=6$:
$$A_6=600(0.8)^{6}$$
$$A_6=600(0.262144)=157.2864$$
$$\boxed{A_6\approx157.3\text{ cm}^2\text{ (1 d.p.)}}$$
(b) Require $A_n<50$:
$$600(0.8)^{n}<50$$
$$(0.8)^{n}<\dfrac{50}{600}=\dfrac{1}{12}$$
Taking logarithms of both sides (note $\ln0.8<0$, so the inequality direction reverses):
$$n\ln0.8<\ln\dfrac{1}{12}$$
$$n>\dfrac{\ln(1/12)}{\ln0.8}\approx\dfrac{-2.4849}{-0.2231}\approx11.14$$
Since $n$ is a positive integer, test $n=11$ and $n=12$:
$$A_{11}=600(0.8)^{11}\approx51.5\text{ cm}^2>50\text{ cm}^2$$
$$A_{12}=600(0.8)^{12}\approx41.2\text{ cm}^2<50\text{ cm}^2$$
$$\boxed{12\text{ copies}}$$
QUESTION 41
7 marks
Medium
In an arithmetic sequence with first term $u_1=a$ and common difference $d$, the 4th term is three times the 1st term, and the 7th term is 2 more than twice the 3rd term.
(a) Show that $a=1.5d$ and find a second equation relating $a$ and $d$. [3]
(b) Find the values of $a$ and $d$. [2]
(c) Find $S_{25}$, the sum of the first 25 terms. [2]
Show complete worked solution
(a) The 4th term is three times the 1st term:
$$u_4=3u_1$$
$$a+3d=3a$$
$$3d=2a$$
$$a=1.5d \quad (1)$$
The 7th term is 2 more than twice the 3rd term:
$$u_7=2u_3+2$$
$$a+6d=2(a+2d)+2$$
$$a+6d=2a+4d+2$$
$$2d-2=a \quad (2)$$
(b) Substituting (1) into (2):
$$1.5d=2d-2$$
$$2=2d-1.5d$$
$$2=0.5d$$
$$d=4$$
Substituting into (1):
$$a=1.5(4)=6$$
$$\boxed{a=6,\ d=4}$$
Check using equation (2): $a+6d=6+24=30$, and $2(a+2d)+2=2(6+8)+2=28+2=30$. Consistent.
(c) Using $S_n=\dfrac{n}{2}\big(2a+(n-1)d\big)$ with $n=25$:
$$S_{25}=\dfrac{25}{2}\big(2(6)+24(4)\big)$$
$$S_{25}=12.5(12+96)=12.5(108)=1350$$
$$\boxed{S_{25}=1350}$$
QUESTION 42
7 marks
Medium
Consider the geometric series $6+18+54+\cdots$, whose first term is 6 and common ratio is 3.
(a) Write down an expression for $S_n$, the sum of the first $n$ terms, using sigma notation. [2]
(b) Find the smallest value of $n$ for which $S_n$ first exceeds one million. [5]
Show complete worked solution
(a) Since $u_r=6(3)^{r-1}$ for $r=1,2,3,\ldots$:
$$\boxed{S_n=\displaystyle\sum_{r=1}^{n}6(3)^{r-1}}$$
(b) Using $S_n=\dfrac{u_1(r^{n}-1)}{r-1}$ with $u_1=6$, $r=3$:
$$S_n=\dfrac{6(3^{n}-1)}{3-1}=3(3^{n}-1)$$
Require $S_n>1\,000\,000$:
$$3(3^{n}-1)>1\,000\,000$$
$$3^{n}-1>333\,333.33\ldots$$
$$3^{n}>333\,334.33\ldots$$
Taking logarithms of both sides:
$$n>\dfrac{\ln(333\,334.33)}{\ln3}\approx\dfrac{12.716}{1.0986}\approx11.58$$
Since $n$ is a positive integer, test $n=11$ and $n=12$:
$$S_{11}=3(3^{11}-1)=3(177\,146)=531\,438<1\,000\,000$$
$$S_{12}=3(3^{12}-1)=3(531\,440)=1\,594\,320>1\,000\,000$$
$$\boxed{n=12}$$
QUESTION 43
7 marks
Medium
A sequence $u_1,u_2,u_3,\ldots$ is defined so that $\log_2u_1,\ \log_2u_2,\ \log_2u_3,\ \ldots$ is an arithmetic sequence with first term 3 and common difference 1.
(a) Show that $\{u_n\}$ is a geometric sequence, and find its common ratio. [4]
(b) Find $u_1$ and hence find $S_{10}$, the sum of the first 10 terms of $\{u_n\}$. [3]
Show complete worked solution
(a) Since $\log_2u_n$ is arithmetic with first term 3 and common difference 1:
$$\log_2u_n=3+(n-1)(1)=n+2$$
Converting from logarithmic to exponential form:
$$u_n=2^{n+2}$$
To test whether $\{u_n\}$ is geometric, consider the ratio of consecutive terms:
$$\dfrac{u_{n+1}}{u_n}=\dfrac{2^{n+3}}{2^{n+2}}=2^{(n+3)-(n+2)}=2^1=2$$
Since this ratio is constant (independent of $n$), $\{u_n\}$ is geometric with common ratio $r=2$.
$$\boxed{r=2}$$
(b) Using $u_n=2^{n+2}$ with $n=1$:
$$u_1=2^{3}=8$$
$$\boxed{u_1=8}$$
Using $S_n=\dfrac{u_1(r^{n}-1)}{r-1}$ with $u_1=8$, $r=2$, $n=10$:
$$S_{10}=\dfrac{8(2^{10}-1)}{2-1}$$
$$S_{10}=8(1024-1)=8(1023)=8184$$
$$\boxed{S_{10}=8184}$$
QUESTION 44
10 marks
Hard
The sum of the first $n$ terms of a sequence is given by $S_n=2n^2-n$.
(a) Find $u_1$. [2]
(b) Show that $u_n=4n-3$, and hence show that the sequence is arithmetic, stating the first term and common difference. [4]
(c) Use the arithmetic series sum formula, together with the values found in part (b), to verify that $S_{30}=1770$, and confirm this agrees with the given formula for $S_n$. [2]
(d) Determine whether 137 is a term of the sequence, and if so, state its position. [2]
Show complete worked solution
(a) Since $S_1=u_1$:
$$u_1=2(1)^2-1=2-1=1$$
$$\boxed{u_1=1}$$
(b) For $n\ge2$, $u_n=S_n-S_{n-1}$. First find $S_{n-1}$:
$$S_{n-1}=2(n-1)^2-(n-1)$$
Expanding $(n-1)^2=n^2-2n+1$:
$$S_{n-1}=2(n^2-2n+1)-n+1=2n^2-4n+2-n+1=2n^2-5n+3$$
So:
$$u_n=S_n-S_{n-1}=(2n^2-n)-(2n^2-5n+3)=4n-3$$
Checking $n=1$: $u_1=4(1)-3=1$, which agrees with part (a), so $u_n=4n-3$ holds for all $n\ge1$.
Since $u_n=4n-3$ is linear in $n$:
$$u_{n+1}-u_n=\big(4(n+1)-3\big)-(4n-3)=4$$
a constant, so the sequence is arithmetic with first term $u_1=1$ and common difference $d=4$.
$$\boxed{u_1=1,\ d=4}$$
(c) Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ with $u_1=1$, $d=4$, $n=30$:
$$S_{30}=\dfrac{30}{2}\big(2(1)+29(4)\big)$$
$$S_{30}=15(2+116)=15(118)=1770$$
Checking against the original formula, $S_n=2n^2-n$ with $n=30$:
$$S_{30}=2(30)^2-30=2(900)-30=1800-30=1770$$
Both methods give the same value, confirming consistency.
$$\boxed{S_{30}=1770}$$
(d) Suppose $u_n=137$:
$$4n-3=137$$
$$4n=140$$
$$n=35$$
Since $n=35$ is a positive integer, 137 is indeed a term of the sequence, at position 35.
$$\boxed{137=u_{35}}$$
QUESTION 45
10 marks
Hard
The first term of an arithmetic sequence is 4, and its common difference is $d$, where $d\neq0$. The first, fourth, and eighth terms of this arithmetic sequence, taken in that order, are also the first three consecutive terms of a geometric sequence.
(a) Show that $d(9d-4)=0$. [5]
(b) Hence find the exact value of $d$, and find the common ratio $r$ of the geometric sequence. [3]
(c) Determine, with a reason, whether the geometric sequence in part (b) has a sum to infinity. [2]
Show complete worked solution
(a) The arithmetic sequence has $u_1=4$ and common difference $d$, so:
$$u_1=4,\qquad u_4=4+3d,\qquad u_8=4+7d$$
These three values, in order, form a geometric sequence, so the ratio between consecutive terms is constant:
$$\dfrac{u_4}{u_1}=\dfrac{u_8}{u_4}$$
$$\dfrac{4+3d}{4}=\dfrac{4+7d}{4+3d}$$
Cross-multiplying:
$$(4+3d)^2=4(4+7d)$$
Expanding the left side:
$$16+24d+9d^2$$
Expanding the right side:
$$16+28d$$
So:
$$16+24d+9d^2=16+28d$$
Rearranging:
$$9d^2+24d-28d=0$$
$$9d^2-4d=0$$
$$d(9d-4)=0$$
as required.
(b) Since $d\neq0$, the factor $9d-4$ must equal zero:
$$9d-4=0$$
$$d=\dfrac{4}{9}$$
$$\boxed{d=\dfrac{4}{9}}$$
The common ratio is found from $r=\dfrac{u_4}{u_1}$:
$$u_4=4+3\left(\dfrac{4}{9}\right)=4+\dfrac{4}{3}=\dfrac{16}{3}$$
$$r=\dfrac{u_4}{u_1}=\dfrac{16/3}{4}=\dfrac{16}{3}\times\dfrac{1}{4}=\dfrac{4}{3}$$
$$\boxed{r=\dfrac{4}{3}}$$
(As a check, $u_8=4+7\left(\dfrac{4}{9}\right)=4+\dfrac{28}{9}=\dfrac{64}{9}$, and $\dfrac{u_8}{u_4}=\dfrac{64/9}{16/3}=\dfrac{64}{9}\times\dfrac{3}{16}=\dfrac{4}{3}=r$, consistent.)
(c) The sum to infinity of a geometric series exists only when $|r|<1$. Here $r=\dfrac{4}{3}$, so:
$$|r|=\dfrac{4}{3}>1$$
Since this condition is not satisfied, the geometric sequence does not have a sum to infinity; the series diverges.
$$\boxed{\text{No sum to infinity exists, since }|r|=\tfrac{4}{3}>1}$$
QUESTION 46
10 marks
Hard
(a) For a geometric sequence with first term $a$ and common ratio $r\neq1$, let $S_n=a+ar+ar^2+\cdots+ar^{n-1}$ denote the sum of the first $n$ terms. By considering $S_n-rS_n$, show that
$$S_n=\dfrac{a(r^{n}-1)}{r-1}$$ [5]
(b) A geometric sequence has first term $a=7$ and common ratio $r=5$. Find the smallest value of $n$ for which $S_n$ first exceeds 500 000. [5]
Show complete worked solution
(a) Write out $S_n$ and $rS_n$ term by term:
$$S_n=a+ar+ar^2+\cdots+ar^{n-2}+ar^{n-1}$$
$$rS_n=ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^{n}$$
Subtracting the second equation from the first, every term $ar, ar^2,\ldots,ar^{n-1}$ cancels, leaving only the first term of $S_n$ and the last term of $rS_n$:
$$S_n-rS_n=a-ar^{n}$$
Factorising the left side and the right side:
$$S_n(1-r)=a(1-r^{n})$$
Since $r\neq1$, divide both sides by $(1-r)$:
$$S_n=\dfrac{a(1-r^{n})}{1-r}$$
Multiplying numerator and denominator by $-1$:
$$S_n=\dfrac{a(r^{n}-1)}{r-1}$$
as required.
(b) Using the formula with $a=7$, $r=5$:
$$S_n=\dfrac{7(5^{n}-1)}{5-1}=\dfrac{7(5^{n}-1)}{4}$$
Require $S_n>500\,000$:
$$\dfrac{7(5^{n}-1)}{4}>500\,000$$
$$5^{n}-1>\dfrac{2\,000\,000}{7}\approx285\,714.29$$
$$5^{n}>285\,715.29$$
Taking logarithms of both sides:
$$n>\dfrac{\ln(285\,715.29)}{\ln5}\approx\dfrac{12.563}{1.6094}\approx7.81$$
Since $n$ is a positive integer, test $n=7$ and $n=8$:
$$S_7=\dfrac{7(5^{7}-1)}{4}=\dfrac{7(78\,124)}{4}=136\,717<500\,000$$
$$S_8=\dfrac{7(5^{8}-1)}{4}=\dfrac{7(390\,624)}{4}=683\,592>500\,000$$
$$\boxed{n=8}$$
QUESTION 47
10 marks
Hard
A square has side length 20 cm. A second square is formed by joining the midpoints of the sides of the first square; this halves the area at each step, so a third square is formed by joining the midpoints of the second square, and so on indefinitely, forming an infinite sequence of nested squares.
(a) Show that the areas of the squares form a geometric sequence with common ratio $r=\dfrac{1}{2}$, and state the first term. [3]
(b) Find $S_\infty$, the sum to infinity of the areas of all the squares in the sequence (including overlaps, i.e. simply adding every square's area). [3]
(c) The "border" region of square $n$ is defined as the area of square $n$ that lies outside square $n+1$, that is, $A_n-A_{n+1}$. Show that the border areas also form a geometric sequence, and show that the sum to infinity of all the border areas equals the area of the original (first) square. [4]
Show complete worked solution
(a) The first square has area $A_1=20^2=400\text{ cm}^2$. Since joining the midpoints of a square exactly halves its area at each stage, each area is $\dfrac{1}{2}$ of the previous one:
$$\dfrac{A_{n+1}}{A_n}=\dfrac{1}{2}\text{ for every }n$$
This constant ratio confirms the areas form a geometric sequence with $r=\dfrac{1}{2}$.
$$\boxed{A_1=400\text{ cm}^2,\ r=\dfrac{1}{2}}$$
(b) Since $|r|=\dfrac{1}{2}<1$, the series converges. Using $S_\infty=\dfrac{u_1}{1-r}$:
$$S_\infty=\dfrac{400}{1-\tfrac{1}{2}}=\dfrac{400}{\tfrac{1}{2}}=800$$
$$\boxed{S_\infty=800\text{ cm}^2}$$
(c) The border area of square $n$ is:
$$B_n=A_n-A_{n+1}$$
Since $A_{n+1}=\dfrac{1}{2}A_n$:
$$B_n=A_n-\dfrac{1}{2}A_n=\dfrac{1}{2}A_n$$
Since $A_n$ is itself geometric with ratio $\dfrac{1}{2}$, and $B_n$ is a constant multiple ($\dfrac{1}{2}$) of $A_n$, $B_n$ is also geometric with the same common ratio $r=\dfrac{1}{2}$. Its first term is:
$$B_1=\dfrac{1}{2}A_1=\dfrac{1}{2}(400)=200$$
Using $S_\infty=\dfrac{u_1}{1-r}$ for the border areas:
$$\displaystyle\sum_{n=1}^{\infty}B_n=\dfrac{200}{1-\tfrac{1}{2}}=\dfrac{200}{\tfrac{1}{2}}=400$$
Since $400\text{ cm}^2$ is exactly the area of the original square, $A_1$, the sum to infinity of all the border areas equals the area of the original square, as required. (This makes geometric sense: as $n\to\infty$, the nested squares shrink toward a single point, so the original square is entirely partitioned, without gaps or overlaps, into the infinite sequence of border regions.)
$$\boxed{\displaystyle\sum_{n=1}^{\infty}B_n=400\text{ cm}^2=A_1}$$
QUESTION 48
12 marks
Hard
Prove by mathematical induction that, for an arithmetic sequence with first term $a$ and common difference $d$, the sum of the first $n$ terms is given by
$$S_n=\dfrac{n}{2}\big(2a+(n-1)d\big)$$
for all positive integers $n$.
Show complete worked solution
Let $P(n)$ be the proposition that $S_n=\dfrac{n}{2}\big(2a+(n-1)d\big)$ for the sum of the first $n$ terms of an arithmetic sequence with first term $a$ and common difference $d$.
Base case ($n=1$):
The left side is $S_1=u_1=a$ (the sum of just the first term).
The right side is:
$$\dfrac{1}{2}\big(2a+(1-1)d\big)=\dfrac{1}{2}(2a)=a$$
Since both sides equal $a$, $P(1)$ is true.
Inductive hypothesis:
Assume $P(k)$ is true for some positive integer $k$, that is, assume:
$$S_k=\dfrac{k}{2}\big(2a+(k-1)d\big)$$
Inductive step:
We must show $P(k+1)$ is true, that is, $S_{k+1}=\dfrac{k+1}{2}\big(2a+kd\big)$.
The sum of the first $k+1$ terms is the sum of the first $k$ terms plus the $(k+1)$th term:
$$S_{k+1}=S_k+u_{k+1}$$
The $(k+1)$th term of the arithmetic sequence is $u_{k+1}=a+kd$. Using the inductive hypothesis for $S_k$:
$$S_{k+1}=\dfrac{k}{2}\big(2a+(k-1)d\big)+\big(a+kd\big)$$
Expand the first term:
$$S_{k+1}=ak+\dfrac{k(k-1)d}{2}+a+kd$$
Group the $a$ terms and the $d$ terms:
$$S_{k+1}=a(k+1)+d\left(\dfrac{k(k-1)}{2}+k\right)$$
Simplify the bracket containing $d$, writing $k$ as $\dfrac{2k}{2}$:
$$\dfrac{k(k-1)}{2}+\dfrac{2k}{2}=\dfrac{k^2-k+2k}{2}=\dfrac{k^2+k}{2}=\dfrac{k(k+1)}{2}$$
So:
$$S_{k+1}=a(k+1)+d\cdot\dfrac{k(k+1)}{2}$$
Factor out $\dfrac{k+1}{2}$:
$$S_{k+1}=\dfrac{k+1}{2}\big(2a+kd\big)$$
This is exactly the statement of $P(k+1)$, so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion:
Since $P(1)$ is true, and $P(k)$ true implies $P(k+1)$ true, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. That is,
$$\boxed{S_n=\dfrac{n}{2}\big(2a+(n-1)d\big)\text{ for all positive integers }n}$$
QUESTION 49
10 marks
Hard
The sum of the first $n$ terms of a geometric sequence is given by $S_n=5(3^n-1)$.
(a) Find $u_1$, and show that $u_n=10(3)^{n-1}$. [5]
(b) Confirm that $\{u_n\}$ is geometric by stating its common ratio, and find the smallest value of $n$ for which $u_n$ first exceeds 100 000. [5]
Show complete worked solution
(a) Since $S_1=u_1$:
$$u_1=5(3^1-1)=5(2)=10$$
$$\boxed{u_1=10}$$
For $n\ge2$, $u_n=S_n-S_{n-1}$:
$$S_{n-1}=5\big(3^{n-1}-1\big)$$
$$u_n=5(3^{n}-1)-5(3^{n-1}-1)=5(3^{n}-3^{n-1})$$
Factor $3^{n-1}$ from the bracket, using $3^n=3\cdot3^{n-1}$:
$$u_n=5\cdot3^{n-1}(3-1)=5\cdot3^{n-1}(2)=10(3)^{n-1}$$
Checking $n=1$: $u_1=10(3)^{0}=10$, which agrees with the value found above, so $u_n=10(3)^{n-1}$ holds for all $n\ge1$, as required.
(b) Writing $u_n=10(3)^{n-1}$ shows that consecutive terms are related by:
$$\dfrac{u_{n+1}}{u_n}=\dfrac{10(3)^{n}}{10(3)^{n-1}}=3$$
a constant ratio, confirming $\{u_n\}$ is geometric with common ratio $r=3$.
Require $u_n>100\,000$:
$$10(3)^{n-1}>100\,000$$
$$3^{n-1}>10\,000$$
Taking logarithms of both sides:
$$(n-1)\ln3>\ln10\,000$$
$$n-1>\dfrac{\ln10\,000}{\ln3}\approx\dfrac{9.2103}{1.0986}\approx8.38$$
$$n>9.38$$
Since $n$ is a positive integer, test $n=9$ and $n=10$:
$$u_9=10(3)^{8}=10(6561)=65\,610<100\,000$$
$$u_{10}=10(3)^{9}=10(19\,683)=196\,830>100\,000$$
$$\boxed{n=10}$$
QUESTION 50
9 marks
Hard
A sequence $\{v_n\}$ is defined by $v_n=u_n+w_n$, where $\{u_n\}$ is an arithmetic sequence with $u_1=2$ and common difference $d=5$, and $\{w_n\}$ is a geometric sequence with $w_1=3$ and common ratio $r=2$.
(a) Find an expression for $v_n$ in terms of $n$. [3]
(b) Find $\displaystyle\sum_{n=1}^{10}v_n$ by writing it as the sum of an arithmetic series and a geometric series. [6]
Show complete worked solution
(a) Using $u_n=u_1+(n-1)d$:
$$u_n=2+(n-1)(5)=5n-3$$
Using $w_n=w_1r^{n-1}$:
$$w_n=3(2)^{n-1}$$
So:
$$\boxed{v_n=5n-3+3(2)^{n-1}}$$
(b) The sum $\displaystyle\sum_{n=1}^{10}v_n$ can be split, since summation distributes over addition:
$$\displaystyle\sum_{n=1}^{10}v_n=\displaystyle\sum_{n=1}^{10}(5n-3)+\displaystyle\sum_{n=1}^{10}3(2)^{n-1}$$
The first sum is an arithmetic series with first term $u_1=5(1)-3=2$, common difference $d=5$, and $n=10$ terms. Using $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$:
$$\displaystyle\sum_{n=1}^{10}(5n-3)=\dfrac{10}{2}\big(2(2)+9(5)\big)=5(4+45)=5(49)=245$$
The second sum is a geometric series with first term $w_1=3$, common ratio $r=2$, and $n=10$ terms. Using $S_n=\dfrac{w_1(r^{n}-1)}{r-1}$:
$$\displaystyle\sum_{n=1}^{10}3(2)^{n-1}=\dfrac{3(2^{10}-1)}{2-1}=3(1024-1)=3(1023)=3069$$
Adding the two results:
$$\displaystyle\sum_{n=1}^{10}v_n=245+3069=3314$$
$$\boxed{\displaystyle\sum_{n=1}^{10}v_n=3314}$$
Financial and Growth Models 50 questions
QUESTION 1
4 marks
Easy
\$2000 is invested at a rate of $5\%$ per annum, compounded annually. Without using a calculator, show that the value of the investment after 2 years is exactly \$2205.
Show complete worked solution
Using the compound interest formula $FV=PV(1+i)^{n}$, with $PV=2000$, $i=0.05$, $n=2$:
$$FV=2000(1.05)^{2}$$
First find $(1.05)^{2}$ by hand:
$$(1.05)^{2}=1.05\times1.05=1.1025$$
Substituting:
$$FV=2000\times1.1025=2205$$
Since this matches the given value,
$$\boxed{FV=\$2205}$$ as required.
QUESTION 2
5 marks
Medium
A car depreciates in value by $20\%$ each year. Its value when new was \$15\,000. Without using a calculator, find its exact value after 3 years.
Show complete worked solution
Depreciation of $20\%$ per year means the value is multiplied by $(1-0.20)=0.8$ each year. Using $FV=PV(1-i)^{n}$ with $PV=15\,000$, $i=0.20$, $n=3$:
$$FV=15\,000(0.8)^{3}$$
Find $(0.8)^{3}$ by hand:
$$(0.8)^{2}=0.64,\qquad (0.8)^{3}=0.64\times0.8=0.512$$
Substituting:
$$FV=15\,000\times0.512=7\,680$$
$$\boxed{FV=\$7680}$$
QUESTION 3
6 marks
Easy
\$5000 is invested at a nominal annual interest rate of $6\%$, compounded monthly, for 4 years. Find:
(a) the value of the investment at the end of the 4 years, correct to 2 decimal places;
(b) the total interest earned, correct to 2 decimal places.
Show complete worked solution
(a) Since interest is compounded monthly, the monthly rate is $\dfrac{0.06}{12}=0.005$, and the number of monthly periods over 4 years is $12\times4=48$. Using $FV=PV\left(1+\dfrac{i}{k}\right)^{kn}$:
$$FV=5000(1.005)^{48}$$
$$(1.005)^{48}\approx1.270489$$
$$FV\approx5000\times1.270489\approx6352.446$$
$$\boxed{FV\approx\$6352.45\text{ (2 d.p.)}}$$
(b) Using $\text{Interest}=FV-PV$:
$$6352.45-5000=1352.45$$
$$\boxed{\text{Interest}\approx\$1352.45}$$
QUESTION 4
7 marks
Medium
A machine was bought for \$80\,000. Using reducing-balance (compound) depreciation, its value falls to \$32\,000 after 5 years.
(a) Find the annual rate of depreciation, correct to 3 significant figures.
(b) Find, to the nearest year, the total time after purchase for the machine's value to first fall below \$10\,000.
Show complete worked solution
(a) Let the annual depreciation rate be $r$ (as a decimal), so the value after $n$ years is $V_n=80\,000(1-r)^{n}$. Substituting $V_5=32\,000$:
$$32\,000=80\,000(1-r)^{5}$$
$$(1-r)^{5}=\dfrac{32\,000}{80\,000}=0.4$$
Taking the 5th root of both sides:
$$1-r=0.4^{1/5}\approx0.83255$$
$$r\approx0.16745$$
$$\boxed{r\approx16.7\%\text{ per annum (3 s.f.)}}$$
(b) Using $1-r\approx0.83255$, require $V_n=80\,000(0.83255)^{n}<10\,000$:
$$(0.83255)^{n}<\dfrac{10\,000}{80\,000}=0.125$$
Taking logarithms of both sides (note $\ln0.83255<0$, so the inequality reverses):
$$n\ln(0.83255)<\ln(0.125)$$
$$n>\dfrac{\ln0.125}{\ln0.83255}\approx\dfrac{-2.0794}{-0.18326}\approx11.35$$
Since $n$ must be a whole number of years, test $n=11$ and $n=12$:
$$V_{11}=80\,000(0.83255)^{11}\approx10\,656>10\,000$$
$$V_{12}=80\,000(0.83255)^{12}\approx8\,872<10\,000$$
$$\boxed{n=12\text{ years}}$$
QUESTION 5
17 marks
Hard
Ali deposits \$500 into a savings account at the end of each year, for a number of years. The account pays a nominal annual interest rate of $4\%$, compounded annually, so that each deposit earns interest for every full year it remains in the account before the next deposit is made.
(a) Show that the amount in the account immediately after the 3rd deposit is \$1560.80. [3]
(b) Show that the amount in the account immediately after the 10th deposit is given by
$$V=500\cdot\dfrac{1.04^{10}-1}{0.04},$$
and calculate this value correct to 2 decimal places. [5]
(c) Find the minimum number of years for which Ali must continue making these annual deposits for the account balance, immediately after a deposit, to first exceed \$8000. [5]
(d) Suppose instead Ali chooses a constant annual deposit $D$ (with the same $4\%$ interest) so that the balance is exactly \$10\,000 immediately after the 10th deposit. Find the value of $D$, correct to 2 decimal places. [4]
Show complete worked solution
(a) After the 1st deposit, the balance is $500$. Each year the balance grows by the annual interest rate before the next deposit is added, so:
$$\text{After year 1 (before 2nd deposit): } 500(1.04)+500=520+500=1020$$
$$\text{After year 2 (before 3rd deposit): } 1020(1.04)+500=1060.80+500=1560.80$$
$$\boxed{\$1560.80}$$ as required.
(b) The 1st deposit earns interest for 9 years by the time of the 10th deposit, growing to $500(1.04)^{9}$; the 2nd deposit earns interest for 8 years, growing to $500(1.04)^{8}$; and so on, down to the 10th deposit, which earns no interest. So the balance immediately after the 10th deposit is
$$V=500(1.04)^{9}+500(1.04)^{8}+\cdots+500(1.04)^{1}+500$$
This is a geometric series with first term $500$, common ratio $1.04$, and 10 terms. Using $S_n=\dfrac{u_1(r^{n}-1)}{r-1}$:
$$V=\dfrac{500\big(1.04^{10}-1\big)}{1.04-1}=500\cdot\dfrac{1.04^{10}-1}{0.04}$$
as required. Now evaluate, using $1.04^{10}\approx1.480244$:
$$V=500\cdot\dfrac{1.480244-1}{0.04}=500\cdot\dfrac{0.480244}{0.04}=500(12.0061)$$
$$\boxed{V\approx\$6003.05\text{ (2 d.p.)}}$$
(c) By the same reasoning as in (b), the balance immediately after the $n$th deposit is
$$V_n=500\cdot\dfrac{1.04^{n}-1}{0.04}=12\,500\big(1.04^{n}-1\big)$$
Require $V_n>8000$:
$$12\,500(1.04^{n}-1)>8000$$
$$1.04^{n}-1>0.64$$
$$1.04^{n}>1.64$$
Taking logarithms of both sides:
$$n>\dfrac{\ln1.64}{\ln1.04}\approx\dfrac{0.4947}{0.03922}\approx12.61$$
Since $n$ is a positive integer, test $n=12$ and $n=13$:
$$V_{12}=12\,500(1.04^{12}-1)\approx12\,500(0.601032)\approx7512.90<8000$$
$$V_{13}=12\,500(1.04^{13}-1)\approx12\,500(0.665074)\approx8313.42>8000$$
$$\boxed{\text{minimum of }13\text{ years}}$$
(d) By the same reasoning as in (b), with deposit $D$ replacing 500:
$$10\,000=D\cdot\dfrac{1.04^{10}-1}{0.04}=D(12.0061)$$
Solving for $D$:
$$D=\dfrac{10\,000}{12.0061}$$
$$\boxed{D\approx\$832.91\text{ (2 d.p.)}}$$
QUESTION 6
4 marks
Easy
\$2500 is invested at a rate of $4\%$ per annum, compounded annually. Without using a calculator, show that the value of the investment after 3 years is exactly \$2812.16.
Show complete worked solution
Using the compound interest formula $FV=PV(1+i)^{n}$, with $PV=2500$, $i=0.04$, $n=3$:
$$FV=2500(1.04)^{3}$$
First find $(1.04)^{3}$ by hand:
$$(1.04)^{2}=1.0816,\qquad (1.04)^{3}=1.0816\times1.04=1.124864$$
Substituting:
$$FV=2500\times1.124864=2812.16$$
Since this matches the given value,
$$\boxed{FV=\$2812.16}$$ as required.
QUESTION 7
4 marks
Easy
A laptop computer is bought for \$1200. Its value depreciates by $15\%$ each year.
Find its value after 4 years, correct to the nearest dollar.
Show complete worked solution
Depreciation of $15\%$ per year means the value is multiplied by $(1-0.15)=0.85$ each year. Using $FV=PV(1-i)^{n}$ with $PV=1200$, $i=0.15$, $n=4$:
$$FV=1200(0.85)^{4}$$
Find $(0.85)^{4}$ by hand:
$$(0.85)^{2}=0.7225,\qquad (0.85)^{4}=(0.7225)^{2}=0.52200625$$
Substituting:
$$FV=1200\times0.52200625=626.4075$$
Rounding to the nearest dollar,
$$\boxed{FV\approx\$626}$$
QUESTION 8
6 marks
Easy
\$2500 is invested at a nominal annual interest rate of $8\%$, compounded quarterly, for 5 years. Find:
(a) the value of the investment at the end of the 5 years, correct to 2 decimal places; [4]
(b) the total interest earned, correct to 2 decimal places. [2]
Show complete worked solution
(a) Since interest is compounded quarterly, the quarterly rate is $\dfrac{0.08}{4}=0.02$, and the number of quarterly periods over 5 years is $4\times5=20$. Using $FV=PV\left(1+\dfrac{i}{k}\right)^{kn}$:
$$FV=2500(1.02)^{20}$$
$$(1.02)^{20}\approx1.485947$$
$$FV\approx2500\times1.485947\approx3714.868490$$
$$\boxed{FV\approx\$3714.87\text{ (2 d.p.)}}$$
(b) Using $\text{Interest}=FV-PV$:
$$3714.868490-2500=1214.868490$$
$$\boxed{\text{Interest}\approx\$1214.87}$$
QUESTION 9
5 marks
Easy
The table shows the population of a small town, which is growing geometrically, over its first two years since Year 0.
(a) Show that these values are consistent with an annual growth rate of $3\%$. [2]
(b) Find the population after 6 years, correct to the nearest 10 people. [3]
| Year | 0 | 1 | 2 |
|---|---|---|---|
| Population | 12\,000 | 12\,360 | 12\,731 |
Show complete worked solution
(a) Check the growth factor between consecutive years:
$$\dfrac{12\,360}{12\,000}=1.03,\qquad \dfrac{12\,731}{12\,360}\approx1.03$$
Since the population is multiplied by $1.03$ each year, the table is consistent with
$$\boxed{\text{an annual growth rate of }3\%}$$
(b) Using $P_n=P_0(1.03)^{n}$ with $P_0=12\,000$, $n=6$:
$$P_6=12\,000(1.03)^{6}$$
$$(1.03)^{6}\approx1.194052$$
$$P_6\approx12\,000\times1.194052\approx14\,328.63$$
Rounding to the nearest 10,
$$\boxed{P_6\approx14\,330\text{ people}}$$
QUESTION 10
6 marks
Easy
A sample of a radioactive isotope has an initial mass of 80 g and a half-life of 5 days.
(a) Find the mass remaining after 15 days. [2]
(b) Find the mass remaining after 12 days, correct to 3 significant figures. [4]
Show complete worked solution
(a) 15 days is exactly $15\div5=3$ half-lives, so the mass is halved 3 times:
$$80\times\left(\dfrac{1}{2}\right)^{3}=80\times\dfrac{1}{8}=10$$
$$\boxed{10\text{ g}}$$
(b) Using $N=N_0\left(\dfrac{1}{2}\right)^{t/h}$ with $N_0=80$, $h=5$, $t=12$:
$$N=80\left(\dfrac{1}{2}\right)^{12/5}=80(0.5)^{2.4}$$
Evaluating $(0.5)^{2.4}$:
$$(0.5)^{2.4}\approx0.189465$$
$$N\approx80\times0.189465\approx15.157$$
$$\boxed{N\approx15.2\text{ g (3 s.f.)}}$$
QUESTION 11
4 marks
Easy
A colony of bacteria starts with 200 cells and doubles in number every 3 hours.
Find the number of bacteria present after 15 hours.
Show complete worked solution
15 hours is exactly $15\div3=5$ doubling periods, so the population is multiplied by $2$ five times. Using $N=N_0(2)^{t/3}$ with $N_0=200$, $t=15$:
$$N=200(2)^{5}$$
$$N=200\times32=6400$$
$$\boxed{N=6400\text{ bacteria}}$$
QUESTION 12
5 marks
Easy
Chen deposits \$300 into a savings account at the end of each year. The account pays interest at $3\%$ per annum, compounded annually, so that each deposit earns interest for every full year it remains in the account before the next deposit is made.
(a) Show that the balance immediately after the 2nd deposit is \$609.00, as shown in the table. [2]
(b) Find the balance immediately after the 3rd deposit, to complete the table. [3]
| Year | 1 | 2 | 3 |
|---|---|---|---|
| Balance immediately after deposit (\$) | 300.00 | 609.00 | ? |
Show complete worked solution
(a) After the 1st deposit, the balance is $300.00$. This grows for one full year before the 2nd deposit is added:
$$300.00\times1.03=309.00$$
$$309.00+300.00=609.00$$
$$\boxed{\$609.00}$$ as required.
(b) The balance of $\$609.00$ grows for one further year, and then the 3rd deposit is added:
$$609.00\times1.03=627.27$$
$$627.27+300.00=927.27$$
$$\boxed{\$927.27}$$
QUESTION 13
4 marks
Easy
Bianca borrows \$2000 at an interest rate of $6\%$ per annum, compounded annually. She repays the loan, together with all interest owed, in a single payment at the end of 3 years.
Find the amount Bianca must repay, correct to 2 decimal places.
Show complete worked solution
Using $FV=PV(1+i)^{n}$ with $PV=2000$, $i=0.06$, $n=3$:
$$FV=2000(1.06)^{3}$$
$$(1.06)^{2}=1.1236,\qquad (1.06)^{3}=1.1236\times1.06=1.191016$$
$$FV=2000\times1.191016=2382.032$$
$$\boxed{FV\approx\$2382.03}$$
QUESTION 14
4 marks
Easy
The value of a house is increasing at a rate of $4\%$ per year. Its current value is \$250\,000.
Find its expected value in 5 years' time, correct to the nearest \$100.
Show complete worked solution
Using $FV=PV(1+i)^{n}$ with $PV=250\,000$, $i=0.04$, $n=5$:
$$FV=250\,000(1.04)^{5}$$
$$(1.04)^{2}=1.0816,\qquad (1.04)^{4}=(1.0816)^{2}=1.16985856$$
$$(1.04)^{5}=1.16985856\times1.04=1.2166529024$$
$$FV=250\,000\times1.2166529024=304\,163.2256$$
Rounding to the nearest \$100,
$$\boxed{FV\approx\$304\,200}$$
QUESTION 15
5 marks
Easy
The temperature, $T\ ^{\circ}\text{C}$, of a cup of coffee $t$ minutes after being poured is modelled by
$$T=20+70(0.85)^{t}$$
(a) Write down the temperature of the coffee at the moment it is poured. [1]
(b) Find the temperature of the coffee after 10 minutes, correct to 1 decimal place. [3]
(c) Write down the temperature that the model predicts the coffee will approach as $t$ becomes very large. [1]
Show complete worked solution
(a) At $t=0$: $T=20+70(0.85)^{0}=20+70(1)=90$.
$$\boxed{90^{\circ}\text{C}}$$
(b) At $t=10$:
$$T=20+70(0.85)^{10}$$
$$(0.85)^{10}\approx0.196874$$
$$T\approx20+70(0.196874)\approx20+13.781\approx33.781$$
$$\boxed{T\approx33.8^{\circ}\text{C (1 d.p.)}}$$
(c) Since $0<0.85<1$, as $t\to\infty$, $(0.85)^{t}\to0$, so $T\to20+0=20$.
$$\boxed{T\to20^{\circ}\text{C}}$$ (the temperature of the surroundings)
QUESTION 16
6 marks
Easy
A car is purchased for \$28\,000. It depreciates in value by $12\%$ each year, using reducing-balance depreciation.
(a) Find the value of the car after 6 years, correct to the nearest dollar. [4]
(b) Find the percentage of the car's original value that remains after 6 years, correct to 1 decimal place. [2]
Show complete worked solution
(a) Using $FV=PV(1-i)^{n}$ with $PV=28\,000$, $i=0.12$, $n=6$:
$$FV=28\,000(0.88)^{6}$$
$$(0.88)^{2}=0.7744,\qquad(0.88)^{3}=0.7744\times0.88=0.681472$$
$$(0.88)^{6}=(0.681472)^{2}\approx0.464404$$
$$FV\approx28\,000\times0.464404\approx13\,003.31$$
$$\boxed{FV\approx\$13\,003\text{ (nearest dollar)}}$$
(b) The fraction of the original value remaining is exactly $(0.88)^{6}\approx0.464404$, so as a percentage:
$$0.464404\times100\%\approx46.4404\%$$
$$\boxed{\approx46.4\%\text{ (1 d.p.)}}$$
QUESTION 17
4 marks
Easy
A rare coin was purchased for \$5000. Its value increases at a rate of $7\%$ per year.
Find the value of the coin after 10 years, correct to 2 decimal places.
Show complete worked solution
Using $FV=PV(1+i)^{n}$ with $PV=5000$, $i=0.07$, $n=10$:
$$FV=5000(1.07)^{10}$$
$$(1.07)^{10}\approx1.967151$$
$$FV\approx5000\times1.967151\approx9835.756786$$
$$\boxed{FV\approx\$9835.76}$$
QUESTION 18
5 marks
Easy
A logging company estimates that the number of trees in a forest is decreasing at a rate of $2\%$ per year due to clearing. The forest currently (Year 0) contains $40\,000$ trees.
(a) Find the number of trees remaining after 4 years, correct to the nearest 100, to complete the table. [2]
(b) Find the number of trees remaining after 8 years, correct to the nearest 100, to complete the table. [3]
| Year | 0 | 4 | 8 |
|---|---|---|---|
| Number of trees | 40\,000 |
Show complete worked solution
(a) Using $N=N_0(1-i)^{t}$ with $N_0=40\,000$, $i=0.02$, $t=4$:
$$N=40\,000(0.98)^{4}$$
$$(0.98)^{2}=0.9604,\qquad(0.98)^{4}=(0.9604)^{2}=0.92236816$$
$$N=40\,000\times0.92236816=36\,894.7264$$
$$\boxed{N\approx36\,900\text{ trees (nearest 100)}}$$
(b) Using $t=8$:
$$N=40\,000(0.98)^{8}=40\,000\left[(0.98)^{4}\right]^{2}$$
$$(0.92236816)^{2}\approx0.850763$$
$$N\approx40\,000\times0.850763\approx34\,030.52$$
$$\boxed{N\approx34\,000\text{ trees (nearest 100)}}$$
QUESTION 19
5 marks
Easy
\$6000 is invested at a nominal annual interest rate of $5\%$, compounded semi-annually (twice per year), for 7 years.
Find the value of the investment at the end of the 7 years, correct to 2 decimal places.
Show complete worked solution
Since interest is compounded semi-annually, the half-yearly rate is $\dfrac{0.05}{2}=0.025$, and the number of half-yearly periods over 7 years is $2\times7=14$. Using $FV=PV\left(1+\dfrac{i}{k}\right)^{kn}$:
$$FV=6000(1.025)^{14}$$
$$(1.025)^{14}\approx1.412974$$
$$FV\approx6000\times1.412974\approx8477.842926$$
$$\boxed{FV\approx\$8477.84}$$
QUESTION 20
4 marks
Easy
Due to inflation of $3\%$ per year, the real value (purchasing power) of a fixed sum of money decreases over time according to $V=V_0(0.97)^{t}$, where $V_0$ is the original value and $t$ is measured in years.
Find the real value, in today's dollars, of \$10\,000 after 4 years of $3\%$ inflation, correct to 2 decimal places.
Show complete worked solution
Using $V=V_0(0.97)^{t}$ with $V_0=10\,000$, $t=4$:
$$V=10\,000(0.97)^{4}$$
$$(0.97)^{2}=0.9409,\qquad(0.97)^{4}=(0.9409)^{2}=0.88529281$$
$$V=10\,000\times0.88529281=8852.9281$$
$$\boxed{V\approx\$8852.93}$$
QUESTION 21
4 marks
Easy
The efficiency of a solar panel decreases by $1.5\%$ each year due to degradation. When new, the panel operates at $100\%$ efficiency.
Find its efficiency after 10 years, correct to 1 decimal place.
Show complete worked solution
Using $E=E_0(1-i)^{t}$ with $E_0=100$, $i=0.015$, $t=10$:
$$E=100(0.985)^{10}$$
$$(0.985)^{10}\approx0.859730$$
$$E\approx100\times0.859730\approx85.9730$$
$$\boxed{E\approx86.0\%\text{ (1 d.p.)}}$$
QUESTION 22
5 marks
Easy
\$1500 is invested at a nominal annual interest rate of $6\%$, compounded weekly (52 weeks per year), for 2 years.
Find the value of the investment at the end of the 2 years, correct to 2 decimal places.
Show complete worked solution
Since interest is compounded weekly, the weekly rate is $\dfrac{0.06}{52}\approx0.0011538$, and the number of weekly periods over 2 years is $52\times2=104$. Using $FV=PV\left(1+\dfrac{i}{k}\right)^{kn}$:
$$FV=1500\left(1+\dfrac{0.06}{52}\right)^{104}$$
$$\left(1+\dfrac{0.06}{52}\right)^{104}\approx1.127419$$
$$FV\approx1500\times1.127419\approx1691.128285$$
$$\boxed{FV\approx\$1691.13}$$
QUESTION 23
4 marks
Easy
An insect population increases by $25\%$ each month. The population is currently 800.
Find the population after 5 months, correct to the nearest whole insect.
Show complete worked solution
Using $N=N_0(1+i)^{t}$ with $N_0=800$, $i=0.25$, $t=5$:
$$N=800(1.25)^{5}$$
$$(1.25)^{2}=1.5625,\qquad(1.25)^{4}=(1.5625)^{2}=2.44140625$$
$$(1.25)^{5}=2.44140625\times1.25=3.0517578125$$
$$N=800\times3.0517578125=2441.40625$$
$$\boxed{N\approx2441\text{ insects}}$$
QUESTION 24
7 marks
Medium
An investment of \$3000 grows to \$4500 in 8 years under annual compound interest.
(a) Find the annual interest rate, correct to 3 significant figures. [4]
(b) Assuming the same annual rate continues to apply, find the value of the investment after a total of 12 years, correct to 2 decimal places. [3]
Show complete worked solution
(a) Using $FV=PV(1+i)^{n}$ with $PV=3000$, $FV=4500$, $n=8$:
$$4500=3000(1+i)^{8}$$
$$(1+i)^{8}=\dfrac{4500}{3000}=1.5$$
Taking the 8th root of both sides:
$$1+i=1.5^{1/8}\approx1.051990$$
$$i\approx0.051990$$
$$\boxed{i\approx5.20\%\text{ per annum (3 s.f.)}}$$
(b) Since $(1+i)^{8}=1.5$, and $12=8\times\dfrac{3}{2}$:
$$(1+i)^{12}=\left[(1+i)^{8}\right]^{3/2}=1.5^{1.5}$$
$$1.5^{1.5}=1.5\times\sqrt{1.5}\approx1.5\times1.224745\approx1.837117$$
$$FV_{12}=3000\times1.837117\approx5511.351921$$
$$\boxed{FV_{12}\approx\$5511.35\text{ (2 d.p.)}}$$
QUESTION 25
6 marks
Medium
\$5000 is invested at $4.5\%$ per annum, compounded annually.
Find the minimum number of whole years required for the value of the investment to first exceed \$8000.
Show complete worked solution
Require $5000(1.045)^{n}>8000$:
$$(1.045)^{n}>\dfrac{8000}{5000}=1.6$$
Taking logarithms of both sides (since $\ln1.045>0$, the inequality direction is unchanged):
$$n\ln1.045>\ln1.6$$
$$n>\dfrac{\ln1.6}{\ln1.045}\approx\dfrac{0.470004}{0.044017}\approx10.678$$
Since $n$ must be a whole number, test $n=10$ and $n=11$:
$$V_{10}=5000(1.045)^{10}\approx7764.85<8000$$
$$V_{11}=5000(1.045)^{11}\approx8114.27>8000$$
$$\boxed{n=11\text{ years}}$$
QUESTION 26
6 marks
Medium
A machine purchased for \$50\,000 depreciates by $18\%$ each year, using reducing-balance depreciation.
Find the minimum number of whole years after purchase for the machine's value to first fall below \$15\,000.
Show complete worked solution
Require $50\,000(0.82)^{n}<15\,000$:
$$(0.82)^{n}<\dfrac{15\,000}{50\,000}=0.3$$
Taking logarithms of both sides (note $\ln0.82<0$, so the inequality reverses):
$$n\ln0.82<\ln0.3$$
$$n>\dfrac{\ln0.3}{\ln0.82}\approx\dfrac{-1.203973}{-0.198451}\approx6.067$$
Since $n$ must be a whole number, test $n=6$ and $n=7$:
$$V_{6}=50\,000(0.82)^{6}\approx15\,200.33>15\,000$$
$$V_{7}=50\,000(0.82)^{7}\approx12\,464.27<15\,000$$
$$\boxed{n=7\text{ years}}$$
QUESTION 27
8 marks
Medium
Elin has \$10\,000 to invest for 3 years and is choosing between two accounts:
Bank A: nominal interest rate of $6\%$ per annum, compounded monthly.
Bank B: interest rate of $6.1\%$ per annum, compounded annually.
(a) Find the value of the investment after 3 years in Bank A, correct to 2 decimal places. [3]
(b) Find the value of the investment after 3 years in Bank B, correct to 2 decimal places. [2]
(c) Determine which bank offers the better return, and find the difference in the final values, correct to 2 decimal places. [3]
Show complete worked solution
(a) The monthly rate is $\dfrac{0.06}{12}=0.005$, and the number of months in 3 years is $36$:
$$FV_A=10\,000(1.005)^{36}$$
$$(1.005)^{36}\approx1.196680$$
$$FV_A\approx10\,000\times1.196680\approx11\,966.81$$
$$\boxed{FV_A\approx\$11\,966.81}$$
(b)
$$FV_B=10\,000(1.061)^{3}$$
$$(1.061)^{2}=1.125721,\qquad(1.061)^{3}=1.125721\times1.061\approx1.194390$$
$$FV_B\approx10\,000\times1.194390\approx11\,943.90$$
$$\boxed{FV_B\approx\$11\,943.90}$$
(c) Comparing the two values:
$$FV_A-FV_B\approx11\,966.81-11\,943.90=22.91$$
$$\boxed{\text{Bank A gives the better return, by approximately }\$22.91}$$
QUESTION 28
7 marks
Medium
At the end of each year, Farah deposits \$800 into a savings account paying $3.5\%$ per annum, compounded annually, so that each deposit earns interest for every full year it remains in the account before the next deposit is made.
(a) Show that the balance immediately after the 6th deposit is given by
$$V=800\cdot\dfrac{1.035^{6}-1}{0.035}$$ [4]
(b) Calculate this value, correct to 2 decimal places. [3]
Show complete worked solution
(a) The 1st deposit earns interest for 5 years by the time of the 6th deposit, growing to $800(1.035)^{5}$; the 2nd deposit earns interest for 4 years, growing to $800(1.035)^{4}$; and so on, down to the 6th deposit, which earns no interest. So the balance immediately after the 6th deposit is
$$V=800(1.035)^{5}+800(1.035)^{4}+\cdots+800(1.035)+800$$
This is a geometric series with first term $800$, common ratio $1.035$, and 6 terms. Using $S_n=\dfrac{u_1(r^{n}-1)}{r-1}$:
$$V=\dfrac{800\big(1.035^{6}-1\big)}{1.035-1}=800\cdot\dfrac{1.035^{6}-1}{0.035}$$
as required.
(b) Now evaluate, using $1.035^{6}\approx1.229255$:
$$V=800\cdot\dfrac{1.229255-1}{0.035}=800\cdot\dfrac{0.229255}{0.035}\approx800(6.550152)$$
$$\boxed{V\approx\$5240.12\text{ (2 d.p.)}}$$
QUESTION 29
7 marks
Medium
A loan of \$20\,000 is to be repaid by 6 equal payments, one made at the end of each year, at an annual interest rate of $5\%$, compounded annually. The size of each payment $D$ satisfies
$$20\,000=D\cdot\dfrac{1-(1.05)^{-6}}{0.05}$$
Find the value of $D$, correct to 2 decimal places.
Show complete worked solution
First evaluate $(1.05)^{-6}$:
$$(1.05)^{6}\approx1.340096,\qquad(1.05)^{-6}\approx\dfrac{1}{1.340096}\approx0.746215$$
So:
$$1-(1.05)^{-6}\approx1-0.746215=0.253785$$
Substituting into the equation:
$$20\,000=D\cdot\dfrac{0.253785}{0.05}=D(5.075697)$$
Rearranging for $D$:
$$D=\dfrac{20\,000}{5.075697}$$
$$\boxed{D\approx\$3940.35\text{ (2 d.p.)}}$$
QUESTION 30
8 marks
Medium
A radioactive substance decays such that $30\%$ of the mass present is lost every 10 years. A sample initially has a mass of 200 g.
(a) Find the mass remaining after 25 years, correct to 2 decimal places. [4]
(b) Find the half-life of the substance, correct to 2 decimal places. [4]
Show complete worked solution
(a) Losing $30\%$ every 10 years means the mass is multiplied by $(1-0.3)=0.7$ every 10 years, so $N=200(0.7)^{t/10}$. At $t=25$:
$$N=200(0.7)^{25/10}=200(0.7)^{2.5}$$
$$(0.7)^{2.5}\approx0.409963$$
$$N\approx200\times0.409963\approx81.99$$
$$\boxed{N\approx81.99\text{ g (2 d.p.)}}$$
(b) The half-life $h$ satisfies $(0.7)^{h/10}=0.5$. Taking logarithms of both sides:
$$\dfrac{h}{10}\ln0.7=\ln0.5$$
$$h=\dfrac{10\ln0.5}{\ln0.7}\approx\dfrac{10(-0.693147)}{-0.356675}$$
$$\boxed{h\approx19.43\text{ years (2 d.p.)}}$$
QUESTION 31
8 marks
Medium
The number of bacteria in a culture is modelled by $N=N_0e^{kt}$, where $N_0$ is the initial number of bacteria, $t$ is the time in hours, and $k$ is a constant. Initially there are 1000 bacteria, and after 6 hours there are 2500 bacteria.
(a) Show that $k=\dfrac{1}{6}\ln2.5$, and find the value of $k$ correct to 4 significant figures. [4]
(b) Find the number of bacteria present after 10 hours, correct to the nearest whole number. [4]
Show complete worked solution
(a) With $N_0=1000$, at $t=6$, $N=2500$:
$$2500=1000e^{6k}$$
$$e^{6k}=2.5$$
Taking natural logarithms of both sides:
$$6k=\ln2.5$$
$$k=\dfrac{1}{6}\ln2.5$$
as required. Numerically, $\ln2.5\approx0.916291$, so
$$\boxed{k\approx0.1527\text{ (4 s.f.)}}$$
(b) Using $N=1000e^{kt}$ with $t=10$:
$$N=1000e^{10(0.152715)}=1000e^{1.527152}$$
$$e^{1.527152}\approx4.605039$$
$$N\approx4605.04$$
$$\boxed{N\approx4605\text{ bacteria (nearest whole number)}}$$
QUESTION 32
9 marks
Medium
An object at a temperature of $90^{\circ}\text{C}$ is placed in a room kept at a constant $20^{\circ}\text{C}$. Its temperature, $T\ ^{\circ}\text{C}$, after $t$ minutes is modelled by Newton's Law of Cooling:
$$T=20+70e^{-kt}$$
where $k$ is a positive constant. After 5 minutes, the temperature of the object is $65^{\circ}\text{C}$.
(a) Find the value of $k$, correct to 4 significant figures. [4]
(b) Find the temperature of the object after 15 minutes, correct to 1 decimal place. [3]
(c) State the temperature that the model predicts the object will approach as $t\to\infty$, and explain why this is a sensible long-term prediction. [2]
Show complete worked solution
(a) At $t=5$, $T=65$:
$$65=20+70e^{-5k}$$
$$45=70e^{-5k}$$
$$e^{-5k}=\dfrac{45}{70}=\dfrac{9}{14}$$
Taking natural logarithms:
$$-5k=\ln\left(\dfrac{9}{14}\right)$$
$$k=-\dfrac{1}{5}\ln\left(\dfrac{9}{14}\right)$$
$$\boxed{k\approx0.08837\text{ (4 s.f.)}}$$
(b) At $t=15$:
$$T=20+70e^{-15(0.08837)}=20+70e^{-1.325499}$$
$$e^{-1.325499}\approx0.265668$$
$$T\approx20+70(0.265668)\approx20+18.597\approx38.597$$
$$\boxed{T\approx38.6^{\circ}\text{C (1 d.p.)}}$$
(c) As $t\to\infty$, $e^{-kt}\to0$ since $k>0$, so $T\to20+0=20$.
$$\boxed{T\to20^{\circ}\text{C}}$$
This is sensible because the object cannot cool below the temperature of its surroundings; the model predicts it gets closer and closer to room temperature, but never below it.
QUESTION 33
7 marks
Medium
Yusuf wants to accumulate \$15\,000 in a savings account after 8 years, by making equal deposits at the end of each year. The account pays $4\%$ per annum, compounded annually. The size of each deposit $D$ satisfies
$$15\,000=D\cdot\dfrac{(1.04)^{8}-1}{0.04}$$
Find the value of $D$, correct to 2 decimal places.
Show complete worked solution
First evaluate $(1.04)^{8}$:
$$(1.04)^{8}\approx1.368569$$
So:
$$(1.04)^{8}-1\approx0.368569$$
Substituting into the equation:
$$15\,000=D\cdot\dfrac{0.368569}{0.04}=D(9.214226)$$
Rearranging for $D$:
$$D=\dfrac{15\,000}{9.214226}$$
$$\boxed{D\approx\$1627.92\text{ (2 d.p.)}}$$
QUESTION 34
9 marks
Medium
The population of Town A is currently $20\,000$ and is growing at $2\%$ per year. The population of Town B is currently $15\,000$ and is growing at $4\%$ per year.
(a) Write down expressions for the population of each town after $t$ years. [2]
(b) Using the values in the table, determine the minimum whole number of years from now for the population of Town B to first exceed the population of Town A. [3]
(c) Find, using logarithms, the value of $t$ (not necessarily a whole number) at which the two populations would be exactly equal, correct to 2 decimal places. [4]
| Years from now, $t$ | 14 | 15 |
|---|---|---|
| Population of Town A | 26\,389.58 | 26\,917.37 |
| Population of Town B | 25\,975.15 | 27\,014.15 |
Show complete worked solution
(a)
$$\boxed{P_A(t)=20\,000(1.02)^{t},\qquad P_B(t)=15\,000(1.04)^{t}}$$
(b) From the table, at $t=14$, Town B's population ($25\,975.15$) is less than Town A's ($26\,389.58$); at $t=15$, Town B's population ($27\,014.15$) is greater than Town A's ($26\,917.37$). So Town B's population first exceeds Town A's during year 15.
$$\boxed{n=15\text{ years}}$$
(c) Setting the populations equal:
$$20\,000(1.02)^{t}=15\,000(1.04)^{t}$$
$$\left(\dfrac{1.04}{1.02}\right)^{t}=\dfrac{20\,000}{15\,000}=\dfrac{4}{3}$$
Taking logarithms of both sides:
$$t\ln\left(\dfrac{1.04}{1.02}\right)=\ln\left(\dfrac{4}{3}\right)$$
$$t=\dfrac{\ln(4/3)}{\ln(1.04/1.02)}\approx\dfrac{0.287682}{0.019418}$$
$$\boxed{t\approx14.82\text{ years (2 d.p.)}}$$
QUESTION 35
8 marks
Medium
A car undergoes reducing-balance depreciation. Its value is \$18\,000 after 2 years, and \$14\,580 after 4 years from purchase.
(a) Find the annual rate of depreciation. [4]
(b) Find the original purchase price of the car. [4]
Show complete worked solution
(a) Let $P_0$ be the original price and $r$ the annual depreciation rate (as a decimal), so the value after $t$ years is $V(t)=P_0(1-r)^{t}$. Then
$$V(2)=P_0(1-r)^{2}=18\,000,\qquad V(4)=P_0(1-r)^{4}=14\,580$$
Dividing the second equation by the first eliminates $P_0$:
$$\dfrac{V(4)}{V(2)}=(1-r)^{2}=\dfrac{14\,580}{18\,000}=0.81$$
Taking the (positive) square root, since $0<1-r<1$:
$$1-r=\sqrt{0.81}=0.9$$
$$\boxed{r=10\%\text{ per annum}}$$
(b) Substituting into $V(2)=P_0(0.9)^{2}=18\,000$:
$$P_0(0.81)=18\,000$$
$$P_0=\dfrac{18\,000}{0.81}$$
$$\boxed{P_0\approx\$22\,222.22\text{ (2 d.p.)}}$$
QUESTION 36
8 marks
Medium
An investment of \$8000 grows at a nominal rate of $6\%$ per annum, compounded annually. At the same time, inflation reduces the purchasing power of money by $2.5\%$ per annum.
(a) Show that the approximate annual real (inflation-adjusted) growth factor of the investment is given by $\dfrac{1.06}{1.025}$, and hence find the real annual growth rate, correct to 3 significant figures. [3]
(b) Hence find the real value of the investment (in today's purchasing power) after 5 years, correct to 2 decimal places. [5]
Show complete worked solution
(a) After 1 year, the nominal value is multiplied by $1.06$. However, due to $2.5\%$ inflation, $\$1$ received in a year's time is worth only $\dfrac{1}{1.025}$ of $\$1$ today. So the real (purchasing-power) growth factor per year is
$$\dfrac{1.06}{1.025}$$
Evaluating:
$$\dfrac{1.06}{1.025}\approx1.034146$$
So the real annual growth rate is approximately $0.034146$.
$$\boxed{\text{real rate}\approx3.41\%\text{ per annum (3 s.f.)}}$$
(b) The real value after 5 years is the nominal future value, deflated by inflation over 5 years:
$$V_{\text{real}}=8000\left(\dfrac{1.06}{1.025}\right)^{5}$$
Using the unrounded ratio $1.06/1.025\approx1.0341463$:
$$\left(1.0341463\right)^{5}\approx1.182796$$
$$V_{\text{real}}\approx8000\times1.182796\approx9462.371316$$
$$\boxed{V_{\text{real}}\approx\$9462.37\text{ (2 d.p.)}}$$
QUESTION 37
8 marks
Medium
At the end of each quarter (3-month period), Priya deposits \$200 into a savings account that pays a nominal annual interest rate of $6\%$, compounded quarterly, so that each deposit earns interest for every full quarter it remains in the account before the next deposit is made.
Find the balance in the account immediately after the 12th deposit (i.e. after 3 years), correct to 2 decimal places.
Show complete worked solution
The quarterly interest rate is $i=\dfrac{0.06}{4}=0.015$. As with an ordinary annuity, the balance immediately after the $n$th deposit is
$$V_n=200\cdot\dfrac{(1.015)^{n}-1}{0.015}$$
With $n=12$:
$$(1.015)^{12}\approx1.195618$$
$$V_{12}=200\cdot\dfrac{1.195618-1}{0.015}=200\cdot\dfrac{0.195618}{0.015}\approx200(13.0412)$$
$$\boxed{V_{12}\approx\$2608.24\text{ (2 d.p.)}}$$
QUESTION 38
7 marks
Medium
A savings account offers a nominal annual interest rate of $9\%$, compounded monthly.
(a) Find the effective annual interest rate, correct to 3 significant figures. [3]
(b) Hence, or otherwise, find the value of a \$4000 investment in this account after 4 years, correct to 2 decimal places. [4]
Show complete worked solution
(a) The monthly interest rate is $\dfrac{0.09}{12}=0.0075$. The effective annual rate satisfies $1+i_{\text{eff}}=(1.0075)^{12}$:
$$(1.0075)^{12}\approx1.093807$$
$$i_{\text{eff}}\approx0.093807$$
$$\boxed{i_{\text{eff}}\approx9.38\%\text{ (3 s.f.)}}$$
(b) Compounding monthly over 4 years ($48$ months):
$$FV=4000(1.0075)^{48}$$
$$(1.0075)^{48}\approx1.431405$$
$$FV\approx4000\times1.431405\approx5725.621333$$
$$\boxed{FV\approx\$5725.62\text{ (2 d.p.)}}$$
QUESTION 39
8 marks
Medium
The number of fish in a lake is decreasing at a rate of $5\%$ per year, due to environmental factors. There are currently $50\,000$ fish in the lake.
(a) Find the number of fish remaining after 10 years, correct to the nearest whole number. [3]
(b) Find the minimum number of whole years for the fish population to first fall below $25\,000$. [5]
Show complete worked solution
(a) Using $N=50\,000(0.95)^{t}$ with $t=10$:
$$N=50\,000(0.95)^{10}$$
$$(0.95)^{10}\approx0.598737$$
$$N\approx50\,000\times0.598737\approx29\,936.85$$
$$\boxed{N\approx29\,937\text{ fish (nearest whole number)}}$$
(b) Require $50\,000(0.95)^{n}<25\,000$:
$$(0.95)^{n}<0.5$$
Taking logarithms of both sides (note $\ln0.95<0$, so the inequality reverses):
$$n\ln0.95<\ln0.5$$
$$n>\dfrac{\ln0.5}{\ln0.95}\approx\dfrac{-0.693147}{-0.051293}\approx13.513$$
Since $n$ must be a whole number, test $n=13$ and $n=14$:
$$N_{13}=50\,000(0.95)^{13}\approx25\,667.10>25\,000$$
$$N_{14}=50\,000(0.95)^{14}\approx24\,383.75<25\,000$$
$$\boxed{n=14\text{ years}}$$
QUESTION 40
9 marks
Medium
Rina borrows \$12\,000 at an interest rate of $4\%$ per annum, compounded annually. At the end of each year, interest is added to the balance and then Rina makes a repayment of \$2500 (except possibly the final repayment, which may differ).
(a) Show that the balance immediately after the 1st repayment is \$9980.00, as shown in the table. [2]
(b) Find the balance immediately after the 5th repayment, to complete the table, correct to 2 decimal places. [4]
(c) Find the single final repayment Rina would need to make at the end of year 6 in order to reduce the balance to exactly zero (i.e. to fully repay the loan). [3]
| Year | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Balance after repayment (\$) | 9980.00 | 7879.20 | 5694.37 | 3422.14 | ? |
Show complete worked solution
(a) The balance grows by $4\%$ interest, then $\$2500$ is subtracted:
$$12\,000(1.04)-2500=12\,480-2500=9980.00$$
$$\boxed{\$9980.00}$$ as required.
(b) Continue iterating, year by year:
$$\text{Year 2: }9980.00(1.04)-2500=10\,379.20-2500=7879.20$$
$$\text{Year 3: }7879.20(1.04)-2500=8194.368-2500=5694.368\approx5694.37$$
$$\text{Year 4: }5694.368(1.04)-2500=5922.14272-2500=3422.14272\approx3422.14$$
$$\text{Year 5: }3422.14272(1.04)-2500=3559.028429-2500=1059.028429$$
$$\boxed{\text{Balance after 5th repayment}\approx\$1059.03\text{ (2 d.p.)}}$$
(c) At the end of year 6, the outstanding balance first grows by $4\%$ interest, and the final repayment must clear it exactly:
$$1059.028429\times1.04\approx1101.389566$$
$$\boxed{\text{Final repayment}\approx\$1101.39\text{ (2 d.p.)}}$$
QUESTION 41
8 marks
Medium
Two investment options are available for a \$5000 deposit:
Option A: simple interest at $8\%$ per annum.
Option B: compound interest at $6\%$ per annum, compounded annually.
(a) Write down a formula for the value of each option after $t$ years. [2]
(b) Using the values in the table, find the minimum whole number of years for Option B to first exceed Option A in value. [3]
(c) Explain, in terms of the two types of growth involved, why Option B must eventually always exceed Option A in value for large enough $t$, regardless of how far behind it is after 10 years. [3]
| Years, $t$ | 10 | 11 |
|---|---|---|
| Option A value (\$) | 9000.00 | 9400.00 |
| Option B value (\$) | 8954.24 | 9491.49 |
Show complete worked solution
(a)
$$\boxed{A(t)=5000(1+0.08t),\qquad B(t)=5000(1.06)^{t}}$$
(b) From the table, at $t=10$, Option B ($8954.24$) is less than Option A ($9000.00$); at $t=11$, Option B ($9491.49$) is greater than Option A ($9400.00$). So Option B first exceeds Option A during year 11.
$$\boxed{n=11\text{ years}}$$
(c) Option A grows linearly (arithmetically): its value increases by the same fixed amount, $5000\times0.08=400$, every year. Option B grows exponentially (geometrically): its value is multiplied by the same factor, $1.06$, every year, so the actual dollar increase itself keeps growing larger each year. Because an exponential function with ratio greater than 1 eventually grows faster than any linear function of $t$, Option B's value must eventually overtake Option A's value, no matter how large the gap is at $t=10$ -- it is only a matter of time.
QUESTION 42
14 marks
Hard
At the end of each year, Marcus deposits \$1200 into an investment fund for his daughter's university education. The fund pays an annual interest rate of $4.5\%$, compounded annually, so that each deposit earns interest for every full year it remains in the fund before the next deposit is made.
(a) Show that the amount in the fund immediately after the 3rd deposit is \$3764.43. [3]
(b) Show that the amount in the fund immediately after the $n$th deposit is given by
$$V_n=1200\cdot\dfrac{1.045^{n}-1}{0.045},$$
and calculate the amount in the fund immediately after the 15th deposit, correct to 2 decimal places. [5]
(c) Find the minimum number of deposits required for the fund balance, immediately after a deposit, to first exceed \$25\,000. [3]
(d) Suppose instead Marcus chooses a constant annual deposit $D$ (with the same $4.5\%$ interest) so that the balance is exactly \$30\,000 immediately after the 12th deposit. Find the value of $D$, correct to 2 decimal places. [3]
Show complete worked solution
(a) After the 1st deposit, the balance is $1200$. Each year the balance grows by $4.5\%$ interest before the next deposit is added:
$$\text{After year 1 (before 2nd deposit): }1200(1.045)+1200=1254+1200=2454$$
$$\text{After year 2 (before 3rd deposit): }2454(1.045)+1200=2564.43+1200=3764.43$$
$$\boxed{\$3764.43}$$ as required.
(b) The 1st deposit earns interest for $(n-1)$ years by the time of the $n$th deposit, growing to $1200(1.045)^{n-1}$; the 2nd deposit earns interest for $(n-2)$ years, growing to $1200(1.045)^{n-2}$; and so on, down to the $n$th deposit, which earns no interest. So
$$V_n=1200(1.045)^{n-1}+1200(1.045)^{n-2}+\cdots+1200(1.045)+1200$$
This is a geometric series with first term $1200$, common ratio $1.045$, and $n$ terms. Using $S_n=\dfrac{u_1(r^{n}-1)}{r-1}$:
$$V_n=\dfrac{1200\big(1.045^{n}-1\big)}{1.045-1}=1200\cdot\dfrac{1.045^{n}-1}{0.045}$$
as required. For $n=15$, using $1.045^{15}\approx1.935282$:
$$V_{15}=1200\cdot\dfrac{1.935282-1}{0.045}=1200\cdot\dfrac{0.935282}{0.045}\approx1200(20.78405)$$
$$\boxed{V_{15}\approx\$24\,940.87\text{ (2 d.p.)}}$$
(c) By the same reasoning, $V_n=1200\cdot\dfrac{1.045^{n}-1}{0.045}$. Require $V_n>25\,000$:
$$1200(1.045^{n}-1)>25\,000(0.045)=1125$$
$$1.045^{n}-1>\dfrac{1125}{1200}=0.9375$$
$$1.045^{n}>1.9375$$
Taking logarithms of both sides:
$$n>\dfrac{\ln1.9375}{\ln1.045}\approx15.026$$
Since $n$ is a positive integer, test $n=15$ and $n=16$:
$$V_{15}\approx24\,940.87<25\,000$$
$$V_{16}=1200\cdot\dfrac{1.045^{16}-1}{0.045}\approx27\,263.20>25\,000$$
$$\boxed{\text{minimum of }16\text{ deposits}}$$
(d) By the same reasoning as in (b), with deposit $D$ replacing 1200, and $n=12$:
$$30\,000=D\cdot\dfrac{1.045^{12}-1}{0.045}$$
Using $1.045^{12}\approx1.695881$:
$$30\,000=D\cdot\dfrac{0.695881}{0.045}=D(15.46403)$$
$$D=\dfrac{30\,000}{15.46403}$$
$$\boxed{D\approx\$1939.99\text{ (2 d.p.)}}$$
QUESTION 43
14 marks
Hard
Diego takes out a loan of \$18\,000 to buy a car. The loan is charged interest at $5.5\%$ per annum, compounded annually, and is to be repaid by 5 equal payments, one made at the end of each year, so that the loan balance is exactly zero immediately after the 5th payment.
(a) Let $D$ be the size of each annual payment. Show that the outstanding balance immediately after the 1st payment is $18\,000(1.055)-D$, and hence show that the outstanding balance immediately after the 5th payment is
$$18\,000(1.055)^{5}-D\big[(1.055)^{4}+(1.055)^{3}+(1.055)^{2}+(1.055)+1\big].$$ [3]
(b) Hence show that
$$D=\dfrac{18\,000(1.055)^{5}}{\dfrac{(1.055)^{5}-1}{0.055}},$$
and calculate the value of $D$, correct to 2 decimal places. [5]
(c) Find the outstanding balance immediately after the 2nd payment, correct to 2 decimal places. [3]
(d) Find the total amount of interest Diego pays over the life of the loan, correct to 2 decimal places. [3]
Show complete worked solution
(a) Immediately after the 1st payment, the loan balance has grown by one year's interest and then $D$ is subtracted:
$$18\,000(1.055)-D$$
as required. This balance then grows by interest and has $D$ subtracted each subsequent year. Following the balance through to the end of year 5 (multiplying the outstanding balance by $1.055$ each year and subtracting a further $D$ at each step) gives
$$18\,000(1.055)^{5}-D\big[(1.055)^{4}+(1.055)^{3}+(1.055)^{2}+(1.055)+1\big]$$
as required, since each of the 5 payments of $D$ is carried forward by interest for a different number of remaining years.
(b) Since the loan is exactly repaid, this balance equals zero:
$$18\,000(1.055)^{5}=D\big[(1.055)^{4}+(1.055)^{3}+(1.055)^{2}+(1.055)+1\big]$$
The bracketed sum is a geometric series with first term $1$, common ratio $1.055$, and 5 terms, so it equals $\dfrac{(1.055)^{5}-1}{0.055}$. Substituting and rearranging for $D$:
$$D=\dfrac{18\,000(1.055)^{5}}{\dfrac{(1.055)^{5}-1}{0.055}}$$
as required. Now evaluate, using $(1.055)^{5}\approx1.306960$:
$$\dfrac{(1.055)^{5}-1}{0.055}=\dfrac{0.306960}{0.055}\approx5.581091$$
$$D=\dfrac{18\,000\times1.306960}{5.581091}=\dfrac{23\,525.28}{5.581091}$$
$$\boxed{D\approx\$4215.18\text{ (2 d.p.)}}$$
(c) Iterating year by year, using the unrounded value of $D$:
$$\text{After payment 1: }18\,000(1.055)-4215.18=18\,990-4215.18=14\,774.82$$
$$\text{After payment 2: }14\,774.82(1.055)-4215.18\approx15\,587.44-4215.18$$
$$\boxed{\approx\$11\,372.26\text{ (2 d.p.)}}$$
(d) The total amount repaid is $5D\approx5\times4215.18=21\,075.88$ (using the more precise value of $D$, $5D\approx21\,075.88$). The interest paid is the total repaid minus the original loan:
$$21\,075.88-18\,000=3075.88$$
$$\boxed{\text{Total interest}\approx\$3075.88\text{ (2 d.p.)}}$$
QUESTION 44
10 marks
Hard
A company buys a piece of equipment for \$60\,000. For the first 2 years, it depreciates by $25\%$ per year (heavier early depreciation). After that, it depreciates by $10\%$ per year (reducing-balance), for as long as the company retains it.
(a) Show that the value of the equipment after the first 2 years is \$33\,750. [2]
(b) Show that the value 3 years into the second depreciation phase (i.e. 5 years after purchase) is \$24\,603.75, confirming the table. [3]
(c) Using the table, determine the minimum number of whole years after purchase for the equipment's value to first fall below \$20\,000. [2]
(d) Find the total percentage decrease in the equipment's value over the first 5 years after purchase (from the original \$60\,000 to the value found in part (b)), correct to 1 decimal place. [3]
| Years since purchase | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|
| Value (\$) | 33\,750.00 | 30\,375.00 | 27\,337.50 | 24\,603.75 | 22\,143.38 | 19\,929.04 |
Show complete worked solution
(a) During the first 2 years, the value is multiplied by $(1-0.25)=0.75$ each year:
$$V_2=60\,000(0.75)^{2}=60\,000(0.5625)=33\,750$$
$$\boxed{\$33\,750}$$ as required.
(b) From the start of the second phase, the value is $\$33\,750$, and now depreciates by $10\%$ per year, i.e. it is multiplied by $0.9$ each year. After 3 more years (5 years after purchase in total):
$$V=33\,750(0.9)^{3}$$
$$(0.9)^{3}=0.729$$
$$V=33\,750\times0.729=24\,603.75$$
$$\boxed{\$24\,603.75}$$ as required, confirming the table.
(c) From the table, at 6 years the value is $\$22\,143.38$ (still above $\$20\,000$); at 7 years it is $\$19\,929.04$ (below $\$20\,000$).
$$\boxed{7\text{ years}}$$
(d) Percentage decrease $=\left(1-\dfrac{24\,603.75}{60\,000}\right)\times100\%$:
$$\dfrac{24\,603.75}{60\,000}=0.4100625$$
$$1-0.4100625=0.5899375$$
$$0.5899375\times100\%=58.99375\%$$
$$\boxed{\approx59.0\%\text{ (1 d.p.)}}$$
QUESTION 45
11 marks
Hard
A sample of a radioactive substance has a half-life of 1600 years. The substance decays according to $N=N_0e^{-kt}$, where $N_0$ is the initial mass, $t$ is the time in years, and $k$ is a positive constant.
(a) Show that $k=\dfrac{\ln2}{1600}$, and find the value of $k$, correct to 4 significant figures. [3]
(b) A particular sample originally had a mass of 50 g. It currently contains only $15\%$ of its original mass. Find the age of the sample, in years, correct to the nearest 10 years. [4]
(c) Find the mass of the sample, correct to 2 decimal places, a further 500 years from now. [3]
(d) Explain why, according to this model, the mass of the substance never reaches exactly zero, no matter how much time passes. [1]
Show complete worked solution
(a) A half-life of 1600 years means $N=\dfrac{N_0}{2}$ when $t=1600$:
$$\dfrac{N_0}{2}=N_0e^{-1600k}$$
$$\dfrac{1}{2}=e^{-1600k}$$
Taking natural logarithms:
$$\ln\left(\dfrac{1}{2}\right)=-1600k$$
$$-\ln2=-1600k$$
$$k=\dfrac{\ln2}{1600}$$
as required. Numerically, $\ln2\approx0.693147$, so
$$\boxed{k\approx0.0004332\text{ per year (4 s.f.)}}$$
(b) $15\%$ remaining means $\dfrac{N}{N_0}=0.15=e^{-kt}$. Taking natural logarithms:
$$\ln0.15=-kt$$
$$t=\dfrac{-\ln0.15}{k}=\dfrac{\ln(1/0.15)}{k}$$
$$\ln(1/0.15)=\ln(6.6\overline{6})\approx1.897120$$
$$t\approx\dfrac{1.897120}{0.00043322}\approx4379.14$$
Rounding to the nearest 10 years,
$$\boxed{t\approx4380\text{ years}}$$
(c) The current mass is $15\%$ of the original $50$ g, i.e. $0.15\times50=7.5$ g. A further 500 years from now, the mass is
$$N=7.5\,e^{-500k}=7.5\,e^{-500(0.00043322)}=7.5\,e^{-0.216608}$$
$$e^{-0.216608}\approx0.805245$$
$$N\approx7.5\times0.805245\approx6.0393$$
$$\boxed{N\approx6.04\text{ g (2 d.p.)}}$$
(d) Since $e^{-kt}>0$ for every finite value of $t$, the model gives $N=N_0e^{-kt}>0$ always -- the mass gets closer and closer to zero as $t$ increases, but the exponential term never actually reaches zero for any finite $t$, so the model never predicts an exact zero mass.
QUESTION 46
11 marks
Hard
The population of a bacteria culture, $t$ hours after being first observed, is modelled by $P=P_0e^{kt}$, where $P_0$ and $k$ are constants. It is known that $P=5000$ when $t=2$, and $P=13\,000$ when $t=5$.
(a) Show that $k=\dfrac{1}{3}\ln2.6$, and find the value of $k$, correct to 4 significant figures. [4]
(b) Find the value of $P_0$, correct to the nearest whole number. [2]
(c) Find the population predicted by the model at $t=8$, correct to the nearest 100. [2]
(d) Find the doubling time of the population (the time taken for the population to double in size), correct to 2 decimal places. [3]
Show complete worked solution
(a) Using the two given conditions:
$$P_0e^{2k}=5000,\qquad P_0e^{5k}=13\,000$$
Dividing the second equation by the first eliminates $P_0$:
$$\dfrac{P_0e^{5k}}{P_0e^{2k}}=\dfrac{13\,000}{5000}$$
$$e^{3k}=2.6$$
Taking natural logarithms:
$$3k=\ln2.6$$
$$k=\dfrac{1}{3}\ln2.6$$
as required. Numerically, $\ln2.6\approx0.955511$, so
$$\boxed{k\approx0.3185\text{ (4 s.f.)}}$$
(b) From $P_0e^{2k}=5000$:
$$P_0=5000e^{-2k}=5000e^{-2(0.318504)}=5000e^{-0.637008}$$
$$e^{-0.637008}\approx0.528873$$
$$P_0\approx5000\times0.528873\approx2644.36$$
$$\boxed{P_0\approx2644\text{ (nearest whole number)}}$$
(c)
$$P(8)=P_0e^{8k}\approx2644.36\,e^{8(0.318504)}=2644.36\,e^{2.548029}$$
$$e^{2.548029}\approx12.7818$$
$$P(8)\approx2644.36\times12.7818\approx33\,800.0$$
$$\boxed{P(8)\approx33\,800\text{ (nearest 100)}}$$
(d) The doubling time $T$ satisfies $e^{kT}=2$:
$$T=\dfrac{\ln2}{k}\approx\dfrac{0.693147}{0.318504}$$
$$\boxed{T\approx2.18\text{ hours (2 d.p.)}}$$
QUESTION 47
12 marks
Hard
Petra wants to have exactly \$50\,000 available in 10 years' time. She is considering two different savings strategies, both earning $5\%$ per annum, compounded annually.
Strategy 1 (lump sum): invest a single amount now, and let it grow for 10 years.
Strategy 2 (annuity): deposit an equal amount at the end of each year for 10 years.
(a) For Strategy 1, find the amount Petra would need to invest now, correct to 2 decimal places. [3]
(b) For Strategy 2, given that the balance immediately after the 10th deposit is $V=D\cdot\dfrac{1.05^{10}-1}{0.05}$, find the value of the annual deposit $D$ required, correct to 2 decimal places. [4]
(c) Find the total amount Petra would pay into the fund under each strategy (i.e. the sum of all deposits made), correct to 2 decimal places for each. [3]
(d) Hence explain, with reference to your answers to part (c), why Strategy 2 requires Petra to pay in more money in total than Strategy 1, even though both strategies reach the same final value of \$50\,000. [2]
Show complete worked solution
(a) Using $PV=\dfrac{FV}{(1+i)^{n}}$ with $FV=50\,000$, $i=0.05$, $n=10$:
$$(1.05)^{10}\approx1.628895$$
$$PV=\dfrac{50\,000}{1.628895}$$
$$\boxed{PV\approx\$30\,695.66\text{ (2 d.p.)}}$$
(b) Using $(1.05)^{10}\approx1.628895$:
$$50\,000=D\cdot\dfrac{1.628895-1}{0.05}=D\cdot\dfrac{0.628895}{0.05}=D(12.57789)$$
$$D=\dfrac{50\,000}{12.57789}$$
$$\boxed{D\approx\$3975.23\text{ (2 d.p.)}}$$
(c) Strategy 1: a single payment now of $\$30\,695.66$ (from part (a)).
Strategy 2: total paid $=10D\approx10\times3975.228748\approx39\,752.29$.
$$\boxed{\text{Strategy 1}\approx\$30\,695.66;\qquad\text{Strategy 2}\approx\$39\,752.29}$$
(d) Under Strategy 1, the entire $\$30\,695.66$ is invested immediately and earns compound interest for the full 10 years, so interest does most of the "work" of reaching $\$50\,000$. Under Strategy 2, each deposit earns interest for a shorter period than the one before it -- the final deposit, made at the end of year 10, earns no interest at all. Because the deposits under Strategy 2 collectively earn less total interest than the single lump sum under Strategy 1, Petra herself must contribute more in total deposits under Strategy 2 to reach the same $\$50\,000$ target.
QUESTION 48
10 marks
Hard
An object at $100^{\circ}\text{C}$ is placed in a room kept at a constant $20^{\circ}\text{C}$. Its temperature, $T\ ^{\circ}\text{C}$, after $t$ minutes follows Newton's Law of Cooling:
$$T=20+80e^{-kt}$$
where $k$ is a positive constant. After 4 minutes, the object's temperature is $70^{\circ}\text{C}$.
(a) Find the value of $k$, correct to 4 significant figures. [4]
(b) Find, correct to the nearest minute, the time at which the object's temperature first reaches $25^{\circ}\text{C}$. [3]
(c) Find the temperature of the object after 20 minutes, correct to 1 decimal place. [2]
(d) State the equation of the horizontal asymptote of the graph of $T$ against $t$ (for $t\ge0$), and explain its physical meaning. [1]
Show complete worked solution
(a) At $t=4$, $T=70$:
$$70=20+80e^{-4k}$$
$$50=80e^{-4k}$$
$$e^{-4k}=\dfrac{50}{80}=0.625$$
Taking natural logarithms:
$$-4k=\ln0.625$$
$$k=-\dfrac{1}{4}\ln0.625$$
$$\boxed{k\approx0.1175\text{ (4 s.f.)}}$$
(b) Require $T=25$:
$$25=20+80e^{-kt}$$
$$5=80e^{-kt}$$
$$e^{-kt}=\dfrac{5}{80}=0.0625$$
$$-kt=\ln0.0625$$
$$t=\dfrac{-\ln0.0625}{k}=\dfrac{\ln16}{k}$$
$$\ln16\approx2.772589$$
$$t\approx\dfrac{2.772589}{0.117501}\approx23.596$$
Rounding to the nearest minute,
$$\boxed{t\approx24\text{ minutes}}$$
(c)
$$T(20)=20+80e^{-20(0.117501)}=20+80e^{-2.350017}$$
$$e^{-2.350017}\approx0.095367$$
$$T\approx20+80(0.095367)\approx20+7.629\approx27.629$$
$$\boxed{T\approx27.6^{\circ}\text{C (1 d.p.)}}$$
(d) As $t\to\infty$, $e^{-kt}\to0$, so $T\to20$.
$$\boxed{T=20}$$
This is the temperature of the room: the object's temperature gets closer and closer to the ambient temperature but, according to the model, never falls below it.
QUESTION 49
13 marks
Hard
A city government issues a bond that will require a single repayment of \$200\,000 in 15 years' time. To prepare for this, the city sets up a sinking fund: it makes an equal deposit at the end of each year for 15 years into an account earning $4.5\%$ per annum, compounded annually, so that the fund reaches exactly \$200\,000 at the end of the 15 years.
(a) Find the required annual deposit, correct to 2 decimal places. [4]
(b) Find the total amount deposited into the fund over the 15 years, and hence find the total interest earned by the fund, both correct to 2 decimal places. [4]
(c) Interest rates fall, and the fund is only able to earn $3.5\%$ per annum instead. Find the new annual deposit that would be required for the fund to still reach \$200\,000 after 15 years, correct to 2 decimal places. [3]
(d) Find the percentage increase in the required annual deposit caused by the fall in interest rate, correct to 1 decimal place. [2]
Show complete worked solution
(a) Using the sinking-fund formula $D=\dfrac{FV\cdot i}{(1+i)^{n}-1}$ with $FV=200\,000$, $i=0.045$, $n=15$:
$$(1.045)^{15}\approx1.935282$$
$$D=\dfrac{200\,000(0.045)}{1.935282-1}=\dfrac{9000}{0.935282}$$
$$\boxed{D\approx\$9622.76\text{ (2 d.p.)}}$$
(b) Total deposited $=15D\approx15\times9622.761623\approx144\,341.42$.
Interest earned $=200\,000-144\,341.42=55\,658.58$.
$$\boxed{\text{Total deposited}\approx\$144\,341.42;\qquad\text{Interest earned}\approx\$55\,658.58}$$
(c) Using $i=0.035$:
$$(1.035)^{15}\approx1.675349$$
$$D_{\text{new}}=\dfrac{200\,000(0.035)}{1.675349-1}=\dfrac{7000}{0.675349}$$
$$\boxed{D_{\text{new}}\approx\$10\,365.01\text{ (2 d.p.)}}$$
(d) Percentage increase $=\dfrac{D_{\text{new}}-D}{D}\times100\%$:
$$\dfrac{10\,365.01-9622.76}{9622.76}\times100\%\approx\dfrac{742.25}{9622.76}\times100\%$$
$$\boxed{\approx7.7\%\text{ (1 d.p.)}}$$
QUESTION 50
15 marks
Hard
A delivery company buys a van for \$45\,000. The van depreciates by $20\%$ per year (reducing-balance). The company plans to replace it after 6 years with a new van expected to cost \$52\,000, using the trade-in (resale) value of the old van plus a replacement fund. The company deposits a fixed amount into the fund at the end of each year for 6 years, earning $3\%$ per annum, compounded annually.
(a) Find the trade-in (resale) value of the old van after 6 years, correct to 2 decimal places. [3]
(b) Hence find the fixed annual deposit into the replacement fund, correct to 2 decimal places, given that the fund balance after 6 years (shown in the table) plus the trade-in value must equal exactly \$52\,000. [4]
(c) Show that the fund balance immediately after the 3rd deposit is \$19\,211.07, confirming the table. [3]
(d) After 3 years, the company decides to increase the annual deposit by $10\%$ for the remaining 3 years (years 4, 5 and 6), while the trade-in value and target purchase price stay as before. Find the new fund balance after 6 years, and determine whether it is enough (together with the trade-in value) to cover the \$52\,000 replacement cost. [5]
| Year | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Fund balance (\$) | 6215.36 | 12\,617.19 | 19\,211.07 | 26\,002.76 | 32\,998.21 | 40\,203.52 |
Show complete worked solution
(a) Using $V=P_0(1-i)^{t}$ with $P_0=45\,000$, $i=0.20$, $t=6$:
$$V=45\,000(0.8)^{6}$$
$$(0.8)^{6}=0.262144$$
$$V=45\,000\times0.262144=11\,796.48$$
$$\boxed{\text{Trade-in value}=\$11\,796.48}$$
(b) The fund needs to make up the difference:
$$52\,000-11\,796.48=40\,203.52$$
Using the sinking-fund formula $D=\dfrac{FV\cdot i}{(1+i)^{n}-1}$ with $FV=40\,203.52$, $i=0.03$, $n=6$:
$$(1.03)^{6}\approx1.194052$$
$$D=\dfrac{40\,203.52(0.03)}{1.194052-1}=\dfrac{1206.1056}{0.194052}$$
$$\boxed{D\approx\$6215.36\text{ (2 d.p.)}}$$
matching the table.
(c) Iterating year by year, using the unrounded value of $D\approx6215.363701$:
$$\text{Year 1: }6215.36$$
$$\text{Year 2: }6215.36(1.03)+6215.36\approx6401.82+6215.36=12\,617.19$$
$$\text{Year 3: }12\,617.19(1.03)+6215.36\approx12\,995.70+6215.36=19\,211.07$$
$$\boxed{\$19\,211.07}$$ as required, confirming the table.
(d) After year 3, the balance is $\$19\,211.07$ (more precisely $19\,211.067664$). The new deposit is $10\%$ larger:
$$D_{\text{new}}=6215.363701\times1.10\approx6836.900071$$
Continuing to iterate for years 4, 5 and 6:
$$\text{Year 4: }19\,211.067664(1.03)+6836.900071\approx19\,787.40+6836.90=26\,624.30$$
$$\text{Year 5: }26\,624.30(1.03)+6836.900071\approx27\,423.03+6836.90=34\,259.93$$
$$\text{Year 6: }34\,259.93(1.03)+6836.900071\approx35\,287.73+6836.90=42\,124.63$$
The total funds available are the new fund balance plus the trade-in value:
$$42\,124.63+11\,796.48=53\,921.11$$
Since $53\,921.11>52\,000$,
$$\boxed{\text{Yes -- the fund is sufficient, with a surplus of approximately }\$1921.11}$$
Exponents & Logarithms 50 questions
QUESTION 1
3 marks
Easy
Without using a calculator, find the value of $(3.2\times10^{5})\times(4\times10^{-2})$, giving your answer in the form $a\times10^{k}$, where $1\le a<10$ and $k\in\mathbb{Z}$.
Show complete worked solution
Using the law $(a\times10^{m})\times(b\times10^{n})=(a\times b)\times10^{m+n}$, multiply the coefficients and the powers of 10 separately.
$$3.2\times4=12.8$$
$$10^{5}\times10^{-2}=10^{3}$$
So $(3.2\times10^{5})\times(4\times10^{-2})=12.8\times10^{3}$.
Since $12.8$ is not between 1 and 10, rewrite $12.8=1.28\times10^{1}$:
$$12.8\times10^{3}=1.28\times10^{1}\times10^{3}=1.28\times10^{4}$$
$$\boxed{1.28\times10^{4}}$$
QUESTION 2
6 marks
Medium
Let $a=2.5\times10^{8}$ and $b=5\times10^{-3}$. Without using a calculator,
(a) find $a\times b$, giving your answer in the form $c\times10^{k}$;
(b) find $\dfrac{a}{b}$, giving your answer in the form $c\times10^{k}$;
(c) find $a+b$, giving your answer correct to 3 significant figures in the form $c\times10^{k}$.
Show complete worked solution
(a) Using the law $(a\times10^{m})\times(b\times10^{n})=(a\times b)\times10^{m+n}$, multiply the coefficients and the powers of 10 separately:
$$a\times b=(2.5\times10^{8})\times(5\times10^{-3})$$
$$2.5\times5=12.5,\qquad 10^{8}\times10^{-3}=10^{5}$$
$$a\times b=12.5\times10^{5}=1.25\times10^{6}$$
$$\boxed{a\times b=1.25\times10^{6}}$$
(b) Using the law $\dfrac{a\times10^{m}}{b\times10^{n}}=\dfrac{a}{b}\times10^{m-n}$, divide the coefficients and the powers of 10 separately:
$$\dfrac{a}{b}=\dfrac{2.5\times10^{8}}{5\times10^{-3}}$$
$$2.5\div5=0.5,\qquad 10^{8}\div10^{-3}=10^{11}$$
$$\dfrac{a}{b}=0.5\times10^{11}=5\times10^{10}$$
$$\boxed{\dfrac{a}{b}=5\times10^{10}}$$
(c) Since $a$ and $b$ have very different orders of magnitude, write both numbers in ordinary decimal form with the same place value before adding:
$$a=2.5\times10^{8}=250\,000\,000,\qquad b=5\times10^{-3}=0.005$$
$$a+b=250\,000\,000.005$$
Since $b$ does not affect $a$ until far beyond the 3rd significant figure, this rounds to
$$\boxed{a+b\approx2.50\times10^{8}}$$ (3 s.f.)
QUESTION 3
4 marks
Easy
A single bacterium has mass $9.6\times10^{-13}\text{ g}$. A colony consists of $4.5\times10^{7}$ bacteria. Find the total mass of the colony in kilograms, giving your answer in the form $a\times10^{k}$, where $1\le a<10$ and $k\in\mathbb{Z}$.
Show complete worked solution
Using $\text{mass}=(\text{mass of one bacterium})\times(\text{number of bacteria})$, multiply:
$$(9.6\times10^{-13})\times(4.5\times10^{7})$$
Multiply the coefficients and powers of 10 separately:
$$9.6\times4.5=43.2,\qquad 10^{-13}\times10^{7}=10^{-6}$$
$$\text{Total mass}=43.2\times10^{-6}\text{ g}=4.32\times10^{-5}\text{ g}$$
Convert grams to kilograms by dividing by $10^{3}$:
$$4.32\times10^{-5}\div10^{3}=4.32\times10^{-8}\text{ kg}$$
$$\boxed{4.32\times10^{-8}\text{ kg}}$$
QUESTION 4
6 marks
Medium
The distance from Earth to a certain star is $4.014\times10^{16}\text{ m}$. Light travels at a speed of $3\times10^{8}\text{ m s}^{-1}$. Given that $1$ year $=3.156\times10^{7}$ seconds, find the time taken for light from the star to reach Earth, in years, correct to 3 significant figures.
Show complete worked solution
Using $\text{time}=\dfrac{\text{distance}}{\text{speed}}$:
$$t=\dfrac{4.014\times10^{16}}{3\times10^{8}}\text{ s}$$
Divide the coefficients and subtract the powers of 10:
$$t=\dfrac{4.014}{3}\times10^{16-8}=1.338\times10^{8}\text{ s}$$
Convert seconds to years by dividing by $3.156\times10^{7}$:
$$t=\dfrac{1.338\times10^{8}}{3.156\times10^{7}}=\dfrac{1.338}{0.3156}\text{ years}$$
$$t=4.2395\ldots\text{ years}$$
$$\boxed{t\approx4.24\text{ years (3 s.f.)}}$$
QUESTION 5
13 marks
Hard
A geologist is studying the sand on a beach.
(a) The mass of a single grain of sand is $m=2.6\times10^{-3}\text{ g}$ and the total number of grains of sand on the beach is estimated as $N=7.5\times10^{12}$. Find the total mass of sand on the beach, in kilograms, giving your answer in the form $a\times10^{k}$, where $1\le a<10$ and $k\in\mathbb{Z}$. [4]
(b) The volume occupied by the sand on the beach is estimated to be $V=1.2\times10^{4}\text{ m}^{3}$. Using your answer to part (a), find the density of the sand, $\rho=\dfrac{\text{mass}}{\text{volume}}$, in $\text{kg m}^{-3}$, correct to 3 significant figures. [4]
(c) The geologist wants to transport all the sand from part (a) using containers, each of which can hold $5.0\times10^{2}\text{ kg}$ of sand. Find the number of containers required to hold all the sand, giving your answer to the nearest whole number. [5]
Show complete worked solution
(a) Using $\text{mass}=m\times N$, multiply:
$$(2.6\times10^{-3})\times(7.5\times10^{12})$$
Multiply the coefficients and powers of 10 separately:
$$2.6\times7.5=19.5,\qquad 10^{-3}\times10^{12}=10^{9}$$
$$\text{Total mass}=19.5\times10^{9}\text{ g}=1.95\times10^{10}\text{ g}$$
Convert to kilograms (divide by $10^{3}$):
$$1.95\times10^{10}\div10^{3}=1.95\times10^{7}\text{ kg}$$
$$\boxed{\text{total mass}=1.95\times10^{7}\text{ kg}}$$
(b) Using $\rho=\dfrac{\text{mass}}{\text{volume}}$ with the result from (a):
$$\rho=\dfrac{1.95\times10^{7}}{1.2\times10^{4}}=\dfrac{1.95}{1.2}\times10^{7-4}$$
$$\rho=1.625\times10^{3}$$
Correct to 3 significant figures:
$$\boxed{\rho\approx1.63\times10^{3}\text{ kg m}^{-3}}$$
(c) Using $\text{number of containers}=\dfrac{\text{total mass}}{\text{capacity of one container}}$:
$$\dfrac{1.95\times10^{7}}{5.0\times10^{2}}=\dfrac{1.95}{5.0}\times10^{7-2}$$
$$=0.39\times10^{5}=3.9\times10^{4}=39\,000$$
Since this is already a whole number,
$$\boxed{39\,000\text{ containers}}$$
QUESTION 6
4 marks
Easy
Without using a calculator, solve for $x$: $3^{x+1}=81$.
Show complete worked solution
Using the strategy of expressing both sides as powers of the same base, write 81 as a power of 3:
$$81=3^{4}$$
So the equation becomes:
$$3^{x+1}=3^{4}$$
Since the bases are equal, the exponents must be equal:
$$x+1=4$$
$$\boxed{x=3}$$
QUESTION 7
6 marks
Medium
Without using a calculator, solve the equation $2^{2x}-5(2^{x})+4=0$ for $x$.
Show complete worked solution
Note that $2^{2x}=(2^{x})^{2}$. Using the substitution $y=2^{x}$ (where $y>0$), the equation becomes a quadratic in $y$:
$$y^{2}-5y+4=0$$
Factorising:
$$(y-1)(y-4)=0$$
$$y=1 \text{ or } y=4$$
Substituting back $y=2^x$:
Case 1: $2^{x}=1=2^{0} \implies x=0$
Case 2: $2^{x}=4=2^{2} \implies x=2$
$$\boxed{x=0\text{ or }x=2}$$
QUESTION 8
5 marks
Easy
Solve the equation $5^{x}=40$ for $x$, giving your answer correct to 3 significant figures.
Show complete worked solution
Taking logarithms of both sides:
$$\log(5^{x})=\log(40)$$
Using the power law of logarithms, $\log(5^{x})=x\log5$:
$$x\log5=\log40$$
Solving for $x$:
$$x=\dfrac{\log40}{\log5}$$
Substituting $\log40\approx1.60206$ and $\log5\approx0.69897$:
$$x=\dfrac{1.60206}{0.69897}\approx2.2920$$
$$\boxed{x\approx2.29\text{ (3 s.f.)}}$$
QUESTION 9
7 marks
Medium
Solve the equation $\log_2(x+3)+\log_2(x-3)=4$ for $x$, explaining why any extraneous solution must be rejected.
Show complete worked solution
Using the product law of logarithms, $\log_2A+\log_2B=\log_2(AB)$:
$$\log_2\big[(x+3)(x-3)\big]=4$$
Converting from logarithmic to exponential form:
$$(x+3)(x-3)=2^{4}=16$$
Expanding using difference of squares:
$$x^{2}-9=16$$
$$x^{2}=25$$
$$x=\pm5$$
Check the domain: both $\log_2(x+3)$ and $\log_2(x-3)$ require positive arguments, so $x+3>0$ and $x-3>0$, i.e. $x>3$.
For $x=-5$: $x-3=-8<0$, so $\log_2(x-3)$ is undefined, and this solution is rejected.
For $x=5$: $x+3=8>0$ and $x-3=2>0$, both valid.
$$\boxed{x=5}$$
QUESTION 10
16 marks
Hard
The sound intensity level $L$, measured in decibels (dB), of a sound with intensity $I$ (in $\text{W m}^{-2}$) is defined by
$$L=10\log_{10}\left(\dfrac{I}{I_0}\right),$$
where $I_0=10^{-12}\text{ W m}^{-2}$ is a reference intensity.
(a) A sound has intensity $I=2.5\times10^{-6}\text{ W m}^{-2}$. Find its intensity level $L$, correct to 1 decimal place. [3]
(b) A second sound has an intensity level of 85 dB. Find its intensity $I$, giving your answer in the form $a\times10^{k}\text{ W m}^{-2}$. [4]
(c) Show that if the intensity of a sound is doubled, its intensity level increases by exactly $10\log_{10}2$ dB, regardless of the original intensity. [4]
(d) Two sounds, with intensity levels $L_1=70$ dB and $L_2=73$ dB respectively, occur together; their intensities add. Find the combined intensity level $L_3$, correct to 1 decimal place. [5]
Show complete worked solution
(a) Using $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$ with $I=2.5\times10^{-6}$, $I_0=10^{-12}$:
$$L=10\log_{10}\left(\dfrac{2.5\times10^{-6}}{10^{-12}}\right)=10\log_{10}(2.5\times10^{6})$$
Using the product law, $\log_{10}(2.5\times10^{6})=\log_{10}2.5+6$, and $\log_{10}2.5\approx0.39794$:
$$L=10(0.39794+6)=10(6.39794)=63.9794$$
$$\boxed{L\approx64.0\text{ dB (1 d.p.)}}$$
(b) Substituting $L=85$ into $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$:
$$85=10\log_{10}\left(\dfrac{I}{10^{-12}}\right)$$
$$\log_{10}\left(\dfrac{I}{10^{-12}}\right)=8.5$$
Converting to exponential form:
$$\dfrac{I}{10^{-12}}=10^{8.5}$$
$$I=10^{8.5}\times10^{-12}=10^{-3.5}$$
Writing this in the form $a\times10^k$: $10^{-3.5}=10^{-4}\times10^{0.5}\approx10^{-4}\times3.16228$:
$$\boxed{I\approx3.16\times10^{-4}\text{ W m}^{-2}}$$
(c) Let a sound have intensity $I$ and level $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$. Doubling the intensity gives a new level:
$$L'=10\log_{10}\left(\dfrac{2I}{I_0}\right)$$
Using the product law of logarithms:
$$L'=10\log_{10}\left(2\cdot\dfrac{I}{I_0}\right)=10\left[\log_{10}2+\log_{10}\left(\dfrac{I}{I_0}\right)\right]$$
$$L'=10\log_{10}2+10\log_{10}\left(\dfrac{I}{I_0}\right)=L+10\log_{10}2$$
So $L'-L=10\log_{10}2$, a constant independent of $I$. $\blacksquare$
(d) First find the individual intensities using $I=I_0\cdot10^{L/10}$:
$$I_1=10^{-12}\times10^{70/10}=10^{-12}\times10^{7}=10^{-5}\text{ W m}^{-2}$$
$$I_2=10^{-12}\times10^{73/10}=10^{-12}\times10^{7.3}$$
Using $10^{7.3}=10^{7}\times10^{0.3}\approx10^{7}\times1.99526$:
$$I_2\approx1.99526\times10^{-5}\text{ W m}^{-2}$$
Since intensities add:
$$I_3=I_1+I_2\approx1\times10^{-5}+1.99526\times10^{-5}=2.99526\times10^{-5}\text{ W m}^{-2}$$
Now find $L_3$ using $L=10\log_{10}(I/I_0)$:
$$L_3=10\log_{10}\left(\dfrac{2.99526\times10^{-5}}{10^{-12}}\right)=10\log_{10}(2.99526\times10^{7})$$
$$=10\big[\log_{10}(2.99526)+7\big]$$
Using $\log_{10}(2.99526)\approx0.47643$:
$$L_3=10(0.47643+7)=10(7.47643)=74.7643$$
$$\boxed{L_3\approx74.8\text{ dB (1 d.p.)}}$$
QUESTION 11
4 marks
Easy
Simplify fully, giving your answer in the form $2^{k}$ where $k \in \mathbb{Q}$: $$\dfrac{8^{2/3} \times 2^{-1}}{4^{1/2}}$$
Show complete worked solution
Write every term as a power of 2, using the law $(a^m)^n=a^{mn}$:
$$8^{2/3}=(2^3)^{2/3}=2^{2},\qquad 4^{1/2}=(2^2)^{1/2}=2^{1}$$
So the expression becomes
$$\dfrac{2^{2}\times2^{-1}}{2^{1}}$$
Applying $a^m\times a^n=a^{m+n}$ to the numerator:
$$2^{2}\times2^{-1}=2^{2-1}=2^{1}$$
Applying $a^m\div a^n=a^{m-n}$:
$$\dfrac{2^{1}}{2^{1}}=2^{1-1}=2^{0}$$
$$\boxed{k=0}$$ (the expression simplifies to $1$).
QUESTION 12
5 marks
Medium
Solve the equation $\log_2(x+3) + \log_2(x-1) = 5$, where $x > 1$.
Show complete worked solution
Using the law $\log_a m+\log_a n=\log_a(mn)$, combine the logarithms:
$$\log_2\big[(x+3)(x-1)\big]=5$$
Convert from logarithmic to exponential form ($\log_2 y=5 \Rightarrow y=2^5$):
$$(x+3)(x-1)=32$$
Expand:
$$x^2+2x-3=32$$
$$x^2+2x-35=0$$
Factorise:
$$(x+7)(x-5)=0$$
$$x=-7 \text{ or } x=5$$
Since the domain requires $x>1$ (so that both $x+3>0$ and $x-1>0$), reject $x=-7$.
$$\boxed{x=5}$$
QUESTION 13
6 marks
Medium
Solve the equation $3^{x+1} = 5^{2x}$, giving your answer correct to 3 significant figures.
Show complete worked solution
Take the natural logarithm of both sides:
$$\ln\left(3^{x+1}\right)=\ln\left(5^{2x}\right)$$
Using the law $\ln(a^m)=m\ln a$:
$$(x+1)\ln3=2x\ln5$$
Expand the left side:
$$x\ln3+\ln3=2x\ln5$$
Collect the $x$ terms on one side:
$$\ln3=2x\ln5-x\ln3=x(2\ln5-\ln3)$$
Solve for $x$:
$$x=\dfrac{\ln3}{2\ln5-\ln3}$$
Substituting $\ln3=1.098612\ldots$, $\ln5=1.609438\ldots$, so $2\ln5-\ln3=3.218876\ldots-1.098612\ldots=2.120264\ldots$:
$$x=\dfrac{1.098612\ldots}{2.120264\ldots}=0.51817\ldots$$
$$\boxed{x=0.518\text{ (3 s.f.)}}$$
QUESTION 14
4 marks
Easy
Given that $\log_a 2 = p$ and $\log_a 3 = q$, express $\log_a 72$ in terms of $p$ and $q$.
Show complete worked solution
Write $72$ as a product of powers of $2$ and $3$:
$$72=8\times9=2^3\times3^2$$
Apply the law $\log_a(mn)=\log_a m+\log_a n$:
$$\log_a 72=\log_a(2^3)+\log_a(3^2)$$
Apply the law $\log_a(m^k)=k\log_a m$:
$$\log_a 72=3\log_a 2+2\log_a 3$$
Substitute $\log_a 2=p$ and $\log_a 3=q$:
$$\boxed{\log_a 72=3p+2q}$$
QUESTION 15
15 marks
Hard
This question explores equations involving exponents and logarithms.
(a) [5 marks] By using the substitution $y = 3^x$, solve the equation $9^{x} - 10(3^{x}) + 9 = 0$ for $x$.
(b) [6 marks] By using the substitution $u = \log_2 x$, solve the equation $(\log_2 x)^2 - \log_2(x^4) - 5 = 0$ for $x > 0$, giving your answers in exact form.
(c) [4 marks] Hence, using the change of base formula, find the exact value of $\log_x 2$ for each of the solutions found in part (b).
Show complete worked solution
(a) Note $9^x=(3^x)^2$, so with $y=3^x$ the equation becomes:
$$y^2-10y+9=0$$
Factorising:
$$(y-1)(y-9)=0 \Rightarrow y=1 \text{ or } y=9$$
Since $y=3^x$:
$$3^x=1 \Rightarrow x=0, \qquad 3^x=9=3^2 \Rightarrow x=2$$
$$\boxed{x=0 \text{ or } x=2}$$
(b) Using the law $\log_2(x^4)=4\log_2 x$, and substituting $u=\log_2 x$:
$$u^2-4u-5=0$$
Factorising:
$$(u-5)(u+1)=0 \Rightarrow u=5 \text{ or } u=-1$$
Since $u=\log_2 x$:
$$\log_2 x=5 \Rightarrow x=2^5=32, \qquad \log_2 x=-1 \Rightarrow x=2^{-1}=\dfrac{1}{2}$$
$$\boxed{x=32 \text{ or } x=\dfrac{1}{2}}$$
(c) By the change of base formula, $\log_x 2=\dfrac{\log_2 2}{\log_2 x}=\dfrac{1}{\log_2 x}=\dfrac{1}{u}$.
For $x=32$ (where $u=5$):
$$\log_{32}2=\dfrac{1}{5}$$
For $x=\dfrac{1}{2}$ (where $u=-1$):
$$\log_{1/2}2=\dfrac{1}{-1}=-1$$
$$\boxed{\log_{32}2=\dfrac{1}{5}, \quad \log_{1/2}2=-1}$$
QUESTION 16
4 marks
Easy
Simplify fully, giving your answer in the form $a\sqrt{3}$: $$\sqrt{75}+\sqrt{12}-\sqrt{27}$$
Show complete worked solution
Write each surd in terms of $\sqrt3$ by extracting the largest square factor from each number under the root sign.
$$\sqrt{75}=\sqrt{25\times3}=5\sqrt3$$
$$\sqrt{12}=\sqrt{4\times3}=2\sqrt3$$
$$\sqrt{27}=\sqrt{9\times3}=3\sqrt3$$
Substituting these into the original expression:
$$5\sqrt3+2\sqrt3-3\sqrt3=(5+2-3)\sqrt3$$
$$\boxed{4\sqrt3}$$
QUESTION 17
4 marks
Easy
Without using a calculator, solve for $x$: $5^{2x-1}=125$.
Show complete worked solution
Using the strategy of expressing both sides as powers of the same base, write 125 as a power of 5:
$$125=5^{3}$$
So the equation becomes:
$$5^{2x-1}=5^{3}$$
Since the bases are equal, the exponents must be equal:
$$2x-1=3$$
$$2x=4$$
$$\boxed{x=2}$$
QUESTION 18
5 marks
Easy
Simplify $\left(\dfrac{a^{3}b^{-2}}{a^{-1}b^{4}}\right)^{2}$, giving your answer in the form $a^{m}b^{n}$.
Show complete worked solution
First simplify inside the brackets using the law $\dfrac{a^{m}}{a^{n}}=a^{m-n}$, applied separately to the powers of $a$ and the powers of $b$:
$$\dfrac{a^{3}b^{-2}}{a^{-1}b^{4}}=a^{3-(-1)}b^{-2-4}=a^{4}b^{-6}$$
Now apply the outer power of 2 using the law $(a^{m}b^{n})^{k}=a^{mk}b^{nk}$:
$$\left(a^{4}b^{-6}\right)^{2}=a^{4\times2}b^{-6\times2}=a^{8}b^{-12}$$
$$\boxed{a^{8}b^{-12}}$$
QUESTION 19
4 marks
Easy
Without using a calculator, evaluate $\log_3 81+\log_2\dfrac{1}{8}$.
Show complete worked solution
Evaluate each logarithm separately by asking what power of the base gives the argument.
Since $3^{4}=81$:
$$\log_3 81=4$$
Since $2^{-3}=\dfrac{1}{8}$:
$$\log_2\dfrac{1}{8}=-3$$
Adding the two results:
$$4+(-3)=1$$
$$\boxed{1}$$
QUESTION 20
5 marks
Easy
Solve the equation $\log_2(3x-1)=5$ for $x$.
Show complete worked solution
Convert the equation from logarithmic to exponential form, using the definition $\log_a y=k \iff y=a^{k}$:
$$3x-1=2^{5}$$
$$3x-1=32$$
$$3x=33$$
$$x=11$$
Check the domain: the argument of the logarithm requires $3x-1>0$, i.e. $x>\dfrac13$. Since $x=11$ satisfies this, the solution is valid.
$$\boxed{x=11}$$
QUESTION 21
5 marks
Easy
Simplify $\dfrac{2^{n+3}-2^{n+1}}{2^{n}}$ to a single integer, where $n\in\mathbb{Z}$.
Show complete worked solution
Factorise $2^{n}$ out of both terms in the numerator, using the law $a^{m+n}=a^{m}\times a^{n}$:
$$2^{n+3}-2^{n+1}=2^{n}\times2^{3}-2^{n}\times2^{1}=2^{n}\left(2^{3}-2^{1}\right)$$
$$=2^{n}(8-2)=2^{n}\times6$$
Substituting this back into the original expression:
$$\dfrac{2^{n}\times6}{2^{n}}=6$$
Since $2^n\neq0$ for all $n\in\mathbb{Z}$, this cancellation is valid for every $n$.
$$\boxed{6}$$
QUESTION 22
5 marks
Easy
Without using a calculator, solve for $x$: $4^{x}=8^{x-1}$.
Show complete worked solution
Using the strategy of expressing both sides as powers of the same base, write 4 and 8 as powers of 2:
$$4^{x}=(2^{2})^{x}=2^{2x}$$
$$8^{x-1}=(2^{3})^{x-1}=2^{3x-3}$$
So the equation becomes:
$$2^{2x}=2^{3x-3}$$
Since the bases are equal, the exponents must be equal:
$$2x=3x-3$$
$$3=x$$
$$\boxed{x=3}$$
QUESTION 23
4 marks
Easy
Write $2\ln x+\ln y-3\ln z$ as a single logarithm, where $x,y,z>0$.
Show complete worked solution
Using the power law $k\ln m=\ln(m^{k})$, rewrite each coefficient as an exponent:
$$2\ln x=\ln(x^{2}),\qquad 3\ln z=\ln(z^{3})$$
So the expression becomes:
$$\ln(x^{2})+\ln y-\ln(z^{3})$$
Using the product law $\ln m+\ln n=\ln(mn)$ on the first two terms, then the quotient law $\ln m-\ln n=\ln\left(\dfrac{m}{n}\right)$:
$$\ln\left(x^{2}y\right)-\ln\left(z^{3}\right)=\ln\left(\dfrac{x^{2}y}{z^{3}}\right)$$
$$\boxed{\ln\left(\dfrac{x^{2}y}{z^{3}}\right)}$$
QUESTION 24
4 marks
Easy
Without using a calculator, solve for $x$: $9^{x}=\dfrac{1}{3}$.
Show complete worked solution
Using the strategy of expressing both sides as powers of the same base, write 9 and $\dfrac13$ as powers of 3:
$$9^{x}=(3^{2})^{x}=3^{2x}$$
$$\dfrac{1}{3}=3^{-1}$$
So the equation becomes:
$$3^{2x}=3^{-1}$$
Since the bases are equal, the exponents must be equal:
$$2x=-1$$
$$\boxed{x=-\dfrac{1}{2}}$$
QUESTION 25
4 marks
Easy
Given that $\log_{10}2\approx0.3010$ and $\log_{10}3\approx0.4771$, find $\log_{10}12$ without using a calculator.
Show complete worked solution
Write 12 as a product of powers of 2 and 3:
$$12=4\times3=2^{2}\times3$$
Apply the law $\log_{10}(mn)=\log_{10}m+\log_{10}n$:
$$\log_{10}12=\log_{10}(2^{2})+\log_{10}3$$
Apply the law $\log_{10}(m^{k})=k\log_{10}m$ to the first term:
$$\log_{10}12=2\log_{10}2+\log_{10}3$$
Substituting the given values:
$$\log_{10}12\approx2(0.3010)+0.4771=0.6020+0.4771$$
$$\boxed{\log_{10}12\approx1.0791}$$
QUESTION 26
4 marks
Easy
Simplify $\sqrt[3]{27x^{6}y^{-3}}$ fully, giving your answer in the form $ax^{m}y^{n}$.
Show complete worked solution
Write the cube root as a power of $\dfrac13$, then apply the law $(a^{m})^{n}=a^{mn}$ to each factor inside:
$$\sqrt[3]{27x^{6}y^{-3}}=\left(27x^{6}y^{-3}\right)^{1/3}=27^{1/3}\times x^{6\times\frac13}\times y^{-3\times\frac13}$$
Since $27^{1/3}=3$ (as $3^{3}=27$):
$$=3\times x^{2}\times y^{-1}$$
$$\boxed{3x^{2}y^{-1}}$$
QUESTION 27
5 marks
Easy
Solve the equation $\log_5(2x+3)=2$ for $x$.
Show complete worked solution
Convert the equation from logarithmic to exponential form, using the definition $\log_a y=k \iff y=a^{k}$:
$$2x+3=5^{2}$$
$$2x+3=25$$
$$2x=22$$
$$x=11$$
Check the domain: the argument requires $2x+3>0$, i.e. $x>-\dfrac32$. Since $x=11$ satisfies this, the solution is valid.
$$\boxed{x=11}$$
QUESTION 28
5 marks
Easy
Rationalise the denominator and simplify fully: $\dfrac{6}{\sqrt5-\sqrt2}$.
Show complete worked solution
Multiply the numerator and denominator by the conjugate of the denominator, $\sqrt5+\sqrt2$, using the difference-of-squares identity $(a-b)(a+b)=a^{2}-b^{2}$ to remove the surds from the denominator:
$$\dfrac{6}{\sqrt5-\sqrt2}\times\dfrac{\sqrt5+\sqrt2}{\sqrt5+\sqrt2}=\dfrac{6(\sqrt5+\sqrt2)}{(\sqrt5)^{2}-(\sqrt2)^{2}}$$
$$=\dfrac{6(\sqrt5+\sqrt2)}{5-2}=\dfrac{6(\sqrt5+\sqrt2)}{3}$$
Dividing every term by 3:
$$\boxed{2\left(\sqrt5+\sqrt2\right)}$$
QUESTION 29
5 marks
Easy
Without using a calculator, evaluate $8^{2/3}+9^{-1/2}$, giving your answer as an exact fraction.
Show complete worked solution
Evaluate each term separately. For $8^{2/3}$, use $a^{m/n}=\left(\sqrt[n]{a}\right)^{m}$:
$$8^{2/3}=\left(\sqrt[3]{8}\right)^{2}=2^{2}=4$$
For $9^{-1/2}$, use $a^{-k}=\dfrac{1}{a^{k}}$ together with $a^{1/2}=\sqrt{a}$:
$$9^{-1/2}=\dfrac{1}{9^{1/2}}=\dfrac{1}{\sqrt9}=\dfrac{1}{3}$$
Adding the two results, using a common denominator of 3:
$$4+\dfrac13=\dfrac{12}{3}+\dfrac13=\dfrac{13}{3}$$
$$\boxed{\dfrac{13}{3}}$$
QUESTION 30
7 marks
Medium
Solve the equation $\log_2 x+\log_4 x=3$ for $x>0$.
Show complete worked solution
The two logarithms have different bases, so first use the change of base formula $\log_4 x=\dfrac{\log_2 x}{\log_2 4}$ to rewrite $\log_4 x$ in base 2:
$$\log_4 x=\dfrac{\log_2 x}{2}$$
Substituting into the original equation:
$$\log_2 x+\dfrac{\log_2 x}{2}=3$$
Combining the like terms on the left:
$$\dfrac{3}{2}\log_2 x=3$$
$$\log_2 x=2$$
Converting to exponential form:
$$x=2^{2}=4$$
Check: $\log_2 4+\log_4 4=2+1=3$, as required, and $x=4>0$ satisfies the domain.
$$\boxed{x=4}$$
QUESTION 31
7 marks
Medium
Solve the simultaneous equations for $x$ and $y$: $$2^{x+y}=32,\qquad 2^{x-y}=2$$
Show complete worked solution
Write the right-hand side of each equation as a power of 2, so that both sides of each equation share the same base.
$$2^{x+y}=2^{5}\ \implies\ x+y=5$$
$$2^{x-y}=2^{1}\ \implies\ x-y=1$$
This gives a linear system in $x$ and $y$. Adding the two equations eliminates $y$:
$$(x+y)+(x-y)=5+1$$
$$2x=6\ \implies\ x=3$$
Substituting $x=3$ into $x+y=5$:
$$3+y=5\ \implies\ y=2$$
Check: $2^{3+2}=2^{5}=32$ and $2^{3-2}=2^{1}=2$, both correct.
$$\boxed{x=3,\ y=2}$$
QUESTION 32
7 marks
Medium
Solve the equation $\log_3(x+5)-\log_3(x-1)=2$ for $x$.
Show complete worked solution
Using the quotient law $\log_a m-\log_a n=\log_a\left(\dfrac{m}{n}\right)$, combine the logarithms:
$$\log_3\left(\dfrac{x+5}{x-1}\right)=2$$
Convert to exponential form:
$$\dfrac{x+5}{x-1}=3^{2}=9$$
Multiply both sides by $(x-1)$:
$$x+5=9(x-1)=9x-9$$
$$5+9=9x-x$$
$$14=8x$$
$$x=\dfrac{14}{8}=\dfrac{7}{4}$$
Check the domain: $\log_3(x+5)$ and $\log_3(x-1)$ require $x>-5$ and $x>1$ respectively, so overall $x>1$. Since $\dfrac74=1.75>1$, the solution is valid.
$$\boxed{x=\dfrac{7}{4}}$$
QUESTION 33
8 marks
Medium
Solve the equation $2^{2x+1}-9(2^{x})+4=0$ for $x$.
Show complete worked solution
Note that $2^{2x+1}=2\times2^{2x}=2\times(2^{x})^{2}$. Using the substitution $y=2^{x}$ (where $y>0$, since an exponential is always positive), the equation becomes a quadratic in $y$:
$$2y^{2}-9y+4=0$$
Factorising (looking for two numbers multiplying to $2\times4=8$ and adding to $-9$, namely $-1$ and $-8$):
$$2y^{2}-y-8y+4=0$$
$$y(2y-1)-4(2y-1)=0$$
$$(2y-1)(y-4)=0$$
$$y=\dfrac12\ \text{ or }\ y=4$$
Both values satisfy $y>0$, so both are valid. Substituting back $y=2^{x}$:
Case 1: $2^{x}=\dfrac12=2^{-1}\implies x=-1$
Case 2: $2^{x}=4=2^{2}\implies x=2$
$$\boxed{x=-1\text{ or }x=2}$$
QUESTION 34
8 marks
Medium
Solve the simultaneous equations for $x$ and $y$, where $x,y>0$: $$\log_2 x+\log_2 y=5,\qquad x-y=4$$
Show complete worked solution
Using the product law, rewrite the first equation without logarithms:
$$\log_2 x+\log_2 y=\log_2(xy)=5\ \implies\ xy=2^{5}=32$$
From the second equation, $x=y+4$. Substitute this into $xy=32$:
$$(y+4)y=32$$
$$y^{2}+4y-32=0$$
Factorising (two numbers multiplying to $-32$ and adding to 4, namely 8 and $-4$):
$$(y+8)(y-4)=0$$
$$y=-8\ \text{or}\ y=4$$
Since the domain requires $y>0$ (for $\log_2 y$ to be defined), reject $y=-8$. So $y=4$, and:
$$x=y+4=8$$
Check: $\log_2 8+\log_2 4=3+2=5$, and $8-4=4$, both correct; also $x=8>0$.
$$\boxed{x=8,\ y=4}$$
QUESTION 35
7 marks
Medium
By using the substitution $y=e^{x}$, solve the equation $e^{2x}-4e^{x}-5=0$ for $x$, giving your answer in exact form.
Show complete worked solution
Note that $e^{2x}=(e^{x})^{2}$. With the substitution $y=e^{x}$ (where $y>0$, since $e^x$ is always positive), the equation becomes a quadratic in $y$:
$$y^{2}-4y-5=0$$
Factorising:
$$(y-5)(y+1)=0$$
$$y=5\ \text{or}\ y=-1$$
Since $y=e^{x}>0$ for all real $x$, the value $y=-1$ is impossible and must be rejected.
Substituting back $y=e^{x}$ for the remaining case:
$$e^{x}=5$$
Taking the natural logarithm of both sides:
$$\boxed{x=\ln5}$$
QUESTION 36
6 marks
Medium
Solve the equation $\log_x 81=4$ for $x$, where $x$ is the base of the logarithm.
Show complete worked solution
Convert the equation from logarithmic to exponential form, using the definition $\log_a y=k \iff y=a^{k}$, with the base $x$ as the unknown:
$$x^{4}=81$$
Take the fourth root of both sides:
$$x=\pm81^{1/4}=\pm3$$
Since $x$ is the base of a logarithm, it must satisfy $x>0$ and $x\neq1$. This rules out $x=-3$.
Check: $\log_3 81=4$ since $3^{4}=81$, as required.
$$\boxed{x=3}$$
QUESTION 37
6 marks
Medium
Express $\log_a\left(\dfrac{a^{2}\sqrt{b}}{c^{3}}\right)$ in terms of $\log_a b$ and $\log_a c$, given that $a,b,c>0$ and $a\neq1$.
Show complete worked solution
Using the quotient law $\log_a\left(\dfrac{m}{n}\right)=\log_a m-\log_a n$:
$$\log_a\left(\dfrac{a^{2}\sqrt{b}}{c^{3}}\right)=\log_a\left(a^{2}\sqrt{b}\right)-\log_a\left(c^{3}\right)$$
Using the product law $\log_a(mn)=\log_a m+\log_a n$ on the first term, and writing $\sqrt{b}=b^{1/2}$:
$$\log_a\left(a^{2}\sqrt{b}\right)=\log_a\left(a^{2}\right)+\log_a\left(b^{1/2}\right)$$
Using the power law $\log_a\left(m^{k}\right)=k\log_a m$ throughout, and $\log_a a=1$:
$$\log_a\left(a^{2}\right)=2\log_a a=2,\qquad \log_a\left(b^{1/2}\right)=\dfrac12\log_a b,\qquad \log_a\left(c^{3}\right)=3\log_a c$$
Combining all the pieces:
$$\boxed{\log_a\left(\dfrac{a^{2}\sqrt{b}}{c^{3}}\right)=2+\dfrac12\log_a b-3\log_a c}$$
QUESTION 38
9 marks
Medium
Solve the simultaneous equations for $x$ and $y$: $$3^{x}\times9^{y}=27,\qquad 2^{x}=4^{y+1}$$
Show complete worked solution
Rewrite every term as a power of a common base within each equation. In the first equation, write $9=3^{2}$ and $27=3^{3}$:
$$3^{x}\times3^{2y}=3^{3}$$
$$3^{x+2y}=3^{3}\implies x+2y=3\qquad(1)$$
In the second equation, write $4=2^{2}$:
$$2^{x}=2^{2(y+1)}$$
$$x=2y+2\qquad(2)$$
Substitute (2) into (1):
$$(2y+2)+2y=3$$
$$4y+2=3$$
$$4y=1\implies y=\dfrac14$$
Substitute back into (2):
$$x=2\left(\dfrac14\right)+2=\dfrac12+2=\dfrac52$$
Check in the original equations: $3^{5/2}\times9^{1/4}=3^{5/2}\times3^{1/2}=3^{3}=27$, and $2^{5/2}=4^{5/4}=2^{5/2}$, both correct.
$$\boxed{x=\dfrac52,\ y=\dfrac14}$$
QUESTION 39
8 marks
Medium
Solve the equation $\log_2(x+4)+\log_2(x-2)=5$ for $x$, giving your answer in exact (surd) form.
Show complete worked solution
Using the product law $\log_a m+\log_a n=\log_a(mn)$, combine the logarithms:
$$\log_2\big[(x+4)(x-2)\big]=5$$
Convert to exponential form:
$$(x+4)(x-2)=2^{5}=32$$
Expand the left-hand side:
$$x^{2}+2x-8=32$$
$$x^{2}+2x-40=0$$
This does not factorise over the integers, so apply the quadratic formula with $a=1,\ b=2,\ c=-40$:
$$x=\dfrac{-2\pm\sqrt{2^{2}-4(1)(-40)}}{2(1)}=\dfrac{-2\pm\sqrt{4+160}}{2}=\dfrac{-2\pm\sqrt{164}}{2}$$
Simplify $\sqrt{164}=\sqrt{4\times41}=2\sqrt{41}$:
$$x=\dfrac{-2\pm2\sqrt{41}}{2}=-1\pm\sqrt{41}$$
Check the domain: $\log_2(x+4)$ and $\log_2(x-2)$ require $x>-4$ and $x>2$ respectively, so overall $x>2$. Since $\sqrt{41}\approx6.40$, the value $-1-\sqrt{41}\approx-7.40$ is negative and must be rejected, while $-1+\sqrt{41}\approx5.40>2$ is valid.
$$\boxed{x=-1+\sqrt{41}}$$
QUESTION 40
6 marks
Medium
Without using a calculator, solve for $x$: $3^{x+2}-3^{x}=24$.
Show complete worked solution
Factorise $3^{x}$ out of both terms on the left-hand side, using the law $a^{m+n}=a^{m}\times a^{n}$:
$$3^{x}\times3^{2}-3^{x}=3^{x}\left(3^{2}-1\right)$$
$$=3^{x}(9-1)=8\times3^{x}$$
So the equation becomes:
$$8\times3^{x}=24$$
$$3^{x}=3$$
Since $3=3^{1}$, and the bases are equal, the exponents must be equal:
$$\boxed{x=1}$$
QUESTION 41
7 marks
Medium
Solve the equation $\log_2(x+6)=1+\log_2(x-2)$ for $x$.
Show complete worked solution
Rearrange so that both logarithmic terms are on the same side:
$$\log_2(x+6)-\log_2(x-2)=1$$
Using the quotient law $\log_a m-\log_a n=\log_a\left(\dfrac{m}{n}\right)$:
$$\log_2\left(\dfrac{x+6}{x-2}\right)=1$$
Convert to exponential form:
$$\dfrac{x+6}{x-2}=2^{1}=2$$
Multiply both sides by $(x-2)$:
$$x+6=2(x-2)=2x-4$$
$$6+4=2x-x$$
$$x=10$$
Check the domain: $\log_2(x+6)$ and $\log_2(x-2)$ require $x>-6$ and $x>2$ respectively, so overall $x>2$. Since $x=10>2$, the solution is valid. As a further check, $\dfrac{16}{8}=2$, as required.
$$\boxed{x=10}$$
QUESTION 42
6 marks
Medium
Given that $2^{x}=5$ and $2^{y}=3$, find the exact value of $2^{2x-y}$, without finding $x$ or $y$ individually.
Show complete worked solution
Using the laws of exponents, rewrite $2^{2x-y}$ in terms of $2^{x}$ and $2^{y}$:
$$2^{2x-y}=\dfrac{2^{2x}}{2^{y}}=\dfrac{\left(2^{x}\right)^{2}}{2^{y}}$$
Substituting the given values $2^{x}=5$ and $2^{y}=3$:
$$2^{2x-y}=\dfrac{5^{2}}{3}=\dfrac{25}{3}$$
$$\boxed{2^{2x-y}=\dfrac{25}{3}}$$
QUESTION 43
7 marks
Medium
Solve the simultaneous equations for $x$ and $y$, where $x,y>0$: $$\log_2 x-\log_2 y=3,\qquad x+y=18$$
Show complete worked solution
Using the quotient law, rewrite the first equation without logarithms:
$$\log_2 x-\log_2 y=\log_2\left(\dfrac{x}{y}\right)=3\ \implies\ \dfrac{x}{y}=2^{3}=8$$
So $x=8y$. Substitute this into the second equation:
$$8y+y=18$$
$$9y=18$$
$$y=2$$
Then:
$$x=8(2)=16$$
Check: $\log_2 16-\log_2 2=4-1=3$, and $16+2=18$, both correct; also $x,y>0$.
$$\boxed{x=16,\ y=2}$$
QUESTION 44
12 marks
Hard
Consider the simultaneous equations $$\log_2(x+y)=4,\qquad \log_2 x-\log_2 y=\log_2 3,$$ where $x,y>0$.
(a) Solve the simultaneous equations to find the exact values of $x$ and $y$. [5]
(b) Hence evaluate $\log_2(xy)$ exactly, giving your answer in the form $a+\log_2 3$, where $a\in\mathbb{Z}$. [3]
(c) Using the change of base formula, express $\log_x y$ exactly in the form $\dfrac{a}{b+\log_2 3}$, where $a,b\in\mathbb{Z}$. [4]
Show complete worked solution
(a) Convert the first equation to exponential form:
$$x+y=2^{4}=16$$
For the second equation, use the quotient law $\log_2 x-\log_2 y=\log_2\left(\dfrac{x}{y}\right)$:
$$\log_2\left(\dfrac{x}{y}\right)=\log_2 3\ \implies\ \dfrac{x}{y}=3\ \implies\ x=3y$$
Substitute $x=3y$ into $x+y=16$:
$$3y+y=16\implies4y=16\implies y=4$$
$$x=3(4)=12$$
Check: $\log_2(12+4)=\log_2 16=4$, and $\log_2 12-\log_2 4=\log_2\left(\dfrac{12}{4}\right)=\log_2 3$, both correct; also $x,y>0$.
$$\boxed{x=12,\ y=4}$$
(b) Using the values found in (a):
$$\log_2(xy)=\log_2(12\times4)=\log_2 48$$
Write $48=16\times3=2^{4}\times3$ and apply the product law:
$$\log_2 48=\log_2\left(2^{4}\right)+\log_2 3$$
$$\boxed{\log_2(xy)=4+\log_2 3}$$
(c) Using the change of base formula $\log_x y=\dfrac{\log_2 y}{\log_2 x}$:
$$\log_x y=\dfrac{\log_2 4}{\log_2 12}$$
Since $\log_2 4=2$, and $\log_2 12=\log_2(4\times3)=2+\log_2 3$ (by the product law):
$$\boxed{\log_x y=\dfrac{2}{2+\log_2 3}}$$
QUESTION 45
15 marks
Hard
This question explores several different types of equations involving exponents and logarithms.
(a) By using the substitution $y=5^{x}$, solve the equation $25^{x}-6(5^{x})+5=0$ for $x$. [5]
(b) Solve the equation $\log_2 x+\log_2(x+2)-\log_2 3=3$ for $x$, stating clearly why any extraneous solution must be rejected. [5]
(c) By using the substitution $t=2^{x}$, solve the equation $4^{x+1}=3(2^{x})+1$ for $x$. [5]
Show complete worked solution
(a) Note that $25^{x}=\left(5^{x}\right)^{2}$. With the substitution $y=5^{x}$ (where $y>0$), the equation becomes a quadratic in $y$:
$$y^{2}-6y+5=0$$
Factorising:
$$(y-1)(y-5)=0$$
$$y=1\ \text{or}\ y=5$$
Both values satisfy $y>0$, so both are valid. Substituting back $y=5^{x}$:
Case 1: $5^{x}=1=5^{0}\implies x=0$
Case 2: $5^{x}=5=5^{1}\implies x=1$
$$\boxed{x=0\text{ or }x=1}$$
(b) First isolate a single logarithm on the left. Using the quotient law on the first and third terms:
$$\log_2\left(\dfrac{x}{3}\right)+\log_2(x+2)=3$$
Now use the product law to combine the two remaining logarithms:
$$\log_2\left[\dfrac{x(x+2)}{3}\right]=3$$
Convert to exponential form:
$$\dfrac{x(x+2)}{3}=2^{3}=8$$
$$x(x+2)=24$$
$$x^{2}+2x-24=0$$
Factorising:
$$(x+6)(x-4)=0$$
$$x=-6\ \text{or}\ x=4$$
Check the domain: $\log_2 x$ requires $x>0$. Since $x=-6<0$, this makes $\log_2(-6)$ undefined, so $x=-6$ must be rejected. Since $x=4>0$ (and $x+2=6>0$), this solution is valid.
$$\boxed{x=4}$$
(c) Note that $4^{x+1}=4\times4^{x}=4\times\left(2^{x}\right)^{2}$. With the substitution $t=2^{x}$ (where $t>0$), the equation becomes a quadratic in $t$:
$$4t^{2}=3t+1$$
$$4t^{2}-3t-1=0$$
Factorising (two numbers multiplying to $4\times(-1)=-4$ and adding to $-3$, namely $-4$ and $1$):
$$4t^{2}-4t+t-1=0$$
$$4t(t-1)+(t-1)=0$$
$$(4t+1)(t-1)=0$$
$$t=-\dfrac14\ \text{or}\ t=1$$
Since $t=2^{x}>0$ for all real $x$, the value $t=-\dfrac14$ is impossible and must be rejected. For the remaining case:
$$2^{x}=1=2^{0}$$
$$\boxed{x=0}$$
QUESTION 46
12 marks
Hard
The population of a town is modelled by $P(t)=P_0e^{kt}$, where $t$ is the time in years after 2010, and $P_0$ and $k$ are constants.
(a) Using the data in the table, show that $P_0=12\,000$ and that $k=\dfrac{1}{10}\ln\left(\dfrac32\right)$. [3]
(b) Hence find the population predicted by the model for the year 2030. [4]
(c) Find, correct to 1 decimal place, the value of $t$ at which the model predicts the population will first reach 30\,000, and hence state the year in which this occurs. [5]
| Year | 2010 | 2020 |
|---|---|---|
| Population $P$ | 12\,000 | 18\,000 |
Show complete worked solution
(a) At $t=0$ (the year 2010), $P(0)=P_0e^{0}=P_0$. Since the population in 2010 is 12\,000:
$$P_0=12\,000$$
At $t=10$ (the year 2020), the population is 18\,000, so:
$$12\,000\,e^{10k}=18\,000$$
$$e^{10k}=\dfrac{18\,000}{12\,000}=\dfrac32$$
Taking the natural logarithm of both sides:
$$10k=\ln\left(\dfrac32\right)$$
$$\boxed{k=\dfrac{1}{10}\ln\left(\dfrac32\right)}$$
(b) The year 2030 corresponds to $t=20$. Using $P(t)=12\,000\,e^{kt}$ and the fact that $e^{10k}=\dfrac32$ from part (a):
$$P(20)=12\,000\,e^{20k}=12\,000\left(e^{10k}\right)^{2}=12\,000\left(\dfrac32\right)^{2}$$
$$=12\,000\times\dfrac94=27\,000$$
$$\boxed{P(2030)=27\,000}$$
(c) Substitute $P(t)=30\,000$ into the model:
$$12\,000\,e^{kt}=30\,000$$
$$e^{kt}=\dfrac{30\,000}{12\,000}=\dfrac{5}{2}$$
Taking the natural logarithm of both sides:
$$kt=\ln\left(\dfrac52\right)$$
$$t=\dfrac{\ln\left(\dfrac52\right)}{k}=\dfrac{\ln\left(\dfrac52\right)}{\dfrac{1}{10}\ln\left(\dfrac32\right)}=\dfrac{10\ln\left(\dfrac52\right)}{\ln\left(\dfrac32\right)}$$
Using $\ln\left(\dfrac52\right)\approx0.91629$ and $\ln\left(\dfrac32\right)\approx0.40546$:
$$t\approx\dfrac{10(0.91629)}{0.40546}\approx\dfrac{9.1629}{0.40546}\approx22.5985$$
$$\boxed{t\approx22.6\text{ (1 d.p.)}}$$
Since $t\approx22.6$ years after 2010 falls between $t=22$ (the year 2032) and $t=23$ (the year 2033), the population first reaches 30\,000 during the year $\boxed{2032}$.
QUESTION 47
10 marks
Hard
Two positive numbers $x$ and $y$, with $x>y$, satisfy $x+y=20$.
(a) Given also that $\log_2 x+\log_2 y=6$, show that $xy=64$. [2]
(b) Solve for $x$ and $y$. [5]
(c) Hence find the exact value of $\log_4 x+\log_4 y$. [3]
Show complete worked solution
(a) Using the product law of logarithms, $\log_2 x+\log_2 y=\log_2(xy)$, the equation becomes:
$$\log_2(xy)=6$$
Converting to exponential form:
$$\boxed{xy=2^{6}=64}$$
(b) Since $x+y=20$ and $xy=64$, $x$ and $y$ are the roots of the quadratic equation $t^{2}-(x+y)t+xy=0$, i.e.
$$t^{2}-20t+64=0$$
Factorising (two numbers multiplying to 64 and adding to 20, namely 16 and 4):
$$(t-16)(t-4)=0$$
$$t=16\ \text{or}\ t=4$$
Since $x>y$, and both values are positive (satisfying the domain of the logarithms):
$$\boxed{x=16,\ y=4}$$
(c) Using the values from (b), and the product law:
$$\log_4 x+\log_4 y=\log_4(xy)=\log_4 64$$
Since $4^{3}=64$:
$$\boxed{\log_4 x+\log_4 y=3}$$
QUESTION 48
10 marks
Hard
(a) Solve the equation $\log_2 x=\log_4(x+6)$ for $x>0$. [4]
(b) Solve the equation $\log_3 x=\log_9(x+10)+1$ for $x>0$. [6]
Show complete worked solution
(a) Use the change of base formula to write the right-hand side in base 2: $\log_4(x+6)=\dfrac{\log_2(x+6)}{\log_2 4}=\dfrac{\log_2(x+6)}{2}$. The equation becomes:
$$\log_2 x=\dfrac{\log_2(x+6)}{2}$$
Multiply both sides by 2:
$$2\log_2 x=\log_2(x+6)$$
Using the power law, $2\log_2 x=\log_2\left(x^{2}\right)$:
$$\log_2\left(x^{2}\right)=\log_2(x+6)$$
Since the logarithms (with the same base) are equal, the arguments must be equal:
$$x^{2}=x+6$$
$$x^{2}-x-6=0$$
$$(x-3)(x+2)=0$$
$$x=3\ \text{or}\ x=-2$$
Since the domain requires $x>0$, reject $x=-2$.
$$\boxed{x=3}$$
(b) Use the change of base formula to write $\log_9(x+10)=\dfrac{\log_3(x+10)}{\log_3 9}=\dfrac{\log_3(x+10)}{2}$. The equation becomes:
$$\log_3 x=\dfrac{\log_3(x+10)}{2}+1$$
Rearrange so the logarithms are on one side:
$$\log_3 x-\dfrac12\log_3(x+10)=1$$
Using the power law, $\dfrac12\log_3(x+10)=\log_3\left[(x+10)^{1/2}\right]$, and then the quotient law:
$$\log_3\left[\dfrac{x}{(x+10)^{1/2}}\right]=1$$
Convert to exponential form:
$$\dfrac{x}{\sqrt{x+10}}=3^{1}=3$$
$$x=3\sqrt{x+10}$$
Square both sides (noting this may introduce extraneous solutions to be checked afterwards):
$$x^{2}=9(x+10)=9x+90$$
$$x^{2}-9x-90=0$$
Using the quadratic formula with $a=1,\ b=-9,\ c=-90$:
$$x=\dfrac{9\pm\sqrt{81+360}}{2}=\dfrac{9\pm\sqrt{441}}{2}=\dfrac{9\pm21}{2}$$
$$x=15\ \text{or}\ x=-6$$
Since the domain requires $x>0$, reject $x=-6$. Check $x=15$ in the squared step: $x=3\sqrt{x+10}$ gives $3\sqrt{25}=3(5)=15$, which matches, so $x=15$ is confirmed (not extraneous).
$$\boxed{x=15}$$
QUESTION 49
10 marks
Hard
(a) Using the change of base formula, prove that $\log_a b\times\log_b a=1$ for all $a,b>0$ with $a,b\neq1$. [4]
(b) Given that $\log_2 7=k$, use the result from part (a) to express $\log_7 32$ in terms of $k$. [3]
(c) Hence show that $\log_x 2\times\log_2\left(x^{2}\right)=2$ for all $x>0,\ x\neq1$. [3]
Show complete worked solution
(a) Using the change of base formula (base 10, or any fixed base), $\log_a b=\dfrac{\log b}{\log a}$ and $\log_b a=\dfrac{\log a}{\log b}$. Multiplying these together:
$$\log_a b\times\log_b a=\dfrac{\log b}{\log a}\times\dfrac{\log a}{\log b}$$
Since $a,b\neq1$, neither $\log a$ nor $\log b$ is zero, so the factors of $\log a$ and $\log b$ cancel:
$$=\dfrac{\log b\times\log a}{\log a\times\log b}=1$$
$$\boxed{\log_a b\times\log_b a=1}\qquad\blacksquare$$
(b) Write $32$ as a power of 2 and apply the power law:
$$\log_7 32=\log_7\left(2^{5}\right)=5\log_7 2$$
By part (a) with $a=2,\ b=7$: $\log_2 7\times\log_7 2=1$, so $\log_7 2=\dfrac{1}{\log_2 7}=\dfrac1k$. Substituting:
$$\log_7 32=5\times\dfrac1k$$
$$\boxed{\log_7 32=\dfrac{5}{k}}$$
(c) Using the power law on the second factor, $\log_2\left(x^{2}\right)=2\log_2 x$:
$$\log_x 2\times\log_2\left(x^{2}\right)=\log_x 2\times2\log_2 x=2\left(\log_x 2\times\log_2 x\right)$$
By part (a), with $a=x,\ b=2$ (valid since $x>0,\ x\neq1$): $\log_x 2\times\log_2 x=1$. Therefore:
$$\log_x 2\times\log_2\left(x^{2}\right)=2\times1=2$$
Since this holds regardless of the specific value of $x$ (for all $x>0,\ x\neq1$), the identity is proven.
$$\boxed{\log_x 2\times\log_2\left(x^{2}\right)=2}\qquad\blacksquare$$
QUESTION 50
12 marks
Hard
Consider the simultaneous equations $$3^{x}\times2^{y}=12,\qquad 3^{x}\times4^{y}=24$$
(a) By dividing the second equation by the first, show that $y=1$. [3]
(b) Hence find the exact value of $x$, giving your answer in the form $1+\log_3 2$. [4]
(c) Using the change of base formula, find the value of $x$ correct to 3 significant figures. [5]
Show complete worked solution
(a) Dividing the second equation by the first:
$$\dfrac{3^{x}\times4^{y}}{3^{x}\times2^{y}}=\dfrac{24}{12}$$
The factor $3^{x}$ cancels, and $\dfrac{4^{y}}{2^{y}}=\left(\dfrac{4}{2}\right)^{y}=2^{y}$:
$$2^{y}=2$$
Since $2=2^{1}$, and the bases are equal, the exponents must be equal:
$$\boxed{y=1}\qquad\blacksquare$$
(b) Substitute $y=1$ into the first equation:
$$3^{x}\times2^{1}=12$$
$$3^{x}=\dfrac{12}{2}=6$$
Write $6=3\times2$, and use the product law of logarithms after converting to logarithmic form. Taking $\log_3$ of both sides:
$$x=\log_3 6=\log_3(3\times2)=\log_3 3+\log_3 2$$
Since $\log_3 3=1$:
$$\boxed{x=1+\log_3 2}$$
(c) Using the change of base formula, $\log_3 2=\dfrac{\ln2}{\ln3}$. Substituting $\ln2\approx0.693147$ and $\ln3\approx1.098612$:
$$\log_3 2\approx\dfrac{0.693147}{1.098612}\approx0.630930$$
So:
$$x\approx1+0.630930=1.630930$$
$$\boxed{x\approx1.63\text{ (3 s.f.)}}$$
The Binomial Theorem 50 questions
QUESTION 1
4 marks
Easy
Find the coefficient of $x^3$ in the expansion of $(2+x)^5$.
Show complete worked solution
By the binomial theorem, the general term in the expansion of $(2+x)^5$ is
$$T_{k+1}=\binom{5}{k}2^{5-k}x^{k}$$
For the term in $x^3$, set $k=3$:
$$T_4=\binom{5}{3}2^{2}x^3$$
Compute $\binom{5}{3}=10$ and $2^2=4$:
$$T_4=10\times4\times x^3=40x^3$$
$$\boxed{\text{Coefficient}=40}$$
QUESTION 2
5 marks
Medium
Use the binomial theorem (or Pascal's triangle) to expand $(x-2y)^4$ fully, simplifying each term.
Show complete worked solution
The row of Pascal's triangle for $n=4$ gives the coefficients $1,4,6,4,1$. Using the binomial theorem with $a=x$, $b=-2y$, $n=4$:
$$(x-2y)^4=\binom{4}{0}x^4+\binom{4}{1}x^3(-2y)+\binom{4}{2}x^2(-2y)^2+\binom{4}{3}x(-2y)^3+\binom{4}{4}(-2y)^4$$
Substitute the coefficients and simplify each term:
$$\binom{4}{0}x^4=x^4$$
$$\binom{4}{1}x^3(-2y)=4x^3(-2y)=-8x^3y$$
$$\binom{4}{2}x^2(-2y)^2=6x^2(4y^2)=24x^2y^2$$
$$\binom{4}{3}x(-2y)^3=4x(-8y^3)=-32xy^3$$
$$\binom{4}{4}(-2y)^4=16y^4$$
Combining:
$$\boxed{(x-2y)^4=x^4-8x^3y+24x^2y^2-32xy^3+16y^4}$$
QUESTION 3
6 marks
Medium
In the expansion of $(1+3x)^n$, where $n \in \mathbb{Z}^+$, the coefficient of the term in $x^2$ is $405$. Find the value of $n$.
Show complete worked solution
The general term of $(1+3x)^n$ is
$$T_{k+1}=\binom{n}{k}(3x)^{k}=\binom{n}{k}3^k x^k$$
For the term in $x^2$, set $k=2$:
$$\binom{n}{2}\times3^2=405$$
$$9\binom{n}{2}=405$$
$$\binom{n}{2}=45$$
Using $\binom{n}{2}=\dfrac{n(n-1)}{2}$:
$$\dfrac{n(n-1)}{2}=45$$
$$n(n-1)=90$$
$$n^2-n-90=0$$
Factorise:
$$(n-10)(n+9)=0$$
$$n=10 \text{ or } n=-9$$
Since $n\in\mathbb{Z}^+$, reject $n=-9$.
$$\boxed{n=10}$$
QUESTION 4
4 marks
Easy
Find the term containing $x^3$ in the expansion of $(1+2x)^7$, simplifying the coefficient fully.
Show complete worked solution
The general term of $(1+2x)^7$ is
$$T_{k+1}=\binom{7}{k}(2x)^k=\binom{7}{k}2^k x^k$$
For the term in $x^3$, set $k=3$:
$$T_4=\binom{7}{3}2^3 x^3$$
Compute $\binom{7}{3}=35$ and $2^3=8$:
$$T_4=35\times8\times x^3=280x^3$$
$$\boxed{280x^3}$$
QUESTION 5
13 marks
Hard
Consider the expansion of $(1+x)^n$, where $n \in \mathbb{Z}^+$.
(a) [3 marks] Write down, in terms of $n$ and $r$, the general term $T_{r+1}$ (the term in $x^r$) in the expansion of $(1+x)^n$.
(b) [4 marks] In this expansion, the coefficients of the terms in $x^3$ and $x^4$ are in the ratio $1:3$. Find the value of $n$.
(c) [6 marks] Prove Pascal's rule, $\dbinom{n}{r} + \dbinom{n}{r+1} = \dbinom{n+1}{r+1}$, for $0 \le r \le n-1$, $n \in \mathbb{Z}^+$. Hence, using the value of $n$ found in part (b), evaluate $\dbinom{15}{3} + \dbinom{15}{4}$.
Show complete worked solution
(a) By the binomial theorem, the term in $x^r$ is
$$\boxed{T_{r+1}=\binom{n}{r}x^r}$$
(b) The coefficient of $x^3$ is $\binom{n}{3}$ and the coefficient of $x^4$ is $\binom{n}{4}$. Using the identity $\binom{n}{r+1}=\binom{n}{r}\times\dfrac{n-r}{r+1}$ with $r=3$:
$$\binom{n}{4}=\binom{n}{3}\times\dfrac{n-3}{4}$$
So
$$\dfrac{\binom{n}{3}}{\binom{n}{4}}=\dfrac{4}{n-3}$$
The given ratio is $\binom{n}{3}:\binom{n}{4}=1:3$, i.e. $\dfrac{\binom{n}{3}}{\binom{n}{4}}=\dfrac{1}{3}$. Therefore:
$$\dfrac{4}{n-3}=\dfrac{1}{3}$$
$$n-3=12$$
$$\boxed{n=15}$$
(c) Starting from the factorial definition:
$$\binom{n}{r}+\binom{n}{r+1}=\dfrac{n!}{r!(n-r)!}+\dfrac{n!}{(r+1)!(n-r-1)!}$$
Factor out $\dfrac{n!}{r!(n-r-1)!}$ from both terms, using $(n-r)!=(n-r)(n-r-1)!$ and $(r+1)!=(r+1)r!$:
$$=\dfrac{n!}{r!(n-r-1)!}\left[\dfrac{1}{n-r}+\dfrac{1}{r+1}\right]$$
Combine the bracket over a common denominator:
$$\dfrac{1}{n-r}+\dfrac{1}{r+1}=\dfrac{(r+1)+(n-r)}{(n-r)(r+1)}=\dfrac{n+1}{(n-r)(r+1)}$$
So:
$$\binom{n}{r}+\binom{n}{r+1}=\dfrac{n!}{r!(n-r-1)!}\times\dfrac{n+1}{(n-r)(r+1)}=\dfrac{(n+1)!}{(r+1)!(n-r)!}=\binom{n+1}{r+1}$$
as required, since $(n-r)(n-r-1)!=(n-r)!$ and $(r+1)r!=(r+1)!$.
Applying this with $n=15$, $r=3$:
$$\binom{15}{3}+\binom{15}{4}=\binom{16}{4}$$
Direct computation:
$$\binom{16}{4}=\dfrac{16\times15\times14\times13}{4!}=\dfrac{43680}{24}=1820$$
$$\boxed{\binom{15}{3}+\binom{15}{4}=1820}$$
QUESTION 6
4 marks
Easy
Find the first three terms, in ascending powers of $x$, in the binomial expansion of $(1+x)^{-2}$.
Show complete worked solution
Using the extended binomial theorem, $(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\ldots$, valid for $|x|<1$, applied here with $n=-2$.
Constant term: $1$
Term in $x$:
$$nx=(-2)x=-2x$$
Term in $x^2$:
$$\dfrac{n(n-1)}{2!}x^2=\dfrac{(-2)(-3)}{2}x^2=\dfrac{6}{2}x^2=3x^2$$
$$\boxed{(1+x)^{-2}\approx1-2x+3x^2}$$
QUESTION 7
6 marks
Medium
Find the first four terms, in ascending powers of $x$, in the binomial expansion of $(1-3x)^{1/3}$. State the range of values of $x$ for which the expansion is valid.
Show complete worked solution
Using $(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\dfrac{n(n-1)(n-2)}{3!}u^3+\ldots$ with $u=-3x$ and $n=\dfrac{1}{3}$:
Term 0: $1$
Term in $u$ (i.e. $x$):
$$nu=\dfrac{1}{3}(-3x)=-x$$
Term in $u^2$ (i.e. $x^2$):
$$\dfrac{n(n-1)}{2}u^2=\dfrac{\tfrac{1}{3}\left(-\tfrac{2}{3}\right)}{2}u^2=-\dfrac{1}{9}u^2$$
Since $u^2=(-3x)^2=9x^2$:
$$-\dfrac{1}{9}(9x^2)=-x^2$$
Term in $u^3$ (i.e. $x^3$):
$$\dfrac{n(n-1)(n-2)}{6}u^3=\dfrac{\tfrac{1}{3}\left(-\tfrac{2}{3}\right)\left(-\tfrac{5}{3}\right)}{6}u^3=\dfrac{\tfrac{10}{27}}{6}u^3=\dfrac{5}{81}u^3$$
Since $u^3=(-3x)^3=-27x^3$:
$$\dfrac{5}{81}(-27x^3)=-\dfrac{5}{3}x^3$$
$$\boxed{(1-3x)^{1/3}\approx1-x-x^2-\dfrac{5}{3}x^3}$$
Validity: the expansion of $(1+u)^n$ for non-integer $n$ is valid for $|u|<1$, i.e. $|-3x|<1$, so
$$\boxed{-\dfrac{1}{3}
QUESTION 8
7 marks
Medium
(i) Find the binomial expansion of $\dfrac{1}{(1+2x)^4}$ in ascending powers of $x$, up to and including the term in $x^3$, and state the coefficient of $x^3$.
(ii) By substituting $x=0.01$, use your expansion to find an approximate value for $\dfrac{1}{(1.02)^4}$, giving your answer to 5 decimal places.
Show complete worked solution
(i) Write $\dfrac{1}{(1+2x)^4}=(1+2x)^{-4}$. Using $(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\dfrac{n(n-1)(n-2)}{3!}u^3+\ldots$ with $u=2x$, $n=-4$:
Term in $x$: $nu=(-4)(2x)=-8x$
Term in $x^2$:
$$\dfrac{n(n-1)}{2}u^2=\dfrac{(-4)(-5)}{2}(2x)^2=10\times4x^2=40x^2$$
Term in $x^3$:
$$\dfrac{n(n-1)(n-2)}{6}u^3=\dfrac{(-4)(-5)(-6)}{6}(2x)^3=(-20)\times8x^3=-160x^3$$
$$(1+2x)^{-4}\approx1-8x+40x^2-160x^3$$
$$\boxed{\text{Coefficient of }x^3=-160}$$
(ii) Substitute $x=0.01$ into the expansion (note $1+2(0.01)=1.02$):
$$1-8(0.01)+40(0.01)^2-160(0.01)^3$$
$$=1-0.08+0.004-0.00016$$
$$=0.92384$$
$$\boxed{\dfrac{1}{(1.02)^4}\approx0.92384\text{ (5 d.p.)}}$$
QUESTION 9
5 marks
Easy
Write down the first three terms, in ascending powers of $x$, of the binomial expansion of $\sqrt{1+x}$, and state the values of $x$ for which this expansion is valid.
Show complete worked solution
Write $\sqrt{1+x}=(1+x)^{1/2}$. Using $(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\ldots$ with $n=\dfrac{1}{2}$:
Term 0: $1$
Term in $x$: $nx=\dfrac{1}{2}x$
Term in $x^2$:
$$\dfrac{n(n-1)}{2}x^2=\dfrac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2}x^2=-\dfrac{1}{8}x^2$$
$$\boxed{\sqrt{1+x}\approx1+\dfrac{1}{2}x-\dfrac{1}{8}x^2}$$
Since $n=\dfrac{1}{2}$ is not a positive integer, the expansion is an infinite series that converges only for $|x|<1$.
$$\boxed{-1
QUESTION 10
15 marks
Hard
Consider the function $f(x) = (1+ax)^{-3}$, where $a$ is a non-zero constant, expanded as a binomial series in ascending powers of $x$.
(a) [4 marks] Show that the first three terms of the expansion are $1 - 3ax + 6a^2x^2$.
(b) [5 marks] Given that the coefficient of $x^2$ in the expansion is $54$, and that $a > 0$, find the value of $a$.
(c) [6 marks] Using the value of $a$ found in part (b) and the first three terms of the expansion, find an approximate value for $f(0.01) = (1.03)^{-3}$, giving your answer to 5 significant figures. State also the range of values of $x$ for which the full expansion of $f(x)$ is valid.
Show complete worked solution
(a) Using $(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\ldots$ with $u=ax$ and $n=-3$:
Term in $x$: $nu=(-3)(ax)=-3ax$
Term in $x^2$:
$$\dfrac{n(n-1)}{2}u^2=\dfrac{(-3)(-4)}{2}(ax)^2=6(ax)^2=6a^2x^2$$
So:
$$(1+ax)^{-3}=1-3ax+6a^2x^2-\ldots$$
as required.
(b) The coefficient of $x^2$ is $6a^2$. Setting this equal to 54:
$$6a^2=54$$
$$a^2=9$$
$$a=\pm3$$
Since $a>0$:
$$\boxed{a=3}$$
(c) With $a=3$, the first three terms become:
$$1-3(3)x+6(3)^2x^2=1-9x+54x^2$$
So $f(x)=(1+3x)^{-3}\approx1-9x+54x^2$. Substitute $x=0.01$ (giving $1+3(0.01)=1.03$):
$$1-9(0.01)+54(0.01)^2=1-0.09+0.0054=0.9154$$
$$\boxed{(1.03)^{-3}\approx0.91540\text{ (5 s.f.)}}$$
Validity: the expansion of $(1+ax)^n$ for non-integer $n$ converges for $|ax|<1$. With $a=3$:
$$|3x|<1 \quad\Rightarrow\quad \boxed{-\dfrac{1}{3}
QUESTION 11
4 marks
Easy
Find the coefficient of $x^4$ in the expansion of $(3+x)^7$.
Show complete worked solution
By the binomial theorem, the general term in the expansion of $(3+x)^7$ is
$$T_{k+1}=\binom{7}{k}3^{7-k}x^{k}$$
For the term in $x^4$, set $k=4$:
$$T_5=\binom{7}{4}3^{3}x^4$$
Compute $\binom{7}{4}=35$ and $3^3=27$:
$$T_5=35\times27\times x^4=945x^4$$
$$\boxed{\text{Coefficient}=945}$$
QUESTION 12
4 marks
Easy
Find the term containing $x^5$ in the expansion of $(1+2x)^8$, giving the coefficient in fully simplified form.
Show complete worked solution
The general term of $(1+2x)^8$ is
$$T_{k+1}=\binom{8}{k}(2x)^{k}=\binom{8}{k}2^kx^k$$
For the term in $x^5$, set $k=5$:
$$T_6=\binom{8}{5}2^5x^5$$
Compute $\binom{8}{5}=56$ and $2^5=32$:
$$T_6=56\times32\times x^5=1792x^5$$
$$\boxed{1792x^5}$$
QUESTION 13
5 marks
Easy
Use the binomial theorem (or Pascal's triangle) to expand $(x+3y)^4$ fully, simplifying each term.
Show complete worked solution
The row of Pascal's triangle for $n=4$ gives the coefficients $1,4,6,4,1$. Using the binomial theorem with $a=x$, $b=3y$, $n=4$:
$$(x+3y)^4=\binom{4}{0}x^4+\binom{4}{1}x^3(3y)+\binom{4}{2}x^2(3y)^2+\binom{4}{3}x(3y)^3+\binom{4}{4}(3y)^4$$
Substitute the coefficients and simplify each term:
$$\binom{4}{0}x^4=x^4$$
$$\binom{4}{1}x^3(3y)=4x^3(3y)=12x^3y$$
$$\binom{4}{2}x^2(3y)^2=6x^2(9y^2)=54x^2y^2$$
$$\binom{4}{3}x(3y)^3=4x(27y^3)=108xy^3$$
$$\binom{4}{4}(3y)^4=81y^4$$
Combining:
$$\boxed{(x+3y)^4=x^4+12x^3y+54x^2y^2+108xy^3+81y^4}$$
QUESTION 14
4 marks
Easy
Expand $(2x-1)^3$ fully using the binomial theorem.
Show complete worked solution
The row of Pascal's triangle for $n=3$ gives coefficients $1,3,3,1$. With $a=2x$, $b=-1$, $n=3$:
$$(2x-1)^3=\binom{3}{0}(2x)^3+\binom{3}{1}(2x)^2(-1)+\binom{3}{2}(2x)(-1)^2+\binom{3}{3}(-1)^3$$
Simplify each term:
$$\binom{3}{0}(2x)^3=8x^3$$
$$\binom{3}{1}(2x)^2(-1)=3(4x^2)(-1)=-12x^2$$
$$\binom{3}{2}(2x)(1)=3(2x)=6x$$
$$\binom{3}{3}(-1)=-1$$
$$\boxed{(2x-1)^3=8x^3-12x^2+6x-1}$$
QUESTION 15
5 marks
Easy
The diagram shows Pascal's triangle for rows $n=0$ to $n=6$. Using the row for $n=6$, find the coefficient of $x^4$ in the expansion of $(2+x)^6$.
Show complete worked solution
From the $n=6$ row of Pascal's triangle shown in the diagram, the binomial coefficients are $1,6,15,20,15,6,1$.
The general term in the expansion of $(2+x)^6$ is
$$T_{k+1}=\binom{6}{k}2^{6-k}x^{k}$$
For the term in $x^4$, set $k=4$. From the triangle, $\binom{6}{4}=15$, and $2^{6-4}=2^2=4$:
$$T_5=15\times4\times x^4=60x^4$$
$$\boxed{\text{Coefficient}=60}$$
QUESTION 16
4 marks
Easy
Find the first three terms, in ascending powers of $x$, in the binomial expansion of $(1+x)^{-3}$.
Show complete worked solution
Using the extended binomial theorem, $(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\ldots$, valid for $|x|<1$, applied here with $n=-3$.
Constant term: $1$
Term in $x$:
$$nx=(-3)x=-3x$$
Term in $x^2$:
$$\dfrac{n(n-1)}{2!}x^2=\dfrac{(-3)(-4)}{2}x^2=\dfrac{12}{2}x^2=6x^2$$
$$\boxed{(1+x)^{-3}\approx1-3x+6x^2}$$
QUESTION 17
5 marks
Easy
Find the first three terms, in ascending powers of $x$, of the binomial expansion of $(1-2x)^{1/2}$, and state the range of values of $x$ for which the expansion is valid.
Show complete worked solution
Using $(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\ldots$ with $u=-2x$ and $n=\dfrac{1}{2}$:
Term 0: $1$
Term in $u$ (i.e. $x$):
$$nu=\dfrac{1}{2}(-2x)=-x$$
Term in $u^2$ (i.e. $x^2$):
$$\dfrac{n(n-1)}{2}u^2=\dfrac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2}u^2=-\dfrac{1}{8}u^2$$
Since $u^2=(-2x)^2=4x^2$:
$$-\dfrac{1}{8}(4x^2)=-\dfrac{1}{2}x^2$$
$$\boxed{(1-2x)^{1/2}\approx1-x-\dfrac{1}{2}x^2}$$
Validity: since $n=\dfrac{1}{2}$ is not a positive integer, the expansion is an infinite series, valid for $|u|<1$, i.e. $|-2x|<1$, so
$$\boxed{-\dfrac{1}{2}
QUESTION 18
6 marks
Easy
In the expansion of $(1+2x)^n$, where $n \in \mathbb{Z}^+$, the coefficient of the term in $x^3$ is $160$. Find the value of $n$.
Show complete worked solution
The general term of $(1+2x)^n$ is
$$T_{k+1}=\binom{n}{k}(2x)^{k}=\binom{n}{k}2^kx^k$$
For the term in $x^3$, set $k=3$:
$$\binom{n}{3}\times2^3=160$$
$$8\binom{n}{3}=160$$
$$\binom{n}{3}=20$$
Using $\binom{n}{3}=\dfrac{n(n-1)(n-2)}{6}$:
$$\dfrac{n(n-1)(n-2)}{6}=20$$
$$n(n-1)(n-2)=120$$
Testing small positive integers, $n=6$ gives $6\times5\times4=120$, which satisfies the equation.
$$\boxed{n=6}$$
QUESTION 19
5 marks
Easy
Consider the expansion of $(2x+5)^9$ in ascending powers of $x$.
(a) [2] Write down the number of terms in the expansion, and state the general term $T_{r+1}$ in terms of $r$.
(b) [3] Hence find the term in $x^6$.
Show complete worked solution
(a) An expansion of $(a+b)^9$ has $9+1=10$ terms. The general term is
$$\boxed{T_{r+1}=\binom{9}{r}(2x)^{9-r}5^{r}}$$
(b) For the term in $x^6$, the power of $x$ is $9-r=6$, so $r=3$:
$$T_4=\binom{9}{3}(2x)^{6}5^{3}$$
Compute $\binom{9}{3}=84$, $(2x)^6=64x^6$, and $5^3=125$:
$$T_4=84\times64x^6\times125=672000x^6$$
$$\boxed{672000x^6}$$
QUESTION 20
4 marks
Easy
Find the ratio of the coefficient of $x^5$ to the coefficient of $x^4$ in the expansion of $(1+x)^{12}$, giving your answer as a fraction in lowest terms.
Show complete worked solution
The coefficient of $x^5$ in $(1+x)^{12}$ is $\binom{12}{5}$ and the coefficient of $x^4$ is $\binom{12}{4}$.
Compute each value:
$$\binom{12}{5}=792 \qquad \binom{12}{4}=495$$
The required ratio is
$$\dfrac{792}{495}$$
Dividing numerator and denominator by their highest common factor, $99$:
$$\dfrac{792}{495}=\dfrac{8}{5}$$
$$\boxed{8:5}$$
QUESTION 21
5 marks
Easy
By expanding $(2+x)^4$ fully and substituting $x=0.01$, find the exact value of $(2.01)^4$.
Show complete worked solution
By the binomial theorem,
$$(2+x)^4=\binom{4}{0}2^4+\binom{4}{1}2^3x+\binom{4}{2}2^2x^2+\binom{4}{3}2x^3+\binom{4}{4}x^4$$
$$=16+32x+24x^2+8x^3+x^4$$
Since $2.01=2+0.01$, substitute $x=0.01$ (this is an exact substitution, not an approximation, because the expansion of an integer power is a finite sum):
$$16+32(0.01)+24(0.01)^2+8(0.01)^3+(0.01)^4$$
$$=16+0.32+0.0024+0.000008+0.00000001$$
$$=16.32240801$$
$$\boxed{(2.01)^4=16.32240801}$$
QUESTION 22
5 marks
Easy
Find the coefficient of $x^3$ in the expansion of $(3-2x)^6$.
Show complete worked solution
The general term of $(3-2x)^6$ is
$$T_{k+1}=\binom{6}{k}3^{6-k}(-2x)^{k}=\binom{6}{k}3^{6-k}(-2)^kx^k$$
For the term in $x^3$, set $k=3$:
$$T_4=\binom{6}{3}3^{3}(-2)^3x^3$$
Compute $\binom{6}{3}=20$, $3^3=27$, and $(-2)^3=-8$:
$$T_4=20\times27\times(-8)\times x^3=-4320x^3$$
$$\boxed{\text{Coefficient}=-4320}$$
QUESTION 23
4 marks
Easy
Find the first three terms, in ascending powers of $x$, of the binomial expansion of $(1+4x)^{-1}$, and state the range of values of $x$ for which the expansion is valid.
Show complete worked solution
Using $(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\ldots$ with $u=4x$ and $n=-1$:
Term 0: $1$
Term in $u$:
$$nu=(-1)(4x)=-4x$$
Term in $u^2$:
$$\dfrac{n(n-1)}{2}u^2=\dfrac{(-1)(-2)}{2}(4x)^2=1\times16x^2=16x^2$$
$$\boxed{(1+4x)^{-1}\approx1-4x+16x^2}$$
Validity: since $n=-1$ is not a positive integer, the series is infinite and converges only for $|u|<1$, i.e. $|4x|<1$, so
$$\boxed{-\dfrac{1}{4}
QUESTION 24
4 marks
Easy
Find the coefficient of the term containing $x^2y^5$ in the expansion of $(x+y)^7$.
Show complete worked solution
The general term in the expansion of $(x+y)^7$ is
$$T_{k+1}=\binom{7}{k}x^{7-k}y^{k}$$
For the term containing $x^2y^5$, the power of $y$ is $k=5$ (which automatically gives the power of $x$ as $7-5=2$, as required):
$$T_6=\binom{7}{5}x^2y^5$$
Compute $\binom{7}{5}=21$:
$$T_6=21x^2y^5$$
$$\boxed{\text{Coefficient}=21}$$
QUESTION 25
5 marks
Easy
In the expansion of $(1+kx)^5$, where $k>0$, the coefficient of the term in $x^2$ is $90$. Find the value of $k$.
Show complete worked solution
The general term of $(1+kx)^5$ is
$$T_{r+1}=\binom{5}{r}(kx)^{r}=\binom{5}{r}k^rx^r$$
For the term in $x^2$, set $r=2$:
$$\binom{5}{2}k^2=90$$
Compute $\binom{5}{2}=10$:
$$10k^2=90$$
$$k^2=9$$
$$k=\pm3$$
Since $k>0$, reject $k=-3$.
$$\boxed{k=3}$$
QUESTION 26
4 marks
Easy
State the number of terms in the expansion of $(1+x)^{10}$, and find the coefficient of the third term (in ascending powers of $x$) in this expansion.
Show complete worked solution
An expansion of $(1+x)^{10}$ has $10+1=11$ terms.
$$\boxed{\text{Number of terms}=11}$$
In ascending powers of $x$, the terms are $T_1,T_2,T_3,\ldots$, where $T_{r+1}=\binom{10}{r}x^r$. The third term corresponds to $r=2$:
$$T_3=\binom{10}{2}x^2$$
Compute $\binom{10}{2}=\dfrac{10\times9}{2}=45$:
$$T_3=45x^2$$
$$\boxed{\text{Coefficient of the third term}=45}$$
QUESTION 27
6 marks
Medium
In the expansion of $(2+ax)^6$, where $a$ is a non-zero constant, the coefficient of the $x^2$ term is $2160$. Find the possible value(s) of $a$.
Show complete worked solution
The general term of $(2+ax)^6$ is
$$T_{k+1}=\binom{6}{k}2^{6-k}(ax)^k=\binom{6}{k}2^{6-k}a^kx^k$$
For the term in $x^2$, set $k=2$:
$$\binom{6}{2}2^4a^2=2160$$
Compute $\binom{6}{2}=15$ and $2^4=16$:
$$15\times16\times a^2=2160$$
$$240a^2=2160$$
$$a^2=9$$
$$\boxed{a=3 \text{ or } a=-3}$$
QUESTION 28
6 marks
Medium
In the expansion of $(1+3x)^n$, where $n \in \mathbb{Z}^+$, the coefficient of the $x^3$ term is three times the coefficient of the $x^2$ term. Find the value of $n$.
Show complete worked solution
The general term of $(1+3x)^n$ is $T_{k+1}=\binom{n}{k}3^kx^k$, so the coefficient of $x^2$ is $9\binom{n}{2}$ and the coefficient of $x^3$ is $27\binom{n}{3}$.
The given condition states
$$27\binom{n}{3}=3\times9\binom{n}{2}$$
$$27\binom{n}{3}=27\binom{n}{2}$$
$$\binom{n}{3}=\binom{n}{2}$$
Using $\binom{n}{3}=\binom{n}{2}\times\dfrac{n-2}{3}$:
$$\binom{n}{2}\times\dfrac{n-2}{3}=\binom{n}{2}$$
Since $\binom{n}{2}\ne0$ for $n\geq2$, divide both sides by $\binom{n}{2}$:
$$\dfrac{n-2}{3}=1$$
$$n-2=3$$
$$\boxed{n=5}$$
Check: with $n=5$, $\binom{5}{2}=10$ and $\binom{5}{3}=10$, so the coefficients are $9(10)=90$ and $27(10)=270$, and indeed $270=3\times90$.
QUESTION 29
7 marks
Medium
Find the term independent of $x$ in the expansion of $\left(3x-\dfrac{2}{x^2}\right)^6$.
Show complete worked solution
The general term of $\left(3x-\dfrac{2}{x^2}\right)^6$ is
$$T_{r+1}=\binom{6}{r}(3x)^{6-r}\left(-\dfrac{2}{x^2}\right)^r=\binom{6}{r}3^{6-r}(-2)^rx^{6-r}x^{-2r}=\binom{6}{r}3^{6-r}(-2)^rx^{6-3r}$$
The term is independent of $x$ when the exponent of $x$ is zero:
$$6-3r=0$$
$$r=2$$
Substitute $r=2$:
$$T_3=\binom{6}{2}3^{4}(-2)^2$$
Compute $\binom{6}{2}=15$, $3^4=81$, and $(-2)^2=4$:
$$T_3=15\times81\times4=4860$$
$$\boxed{\text{Term independent of }x=4860}$$
QUESTION 30
7 marks
Medium
Find the first four terms, in ascending powers of $x$, in the binomial expansion of $(1+5x)^{-2}$. State the range of values of $x$ for which the expansion is valid.
Show complete worked solution
Using $(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\dfrac{n(n-1)(n-2)}{3!}u^3+\ldots$ with $u=5x$ and $n=-2$:
Term 0: $1$
Term in $u$:
$$nu=(-2)(5x)=-10x$$
Term in $u^2$:
$$\dfrac{n(n-1)}{2}u^2=\dfrac{(-2)(-3)}{2}(5x)^2=3\times25x^2=75x^2$$
Term in $u^3$:
$$\dfrac{n(n-1)(n-2)}{6}u^3=\dfrac{(-2)(-3)(-4)}{6}(5x)^3=(-4)\times125x^3=-500x^3$$
$$\boxed{(1+5x)^{-2}\approx1-10x+75x^2-500x^3}$$
Validity: the expansion of $(1+u)^n$ for non-integer $n$ converges for $|u|<1$, i.e. $|5x|<1$, so
$$\boxed{-\dfrac{1}{5}
QUESTION 31
7 marks
Medium
Find the first four terms, in ascending powers of $x$, of the binomial expansion of $\dfrac{1}{(1-x)^3}$. State the range of values of $x$ for which the expansion is valid.
Show complete worked solution
Write $\dfrac{1}{(1-x)^3}=(1-x)^{-3}$. Using $(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\dfrac{n(n-1)(n-2)}{3!}u^3+\ldots$ with $u=-x$ and $n=-3$:
Term 0: $1$
Term in $u$:
$$nu=(-3)(-x)=3x$$
Term in $u^2$:
$$\dfrac{n(n-1)}{2}u^2=\dfrac{(-3)(-4)}{2}(-x)^2=6x^2$$
Term in $u^3$:
$$\dfrac{n(n-1)(n-2)}{6}u^3=\dfrac{(-3)(-4)(-5)}{6}(-x)^3=(-10)(-x^3)=10x^3$$
$$\boxed{(1-x)^{-3}\approx1+3x+6x^2+10x^3}$$
Validity: for non-integer $n$, the series converges for $|u|<1$, i.e. $|-x|<1$, so
$$\boxed{-1
QUESTION 32
7 marks
Medium
Use the first four terms of the binomial expansion of $(2-x)^7$ to find an approximate value of $(1.98)^7$, giving your answer correct to 2 decimal places.
Show complete worked solution
Write $1.98=2-0.02$, so $(1.98)^7=(2-x)^7$ with $x=0.02$. By the binomial theorem, the first four terms of $(2-x)^7$ are
$$\binom{7}{0}2^7-\binom{7}{1}2^6x+\binom{7}{2}2^5x^2-\binom{7}{3}2^4x^3$$
Compute the coefficients: $\binom{7}{0}2^7=128$, $\binom{7}{1}2^6=7\times64=448$, $\binom{7}{2}2^5=21\times32=672$, $\binom{7}{3}2^4=35\times16=560$:
$$128-448x+672x^2-560x^3$$
Since $x=0.02$ is small, the remaining terms (in $x^4,x^5,\ldots$) are negligible at this level of accuracy. Substitute $x=0.02$:
$$128-448(0.02)+672(0.02)^2-560(0.02)^3$$
$$=128-8.96+0.2688-0.00448$$
$$=119.30432$$
$$\boxed{(1.98)^7\approx119.30\text{ (2 d.p.)}}$$
QUESTION 33
8 marks
Medium
Find the coefficient of $x^2$ in the expansion of $(1+x)^5(1-2x)^4$.
Show complete worked solution
To find the coefficient of $x^2$ in the product, expand each factor as far as the $x^2$ term and combine.
By the binomial theorem, the first three terms of $(1+x)^5$ are
$$\binom{5}{0}+\binom{5}{1}x+\binom{5}{2}x^2=1+5x+10x^2$$
and the first three terms of $(1-2x)^4$ are
$$\binom{4}{0}+\binom{4}{1}(-2x)+\binom{4}{2}(-2x)^2=1-8x+24x^2$$
The coefficient of $x^2$ in the product $(1+x)^5(1-2x)^4$ comes from pairing terms whose powers of $x$ sum to $2$:
$$[x^0\text{ of }(1+x)^5]\times[x^2\text{ of }(1-2x)^4]+[x^1]\times[x^1]+[x^2]\times[x^0]$$
$$=(1)(24)+(5)(-8)+(10)(1)$$
$$=24-40+10$$
$$=-6$$
$$\boxed{\text{Coefficient of }x^2=-6}$$
QUESTION 34
6 marks
Medium
In the expansion of $(1+x)^n$, the coefficients of the terms in $x^3$ and $x^5$ are equal. Find the value of $n$, given that $n\ne8$ is not required (i.e. state the smallest possible value of $n$).
Show complete worked solution
The coefficient of $x^3$ is $\binom{n}{3}$ and the coefficient of $x^5$ is $\binom{n}{5}$. These are given to be equal:
$$\binom{n}{3}=\binom{n}{5}$$
By the symmetry property of binomial coefficients, $\binom{n}{r}=\binom{n}{n-r}$, so $\binom{n}{3}=\binom{n}{n-3}$. Comparing with $\binom{n}{5}$, either $n-3=5$ (giving a genuinely different pair of indices) or $3=5$ (impossible). Hence:
$$n-3=5$$
$$\boxed{n=8}$$
Check: $\binom{8}{3}=56$ and $\binom{8}{5}=56$, confirming the coefficients are equal.
QUESTION 35
7 marks
Medium
By writing $\sqrt{1.1}$ as $(1+0.1)^{1/2}$ and using the first four terms of the binomial expansion of $(1+x)^{1/2}$, find an approximate value of $\sqrt{1.1}$, giving your answer correct to 4 decimal places.
Show complete worked solution
Using $(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\ldots$ with $n=\dfrac{1}{2}$:
Term in $x$: $\dfrac{1}{2}x$
Term in $x^2$:
$$\dfrac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2}x^2=-\dfrac{1}{8}x^2$$
Term in $x^3$:
$$\dfrac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{6}x^3=\dfrac{3/8}{6}x^3=\dfrac{1}{16}x^3$$
So
$$(1+x)^{1/2}\approx1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3$$
Substitute $x=0.1$:
$$1+\dfrac{1}{2}(0.1)-\dfrac{1}{8}(0.1)^2+\dfrac{1}{16}(0.1)^3$$
$$=1+0.05-0.00125+0.0000625$$
$$=1.0488125$$
$$\boxed{\sqrt{1.1}\approx1.0488\text{ (4 d.p.)}}$$
QUESTION 36
7 marks
Medium
Find the constant term in the expansion of $\left(2x^2+\dfrac{1}{x}\right)^6$.
Show complete worked solution
The general term of $\left(2x^2+\dfrac{1}{x}\right)^6$ is
$$T_{r+1}=\binom{6}{r}(2x^2)^{6-r}\left(\dfrac{1}{x}\right)^r=\binom{6}{r}2^{6-r}x^{12-2r}x^{-r}=\binom{6}{r}2^{6-r}x^{12-3r}$$
The term is constant when the exponent of $x$ is zero:
$$12-3r=0$$
$$r=4$$
Substitute $r=4$:
$$T_5=\binom{6}{4}2^{2}$$
Compute $\binom{6}{4}=15$ and $2^2=4$:
$$T_5=15\times4=60$$
$$\boxed{\text{Constant term}=60}$$
QUESTION 37
7 marks
Medium
In the expansion of $(1+x)^n$, the coefficient of the term in $x^4$ is four times the coefficient of the term in $x^3$. Find the value of $n$.
Show complete worked solution
The coefficient of $x^3$ is $\binom{n}{3}$ and the coefficient of $x^4$ is $\binom{n}{4}$. The condition states
$$\binom{n}{4}=4\binom{n}{3}$$
Using $\binom{n}{4}=\binom{n}{3}\times\dfrac{n-3}{4}$:
$$\binom{n}{3}\times\dfrac{n-3}{4}=4\binom{n}{3}$$
Since $\binom{n}{3}\ne0$ for $n\geq3$, divide both sides by $\binom{n}{3}$:
$$\dfrac{n-3}{4}=4$$
$$n-3=16$$
$$\boxed{n=19}$$
Check: $\binom{19}{3}=969$ and $\binom{19}{4}=3876=4\times969$, confirming the result.
QUESTION 38
7 marks
Medium
Consider the expansion of $(1+x)^n$, where $n \in \mathbb{Z}^+$.
(a) [4] Given that the third term (in ascending powers of $x$) is $45x^2$, find the value of $n$.
(b) [3] Hence find the coefficient of $x^7$ in this expansion.
Show complete worked solution
(a) In ascending powers of $x$, the third term corresponds to $r=2$ in $T_{r+1}=\binom{n}{r}x^r$:
$$T_3=\binom{n}{2}x^2$$
Setting this equal to $45x^2$:
$$\binom{n}{2}=45$$
$$\dfrac{n(n-1)}{2}=45$$
$$n(n-1)=90$$
$$n^2-n-90=0$$
Factorise:
$$(n-10)(n+9)=0$$
$$n=10 \text{ or } n=-9$$
Since $n\in\mathbb{Z}^+$, reject $n=-9$.
$$\boxed{n=10}$$
(b) The coefficient of $x^7$ in $(1+x)^{10}$ is $\binom{10}{7}$:
$$\binom{10}{7}=\dfrac{10!}{7!\,3!}=120$$
$$\boxed{\text{Coefficient of }x^7=120}$$
QUESTION 39
8 marks
Medium
Find the first three terms, in ascending powers of $x$, of the binomial expansion of $(8+x)^{1/3}$, by first writing $(8+x)^{1/3}=8^{1/3}\left(1+\dfrac{x}{8}\right)^{1/3}$. State the range of values of $x$ for which the expansion is valid.
Show complete worked solution
Since $8^{1/3}=2$, write
$$(8+x)^{1/3}=2\left(1+\dfrac{x}{8}\right)^{1/3}$$
Using $(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\ldots$ with $u=\dfrac{x}{8}$ and $n=\dfrac{1}{3}$:
Term 0: $1$
Term in $u$:
$$nu=\dfrac{1}{3}\times\dfrac{x}{8}=\dfrac{x}{24}$$
Term in $u^2$:
$$\dfrac{n(n-1)}{2}u^2=\dfrac{\tfrac{1}{3}\left(-\tfrac{2}{3}\right)}{2}\left(\dfrac{x}{8}\right)^2=-\dfrac{1}{9}\times\dfrac{x^2}{64}=-\dfrac{x^2}{576}$$
So
$$\left(1+\dfrac{x}{8}\right)^{1/3}\approx1+\dfrac{x}{24}-\dfrac{x^2}{576}$$
Multiplying through by the factor of $2$:
$$\boxed{(8+x)^{1/3}\approx2+\dfrac{x}{12}-\dfrac{x^2}{288}}$$
Validity: the inner expansion is valid for $|u|<1$, i.e. $\left|\dfrac{x}{8}\right|<1$, so
$$\boxed{-8
QUESTION 40
7 marks
Medium
In the expansion of $\left(x+\dfrac{1}{x}\right)^8$:
(a) [4] Find the term independent of $x$.
(b) [3] Find the coefficient of $x^4$.
Show complete worked solution
The general term of $\left(x+\dfrac{1}{x}\right)^8$ is
$$T_{r+1}=\binom{8}{r}x^{8-r}\left(\dfrac{1}{x}\right)^r=\binom{8}{r}x^{8-2r}$$
(a) The term is independent of $x$ when $8-2r=0$, i.e. $r=4$:
$$T_5=\binom{8}{4}=70$$
$$\boxed{\text{Term independent of }x=70}$$
(b) The term in $x^4$ requires $8-2r=4$, i.e. $r=2$:
$$T_3=\binom{8}{2}x^4$$
Compute $\binom{8}{2}=28$:
$$T_3=28x^4$$
$$\boxed{\text{Coefficient of }x^4=28}$$
QUESTION 41
8 marks
Medium
Find the first three terms, in ascending powers of $x$, of the binomial expansion of $(4-x)^{-1/2}$, by first writing $(4-x)^{-1/2}=4^{-1/2}\left(1-\dfrac{x}{4}\right)^{-1/2}$. State the range of values of $x$ for which the expansion is valid.
Show complete worked solution
Since $4^{-1/2}=\dfrac{1}{2}$, write
$$(4-x)^{-1/2}=\dfrac{1}{2}\left(1-\dfrac{x}{4}\right)^{-1/2}$$
Using $(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\ldots$ with $u=-\dfrac{x}{4}$ and $n=-\dfrac{1}{2}$:
Term 0: $1$
Term in $u$:
$$nu=\left(-\dfrac{1}{2}\right)\left(-\dfrac{x}{4}\right)=\dfrac{x}{8}$$
Term in $u^2$:
$$\dfrac{n(n-1)}{2}u^2=\dfrac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{2}\left(\dfrac{x}{4}\right)^2=\dfrac{3}{8}\times\dfrac{x^2}{16}=\dfrac{3x^2}{128}$$
So
$$\left(1-\dfrac{x}{4}\right)^{-1/2}\approx1+\dfrac{x}{8}+\dfrac{3x^2}{128}$$
Multiplying through by the factor of $\dfrac{1}{2}$:
$$\boxed{(4-x)^{-1/2}\approx\dfrac{1}{2}+\dfrac{x}{16}+\dfrac{3x^2}{256}}$$
Validity: the inner expansion is valid for $|u|<1$, i.e. $\left|\dfrac{x}{4}\right|<1$, so
$$\boxed{-4
QUESTION 42
8 marks
Medium
The first three terms, in ascending powers of $x$, of the binomial expansion of $(1+ax)^n$, where $n\in\mathbb{Z}^+$ and $a>0$, are $1+24x+252x^2+\ldots$. Find the values of $a$ and $n$.
Show complete worked solution
By the binomial theorem, $(1+ax)^n=1+nax+\dbinom{n}{2}a^2x^2+\ldots$. Comparing with the given expansion:
$$na=24 \quad \text{(i)}$$
$$\binom{n}{2}a^2=252 \quad \text{(ii)}$$
From (i), $a=\dfrac{24}{n}$. Substitute into (ii), using $\binom{n}{2}=\dfrac{n(n-1)}{2}$:
$$\dfrac{n(n-1)}{2}\times\dfrac{576}{n^2}=252$$
$$\dfrac{288(n-1)}{n}=252$$
$$288(n-1)=252n$$
$$288n-288=252n$$
$$36n=288$$
$$n=8$$
Substitute back into (i):
$$a=\dfrac{24}{8}=3$$
$$\boxed{a=3,\ n=8}$$
Check: $\binom{8}{2}(3)^2=28\times9=252$, confirming the value of the $x^2$ coefficient.
QUESTION 43
10 marks
Hard
In the expansion of $(1+x)^n$, three consecutive binomial coefficients $\dbinom{n}{r-1}$, $\dbinom{n}{r}$, $\dbinom{n}{r+1}$ satisfy
$$\binom{n}{r-1}:\binom{n}{r}:\binom{n}{r+1}=1:3:5$$
(a) [5] Using the identity $\dbinom{n}{r}=\dbinom{n}{r-1}\times\dfrac{n-r+1}{r}$ (and its analogue for $\dbinom{n}{r+1}$ in terms of $\dbinom{n}{r}$), show that $n=4r-1$ and $3n=8r+5$. Hence find the values of $n$ and $r$.
(b) [5] Hence find the value of $\dbinom{n}{r-1}+\dbinom{n}{r}+\dbinom{n}{r+1}$.
Show complete worked solution
(a) Since $\dbinom{n}{r}=\dbinom{n}{r-1}\times\dfrac{n-r+1}{r}$, the ratio $\dbinom{n}{r-1}:\dbinom{n}{r}=1:3$ gives
$$\dfrac{\binom{n}{r}}{\binom{n}{r-1}}=\dfrac{n-r+1}{r}=3$$
$$n-r+1=3r$$
$$n=4r-1 \quad \text{(i)}$$
Similarly, $\dbinom{n}{r+1}=\dbinom{n}{r}\times\dfrac{n-r}{r+1}$, so the ratio $\dbinom{n}{r}:\dbinom{n}{r+1}=3:5$ gives
$$\dfrac{\binom{n}{r+1}}{\binom{n}{r}}=\dfrac{n-r}{r+1}=\dfrac{5}{3}$$
$$3(n-r)=5(r+1)$$
$$3n-3r=5r+5$$
$$3n=8r+5 \quad \text{(ii)}$$
as required. Substituting (i) into (ii):
$$3(4r-1)=8r+5$$
$$12r-3=8r+5$$
$$4r=8$$
$$r=2$$
Then from (i), $n=4(2)-1=7$.
$$\boxed{n=7,\ r=2}$$
(b) With $n=7$, $r=2$:
$$\binom{7}{1}=7,\qquad\binom{7}{2}=21,\qquad\binom{7}{3}=35$$
(Check: $7:21:35=1:3:5$, confirming the values found in part (a).)
$$\binom{7}{1}+\binom{7}{2}+\binom{7}{3}=7+21+35=63$$
$$\boxed{63}$$
QUESTION 44
11 marks
Hard
(a) [3] By substituting $x=1$ into the binomial expansion of $(1+x)^n$, show that
$$\sum_{r=0}^{n}\binom{n}{r}=2^n$$
(b) [3] By substituting an appropriate value of $x$ into the same expansion, show that, for $n\geq1$,
$$\sum_{r=0}^{n}(-1)^r\binom{n}{r}=0$$
(c) [5] Hence, using the results of parts (a) and (b) with $n=10$, find the value of
$$\binom{10}{0}+\binom{10}{2}+\binom{10}{4}+\binom{10}{6}+\binom{10}{8}+\binom{10}{10}$$
(the sum of the binomial coefficients at even positions).
Show complete worked solution
(a) By the binomial theorem, $(1+x)^n=\displaystyle\sum_{r=0}^{n}\binom{n}{r}x^r$. Substituting $x=1$:
$$(1+1)^n=\sum_{r=0}^{n}\binom{n}{r}(1)^r$$
$$2^n=\sum_{r=0}^{n}\binom{n}{r}$$
as required, so $\boxed{\displaystyle\sum_{r=0}^{n}\binom{n}{r}=2^n}$.
(b) Substituting $x=-1$ into the same expansion:
$$(1+(-1))^n=\sum_{r=0}^{n}\binom{n}{r}(-1)^r$$
$$0^n=\sum_{r=0}^{n}(-1)^r\binom{n}{r}$$
For $n\geq1$, $0^n=0$, so $\boxed{\displaystyle\sum_{r=0}^{n}(-1)^r\binom{n}{r}=0}$, as required.
(c) Let $E$ denote the sum of the coefficients at even positions ($r=0,2,4,6,8,10$) and $O$ denote the sum at odd positions ($r=1,3,5,7,9$), for $n=10$.
From part (a) with $n=10$: $E+O=2^{10}=1024$.
From part (b) with $n=10$: $E-O=0$ (since the alternating sum groups exactly the even-position terms as positive and odd-position terms as negative).
Adding these two equations:
$$2E=1024$$
$$E=512$$
$$\boxed{\binom{10}{0}+\binom{10}{2}+\binom{10}{4}+\binom{10}{6}+\binom{10}{8}+\binom{10}{10}=512}$$
QUESTION 45
11 marks
Hard
(a) [7] Find the first four terms, in ascending powers of $x$, of the binomial expansion of $(1-2x)^{-1/2}$. State the range of values of $x$ for which the expansion is valid.
(b) [4] By substituting an appropriate value of $x$ into your expansion, find an approximate value of $\dfrac{1}{\sqrt{1.02}}$, giving your answer correct to 5 decimal places.
Show complete worked solution
(a) Using $(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\dfrac{n(n-1)(n-2)}{3!}u^3+\ldots$ with $u=-2x$ and $n=-\dfrac{1}{2}$:
Term 0: $1$
Term in $u$:
$$nu=\left(-\dfrac{1}{2}\right)(-2x)=x$$
Term in $u^2$:
$$\dfrac{n(n-1)}{2}u^2=\dfrac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{2}(-2x)^2=\dfrac{3}{8}\times4x^2=\dfrac{3}{2}x^2$$
Term in $u^3$:
$$\dfrac{n(n-1)(n-2)}{6}u^3=\dfrac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)\left(-\tfrac{5}{2}\right)}{6}(-2x)^3=\left(-\dfrac{5}{16}\right)(-8x^3)=\dfrac{5}{2}x^3$$
$$\boxed{(1-2x)^{-1/2}\approx1+x+\dfrac{3}{2}x^2+\dfrac{5}{2}x^3}$$
Validity: for non-integer $n$, the series converges for $|u|<1$, i.e. $|-2x|<1$, so
$$\boxed{-\dfrac{1}{2}
QUESTION 46
13 marks
Hard
Consider the function $f(x)=\dfrac{\sqrt{1+x}}{1-2x}$.
(a) [5] Find the binomial expansion of $\sqrt{1+x}$, up to and including the term in $x^2$, and separately the binomial expansion of $(1-2x)^{-1}$, up to and including the term in $x^2$.
(b) [4] Hence show that the first three terms, in ascending powers of $x$, of the expansion of $f(x)$ are $1+\dfrac{5}{2}x+\dfrac{39}{8}x^2$.
(c) [4] State, with justification, the range of values of $x$ for which the full expansion of $f(x)$ is valid.
Show complete worked solution
(a) Write $\sqrt{1+x}=(1+x)^{1/2}$. Using $(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\ldots$ with $n=\dfrac{1}{2}$:
$$\sqrt{1+x}\approx1+\dfrac{1}{2}x+\dfrac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2}x^2=1+\dfrac{1}{2}x-\dfrac{1}{8}x^2$$
Write $(1-2x)^{-1}$. Using the same series with $u=-2x$, $n=-1$:
$$(1-2x)^{-1}\approx1+(-1)(-2x)+\dfrac{(-1)(-2)}{2}(-2x)^2=1+2x+4x^2$$
$$\boxed{\sqrt{1+x}\approx1+\dfrac{1}{2}x-\dfrac{1}{8}x^2,\qquad(1-2x)^{-1}\approx1+2x+4x^2}$$
(b) Multiply the two expansions and collect terms up to $x^2$, discarding any term of higher degree:
$$\left(1+\dfrac{1}{2}x-\dfrac{1}{8}x^2\right)\left(1+2x+4x^2\right)$$
The constant term: $1\times1=1$.
The $x$ term: $1\times2x+\dfrac{1}{2}x\times1=2x+\dfrac{1}{2}x=\dfrac{5}{2}x$.
The $x^2$ term: $1\times4x^2+\dfrac{1}{2}x\times2x+\left(-\dfrac{1}{8}x^2\right)\times1=4x^2+x^2-\dfrac{1}{8}x^2=\dfrac{39}{8}x^2$.
Combining:
$$\boxed{f(x)\approx1+\dfrac{5}{2}x+\dfrac{39}{8}x^2}$$
as required.
(c) The expansion of $\sqrt{1+x}=(1+x)^{1/2}$ is valid for $|x|<1$. The expansion of $(1-2x)^{-1}$ is valid for $|-2x|<1$, i.e. $|x|<\dfrac{1}{2}$. Since $f(x)$ is the product of both series, its expansion is valid only where **both** individual expansions converge, i.e. where the more restrictive condition holds:
$$\boxed{-\dfrac{1}{2}
QUESTION 47
10 marks
Hard
(a) [6] Show that, for $0\leq r\leq9$, the ratio of consecutive binomial coefficients in the expansion of $(1+x)^{10}$ satisfies
$$\dfrac{\binom{10}{r+1}}{\binom{10}{r}}=\dfrac{10-r}{r+1}$$
and hence find the set of values of $r$ for which $\binom{10}{r+1}>\binom{10}{r}$ (i.e. the coefficients are strictly increasing).
(b) [4] Hence determine the term with the greatest binomial coefficient in the expansion of $(1+x)^{10}$, and state this coefficient.
Show complete worked solution
(a) Using the factorial definition,
$$\dfrac{\binom{10}{r+1}}{\binom{10}{r}}=\dfrac{\dfrac{10!}{(r+1)!(9-r)!}}{\dfrac{10!}{r!(10-r)!}}=\dfrac{r!(10-r)!}{(r+1)!(9-r)!}$$
Since $(r+1)!=(r+1)r!$ and $(10-r)!=(10-r)(9-r)!$:
$$=\dfrac{r!\times(10-r)(9-r)!}{(r+1)r!\times(9-r)!}=\dfrac{10-r}{r+1}$$
as required. The coefficients are strictly increasing (moving from $\binom{10}{r}$ to $\binom{10}{r+1}$) exactly when this ratio exceeds $1$:
$$\dfrac{10-r}{r+1}>1$$
Since $r+1>0$ for all valid $r$, this is equivalent to
$$10-r>r+1$$
$$9>2r$$
$$r<4.5$$
Since $r$ is a non-negative integer, the coefficients strictly increase for
$$\boxed{r=0,1,2,3,4}$$
(that is, $\binom{10}{1}>\binom{10}{0}$, $\binom{10}{2}>\binom{10}{1}$, ..., $\binom{10}{5}>\binom{10}{4}$).
(b) Since the coefficients increase up to and including $\binom{10}{5}$ (reached when $r=4$ gives $\binom{10}{5}>\binom{10}{4}$), and by the symmetry property $\binom{10}{r}=\binom{10}{10-r}$ the coefficients must decrease again after $r=5$, the greatest binomial coefficient occurs at $r=5$, i.e. in the term $T_6$, the term in $x^5$.
$$\binom{10}{5}=\dfrac{10!}{5!\,5!}=252$$
$$\boxed{\text{Greatest coefficient}=252,\text{ occurring in the term in }x^5}$$
QUESTION 48
10 marks
Hard
(a) [4] Find the coefficient of $x^3$ in the binomial expansion of $(1+x)^{-4}$, in ascending powers of $x$.
(b) [3] Given that, for $k>0$, the coefficient of $x^3$ in the expansion of $(1+kx)^{-4}$ is $-160$, find the exact value of $k$.
(c) [3] State the range of values of $x$ for which the expansion of $(1+kx)^{-4}$, using this value of $k$, is valid.
Show complete worked solution
(a) Using $(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\ldots$ with $n=-4$, the coefficient of $x^3$ is
$$\dfrac{n(n-1)(n-2)}{6}=\dfrac{(-4)(-5)(-6)}{6}=\dfrac{-120}{6}=-20$$
$$\boxed{\text{Coefficient of }x^3\text{ in }(1+x)^{-4}=-20}$$
(b) In the expansion of $(1+kx)^{-4}$, replacing $x$ by $kx$ in the result of part (a), the coefficient of $x^3$ is $-20k^3$ (since each power of $x$ picks up a corresponding power of $k$). This is given to equal $-160$:
$$-20k^3=-160$$
$$k^3=8$$
$$\boxed{k=2}$$
(c) The expansion of $(1+kx)^{-4}$ is valid for $|kx|<1$. With $k=2$:
$$|2x|<1$$
$$\boxed{-\dfrac{1}{2}
QUESTION 49
13 marks
Hard
In the expansion of $(a+bx)^5$, where $a$ and $b$ are positive constants, the coefficient of the $x^2$ term is $720$ and the coefficient of the $x^3$ term is $1080$.
(a) [6] Form two equations in $a$ and $b$ from this information, and hence show that $b=\dfrac{3a}{2}$.
(b) [4] Hence find the exact values of $a$ and $b$.
(c) [3] Using these values, find the term in $x^4$ in the expansion.
Show complete worked solution
(a) The general term of $(a+bx)^5$ is $T_{r+1}=\dbinom{5}{r}a^{5-r}b^rx^r$. The coefficient of $x^2$ (set $r=2$) is $\dbinom{5}{2}a^3b^2=10a^3b^2$, and the coefficient of $x^3$ (set $r=3$) is $\dbinom{5}{3}a^2b^3=10a^2b^3$. This gives the two equations:
$$10a^3b^2=720 \quad\Rightarrow\quad a^3b^2=72 \quad \text{(i)}$$
$$10a^2b^3=1080 \quad\Rightarrow\quad a^2b^3=108 \quad \text{(ii)}$$
Dividing (ii) by (i):
$$\dfrac{a^2b^3}{a^3b^2}=\dfrac{108}{72}$$
$$\dfrac{b}{a}=\dfrac{3}{2}$$
$$\boxed{b=\dfrac{3a}{2}}$$
as required.
(b) Substitute $b=\dfrac{3a}{2}$ into equation (i):
$$a^3\left(\dfrac{3a}{2}\right)^2=72$$
$$a^3\times\dfrac{9a^2}{4}=72$$
$$\dfrac{9a^5}{4}=72$$
$$a^5=32$$
$$a=2$$
Then
$$b=\dfrac{3(2)}{2}=3$$
$$\boxed{a=2,\ b=3}$$
Check: $a^3b^2=8\times9=72$, correct, and $a^2b^3=4\times27=108$, also correct.
(c) The term in $x^4$ corresponds to $r=4$:
$$T_5=\binom{5}{4}a^1b^4x^4=5\times2\times3^4x^4$$
Compute $3^4=81$:
$$T_5=5\times2\times81\times x^4=810x^4$$
$$\boxed{810x^4}$$
QUESTION 50
9 marks
Hard
(a) [3] Using the identity $\dbinom{n}{r+1}=\dbinom{n}{r}\times\dfrac{n-r}{r+1}$, express $\dbinom{14}{5}$ and $\dbinom{14}{6}$ each in terms of $\dbinom{14}{4}$.
(b) [4] Hence show that $\dbinom{14}{4}$, $\dbinom{14}{5}$, $\dbinom{14}{6}$ (three consecutive binomial coefficients in the expansion of $(1+x)^{14}$) form an arithmetic sequence, and state the common difference.
(c) [2] Write down the term in $x^5$ in the expansion of $(1+x)^{14}$.
Show complete worked solution
(a) Using $\dbinom{n}{r+1}=\dbinom{n}{r}\times\dfrac{n-r}{r+1}$ with $n=14$, $r=4$:
$$\binom{14}{5}=\binom{14}{4}\times\dfrac{14-4}{5}=\binom{14}{4}\times\dfrac{10}{5}=2\binom{14}{4}$$
With $n=14$, $r=5$:
$$\binom{14}{6}=\binom{14}{5}\times\dfrac{14-5}{6}=\binom{14}{5}\times\dfrac{9}{6}=\dfrac{3}{2}\binom{14}{5}=\dfrac{3}{2}\times2\binom{14}{4}=3\binom{14}{4}$$
$$\boxed{\binom{14}{5}=2\binom{14}{4},\qquad\binom{14}{6}=3\binom{14}{4}}$$
(b) From part (a), if we let $c=\dbinom{14}{4}$, then the three consecutive coefficients are $c$, $2c$, $3c$. The differences between consecutive terms are
$$2c-c=c \qquad \text{and} \qquad 3c-2c=c$$
Since both differences equal $c$, the three coefficients form an arithmetic sequence with common difference $c=\dbinom{14}{4}$.
Computing $\dbinom{14}{4}=\dfrac{14!}{4!\,10!}=\dfrac{14\times13\times12\times11}{24}=1001$:
$$\binom{14}{4}=1001,\qquad\binom{14}{5}=2002,\qquad\binom{14}{6}=3003$$
Indeed, $2002-1001=1001$ and $3003-2002=1001$.
$$\boxed{\text{Common difference}=1001}$$
(c) The term in $x^5$ in $(1+x)^{14}$ is $\dbinom{14}{5}x^5$:
$$\boxed{2002x^5}$$
Counting Principles 50 questions
QUESTION 1
4 marks
Easy
A committee of 4 people is to be chosen from a group of 9 people. Find the number of different committees that can be formed.
Show complete worked solution
Since the order in which committee members are chosen does not matter, this is a combinations problem. The number of ways to choose 4 people from 9 is
$$\binom{9}{4}=\dfrac{9!}{4!\,5!}$$
Compute:
$$\binom{9}{4}=\dfrac{9\times8\times7\times6}{4\times3\times2\times1}=\dfrac{3024}{24}=126$$
$$\boxed{126\text{ committees}}$$
QUESTION 2
5 marks
Medium
The word "MATHS" has 5 distinct letters. (i) Find the number of different 5-letter arrangements that can be made using each letter exactly once. (ii) Find the number of these arrangements that begin with a vowel.
Show complete worked solution
(i) All 5 letters (M, A, T, H, S) are distinct, so the number of arrangements of all 5 letters is
$$5!=5\times4\times3\times2\times1=120$$
$$\boxed{120\text{ arrangements}}$$
(ii) The only vowel among the letters is A. Fixing A in the first position, the remaining 4 letters (M, T, H, S) can be arranged in the remaining 4 positions in
$$4!=4\times3\times2\times1=24$$
ways.
$$\boxed{24\text{ arrangements begin with a vowel}}$$
QUESTION 3
6 marks
Medium
A security code consists of 4 digits (chosen from 0-9) followed by 3 letters (chosen from the 26 letters of the alphabet). No digit may be repeated within the code, but letters may be repeated. Find the total number of different codes that can be formed.
Show complete worked solution
The digit part and the letter part are independent, so by the multiplication principle the total number of codes is the product of the two counts.
Digits (no repetition): the first digit can be chosen in 10 ways, the second in 9 remaining ways, the third in 8 ways, and the fourth in 7 ways:
$$10\times9\times8\times7=5040$$
Letters (repetition allowed): each of the 3 letters can independently be any of the 26 letters:
$$26^3=17576$$
By the multiplication principle, the total number of codes is:
$$5040\times17576=88\,583\,040$$
$$\boxed{88\,583\,040\text{ codes}}$$
QUESTION 4
5 marks
Easy
In how many ways can 6 different books be arranged in a row on a shelf if two particular books must be placed next to each other?
Show complete worked solution
Treat the two particular books that must stay together as a single combined unit. Together with the remaining 4 books, this gives 5 units to arrange in a row:
$$5!=120\text{ ways}$$
Within the combined unit, the two particular books can be arranged in $2!=2$ different orders.
By the multiplication principle, the total number of arrangements is
$$5!\times2!=120\times2=240$$
$$\boxed{240\text{ arrangements}}$$
QUESTION 5
15 marks
Hard
A school has 12 students available for a committee: 7 boys and 5 girls.
(a) [3 marks] Find the number of ways to select a committee of 5 students from the 12, with no restrictions on composition.
(b) [4 marks] Find the number of ways to select a committee of 5 students that contains exactly 3 boys and 2 girls.
(c) [4 marks] Find the number of ways to select a committee of 5 students that contains at least 1 girl.
(d) [4 marks] Once an (unrestricted) committee of 5 has been chosen from the 12 students, one member is then assigned to be chairperson and a different member is assigned to be secretary. Find the total number of ways to first select the committee of 5 from the 12 students, and then assign the chairperson and secretary roles.
Show complete worked solution
(a) Since order does not matter, this is a combinations problem:
$$\binom{12}{5}=\dfrac{12!}{5!\,7!}=\dfrac{12\times11\times10\times9\times8}{5\times4\times3\times2\times1}=\dfrac{95040}{120}=792$$
$$\boxed{792}$$
(b) Choose 3 boys from 7, and independently choose 2 girls from 5. By the multiplication principle, multiply the two counts:
$$\binom{7}{3}\times\binom{5}{2}=35\times10=350$$
$$\boxed{350}$$
(c) Use the complement: (committees with at least 1 girl) = (all committees) - (committees with no girls, i.e. all 5 members are boys).
Number with no girls: $\binom{7}{5}=21$ (choosing all 5 members from the 7 boys).
$$\binom{12}{5}-\binom{7}{5}=792-21=771$$
$$\boxed{771}$$
(d) First choose the unrestricted committee of 5 from 12: $\binom{12}{5}=792$ ways.
Then, from within that committee of 5, assign 2 distinct roles (chairperson, secretary) to 2 different members. This is a permutation of 2 objects chosen from 5:
$$P(5,2)=5\times4=20$$
By the multiplication principle, the total number of ways is
$$792\times20=15\,840$$
$$\boxed{15\,840}$$
QUESTION 6
4 marks
Easy
A sandwich shop offers a build-your-own combo: a choice of 4 different types of bread, 3 different fillings, and 2 different sauces. A customer selects exactly one bread, one filling, and one sauce. Find the number of different sandwich combinations possible.
Show complete worked solution
Since the customer makes three independent choices - one bread, one filling, and one sauce - the total number of combinations is found using the multiplication principle: multiply the number of options for each choice together.
$$4\times3\times2=24$$
$$\boxed{24\text{ sandwich combinations}}$$
QUESTION 7
4 marks
Easy
A café offers 5 different cakes and 7 different ice-cream flavours. A customer will choose exactly one dessert, either a cake or an ice-cream, but not both. Find the number of different desserts the customer could choose.
Show complete worked solution
Choosing a cake and choosing an ice-cream are mutually exclusive options - the customer picks from one group or the other, not both. By the addition principle, the total number of choices is the sum of the number of options in each group.
$$5+7=12$$
$$\boxed{12\text{ possible desserts}}$$
QUESTION 8
4 marks
Easy
An art gallery has 7 different paintings to hang in a single row along one wall. Find the number of different ways the paintings can be arranged.
Show complete worked solution
All 7 paintings are different and every position in the row is distinct, so this is a permutation of 7 distinct objects. The number of arrangements is
$$7!=7\times6\times5\times4\times3\times2\times1$$
Compute:
$$7!=5040$$
$$\boxed{5040\text{ arrangements}}$$
QUESTION 9
4 marks
Easy
A website requires users to create a 5-character password using only the digits 0-9, where digits may be repeated any number of times. Find the total number of different passwords possible.
Show complete worked solution
Each of the 5 character positions can independently be any of the 10 digits, since repetition is allowed. By the multiplication principle, the total number of passwords is
$$10\times10\times10\times10\times10=10^5$$
$$\boxed{100\,000\text{ passwords}}$$
QUESTION 10
5 marks
Easy
10 horses compete in a race. Assuming there are no ties, find the number of different ways the 1st, 2nd, 3rd and 4th place finishers can be determined.
Show complete worked solution
The order in which horses finish matters, and no horse can finish in more than one place, so this is a permutation of 4 positions chosen from 10 horses.
$$P(10,4)=\dfrac{10!}{(10-4)!}=10\times9\times8\times7$$
Compute:
$$10\times9\times8\times7=5040$$
$$\boxed{5040\text{ ways}}$$
QUESTION 11
5 marks
Easy
A music app allows a user to select 6 songs from a list of 18 songs to create a playlist. The order in which the songs are selected does not matter. Find the number of different playlists that can be formed.
Show complete worked solution
Since the order of selection does not matter, this is a combinations problem. The number of ways to choose 6 songs from 18 is
$$\binom{18}{6}=\dfrac{18!}{6!\,12!}$$
Compute:
$$\binom{18}{6}=\dfrac{18\times17\times16\times15\times14\times13}{6\times5\times4\times3\times2\times1}=\dfrac{13\,366\,080}{720}=18\,564$$
$$\boxed{18\,564\text{ playlists}}$$
QUESTION 12
4 marks
Easy
A fruit salad is made by choosing 3 different kinds of fruit from 9 kinds available at a market. The order of selection does not matter. Find the number of different fruit salads that can be made.
Show complete worked solution
Since the order of selection does not matter, this is a combinations problem.
$$\binom{9}{3}=\dfrac{9!}{3!\,6!}=\dfrac{9\times8\times7}{3\times2\times1}=\dfrac{504}{6}=84$$
$$\boxed{84\text{ fruit salads}}$$
QUESTION 13
5 marks
Easy
8 friends sit down for dinner at a circular table. Two seating arrangements are considered the same if one can be obtained from the other by rotating everyone around the table. Find the number of different seating arrangements.
Show complete worked solution
For a circular arrangement of $n$ distinct objects where rotations are considered identical, the number of arrangements is $(n-1)!$ rather than $n!$, since fixing one person's seat removes the rotational symmetry and the remaining $n-1$ people can be arranged in the remaining seats in any order.
Here $n=8$, so the number of seating arrangements is
$$(8-1)!=7!=7\times6\times5\times4\times3\times2\times1=5040$$
$$\boxed{5040\text{ seating arrangements}}$$
QUESTION 14
5 marks
Easy
Find the number of distinct arrangements of all the letters of the word "LETTER".
Show complete worked solution
The word LETTER has 6 letters in total, but they are not all different: the letter E occurs twice and the letter T occurs twice (L and R occur once each). When arranging a multiset of objects, the number of distinct arrangements is the total factorial divided by the factorial of each repeated letter's count, since swapping identical letters with each other does not create a new arrangement.
$$\dfrac{6!}{2!\,2!}=\dfrac{720}{4}=180$$
$$\boxed{180\text{ distinct arrangements}}$$
QUESTION 15
4 marks
Easy
A code is formed by choosing 3 digits from 1 to 7, with no digit repeated, and arranging them in a row. Find the number of different codes that can be formed.
Show complete worked solution
Since no digit may repeat and the order of the digits matters, this is a permutation of 3 positions chosen from 7 digits.
$$P(7,3)=7\times6\times5=210$$
$$\boxed{210\text{ codes}}$$
QUESTION 16
4 marks
Easy
Three distinguishable six-sided dice - a red die, a blue die, and a green die - are rolled once. Find the number of different possible outcomes, where an outcome records the number shown on each die.
Show complete worked solution
Each die independently shows one of 6 possible numbers, and the three results are recorded separately since the dice are distinguishable. By the multiplication principle,
$$6\times6\times6=6^3=216$$
$$\boxed{216\text{ outcomes}}$$
QUESTION 17
5 marks
Easy
A hand of 4 cards is dealt from a standard, well-shuffled deck of 52 playing cards. Find the number of different 4-card hands possible, where the order in which the cards are dealt does not matter.
Show complete worked solution
Since the order of the cards does not matter, this is a combinations problem.
$$\binom{52}{4}=\dfrac{52\times51\times50\times49}{4\times3\times2\times1}=\dfrac{6\,497\,400}{24}=270\,725$$
$$\boxed{270\,725\text{ hands}}$$
QUESTION 18
4 marks
Easy
9 athletes compete in a 100m sprint final. Assuming there are no ties, find the number of different ways gold, silver, and bronze medals can be awarded.
Show complete worked solution
The order in which the medals are awarded matters, and no athlete can receive more than one medal, so this is a permutation of 3 positions chosen from 9 athletes.
$$P(9,3)=9\times8\times7=504$$
$$\boxed{504\text{ ways}}$$
QUESTION 19
5 marks
Easy
A company selects 5 employees from a department of 16 employees to attend a training course. The order of selection does not matter. Find the number of different groups of 5 employees that could be selected.
Show complete worked solution
Since the order of selection does not matter, this is a combinations problem.
$$\binom{16}{5}=\dfrac{16\times15\times14\times13\times12}{5\times4\times3\times2\times1}=\dfrac{524\,160}{120}=4368$$
$$\boxed{4368\text{ groups}}$$
QUESTION 20
4 marks
Easy
A student has 6 different shirts and 5 different pairs of trousers. Find the number of different outfits, one shirt and one pair of trousers, the student could wear.
Show complete worked solution
The shirt and the trousers are chosen independently, so by the multiplication principle the number of outfits is
$$6\times5=30$$
$$\boxed{30\text{ outfits}}$$
QUESTION 21
4 marks
Easy
6 different flags are to be arranged in a row on a display stand, one flag per position. Find the number of different arrangements possible.
Show complete worked solution
All 6 flags are different and every position is distinct, so this is a permutation of 6 distinct objects.
$$6!=6\times5\times4\times3\times2\times1=720$$
$$\boxed{720\text{ arrangements}}$$
QUESTION 22
6 marks
Easy
In a lottery game, a player chooses 5 different numbers from the integers 1 to 40. The order in which the numbers are chosen does not matter. Find the number of different selections of 5 numbers that are possible.
Show complete worked solution
Since the order of selection does not matter, this is a combinations problem.
$$\binom{40}{5}=\dfrac{40\times39\times38\times37\times36}{5\times4\times3\times2\times1}=\dfrac{78\,960\,960}{120}=658\,008$$
$$\boxed{658\,008\text{ selections}}$$
QUESTION 23
5 marks
Easy
6 different keys are placed on a circular keyring. Two arrangements of keys on the ring are considered the same if one can be obtained from the other either by rotating the ring or by flipping it over, since the ring can be viewed from either side. Find the number of distinct arrangements of the 6 keys on the ring.
Show complete worked solution
For a circular arrangement of $n$ distinct objects, rotations alone give $(n-1)!$ distinct arrangements. Since the ring may also be flipped over, each arrangement and its mirror-image (reflection) look the same, so the count of rotation-only arrangements must be further divided by 2.
With $n=6$:
$$\dfrac{(6-1)!}{2}=\dfrac{120}{2}=60$$
$$\boxed{60\text{ distinct arrangements}}$$
QUESTION 24
6 marks
Medium
The word "TRIANGLE" has 8 letters, all different.
(a) [2] Find the number of different arrangements of all 8 letters.
(b) [4] Find the number of these arrangements in which the three vowels (I, A, E) are all next to each other.
Show complete worked solution
(a) All 8 letters are distinct, so the number of arrangements of all 8 letters is
$$8!=8\times7\times6\times5\times4\times3\times2\times1=40\,320$$
$$\boxed{40\,320}$$
(b) Treat the three vowels I, A, E as a single block that must stay together. This block, together with the 5 consonants (T, R, N, G, L), gives 6 units to arrange in a row:
$$6!=720$$
Within the block, the 3 vowels can be arranged in $3!=6$ different orders.
By the multiplication principle, the number of arrangements with the vowels together is
$$6!\times3!=720\times6=4320$$
$$\boxed{4320}$$
QUESTION 25
6 marks
Medium
How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 (each digit used at most once) if the number formed must be even?
Show complete worked solution
For the number to be even, the last (units) digit must be one of 2, 4, or 6, so there are 3 choices for the last digit.
Once the last digit is fixed, 5 digits remain, and the first three digits of the number must be filled from these 5 digits without repetition, so there are $P(5,3)=5\times4\times3=60$ ways to fill the remaining three positions in order.
By the multiplication principle, the total number of even 4-digit numbers is
$$3\times60=180$$
$$\boxed{180\text{ numbers}}$$
QUESTION 26
6 marks
Medium
A sports club has 8 men and 6 women. A mixed team of 5 players is to be selected that contains exactly 2 women and 3 men. Find the number of different teams that can be selected.
Show complete worked solution
The 2 women and 3 men are chosen independently, and within each group the order of selection does not matter, so both selections are combinations. Choose 2 women from 6, and choose 3 men from 8:
$$\binom{6}{2}=\dfrac{6\times5}{2\times1}=15\qquad\binom{8}{3}=\dfrac{8\times7\times6}{3\times2\times1}=56$$
By the multiplication principle, the total number of teams is
$$15\times56=840$$
$$\boxed{840\text{ teams}}$$
QUESTION 27
6 marks
Medium
10 cards, numbered 1 to 10, lie on a table. A hand of 4 cards is chosen from the 10 (order does not matter). Find the number of different hands that contain at least one of card 1 or card 2.
Show complete worked solution
Use the complement: (hands with at least one of card 1 or card 2) = (all possible hands) $-$ (hands with neither card 1 nor card 2).
All possible hands of 4 from the 10 cards:
$$\binom{10}{4}=\dfrac{10\times9\times8\times7}{4\times3\times2\times1}=210$$
Hands containing neither card 1 nor card 2 must be chosen entirely from the other 8 cards:
$$\binom{8}{4}=\dfrac{8\times7\times6\times5}{4\times3\times2\times1}=70$$
Subtract:
$$210-70=140$$
$$\boxed{140\text{ hands}}$$
QUESTION 28
7 marks
Medium
7 people, including two particular friends X and Y, are seated around a circular table. Rotations of a seating arrangement are considered identical. Find the number of arrangements in which X and Y sit next to each other.
Show complete worked solution
Treat X and Y, who must sit together, as a single combined unit. Together with the other 5 people, this gives 6 units to arrange around the circular table. For a circular arrangement of 6 distinct units (rotations identical), the number of arrangements is $(6-1)!=5!=120$.
Within the combined unit, X and Y can be arranged in $2!=2$ different orders (X to the left of Y, or Y to the left of X).
By the multiplication principle, the total number of arrangements is
$$5!\times2!=120\times2=240$$
$$\boxed{240\text{ arrangements}}$$
QUESTION 29
6 marks
Medium
The word "ASSESS" consists of 6 letters: A occurs once, S occurs four times, and E occurs once.
(a) [3] Find the number of distinct arrangements of all 6 letters.
(b) [3] Find the number of these arrangements in which all four S's are next to each other.
Show complete worked solution
(a) Since the four S's are identical, the number of distinct arrangements of the multiset of letters is the total factorial divided by the factorial of the repeated letter's count:
$$\dfrac{6!}{4!}=\dfrac{720}{24}=30$$
$$\boxed{30}$$
(b) Treat the four identical S's as a single block (there is only one way to arrange identical letters within the block). This block, together with the letters A and E, gives 3 units to arrange in a row:
$$3!=6$$
$$\boxed{6}$$
QUESTION 30
6 marks
Medium
A password consists of 2 different letters (from the 26 letters of the alphabet, no letter repeated) followed by 3 digits (from 0-9, digits may repeat). The first letter of the password must be a vowel (A, E, I, O, or U). Find the total number of different passwords possible.
Show complete worked solution
Build the password position by position.
First letter: must be a vowel, so there are 5 choices.
Second letter: must be different from the first letter, so there are 25 remaining choices (any of the 26 letters except the one already used).
Each of the 3 digit positions: repetition is allowed, so there are 10 choices for each digit, giving $10\times10\times10=10^3=1000$ possibilities for the digit part.
By the multiplication principle, the total number of passwords is
$$5\times25\times1000=125\,000$$
$$\boxed{125\,000\text{ passwords}}$$
QUESTION 31
7 marks
Medium
A working group of 5 people is to be chosen from a department of 4 teachers and 9 students, such that the group contains at least 3 teachers. Find the number of different groups that can be chosen.
Show complete worked solution
"At least 3 teachers" in a group of 5 means either exactly 3 teachers (and 2 students) or exactly 4 teachers (and 1 student); these two cases are mutually exclusive, so their counts are added by the addition principle.
Exactly 3 teachers and 2 students:
$$\binom{4}{3}\times\binom{9}{2}=4\times36=144$$
Exactly 4 teachers and 1 student:
$$\binom{4}{4}\times\binom{9}{1}=1\times9=9$$
Add the two cases:
$$144+9=153$$
$$\boxed{153\text{ groups}}$$
QUESTION 32
6 marks
Medium
5 different mathematics books and 3 different physics books are to be arranged in a row on a shelf, with all 3 physics books next to each other. Find the number of different arrangements possible.
Show complete worked solution
Treat the 3 physics books, which must stay together, as a single combined block. Together with the 5 mathematics books, this gives 6 units to arrange in a row:
$$6!=720$$
Within the block, the 3 physics books can be arranged among themselves in $3!=6$ different orders.
By the multiplication principle, the total number of arrangements is
$$6!\times3!=720\times6=4320$$
$$\boxed{4320\text{ arrangements}}$$
QUESTION 33
8 marks
Medium
4 boys and 3 girls are to be arranged in a row of 7 seats so that no two girls sit next to each other. Find the number of different arrangements possible.
Show complete worked solution
First arrange the 4 boys in a row; since they are all different, this can be done in $4!=24$ ways. Arranging the boys creates 5 possible gaps where a girl could be placed: one gap before the first boy, one gap between each pair of adjacent boys, and one gap after the last boy - that is, $4+1=5$ gaps in total.
To ensure no two girls are adjacent, place the 3 girls into 3 different gaps (each gap holding at most one girl), and arrange the 3 girls within the chosen gaps in order:
$$P(5,3)=5\times4\times3=60$$
By the multiplication principle, the total number of arrangements is
$$4!\times P(5,3)=24\times60=1440$$
$$\boxed{1440\text{ arrangements}}$$
QUESTION 34
7 marks
Medium
A fruit bowl contains 5 different varieties of apple and 6 different varieties of orange. A selection of 4 pieces of fruit is chosen from the 11 varieties, with at most 2 apple varieties included. Find the number of different selections possible.
Show complete worked solution
"At most 2 apple varieties" means the selection contains 0, 1, or 2 apple varieties (with the rest oranges); these cases are mutually exclusive, so their counts are added.
0 apples, 4 oranges: $\binom{5}{0}\times\binom{6}{4}=1\times15=15$
1 apple, 3 oranges: $\binom{5}{1}\times\binom{6}{3}=5\times20=100$
2 apples, 2 oranges: $\binom{5}{2}\times\binom{6}{2}=10\times15=150$
Add the three cases:
$$15+100+150=265$$
$$\boxed{265\text{ selections}}$$
QUESTION 35
7 marks
Medium
6 people are seated around a circular table, with rotations of a seating arrangement considered identical. Two of the people, P and Q, refuse to sit next to each other. Find the number of seating arrangements in which P and Q do not sit together.
Show complete worked solution
Use the complement: (arrangements with P, Q not together) = (all arrangements) $-$ (arrangements with P, Q together).
All circular arrangements of the 6 people:
$$(6-1)!=5!=120$$
Arrangements with P and Q together: treat them as a single block, giving 5 units to arrange around the table, $(5-1)!=4!=24$; within the block P and Q can be arranged in $2!=2$ orders, giving $24\times2=48$.
Subtract:
$$120-48=72$$
$$\boxed{72\text{ arrangements}}$$
QUESTION 36
6 marks
Medium
The word "COMMITTEE" has 9 letters: C occurs once, O once, M twice, I once, T twice, and E twice. Find the number of distinct arrangements of all 9 letters that begin with the letter C.
Show complete worked solution
Fix the letter C in the first position. The remaining 8 letters - O, M, M, I, T, T, E, E - must be arranged in the remaining 8 positions. Among these 8 letters, M occurs twice, T occurs twice, and E occurs twice, so the number of distinct arrangements of these 8 letters is
$$\dfrac{8!}{2!\,2!\,2!}=\dfrac{40\,320}{8}=5040$$
$$\boxed{5040\text{ arrangements}}$$
QUESTION 37
7 marks
Medium
A quiz team of 6 students is chosen from a class of 15 students, and then one of the 6 team members is chosen to be team captain. Find the total number of different ways the team and captain can be chosen.
Show complete worked solution
First choose the team of 6 from the 15 students; since order does not matter, this is a combination:
$$\binom{15}{6}=5005$$
Then, from within the chosen team of 6, one student is selected to be captain; this can be done in 6 ways.
By the multiplication principle, the total number of ways is
$$5005\times6=30\,030$$
$$\boxed{30\,030\text{ ways}}$$
QUESTION 38
7 marks
Medium
The digits 1, 2, 3, 4, 5, 6, 7 are each used exactly once to form a 7-digit number. Find the number of such numbers in which the digit 4 is not in the first position and not in the last position.
Show complete worked solution
Use the complement: (numbers with 4 not first or last) = (all numbers) $-$ (numbers with 4 first or last).
All arrangements of the 7 digits: $7!=5040$.
Numbers with 4 in the first position: the remaining 6 digits fill the other 6 positions in any order, giving $6!=720$ numbers. Similarly, numbers with 4 in the last position also give $6!=720$ numbers. These two cases cannot overlap (4 cannot occupy both the first and last position at once), so by the addition principle there are $720+720=1440$ numbers with 4 first or last.
Subtract:
$$5040-1440=3600$$
$$\boxed{3600\text{ numbers}}$$
QUESTION 39
7 marks
Medium
A charity's fundraising panel of 4 people is to be selected from a pool of 10 volunteers, two of whom, Priya and Sam, are married to each other. Find the number of different panels in which Priya and Sam are either both selected or both left out.
Show complete worked solution
The two cases are mutually exclusive, so their counts are added by the addition principle.
Both included: the couple take 2 of the 4 places, so the remaining 2 places are filled from the other 8 people:
$$\binom{8}{2}=28$$
Both excluded: all 4 places are filled from the other 8 people:
$$\binom{8}{4}=70$$
Add the two cases:
$$28+70=98$$
$$\boxed{98\text{ committees}}$$
QUESTION 40
6 marks
Medium
9 balls are arranged in a row: 3 identical red balls, 2 identical blue balls, and 4 identical green balls (balls of the same colour cannot be told apart). Find the number of distinguishable arrangements of the 9 balls in the row.
Show complete worked solution
This is an arrangement of a multiset of objects. The number of distinguishable arrangements is the total factorial divided by the factorial of each colour's count, since swapping identical balls of the same colour with each other does not create a new arrangement:
$$\dfrac{9!}{3!\,2!\,4!}=\dfrac{362\,880}{6\times2\times24}=\dfrac{362\,880}{288}=1260$$
$$\boxed{1260\text{ arrangements}}$$
QUESTION 41
8 marks
Medium
A box contains 12 distinct cards: 4 red, 4 blue, and 4 yellow. A hand of 3 cards is drawn from the box (order does not matter). Find the number of different hands in which exactly two of the three colours are represented.
Show complete worked solution
Use the complement within all possible hands: (hands with exactly two colours) = (all hands) $-$ (hands with only one colour) $-$ (hands with all three colours).
All hands of 3 from the 12 cards:
$$\binom{12}{3}=220$$
Hands with only one colour: all 3 cards must come from the same colour group of 4 cards, and there are 3 colour groups:
$$3\times\binom{4}{3}=3\times4=12$$
Hands with all three colours represented: since there are exactly 3 colours and 3 cards are drawn, this means exactly one card of each colour:
$$\binom{4}{1}\times\binom{4}{1}\times\binom{4}{1}=4\times4\times4=64$$
Subtract both cases from the total:
$$220-12-64=144$$
$$\boxed{144\text{ hands}}$$
QUESTION 42
10 marks
Hard
A gym issues membership card numbers made up of 2 letters (from the 26 letters of the alphabet) followed by 4 digits (from 0-9).
(a) [2] Given that letters and digits may each be repeated, find the total number of possible card numbers.
(b) [3] Find the number of card numbers in which the letters may repeat but no digit is repeated.
(c) [3] Find the number of card numbers in which the two letters must be different from each other and no digit is repeated.
(d) [2] Find the number of card numbers in which the two letters are the same as each other and all four digits are the same as each other.
Show complete worked solution
(a) The 2 letters and 4 digits are chosen independently, and repetition is allowed throughout, so by the multiplication principle:
$$26\times26\times10\times10\times10\times10=26^2\times10^4=676\times10\,000=6\,760\,000$$
$$\boxed{6\,760\,000}$$
(b) The letter part is unchanged from part (a): $26^2=676$ ways (repetition allowed). For the digit part, no digit may repeat, so this is a permutation of 4 positions chosen from 10 digits:
$$P(10,4)=10\times9\times8\times7=5040$$
By the multiplication principle:
$$676\times5040=3\,407\,040$$
$$\boxed{3\,407\,040}$$
(c) Now the two letters must also be different from each other: the first letter has 26 choices, and the second letter has 25 remaining choices, giving $26\times25=650$ ways for the letters. The digit part is unchanged from part (b): $P(10,4)=5040$.
By the multiplication principle:
$$650\times5040=3\,276\,000$$
$$\boxed{3\,276\,000}$$
(d) If the two letters are the same as each other, there are 26 ways to choose that repeated letter (it can be any one of the 26 letters, used in both positions). If all four digits are the same as each other, there are 10 ways to choose that repeated digit (it can be any one of the 10 digits, used in all four positions). By the multiplication principle, the total number of such card numbers is
$$26\times10=260$$
$$\boxed{260}$$
QUESTION 43
11 marks
Hard
8 people, including a couple X and Y, and two other people P and Q who dislike each other, are seated around a circular table (rotations of an arrangement are considered identical).
(a) [2] Find the total number of seating arrangements with no restrictions.
(b) [3] Find the number of arrangements in which X and Y sit next to each other.
(c) [3] Find the number of arrangements in which X and Y sit next to each other, and P and Q also sit next to each other.
(d) [3] Hence find the number of arrangements in which X and Y sit next to each other, but P and Q do not sit next to each other.
Show complete worked solution
(a) For a circular arrangement of 8 distinct people (rotations identical):
$$(8-1)!=7!=5040$$
$$\boxed{5040}$$
(b) Treat X and Y as a single block. Together with the other 6 people, this gives 7 units to arrange around the table: $(7-1)!=6!=720$. Within the block, X and Y can be ordered in $2!=2$ ways.
$$6!\times2!=720\times2=1440$$
$$\boxed{1440}$$
(c) Treat X,Y as one block and P,Q as another block. Together with the other 4 people, this gives 6 units to arrange around the table: $(6-1)!=5!=120$. Within the X,Y block there are $2!=2$ orderings, and within the P,Q block there are $2!=2$ orderings.
$$5!\times2!\times2!=120\times2\times2=480$$
$$\boxed{480}$$
(d) Use the complement within the arrangements found in part (b): (X,Y together, P,Q not together) = (X,Y together) $-$ (X,Y together AND P,Q together).
From (b), the number with X,Y together is 1440. From (c), the number with X,Y together and P,Q together is 480.
$$1440-480=960$$
$$\boxed{960}$$
QUESTION 44
15 marks
Hard
A company has 6 engineers, 5 designers, and 4 managers, as summarised in the table below.
A project team of 7 people is to be formed from these 15 people.
(a) [3] Find the number of ways to form the team with no restrictions on its composition.
(b) [4] Find the number of ways to form a team containing exactly 3 engineers, 2 designers, and 2 managers.
(c) [4] Find the number of ways to form a team containing at least 5 engineers.
(d) [4] Find the number of ways to form a team containing at most 1 manager.
| Role | Number of people |
|---|---|
| Engineers | 6 |
| Designers | 5 |
| Managers | 4 |
Show complete worked solution
(a) Since order does not matter, this is a combinations problem:
$$\binom{15}{7}=6435$$
$$\boxed{6435}$$
(b) Choose 3 engineers from 6, 2 designers from 5, and 2 managers from 4; these three choices are independent, so multiply the counts.
$$\binom{6}{3}\times\binom{5}{2}\times\binom{4}{2}=20\times10\times6=1200$$
$$\boxed{1200}$$
(c) "At least 5 engineers" in a team of 7 means either exactly 5 engineers (and 2 non-engineers) or exactly 6 engineers (and 1 non-engineer), since there are only 6 engineers available; these cases are mutually exclusive, so add their counts. Non-engineers total $5+4=9$ people.
Exactly 5 engineers: $\binom{6}{5}\times\binom{9}{2}=6\times36=216$
Exactly 6 engineers: $\binom{6}{6}\times\binom{9}{1}=1\times9=9$
Add:
$$216+9=225$$
$$\boxed{225}$$
(d) "At most 1 manager" means either 0 managers or exactly 1 manager; these cases are mutually exclusive, so add their counts. Non-managers total $6+5=11$ people.
0 managers (all 7 from the 11 non-managers): $\binom{11}{7}=330$
Exactly 1 manager (and 6 from the 11 non-managers): $\binom{4}{1}\times\binom{11}{6}=4\times462=1848$
Add:
$$330+1848=2178$$
$$\boxed{2178}$$
QUESTION 45
11 marks
Hard
The word "MISSISSIPPI" has 11 letters: M occurs once, I occurs four times, S occurs four times, and P occurs twice.
(a) [3] Find the number of distinct arrangements of all 11 letters.
(b) [4] Find the number of these arrangements in which all four I's are next to each other.
(c) [4] Find the number of these arrangements that begin with the letter S and end with the letter S.
Show complete worked solution
(a) This is an arrangement of a multiset of 11 letters with repeats of I (4), S (4), and P (2):
$$\dfrac{11!}{4!\,4!\,2!}=\dfrac{39\,916\,800}{24\times24\times2}=\dfrac{39\,916\,800}{1152}=34\,650$$
$$\boxed{34\,650}$$
(b) Treat the four I's as a single block (they are identical, so there is only one way to arrange them within the block). This block, together with the remaining 7 letters (S,S,S,S,P,P,M), gives 8 units to arrange, where the 4 S's are identical and the 2 P's are identical:
$$\dfrac{8!}{4!\,2!}=\dfrac{40\,320}{48}=840$$
$$\boxed{840}$$
(c) Fix an S at the first position and an S at the last position, using up 2 of the 4 S's. The remaining 9 letters to place in the 9 middle positions are I,I,I,I,S,S,P,P,M - that is, 4 I's, 2 remaining S's, 2 P's, and 1 M:
$$\dfrac{9!}{4!\,2!\,2!}=\dfrac{362\,880}{24\times2\times2}=\dfrac{362\,880}{96}=3780$$
$$\boxed{3780}$$
QUESTION 46
16 marks
Hard
A 5-card hand is dealt from a standard, well-shuffled deck of 52 playing cards (13 ranks in each of 4 suits).
(a) [3] Find the total number of different 5-card hands possible.
(b) [4] A "flush" is a hand in which all 5 cards belong to the same suit. Find the number of different flush hands possible.
(c) [5] Find the number of hands that contain exactly 3 kings and exactly 2 queens.
(d) [4] Find the number of hands that contain at least 1 ace.
Show complete worked solution
(a) Since the order of the cards does not matter, this is a combinations problem:
$$\binom{52}{5}=2\,598\,960$$
$$\boxed{2\,598\,960}$$
(b) There are 4 suits, and for each suit, the 5 cards of the flush must be chosen from that suit's 13 cards:
$$\binom{13}{5}=1287$$
Since the suit can be any one of the 4 suits, by the multiplication principle:
$$4\times1287=5148$$
$$\boxed{5148}$$
(c) There are exactly 4 kings and exactly 4 queens in the deck. Choose 3 of the 4 kings and 2 of the 4 queens; these two choices are independent, so multiply the counts:
$$\binom{4}{3}\times\binom{4}{2}=4\times6=24$$
$$\boxed{24}$$
(d) Use the complement: (hands with at least 1 ace) = (all hands) $-$ (hands with no ace).
Hands with no ace are chosen entirely from the other 48 cards:
$$\binom{48}{5}=1\,712\,304$$
Subtract from the total found in (a):
$$2\,598\,960-1\,712\,304=886\,656$$
$$\boxed{886\,656}$$
QUESTION 47
12 marks
Hard
A quiz league has 12 players available: 7 experienced players and 5 novice players. A team of 5 players is selected, then seated around a circular table during the competition (rotations of a seating arrangement are considered identical), and then a captain and a different vice-captain are appointed from among the 5 team members.
(a) [3] Find the number of ways to select a team of exactly 3 experienced players and 2 novice players.
(b) [3] Find the number of ways to seat 5 people around the circular table.
(c) [3] Find the number of ways to appoint a captain and a different vice-captain from 5 people.
(d) [3] Hence find the total number of ways to select a team of exactly 3 experienced and 2 novice players, seat the team around the table, and appoint the captain and vice-captain.
Show complete worked solution
(a) Choose 3 experienced players from 7, and 2 novice players from 5; these choices are independent, so multiply:
$$\binom{7}{3}\times\binom{5}{2}=35\times10=350$$
$$\boxed{350}$$
(b) For a circular arrangement of 5 distinct people (rotations identical):
$$(5-1)!=4!=24$$
$$\boxed{24}$$
(c) Appointing a captain and a different vice-captain from 5 people is a permutation of 2 positions chosen from 5:
$$P(5,2)=5\times4=20$$
$$\boxed{20}$$
(d) Selecting the team, seating the team, and appointing the two roles are three independent stages, so by the multiplication principle the total number of ways is the product of the answers to (a), (b), and (c):
$$350\times24\times20=168\,000$$
$$\boxed{168\,000\text{ ways}}$$
QUESTION 48
9 marks
Hard
New license plates consist of 3 letters (from the 26 letters of the alphabet) followed by 4 digits (from 0-9). To avoid confusion with the digits 0 and 1, the first letter of the plate may not be O or I (the other two letters have no such restriction, and letters may repeat).
(a) [3] Given that digits may repeat, find the number of valid plates with the first-letter restriction applied.
(b) [3] Find the number of valid plates in which, in addition, no digit is repeated.
(c) [3] Hence find the number of valid plates in which at least one digit is repeated.
Show complete worked solution
(a) The first letter has $26-2=24$ choices (excluding O and I), and each of the other two letters has 26 choices (repetition allowed). The 4 digits each have 10 choices independently (repetition allowed).
$$24\times26\times26\times10^4=24\times676\times10\,000=162\,240\,000$$
$$\boxed{162\,240\,000}$$
(b) The letter part is unchanged: $24\times26\times26=16\,224$ ways. For the digits, no repetition is allowed, so this is a permutation of 4 positions from 10 digits:
$$P(10,4)=10\times9\times8\times7=5040$$
By the multiplication principle:
$$16\,224\times5040=81\,768\,960$$
$$\boxed{81\,768\,960}$$
(c) Within the valid plates counted in part (a), every plate either has no digit repeated (counted in part (b)) or has at least one digit repeated; these two cases partition all the valid plates, so
$$(\text{at least one digit repeated})=(\text{a})-(\text{b})=162\,240\,000-81\,768\,960=80\,471\,040$$
$$\boxed{80\,471\,040}$$
QUESTION 49
11 marks
Hard
9 delegates, including three delegates A, B, and C from the same country, are seated around a circular table (rotations of a seating arrangement are considered identical).
(a) [3] Find the total number of seating arrangements with no restrictions.
(b) [4] Find the number of arrangements in which A, B, and C all sit next to each other, as a single block, in any order.
(c) [4] Find the number of arrangements in which no two of A, B, and C sit next to each other.
Show complete worked solution
(a) For a circular arrangement of 9 distinct people (rotations identical):
$$(9-1)!=8!=40\,320$$
$$\boxed{40\,320}$$
(b) Treat A, B, and C as a single block. Together with the other 6 delegates, this gives 7 units to arrange around the table: $(7-1)!=6!=720$. Within the block, A, B, and C can be ordered in $3!=6$ ways.
$$6!\times3!=720\times6=4320$$
$$\boxed{4320}$$
(c) First seat the other 6 delegates around the table: $(6-1)!=5!=120$ ways. This creates exactly 6 gaps around the table, one between each pair of adjacent seated delegates. To ensure no two of A, B, C are adjacent to each other, place A, B, and C into 3 different gaps (at most one of them per gap) and arrange them in order:
$$P(6,3)=6\times5\times4=120$$
By the multiplication principle, the total number of arrangements is
$$5!\times P(6,3)=120\times120=14\,400$$
$$\boxed{14\,400\text{ arrangements}}$$
QUESTION 50
13 marks
Hard
A survey panel of 6 people is selected from 9 scientists and 7 engineers (16 people in total).
(a) [3] Find the total number of ways to select the panel with no restrictions.
(b) [4] Find the number of ways to select a panel containing exactly 4 scientists and 2 engineers.
(c) [3] Find the number of ways to select a panel containing at least 1 engineer.
(d) [3] Find the number of ways to select a panel containing at most 3 scientists.
Show complete worked solution
(a) $$\binom{16}{6}=8008$$
$$\boxed{8008}$$
(b) Choose 4 scientists from 9 and 2 engineers from 7; the choices are independent, so multiply:
$$\binom{9}{4}\times\binom{7}{2}=126\times21=2646$$
$$\boxed{2646}$$
(c) Use the complement: (at least 1 engineer) = (all panels) $-$ (panels with no engineer, i.e. all 6 members are scientists).
Panels with no engineer: $\binom{9}{6}=84$.
$$8008-84=7924$$
$$\boxed{7924}$$
(d) "At most 3 scientists" means the panel has 0, 1, 2, or 3 scientists (with the rest engineers); these cases are mutually exclusive, so add their counts.
0 scientists, 6 engineers: $\binom{9}{0}\times\binom{7}{6}=1\times7=7$
1 scientist, 5 engineers: $\binom{9}{1}\times\binom{7}{5}=9\times21=189$
2 scientists, 4 engineers: $\binom{9}{2}\times\binom{7}{4}=36\times35=1260$
3 scientists, 3 engineers: $\binom{9}{3}\times\binom{7}{3}=84\times35=2940$
Add all four cases:
$$7+189+1260+2940=4396$$
$$\boxed{4396\text{ ways}}$$
Complex Numbers: Algebraic Operations 50 questions
QUESTION 1
4 marks
Easy
Express $(2+3i)(1-4i)$ in the form $a+bi$, where $a, b \in \mathbb{Z}$.
Show complete worked solution
Using the distributive law to expand the brackets:
$$(2+3i)(1-4i)=2(1)+2(-4i)+3i(1)+3i(-4i)$$
$$=2-8i+3i-12i^2$$
Since $i^2=-1$:
$$=2-8i+3i+12$$
Combine real and imaginary parts:
$$=14-5i$$
$$\boxed{(2+3i)(1-4i)=14-5i}$$
QUESTION 2
5 marks
Easy
Find real numbers $p$ and $q$ such that $(p+qi)(2-i) = 7-i$.
Show complete worked solution
Expand the left-hand side using the distributive law:
$$(p+qi)(2-i)=2p-pi+2qi-qi^2$$
Since $i^2=-1$:
$$=2p-pi+2qi+q=(2p+q)+(2q-p)i$$
Equating this to $7-i$, and equating real and imaginary parts separately (since $p,q$ are real):
$$\text{Real: } 2p+q=7 \quad (1)$$
$$\text{Imaginary: } 2q-p=-1 \quad (2)$$
From (1), $q=7-2p$. Substitute into (2):
$$2(7-2p)-p=-1$$
$$14-4p-p=-1$$
$$14-5p=-1$$
$$-5p=-15$$
$$p=3$$
Then $q=7-2(3)=1$.
Check: $(3+i)(2-i)=6-3i+2i-i^2=6-i+1=7-i$ $\checkmark$
$$\boxed{p=3,\ q=1}$$
QUESTION 3
6 marks
Medium
The complex numbers $z_1 = 4+i$ and $z_2 = -2+5i$ are represented by the points $A$ and $B$ respectively on an Argand diagram. Find, in the form $y = mx+c$, the Cartesian equation of the locus of points $P$ representing complex numbers $z$ such that $|z-z_1| = |z-z_2|$.
Show complete worked solution
The locus $|z-z_1|=|z-z_2|$ is the set of points equidistant from $A(4,1)$ and $B(-2,5)$, i.e. the perpendicular bisector of $AB$.
Let $z=x+iy$. Squaring both sides of $|z-z_1|=|z-z_2|$:
$$|z-z_1|^2=|z-z_2|^2$$
$$(x-4)^2+(y-1)^2=(x+2)^2+(y-5)^2$$
Expand both sides:
$$x^2-8x+16+y^2-2y+1=x^2+4x+4+y^2-10y+25$$
Cancel $x^2$ and $y^2$ from both sides:
$$-8x+16-2y+1=4x+4-10y+25$$
$$-8x-2y+17=4x-10y+29$$
Collect terms:
$$-8x-4x-2y+10y+17-29=0$$
$$-12x+8y-12=0$$
Divide through by $-4$:
$$3x-2y+3=0$$
$$2y=3x+3$$
$$y=\dfrac{3}{2}x+\dfrac{3}{2}$$
$$\boxed{y=\dfrac{3}{2}x+\dfrac{3}{2}}$$
QUESTION 4
6 marks
Medium
Find the complex number $z = x+iy$, where $x, y \in \mathbb{R}$, that satisfies the equation $z + 3\bar{z} = 12 - 8i$.
Show complete worked solution
Let $z=x+iy$, so $\bar{z}=x-iy$. Substituting into the equation:
$$z+3\bar{z}=(x+iy)+3(x-iy)$$
$$=x+iy+3x-3iy$$
$$=4x-2iy$$
So:
$$4x-2iy=12-8i$$
Equating real and imaginary parts:
$$\text{Real: } 4x=12 \Rightarrow x=3$$
$$\text{Imaginary: } -2y=-8 \Rightarrow y=4$$
$$\boxed{z=3+4i}$$
Check: $z+3\bar{z}=(3+4i)+3(3-4i)=3+4i+9-12i=12-8i$ $\checkmark$
QUESTION 5
14 marks
Hard
The complex number $z = x+iy$ satisfies $|z-(3+4i)| = 5$.
(a) Show that the Cartesian equation of the locus of $z$ is $(x-3)^2 + (y-4)^2 = 25$.
(b) Find the two points at which this locus crosses the real axis (i.e. where $y=0$), giving your answers as complex numbers.
(c) Two points $A$ and $B$ on the locus both have imaginary part equal to $8$. Find the complex numbers represented by $A$ and $B$, and hence find the distance $AB$.
(d) Find the complex number on the locus with the greatest possible imaginary part, giving your answer in Cartesian form.
Show complete worked solution
(a) $|z-(3+4i)|=5$ means the distance from $z=x+iy$ to the point $3+4i$ is 5:
$$|(x-3)+i(y-4)|=5$$
$$\sqrt{(x-3)^2+(y-4)^2}=5$$
Squaring both sides:
$$(x-3)^2+(y-4)^2=25$$
This is the Cartesian equation of a circle, centre $(3,4)$, radius 5, as required.
(b) Setting $y=0$:
$$(x-3)^2+(0-4)^2=25$$
$$(x-3)^2+16=25$$
$$(x-3)^2=9$$
$$x-3=\pm3$$
$$x=6 \text{ or } x=0$$
So the locus crosses the real axis at the points representing
$$\boxed{z=6 \text{ and } z=0}$$
(c) Setting $y=8$:
$$(x-3)^2+(8-4)^2=25$$
$$(x-3)^2+16=25$$
$$(x-3)^2=9$$
$$x=6 \text{ or } x=0$$
So $A$ and $B$ represent $z_A=6+8i$ and $z_B=0+8i=8i$ (in either order). Since $A$ and $B$ have the same imaginary part, $AB$ is a horizontal segment:
$$AB=|6-0|=6$$
$$\boxed{z_A=6+8i,\ z_B=8i,\ AB=6}$$
(d) The imaginary part $y$ is maximised when the point is at the top of the circle, directly above the centre, where $x=3$:
$$(3-3)^2+(y-4)^2=25$$
$$(y-4)^2=25$$
$$y-4=\pm5$$
$$y=9 \text{ or } y=-1$$
The greatest value is $y=9$, occurring at $x=3$.
$$\boxed{z=3+9i}$$
QUESTION 6
5 marks
Easy
Simplify $(5-2i)+(3+7i)-(1-4i)$, giving your answer in the form $a+bi$.
Show complete worked solution
Removing the brackets, taking care with the signs (in particular the minus sign in front of the third bracket):
$$(5-2i)+(3+7i)-(1-4i)=5-2i+3+7i-1+4i$$
Collect the real parts and the imaginary parts separately:
$$\text{Real: } 5+3-1=7$$
$$\text{Imaginary: } -2+7+4=9$$
$$\boxed{(5-2i)+(3+7i)-(1-4i)=7+9i}$$
QUESTION 7
5 marks
Easy
Expand and simplify $(4+3i)(2-5i)$, giving your answer in the form $a+bi$, where $a, b \in \mathbb{Z}$.
Show complete worked solution
Using the distributive law to expand the brackets:
$$(4+3i)(2-5i)=4(2)+4(-5i)+3i(2)+3i(-5i)$$
$$=8-20i+6i-15i^2$$
Since $i^2=-1$:
$$=8-20i+6i+15$$
Collecting the real and imaginary parts:
$$=23-14i$$
$$\boxed{(4+3i)(2-5i)=23-14i}$$
QUESTION 8
5 marks
Easy
Expand and simplify $(1+2i)^2$, giving your answer in the form $a+bi$.
Show complete worked solution
Writing $(1+2i)^2=(1+2i)(1+2i)$ and expanding using the distributive law:
$$(1+2i)(1+2i)=1(1)+1(2i)+2i(1)+2i(2i)$$
$$=1+2i+2i+4i^2$$
Since $i^2=-1$:
$$=1+4i-4$$
$$\boxed{(1+2i)^2=-3+4i}$$
QUESTION 9
5 marks
Easy
Express $\dfrac{6+2i}{1-i}$ in the form $a+bi$, where $a, b \in \mathbb{Z}$.
Show complete worked solution
To divide by the complex number $1-i$, multiply the numerator and denominator by its conjugate, $1+i$:
$$\frac{6+2i}{1-i}=\frac{(6+2i)(1+i)}{(1-i)(1+i)}$$
Expand the numerator:
$$(6+2i)(1+i)=6+6i+2i+2i^2=6+8i-2=4+8i$$
Expand the denominator, which is a difference of two squares:
$$(1-i)(1+i)=1^2-i^2=1-(-1)=2$$
So:
$$\frac{6+2i}{1-i}=\frac{4+8i}{2}=2+4i$$
$$\boxed{\dfrac{6+2i}{1-i}=2+4i}$$
QUESTION 10
4 marks
Easy
Let $z=3-5i$. Find $z+\bar z$ and $z\bar z$, showing that both are real numbers.
Show complete worked solution
The conjugate of $z=3-5i$ is obtained by reversing the sign of the imaginary part:
$$\bar z=3+5i$$
Sum:
$$z+\bar z=(3-5i)+(3+5i)=6+0i=6$$
which is real, since the imaginary parts cancel.
Product:
$$z\bar z=(3-5i)(3+5i)=3^2-(5i)^2=9-25i^2=9+25=34$$
which is real, since this is a difference of two squares and $i^2=-1$ turns the second term positive.
$$\boxed{z+\bar z=6,\quad z\bar z=34}$$
QUESTION 11
5 marks
Easy
Find real numbers $p$ and $q$ such that $(p+qi)(3+2i) = 11+3i$.
Show complete worked solution
Expand the left-hand side using the distributive law:
$$(p+qi)(3+2i)=3p+2pi+3qi+2qi^2$$
Since $i^2=-1$:
$$=3p+2pi+3qi-2q=(3p-2q)+(2p+3q)i$$
Equating this to $11+3i$, and equating real and imaginary parts separately (since $p,q$ are real):
$$\text{Real: } 3p-2q=11 \quad (1)$$
$$\text{Imaginary: } 2p+3q=3 \quad (2)$$
From (1), $2q=3p-11$, so $q=\dfrac{3p-11}{2}$. Substitute into (2):
$$2p+3\left(\frac{3p-11}{2}\right)=3$$
Multiply through by 2 to clear the fraction:
$$4p+3(3p-11)=6$$
$$4p+9p-33=6$$
$$13p=39$$
$$p=3$$
Then $q=\dfrac{3(3)-11}{2}=\dfrac{-2}{2}=-1$.
Check: $(3-i)(3+2i)=9+6i-3i-2i^2=9+3i+2=11+3i$ $\checkmark$
$$\boxed{p=3,\ q=-1}$$
QUESTION 12
5 marks
Easy
Solve the equation $z^2-4z+13=0$, giving your answers in the form $a+bi$.
Show complete worked solution
Using the quadratic formula with $a=1$, $b=-4$, $c=13$:
$$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\pm\sqrt{(-4)^2-4(1)(13)}}{2(1)}$$
Compute the discriminant:
$$(-4)^2-4(1)(13)=16-52=-36$$
Since the discriminant is negative, the roots are non-real. Write $-36=36i^2$, so:
$$\sqrt{-36}=\sqrt{36i^2}=6i$$
Substituting back:
$$z=\frac{4\pm6i}{2}=2\pm3i$$
$$\boxed{z=2+3i \text{ or } z=2-3i}$$
QUESTION 13
5 marks
Easy
Solve the equation $2z^2+2z+5=0$, giving your answers in the form $a+bi$.
Show complete worked solution
Using the quadratic formula with $a=2$, $b=2$, $c=5$:
$$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-2\pm\sqrt{2^2-4(2)(5)}}{2(2)}$$
Compute the discriminant:
$$2^2-4(2)(5)=4-40=-36$$
Since the discriminant is negative:
$$\sqrt{-36}=6i$$
Substituting back:
$$z=\frac{-2\pm6i}{4}$$
Divide every term by the common factor 2:
$$z=\frac{-1\pm3i}{2}=-\frac{1}{2}\pm\frac{3}{2}i$$
$$\boxed{z=-\dfrac{1}{2}+\dfrac{3}{2}i \text{ or } z=-\dfrac{1}{2}-\dfrac{3}{2}i}$$
QUESTION 14
5 marks
Easy
A quadratic equation with real coefficients, $z^2+bz+c=0$ where $b,c\in\mathbb{R}$, has $z=4-i$ as one of its roots.
(a) Write down the other root. [1]
(b) Find the values of $b$ and $c$. [4]
Show complete worked solution
(a) Since the coefficients of the quadratic are real, non-real roots occur in conjugate pairs. As $4-i$ is a root, its conjugate must also be a root:
$$\boxed{z=4+i}$$
(b) For a monic quadratic $z^2+bz+c=0$ with roots $z_1,z_2$, the sum of the roots is $-b$ and the product of the roots is $c$.
Sum of roots:
$$z_1+z_2=(4-i)+(4+i)=8$$
So $-b=8$, giving $b=-8$.
Product of roots:
$$z_1z_2=(4-i)(4+i)=4^2-i^2=16-(-1)=17$$
So $c=17$.
Check by substitution: $(4-i)^2-8(4-i)+17=16-8i+i^2-32+8i+17=16-8i-1-32+8i+17=0$ $\checkmark$
$$\boxed{b=-8,\ c=17}$$
QUESTION 15
5 marks
Easy
Find the Cartesian equation of the locus of points in the complex plane representing the complex number $z$ satisfying $|z-(2-3i)|=4$.
Show complete worked solution
Let $z=x+iy$, where $x,y\in\mathbb{R}$. The condition $|z-(2-3i)|=4$ states that the distance from the point representing $z$ to the fixed point $2-3i$ is always $4$, so the locus is a circle of radius $4$ centred at $(2,-3)$.
Writing this out algebraically:
$$|z-(2-3i)|=|(x-2)+i(y+3)|=4$$
$$\sqrt{(x-2)^2+(y+3)^2}=4$$
Squaring both sides:
$$\boxed{(x-2)^2+(y+3)^2=16}$$
QUESTION 16
6 marks
Easy
The points $A$ and $B$ on an Argand diagram represent the complex numbers $z_1=1+2i$ and $z_2=5-4i$ respectively. Find, in the form $y=mx+c$, the Cartesian equation of the locus of points $P$ representing complex numbers $z$ such that $|z-z_1|=|z-z_2|$.
Show complete worked solution
The locus $|z-z_1|=|z-z_2|$ is the set of points equidistant from $A(1,2)$ and $B(5,-4)$, i.e. the perpendicular bisector of $AB$.
Let $z=x+iy$. Squaring both sides of $|z-z_1|=|z-z_2|$:
$$|z-z_1|^2=|z-z_2|^2$$
$$(x-1)^2+(y-2)^2=(x-5)^2+(y+4)^2$$
Expand both sides:
$$x^2-2x+1+y^2-4y+4=x^2-10x+25+y^2+8y+16$$
Cancel $x^2$ and $y^2$ from both sides:
$$-2x+1-4y+4=-10x+25+8y+16$$
$$-2x-4y+5=-10x+8y+41$$
Collect terms onto one side:
$$-2x+10x-4y-8y+5-41=0$$
$$8x-12y-36=0$$
Divide through by $4$:
$$2x-3y-9=0$$
$$3y=2x-9$$
$$\boxed{y=\dfrac{2}{3}x-3}$$
QUESTION 17
4 marks
Easy
Simplify $(7-i)(7+i)$, giving your answer as a real number.
Show complete worked solution
This product has the form $(a-b)(a+b)=a^2-b^2$, a difference of two squares, with $a=7$ and $b=i$:
$$(7-i)(7+i)=7^2-i^2$$
Since $i^2=-1$:
$$=49-(-1)=49+1=50$$
$$\boxed{(7-i)(7+i)=50}$$
QUESTION 18
5 marks
Easy
Express $\dfrac{3-4i}{2+i}$ in the form $a+bi$, where $a, b \in \mathbb{Q}$.
Show complete worked solution
Multiply the numerator and denominator by the conjugate of the denominator, $2-i$:
$$\frac{3-4i}{2+i}=\frac{(3-4i)(2-i)}{(2+i)(2-i)}$$
Expand the numerator:
$$(3-4i)(2-i)=6-3i-8i+4i^2=6-11i-4=2-11i$$
Expand the denominator:
$$(2+i)(2-i)=2^2-i^2=4+1=5$$
So:
$$\frac{3-4i}{2+i}=\frac{2-11i}{5}$$
$$\boxed{\dfrac{3-4i}{2+i}=\dfrac{2}{5}-\dfrac{11}{5}i}$$
QUESTION 19
5 marks
Easy
Find the complex number $z=x+iy$, where $x,y\in\mathbb{R}$, that satisfies the equation $z-3\bar z=-8+12i$.
Show complete worked solution
Let $z=x+iy$, so $\bar z=x-iy$. Substituting into the equation:
$$z-3\bar z=(x+iy)-3(x-iy)$$
$$=x+iy-3x+3iy$$
$$=-2x+4iy$$
So:
$$-2x+4iy=-8+12i$$
Equating real and imaginary parts:
$$\text{Real: } -2x=-8 \Rightarrow x=4$$
$$\text{Imaginary: } 4y=12 \Rightarrow y=3$$
$$\boxed{z=4+3i}$$
Check: $z-3\bar z=(4+3i)-3(4-3i)=4+3i-12+9i=-8+12i$ $\checkmark$
QUESTION 20
5 marks
Easy
Solve the equation $3z^2-6z+15=0$, giving your answers in the form $a+bi$.
Show complete worked solution
First divide the whole equation by the common factor $3$:
$$z^2-2z+5=0$$
Using the quadratic formula with $a=1$, $b=-2$, $c=5$:
$$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2\pm\sqrt{(-2)^2-4(1)(5)}}{2}$$
Compute the discriminant:
$$(-2)^2-4(1)(5)=4-20=-16$$
Since the discriminant is negative:
$$\sqrt{-16}=4i$$
Substituting back:
$$z=\frac{2\pm4i}{2}=1\pm2i$$
$$\boxed{z=1+2i \text{ or } z=1-2i}$$
QUESTION 21
6 marks
Easy
Find real numbers $x$ and $y$ such that $(x+yi)(1+3i)=-5+5i$.
Show complete worked solution
Expand the left-hand side using the distributive law:
$$(x+yi)(1+3i)=x+3xi+yi+3yi^2$$
Since $i^2=-1$:
$$=x+3xi+yi-3y=(x-3y)+(3x+y)i$$
Equating this to $-5+5i$, and equating real and imaginary parts separately:
$$\text{Real: } x-3y=-5 \quad (1)$$
$$\text{Imaginary: } 3x+y=5 \quad (2)$$
From (2), $y=5-3x$. Substitute into (1):
$$x-3(5-3x)=-5$$
$$x-15+9x=-5$$
$$10x=10$$
$$x=1$$
Then $y=5-3(1)=2$.
Check: $(1+2i)(1+3i)=1+3i+2i+6i^2=1+5i-6=-5+5i$ $\checkmark$
$$\boxed{x=1,\ y=2}$$
QUESTION 22
5 marks
Easy
Express $(2-i)^3$ in the form $a+bi$, where $a,b\in\mathbb{Z}$.
Show complete worked solution
First find $(2-i)^2$:
$$(2-i)^2=(2-i)(2-i)=4-2i-2i+i^2=4-4i-1=3-4i$$
Now multiply this result by a further factor of $(2-i)$:
$$(2-i)^3=(3-4i)(2-i)$$
$$=3(2)+3(-i)+(-4i)(2)+(-4i)(-i)$$
$$=6-3i-8i+4i^2$$
Since $i^2=-1$:
$$=6-11i-4$$
$$\boxed{(2-i)^3=2-11i}$$
QUESTION 23
6 marks
Easy
The points $A$ and $B$ on an Argand diagram represent the complex numbers $z_1=-1+3i$ and $z_2=5+3i$. Find the Cartesian equation of the locus of points $P$ representing $z$ such that $|z-z_1|=|z-z_2|$.
Show complete worked solution
The locus is the perpendicular bisector of the segment $AB$, where $A(-1,3)$ and $B(5,3)$.
Let $z=x+iy$. Squaring both sides of $|z-z_1|=|z-z_2|$:
$$(x+1)^2+(y-3)^2=(x-5)^2+(y-3)^2$$
Since the term $(y-3)^2$ appears identically on both sides, it cancels:
$$(x+1)^2=(x-5)^2$$
Expand both sides:
$$x^2+2x+1=x^2-10x+25$$
Cancel $x^2$ from both sides:
$$2x+1=-10x+25$$
$$12x=24$$
$$x=2$$
This makes sense geometrically: since $A$ and $B$ have the same imaginary part, $AB$ is horizontal, so its perpendicular bisector is the vertical line through the midpoint $(2,3)$.
$$\boxed{x=2}$$
QUESTION 24
7 marks
Medium
Given that $\dfrac{(3+2i)(1-i)}{2+i}=a+bi$, find the values of $a$ and $b$.
Show complete worked solution
First simplify the numerator by expanding the product:
$$(3+2i)(1-i)=3-3i+2i-2i^2=3-i+2=5-i$$
So the expression becomes:
$$\frac{5-i}{2+i}$$
To divide by $2+i$, multiply the numerator and denominator by its conjugate, $2-i$:
$$\frac{5-i}{2+i}=\frac{(5-i)(2-i)}{(2+i)(2-i)}$$
Expand the numerator:
$$(5-i)(2-i)=10-5i-2i+i^2=10-7i-1=9-7i$$
Expand the denominator:
$$(2+i)(2-i)=4-i^2=4+1=5$$
So:
$$\frac{5-i}{2+i}=\frac{9-7i}{5}=\frac{9}{5}-\frac{7}{5}i$$
$$\boxed{a=\dfrac{9}{5},\ b=-\dfrac{7}{5}}$$
QUESTION 25
8 marks
Medium
Solve the equation $z^2+2z+5=0$, giving your answers in the form $a+bi$. Hence verify that the sum of the roots is $-2$ and the product of the roots is $5$.
Show complete worked solution
Using the quadratic formula with $a=1$, $b=2$, $c=5$:
$$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-2\pm\sqrt{2^2-4(1)(5)}}{2}$$
Compute the discriminant:
$$2^2-4(1)(5)=4-20=-16$$
Since the discriminant is negative:
$$\sqrt{-16}=4i$$
Substituting back:
$$z=\frac{-2\pm4i}{2}=-1\pm2i$$
So the roots are $z_1=-1+2i$ and $z_2=-1-2i$.
Verifying the sum:
$$z_1+z_2=(-1+2i)+(-1-2i)=-2$$
which matches $-\dfrac{b}{a}=-\dfrac{2}{1}=-2$ $\checkmark$
Verifying the product:
$$z_1z_2=(-1+2i)(-1-2i)=(-1)^2-(2i)^2=1-4i^2=1+4=5$$
which matches $\dfrac{c}{a}=\dfrac{5}{1}=5$ $\checkmark$
$$\boxed{z=-1+2i \text{ or } z=-1-2i}$$
QUESTION 26
7 marks
Medium
The equation $z^2-6z+k=0$, where $k\in\mathbb{R}$, has a root $z=3+2i$.
(a) Write down the other root. [1]
(b) Find the value of $k$. [3]
(c) Verify your value of $k$ by substituting $z=3+2i$ into the original equation. [3]
Show complete worked solution
(a) The coefficients of the quadratic are real, so non-real roots occur in conjugate pairs:
$$\boxed{z=3-2i}$$
(b) For $z^2-6z+k=0$, the product of the roots equals $k$ (the constant term, since the coefficient of $z^2$ is $1$):
$$k=(3+2i)(3-2i)=3^2-(2i)^2=9-4i^2=9+4=13$$
$$\boxed{k=13}$$
(c) Substitute $z=3+2i$ into $z^2-6z+13$:
$$(3+2i)^2-6(3+2i)+13$$
First expand $(3+2i)^2$:
$$(3+2i)^2=9+12i+4i^2=9+12i-4=5+12i$$
Then:
$$5+12i-6(3+2i)+13=5+12i-18-12i+13$$
Collecting real and imaginary parts:
$$=(5-18+13)+(12i-12i)=0+0i=0$$
$$\boxed{0}$$ as required, confirming $k=13$.
QUESTION 27
8 marks
Medium
The cubic equation $z^3-4z^2+6z-4=0$ has one real root.
(a) Show that $z=2$ is a root of the equation. [2]
(b) Hence find the other two roots. [6]
Show complete worked solution
(a) Substitute $z=2$ into the left-hand side:
$$2^3-4(2)^2+6(2)-4=8-16+12-4=0$$
Since the equation is satisfied, $z=2$ is a root.
(b) By the factor theorem, $(z-2)$ is a factor of $z^3-4z^2+6z-4$. Write:
$$z^3-4z^2+6z-4=(z-2)(z^2+pz+q)$$
Expanding the right-hand side:
$$(z-2)(z^2+pz+q)=z^3+pz^2+qz-2z^2-2pz-2q=z^3+(p-2)z^2+(q-2p)z-2q$$
Matching to $z^3-4z^2+6z-4$: from the constant term, $-2q=-4$, so $q=2$. From the $z^2$ term, $p-2=-4$, so $p=-2$. Checking the $z$ term: $q-2p=2-2(-2)=2+4=6$ $\checkmark$
So:
$$z^3-4z^2+6z-4=(z-2)(z^2-2z+2)$$
Setting the quadratic factor to zero:
$$z^2-2z+2=0$$
Using the quadratic formula with $a=1$, $b=-2$, $c=2$:
$$z=\frac{2\pm\sqrt{(-2)^2-4(1)(2)}}{2}=\frac{2\pm\sqrt{4-8}}{2}=\frac{2\pm\sqrt{-4}}{2}$$
Since $\sqrt{-4}=2i$:
$$z=\frac{2\pm2i}{2}=1\pm i$$
$$\boxed{z=2,\ z=1+i,\ z=1-i}$$
QUESTION 28
8 marks
Medium
The cubic equation $z^3-5z^2+9z-5=0$, where all coefficients are real, has $z=2+i$ as one of its roots.
(a) Write down a second root. [1]
(b) Find the third (real) root, using the fact that for a cubic $z^3+bz^2+cz+d=0$ the sum of all three roots equals $-b$. [4]
(c) Verify the third root by direct substitution. [3]
Show complete worked solution
(a) Since the cubic has real coefficients, complex roots occur in conjugate pairs, so:
$$\boxed{z=2-i}$$
(b) Let the real root be $r$. For $z^3-5z^2+9z-5=0$ (so $b=-5$), the sum of all three roots is $-b=5$:
$$(2+i)+(2-i)+r=5$$
$$4+r=5$$
$$\boxed{r=1}$$
(c) Substitute $z=1$ into $z^3-5z^2+9z-5$:
$$1^3-5(1)^2+9(1)-5=1-5+9-5=0$$
$$\boxed{0}$$ as required, confirming that $z=1$ is the third root.
So the three roots of the equation are $z=1$, $z=2+i$, $z=2-i$.
QUESTION 29
8 marks
Medium
Solve the simultaneous equations for the complex numbers $z$ and $w$:
$$z+2w=5$$
$$2z-w=5+5i$$
Show complete worked solution
From the first equation, make $z$ the subject:
$$z=5-2w \quad (1)$$
Substitute (1) into the second equation:
$$2(5-2w)-w=5+5i$$
$$10-4w-w=5+5i$$
$$10-5w=5+5i$$
$$-5w=-5+5i$$
$$w=1-i$$
Substitute $w=1-i$ back into (1):
$$z=5-2(1-i)=5-2+2i=3+2i$$
Check in the second equation: $2(3+2i)-(1-i)=6+4i-1+i=5+5i$ $\checkmark$
$$\boxed{z=3+2i,\ w=1-i}$$
QUESTION 30
8 marks
Medium
Solve the simultaneous equations for the complex numbers $z$ and $w$:
$$(1+i)z+w=2+i$$
$$z-iw=-i$$
Show complete worked solution
From the second equation, make $z$ the subject:
$$z=-i+iw \quad (1)$$
Substitute (1) into the first equation:
$$(1+i)(-i+iw)+w=2+i$$
Expand $(1+i)(-i)$:
$$(1+i)(-i)=-i-i^2=-i+1=1-i$$
Expand $(1+i)(iw)$:
$$(1+i)(iw)=(i+i^2)w=(i-1)w$$
So the equation becomes:
$$(1-i)+(i-1)w+w=2+i$$
Combine the $w$ terms: $(i-1)w+w=(i-1+1)w=iw$:
$$(1-i)+iw=2+i$$
$$iw=2+i-1+i=1+2i$$
Divide both sides by $i$ (equivalently, multiply by $-i$, since $\frac{1}{i}=-i$):
$$w=\frac{1+2i}{i}=(1+2i)(-i)=-i-2i^2=-i+2=2-i$$
Substitute $w=2-i$ back into (1):
$$z=-i+i(2-i)=-i+2i-i^2=-i+2i+1=1+i$$
Check in the first equation: $(1+i)(1+i)+(2-i)=(1+2i+i^2)+(2-i)=(2i)+(2-i)=2+i$ $\checkmark$
$$\boxed{z=1+i,\ w=2-i}$$
QUESTION 31
7 marks
Medium
Points $A$ and $B$ on an Argand diagram represent the complex numbers $z_1=1+4i$ and $z_2=-3-2i$. Find the Cartesian equation of the locus of points $P$ representing $z$ such that $|z-z_1|=|z-z_2|$, giving your answer in the form $ax+by=c$, where $a,b,c\in\mathbb{Z}$.
Show complete worked solution
The locus is the perpendicular bisector of $AB$, where $A(1,4)$ and $B(-3,-2)$.
Let $z=x+iy$. Squaring both sides of $|z-z_1|=|z-z_2|$:
$$(x-1)^2+(y-4)^2=(x+3)^2+(y+2)^2$$
Expand both sides:
$$x^2-2x+1+y^2-8y+16=x^2+6x+9+y^2+4y+4$$
Cancel $x^2$ and $y^2$:
$$-2x+1-8y+16=6x+9+4y+4$$
$$-2x-8y+17=6x+4y+13$$
Collect terms:
$$-2x-6x-8y-4y+17-13=0$$
$$-8x-12y+4=0$$
Divide through by $-4$:
$$2x+3y-1=0$$
$$\boxed{2x+3y=1}$$
QUESTION 32
8 marks
Medium
Consider the quadratic equation $2z^2-4z+10=0$.
(a) Show that the equation has no real roots. [2]
(b) Solve the equation, giving your answers in the form $a+bi$. [4]
(c) Verify that the sum of your two roots equals $2$. [2]
Show complete worked solution
(a) For $2z^2-4z+10=0$, we have $a=2$, $b=-4$, $c=10$. The discriminant is:
$$b^2-4ac=(-4)^2-4(2)(10)=16-80=-64$$
Since the discriminant is negative, the equation has no real roots.
(b) Divide the whole equation by the common factor $2$:
$$z^2-2z+5=0$$
Using the quadratic formula with $a=1$, $b=-2$, $c=5$:
$$z=\frac{2\pm\sqrt{(-2)^2-4(1)(5)}}{2}=\frac{2\pm\sqrt{4-20}}{2}=\frac{2\pm\sqrt{-16}}{2}$$
Since $\sqrt{-16}=4i$:
$$z=\frac{2\pm4i}{2}=1\pm2i$$
$$\boxed{z=1+2i \text{ or } z=1-2i}$$
(c) Sum of the roots:
$$(1+2i)+(1-2i)=2$$
$$\boxed{2}$$ as required, which agrees with $-\dfrac{b}{a}=\dfrac{4}{2}=2$ for the original equation.
QUESTION 33
8 marks
Medium
Find the real numbers $x$ and $y$ satisfying the equation $\dfrac{3+xi}{1+yi}=1-2i$.
Show complete worked solution
Multiply both sides by $(1+yi)$ to clear the denominator:
$$3+xi=(1-2i)(1+yi)$$
Expand the right-hand side:
$$(1-2i)(1+yi)=1+yi-2i-2yi^2=1+yi-2i+2y=(1+2y)+(y-2)i$$
So:
$$3+xi=(1+2y)+(y-2)i$$
Equating real and imaginary parts (since $x,y$ are real):
$$\text{Real: } 3=1+2y \Rightarrow y=1$$
$$\text{Imaginary: } x=y-2=1-2=-1$$
Check: $\dfrac{3-i}{1+i}=\dfrac{(3-i)(1-i)}{(1+i)(1-i)}=\dfrac{3-3i-i+i^2}{2}=\dfrac{3-4i-1}{2}=\dfrac{2-4i}{2}=1-2i$ $\checkmark$
$$\boxed{x=-1,\ y=1}$$
QUESTION 34
7 marks
Medium
The cubic equation $z^3+z^2+3z-5=0$ has one real root.
(a) By testing values, find the real root. [3]
(b) Hence find the other two roots. [4]
Show complete worked solution
(a) Since the leading and constant coefficients are $1$ and $-5$, any rational root must divide $5$, so possible values are $\pm1,\pm5$. Testing $z=1$:
$$1^3+1^2+3(1)-5=1+1+3-5=0$$
So $z=1$ is the real root.
(b) By the factor theorem, $(z-1)$ is a factor. Write:
$$z^3+z^2+3z-5=(z-1)(z^2+pz+q)$$
Expanding the right-hand side:
$$(z-1)(z^2+pz+q)=z^3+pz^2+qz-z^2-pz-q=z^3+(p-1)z^2+(q-p)z-q$$
Matching constant terms: $-q=-5$, so $q=5$. Matching $z^2$ coefficients: $p-1=1$, so $p=2$. Check the $z$ coefficient: $q-p=5-2=3$ $\checkmark$
So:
$$z^3+z^2+3z-5=(z-1)(z^2+2z+5)$$
Setting the quadratic factor to zero:
$$z^2+2z+5=0$$
Using the quadratic formula with $a=1$, $b=2$, $c=5$:
$$z=\frac{-2\pm\sqrt{4-20}}{2}=\frac{-2\pm\sqrt{-16}}{2}=\frac{-2\pm4i}{2}=-1\pm2i$$
$$\boxed{z=1,\ z=-1+2i,\ z=-1-2i}$$
QUESTION 35
8 marks
Medium
The locus of a point $P$ representing the complex number $z$ is given by $|z-6i|=|z+4|$.
(a) Find the Cartesian equation of the locus, giving your answer in the form $y=mx+c$. [5]
(b) Find the point where the locus crosses the real axis. [3]
Show complete worked solution
(a) The point $6i$ corresponds to $A(0,6)$, and $-4$ corresponds to $B(-4,0)$. The locus $|z-6i|=|z+4|$ is the perpendicular bisector of $AB$.
Let $z=x+iy$. Squaring both sides:
$$x^2+(y-6)^2=(x+4)^2+y^2$$
Expand both sides:
$$x^2+y^2-12y+36=x^2+8x+16+y^2$$
Cancel $x^2$ and $y^2$:
$$-12y+36=8x+16$$
$$-12y=8x-20$$
$$y=\frac{8x-20}{-12}=\frac{-8x+20}{12}$$
Simplify by dividing numerator and denominator by $4$:
$$\boxed{y=-\dfrac{2}{3}x+\dfrac{5}{3}}$$
(b) The locus crosses the real axis where $y=0$:
$$0=-\frac{2}{3}x+\frac{5}{3}$$
$$\frac{2}{3}x=\frac{5}{3}$$
$$x=\frac{5}{2}$$
$$\boxed{z=\dfrac{5}{2}}$$
QUESTION 36
8 marks
Medium
Solve the equation $z^2=-8+6i$, giving your answers in the form $a+bi$, where $a,b\in\mathbb{R}$.
Show complete worked solution
Let $z=x+iy$, where $x,y\in\mathbb{R}$. Then:
$$z^2=(x+iy)^2=x^2-y^2+2xyi$$
Equating this to $-8+6i$ and comparing real and imaginary parts:
$$\text{Real: } x^2-y^2=-8 \quad (1)$$
$$\text{Imaginary: } 2xy=6 \Rightarrow xy=3 \Rightarrow y=\frac{3}{x} \quad (2)$$
Substitute (2) into (1):
$$x^2-\left(\frac{3}{x}\right)^2=-8$$
$$x^2-\frac{9}{x^2}=-8$$
Multiply through by $x^2$ (note $x\ne0$, since $xy=3$):
$$x^4-9=-8x^2$$
$$x^4+8x^2-9=0$$
This is quadratic in $x^2$. Let $u=x^2$:
$$u^2+8u-9=0$$
$$(u+9)(u-1)=0$$
$$u=-9 \text{ or } u=1$$
Since $u=x^2\ge0$ for real $x$, reject $u=-9$. So $x^2=1$, giving $x=1$ or $x=-1$.
Using $y=3/x$: if $x=1$, $y=3$; if $x=-1$, $y=-3$.
Check: $(1+3i)^2=1+6i+9i^2=1+6i-9=-8+6i$ $\checkmark$
$$\boxed{z=1+3i \text{ or } z=-1-3i}$$
QUESTION 37
8 marks
Medium
The complex numbers $z$ and $w$ satisfy the simultaneous equations
$$z+2w=9-i$$
$$\bar z-w=3+4i$$
Find $z$ and $w$.
Show complete worked solution
Let $z=x+iy$, where $x,y\in\mathbb{R}$, so $\bar z=x-iy$.
From the first equation:
$$2w=9-i-z=9-i-(x+iy)=(9-x)-i(1+y)$$
$$w=\frac{9-x}{2}-\frac{1+y}{2}i \quad (1)$$
From the second equation:
$$w=\bar z-(3+4i)=(x-iy)-3-4i=(x-3)-i(y+4) \quad (2)$$
Equate the real parts of (1) and (2):
$$\frac{9-x}{2}=x-3$$
$$9-x=2x-6$$
$$15=3x$$
$$x=5$$
Equate the imaginary parts of (1) and (2):
$$-\frac{1+y}{2}=-(y+4)$$
$$1+y=2(y+4)$$
$$1+y=2y+8$$
$$-7=y$$
So $z=5-7i$. Substitute into (2):
$$w=(5-3)-i(-7+4)=2-i(-3)=2+3i$$
Check in the first equation: $z+2w=(5-7i)+2(2+3i)=5-7i+4+6i=9-i$ $\checkmark$
Check in the second equation: $\bar z-w=(5+7i)-(2+3i)=3+4i$ $\checkmark$
$$\boxed{z=5-7i,\ w=2+3i}$$
QUESTION 38
7 marks
Medium
Simplify $\dfrac{1}{2+i}+\dfrac{1}{2-i}$, giving your answer as a real number.
Show complete worked solution
Write the sum of the two fractions over a common denominator:
$$\frac{1}{2+i}+\frac{1}{2-i}=\frac{(2-i)+(2+i)}{(2+i)(2-i)}$$
Simplify the numerator (the imaginary parts cancel):
$$(2-i)+(2+i)=4$$
Simplify the denominator, a difference of two squares:
$$(2+i)(2-i)=2^2-i^2=4+1=5$$
So:
$$\frac{1}{2+i}+\frac{1}{2-i}=\frac{4}{5}$$
$$\boxed{\dfrac{1}{2+i}+\dfrac{1}{2-i}=\dfrac{4}{5}}$$
QUESTION 39
7 marks
Medium
Given that $z=3-i$ is a root of the equation $2z^2+bz+c=0$, where $b,c\in\mathbb{R}$, find the values of $b$ and $c$.
Show complete worked solution
Since the coefficients of the quadratic are real, the other root must be the conjugate, $z=3+i$.
For $2z^2+bz+c=0$, dividing through by $2$ gives $z^2+\dfrac{b}{2}z+\dfrac{c}{2}=0$, so the sum of the roots is $-\dfrac{b}{2}$ and the product of the roots is $\dfrac{c}{2}$.
Sum of roots:
$$(3-i)+(3+i)=6$$
So $-\dfrac{b}{2}=6$, giving $b=-12$.
Product of roots:
$$(3-i)(3+i)=3^2-i^2=9+1=10$$
So $\dfrac{c}{2}=10$, giving $c=20$.
Check by substitution into $2z^2-12z+20$ at $z=3-i$: first $(3-i)^2=9-6i+i^2=8-6i$, so $2(8-6i)-12(3-i)+20=16-12i-36+12i+20=0$ $\checkmark$
$$\boxed{b=-12,\ c=20}$$
QUESTION 40
7 marks
Medium
Find the set of values of the real number $k$ for which the equation $z^2+kz+(k+3)=0$ has two distinct non-real roots.
Show complete worked solution
A quadratic $az^2+bz+c=0$ with real coefficients has two distinct non-real (complex conjugate) roots exactly when its discriminant is negative.
Here $a=1$, $b=k$, $c=k+3$, so the discriminant is:
$$b^2-4ac=k^2-4(1)(k+3)=k^2-4k-12$$
We require:
$$k^2-4k-12<0$$
Factorise the quadratic in $k$:
$$k^2-4k-12=(k-6)(k+2)$$
So we need:
$$(k-6)(k+2)<0$$
This product is negative when the two factors have opposite signs, which occurs between the roots $k=-2$ and $k=6$ of $(k-6)(k+2)=0$:
$$\boxed{-2
QUESTION 41
7 marks
Medium
(a) Find the Cartesian equation of the locus of points representing the complex number $z=x+iy$ satisfying $|z-3+i|=2$, stating its centre and radius. [3]
(b) Determine whether the point representing $z=5-i$ lies on, inside, or outside this locus. [4]
Show complete worked solution
(a) Rewrite $|z-3+i|$ as $|z-(3-i)|$, so the condition is $|z-(3-i)|=2$: the distance from $z$ to the fixed point $3-i$ is always $2$.
$$|(x-3)+i(y+1)|=2$$
$$\sqrt{(x-3)^2+(y+1)^2}=2$$
Squaring both sides:
$$\boxed{(x-3)^2+(y+1)^2=4}$$
This is a circle with centre $(3,-1)$ and radius $2$.
(b) The point representing $z=5-i$ is $(5,-1)$. Substitute into the left-hand side of the locus equation:
$$(5-3)^2+(-1+1)^2=2^2+0^2=4$$
Since this equals the right-hand side, $4=4$, the point satisfies the equation exactly, so it lies exactly on the circle.
$$\boxed{\text{The point } z=5-i \text{ lies on the locus.}}$$
QUESTION 42
13 marks
Hard
The quartic equation $z^4-6z^3+18z^2-30z+25=0$ has $z=1+2i$ as one of its roots.
(a) Write down a second root, and hence state a quadratic factor of the quartic with real coefficients. [4]
(b) By dividing the quartic by this quadratic factor, find the other two roots. [8]
(c) Hence write the quartic as a product of two quadratic factors with real coefficients. [1]
Show complete worked solution
(a) Since the quartic has real coefficients, non-real roots occur in conjugate pairs, so a second root is:
$$z=1-2i$$
The quadratic with these two roots has sum $(1+2i)+(1-2i)=2$ and product $(1+2i)(1-2i)=1-4i^2=1+4=5$, so it is:
$$\boxed{z^2-2z+5}$$
(b) Divide $z^4-6z^3+18z^2-30z+25$ by $z^2-2z+5$ using polynomial long division.
Divide the leading term $z^4$ by $z^2$ to get $z^2$. Multiply the divisor by $z^2$: $z^4-2z^3+5z^2$. Subtract this from the quartic:
$$(z^4-6z^3+18z^2)-(z^4-2z^3+5z^2)=-4z^3+13z^2$$
Bring down the next term, $-30z$, giving $-4z^3+13z^2-30z$.
Divide $-4z^3$ by $z^2$ to get $-4z$. Multiply the divisor by $-4z$: $-4z^3+8z^2-20z$. Subtract:
$$(-4z^3+13z^2-30z)-(-4z^3+8z^2-20z)=5z^2-10z$$
Bring down the next term, $+25$, giving $5z^2-10z+25$.
Divide $5z^2$ by $z^2$ to get $5$. Multiply the divisor by $5$: $5z^2-10z+25$. Subtract:
$$(5z^2-10z+25)-(5z^2-10z+25)=0$$
The division is exact, with quotient $z^2-4z+5$. So:
$$z^4-6z^3+18z^2-30z+25=(z^2-2z+5)(z^2-4z+5)$$
Setting the quotient to zero:
$$z^2-4z+5=0$$
Using the quadratic formula with $a=1$, $b=-4$, $c=5$:
$$z=\frac{4\pm\sqrt{(-4)^2-4(1)(5)}}{2}=\frac{4\pm\sqrt{16-20}}{2}=\frac{4\pm\sqrt{-4}}{2}$$
Since $\sqrt{-4}=2i$:
$$z=\frac{4\pm2i}{2}=2\pm i$$
$$\boxed{z=2+i \text{ or } z=2-i}$$
(c) Combining the two quadratic factors found above:
$$\boxed{z^4-6z^3+18z^2-30z+25=(z^2-2z+5)(z^2-4z+5)}$$
QUESTION 43
11 marks
Hard
Consider the cubic equation $z^3-2z^2+9z-18=0$.
(a) By testing factors of $18$, show that $z=2$ is a root of the equation. [2]
(b) Hence find the other two roots, which are purely imaginary. [6]
(c) Verify that the sum of all three roots is $2$, and that the product of all three roots is $18$. [3]
Show complete worked solution
(a) Since the equation has integer coefficients with leading coefficient $1$ and constant term $-18$, any rational root must be a factor of $18$: possible values include $\pm1,\pm2,\pm3,\pm6,\pm9,\pm18$. Testing $z=2$:
$$2^3-2(2)^2+9(2)-18=8-8+18-18=0$$
Since the equation is satisfied, $z=2$ is a root.
(b) By the factor theorem, $(z-2)$ is a factor. Write:
$$z^3-2z^2+9z-18=(z-2)(z^2+pz+q)$$
Expanding the right-hand side:
$$(z-2)(z^2+pz+q)=z^3+pz^2+qz-2z^2-2pz-2q=z^3+(p-2)z^2+(q-2p)z-2q$$
Matching constant terms: $-2q=-18$, so $q=9$. Matching $z^2$ coefficients: $p-2=-2$, so $p=0$. Check the $z$ coefficient: $q-2p=9-0=9$ $\checkmark$
So:
$$z^3-2z^2+9z-18=(z-2)(z^2+9)$$
Setting the quadratic factor to zero:
$$z^2+9=0$$
$$z^2=-9$$
$$z=\pm\sqrt{-9}=\pm3i$$
$$\boxed{z=2,\ z=3i,\ z=-3i}$$
(c) Sum of roots:
$$2+3i+(-3i)=2$$
which agrees with $-\dfrac{b}{a}=-\dfrac{-2}{1}=2$ for the original cubic $\checkmark$
Product of roots:
$$2\times3i\times(-3i)=2\times(-9i^2)=2\times9=18$$
which agrees with $-\dfrac{d}{a}=-\dfrac{-18}{1}=18$ $\checkmark$
$$\boxed{\text{Sum}=2,\ \text{Product}=18}$$
QUESTION 44
14 marks
Hard
The complex number $z=x+iy$ satisfies $|z-(5-2i)|=4$.
(a) Show that the Cartesian equation of the locus of $z$ is $(x-5)^2+(y+2)^2=16$. [4]
(b) Find the two points at which this locus crosses the line $y=-2$ (the horizontal line through the centre), giving your answers as complex numbers. [3]
(c) Determine whether the point representing $z=8+2i$ lies inside, on, or outside the locus. [3]
(d) Find the complex number on the locus with the least possible real part, giving the answer in Cartesian form. [4]
Show complete worked solution
(a) $|z-(5-2i)|=4$ means the distance from $z=x+iy$ to the point $5-2i$ is $4$:
$$|(x-5)+i(y+2)|=4$$
$$\sqrt{(x-5)^2+(y+2)^2}=4$$
Squaring both sides:
$$(x-5)^2+(y+2)^2=16$$
This is the Cartesian equation of a circle, centre $(5,-2)$, radius $4$, as required.
$$\boxed{(x-5)^2+(y+2)^2=16}$$
(b) Setting $y=-2$:
$$(x-5)^2+(-2+2)^2=16$$
$$(x-5)^2=16$$
$$x-5=\pm4$$
$$x=9 \text{ or } x=1$$
So the locus crosses the line $y=-2$ at the points representing
$$\boxed{z=9-2i \text{ and } z=1-2i}$$
(c) Substitute $x=8,y=2$ into the left-hand side of the locus equation:
$$(8-5)^2+(2+2)^2=3^2+4^2=9+16=25$$
Since $25>16$, the point representing $z=8+2i$ lies outside the locus.
$$\boxed{\text{Outside}}$$
(d) The real part $x$ is minimised at the leftmost point of the circle, directly to the left of the centre, where $y=-2$ (the same horizontal line as in part (b)):
$$(x-5)^2+0=16$$
$$x-5=\pm4$$
$$x=9 \text{ or } x=1$$
The least value is $x=1$, occurring at $y=-2$ (this is the same point found in part (b)).
$$\boxed{z=1-2i}$$
QUESTION 45
10 marks
Hard
Solve the simultaneous equations for the complex numbers $z$ and $w$:
$$(2-i)z+3w=1-3i$$
$$iz-w=-1+3i$$
Show complete worked solution
From the second equation, make $w$ the subject:
$$w=iz-(-1+3i)=iz+1-3i \quad (1)$$
Substitute (1) into the first equation:
$$(2-i)z+3(iz+1-3i)=1-3i$$
Expand:
$$(2-i)z+3iz+3-9i=1-3i$$
Combine the $z$ terms:
$$[(2-i)+3i]z=1-3i-3+9i$$
$$(2+2i)z=-2+6i$$
Divide both sides by $(2+2i)$, and multiply numerator and denominator by the conjugate $(2-2i)$:
$$z=\frac{-2+6i}{2+2i}=\frac{(-2+6i)(2-2i)}{(2+2i)(2-2i)}$$
Expand the numerator:
$$(-2+6i)(2-2i)=-4+4i+12i-12i^2=-4+16i+12=8+16i$$
Expand the denominator:
$$(2+2i)(2-2i)=2^2-(2i)^2=4-4i^2=4+4=8$$
So:
$$z=\frac{8+16i}{8}=1+2i$$
Substitute $z=1+2i$ back into (1):
$$w=i(1+2i)+1-3i=i+2i^2+1-3i=i-2+1-3i=-1-2i$$
Check in the first equation: $(2-i)(1+2i)+3(-1-2i)=(2+4i-i-2i^2)+(-3-6i)=(2+3i+2)+(-3-6i)=(4+3i)+(-3-6i)=1-3i$ $\checkmark$
$$\boxed{z=1+2i,\ w=-1-2i}$$
QUESTION 46
11 marks
Hard
(a) Find the two square roots of $-5+12i$, giving your answers in the form $a+bi$, where $a,b\in\mathbb{R}$. [6]
(b) Hence solve the equation $z^2-(4-i)z+(5-5i)=0$, giving your answers in the form $a+bi$. [5]
Show complete worked solution
(a) Let $a+bi$, where $a,b\in\mathbb{R}$, be a square root of $-5+12i$, so:
$$(a+bi)^2=-5+12i$$
$$a^2-b^2+2abi=-5+12i$$
Equating real and imaginary parts:
$$a^2-b^2=-5 \quad (1)$$
$$2ab=12 \Rightarrow ab=6 \Rightarrow b=\frac{6}{a} \quad (2)$$
Substitute (2) into (1):
$$a^2-\left(\frac{6}{a}\right)^2=-5$$
$$a^2-\frac{36}{a^2}=-5$$
Multiply through by $a^2$:
$$a^4-36=-5a^2$$
$$a^4+5a^2-36=0$$
Let $u=a^2$:
$$u^2+5u-36=0$$
$$(u+9)(u-4)=0$$
$$u=-9 \text{ or } u=4$$
Since $u=a^2\ge0$, reject $u=-9$. So $a^2=4$, giving $a=2$ or $a=-2$.
Using $b=6/a$: if $a=2$, $b=3$; if $a=-2$, $b=-3$.
Check: $(2+3i)^2=4+12i+9i^2=4+12i-9=-5+12i$ $\checkmark$
$$\boxed{\text{Square roots of } -5+12i \text{ are } \pm(2+3i)}$$
(b) Using the quadratic formula on $z^2-(4-i)z+(5-5i)=0$, with $a=1$, $b=-(4-i)$, $c=5-5i$:
$$z=\frac{(4-i)\pm\sqrt{(4-i)^2-4(5-5i)}}{2}$$
First compute $(4-i)^2$:
$$(4-i)^2=16-8i+i^2=16-8i-1=15-8i$$
Then compute $4(5-5i)=20-20i$, so the discriminant is:
$$(15-8i)-(20-20i)=15-8i-20+20i=-5+12i$$
By part (a), the square roots of $-5+12i$ are $\pm(2+3i)$. Substituting:
$$z=\frac{(4-i)\pm(2+3i)}{2}$$
Taking the $+$ sign:
$$z=\frac{(4-i)+(2+3i)}{2}=\frac{6+2i}{2}=3+i$$
Taking the $-$ sign:
$$z=\frac{(4-i)-(2+3i)}{2}=\frac{2-4i}{2}=1-2i$$
$$\boxed{z=3+i \text{ or } z=1-2i}$$
QUESTION 47
10 marks
Hard
Given that $z=1-i$ and $z=-2+3i$ are two roots of the quartic equation $z^4+bz^3+cz^2+dz+e=0$, where $b,c,d,e\in\mathbb{R}$:
(a) Write down the other two roots. [2]
(b) Find the two quadratic factors, with real coefficients, corresponding to each conjugate pair of roots. [4]
(c) Hence find the values of $b$, $c$, $d$ and $e$. [4]
Show complete worked solution
(a) Since the quartic has real coefficients, non-real roots occur in conjugate pairs:
$$\boxed{z=1+i \text{ and } z=-2-3i}$$
(b) The quadratic with roots $1-i$ and $1+i$ has sum $2$ and product $(1-i)(1+i)=1-i^2=1+1=2$:
$$z^2-2z+2$$
The quadratic with roots $-2+3i$ and $-2-3i$ has sum $-4$ and product $(-2+3i)(-2-3i)=(-2)^2-(3i)^2=4-9i^2=4+9=13$:
$$z^2+4z+13$$
$$\boxed{z^2-2z+2 \text{ and } z^2+4z+13}$$
(c) The quartic is the product of these two quadratic factors:
$$z^4+bz^3+cz^2+dz+e=(z^2-2z+2)(z^2+4z+13)$$
Expanding the product term by term:
$$(z^2-2z+2)(z^2+4z+13)$$
$$=z^2(z^2+4z+13)-2z(z^2+4z+13)+2(z^2+4z+13)$$
$$=(z^4+4z^3+13z^2)+(-2z^3-8z^2-26z)+(2z^2+8z+26)$$
Collecting like powers of $z$:
$$z^4 \text{ term: } z^4$$
$$z^3 \text{ terms: } 4z^3-2z^3=2z^3$$
$$z^2 \text{ terms: } 13z^2-8z^2+2z^2=7z^2$$
$$z \text{ terms: } -26z+8z=-18z$$
$$\text{constant: } 26$$
So:
$$z^4+bz^3+cz^2+dz+e=z^4+2z^3+7z^2-18z+26$$
$$\boxed{b=2,\ c=7,\ d=-18,\ e=26}$$
QUESTION 48
10 marks
Hard
Two loci in the complex plane are defined by $|z-3|=5$ and $|z-3-8i|=5$.
(a) Write down the Cartesian equation of each locus. [3]
(b) Find the point(s) of intersection of the two loci. [7]
Show complete worked solution
(a) The first locus, $|z-3|=5$, is a circle of radius $5$ centred at $(3,0)$:
$$(x-3)^2+y^2=25 \quad (1)$$
The second locus, $|z-3-8i|=|z-(3+8i)|=5$, is a circle of radius $5$ centred at $(3,8)$:
$$(x-3)^2+(y-8)^2=25 \quad (2)$$
$$\boxed{(x-3)^2+y^2=25 \text{ and } (x-3)^2+(y-8)^2=25}$$
(b) Subtract equation (1) from equation (2) to eliminate the $(x-3)^2$ term:
$$(y-8)^2-y^2=25-25$$
$$(y-8)^2-y^2=0$$
Expand $(y-8)^2=y^2-16y+64$:
$$y^2-16y+64-y^2=0$$
$$-16y+64=0$$
$$y=4$$
Substitute $y=4$ into equation (1):
$$(x-3)^2+4^2=25$$
$$(x-3)^2=25-16=9$$
$$x-3=\pm3$$
$$x=6 \text{ or } x=0$$
So the two circles intersect at the points $(6,4)$ and $(0,4)$, representing
$$\boxed{z=6+4i \text{ and } z=4i}$$
QUESTION 49
10 marks
Hard
The cubic equation $z^3+pz^2+37z+q=0$, where $p,q\in\mathbb{R}$, has $z=2+3i$ as a root.
(a) Write down a second root. [1]
(b) Given that the third root is real, use the fact that the sum of the products of the roots taken two at a time equals $37$ (the coefficient of $z$, since the equation is monic) to find this real root. [5]
(c) Hence find the values of $p$ and $q$. [4]
Show complete worked solution
(a) Since the cubic has real coefficients, complex roots occur in conjugate pairs:
$$\boxed{z=2-3i}$$
(b) Let the three roots be $2+3i$, $2-3i$, and a real root $r$. For a monic cubic $z^3+pz^2+37z+q=0$ with roots $\alpha,\beta,\gamma$, the sum of the products of pairs of roots is $\alpha\beta+\beta\gamma+\gamma\alpha=37$.
$$\alpha\beta=(2+3i)(2-3i)=2^2-(3i)^2=4-9i^2=4+9=13$$
$$\beta\gamma+\gamma\alpha=r\big[(2-3i)+(2+3i)\big]=r(4)=4r$$
So:
$$13+4r=37$$
$$4r=24$$
$$\boxed{r=6}$$
(c) The sum of all three roots is $-p$:
$$(2+3i)+(2-3i)+6=10$$
$$-p=10 \Rightarrow p=-10$$
The product of all three roots is $-q$:
$$(2+3i)(2-3i)(6)=13\times6=78$$
$$-q=78 \Rightarrow q=-78$$
Check by substituting the real root $z=6$ into $z^3-10z^2+37z-78$: $216-360+222-78=0$ $\checkmark$
$$\boxed{p=-10,\ q=-78}$$
QUESTION 50
10 marks
Hard
The complex numbers $z$ and $w$ satisfy the simultaneous equations
$$z+iw=2$$
$$2z-w=7-4i$$
(a) Solve the system to find $z$ and $w$. [7]
(b) Hence find $z\bar w$, giving your answer in the form $a+bi$. [3]
Show complete worked solution
(a) From the second equation, make $w$ the subject:
$$w=2z-(7-4i)=2z-7+4i \quad (1)$$
Substitute (1) into the first equation:
$$z+i(2z-7+4i)=2$$
Expand:
$$z+2iz-7i+4i^2=2$$
$$z+2iz-7i-4=2$$
Combine the $z$ terms:
$$z(1+2i)=2+7i+4=6+7i$$
$$z=\frac{6+7i}{1+2i}$$
Multiply the numerator and denominator by the conjugate $1-2i$:
$$z=\frac{(6+7i)(1-2i)}{(1+2i)(1-2i)}$$
Expand the numerator:
$$(6+7i)(1-2i)=6-12i+7i-14i^2=6-5i+14=20-5i$$
Expand the denominator:
$$(1+2i)(1-2i)=1-4i^2=1+4=5$$
So:
$$z=\frac{20-5i}{5}=4-i$$
Substitute $z=4-i$ back into (1):
$$w=2(4-i)-7+4i=8-2i-7+4i=1+2i$$
Check in the first equation: $z+iw=(4-i)+i(1+2i)=4-i+i+2i^2=4-i+i-2=2$ $\checkmark$
$$\boxed{z=4-i,\ w=1+2i}$$
(b) Since $w=1+2i$, its conjugate is $\bar w=1-2i$. Then:
$$z\bar w=(4-i)(1-2i)=4-8i-i+2i^2=4-9i-2$$
$$\boxed{z\bar w=2-9i}$$
Complex Numbers: Geometric Representations 50 questions
QUESTION 1
4 marks
Easy
Express $z = -1+i\sqrt{3}$ in the modulus-argument form $z = r(\cos\theta + i\sin\theta)$, where $r>0$ and $-\pi < \theta \le \pi$, stating the exact values of $r$ and $\theta$.
Show complete worked solution
Using $r=|z|$ and finding the argument from the quadrant of $z$:
$$r=|z|=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=\sqrt4=2$$
Since $z$ has negative real part and positive imaginary part, $z$ lies in the second quadrant. The reference angle is
$$\arctan\left(\dfrac{\sqrt3}{1}\right)=\dfrac{\pi}{3}$$
Since $z$ is in the second quadrant:
$$\theta=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}$$
$$\boxed{z=2\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)}$$
QUESTION 2
4 marks
Easy
Write $z = 3e^{-i\pi/4}$ in Cartesian form $a+bi$, giving $a$ and $b$ as exact values.
Show complete worked solution
Using Euler's formula, $e^{i\theta}=\cos\theta+i\sin\theta$:
$$z=3e^{-i\pi/4}=3\left(\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right)\right)$$
Since cosine is even and sine is odd:
$$\cos\left(-\dfrac{\pi}{4}\right)=\cos\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}, \qquad \sin\left(-\dfrac{\pi}{4}\right)=-\sin\dfrac{\pi}{4}=-\dfrac{\sqrt2}{2}$$
Substituting:
$$z=3\left(\dfrac{\sqrt2}{2}-i\dfrac{\sqrt2}{2}\right)=\dfrac{3\sqrt2}{2}-\dfrac{3\sqrt2}{2}i$$
$$\boxed{z=\dfrac{3\sqrt2}{2}-\dfrac{3\sqrt2}{2}i}$$
QUESTION 3
6 marks
Medium
Let $z_1 = 4e^{i\pi/3}$ and $z_2 = 2e^{i\pi/6}$. Find $z_1z_2$ and $\dfrac{z_1}{z_2}$, giving each answer in the form $re^{i\theta}$. Hence express $z_1z_2$ in Cartesian form.
Show complete worked solution
To multiply complex numbers in Euler form, multiply the moduli and add the arguments:
$$z_1z_2=(4)(2)\,e^{i(\pi/3+\pi/6)}=8e^{i(2\pi/6+\pi/6)}=8e^{i\pi/2}$$
To divide, divide the moduli and subtract the arguments:
$$\dfrac{z_1}{z_2}=\dfrac{4}{2}e^{i(\pi/3-\pi/6)}=2e^{i(2\pi/6-\pi/6)}=2e^{i\pi/6}$$
Converting $z_1z_2$ to Cartesian form using $e^{i\theta}=\cos\theta+i\sin\theta$:
$$z_1z_2=8e^{i\pi/2}=8\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)=8(0+i)=8i$$
$$\boxed{z_1z_2=8e^{i\pi/2}=8i, \qquad \dfrac{z_1}{z_2}=2e^{i\pi/6}}$$
QUESTION 4
7 marks
Medium
Let $z = \dfrac{(1+i)^3}{1-i\sqrt3}$.
(a) Show that $(1+i)^3 = -2+2i$.
(b) Express $(1+i)^3$ and $1-i\sqrt3$ each in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$.
(c) Hence express $z$ in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$.
Show complete worked solution
(a) Using $(1+i)^2=1+2i+i^2$ and $i^2=-1$:
$$(1+i)^2=1+2i-1=2i$$
So:
$$(1+i)^3=(1+i)^2(1+i)=2i(1+i)=2i+2i^2=2i-2=-2+2i$$
as required.
(b) For $(1+i)^3=-2+2i$: using $r=|z|$ and the quadrant of $z$:
$$r_1=\sqrt{(-2)^2+2^2}=\sqrt{4+4}=2\sqrt2$$
The point $(-2,2)$ lies in the second quadrant; the reference angle is $\arctan(2/2)=\pi/4$, so:
$$\theta_1=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$$
$$(1+i)^3=2\sqrt2\,e^{i3\pi/4}$$
For $1-i\sqrt3$:
$$r_2=\sqrt{1^2+(\sqrt3)^2}=\sqrt{1+3}=2$$
The point $(1,-\sqrt3)$ lies in the fourth quadrant; the reference angle is $\arctan(\sqrt3/1)=\pi/3$, so:
$$\theta_2=-\dfrac{\pi}{3}$$
$$\boxed{(1+i)^3=2\sqrt2\,e^{i3\pi/4}, \qquad 1-i\sqrt3=2e^{-i\pi/3}}$$
(c) Dividing, using the rule for division in Euler form:
$$z=\dfrac{2\sqrt2\,e^{i3\pi/4}}{2e^{-i\pi/3}}=\sqrt2\,e^{i(3\pi/4-(-\pi/3))}=\sqrt2\,e^{i(3\pi/4+\pi/3)}$$
Using a common denominator of 12: $3\pi/4=9\pi/12$ and $\pi/3=4\pi/12$, so the argument is $13\pi/12$.
Since $13\pi/12>\pi$, subtract $2\pi$ to bring it into the required range $(-\pi,\pi]$:
$$\dfrac{13\pi}{12}-2\pi=\dfrac{13\pi}{12}-\dfrac{24\pi}{12}=-\dfrac{11\pi}{12}$$
$$\boxed{z=\sqrt2\,e^{-i11\pi/12}}$$
QUESTION 5
16 marks
Hard
(a) Using Euler's formula $e^{i\theta} = \cos\theta+i\sin\theta$, show that $\cos\theta = \dfrac{e^{i\theta}+e^{-i\theta}}{2}$ and $\sin\theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}$.
(b) Hence show that $\cos A\cos B = \dfrac{1}{2}\left[\cos(A-B)+\cos(A+B)\right]$.
(c) Using the result from part (b), find the exact value of $\cos75^{\circ}\cos15^{\circ}$.
(d) By writing $e^{i\alpha}+e^{i\beta} = e^{i(\alpha+\beta)/2}\left(e^{i(\alpha-\beta)/2}+e^{-i(\alpha-\beta)/2}\right)$, show that $e^{i\alpha}+e^{i\beta} = 2\cos\left(\dfrac{\alpha-\beta}{2}\right)e^{i(\alpha+\beta)/2}$. Hence find the modulus and argument of $w = e^{i\pi/3}+e^{i2\pi/3}$.
Show complete worked solution
(a) By Euler's formula:
$$e^{i\theta}=\cos\theta+i\sin\theta, \qquad e^{-i\theta}=\cos(-\theta)+i\sin(-\theta)=\cos\theta-i\sin\theta$$
(using $\cos$ even, $\sin$ odd). Adding these two equations:
$$e^{i\theta}+e^{-i\theta}=2\cos\theta \quad \Rightarrow \quad \cos\theta=\dfrac{e^{i\theta}+e^{-i\theta}}{2}$$
Subtracting them:
$$e^{i\theta}-e^{-i\theta}=2i\sin\theta \quad \Rightarrow \quad \sin\theta=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}$$
$$\boxed{\cos\theta=\dfrac{e^{i\theta}+e^{-i\theta}}{2}, \qquad \sin\theta=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}}$$
(b) Using the results of part (a):
$$\cos A\cos B=\dfrac{e^{iA}+e^{-iA}}{2}\cdot\dfrac{e^{iB}+e^{-iB}}{2}=\dfrac{1}{4}\left(e^{i(A+B)}+e^{i(A-B)}+e^{-i(A-B)}+e^{-i(A+B)}\right)$$
Group into conjugate pairs and use part (a) in reverse:
$$=\dfrac{1}{4}\left[\left(e^{i(A+B)}+e^{-i(A+B)}\right)+\left(e^{i(A-B)}+e^{-i(A-B)}\right)\right]$$
$$=\dfrac{1}{4}\left[2\cos(A+B)+2\cos(A-B)\right]$$
$$=\dfrac{1}{2}\left[\cos(A-B)+\cos(A+B)\right]$$
as required.
(c) Let $A=75^{\circ}$, $B=15^{\circ}$. Substituting into the result of part (b):
$$\cos75^{\circ}\cos15^{\circ}=\dfrac{1}{2}\left[\cos(60^{\circ})+\cos(90^{\circ})\right]=\dfrac{1}{2}\left[\dfrac{1}{2}+0\right]=\dfrac{1}{4}$$
$$\boxed{\cos75^{\circ}\cos15^{\circ}=\dfrac{1}{4}}$$
(d) Factoring out $e^{i(\alpha+\beta)/2}$:
$$e^{i\alpha}+e^{i\beta}=e^{i(\alpha+\beta)/2}\left(e^{i(\alpha-\beta)/2}+e^{-i(\alpha-\beta)/2}\right)$$
By part (a), applied with $\theta=(\alpha-\beta)/2$:
$$e^{i(\alpha-\beta)/2}+e^{-i(\alpha-\beta)/2}=2\cos\left(\dfrac{\alpha-\beta}{2}\right)$$
So:
$$e^{i\alpha}+e^{i\beta}=2\cos\left(\dfrac{\alpha-\beta}{2}\right)e^{i(\alpha+\beta)/2}$$
as required. Applying this with $\alpha=\pi/3$, $\beta=2\pi/3$:
$$\dfrac{\alpha-\beta}{2}=\dfrac{\pi/3-2\pi/3}{2}=\dfrac{-\pi/3}{2}=-\dfrac{\pi}{6}, \qquad \dfrac{\alpha+\beta}{2}=\dfrac{\pi}{2}$$
$$w=2\cos\left(-\dfrac{\pi}{6}\right)e^{i\pi/2}=2\cdot\dfrac{\sqrt3}{2}\,e^{i\pi/2}=\sqrt3\,e^{i\pi/2}$$
Check directly: $e^{i\pi/3}+e^{i2\pi/3}=\left(\tfrac12+i\tfrac{\sqrt3}{2}\right)+\left(-\tfrac12+i\tfrac{\sqrt3}{2}\right)=i\sqrt3$, which has modulus $\sqrt3$ and argument $\pi/2$. $\checkmark$
$$\boxed{|w|=\sqrt3, \ \arg(w)=\dfrac{\pi}{2}}$$
QUESTION 6
4 marks
Easy
Use De Moivre's theorem to find the exact value of $(1+i)^8$, giving your answer in Cartesian form.
Show complete worked solution
First write $1+i$ in modulus-argument form:
$$r=|1+i|=\sqrt{1^2+1^2}=\sqrt2, \qquad \theta=\arg(1+i)=\dfrac{\pi}{4}$$
So:
$$1+i=\sqrt2\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$$
By De Moivre's theorem:
$$(1+i)^8=(\sqrt2)^8\left(\cos\dfrac{8\pi}{4}+i\sin\dfrac{8\pi}{4}\right)=16\left(\cos2\pi+i\sin2\pi\right)$$
Since $\cos2\pi=1$ and $\sin2\pi=0$:
$$(1+i)^8=16(1+0i)=16$$
$$\boxed{(1+i)^8=16}$$
QUESTION 7
5 marks
Easy
The polynomial $p(x) = x^3 - 6x^2 + 21x - 26$ has real coefficients. Given that $z=2-3i$ is a root of $p(x)=0$, find the other two roots.
Show complete worked solution
Since $p(x)$ has real coefficients, complex roots occur in conjugate pairs. As $z=2-3i$ is a root, its conjugate $z^*=2+3i$ is also a root.
Let the third (real) root be $r$. For a cubic $x^3+bx^2+cx+d$ with roots $\alpha,\beta,\gamma$, the sum of roots is $-b$. Here $b=-6$, so:
$$(2-3i)+(2+3i)+r=-(-6)=6$$
$$4+r=6$$
$$r=2$$
Check using the product of roots ($=-d=26$ for this cubic):
$$(2-3i)(2+3i)(2)=(4+9)(2)=13\times2=26 \checkmark$$
$$\boxed{\text{the other two roots are } z=2+3i \text{ and } z=2}$$
QUESTION 8
6 marks
Medium
Use De Moivre's theorem to prove that $\cos3\theta = 4\cos^3\theta - 3\cos\theta$.
Show complete worked solution
By De Moivre's theorem:
$$(\cos\theta+i\sin\theta)^3=\cos3\theta+i\sin3\theta$$
Expand the left-hand side using the binomial theorem:
$$(\cos\theta+i\sin\theta)^3=\cos^3\theta+3\cos^2\theta(i\sin\theta)+3\cos\theta(i\sin\theta)^2+(i\sin\theta)^3$$
$$=\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta$$
Group real and imaginary parts:
$$=\left(\cos^3\theta-3\cos\theta\sin^2\theta\right)+i\left(3\cos^2\theta\sin\theta-\sin^3\theta\right)$$
Equating real parts with $\cos3\theta+i\sin3\theta$:
$$\cos3\theta=\cos^3\theta-3\cos\theta\sin^2\theta$$
Using $\sin^2\theta=1-\cos^2\theta$:
$$\cos3\theta=\cos^3\theta-3\cos\theta(1-\cos^2\theta)$$
$$=\cos^3\theta-3\cos\theta+3\cos^3\theta$$
$$=4\cos^3\theta-3\cos\theta$$
as required. $\blacksquare$
QUESTION 9
8 marks
Medium
Find the three cube roots of $z=-8i$. Give each root in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$, and hence express each root in Cartesian form $a+bi$. Verify your answers by showing that the sum of the three roots is zero.
Show complete worked solution
First write $z=-8i$ in polar form:
$$r=|-8i|=8$$
Since $-8i$ lies on the negative imaginary axis:
$$\theta=-\dfrac{\pi}{2}$$
So $z=8e^{-i\pi/2}$.
The cube roots have modulus $8^{1/3}=2$ and arguments
$$\theta_k=\dfrac{-\pi/2+2k\pi}{3}, \quad k=0,1,2$$
$$k=0: \ \theta_0=-\dfrac{\pi}{6}$$
$$k=1: \ \theta_1=\dfrac{-\pi/2+2\pi}{3}=\dfrac{3\pi/2}{3}=\dfrac{\pi}{2}$$
$$k=2: \ \theta_2=\dfrac{-\pi/2+4\pi}{3}=\dfrac{7\pi/2}{3}=\dfrac{7\pi}{6}$$
Since $\dfrac{7\pi}{6}>\pi$, subtract $2\pi$ to bring it into range: $\dfrac{7\pi}{6}-2\pi=-\dfrac{5\pi}{6}$
So the three roots in polar form are:
$$w_0=2e^{-i\pi/6}, \quad w_1=2e^{i\pi/2}, \quad w_2=2e^{-i5\pi/6}$$
Converting to Cartesian form using $\cos\theta+i\sin\theta$:
$$w_0=2\left(\cos(-30^\circ)+i\sin(-30^\circ)\right)=2\left(\dfrac{\sqrt3}{2}-\dfrac{1}{2}i\right)=\sqrt3-i$$
$$w_1=2\left(\cos90^\circ+i\sin90^\circ\right)=2(0+i)=2i$$
$$w_2=2\left(\cos(-150^\circ)+i\sin(-150^\circ)\right)=2\left(-\dfrac{\sqrt3}{2}-\dfrac{1}{2}i\right)=-\sqrt3-i$$
Sum check:
$$(\sqrt3-i)+(2i)+(-\sqrt3-i)=(\sqrt3-\sqrt3)+i(-1+2-1)=0+0i=0 \checkmark$$
(This is expected since the roots satisfy $w^3+8i=0$, a cubic in $w$ with zero coefficient of $w^2$, so the sum of roots is 0.)
$$\boxed{w_0=\sqrt3-i,\ \ w_1=2i,\ \ w_2=-\sqrt3-i}$$
QUESTION 10
14 marks
Hard
Let $z = 1+i\sqrt3$.
(a) Express $z$ in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$.
(b) Use De Moivre's theorem to find the value of $z^6$, giving your answer in Cartesian form.
(c) Find all solutions $w$ to the equation $w^5=z$, giving the modulus and the exact argument (in the range $-\pi<\theta\le\pi$) of each solution.
(d) State the sum of the five solutions found in part (c), and justify your answer.
Show complete worked solution
(a) Using $r=|z|$ and the quadrant of $z$:
$$r=|z|=\sqrt{1^2+(\sqrt3)^2}=\sqrt{1+3}=2$$
$(1,\sqrt3)$ lies in the first quadrant:
$$\theta=\arctan(\sqrt3/1)=\dfrac{\pi}{3}$$
$$z=2e^{i\pi/3}$$
(b) By De Moivre's theorem:
$$z^6=2^6\,e^{i6(\pi/3)}=64\,e^{i2\pi}=64(\cos2\pi+i\sin2\pi)=64(1+0i)$$
$$\boxed{z^6=64}$$
(c) We need $w^5=2e^{i\pi/3}$. Write $w=\rho e^{i\phi}$. Then $\rho^5=2 \Rightarrow \rho=2^{1/5}$, and:
$$5\phi=\dfrac{\pi}{3}+2k\pi \ \Rightarrow\ \phi_k=\dfrac{\pi/3+2k\pi}{5}, \quad k=0,1,2,3,4$$
Writing all values with common denominator 15 ($\pi=15\pi/15$):
$$k=0: \ \phi_0=\dfrac{\pi}{15}$$
$$k=1: \ \phi_1=\dfrac{7\pi}{15}$$
$$k=2: \ \phi_2=\dfrac{13\pi}{15}$$
$$k=3: \ \phi_3=\dfrac{19\pi}{15} \ \to\ \text{subtract } 2\pi=\dfrac{30\pi}{15}: \ \phi_3=-\dfrac{11\pi}{15}$$
$$k=4: \ \phi_4=\dfrac{25\pi}{15} \ \to\ \text{subtract } 2\pi: \ \phi_4=-\dfrac{5\pi}{15}=-\dfrac{\pi}{3}$$
(These are equally spaced by $\dfrac{2\pi}{5}=\dfrac{6\pi}{15}$, confirming: $-\dfrac{11\pi}{15}, -\dfrac{\pi}{3}\left(=-\dfrac{5\pi}{15}\right), \dfrac{\pi}{15}, \dfrac{7\pi}{15}, \dfrac{13\pi}{15}$.)
The five solutions all have modulus $2^{1/5}$, with arguments
$$\boxed{-\dfrac{11\pi}{15},\ -\dfrac{\pi}{3},\ \dfrac{\pi}{15},\ \dfrac{7\pi}{15},\ \dfrac{13\pi}{15}}$$
(d) The five roots of $w^5-z=0$ are equally spaced points on a circle of radius $2^{1/5}$ centred at the origin (vertices of a regular pentagon). By symmetry, their sum is 0. Algebraically, this follows because $w^5-z=0$ has zero coefficient of $w^4$, so the sum of all five roots equals $-(\text{coefficient of }w^4)=0$.
$$\boxed{\text{sum}=0}$$
QUESTION 11
4 marks
Easy
Express $z = 3-3i$ in the modulus-argument form $z = r(\cos\theta + i\sin\theta)$, where $r>0$ and $-\pi < \theta \le \pi$, stating the exact values of $r$ and $\theta$.
Show complete worked solution
Using $r=|z|$ and finding the argument from the quadrant of $z$:
$$r=|z|=\sqrt{3^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt2$$
Since $z$ has positive real part and negative imaginary part, $z$ lies in the fourth quadrant. The reference angle is
$$\arctan\left(\dfrac{3}{3}\right)=\dfrac{\pi}{4}$$
Since $z$ is in the fourth quadrant:
$$\theta=-\dfrac{\pi}{4}$$
$$\boxed{z=3\sqrt2\left(\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right)\right)}$$
QUESTION 12
4 marks
Easy
Write $z = 5e^{i2\pi/3}$ in Cartesian form $a+bi$, giving $a$ and $b$ as exact values.
Show complete worked solution
Using Euler's formula, $e^{i\theta}=\cos\theta+i\sin\theta$:
$$z=5e^{i2\pi/3}=5\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)$$
The angle $\dfrac{2\pi}{3}$ has reference angle $\dfrac{\pi}{3}$ and lies in the second quadrant, so:
$$\cos\dfrac{2\pi}{3}=-\dfrac{1}{2}, \qquad \sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}$$
Substituting:
$$z=5\left(-\dfrac{1}{2}+i\dfrac{\sqrt3}{2}\right)=-\dfrac{5}{2}+\dfrac{5\sqrt3}{2}i$$
$$\boxed{z=-\dfrac{5}{2}+\dfrac{5\sqrt3}{2}i}$$
QUESTION 13
4 marks
Easy
Find the modulus and argument of $z = -4+4i$, giving the argument exactly in the range $-\pi < \theta \le \pi$.
Show complete worked solution
Using $r=|z|$:
$$r=|z|=\sqrt{(-4)^2+4^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt2$$
The point $(-4,4)$ lies in the second quadrant. The reference angle is
$$\arctan\left(\dfrac{4}{4}\right)=\dfrac{\pi}{4}$$
Since $z$ is in the second quadrant:
$$\theta=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$$
$$\boxed{|z|=4\sqrt2, \qquad \arg(z)=\dfrac{3\pi}{4}}$$
QUESTION 14
5 marks
Easy
Use De Moivre's theorem to find the exact value of $(\sqrt3+i)^4$, giving your answer in Cartesian form.
Show complete worked solution
First write $\sqrt3+i$ in modulus-argument form. Using $r=|z|$ and the quadrant of $z$:
$$r=\sqrt{(\sqrt3)^2+1^2}=\sqrt{3+1}=2$$
The point $(\sqrt3,1)$ lies in the first quadrant:
$$\theta=\arctan\left(\dfrac{1}{\sqrt3}\right)=\dfrac{\pi}{6}$$
So $\sqrt3+i=2\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)$.
By De Moivre's theorem:
$$(\sqrt3+i)^4=2^4\left(\cos\dfrac{4\pi}{6}+i\sin\dfrac{4\pi}{6}\right)=16\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)$$
Since $\cos\dfrac{2\pi}{3}=-\dfrac{1}{2}$ and $\sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}$:
$$(\sqrt3+i)^4=16\left(-\dfrac{1}{2}+i\dfrac{\sqrt3}{2}\right)=-8+8\sqrt3\,i$$
$$\boxed{(\sqrt3+i)^4=-8+8\sqrt3\,i}$$
QUESTION 15
4 marks
Easy
Express the purely imaginary number $z=-6i$ in the form $re^{i\theta}$, where $r>0$ and $-\pi < \theta \le \pi$.
Show complete worked solution
The point representing $z=-6i$ on the Argand diagram is $(0,-6)$, which lies on the negative imaginary axis.
The modulus is
$$r=|z|=\sqrt{0^2+(-6)^2}=6$$
A point on the negative imaginary axis has argument $-\dfrac{\pi}{2}$ (a quarter turn clockwise from the positive real axis), so
$$\theta=-\dfrac{\pi}{2}$$
$$\boxed{z=6e^{-i\pi/2}}$$
QUESTION 16
5 marks
Easy
Write $z=2e^{i5\pi/6}$ in Cartesian form $a+bi$, giving $a$ and $b$ as exact values.
Show complete worked solution
Using Euler's formula, $e^{i\theta}=\cos\theta+i\sin\theta$:
$$z=2e^{i5\pi/6}=2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)$$
The angle $\dfrac{5\pi}{6}$ has reference angle $\dfrac{\pi}{6}$ and lies in the second quadrant, so:
$$\cos\dfrac{5\pi}{6}=-\dfrac{\sqrt3}{2}, \qquad \sin\dfrac{5\pi}{6}=\dfrac{1}{2}$$
Substituting:
$$z=2\left(-\dfrac{\sqrt3}{2}+i\dfrac{1}{2}\right)=-\sqrt3+i$$
$$\boxed{z=-\sqrt3+i}$$
QUESTION 17
5 marks
Easy
Sketch the locus of points $z$ satisfying $|z-3|=2$ on an Argand diagram, and find its Cartesian equation.
Show complete worked solution
The expression $|z-3|$ is the distance from the point representing $z$ to the point representing $3$ (that is, $(3,0)$). The equation $|z-3|=2$ therefore describes the set of points a constant distance $2$ from $(3,0)$, which is a circle of radius $2$ centred at $(3,0)$.
The sketch is shown in the diagram: a circle of radius $2$ centred at the point $3$ on the real axis.
To find the Cartesian equation, let $z=x+iy$:
$$|z-3|=|(x-3)+iy|=\sqrt{(x-3)^2+y^2}$$
Setting this equal to $2$ and squaring both sides:
$$(x-3)^2+y^2=4$$
$$\boxed{(x-3)^2+y^2=4 \text{, a circle of centre } (3,0) \text{ and radius } 2}$$
QUESTION 18
4 marks
Easy
State the centre and radius of the locus $|z|=5$, and find its Cartesian equation.
Show complete worked solution
The expression $|z|$ is the distance from the point representing $z$ to the origin. The equation $|z|=5$ describes the set of points a constant distance $5$ from the origin, so the locus is a circle centred at the origin with radius $5$.
Letting $z=x+iy$:
$$|z|=\sqrt{x^2+y^2}=5$$
Squaring both sides:
$$x^2+y^2=25$$
$$\boxed{\text{centre } (0,0), \text{ radius } 5, \text{ Cartesian equation } x^2+y^2=25}$$
QUESTION 19
4 marks
Easy
Let $z_1=2e^{i\pi/4}$ and $z_2=3e^{i\pi/3}$. Find $z_1z_2$, giving your answer in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$.
Show complete worked solution
To multiply complex numbers in Euler form, multiply the moduli and add the arguments:
$$z_1z_2=(2)(3)\,e^{i(\pi/4+\pi/3)}$$
Using a common denominator of $12$: $\dfrac{\pi}{4}=\dfrac{3\pi}{12}$ and $\dfrac{\pi}{3}=\dfrac{4\pi}{12}$, so
$$\dfrac{\pi}{4}+\dfrac{\pi}{3}=\dfrac{3\pi}{12}+\dfrac{4\pi}{12}=\dfrac{7\pi}{12}$$
Since $\dfrac{7\pi}{12}$ already lies in $(-\pi,\pi]$, no adjustment is needed.
$$\boxed{z_1z_2=6e^{i7\pi/12}}$$
QUESTION 20
4 marks
Easy
Let $z_1=2e^{i\pi/4}$ and $z_2=3e^{i\pi/3}$. Find $\dfrac{z_1}{z_2}$, giving your answer in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$.
Show complete worked solution
To divide complex numbers in Euler form, divide the moduli and subtract the arguments:
$$\dfrac{z_1}{z_2}=\dfrac{2}{3}\,e^{i(\pi/4-\pi/3)}$$
Using a common denominator of $12$: $\dfrac{\pi}{4}=\dfrac{3\pi}{12}$ and $\dfrac{\pi}{3}=\dfrac{4\pi}{12}$, so
$$\dfrac{\pi}{4}-\dfrac{\pi}{3}=\dfrac{3\pi}{12}-\dfrac{4\pi}{12}=-\dfrac{\pi}{12}$$
Since $-\dfrac{\pi}{12}$ already lies in $(-\pi,\pi]$, no adjustment is needed.
$$\boxed{\dfrac{z_1}{z_2}=\dfrac{2}{3}e^{-i\pi/12}}$$
QUESTION 21
5 marks
Easy
Use De Moivre's theorem to find the exact value of $(1-i)^6$, giving your answer in Cartesian form.
Show complete worked solution
First write $1-i$ in modulus-argument form:
$$r=|1-i|=\sqrt{1^2+(-1)^2}=\sqrt2, \qquad \theta=\arg(1-i)=-\dfrac{\pi}{4}$$
So $1-i=\sqrt2\left(\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right)\right)$.
By De Moivre's theorem:
$$(1-i)^6=(\sqrt2)^6\left(\cos\left(-\dfrac{6\pi}{4}\right)+i\sin\left(-\dfrac{6\pi}{4}\right)\right)=8\left(\cos\left(-\dfrac{3\pi}{2}\right)+i\sin\left(-\dfrac{3\pi}{2}\right)\right)$$
Since $-\dfrac{3\pi}{2}$ is outside $(-\pi,\pi]$, add $2\pi$:
$$-\dfrac{3\pi}{2}+2\pi=\dfrac{\pi}{2}$$
$$(1-i)^6=8\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)=8(0+i)=8i$$
$$\boxed{(1-i)^6=8i}$$
QUESTION 22
6 marks
Easy
Find all four solutions of $z^4=1$ (the fourth roots of unity), giving each in Cartesian form. Describe their geometric arrangement on an Argand diagram.
Show complete worked solution
Write $1=1e^{i0}$ (modulus $1$, argument $0$). By the $n$th roots formula, the fourth roots have modulus $1^{1/4}=1$ and arguments
$$\theta_k=\dfrac{0+2k\pi}{4}=\dfrac{k\pi}{2}, \quad k=0,1,2,3$$
$$k=0:\ \theta_0=0 \qquad k=1:\ \theta_1=\dfrac{\pi}{2} \qquad k=2:\ \theta_2=\pi \qquad k=3:\ \theta_3=\dfrac{3\pi}{2}\left(\text{equivalently } -\dfrac{\pi}{2}\right)$$
Converting each to Cartesian form using $\cos\theta+i\sin\theta$:
$$w_0=\cos0+i\sin0=1, \qquad w_1=\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}=i$$
$$w_2=\cos\pi+i\sin\pi=-1, \qquad w_3=\cos\left(-\dfrac{\pi}{2}\right)+i\sin\left(-\dfrac{\pi}{2}\right)=-i$$
All four roots have modulus $1$, so they lie on the unit circle, and their arguments are equally spaced by $\dfrac{2\pi}{4}=\dfrac{\pi}{2}$. They are therefore the vertices of a square inscribed in the unit circle, as shown in the diagram.
$$\boxed{z=1,\ i,\ -1,\ -i}$$
QUESTION 23
5 marks
Easy
Sketch the locus of points $z$ (with $z\ne0$) satisfying $\arg(z)=\dfrac{\pi}{4}$ on an Argand diagram, and describe it geometrically.
Show complete worked solution
The condition $\arg(z)=\dfrac{\pi}{4}$ fixes the angle that the line segment from the origin to $z$ makes with the positive real axis, but places no restriction on $|z|$ (except $z\ne0$, since the origin itself has no defined argument).
The locus is therefore a half-line (ray) starting from, but not including, the origin, making an angle of $\dfrac{\pi}{4}$ (that is, $45^{\circ}$) with the positive real axis, as shown by the dashed ray in the diagram - the open circle at the origin indicates that the point $z=0$ is excluded.
Every point on this ray has the form $z=t\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)=t\left(\dfrac{\sqrt2}{2}+i\dfrac{\sqrt2}{2}\right)$ for some $t>0$, that is, points of the form $y=x$ with $x>0$.
$$\boxed{\text{a ray from the origin (excluded) at angle } \dfrac{\pi}{4} \text{ to the positive real axis, i.e. } y=x,\ x>0}$$
QUESTION 24
5 marks
Easy
Express $z=-2-2\sqrt3\,i$ in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$.
Show complete worked solution
Using $r=|z|$:
$$r=|z|=\sqrt{(-2)^2+(-2\sqrt3)^2}=\sqrt{4+12}=\sqrt{16}=4$$
The point $(-2,-2\sqrt3)$ lies in the third quadrant (both coordinates negative). The reference angle is
$$\arctan\left(\dfrac{2\sqrt3}{2}\right)=\arctan(\sqrt3)=\dfrac{\pi}{3}$$
Since $z$ is in the third quadrant, measuring the argument as the angle in $(-\pi,\pi]$ means going clockwise from the negative real axis:
$$\theta=-\left(\pi-\dfrac{\pi}{3}\right)=-\dfrac{2\pi}{3}$$
$$\boxed{z=4e^{-i2\pi/3}}$$
QUESTION 25
5 marks
Easy
The point $z=3+4i$ is rotated $90^{\circ}$ anticlockwise about the origin on an Argand diagram. Find the Cartesian form of the image point, and justify your method.
Show complete worked solution
Multiplying a complex number by $e^{i\pi/2}=i$ rotates its point on the Argand diagram by $90^{\circ}$ ($\dfrac{\pi}{2}$ radians) anticlockwise about the origin, since multiplication by $e^{i\pi/2}$ adds $\dfrac{\pi}{2}$ to the argument while leaving the modulus unchanged.
The image is therefore
$$w=iz=i(3+4i)=3i+4i^2=3i-4=-4+3i$$
As a check, $|z|=\sqrt{3^2+4^2}=5$ and $|w|=\sqrt{(-4)^2+3^2}=\sqrt{16+9}=5$, confirming the modulus is unchanged, as expected for a rotation. The vectors are shown in the diagram.
$$\boxed{w=-4+3i}$$
QUESTION 26
4 marks
Easy
The point $z=5-2i$ is reflected in the real axis on an Argand diagram. State the Cartesian form of the image point and explain the connection with complex conjugation.
Show complete worked solution
Reflecting a point $(a,b)$ in the real axis (the $x$-axis) sends it to $(a,-b)$, that is, it negates the imaginary part while leaving the real part unchanged.
For $z=5-2i=(5,-2)$, the reflected image is therefore
$$w=5+2i$$
This is exactly the complex conjugate of $z$: $\bar z = 5+2i = w$. In general, reflecting any point $z$ in the real axis on an Argand diagram produces its conjugate $\bar z$, since conjugation is defined precisely as negating the imaginary part.
$$\boxed{w=5+2i}$$
QUESTION 27
7 marks
Medium
Let $z=\sqrt3+i$.
(a) Express $z$ in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$. [2]
(b) Use De Moivre's theorem to find $z^5$, giving your answer first in the form $re^{i\theta}$ and then in Cartesian form. [5]
Show complete worked solution
(a) Using $r=|z|$ and the quadrant of $z$:
$$r=\sqrt{(\sqrt3)^2+1^2}=\sqrt{3+1}=2$$
The point $(\sqrt3,1)$ lies in the first quadrant:
$$\theta=\arctan\left(\dfrac{1}{\sqrt3}\right)=\dfrac{\pi}{6}$$
$$\boxed{z=2e^{i\pi/6}}$$
(b) By De Moivre's theorem, $z^5=r^5e^{i5\theta}$:
$$r^5=2^5=32, \qquad 5\theta=5\cdot\dfrac{\pi}{6}=\dfrac{5\pi}{6}$$
Since $\dfrac{5\pi}{6}$ already lies in $(-\pi,\pi]$:
$$z^5=32e^{i5\pi/6}$$
Converting to Cartesian form: $\cos\dfrac{5\pi}{6}=-\dfrac{\sqrt3}{2}$ and $\sin\dfrac{5\pi}{6}=\dfrac{1}{2}$, so
$$z^5=32\left(-\dfrac{\sqrt3}{2}+i\dfrac12\right)=-16\sqrt3+16i$$
$$\boxed{z^5=32e^{i5\pi/6}=-16\sqrt3+16i}$$
QUESTION 28
7 marks
Medium
Use De Moivre's theorem to prove that $\sin3\theta=3\sin\theta-4\sin^3\theta$.
Show complete worked solution
By De Moivre's theorem:
$$(\cos\theta+i\sin\theta)^3=\cos3\theta+i\sin3\theta$$
Expand the left-hand side using the binomial theorem:
$$(\cos\theta+i\sin\theta)^3=\cos^3\theta+3\cos^2\theta(i\sin\theta)+3\cos\theta(i\sin\theta)^2+(i\sin\theta)^3$$
$$=\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta$$
Group real and imaginary parts:
$$=\left(\cos^3\theta-3\cos\theta\sin^2\theta\right)+i\left(3\cos^2\theta\sin\theta-\sin^3\theta\right)$$
Equating imaginary parts with $\cos3\theta+i\sin3\theta$:
$$\sin3\theta=3\cos^2\theta\sin\theta-\sin^3\theta$$
Using $\cos^2\theta=1-\sin^2\theta$:
$$\sin3\theta=3(1-\sin^2\theta)\sin\theta-\sin^3\theta$$
$$=3\sin\theta-3\sin^3\theta-\sin^3\theta$$
$$=3\sin\theta-4\sin^3\theta$$
as required. $\blacksquare$
QUESTION 29
8 marks
Medium
Find all four fourth roots of $z=-16$. Give each root in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$, and hence in Cartesian form.
Show complete worked solution
First write $z=-16$ in polar form. Since $-16$ is a negative real number:
$$r=|-16|=16, \qquad \theta=\pi$$
so $z=16e^{i\pi}$.
The fourth roots have modulus $16^{1/4}=2$ and arguments
$$\theta_k=\dfrac{\pi+2k\pi}{4}, \quad k=0,1,2,3$$
$$k=0:\ \theta_0=\dfrac{\pi}{4} \qquad k=1:\ \theta_1=\dfrac{3\pi}{4}$$
$$k=2:\ \theta_2=\dfrac{5\pi}{4}\ \to\ \text{subtract } 2\pi:\ \theta_2=-\dfrac{3\pi}{4}$$
$$k=3:\ \theta_3=\dfrac{7\pi}{4}\ \to\ \text{subtract } 2\pi:\ \theta_3=-\dfrac{\pi}{4}$$
So the four roots in polar form are:
$$w_0=2e^{i\pi/4},\quad w_1=2e^{i3\pi/4},\quad w_2=2e^{-i3\pi/4},\quad w_3=2e^{-i\pi/4}$$
Converting to Cartesian form using $\cos\dfrac{\pi}{4}=\sin\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}$:
$$w_0=2\left(\dfrac{\sqrt2}{2}+i\dfrac{\sqrt2}{2}\right)=\sqrt2+\sqrt2\,i, \qquad w_1=-\sqrt2+\sqrt2\,i$$
$$w_2=-\sqrt2-\sqrt2\,i, \qquad w_3=\sqrt2-\sqrt2\,i$$
These four points, shown in the diagram, are equally spaced by $\dfrac{\pi}{2}$ around the circle $|z|=2$.
$$\boxed{w_0=\sqrt2+\sqrt2\,i,\ w_1=-\sqrt2+\sqrt2\,i,\ w_2=-\sqrt2-\sqrt2\,i,\ w_3=\sqrt2-\sqrt2\,i}$$
QUESTION 30
8 marks
Medium
Find all three cube roots of $z=-4\sqrt2+4\sqrt2\,i$. Give each root in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$.
Show complete worked solution
First write $z$ in polar form. Using $r=|z|$:
$$r=\sqrt{(-4\sqrt2)^2+(4\sqrt2)^2}=\sqrt{32+32}=\sqrt{64}=8$$
The point $(-4\sqrt2,4\sqrt2)$ lies in the second quadrant; the reference angle is
$$\arctan\left(\dfrac{4\sqrt2}{4\sqrt2}\right)=\arctan(1)=\dfrac{\pi}{4}$$
so
$$\theta=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$$
Thus $z=8e^{i3\pi/4}$.
The cube roots have modulus $8^{1/3}=2$ and arguments
$$\theta_k=\dfrac{3\pi/4+2k\pi}{3}, \quad k=0,1,2$$
$$k=0:\ \theta_0=\dfrac{\pi}{4}$$
$$k=1:\ \theta_1=\dfrac{3\pi/4+2\pi}{3}=\dfrac{11\pi/4}{3}=\dfrac{11\pi}{12}$$
$$k=2:\ \theta_2=\dfrac{3\pi/4+4\pi}{3}=\dfrac{19\pi/4}{3}=\dfrac{19\pi}{12}\ \to\ \text{subtract } 2\pi=\dfrac{24\pi}{12}:\ \theta_2=-\dfrac{5\pi}{12}$$
The three roots (shown equally spaced by $\dfrac{2\pi}{3}$ on the circle $|z|=2$ in the diagram) are:
$$\boxed{w_0=2e^{i\pi/4},\quad w_1=2e^{i11\pi/12},\quad w_2=2e^{-i5\pi/12}}$$
QUESTION 31
7 marks
Medium
Find the Cartesian equation of the locus of points $z$ satisfying $|z-2i|=|z-4|$, and describe it geometrically.
Show complete worked solution
The equation $|z-2i|=|z-4|$ states that the point representing $z$ is equidistant from the point representing $2i$ (that is, $(0,2)$) and the point representing $4$ (that is, $(4,0)$). This describes the perpendicular bisector of the line segment joining $(0,2)$ and $(4,0)$.
Let $z=x+iy$. Then:
$$|z-2i|=\sqrt{x^2+(y-2)^2}, \qquad |z-4|=\sqrt{(x-4)^2+y^2}$$
Setting these equal and squaring both sides:
$$x^2+(y-2)^2=(x-4)^2+y^2$$
$$x^2+y^2-4y+4=x^2-8x+16+y^2$$
Cancel $x^2$ and $y^2$ from both sides:
$$-4y+4=-8x+16$$
$$8x-4y=12$$
$$\boxed{y=2x-3 \text{ , the perpendicular bisector of the segment joining } 2i \text{ and } 4}$$
QUESTION 32
6 marks
Medium
(a) Sketch and describe the locus $|z+2|<4$ on an Argand diagram. [3]
(b) Determine, with justification, whether the point $z=1+3i$ lies inside this region. [3]
Show complete worked solution
(a) The expression $|z+2|=|z-(-2)|$ is the distance from $z$ to the point $(-2,0)$. The inequality $|z+2|<4$ describes all points whose distance from $(-2,0)$ is less than $4$, so the locus is the open disc (interior only, boundary excluded) of radius $4$ centred at $(-2,0)$, shown dashed in the diagram.
(b) Compute the distance from $z=1+3i$ to the centre $-2$:
$$|z-(-2)|=|1+3i+2|=|3+3i|=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt2$$
Since $3\sqrt2\approx4.24$, we have $3\sqrt2>4$, so $z=1+3i$ lies outside the disc.
$$\boxed{|z+2|<4 \text{ is the open disc of radius } 4 \text{ centred at } -2; \quad z=1+3i \text{ lies outside it, since } |3+3i|=3\sqrt2>4}$$
QUESTION 33
7 marks
Medium
The transformation $w=(1+i)z$ is applied to points on an Argand diagram.
(a) Express $1+i$ in the form $re^{i\theta}$, and hence describe the geometric effect of this transformation. [3]
(b) Find the image $w$ of the point $z=2+i$ under this transformation. [3]
Show complete worked solution
(a) Using $r=|1+i|$ and the quadrant:
$$r=\sqrt{1^2+1^2}=\sqrt2, \qquad \theta=\arctan\left(\dfrac{1}{1}\right)=\dfrac{\pi}{4}$$
so $1+i=\sqrt2\,e^{i\pi/4}$.
Multiplying $z$ by $\sqrt2\,e^{i\pi/4}$ multiplies its modulus by $\sqrt2$ and adds $\dfrac{\pi}{4}$ to its argument. Geometrically, the transformation $w=(1+i)z$ is therefore an enlargement (dilation) with scale factor $\sqrt2$ about the origin, combined with a rotation of $\dfrac{\pi}{4}$ ($45^{\circ}$) anticlockwise about the origin.
(b) Substituting $z=2+i$:
$$w=(1+i)(2+i)=2+i+2i+i^2=2+3i-1=1+3i$$
As a check: $|z|=\sqrt{4+1}=\sqrt5$ and $|w|=\sqrt{1+9}=\sqrt{10}=\sqrt2\cdot\sqrt5$, confirming the modulus scaled by $\sqrt2$, as shown in the diagram.
$$\boxed{1+i=\sqrt2\,e^{i\pi/4} \text{ (rotation } \tfrac{\pi}{4}\text{, enlargement } \sqrt2\text{)}, \qquad w=1+3i}$$
QUESTION 34
7 marks
Medium
Use the modulus-argument method to find both square roots of $z=8i$, and verify your answers by squaring.
Show complete worked solution
First write $z=8i$ in polar form. Since $8i$ lies on the positive imaginary axis:
$$r=|8i|=8, \qquad \theta=\dfrac{\pi}{2}$$
so $z=8e^{i\pi/2}$.
The square roots have modulus $8^{1/2}=2\sqrt2$ and arguments
$$\phi_k=\dfrac{\pi/2+2k\pi}{2}, \quad k=0,1$$
$$k=0:\ \phi_0=\dfrac{\pi}{4} \qquad\qquad k=1:\ \phi_1=\dfrac{\pi}{4}+\pi=\dfrac{5\pi}{4}\ \left(\text{equivalently } -\dfrac{3\pi}{4}\right)$$
Converting to Cartesian form: $\cos\dfrac{\pi}{4}=\sin\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}$, so
$$w_0=2\sqrt2\left(\dfrac{\sqrt2}{2}+i\dfrac{\sqrt2}{2}\right)=2+2i$$
and $w_1=-w_0$ (since the two square roots are diametrically opposite, differing in argument by $\pi$):
$$w_1=-2-2i$$
Verification by squaring:
$$w_0^2=(2+2i)^2=4+8i+4i^2=4+8i-4=8i \ \checkmark$$
$$w_1^2=(-2-2i)^2=(-1)^2(2+2i)^2=8i \ \checkmark$$
$$\boxed{w_0=2+2i, \qquad w_1=-2-2i}$$
QUESTION 35
6 marks
Medium
Using Euler's formula, show that $\cos^2\theta=\dfrac{1+\cos2\theta}{2}$.
Show complete worked solution
By Euler's formula, $\cos\theta=\dfrac{e^{i\theta}+e^{-i\theta}}{2}$ (obtained by adding $e^{i\theta}=\cos\theta+i\sin\theta$ and $e^{-i\theta}=\cos\theta-i\sin\theta$).
Squaring both sides:
$$\cos^2\theta=\left(\dfrac{e^{i\theta}+e^{-i\theta}}{2}\right)^2=\dfrac{e^{2i\theta}+2e^{i\theta}e^{-i\theta}+e^{-2i\theta}}{4}$$
Since $e^{i\theta}e^{-i\theta}=e^0=1$:
$$\cos^2\theta=\dfrac{e^{2i\theta}+2+e^{-2i\theta}}{4}=\dfrac{\left(e^{i2\theta}+e^{-i2\theta}\right)+2}{4}$$
Using Euler's formula again, this time with $2\theta$ in place of $\theta$: $e^{i2\theta}+e^{-i2\theta}=2\cos2\theta$, so:
$$\cos^2\theta=\dfrac{2\cos2\theta+2}{4}=\dfrac{\cos2\theta+1}{2}=\dfrac{1+\cos2\theta}{2}$$
as required. $\blacksquare$
QUESTION 36
7 marks
Medium
The point $A$ represents $0$ and the point $B$ represents $4$ on an Argand diagram. The point $C$ represents $4e^{i\pi/3}$.
(a) Explain why multiplying $4$ by $e^{i\pi/3}$ corresponds to a rotation of $B$ about $A$ through $\dfrac{\pi}{3}$. [2]
(b) Show, using the moduli $|AB|$, $|AC|$ and $|BC|$, that triangle $ABC$ is equilateral. [5]
Show complete worked solution
(a) Multiplying a complex number by $e^{i\pi/3}$ leaves the modulus unchanged (since $|e^{i\pi/3}|=1$) and adds $\dfrac{\pi}{3}$ to the argument. Since $A$ is the origin, multiplying $B=4$ by $e^{i\pi/3}$ therefore rotates the point $B$ about $A$ through angle $\dfrac{\pi}{3}$ anticlockwise, producing $C$.
(b) The side length $|AB|$ is:
$$|AB|=|4-0|=4$$
Since $C=4e^{i\pi/3}$ is obtained from $B$ by a rotation about $A$ (which preserves distance from $A$):
$$|AC|=|4e^{i\pi/3}-0|=4|e^{i\pi/3}|=4(1)=4$$
For $|BC|$, first find $C$ in Cartesian form: $\cos\dfrac{\pi}{3}=\dfrac12$, $\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}$, so
$$C=4\left(\dfrac12+i\dfrac{\sqrt3}{2}\right)=2+2\sqrt3\,i$$
Then:
$$|BC|=|C-B|=|(2+2\sqrt3\,i)-4|=|-2+2\sqrt3\,i|=\sqrt{(-2)^2+(2\sqrt3)^2}=\sqrt{4+12}=\sqrt{16}=4$$
Since $|AB|=|AC|=|BC|=4$ (shown in the diagram), triangle $ABC$ is equilateral. $\blacksquare$
$$\boxed{|AB|=|AC|=|BC|=4, \text{ so } ABC \text{ is equilateral}}$$
QUESTION 37
7 marks
Medium
Find all six sixth roots of unity ($z^6=1$), giving each in Cartesian form, and show that their sum is zero.
Show complete worked solution
Write $1=1e^{i0}$. The sixth roots have modulus $1^{1/6}=1$ and arguments
$$\theta_k=\dfrac{2k\pi}{6}=\dfrac{k\pi}{3}, \quad k=0,1,2,3,4,5$$
These give arguments $0,\ \dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \pi,\ \dfrac{4\pi}{3}\left(=-\dfrac{2\pi}{3}\right),\ \dfrac{5\pi}{3}\left(=-\dfrac{\pi}{3}\right)$.
Converting each to Cartesian form (using $\cos\dfrac{\pi}{3}=\dfrac12$, $\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}$):
$$w_0=1, \quad w_1=\dfrac12+\dfrac{\sqrt3}{2}i, \quad w_2=-\dfrac12+\dfrac{\sqrt3}{2}i$$
$$w_3=-1, \quad w_4=-\dfrac12-\dfrac{\sqrt3}{2}i, \quad w_5=\dfrac12-\dfrac{\sqrt3}{2}i$$
These six points, shown in the diagram, are equally spaced around the unit circle (a regular hexagon).
To find the sum, add real and imaginary parts separately. The real parts are $1,\ \tfrac12,\ -\tfrac12,\ -1,\ -\tfrac12,\ \tfrac12$, which sum to $0$. The imaginary parts are $0,\ \tfrac{\sqrt3}{2},\ \tfrac{\sqrt3}{2},\ 0,\ -\tfrac{\sqrt3}{2},\ -\tfrac{\sqrt3}{2}$, which also sum to $0$.
(This is expected: the roots satisfy $z^6-1=0$, a polynomial in which the coefficient of $z^5$ is $0$, so by Vieta's formulas the sum of all six roots is $0$.)
$$\boxed{w_0+w_1+w_2+w_3+w_4+w_5=0}$$
QUESTION 38
6 marks
Medium
Let $z_1=4e^{i\pi/5}$ and $z_2=2e^{i3\pi/10}$. Find the modulus and argument of $z_1^2z_2$, giving the argument exactly in the range $-\pi<\theta\le\pi$.
Show complete worked solution
By De Moivre's theorem, $z_1^2$ has modulus $4^2=16$ and argument $2\cdot\dfrac{\pi}{5}=\dfrac{2\pi}{5}$, so $z_1^2=16e^{i2\pi/5}$.
Multiplying by $z_2$: multiply moduli and add arguments:
$$|z_1^2z_2|=16\times2=32$$
$$\arg(z_1^2z_2)=\dfrac{2\pi}{5}+\dfrac{3\pi}{10}$$
Using a common denominator of $10$: $\dfrac{2\pi}{5}=\dfrac{4\pi}{10}$, so
$$\arg(z_1^2z_2)=\dfrac{4\pi}{10}+\dfrac{3\pi}{10}=\dfrac{7\pi}{10}$$
Since $\dfrac{7\pi}{10}$ already lies in $(-\pi,\pi]$, no adjustment is needed.
$$\boxed{|z_1^2z_2|=32, \qquad \arg(z_1^2z_2)=\dfrac{7\pi}{10}}$$
QUESTION 39
6 marks
Medium
The point $z=3+2i$ is reflected in the line $y=x$ on an Argand diagram.
(a) Show that this reflection is given by $w=i\bar z$. [3]
(b) Hence find the Cartesian form of the image of $z=3+2i$. [3]
Show complete worked solution
(a) Reflecting a point $(a,b)$ in the line $y=x$ produces the point $(b,a)$ (the coordinates swap). In complex form, this sends $z=a+bi$ to $w=b+ai$.
Starting from $\bar z=a-bi$ and multiplying by $i$:
$$i\bar z=i(a-bi)=ai-bi^2=b+ai$$
This is exactly $w=b+ai$, the reflected point. So the reflection in $y=x$ is given by $w=i\bar z$, as required.
(b) For $z=3+2i$: $\bar z=3-2i$, so
$$w=i(3-2i)=3i-2i^2=2+3i$$
This matches the coordinate-swap rule directly: $(3,2)\to(2,3)$, shown by the reflection across the dashed line $y=x$ in the diagram.
$$\boxed{w=2+3i}$$
QUESTION 40
6 marks
Medium
Sketch the locus of points $z$ (with $z\ne3$) satisfying $\arg(z-3)=\dfrac{\pi}{3}$ on an Argand diagram, and describe it geometrically.
Show complete worked solution
The expression $z-3$ represents the vector from the point $3$ (that is, $(3,0)$) to the point $z$. The condition $\arg(z-3)=\dfrac{\pi}{3}$ fixes the direction of this vector to be $\dfrac{\pi}{3}$ ($60^{\circ}$) from the positive real direction, with no restriction on $|z-3|$ other than it being positive (since $z\ne3$).
The locus is therefore a half-line (ray) starting from, but not including, the point $3$ on the real axis, making an angle of $\dfrac{\pi}{3}$ with the positive real direction, as shown by the dashed ray in the diagram (the open circle marks the excluded starting point).
Writing $z=3+t\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)=3+t\left(\dfrac12+i\dfrac{\sqrt3}{2}\right)$ for $t>0$ gives every point on the ray.
$$\boxed{\text{a ray from the point } 3 \text{ (excluded), at angle } \dfrac{\pi}{3} \text{ to the positive real direction}}$$
QUESTION 41
7 marks
Medium
A square on an Argand diagram has vertices $0$, $2$, $2+2i$ and $2i$. The transformation $w=iz$ is applied to each vertex.
(a) Find the image of each vertex under this transformation. [3]
(b) Describe the geometric effect of the transformation $w=iz$ on the square. [3]
Show complete worked solution
(a) Multiplying each vertex by $i$:
$$0 \to i(0)=0$$
$$2 \to i(2)=2i$$
$$2+2i \to i(2+2i)=2i+2i^2=-2+2i$$
$$2i \to i(2i)=2i^2=-2$$
So the image vertices are $0,\ 2i,\ -2+2i,\ -2$, shown in amber in the diagram alongside the original square in green.
(b) Since $i=e^{i\pi/2}$ has modulus $1$, multiplying by $i$ leaves all distances from the origin unchanged and adds $\dfrac{\pi}{2}$ to every argument. The transformation $w=iz$ is therefore a rotation of $90^{\circ}$ ($\dfrac{\pi}{2}$ radians) anticlockwise about the origin - it maps the original square onto a congruent square, rotated a quarter-turn about $0$ (one of its own vertices, which is why that vertex maps to itself).
$$\boxed{\text{images: } 0,\ 2i,\ -2+2i,\ -2 \text{; the map } w=iz \text{ is a } 90^{\circ} \text{ rotation about the origin}}$$
QUESTION 42
8 marks
Medium
One cube root of a complex number $z$ is known to be $w_0=2e^{i\pi/9}$.
(a) Find $z$ in the form $re^{i\theta}$. [3]
(b) Find the other two cube roots of $z$, giving each in the form $re^{i\theta}$ with $-\pi<\theta\le\pi$. [5]
Show complete worked solution
(a) Since $w_0$ is a cube root of $z$, we have $z=w_0^3$. By De Moivre's theorem:
$$z=w_0^3=2^3e^{i3(\pi/9)}=8e^{i\pi/3}$$
$$\boxed{z=8e^{i\pi/3}}$$
(b) The three cube roots of $z$ all have the same modulus, $8^{1/3}=2$, and their arguments are equally spaced by $\dfrac{2\pi}{3}$ around the circle $|w|=2$. Since $w_0=2e^{i\pi/9}$ is one of them, the other two are found by adding $\dfrac{2\pi}{3}$ and $\dfrac{4\pi}{3}$ to its argument:
$$w_1: \ \phi_1=\dfrac{\pi}{9}+\dfrac{2\pi}{3}=\dfrac{\pi}{9}+\dfrac{6\pi}{9}=\dfrac{7\pi}{9}$$
$$w_2: \ \phi_2=\dfrac{\pi}{9}+\dfrac{4\pi}{3}=\dfrac{\pi}{9}+\dfrac{12\pi}{9}=\dfrac{13\pi}{9}\ \to\ \text{subtract } 2\pi=\dfrac{18\pi}{9}:\ \phi_2=-\dfrac{5\pi}{9}$$
Both $\dfrac{7\pi}{9}$ and $-\dfrac{5\pi}{9}$ lie in $(-\pi,\pi]$. The three roots are shown, equally spaced, on the circle $|w|=2$ in the diagram.
$$\boxed{w_1=2e^{i7\pi/9}, \qquad w_2=2e^{-i5\pi/9}}$$
QUESTION 43
13 marks
Hard
(a) Use De Moivre's theorem and the binomial theorem to show that $\cos4\theta=8\cos^4\theta-8\cos^2\theta+1$. [7]
(b) Hence, by writing $x=\cos^2\theta$, show that $\cos^2\dfrac{3\pi}{8}=\dfrac{2-\sqrt2}{4}$. [4]
(c) Hence find the exact value of $\cos\dfrac{3\pi}{8}$, given that $\cos\dfrac{3\pi}{8}>0$. [2]
Show complete worked solution
(a) By De Moivre's theorem:
$$(\cos\theta+i\sin\theta)^4=\cos4\theta+i\sin4\theta$$
Expand the left-hand side using the binomial theorem, writing $c=\cos\theta$ and $s=\sin\theta$ for brevity:
$$(c+is)^4=c^4+4c^3(is)+6c^2(is)^2+4c(is)^3+(is)^4$$
$$=c^4+4ic^3s-6c^2s^2-4ics^3+s^4$$
Group real and imaginary parts:
$$=\left(c^4-6c^2s^2+s^4\right)+i\left(4c^3s-4cs^3\right)$$
Equating real parts with $\cos4\theta+i\sin4\theta$:
$$\cos4\theta=c^4-6c^2s^2+s^4=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta$$
Using $\sin^2\theta=1-\cos^2\theta$, replace every $s^2$ with $1-c^2$:
$$\cos4\theta=c^4-6c^2(1-c^2)+(1-c^2)^2$$
$$=c^4-6c^2+6c^4+\left(1-2c^2+c^4\right)$$
$$=c^4-6c^2+6c^4+1-2c^2+c^4$$
Collecting like powers of $c$:
$$=(c^4+6c^4+c^4)+(-6c^2-2c^2)+1=8c^4-8c^2+1$$
$$\cos4\theta=8\cos^4\theta-8\cos^2\theta+1$$
as required. $\blacksquare$
(b) Let $\theta=\dfrac{3\pi}{8}$, so $4\theta=\dfrac{3\pi}{2}$, and $\cos\dfrac{3\pi}{2}=0$. Substituting into the identity from part (a):
$$0=8\cos^4\theta-8\cos^2\theta+1$$
Let $x=\cos^2\theta$. This becomes a quadratic in $x$:
$$8x^2-8x+1=0$$
By the quadratic formula:
$$x=\dfrac{8\pm\sqrt{64-32}}{16}=\dfrac{8\pm\sqrt{32}}{16}=\dfrac{8\pm4\sqrt2}{16}=\dfrac{2\pm\sqrt2}{4}$$
Since $\theta=\dfrac{3\pi}{8}=67.5^{\circ}$ is close to $90^{\circ}$, $\cos\theta$ is small and positive, so $x=\cos^2\theta$ must be the smaller root:
$$x=\dfrac{2-\sqrt2}{4}$$
$$\boxed{\cos^2\dfrac{3\pi}{8}=\dfrac{2-\sqrt2}{4}}$$
(c) Taking the positive square root (since $\cos\dfrac{3\pi}{8}>0$, given):
$$\cos\dfrac{3\pi}{8}=\sqrt{\dfrac{2-\sqrt2}{4}}=\dfrac{\sqrt{2-\sqrt2}}{2}$$
$$\boxed{\cos\dfrac{3\pi}{8}=\dfrac{\sqrt{2-\sqrt2}}{2}}$$
QUESTION 44
14 marks
Hard
(a) Find all five solutions of $z^5=1$, giving each in the form $e^{i\theta}$ with $-\pi<\theta\le\pi$. [4]
(b) By using the formula for the sum of a finite geometric series, show that $1+\omega+\omega^2+\omega^3+\omega^4=0$, where $\omega=e^{i2\pi/5}$. [4]
(c) Hence show that $\cos\dfrac{2\pi}{5}+\cos\dfrac{4\pi}{5}=-\dfrac12$. [6]
Show complete worked solution
(a) Write $1=e^{i0}$. The five fifth roots of unity have modulus $1^{1/5}=1$ and arguments
$$\theta_k=\dfrac{2k\pi}{5}, \quad k=0,1,2,3,4$$
This gives $\theta_0=0$, $\theta_1=\dfrac{2\pi}{5}$, $\theta_2=\dfrac{4\pi}{5}$, $\theta_3=\dfrac{6\pi}{5}$, $\theta_4=\dfrac{8\pi}{5}$. The last two exceed $\pi$, so subtract $2\pi=\dfrac{10\pi}{5}$:
$$\theta_3=\dfrac{6\pi}{5}-\dfrac{10\pi}{5}=-\dfrac{4\pi}{5}, \qquad \theta_4=\dfrac{8\pi}{5}-\dfrac{10\pi}{5}=-\dfrac{2\pi}{5}$$
$$\boxed{z=1,\ e^{i2\pi/5},\ e^{i4\pi/5},\ e^{-i4\pi/5},\ e^{-i2\pi/5}}$$
(b) Let $\omega=e^{i2\pi/5}$. Since $\omega\ne1$, the sum $1+\omega+\omega^2+\omega^3+\omega^4$ is a geometric series with first term $1$, common ratio $\omega$, and $5$ terms, so by the geometric series formula:
$$1+\omega+\omega^2+\omega^3+\omega^4=\dfrac{\omega^5-1}{\omega-1}$$
Since $\omega=e^{i2\pi/5}$ is a fifth root of unity, $\omega^5=\left(e^{i2\pi/5}\right)^5=e^{i2\pi}=1$. So the numerator is $\omega^5-1=1-1=0$, and since $\omega\ne1$ the denominator $\omega-1\ne0$:
$$1+\omega+\omega^2+\omega^3+\omega^4=\dfrac{0}{\omega-1}=0$$
as required. $\blacksquare$
(c) By part (a), the powers $\omega,\omega^2,\omega^3,\omega^4$ are $e^{i2\pi/5},\ e^{i4\pi/5},\ e^{i6\pi/5}\left(=e^{-i4\pi/5}\right),\ e^{i8\pi/5}\left(=e^{-i2\pi/5}\right)$ - the four nontrivial fifth roots of unity found in part (a). Taking the real part of the identity in part (b):
$$\operatorname{Re}(1)+\operatorname{Re}(\omega)+\operatorname{Re}(\omega^2)+\operatorname{Re}(\omega^3)+\operatorname{Re}(\omega^4)=0$$
$$1+\cos\dfrac{2\pi}{5}+\cos\dfrac{4\pi}{5}+\cos\left(-\dfrac{4\pi}{5}\right)+\cos\left(-\dfrac{2\pi}{5}\right)=0$$
Since cosine is an even function, $\cos\left(-\dfrac{4\pi}{5}\right)=\cos\dfrac{4\pi}{5}$ and $\cos\left(-\dfrac{2\pi}{5}\right)=\cos\dfrac{2\pi}{5}$:
$$1+\cos\dfrac{2\pi}{5}+\cos\dfrac{4\pi}{5}+\cos\dfrac{4\pi}{5}+\cos\dfrac{2\pi}{5}=0$$
$$1+2\cos\dfrac{2\pi}{5}+2\cos\dfrac{4\pi}{5}=0$$
$$2\left(\cos\dfrac{2\pi}{5}+\cos\dfrac{4\pi}{5}\right)=-1$$
$$\cos\dfrac{2\pi}{5}+\cos\dfrac{4\pi}{5}=-\dfrac12$$
as required. $\blacksquare$
$$\boxed{\cos\dfrac{2\pi}{5}+\cos\dfrac{4\pi}{5}=-\dfrac12}$$
QUESTION 45
12 marks
Hard
Consider the locus of points $z$ satisfying $|z-8|=2|z-2|$.
(a) By writing $z=x+iy$, show that this locus is the circle $x^2+y^2=16$. [6]
(b) State the centre and radius of this circle, and determine, showing your reasoning, whether each of the points $z=8$ and $z=-4$ lies on this locus. [3]
(c) Find, in Cartesian form, the point $P$ on this locus for which $\arg(z)=\dfrac{\pi}{3}$. [4]
Show complete worked solution
(a) Let $z=x+iy$. Then $z-8=(x-8)+iy$ and $z-2=(x-2)+iy$, so:
$$|z-8|=\sqrt{(x-8)^2+y^2}, \qquad |z-2|=\sqrt{(x-2)^2+y^2}$$
The equation $|z-8|=2|z-2|$ becomes:
$$\sqrt{(x-8)^2+y^2}=2\sqrt{(x-2)^2+y^2}$$
Squaring both sides:
$$(x-8)^2+y^2=4\left[(x-2)^2+y^2\right]$$
Expand both sides:
$$x^2-16x+64+y^2=4\left(x^2-4x+4+y^2\right)=4x^2-16x+16+4y^2$$
Bring all terms to one side:
$$x^2-16x+64+y^2-4x^2+16x-16-4y^2=0$$
The $-16x$ and $+16x$ terms cancel:
$$(x^2-4x^2)+(y^2-4y^2)+(64-16)=0$$
$$-3x^2-3y^2+48=0$$
Divide by $-3$:
$$x^2+y^2=16$$
as required. $\blacksquare$
(b) Since $x^2+y^2=16$ is of the form $x^2+y^2=r^2$, the locus is a circle with centre $(0,0)$ (the origin) and radius $4$.
Checking $z=8$: this corresponds to $(8,0)$, and $8^2+0^2=64\ne16$, so $z=8$ does **not** lie on the circle (indeed, substituting into the original equation gives $|8-8|=0$ on the left but $2|8-2|=12$ on the right, confirming it is not a solution).
Checking $z=-4$: this corresponds to $(-4,0)$, and $(-4)^2+0^2=16$, so $z=-4$ **does** lie on the circle. Substituting into the original equation: $|-4-8|=|-12|=12$ and $2|-4-2|=2|-6|=12$, which agree. $\checkmark$
$$\boxed{\text{centre } (0,0), \text{ radius } 4; \ z=8 \text{ is not on the locus, } z=-4 \text{ is}}$$
(c) A point on the circle $|z|=4$ with $\arg(z)=\dfrac{\pi}{3}$ can be written as $P=4e^{i\pi/3}$. Converting to Cartesian form using $\cos\dfrac{\pi}{3}=\dfrac12$ and $\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}$:
$$P=4\left(\dfrac12+i\dfrac{\sqrt3}{2}\right)=2+2\sqrt3\,i$$
Check: $|P-8|=|2+2\sqrt3i-8|=|-6+2\sqrt3i|=\sqrt{36+12}=\sqrt{48}=4\sqrt3$, and $2|P-2|=2|2\sqrt3i|=2(2\sqrt3)=4\sqrt3$. These agree, confirming $P$ lies on the locus, as shown in the diagram. $\checkmark$
$$\boxed{P=2+2\sqrt3\,i}$$
QUESTION 46
13 marks
Hard
(a) Find all five fifth roots of $z=32$, giving each in the form $re^{i\theta}$ with $-\pi<\theta\le\pi$, and describe their geometric arrangement. [6]
(b) Without evaluating any trigonometric function, state the sum of the five roots found in part (a), justifying your answer using the coefficients of the equation $w^5-32=0$. [3]
(c) Show that the product of the five roots equals $32$. [4]
Show complete worked solution
(a) Write $z=32=32e^{i0}$. The fifth roots have modulus $32^{1/5}=2$ (since $2^5=32$) and arguments
$$\theta_k=\dfrac{2k\pi}{5}, \quad k=0,1,2,3,4$$
This gives $\theta_0=0$, $\theta_1=\dfrac{2\pi}{5}$, $\theta_2=\dfrac{4\pi}{5}$, $\theta_3=\dfrac{6\pi}{5}$, $\theta_4=\dfrac{8\pi}{5}$. The last two lie outside $(-\pi,\pi]$, so subtract $2\pi=\dfrac{10\pi}{5}$:
$$\theta_3=\dfrac{6\pi}{5}-\dfrac{10\pi}{5}=-\dfrac{4\pi}{5}, \qquad \theta_4=\dfrac{8\pi}{5}-\dfrac{10\pi}{5}=-\dfrac{2\pi}{5}$$
$$\boxed{w_0=2e^{i0},\ w_1=2e^{i2\pi/5},\ w_2=2e^{i4\pi/5},\ w_3=2e^{-i4\pi/5},\ w_4=2e^{-i2\pi/5}}$$
All five roots have the same modulus $2$, so they lie on the circle $|w|=2$, and their arguments are equally spaced by $\dfrac{2\pi}{5}$. They are therefore the vertices of a regular pentagon inscribed in this circle, with one vertex on the positive real axis, as shown in the diagram.
(b) The five roots are precisely the roots of $w^5-32=0$, a degree-$5$ polynomial equation $w^5+0w^4+0w^3+0w^2+0w-32=0$. By Vieta's formulas, the sum of the roots of $w^5+c_4w^4+\dots+c_0=0$ equals $-c_4$. Here the coefficient of $w^4$ is $0$, so:
$$\boxed{w_0+w_1+w_2+w_3+w_4=0}$$
(No trigonometric evaluation is needed - this follows purely from the missing $w^4$ term.)
(c) By Vieta's formulas, the product of the roots of $w^5+c_4w^4+c_3w^3+c_2w^2+c_1w+c_0=0$ equals $(-1)^5c_0$. Here $c_0=-32$, so:
$$w_0w_1w_2w_3w_4=(-1)^5(-32)=32$$
This can also be verified directly using the moduli and arguments: the product of the five roots has modulus $2\times2\times2\times2\times2=2^5=32$, and argument
$$0+\dfrac{2\pi}{5}+\dfrac{4\pi}{5}-\dfrac{4\pi}{5}-\dfrac{2\pi}{5}=0$$
so the product is $32e^{i0}=32$, confirming the result. $\blacksquare$
$$\boxed{w_0w_1w_2w_3w_4=32}$$
QUESTION 47
14 marks
Hard
(a) By writing $2i\sin\theta=e^{i\theta}-e^{-i\theta}$ and expanding $(2i\sin\theta)^5$ using the binomial theorem, show that
$$\sin^5\theta=\dfrac{1}{16}\left(\sin5\theta-5\sin3\theta+10\sin\theta\right)$$ [9]
(b) Hence find the exact value of $\sin^5\dfrac{\pi}{6}$ using the identity from part (a), and verify your answer by direct computation of $\left(\sin\dfrac{\pi}{6}\right)^5$. [5]
Show complete worked solution
(a) From Euler's formula, $e^{i\theta}=\cos\theta+i\sin\theta$ and $e^{-i\theta}=\cos\theta-i\sin\theta$, so subtracting gives $e^{i\theta}-e^{-i\theta}=2i\sin\theta$.
Raise both sides to the fifth power:
$$(2i\sin\theta)^5=\left(e^{i\theta}-e^{-i\theta}\right)^5$$
Expand the right-hand side using the binomial theorem, with $a=e^{i\theta}$, $b=e^{-i\theta}$:
$$\left(e^{i\theta}-e^{-i\theta}\right)^5=e^{i5\theta}-5e^{i3\theta}+10e^{i\theta}-10e^{-i\theta}+5e^{-i3\theta}-e^{-i5\theta}$$
(using binomial coefficients $\binom{5}{0},\binom{5}{1},\dots,\binom{5}{5}=1,5,10,10,5,1$ with alternating signs, and simplifying exponents, e.g. $e^{i\theta}\cdot e^{-4i\theta}=e^{-3i\theta}$ etc.)
Group into conjugate pairs:
$$=\left(e^{i5\theta}-e^{-i5\theta}\right)-5\left(e^{i3\theta}-e^{-i3\theta}\right)+10\left(e^{i\theta}-e^{-i\theta}\right)$$
Using $e^{i n\theta}-e^{-in\theta}=2i\sin(n\theta)$ for each bracket:
$$=2i\sin5\theta-5(2i\sin3\theta)+10(2i\sin\theta)$$
$$=2i\left(\sin5\theta-5\sin3\theta+10\sin\theta\right)$$
So:
$$(2i\sin\theta)^5=2i\left(\sin5\theta-5\sin3\theta+10\sin\theta\right)$$
The left-hand side is $(2i)^5\sin^5\theta=32i^5\sin^5\theta$. Since $i^5=i^4\cdot i=1\cdot i=i$, this is $32i\sin^5\theta$:
$$32i\sin^5\theta=2i\left(\sin5\theta-5\sin3\theta+10\sin\theta\right)$$
Divide both sides by $32i$:
$$\sin^5\theta=\dfrac{2}{32}\left(\sin5\theta-5\sin3\theta+10\sin\theta\right)=\dfrac{1}{16}\left(\sin5\theta-5\sin3\theta+10\sin\theta\right)$$
as required. $\blacksquare$
(b) Let $\theta=\dfrac{\pi}{6}$. Then $3\theta=\dfrac{\pi}{2}$ and $5\theta=\dfrac{5\pi}{6}$. Using the identity from part (a):
$$\sin^5\dfrac{\pi}{6}=\dfrac{1}{16}\left(\sin\dfrac{5\pi}{6}-5\sin\dfrac{\pi}{2}+10\sin\dfrac{\pi}{6}\right)$$
Now $\sin\dfrac{5\pi}{6}=\dfrac12$ (reference angle $\dfrac{\pi}{6}$, second quadrant), $\sin\dfrac{\pi}{2}=1$, and $\sin\dfrac{\pi}{6}=\dfrac12$:
$$\sin^5\dfrac{\pi}{6}=\dfrac{1}{16}\left(\dfrac12-5(1)+10\left(\dfrac12\right)\right)=\dfrac{1}{16}\left(\dfrac12-5+5\right)=\dfrac{1}{16}\cdot\dfrac12=\dfrac{1}{32}$$
Direct verification: $\sin\dfrac{\pi}{6}=\dfrac12$, so
$$\left(\sin\dfrac{\pi}{6}\right)^5=\left(\dfrac12\right)^5=\dfrac{1}{32}$$
This matches the value found using the identity. $\checkmark$
$$\boxed{\sin^5\dfrac{\pi}{6}=\dfrac{1}{32}}$$
QUESTION 48
13 marks
Hard
Triangle $ABC$ on an Argand diagram has vertices $A=0$, $B=2$ and $C=1+\sqrt3\,i$.
(a) Show that triangle $ABC$ is equilateral. [4]
(b) The transformation $w=e^{i\pi/3}z+2$ is applied to each vertex. Find the images $A'$, $B'$, $C'$ of $A$, $B$, $C$ under this transformation, giving each in Cartesian form. [6]
(c) Show that triangle $A'B'C'$ is also equilateral with the same side length as triangle $ABC$, and explain why this must be true without further calculation, based on the nature of the transformation. [3]
Show complete worked solution
(a) Compute each side length:
$$|AB|=|2-0|=2$$
$$|AC|=|1+\sqrt3i-0|=\sqrt{1^2+(\sqrt3)^2}=\sqrt{1+3}=2$$
$$|BC|=|(1+\sqrt3i)-2|=|-1+\sqrt3i|=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=2$$
Since $|AB|=|AC|=|BC|=2$, triangle $ABC$ is equilateral with side length $2$. $\blacksquare$
(b) Note that $e^{i\pi/3}=\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}=\dfrac12+\dfrac{\sqrt3}{2}i$.
For $A=0$:
$$A'=e^{i\pi/3}(0)+2=2$$
For $B=2$:
$$B'=e^{i\pi/3}(2)+2=2\left(\dfrac12+\dfrac{\sqrt3}{2}i\right)+2=(1+\sqrt3i)+2=3+\sqrt3i$$
For $C=1+\sqrt3i$:
$$e^{i\pi/3}(1+\sqrt3i)=\left(\dfrac12+\dfrac{\sqrt3}{2}i\right)(1+\sqrt3i)=\dfrac12+\dfrac{\sqrt3}{2}i+\dfrac{\sqrt3}{2}i+\dfrac{3}{2}i^2$$
$$=\dfrac12+\sqrt3i-\dfrac32=-1+\sqrt3i$$
so
$$C'=(-1+\sqrt3i)+2=1+\sqrt3i$$
$$\boxed{A'=2, \qquad B'=3+\sqrt3\,i, \qquad C'=1+\sqrt3\,i}$$
(c) Computing the side lengths of the image triangle:
$$|A'B'|=|(3+\sqrt3i)-2|=|1+\sqrt3i|=\sqrt{1+3}=2$$
$$|A'C'|=|(1+\sqrt3i)-2|=|-1+\sqrt3i|=\sqrt{1+3}=2$$
$$|B'C'|=|(1+\sqrt3i)-(3+\sqrt3i)|=|-2|=2$$
Since $|A'B'|=|A'C'|=|B'C'|=2$, triangle $A'B'C'$ is equilateral with the same side length $2$ as triangle $ABC$, as shown in the diagram (original triangle in green, image in amber). $\blacksquare$
This must be true without further calculation because the transformation $w=e^{i\pi/3}z+2$ consists of a rotation about the origin through $\dfrac{\pi}{3}$ (multiplication by $e^{i\pi/3}$, which has modulus $1$ and so preserves all distances) followed by a translation by $2$ (which also preserves all distances between points, merely repositioning the whole figure). Since both steps are isometries (distance-preserving maps), every distance between vertices - and hence the shape and size of the triangle - is preserved by their composition.
$$\boxed{\text{triangle } A'B'C' \text{ is equilateral with side length } 2, \text{ since } w=e^{i\pi/3}z+2 \text{ is a rotation followed by a translation, both isometries}}$$
QUESTION 49
15 marks
Hard
Let $z_0$ be any fixed complex number (a "centre"), and let $z$ be any complex number with $z\ne z_0$. Define
$$P_1=z_0+(z-z_0)e^{i2\pi/3}, \qquad P_2=z_0+(z-z_0)e^{i4\pi/3}$$
(a) Show that $z$, $P_1$ and $P_2$ are all the same distance from $z_0$. [3]
(b) Show that $|P_1-P_2|=\sqrt3\,|z-z_0|$, and hence show that $z$, $P_1$ and $P_2$ are the vertices of an equilateral triangle for any choice of $z\ne z_0$. [7]
(c) Apply this result with $z_0=1+i$ and $z=4+i$ to find $P_1$ and $P_2$ in Cartesian form. [5]
Show complete worked solution
(a) The distance from $z_0$ to $z$ is $|z-z_0|$. The distance from $z_0$ to $P_1$ is:
$$|P_1-z_0|=\left|(z-z_0)e^{i2\pi/3}\right|=|z-z_0|\left|e^{i2\pi/3}\right|=|z-z_0|(1)=|z-z_0|$$
since $\left|e^{i2\pi/3}\right|=1$ for any real angle. Similarly:
$$|P_2-z_0|=\left|(z-z_0)e^{i4\pi/3}\right|=|z-z_0|$$
So all three points $z$, $P_1$, $P_2$ are the same distance, $|z-z_0|$, from $z_0$. $\blacksquare$
(b) Compute $P_1-P_2$:
$$P_1-P_2=(z-z_0)e^{i2\pi/3}-(z-z_0)e^{i4\pi/3}=(z-z_0)\left(e^{i2\pi/3}-e^{i4\pi/3}\right)$$
So:
$$|P_1-P_2|=|z-z_0|\left|e^{i2\pi/3}-e^{i4\pi/3}\right|$$
Now evaluate $e^{i2\pi/3}-e^{i4\pi/3}$ using $\cos\dfrac{2\pi}{3}=-\dfrac12,\ \sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}$ and $\cos\dfrac{4\pi}{3}=-\dfrac12,\ \sin\dfrac{4\pi}{3}=-\dfrac{\sqrt3}{2}$:
$$e^{i2\pi/3}-e^{i4\pi/3}=\left(-\dfrac12+\dfrac{\sqrt3}{2}i\right)-\left(-\dfrac12-\dfrac{\sqrt3}{2}i\right)=\sqrt3\,i$$
So $\left|e^{i2\pi/3}-e^{i4\pi/3}\right|=|\sqrt3i|=\sqrt3$, and therefore:
$$|P_1-P_2|=\sqrt3\,|z-z_0|$$
as required. Now, using part (a), $|z-P_1|$: since $P_1$ is obtained by rotating $z$ about $z_0$ through $\dfrac{2\pi}{3}$, and rotation is an isometry about $z_0$, the chord length between two points on a circle of radius $\rho=|z-z_0|$ separated by angle $\dfrac{2\pi}{3}$ is $2\rho\sin\left(\dfrac{2\pi/3}{2}\right)=2\rho\sin\dfrac{\pi}{3}=2\rho\cdot\dfrac{\sqrt3}{2}=\rho\sqrt3$. So $|z-P_1|=\sqrt3\,|z-z_0|$. By an identical argument (the angle between $z$ and $P_2$ about $z_0$ is $\dfrac{4\pi}{3}$, equivalent to $-\dfrac{2\pi}{3}$, giving the same chord length), $|z-P_2|=\sqrt3\,|z-z_0|$ also.
So $|z-P_1|=|z-P_2|=|P_1-P_2|=\sqrt3\,|z-z_0|$: all three sides of triangle $zP_1P_2$ are equal, so it is equilateral for any $z\ne z_0$. $\blacksquare$
(c) With $z_0=1+i$ and $z=4+i$:
$$z-z_0=(4+i)-(1+i)=3$$
Using $e^{i2\pi/3}=-\dfrac12+\dfrac{\sqrt3}{2}i$:
$$P_1=z_0+3e^{i2\pi/3}=(1+i)+3\left(-\dfrac12+\dfrac{\sqrt3}{2}i\right)=(1+i)+\left(-\dfrac32+\dfrac{3\sqrt3}{2}i\right)$$
$$=\left(1-\dfrac32\right)+\left(1+\dfrac{3\sqrt3}{2}\right)i=-\dfrac12+\left(1+\dfrac{3\sqrt3}{2}\right)i$$
Using $e^{i4\pi/3}=-\dfrac12-\dfrac{\sqrt3}{2}i$:
$$P_2=z_0+3e^{i4\pi/3}=(1+i)+3\left(-\dfrac12-\dfrac{\sqrt3}{2}i\right)=(1+i)+\left(-\dfrac32-\dfrac{3\sqrt3}{2}i\right)$$
$$=-\dfrac12+\left(1-\dfrac{3\sqrt3}{2}\right)i$$
The three points $z=4+i$, $P_1$ and $P_2$ (shown with centre $z_0$ in the diagram) form an equilateral triangle of side $3\sqrt3$, by part (b).
$$\boxed{P_1=-\dfrac12+\left(1+\dfrac{3\sqrt3}{2}\right)i, \qquad P_2=-\dfrac12+\left(1-\dfrac{3\sqrt3}{2}\right)i}$$
QUESTION 50
16 marks
Hard
(a) Find all six solutions of $z^6=-64$, giving each in Cartesian form. [7]
(b) By pairing each root with its complex conjugate, show that
$$z^6+64=\left(z^2-2\sqrt3z+4\right)\left(z^2+2\sqrt3z+4\right)\left(z^2+4\right)$$ [9]
Show complete worked solution
(a) Write $z=-64=64e^{i\pi}$ (since $-64$ is a negative real number, with modulus $64$ and argument $\pi$).
The sixth roots have modulus $64^{1/6}=2$ (since $2^6=64$) and arguments
$$\theta_k=\dfrac{\pi+2k\pi}{6}, \quad k=0,1,2,3,4,5$$
$$k=0:\ \theta_0=\dfrac{\pi}{6} \qquad k=1:\ \theta_1=\dfrac{3\pi}{6}=\dfrac{\pi}{2} \qquad k=2:\ \theta_2=\dfrac{5\pi}{6}$$
$$k=3:\ \theta_3=\dfrac{7\pi}{6}\ \to\ \text{subtract } 2\pi:\ \theta_3=-\dfrac{5\pi}{6}$$
$$k=4:\ \theta_4=\dfrac{9\pi}{6}=\dfrac{3\pi}{2}\ \to\ \text{subtract } 2\pi:\ \theta_4=-\dfrac{\pi}{2}$$
$$k=5:\ \theta_5=\dfrac{11\pi}{6}\ \to\ \text{subtract } 2\pi:\ \theta_5=-\dfrac{\pi}{6}$$
Converting each to Cartesian form (modulus $2$ throughout), using $\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2},\ \sin\dfrac{\pi}{6}=\dfrac12$:
$$w_0=2e^{i\pi/6}=\sqrt3+i, \qquad w_1=2e^{i\pi/2}=2i, \qquad w_2=2e^{i5\pi/6}=-\sqrt3+i$$
$$w_3=2e^{-i5\pi/6}=-\sqrt3-i, \qquad w_4=2e^{-i\pi/2}=-2i, \qquad w_5=2e^{-i\pi/6}=\sqrt3-i$$
These six points are equally spaced by $\dfrac{\pi}{3}$ ($60^{\circ}$) around the circle $|z|=2$, forming a regular hexagon, as shown in the diagram.
$$\boxed{z=\sqrt3+i,\ 2i,\ -\sqrt3+i,\ -\sqrt3-i,\ -2i,\ \sqrt3-i}$$
(b) Since $z^6+64$ has real coefficients, its roots occur in conjugate pairs; grouping the six roots from part (a) into conjugate pairs:
$$w_0=\sqrt3+i \text{ and } w_5=\sqrt3-i=\overline{w_0}$$
$$w_2=-\sqrt3+i \text{ and } w_3=-\sqrt3-i=\overline{w_2}$$
$$w_1=2i \text{ and } w_4=-2i=\overline{w_1}$$
For each conjugate pair $\alpha,\bar\alpha$, the quadratic factor is $(z-\alpha)(z-\bar\alpha)=z^2-(\alpha+\bar\alpha)z+\alpha\bar\alpha=z^2-2\operatorname{Re}(\alpha)z+|\alpha|^2$.
For the pair $\sqrt3\pm i$: $\operatorname{Re}(\alpha)=\sqrt3$ and $|\alpha|^2=(\sqrt3)^2+1^2=4$, giving the factor
$$z^2-2\sqrt3z+4$$
For the pair $-\sqrt3\pm i$: $\operatorname{Re}(\alpha)=-\sqrt3$ and $|\alpha|^2=(-\sqrt3)^2+1^2=4$, giving the factor
$$z^2+2\sqrt3z+4$$
For the pair $\pm2i$: $\operatorname{Re}(\alpha)=0$ and $|\alpha|^2=0^2+2^2=4$, giving the factor
$$z^2+4$$
Since $z^6+64$ is a monic (leading coefficient $1$) degree-$6$ polynomial whose six roots are exactly $w_0,\dots,w_5$, it equals the product of these three quadratic factors:
$$z^6+64=\left(z^2-2\sqrt3z+4\right)\left(z^2+2\sqrt3z+4\right)\left(z^2+4\right)$$
As a check, multiply the first two factors: $\left(z^2+4-2\sqrt3z\right)\left(z^2+4+2\sqrt3z\right)=(z^2+4)^2-\left(2\sqrt3z\right)^2=z^4+8z^2+16-12z^2=z^4-4z^2+16$.
Multiplying by the remaining factor $(z^2+4)$:
$$\left(z^4-4z^2+16\right)(z^2+4)=z^6+4z^4-4z^4-16z^2+16z^2+64=z^6+64 \ \checkmark$$
as required. $\blacksquare$
Proof by Deduction, Contradiction, and Counterexample 50 questions
QUESTION 1
4 marks
Easy
Prove that $(x+2)^{2}-(x-2)^{2}=8x$ for all real numbers $x$.
Show complete worked solution
Using a direct algebraic proof, start from the left-hand side (LHS) and work towards the right-hand side (RHS):
$$\text{LHS}=(x+2)^{2}-(x-2)^{2}$$
Expand each square:
$$(x+2)^{2}=x^{2}+4x+4,\qquad (x-2)^{2}=x^{2}-4x+4$$
Substituting back:
$$\text{LHS}=(x^{2}+4x+4)-(x^{2}-4x+4)$$
Distributing the negative sign and combining like terms:
$$=x^{2}+4x+4-x^{2}+4x-4=8x$$
$$=\text{RHS}$$
Since $\text{LHS}=\text{RHS}$ for all real $x$, the identity is proven. $\blacksquare$
QUESTION 2
5 marks
Medium
Prove that $\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1}{n(n+1)}$ for all positive integers $n$.
Show complete worked solution
Starting from the left-hand side and combining the fractions over a common denominator:
$$\text{LHS}=\dfrac{1}{n}-\dfrac{1}{n+1}$$
Write both fractions with the common denominator $n(n+1)$:
$$=\dfrac{n+1}{n(n+1)}-\dfrac{n}{n(n+1)}$$
Combine the numerators over the shared denominator:
$$=\dfrac{(n+1)-n}{n(n+1)}$$
Simplify the numerator:
$$=\dfrac{1}{n(n+1)}=\text{RHS}$$
Since $\text{LHS}=\text{RHS}$ for all positive integers $n$ (so $n\neq0$ and $n+1\neq0$ automatically), the identity is proven. $\blacksquare$
QUESTION 3
6 marks
Easy
(a) Show that the sum of the squares of two consecutive integers $n$ and $n+1$ can be written as $2n^{2}+2n+1$.
(b) Hence prove that this sum is always an odd number, for any integer $n$.
Show complete worked solution
(a) Starting from the left-hand side and expanding:
$$\text{LHS}=n^{2}+(n+1)^{2}$$
Expand $(n+1)^2$:
$$=n^{2}+n^{2}+2n+1$$
Combine like terms:
$$=2n^{2}+2n+1=\text{RHS}$$
Since $\text{LHS}=\text{RHS}$ for all integers $n$, the identity is proven. $\blacksquare$
(b) From part (a), the sum is $2n^{2}+2n+1=2(n^{2}+n)+1$.
Since $n^{2}+n$ is an integer (as $n$ is an integer), $2(n^{2}+n)$ is an even integer.
An even integer plus 1 is, by definition, an odd integer.
Therefore $2n^{2}+2n+1$ is always odd, for any integer $n$. $\blacksquare$
QUESTION 4
6 marks
Medium
Prove, using the formula for the sum of an arithmetic series, that
$$\sum_{r=1}^{n}(2r-1)=n^{2}$$
for all positive integers $n$.
Show complete worked solution
The terms $2r-1$ for $r=1,2,3,\ldots,n$ are $1,3,5,\ldots,(2n-1)$, the first $n$ odd numbers - an arithmetic sequence with first term $u_1=1$ and common difference $d=2$.
Starting from the left-hand side and applying the sum formula for an arithmetic series, $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$:
$$\text{LHS}=\sum_{r=1}^{n}(2r-1)=\dfrac{n}{2}\big(2(1)+(n-1)(2)\big)$$
Simplify inside the brackets:
$$=\dfrac{n}{2}\big(2+2n-2\big)=\dfrac{n}{2}(2n)=n^{2}=\text{RHS}$$
Since $\text{LHS}=\text{RHS}$ for all positive integers $n$, the identity is proven. $\blacksquare$
QUESTION 5
14 marks
Hard
(a) Prove that $(2n+1)^{2}-(2n-1)^{2}=8n$ for all integers $n$. [3]
(b) Hence show that the difference between the squares of any two consecutive odd positive integers is always a multiple of 8. [3]
(c) Prove that $a^{2}+b^{2}\ge2ab$ for all real numbers $a$ and $b$, and hence deduce that $\dfrac{a^{2}+b^{2}}{2}\ge ab$. [4]
(d) Using the result of part (c), prove that $x+\dfrac{1}{x}\ge2$ for all positive real numbers $x$. [4]
Show complete worked solution
(a) Starting from the left-hand side:
$$\text{LHS}=(2n+1)^{2}-(2n-1)^{2}$$
Expand each square:
$$(2n+1)^{2}=4n^{2}+4n+1,\qquad (2n-1)^{2}=4n^{2}-4n+1$$
Subtract:
$$\text{LHS}=(4n^{2}+4n+1)-(4n^{2}-4n+1)=8n=\text{RHS}$$
Since $\text{LHS}=\text{RHS}$ for all integers $n$, the identity is proven. $\blacksquare$
(b) Any two consecutive odd positive integers can be written as $2k-1$ and $2k+1$ for some positive integer $k$.
The difference of their squares is $(2k+1)^{2}-(2k-1)^{2}$, exactly the identity proven in part (a) with $n=k$:
$$(2k+1)^{2}-(2k-1)^{2}=8k$$
Since $k$ is a positive integer, $8k$ is a multiple of 8. Therefore the difference between the squares of any two consecutive odd positive integers is always a multiple of 8. $\blacksquare$
(c) Consider $(a-b)^{2}$. Since it is the square of a real number, it can never be negative:
$$(a-b)^{2}\ge0 \text{ for all real } a,b$$
Expand the left-hand side:
$$a^{2}-2ab+b^{2}\ge0$$
Add $2ab$ to both sides:
$$a^{2}+b^{2}\ge2ab$$
This proves the inequality for all real $a$ and $b$. $\blacksquare$
Dividing both sides by 2 (which does not change the direction of the inequality, since 2 is positive):
$$\dfrac{a^{2}+b^{2}}{2}\ge ab$$
as required.
(d) Let $x$ be a positive real number, and set $a=\sqrt{x}$ and $b=\dfrac{1}{\sqrt{x}}$ (both real and well-defined, since $x>0$). By the result of part (c):
$$a^{2}+b^{2}\ge2ab$$
Substitute $a=\sqrt{x}$, $b=\dfrac{1}{\sqrt{x}}$, so $a^{2}=x$, $b^{2}=\dfrac{1}{x}$, $2ab=2\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}=2$:
$$x+\dfrac{1}{x}\ge2$$
for all positive real numbers $x$. $\blacksquare$
QUESTION 6
3 marks
Easy
Prove that the sum of any two even integers is always even.
Show complete worked solution
Let $m$ and $n$ be any two even integers. By the definition of an even number, $m$ and $n$ can be written as
$$m=2a,\qquad n=2b$$
for some integers $a$ and $b$.
Then their sum is
$$m+n=2a+2b=2(a+b)$$
Since $a$ and $b$ are integers, $a+b$ is also an integer, so $m+n$ is $2$ times an integer.
Therefore $m+n$ is even, for any two even integers $m$ and $n$. $\blacksquare$
QUESTION 7
4 marks
Easy
Prove that the sum of any two odd integers is always even.
Show complete worked solution
Let $m$ and $n$ be any two odd integers. By the definition of an odd number, $m$ and $n$ can be written as
$$m=2a+1,\qquad n=2b+1$$
for some integers $a$ and $b$.
Then their sum is
$$m+n=(2a+1)+(2b+1)=2a+2b+2=2(a+b+1)$$
Since $a$ and $b$ are integers, $a+b+1$ is also an integer, so $m+n$ is $2$ times an integer.
Therefore $m+n$ is even, for any two odd integers $m$ and $n$. $\blacksquare$
QUESTION 8
4 marks
Easy
Prove that the product of any two odd integers is always odd.
Show complete worked solution
Let $m$ and $n$ be any two odd integers, so
$$m=2a+1,\qquad n=2b+1$$
for some integers $a$ and $b$.
Then their product is
$$mn=(2a+1)(2b+1)=4ab+2a+2b+1=2(2ab+a+b)+1$$
Since $a$ and $b$ are integers, $2ab+a+b$ is also an integer, so $mn$ is of the form $2k+1$ where $k=2ab+a+b$ is an integer.
Therefore $mn$ is odd, for any two odd integers $m$ and $n$. $\blacksquare$
QUESTION 9
4 marks
Easy
Prove that if $n$ is an odd integer, then $n^{2}$ is also odd.
Show complete worked solution
Let $n$ be an odd integer, so $n=2k+1$ for some integer $k$.
Then
$$n^{2}=(2k+1)^{2}=4k^{2}+4k+1=2(2k^{2}+2k)+1$$
Since $k$ is an integer, $2k^{2}+2k$ is also an integer, so $n^2$ is of the form $2j+1$ where $j=2k^{2}+2k$ is an integer.
Therefore $n^2$ is odd whenever $n$ is odd. $\blacksquare$
QUESTION 10
3 marks
Easy
Prove that $n(n+1)$ is always an even number, for any integer $n$.
Show complete worked solution
Any integer $n$ is either even or odd.
Case 1: $n$ is even, so $n=2k$ for some integer $k$. Then $n(n+1)=2k(n+1)$, which is $2$ times an integer, so it is even.
Case 2: $n$ is odd, so $n+1$ is even, and $n+1=2k$ for some integer $k$. Then $n(n+1)=n(2k)=2(nk)$, which is $2$ times an integer, so it is even.
In both possible cases $n(n+1)$ is even. Since every integer $n$ is either even or odd, $n(n+1)$ is always even. $\blacksquare$
QUESTION 11
4 marks
Easy
Prove that $n^{2}+n+2$ is even for every integer $n$.
Show complete worked solution
Note that $n^{2}+n=n(n+1)$, the product of two consecutive integers.
Of any two consecutive integers, one must be even, so their product $n(n+1)$ is always even; that is, $n(n+1)=2k$ for some integer $k$.
Then
$$n^{2}+n+2=n(n+1)+2=2k+2=2(k+1)$$
Since $k$ is an integer, $k+1$ is also an integer, so $n^2+n+2$ is $2$ times an integer.
Therefore $n^2+n+2$ is even for every integer $n$. $\blacksquare$
QUESTION 12
3 marks
Easy
Prove that the product of an even integer and any other integer is always even.
Show complete worked solution
Let $m$ be an even integer and $n$ be any integer. Since $m$ is even, $m=2k$ for some integer $k$.
Then
$$mn=(2k)n=2(kn)$$
Since $k$ and $n$ are both integers, $kn$ is also an integer, so $mn$ is $2$ times an integer.
Therefore $mn$ is even, for any even integer $m$ and any integer $n$. $\blacksquare$
QUESTION 13
4 marks
Easy
Prove that if $a$, $b$ and $c$ are integers with $a\mid b$ and $a\mid c$, then $a\mid(b+c)$.
Show complete worked solution
Since $a\mid b$, there exists an integer $p$ such that
$$b=ap$$
Since $a\mid c$, there exists an integer $q$ such that
$$c=aq$$
Adding these two equations:
$$b+c=ap+aq=a(p+q)$$
Since $p$ and $q$ are integers, $p+q$ is also an integer, so $b+c$ is $a$ times an integer.
Therefore $a\mid(b+c)$. $\blacksquare$
QUESTION 14
4 marks
Easy
Prove that the difference between the squares of any two consecutive positive integers is always odd.
Show complete worked solution
Let $n$ and $n+1$ be two consecutive positive integers. The difference between their squares is
$$(n+1)^{2}-n^{2}$$
Expanding $(n+1)^2$:
$$=n^{2}+2n+1-n^{2}=2n+1$$
Since $n$ is a positive integer, $2n$ is an even integer, so $2n+1$ is of the form (even) $+1$, which is odd.
Therefore the difference between the squares of any two consecutive positive integers, $(n+1)^{2}-n^{2}=2n+1$, is always odd. $\blacksquare$
QUESTION 15
3 marks
Easy
A student claims that "the sum of any two prime numbers is always an even number." Show, by means of a counterexample, that this statement is false.
Show complete worked solution
The general claim is that $p+q$ is even for any two prime numbers $p$ and $q$.
Consider the prime numbers $p=2$ and $q=3$. Both $2$ and $3$ are prime numbers.
Their sum is
$$p+q=2+3=5$$
Since $5$ is odd, not even, this is a counterexample to the claim.
Therefore the statement "the sum of any two prime numbers is always even" is false. $\blacksquare$
QUESTION 16
3 marks
Easy
Disprove the statement: "For all real numbers $a$ and $b$, $(a+b)^{2}=a^{2}+b^{2}$."
Show complete worked solution
Consider $a=1$ and $b=1$, both real numbers.
The left-hand side gives
$$(a+b)^{2}=(1+1)^{2}=2^{2}=4$$
The right-hand side gives
$$a^{2}+b^{2}=1^{2}+1^{2}=1+1=2$$
Since $4\neq2$, the equation $(a+b)^2=a^2+b^2$ fails for $a=b=1$.
Therefore the statement "$(a+b)^2=a^2+b^2$ for all real numbers $a$ and $b$" is false. $\blacksquare$
QUESTION 17
3 marks
Easy
Disprove the statement: "Every prime number is odd."
Show complete worked solution
Consider the number $2$.
By definition, $2$ is a prime number, since its only positive divisors are $1$ and $2$ itself.
However, $2=2(1)$, so $2$ is even, not odd.
This is a counterexample: $2$ is a prime number that is not odd.
Therefore the statement "every prime number is odd" is false. $\blacksquare$
QUESTION 18
4 marks
Easy
Disprove the statement: "For all real numbers $x$, $\sqrt{x^{2}}=x$."
Show complete worked solution
Consider $x=-3$, a real number.
The left-hand side gives
$$\sqrt{x^{2}}=\sqrt{(-3)^{2}}=\sqrt{9}=3$$
since the symbol $\sqrt{\ }$ always denotes the non-negative square root.
The right-hand side gives
$$x=-3$$
Since $3\neq-3$, the equation $\sqrt{x^2}=x$ fails for $x=-3$.
Therefore the statement "$\sqrt{x^2}=x$ for all real numbers $x$" is false. $\blacksquare$
QUESTION 19
4 marks
Easy
Disprove the statement: "For all integers $a$ and $b$, if $a^{2}=b^{2}$ then $a=b$."
Show complete worked solution
Consider $a=3$ and $b=-3$, both integers.
Then
$$a^{2}=3^{2}=9,\qquad b^{2}=(-3)^{2}=9$$
so $a^2=b^2$ is satisfied.
However,
$$a=3,\qquad b=-3$$
and clearly $3\neq-3$, so $a\neq b$.
This is a counterexample: $a^2=b^2$ holds but $a=b$ does not.
Therefore the statement "if $a^2=b^2$ then $a=b$" is false. $\blacksquare$
QUESTION 20
5 marks
Easy
Prove, by contradiction, that there is no smallest positive rational number.
Show complete worked solution
Using proof by contradiction, suppose that there is a smallest positive rational number, and call it $q$.
Since $q$ is rational and non-zero, $\dfrac{q}{2}$ is also rational, being the ratio of the rational number $q$ to the integer $2$.
Since $q>0$, dividing by $2$ gives $\dfrac{q}{2}>0$, so $\dfrac{q}{2}$ is a positive rational number.
Also, since $q>0$,
$$\dfrac{q}{2}
QUESTION 21
5 marks
Easy
Prove, by contradiction, that if $n^{2}$ is even, then $n$ is even, for any integer $n$.
Show complete worked solution
Using proof by contradiction, suppose that $n^{2}$ is even but $n$ is not even, i.e. $n$ is odd.
Since $n$ is odd, $n=2k+1$ for some integer $k$. Then
$$n^{2}=(2k+1)^{2}=4k^{2}+4k+1=2(2k^{2}+2k)+1$$
Since $k$ is an integer, $2k^{2}+2k$ is an integer, so $n^2$ is of the form (even)$+1$, which means $n^2$ is odd.
This contradicts the assumption that $n^2$ is even.
Therefore the assumption that $n$ is odd must be false, so $n$ must be even whenever $n^2$ is even. $\blacksquare$
QUESTION 22
4 marks
Easy
Prove that the sum of any three consecutive integers is always divisible by $3$.
Show complete worked solution
Let the three consecutive integers be $n$, $n+1$ and $n+2$, for some integer $n$. Their sum is
$$n+(n+1)+(n+2)=3n+3=3(n+1)$$
Since $n$ is an integer, $n+1$ is also an integer, so the sum is $3$ times an integer.
Therefore the sum of any three consecutive integers is always divisible by $3$. $\blacksquare$
QUESTION 23
3 marks
Easy
Disprove the statement: "If $p$ is a prime number, then $p+2$ is also a prime number."
Show complete worked solution
Consider $p=7$, which is a prime number, since its only positive divisors are $1$ and $7$.
Then
$$p+2=7+2=9$$
However, $9=3\times3$, so $9$ has a divisor other than $1$ and itself, meaning $9$ is not prime.
This is a counterexample: $p=7$ is prime, but $p+2=9$ is not prime.
Therefore the statement "if $p$ is prime, then $p+2$ is also prime" is false. $\blacksquare$
QUESTION 24
6 marks
Medium
Prove, by contradiction, that $\sqrt{2}$ is irrational.
Show complete worked solution
Using proof by contradiction, suppose that $\sqrt{2}$ is rational.
Then $\sqrt{2}$ can be written as $\sqrt{2}=\dfrac{p}{q}$, where $p$ and $q$ are integers with no common factor other than $1$ (i.e. the fraction is in lowest terms) and $q\neq0$.
Squaring both sides:
$$2=\dfrac{p^{2}}{q^{2}}$$
$$p^{2}=2q^{2}$$
Since $q^2$ is an integer, $p^2$ is $2$ times an integer, so $p^2$ is even. Since the square of an odd number is always odd, $p^2$ being even means $p$ itself must be even, so $p=2k$ for some integer $k$.
Substituting $p=2k$ into $p^2=2q^2$:
$$(2k)^{2}=2q^{2}$$
$$4k^{2}=2q^{2}$$
$$q^{2}=2k^{2}$$
Since $k^2$ is an integer, $q^2$ is $2$ times an integer, so $q^2$ is even, and therefore $q$ is also even.
But $p$ and $q$ are both even, so they share the common factor $2$.
This contradicts the assumption that $p$ and $q$ have no common factor other than $1$.
Therefore the assumption that $\sqrt{2}$ is rational must be false, so $\sqrt{2}$ is irrational. $\blacksquare$
QUESTION 25
6 marks
Medium
Prove, by contradiction, that $\sqrt{3}$ is irrational.
Show complete worked solution
Using proof by contradiction, suppose that $\sqrt{3}$ is rational.
Then $\sqrt{3}=\dfrac{p}{q}$, where $p$ and $q$ are integers with no common factor other than $1$, and $q\neq0$.
Squaring both sides:
$$3=\dfrac{p^{2}}{q^{2}}$$
$$p^{2}=3q^{2}$$
Since $q^2$ is an integer, $p^2$ is $3$ times an integer, so $3\mid p^2$. Since $3$ is prime, this means $3\mid p$ (if $p$ were not divisible by $3$, then $p^2$ would not be divisible by $3$ either), so $p=3k$ for some integer $k$.
Substituting $p=3k$ into $p^2=3q^2$:
$$(3k)^{2}=3q^{2}$$
$$9k^{2}=3q^{2}$$
$$q^{2}=3k^{2}$$
Since $k^2$ is an integer, $q^2$ is $3$ times an integer, so $3\mid q^2$, and by the same reasoning $3\mid q$.
But $p$ and $q$ are both divisible by $3$, so they share the common factor $3$.
This contradicts the assumption that $p$ and $q$ have no common factor other than $1$.
Therefore the assumption that $\sqrt{3}$ is rational must be false, so $\sqrt{3}$ is irrational. $\blacksquare$
QUESTION 26
6 marks
Medium
Prove, by contradiction, that $\sqrt{5}$ is irrational.
Show complete worked solution
Using proof by contradiction, suppose that $\sqrt{5}$ is rational.
Then $\sqrt{5}=\dfrac{p}{q}$, where $p$ and $q$ are integers with no common factor other than $1$, and $q\neq0$.
Squaring both sides:
$$5=\dfrac{p^{2}}{q^{2}}$$
$$p^{2}=5q^{2}$$
Since $q^2$ is an integer, $5\mid p^2$. Since $5$ is prime, this means $5\mid p$, so $p=5k$ for some integer $k$.
Substituting $p=5k$ into $p^2=5q^2$:
$$(5k)^{2}=5q^{2}$$
$$25k^{2}=5q^{2}$$
$$q^{2}=5k^{2}$$
Since $k^2$ is an integer, $5\mid q^2$, and since $5$ is prime, $5\mid q$.
But $p$ and $q$ are both divisible by $5$, so they share the common factor $5$.
This contradicts the assumption that $p$ and $q$ have no common factor other than $1$.
Therefore the assumption that $\sqrt{5}$ is rational must be false, so $\sqrt{5}$ is irrational. $\blacksquare$
QUESTION 27
7 marks
Medium
Let $p$ be a prime number. Prove, by contradiction, that $\sqrt{p}$ is irrational.
Show complete worked solution
Using proof by contradiction, suppose that $\sqrt{p}$ is rational, where $p$ is a prime number.
Then $\sqrt{p}=\dfrac{a}{b}$, where $a$ and $b$ are integers with no common factor other than $1$, and $b\neq0$.
Squaring both sides:
$$p=\dfrac{a^{2}}{b^{2}}$$
$$a^{2}=pb^{2}$$
Since $b^2$ is an integer, $p\mid a^2$. Because $p$ is prime, if $p$ divides a product it must divide at least one of the factors; applying this to $a^2=a\times a$ gives $p\mid a$. So $a=pk$ for some integer $k$.
Substituting $a=pk$ into $a^2=pb^2$:
$$(pk)^{2}=pb^{2}$$
$$p^{2}k^{2}=pb^{2}$$
Dividing both sides by $p$ (valid since $p\neq0$):
$$pk^{2}=b^{2}$$
Since $k^2$ is an integer, $p\mid b^2$, and since $p$ is prime, $p\mid b$ by the same reasoning as above.
But then $a$ and $b$ are both divisible by $p$, so they share the common factor $p$.
This contradicts the assumption that $a$ and $b$ have no common factor other than $1$.
Therefore the assumption that $\sqrt{p}$ is rational must be false, so $\sqrt{p}$ is irrational for every prime number $p$. $\blacksquare$
QUESTION 28
7 marks
Medium
Prove that $n^{3}-n$ is divisible by $6$ for every integer $n$.
Show complete worked solution
Factorise the expression:
$$n^{3}-n=n(n^{2}-1)=n(n-1)(n+1)=(n-1)n(n+1)$$
This is the product of three consecutive integers $n-1$, $n$ and $n+1$.
Divisibility by $2$: among any two consecutive integers, at least one is even. In particular, among $n-1$ and $n$, at least one is even, so the product $(n-1)n(n+1)$ is even, i.e. divisible by $2$.
Divisibility by $3$: among any three consecutive integers, exactly one is divisible by $3$ (since every third integer is a multiple of $3$). So one of $n-1$, $n$, $n+1$ is divisible by $3$, and therefore the product $(n-1)n(n+1)$ is divisible by $3$.
Since the product $(n-1)n(n+1)$ is divisible by both $2$ and $3$, and $2$ and $3$ share no common factor other than $1$, the product is divisible by $2\times3=6$.
Therefore $n^3-n$ is divisible by $6$ for every integer $n$. $\blacksquare$
QUESTION 29
6 marks
Medium
Prove that the square of any integer $n$ is either a multiple of $4$, or one more than a multiple of $4$.
Show complete worked solution
Every integer $n$ is either even or odd.
Case 1: $n$ is even, so $n=2k$ for some integer $k$. Then
$$n^{2}=4k^{2}$$
which is a multiple of $4$.
Case 2: $n$ is odd, so $n=2k+1$ for some integer $k$. Then
$$n^{2}=(2k+1)^{2}=4k^{2}+4k+1=4(k^{2}+k)+1$$
Since $k$ is an integer, $k^2+k$ is also an integer, so $n^2$ is one more than a multiple of $4$.
Since every integer $n$ falls into exactly one of these two cases, $n^2$ is always either a multiple of $4$, or one more than a multiple of $4$. $\blacksquare$
QUESTION 30
7 marks
Medium
The $n$th triangular number is given by $T_{n}=\dfrac{n(n+1)}{2}$. Prove that the sum of any two consecutive triangular numbers, $T_{n}+T_{n+1}$, is always a perfect square.
Show complete worked solution
Using the formula for the $n$th triangular number:
$$T_{n}+T_{n+1}=\dfrac{n(n+1)}{2}+\dfrac{(n+1)(n+2)}{2}$$
Combine the fractions over the common denominator $2$:
$$=\dfrac{n(n+1)+(n+1)(n+2)}{2}$$
Factor out the common factor $(n+1)$ from the numerator:
$$=\dfrac{(n+1)\big[n+(n+2)\big]}{2}=\dfrac{(n+1)(2n+2)}{2}$$
Factor $2$ from $(2n+2)$:
$$=\dfrac{(n+1)\times2(n+1)}{2}=\dfrac{2(n+1)^{2}}{2}=(n+1)^{2}$$
Since $n$ is an integer, $n+1$ is also an integer, so $T_n+T_{n+1}=(n+1)^2$ is the square of an integer, i.e. a perfect square.
Therefore the sum of any two consecutive triangular numbers is always a perfect square. $\blacksquare$
QUESTION 31
6 marks
Medium
Prove, by contradiction, that if $x$ is a rational number and $y$ is an irrational number, then $x+y$ is irrational.
Show complete worked solution
Using proof by contradiction, suppose that $x$ is rational, $y$ is irrational, but $x+y$ is rational.
Since $x$ is rational, $x=\dfrac{a}{b}$ for some integers $a,b$ with $b\neq0$. Since $x+y$ is assumed rational, $x+y=\dfrac{c}{d}$ for some integers $c,d$ with $d\neq0$.
Rearranging:
$$y=(x+y)-x=\dfrac{c}{d}-\dfrac{a}{b}$$
Writing this over a common denominator:
$$y=\dfrac{cb-ad}{db}$$
Since $a,b,c,d$ are integers, $cb-ad$ and $db$ are also integers, and $db\neq0$ (as $b\neq0$ and $d\neq0$). So $y$ is expressed as a ratio of two integers, meaning $y$ is rational.
This contradicts the given fact that $y$ is irrational.
Therefore the assumption that $x+y$ is rational must be false, so $x+y$ is irrational whenever $x$ is rational and $y$ is irrational. $\blacksquare$
QUESTION 32
6 marks
Medium
Prove, by contradiction, that if $x$ is a non-zero rational number and $y$ is an irrational number, then $xy$ is irrational.
Show complete worked solution
Using proof by contradiction, suppose that $x$ is a non-zero rational number, $y$ is irrational, but $xy$ is rational.
Since $x$ is rational and non-zero, $x=\dfrac{a}{b}$ for some non-zero integers $a,b$ with $b\neq0$. Since $xy$ is assumed rational, $xy=\dfrac{c}{d}$ for some integers $c,d$ with $d\neq0$.
Since $x\neq0$, we can divide both sides of $xy=\dfrac{c}{d}$ by $x$:
$$y=\dfrac{xy}{x}=\dfrac{c}{d}\div\dfrac{a}{b}=\dfrac{c}{d}\times\dfrac{b}{a}=\dfrac{cb}{da}$$
Since $a,b,c,d$ are integers with $a\neq0$ and $d\neq0$, both $cb$ and $da$ are integers with $da\neq0$. So $y$ is expressed as a ratio of two integers, meaning $y$ is rational.
This contradicts the given fact that $y$ is irrational.
Therefore the assumption that $xy$ is rational must be false, so $xy$ is irrational whenever $x$ is a non-zero rational number and $y$ is irrational. $\blacksquare$
QUESTION 33
6 marks
Medium
(a) Show that $n^{2}-n=n(n-1)$ for all integers $n$.
(b) Hence prove that $n^{2}-n$ is always an even number, for any integer $n$.
Show complete worked solution
(a) Starting from the right-hand side and expanding:
$$\text{RHS}=n(n-1)=n^{2}-n=\text{LHS}$$
Since $\text{LHS}=\text{RHS}$ for all integers $n$, the identity is proven. $\blacksquare$
(b) From part (a), $n^2-n=n(n-1)$, the product of two consecutive integers $n-1$ and $n$.
Of any two consecutive integers, one must be even, so their product $n(n-1)$ is always even.
Therefore $n^2-n$ is always an even number, for any integer $n$. $\blacksquare$
QUESTION 34
6 marks
Medium
A student conjectures that "the expression $n^{2}+n+41$ gives a prime number for every positive integer $n$." Show, by means of a counterexample, that this conjecture is false.
Show complete worked solution
The general claim is that $n^2+n+41$ is prime for every positive integer $n$.
Consider $n=40$.
$$n^{2}+n+41=40^{2}+40+41=1600+40+41=1681$$
Now check whether $1681$ is prime. Notice that
$$1681=41^{2}=41\times41$$
Since $1681$ has a divisor ($41$) other than $1$ and itself, $1681$ is not prime.
This is a counterexample: for $n=40$, the expression $n^2+n+41=1681$ is not a prime number.
Therefore the conjecture "$n^2+n+41$ is prime for every positive integer $n$" is false. $\blacksquare$
QUESTION 35
6 marks
Medium
Disprove the statement: "If $a$, $b$ and $c$ are integers with $a\mid bc$, then $a\mid b$ or $a\mid c$."
Show complete worked solution
Consider $a=6$, $b=4$ and $c=3$.
Then
$$bc=4\times3=12$$
and $12=6\times2$, so $6\mid bc$, meaning $a\mid bc$ is satisfied.
However, $4\div6$ is not an integer, so $a\nmid b$; and $3\div6$ is not an integer, so $a\nmid c$.
This is a counterexample: $a\mid bc$ holds, but neither $a\mid b$ nor $a\mid c$ holds.
Therefore the statement "if $a\mid bc$ then $a\mid b$ or $a\mid c$" is false. $\blacksquare$
QUESTION 36
7 marks
Medium
A student claims that "$2^{n}-1$ is always a prime number whenever $n$ is a prime number." Show, by means of a counterexample, that this claim is false.
Show complete worked solution
The general claim is that $2^n-1$ is prime whenever $n$ is prime.
Consider $n=11$. Since $11$ has only $1$ and itself as positive divisors, $11$ is prime.
$$2^{11}-1=2048-1=2047$$
Now check whether $2047$ is prime. Testing division by $23$:
$$2047\div23=89$$
Since $23\times89=2047$, this confirms $2047=23\times89$.
Since $2047$ has a divisor ($23$) other than $1$ and itself, $2047$ is not prime.
This is a counterexample: $n=11$ is prime, but $2^{11}-1=2047$ is not prime.
Therefore the claim "$2^n-1$ is always prime whenever $n$ is prime" is false. $\blacksquare$
QUESTION 37
6 marks
Medium
Disprove the statement: "$n^{2}-n+11$ is a prime number for every positive integer $n$."
Show complete worked solution
Consider $n=11$.
$$n^{2}-n+11=11^{2}-11+11=121-11+11=121$$
Now check whether $121$ is prime. Notice that
$$121=11^{2}=11\times11$$
Since $121$ has a divisor ($11$) other than $1$ and itself, $121$ is not prime.
This is a counterexample: for $n=11$, the expression $n^2-n+11=121$ is not a prime number.
Therefore the statement "$n^2-n+11$ is prime for every positive integer $n$" is false. $\blacksquare$
QUESTION 38
7 marks
Medium
Prove that the square of any odd integer can be written in the form $8k+1$, for some integer $k$.
Show complete worked solution
Let $n$ be an odd integer, so $n=2m+1$ for some integer $m$. Then
$$n^{2}=(2m+1)^{2}=4m^{2}+4m+1=4m(m+1)+1$$
Now consider $m(m+1)$, the product of two consecutive integers $m$ and $m+1$. Of any two consecutive integers, one is always even, so $m(m+1)$ is even, and can be written as $m(m+1)=2k$ for some integer $k$.
Substituting this back:
$$n^{2}=4\times2k+1=8k+1$$
Since $k$ is an integer, $n^2$ is of the form $8k+1$.
Therefore the square of any odd integer can always be written in the form $8k+1$, for some integer $k$. $\blacksquare$
QUESTION 39
6 marks
Medium
Prove that the product of any two consecutive even integers is divisible by $8$.
Show complete worked solution
Let the two consecutive even integers be $2n$ and $2n+2$, for some integer $n$. Their product is
$$2n(2n+2)=4n(n+1)$$
Now consider $n(n+1)$, the product of two consecutive integers $n$ and $n+1$. Of any two consecutive integers, one is always even, so $n(n+1)=2k$ for some integer $k$.
Substituting this back:
$$2n(2n+2)=4\times2k=8k$$
Since $k$ is an integer, the product $2n(2n+2)$ is $8$ times an integer.
Therefore the product of any two consecutive even integers is always divisible by $8$. $\blacksquare$
QUESTION 40
5 marks
Medium
Disprove the statement: "If $n$ is a composite number, then $n+1$ is a prime number."
Show complete worked solution
Consider $n=8$. Since $8=2\times4$, $8$ has divisors other than $1$ and itself, so $8$ is composite.
Then
$$n+1=8+1=9$$
However, $9=3\times3$, so $9$ has a divisor other than $1$ and itself, meaning $9$ is not prime.
This is a counterexample: $n=8$ is composite, but $n+1=9$ is not prime.
Therefore the statement "if $n$ is composite, then $n+1$ is prime" is false. $\blacksquare$
QUESTION 41
6 marks
Medium
Prove, by contradiction, that $\sqrt{6}$ is irrational.
Show complete worked solution
Using proof by contradiction, suppose that $\sqrt{6}$ is rational.
Then $\sqrt{6}=\dfrac{p}{q}$, where $p$ and $q$ are integers with no common factor other than $1$, and $q\neq0$.
Squaring both sides:
$$6=\dfrac{p^{2}}{q^{2}}$$
$$p^{2}=6q^{2}$$
Since $q^2$ is an integer, $p^2=2(3q^{2})$ is $2$ times an integer, so $p^2$ is even, and therefore $p$ is even. So $p=2k$ for some integer $k$.
Substituting $p=2k$ into $p^2=6q^2$:
$$(2k)^{2}=6q^{2}$$
$$4k^{2}=6q^{2}$$
$$2k^{2}=3q^{2}$$
Since the left-hand side, $2k^2$, is even, the right-hand side $3q^2$ must also be even. Since $3$ is odd, this means $q^2$ must be even (an odd number times an odd number is always odd), and therefore $q$ is also even.
But $p$ and $q$ are both even, so they share the common factor $2$.
This contradicts the assumption that $p$ and $q$ have no common factor other than $1$.
Therefore the assumption that $\sqrt{6}$ is rational must be false, so $\sqrt{6}$ is irrational. $\blacksquare$
QUESTION 42
9 marks
Hard
Prove, by contradiction, that there are infinitely many prime numbers.
Show complete worked solution
Using proof by contradiction, suppose that there are only finitely many prime numbers. Then they can be listed in full as
$$p_{1},\ p_{2},\ p_{3},\ \ldots,\ p_{n}$$
for some positive integer $n$, where $p_1,p_2,\ldots,p_n$ is the complete list of every prime number that exists.
Consider the number
$$N=p_{1}p_{2}p_{3}\cdots p_{n}+1$$
formed by multiplying every prime in the list together and adding $1$.
Since $N$ is an integer greater than $1$, $N$ must have at least one prime factor; call it $q$.
Because $p_1,p_2,\ldots,p_n$ is assumed to be the complete list of all prime numbers, $q$ must be equal to one of $p_1,p_2,\ldots,p_n$; say $q=p_i$ for some $i$.
Since $q=p_i$ divides both $N$ and the product $p_1p_2\cdots p_n$ (as $p_i$ is one of the factors in that product), $q$ must divide their difference:
$$N-p_{1}p_{2}\cdots p_{n}=1$$
So $q\mid1$. But the only positive integer that divides $1$ is $1$ itself, and $q$ is prime, so $q\ge2$.
This contradicts $q\mid1$.
Therefore the assumption that there are only finitely many prime numbers must be false, so there are infinitely many prime numbers. $\blacksquare$
QUESTION 43
10 marks
Hard
(a) Prove, by contradiction, that $\sqrt{6}$ is irrational. [5]
(b) Hence, using proof by contradiction, prove that $\sqrt{2}+\sqrt{3}$ is irrational. [5]
Show complete worked solution
(a) Using proof by contradiction, suppose that $\sqrt{6}$ is rational, so $\sqrt{6}=\dfrac{p}{q}$ for integers $p,q$ with no common factor other than $1$, and $q\neq0$.
Squaring both sides:
$$p^{2}=6q^{2}$$
Since $p^2=2(3q^2)$ is even, $p$ is even, so $p=2k$ for some integer $k$. Substituting:
$$4k^{2}=6q^{2}\quad\Rightarrow\quad2k^{2}=3q^{2}$$
Since $2k^2$ is even, $3q^2$ is even, and since $3$ is odd, $q^2$ must be even, so $q$ is even.
But then $p$ and $q$ are both even, contradicting the assumption that they share no common factor other than $1$.
Therefore the assumption that $\sqrt{6}$ is rational must be false, so $\sqrt{6}$ is irrational. $\blacksquare$
(b) Using proof by contradiction, suppose that $\sqrt{2}+\sqrt{3}$ is rational, and call it $r$, so
$$\sqrt{2}+\sqrt{3}=r$$
Squaring both sides:
$$(\sqrt{2}+\sqrt{3})^{2}=r^{2}$$
$$2+2\sqrt{2}\sqrt{3}+3=r^{2}$$
$$5+2\sqrt{6}=r^{2}$$
Rearranging to isolate $\sqrt{6}$:
$$\sqrt{6}=\dfrac{r^{2}-5}{2}$$
Since $r$ is assumed rational, $r^2$ is rational, so $r^2-5$ is rational, and dividing a rational number by $2$ gives a rational number. So the right-hand side, $\dfrac{r^2-5}{2}$, is rational, which means $\sqrt{6}$ is rational.
This contradicts the result of part (a), that $\sqrt{6}$ is irrational.
Therefore the assumption that $\sqrt{2}+\sqrt{3}$ is rational must be false, so $\sqrt{2}+\sqrt{3}$ is irrational. $\blacksquare$
QUESTION 44
10 marks
Hard
(a) Prove that every prime number $p>3$ can be written in the form $6k+1$ or $6k-1$, for some integer $k$. [4]
(b) Hence prove that $p^{2}-1$ is divisible by $24$ for every prime number $p>3$. [6]
Show complete worked solution
(a) Every integer can be written in one of the six forms $6k$, $6k+1$, $6k+2$, $6k+3$, $6k+4$ or $6k+5$, for some integer $k$, according to its remainder upon division by $6$.
Consider a prime number $p>3$. Since $p>3$ is prime, $p$ is not divisible by $2$ and not divisible by $3$.
$6k$, $6k+2$ and $6k+4$ are all even (each is $2$ times an integer), so $p$ cannot take these forms.
$6k+3=3(2k+1)$ is divisible by $3$, so $p$ cannot take this form either.
This leaves only $6k+1$ and $6k+5$. Since $6k+5=6(k+1)-1$, this form can be written as $6k'-1$ with $k'=k+1$.
Therefore every prime number $p>3$ must be of the form $6k+1$ or $6k-1$, for some integer $k$. $\blacksquare$
(b) From part (a), $p=6k\pm1$ for some integer $k$. Then
$$p^{2}=(6k\pm1)^{2}=36k^{2}\pm12k+1$$
So
$$p^{2}-1=36k^{2}\pm12k=12k(3k\pm1)$$
Now consider the factor $k(3k\pm1)$. If $k$ is even, then $k(3k\pm1)$ is even. If $k$ is odd, then $3k$ is odd, so $3k\pm1$ is even, and again $k(3k\pm1)$ is even.
In either case, $k(3k\pm1)$ is even, so $k(3k\pm1)=2m$ for some integer $m$.
Substituting back:
$$p^{2}-1=12\times2m=24m$$
Since $m$ is an integer, $p^2-1$ is $24$ times an integer.
Therefore $p^2-1$ is divisible by $24$ for every prime number $p>3$. $\blacksquare$
QUESTION 45
8 marks
Hard
Prove, by contradiction, that $\log_{2}3$ is irrational.
Show complete worked solution
Using proof by contradiction, suppose that $\log_{2}3$ is rational.
Since $3>1$ and the base $2>1$, $\log_2 3>0$, so it can be written as a ratio of two positive integers:
$$\log_{2}3=\dfrac{p}{q}$$
where $p$ and $q$ are positive integers.
By the definition of a logarithm, $\log_2 3=\dfrac{p}{q}$ means
$$2^{p/q}=3$$
Raising both sides to the power $q$:
$$2^{p}=3^{q}$$
Since $p$ is a positive integer, the left-hand side $2^p$ is an even number (it is $2$ multiplied by itself $p$ times, so it is divisible by $2$).
Since $q$ is a positive integer, the right-hand side $3^q$ is a product of odd numbers ($3\times3\times\cdots\times3$), so $3^q$ is odd.
So the equation $2^p=3^q$ says that an even number equals an odd number, which is impossible.
This is a contradiction.
Therefore the assumption that $\log_2 3$ is rational must be false, so $\log_2 3$ is irrational. $\blacksquare$
QUESTION 46
9 marks
Hard
Prove, by contradiction, that $\sqrt[3]{2}$ is irrational.
Show complete worked solution
Using proof by contradiction, suppose that $\sqrt[3]{2}$ is rational.
Then $\sqrt[3]{2}=\dfrac{p}{q}$, where $p$ and $q$ are integers with no common factor other than $1$, and $q\neq0$.
Cubing both sides:
$$2=\dfrac{p^{3}}{q^{3}}$$
$$p^{3}=2q^{3}$$
Since $q^3$ is an integer, $p^3$ is $2$ times an integer, so $p^3$ is even. Since the cube of an odd number is always odd (an odd number multiplied by itself any number of times remains odd), $p^3$ being even means $p$ itself must be even, so $p=2k$ for some integer $k$.
Substituting $p=2k$ into $p^3=2q^3$:
$$(2k)^{3}=2q^{3}$$
$$8k^{3}=2q^{3}$$
$$q^{3}=4k^{3}$$
Since $k^3$ is an integer, $q^3=2(2k^3)$ is $2$ times an integer, so $q^3$ is even, and by the same reasoning as above, $q$ must also be even.
But $p$ and $q$ are both even, so they share the common factor $2$.
This contradicts the assumption that $p$ and $q$ have no common factor other than $1$.
Therefore the assumption that $\sqrt[3]{2}$ is rational must be false, so $\sqrt[3]{2}$ is irrational. $\blacksquare$
QUESTION 47
8 marks
Hard
(a) Show that $n(n+1)(n+2)(n+3)+1=(n^{2}+3n+1)^{2}$ for all integers $n$. [4]
(b) Hence prove that the product of any four consecutive positive integers, increased by $1$, is always a perfect square. [4]
Show complete worked solution
(a) Starting from the left-hand side, group the four factors into two pairs and expand each pair:
$$\text{LHS}=n(n+1)(n+2)(n+3)+1=\big[n(n+3)\big]\big[(n+1)(n+2)\big]+1$$
Expanding each bracket:
$$n(n+3)=n^{2}+3n,\qquad(n+1)(n+2)=n^{2}+3n+2$$
Let $u=n^2+3n$, so the product becomes
$$\text{LHS}=u(u+2)+1=u^{2}+2u+1=(u+1)^{2}$$
Substituting back $u=n^2+3n$:
$$\text{LHS}=(n^{2}+3n+1)^{2}=\text{RHS}$$
Since $\text{LHS}=\text{RHS}$ for all integers $n$, the identity is proven. $\blacksquare$
(b) Let $n$, $n+1$, $n+2$, $n+3$ be any four consecutive positive integers. By part (a), their product increased by $1$ is
$$n(n+1)(n+2)(n+3)+1=(n^{2}+3n+1)^{2}$$
Since $n$ is a positive integer, $n^2+3n+1$ is also an integer, so the right-hand side is the square of an integer, i.e. a perfect square.
Therefore the product of any four consecutive positive integers, increased by $1$, is always a perfect square. $\blacksquare$
QUESTION 48
8 marks
Hard
Prove, by contradiction, that there do not exist positive integers $x$ and $y$, with $y\ge1$, such that $x^{2}-y^{2}=1$.
Show complete worked solution
Using proof by contradiction, suppose that there do exist positive integers $x$ and $y$, with $y\ge1$, such that
$$x^{2}-y^{2}=1$$
Factorising the left-hand side as a difference of two squares:
$$(x-y)(x+y)=1$$
Since $y\ge1$ and $x^{2}=y^{2}+1>y^{2}$, and $x$ is a positive integer, it follows that $x>y$, so $x-y$ is a positive integer, meaning $x-y\ge1$.
Also, since $x$ and $y$ are both positive integers, $x+y$ is a positive integer with $x+y\ge2$.
So $(x-y)$ and $(x+y)$ are positive integers with $(x-y)(x+y)=1$ and $x+y\ge2$. But the only pair of positive integers whose product is $1$ is $1\times1$, which would require $x+y=1$.
This contradicts $x+y\ge2$.
Therefore the assumption that such positive integers $x$ and $y$ exist must be false, so there do not exist positive integers $x$ and $y$, with $y\ge1$, satisfying $x^2-y^2=1$. $\blacksquare$
QUESTION 49
9 marks
Hard
Prove, by contradiction, that there is no integer $n$ for which $n^{2}+1$ is divisible by $4$.
Show complete worked solution
Using proof by contradiction, suppose that there exists an integer $n$ such that $n^{2}+1$ is divisible by $4$, i.e.
$$n^{2}+1=4m$$
for some integer $m$, so $n^2\equiv-1\equiv3\pmod{4}$.
Every integer $n$ is either even or odd.
Case 1: $n$ is even, so $n=2k$ for some integer $k$. Then
$$n^{2}=4k^{2}$$
so $n^2$ is a multiple of $4$, meaning $n^2\equiv0\pmod{4}$.
Case 2: $n$ is odd, so $n=2k+1$ for some integer $k$. Then
$$n^{2}=4k^{2}+4k+1=4(k^{2}+k)+1$$
so $n^2$ leaves remainder $1$ upon division by $4$, meaning $n^2\equiv1\pmod{4}$.
In either case, $n^2\equiv0\pmod{4}$ or $n^2\equiv1\pmod{4}$; there is no integer $n$ for which $n^2\equiv3\pmod{4}$.
This contradicts the assumption that $n^2\equiv3\pmod{4}$.
Therefore the assumption that such an integer $n$ exists must be false, so there is no integer $n$ for which $n^2+1$ is divisible by $4$. $\blacksquare$
QUESTION 50
10 marks
Hard
(a) Prove that if $n$ is an integer and $7\mid n^{2}$, then $7\mid n$. [5]
(b) Hence, using proof by contradiction, prove that $\sqrt{7}$ is irrational. [5]
Show complete worked solution
(a) Using proof by contrapositive, it is equivalent to prove that if $7\nmid n$, then $7\nmid n^{2}$.
If $7\nmid n$, then $n$ leaves a remainder of $1,2,3,4,5$ or $6$ upon division by $7$, so $n=7t+r$ for some integer $t$ and some $r\in\{1,2,3,4,5,6\}$.
Then
$$n^{2}=(7t+r)^{2}=49t^{2}+14tr+r^{2}=7(7t^{2}+2tr)+r^{2}$$
So $n^2$ leaves the same remainder as $r^2$ upon division by $7$. Checking each possible value of $r$:
$$r=1:r^{2}=1,\quad r=2:r^{2}=4,\quad r=3:r^{2}=9\equiv2,\quad r=4:r^{2}=16\equiv2,\quad r=5:r^{2}=25\equiv4,\quad r=6:r^{2}=36\equiv1\pmod{7}$$
In every case, $r^2\pmod{7}\in\{1,2,4\}$, never $0$. So $n^2$ is never divisible by $7$ when $n$ is not divisible by $7$.
By the contrapositive, if $7\mid n^2$, then $7\mid n$. $\blacksquare$
(b) Using proof by contradiction, suppose that $\sqrt{7}$ is rational.
Then $\sqrt{7}=\dfrac{p}{q}$, where $p$ and $q$ are integers with no common factor other than $1$, and $q\neq0$.
Squaring both sides:
$$7=\dfrac{p^{2}}{q^{2}}\quad\Rightarrow\quad p^{2}=7q^{2}$$
Since $q^2$ is an integer, $7\mid p^2$, and by part (a), $7\mid p$, so $p=7k$ for some integer $k$.
Substituting $p=7k$ into $p^2=7q^2$:
$$(7k)^{2}=7q^{2}\quad\Rightarrow\quad49k^{2}=7q^{2}\quad\Rightarrow\quad q^{2}=7k^{2}$$
Since $k^2$ is an integer, $7\mid q^2$, and by part (a) again, $7\mid q$.
But then $p$ and $q$ are both divisible by $7$, so they share the common factor $7$.
This contradicts the assumption that $p$ and $q$ have no common factor other than $1$.
Therefore the assumption that $\sqrt{7}$ is rational must be false, so $\sqrt{7}$ is irrational. $\blacksquare$
Proof by Induction: Series, Divisibility, and Inequalities 5 questions
QUESTION 1
4 marks
Easy
Prove by contradiction that if n^2 is even, then n is even, for n \in \mathbb{Z}.
Show complete worked solution
Using proof by contradiction, suppose that $n^2$ is even but $n$ is odd.
Since $n$ is odd, $n=2k+1$ for some integer $k$. Then:
$$n^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1$$
Since $2k^2+2k$ is an integer, $n^2=2(2k^2+2k)+1$ is odd.
This contradicts the assumption that $n^2$ is even.
Therefore the assumption that $n$ is odd must be false, so $n$ is even. $\blacksquare$
QUESTION 2
5 marks
Easy
Prove by mathematical induction that 5^n - 1 is divisible by 4, for all positive integers n.
Show complete worked solution
Let $P(n)$ be the proposition that $5^n-1$ is divisible by 4. Using proof by mathematical induction:
Base case ($n=1$):
$$5^1-1=4=4(1)$$
which is divisible by 4, so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$5^k-1=4m \text{ for some integer } m, \quad \text{so } 5^k=4m+1$$
Consider $n=k+1$:
$$5^{k+1}-1=5\cdot5^k-1=5(4m+1)-1=20m+5-1=20m+4=4(5m+1)$$
Since $5m+1$ is an integer, $5^{k+1}-1$ is divisible by 4, so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 3
7 marks
Medium
Prove by mathematical induction that for all positive integers n,
\sum_{r=1}^{n} r\cdot2^r = (n-1)2^{n+1}+2.
Show complete worked solution
Let $P(n)$ be the statement $\displaystyle\sum_{r=1}^{n} r\cdot2^r=(n-1)2^{n+1}+2$. Using proof by mathematical induction:
Base case ($n=1$):
$$\text{LHS}=1\cdot2^1=2, \qquad \text{RHS}=(1-1)2^{2}+2=0+2=2$$
$\text{LHS}=\text{RHS}$, so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\sum_{r=1}^{k} r\cdot2^r=(k-1)2^{k+1}+2$$
Consider $n=k+1$:
$$\sum_{r=1}^{k+1} r\cdot2^r=\sum_{r=1}^{k} r\cdot2^r+(k+1)2^{k+1}$$
Using the inductive hypothesis:
$$=\left[(k-1)2^{k+1}+2\right]+(k+1)2^{k+1}$$
$$=2^{k+1}\left[(k-1)+(k+1)\right]+2$$
$$=2^{k+1}(2k)+2$$
$$=k\cdot2^{k+2}+2$$
Now check this equals the right-hand side of $P(k+1)$, i.e. $\left((k+1)-1\right)2^{(k+1)+1}+2=k\cdot2^{k+2}+2$.
This matches exactly, so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 4
7 marks
Medium
Prove by contradiction that \sqrt3 is an irrational number. You may use the fact that if 3 divides p^2 (for an integer p), then 3 divides p.
Show complete worked solution
Using proof by contradiction, suppose that $\sqrt3$ is rational.
Then $\sqrt3$ can be written as $\sqrt3=\dfrac{p}{q}$, where $p,q\in\mathbb{Z}$, $q\neq0$, and $\dfrac{p}{q}$ is in lowest terms (i.e. $\gcd(p,q)=1$).
Squaring both sides:
$$3=\dfrac{p^2}{q^2} \ \Rightarrow\ p^2=3q^2$$
So $p^2$ is divisible by 3. By the given fact, this means $p$ is divisible by 3, so write $p=3m$ for some integer $m$.
Substituting back:
$$(3m)^2=3q^2$$
$$9m^2=3q^2$$
$$q^2=3m^2$$
So $q^2$ is also divisible by 3, which means (by the same fact) $q$ is divisible by 3.
But then both $p$ and $q$ are divisible by 3, so $\gcd(p,q)\ge3$, contradicting the assumption that $\dfrac{p}{q}$ is in lowest terms ($\gcd(p,q)=1$).
This contradiction shows the original assumption is false. Therefore $\sqrt3$ is irrational. $\blacksquare$
QUESTION 5
16 marks
Hard
(a) Prove by mathematical induction that \displaystyle\sum_{r=1}^{n} r^3 = \left(\dfrac{n(n+1)}{2}\right)^2 for all positive integers n.
(b) Prove by mathematical induction that 4^n > n^4 for all integers n \ge 5.
(c) A student conjectures that n^2-n+41 is a prime number for every positive integer n. Show that this conjecture is false by giving a counterexample.
Show complete worked solution
(a) Let $P(n)$ be the statement $\displaystyle\sum_{r=1}^{n} r^3=\left(\dfrac{n(n+1)}{2}\right)^2$. Using proof by mathematical induction:
Base case ($n=1$):
$$\text{LHS}=1^3=1, \qquad \text{RHS}=\left(\dfrac{1\cdot2}{2}\right)^2=1^2=1$$
So $P(1)$ is true.
Inductive step: assume $P(k)$ is true, i.e.
$$\sum_{r=1}^{k} r^3=\left(\dfrac{k(k+1)}{2}\right)^2$$
Consider $n=k+1$:
$$\sum_{r=1}^{k+1} r^3=\left(\dfrac{k(k+1)}{2}\right)^2+(k+1)^3=\dfrac{k^2(k+1)^2}{4}+(k+1)^3$$
Factor out $(k+1)^2$:
$$=(k+1)^2\left[\dfrac{k^2}{4}+(k+1)\right]=(k+1)^2\left[\dfrac{k^2+4k+4}{4}\right]=(k+1)^2\cdot\dfrac{(k+2)^2}{4}=\left(\dfrac{(k+1)(k+2)}{2}\right)^2$$
This is exactly the right-hand side of $P(k+1)$, so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
(b) Let $P(n)$ be the statement $4^n>n^4$. Using proof by mathematical induction:
Base case ($n=5$):
$$4^5=1024, \qquad 5^4=625$$
Since $1024>625$, $P(5)$ is true.
Inductive step: assume $P(k)$ is true for some integer $k\ge5$, i.e. $4^k>k^4$. We want to show $4^{k+1}>(k+1)^4$.
First, for $k\ge5$:
$$\dfrac{(k+1)^4}{k^4}=\left(1+\dfrac1k\right)^4 \le \left(1+\dfrac15\right)^4=\left(\dfrac65\right)^4=\dfrac{1296}{625}=2.0736<4$$
(since $1+1/k$ is a decreasing function of $k$, its maximum on $k\ge5$ occurs at $k=5$). So $(k+1)^4<4k^4$ for all $k\ge5$.
Now, using the inductive hypothesis:
$$4^{k+1}=4\cdot4^k>4k^4>(k+1)^4$$
So $4^{k+1}>(k+1)^4$, i.e. $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(5)$ is true, and $P(k)\Rightarrow P(k+1)$ for all integers $k\ge5$, by the principle of mathematical induction, $4^n>n^4$ for all integers $n\ge5$. $\blacksquare$
(c) Using proof by counterexample, testing $n=41$:
$$41^2-41+41=41^2=1681=41\times41$$
Since 1681 has 41 as a factor (other than 1 and itself), it is not prime.
So $n=41$ is a counterexample: $n^2-n+41=1681$, which is not prime. This disproves the conjecture. $\blacksquare$
Proof by Induction: Differentiation & Integration 50 questions
QUESTION 1
5 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(e^{4x}\right) = 4^n e^{4x}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(e^{4x}\right) = 4^n e^{4x}$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}\left(e^{4x}\right) = 4e^{4x} = 4^1e^{4x}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(e^{4x}\right) = 4^k e^{4x}$$
Consider $n=k+1$. Differentiating both sides of the assumption with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(e^{4x}\right) = \frac{d}{dx}\left(4^ke^{4x}\right) = 4^k\cdot4e^{4x} = 4^{k+1}e^{4x}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 2
6 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(e^{-2x}\right) = (-2)^n e^{-2x}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(e^{-2x}\right) = (-2)^n e^{-2x}$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}\left(e^{-2x}\right) = -2e^{-2x} = (-2)^1e^{-2x}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(e^{-2x}\right) = (-2)^k e^{-2x}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(e^{-2x}\right) = \frac{d}{dx}\left((-2)^ke^{-2x}\right) = (-2)^k\cdot(-2)e^{-2x} = (-2)^{k+1}e^{-2x}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 3
7 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}(\sin x) = \sin\left(x+\dfrac{n\pi}{2}\right)$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}(\sin x) = \sin\left(x+\dfrac{n\pi}{2}\right)$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}(\sin x) = \cos x$$
and
$$\sin\left(x+\frac{\pi}{2}\right) = \sin x\cos\frac{\pi}{2}+\cos x\sin\frac{\pi}{2} = \cos x$$
so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}(\sin x) = \sin\left(x+\frac{k\pi}{2}\right)$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}(\sin x) = \frac{d}{dx}\left[\sin\left(x+\frac{k\pi}{2}\right)\right] = \cos\left(x+\frac{k\pi}{2}\right)$$
Using $\cos\theta = \sin\left(\theta+\dfrac{\pi}{2}\right)$ with $\theta = x+\dfrac{k\pi}{2}$:
$$\cos\left(x+\frac{k\pi}{2}\right) = \sin\left(x+\frac{k\pi}{2}+\frac{\pi}{2}\right) = \sin\left(x+\frac{(k+1)\pi}{2}\right)$$
so $\dfrac{d^{k+1}}{dx^{k+1}}(\sin x) = \sin\left(x+\dfrac{(k+1)\pi}{2}\right)$, meaning $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 4
7 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}(\cos x) = \cos\left(x+\dfrac{n\pi}{2}\right)$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}(\cos x) = \cos\left(x+\dfrac{n\pi}{2}\right)$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}(\cos x) = -\sin x$$
and
$$\cos\left(x+\frac{\pi}{2}\right) = \cos x\cos\frac{\pi}{2}-\sin x\sin\frac{\pi}{2} = -\sin x$$
so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}(\cos x) = \cos\left(x+\frac{k\pi}{2}\right)$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}(\cos x) = \frac{d}{dx}\left[\cos\left(x+\frac{k\pi}{2}\right)\right] = -\sin\left(x+\frac{k\pi}{2}\right)$$
Using $-\sin\theta = \cos\left(\theta+\dfrac{\pi}{2}\right)$ with $\theta = x+\dfrac{k\pi}{2}$:
$$-\sin\left(x+\frac{k\pi}{2}\right) = \cos\left(x+\frac{k\pi}{2}+\frac{\pi}{2}\right) = \cos\left(x+\frac{(k+1)\pi}{2}\right)$$
so $\dfrac{d^{k+1}}{dx^{k+1}}(\cos x) = \cos\left(x+\dfrac{(k+1)\pi}{2}\right)$, meaning $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 5
8 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{1-x}\right) = \dfrac{n!}{(1-x)^{n+1}}$ for all $n \in \mathbb{Z}^+$, where $x \neq 1$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{1-x}\right) = \dfrac{n!}{(1-x)^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{d}{dx}\left[(1-x)^{-1}\right] = -(1-x)^{-2}\cdot(-1) = \frac{1}{(1-x)^2} = \frac{1!}{(1-x)^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{1}{1-x}\right) = k!\,(1-x)^{-(k+1)}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the chain rule:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{1-x}\right) = k!\cdot\left[-(k+1)\right](1-x)^{-(k+2)}\cdot(-1) = k!(k+1)(1-x)^{-(k+2)}$$
Since $k!(k+1) = (k+1)!$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{1-x}\right) = \frac{(k+1)!}{(1-x)^{k+2}} = \frac{(k+1)!}{(1-x)^{(k+1)+1}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq1$. $\blacksquare$
QUESTION 6
8 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}(\ln x) = \dfrac{(-1)^{n-1}(n-1)!}{x^n}$ for all $n \in \mathbb{Z}^+$, where $x>0$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}(\ln x) = \dfrac{(-1)^{n-1}(n-1)!}{x^n}$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}(\ln x) = \frac{1}{x}$$
and
$$\frac{(-1)^{0}(0)!}{x^1} = \frac{1}{x}$$
so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}(\ln x) = (-1)^{k-1}(k-1)!\,x^{-k}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}(\ln x) = (-1)^{k-1}(k-1)!\cdot(-k)x^{-k-1} = (-1)^{k}\,k!\,x^{-(k+1)}$$
since $(-1)^{k-1}\cdot(-k) = (-1)^k\cdot k$ and $k\cdot(k-1)! = k!$. This gives:
$$\frac{d^{k+1}}{dx^{k+1}}(\ln x) = \frac{(-1)^{(k+1)-1}\left((k+1)-1\right)!}{x^{k+1}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x>0$. $\blacksquare$
QUESTION 7
10 marks
Hard
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(xe^x\right) = (x+n)e^x$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(xe^x\right) = (x+n)e^x$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left(xe^x\right) = (1)e^x+xe^x = (x+1)e^x$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(xe^x\right) = (x+k)e^x$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the product rule on the right-hand side:
$$\frac{d^{k+1}}{dx^{k+1}}\left(xe^x\right) = \frac{d}{dx}\left[(x+k)e^x\right] = (1)e^x+(x+k)e^x = (x+k+1)e^x = \left(x+(k+1)\right)e^x$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 8
11 marks
Hard
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(x^2e^x\right) = \left(x^2+2nx+n(n-1)\right)e^x$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(x^2e^x\right) = \left(x^2+2nx+n(n-1)\right)e^x$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left(x^2e^x\right) = 2xe^x+x^2e^x = (x^2+2x)e^x$$
and the formula gives $\left(x^2+2(1)x+1(0)\right)e^x = (x^2+2x)e^x$, so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(x^2e^x\right) = \left(x^2+2kx+k(k-1)\right)e^x$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the product rule on the right-hand side:
$$\frac{d^{k+1}}{dx^{k+1}}\left(x^2e^x\right) = (2x+2k)e^x+\left(x^2+2kx+k(k-1)\right)e^x$$
$$= \left(x^2+2kx+2x+2k+k(k-1)\right)e^x = \left(x^2+2(k+1)x+\left[2k+k(k-1)\right]\right)e^x$$
Since $2k+k(k-1) = k(2+k-1) = k(k+1) = (k+1)\left((k+1)-1\right)$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(x^2e^x\right) = \left(x^2+2(k+1)x+(k+1)\left((k+1)-1\right)\right)e^x$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 9
12 marks
Hard
Let $I_n = \displaystyle\int_0^{\infty} x^n e^{-x}\,dx$ for integers $n\ge0$, given that $I_0=1$. Use integration by parts to show that $I_n = nI_{n-1}$ for $n\ge1$, and hence prove by mathematical induction that $I_n = n!$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
First, use integration by parts on $I_n = \displaystyle\int_0^{\infty} x^n e^{-x}\,dx$ for $n\ge1$, with $u=x^n$, $dv=e^{-x}dx$, so $du=nx^{n-1}dx$ and $v=-e^{-x}$:
$$I_n = \left[-x^ne^{-x}\right]_0^{\infty} + n\int_0^{\infty}x^{n-1}e^{-x}\,dx$$
As $x\to\infty$, $x^ne^{-x}\to0$ since exponential decay dominates any power of $x$, and at $x=0$ the term $x^ne^{-x}=0$ for $n\ge1$, so the boundary term vanishes:
$$I_n = n\int_0^{\infty}x^{n-1}e^{-x}\,dx = nI_{n-1}$$
Now let $P(n)$ be the proposition that $I_n = n!$. Using proof by mathematical induction:
Base case ($n=1$):
$$I_1 = 1\cdot I_0 = 1\cdot1 = 1 = 1!$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e. $I_k=k!$. Consider $n=k+1$. Using the recurrence relation established above:
$$I_{k+1} = (k+1)I_k = (k+1)\cdot k! = (k+1)!$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $I_n=n!$ for all positive integers $n$. $\blacksquare$
QUESTION 10
10 marks
Hard
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(xe^{-x}\right) = (-1)^n(x-n)e^{-x}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(xe^{-x}\right) = (-1)^n(x-n)e^{-x}$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left(xe^{-x}\right) = (1)e^{-x}+x\left(-e^{-x}\right) = (1-x)e^{-x} = -(x-1)e^{-x} = (-1)^1(x-1)e^{-x}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(xe^{-x}\right) = (-1)^k(x-k)e^{-x}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the product rule on the right-hand side:
$$\frac{d^{k+1}}{dx^{k+1}}\left(xe^{-x}\right) = (-1)^k\left[(1)e^{-x}+(x-k)\left(-e^{-x}\right)\right] = (-1)^k\left[1-(x-k)\right]e^{-x}$$
$$= (-1)^k\left[(k+1)-x\right]e^{-x} = (-1)^{k+1}\left(x-(k+1)\right)e^{-x}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 11
5 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(e^{5x}\right) = 5^n e^{5x}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(e^{5x}\right) = 5^n e^{5x}$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}\left(e^{5x}\right) = 5e^{5x} = 5^1e^{5x}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(e^{5x}\right) = 5^k e^{5x}$$
Consider $n=k+1$. Differentiating both sides of the assumption with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(e^{5x}\right) = \frac{d}{dx}\left(5^ke^{5x}\right) = 5^k\cdot5e^{5x} = 5^{k+1}e^{5x}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 12
6 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(e^{-x/3}\right) = \left(-\dfrac13\right)^n e^{-x/3}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(e^{-x/3}\right) = \left(-\dfrac13\right)^n e^{-x/3}$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}\left(e^{-x/3}\right) = -\frac13e^{-x/3} = \left(-\frac13\right)^1e^{-x/3}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(e^{-x/3}\right) = \left(-\frac13\right)^k e^{-x/3}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(e^{-x/3}\right) = \frac{d}{dx}\left[\left(-\frac13\right)^ke^{-x/3}\right] = \left(-\frac13\right)^k\cdot\left(-\frac13\right)e^{-x/3} = \left(-\frac13\right)^{k+1}e^{-x/3}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 13
5 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(e^{2x+3}\right) = 2^n e^{2x+3}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(e^{2x+3}\right) = 2^n e^{2x+3}$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(e^{2x+3}\right) = 2e^{2x+3} = 2^1e^{2x+3}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(e^{2x+3}\right) = 2^k e^{2x+3}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(e^{2x+3}\right) = \frac{d}{dx}\left(2^ke^{2x+3}\right) = 2^k\cdot2e^{2x+3} = 2^{k+1}e^{2x+3}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 14
6 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\sin 3x\right) = 3^n\sin\left(3x+\dfrac{n\pi}{2}\right)$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\sin 3x\right) = 3^n\sin\left(3x+\dfrac{n\pi}{2}\right)$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(\sin 3x\right) = 3\cos 3x$$
and
$$3^1\sin\left(3x+\frac{\pi}{2}\right) = 3\left[\sin 3x\cos\frac{\pi}{2}+\cos 3x\sin\frac{\pi}{2}\right] = 3\cos 3x$$
so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\sin 3x\right) = 3^k\sin\left(3x+\frac{k\pi}{2}\right)$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\sin 3x\right) = \frac{d}{dx}\left[3^k\sin\left(3x+\frac{k\pi}{2}\right)\right] = 3^k\cdot3\cos\left(3x+\frac{k\pi}{2}\right) = 3^{k+1}\cos\left(3x+\frac{k\pi}{2}\right)$$
Using $\cos\theta = \sin\left(\theta+\dfrac{\pi}{2}\right)$ with $\theta = 3x+\dfrac{k\pi}{2}$:
$$3^{k+1}\cos\left(3x+\frac{k\pi}{2}\right) = 3^{k+1}\sin\left(3x+\frac{k\pi}{2}+\frac{\pi}{2}\right) = 3^{k+1}\sin\left(3x+\frac{(k+1)\pi}{2}\right)$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 15
6 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\cos 4x\right) = 4^n\cos\left(4x+\dfrac{n\pi}{2}\right)$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\cos 4x\right) = 4^n\cos\left(4x+\dfrac{n\pi}{2}\right)$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(\cos 4x\right) = -4\sin 4x$$
and
$$4^1\cos\left(4x+\frac{\pi}{2}\right) = 4\left[\cos 4x\cos\frac{\pi}{2}-\sin 4x\sin\frac{\pi}{2}\right] = -4\sin 4x$$
so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\cos 4x\right) = 4^k\cos\left(4x+\frac{k\pi}{2}\right)$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\cos 4x\right) = \frac{d}{dx}\left[4^k\cos\left(4x+\frac{k\pi}{2}\right)\right] = -4^k\cdot4\sin\left(4x+\frac{k\pi}{2}\right) = -4^{k+1}\sin\left(4x+\frac{k\pi}{2}\right)$$
Using $-\sin\theta = \cos\left(\theta+\dfrac{\pi}{2}\right)$ with $\theta = 4x+\dfrac{k\pi}{2}$:
$$-4^{k+1}\sin\left(4x+\frac{k\pi}{2}\right) = 4^{k+1}\cos\left(4x+\frac{k\pi}{2}+\frac{\pi}{2}\right) = 4^{k+1}\cos\left(4x+\frac{(k+1)\pi}{2}\right)$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 16
6 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{x+2}\right) = \dfrac{(-1)^n n!}{(x+2)^{n+1}}$ for all $n \in \mathbb{Z}^+$, where $x\neq-2$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{x+2}\right) = \dfrac{(-1)^n n!}{(x+2)^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}\left(\frac{1}{x+2}\right) = \frac{d}{dx}\left[(x+2)^{-1}\right] = -(x+2)^{-2} = \frac{(-1)^1\,1!}{(x+2)^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{1}{x+2}\right) = (-1)^k k!\,(x+2)^{-(k+1)}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{x+2}\right) = (-1)^k k!\cdot\left[-(k+1)\right](x+2)^{-(k+2)} = (-1)^{k+1}(k+1)!\,(x+2)^{-(k+2)}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq-2$. $\blacksquare$
QUESTION 17
6 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{3-x}\right) = \dfrac{n!}{(3-x)^{n+1}}$ for all $n \in \mathbb{Z}^+$, where $x\neq3$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{3-x}\right) = \dfrac{n!}{(3-x)^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}\left(\frac{1}{3-x}\right) = \frac{d}{dx}\left[(3-x)^{-1}\right] = -(3-x)^{-2}\cdot(-1) = \frac{1}{(3-x)^{2}} = \frac{1!}{(3-x)^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{1}{3-x}\right) = k!\,(3-x)^{-(k+1)}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the chain rule:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{3-x}\right) = k!\cdot\left[-(k+1)\right](3-x)^{-(k+2)}\cdot(-1) = k!(k+1)(3-x)^{-(k+2)} = (k+1)!\,(3-x)^{-(k+2)}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq3$. $\blacksquare$
QUESTION 18
7 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{2x+1}\right) = \dfrac{(-1)^n 2^n n!}{(2x+1)^{n+1}}$ for all $n \in \mathbb{Z}^+$, where $x\neq-\dfrac12$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{2x+1}\right) = \dfrac{(-1)^n 2^n n!}{(2x+1)^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(\frac{1}{2x+1}\right) = -2(2x+1)^{-2} = \frac{(-1)^1\,2^1\,1!}{(2x+1)^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{1}{2x+1}\right) = (-1)^k 2^k k!\,(2x+1)^{-(k+1)}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the chain rule:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{2x+1}\right) = (-1)^k 2^k k!\cdot\left[-(k+1)\right](2x+1)^{-(k+2)}\cdot2 = (-1)^{k+1}2^{k+1}(k+1)!\,(2x+1)^{-(k+2)}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq-\dfrac12$. $\blacksquare$
QUESTION 19
5 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{x}\right) = \dfrac{(-1)^n n!}{x^{n+1}}$ for all $n \in \mathbb{Z}^+$, where $x\neq0$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{x}\right) = \dfrac{(-1)^n n!}{x^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} = \frac{(-1)^1\,1!}{x^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{1}{x}\right) = (-1)^k k!\,x^{-(k+1)}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{x}\right) = (-1)^k k!\cdot\left[-(k+1)\right]x^{-(k+2)} = (-1)^{k+1}(k+1)!\,x^{-(k+2)}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq0$. $\blacksquare$
QUESTION 20
5 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\ln 3x\right) = \dfrac{(-1)^{n-1}(n-1)!}{x^n}$ for all $n \in \mathbb{Z}^+$, where $x>0$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\ln 3x\right) = \dfrac{(-1)^{n-1}(n-1)!}{x^n}$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(\ln 3x\right) = \frac{1}{3x}\cdot3 = \frac{1}{x}$$
and
$$\frac{(-1)^{0}(0)!}{x^1} = \frac{1}{x}$$
so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\ln 3x\right) = (-1)^{k-1}(k-1)!\,x^{-k}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\ln 3x\right) = (-1)^{k-1}(k-1)!\cdot(-k)x^{-k-1} = (-1)^{k}\,k!\,x^{-(k+1)}$$
since $(-1)^{k-1}\cdot(-k) = (-1)^k\cdot k$ and $k\cdot(k-1)! = k!$. This gives:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\ln 3x\right) = \frac{(-1)^{(k+1)-1}\left((k+1)-1\right)!}{x^{k+1}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x>0$. $\blacksquare$
QUESTION 21
6 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\ln(x+4)\right) = \dfrac{(-1)^{n-1}(n-1)!}{(x+4)^n}$ for all $n \in \mathbb{Z}^+$, where $x>-4$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\ln(x+4)\right) = \dfrac{(-1)^{n-1}(n-1)!}{(x+4)^n}$. Using proof by mathematical induction:
Base case ($n=1$):
$$\frac{d}{dx}\left(\ln(x+4)\right) = \frac{1}{x+4}$$
and
$$\frac{(-1)^{0}(0)!}{(x+4)^1} = \frac{1}{x+4}$$
so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\ln(x+4)\right) = (-1)^{k-1}(k-1)!\,(x+4)^{-k}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\ln(x+4)\right) = (-1)^{k-1}(k-1)!\cdot(-k)(x+4)^{-k-1} = (-1)^{k}\,k!\,(x+4)^{-(k+1)}$$
since $(-1)^{k-1}\cdot(-k) = (-1)^k\cdot k$ and $k\cdot(k-1)! = k!$. This gives:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\ln(x+4)\right) = \frac{(-1)^{(k+1)-1}\left((k+1)-1\right)!}{(x+4)^{k+1}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x>-4$. $\blacksquare$
QUESTION 22
7 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{(x-1)^2}\right) = \dfrac{(-1)^n(n+1)!}{(x-1)^{n+2}}$ for all $n \in \mathbb{Z}^+$, where $x\neq1$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{(x-1)^2}\right) = \dfrac{(-1)^n(n+1)!}{(x-1)^{n+2}}$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(\frac{1}{(x-1)^2}\right) = -2(x-1)^{-3} = \frac{(-1)^1\,2!}{(x-1)^{3}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{1}{(x-1)^2}\right) = (-1)^k(k+1)!\,(x-1)^{-(k+2)}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{(x-1)^2}\right) = (-1)^k(k+1)!\cdot\left[-(k+2)\right](x-1)^{-(k+3)} = (-1)^{k+1}(k+1)!(k+2)\,(x-1)^{-(k+3)}$$
Since $(k+1)!(k+2) = (k+2)!$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{(x-1)^2}\right) = \frac{(-1)^{k+1}(k+2)!}{(x-1)^{k+3}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq1$. $\blacksquare$
QUESTION 23
7 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{x}{1-x}\right) = \dfrac{n!}{(1-x)^{n+1}}$ for all $n \in \mathbb{Z}^+$, where $x\neq1$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{x}{1-x}\right) = \dfrac{n!}{(1-x)^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$): using the quotient rule,
$$\frac{d}{dx}\left(\frac{x}{1-x}\right) = \frac{(1)(1-x)-x(-1)}{(1-x)^2} = \frac{(1-x)+x}{(1-x)^2} = \frac{1}{(1-x)^2} = \frac{1!}{(1-x)^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{x}{1-x}\right) = k!\,(1-x)^{-(k+1)}$$
Consider $n=k+1$. For $k\ge1$ the right-hand side no longer contains $x$ except inside the power of $(1-x)$, so differentiating both sides with respect to $x$ using the chain rule:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{x}{1-x}\right) = k!\cdot\left[-(k+1)\right](1-x)^{-(k+2)}\cdot(-1) = k!(k+1)(1-x)^{-(k+2)} = (k+1)!\,(1-x)^{-(k+2)}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq1$. $\blacksquare$
QUESTION 24
6 marks
Easy
Let $I_n = \displaystyle\int_0^{\infty} x^n e^{-2x}\,dx$ for integers $n\ge0$, given that $I_0=\dfrac12$. Use integration by parts to show that $I_n = \dfrac{n}{2}I_{n-1}$ for $n\ge1$, and hence prove by mathematical induction that $I_n = \dfrac{n!}{2^{n+1}}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
First, use integration by parts on $I_n = \displaystyle\int_0^{\infty} x^n e^{-2x}\,dx$ for $n\ge1$, with $u=x^n$, $dv=e^{-2x}dx$, so $du=nx^{n-1}dx$ and $v=-\dfrac12e^{-2x}$:
$$I_n = \left[-\frac12x^ne^{-2x}\right]_0^{\infty} + \frac{n}{2}\int_0^{\infty}x^{n-1}e^{-2x}\,dx$$
As $x\to\infty$, $x^ne^{-2x}\to0$ since exponential decay dominates any power of $x$, and at $x=0$ the term $x^ne^{-2x}=0$ for $n\ge1$, so the boundary term vanishes:
$$I_n = \frac{n}{2}\int_0^{\infty}x^{n-1}e^{-2x}\,dx = \frac{n}{2}I_{n-1}$$
Now let $P(n)$ be the proposition that $I_n = \dfrac{n!}{2^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$):
$$I_1 = \frac12\cdot I_0 = \frac12\cdot\frac12 = \frac14 = \frac{1!}{2^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e. $I_k=\dfrac{k!}{2^{k+1}}$. Consider $n=k+1$. Using the recurrence relation established above:
$$I_{k+1} = \frac{k+1}{2}I_k = \frac{k+1}{2}\cdot\frac{k!}{2^{k+1}} = \frac{(k+1)!}{2^{k+2}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $I_n=\dfrac{n!}{2^{n+1}}$ for all positive integers $n$. $\blacksquare$
QUESTION 25
6 marks
Easy
Let $J_n = \displaystyle\int_0^{\infty} x^n e^{-3x}\,dx$ for integers $n\ge0$, given that $J_0=\dfrac13$. Use integration by parts to show that $J_n = \dfrac{n}{3}J_{n-1}$ for $n\ge1$, and hence prove by mathematical induction that $J_n = \dfrac{n!}{3^{n+1}}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
First, use integration by parts on $J_n = \displaystyle\int_0^{\infty} x^n e^{-3x}\,dx$ for $n\ge1$, with $u=x^n$, $dv=e^{-3x}dx$, so $du=nx^{n-1}dx$ and $v=-\dfrac13e^{-3x}$:
$$J_n = \left[-\frac13x^ne^{-3x}\right]_0^{\infty} + \frac{n}{3}\int_0^{\infty}x^{n-1}e^{-3x}\,dx$$
As $x\to\infty$, $x^ne^{-3x}\to0$ since exponential decay dominates any power of $x$, and at $x=0$ the term $x^ne^{-3x}=0$ for $n\ge1$, so the boundary term vanishes:
$$J_n = \frac{n}{3}\int_0^{\infty}x^{n-1}e^{-3x}\,dx = \frac{n}{3}J_{n-1}$$
Now let $P(n)$ be the proposition that $J_n = \dfrac{n!}{3^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$):
$$J_1 = \frac13\cdot J_0 = \frac13\cdot\frac13 = \frac19 = \frac{1!}{3^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e. $J_k=\dfrac{k!}{3^{k+1}}$. Consider $n=k+1$. Using the recurrence relation established above:
$$J_{k+1} = \frac{k+1}{3}J_k = \frac{k+1}{3}\cdot\frac{k!}{3^{k+1}} = \frac{(k+1)!}{3^{k+2}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $J_n=\dfrac{n!}{3^{n+1}}$ for all positive integers $n$. $\blacksquare$
QUESTION 26
6 marks
Easy
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\sin\dfrac{x}{2}\right) = \left(\dfrac12\right)^n\sin\left(\dfrac{x}{2}+\dfrac{n\pi}{2}\right)$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\sin\dfrac{x}{2}\right) = \left(\dfrac12\right)^n\sin\left(\dfrac{x}{2}+\dfrac{n\pi}{2}\right)$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(\sin\frac{x}{2}\right) = \frac12\cos\frac{x}{2}$$
and
$$\left(\frac12\right)^1\sin\left(\frac{x}{2}+\frac{\pi}{2}\right) = \frac12\left[\sin\frac{x}{2}\cos\frac{\pi}{2}+\cos\frac{x}{2}\sin\frac{\pi}{2}\right] = \frac12\cos\frac{x}{2}$$
so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\sin\frac{x}{2}\right) = \left(\frac12\right)^k\sin\left(\frac{x}{2}+\frac{k\pi}{2}\right)$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\sin\frac{x}{2}\right) = \left(\frac12\right)^k\cdot\frac12\cos\left(\frac{x}{2}+\frac{k\pi}{2}\right) = \left(\frac12\right)^{k+1}\cos\left(\frac{x}{2}+\frac{k\pi}{2}\right)$$
Using $\cos\theta = \sin\left(\theta+\dfrac{\pi}{2}\right)$ with $\theta = \dfrac{x}{2}+\dfrac{k\pi}{2}$:
$$\left(\frac12\right)^{k+1}\cos\left(\frac{x}{2}+\frac{k\pi}{2}\right) = \left(\frac12\right)^{k+1}\sin\left(\frac{x}{2}+\frac{k\pi}{2}+\frac{\pi}{2}\right) = \left(\frac12\right)^{k+1}\sin\left(\frac{x}{2}+\frac{(k+1)\pi}{2}\right)$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 27
8 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(xe^{2x}\right) = 2^{n-1}(2x+n)e^{2x}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(xe^{2x}\right) = 2^{n-1}(2x+n)e^{2x}$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left(xe^{2x}\right) = e^{2x}+2xe^{2x} = (2x+1)e^{2x}$$
and the formula gives $2^{0}(2x+1)e^{2x} = (2x+1)e^{2x}$, so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(xe^{2x}\right) = 2^{k-1}(2x+k)e^{2x}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the product rule on the right-hand side:
$$\frac{d^{k+1}}{dx^{k+1}}\left(xe^{2x}\right) = 2^{k-1}\left[2e^{2x}+(2x+k)\cdot2e^{2x}\right] = 2^{k-1}\cdot2\left[1+(2x+k)\right]e^{2x} = 2^{k}(2x+k+1)e^{2x}$$
Since $k+1=(k+1)$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(xe^{2x}\right) = 2^{(k+1)-1}\left(2x+(k+1)\right)e^{2x}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 28
8 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(xe^{-3x}\right) = (-3)^{n-1}(n-3x)e^{-3x}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(xe^{-3x}\right) = (-3)^{n-1}(n-3x)e^{-3x}$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left(xe^{-3x}\right) = e^{-3x}-3xe^{-3x} = (1-3x)e^{-3x}$$
and the formula gives $(-3)^{0}(1-3x)e^{-3x} = (1-3x)e^{-3x}$, so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(xe^{-3x}\right) = (-3)^{k-1}(k-3x)e^{-3x}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the product rule on the right-hand side:
$$\frac{d^{k+1}}{dx^{k+1}}\left(xe^{-3x}\right) = (-3)^{k-1}\left[-3e^{-3x}+(k-3x)(-3)e^{-3x}\right] = (-3)^{k-1}\cdot(-3)\left[1+(k-3x)\right]e^{-3x} = (-3)^{k}(k+1-3x)e^{-3x}$$
so:
$$\frac{d^{k+1}}{dx^{k+1}}\left(xe^{-3x}\right) = (-3)^{(k+1)-1}\left((k+1)-3x\right)e^{-3x}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 29
8 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{(x+1)^3}\right) = \dfrac{(-1)^n(n+2)!}{2(x+1)^{n+3}}$ for all $n \in \mathbb{Z}^+$, where $x\neq-1$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{(x+1)^3}\right) = \dfrac{(-1)^n(n+2)!}{2(x+1)^{n+3}}$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(\frac{1}{(x+1)^3}\right) = -3(x+1)^{-4} = \frac{(-1)^1\,3!}{2(x+1)^{4}}$$
since $3!/2=3$, so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{1}{(x+1)^3}\right) = \frac{(-1)^k(k+2)!}{2}\,(x+1)^{-(k+3)}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{(x+1)^3}\right) = \frac{(-1)^k(k+2)!}{2}\cdot\left[-(k+3)\right](x+1)^{-(k+4)} = \frac{(-1)^{k+1}(k+2)!(k+3)}{2}\,(x+1)^{-(k+4)}$$
Since $(k+2)!(k+3) = (k+3)!$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{(x+1)^3}\right) = \frac{(-1)^{k+1}(k+3)!}{2(x+1)^{k+4}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq-1$. $\blacksquare$
QUESTION 30
8 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{(2-x)^2}\right) = \dfrac{(n+1)!}{(2-x)^{n+2}}$ for all $n \in \mathbb{Z}^+$, where $x\neq2$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{(2-x)^2}\right) = \dfrac{(n+1)!}{(2-x)^{n+2}}$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(\frac{1}{(2-x)^2}\right) = -2(2-x)^{-3}\cdot(-1) = \frac{2}{(2-x)^{3}} = \frac{2!}{(2-x)^{3}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{1}{(2-x)^2}\right) = (k+1)!\,(2-x)^{-(k+2)}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the chain rule:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{(2-x)^2}\right) = (k+1)!\cdot\left[-(k+2)\right](2-x)^{-(k+3)}\cdot(-1) = (k+1)!(k+2)\,(2-x)^{-(k+3)}$$
Since $(k+1)!(k+2) = (k+2)!$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{(2-x)^2}\right) = \frac{(k+2)!}{(2-x)^{k+3}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq2$. $\blacksquare$
QUESTION 31
7 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\sin(2x+1)\right) = 2^n\sin\left(2x+1+\dfrac{n\pi}{2}\right)$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\sin(2x+1)\right) = 2^n\sin\left(2x+1+\dfrac{n\pi}{2}\right)$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(\sin(2x+1)\right) = 2\cos(2x+1)$$
and
$$2^1\sin\left(2x+1+\frac{\pi}{2}\right) = 2\left[\sin(2x+1)\cos\frac{\pi}{2}+\cos(2x+1)\sin\frac{\pi}{2}\right] = 2\cos(2x+1)$$
so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\sin(2x+1)\right) = 2^k\sin\left(2x+1+\frac{k\pi}{2}\right)$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\sin(2x+1)\right) = 2^k\cdot2\cos\left(2x+1+\frac{k\pi}{2}\right) = 2^{k+1}\cos\left(2x+1+\frac{k\pi}{2}\right)$$
Using $\cos\theta = \sin\left(\theta+\dfrac{\pi}{2}\right)$ with $\theta = 2x+1+\dfrac{k\pi}{2}$:
$$2^{k+1}\cos\left(2x+1+\frac{k\pi}{2}\right) = 2^{k+1}\sin\left(2x+1+\frac{k\pi}{2}+\frac{\pi}{2}\right) = 2^{k+1}\sin\left(2x+1+\frac{(k+1)\pi}{2}\right)$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 32
7 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\cos(3x-2)\right) = 3^n\cos\left(3x-2+\dfrac{n\pi}{2}\right)$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\cos(3x-2)\right) = 3^n\cos\left(3x-2+\dfrac{n\pi}{2}\right)$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(\cos(3x-2)\right) = -3\sin(3x-2)$$
and
$$3^1\cos\left(3x-2+\frac{\pi}{2}\right) = 3\left[\cos(3x-2)\cos\frac{\pi}{2}-\sin(3x-2)\sin\frac{\pi}{2}\right] = -3\sin(3x-2)$$
so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\cos(3x-2)\right) = 3^k\cos\left(3x-2+\frac{k\pi}{2}\right)$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\cos(3x-2)\right) = -3^k\cdot3\sin\left(3x-2+\frac{k\pi}{2}\right) = -3^{k+1}\sin\left(3x-2+\frac{k\pi}{2}\right)$$
Using $-\sin\theta = \cos\left(\theta+\dfrac{\pi}{2}\right)$ with $\theta = 3x-2+\dfrac{k\pi}{2}$:
$$-3^{k+1}\sin\left(3x-2+\frac{k\pi}{2}\right) = 3^{k+1}\cos\left(3x-2+\frac{k\pi}{2}+\frac{\pi}{2}\right) = 3^{k+1}\cos\left(3x-2+\frac{(k+1)\pi}{2}\right)$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 33
8 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{x}{x+1}\right) = \dfrac{(-1)^{n+1}n!}{(x+1)^{n+1}}$ for all $n \in \mathbb{Z}^+$, where $x\neq-1$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{x}{x+1}\right) = \dfrac{(-1)^{n+1}n!}{(x+1)^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$): using the quotient rule,
$$\frac{d}{dx}\left(\frac{x}{x+1}\right) = \frac{(1)(x+1)-x(1)}{(x+1)^2} = \frac{1}{(x+1)^2} = \frac{(-1)^{2}\,1!}{(x+1)^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{x}{x+1}\right) = (-1)^{k+1}k!\,(x+1)^{-(k+1)}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{x}{x+1}\right) = (-1)^{k+1}k!\cdot\left[-(k+1)\right](x+1)^{-(k+2)} = (-1)^{k+2}(k+1)!\,(x+1)^{-(k+2)}$$
so:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{x}{x+1}\right) = \frac{(-1)^{(k+1)+1}(k+1)!}{(x+1)^{(k+1)+1}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq-1$. $\blacksquare$
QUESTION 34
9 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{x^2-1}\right) = \dfrac{(-1)^n n!}{2}\left[\dfrac{1}{(x-1)^{n+1}}-\dfrac{1}{(x+1)^{n+1}}\right]$ for all $n \in \mathbb{Z}^+$, where $x\neq\pm1$.
Show complete worked solution
First write $\dfrac{1}{x^2-1}$ in partial fractions:
$$\frac{1}{x^2-1} = \frac{1}{(x-1)(x+1)} = \frac12\left[\frac{1}{x-1}-\frac{1}{x+1}\right]$$
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{x^2-1}\right) = \dfrac{(-1)^n n!}{2}\left[\dfrac{1}{(x-1)^{n+1}}-\dfrac{1}{(x+1)^{n+1}}\right]$. Using proof by mathematical induction:
Base case ($n=1$): differentiating the partial fraction form,
$$\frac{d}{dx}\left(\frac{1}{x^2-1}\right) = \frac12\left[-\frac{1}{(x-1)^2}+\frac{1}{(x+1)^2}\right] = \frac{(-1)^1\,1!}{2}\left[\frac{1}{(x-1)^2}-\frac{1}{(x+1)^2}\right]$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{1}{x^2-1}\right) = \frac{(-1)^k k!}{2}\left[(x-1)^{-(k+1)}-(x+1)^{-(k+1)}\right]$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{x^2-1}\right) = \frac{(-1)^k k!}{2}\left[-(k+1)(x-1)^{-(k+2)}+(k+1)(x+1)^{-(k+2)}\right]$$
$$= \frac{(-1)^k k!(k+1)}{2}\left[-(x-1)^{-(k+2)}+(x+1)^{-(k+2)}\right] = \frac{(-1)^{k+1}(k+1)!}{2}\left[(x-1)^{-(k+2)}-(x+1)^{-(k+2)}\right]$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq\pm1$. $\blacksquare$
QUESTION 35
8 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left((2x-1)e^x\right) = (2x-1+2n)e^x$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left((2x-1)e^x\right) = (2x-1+2n)e^x$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left((2x-1)e^x\right) = 2e^x+(2x-1)e^x = (2x+1)e^x$$
and the formula gives $(2x-1+2)e^x = (2x+1)e^x$, so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left((2x-1)e^x\right) = (2x-1+2k)e^x$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the product rule on the right-hand side:
$$\frac{d^{k+1}}{dx^{k+1}}\left((2x-1)e^x\right) = 2e^x+(2x-1+2k)e^x = (2x+1+2k)e^x = \left(2x-1+2(k+1)\right)e^x$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 36
8 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left((3x+2)e^{-x}\right) = (-1)^n(3x+2-3n)e^{-x}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left((3x+2)e^{-x}\right) = (-1)^n(3x+2-3n)e^{-x}$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left((3x+2)e^{-x}\right) = 3e^{-x}-(3x+2)e^{-x} = (1-3x)e^{-x} = -(3x-1)e^{-x} = (-1)^1(3x+2-3)e^{-x}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left((3x+2)e^{-x}\right) = (-1)^k(3x+2-3k)e^{-x}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the product rule on the right-hand side:
$$\frac{d^{k+1}}{dx^{k+1}}\left((3x+2)e^{-x}\right) = (-1)^k\left[3e^{-x}+(3x+2-3k)(-1)e^{-x}\right] = (-1)^k\left[3-(3x+2-3k)\right]e^{-x}$$
$$= (-1)^k\left(3k+1-3x\right)e^{-x} = (-1)^{k+1}\left(3x-3k-1\right)e^{-x} = (-1)^{k+1}\left(3x+2-3(k+1)\right)e^{-x}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 37
8 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{x+3}{x+1}\right) = \dfrac{2(-1)^n n!}{(x+1)^{n+1}}$ for all $n \in \mathbb{Z}^+$, where $x\neq-1$.
Show complete worked solution
First write $\dfrac{x+3}{x+1} = 1+\dfrac{2}{x+1}$, since $(x+1)+2=x+3$.
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{x+3}{x+1}\right) = \dfrac{2(-1)^n n!}{(x+1)^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$): using the quotient rule,
$$\frac{d}{dx}\left(\frac{x+3}{x+1}\right) = \frac{(1)(x+1)-(x+3)(1)}{(x+1)^2} = \frac{-2}{(x+1)^2} = \frac{2(-1)^1\,1!}{(x+1)^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{x+3}{x+1}\right) = 2(-1)^k k!\,(x+1)^{-(k+1)}$$
Consider $n=k+1$. Since the derivative of the constant term $1$ vanishes, differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{x+3}{x+1}\right) = 2(-1)^k k!\cdot\left[-(k+1)\right](x+1)^{-(k+2)} = 2(-1)^{k+1}(k+1)!\,(x+1)^{-(k+2)}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq-1$. $\blacksquare$
QUESTION 38
8 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{2x-1}{x-2}\right) = \dfrac{3(-1)^n n!}{(x-2)^{n+1}}$ for all $n \in \mathbb{Z}^+$, where $x\neq2$.
Show complete worked solution
First write $\dfrac{2x-1}{x-2} = 2+\dfrac{3}{x-2}$, since $2(x-2)+3=2x-1$.
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{2x-1}{x-2}\right) = \dfrac{3(-1)^n n!}{(x-2)^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$): using the quotient rule,
$$\frac{d}{dx}\left(\frac{2x-1}{x-2}\right) = \frac{(2)(x-2)-(2x-1)(1)}{(x-2)^2} = \frac{-3}{(x-2)^2} = \frac{3(-1)^1\,1!}{(x-2)^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{2x-1}{x-2}\right) = 3(-1)^k k!\,(x-2)^{-(k+1)}$$
Consider $n=k+1$. Since the derivative of the constant term $2$ vanishes, differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{2x-1}{x-2}\right) = 3(-1)^k k!\cdot\left[-(k+1)\right](x-2)^{-(k+2)} = 3(-1)^{k+1}(k+1)!\,(x-2)^{-(k+2)}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq2$. $\blacksquare$
QUESTION 39
9 marks
Medium
Let $D_n = \displaystyle\int_0^{1} (\ln x)^n\,dx$ for integers $n\ge0$, given that $D_0=1$. Use integration by parts to show that $D_n = -nD_{n-1}$ for $n\ge1$, and hence prove by mathematical induction that $D_n = (-1)^n n!$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
First, use integration by parts on $D_n = \displaystyle\int_0^{1} (\ln x)^n\,dx$ for $n\ge1$, with $u=(\ln x)^n$, $dv=dx$, so $du=n(\ln x)^{n-1}\cdot\dfrac1x\,dx$ and $v=x$:
$$D_n = \left[x(\ln x)^n\right]_0^{1} - \int_0^{1}x\cdot n(\ln x)^{n-1}\cdot\frac1x\,dx = \left[x(\ln x)^n\right]_0^{1} - n\int_0^{1}(\ln x)^{n-1}\,dx$$
At $x=1$, $(\ln1)^n=0$; as $x\to0^+$, $x(\ln x)^n\to0$ since $x$ dominates any power of $\ln x$. So the boundary term vanishes:
$$D_n = -n\int_0^{1}(\ln x)^{n-1}\,dx = -nD_{n-1}$$
Now let $P(n)$ be the proposition that $D_n = (-1)^n n!$. Using proof by mathematical induction:
Base case ($n=1$):
$$D_1 = -1\cdot D_0 = -1\cdot1 = -1 = (-1)^1\,1!$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e. $D_k=(-1)^k k!$. Consider $n=k+1$. Using the recurrence relation established above:
$$D_{k+1} = -(k+1)D_k = -(k+1)\cdot(-1)^k k! = (-1)^{k+1}(k+1)!$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $D_n=(-1)^n n!$ for all positive integers $n$. $\blacksquare$
QUESTION 40
8 marks
Medium
Let $\lambda>0$ be a constant, and let $I_n = \displaystyle\int_0^{\infty} x^n e^{-\lambda x}\,dx$ for integers $n\ge0$, given that $I_0=\dfrac1\lambda$. Use integration by parts to show that $I_n = \dfrac{n}{\lambda}I_{n-1}$ for $n\ge1$, and hence prove by mathematical induction that $I_n = \dfrac{n!}{\lambda^{n+1}}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
First, use integration by parts on $I_n = \displaystyle\int_0^{\infty} x^n e^{-\lambda x}\,dx$ for $n\ge1$, with $u=x^n$, $dv=e^{-\lambda x}dx$, so $du=nx^{n-1}dx$ and $v=-\dfrac1\lambda e^{-\lambda x}$:
$$I_n = \left[-\frac1\lambda x^ne^{-\lambda x}\right]_0^{\infty} + \frac{n}{\lambda}\int_0^{\infty}x^{n-1}e^{-\lambda x}\,dx$$
Since $\lambda>0$, exponential decay dominates any power of $x$, so $x^ne^{-\lambda x}\to0$ as $x\to\infty$, and at $x=0$ the term $x^ne^{-\lambda x}=0$ for $n\ge1$. The boundary term vanishes:
$$I_n = \frac{n}{\lambda}\int_0^{\infty}x^{n-1}e^{-\lambda x}\,dx = \frac{n}{\lambda}I_{n-1}$$
Now let $P(n)$ be the proposition that $I_n = \dfrac{n!}{\lambda^{n+1}}$. Using proof by mathematical induction:
Base case ($n=1$):
$$I_1 = \frac1\lambda\cdot I_0 = \frac1\lambda\cdot\frac1\lambda = \frac{1}{\lambda^2} = \frac{1!}{\lambda^{2}}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e. $I_k=\dfrac{k!}{\lambda^{k+1}}$. Consider $n=k+1$. Using the recurrence relation established above:
$$I_{k+1} = \frac{k+1}{\lambda}I_k = \frac{k+1}{\lambda}\cdot\frac{k!}{\lambda^{k+1}} = \frac{(k+1)!}{\lambda^{k+2}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $I_n=\dfrac{n!}{\lambda^{n+1}}$ for all positive integers $n$. $\blacksquare$
QUESTION 41
9 marks
Medium
Let $H_n = \displaystyle\int_0^{\infty} x^{2n+1} e^{-x^2}\,dx$ for integers $n\ge0$, given that $H_0=\dfrac12$. Use integration by parts to show that $H_n = nH_{n-1}$ for $n\ge1$, and hence prove by mathematical induction that $H_n = \dfrac{n!}{2}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
First, use integration by parts on $H_n = \displaystyle\int_0^{\infty} x^{2n+1} e^{-x^2}\,dx$ for $n\ge1$. Write $x^{2n+1}e^{-x^2} = x^{2n}\cdot xe^{-x^2}$ and let $u=x^{2n}$, $dv=xe^{-x^2}dx$, so $du=2nx^{2n-1}dx$ and $v=-\dfrac12e^{-x^2}$ (since $\dfrac{d}{dx}\left(-\dfrac12e^{-x^2}\right)=xe^{-x^2}$):
$$H_n = \left[-\frac12x^{2n}e^{-x^2}\right]_0^{\infty} + n\int_0^{\infty}x^{2n-1}e^{-x^2}\,dx$$
As $x\to\infty$, $x^{2n}e^{-x^2}\to0$ since $e^{-x^2}$ decays faster than any power of $x$ grows, and at $x=0$ the term $x^{2n}e^{-x^2}=0$ for $n\ge1$. The boundary term vanishes:
$$H_n = n\int_0^{\infty}x^{2n-1}e^{-x^2}\,dx = nH_{n-1}$$
Now let $P(n)$ be the proposition that $H_n = \dfrac{n!}{2}$. Using proof by mathematical induction:
Base case ($n=1$):
$$H_1 = 1\cdot H_0 = 1\cdot\frac12 = \frac12 = \frac{1!}{2}$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e. $H_k=\dfrac{k!}{2}$. Consider $n=k+1$. Using the recurrence relation established above:
$$H_{k+1} = (k+1)H_k = (k+1)\cdot\frac{k!}{2} = \frac{(k+1)!}{2}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $H_n=\dfrac{n!}{2}$ for all positive integers $n$. $\blacksquare$
QUESTION 42
9 marks
Medium
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{x(x+1)}\right) = (-1)^n n!\left[\dfrac{1}{x^{n+1}}-\dfrac{1}{(x+1)^{n+1}}\right]$ for all $n \in \mathbb{Z}^+$, where $x\neq0,-1$.
Show complete worked solution
First write $\dfrac{1}{x(x+1)}$ in partial fractions:
$$\frac{1}{x(x+1)} = \frac{1}{x}-\frac{1}{x+1}$$
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{x(x+1)}\right) = (-1)^n n!\left[\dfrac{1}{x^{n+1}}-\dfrac{1}{(x+1)^{n+1}}\right]$. Using proof by mathematical induction:
Base case ($n=1$): differentiating the partial fraction form,
$$\frac{d}{dx}\left(\frac{1}{x(x+1)}\right) = -\frac{1}{x^2}+\frac{1}{(x+1)^2} = (-1)^1\,1!\left[\frac{1}{x^2}-\frac{1}{(x+1)^2}\right]$$
so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{1}{x(x+1)}\right) = (-1)^k k!\left[x^{-(k+1)}-(x+1)^{-(k+1)}\right]$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{x(x+1)}\right) = (-1)^k k!\left[-(k+1)x^{-(k+2)}+(k+1)(x+1)^{-(k+2)}\right]$$
$$= (-1)^k k!(k+1)\left[-x^{-(k+2)}+(x+1)^{-(k+2)}\right] = (-1)^{k+1}(k+1)!\left[x^{-(k+2)}-(x+1)^{-(k+2)}\right]$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for $x\neq0,-1$. $\blacksquare$
QUESTION 43
11 marks
Hard
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(x^2e^{-x}\right) = (-1)^n\left(x^2-2nx+n(n-1)\right)e^{-x}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(x^2e^{-x}\right) = (-1)^n\left(x^2-2nx+n(n-1)\right)e^{-x}$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left(x^2e^{-x}\right) = 2xe^{-x}-x^2e^{-x} = -(x^2-2x)e^{-x}$$
and the formula gives $(-1)^1\left(x^2-2(1)x+1(0)\right)e^{-x} = -(x^2-2x)e^{-x}$, so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(x^2e^{-x}\right) = (-1)^k\left(x^2-2kx+k(k-1)\right)e^{-x}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$, using the product rule on the right-hand side:
$$\frac{d^{k+1}}{dx^{k+1}}\left(x^2e^{-x}\right) = (-1)^k\left\{(2x-2k)e^{-x}+\left(x^2-2kx+k(k-1)\right)(-1)e^{-x}\right\}$$
$$= (-1)^k\left\{(2x-2k)-\left(x^2-2kx+k(k-1)\right)\right\}e^{-x} = (-1)^{k+1}\left\{x^2-2kx+k(k-1)-2x+2k\right\}e^{-x}$$
$$= (-1)^{k+1}\left\{x^2-2(k+1)x+\left[k(k-1)+2k\right]\right\}e^{-x}$$
Since $k(k-1)+2k = k^2+k = k(k+1) = (k+1)\left((k+1)-1\right)$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(x^2e^{-x}\right) = (-1)^{k+1}\left(x^2-2(k+1)x+(k+1)\left((k+1)-1\right)\right)e^{-x}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 44
12 marks
Hard
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(x^3e^{x}\right) = \left(x^3+3nx^2+3n(n-1)x+n(n-1)(n-2)\right)e^{x}$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(x^3e^{x}\right) = \left(x^3+3nx^2+3n(n-1)x+n(n-1)(n-2)\right)e^{x}$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left(x^3e^{x}\right) = 3x^2e^x+x^3e^x = (x^3+3x^2)e^x$$
and the formula gives $\left(x^3+3(1)x^2+3(1)(0)x+1(0)(-1)\right)e^x = (x^3+3x^2)e^x$, so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(x^3e^{x}\right) = \left(x^3+3kx^2+3k(k-1)x+k(k-1)(k-2)\right)e^{x}$$
Consider $n=k+1$. Write the bracket as $B=x^3+3kx^2+3k(k-1)x+k(k-1)(k-2)$, so $\dfrac{d^k}{dx^k}\left(x^3e^x\right)=Be^x$. Differentiating both sides with respect to $x$, using the product rule:
$$\frac{d^{k+1}}{dx^{k+1}}\left(x^3e^{x}\right) = B'e^x+Be^x = (B'+B)e^x$$
where $B' = 3x^2+6kx+3k(k-1)$. Adding term by term:
$$B'+B = x^3+\left(3+3k\right)x^2+\left(6k+3k(k-1)\right)x+\left(3k(k-1)+k(k-1)(k-2)\right)$$
The coefficient of $x^2$ is $3+3k=3(k+1)$. The coefficient of $x$ is $6k+3k(k-1)=3k^2+3k=3k(k+1)=3(k+1)\left((k+1)-1\right)$. The constant term is:
$$3k(k-1)+k(k-1)(k-2) = k(k-1)\left[3+(k-2)\right] = k(k-1)(k+1) = (k+1)\,k\,(k-1) = (k+1)\left((k+1)-1\right)\left((k+1)-2\right)$$
so:
$$\frac{d^{k+1}}{dx^{k+1}}\left(x^3e^{x}\right) = \left(x^3+3(k+1)x^2+3(k+1)k\,x+(k+1)k(k-1)\right)e^x$$
which is exactly the formula for $n=k+1$, so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 45
12 marks
Hard
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(x\sin x\right) = x\sin\left(x+\dfrac{n\pi}{2}\right)-n\cos\left(x+\dfrac{n\pi}{2}\right)$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(x\sin x\right) = x\sin\left(x+\dfrac{n\pi}{2}\right)-n\cos\left(x+\dfrac{n\pi}{2}\right)$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left(x\sin x\right) = \sin x+x\cos x$$
and the formula gives $x\sin\left(x+\dfrac{\pi}{2}\right)-1\cdot\cos\left(x+\dfrac{\pi}{2}\right) = x\cos x-(-\sin x) = x\cos x+\sin x$, using $\sin\left(x+\dfrac\pi2\right)=\cos x$ and $\cos\left(x+\dfrac\pi2\right)=-\sin x$. The two sides agree, so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e., writing $\theta=x+\dfrac{k\pi}{2}$,
$$\frac{d^k}{dx^k}\left(x\sin x\right) = x\sin\theta-k\cos\theta$$
Consider $n=k+1$. Differentiating both sides with respect to $x$ (noting $\dfrac{d\theta}{dx}=1$), using the product rule on the first term:
$$\frac{d^{k+1}}{dx^{k+1}}\left(x\sin x\right) = \left[\sin\theta+x\cos\theta\right]+k\sin\theta = x\cos\theta+(k+1)\sin\theta$$
Now use $\cos\theta = \sin\left(\theta+\dfrac{\pi}{2}\right)$ and $\sin\theta = -\cos\left(\theta+\dfrac{\pi}{2}\right)$:
$$x\cos\theta+(k+1)\sin\theta = x\sin\left(\theta+\frac{\pi}{2}\right)-(k+1)\cos\left(\theta+\frac{\pi}{2}\right)$$
Since $\theta+\dfrac{\pi}{2}=x+\dfrac{(k+1)\pi}{2}$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(x\sin x\right) = x\sin\left(x+\frac{(k+1)\pi}{2}\right)-(k+1)\cos\left(x+\frac{(k+1)\pi}{2}\right)$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 46
12 marks
Hard
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(x\cos x\right) = x\cos\left(x+\dfrac{n\pi}{2}\right)+n\sin\left(x+\dfrac{n\pi}{2}\right)$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(x\cos x\right) = x\cos\left(x+\dfrac{n\pi}{2}\right)+n\sin\left(x+\dfrac{n\pi}{2}\right)$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left(x\cos x\right) = \cos x-x\sin x$$
and the formula gives $x\cos\left(x+\dfrac{\pi}{2}\right)+1\cdot\sin\left(x+\dfrac{\pi}{2}\right) = x(-\sin x)+\cos x = \cos x-x\sin x$, using $\cos\left(x+\dfrac\pi2\right)=-\sin x$ and $\sin\left(x+\dfrac\pi2\right)=\cos x$. The two sides agree, so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e., writing $\theta=x+\dfrac{k\pi}{2}$,
$$\frac{d^k}{dx^k}\left(x\cos x\right) = x\cos\theta+k\sin\theta$$
Consider $n=k+1$. Differentiating both sides with respect to $x$ (noting $\dfrac{d\theta}{dx}=1$), using the product rule on the first term:
$$\frac{d^{k+1}}{dx^{k+1}}\left(x\cos x\right) = \left[\cos\theta-x\sin\theta\right]+k\cos\theta = -x\sin\theta+(k+1)\cos\theta$$
Now use $-\sin\theta = \cos\left(\theta+\dfrac{\pi}{2}\right)$ and $\cos\theta = \sin\left(\theta+\dfrac{\pi}{2}\right)$:
$$-x\sin\theta+(k+1)\cos\theta = x\cos\left(\theta+\frac{\pi}{2}\right)+(k+1)\sin\left(\theta+\frac{\pi}{2}\right)$$
Since $\theta+\dfrac{\pi}{2}=x+\dfrac{(k+1)\pi}{2}$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(x\cos x\right) = x\cos\left(x+\frac{(k+1)\pi}{2}\right)+(k+1)\sin\left(x+\frac{(k+1)\pi}{2}\right)$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 47
11 marks
Hard
Let $m$ be a fixed positive integer and $a$ a real constant. Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{(x+a)^m}\right) = \dfrac{(-1)^n(m+n-1)!}{(m-1)!\,(x+a)^{m+n}}$ for all $n \in \mathbb{Z}^+$, where $x\neq-a$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(\dfrac{1}{(x+a)^m}\right) = \dfrac{(-1)^n(m+n-1)!}{(m-1)!\,(x+a)^{m+n}}$, for the fixed positive integer $m$. Using proof by mathematical induction:
Base case ($n=1$): using the chain rule,
$$\frac{d}{dx}\left(\frac{1}{(x+a)^m}\right) = -m(x+a)^{-(m+1)}$$
and the formula gives $\dfrac{(-1)^1\,m!}{(m-1)!}(x+a)^{-(m+1)} = -m(x+a)^{-(m+1)}$, since $\dfrac{m!}{(m-1)!}=m$. The two sides agree, so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e.
$$\frac{d^k}{dx^k}\left(\frac{1}{(x+a)^m}\right) = \frac{(-1)^k(m+k-1)!}{(m-1)!}\,(x+a)^{-(m+k)}$$
Consider $n=k+1$. Differentiating both sides with respect to $x$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{(x+a)^m}\right) = \frac{(-1)^k(m+k-1)!}{(m-1)!}\cdot\left[-(m+k)\right](x+a)^{-(m+k+1)} = \frac{(-1)^{k+1}(m+k-1)!(m+k)}{(m-1)!}\,(x+a)^{-(m+k+1)}$$
Since $(m+k-1)!(m+k) = (m+k)!$:
$$\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{(x+a)^m}\right) = \frac{(-1)^{k+1}(m+k)!}{(m-1)!\,(x+a)^{m+k+1}} = \frac{(-1)^{k+1}\left(m+(k+1)-1\right)!}{(m-1)!\,(x+a)^{m+(k+1)}}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$, for each fixed positive integer $m$ and $x\neq-a$. $\blacksquare$
QUESTION 48
13 marks
Hard
Prove by mathematical induction that $\dfrac{d^n}{dx^n}\left(e^{x}\cos x\right) = \left(\sqrt2\right)^n e^{x}\cos\left(x+\dfrac{n\pi}{4}\right)$ for all $n \in \mathbb{Z}^+$.
Show complete worked solution
Let $P(n)$ be the proposition that $\dfrac{d^n}{dx^n}\left(e^{x}\cos x\right) = \left(\sqrt2\right)^n e^{x}\cos\left(x+\dfrac{n\pi}{4}\right)$. Using proof by mathematical induction:
Base case ($n=1$): using the product rule,
$$\frac{d}{dx}\left(e^x\cos x\right) = e^x\cos x-e^x\sin x = e^x(\cos x-\sin x)$$
and, using $\cos\theta-\sin\theta = \sqrt2\left[\cos\theta\cos\dfrac\pi4-\sin\theta\sin\dfrac\pi4\right] = \sqrt2\cos\left(\theta+\dfrac\pi4\right)$ with $\theta=x$:
$$\left(\sqrt2\right)^1 e^x\cos\left(x+\frac{\pi}{4}\right) = \sqrt2\,e^x\cos\left(x+\frac\pi4\right) = e^x(\cos x-\sin x)$$
so the two sides agree and $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e., writing $\theta=x+\dfrac{k\pi}{4}$,
$$\frac{d^k}{dx^k}\left(e^x\cos x\right) = \left(\sqrt2\right)^k e^x\cos\theta$$
Consider $n=k+1$. Differentiating both sides with respect to $x$ (noting $\dfrac{d\theta}{dx}=1$), using the product rule:
$$\frac{d^{k+1}}{dx^{k+1}}\left(e^x\cos x\right) = \left(\sqrt2\right)^k\left[e^x\cos\theta-e^x\sin\theta\right] = \left(\sqrt2\right)^k e^x\left(\cos\theta-\sin\theta\right)$$
Using the identity $\cos\theta-\sin\theta = \sqrt2\cos\left(\theta+\dfrac\pi4\right)$ established above:
$$\left(\sqrt2\right)^k e^x\left(\cos\theta-\sin\theta\right) = \left(\sqrt2\right)^k\cdot\sqrt2\,e^x\cos\left(\theta+\frac{\pi}{4}\right) = \left(\sqrt2\right)^{k+1}e^x\cos\left(x+\frac{(k+1)\pi}{4}\right)$$
since $\theta+\dfrac\pi4 = x+\dfrac{k\pi}{4}+\dfrac{\pi}{4} = x+\dfrac{(k+1)\pi}{4}$, so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$. $\blacksquare$
QUESTION 49
14 marks
Hard
Let $W_n = \displaystyle\int_0^{\pi/2}\sin^n x\,dx$ for integers $n\ge0$, given that $W_0=\dfrac{\pi}{2}$. Use integration by parts to show that $W_n = \dfrac{n-1}{n}W_{n-2}$ for $n\ge2$, and hence prove by mathematical induction that $W_{2m} = \dfrac{\pi}{2}\cdot\dfrac{(2m)!}{4^m(m!)^2}$ for all $m \in \mathbb{Z}^+$.
Show complete worked solution
First, use integration by parts on $W_n = \displaystyle\int_0^{\pi/2}\sin^n x\,dx = \int_0^{\pi/2}\sin^{n-1}x\cdot\sin x\,dx$ for $n\ge2$, with $u=\sin^{n-1}x$, $dv=\sin x\,dx$, so $du=(n-1)\sin^{n-2}x\cos x\,dx$ and $v=-\cos x$:
$$W_n = \left[-\sin^{n-1}x\cos x\right]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x\,dx$$
At $x=\dfrac{\pi}{2}$, $\cos x=0$; at $x=0$, $\sin^{n-1}x=0$ for $n\ge2$. So the boundary term vanishes. Using $\cos^2x=1-\sin^2x$:
$$W_n = (n-1)\int_0^{\pi/2}\sin^{n-2}x\left(1-\sin^2x\right)dx = (n-1)\left(W_{n-2}-W_n\right)$$
Rearranging: $W_n+(n-1)W_n = (n-1)W_{n-2}$, so $nW_n = (n-1)W_{n-2}$, giving
$$W_n = \frac{n-1}{n}W_{n-2}$$
Now let $P(m)$ be the proposition that $W_{2m} = \dfrac{\pi}{2}\cdot\dfrac{(2m)!}{4^m(m!)^2}$. Using proof by mathematical induction:
Base case ($m=1$): applying the recurrence with $n=2$,
$$W_2 = \frac{2-1}{2}W_0 = \frac12\cdot\frac{\pi}{2} = \frac{\pi}{4}$$
and the formula gives $\dfrac{\pi}{2}\cdot\dfrac{2!}{4^1(1!)^2} = \dfrac{\pi}{2}\cdot\dfrac24 = \dfrac{\pi}{4}$, so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e. $W_{2k} = \dfrac{\pi}{2}\cdot\dfrac{(2k)!}{4^k(k!)^2}$. Consider $m=k+1$. Applying the recurrence with $n=2k+2$:
$$W_{2k+2} = \frac{2k+1}{2k+2}W_{2k} = \frac{2k+1}{2k+2}\cdot\frac{\pi}{2}\cdot\frac{(2k)!}{4^k(k!)^2}$$
Now simplify the target: since $(2k+2)! = (2k+2)(2k+1)(2k)!$ and $4^{k+1}\left((k+1)!\right)^2 = 4\cdot4^k(k+1)^2(k!)^2$,
$$\frac{\pi}{2}\cdot\frac{(2k+2)!}{4^{k+1}\left((k+1)!\right)^2} = \frac{\pi}{2}\cdot\frac{(2k)!}{4^k(k!)^2}\cdot\frac{(2k+2)(2k+1)}{4(k+1)^2}$$
Since $\dfrac{2k+2}{k+1}=2$, the last fraction is $\dfrac{2(k+1)(2k+1)}{4(k+1)^2} = \dfrac{2k+1}{2(k+1)} = \dfrac{2k+1}{2k+2}$, so
$$\frac{\pi}{2}\cdot\frac{(2k+2)!}{4^{k+1}\left((k+1)!\right)^2} = \frac{2k+1}{2k+2}\cdot\frac{\pi}{2}\cdot\frac{(2k)!}{4^k(k!)^2} = W_{2k+2}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $W_{2m}=\dfrac{\pi}{2}\cdot\dfrac{(2m)!}{4^m(m!)^2}$ for all positive integers $m$. $\blacksquare$
QUESTION 50
14 marks
Hard
Let $W_n = \displaystyle\int_0^{\pi/2}\sin^n x\,dx$ for integers $n\ge0$, given that $W_1=1$. Use integration by parts to show that $W_n = \dfrac{n-1}{n}W_{n-2}$ for $n\ge2$, and hence prove by mathematical induction that $W_{2m+1} = \dfrac{4^m(m!)^2}{(2m+1)!}$ for all $m \in \mathbb{Z}^+$.
Show complete worked solution
First, use integration by parts on $W_n = \displaystyle\int_0^{\pi/2}\sin^n x\,dx = \int_0^{\pi/2}\sin^{n-1}x\cdot\sin x\,dx$ for $n\ge2$, with $u=\sin^{n-1}x$, $dv=\sin x\,dx$, so $du=(n-1)\sin^{n-2}x\cos x\,dx$ and $v=-\cos x$:
$$W_n = \left[-\sin^{n-1}x\cos x\right]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x\,dx$$
At $x=\dfrac{\pi}{2}$, $\cos x=0$; at $x=0$, $\sin^{n-1}x=0$ for $n\ge2$. So the boundary term vanishes. Using $\cos^2x=1-\sin^2x$:
$$W_n = (n-1)\int_0^{\pi/2}\sin^{n-2}x\left(1-\sin^2x\right)dx = (n-1)\left(W_{n-2}-W_n\right)$$
Rearranging: $nW_n = (n-1)W_{n-2}$, giving
$$W_n = \frac{n-1}{n}W_{n-2}$$
Now let $P(m)$ be the proposition that $W_{2m+1} = \dfrac{4^m(m!)^2}{(2m+1)!}$. Using proof by mathematical induction:
Base case ($m=1$): applying the recurrence with $n=3$,
$$W_3 = \frac{3-1}{3}W_1 = \frac23\cdot1 = \frac23$$
and the formula gives $\dfrac{4^1(1!)^2}{3!} = \dfrac{4}{6} = \dfrac23$, so $P(1)$ is true.
Inductive step: assume $P(k)$ is true for some positive integer $k$, i.e. $W_{2k+1} = \dfrac{4^k(k!)^2}{(2k+1)!}$. Consider $m=k+1$. Applying the recurrence with $n=2k+3$:
$$W_{2k+3} = \frac{2k+2}{2k+3}W_{2k+1} = \frac{2k+2}{2k+3}\cdot\frac{4^k(k!)^2}{(2k+1)!}$$
Now simplify the target: since $(2k+3)! = (2k+3)(2k+2)(2k+1)!$ and $4^{k+1}\left((k+1)!\right)^2 = 4\cdot4^k(k+1)^2(k!)^2$,
$$\frac{4^{k+1}\left((k+1)!\right)^2}{(2k+3)!} = \frac{4^k(k!)^2}{(2k+1)!}\cdot\frac{4(k+1)^2}{(2k+3)(2k+2)}$$
Since $2k+2=2(k+1)$, the last fraction is $\dfrac{4(k+1)^2}{(2k+3)\cdot2(k+1)} = \dfrac{2(k+1)}{2k+3} = \dfrac{2k+2}{2k+3}$, so
$$\frac{4^{k+1}\left((k+1)!\right)^2}{(2k+3)!} = \frac{2k+2}{2k+3}\cdot\frac{4^k(k!)^2}{(2k+1)!} = W_{2k+3}$$
so $P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: since $P(1)$ is true, and $P(k)\Rightarrow P(k+1)$ for all positive integers $k$, by the principle of mathematical induction, $W_{2m+1}=\dfrac{4^m(m!)^2}{(2m+1)!}$ for all positive integers $m$. $\blacksquare$
Systems of Linear Equations 50 questions
QUESTION 1
4 marks
Easy
Express $\dfrac{7x-4}{(x-1)(x+2)}$ in partial fractions.
Show complete worked solution
Since the denominator has two distinct linear factors, write
$$\dfrac{7x-4}{(x-1)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}$$
Multiply both sides by $(x-1)(x+2)$:
$$7x-4=A(x+2)+B(x-1)$$
Find $A$: substitute $x=1$ (which eliminates $B$):
$$7(1)-4=A(1+2)$$
$$3=3A \quad\Rightarrow\quad A=1$$
Find $B$: substitute $x=-2$ (which eliminates $A$):
$$7(-2)-4=B(-2-1)$$
$$-18=-3B \quad\Rightarrow\quad B=6$$
$$\boxed{\dfrac{7x-4}{(x-1)(x+2)}=\dfrac{1}{x-1}+\dfrac{6}{x+2}}$$
QUESTION 2
6 marks
Medium
Express $\dfrac{x+7}{(x+1)^2(x-2)}$ in partial fractions.
Show complete worked solution
Since $(x+1)$ is a repeated linear factor, write
$$\dfrac{x+7}{(x+1)^2(x-2)}=\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}+\dfrac{C}{x-2}$$
Multiply both sides by $(x+1)^2(x-2)$:
$$x+7=A(x+1)(x-2)+B(x-2)+C(x+1)^2$$
Find $B$: substitute $x=-1$ (eliminates $A$ and $C$):
$$-1+7=B(-1-2)$$
$$6=-3B \quad\Rightarrow\quad B=-2$$
Find $C$: substitute $x=2$ (eliminates $A$ and $B$):
$$2+7=C(2+1)^2$$
$$9=9C \quad\Rightarrow\quad C=1$$
Find $A$: compare coefficients of $x^2$ on both sides. On the right, the $x^2$ terms come from $A(x+1)(x-2)$ and $C(x+1)^2$, giving coefficient $A+C$; the left side has no $x^2$ term, so:
$$0=A+C \quad\Rightarrow\quad A=-C=-1$$
Check using $x=0$: LHS $=7$. RHS $=A(1)(-2)+B(-2)+C(1)=-2A-2B+C=-2(-1)-2(-2)+1=2+4+1=7$. $\checkmark$
$$\boxed{\dfrac{x+7}{(x+1)^2(x-2)}=-\dfrac{1}{x+1}-\dfrac{2}{(x+1)^2}+\dfrac{1}{x-2}}$$
QUESTION 3
6 marks
Medium
Express $\dfrac{2x^2+3x-1}{(x-1)(x+1)(x+3)}$ in partial fractions.
Show complete worked solution
Since the denominator has three distinct linear factors, write
$$\dfrac{2x^2+3x-1}{(x-1)(x+1)(x+3)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{C}{x+3}$$
Multiply both sides by $(x-1)(x+1)(x+3)$:
$$2x^2+3x-1=A(x+1)(x+3)+B(x-1)(x+3)+C(x-1)(x+1)$$
Find $A$: substitute $x=1$:
$$2(1)+3(1)-1=A(2)(4)$$
$$4=8A \quad\Rightarrow\quad A=\dfrac{1}{2}$$
Find $B$: substitute $x=-1$:
$$2(1)+3(-1)-1=B(-2)(2)$$
$$-2=-4B \quad\Rightarrow\quad B=\dfrac{1}{2}$$
Find $C$: substitute $x=-3$:
$$2(9)+3(-3)-1=C(-4)(-2)$$
$$18-9-1=8C$$
$$8=8C \quad\Rightarrow\quad C=1$$
Check using $x=0$: LHS $=-1$. RHS $=A(1)(3)+B(-1)(3)+C(-1)(1)=3A-3B-C=1.5-1.5-1=-1$. $\checkmark$
$$\boxed{\dfrac{2x^2+3x-1}{(x-1)(x+1)(x+3)}=\dfrac{1}{2(x-1)}+\dfrac{1}{2(x+1)}+\dfrac{1}{x+3}}$$
QUESTION 4
5 marks
Easy
Express $\dfrac{x^2+1}{x^2-1}$ in partial fractions.
Show complete worked solution
Since the numerator and denominator have the same degree (both degree 2), the fraction is improper. Perform algebraic division first, by writing the numerator in terms of the denominator:
$$x^2+1=(x^2-1)+2$$
So:
$$\dfrac{x^2+1}{x^2-1}=\dfrac{(x^2-1)+2}{x^2-1}=1+\dfrac{2}{x^2-1}=1+\dfrac{2}{(x-1)(x+1)}$$
Now decompose the proper fraction $\dfrac{2}{(x-1)(x+1)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}$:
$$2=A(x+1)+B(x-1)$$
Find $A$: substitute $x=1$:
$$2=2A \Rightarrow A=1$$
Find $B$: substitute $x=-1$:
$$2=-2B \Rightarrow B=-1$$
$$\boxed{\dfrac{x^2+1}{x^2-1}=1+\dfrac{1}{x-1}-\dfrac{1}{x+1}}$$
QUESTION 5
15 marks
Hard
Consider the function $f(x) = \dfrac{8+x}{(1-x)(2+x)}$.
(a) [5 marks] Express $f(x)$ in the form $\dfrac{A}{1-x} + \dfrac{B}{2+x}$, finding the values of the constants $A$ and $B$.
(b) [6 marks] By expanding each partial fraction from part (a) as a binomial series, show that
$$f(x) = 4 + \dfrac{5}{2}x + \dfrac{13}{4}x^2 + \dfrac{23}{8}x^3 + \ldots$$
(c) [4 marks] State the range of values of $x$ for which this series expansion of $f(x)$ is valid, justifying your answer by considering the validity of each partial fraction's expansion separately.
Show complete worked solution
(a) Multiply both sides by $(1-x)(2+x)$:
$$8+x=A(2+x)+B(1-x)$$
Find $A$: substitute $x=1$:
$$8+1=A(2+1)$$
$$9=3A \quad\Rightarrow\quad A=3$$
Find $B$: substitute $x=-2$:
$$8-2=B(1-(-2))$$
$$6=3B \quad\Rightarrow\quad B=2$$
$$\boxed{f(x)=\dfrac{3}{1-x}+\dfrac{2}{2+x}, \quad A=3,\ B=2}$$
(b) Expand $\dfrac{3}{1-x}=3(1-x)^{-1}$ using $(1-x)^{-1}=1+x+x^2+x^3+\ldots$ (valid for $|x|<1$):
$$3(1-x)^{-1}=3+3x+3x^2+3x^3+\ldots$$
Expand $\dfrac{2}{2+x}$. First factor out 2 from the denominator:
$$\dfrac{2}{2+x}=\dfrac{2}{2\left(1+\tfrac{x}{2}\right)}=\left(1+\dfrac{x}{2}\right)^{-1}$$
Using $(1+u)^{-1}=1-u+u^2-u^3+\ldots$ with $u=\dfrac{x}{2}$:
$$\left(1+\dfrac{x}{2}\right)^{-1}=1-\dfrac{x}{2}+\dfrac{x^2}{4}-\dfrac{x^3}{8}+\ldots$$
Add the two series term by term:
Constant: $3+1=4$
$x$: $3-\dfrac{1}{2}=\dfrac{5}{2}$
$x^2$: $3+\dfrac{1}{4}=\dfrac{13}{4}$
$x^3$: $3-\dfrac{1}{8}=\dfrac{23}{8}$
$$\boxed{f(x)=4+\dfrac{5}{2}x+\dfrac{13}{4}x^2+\dfrac{23}{8}x^3+\ldots}$$
as required.
(c) The expansion of $\dfrac{3}{1-x}=3(1-x)^{-1}$ is valid for $|x|<1$.
The expansion of $\left(1+\dfrac{x}{2}\right)^{-1}$ is valid for $\left|\dfrac{x}{2}\right|<1$, i.e. $|x|<2$.
The combined series for $f(x)$ is valid only where both individual expansions converge, i.e. the intersection of $|x|<1$ and $|x|<2$:
$$\boxed{-1
QUESTION 6
4 marks
Easy
Solve the simultaneous equations:
2x+y=7
x-y=2
Show complete worked solution
Using the elimination method:
$$2x+y=7 \quad (1)$$
$$x-y=2 \quad (2)$$
Add (1) and (2) to eliminate $y$:
$$(2x+y)+(x-y)=7+2$$
$$3x=9$$
$$x=3$$
Substitute $x=3$ into (2):
$$3-y=2$$
$$y=1$$
Check in (1): $2(3)+1=7$ $\checkmark$
$$\boxed{x=3,\ y=1}$$
QUESTION 7
4 marks
Easy
Solve the system of equations:
x+y+z=9
y+z=5
z=2
Show complete worked solution
The system is already in triangular form, so solve by back-substitution.
From the third equation:
$$z=2$$
Substitute into the second equation:
$$y+2=5$$
$$y=3$$
Substitute $y=3$, $z=2$ into the first equation:
$$x+3+2=9$$
$$x=4$$
$$\boxed{x=4,\ y=3,\ z=2}$$
QUESTION 8
7 marks
Medium
Solve the system of equations:
x+y+2z=9
2x+4y-3z=1
3x+6y-5z=0
Show complete worked solution
Using Gaussian elimination:
$$x+y+2z=9 \quad (1)$$
$$2x+4y-3z=1 \quad (2)$$
$$3x+6y-5z=0 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x+4y-3z)-2(x+y+2z)=1-2(9)$$
$$2x+4y-3z-2x-2y-4z=1-18$$
$$2y-7z=-17 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-\,3\times$(1):
$$(3x+6y-5z)-3(x+y+2z)=0-3(9)$$
$$3x+6y-5z-3x-3y-6z=-27$$
$$3y-11z=-27 \quad (5)$$
Now eliminate $y$ from (4) and (5): compute $3\times$(4) $-\,2\times$(5):
$$3(2y-7z)-2(3y-11z)=3(-17)-2(-27)$$
$$6y-21z-6y+22z=-51+54$$
$$z=3$$
Substitute $z=3$ into (4):
$$2y-7(3)=-17$$
$$2y-21=-17$$
$$2y=4$$
$$y=2$$
Substitute $y=2$, $z=3$ into (1):
$$x+2+6=9$$
$$x=1$$
Check in (3): $3(1)+6(2)-5(3)=3+12-15=0$ $\checkmark$
$$\boxed{x=1,\ y=2,\ z=3}$$
QUESTION 9
7 marks
Medium
A bakery sells croissants, muffins and scones at prices \$c, \$m and \$s respectively. On one morning:
2 croissants + 1 muffin + 1 scone cost \$12
1 croissant + 2 muffins + 1 scone cost \$11
1 croissant + 1 muffin + 2 scones cost \$13
Find the price of each item.
Show complete worked solution
Let the equations be:
$$2c+m+s=12 \quad (1)$$
$$c+2m+s=11 \quad (2)$$
$$c+m+2s=13 \quad (3)$$
Subtract (2) from (1) to eliminate $s$:
$$(2c+m+s)-(c+2m+s)=12-11$$
$$c-m=1 \quad (4)$$
Subtract (3) from (2) to eliminate $c$:
$$(c+2m+s)-(c+m+2s)=11-13$$
$$m-s=-2 \quad (5)$$
Add (1)+(2)+(3):
$$4c+4m+4s=12+11+13=36$$
$$c+m+s=9 \quad (6)$$
From (4), $c=m+1$. From (5), $m=s-2$, so $c=(s-2)+1=s-1$.
Substitute $c=s-1$ and $m=s-2$ into (6):
$$(s-1)+(s-2)+s=9$$
$$3s-3=9$$
$$3s=12$$
$$s=4$$
Then $m=s-2=2$, and $c=s-1=3$.
Check in (1): $2(3)+2+4=6+2+4=12$ $\checkmark$
$$\boxed{\text{croissant}=\$3,\ \text{muffin}=\$2,\ \text{scone}=\$4}$$
QUESTION 10
14 marks
Hard
Consider the system of equations
x+2y-z=1
2x-y+3z=4
3x+y+\lambda z=\mu
where \lambda,\mu\in\mathbb{R}.
(a) Show that the determinant of the coefficient matrix is -5\lambda+10, and hence state the value of \lambda for which the system does not have a unique solution.
(b) For this value of \lambda, find the value of \mu for which the system is consistent (i.e. has infinitely many solutions), and find the general solution in this case.
(c) For the same value of \lambda, but with \mu not equal to the value found in part (b), explain why the system has no solution. Briefly describe the geometric configuration of the three planes in each of the three cases: (i) unique solution, (ii) infinitely many solutions, (iii) no solution.
Show complete worked solution
(a) The coefficient matrix is
$$\begin{pmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 3 & 1 & \lambda \end{pmatrix}$$
Expanding the determinant along the first row:
$$\det = 1\left[(-1)(\lambda)-(3)(1)\right] - 2\left[(2)(\lambda)-(3)(3)\right] + (-1)\left[(2)(1)-(-1)(3)\right]$$
$$=1(-\lambda-3) - 2(2\lambda-9) - 1(2+3)$$
$$=(-\lambda-3) - (4\lambda-18) - 5$$
$$=-\lambda-3-4\lambda+18-5$$
$$=-5\lambda+10$$
as required. The system fails to have a unique solution when the determinant is zero:
$$-5\lambda+10=0 \ \Rightarrow\ \boxed{\lambda=2}$$
(b) Set $\lambda=2$. The system becomes:
$$x+2y-z=1 \quad (1)$$
$$2x-y+3z=4 \quad (2)$$
$$3x+y+2z=\mu \quad (3)$$
Since $\det=0$, row (3) must be a linear combination of rows (1) and (2) for consistency. Find $a,b$ such that $a\times$(1)$+b\times$(2) reproduces the coefficients of (3):
$$x: a+2b=3$$
$$y: 2a-b=1$$
$$z: -a+3b=2$$
From the $x$-equation, $a=3-2b$. Substitute into the $y$-equation:
$$2(3-2b)-b=1$$
$$6-4b-b=1$$
$$6-5b=1$$
$$b=1, \text{ so } a=3-2(1)=1$$
Check the $z$-equation: $-1+3(1)=2$ $\checkmark$ (consistent with the coefficient of $z$ in (3))
So (3) $=1\times$(1)$+1\times$(2) in terms of the left-hand-side coefficients. For the system to be consistent, the right-hand side must satisfy the same relation:
$$\mu=1(1)+1(4)=5$$
So the system has infinitely many solutions when $\mu=5$.
For the general solution, let $z=t$. From (1):
$$x=1-2y+t$$
Substitute into (2):
$$2(1-2y+t)-y+3t=4$$
$$2-4y+2t-y+3t=4$$
$$2-5y+5t=4$$
$$-5y=2-5t$$
$$y=t-\dfrac25$$
Then:
$$x=1-2\left(t-\dfrac25\right)+t=1-2t+\dfrac45+t=\dfrac95-t$$
Check in (3) with $\mu=5$: $3\left(\dfrac95-t\right)+\left(t-\dfrac25\right)+2t=\dfrac{27}{5}-3t+t-\dfrac25+2t=\dfrac{25}{5}+0t=5$ $\checkmark$
$$\boxed{\mu=5; \quad x=\dfrac95-t,\ \ y=t-\dfrac25,\ \ z=t,\ \ t\in\mathbb{R}}$$
(c) When $\lambda=2$ and $\mu\ne5$, the same elimination shows that equation (3) reduces to $0=\mu-5\ne0$ after eliminating $x,y,z$ using (1) and (2) - a contradiction. So no values of $x,y,z$ can satisfy all three equations simultaneously, and the system has no solution.
Geometric interpretation:
(i) When $\lambda\ne2$, the determinant is non-zero, so the three planes intersect at exactly one point (unique solution).
(ii) When $\lambda=2$ and $\mu=5$, the three planes share a common line of intersection (infinitely many solutions lying on that line).
(iii) When $\lambda=2$ and $\mu\ne5$, the planes have no common point - their pairwise lines of intersection are parallel but distinct, so the three planes form a triangular prism, with no point lying on all three planes.
$$\boxed{\text{(i) unique point; (ii) common line; (iii) no common point (triangular prism)}}$$
QUESTION 11
4 marks
Easy
Solve the simultaneous equations:
2x+3y=16
x-y=3
Show complete worked solution
Using the elimination method:
$$2x+3y=16 \quad (1)$$
$$x-y=3 \quad (2)$$
Multiply (2) by 3 so that the $y$-terms will cancel when added to (1):
$$3x-3y=9 \quad (3)$$
Add (1) and (3):
$$(2x+3y)+(3x-3y)=16+9$$
$$5x=25$$
$$x=5$$
Substitute $x=5$ into (2):
$$5-y=3$$
$$y=2$$
Check in (1): $2(5)+3(2)=10+6=16$ $\checkmark$
$$\boxed{x=5,\ y=2}$$
QUESTION 12
4 marks
Easy
Solve the simultaneous equations:
3x+2y=6
x+3y=-5
Show complete worked solution
Using the elimination method:
$$3x+2y=6 \quad (1)$$
$$x+3y=-5 \quad (2)$$
Multiply (2) by 3 so that the $x$-coefficients match:
$$3x+9y=-15 \quad (3)$$
Subtract (1) from (3) to eliminate $x$:
$$(3x+9y)-(3x+2y)=-15-6$$
$$7y=-21$$
$$y=-3$$
Substitute $y=-3$ into (2):
$$x+3(-3)=-5$$
$$x-9=-5$$
$$x=4$$
Check in (1): $3(4)+2(-3)=12-6=6$ $\checkmark$
$$\boxed{x=4,\ y=-3}$$
QUESTION 13
4 marks
Easy
Solve the simultaneous equations:
y=2x-3
3x+y=22
Show complete worked solution
Using the substitution method, label the equations:
$$y=2x-3 \quad (1)$$
$$3x+y=22 \quad (2)$$
Substitute (1) into (2):
$$3x+(2x-3)=22$$
$$5x-3=22$$
$$5x=25$$
$$x=5$$
Substitute $x=5$ into (1):
$$y=2(5)-3=7$$
Check in (2): $3(5)+7=15+7=22$ $\checkmark$
$$\boxed{x=5,\ y=7}$$
QUESTION 14
5 marks
Easy
Solve the simultaneous equations:
4x+y=-1
2x-3y=-25
Show complete worked solution
Using the elimination method:
$$4x+y=-1 \quad (1)$$
$$2x-3y=-25 \quad (2)$$
Multiply (1) by 3 so that the $y$-terms will cancel when added to (2):
$$12x+3y=-3 \quad (3)$$
Add (2) and (3):
$$(2x-3y)+(12x+3y)=-25+(-3)$$
$$14x=-28$$
$$x=-2$$
Substitute $x=-2$ into (1):
$$4(-2)+y=-1$$
$$-8+y=-1$$
$$y=7$$
Check in (2): $2(-2)-3(7)=-4-21=-25$ $\checkmark$
$$\boxed{x=-2,\ y=7}$$
QUESTION 15
4 marks
Easy
The sum of two numbers is 26. Their difference is 8. Find the two numbers.
Show complete worked solution
Let the two numbers be $x$ and $y$, with $x>y$.
$$x+y=26 \quad (1)$$
$$x-y=8 \quad (2)$$
Add (1) and (2) to eliminate $y$:
$$(x+y)+(x-y)=26+8$$
$$2x=34$$
$$x=17$$
Substitute $x=17$ into (1):
$$17+y=26$$
$$y=9$$
Check in (2): $17-9=8$ $\checkmark$
$$\boxed{\text{The two numbers are }17\text{ and }9.}$$
QUESTION 16
5 marks
Easy
At a market stall, apples cost \$a each and oranges cost \$o each.
3 apples and 2 oranges cost \$13.
2 apples and 5 oranges cost \$16.
Find the price of an apple and the price of an orange.
Show complete worked solution
Let the equations be:
$$3a+2o=13 \quad (1)$$
$$2a+5o=16 \quad (2)$$
Eliminate $a$: multiply (1) by 2 and (2) by 3:
$$6a+4o=26 \quad (3)$$
$$6a+15o=48 \quad (4)$$
Subtract (3) from (4):
$$(6a+15o)-(6a+4o)=48-26$$
$$11o=22$$
$$o=2$$
Substitute $o=2$ into (1):
$$3a+2(2)=13$$
$$3a+4=13$$
$$3a=9$$
$$a=3$$
Check in (2): $2(3)+5(2)=6+10=16$ $\checkmark$
$$\boxed{\text{apple}=\$3,\ \text{orange}=\$2}$$
QUESTION 17
4 marks
Easy
Solve the system of equations:
x+y+z=15
y+z=11
z=4
Show complete worked solution
The system is already in triangular form, so solve by back-substitution.
From the third equation:
$$z=4$$
Substitute into the second equation:
$$y+4=11$$
$$y=7$$
Substitute $y=7$, $z=4$ into the first equation:
$$x+7+4=15$$
$$x=4$$
$$\boxed{x=4,\ y=7,\ z=4}$$
QUESTION 18
4 marks
Easy
Solve the system of equations:
x+y+z=9
y-2z=5
z=-1
Show complete worked solution
The system is already in triangular form, so solve by back-substitution.
From the third equation:
$$z=-1$$
Substitute into the second equation:
$$y-2(-1)=5$$
$$y+2=5$$
$$y=3$$
Substitute $y=3$, $z=-1$ into the first equation:
$$x+3+(-1)=9$$
$$x+2=9$$
$$x=7$$
$$\boxed{x=7,\ y=3,\ z=-1}$$
QUESTION 19
5 marks
Easy
Solve the system of equations:
x-y+z=8
y+3z=17
2z=10
Show complete worked solution
The system is already in triangular form, so solve by back-substitution.
From the third equation:
$$2z=10$$
$$z=5$$
Substitute into the second equation:
$$y+3(5)=17$$
$$y+15=17$$
$$y=2$$
Substitute $y=2$, $z=5$ into the first equation:
$$x-2+5=8$$
$$x+3=8$$
$$x=5$$
$$\boxed{x=5,\ y=2,\ z=5}$$
QUESTION 20
5 marks
Easy
Solve the system of equations:
2x+y-z=11
3y-z=13
z=2
Show complete worked solution
The system is already in triangular form, so solve by back-substitution.
From the third equation:
$$z=2$$
Substitute into the second equation:
$$3y-2=13$$
$$3y=15$$
$$y=5$$
Substitute $y=5$, $z=2$ into the first equation:
$$2x+5-2=11$$
$$2x+3=11$$
$$2x=8$$
$$x=4$$
$$\boxed{x=4,\ y=5,\ z=2}$$
QUESTION 21
6 marks
Easy
A jar contains only nickels (5 cents), dimes (10 cents) and quarters (25 cents). There are 30 coins in total, worth 430 cents altogether. There are 2 more quarters than nickels.
Let $n$, $d$ and $q$ be the numbers of nickels, dimes and quarters. Find $n$, $d$ and $q$.
Show complete worked solution
Let the equations be:
$$n+d+q=30 \quad (1)$$
$$5n+10d+25q=430 \quad (2)$$
$$q=n+2 \quad (3)$$
Rearrange (3): $n-q=-2 \quad (3')$
Divide (2) by 5:
$$n+2d+5q=86 \quad (4)$$
Subtract (1) from (4) to eliminate $n$:
$$(n+2d+5q)-(n+d+q)=86-30$$
$$d+4q=56 \quad (5)$$
From (3'), $n=q-2$. Substitute into (1):
$$(q-2)+d+q=30$$
$$d+2q=32 \quad (6)$$
Subtract (6) from (5):
$$(d+4q)-(d+2q)=56-32$$
$$2q=24$$
$$q=12$$
Substitute $q=12$ into (6):
$$d+24=32$$
$$d=8$$
Substitute $q=12$ into (3): $n=12-2=10$
Check in (2): $5(10)+10(8)+25(12)=50+80+300=430$ $\checkmark$
$$\boxed{n=10,\ d=8,\ q=12}$$
QUESTION 22
6 marks
Easy
In a family, the mother's age $m$ is three times the son's age $s$. The daughter's age $d$ is 4 years more than the son's age. The three ages add up to 64 years.
Find $s$, $d$ and $m$.
Show complete worked solution
Let the equations be:
$$3s-m=0 \quad (1)$$
$$s+d+m=64 \quad (2)$$
$$s-d=-4 \quad (3)$$
From (1), $m=3s$. Substitute into (2):
$$s+d+3s=64$$
$$4s+d=64 \quad (4)$$
From (3), $d=s+4$. Substitute into (4):
$$4s+(s+4)=64$$
$$5s+4=64$$
$$5s=60$$
$$s=12$$
Substitute $s=12$ into (3): $d=12+4=16$
Substitute $s=12$ into (1): $m=3(12)=36$
Check in (2): $12+16+36=64$ $\checkmark$
$$\boxed{s=12,\ d=16,\ m=36}$$
QUESTION 23
4 marks
Easy
Solve the simultaneous equations:
2x-y=-2
-3x+2y=1
Show complete worked solution
Using the elimination method:
$$2x-y=-2 \quad (1)$$
$$-3x+2y=1 \quad (2)$$
Multiply (1) by 2 so that the $y$-terms will cancel when added to (2):
$$4x-2y=-4 \quad (3)$$
Add (2) and (3):
$$(-3x+2y)+(4x-2y)=1+(-4)$$
$$x=-3$$
Substitute $x=-3$ into (1):
$$2(-3)-y=-2$$
$$-6-y=-2$$
$$-y=4$$
$$y=-4$$
Check in (2): $-3(-3)+2(-4)=9-8=1$ $\checkmark$
$$\boxed{x=-3,\ y=-4}$$
QUESTION 24
5 marks
Easy
A chemist mixes a 10\% acid solution with a 50\% acid solution to make 40 litres of a 30\% acid solution. Let $x$ be the number of litres of the 10\% solution used and $y$ the number of litres of the 50\% solution used.
Find $x$ and $y$.
Show complete worked solution
The total volume gives:
$$x+y=40 \quad (1)$$
The total amount of pure acid gives:
$$0.1x+0.5y=0.3(40) \quad (2)$$
which simplifies to
$$0.1x+0.5y=12 \quad (2')$$
From (1), $y=40-x$. Substitute into (2'):
$$0.1x+0.5(40-x)=12$$
$$0.1x+20-0.5x=12$$
$$-0.4x=-8$$
$$x=20$$
Substitute $x=20$ into (1):
$$y=40-20=20$$
Check in (2): $0.1(20)+0.5(20)=2+10=12$ $\checkmark$
$$\boxed{x=20\text{ litres},\ y=20\text{ litres}}$$
QUESTION 25
5 marks
Easy
Solve the simultaneous equations:
3x+4y=13
5x-2y=39
Show complete worked solution
Using the elimination method:
$$3x+4y=13 \quad (1)$$
$$5x-2y=39 \quad (2)$$
Multiply (2) by 2 so that the $y$-terms will cancel when added to (1):
$$10x-4y=78 \quad (3)$$
Add (1) and (3):
$$(3x+4y)+(10x-4y)=13+78$$
$$13x=91$$
$$x=7$$
Substitute $x=7$ into (2):
$$5(7)-2y=39$$
$$35-2y=39$$
$$-2y=4$$
$$y=-2$$
Check in (1): $3(7)+4(-2)=21-8=13$ $\checkmark$
$$\boxed{x=7,\ y=-2}$$
QUESTION 26
6 marks
Easy
Solve the system of equations:
x+y+z=6
2x+y=7
z=1
Show complete worked solution
From the third equation:
$$z=1$$
Substitute into the first equation:
$$x+y+1=6$$
$$x+y=5 \quad (1)$$
The second equation is:
$$2x+y=7 \quad (2)$$
Subtract (1) from (2) to eliminate $y$:
$$(2x+y)-(x+y)=7-5$$
$$x=2$$
Substitute $x=2$ into (1):
$$2+y=5$$
$$y=3$$
Check in (2): $2(2)+3=4+3=7$ $\checkmark$
$$\boxed{x=2,\ y=3,\ z=1}$$
QUESTION 27
7 marks
Medium
Solve the system of equations:
x+y+z=8
2x-y+3z=16
x+2y-z=3
Show complete worked solution
Using Gaussian elimination:
$$x+y+z=8 \quad (1)$$
$$2x-y+3z=16 \quad (2)$$
$$x+2y-z=3 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x-y+3z)-2(x+y+z)=16-2(8)$$
$$-3y+z=0 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-$ (1):
$$(x+2y-z)-(x+y+z)=3-8$$
$$y-2z=-5 \quad (5)$$
Now eliminate $y$ from (4) and (5): compute (4) $+\,3\times$(5):
$$(-3y+z)+3(y-2z)=0+3(-5)$$
$$-5z=-15$$
$$z=3$$
Substitute $z=3$ into (4):
$$-3y+3=0$$
$$y=1$$
Substitute $y=1$, $z=3$ into (1):
$$x+1+3=8$$
$$x=4$$
Check in (2): $2(4)-1+3(3)=8-1+9=16$ $\checkmark$
$$\boxed{x=4,\ y=1,\ z=3}$$
QUESTION 28
7 marks
Medium
Solve the system of equations:
x+2y+3z=9
2x+3y-z=-2
3x-y+2z=13
Show complete worked solution
Using Gaussian elimination:
$$x+2y+3z=9 \quad (1)$$
$$2x+3y-z=-2 \quad (2)$$
$$3x-y+2z=13 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x+3y-z)-2(x+2y+3z)=-2-2(9)$$
$$-y-7z=-20 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-\,3\times$(1):
$$(3x-y+2z)-3(x+2y+3z)=13-3(9)$$
$$-7y-7z=-14 \quad (5)$$
Now eliminate $y$ from (4) and (5): compute (5) $-\,7\times$(4):
$$(-7y-7z)-7(-y-7z)=-14-7(-20)$$
$$42z=126$$
$$z=3$$
Substitute $z=3$ into (4):
$$-y-7(3)=-20$$
$$-y-21=-20$$
$$y=-1$$
Substitute $y=-1$, $z=3$ into (1):
$$x+2(-1)+3(3)=9$$
$$x-2+9=9$$
$$x=2$$
Check in (3): $3(2)-(-1)+2(3)=6+1+6=13$ $\checkmark$
$$\boxed{x=2,\ y=-1,\ z=3}$$
QUESTION 29
7 marks
Medium
Solve the system of equations:
x-2y+z=0
3x+y-2z=19
2x+3y+z=15
Show complete worked solution
Using Gaussian elimination:
$$x-2y+z=0 \quad (1)$$
$$3x+y-2z=19 \quad (2)$$
$$2x+3y+z=15 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,3\times$(1):
$$(3x+y-2z)-3(x-2y+z)=19-3(0)$$
$$7y-5z=19 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-\,2\times$(1):
$$(2x+3y+z)-2(x-2y+z)=15-2(0)$$
$$7y-z=15 \quad (5)$$
Now eliminate $y$ from (4) and (5): compute (5) $-$ (4):
$$(7y-z)-(7y-5z)=15-19$$
$$4z=-4$$
$$z=-1$$
Substitute $z=-1$ into (5):
$$7y-(-1)=15$$
$$7y=14$$
$$y=2$$
Substitute $y=2$, $z=-1$ into (1):
$$x-2(2)+(-1)=0$$
$$x-5=0$$
$$x=5$$
Check in (2): $3(5)+2-2(-1)=15+2+2=19$ $\checkmark$
$$\boxed{x=5,\ y=2,\ z=-1}$$
QUESTION 30
7 marks
Medium
Solve the system of equations:
x+y-2z=3
2x-3y+z=1
3x+y+4z=15
Show complete worked solution
Using Gaussian elimination:
$$x+y-2z=3 \quad (1)$$
$$2x-3y+z=1 \quad (2)$$
$$3x+y+4z=15 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x-3y+z)-2(x+y-2z)=1-2(3)$$
$$-5y+5z=-5 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-\,3\times$(1):
$$(3x+y+4z)-3(x+y-2z)=15-3(3)$$
$$-2y+10z=6 \quad (5)$$
Now eliminate $y$ from (4) and (5): compute $5\times$(5) $-\,2\times$(4):
$$5(-2y+10z)-2(-5y+5z)=5(6)-2(-5)$$
$$40z=40$$
$$z=1$$
Substitute $z=1$ into (4):
$$-5y+5(1)=-5$$
$$-5y=-10$$
$$y=2$$
Substitute $y=2$, $z=1$ into (1):
$$x+2-2(1)=3$$
$$x=3$$
Check in (3): $3(3)+2+4(1)=9+2+4=15$ $\checkmark$
$$\boxed{x=3,\ y=2,\ z=1}$$
QUESTION 31
7 marks
Medium
Solve the system of equations:
x+3y-z=15
2x+y+3z=0
-x+2y+2z=3
Show complete worked solution
Using Gaussian elimination:
$$x+3y-z=15 \quad (1)$$
$$2x+y+3z=0 \quad (2)$$
$$-x+2y+2z=3 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x+y+3z)-2(x+3y-z)=0-2(15)$$
$$-5y+5z=-30 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $+$ (1):
$$(-x+2y+2z)+(x+3y-z)=3+15$$
$$5y+z=18 \quad (5)$$
Now eliminate $y$ from (4) and (5): compute (5) $+$ (4):
$$(5y+z)+(-5y+5z)=18+(-30)$$
$$6z=-12$$
$$z=-2$$
Substitute $z=-2$ into (5):
$$5y+(-2)=18$$
$$5y=20$$
$$y=4$$
Substitute $y=4$, $z=-2$ into (1):
$$x+3(4)-(-2)=15$$
$$x+14=15$$
$$x=1$$
Check in (2): $2(1)+4+3(-2)=2+4-6=0$ $\checkmark$
$$\boxed{x=1,\ y=4,\ z=-2}$$
QUESTION 32
7 marks
Medium
Solve the system of equations:
x-y+2z=1
3x+2y-z=11
2x+3y+z=14
Show complete worked solution
Using Gaussian elimination:
$$x-y+2z=1 \quad (1)$$
$$3x+2y-z=11 \quad (2)$$
$$2x+3y+z=14 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,3\times$(1):
$$(3x+2y-z)-3(x-y+2z)=11-3(1)$$
$$5y-7z=8 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-\,2\times$(1):
$$(2x+3y+z)-2(x-y+2z)=14-2(1)$$
$$5y-3z=12 \quad (5)$$
Now eliminate $y$ from (4) and (5): compute (5) $-$ (4):
$$(5y-3z)-(5y-7z)=12-8$$
$$4z=4$$
$$z=1$$
Substitute $z=1$ into (5):
$$5y-3(1)=12$$
$$5y=15$$
$$y=3$$
Substitute $y=3$, $z=1$ into (1):
$$x-3+2(1)=1$$
$$x-1=1$$
$$x=2$$
Check in (3): $2(2)+3(3)+1=4+9+1=14$ $\checkmark$
$$\boxed{x=2,\ y=3,\ z=1}$$
QUESTION 33
7 marks
Medium
Solve the system of equations:
x+2y-z=3
3x-y+2z=-2
2x+y+3z=7
Show complete worked solution
Using Gaussian elimination:
$$x+2y-z=3 \quad (1)$$
$$3x-y+2z=-2 \quad (2)$$
$$2x+y+3z=7 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,3\times$(1):
$$(3x-y+2z)-3(x+2y-z)=-2-3(3)$$
$$-7y+5z=-11 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-\,2\times$(1):
$$(2x+y+3z)-2(x+2y-z)=7-2(3)$$
$$-3y+5z=1 \quad (5)$$
Now eliminate $y$ from (4) and (5): compute $3\times$(4) $-\,7\times$(5):
$$3(-7y+5z)-7(-3y+5z)=3(-11)-7(1)$$
$$-20z=-40$$
$$z=2$$
Substitute $z=2$ into (5):
$$-3y+5(2)=1$$
$$-3y=-9$$
$$y=3$$
Substitute $y=3$, $z=2$ into (1):
$$x+2(3)-2=3$$
$$x+4=3$$
$$x=-1$$
Check in (2): $3(-1)-3+2(2)=-3-3+4=-2$ $\checkmark$
$$\boxed{x=-1,\ y=3,\ z=2}$$
QUESTION 34
7 marks
Medium
Solve the system of equations:
x+y+2z=7
2x-y+z=11
3x+2y-z=-4
Show complete worked solution
Using Gaussian elimination:
$$x+y+2z=7 \quad (1)$$
$$2x-y+z=11 \quad (2)$$
$$3x+2y-z=-4 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x-y+z)-2(x+y+2z)=11-2(7)$$
$$-3y-3z=-3 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-\,3\times$(1):
$$(3x+2y-z)-3(x+y+2z)=-4-3(7)$$
$$-y-7z=-25 \quad (5)$$
Now eliminate $y$ from (4) and (5): compute (4) $-\,3\times$(5):
$$(-3y-3z)-3(-y-7z)=-3-3(-25)$$
$$18z=72$$
$$z=4$$
Substitute $z=4$ into (5):
$$-y-7(4)=-25$$
$$-y-28=-25$$
$$y=-3$$
Substitute $y=-3$, $z=4$ into (1):
$$x+(-3)+2(4)=7$$
$$x+5=7$$
$$x=2$$
Check in (3): $3(2)+2(-3)-4=6-6-4=-4$ $\checkmark$
$$\boxed{x=2,\ y=-3,\ z=4}$$
QUESTION 35
8 marks
Medium
An investor divides \$10000 between three accounts paying simple annual interest of 5\%, 4\% and 6\% respectively. Let \$x, \$y and \$z be the amounts invested at each rate. The total interest earned in one year is \$490. The amount invested at 5\% is \$1000 less than twice the amount invested at 4\%.
Find $x$, $y$ and $z$.
Show complete worked solution
Let the equations be:
$$x+y+z=10000 \quad (1)$$
$$0.05x+0.04y+0.06z=490 \quad (2)$$
$$x=2y-1000 \quad (3)$$
Rearrange (3): $x-2y=-1000 \quad (3')$
Eliminate $x$ from (2) using (1): compute (2) $-\,0.05\times$(1):
$$(0.05x+0.04y+0.06z)-0.05(x+y+z)=490-0.05(10000)$$
$$-0.01y+0.01z=-10$$
Multiply through by $-100$:
$$y-z=1000 \quad (4)$$
Eliminate $x$ from (3') using (1): compute (1) $-$ (3'):
$$(x+y+z)-(x-2y)=10000-(-1000)$$
$$3y+z=11000 \quad (5)$$
Now eliminate $z$ from (4) and (5): compute (4) $+$ (5):
$$(y-z)+(3y+z)=1000+11000$$
$$4y=12000$$
$$y=3000$$
Substitute $y=3000$ into (4):
$$3000-z=1000$$
$$z=2000$$
Substitute $y=3000$, $z=2000$ into (1):
$$x+3000+2000=10000$$
$$x=5000$$
Check in (2): $0.05(5000)+0.04(3000)+0.06(2000)=250+120+120=490$ $\checkmark$
$$\boxed{x=\$5000,\ y=\$3000,\ z=\$2000}$$
QUESTION 36
8 marks
Medium
At a concert, tickets are sold in three categories: adult (\$12), child (\$8) and senior (\$6). A total of 80 tickets are sold, raising \$770. The number of adult tickets sold is 15 more than the number of child tickets sold.
Let $a$, $c$ and $s$ be the numbers of adult, child and senior tickets sold. Find $a$, $c$ and $s$.
Show complete worked solution
Let the equations be:
$$a+c+s=80 \quad (1)$$
$$12a+8c+6s=770 \quad (2)$$
$$a-c=15 \quad (3)$$
Eliminate $a$ from (2) using (1): compute (2) $-\,12\times$(1):
$$(12a+8c+6s)-12(a+c+s)=770-12(80)$$
$$-4c-6s=-190$$
Divide by $-2$:
$$2c+3s=95 \quad (4)$$
Eliminate $a$ from (3) using (1): compute (1) $-$ (3):
$$(a+c+s)-(a-c)=80-15$$
$$2c+s=65 \quad (5)$$
From (5): $s=65-2c$. Substitute into (4):
$$2c+3(65-2c)=95$$
$$2c+195-6c=95$$
$$-4c=-100$$
$$c=25$$
Substitute $c=25$ into (5):
$$s=65-2(25)=15$$
Substitute $c=25$, $s=15$ into (1):
$$a+25+15=80$$
$$a=40$$
Check in (2): $12(40)+8(25)+6(15)=480+200+90=770$ $\checkmark$
$$\boxed{a=40,\ c=25,\ s=15}$$
QUESTION 37
8 marks
Medium
A pharmacist combines three saline solutions with concentrations 10\%, 20\% and 50\% to produce 100 litres of solution containing 23\% salt. Let $x$, $y$ and $z$ be the number of litres of the 10\%, 20\% and 50\% solutions used. Twice the amount of the 10\% solution is 10 litres more than the amount of the 20\% solution.
Find $x$, $y$ and $z$.
Show complete worked solution
Let the equations be:
$$x+y+z=100 \quad (1)$$
$$0.10x+0.20y+0.50z=23 \quad (2)$$
$$2x=y-10 \quad (3)$$
Rearrange (3): $-2x+y=-10 \quad (3')$
Eliminate $x$ from (2) using (1): compute (2) $-\,0.1\times$(1):
$$(0.10x+0.20y+0.50z)-0.1(x+y+z)=23-0.1(100)$$
$$0.1y+0.4z=13$$
Multiply through by 10:
$$y+4z=130 \quad (4)$$
Eliminate $x$ from (3') using (1): compute (3') $+\,2\times$(1):
$$(-2x+y)+2(x+y+z)=-10+2(100)$$
$$3y+2z=190 \quad (5)$$
From (4): $y=130-4z$. Substitute into (5):
$$3(130-4z)+2z=190$$
$$390-12z+2z=190$$
$$-10z=-200$$
$$z=20$$
Substitute $z=20$ into (4):
$$y=130-4(20)=50$$
Substitute $y=50$, $z=20$ into (1):
$$x+50+20=100$$
$$x=30$$
Check in (2): $0.10(30)+0.20(50)+0.50(20)=3+10+10=23$ $\checkmark$
$$\boxed{x=30\text{ L},\ y=50\text{ L},\ z=20\text{ L}}$$
QUESTION 38
7 marks
Medium
The three interior angles of a triangle are $A$, $B$ and $C$ degrees. Angle $B$ exceeds angle $A$ by $20\degree$, and angle $B$ exceeds angle $C$ by $10\degree$.
Find $A$, $B$ and $C$.
Show complete worked solution
The angle sum of a triangle gives:
$$A+B+C=180 \quad (1)$$
$$A-B=-20 \quad (2)$$
$$B-C=10 \quad (3)$$
Eliminate $A$ from (2) using (1): compute (2) $-$ (1):
$$(A-B)-(A+B+C)=-20-180$$
$$-2B-C=-200$$
Multiply by $-1$:
$$2B+C=200 \quad (4)$$
Now eliminate $C$ from (4) and (3): compute (4) $+$ (3):
$$(2B+C)+(B-C)=200+10$$
$$3B=210$$
$$B=70$$
Substitute $B=70$ into (3):
$$70-C=10$$
$$C=60$$
Substitute $B=70$, $C=60$ into (1):
$$A+70+60=180$$
$$A=50$$
Check in (2): $50-70=-20$ $\checkmark$
$$\boxed{A=50\degree,\ B=70\degree,\ C=60\degree}$$
QUESTION 39
8 marks
Medium
Solve the system of equations:
x+2y+3z=11
2x-y+6z=17
x+3y-3z=-2
Show complete worked solution
Using Gaussian elimination:
$$x+2y+3z=11 \quad (1)$$
$$2x-y+6z=17 \quad (2)$$
$$x+3y-3z=-2 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x-y+6z)-2(x+2y+3z)=17-2(11)$$
$$-5y=-5 \quad (4)$$
Since (4) has no $z$-term, solve directly:
$$y=1$$
Eliminate $x$ from (3) using (1): compute (3) $-$ (1):
$$(x+3y-3z)-(x+2y+3z)=-2-11$$
$$y-6z=-13 \quad (5)$$
Substitute $y=1$ into (5):
$$1-6z=-13$$
$$-6z=-14$$
$$z=\dfrac{14}{6}=\dfrac{7}{3}$$
Substitute $y=1$, $z=\dfrac{7}{3}$ into (1):
$$x+2(1)+3\left(\dfrac{7}{3}\right)=11$$
$$x+2+7=11$$
$$x=2$$
Check in (2): $2(2)-1+6\left(\dfrac{7}{3}\right)=4-1+14=17$ $\checkmark$
$$\boxed{(x,y,z)=\left(2,\ 1,\ \dfrac{7}{3}\right)}$$
QUESTION 40
8 marks
Medium
Solve the system of equations:
x+5y+2z=9
2x-5y-z=-1
x+10y+3z=16
Show complete worked solution
Using Gaussian elimination:
$$x+5y+2z=9 \quad (1)$$
$$2x-5y-z=-1 \quad (2)$$
$$x+10y+3z=16 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x-5y-z)-2(x+5y+2z)=-1-2(9)$$
$$-15y-5z=-19 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-$ (1):
$$(x+10y+3z)-(x+5y+2z)=16-9$$
$$5y+z=7 \quad (5)$$
Now eliminate $z$ from (4) and (5): compute (4) $+\,5\times$(5):
$$(-15y-5z)+5(5y+z)=-19+5(7)$$
$$10y=16$$
$$y=\dfrac{16}{10}=\dfrac{8}{5}$$
Substitute $y=\dfrac{8}{5}$ into (5):
$$5\left(\dfrac{8}{5}\right)+z=7$$
$$8+z=7$$
$$z=-1$$
Substitute $y=\dfrac{8}{5}$, $z=-1$ into (1):
$$x+5\left(\dfrac{8}{5}\right)+2(-1)=9$$
$$x+8-2=9$$
$$x=3$$
Check in (2): $2(3)-5\left(\dfrac{8}{5}\right)-(-1)=6-8+1=-1$ $\checkmark$
$$\boxed{(x,y,z)=\left(3,\ \dfrac{8}{5},\ -1\right)}$$
QUESTION 41
9 marks
Medium
Consider the three planes
$$\pi_1: x+y-z=1$$
$$\pi_2: 2x+3y+z=4$$
$$\pi_3: x-2y+3z=8$$
(a) [7] Solve the system of equations to find the point of intersection of $\pi_1$, $\pi_2$ and $\pi_3$.
(b) [2] State what your answer to part (a) tells you about the geometric relationship between the three planes.
Show complete worked solution
(a) Label the equations:
$$x+y-z=1 \quad (1)$$
$$2x+3y+z=4 \quad (2)$$
$$x-2y+3z=8 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x+3y+z)-2(x+y-z)=4-2(1)$$
$$y+3z=2 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-$ (1):
$$(x-2y+3z)-(x+y-z)=8-1$$
$$-3y+4z=7 \quad (5)$$
Now eliminate $y$ from (4) and (5): compute (5) $+\,3\times$(4):
$$(-3y+4z)+3(y+3z)=7+3(2)$$
$$13z=13$$
$$z=1$$
Substitute $z=1$ into (4):
$$y+3(1)=2$$
$$y=-1$$
Substitute $y=-1$, $z=1$ into (1):
$$x+(-1)-1=1$$
$$x=3$$
Check in (3): $3-2(-1)+3(1)=3+2+3=8$ $\checkmark$
$$\boxed{(x,y,z)=(3,-1,1)}$$
(b) Since the system has exactly one solution, the three planes $\pi_1$, $\pi_2$ and $\pi_3$ intersect at the single point $\boxed{(3,-1,1)}$ -- this is the only point that lies on all three planes simultaneously.
QUESTION 42
9 marks
Medium
Consider the three planes
$$\pi_1: x-3y+2z=-5$$
$$\pi_2: 2x+y-z=7$$
$$\pi_3: 3x+2y+z=6$$
(a) [7] Solve the system of equations to find the point of intersection of $\pi_1$, $\pi_2$ and $\pi_3$.
(b) [2] Explain, without further calculation, why a system of three linear equations in three unknowns cannot have exactly two solutions.
Show complete worked solution
(a) Label the equations:
$$x-3y+2z=-5 \quad (1)$$
$$2x+y-z=7 \quad (2)$$
$$3x+2y+z=6 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x+y-z)-2(x-3y+2z)=7-2(-5)$$
$$7y-5z=17 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-\,3\times$(1):
$$(3x+2y+z)-3(x-3y+2z)=6-3(-5)$$
$$11y-5z=21 \quad (5)$$
Now eliminate $z$ from (4) and (5): compute (5) $-$ (4):
$$(11y-5z)-(7y-5z)=21-17$$
$$4y=4$$
$$y=1$$
Substitute $y=1$ into (4):
$$7(1)-5z=17$$
$$-5z=10$$
$$z=-2$$
Substitute $y=1$, $z=-2$ into (1):
$$x-3(1)+2(-2)=-5$$
$$x-3-4=-5$$
$$x=2$$
Check in (3): $3(2)+2(1)+(-2)=6+2-2=6$ $\checkmark$
$$\boxed{(x,y,z)=(2,1,-2)}$$
(b) Each equation represents a plane in three-dimensional space, and each row operation used to solve the system produces an equivalent linear equation (a line, in the reduced 2-variable stage, or a single value once fully reduced). A system of linear equations always reduces to exactly one of three outcomes: a unique solution (the coefficient determinant is non-zero, as here), infinitely many solutions (the equations reduce to fewer independent constraints than unknowns, giving a common line or plane of solutions), or no solution (the reduction produces a contradiction such as $0=k$ for some $k\ne0$). There is no way for consistent linear equations to intersect in exactly two isolated points -- two distinct planes either do not meet, meet in a whole line, or (if identical) meet in a whole plane, and a line together with a third plane either meets it in one point, lies entirely in it, or misses it entirely. $\boxed{\text{A linear system has 0, 1, or infinitely many solutions -- never exactly 2.}}$
QUESTION 43
14 marks
Hard
Consider the system of equations
2x+y-z=4
x-y+3z=-2
x+2y+\lambda z=\mu
where $\lambda,\mu\in\mathbb{R}$.
(a) [4] Show that the determinant of the coefficient matrix is $-3\lambda-12$, and hence state the value of $\lambda$ for which the system does not have a unique solution.
(b) [6] For this value of $\lambda$, find the value of $\mu$ for which the system is consistent, and find the general solution in this case.
(c) [4] For the same value of $\lambda$, but with $\mu$ not equal to the value found in part (b), explain why the system has no solution, and describe the geometric configuration of the three planes in each of the cases (i) $\lambda\ne-4$, (ii) $\lambda=-4,\ \mu$ as in (b), (iii) $\lambda=-4$, $\mu$ otherwise.
Show complete worked solution
(a) The coefficient matrix is
$$\begin{pmatrix} 2 & 1 & -1 \\ 1 & -1 & 3 \\ 1 & 2 & \lambda \end{pmatrix}$$
Expanding the determinant along the first row:
$$\det = 2\left[(-1)(\lambda)-(3)(2)\right] - 1\left[(1)(\lambda)-(3)(1)\right] + (-1)\left[(1)(2)-(-1)(1)\right]$$
$$=2(-\lambda-6) - 1(\lambda-3) - 1(2+1)$$
$$=(-2\lambda-12) - (\lambda-3) - 3$$
$$=-2\lambda-12-\lambda+3-3$$
$$=-3\lambda-12$$
as required. The system fails to have a unique solution when the determinant is zero:
$$-3\lambda-12=0 \ \Rightarrow\ \boxed{\lambda=-4}$$
(b) Set $\lambda=-4$. The system becomes:
$$2x+y-z=4 \quad (1)$$
$$x-y+3z=-2 \quad (2)$$
$$x+2y-4z=\mu \quad (3)$$
Add (1) and (2) to eliminate $y$:
$$(2x+y-z)+(x-y+3z)=4+(-2)$$
$$3x+2z=2 \quad (4)$$
Let $z=t$. From (4):
$$3x=2-2t \ \Rightarrow\ x=\dfrac{2-2t}{3}$$
Substitute into (2) to find $y$:
$$\dfrac{2-2t}{3}-y+3t=-2$$
$$y=\dfrac{2-2t}{3}+3t+2=\dfrac{2-2t+9t+6}{3}=\dfrac{8+7t}{3}$$
So the general solution, for any real $t$, is $x=\dfrac{2-2t}{3}$, $y=\dfrac{8+7t}{3}$, $z=t$. For the system to be consistent, this must also satisfy (3) for every value of $t$:
$$\dfrac{2-2t}{3}+2\left(\dfrac{8+7t}{3}\right)-4t=\dfrac{(2-2t)+2(8+7t)}{3}-4t=\dfrac{2-2t+16+14t}{3}-4t=\dfrac{18+12t}{3}-4t=6+4t-4t=6$$
Since this equals $6$ regardless of $t$, the system is consistent precisely when $\mu=6$.
$$\boxed{\mu=6; \quad x=\dfrac{2-2t}{3},\ \ y=\dfrac{8+7t}{3},\ \ z=t,\ \ t\in\mathbb{R}}$$
(c) When $\lambda=-4$ and $\mu\ne6$, the working in part (b) shows that equations (1) and (2) force $x+2y-4z$ to equal $6$ identically (for every value of $t$) -- but equation (3) requires $x+2y-4z=\mu\ne6$. No values of $x,y,z$ can satisfy this contradiction, so the system has no solution.
Geometric interpretation:
(i) When $\lambda\ne-4$, the determinant is non-zero, so the three planes intersect at exactly one point (unique solution).
(ii) When $\lambda=-4$ and $\mu=6$, the three planes share a common line of intersection, given by the general solution in part (b) (infinitely many solutions lying on that line).
(iii) When $\lambda=-4$ and $\mu\ne6$, the planes have no common point: their pairwise lines of intersection are parallel but distinct, so the three planes form a triangular prism.
$$\boxed{\text{(i) unique point; (ii) common line; (iii) no common point (triangular prism)}}$$
QUESTION 44
16 marks
Hard
Consider the system of equations
x+y+\lambda z=1
x+\lambda y+z=1
\lambda x+y+z=1
where $\lambda\in\mathbb{R}$.
(a) [6] Show that the determinant of the coefficient matrix can be written as $-(\lambda-1)^2(\lambda+2)$, and hence find all values of $\lambda$ for which the system does not have a unique solution.
(b) [5] Solve the system when $\lambda=1$, and describe geometrically the set of all solutions.
(c) [5] Show that the system has no solution when $\lambda=-2$, and describe the geometric configuration of the three planes in this case.
Show complete worked solution
(a) The coefficient matrix is
$$\begin{pmatrix} 1 & 1 & \lambda \\ 1 & \lambda & 1 \\ \lambda & 1 & 1 \end{pmatrix}$$
Expanding the determinant along the first row:
$$\det = 1\left[(\lambda)(1)-(1)(1)\right] - 1\left[(1)(1)-(1)(\lambda)\right] + \lambda\left[(1)(1)-(\lambda)(\lambda)\right]$$
$$=(\lambda-1) - (1-\lambda) + \lambda(1-\lambda^2)$$
Since $1-\lambda=-(\lambda-1)$, the first two terms combine:
$$(\lambda-1)-(1-\lambda)=(\lambda-1)+(\lambda-1)=2(\lambda-1)$$
and $1-\lambda^2=(1-\lambda)(1+\lambda)=-(\lambda-1)(\lambda+1)$, so:
$$\det=2(\lambda-1)-\lambda(\lambda-1)(\lambda+1)=(\lambda-1)\left[2-\lambda(\lambda+1)\right]=(\lambda-1)(2-\lambda-\lambda^2)$$
Factorising the quadratic $2-\lambda-\lambda^2=-(\lambda^2+\lambda-2)=-(\lambda-1)(\lambda+2)$:
$$\det=(\lambda-1)\cdot\left[-(\lambda-1)(\lambda+2)\right]=-(\lambda-1)^2(\lambda+2)$$
as required. The system fails to have a unique solution when $\det=0$:
$$-(\lambda-1)^2(\lambda+2)=0 \ \Rightarrow\ \boxed{\lambda=1\ \text{(repeated) or}\ \lambda=-2}$$
(b) Set $\lambda=1$. All three equations become identical:
$$x+y+z=1$$
$$x+y+z=1$$
$$x+y+z=1$$
Since the three equations are the same single equation, any point $(x,y,z)$ satisfying $x+y+z=1$ is a solution. Letting $y=s$ and $z=t$ be free parameters:
$$\boxed{x=1-s-t,\ \ y=s,\ \ z=t,\ \ s,t\in\mathbb{R}}$$
Geometrically, when $\lambda=1$ the three planes are not just parallel but identical -- all three equations describe the same plane $x+y+z=1$. The solution set is therefore this entire plane (a two-parameter family of solutions), not merely a line.
(c) Set $\lambda=-2$. The system becomes:
$$x+y-2z=1 \quad (1)$$
$$x-2y+z=1 \quad (2)$$
$$-2x+y+z=1 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-$ (1):
$$(x-2y+z)-(x+y-2z)=1-1$$
$$-3y+3z=0 \ \Rightarrow\ y-z=0 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $+\,2\times$(1):
$$(-2x+y+z)+2(x+y-2z)=1+2(1)$$
$$3y-3z=3 \ \Rightarrow\ y-z=1 \quad (5)$$
Equations (4) and (5) both express $y-z$, but give the contradictory values $0$ and $1$. Since $0\ne1$, no values of $x,y,z$ can satisfy the system.
$$\boxed{\text{No solution when }\lambda=-2}$$
Geometric interpretation: since the coefficient determinant is zero but the equations are inconsistent, the three planes do not share a common point. By the symmetry of the coefficients, no two of the three planes are parallel (their normal vectors $(1,1,-2)$, $(1,-2,1)$, $(-2,1,1)$ are pairwise non-proportional), so each pair of planes intersects in its own line; equations (4) and (5) show these three lines of intersection are parallel to each other but distinct. The three planes therefore form a triangular prism, with no point lying on all three.
QUESTION 45
11 marks
Hard
Consider the three planes
$$\pi_1: x+2y-z=4$$
$$\pi_2: 2x+4y-2z=9$$
$$\pi_3: x-y+z=1$$
(a) [5] Write down the augmented matrix for this system, and use elementary row operations to begin reducing it to row-echelon form. State clearly what your reduction shows about the system.
(b) [3] By comparing the coefficients of $\pi_1$ and $\pi_2$ directly (without further row reduction), explain algebraically why the system is inconsistent.
(c) [3] Describe the geometric relationship between $\pi_1$, $\pi_2$ and $\pi_3$.
Show complete worked solution
(a) The augmented matrix is
$$\left(\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 2 & 4 & -2 & 9 \\ 1 & -1 & 1 & 1 \end{array}\right)$$
Apply $R_2 \to R_2-2R_1$ to eliminate $x$ from row 2:
$$R_2-2R_1 = (2-2(1),\ 4-2(2),\ -2-2(-1),\ 9-2(4)) = (0,\ 0,\ 0,\ 1)$$
$$\left(\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 0 & 0 & 0 & 1 \\ 1 & -1 & 1 & 1 \end{array}\right)$$
Row 2 now reads $0x+0y+0z=1$, i.e. $0=1$, which is never true. This is already a contradiction, so the system is inconsistent -- there is no need to continue reducing row 3, since no values of $x,y,z$ can satisfy row 2.
$$\boxed{\text{The system is inconsistent: no solution exists.}}$$
(b) Comparing the coefficients of $\pi_1$ and $\pi_2$ directly: the coefficients of $\pi_2$ are exactly twice those of $\pi_1$, since $(2,4,-2)=2(1,2,-1)$. This means $\pi_1$ and $\pi_2$ have parallel (in fact, scalar-multiple) normal vectors, so $\pi_1$ and $\pi_2$ are parallel planes. If $\pi_2$ were simply $2\times\pi_1$, its constant term would be $2(4)=8$; but the actual constant term of $\pi_2$ is $9\ne8$. Hence $\pi_1$ and $\pi_2$ are parallel and distinct planes, which never intersect -- so no point can lie on both, and the full system (which requires a point on all three planes) has no solution regardless of $\pi_3$.
(c) $\pi_1$ and $\pi_2$ are parallel, distinct planes (they never meet). $\pi_3$ has normal vector $(1,-1,1)$, which is not a scalar multiple of $(1,2,-1)$, so $\pi_3$ is not parallel to $\pi_1$ or $\pi_2$; it cuts each of them in its own line. Because $\pi_1\parallel\pi_2$, these two lines of intersection ($\pi_1\cap\pi_3$ and $\pi_2\cap\pi_3$) are themselves parallel to each other. So $\pi_3$ crosses two parallel "sheets" $\pi_1,\pi_2$ in two parallel lines, and there is no point common to all three planes.
$$\boxed{\pi_1\parallel\pi_2\text{ (distinct)}; \ \pi_3\text{ meets each in a line; the three planes share no common point.}}$$
QUESTION 46
12 marks
Hard
Consider the system of equations
x+y+z=1
2x+3y+4z=3
3x+4y+5z=\lambda
where $\lambda\in\mathbb{R}$.
(a) [3] Show, using a row operation, that the determinant of the coefficient matrix is zero for every value of $\lambda$.
(b) [6] Find the value of $\lambda$ for which the system is consistent, and find the general solution in this case.
(c) [3] Explain why, for this system, no value of $\lambda$ can ever produce a unique solution, and describe the geometric configuration of the three planes for (i) $\lambda$ equal to the value found in part (b), and (ii) any other value of $\lambda$.
Show complete worked solution
(a) The coefficient matrix, which does not involve $\lambda$ at all (since $\lambda$ only appears in the constant term of equation 3), is
$$\begin{pmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix}$$
Apply the row operation $R_3\to R_3-R_2-R_1$:
$$R_3-R_2-R_1=(3-2-1,\ 4-3-1,\ 5-4-1)=(0,0,0)$$
Since row 3 becomes entirely zero, the rows of the matrix are linearly dependent, so:
$$\boxed{\det=0 \text{ for every value of }\lambda}$$
(b) Label the equations:
$$x+y+z=1 \quad (1)$$
$$2x+3y+4z=3 \quad (2)$$
$$3x+4y+5z=\lambda \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x+3y+4z)-2(x+y+z)=3-2(1)$$
$$y+2z=1 \quad (4)$$
From (4), $y=1-2z$. Let $z=t$. Substitute into (1):
$$x+(1-2t)+t=1$$
$$x=t$$
So the general solution has the form $x=t$, $y=1-2t$, $z=t$. Since row 3 was shown in part (a) to equal row1+row2 for the coefficients, consistency requires the same relation for the constants:
$$\lambda = 1+3=4$$
Check in (3) directly: $3(t)+4(1-2t)+5(t)=3t+4-8t+5t=4+0t=4=\lambda$ $\checkmark$ (true for every $t$, confirming consistency exactly when $\lambda=4$).
$$\boxed{\lambda=4; \quad x=t,\ \ y=1-2t,\ \ z=t,\ \ t\in\mathbb{R}}$$
(c) Because the coefficient matrix has determinant zero for every value of $\lambda$ (part (a)) -- the coefficients of $\lambda$-containing equation 3 are fixed and always equal to row1+row2 -- the three plane normals are always linearly dependent, regardless of $\lambda$. A unique solution requires a non-zero coefficient determinant, so this system can never have a unique solution, for any value of $\lambda$.
(i) When $\lambda=4$, the system is consistent with infinitely many solutions: the three planes share the common line found in part (b).
(ii) When $\lambda\ne4$, exactly as in part (a) of a typical parameter question, row 3 minus row 2 minus row 1 gives $0=\lambda-4\ne0$, a contradiction, so the system has no solution -- the three planes form a triangular prism (pairwise intersections are three parallel, distinct lines).
$$\boxed{\text{Never a unique point; (i) common line when }\lambda=4\text{; (ii) triangular prism otherwise.}}$$
QUESTION 47
11 marks
Hard
Consider the system of equations
2x+y-z=-3
x-3y+2z=13
3x+2y+z=2
(a) [7] Write the system as an augmented matrix and reduce it to row-echelon form, stating each row operation used. Hence solve the system.
(b) [4] Verify your solution by substitution into all three original equations, and state what your solution represents geometrically.
Show complete worked solution
(a) The augmented matrix is
$$\left(\begin{array}{ccc|c} 2 & 1 & -1 & -3 \\ 1 & -3 & 2 & 13 \\ 3 & 2 & 1 & 2 \end{array}\right)$$
Apply $R_2\to 2R_2-R_1$ to eliminate $x$ from row 2:
$$2R_2-R_1=(2(1)-2,\ 2(-3)-1,\ 2(2)-(-1),\ 2(13)-(-3))=(0,-7,5,29)$$
Apply $R_3\to 2R_3-3R_1$ to eliminate $x$ from row 3:
$$2R_3-3R_1=(2(3)-3(2),\ 2(2)-3(1),\ 2(1)-3(-1),\ 2(2)-3(-3))=(0,1,5,13)$$
The matrix is now:
$$\left(\begin{array}{ccc|c} 2 & 1 & -1 & -3 \\ 0 & -7 & 5 & 29 \\ 0 & 1 & 5 & 13 \end{array}\right)$$
Apply $R_3\to 7R_3+R_2$ to eliminate $y$ from row 3:
$$7R_3+R_2=(0,\ 7(1)+(-7),\ 7(5)+5,\ 7(13)+29)=(0,0,40,120)$$
This gives the row-echelon form:
$$\left(\begin{array}{ccc|c} 2 & 1 & -1 & -3 \\ 0 & -7 & 5 & 29 \\ 0 & 0 & 40 & 120 \end{array}\right)$$
From row 3: $40z=120 \Rightarrow z=3$.
Back-substitute into row 2: $-7y+5(3)=29 \Rightarrow -7y=14 \Rightarrow y=-2$.
Back-substitute into row 1: $2x+(-2)-3=-3 \Rightarrow 2x=2 \Rightarrow x=1$.
$$\boxed{(x,y,z)=(1,-2,3)}$$
(b) Check in the original equations:
Equation 1: $2(1)+(-2)-3=2-2-3=-3$ $\checkmark$
Equation 2: $1-3(-2)+2(3)=1+6+6=13$ $\checkmark$
Equation 3: $3(1)+2(-2)+3=3-4+3=2$ $\checkmark$
All three equations are satisfied, confirming the solution. Since a unique triple $(x,y,z)$ satisfies all three equations, the three planes represented by the equations intersect at exactly the single point $\boxed{(1,-2,3)}$ in three-dimensional space -- this is the only point common to all three planes.
QUESTION 48
13 marks
Hard
A factory blends three raw materials, of masses $x$, $y$ and $z$ kilograms, subject to the constraints:
Total mass: $x+y+z=10$
Cost constraint: $2x+3y+z=22$
Quality constraint: $3x+\lambda y+2z=Q$
where $\lambda$ is a sensitivity parameter of the blending process and $Q$ is a target quality score.
(a) [4] Show that the determinant of the coefficient matrix is $\lambda-4$, and hence state the value of $\lambda$ for which the blend masses are not uniquely determined by the constraints.
(b) [4] For this value of $\lambda$ and a target quality score of $Q=30$, show that no blend $(x,y,z)$ can satisfy all three constraints simultaneously.
(c) [5] For the same value of $\lambda$, find the target quality score $Q$ for which infinitely many blends satisfy the constraints, and find the general solution in terms of a parameter $t$.
Show complete worked solution
(a) The coefficient matrix is
$$\begin{pmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 3 & \lambda & 2 \end{pmatrix}$$
Expanding the determinant along the first row:
$$\det = 1\left[(3)(2)-(1)(\lambda)\right] - 1\left[(2)(2)-(1)(3)\right] + 1\left[(2)(\lambda)-(3)(3)\right]$$
$$=1(6-\lambda) - 1(4-3) + 1(2\lambda-9)$$
$$=(6-\lambda) - 1 + (2\lambda-9)$$
$$=\lambda-4$$
as required. The blend masses fail to be uniquely determined when the determinant is zero:
$$\lambda-4=0 \ \Rightarrow\ \boxed{\lambda=4}$$
(b) Set $\lambda=4$, $Q=30$. The system becomes:
$$x+y+z=10 \quad (1)$$
$$2x+3y+z=22 \quad (2)$$
$$3x+4y+2z=30 \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-\,2\times$(1):
$$(2x+3y+z)-2(x+y+z)=22-2(10)$$
$$y-z=2 \quad (4)$$
Eliminate $x$ from (3) using (1): compute (3) $-\,3\times$(1):
$$(3x+4y+2z)-3(x+y+z)=30-3(10)$$
$$y-z=0 \quad (5)$$
Equations (4) and (5) both give a value for $y-z$, but (4) requires $y-z=2$ while (5) requires $y-z=0$. Since $2\ne0$, this is a contradiction.
$$\boxed{\text{No blend satisfies all three constraints when }Q=30.}$$
(c) Repeat the elimination with a general target $Q$ in place of $30$: compute (3) $-\,3\times$(1) again:
$$(3x+4y+2z)-3(x+y+z)=Q-3(10)$$
$$y-z=Q-30 \quad (5')$$
For consistency with (4) $\left(y-z=2\right)$, we need:
$$Q-30=2 \ \Rightarrow\ Q=32$$
With $Q=32$, both (4) and (5') read $y-z=2$. Let $z=t$. Then $y=2+t$, and from (1):
$$x=10-y-z=10-(2+t)-t=8-2t$$
Check in (2): $2(8-2t)+3(2+t)+t=16-4t+6+3t+t=22+0t=22$ $\checkmark$ (constant, confirming consistency for every $t$).
$$\boxed{Q=32; \quad x=8-2t,\ \ y=2+t,\ \ z=t,\ \ t\in\mathbb{R},\ 0\le t\le4\text{ (for non-negative masses)}}$$
QUESTION 49
12 marks
Hard
Consider the system of equations
x+2y+\lambda z=3
2x-y+z=4
x+2y+z=5
where $\lambda\in\mathbb{R}$.
(a) [4] Show that the determinant of the coefficient matrix is $5\lambda-5$, and hence find the value of $\lambda$ for which the system does not have a unique solution.
(b) [4] For this value of $\lambda$, show algebraically -- by comparing equations directly, and by row reduction -- that the system has no solution.
(c) [4] Explain, in terms of the normal vectors of the planes, why this particular value of $\lambda$ produces no solution, and contrast the geometric configuration of the three planes for this value of $\lambda$ with the configuration for any other value of $\lambda$.
Show complete worked solution
(a) The coefficient matrix is
$$\begin{pmatrix} 1 & 2 & \lambda \\ 2 & -1 & 1 \\ 1 & 2 & 1 \end{pmatrix}$$
Expanding the determinant along the first row:
$$\det = 1\left[(-1)(1)-(1)(2)\right] - 2\left[(2)(1)-(1)(1)\right] + \lambda\left[(2)(2)-(-1)(1)\right]$$
$$=1(-1-2) - 2(2-1) + \lambda(4+1)$$
$$=-3-2+5\lambda$$
$$=5\lambda-5$$
as required. The system fails to have a unique solution when the determinant is zero:
$$5\lambda-5=0 \ \Rightarrow\ \boxed{\lambda=1}$$
(b) Set $\lambda=1$. The system becomes:
$$x+2y+z=3 \quad (1)$$
$$2x-y+z=4 \quad (2)$$
$$x+2y+z=5 \quad (3)$$
Comparing (1) and (3) directly: both equations have identical coefficients $(1,2,1)$ on the left-hand side, but different constants ($3$ and $5$). By row reduction, compute (3) $-$ (1):
$$(x+2y+z)-(x+2y+z)=5-3$$
$$0=2$$
This is a contradiction, so the system has no solution, regardless of equation (2).
$$\boxed{\text{No solution when }\lambda=1.}$$
(c) The normal vector of the plane in equation (1) is $(1,2,1)$, and the normal vector of the plane in equation (3) is also $(1,2,1)$ -- exactly the same vector. Two planes with identical (or scalar-multiple) normal vectors are parallel. Since the constants on the right-hand side differ ($3\ne5$), the two planes are parallel but distinct, so they never intersect. As no point can lie on both of these two planes, no point can lie on all three planes, and the system has no solution however equation (2) is chosen.
For any $\lambda\ne1$, the $z$-coefficient in equation (1) is different from that in equation (3), so their normal vectors $(1,2,\lambda)$ and $(1,2,1)$ are no longer parallel; the determinant is then non-zero and the three planes intersect at a single unique point.
$$\boxed{\lambda=1:\ \pi_1\parallel\pi_3\text{ (distinct), no solution;} \quad \lambda\ne1:\ \text{unique point of intersection.}}$$
QUESTION 50
15 marks
Hard
Consider the system of equations
x+y+z=6
x-y+2z=5
2x+\lambda y+3z=\mu
where $\lambda,\mu\in\mathbb{R}$.
(a) [3] Show that the determinant of the coefficient matrix is $-\lambda$, and hence state the value of $\lambda$ for which the system does not have a unique solution.
(b) [4] For this value of $\lambda$, find the value of $\mu$ for which the system is consistent, and find the general solution as coordinates in terms of a parameter $t$.
(c) [4] Write the solution set from part (b) as a vector equation of a line, $\mathbf{r}=\mathbf{a}+t\mathbf{d}$, and verify that the direction vector $\mathbf{d}$ is parallel to the cross product of the normal vectors of the first two planes.
(d) [4] For the same value of $\lambda$, but with $\mu$ not equal to the value found in part (b), describe the geometric configuration of the three planes, explaining your reasoning.
Show complete worked solution
(a) The coefficient matrix is
$$\begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ 2 & \lambda & 3 \end{pmatrix}$$
Expanding the determinant along the first row:
$$\det = 1\left[(-1)(3)-(2)(\lambda)\right] - 1\left[(1)(3)-(2)(2)\right] + 1\left[(1)(\lambda)-(-1)(2)\right]$$
$$=1(-3-2\lambda) - 1(3-4) + 1(\lambda+2)$$
$$=(-3-2\lambda) - (-1) + (\lambda+2)$$
$$=-3-2\lambda+1+\lambda+2$$
$$=-\lambda$$
as required. The system fails to have a unique solution when the determinant is zero:
$$-\lambda=0 \ \Rightarrow\ \boxed{\lambda=0}$$
(b) Set $\lambda=0$. The system becomes:
$$x+y+z=6 \quad (1)$$
$$x-y+2z=5 \quad (2)$$
$$2x+3z=\mu \quad (3)$$
Eliminate $x$ from (2) using (1): compute (2) $-$ (1):
$$(x-y+2z)-(x+y+z)=5-6$$
$$-2y+z=-1 \ \Rightarrow\ z=2y-1 \quad (4)$$
Let $y=t$. Then $z=2t-1$. Substitute into (1):
$$x+t+(2t-1)=6$$
$$x=7-3t$$
For consistency, check in (3) for every $t$:
$$2(7-3t)+3(2t-1)=14-6t+6t-3=11$$
Since this is constant ($=11$) regardless of $t$, the system is consistent precisely when $\mu=11$.
$$\boxed{\mu=11; \quad x=7-3t,\ \ y=t,\ \ z=2t-1,\ \ t\in\mathbb{R}}$$
(c) Writing the solution as a position vector plus a multiple of a direction vector, using the constant terms (at $t=0$) as the point $\mathbf{a}$ and the coefficients of $t$ as the direction $\mathbf{d}$:
$$\boxed{\mathbf{r}=\begin{pmatrix}7\\0\\-1\end{pmatrix}+t\begin{pmatrix}-3\\1\\2\end{pmatrix}}$$
To verify, the line of intersection of planes (1) and (2) must be parallel to the cross product of their normal vectors $\mathbf{n_1}=(1,1,1)$ and $\mathbf{n_2}=(1,-1,2)$:
$$\mathbf{n_1}\times\mathbf{n_2}=\begin{pmatrix}(1)(2)-(1)(-1)\\(1)(1)-(1)(2)\\(1)(-1)-(1)(1)\end{pmatrix}=\begin{pmatrix}2+1\\1-2\\-1-1\end{pmatrix}=\begin{pmatrix}3\\-1\\-2\end{pmatrix}$$
This is $-1$ times the direction vector $\mathbf{d}=(-3,1,2)$ found above, i.e. $\mathbf{n_1}\times\mathbf{n_2}=-\mathbf{d}$. Since a direction vector and its negative describe the same line, $\mathbf{d}$ is indeed parallel to $\mathbf{n_1}\times\mathbf{n_2}$, confirming the result. $\checkmark$
(d) When $\lambda=0$ and $\mu\ne11$, planes (1) and (2) still intersect in the same line $\mathbf{r}=(7,0,-1)+t(-3,1,2)$ found in part (b), since this line does not depend on $\mu$. However, plane (3), $2x+3z=\mu$, does not contain this line unless $\mu=11$ (shown in part (b)); since the line's direction vector $(-3,1,2)$ satisfies $2(-3)+3(2)=-6+6=0$, the line is parallel to plane (3)'s normal-vector condition (the line direction is perpendicular to the normal $(2,0,3)$, confirming the line is parallel to plane (3)), but it lies outside plane (3) when $\mu\ne11$. So plane (3) is parallel to the line of intersection of (1) and (2) but does not contain it: the three planes have no common point, and they form a triangular prism.
$$\boxed{\mu\ne11: \text{plane 3 is parallel to the line }\pi_1\cap\pi_2\text{ but does not contain it -- triangular prism, no solution.}}$$