DP (Grade 11 & 12) · Maths AI HL
Functions I
350 questions across 7 sub-topics
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Straight Line Equations
Linear Functions
Quadratic Functions
Cubic Functions
Direct & Inverse Variation
Exponential Functions
Logarithmic Functions
Straight Line Equations 50 questions
QUESTION 1
5 marks
Easy
A small workshop manufactures custom phone cases. The total weekly cost $C$ (in USD) to
produce $n$ cases is modelled by a linear function of $n$. Records show that producing
$n=50$ cases costs $C=\$620$, and producing $n=200$ cases costs $C=\$1400$.
(a) Find the gradient of the line through these two data points, and interpret its
meaning in this context. [2]
(b) Find the equation of the linear cost model in the form $C = mn+c$. [2]
(c) Use your model to predict the cost of producing $350$ cases. [1]
Show complete worked solution
(a) Using the gradient formula between two points $(n_1,C_1)$ and $(n_2,C_2)$ on the line:
$$m = \dfrac{C_2 - C_1}{n_2 - n_1}$$
Substituting the given data points $(50,620)$ and $(200,1400)$:
$$m = \dfrac{1400-620}{200-50}$$
$$m = \dfrac{780}{150} = 5.2$$
Interpretation: each additional case produced increases the weekly cost by $\$5.20$; this is the marginal (variable) cost per case.
$$\boxed{m = 5.2}$$
(b) Using the point-gradient form of a line with gradient $m=5.2$ and the point $(50,620)$:
$$C - 620 = 5.2(n-50)$$
Expanding and simplifying:
$$C = 5.2n - 260 + 620$$
$$C = 5.2n + 360$$
$$\boxed{C(n) = 5.2n + 360}$$
(The intercept $360$ represents the fixed weekly cost, e.g. rent and equipment, incurred even when $n=0$.)
(c) Substituting $n=350$ into the model from part (b):
$$C(350) = 5.2(350)+360$$
$$C(350) = 1820+360$$
$$\boxed{C(350) = \$2180}$$
QUESTION 2
4 marks
Easy
A city planning office is drawing a proposed new access road that must run parallel to
Elm Street. On the city's coordinate grid (units in km), Elm Street's centreline
satisfies the equation
$$3x - 2y = 8.$$
The new road must pass through the point $(4,-1)$.
(a) Write Elm Street's equation in the form $y=mx+c$ and state its gradient. [2]
(b) Find the equation of the new road, giving your answer in the form $y=mx+c$. [2]
Show complete worked solution
(a) Rearranging Elm Street's equation into the form $y=mx+c$:
$$3x-2y=8$$
$$-2y = -3x+8$$
$$y = \dfrac{3}{2}x - 4$$
Gradient of Elm Street: $m = 1.5$.
$$\boxed{y = \dfrac{3}{2}x - 4,\ \ m=1.5}$$
(b) Since the new road is parallel to Elm Street, it has the same gradient, $m=1.5$.
Using the point-gradient form with $m=1.5$ and the point $(4,-1)$:
$$y-(-1) = 1.5(x-4)$$
$$y+1 = 1.5x-6$$
Simplifying:
$$y = 1.5x-7$$
$$\boxed{y = 1.5x - 7}$$
QUESTION 3
6 marks
Medium
A delivery drone's flight path over level ground is modelled as a straight line $L_1$
passing through the points $A(2,5)$ and $B(8,17)$, where $x$ is horizontal distance (m)
and $y$ is height above the ground (m).
(a) Find the gradient of $L_1$ and its equation in the form $y=mx+c$. [3]
(b) A safety-corridor boundary is modelled by a line $L_2$ that is perpendicular to
$L_1$ and passes through the point $(10,3)$. Find the equation of $L_2$, giving your
answer in the form $ax+by+d=0$ where $a,b,d \in \mathbb{Z}$. [3]
Show complete worked solution
(a) Using the gradient formula for the line through $A(2,5)$ and $B(8,17)$:
$$m_1 = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{17-5}{8-2}$$
$$m_1 = \dfrac{12}{6} = 2$$
Using the point-gradient form with $m_1=2$ and point $A(2,5)$:
$$y-5 = 2(x-2)$$
$$y = 2x-4+5 = 2x+1$$
$$\boxed{L_1:\ y = 2x+1}$$
(b) Since $L_2 \perp L_1$, the gradient of $L_2$ is the negative reciprocal of $m_1$:
$$m_2 = -\dfrac{1}{m_1} = -\dfrac{1}{2}$$
Using the point-gradient form with $m_2=-\dfrac{1}{2}$ and the point $(10,3)$:
$$y-3 = -\dfrac{1}{2}(x-10)$$
$$y = 3-\dfrac{1}{2}x+5$$
$$y = -\dfrac{1}{2}x+8$$
Multiplying through by $2$ to clear the fraction, and rearranging into the form $ax+by+d=0$:
$$2y = -x+16$$
$$x+2y-16=0$$
$$\boxed{L_2:\ x + 2y - 16 = 0}$$
QUESTION 4
7 marks
Medium
Two mobile-phone relay towers are located at $A(-3,4)$ and $B(9,-2)$, with coordinates
measured in kilometres on a regional grid. A new booster tower is to be placed
somewhere on the perpendicular bisector of $[AB]$, so that it is always equidistant
from both existing towers.
(a) Find the midpoint of $[AB]$ and the gradient of $[AB]$. [3]
(b) Find the equation of the perpendicular bisector of $[AB]$, giving your answer in
the form $ax+by=c$. [2]
(c) Determine, with justification, whether the point $(5,5)$ could be a valid site for
the new booster tower. [2]
Show complete worked solution
(a) Using the midpoint formula for $A(-3,4)$ and $B(9,-2)$:
$$M = \left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$$
$$M = \left(\dfrac{-3+9}{2},\ \dfrac{4+(-2)}{2}\right)$$
$$M = (3,1)$$
Using the gradient formula for $A$ and $B$:
$$m_{AB} = \dfrac{-2-4}{9-(-3)} = \dfrac{-6}{12} = -0.5$$
$$\boxed{M=(3,1),\ \ m_{AB} = -0.5}$$
(b) The perpendicular bisector has gradient equal to the negative reciprocal of $m_{AB}$:
$$m_{\perp} = -\dfrac{1}{-0.5} = 2$$
Using the point-gradient form through the midpoint $M(3,1)$ with gradient $2$:
$$y-1 = 2(x-3)$$
$$y = 2x-6+1 = 2x-5$$
Rearranging into the form $ax+by=c$:
$$\boxed{2x - y = 5}$$
(c) Substituting $(5,5)$ into $2x-y=5$:
$$2(5)-5 = 5$$
This equals the right-hand side, so $(5,5)$ satisfies the equation and lies on the perpendicular bisector.
Confirming with the distance formula, $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$:
$$AP = \sqrt{(5-(-3))^2+(5-4)^2} = \sqrt{64+1} = \sqrt{65} \approx 8.06\text{ km}$$
$$BP = \sqrt{(5-9)^2+(5-(-2))^2} = \sqrt{16+49} = \sqrt{65} \approx 8.06\text{ km}$$
Since $AP=BP$, the point $(5,5)$ is equidistant from both towers.
$$\boxed{\text{Yes; } (5,5)\text{ is equidistant (}AP=BP=\sqrt{65}\approx 8.06\text{ km), so it is a valid site.}}$$
QUESTION 5
15 marks
Hard
An engineering firm is designing a network of straight elevated pedestrian bridges in a
park, modelled on a coordinate grid with units in metres.
Bridge $A$ is a straight segment connecting $P(0,40)$ to $Q(120,100)$.
Bridge $B$ is a second straight walkway that meets Bridge $A$ at exactly the point $Q$,
crossing it at right angles, and continues from $Q$ towards a viewing platform $R$.
(a) Find the gradient of Bridge $A$ and its equation in the form $y=mx+c$. [4]
(b) Find the equation of the line containing Bridge $B$, given that it is perpendicular
to Bridge $A$ and passes through $Q$. [5]
(c) The viewing platform $R$ lies exactly $130$ m along Bridge $B$ from $Q$, in the
direction for which the $x$-coordinate is smaller than that of $Q$ (i.e. away from the
park entrance). Find the coordinates of $R$, giving each coordinate to 3 significant
figures. [6]
Show complete worked solution
(a) Using the gradient formula for Bridge $A$ through $P(0,40)$ and $Q(120,100)$:
$$m_A = \dfrac{100-40}{120-0} = \dfrac{60}{120} = 0.5$$
Using the point-gradient form with $m_A=0.5$ and point $P(0,40)$:
$$y-40 = 0.5(x-0)$$
$$y = 0.5x+40$$
$$\boxed{\text{Bridge } A:\ y = 0.5x + 40}$$
(b) Since Bridge $B \perp$ Bridge $A$, its gradient is the negative reciprocal of $m_A$:
$$m_B = -\dfrac{1}{0.5} = -2$$
Using the point-gradient form with $m_B=-2$ and point $Q(120,100)$:
$$y-100 = -2(x-120)$$
$$y = -2x+240+100$$
$$y = -2x+340$$
$$\boxed{\text{Bridge } B:\ y = -2x+340}$$
(c) A direction vector along Bridge $B$ (gradient $-2$, i.e. a run of $1$ for a rise of $-2$) is
$$\begin{pmatrix}1\\-2\end{pmatrix}$$
with magnitude
$$\left|\begin{pmatrix}1\\-2\end{pmatrix}\right| = \sqrt{1^2+(-2)^2} = \sqrt5$$
Since $R$ must have a smaller $x$-coordinate than $Q$, the required unit vector points in the opposite direction to $\begin{pmatrix}1\\-2\end{pmatrix}$:
$$\hat{u} = \left(-\dfrac{1}{\sqrt5},\ \dfrac{2}{\sqrt5}\right)$$
Moving $130$ m from $Q$ along $\hat u$:
$$R = Q + 130\hat{u} = \left(120 - \dfrac{130}{\sqrt5},\ 100+\dfrac{260}{\sqrt5}\right)$$
Evaluating the two terms:
$$\dfrac{130}{\sqrt5} = 58.1378\ldots, \qquad \dfrac{260}{\sqrt5} = 116.276\ldots$$
So:
$$R_x = 120-58.1378\ldots = 61.862\ldots \approx 61.9$$
$$R_y = 100+116.276\ldots = 216.276\ldots \approx 216$$
Check: the distance $QR = \sqrt{58.14^2+116.28^2}\approx 130$ m, and $R$ satisfies $y=-2x+340$ since $-2(61.9)+340 = 216.2$, confirming $R$ lies on Bridge $B$.
$$\boxed{R \approx (61.9,\ 216)}$$
QUESTION 6
4 marks
Easy
A ski resort is installing a new chairlift. On a coordinate grid where $x$ is the horizontal
distance from the base station (in metres) and $y$ is the height above the base station (in
metres), the base tower is located at $A(0,20)$ and the summit tower is located at $B(300,260)$.
The chairlift cable runs in a straight line between the two towers.
(a) Find the gradient of the chairlift cable. [2]
(b) Find the equation of the cable line in the form $y=mx+c$. [2]
Show complete worked solution
(a) Using the gradient formula between $A(0,20)$ and $B(300,260)$:
$$m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{260-20}{300-0}$$
$$m = \dfrac{240}{300} = 0.8$$
$$\boxed{m=0.8}$$
(b) Using the point-gradient form with $m=0.8$ and the point $A(0,20)$:
$$y-20 = 0.8(x-0)$$
$$y = 0.8x+20$$
$$\boxed{y = 0.8x+20}$$
QUESTION 7
4 marks
Easy
Two irrigation canals run in straight lines across a farm's coordinate map (units in km).
Canal $A$ has equation $y=2x-5$. Canal $B$ passes through the points $(1,3)$ and $(4,9)$.
(a) Find the gradient of Canal $B$. [2]
(b) Determine, with a reason, whether Canal $A$ and Canal $B$ are parallel. [2]
Show complete worked solution
(a) Using the gradient formula for the two points on Canal $B$, $(1,3)$ and $(4,9)$:
$$m_B = \dfrac{9-3}{4-1} = \dfrac{6}{3} = 2$$
$$\boxed{m_B = 2}$$
(b) Canal $A$ has equation $y=2x-5$, so its gradient is $m_A=2$.
Since $m_A = m_B = 2$, the two lines have equal gradients.
$$\boxed{\text{Canal } A \text{ and Canal } B \text{ are parallel, since } m_A=m_B=2}$$
QUESTION 8
4 marks
Easy
A forest adventure park is installing a zip-line. On the park's coordinate map (units in
metres), the launch tower is at $(5,45)$ and the landing platform is at $(85,5)$.
(a) Find the gradient of the zip-line. [1]
(b) Find the equation of the zip-line, giving your answer in the form $ax+by=d$
where $a,b,d \in \mathbb{Z}$. [3]
Show complete worked solution
(a) Using the gradient formula between $(5,45)$ and $(85,5)$:
$$m = \dfrac{5-45}{85-5} = \dfrac{-40}{80} = -0.5$$
$$\boxed{m=-0.5}$$
(b) Using the point-gradient form with $m=-0.5$ and the point $(5,45)$:
$$y-45 = -0.5(x-5)$$
$$y = -0.5x+2.5+45$$
$$y = -0.5x+47.5$$
Multiplying every term by $2$ to clear the fraction and rearranging into the form $ax+by=d$:
$$2y = -x+95$$
$$\boxed{x+2y=95}$$
QUESTION 9
3 marks
Easy
Two GPS waypoints on a hiking trail are recorded as $A(12.4,7.8)$ and $B(15.8,3.2)$, with
coordinates measured in kilometres on a regional map.
Find the coordinates of the midpoint of $[AB]$, which the trail organisers will use as the
location of a new rest bench. [3]
Show complete worked solution
Using the midpoint formula for $A(12.4,7.8)$ and $B(15.8,3.2)$:
$$M = \left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$$
$$M = \left(\dfrac{12.4+15.8}{2},\ \dfrac{7.8+3.2}{2}\right)$$
$$M = \left(\dfrac{28.2}{2},\ \dfrac{11}{2}\right)$$
$$\boxed{M = (14.1,\ 5.5)}$$
QUESTION 10
3 marks
Easy
Two lighthouses are marked on a nautical chart at $A(3,2)$ and $B(15,11)$, with coordinates
measured in kilometres.
Find the straight-line distance between the two lighthouses. [3]
Show complete worked solution
Using the distance formula for $A(3,2)$ and $B(15,11)$:
$$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
$$d = \sqrt{(15-3)^2+(11-2)^2}$$
$$d = \sqrt{12^2+9^2} = \sqrt{144+81} = \sqrt{225}$$
$$\boxed{d = 15\text{ km}}$$
QUESTION 11
4 marks
Easy
A straight fence line on a farm's coordinate map (units in metres) has equation
$$4x - 3y = 12.$$
(a) Rearrange this equation into the form $y=mx+c$. [3]
(b) State the gradient of the fence. [1]
Show complete worked solution
(a) Rearranging $4x-3y=12$:
$$-3y = -4x+12$$
Dividing every term by $-3$:
$$y = \dfrac{4}{3}x - 4$$
$$\boxed{y = \dfrac{4}{3}x - 4}$$
(b) Comparing with $y=mx+c$, the gradient is the coefficient of $x$:
$$\boxed{m = \dfrac{4}{3} \approx 1.33}$$
QUESTION 12
4 marks
Easy
On a city coordinate grid, Road $P$ has gradient $\dfrac{3}{4}$. Road $Q$ passes through the
points $(0,0)$ and $(8,-6)$.
(a) Find the gradient of Road $Q$. [2]
(b) Determine, with justification, whether Road $P$ and Road $Q$ are perpendicular. [2]
Show complete worked solution
(a) Using the gradient formula for the two points on Road $Q$, $(0,0)$ and $(8,-6)$:
$$m_Q = \dfrac{-6-0}{8-0} = \dfrac{-6}{8} = -\dfrac{3}{4}$$
$$\boxed{m_Q = -\dfrac{3}{4}}$$
(b) Two lines are perpendicular only if the product of their gradients equals $-1$. Testing:
$$m_P \times m_Q = \dfrac{3}{4}\times\left(-\dfrac{3}{4}\right) = -\dfrac{9}{16}$$
Since $-\dfrac{9}{16}\neq -1$, the roads are not perpendicular.
$$\boxed{\text{Not perpendicular, since } m_P m_Q = -\dfrac{9}{16}\neq -1}$$
QUESTION 13
4 marks
Easy
A city is painting a new dedicated bike path that must run parallel to an existing footpath.
On the city's coordinate grid (units in metres), the footpath has equation $y=-2x+7$. The new
bike path must pass through the point $(3,-1)$.
(a) State the gradient the bike path must have. [1]
(b) Find the equation of the bike path, giving your answer in the form $y=mx+c$. [3]
Show complete worked solution
(a) Since the bike path is parallel to the footpath, it must have the same gradient as
$y=-2x+7$:
$$\boxed{m = -2}$$
(b) Using the point-gradient form with $m=-2$ and the point $(3,-1)$:
$$y-(-1) = -2(x-3)$$
$$y+1 = -2x+6$$
$$y = -2x+5$$
$$\boxed{y = -2x+5}$$
QUESTION 14
5 marks
Easy
On a stadium's coordinate grid (units in metres), a sprint lane's centreline runs in a
straight line from the start marker at $(10,5)$ to the finish marker at $(110,45)$.
(a) Find the gradient of the sprint lane's centreline. [2]
(b) Find the equation of the centreline in the form $y=mx+c$. [3]
Show complete worked solution
(a) Using the gradient formula between $(10,5)$ and $(110,45)$:
$$m = \dfrac{45-5}{110-10} = \dfrac{40}{100} = 0.4$$
$$\boxed{m=0.4}$$
(b) Using the point-gradient form with $m=0.4$ and the point $(10,5)$:
$$y-5 = 0.4(x-10)$$
$$y = 0.4x-4+5$$
$$y = 0.4x+1$$
$$\boxed{y = 0.4x+1}$$
QUESTION 15
5 marks
Easy
Two power-line pylons stand on a coordinate map of a utility corridor (units in metres),
where $x$ is horizontal distance and $y$ is the height of the cable above the ground. Pylon $A$
supports the cable at $(20,150)$ and Pylon $B$ supports the cable at $(170,180)$. Between the
pylons the cable is modelled as a straight line.
(a) Find the gradient of the cable between the pylons. [2]
(b) Find the equation of the cable line in the form $y=mx+c$. [3]
Show complete worked solution
(a) Using the gradient formula between $A(20,150)$ and $B(170,180)$:
$$m = \dfrac{180-150}{170-20} = \dfrac{30}{150} = 0.2$$
$$\boxed{m=0.2}$$
(b) Using the point-gradient form with $m=0.2$ and the point $A(20,150)$:
$$y-150 = 0.2(x-20)$$
$$y = 0.2x-4+150$$
$$y = 0.2x+146$$
$$\boxed{y = 0.2x+146}$$
QUESTION 16
3 marks
Easy
Two navigation buoys are marked on a harbour chart at $A(-4,6)$ and $B(5,-6)$, with
coordinates measured in nautical miles.
Find the distance between the two buoys. [3]
Show complete worked solution
Using the distance formula for $A(-4,6)$ and $B(5,-6)$:
$$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
$$d = \sqrt{(5-(-4))^2+(-6-6)^2}$$
$$d = \sqrt{9^2+(-12)^2} = \sqrt{81+144} = \sqrt{225}$$
$$\boxed{d = 15\text{ nautical miles}}$$
QUESTION 17
4 marks
Easy
A straight irrigation channel on a farm's coordinate map (units in metres) has equation
$$5x+2y=20.$$
(a) Find the point where the channel crosses the $x$-axis. [2]
(b) Find the point where the channel crosses the $y$-axis. [2]
Show complete worked solution
(a) The channel crosses the $x$-axis where $y=0$. Substituting into $5x+2y=20$:
$$5x+2(0)=20$$
$$5x = 20$$
$$x = 4$$
$$\boxed{(4,0)}$$
(b) The channel crosses the $y$-axis where $x=0$. Substituting into $5x+2y=20$:
$$5(0)+2y=20$$
$$2y = 20$$
$$y = 10$$
$$\boxed{(0,10)}$$
QUESTION 18
4 marks
Easy
A new irrigation line must run parallel to an existing boundary fence. On the farm's
coordinate map (units in metres) the fence has equation $2x-y=7$. The irrigation line must pass
through the point $(3,4)$.
(a) Find the gradient of the fence. [2]
(b) Find the equation of the irrigation line, in the form $y=mx+c$. [2]
Show complete worked solution
(a) Rearranging the fence equation $2x-y=7$ into the form $y=mx+c$:
$$-y = -2x+7$$
$$y = 2x-7$$
The gradient of the fence is:
$$\boxed{m=2}$$
(b) Since the irrigation line is parallel to the fence, it has the same gradient, $m=2$. Using
the point-gradient form with the point $(3,4)$:
$$y-4 = 2(x-3)$$
$$y = 2x-6+4$$
$$y = 2x-2$$
$$\boxed{y = 2x-2}$$
QUESTION 19
3 marks
Easy
On a city coordinate grid (units in km, with $x$ measured east and $y$ measured north), a
short access road runs due north-south (i.e. parallel to the $y$-axis) and passes through the
point $(35,12)$.
(a) Write down the equation of this road. [2]
(b) Explain why the gradient of this road cannot be calculated using the usual
gradient formula. [1]
Show complete worked solution
(a) A line running due north-south (parallel to the $y$-axis) has every point on it sharing
the same $x$-coordinate. Since the road passes through $(35,12)$:
$$\boxed{x = 35}$$
(b) The gradient formula is $m=\dfrac{y_2-y_1}{x_2-x_1}$. Every point on this road has the same
$x$-coordinate, so $x_2-x_1=0$ for any two points on it, which would require dividing by zero.
$$\boxed{\text{The gradient is undefined, since it would require division by } x_2-x_1=0}$$
QUESTION 20
5 marks
Easy
Two lamp posts in a park are located at $A(2,3)$ and $B(10,9)$, with coordinates measured in
metres on the park's layout grid. The park manager wants to place a new bench exactly halfway
between the two lamp posts.
(a) Find the coordinates of the midpoint of $[AB]$, where the bench will be
placed. [3]
(b) Find the distance between the two lamp posts. [2]
Show complete worked solution
(a) Using the midpoint formula for $A(2,3)$ and $B(10,9)$:
$$M = \left(\dfrac{2+10}{2},\ \dfrac{3+9}{2}\right)$$
$$M = \left(\dfrac{12}{2},\ \dfrac{12}{2}\right)$$
$$\boxed{M = (6,6)}$$
(b) Using the distance formula for $A(2,3)$ and $B(10,9)$:
$$d = \sqrt{(10-2)^2+(9-3)^2} = \sqrt{8^2+6^2} = \sqrt{64+36} = \sqrt{100}$$
$$\boxed{d = 10\text{ m}}$$
QUESTION 21
4 marks
Easy
A ranger is checking whether three trail markers, $A(1,2)$, $B(4,8)$ and $C(6,12)$, lie on
the same straight section of a hiking trail. Coordinates are measured in hundreds of metres on
the trail map.
(a) Find the gradient of $[AB]$. [2]
(b) Find the gradient of $[BC]$, and hence determine whether the three markers
are collinear (lie on the same straight line). [2]
Show complete worked solution
(a) Using the gradient formula for $A(1,2)$ and $B(4,8)$:
$$m_{AB} = \dfrac{8-2}{4-1} = \dfrac{6}{3} = 2$$
$$\boxed{m_{AB} = 2}$$
(b) Using the gradient formula for $B(4,8)$ and $C(6,12)$:
$$m_{BC} = \dfrac{12-8}{6-4} = \dfrac{4}{2} = 2$$
Since $m_{AB} = m_{BC} = 2$ and both segments share the common point $B$, the three points lie on
the same straight line.
$$\boxed{\text{Yes, } A,\ B,\ C \text{ are collinear, since } m_{AB}=m_{BC}=2}$$
QUESTION 22
5 marks
Easy
Two mobile-phone towers are located at $A(0,0)$ and $B(8,6)$, with coordinates measured in
kilometres on a regional grid.
(a) Find the midpoint of $[AB]$. [2]
(b) A new relay mast is proposed for the point $P(7,-1)$. Use the distance
formula to determine whether $P$ is equidistant from both towers. [3]
Show complete worked solution
(a) Using the midpoint formula for $A(0,0)$ and $B(8,6)$:
$$M = \left(\dfrac{0+8}{2},\ \dfrac{0+6}{2}\right)$$
$$\boxed{M = (4,3)}$$
(b) Using the distance formula, $PA$ is the distance from $P(7,-1)$ to $A(0,0)$:
$$PA = \sqrt{(7-0)^2+(-1-0)^2} = \sqrt{49+1} = \sqrt{50}$$
$PB$ is the distance from $P(7,-1)$ to $B(8,6)$:
$$PB = \sqrt{(7-8)^2+(-1-6)^2} = \sqrt{1+49} = \sqrt{50}$$
Since $PA = PB = \sqrt{50}\approx 7.07\text{ km}$, $P$ is equidistant from both towers.
$$\boxed{\text{Yes; } PA=PB=\sqrt{50}\approx 7.07\text{ km}}$$
QUESTION 23
4 marks
Easy
Two aircraft flight paths are modelled on a radar coordinate grid (units in km) by the
equations
$$L_1:\ y = 2x-3 \qquad\qquad L_2:\ y = -x+9$$
(a) By setting the two expressions for $y$ equal, find the $x$-coordinate where
the flight paths cross. [2]
(b) Find the $y$-coordinate of the crossing point, and state the coordinates of
the point where the flight paths intersect. [2]
Show complete worked solution
(a) At the point where the paths cross, both equations give the same $y$-value, so:
$$2x-3 = -x+9$$
$$2x+x = 9+3$$
$$3x = 12$$
$$\boxed{x=4}$$
(b) Substituting $x=4$ into $L_1$: $y=2x-3$:
$$y = 2(4)-3 = 8-3 = 5$$
Check using $L_2$: $y=-x+9 = -4+9=5$. Both agree.
$$\boxed{\text{The flight paths cross at } (4,5)}$$
QUESTION 24
7 marks
Medium
A delivery drone descends along a straight-line flight path over level ground. On a
coordinate grid where $x$ is horizontal distance (m) and $y$ is height above the ground (m), the
drone is at $(2,50)$ and later at $(10,10)$.
(a) Find the gradient of the flight path. [2]
(b) Find the equation of the flight path in the form $y=mx+c$. [3]
(c) Find the horizontal distance $x$ at which the drone reaches the ground
($y=0$), and interpret this value in context. [2]
Show complete worked solution
(a) Using the gradient formula between $(2,50)$ and $(10,10)$:
$$m = \dfrac{10-50}{10-2} = \dfrac{-40}{8} = -5$$
$$\boxed{m=-5}$$
(b) Using the point-gradient form with $m=-5$ and the point $(2,50)$:
$$y-50 = -5(x-2)$$
$$y = -5x+10+50$$
$$y = -5x+60$$
$$\boxed{y=-5x+60}$$
(c) The drone reaches the ground when $y=0$. Substituting into the equation from part (b):
$$0 = -5x+60$$
$$5x = 60$$
$$x = 12$$
This means the drone reaches ground level $12$ m (horizontally) from the origin of the flight
path being modelled.
$$\boxed{x=12\text{ m}}$$
QUESTION 25
7 marks
Medium
Two straight streets are marked on a city planning grid (units in metres). Street $A$ has
equation $x+y=10$. Street $B$ passes through the points $(-2,3)$ and $(10,9)$.
(a) Find the equation of Street $B$, giving your answer in the form $y=mx+c$. [3]
(b) Find the coordinates of the crossroads where Street $A$ and Street $B$
intersect, by solving the two equations simultaneously. [4]
Show complete worked solution
(a) Using the gradient formula for the two given points on Street $B$:
$$m_B = \dfrac{9-3}{10-(-2)} = \dfrac{6}{12} = 0.5$$
Using the point-gradient form with $m_B=0.5$ and the point $(10,9)$:
$$y-9 = 0.5(x-10)$$
$$y = 0.5x-5+9$$
$$y = 0.5x+4$$
$$\boxed{\text{Street } B:\ y=0.5x+4}$$
(b) Rearranging Street $A$'s equation $x+y=10$ into the form $y=mx+c$:
$$y = -x+10$$
At the crossroads, both equations give the same $y$-value, so:
$$0.5x+4 = -x+10$$
$$0.5x+x = 10-4$$
$$1.5x = 6$$
$$x = 4$$
Substituting $x=4$ into $y=-x+10$:
$$y = -4+10 = 6$$
Check using Street $B$: $y=0.5(4)+4 = 2+4=6$. Both agree.
$$\boxed{\text{Crossroads at } (4,6)}$$
QUESTION 26
8 marks
Medium
A water utility company is laying a main pipeline in a straight line through the points
$(0,10)$ and $(20,30)$ on a coordinate map of the district (units in metres). A spur pipeline
must run perpendicular to the main pipeline, starting at a pumping station located at $(16,2)$.
(a) Find the gradient of the main pipeline, and hence its equation in the form
$y=mx+c$. [2]
(b) Find the equation of the spur pipeline. [3]
(c) Find the coordinates of the junction where the spur pipeline meets the main
pipeline. [2]
(d) Find the length of the spur pipeline from the pumping station to the
junction. [1]
Show complete worked solution
(a) Using the gradient formula between $(0,10)$ and $(20,30)$:
$$m_{\text{main}} = \dfrac{30-10}{20-0} = \dfrac{20}{20} = 1$$
Using the point-gradient form with $m=1$ and the point $(0,10)$:
$$y-10 = 1(x-0)$$
$$\boxed{\text{Main pipeline: } y=x+10}$$
(b) Since the spur is perpendicular to the main pipeline, its gradient is the negative reciprocal
of $m_{\text{main}}=1$:
$$m_{\text{spur}} = -\dfrac{1}{1} = -1$$
Using the point-gradient form with $m=-1$ and the point $(16,2)$:
$$y-2 = -1(x-16)$$
$$y = -x+16+2$$
$$\boxed{\text{Spur pipeline: } y=-x+18}$$
(c) At the junction, both pipeline equations give the same $y$-value:
$$x+10 = -x+18$$
$$2x = 8$$
$$x = 4$$
$$y = 4+10 = 14$$
$$\boxed{\text{Junction at } (4,14)}$$
(d) Using the distance formula between the pumping station $(16,2)$ and the junction $(4,14)$:
$$d = \sqrt{(16-4)^2+(2-14)^2} = \sqrt{12^2+(-12)^2} = \sqrt{144+144} = \sqrt{288}$$
$$\boxed{d = 12\sqrt2 \approx 17.0\text{ m}}$$
QUESTION 27
7 marks
Medium
A railway platform edge is modelled on a station's coordinate plan (units in metres) by the
equation $5x-2y=10$. A new section of railway track must be laid parallel to the platform edge,
passing through the point $(4,3)$.
(a) Find the gradient of the platform edge. [2]
(b) Find the equation of the new track, in the form $y=mx+c$. [3]
(c) Find the point where the new track crosses the $y$-axis, and state its
coordinates. [2]
Show complete worked solution
(a) Rearranging the platform edge equation $5x-2y=10$ into the form $y=mx+c$:
$$-2y = -5x+10$$
$$y = 2.5x-5$$
$$\boxed{m = 2.5}$$
(b) Since the new track is parallel to the platform edge, it has the same gradient, $m=2.5$.
Using the point-gradient form with the point $(4,3)$:
$$y-3 = 2.5(x-4)$$
$$y = 2.5x-10+3$$
$$\boxed{y = 2.5x-7}$$
(c) The track crosses the $y$-axis where $x=0$:
$$y = 2.5(0)-7 = -7$$
$$\boxed{(0,-7)}$$
QUESTION 28
6 marks
Medium
A cable car system runs on a straight cable between a base tower at $A(0,0)$ and a summit
tower at $B(240,180)$, with coordinates in metres on the mountain's coordinate grid. A support
pulley station is to be installed at the midpoint of the cable.
(a) Find the coordinates of the midpoint of $[AB]$, where the pulley station
will be installed. [2]
(b) Find the total length of the cable, $AB$. [2]
(c) Find the length of cable between the pulley station and the summit tower
$B$. [2]
Show complete worked solution
(a) Using the midpoint formula for $A(0,0)$ and $B(240,180)$:
$$M = \left(\dfrac{0+240}{2},\ \dfrac{0+180}{2}\right)$$
$$\boxed{M = (120,90)}$$
(b) Using the distance formula for $A(0,0)$ and $B(240,180)$:
$$AB = \sqrt{(240-0)^2+(180-0)^2} = \sqrt{57600+32400} = \sqrt{90000}$$
$$\boxed{AB = 300\text{ m}}$$
(c) Using the distance formula for $M(120,90)$ and $B(240,180)$:
$$MB = \sqrt{(240-120)^2+(180-90)^2} = \sqrt{120^2+90^2} = \sqrt{14400+8100} = \sqrt{22500}$$
$$\boxed{MB = 150\text{ m}}$$
(Since $M$ is the midpoint of $[AB]$, this is exactly half of the total cable length found in
part (b), confirming $MB=\tfrac12(300)=150$ m.)
QUESTION 29
7 marks
Medium
A surveyor records three corner markers of a triangular plot of land on a coordinate grid
(units in metres): $A(1,1)$, $B(7,3)$ and $C(5,9)$.
(a) Find the gradient of $[AB]$. [2]
(b) Find the gradient of $[BC]$. [2]
(c) Show that the plot has a right angle at $B$, and find the equation of the
line through $A$ and $B$, in the form $ax+by=c$ where $a,b,c \in \mathbb{Z}$. [3]
Show complete worked solution
(a) Using the gradient formula for $A(1,1)$ and $B(7,3)$:
$$m_{AB} = \dfrac{3-1}{7-1} = \dfrac{2}{6} = \dfrac{1}{3}$$
$$\boxed{m_{AB} = \dfrac13}$$
(b) Using the gradient formula for $B(7,3)$ and $C(5,9)$:
$$m_{BC} = \dfrac{9-3}{5-7} = \dfrac{6}{-2} = -3$$
$$\boxed{m_{BC} = -3}$$
(c) Testing whether $[AB]\perp[BC]$ by multiplying the gradients:
$$m_{AB}\times m_{BC} = \dfrac13\times(-3) = -1$$
Since the product equals $-1$, the sides $[AB]$ and $[BC]$ are perpendicular, so the plot has a
right angle at $B$.
Using the point-gradient form for the line through $A$ and $B$, with $m_{AB}=\tfrac13$ and point
$A(1,1)$:
$$y-1 = \dfrac13(x-1)$$
Multiplying through by $3$:
$$3y-3 = x-1$$
$$\boxed{x - 3y = -2}$$
QUESTION 30
7 marks
Medium
Two aircraft flight paths are tracked over a $40$ km wide air corridor ($0\le x\le 40$) on a
radar coordinate grid (units in km). Flight Path 1 has equation $y=1.5x-2$. Flight Path 2 passes
through the points $(0,25)$ and $(20,5)$.
(a) Find the equation of Flight Path 2, in the form $y=mx+c$. [2]
(b) Find the coordinates of the point where the two flight paths cross. [3]
(c) State, with a reason, whether this crossing point lies within the air
corridor. [2]
Show complete worked solution
(a) Using the gradient formula for the two given points on Flight Path 2:
$$m = \dfrac{5-25}{20-0} = \dfrac{-20}{20} = -1$$
Using the point-gradient form with $m=-1$ and the point $(0,25)$:
$$y-25 = -1(x-0)$$
$$\boxed{\text{Flight Path 2: } y = -x+25}$$
(b) At the crossing point, both equations give the same $y$-value:
$$1.5x-2 = -x+25$$
$$1.5x+x = 25+2$$
$$2.5x = 27$$
$$x = 10.8$$
Substituting into $y=-x+25$:
$$y = -10.8+25 = 14.2$$
$$\boxed{\text{Crossing point } (10.8,\ 14.2)}$$
(c) The corridor requires $0\le x\le 40$. Since $x=10.8$ satisfies $0\le 10.8\le 40$:
$$\boxed{\text{Yes, the crossing point lies within the corridor, since } 0\le 10.8\le 40}$$
QUESTION 31
6 marks
Medium
A suspension bridge deck is modelled as a straight line on a coordinate grid (units in
metres), where $x$ is horizontal distance from the left pier and $y$ is the height of the deck
above the water. The left pier supports the deck at $(0,12)$ and the right pier supports the
deck at $(150,42)$.
(a) Find the gradient of the deck, and hence its equation in the form $y=mx+c$. [4]
(b) A sensor is mounted on the deck at $x=90$. Find the height of the deck at
this point. [2]
Show complete worked solution
(a) Using the gradient formula between $(0,12)$ and $(150,42)$:
$$m = \dfrac{42-12}{150-0} = \dfrac{30}{150} = 0.2$$
Using the point-gradient form with $m=0.2$ and the point $(0,12)$:
$$y-12 = 0.2(x-0)$$
$$\boxed{y = 0.2x+12}$$
(b) Substituting $x=90$ into the deck equation:
$$y = 0.2(90)+12 = 18+12$$
$$\boxed{y = 30\text{ m}}$$
QUESTION 32
6 marks
Medium
A campus network technician wants to place a new Wi-Fi router exactly halfway between an
existing router $A$ and a second existing router $B$, on the campus coordinate map (units in
metres). Router $A$ is at $(5,8)$, and the midpoint of $A$ and $B$ has already been surveyed and
marked at $M(11,17)$.
(a) Write down the midpoint formula equations relating $A$, $B$ and $M$. [2]
(b) Find the coordinates of router $B$. [4]
Show complete worked solution
(a) Let $B=(x_B,y_B)$. Since $M(11,17)$ is the midpoint of $A(5,8)$ and $B(x_B,y_B)$:
$$11 = \dfrac{5+x_B}{2} \qquad\qquad 17 = \dfrac{8+y_B}{2}$$
(b) Solving the first equation for $x_B$:
$$22 = 5+x_B$$
$$x_B = 17$$
Solving the second equation for $y_B$:
$$34 = 8+y_B$$
$$y_B = 26$$
$$\boxed{B = (17,26)}$$
Check: midpoint of $A(5,8)$ and $B(17,26)$ is $\left(\dfrac{5+17}{2},\dfrac{8+26}{2}\right) =
(11,17) = M$, as required.
QUESTION 33
8 marks
Medium
Two fire stations are located at $A(2,2)$ and $B(10,8)$, with coordinates measured in
kilometres on a regional grid. The boundary between the two stations' service areas is the
perpendicular bisector of $[AB]$: every point on this boundary is equidistant from both stations.
(a) Find the midpoint of $[AB]$ and the gradient of $[AB]$. [3]
(b) Find the equation of the service-area boundary, giving your answer in the
form $ax+by=c$. [3]
(c) The town hall is located at $(12,3)$. By finding the distance from the town
hall to each fire station, determine which station should respond to a call from the town hall. [2]
Show complete worked solution
(a) Using the midpoint formula for $A(2,2)$ and $B(10,8)$:
$$M = \left(\dfrac{2+10}{2},\ \dfrac{2+8}{2}\right) = (6,5)$$
Using the gradient formula for $A$ and $B$:
$$m_{AB} = \dfrac{8-2}{10-2} = \dfrac{6}{8} = 0.75$$
$$\boxed{M=(6,5),\ \ m_{AB}=0.75}$$
(b) The boundary is perpendicular to $[AB]$, so its gradient is the negative reciprocal of
$m_{AB}$:
$$m_\perp = -\dfrac{1}{0.75} = -\dfrac{4}{3}$$
Using the point-gradient form through $M(6,5)$ with gradient $-\dfrac43$:
$$y-5 = -\dfrac43(x-6)$$
$$y = -\dfrac43x+8+5$$
$$y = -\dfrac43x+13$$
Multiplying through by $3$ and rearranging into the form $ax+by=c$:
$$3y = -4x+39$$
$$\boxed{4x+3y=39}$$
(c) Using the distance formula, the distance from the town hall $(12,3)$ to station $A(2,2)$:
$$TA = \sqrt{(12-2)^2+(3-2)^2} = \sqrt{100+1} = \sqrt{101}\approx 10.05\text{ km}$$
The distance from the town hall to station $B(10,8)$:
$$TB = \sqrt{(12-10)^2+(3-8)^2} = \sqrt{4+25} = \sqrt{29}\approx 5.39\text{ km}$$
Since $TB < TA$, the town hall is closer to station $B$.
$$\boxed{\text{Station } B \text{ should respond, since } TB\approx5.39\text{ km} < TA\approx10.05\text{ km}}$$
QUESTION 34
7 marks
Medium
A landscape designer is laying out a parallelogram-shaped garden bed $PQRS$ on a coordinate
plan (units in metres). Three of the vertices are known: $P(1,2)$, $Q(7,2)$ and $R(9,6)$. In a
parallelogram $PQRS$, the diagonals $[PR]$ and $[QS]$ bisect each other (share the same
midpoint).
(a) Find the midpoint of the diagonal $[PR]$. [2]
(b) Use the fact that $[QS]$ has the same midpoint to find the coordinates of
$S$. [3]
(c) Verify your answer by showing that $[PS]$ and $[QR]$ are parallel. [2]
Show complete worked solution
(a) Using the midpoint formula for $P(1,2)$ and $R(9,6)$:
$$M = \left(\dfrac{1+9}{2},\ \dfrac{2+6}{2}\right)$$
$$\boxed{M = (5,4)}$$
(b) Since $[QS]$ has the same midpoint $M(5,4)$, and $Q=(7,2)$, let $S=(x_S,y_S)$:
$$5 = \dfrac{7+x_S}{2} \implies x_S = 3$$
$$4 = \dfrac{2+y_S}{2} \implies y_S = 6$$
$$\boxed{S = (3,6)}$$
(c) Using the gradient formula for $Q(7,2)$ and $R(9,6)$:
$$m_{QR} = \dfrac{6-2}{9-7} = \dfrac{4}{2} = 2$$
Using the gradient formula for $P(1,2)$ and $S(3,6)$:
$$m_{PS} = \dfrac{6-2}{3-1} = \dfrac{4}{2} = 2$$
Since $m_{QR}=m_{PS}=2$, the sides $[PS]$ and $[QR]$ are parallel, confirming $PQRS$ is a
parallelogram.
$$\boxed{S=(3,6);\ \ m_{QR}=m_{PS}=2 \text{ confirms the parallelogram}}$$
QUESTION 35
6 marks
Medium
A tunnel boring machine follows a straight-line path on an engineer's coordinate cross-section
(units in metres), where $x$ is horizontal distance bored and $y$ is elevation relative to a
reference datum. The machine's path passes through $(0,-30)$ and $(400,170)$. A property boundary
crosses the tunnel corridor vertically at $x=250$.
(a) Find the gradient of the machine's path, and hence its equation in the form
$y=mx+c$. [4]
(b) Find the elevation $y$ of the machine's path where it crosses the property
boundary at $x=250$. [2]
Show complete worked solution
(a) Using the gradient formula between $(0,-30)$ and $(400,170)$:
$$m = \dfrac{170-(-30)}{400-0} = \dfrac{200}{400} = 0.5$$
Using the point-gradient form with $m=0.5$ and the point $(0,-30)$:
$$y-(-30) = 0.5(x-0)$$
$$\boxed{y = 0.5x-30}$$
(b) Substituting $x=250$ into the equation:
$$y = 0.5(250)-30 = 125-30$$
$$\boxed{y = 95\text{ m}}$$
QUESTION 36
7 marks
Medium
A ski lift's main cable runs in a straight line from the base station $(0,0)$ to the top
station $(100,60)$ on the mountain's coordinate grid (units in metres). A separate cross-cable
runs from the top station $(100,60)$ to an anchor point at $(115,35)$.
(a) Find the gradient of the main cable. [2]
(b) Find the gradient of the cross-cable. [2]
(c) Determine, using the product of gradients, whether the cross-cable is
perpendicular to the main cable, and find the equation of the cross-cable in the form $y=mx+c$. [3]
Show complete worked solution
(a) Using the gradient formula between $(0,0)$ and $(100,60)$:
$$m_{\text{main}} = \dfrac{60-0}{100-0} = \dfrac{60}{100} = 0.6$$
$$\boxed{m_{\text{main}} = 0.6}$$
(b) Using the gradient formula between $(100,60)$ and $(115,35)$:
$$m_{\text{cross}} = \dfrac{35-60}{115-100} = \dfrac{-25}{15} = -\dfrac53$$
$$\boxed{m_{\text{cross}} = -\dfrac53}$$
(c) Testing the product of gradients:
$$m_{\text{main}}\times m_{\text{cross}} = 0.6\times\left(-\dfrac53\right) = -1$$
Since the product equals $-1$, the cross-cable is perpendicular to the main cable.
Using the point-gradient form with $m=-\dfrac53$ and the point $(100,60)$:
$$y-60 = -\dfrac53(x-100)$$
$$y = -\dfrac53x+\dfrac{500}{3}+60$$
$$\boxed{y = -\dfrac53x+\dfrac{680}{3}\ \ \left(\text{perpendicular, since } m_{\text{main}}m_{\text{cross}}=-1\right)}$$
QUESTION 37
6 marks
Medium
A satellite dish is mounted on a support rail. On the installation's coordinate diagram
(units in cm), the rail has equation $6x-4y=16$. A support arm must be attached parallel to the
rail, anchored at the point $(10,-2)$.
(a) Find the gradient of the rail from its general-form equation. [3]
(b) Find the equation of the support arm, in the form $y=mx+c$. [3]
Show complete worked solution
(a) Rearranging the rail equation $6x-4y=16$ into the form $y=mx+c$:
$$-4y = -6x+16$$
$$y = 1.5x-4$$
$$\boxed{m = 1.5}$$
(b) Since the support arm is parallel to the rail, it has the same gradient, $m=1.5$. Using the
point-gradient form with the point $(10,-2)$:
$$y-(-2) = 1.5(x-10)$$
$$y+2 = 1.5x-15$$
$$y = 1.5x-17$$
$$\boxed{y = 1.5x-17}$$
QUESTION 38
6 marks
Medium
A construction crane's jib is designed to run in a single straight line. Three fixing points
along the jib are recorded on the site's coordinate plan (units in metres) as $A(2,5)$,
$B(6,k)$ and $C(10,17)$, where $k$ is unknown. For the jib to be straight, $A$, $B$ and $C$ must
be collinear.
(a) Write an equation stating that the gradient of $[AB]$ equals the gradient of
$[BC]$. [3]
(b) Solve your equation to find the value of $k$. [3]
Show complete worked solution
(a) The gradient of $[AB]$ is $\dfrac{k-5}{6-2}$, and the gradient of $[BC]$ is
$\dfrac{17-k}{10-6}$. Setting these equal:
$$\dfrac{k-5}{4} = \dfrac{17-k}{4}$$
(b) Since both denominators are equal to $4$, the numerators must also be equal:
$$k-5 = 17-k$$
$$2k = 22$$
$$\boxed{k = 11}$$
Check: gradient of $[AB]=\dfrac{11-5}{4}=1.5$; gradient of $[BC]=\dfrac{17-11}{4}=1.5$. Equal, so
$A(2,5)$, $B(6,11)$ and $C(10,17)$ are indeed collinear.
QUESTION 39
8 marks
Medium
A mountain cabin is located at $C(6,2)$ on a coordinate map (units in hundreds of metres). A
straight forest road has equation $y=2x+1$. Rangers want to build the shortest possible access
path from the cabin to the road, which meets the road at a right angle.
(a) Find the gradient of the road and the gradient of the access path (which is
perpendicular to the road). [2]
(b) Find the equation of the access path. [2]
(c) Find the coordinates of the point $F$ where the access path meets the road
(the foot of the perpendicular), by solving the two equations simultaneously. [3]
(d) Find the length of the access path, $CF$. [1]
Show complete worked solution
(a) The road $y=2x+1$ has gradient $2$. Since the access path is perpendicular to the road,
its gradient is the negative reciprocal:
$$m_{\perp} = -\dfrac12$$
$$\boxed{m_{\text{road}}=2,\ \ m_{\text{path}}=-\dfrac12}$$
(b) Using the point-gradient form with $m=-\dfrac12$ and the point $C(6,2)$:
$$y-2 = -\dfrac12(x-6)$$
$$y = -\dfrac12x+3+2$$
$$\boxed{y = -\dfrac12x+5}$$
(c) At the foot of the perpendicular $F$, both equations give the same $y$-value:
$$2x+1 = -\dfrac12x+5$$
$$2x+\dfrac12x = 5-1$$
$$2.5x = 4$$
$$x = 1.6$$
Substituting into $y=2x+1$:
$$y = 2(1.6)+1 = 3.2+1 = 4.2$$
$$\boxed{F = (1.6,4.2)}$$
(d) Using the distance formula between $C(6,2)$ and $F(1.6,4.2)$:
$$CF = \sqrt{(6-1.6)^2+(2-4.2)^2} = \sqrt{4.4^2+(-2.2)^2} = \sqrt{19.36+4.84} = \sqrt{24.2}$$
$$\boxed{CF \approx 4.92\ \text{(hundred metres)}}$$
QUESTION 40
7 marks
Medium
Two ships' straight-line courses are plotted on a navigation coordinate grid (units in km,
with $x$ measured east and $y$ measured north of a reference point). Ship 1's course passes
through the origin with equation $y=0.75x$ (increasing $x$ is the direction the ship is heading).
Ship 2's course passes through the points $(20,0)$ and $(12,9)$.
(a) Find the equation of Ship 2's course, in the form $y=mx+c$. [3]
(b) Find the coordinates of the point where the two courses cross. [3]
(c) Ship 1 is currently at the origin and travels in the direction of
increasing $x$. State, with a reason, whether the crossing point lies ahead of Ship 1 (in the
direction it is travelling). [1]
Show complete worked solution
(a) Using the gradient formula for the two given points on Ship 2's course:
$$m = \dfrac{9-0}{12-20} = \dfrac{9}{-8} = -\dfrac98$$
Using the point-gradient form with $m=-\dfrac98$ and the point $(20,0)$:
$$y-0 = -\dfrac98(x-20)$$
$$y = -\dfrac98x+\dfrac{180}{8}$$
$$\boxed{y = -\dfrac98x+22.5}$$
(b) At the crossing point, both equations give the same $y$-value:
$$0.75x = -1.125x+22.5$$
$$0.75x+1.125x = 22.5$$
$$1.875x = 22.5$$
$$x = 12$$
Substituting into $y=0.75x$:
$$y = 0.75(12) = 9$$
$$\boxed{\text{Crossing point } (12,9)}$$
(c) Since Ship 1 starts at $x=0$ and travels in the direction of increasing $x$, and the crossing
point has $x=12>0$:
$$\boxed{\text{Yes, the crossing point lies ahead of Ship 1, since its } x\text{-coordinate } (12) \text{ is greater than } 0}$$
QUESTION 41
6 marks
Medium
A quality-control surveyor is checking a newly built fence corner. On the site's coordinate
plan (units in metres), Fence 1 runs from $D(0,0)$ to $E(8,4)$, and Fence 2 (built by the
contractor) runs from $E(8,4)$ to $F(10,0)$. The design specifies that Fence 1 and Fence 2
should meet at a right angle at $E$.
(a) Find the gradient of Fence 1. [2]
(b) Find the gradient of Fence 2. [2]
(c) Using the product of the two gradients, determine whether the contractor
built the fences to meet at a right angle. [2]
Show complete worked solution
(a) Using the gradient formula for $D(0,0)$ and $E(8,4)$:
$$m_{DE} = \dfrac{4-0}{8-0} = \dfrac{4}{8} = 0.5$$
$$\boxed{m_{DE} = 0.5}$$
(b) Using the gradient formula for $E(8,4)$ and $F(10,0)$:
$$m_{EF} = \dfrac{0-4}{10-8} = \dfrac{-4}{2} = -2$$
$$\boxed{m_{EF} = -2}$$
(c) Testing the product of the gradients:
$$m_{DE}\times m_{EF} = 0.5\times(-2) = -1$$
Since the product equals $-1$, Fence 1 and Fence 2 are indeed perpendicular.
$$\boxed{\text{Yes, the fences meet at a right angle, since } m_{DE}\,m_{EF}=-1}$$
QUESTION 42
11 marks
Hard
A surveyor records the three corner markers of a triangular plot of land on a coordinate
grid (units in metres): $A(0,0)$, $B(8,6)$ and $C(17,-6)$.
(a) Find the gradient of $[AB]$ and the length of $[AB]$. [3]
(b) Find the gradient of $[BC]$ and the length of $[BC]$. [3]
(c) Show that the plot has a right angle at $B$. [2]
(d) Find the length of $[AC]$, and hence find the perimeter of the plot. [3]
Show complete worked solution
(a) Using the gradient formula for $A(0,0)$ and $B(8,6)$:
$$m_{AB} = \dfrac{6-0}{8-0} = \dfrac{6}{8} = 0.75$$
Using the distance formula:
$$AB = \sqrt{(8-0)^2+(6-0)^2} = \sqrt{64+36} = \sqrt{100}$$
$$\boxed{m_{AB}=0.75,\ \ AB = 10\text{ m}}$$
(b) Using the gradient formula for $B(8,6)$ and $C(17,-6)$:
$$m_{BC} = \dfrac{-6-6}{17-8} = \dfrac{-12}{9} = -\dfrac43$$
Using the distance formula:
$$BC = \sqrt{(17-8)^2+(-6-6)^2} = \sqrt{81+144} = \sqrt{225}$$
$$\boxed{m_{BC}=-\dfrac43,\ \ BC = 15\text{ m}}$$
(c) Testing the product of the gradients found in (a) and (b):
$$m_{AB}\times m_{BC} = 0.75\times\left(-\dfrac43\right) = -1$$
Since the product equals $-1$, $[AB]$ and $[BC]$ are perpendicular, confirming a right angle at
$B$.
$$\boxed{\text{Right angle at } B,\text{ since } m_{AB}\,m_{BC}=-1}$$
(d) Using the distance formula between $A(0,0)$ and $C(17,-6)$:
$$AC = \sqrt{(17-0)^2+(-6-0)^2} = \sqrt{289+36} = \sqrt{325} \approx 18.03\text{ m}$$
The perimeter is the sum of all three side lengths:
$$P = AB+BC+AC = 10+15+18.03$$
$$\boxed{P \approx 43.0\text{ m}}$$
QUESTION 43
10 marks
Hard
An air-traffic control tower is located at the origin $O(0,0)$ on a radar coordinate grid
(units in km). Corridor Line 1 is a straight flight corridor passing through $(0,20)$ and
$(40,60)$. Corridor Line 2 must be perpendicular to Line 1 and must pass through the point
$(50,0)$. The two corridors cross at a mandatory checkpoint.
(a) Find the equation of Corridor Line 1, in the form $y=mx+c$. [2]
(b) Find the equation of Corridor Line 2. [3]
(c) Find the coordinates of the checkpoint where the two corridors cross. [3]
(d) Find the straight-line distance from the control tower $O$ to the
checkpoint. [2]
Show complete worked solution
(a) Using the gradient formula between $(0,20)$ and $(40,60)$:
$$m_1 = \dfrac{60-20}{40-0} = \dfrac{40}{40} = 1$$
Using the point-gradient form with $m_1=1$ and the point $(0,20)$:
$$y-20 = 1(x-0)$$
$$\boxed{\text{Line 1: } y = x+20}$$
(b) Since Line 2 is perpendicular to Line 1, its gradient is the negative reciprocal of $m_1=1$:
$$m_2 = -\dfrac{1}{1} = -1$$
Using the point-gradient form with $m_2=-1$ and the point $(50,0)$:
$$y-0 = -1(x-50)$$
$$\boxed{\text{Line 2: } y = -x+50}$$
(c) At the checkpoint, both equations give the same $y$-value:
$$x+20 = -x+50$$
$$2x = 30$$
$$x = 15$$
$$y = 15+20 = 35$$
$$\boxed{\text{Checkpoint at } (15,35)}$$
(d) Using the distance formula between $O(0,0)$ and the checkpoint $(15,35)$:
$$d = \sqrt{(15-0)^2+(35-0)^2} = \sqrt{225+1225} = \sqrt{1450}$$
$$\boxed{d \approx 38.1\text{ km}}$$
QUESTION 44
10 marks
Hard
An engineering firm is inspecting a straight suspension cable on a dam wall, modelled on a
coordinate diagram (units in metres) as the line through anchor points $A(20,10)$ and
$B(60,40)$. An inspection drone starts at $B$ and can travel a further $25$ m along the line of
the cable, in either direction from $B$.
(a) Find the gradient of the cable, and hence its equation in the form
$y=mx+c$. [3]
(b) Find a unit vector in the direction from $A$ towards $B$. [3]
(c) Find the coordinates of both possible positions of the drone, $25$ m from
$B$ along the line of the cable. [4]
Show complete worked solution
(a) Using the gradient formula between $A(20,10)$ and $B(60,40)$:
$$m = \dfrac{40-10}{60-20} = \dfrac{30}{40} = 0.75$$
Using the point-gradient form with $m=0.75$ and the point $A(20,10)$:
$$y-10 = 0.75(x-20)$$
$$y = 0.75x-15+10$$
$$\boxed{y = 0.75x-5}$$
(b) A direction vector from $A$ to $B$ is:
$$\overrightarrow{AB} = \begin{pmatrix}60-20\\40-10\end{pmatrix} = \begin{pmatrix}40\\30\end{pmatrix}$$
with magnitude
$$|\overrightarrow{AB}| = \sqrt{40^2+30^2} = \sqrt{1600+900} = \sqrt{2500} = 50$$
Dividing by the magnitude gives the unit vector:
$$\hat{u} = \left(\dfrac{40}{50},\ \dfrac{30}{50}\right) = (0.8,\ 0.6)$$
$$\boxed{\hat u = (0.8,\ 0.6)}$$
(c) Moving $25$ m from $B(60,40)$ in the direction of $\hat u$ (continuing past $B$, away from
$A$):
$$P_1 = B+25\hat u = (60+25(0.8),\ 40+25(0.6)) = (60+20,\ 40+15)$$
$$\boxed{P_1 = (80,55)}$$
Moving $25$ m from $B(60,40)$ in the opposite direction (back towards $A$ and beyond):
$$P_2 = B-25\hat u = (60-20,\ 40-15)$$
$$\boxed{P_2 = (40,25)}$$
Check: $A(20,10)$, $P_2(40,25)$ and $B(60,40)$ all satisfy $y=0.75x-5$
(e.g. $0.75(40)-5=25$, which matches), confirming both points lie on the cable's line.
QUESTION 45
11 marks
Hard
A quadrilateral plot of land has vertices $W(1,1)$, $X(6,3)$, $Y(9,9)$ and $Z(4,7)$ on a
coordinate survey map (units in metres), listed in order around the boundary.
(a) Find the gradients of $[WX]$ and $[ZY]$, and hence show that $[WX]$ and
$[ZY]$ are parallel. [3]
(b) Find the gradients of $[WZ]$ and $[XY]$, and hence show that $[WZ]$ and
$[XY]$ are parallel. [3]
(c) Find the midpoints of the two diagonals, $[WY]$ and $[XZ]$, and comment on
what your results confirm about the quadrilateral. [3]
(d) Find the lengths of both diagonals, $[WY]$ and $[XZ]$. [2]
Show complete worked solution
(a) Using the gradient formula for $W(1,1)$ and $X(6,3)$:
$$m_{WX} = \dfrac{3-1}{6-1} = \dfrac{2}{5} = 0.4$$
Using the gradient formula for $Z(4,7)$ and $Y(9,9)$:
$$m_{ZY} = \dfrac{9-7}{9-4} = \dfrac{2}{5} = 0.4$$
Since $m_{WX}=m_{ZY}=0.4$:
$$\boxed{[WX]\parallel[ZY]}$$
(b) Using the gradient formula for $W(1,1)$ and $Z(4,7)$:
$$m_{WZ} = \dfrac{7-1}{4-1} = \dfrac{6}{3} = 2$$
Using the gradient formula for $X(6,3)$ and $Y(9,9)$:
$$m_{XY} = \dfrac{9-3}{9-6} = \dfrac{6}{3} = 2$$
Since $m_{WZ}=m_{XY}=2$:
$$\boxed{[WZ]\parallel[XY]}$$
(c) Using the midpoint formula for the diagonal $[WY]$, $W(1,1)$ and $Y(9,9)$:
$$M_{WY} = \left(\dfrac{1+9}{2},\ \dfrac{1+9}{2}\right) = (5,5)$$
Using the midpoint formula for the diagonal $[XZ]$, $X(6,3)$ and $Z(4,7)$:
$$M_{XZ} = \left(\dfrac{6+4}{2},\ \dfrac{3+7}{2}\right) = (5,5)$$
Since both diagonals share the same midpoint $(5,5)$, they bisect each other.
$$\boxed{M_{WY}=M_{XZ}=(5,5)\text{, confirming } WXYZ \text{ is a parallelogram (diagonals bisect each other)}}$$
(d) Using the distance formula between $W(1,1)$ and $Y(9,9)$:
$$WY = \sqrt{(9-1)^2+(9-1)^2} = \sqrt{64+64} = \sqrt{128} = 8\sqrt2 \approx 11.31\text{ m}$$
Using the distance formula between $X(6,3)$ and $Z(4,7)$:
$$XZ = \sqrt{(4-6)^2+(7-3)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt5 \approx 4.47\text{ m}$$
$$\boxed{WY \approx 11.31\text{ m},\ \ XZ \approx 4.47\text{ m}}$$
QUESTION 46
11 marks
Hard
Three relay stations are located at $A(0,0)$, $B(8,0)$ and $C(2,6)$ on a coordinate grid
(units in km). Engineers want to find the location of a new control hub that is equidistant from
all three stations.
(a) Find the equation of the perpendicular bisector of $[AB]$. [3]
(b) Find the equation of the perpendicular bisector of $[AC]$. [3]
(c) By solving your two equations simultaneously, find the coordinates of the
control hub. [3]
(d) Verify that the control hub is equidistant from all three stations. [2]
Show complete worked solution
(a) The midpoint of $[AB]$, where $A(0,0)$ and $B(8,0)$, is:
$$M_{AB} = \left(\dfrac{0+8}{2},\ \dfrac{0+0}{2}\right) = (4,0)$$
Since $[AB]$ is horizontal (gradient $0$), the perpendicular bisector is vertical, passing
through $x=4$:
$$\boxed{\text{Perpendicular bisector of } [AB]:\ x=4}$$
(b) The midpoint of $[AC]$, where $A(0,0)$ and $C(2,6)$, is:
$$M_{AC} = \left(\dfrac{0+2}{2},\ \dfrac{0+6}{2}\right) = (1,3)$$
The gradient of $[AC]$ is:
$$m_{AC} = \dfrac{6-0}{2-0} = 3$$
The perpendicular bisector has gradient $-\dfrac13$. Using the point-gradient form through
$M_{AC}(1,3)$:
$$y-3 = -\dfrac13(x-1)$$
$$y = -\dfrac13x+\dfrac13+3$$
$$\boxed{\text{Perpendicular bisector of } [AC]:\ y = -\dfrac13x+\dfrac{10}{3}}$$
(c) Substituting $x=4$ from part (a) into the equation from part (b):
$$y = -\dfrac13(4)+\dfrac{10}{3} = -\dfrac43+\dfrac{10}{3} = \dfrac{6}{3}$$
$$y=2$$
$$\boxed{\text{Control hub at } (4,2)}$$
(d) Using the distance formula from the control hub $(4,2)$ to each station:
$$\text{to } A(0,0):\quad \sqrt{4^2+2^2} = \sqrt{16+4} = \sqrt{20}$$
$$\text{to } B(8,0):\quad \sqrt{(4-8)^2+2^2} = \sqrt{16+4} = \sqrt{20}$$
$$\text{to } C(2,6):\quad \sqrt{(4-2)^2+(2-6)^2} = \sqrt{4+16} = \sqrt{20}$$
All three distances equal $\sqrt{20}\approx4.47$ km, confirming the control hub is equidistant
from all three stations.
$$\boxed{\text{Equidistant, each distance} = \sqrt{20}\approx 4.47\text{ km}}$$
QUESTION 47
10 marks
Hard
A city road runs in a straight line through $(0,5)$ and $(10,15)$ on the city's coordinate
grid (units in metres). A cross street runs through the same point $(10,15)$ and also through
$(25,0)$. For a construction project, the original road must be realigned to a new, parallel
route passing through the point $(2,20)$.
(a) Find the equation of the original road. [2]
(b) Find the equation of the new (realigned) road, which is parallel to the
original road and passes through $(2,20)$. [2]
(c) Find the equation of the cross street. [2]
(d) Find the coordinates of the new crossroads, where the new road meets the
cross street. [2]
(e) Find the distance between the original crossroads at $(10,15)$ and the new
crossroads found in part (d). [2]
Show complete worked solution
(a) Using the gradient formula between $(0,5)$ and $(10,15)$:
$$m = \dfrac{15-5}{10-0} = \dfrac{10}{10} = 1$$
Using the point-gradient form with $m=1$ and the point $(0,5)$:
$$y-5 = 1(x-0)$$
$$\boxed{\text{Original road: } y = x+5}$$
(b) Since the new road is parallel to the original road, it has the same gradient, $m=1$. Using
the point-gradient form with the point $(2,20)$:
$$y-20 = 1(x-2)$$
$$\boxed{\text{New road: } y = x+18}$$
(c) Using the gradient formula for the two given points on the cross street, $(10,15)$ and
$(25,0)$:
$$m = \dfrac{0-15}{25-10} = \dfrac{-15}{15} = -1$$
Using the point-gradient form with $m=-1$ and the point $(10,15)$:
$$y-15 = -1(x-10)$$
$$\boxed{\text{Cross street: } y = -x+25}$$
(d) At the new crossroads, the new road and the cross street give the same $y$-value:
$$x+18 = -x+25$$
$$2x = 7$$
$$x = 3.5$$
$$y = 3.5+18 = 21.5$$
$$\boxed{\text{New crossroads at } (3.5,21.5)}$$
(e) Using the distance formula between the original crossroads $(10,15)$ and the new crossroads
$(3.5,21.5)$:
$$d = \sqrt{(10-3.5)^2+(15-21.5)^2} = \sqrt{6.5^2+(-6.5)^2} = \sqrt{42.25+42.25} = \sqrt{84.5}$$
$$\boxed{d \approx 9.19\text{ m}}$$
QUESTION 48
11 marks
Hard
A hiking trail consists of two straight segments on a coordinate map (units in metres): from
the trailhead $A(0,0)$ to a rest point $B(30,40)$, and from $B(30,40)$ to the summit
$C(70,10)$.
(a) Find the equation of segment $[AB]$. [3]
(b) Find the equation of segment $[BC]$. [3]
(c) Find the lengths of $[AB]$ and $[BC]$, and hence the total length of the
trail from $A$ to $C$ via $B$. [2]
(d) Find the direct (straight-line) distance from $A$ to $C$, and compare it to
the total trail length found in part (c). [3]
Show complete worked solution
(a) Using the gradient formula between $A(0,0)$ and $B(30,40)$:
$$m_{AB} = \dfrac{40-0}{30-0} = \dfrac{40}{30} = \dfrac43$$
Using the point-gradient form with $m_{AB}=\dfrac43$ and the point $A(0,0)$:
$$y-0 = \dfrac43(x-0)$$
$$\boxed{[AB]:\ y = \dfrac43x}$$
(b) Using the gradient formula between $B(30,40)$ and $C(70,10)$:
$$m_{BC} = \dfrac{10-40}{70-30} = \dfrac{-30}{40} = -\dfrac34$$
Using the point-gradient form with $m_{BC}=-\dfrac34$ and the point $B(30,40)$:
$$y-40 = -\dfrac34(x-30)$$
$$y = -0.75x+22.5+40$$
$$\boxed{[BC]:\ y = -0.75x+62.5}$$
(c) Using the distance formula for $[AB]$:
$$AB = \sqrt{(30-0)^2+(40-0)^2} = \sqrt{900+1600} = \sqrt{2500} = 50\text{ m}$$
Using the distance formula for $[BC]$:
$$BC = \sqrt{(70-30)^2+(10-40)^2} = \sqrt{1600+900} = \sqrt{2500} = 50\text{ m}$$
Total trail length:
$$AB+BC = 50+50 = 100\text{ m}$$
$$\boxed{\text{Total trail length} = 100\text{ m}}$$
(d) Using the distance formula for $A(0,0)$ and $C(70,10)$:
$$AC = \sqrt{(70-0)^2+(10-0)^2} = \sqrt{4900+100} = \sqrt{5000} \approx 70.71\text{ m}$$
Since $AC \approx 70.71\text{ m} < 100\text{ m}$, the direct distance from $A$ to $C$ is shorter
than the trail via $B$, as expected since $B$ lies off the direct line from $A$ to $C$.
$$\boxed{AC\approx70.71\text{ m}, \text{ which is less than the } 100\text{ m trail via } B}$$
QUESTION 49
10 marks
Hard
Two towns are located at $P(0,0)$ and $Q(12,16)$ on a regional coordinate grid (units in
km). A proposed emergency-service boundary is the perpendicular bisector of $[PQ]$, so that
locations on either side are closer to one town or the other. A highway runs through the region
with equation $y=0.25x+2$.
(a) Find the midpoint of $[PQ]$ and the gradient of $[PQ]$. [3]
(b) Find the equation of the emergency-service boundary. [3]
(c) Find the coordinates of the point where the boundary crosses the highway. [3]
(d) State what this crossing point represents in context. [1]
Show complete worked solution
(a) Using the midpoint formula for $P(0,0)$ and $Q(12,16)$:
$$M = \left(\dfrac{0+12}{2},\ \dfrac{0+16}{2}\right) = (6,8)$$
Using the gradient formula for $P$ and $Q$:
$$m_{PQ} = \dfrac{16-0}{12-0} = \dfrac{16}{12} = \dfrac43$$
$$\boxed{M=(6,8),\ \ m_{PQ}=\dfrac43}$$
(b) The boundary is perpendicular to $[PQ]$, so its gradient is:
$$m_\perp = -\dfrac{1}{4/3} = -\dfrac34$$
Using the point-gradient form through $M(6,8)$ with gradient $-\dfrac34$:
$$y-8 = -\dfrac34(x-6)$$
$$y = -0.75x+4.5+8$$
$$\boxed{\text{Boundary: } y = -0.75x+12.5}$$
(c) At the crossing point, the boundary and highway equations give the same $y$-value:
$$-0.75x+12.5 = 0.25x+2$$
$$12.5-2 = 0.25x+0.75x$$
$$10.5 = x$$
$$x = 10.5$$
Substituting into the highway equation:
$$y = 0.25(10.5)+2 = 2.625+2 = 4.625$$
$$\boxed{\text{Crossing point at } (10.5,\ 4.625)}$$
(d) This point represents the exact location, along the highway, where a driver is equally close
to both towns $P$ and $Q$: beyond this point in one direction the highway is closer to $P$, and
beyond it in the other direction the highway is closer to $Q$.
$$\boxed{\text{The point where the highway is equidistant from towns } P \text{ and } Q}$$
QUESTION 50
11 marks
Hard
A tower crane's jib runs in a straight line through the base $(0,0)$ and the point
$(24,18)$ on a coordinate diagram of the construction site (units in metres). A fixed anchor
point for a support wire is located at $(5,25)$. The support wire must run from the anchor point
to the jib, meeting it at a right angle (the shortest possible connection).
(a) Find the equation of the jib's line. [2]
(b) Find the equation of the support wire, given that it is perpendicular to the
jib and passes through the anchor point $(5,25)$. [3]
(c) Find the coordinates of the point $F$ where the support wire meets the jib
(the foot of the perpendicular). [4]
(d) Find the length of the support wire needed, from the anchor point to $F$. [2]
Show complete worked solution
(a) Using the gradient formula between $(0,0)$ and $(24,18)$:
$$m_{\text{jib}} = \dfrac{18-0}{24-0} = \dfrac{18}{24} = 0.75$$
Using the point-gradient form with $m=0.75$ and the point $(0,0)$:
$$\boxed{\text{Jib: } y = 0.75x}$$
(b) Since the support wire is perpendicular to the jib, its gradient is the negative reciprocal
of $m_{\text{jib}}=0.75$:
$$m_{\text{wire}} = -\dfrac{1}{0.75} = -\dfrac43$$
Using the point-gradient form with $m=-\dfrac43$ and the point $(5,25)$:
$$y-25 = -\dfrac43(x-5)$$
$$y = -\dfrac43x+\dfrac{20}{3}+25$$
$$\boxed{\text{Support wire: } y = -\dfrac43x+\dfrac{95}{3}}$$
(c) At the foot of the perpendicular $F$, both equations give the same $y$-value:
$$0.75x = -\dfrac43x+\dfrac{95}{3}$$
Multiplying every term by $12$ to clear denominators:
$$9x = -16x+380$$
$$25x = 380$$
$$x = 15.2$$
Substituting into $y=0.75x$:
$$y = 0.75(15.2) = 11.4$$
$$\boxed{F = (15.2,\ 11.4)}$$
(d) Using the distance formula between the anchor point $(5,25)$ and $F(15.2,11.4)$:
$$d = \sqrt{(5-15.2)^2+(25-11.4)^2} = \sqrt{(-10.2)^2+13.6^2} = \sqrt{104.04+184.96} = \sqrt{289}$$
$$\boxed{d = 17\text{ m}}$$
Linear Functions 50 questions
QUESTION 1
5 marks
Easy
The speed of a certain catalysed chemical reaction, $s$ (in cm$^3$ per second), is
modelled as a function of temperature $x$ (in $^{\circ}$C above a threshold value) by
$$f(x) = \sqrt{x-3}+2.$$
(a) State the largest possible domain of $f$ for which $f(x)$ is defined and gives a
real value. [2]
(b) State the range of $f$ over this domain. [2]
(c) Find the value of $f(12)$. [1]
Show complete worked solution
(a) The expression $f(x) = \sqrt{x-3}+2$ requires the quantity under the square root to be non-negative:
$$x-3 \ge 0$$
$$x \ge 3$$
$$\boxed{\text{Domain: } x \ge 3}$$
(b) Since $\sqrt{x-3} \ge 0$ for every $x$ in the domain, adding $2$ to both sides gives:
$$f(x) = \sqrt{x-3}+2 \ge 2$$
The minimum value $2$ is attained at $x=3$ (where $\sqrt{x-3}=0$), and as $x$ increases without bound, $\sqrt{x-3}$ also increases without bound, so $f(x)$ takes every value from $2$ upwards.
$$\boxed{\text{Range: } f(x) \ge 2}$$
(c) Substituting $x=12$ into $f(x)$:
$$f(12) = \sqrt{12-3}+2$$
$$f(12) = \sqrt{9}+2$$
$$f(12) = 3+2 = 5$$
$$\boxed{f(12) = 5\text{ cm}^3\text{s}^{-1}}$$
QUESTION 2
6 marks
Medium
A courier company charges a delivery fee $f(w)$, in dollars, for a package of mass $w$
kg according to
$$f(w) = 8 + 2.5w, \qquad 0 < w \le 20$$
Packages heavier than 20 kg are not accepted by this service.
(a) State the domain and range of $f$. [3]
(b) Find $f(7.4)$ and interpret this value in context. [2]
(c) Find the mass $w$ of a package for which the delivery fee is $\$53$. [1]
Show complete worked solution
(a) The domain is given directly by the restriction on the model:
$$\boxed{\text{Domain: } 0 < w \le 20}$$
To find the range, consider the behaviour of $f(w)=8+2.5w$ at the ends of the domain. As $w \to 0^+$ (but $w\ne0$, since a package must have positive mass), $f(w) \to 8$, but this value is never actually reached. At $w=20$, which is included in the domain:
$$f(20) = 8+2.5(20) = 8+50 = 58$$
Since $f$ is linear and increasing (gradient $2.5>0$), it takes every value between these two ends, with $8$ excluded and $58$ included.
$$\boxed{\text{Range: } 8 < f(w) \le 58}$$
(b) Substituting $w=7.4$ into $f(w)$:
$$f(7.4) = 8+2.5(7.4)$$
$$f(7.4) = 8+18.5 = 26.5$$
$$\boxed{f(7.4) = \$26.50}$$
This means a $7.4$ kg package costs $\$26.50$ to deliver.
(c) Setting $f(w)=53$ and solving for $w$:
$$53 = 8+2.5w$$
$$2.5w = 45$$
$$w = 18$$
$$\boxed{w = 18\text{ kg}}$$
QUESTION 3
5 marks
Easy
A tutor records the number of hours $x$ a student studies for a test and the resulting
test score, $f(x)$ (out of 100):
| Hours studied, $x$ | 2 | 4 | 6 | 8 |
|---|---|---|---|---|
| Score, $f(x)$ | 54 | 62 | 70 | 78 |
The data follow an exact linear pattern.
(a) Find $f(x)$ in the form $f(x)=mx+c$. [3]
(b) Use $f$ to estimate the score of a student who studies for 5 hours. [1]
(c) Solve $f(x)=90$, and comment on whether this prediction can be trusted, given a
sensible domain for the model such as $0 \le x \le 10$. [1]
Show complete worked solution
(a) Using the gradient formula between the first two data points $(2,54)$ and $(4,62)$:
$$m = \dfrac{62-54}{4-2} = \dfrac{8}{2} = 4$$
Using the point-gradient form with $m=4$ and point $(2,54)$:
$$f(x)-54 = 4(x-2)$$
$$f(x) = 4x-8+54$$
$$f(x) = 4x+46$$
$$\boxed{f(x) = 4x+46}$$
(b) Substituting $x=5$ into $f(x)$:
$$f(5) = 4(5)+46$$
$$f(5) = 20+46 = 66$$
$$\boxed{f(5) \approx 66}$$
(c) Setting $f(x)=90$ and solving for $x$:
$$90 = 4x+46$$
$$4x = 44$$
$$x = 11$$
Since the sensible domain for the model is $0 \le x \le 10$, the value $x=11$ lies just outside this domain, so this prediction is an extrapolation. The true relationship between study time and score may not remain linear that far beyond the observed data, so the prediction should be treated with caution.
$$\boxed{x=11\text{ hours (extrapolation - unreliable)}}$$
QUESTION 4
7 marks
Medium
A taxi company's fare, $f(d)$ dollars, for a journey of distance $d$ km is calculated
using the piecewise function
$$f(d) = \begin{cases} 3.50+1.20d, & 0 \le d \le 10 \\ 15.50+1.60(d-10), & 10 < d \le 40 \end{cases}$$
(the service does not operate for journeys longer than 40 km).
(a) Find $f(6)$ and $f(15)$. [3]
(b) State the range of $f$ over its full domain. [2]
(c) A customer is charged $\$30$. Find the distance travelled, giving your answer to 3
significant figures. [2]
Show complete worked solution
(a) Since $0 \le 6 \le 10$, use the first branch, $f(d)=3.50+1.20d$:
$$f(6) = 3.50+1.20(6)$$
$$f(6) = 3.50+7.20 = 10.70$$
$$\boxed{f(6) = \$10.70}$$
Since $10 < 15 \le 40$, use the second branch, $f(d)=15.50+1.60(d-10)$:
$$f(15) = 15.50+1.60(15-10)$$
$$f(15) = 15.50+1.60(5)$$
$$f(15) = 15.50+8.00 = 23.50$$
$$\boxed{f(15) = \$23.50}$$
(b) On $[0,10]$, $f$ is linear with positive gradient $1.20$, so it increases from
$$f(0) = 3.50+1.20(0) = 3.50$$
to
$$f(10) = 3.50+1.20(10) = 3.50+12.00 = 15.50$$
On $(10,40]$, $f$ is linear with positive gradient $1.60$, so it continues increasing from just above $15.50$ up to
$$f(40) = 15.50+1.60(40-10) = 15.50+48.00 = 63.50$$
Since $f$ is continuous at $d=10$ (both branches give $15.50$ there) and increasing throughout, the overall range is:
$$\boxed{3.50 \le f(d) \le 63.50}$$
(c) Since the charge $\$30 > 15.50$, the journey must be on the second branch, so set $f(d)=30$:
$$30 = 15.50+1.60(d-10)$$
$$14.50 = 1.60(d-10)$$
$$d-10 = \dfrac{14.50}{1.60} = 9.0625$$
$$d = 10+9.0625 = 19.0625$$
$$\boxed{d \approx 19.1\text{ km}}$$
QUESTION 5
15 marks
Hard
A furniture company manufactures wooden chairs. Because of factory capacity, at most
400 chairs can be made per month, so all functions below have domain $0 \le x \le 400$,
where $x$ is the number of chairs sold in a month.
The company's monthly revenue (in dollars) from selling $x$ chairs at $\$85$ each is
$$R(x) = 85x.$$
The accounts department reports total monthly cost data:
when $x=100$ chairs are made, the total cost is $\$8500$;
when $x=250$ chairs are made, the total cost is $\$16750$.
The cost function $C(x)$ is known to be linear.
(a) Find $C(x)$, and state the values of the fixed monthly cost and the variable
(marginal) cost per chair. [4]
(b) Define the monthly profit function $P(x)=R(x)-C(x)$. Find $P(x)$ in the form
$P(x)=ax+b$, and state the range of $P$ over the given domain. [6]
(c) Find the break-even production level (the value of $x$ for which $P(x)=0$), and
explain, with reference to the domain, why this is the only break-even point for this
model. [5]
Show complete worked solution
(a) Using the gradient formula for $C$ between $(100,8500)$ and $(250,16750)$:
$$m_C = \dfrac{16750-8500}{250-100} = \dfrac{8250}{150} = 55$$
Using the point-gradient form with $m_C=55$ and point $(100,8500)$:
$$C(x)-8500 = 55(x-100)$$
$$C(x) = 55x-5500+8500$$
$$C(x) = 55x+3000$$
$$\boxed{C(x) = 55x+3000}$$
The fixed monthly cost is the value of $C$ at $x=0$, i.e. $\$3000$ (costs incurred even with no chairs produced, such as rent). The variable (marginal) cost per chair is the gradient, $\$55$ per chair.
(b) Substituting $R(x)=85x$ and $C(x)=55x+3000$ into $P(x)=R(x)-C(x)$:
$$P(x) = 85x-(55x+3000)$$
$$P(x) = 30x-3000$$
$$\boxed{P(x) = 30x-3000}$$
Since $P$ is linear with positive gradient $30>0$, $P$ is increasing on $[0,400]$, so its minimum and maximum values on the domain occur at the endpoints:
$$P(0) = 30(0)-3000 = -3000$$
$$P(400) = 30(400)-3000 = 12000-3000 = 9000$$
$$\boxed{\text{Range: } -3000 \le P(x) \le 9000}$$
(c) Setting $P(x)=0$ and solving for $x$:
$$30x-3000 = 0$$
$$30x = 3000$$
$$x = 100$$
$$\boxed{x = 100\text{ chairs}}$$
Since $P(x)=30x-3000$ is linear with a non-zero gradient, its graph is a straight line that can cross the horizontal axis (i.e. satisfy $P(x)=0$) at most once. As $x=100$ lies within the domain $[0,400]$, this is the unique, valid break-even point: the company needs to sell at least $100$ chairs per month to avoid a loss.
QUESTION 6
6 marks
Easy
A cylindrical water tank is draining at a constant rate. The depth of water, $h$ cm,
is recorded at two times: at $t=2$ minutes, $h=54$ cm; at $t=9$ minutes, $h=19$ cm.
(a) Find a linear model $h(t)$ for the depth of water, $t$ minutes after the tank
began draining. [3]
(b) Interpret the gradient of your model in context. [1]
(c) Find the time at which the tank becomes empty, and state a suitable domain for
this model. [2]
Show complete worked solution
(a) Using the gradient formula between $(2,54)$ and $(9,19)$:
$$m = \dfrac{19-54}{9-2} = \dfrac{-35}{7} = -5$$
Using the point-gradient form with $m=-5$ and point $(2,54)$:
$$h(t)-54 = -5(t-2)$$
$$h(t) = -5t+10+54$$
$$h(t) = 64-5t$$
$$\boxed{h(t) = 64-5t}$$
(b) The gradient $-5$ means the water depth decreases by $5$ cm every minute - the tank drains at a constant rate of $5$ cm/min.
(c) Setting $h(t)=0$ and solving for $t$:
$$64-5t = 0$$
$$5t = 64$$
$$t = 12.8$$
$$\boxed{t = 12.8\text{ minutes}}$$
Since the depth of water cannot be negative, a suitable domain for this model is $0 \le t \le 12.8$.
QUESTION 7
6 marks
Easy
A ball is thrown into the air. Its height above the ground, $h$ metres, is recorded at
various times $t$ seconds after being thrown:
| $t$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $h$ | 1 | 16 | 21 | 16 | 1 |
(a) Use quadratic regression on your GDC to find a model $h(t) = at^2+bt+c$ for this
data. [3]
(b) Find the maximum height reached by the ball, and the time at which it occurs. [1]
(c) Find the positive time at which the ball hits the ground ($h=0$), to 3 significant
figures. [2]
Show complete worked solution
(a) Entering the five data points into the GDC's quadratic regression feature (model $h(t)=at^2+bt+c$) gives:
$$\boxed{h(t) = -5t^2+20t+1}$$
This can be checked against the table, e.g. $h(1) = -5(1)+20(1)+1 = -5+20+1 = 16$, and $h(3) = -5(9)+20(3)+1 = -45+60+1=16$, both of which match the given data.
(b) Since the model is quadratic with $a=-5<0$, the graph is a downward-opening parabola, so its maximum occurs at the vertex. Using the vertex formula:
$$t = -\dfrac{b}{2a} = -\dfrac{20}{2(-5)} = 2$$
Substituting $t=2$ into $h(t)$:
$$h(2) = -5(2)^2+20(2)+1$$
$$h(2) = -20+40+1 = 21$$
$$\boxed{\text{Maximum height} = 21\text{ m at } t=2\text{ s}}$$
(c) Setting $h(t)=0$:
$$-5t^2+20t+1 = 0$$
Multiplying by $-1$:
$$5t^2-20t-1 = 0$$
Using the quadratic formula, $t = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$, with $a=5$, $b=-20$, $c=-1$:
$$t = \dfrac{20\pm\sqrt{(-20)^2-4(5)(-1)}}{2(5)}$$
$$t = \dfrac{20\pm\sqrt{400+20}}{10} = \dfrac{20\pm\sqrt{420}}{10}$$
$$t = \dfrac{20\pm20.4939\ldots}{10}$$
Taking the positive root, since time must be positive:
$$t = \dfrac{20+20.4939\ldots}{10} = 4.0494\ldots$$
$$\boxed{t \approx 4.05\text{ s}}$$
QUESTION 8
7 marks
Medium
An investment portfolio is worth $\$15\,000$ initially ($t=0$), and $\$21\,400$ after
5 years. The value $V(t)$, in dollars, $t$ years after the initial investment, is
modelled by an exponential function
$$V(t) = V_0e^{kt}.$$
(a) Show that $V_0 = 15000$, and find the value of $k$, to 4 significant figures. [4]
(b) Using your model, find the number of years it will take for the investment to
reach $\$50\,000$, to 3 significant figures. [3]
Show complete worked solution
(a) Substituting $t=0$ into $V(t)=V_0e^{kt}$:
$$V(0) = V_0e^{0} = V_0$$
Since $V(0)=15000$:
$$\boxed{V_0 = 15000}$$
Substituting $t=5$, $V(5)=21400$ and $V_0=15000$:
$$21400 = 15000e^{5k}$$
Dividing both sides by $15000$:
$$e^{5k} = \dfrac{21400}{15000} = 1.42\overline{6}$$
Taking the natural logarithm of both sides:
$$5k = \ln(1.42\overline{6}) = 0.355374\ldots$$
$$k = 0.0710748\ldots$$
$$\boxed{k \approx 0.07107}$$
(b) The model is $V(t) = 15000e^{0.07107t}$. Setting $V(t)=50000$:
$$50000 = 15000e^{0.07107t}$$
Dividing both sides by $15000$:
$$e^{0.07107t} = \dfrac{50000}{15000} = 3.3\overline{3}$$
Taking the natural logarithm of both sides:
$$0.07107t = \ln(3.3\overline{3}) = 1.20397\ldots$$
$$t = \dfrac{1.20397\ldots}{0.07107} = 16.9411\ldots$$
$$\boxed{t \approx 16.9\text{ years}}$$
QUESTION 9
11 marks
Medium
A bakery finds that the weekly number of cakes sold, $n$, depends on the price $p$
dollars charged per cake. Two data points from its sales records:
when $p=\$10$, $n=120$ cakes are sold per week;
when $p=\$15$, $n=80$ cakes are sold per week.
The relationship between $n$ and $p$ is modelled as linear.
(a) Find the linear demand function $n(p)$. [3]
(b) The weekly revenue is $R(p) = p \times n(p)$. Show that
$R(p) = 200p-8p^2$, and use technology to find the price $p$ that maximizes weekly
revenue, and the maximum weekly revenue. [4]
(c) Find the range of prices $p$ for which the weekly revenue is at least $\$1200$. [4]
Show complete worked solution
(a) Using the gradient formula between $(10,120)$ and $(15,80)$:
$$m = \dfrac{80-120}{15-10} = \dfrac{-40}{5} = -8$$
Using the point-gradient form with $m=-8$ and point $(10,120)$:
$$n(p)-120 = -8(p-10)$$
$$n(p) = -8p+80+120$$
$$n(p) = 200-8p$$
$$\boxed{n(p) = 200-8p}$$
(b) Substituting $n(p)=200-8p$ into $R(p)=p\times n(p)$:
$$R(p) = p(200-8p)$$
$$R(p) = 200p-8p^2 \qquad \text{as required.}$$
This is a quadratic in $p$ with $a=-8<0$, so its graph is a downward-opening parabola and its maximum occurs at the vertex. Using the vertex formula (or technology):
$$p^{*} = -\dfrac{200}{2(-8)} = 12.5$$
Substituting $p=12.5$ back into $R(p)$:
$$R(12.5) = 200(12.5)-8(12.5)^2$$
$$R(12.5) = 2500-1250 = 1250$$
$$\boxed{\text{Maximum revenue} = \$1250\text{ at } p=\$12.50}$$
(c) Setting $R(p) \ge 1200$:
$$200p-8p^2 \ge 1200$$
Rearranging into standard quadratic form:
$$-8p^2+200p-1200 \ge 0$$
Dividing both sides by $-8$ (reversing the inequality):
$$p^2-25p+150 \le 0$$
Using the quadratic formula to find the boundary values, $p=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $a=1,b=-25,c=150$:
$$p = \dfrac{25\pm\sqrt{625-600}}{2} = \dfrac{25\pm\sqrt{25}}{2} = \dfrac{25\pm5}{2}$$
$$p = 15 \quad \text{or} \quad p = 10$$
Since the parabola $p^2-25p+150$ opens upward, the expression is $\le 0$ between its roots:
$$\boxed{10 \le p \le 15}$$
(i.e. the price must be between $\$10$ and $\$15$ inclusive for the weekly revenue to be at least $\$1200$.)
QUESTION 10
14 marks
Hard
A phone retailer records the resale value, $V$ dollars, of a particular smartphone
model as it ages:
when new ($t=0$ years), $V=\$900$;
after $t=2$ years, $V=\$324$.
(a) Assuming an exponential decay model $V(t) = 900r^t$, find the value of $r$. [3]
(b) Find the annual percentage depreciation rate implied by this model, and use the
model to predict the resale value after 5 years, to the nearest dollar. [3]
(c) A trade-in store guarantees to pay at least $\$150$ for this phone, for phones up
to 6 years old. Using the model, find the first whole number of years, $t$, after
which the model-predicted resale value falls below $\$150$. [4]
(d) A competing, simpler model assumes the value depreciates linearly, using a
straight line through the same two data points $(0,900)$ and $(2,324)$. Find this
linear model $L(t)$, and evaluate $L(5)$. Comparing $L(5)$ with your answer to part
(b), comment on which model (exponential or linear) is more realistic for predicting
long-term resale value, and why. [4]
Show complete worked solution
(a) Substituting $t=2$, $V=324$ into the model $V(t)=900r^t$:
$$324 = 900r^2$$
Dividing both sides by $900$:
$$r^2 = \dfrac{324}{900} = 0.36$$
Taking the positive square root, since $r>0$ for a depreciation factor:
$$r = \sqrt{0.36} = 0.6$$
$$\boxed{r = 0.6}$$
So the model is $V(t) = 900(0.6)^t$.
(b) The annual depreciation rate is found from the decay factor $r$:
$$\text{depreciation rate} = 1-r = 1-0.6 = 0.4 = 40\%\text{ per year}$$
Substituting $t=5$ into the model:
$$V(5) = 900(0.6)^5$$
$$V(5) = 900(0.07776)$$
$$V(5) = 69.984$$
$$\boxed{V(5) \approx \$70}$$
(c) Setting up the inequality $V(t) < 150$:
$$900(0.6)^t < 150$$
Dividing both sides by $900$:
$$(0.6)^t < \dfrac{150}{900} = 0.1\overline{6}$$
Taking the natural logarithm of both sides:
$$t\ln(0.6) < \ln(0.1\overline{6})$$
Since $\ln(0.6)<0$, dividing both sides by $\ln(0.6)$ reverses the inequality:
$$t > \dfrac{\ln(0.1\overline{6})}{\ln(0.6)} = \dfrac{-1.79176\ldots}{-0.51083\ldots} = 3.5075\ldots$$
So the model-predicted value falls below $\$150$ once $t>3.51$ (approx). Checking whole years directly:
$$V(3) = 900(0.6)^3 = 900(0.216) = 194.4 > 150$$
$$V(4) = 900(0.6)^4 = 900(0.1296) = 116.64 < 150$$
$$\boxed{t=4\text{ years}}$$
This is the first whole number of years for which the predicted resale value falls below $\$150$, and it lies within the 6-year guarantee window.
(d) Using the gradient formula between $(0,900)$ and $(2,324)$:
$$m = \dfrac{324-900}{2-0} = \dfrac{-576}{2} = -288$$
Using the point-gradient form with $m=-288$ and point $(0,900)$:
$$L(t) = 900-288t$$
Substituting $t=5$:
$$L(5) = 900-288(5)$$
$$L(5) = 900-1440 = -540$$
$$\boxed{L(t)=900-288t,\quad L(5)=-\$540}$$
Comment: the linear model predicts a resale value of $-\$540$ after 5 years, which is impossible since a value cannot be negative, whereas the exponential model predicts a small but positive value of about $\$70$. This shows the exponential model is more realistic for long-term depreciation, since resale value naturally decreases towards (but never reaches) zero - matching the exponential model's horizontal asymptote at $V=0$ - whereas the linear model eventually gives unrealistic negative values.
QUESTION 11
5 marks
Easy
A local gym charges a one-time sign-up fee of \$45, plus a monthly membership rate of \$32. The total cost, $f(x)$ dollars, of $x$ months of membership is modelled by
$$f(x) = 45+32x, \qquad 0 \le x \le 12.$$
(a) Find $f(9)$, and interpret this value in context. [2]
(b) State the domain of $f$, and hence find its range. [2]
(c) Find the number of months of membership that can be purchased for exactly \$333. [1]
Show complete worked solution
(a) Substituting $x=9$ into $f(x)$:
$$f(9) = 45+32(9)$$
$$f(9) = 45+288 = 333$$
$$\boxed{f(9) = \$333}$$
This means that 9 months of gym membership costs \$333 in total.
(b) The domain is given directly by the model:
$$\boxed{\text{Domain: } 0 \le x \le 12}$$
Since $f$ is linear with positive gradient $32>0$, $f$ is increasing on this domain, so its minimum and maximum occur at the endpoints:
$$f(0) = 45+32(0) = 45$$
$$f(12) = 45+32(12) = 45+384 = 429$$
$$\boxed{\text{Range: } 45 \le f(x) \le 429}$$
(c) Setting $f(x)=333$ and solving for $x$ (this matches the value already found in part (a)):
$$333 = 45+32x$$
$$32x = 288$$
$$x = 9$$
$$\boxed{x = 9\text{ months}}$$
QUESTION 12
5 marks
Easy
A mobile network charges a fixed monthly fee of \$18, plus \$2.50 for every gigabyte (GB) of data used. The monthly cost, $f(x)$ dollars, for using $x$ GB of data is
$$f(x) = 18+2.5x, \qquad 0 \le x \le 50,$$
since the plan is capped at 50 GB per month.
(a) Find $f(12)$. [2]
(b) State the domain and range of $f$. [2]
(c) Find the value of $x$ for which $f(x)=68$. [1]
Show complete worked solution
(a) Substituting $x=12$ into $f(x)$:
$$f(12) = 18+2.5(12)$$
$$f(12) = 18+30 = 48$$
$$\boxed{f(12) = \$48}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le x \le 50}$$
Since $f$ is linear with positive gradient $2.5>0$, $f$ is increasing, so the range is found from the endpoint values:
$$f(0) = 18, \qquad f(50) = 18+2.5(50) = 18+125 = 143$$
$$\boxed{\text{Range: } 18 \le f(x) \le 143}$$
(c) Setting $f(x)=68$ and solving for $x$:
$$68 = 18+2.5x$$
$$2.5x = 50$$
$$x = 20$$
$$\boxed{x = 20\text{ GB}}$$
QUESTION 13
5 marks
Easy
A bike-hire shop charges a fixed deposit of \$8, plus \$3.50 for each hour the bike is hired. The total cost, $f(h)$ dollars, for hiring a bike for $h$ hours is
$$f(h) = 8+3.5h,$$
and bikes may be hired for at most 10 hours per day.
(a) State a suitable domain for $f$, and hence find its range. [3]
(b) Find $f(4.5)$. [1]
(c) Find the number of hours of hire that would cost exactly \$29. [1]
Show complete worked solution
(a) Since hire time cannot be negative and the shop allows at most 10 hours of hire per day:
$$\boxed{\text{Domain: } 0 \le h \le 10}$$
Since $f$ is linear with positive gradient $3.5>0$, $f$ is increasing, so the range comes from the endpoint values:
$$f(0) = 8+3.5(0) = 8$$
$$f(10) = 8+3.5(10) = 8+35 = 43$$
$$\boxed{\text{Range: } 8 \le f(h) \le 43}$$
(b) Substituting $h=4.5$ into $f(h)$:
$$f(4.5) = 8+3.5(4.5)$$
$$f(4.5) = 8+15.75 = 23.75$$
$$\boxed{f(4.5) = \$23.75}$$
(c) Setting $f(h)=29$ and solving for $h$:
$$29 = 8+3.5h$$
$$3.5h = 21$$
$$h = 6$$
$$\boxed{h = 6\text{ hours}}$$
QUESTION 14
5 marks
Easy
A print shop charges a fixed setup fee of \$15 for a print job, plus \$0.08 per page printed. The total cost, $f(x)$ dollars, for a job of $x$ pages is
$$f(x) = 15+0.08x, \qquad 0 \le x \le 1000.$$
(a) Find $f(250)$. [2]
(b) State the domain and range of $f$. [2]
(c) Find the number of pages printed if the total cost was \$47. [1]
Show complete worked solution
(a) Substituting $x=250$ into $f(x)$:
$$f(250) = 15+0.08(250)$$
$$f(250) = 15+20 = 35$$
$$\boxed{f(250) = \$35}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le x \le 1000}$$
Since $f$ is linear and increasing (gradient $0.08>0$), the range comes from the endpoint values:
$$f(0) = 15, \qquad f(1000) = 15+0.08(1000) = 15+80 = 95$$
$$\boxed{\text{Range: } 15 \le f(x) \le 95}$$
(c) Setting $f(x)=47$ and solving for $x$:
$$47 = 15+0.08x$$
$$0.08x = 32$$
$$x = 400$$
$$\boxed{x = 400\text{ pages}}$$
QUESTION 15
5 marks
Easy
A car park charges a flat entry fee of \$4, plus \$2.25 for every hour (or part hour) the car is parked. The total cost, $f(h)$ dollars, of parking for $h$ hours is
$$f(h) = 4+2.25h,$$
and the car park does not allow parking for more than 8 hours.
(a) State a suitable domain for this model, and find the corresponding range. [3]
(b) Find $f(3.5)$. [1]
(c) Find the value of $h$ for which $f(h)=15.25$. [1]
Show complete worked solution
(a) Since parking time cannot be negative, and the maximum stay allowed is 8 hours:
$$\boxed{\text{Domain: } 0 \le h \le 8}$$
Since $f$ is linear with positive gradient $2.25>0$, $f$ is increasing, so:
$$f(0) = 4, \qquad f(8) = 4+2.25(8) = 4+18 = 22$$
$$\boxed{\text{Range: } 4 \le f(h) \le 22}$$
(b) Substituting $h=3.5$ into $f(h)$:
$$f(3.5) = 4+2.25(3.5)$$
$$f(3.5) = 4+7.875 = 11.875$$
$$\boxed{f(3.5) = \$11.875}$$
(c) Setting $f(h)=15.25$ and solving for $h$:
$$15.25 = 4+2.25h$$
$$2.25h = 11.25$$
$$h = 5$$
$$\boxed{h = 5\text{ hours}}$$
QUESTION 16
6 marks
Easy
The temperature in degrees Fahrenheit, $F$, is related to the temperature in degrees Celsius, $C$, by the linear function
$$F(C) = \dfrac{9}{5}C+32, \qquad -20 \le C \le 40.$$
(a) Find $F(20)$. [1]
(b) State the range of $F$ over the given domain. [2]
(c) Find the inverse function $C(F)$, and use it to convert $98.6^{\circ}$F (normal human body temperature) into degrees Celsius. [3]
Show complete worked solution
(a) Substituting $C=20$ into $F(C)$:
$$F(20) = \dfrac{9}{5}(20)+32$$
$$F(20) = 36+32 = 68$$
$$\boxed{F(20) = 68^{\circ}\text{F}}$$
(b) Since $F$ is linear with positive gradient $\dfrac{9}{5}>0$, $F$ is increasing on $-20 \le C \le 40$, so:
$$F(-20) = \dfrac{9}{5}(-20)+32 = -36+32 = -4$$
$$F(40) = \dfrac{9}{5}(40)+32 = 72+32 = 104$$
$$\boxed{\text{Range: } -4 \le F(C) \le 104}$$
(c) Starting from $F=\dfrac{9}{5}C+32$ and making $C$ the subject:
$$F-32 = \dfrac{9}{5}C$$
$$C = \dfrac{5}{9}(F-32)$$
$$\boxed{C(F) = \dfrac{5}{9}(F-32)}$$
Substituting $F=98.6$:
$$C(98.6) = \dfrac{5}{9}(98.6-32)$$
$$C(98.6) = \dfrac{5}{9}(66.6)$$
$$C(98.6) = \dfrac{333}{9} = 37.0$$
$$\boxed{C(98.6) = 37.0^{\circ}\text{C}}$$
QUESTION 17
5 marks
Easy
A plumber charges a call-out fee of \$60, plus \$45 for every hour (or part hour) of work. The total charge, $f(h)$ dollars, for a job lasting $h$ hours is
$$f(h) = 60+45h, \qquad 0 \le h \le 6.$$
(a) Find $f(2.5)$. [2]
(b) State the domain and range of $f$. [2]
(c) Find the value of $h$ for which $f(h)=195$. [1]
Show complete worked solution
(a) Substituting $h=2.5$ into $f(h)$:
$$f(2.5) = 60+45(2.5)$$
$$f(2.5) = 60+112.5 = 172.5$$
$$\boxed{f(2.5) = \$172.50}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le h \le 6}$$
Since $f$ is linear and increasing (gradient $45>0$), the range comes from the endpoint values:
$$f(0) = 60, \qquad f(6) = 60+45(6) = 60+270 = 330$$
$$\boxed{\text{Range: } 60 \le f(h) \le 330}$$
(c) Setting $f(h)=195$ and solving for $h$:
$$195 = 60+45h$$
$$45h = 135$$
$$h = 3$$
$$\boxed{h = 3\text{ hours}}$$
QUESTION 18
5 marks
Easy
A household water utility charges a fixed service fee of \$15 per month, plus \$2.10 for every cubic metre (m$^3$) of water used. The monthly bill, $f(x)$ dollars, for using $x$ m$^3$ of water is
$$f(x) = 15+2.1x, \qquad 0 \le x \le 40.$$
(a) Find $f(24)$. [2]
(b) State the domain and range of $f$. [2]
(c) Find the value of $x$ for which the bill is exactly \$78. [1]
Show complete worked solution
(a) Substituting $x=24$ into $f(x)$:
$$f(24) = 15+2.1(24)$$
$$f(24) = 15+50.4 = 65.4$$
$$\boxed{f(24) = \$65.40}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le x \le 40}$$
Since $f$ is linear and increasing (gradient $2.1>0$), the range comes from the endpoint values:
$$f(0) = 15, \qquad f(40) = 15+2.1(40) = 15+84 = 99$$
$$\boxed{\text{Range: } 15 \le f(x) \le 99}$$
(c) Setting $f(x)=78$ and solving for $x$:
$$78 = 15+2.1x$$
$$2.1x = 63$$
$$x = 30$$
$$\boxed{x = 30\text{ m}^3}$$
QUESTION 19
5 marks
Easy
A currency exchange kiosk charges a flat handling fee of \$5, plus a rate of \$0.90 for every unit of foreign currency exchanged. The total cost in dollars, $f(x)$, of exchanging $x$ units of foreign currency is
$$f(x) = 5+0.9x, \qquad 0 \le x \le 500.$$
(a) Find $f(150)$. [2]
(b) State the domain and range of $f$. [2]
(c) Find the value of $x$ for which $f(x)=95$. [1]
Show complete worked solution
(a) Substituting $x=150$ into $f(x)$:
$$f(150) = 5+0.9(150)$$
$$f(150) = 5+135 = 140$$
$$\boxed{f(150) = \$140}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le x \le 500}$$
Since $f$ is linear and increasing (gradient $0.9>0$), the range comes from the endpoint values:
$$f(0) = 5, \qquad f(500) = 5+0.9(500) = 5+450 = 455$$
$$\boxed{\text{Range: } 5 \le f(x) \le 455}$$
(c) Setting $f(x)=95$ and solving for $x$:
$$95 = 5+0.9x$$
$$0.9x = 90$$
$$x = 100$$
$$\boxed{x = 100\text{ units}}$$
QUESTION 20
5 marks
Easy
A school hires a bus for a day trip. The hire company charges a fixed fee of \$220, plus \$6 for every student on the trip. The total cost, $f(x)$ dollars, for $x$ students is
$$f(x) = 220+6x, \qquad 10 \le x \le 60,$$
since the bus requires a minimum of 10 students and has a maximum capacity of 60 students.
(a) Find $f(45)$. [2]
(b) State the domain and range of $f$. [2]
(c) Find the number of students on the trip if the total cost was \$430. [1]
Show complete worked solution
(a) Substituting $x=45$ into $f(x)$:
$$f(45) = 220+6(45)$$
$$f(45) = 220+270 = 490$$
$$\boxed{f(45) = \$490}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 10 \le x \le 60}$$
Since $f$ is linear and increasing (gradient $6>0$), the range comes from the endpoint values:
$$f(10) = 220+6(10) = 220+60 = 280$$
$$f(60) = 220+6(60) = 220+360 = 580$$
$$\boxed{\text{Range: } 280 \le f(x) \le 580}$$
(c) Setting $f(x)=430$ and solving for $x$:
$$430 = 220+6x$$
$$6x = 210$$
$$x = 35$$
$$\boxed{x = 35\text{ students}}$$
Since $10 \le 35 \le 60$, this value lies within the domain of the model.
QUESTION 21
5 marks
Easy
An internet provider's plan includes a base monthly charge of \$40. If a customer uses more data than the plan includes, an extra charge of \$3 is added for every additional gigabyte (GB) used beyond the included amount. The total monthly cost, $f(y)$ dollars, when $y$ extra GB are used is
$$f(y) = 40+3y, \qquad 0 \le y \le 30.$$
(a) Find $f(8)$. [2]
(b) State the domain and range of $f$. [2]
(c) Find the value of $y$ for which $f(y)=94$. [1]
Show complete worked solution
(a) Substituting $y=8$ into $f(y)$:
$$f(8) = 40+3(8)$$
$$f(8) = 40+24 = 64$$
$$\boxed{f(8) = \$64}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le y \le 30}$$
Since $f$ is linear and increasing (gradient $3>0$), the range comes from the endpoint values:
$$f(0) = 40, \qquad f(30) = 40+3(30) = 40+90 = 130$$
$$\boxed{\text{Range: } 40 \le f(y) \le 130}$$
(c) Setting $f(y)=94$ and solving for $y$:
$$94 = 40+3y$$
$$3y = 54$$
$$y = 18$$
$$\boxed{y = 18\text{ GB}}$$
QUESTION 22
5 marks
Easy
A custom T-shirt printing company charges a fixed design/setup fee of \$80, plus \$6.50 for every T-shirt printed. The total cost, $f(n)$ dollars, of an order of $n$ T-shirts is
$$f(n) = 80+6.5n, \qquad 0 \le n \le 200.$$
(a) Find $f(40)$. [2]
(b) State the domain and range of $f$. [2]
(c) Find the value of $n$ for which $f(n)=353$. [1]
Show complete worked solution
(a) Substituting $n=40$ into $f(n)$:
$$f(40) = 80+6.5(40)$$
$$f(40) = 80+260 = 340$$
$$\boxed{f(40) = \$340}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le n \le 200}$$
Since $f$ is linear and increasing (gradient $6.5>0$), the range comes from the endpoint values:
$$f(0) = 80, \qquad f(200) = 80+6.5(200) = 80+1300 = 1380$$
$$\boxed{\text{Range: } 80 \le f(n) \le 1380}$$
(c) Setting $f(n)=353$ and solving for $n$:
$$353 = 80+6.5n$$
$$6.5n = 273$$
$$n = 42$$
$$\boxed{n = 42\text{ T-shirts}}$$
QUESTION 23
5 marks
Easy
In the lower atmosphere, air temperature decreases at a roughly constant rate as altitude increases. At sea level ($a=0$ km) the temperature is $15^{\circ}$C, and it decreases by $6.5^{\circ}$C for every kilometre of altitude gained. This is modelled by the linear function
$$T(a) = 15-6.5a, \qquad 0 \le a \le 5,$$
where $T(a)$ is the temperature in degrees Celsius at altitude $a$ km.
(a) Find $T(3)$. [2]
(b) State the domain and range of $T$. [2]
(c) Find the altitude at which the temperature is $-1.25^{\circ}$C. [1]
Show complete worked solution
(a) Substituting $a=3$ into $T(a)$:
$$T(3) = 15-6.5(3)$$
$$T(3) = 15-19.5 = -4.5$$
$$\boxed{T(3) = -4.5^{\circ}\text{C}}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le a \le 5}$$
Since $T$ is linear with negative gradient $-6.5<0$, $T$ is decreasing, so the range comes from the endpoint values:
$$T(0) = 15, \qquad T(5) = 15-6.5(5) = 15-32.5 = -17.5$$
$$\boxed{\text{Range: } -17.5 \le T(a) \le 15}$$
(c) Setting $T(a)=-1.25$ and solving for $a$:
$$-1.25 = 15-6.5a$$
$$6.5a = 16.25$$
$$a = 2.5$$
$$\boxed{a = 2.5\text{ km}}$$
QUESTION 24
5 marks
Easy
A candle is lit. Its height, $h(t)$ cm, $t$ minutes after being lit, is modelled by the linear function
$$h(t) = 20-0.4t.$$
(a) Find $h(15)$, and interpret this value in context. [2]
(b) Find the time at which the candle burns out completely (that is, when $h(t)=0$). [2]
(c) Hence state a suitable domain for $h$, and the corresponding range. [1]
Show complete worked solution
(a) Substituting $t=15$ into $h(t)$:
$$h(15) = 20-0.4(15)$$
$$h(15) = 20-6 = 14$$
$$\boxed{h(15) = 14\text{ cm}}$$
This means that 15 minutes after being lit, the candle is 14 cm tall.
(b) Setting $h(t)=0$ and solving for $t$:
$$0 = 20-0.4t$$
$$0.4t = 20$$
$$t = 50$$
$$\boxed{t = 50\text{ minutes}}$$
(c) Since the candle's height cannot be negative, and it burns out completely at $t=50$:
$$\boxed{\text{Domain: } 0 \le t \le 50}$$
Since $h$ is linear with negative gradient $-0.4<0$, $h$ is decreasing from $h(0)=20$ down to $h(50)=0$:
$$\boxed{\text{Range: } 0 \le h(t) \le 20}$$
QUESTION 25
5 marks
Easy
A postal service charges \$3.20 for the first 20 grams of a parcel, plus \$0.15 for every additional gram beyond 20 grams. Writing $x$ for the number of grams by which the parcel's mass exceeds 20 g, the cost in dollars is modelled by
$$f(x) = 3.20+0.15x, \qquad 0 \le x \le 480$$
(the service does not accept parcels heavier than 500 g).
(a) Find $f(80)$, and state the total mass of the parcel in this case. [2]
(b) State the domain and range of $f$. [2]
(c) A parcel costs \$14.30 to post. Find its total mass in grams. [1]
Show complete worked solution
(a) Substituting $x=80$ into $f(x)$:
$$f(80) = 3.20+0.15(80)$$
$$f(80) = 3.20+12 = 15.20$$
$$\boxed{f(80) = \$15.20}$$
Since $x=80$ represents 80 g beyond the initial 20 g, the total mass of the parcel is $20+80=100$ g.
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le x \le 480}$$
Since $f$ is linear and increasing (gradient $0.15>0$), the range comes from the endpoint values:
$$f(0) = 3.20, \qquad f(480) = 3.20+0.15(480) = 3.20+72 = 75.20$$
$$\boxed{\text{Range: } 3.20 \le f(x) \le 75.20}$$
(c) Setting $f(x)=14.30$ and solving for $x$:
$$14.30 = 3.20+0.15x$$
$$0.15x = 11.10$$
$$x = 74$$
Since $x$ is the mass beyond the initial 20 g, the total mass of the parcel is:
$$20+74 = 94$$
$$\boxed{\text{Total mass} = 94\text{ g}}$$
QUESTION 26
5 marks
Easy
A yoga studio offers a term package: a fixed enrolment fee of \$25, plus \$9 for every class attended. The total cost, $f(n)$ dollars, for attending $n$ classes in the term is
$$f(n) = 25+9n, \qquad 0 \le n \le 30.$$
(a) Find $f(12)$. [2]
(b) State the domain and range of $f$. [2]
(c) Find the value of $n$ for which $f(n)=169$. [1]
Show complete worked solution
(a) Substituting $n=12$ into $f(n)$:
$$f(12) = 25+9(12)$$
$$f(12) = 25+108 = 133$$
$$\boxed{f(12) = \$133}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le n \le 30}$$
Since $f$ is linear and increasing (gradient $9>0$), the range comes from the endpoint values:
$$f(0) = 25, \qquad f(30) = 25+9(30) = 25+270 = 295$$
$$\boxed{\text{Range: } 25 \le f(n) \le 295}$$
(c) Setting $f(n)=169$ and solving for $n$:
$$169 = 25+9n$$
$$9n = 144$$
$$n = 16$$
$$\boxed{n = 16\text{ classes}}$$
QUESTION 27
8 marks
Medium
Two ride-share apps charge fares based on distance travelled, $d$ km:
$$\text{App A: } f(d) = 2+1.10d, \qquad \text{App B: } g(d) = 3.5+0.85d, \qquad 0 \le d \le 30.$$
(a) Find $f(10)$ and $g(10)$, and state which app is cheaper for a 10 km trip. [3]
(b) Find the distance at which the two apps charge exactly the same fare (the break-even distance). [3]
(c) With reference to the gradients of $f$ and $g$, explain which app is cheaper for distances greater than the break-even distance. [2]
Show complete worked solution
(a) Substituting $d=10$ into each function:
$$f(10) = 2+1.10(10) = 2+11 = 13$$
$$g(10) = 3.5+0.85(10) = 3.5+8.5 = 12$$
$$\boxed{f(10) = \$13, \quad g(10) = \$12}$$
Since $g(10)0.85$, App A's fare increases more quickly with distance than App B's fare. Once the break-even distance $d=6$ km is passed, App B's slower rate of increase means its fare remains lower.
$$\boxed{\text{App B is cheaper for } d>6\text{ km}}$$
QUESTION 28
9 marks
Medium
A student is comparing two gym memberships for a term of up to 24 months. Gym X charges a one-time joining fee of \$60 plus \$25 per month; Gym Y has no joining fee but charges \$40 per month. The monthly costs, in dollars, are
$$X(m) = 60+25m, \qquad Y(m) = 40m, \qquad 0 \le m \le 24.$$
(a) Find the total cost of a 3-month membership at each gym, and state which is cheaper. [3]
(b) Find the number of months, $m$, at which the two gyms have the same total cost. [3]
(c) Let $D(m) = X(m)-Y(m)$ be the cost difference between the two gyms (Gym X minus Gym Y). Find $D(m)$ in the form $D(m)=am+b$, and interpret the values of $D(0)$ and $D(10)$ in context. [3]
Show complete worked solution
(a) Substituting $m=3$ into each function:
$$X(3) = 60+25(3) = 60+75 = 135$$
$$Y(3) = 40(3) = 120$$
$$\boxed{X(3) = \$135, \quad Y(3) = \$120}$$
Since $Y(3)
QUESTION 29
7 marks
Medium
Two companies produce custom-printed mugs. Company P charges a \$50 setup fee plus \$4 per mug; Company Q charges a \$20 setup fee plus \$5.50 per mug. For an order of $n$ mugs, $0 \le n \le 100$, the total costs in dollars are
$$P(n) = 50+4n, \qquad Q(n) = 20+5.5n.$$
(a) Find $P(10)$ and $Q(10)$. [2]
(b) Find the order size $n$ at which both companies charge the same total price (the break-even order size). [3]
(c) For orders of more than 20 mugs, determine, with a reason based on the gradients of $P$ and $Q$, which company is cheaper. [2]
Show complete worked solution
(a) Substituting $n=10$ into each function:
$$P(10) = 50+4(10) = 50+40 = 90$$
$$Q(10) = 20+5.5(10) = 20+55 = 75$$
$$\boxed{P(10) = \$90, \quad Q(10) = \$75}$$
(b) Setting $P(n)=Q(n)$ and solving for $n$:
$$50+4n = 20+5.5n$$
$$50-20 = 5.5n-4n$$
$$30 = 1.5n$$
$$n = 20$$
Checking: $P(20)=50+80=130$ and $Q(20)=20+110=130$, which agree.
$$\boxed{n = 20\text{ mugs}}$$
(c) The gradient of $P$ is $4$ dollars/mug, while the gradient of $Q$ is $5.5$ dollars/mug. Since $4<5.5$, Company P's cost increases more slowly per additional mug. For order sizes beyond the break-even point $n=20$, this slower rate of increase means Company P becomes the cheaper option.
$$\boxed{\text{Company P is cheaper for } n>20\text{ mugs}}$$
QUESTION 30
8 marks
Medium
Two mobile phone plans model monthly cost, in dollars, as a function of minutes used, $x$:
$$\text{Plan A: } A(x) = 15+0.08x, \qquad \text{Plan B: } B(x) = 30+0.03x, \qquad 0 \le x \le 1000.$$
(a) Find $A(200)$ and $B(200)$, and state which plan is cheaper for 200 minutes of use. [3]
(b) Find the break-even number of minutes, $x$, at which both plans cost the same. [3]
(c) A customer uses an average of 450 minutes per month. Using your result from part (b), determine which plan this customer should choose, and justify your answer. [2]
Show complete worked solution
(a) Substituting $x=200$ into each function:
$$A(200) = 15+0.08(200) = 15+16 = 31$$
$$B(200) = 30+0.03(200) = 30+6 = 36$$
$$\boxed{A(200) = \$31, \quad B(200) = \$36}$$
Since $A(200)
QUESTION 31
9 marks
Medium
A small manufacturer's total cost, $C(x)$ dollars, of producing $x$ units is modelled by the linear function
$$C(x) = 250+18x, \qquad 0 \le x \le 500.$$
(a) Find $C(120)$. [2]
(b) State the domain and range of $C$. [2]
(c) Find the inverse function $C^{-1}(y)$, and explain what $C^{-1}(y)$ represents in this context. [3]
(d) Use $C^{-1}$ to find the maximum number of units that can be produced with a budget of \$4750. [2]
Show complete worked solution
(a) Substituting $x=120$ into $C(x)$:
$$C(120) = 250+18(120)$$
$$C(120) = 250+2160 = 2410$$
$$\boxed{C(120) = \$2410}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le x \le 500}$$
Since $C$ is linear and increasing (gradient $18>0$), the range comes from the endpoint values:
$$C(0) = 250, \qquad C(500) = 250+18(500) = 250+9000 = 9250$$
$$\boxed{\text{Range: } 250 \le C(x) \le 9250}$$
(c) Starting from $y=250+18x$ and making $x$ the subject:
$$y-250 = 18x$$
$$x = \dfrac{y-250}{18}$$
$$\boxed{C^{-1}(y) = \dfrac{y-250}{18}}$$
Since $C(x)$ gives the total cost of producing $x$ units, its inverse $C^{-1}(y)$ takes a total budget $y$ (in dollars) and returns the number of units, $x$, that can be produced for that budget.
(d) Substituting $y=4750$ into $C^{-1}(y)$:
$$C^{-1}(4750) = \dfrac{4750-250}{18}$$
$$C^{-1}(4750) = \dfrac{4500}{18} = 250$$
$$\boxed{\text{Maximum } 250\text{ units can be produced}}$$
QUESTION 32
8 marks
Medium
In an online game, players convert in-game Credits into Gems for trading. The number of Gems received, $g(c)$, for converting $c$ Credits is modelled by the linear function
$$g(c) = 0.75c-15, \qquad 20 \le c \le 400$$
(the game requires a minimum conversion of 20 Credits, and a maximum of 400 Credits at one time; the $-15$ represents a flat conversion fee charged in Gems).
(a) Find $g(100)$. [2]
(b) State the domain and range of $g$. [3]
(c) Find the inverse function $g^{-1}(x)$, and use it to find how many Credits a player needs to convert in order to receive exactly 180 Gems. [3]
Show complete worked solution
(a) Substituting $c=100$ into $g(c)$:
$$g(100) = 0.75(100)-15$$
$$g(100) = 75-15 = 60$$
$$\boxed{g(100) = 60\text{ Gems}}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 20 \le c \le 400}$$
Since $g$ is linear and increasing (gradient $0.75>0$), the range comes from the endpoint values:
$$g(20) = 0.75(20)-15 = 15-15 = 0$$
$$g(400) = 0.75(400)-15 = 300-15 = 285$$
$$\boxed{\text{Range: } 0 \le g(c) \le 285}$$
(c) Starting from $x=0.75c-15$ and making $c$ the subject:
$$x+15 = 0.75c$$
$$c = \dfrac{x+15}{0.75}$$
$$\boxed{g^{-1}(x) = \dfrac{x+15}{0.75}}$$
Substituting $x=180$:
$$g^{-1}(180) = \dfrac{180+15}{0.75} = \dfrac{195}{0.75} = 260$$
Checking: $g(260)=0.75(260)-15 = 195-15=180$, which agrees, and $260$ lies within the domain $[20,400]$.
$$\boxed{c = 260\text{ Credits}}$$
QUESTION 33
8 marks
Medium
A homeowner installs solar panels at an upfront cost of \$6000. The panels are expected to save \$85 per month on electricity bills. The net amount still owed on the initial investment, $N(m)$ dollars, after $m$ months is modelled by
$$N(m) = 6000-85m, \qquad 0 \le m \le 80.$$
(a) Find $N(24)$, and interpret this value in context. [3]
(b) State the domain and range of $N$. [2]
(c) The panels have "paid for themselves" once $N(m) \le 0$. Solve $N(m)=0$, and hence find the first whole number of months by which the panels have paid for themselves. [3]
Show complete worked solution
(a) Substituting $m=24$ into $N(m)$:
$$N(24) = 6000-85(24)$$
$$N(24) = 6000-2040 = 3960$$
$$\boxed{N(24) = \$3960}$$
This means that after 24 months, \$3960 of the original \$6000 investment has not yet been recovered through savings.
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le m \le 80}$$
Since $N$ is linear with negative gradient $-85<0$, $N$ is decreasing, so the range comes from the endpoint values:
$$N(0) = 6000, \qquad N(80) = 6000-85(80) = 6000-6800 = -800$$
$$\boxed{\text{Range: } -800 \le N(m) \le 6000}$$
(c) Setting $N(m)=0$ and solving for $m$:
$$0 = 6000-85m$$
$$85m = 6000$$
$$m = \dfrac{6000}{85} = 70.588\ldots$$
Since $m=70.59$ (to 2 d.p.) is not a whole number, and the investment has "paid for itself" only once $N(m) \le 0$, the panels are not fully paid off until the end of the 71st month:
$$\boxed{m = 71\text{ months}}$$
QUESTION 34
7 marks
Medium
A catering company's records give the total cost of preparing a meal for $x$ guests: for 20 guests, the total cost is \$650; for 50 guests, the total cost is \$1400. The cost function $C(x)$ is known to be linear.
(a) Find $C(x)$ in the form $C(x)=mx+c$. [3]
(b) State the domain and range of $C$ for $10 \le x \le 200$ guests. [2]
(c) A client has a budget of exactly \$2000. Find the maximum number of guests they can invite. [2]
Show complete worked solution
(a) Using the gradient formula between $(20,650)$ and $(50,1400)$:
$$m = \dfrac{1400-650}{50-20} = \dfrac{750}{30} = 25$$
Using the point-gradient form with $m=25$ and point $(20,650)$:
$$C(x)-650 = 25(x-20)$$
$$C(x) = 25x-500+650$$
$$C(x) = 25x+150$$
$$\boxed{C(x) = 25x+150}$$
(b) The domain is given as $10 \le x \le 200$. Since $C$ is linear and increasing (gradient $25>0$), the range comes from the endpoint values:
$$C(10) = 25(10)+150 = 250+150 = 400$$
$$C(200) = 25(200)+150 = 5000+150 = 5150$$
$$\boxed{\text{Domain: } 10 \le x \le 200, \qquad \text{Range: } 400 \le C(x) \le 5150}$$
(c) Setting $C(x) \le 2000$ and solving for $x$:
$$25x+150 \le 2000$$
$$25x \le 1850$$
$$x \le 74$$
Since $x$ must be a whole number of guests, and $C(74)=25(74)+150=1850+150=2000$ exactly:
$$\boxed{x = 74\text{ guests}}$$
QUESTION 35
8 marks
Medium
A city is compared using two delivery services for local parcels: a drone service and a van courier. Their costs, in dollars, for a delivery of $x$ km are modelled by
$$\text{Drone: } D(x) = 6+0.5x, \qquad \text{Van: } V(x) = 10+0.3x, \qquad 0 \le x \le 40.$$
(a) Complete the table below, giving the cost of each service at the given distances.
[3]
(b) Find the break-even distance at which both services cost the same. [3]
(c) A customer needs a parcel delivered 35 km away. Using your result from part (b), determine, with justification, which service is cheaper. [2]
| Distance, $x$ (km) | Drone cost, $D(x)$ (\$) | Van cost, $V(x)$ (\$) |
|---|---|---|
| 5 | ? | ? |
| 15 | ? | ? |
| 25 | ? | ? |
Show complete worked solution
(a) Substituting each value of $x$ into $D(x)$ and $V(x)$:
$$D(5)=6+2.5=8.5, \quad V(5)=10+1.5=11.5$$
$$D(15)=6+7.5=13.5, \quad V(15)=10+4.5=14.5$$
$$D(25)=6+12.5=18.5, \quad V(25)=10+7.5=17.5$$
(b) Setting $D(x)=V(x)$ and solving for $x$:
$$6+0.5x = 10+0.3x$$
$$0.2x = 4$$
$$x = 20$$
Checking: $D(20)=6+10=16$ and $V(20)=10+6=16$, which agrees, and is consistent with the table (the Drone is cheaper at $x=5,15$, but the Van becomes cheaper by $x=25$).
$$\boxed{x = 20\text{ km}}$$
(c) Since the delivery distance of 35 km exceeds the break-even distance of 20 km found in part (b), and the Van's gradient ($0.3$) is smaller than the Drone's gradient ($0.5$), the Van is the cheaper option beyond the break-even point. Checking directly:
$$D(35) = 6+0.5(35) = 6+17.5 = 23.5$$
$$V(35) = 10+0.3(35) = 10+10.5 = 20.5$$
Since $V(35)
| Distance, $x$ (km) | Drone cost, $D(x)$ (\$) | Van cost, $V(x)$ (\$) |
|---|---|---|
| 5 | 8.50 | 11.50 |
| 15 | 13.50 | 14.50 |
| 25 | 18.50 | 17.50 |
QUESTION 36
9 marks
Medium
The population of a small town, $P(t)$, is recorded $t$ years after 2010. In 2010 ($t=0$), the population was 8200; in 2015 ($t=5$), the population was 9450. Assuming the population grows linearly, and that this model remains valid until 2030,
(a) find $P(t)$ in the form $P(t)=mt+c$. [3]
(b) find $P(12)$, and interpret this value in context. [2]
(c) state the domain and range of $P$ for the given model, and find the year in which the population is predicted to reach 11000. [4]
Show complete worked solution
(a) Using the gradient formula between $(0,8200)$ and $(5,9450)$:
$$m = \dfrac{9450-8200}{5-0} = \dfrac{1250}{5} = 250$$
Since $t=0$ gives $P(0)=8200$, the $y$-intercept is $c=8200$:
$$\boxed{P(t) = 8200+250t}$$
(b) Substituting $t=12$ into $P(t)$:
$$P(12) = 8200+250(12)$$
$$P(12) = 8200+3000 = 11200$$
$$\boxed{P(12) = 11200}$$
This means the model predicts the town's population to be $11200$ in the year $2010+12=2022$.
(c) Since the model is valid from 2010 ($t=0$) until 2030 ($t=20$):
$$\boxed{\text{Domain: } 0 \le t \le 20}$$
Since $P$ is linear and increasing (gradient $250>0$), the range comes from the endpoint values:
$$P(0) = 8200, \qquad P(20) = 8200+250(20) = 8200+5000 = 13200$$
$$\boxed{\text{Range: } 8200 \le P(t) \le 13200}$$
Setting $P(t)=11000$ and solving for $t$:
$$11000 = 8200+250t$$
$$250t = 2800$$
$$t = 11.2$$
Since $t=11.2$ lies within the valid domain $[0,20]$, the corresponding year is $2010+11.2=2021.2$, i.e. partway through 2021:
$$\boxed{\text{The population reaches 11000 during the year 2021}}$$
QUESTION 37
8 marks
Medium
A student is deciding whether to rent or buy a projector for a semester-long project lasting up to 30 days. Renting costs \$45 per day. Buying costs a one-time fee of \$900 (with no daily charge). The cost of renting for $d$ days, in dollars, is
$$R(d) = 45d, \qquad 0 \le d \le 30.$$
(a) Find $R(10)$. [2]
(b) Find the number of days, $d$, at which the cost of renting equals the cost of buying. [3]
(c) State the domain and range of $R$. Hence, for an 8-day project, determine which option (renting or buying) is cheaper. [3]
Show complete worked solution
(a) Substituting $d=10$ into $R(d)$:
$$R(10) = 45(10) = 450$$
$$\boxed{R(10) = \$450}$$
(b) Setting $R(d)=900$ (the fixed cost of buying) and solving for $d$:
$$45d = 900$$
$$d = 20$$
$$\boxed{d = 20\text{ days}}$$
(c) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le d \le 30}$$
Since $R$ is linear and increasing (gradient $45>0$), the range comes from the endpoint values:
$$R(0) = 0, \qquad R(30) = 45(30) = 1350$$
$$\boxed{\text{Range: } 0 \le R(d) \le 1350}$$
For an 8-day project, substituting $d=8$ into $R(d)$:
$$R(8) = 45(8) = 360$$
Since $R(8)=\$360$ is less than the \$900 cost of buying, and $d=8$ is less than the break-even value $d=20$ found in part (b):
$$\boxed{\text{Renting is cheaper for an 8-day project}}$$
QUESTION 38
6 marks
Medium
The cost of filling a car's fuel tank, $F(x)$ dollars, is directly proportional to the number of litres, $x$, purchased, at a price of \$1.85 per litre:
$$F(x) = 1.85x, \qquad 0 \le x \le 60$$
(the tank has a capacity of 60 litres).
(a) Find $F(35)$. [2]
(b) State the domain and range of $F$. [2]
(c) Find the inverse function $F^{-1}(x)$, and use it to find how many litres can be purchased with exactly \$74. [2]
Show complete worked solution
(a) Substituting $x=35$ into $F(x)$:
$$F(35) = 1.85(35)$$
$$F(35) = 64.75$$
$$\boxed{F(35) = \$64.75}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le x \le 60}$$
Since $F$ is linear and increasing (gradient $1.85>0$), the range comes from the endpoint values:
$$F(0) = 0, \qquad F(60) = 1.85(60) = 111$$
$$\boxed{\text{Range: } 0 \le F(x) \le 111}$$
(c) Starting from $y=1.85x$ and making $x$ the subject:
$$x = \dfrac{y}{1.85}$$
$$\boxed{F^{-1}(y) = \dfrac{y}{1.85}}$$
Substituting $y=74$:
$$F^{-1}(74) = \dfrac{74}{1.85} = 40$$
$$\boxed{40\text{ litres can be purchased for } \$74}$$
QUESTION 39
8 marks
Medium
Two internet providers are compared for monthly cost, in dollars, based on data usage $g$ GB per month, $0 \le g \le 50$:
$$\text{Provider A: } A(g) = 25+1.5g, \qquad \text{Provider B: charges a flat rate of } \$55\text{, regardless of usage.}$$
(a) Find $A(12)$. [2]
(b) Find the usage, $g$, at which Provider A's cost equals Provider B's flat rate of \$55. [3]
(c) State the domain and range of $A$. A user typically uses 35 GB per month; recommend a provider for this user, giving a reason. [3]
Show complete worked solution
(a) Substituting $g=12$ into $A(g)$:
$$A(12) = 25+1.5(12)$$
$$A(12) = 25+18 = 43$$
$$\boxed{A(12) = \$43}$$
(b) Setting $A(g)=55$ and solving for $g$:
$$55 = 25+1.5g$$
$$1.5g = 30$$
$$g = 20$$
$$\boxed{g = 20\text{ GB}}$$
(c) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le g \le 50}$$
Since $A$ is linear and increasing (gradient $1.5>0$), the range comes from the endpoint values:
$$A(0) = 25, \qquad A(50) = 25+1.5(50) = 25+75 = 100$$
$$\boxed{\text{Range: } 25 \le A(g) \le 100}$$
For a user with $g=35$ GB (which exceeds the break-even usage $g=20$ found in part (b)), substituting into $A(g)$:
$$A(35) = 25+1.5(35) = 25+52.5 = 77.5$$
Since $A(35)=\$77.50$ is greater than Provider B's flat rate of \$55:
$$\boxed{\text{Provider B is recommended for this user}}$$
QUESTION 40
7 marks
Medium
A 3D printing service charges customers based on the length of time their model takes to print. The total cost, $C(t)$ dollars, for a print job lasting $t$ hours is
$$C(t) = 12+3.5t, \qquad 0 \le t \le 48$$
(48 hours is the longest continuous print job the service allows).
(a) Find $C(6)$. [2]
(b) State the domain and range of $C$. [2]
(c) Find the inverse function $C^{-1}(x)$, and use it to find the maximum print time affordable with a \$180 budget, giving your answer to the nearest tenth of an hour. [3]
Show complete worked solution
(a) Substituting $t=6$ into $C(t)$:
$$C(6) = 12+3.5(6)$$
$$C(6) = 12+21 = 33$$
$$\boxed{C(6) = \$33}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le t \le 48}$$
Since $C$ is linear and increasing (gradient $3.5>0$), the range comes from the endpoint values:
$$C(0) = 12, \qquad C(48) = 12+3.5(48) = 12+168 = 180$$
$$\boxed{\text{Range: } 12 \le C(t) \le 180}$$
(c) Starting from $x=12+3.5t$ and making $t$ the subject:
$$x-12 = 3.5t$$
$$t = \dfrac{x-12}{3.5}$$
$$\boxed{C^{-1}(x) = \dfrac{x-12}{3.5}}$$
Substituting $x=180$:
$$C^{-1}(180) = \dfrac{180-12}{3.5} = \dfrac{168}{3.5} = 48.0$$
$$\boxed{\text{Maximum print time} = 48.0\text{ hours}}$$
(This equals the upper end of the domain, confirming that a \$180 budget exactly covers the longest job the service allows.)
QUESTION 41
7 marks
Medium
A monthly subscription box service offers two payment options. Option 1 is a flat subscription of \$18 per month for unlimited boxes. Option 2 has no monthly fee, but charges \$6 per individual box ordered, for $n$ boxes, $0 \le n \le 10$:
$$P(n) = 6n.$$
(a) Find $P(2)$, and state which option is cheaper for a customer who orders 2 boxes in a month. [2]
(b) Find the number of boxes, $n$, at which Option 2 costs the same as Option 1's flat \$18 fee. [3]
(c) State the domain and range of $P$. A customer typically orders 6 boxes per month; recommend an option for this customer, giving a reason. [2]
Show complete worked solution
(a) Substituting $n=2$ into $P(n)$:
$$P(2) = 6(2) = 12$$
$$\boxed{P(2) = \$12}$$
Since $P(2)=\$12$ is less than Option 1's flat \$18 fee, Option 2 (pay-per-box) is cheaper for a customer ordering 2 boxes.
(b) Setting $P(n)=18$ and solving for $n$:
$$6n = 18$$
$$n = 3$$
$$\boxed{n = 3\text{ boxes}}$$
(c) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le n \le 10}$$
Since $P$ is linear and increasing (gradient $6>0$), the range comes from the endpoint values:
$$P(0) = 0, \qquad P(10) = 6(10) = 60$$
$$\boxed{\text{Range: } 0 \le P(n) \le 60}$$
For a customer ordering $n=6$ boxes per month (which exceeds the break-even value $n=3$ found in part (b)), substituting into $P(n)$:
$$P(6) = 6(6) = 36$$
Since $P(6)=\$36$ is greater than Option 1's flat \$18 fee:
$$\boxed{\text{Option 1 (the flat subscription) is recommended for this customer}}$$
QUESTION 42
8 marks
Medium
Two mountain weather stations model air temperature as a linear function of altitude, $a$ km, over the domain $0 \le a \le 6$:
$$\text{Station A: } T_A(a) = 22-6.5a, \qquad \text{Station B: } T_B(a) = 15-3a.$$
(a) Find $T_A(1)$ and $T_B(1)$, and state which station's model predicts the colder temperature at an altitude of 1 km. [3]
(b) Find the altitude, $a$, at which both models predict the same temperature. [3]
(c) State the domain and range of $T_A$ over the given interval. [2]
Show complete worked solution
(a) Substituting $a=1$ into each function:
$$T_A(1) = 22-6.5(1) = 22-6.5 = 15.5$$
$$T_B(1) = 15-3(1) = 15-3 = 12$$
$$\boxed{T_A(1) = 15.5^{\circ}\text{C}, \quad T_B(1) = 12^{\circ}\text{C}}$$
Since $T_B(1)
QUESTION 43
14 marks
Hard
An electric scooter manufacturer's factory can produce at most 300 scooters per month, so all functions below have domain $0 \le x \le 300$, where $x$ is the number of scooters produced (and sold) in a month.
Each scooter sells for \$220, so the monthly revenue, in dollars, is
$$R(x) = 220x.$$
The accounts department reports: when $x=50$ scooters are made, the total monthly cost is \$14000; when $x=150$ scooters are made, the total monthly cost is \$30000. The cost function $C(x)$ is known to be linear.
(a) Find $C(x)$, and state the fixed monthly cost and the marginal (variable) cost per scooter. [4]
(b) Define the monthly profit function $P(x)=R(x)-C(x)$. Find $P(x)$ in the form $P(x)=ax+b$, and find the range of $P$ over the given domain. [4]
(c) Find the break-even production level, and state whether the manufacturer makes a profit or a loss when $x<100$. [3]
(d) Find the inverse function $P^{-1}(y)$, and use it to find the production level required for a monthly profit of \$7800. [3]
Show complete worked solution
(a) Using the gradient formula for $C$ between $(50,14000)$ and $(150,30000)$:
$$m_C = \dfrac{30000-14000}{150-50} = \dfrac{16000}{100} = 160$$
Using the point-gradient form with $m_C=160$ and point $(50,14000)$:
$$C(x)-14000 = 160(x-50)$$
$$C(x) = 160x-8000+14000$$
$$C(x) = 160x+6000$$
$$\boxed{C(x) = 160x+6000}$$
The fixed monthly cost is $C(0)=\$6000$ (costs incurred even if no scooters are made). The marginal cost per scooter is the gradient, $\$160$.
(b) Substituting $R(x)=220x$ and $C(x)=160x+6000$ into $P(x)=R(x)-C(x)$:
$$P(x) = 220x-(160x+6000)$$
$$P(x) = 60x-6000$$
$$\boxed{P(x) = 60x-6000}$$
Since $P$ is linear with positive gradient $60>0$, $P$ is increasing on $[0,300]$, so the range comes from the endpoint values:
$$P(0) = 60(0)-6000 = -6000$$
$$P(300) = 60(300)-6000 = 18000-6000 = 12000$$
$$\boxed{\text{Range: } -6000 \le P(x) \le 12000}$$
(c) Setting $P(x)=0$ and solving for $x$:
$$60x-6000 = 0$$
$$60x = 6000$$
$$x = 100$$
$$\boxed{x = 100\text{ scooters}}$$
Since $P$ is increasing (gradient $60>0$) and $P(100)=0$, for $x<100$ the profit function is negative:
$$\boxed{\text{The manufacturer makes a loss when } x<100}$$
(d) Starting from $y=60x-6000$ and making $x$ the subject:
$$y+6000 = 60x$$
$$x = \dfrac{y+6000}{60}$$
$$\boxed{P^{-1}(y) = \dfrac{y+6000}{60}}$$
Substituting $y=7800$:
$$P^{-1}(7800) = \dfrac{7800+6000}{60} = \dfrac{13800}{60} = 230$$
$$\boxed{x = 230\text{ scooters}}$$
QUESTION 44
11 marks
Hard
Two mobile carriers model total monthly cost, in dollars, as a function of minutes used, $m$, over the domain $0 \le m \le 2000$:
$$\text{Carrier X: } x(m) = 22+0.04m, \qquad \text{Carrier Y: } y(m) = 10+0.055m.$$
(a) Complete the table below.
[3]
(b) Find the break-even number of minutes, $m$. [3]
(c) State, with a reason based on the gradients of $x$ and $y$, which carrier is cheaper for a user averaging 1500 minutes per month. [2]
(d) Find the inverse function $x^{-1}(b)$, interpret it in context, and use it to find the maximum number of minutes a Carrier X user can use on a \$58 monthly budget. [3]
| Minutes, $m$ | Carrier X cost, $x(m)$ (\$) | Carrier Y cost, $y(m)$ (\$) |
|---|---|---|
| 200 | ? | ? |
| 600 | ? | ? |
| 1000 | ? | ? |
Show complete worked solution
(a) Substituting each value of $m$ into $x(m)$ and $y(m)$:
$$x(200)=22+8=30, \quad y(200)=10+11=21$$
$$x(600)=22+24=46, \quad y(600)=10+33=43$$
$$x(1000)=22+40=62, \quad y(1000)=10+55=65$$
(b) Setting $x(m)=y(m)$ and solving for $m$:
$$22+0.04m = 10+0.055m$$
$$22-10 = 0.055m-0.04m$$
$$12 = 0.015m$$
$$m = 800$$
Checking: $x(800)=22+32=54$ and $y(800)=10+44=54$, which agree, and this is consistent with the table (Carrier Y is cheaper at $m=200,600$, but Carrier X becomes cheaper by $m=1000$).
$$\boxed{m = 800\text{ minutes}}$$
(c) Carrier X's gradient is $0.04$ dollars/minute, while Carrier Y's gradient is $0.055$ dollars/minute. Since $0.04<0.055$, Carrier X's cost increases more slowly with usage. For a user averaging $1500$ minutes, which exceeds the break-even point of $800$ minutes, Carrier X is therefore cheaper. Checking directly:
$$x(1500) = 22+0.04(1500) = 22+60 = 82$$
$$y(1500) = 10+0.055(1500) = 10+82.5 = 92.5$$
Since $x(1500)
| Minutes, $m$ | Carrier X cost, $x(m)$ (\$) | Carrier Y cost, $y(m)$ (\$) |
|---|---|---|
| 200 | 30 | 21 |
| 600 | 46 | 43 |
| 1000 | 62 | 65 |
QUESTION 45
12 marks
Hard
A student compares renting a projector to buying one on an instalment plan, for a project lasting up to 36 months.
Renting costs \$45 per month, so the total rental cost, in dollars, after $m$ months is
$$R(m) = 45m, \qquad 0 \le m \le 36.$$
Buying on the instalment plan requires a down payment of \$150, plus \$30 per month, so the total amount paid, in dollars, after $m$ months is
$$B(m) = 150+30m, \qquad 0 \le m \le 36.$$
(a) Find $R(6)$ and $B(6)$, and state which option is cheaper after 6 months. [3]
(b) Find the break-even number of months, $m$, at which both options have cost the same total amount. [3]
(c) Let $D(m)=R(m)-B(m)$. Find $D(m)$ in the form $D(m)=am+b$, and find the range of $D$ over $0 \le m \le 36$. Interpret the values of $D(0)$ and $D(36)$ in context. [3]
(d) Find the inverse function $B^{-1}(y)$, and use it to find the number of months after which a total of \$780 has been paid under the buying plan. [3]
Show complete worked solution
(a) Substituting $m=6$ into each function:
$$R(6) = 45(6) = 270$$
$$B(6) = 150+30(6) = 150+180 = 330$$
$$\boxed{R(6) = \$270, \quad B(6) = \$330}$$
Since $R(6)0$), the range over $0 \le m \le 36$ comes from the endpoint values:
$$D(0) = 15(0)-150 = -150$$
$$D(36) = 15(36)-150 = 540-150 = 390$$
$$\boxed{\text{Range: } -150 \le D(m) \le 390}$$
$D(0)=-150$ means that at the very start ($m=0$), renting has cost \$0 while buying already requires the \$150 down payment, so renting is \$150 cheaper at that point. $D(36)=390$ is positive, meaning that after 36 months, renting has cost \$390 more than buying overall - consistent with $m=36$ being well beyond the break-even point $m=10$ found in part (b), after which renting's higher monthly rate makes buying the better long-term value.
(d) Starting from $y=150+30m$ and making $m$ the subject:
$$y-150 = 30m$$
$$m = \dfrac{y-150}{30}$$
$$\boxed{B^{-1}(y) = \dfrac{y-150}{30}}$$
Substituting $y=780$:
$$B^{-1}(780) = \dfrac{780-150}{30} = \dfrac{630}{30} = 21$$
$$\boxed{m = 21\text{ months}}$$
QUESTION 46
12 marks
Hard
A customer sends money abroad using one of two remittance providers. The total fee charged, in dollars, for sending $x$ dollars is modelled, for $0 \le x \le 5000$, by
$$\text{Provider P: } C_P(x) = 4+0.015x, \qquad \text{Provider Q: } C_Q(x) = 12+0.007x.$$
(a) Find $C_P(500)$ and $C_Q(500)$, and state which provider is cheaper for sending \$500. [3]
(b) Find the break-even transfer amount, $x$, at which both providers charge the same total fee. [3]
(c) State the range of $C_P$ over the given domain. Find the inverse function $C_P^{-1}(y)$, and use it to find the maximum amount that can be sent through Provider P for a total fee budget of \$40. [4]
(d) A customer regularly sends \$3000 each month. Using your result from part (b), determine, with justification, which provider is cheaper for this customer. [2]
Show complete worked solution
(a) Substituting $x=500$ into each function:
$$C_P(500) = 4+0.015(500) = 4+7.5 = 11.5$$
$$C_Q(500) = 12+0.007(500) = 12+3.5 = 15.5$$
$$\boxed{C_P(500) = \$11.50, \quad C_Q(500) = \$15.50}$$
Since $C_P(500)0$) on $0 \le x \le 5000$, the range comes from the endpoint values:
$$C_P(0) = 4, \qquad C_P(5000) = 4+0.015(5000) = 4+75 = 79$$
$$\boxed{\text{Range: } 4 \le C_P(x) \le 79}$$
Starting from $y=4+0.015x$ and making $x$ the subject:
$$y-4 = 0.015x$$
$$x = \dfrac{y-4}{0.015}$$
$$\boxed{C_P^{-1}(y) = \dfrac{y-4}{0.015}}$$
Substituting $y=40$:
$$C_P^{-1}(40) = \dfrac{40-4}{0.015} = \dfrac{36}{0.015} = 2400$$
$$\boxed{\$2400\text{ can be sent through Provider P for a \$40 fee}}$$
(d) The transfer amount of \$3000 exceeds the break-even amount $x=1000$ found in part (b). Since Provider P's gradient ($0.015$) is larger than Provider Q's gradient ($0.007$), Provider P's fee increases more quickly with transfer amount, so beyond the break-even point Provider Q becomes cheaper. Checking directly:
$$C_P(3000) = 4+0.015(3000) = 4+45 = 49$$
$$C_Q(3000) = 12+0.007(3000) = 12+21 = 33$$
Since $C_Q(3000)
QUESTION 47
12 marks
Hard
A student compares two moving companies for a local house move lasting between 1 and 8 hours.
Swift Movers' records show: a 2-hour move costs \$260; a 5-hour move costs \$470. Swift Movers' cost, $S(h)$ dollars, is known to be a linear function of the number of hours, $h$.
ValueVan charges according to $V(h) = 90+60h$.
(a) Find $S(h)$ in the form $S(h)=mh+c$. [3]
(b) Complete the table below.
[3]
(c) Find the value of $h$ for which $S(h)=V(h)$ algebraically, and hence explain, with reference to the domain $1 \le h \le 8$, why ValueVan is cheaper than Swift Movers throughout this domain. [3]
(d) Find the inverse function $S^{-1}(x)$, and use it to find how many hours of Swift Movers' service a budget of \$470 affords. State whether this lies within the model's domain. [3]
| Hours, $h$ | Swift Movers, $S(h)$ (\$) | ValueVan, $V(h)$ (\$) |
|---|---|---|
| 1 | ? | ? |
| 4 | ? | ? |
| 7 | ? | ? |
Show complete worked solution
(a) Using the gradient formula between $(2,260)$ and $(5,470)$:
$$m = \dfrac{470-260}{5-2} = \dfrac{210}{3} = 70$$
Using the point-gradient form with $m=70$ and point $(2,260)$:
$$S(h)-260 = 70(h-2)$$
$$S(h) = 70h-140+260$$
$$S(h) = 70h+120$$
$$\boxed{S(h) = 70h+120}$$
(b) Substituting each value of $h$ into $S(h)$ and $V(h)$:
$$S(1)=70+120=190, \quad V(1)=90+60=150$$
$$S(4)=280+120=400, \quad V(4)=90+240=330$$
$$S(7)=490+120=610, \quad V(7)=90+420=510$$
(c) Setting $S(h)=V(h)$ and solving for $h$:
$$70h+120 = 90+60h$$
$$70h-60h = 90-120$$
$$10h = -30$$
$$h = -3$$
$$\boxed{h = -3}$$
This solution is negative, and therefore lies outside the domain $1 \le h \le 8$. Since $S(h)-V(h) = (70h+120)-(90+60h) = 10h+30$, which is positive for every $h$ in the domain $1 \le h \le 8$ (indeed for every $h \ge -3$), Swift Movers always costs more than ValueVan on this domain, confirming the table's pattern that ValueVan is cheaper at every value tested.
$$\boxed{\text{ValueVan is cheaper than Swift Movers for all } 1 \le h \le 8}$$
(d) Starting from $x=70h+120$ and making $h$ the subject:
$$x-120 = 70h$$
$$h = \dfrac{x-120}{70}$$
$$\boxed{S^{-1}(x) = \dfrac{x-120}{70}}$$
Substituting $x=470$:
$$S^{-1}(470) = \dfrac{470-120}{70} = \dfrac{350}{70} = 5$$
$$\boxed{5\text{ hours}}$$
Since $1 \le 5 \le 8$, this value lies within the model's domain.
| Hours, $h$ | Swift Movers, $S(h)$ (\$) | ValueVan, $V(h)$ (\$) |
|---|---|---|
| 1 | 190 | 150 |
| 4 | 400 | 330 |
| 7 | 610 | 510 |
QUESTION 48
11 marks
Hard
A ride-share driver compares two platforms' weekly net earnings (after costs such as fuel are accounted for) as a function of hours driven per week, $h$, over the domain $5 \le h \le 50$ (the driver must drive at least 5 hours to qualify for either platform, and drives at most 50 hours per week):
$$\text{Platform Alpha: } A(h) = 15h-40, \qquad \text{Platform Beta: } B(h) = 12h+20.$$
(a) Find $A(10)$ and $B(10)$, and state which platform gives higher net earnings for 10 hours of driving. [3]
(b) Find the break-even number of hours, $h$, at which both platforms give the same net earnings. [3]
(c) State the domain and range of $A$ over the given interval. [2]
(d) Find the inverse function $A^{-1}(x)$, and use it to find the number of hours the driver would need to work on Platform Alpha to earn net \$410 in a week. [3]
Show complete worked solution
(a) Substituting $h=10$ into each function:
$$A(10) = 15(10)-40 = 150-40 = 110$$
$$B(10) = 12(10)+20 = 120+20 = 140$$
$$\boxed{A(10) = \$110, \quad B(10) = \$140}$$
Since $B(10)>A(10)$, Platform Beta gives higher net earnings for 10 hours of driving.
(b) Setting $A(h)=B(h)$ and solving for $h$:
$$15h-40 = 12h+20$$
$$15h-12h = 20+40$$
$$3h = 60$$
$$h = 20$$
Checking: $A(20)=300-40=260$ and $B(20)=240+20=260$, which agree.
$$\boxed{h = 20\text{ hours}}$$
(c) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 5 \le h \le 50}$$
Since $A$ is linear and increasing (gradient $15>0$), the range comes from the endpoint values:
$$A(5) = 15(5)-40 = 75-40 = 35$$
$$A(50) = 15(50)-40 = 750-40 = 710$$
$$\boxed{\text{Range: } 35 \le A(h) \le 710}$$
(d) Starting from $x=15h-40$ and making $h$ the subject:
$$x+40 = 15h$$
$$h = \dfrac{x+40}{15}$$
$$\boxed{A^{-1}(x) = \dfrac{x+40}{15}}$$
Substituting $x=410$:
$$A^{-1}(410) = \dfrac{410+40}{15} = \dfrac{450}{15} = 30$$
$$\boxed{h = 30\text{ hours}}$$
Since $5 \le 30 \le 50$, this value lies within the domain of the model.
QUESTION 49
11 marks
Hard
Underwater pressure increases roughly linearly with depth. A dive computer records: at a depth of $d=10$ m, the absolute pressure is $150$ kPa; at $d=30$ m, the absolute pressure is $350$ kPa. The pressure, $P(d)$ kPa, is modelled as a linear function of depth $d$ metres, for $0 \le d \le 40$ (the recreational diving depth limit).
(a) Find $P(d)$ in the form $P(d)=md+c$. [3]
(b) State the domain and range of $P$. [2]
(c) A diver's equipment can safely withstand a pressure of up to 380 kPa. Find the maximum safe depth according to this model, and state whether it lies within the recreational limit of 40 m. [3]
(d) Find the inverse function $P^{-1}(x)$, interpret it in context, and use it to find the depth corresponding to a pressure reading of 275 kPa. [3]
Show complete worked solution
(a) Using the gradient formula between $(10,150)$ and $(30,350)$:
$$m = \dfrac{350-150}{30-10} = \dfrac{200}{20} = 10$$
Using the point-gradient form with $m=10$ and point $(10,150)$:
$$P(d)-150 = 10(d-10)$$
$$P(d) = 10d-100+150$$
$$P(d) = 10d+50$$
$$\boxed{P(d) = 10d+50}$$
(b) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le d \le 40}$$
Since $P$ is linear and increasing (gradient $10>0$), the range comes from the endpoint values:
$$P(0) = 50, \qquad P(40) = 10(40)+50 = 400+50 = 450$$
$$\boxed{\text{Range: } 50 \le P(d) \le 450}$$
(c) Setting $P(d)=380$ and solving for $d$:
$$380 = 10d+50$$
$$10d = 330$$
$$d = 33$$
$$\boxed{d = 33\text{ m}}$$
Since $33 \le 40$, this maximum safe depth lies within the recreational diving limit of 40 m.
(d) Starting from $x=10d+50$ and making $d$ the subject:
$$x-50 = 10d$$
$$d = \dfrac{x-50}{10}$$
$$\boxed{P^{-1}(x) = \dfrac{x-50}{10}}$$
Since $P(d)$ gives the pressure at depth $d$, its inverse $P^{-1}(x)$ takes a pressure reading $x$ and returns the corresponding depth.
Substituting $x=275$:
$$P^{-1}(275) = \dfrac{275-50}{10} = \dfrac{225}{10} = 22.5$$
$$\boxed{d = 22.5\text{ m}}$$
QUESTION 50
11 marks
Hard
Two ride-hailing services, Zip and Roam, charge fares based on distance travelled, $d$ km, over the domain $0 \le d \le 25$:
$$\text{Zip: } Z(d) = 3+1.75d, \qquad \text{Roam: } Ro(d) = 6+1.25d.$$
(a) Complete the table below.
[3]
(b) Find the break-even distance, $d$, at which both services charge the same fare. [3]
(c) State the domain and range of $Z$. [2]
(d) Find the inverse function $Ro^{-1}(y)$, and use it to find the distance corresponding to a Roam fare of \$23.50. [3]
| Distance, $d$ (km) | Zip fare, $Z(d)$ (\$) | Roam fare, $Ro(d)$ (\$) |
|---|---|---|
| 4 | ? | ? |
| 10 | ? | ? |
| 16 | ? | ? |
Show complete worked solution
(a) Substituting each value of $d$ into $Z(d)$ and $Ro(d)$:
$$Z(4)=3+7=10, \quad Ro(4)=6+5=11$$
$$Z(10)=3+17.5=20.5, \quad Ro(10)=6+12.5=18.5$$
$$Z(16)=3+28=31, \quad Ro(16)=6+20=26$$
(b) Setting $Z(d)=Ro(d)$ and solving for $d$:
$$3+1.75d = 6+1.25d$$
$$1.75d-1.25d = 6-3$$
$$0.5d = 3$$
$$d = 6$$
Checking: $Z(6)=3+10.5=13.5$ and $Ro(6)=6+7.5=13.5$, which agree, consistent with the table (Zip is cheaper at $d=4$, but Roam becomes cheaper by $d=10$).
$$\boxed{d = 6\text{ km}}$$
(c) The domain is stated directly in the model:
$$\boxed{\text{Domain: } 0 \le d \le 25}$$
Since $Z$ is linear and increasing (gradient $1.75>0$), the range comes from the endpoint values:
$$Z(0) = 3, \qquad Z(25) = 3+1.75(25) = 3+43.75 = 46.75$$
$$\boxed{\text{Range: } 3 \le Z(d) \le 46.75}$$
(d) Starting from $y=6+1.25d$ and making $d$ the subject:
$$y-6 = 1.25d$$
$$d = \dfrac{y-6}{1.25}$$
$$\boxed{Ro^{-1}(y) = \dfrac{y-6}{1.25}}$$
Substituting $y=23.5$:
$$Ro^{-1}(23.5) = \dfrac{23.5-6}{1.25} = \dfrac{17.5}{1.25} = 14$$
$$\boxed{d = 14\text{ km}}$$
| Distance, $d$ (km) | Zip fare, $Z(d)$ (\$) | Roam fare, $Ro(d)$ (\$) |
|---|---|---|
| 4 | 10.00 | 11.00 |
| 10 | 20.50 | 18.50 |
| 16 | 31.00 | 26.00 |
Quadratic Functions 50 questions
QUESTION 1
4 marks
Easy
Expand and simplify $(x-3)(x+7)$, giving your answer in the form $ax^2+bx+c$.
Show complete worked solution
Expanding the product using the distributive law:
$$(x-3)(x+7) = x(x+7) - 3(x+7)$$
$$= x^2+7x-3x-21$$
$$= x^2+4x-21$$
$$\boxed{x^2+4x-21}$$
QUESTION 2
4 marks
Easy
Factorise $f(x)=x^2-5x+6$ completely, and hence write down the $x$-intercepts of the graph of $y=f(x)$.
Show complete worked solution
We require two numbers that multiply to give $6$ and add to give $-5$. These numbers are $-2$ and $-3$.
$$f(x) = x^2-5x+6 = (x-2)(x-3)$$
The $x$-intercepts occur where $f(x)=0$:
$$(x-2)(x-3)=0 \implies x=2 \text{ or } x=3$$
$$\boxed{f(x)=(x-2)(x-3), \text{ with } x\text{-intercepts at } (2,0) \text{ and } (3,0)}$$
QUESTION 3
5 marks
Easy
Write $f(x)=2(x-3)^2+1$ in expanded (standard) form $ax^2+bx+c$.
Show complete worked solution
Expand the squared bracket first:
$$(x-3)^2 = x^2-6x+9$$
Multiply through by $a=2$:
$$2(x-3)^2 = 2x^2-12x+18$$
Add the constant term $1$:
$$f(x) = 2x^2-12x+18+1 = 2x^2-12x+19$$
$$\boxed{f(x)=2x^2-12x+19}$$
QUESTION 4
4 marks
Easy
The graph of $g(x)=-3(x+4)^2+5$ is a parabola in vertex form.
(a) Write down the coordinates of the vertex. [2]
(b) State whether the vertex is a maximum or a minimum point, giving a reason. [2]
Show complete worked solution
(a) The vertex form $y=a(x-h)^2+k$ has vertex $(h,k)$. Here $g(x)=-3(x-(-4))^2+5$, so $h=-4$ and $k=5$.
$$\boxed{\text{Vertex } = (-4,5)}$$
(b) The coefficient of the squared term is $a=-3$, which is negative. A negative leading coefficient means the parabola opens downward, so the vertex is the highest point on the graph.
$$\boxed{\text{The vertex is a maximum point, since } a=-3<0}$$
QUESTION 5
5 marks
Easy
Find the coordinates of the vertex of $f(x)=x^2-6x+8$ using the formula $x=-\dfrac{b}{2a}$, and state the minimum value of $f$.
Show complete worked solution
For $f(x)=x^2-6x+8$ we have $a=1$, $b=-6$, $c=8$.
The $x$-coordinate of the vertex is
$$x = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3$$
Substitute $x=3$ into $f(x)$ to find the $y$-coordinate:
$$f(3) = (3)^2-6(3)+8 = 9-18+8 = -1$$
$$\boxed{\text{Vertex } = (3,-1)}$$
Since $a=1>0$, the parabola opens upward, so the vertex is a minimum point.
$$\boxed{\text{Minimum value of } f \text{ is } -1}$$
QUESTION 6
5 marks
Easy
Consider $f(x)=2x^2+12x-7$.
(a) Find the equation of the axis of symmetry. [2]
(b) Write down the $y$-intercept of the graph of $y=f(x)$. [1]
Show complete worked solution
(a) For $f(x)=ax^2+bx+c$, the axis of symmetry is the vertical line $x=-\dfrac{b}{2a}$. Here $a=2$, $b=12$:
$$x = -\frac{12}{2(2)} = -\frac{12}{4} = -3$$
$$\boxed{x=-3}$$
(b) The $y$-intercept occurs at $x=0$:
$$f(0) = 2(0)^2+12(0)-7 = -7$$
$$\boxed{(0,-7)}$$
QUESTION 7
5 marks
Easy
Consider $f(x)=x^2-2x-8$.
(a) Factorise $f(x)$. [2]
(b) Hence find the $x$-intercepts and the $y$-intercept of the graph of $y=f(x)$. [3]
Show complete worked solution
(a) We need two numbers that multiply to $-8$ and add to $-2$: these are $-4$ and $2$.
$$f(x) = x^2-2x-8 = (x-4)(x+2)$$
(b) The $x$-intercepts occur where $f(x)=0$:
$$(x-4)(x+2)=0 \implies x=4 \text{ or } x=-2$$
$$\boxed{x\text{-intercepts: } (4,0) \text{ and } (-2,0)}$$
The $y$-intercept occurs at $x=0$:
$$f(0)=0^2-2(0)-8=-8$$
$$\boxed{y\text{-intercept: } (0,-8)}$$
QUESTION 8
4 marks
Easy
Solve the equation $x^2+2x-15=0$ by factorisation.
Show complete worked solution
We need two numbers that multiply to $-15$ and add to $2$: these are $5$ and $-3$.
$$x^2+2x-15 = (x+5)(x-3)$$
Setting the factored form equal to zero:
$$(x+5)(x-3)=0$$
$$x+5=0 \quad \text{or} \quad x-3=0$$
$$\boxed{x=-5 \text{ or } x=3}$$
QUESTION 9
5 marks
Easy
Solve the equation $2x^2-7x-4=0$ using the quadratic formula. Give exact values.
Show complete worked solution
The quadratic formula states that for $ax^2+bx+c=0$,
$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Here $a=2$, $b=-7$, $c=-4$. First compute the discriminant:
$$b^2-4ac = (-7)^2-4(2)(-4) = 49+32 = 81$$
Since $\sqrt{81}=9$:
$$x = \frac{-(-7)\pm 9}{2(2)} = \frac{7\pm 9}{4}$$
$$x = \frac{7+9}{4} = 4 \qquad \text{or} \qquad x = \frac{7-9}{4} = -\frac{1}{2}$$
$$\boxed{x=4 \text{ or } x=-\frac{1}{2}}$$
QUESTION 10
4 marks
Easy
Find the value of the discriminant of $3x^2-2x+5$ and use it to state the number of real roots of $3x^2-2x+5=0$.
Show complete worked solution
For $ax^2+bx+c$ with $a=3$, $b=-2$, $c=5$, the discriminant is
$$\Delta = b^2-4ac = (-2)^2-4(3)(5) = 4-60 = -56$$
Since $\Delta<0$, the equation has no real roots (the graph of $y=3x^2-2x+5$ does not cross the $x$-axis).
$$\boxed{\Delta=-56 < 0 \implies \text{no real roots}}$$
QUESTION 11
5 marks
Easy
Consider the equation $x^2-4x+4=0$.
(a) Show that the discriminant equals zero. [2]
(b) Hence find the repeated root of the equation. [3]
Show complete worked solution
(a) For $x^2-4x+4=0$, $a=1$, $b=-4$, $c=4$:
$$\Delta = b^2-4ac = (-4)^2-4(1)(4) = 16-16 = 0$$
$$\boxed{\Delta=0}$$
(b) Since $\Delta=0$, the quadratic formula gives a single repeated root:
$$x = \frac{-b\pm\sqrt{\Delta}}{2a} = \frac{4\pm 0}{2} = 2$$
This can be confirmed by factorising: $x^2-4x+4=(x-2)^2=0 \implies x=2$.
$$\boxed{x=2 \text{ (repeated root)}}$$
QUESTION 12
5 marks
Easy
Write $f(x)=x^2+8x+5$ in vertex form by completing the square, and hence state the coordinates of the vertex.
Show complete worked solution
Start with the $x^2$ and $x$ terms and halve the coefficient of $x$: half of $8$ is $4$.
$$f(x) = x^2+8x+5$$
$$= (x+4)^2-4^2+5$$
$$= (x+4)^2-16+5$$
$$= (x+4)^2-11$$
This is now in vertex form $y=a(x-h)^2+k$ with $a=1$, $h=-4$, $k=-11$.
$$\boxed{f(x)=(x+4)^2-11, \text{ vertex } = (-4,-11)}$$
QUESTION 13
4 marks
Easy
A ball is thrown vertically into the air. Its height above the ground, in metres, $t$ seconds after being thrown is modelled by
$$h(t) = -5t^2+20t+2$$
Find the height of the ball at the instant it is thrown (i.e. at $t=0$).
Show complete worked solution
The height at the instant the ball is thrown corresponds to $t=0$. Substitute $t=0$ into the model:
$$h(0) = -5(0)^2+20(0)+2 = 0+0+2 = 2$$
$$\boxed{h(0)=2 \text{ metres}}$$
This makes sense: the constant term of a height-time quadratic model always represents the initial height (e.g. the height of the thrower's hand above the ground).
QUESTION 14
5 marks
Easy
A javelin's height above the ground, in metres, $t$ seconds after release is modelled by
$$h(t) = -4.9t^2+14t+1$$
Find the time at which the javelin reaches its maximum height, and find this maximum height.
Show complete worked solution
The model $h(t)=-4.9t^2+14t+1$ is a quadratic with $a=-4.9$, $b=14$, $c=1$. Since $a<0$, the graph opens downward and the vertex is a maximum.
The time at which the maximum occurs is
$$t = -\frac{b}{2a} = -\frac{14}{2(-4.9)} = \frac{14}{9.8} \approx 1.43 \text{ s}$$
Substituting back (using the GDC or exact working):
$$h(1.42857\ldots) = -4.9(1.42857\ldots)^2+14(1.42857\ldots)+1 = -10+20+1=11$$
$$\boxed{\text{Maximum height} = 11 \text{ m, reached at } t\approx 1.43 \text{ s}}$$
QUESTION 15
5 marks
Easy
A shop's weekly revenue, in dollars, from selling $x$ units of a product is modelled by
$$R(x) = -2x^2+40x$$
(a) Find the revenue when $x=5$ units are sold. [2]
(b) Find the number of units that must be sold to maximise revenue, and the maximum revenue. [3]
Show complete worked solution
(a) Substitute $x=5$ into the model:
$$R(5) = -2(5)^2+40(5) = -50+200 = 150$$
$$\boxed{R(5)=\$150}$$
(b) Since $R(x)=-2x^2+40x$ has $a=-2<0$, the vertex is a maximum. The $x$-coordinate of the vertex is
$$x = -\frac{b}{2a} = -\frac{40}{2(-2)} = 10$$
The maximum revenue is
$$R(10) = -2(10)^2+40(10) = -200+400 = 200$$
$$\boxed{\text{Sell 10 units for a maximum revenue of } \$200}$$
QUESTION 16
5 marks
Easy
A parabola has $x$-intercepts at $x=-3$ and $x=5$, and its equation has leading coefficient $a=2$.
(a) Write the equation of the parabola in factored form. [2]
(b) Expand your answer to give the equation in standard form. [3]
Show complete worked solution
(a) A parabola with $x$-intercepts $p$ and $q$, and leading coefficient $a$, has factored form $y=a(x-p)(x-q)$. With $p=-3$, $q=5$, $a=2$:
$$\boxed{y = 2(x+3)(x-5)}$$
(b) Expand $(x+3)(x-5)$ first:
$$(x+3)(x-5) = x^2-5x+3x-15 = x^2-2x-15$$
Now multiply by $2$:
$$y = 2(x^2-2x-15) = 2x^2-4x-30$$
$$\boxed{y=2x^2-4x-30}$$
QUESTION 17
5 marks
Easy
The graph of $y=f(x)$ shown below is a parabola passing through the points $(1,0)$, $(5,0)$ and $(3,-4)$.
(a) Write down the equation of the axis of symmetry. [1]
(b) Write down the coordinates of the vertex. [1]
(c) Write $f(x)$ in factored form. [3]
Show complete worked solution
(a) The two $x$-intercepts, $x=1$ and $x=5$, are symmetric about the axis of symmetry, so the axis lies midway between them:
$$x = \frac{1+5}{2}=3$$
$$\boxed{x=3}$$
(b) The point $(3,-4)$ lies on the axis of symmetry $x=3$, and it is the lowest point shown on the graph, so it is the vertex.
$$\boxed{\text{Vertex} = (3,-4)}$$
(c) Since the graph has $x$-intercepts at $x=1$ and $x=5$ and passes through $(3,-4)$ with leading coefficient $a$:
$$f(x) = a(x-1)(x-5)$$
Substitute $(3,-4)$: $-4 = a(3-1)(3-5) = a(2)(-2) = -4a \implies a=1$.
$$\boxed{f(x)=(x-1)(x-5)}$$
QUESTION 18
4 marks
Easy
Use your GDC to solve $1.5x^2-4x-3=0$. Give your answers correct to 3 significant figures.
Show complete worked solution
Enter $y=1.5x^2-4x-3$ into the GDC's equation solver (or graph the function and find its zeros).
Algebraically, using the quadratic formula with $a=1.5$, $b=-4$, $c=-3$:
$$\Delta = (-4)^2-4(1.5)(-3) = 16+18 = 34$$
$$x = \frac{4\pm\sqrt{34}}{3} = \frac{4\pm 5.83095\ldots}{3}$$
$$x = \frac{4+5.83095\ldots}{3} = 3.27698\ldots \qquad \text{or} \qquad x=\frac{4-5.83095\ldots}{3}=-0.61032\ldots$$
$$\boxed{x=3.28 \text{ or } x=-0.610 \text{ (3 s.f.)}}$$
QUESTION 19
4 marks
Easy
A parabola has vertex at the origin and passes through the point $(2,8)$. Find the equation of the parabola in the form $y=ax^2$.
Show complete worked solution
Since the vertex is at the origin, the equation has the form $y=ax^2$ (a vertex-form parabola with $h=0$, $k=0$).
Substitute the point $(2,8)$:
$$8 = a(2)^2 = 4a$$
$$a = \frac{8}{4}=2$$
$$\boxed{y=2x^2}$$
QUESTION 20
4 marks
Easy
A square garden bed originally had side length $x$ metres. A gardener extends two adjacent sides by $3$ m each, forming a rectangle of dimensions $(x+3)$ by $(x+3)$, i.e. the bed remains square with new side length $(x+3)$.
(a) Write an expression for the new area, $A(x)$, in expanded quadratic form. [2]
(b) Find the new area when $x=2$. [2]
Show complete worked solution
(a) The new area is side length squared:
$$A(x) = (x+3)^2 = x^2+6x+9$$
$$\boxed{A(x)=x^2+6x+9}$$
(b) Substitute $x=2$:
$$A(2) = (2+3)^2 = 5^2=25$$
Checking against the expanded form: $A(2)=2^2+6(2)+9=4+12+9=25$. Both methods agree.
$$\boxed{A(2)=25 \text{ m}^2}$$
QUESTION 21
6 marks
Medium
Consider $f(x)=2x^2-8x+3$.
(a) Write $f(x)$ in vertex form by completing the square. [4]
(b) State the coordinates of the vertex, and whether it is a maximum or a minimum. [2]
Show complete worked solution
(a) Factor out the coefficient of $x^2$ from the first two terms:
$$f(x) = 2x^2-8x+3 = 2(x^2-4x)+3$$
Complete the square inside the bracket: half of $-4$ is $-2$, and $(-2)^2=4$.
$$= 2\big[(x-2)^2-4\big]+3$$
$$= 2(x-2)^2-8+3$$
$$= 2(x-2)^2-5$$
$$\boxed{f(x)=2(x-2)^2-5}$$
(b) The vertex form $y=a(x-h)^2+k$ gives vertex $(h,k)=(2,-5)$. Since $a=2>0$, the parabola opens upward, so the vertex is a minimum.
$$\boxed{\text{Vertex} = (2,-5), \text{ a minimum}}$$
QUESTION 22
6 marks
Medium
Consider $f(x)=-\dfrac{1}{2}(x+2)^2+6$.
(a) Expand $f(x)$ to write it in standard form $ax^2+bx+c$. [4]
(b) Hence state the $y$-intercept of the graph of $y=f(x)$. [2]
Show complete worked solution
(a) Expand the squared bracket:
$$(x+2)^2 = x^2+4x+4$$
Multiply by $-\dfrac{1}{2}$:
$$-\frac{1}{2}(x+2)^2 = -\frac{1}{2}x^2-2x-2$$
Add the constant $6$:
$$f(x) = -\frac{1}{2}x^2-2x-2+6 = -\frac{1}{2}x^2-2x+4$$
$$\boxed{f(x)=-\frac{1}{2}x^2-2x+4}$$
(b) The $y$-intercept occurs at $x=0$. From either form:
$$f(0) = -\frac{1}{2}(0)^2-2(0)+4 = 4$$
$$\boxed{(0,4)}$$
QUESTION 23
7 marks
Medium
A parabola has vertex $(3,-4)$ and passes through the point $(5,4)$. Find the equation of the parabola in standard form $ax^2+bx+c$.
Show complete worked solution
Since the vertex is $(3,-4)$, write the equation in vertex form:
$$y = a(x-3)^2-4$$
Substitute the point $(5,4)$ to find $a$:
$$4 = a(5-3)^2-4$$
$$4 = 4a-4$$
$$8 = 4a$$
$$a = 2$$
So the vertex form is $y=2(x-3)^2-4$. Expand to standard form:
$$y = 2(x^2-6x+9)-4 = 2x^2-12x+18-4 = 2x^2-12x+14$$
$$\boxed{y=2x^2-12x+14}$$
Check: at $x=5$, $y=2(25)-12(5)+14=50-60+14=4$. Correct.
QUESTION 24
7 marks
Medium
A parabola $f(x)=ax^2+bx+c$ passes through the points $(0,1)$, $(1,4)$ and $(2,11)$. Find the values of $a$, $b$ and $c$.
Show complete worked solution
Substitute each point into $f(x)=ax^2+bx+c$ to form a system of equations.
From $(0,1)$: $\quad c = 1$
From $(1,4)$: $\quad a+b+c=4 \implies a+b=3$
From $(2,11)$: $\quad 4a+2b+c=11 \implies 4a+2b=10 \implies 2a+b=5$
Subtract the first reduced equation from the second:
$$(2a+b)-(a+b) = 5-3$$
$$a = 2$$
Substitute back into $a+b=3$: $b=3-2=1$.
$$\boxed{a=2,\ b=1,\ c=1, \text{ so } f(x)=2x^2+x+1}$$
Check: $f(2)=2(4)+2+1=8+2+1=11$. Correct.
QUESTION 25
7 marks
Medium
A model rocket is launched from a platform. Its height above the ground, in metres, $t$ seconds after launch is
$$h(t) = -4.9t^2+18t+1.2$$
(a) Find the initial height of the rocket. [2]
(b) Find the maximum height reached by the rocket, and the time at which it occurs, correct to 3 significant figures. [5]
Show complete worked solution
(a) The initial height occurs at $t=0$:
$$h(0) = -4.9(0)^2+18(0)+1.2 = 1.2$$
$$\boxed{\text{Initial height} = 1.2 \text{ m}}$$
(b) Since $a=-4.9<0$, the vertex of the parabola is a maximum. Its time-coordinate is
$$t = -\frac{b}{2a} = -\frac{18}{2(-4.9)} = \frac{18}{9.8} = 1.83673\ldots \approx 1.84 \text{ s}$$
Substituting into $h(t)$ (using the GDC or direct calculation):
$$h(1.83673\ldots) = -4.9(1.83673\ldots)^2+18(1.83673\ldots)+1.2 \approx 17.7306\ldots$$
$$\boxed{\text{Maximum height} \approx 17.7 \text{ m at } t\approx 1.84 \text{ s (3 s.f.)}}$$
QUESTION 26
7 marks
Medium
A company's weekly revenue, in dollars, from selling a product at price $\$p$ per unit is modelled by
$$R(p) = -5p^2+300p$$
(a) Find the price that maximises revenue. [3]
(b) Find the maximum weekly revenue. [4]
Show complete worked solution
(a) $R(p)=-5p^2+300p$ has $a=-5$, $b=300$. Since $a<0$, the vertex is a maximum, occurring at
$$p = -\frac{b}{2a} = -\frac{300}{2(-5)} = \frac{300}{10}=30$$
$$\boxed{p=\$30}$$
(b) Substitute $p=30$ into $R(p)$:
$$R(30) = -5(30)^2+300(30) = -5(900)+9000 = -4500+9000=4500$$
$$\boxed{\text{Maximum weekly revenue} = \$4500}$$
QUESTION 27
7 marks
Medium
A farmer has 80 m of fencing to enclose a rectangular pen. Let $x$ metres be the width of the pen.
(a) Show that the length of the pen is $(40-x)$ metres, and hence write the area of the pen as $A(x)=x(40-x)$. [3]
(b) Find the dimensions that maximise the area, and state the maximum area. [4]
Show complete worked solution
(a) Let the width be $x$ and the length be $y$. Since all 80 m of fencing is used for the perimeter of the rectangle:
$$2x+2y=80 \implies x+y=40 \implies y=40-x$$
The area is width times length:
$$A(x) = x(40-x)$$
$$\boxed{A(x)=x(40-x) = 40x-x^2}$$
(b) Writing $A(x)=-x^2+40x$, we have $a=-1$, $b=40$. Since $a<0$, the vertex is a maximum, occurring at
$$x = -\frac{b}{2a} = -\frac{40}{2(-1)} = 20$$
Then $y=40-20=20$, and the maximum area is
$$A(20) = 20(40-20) = 20(20) = 400$$
$$\boxed{\text{The pen should be } 20\text{ m by } 20\text{ m, giving a maximum area of } 400 \text{ m}^2}$$
QUESTION 28
8 marks
Medium
Consider $f(x)=x^2-4x+1$ for the restricted domain $-1\le x\le 5$. Find the range of $f$ over this domain.
Show complete worked solution
First find the vertex of $f(x)=x^2-4x+1$, since $a=1>0$ means it is a minimum:
$$x = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2$$
$$f(2) = (2)^2-4(2)+1 = 4-8+1=-3$$
Since $x=2$ lies within the domain $[-1,5]$, the minimum value of $f$ on this domain is $-3$.
Next check both endpoints, since the maximum on a restricted domain occurs at an endpoint:
$$f(-1) = (-1)^2-4(-1)+1 = 1+4+1=6$$
$$f(5) = (5)^2-4(5)+1 = 25-20+1=6$$
Both endpoints give the same value, $6$, which is the maximum on this domain (this happens because $-1$ and $5$ are symmetric about the axis of symmetry $x=2$).
$$\boxed{\text{Range: } -3\le f(x)\le 6}$$
QUESTION 29
6 marks
Medium
Find the values of $k$ for which the equation $x^2+kx+9=0$ has exactly one (repeated) real root.
Show complete worked solution
A quadratic has exactly one repeated real root when its discriminant equals zero. For $x^2+kx+9=0$, $a=1$, $b=k$, $c=9$:
$$\Delta = k^2-4(1)(9) = k^2-36$$
Set $\Delta=0$:
$$k^2-36=0$$
$$k^2=36$$
$$k=\pm 6$$
$$\boxed{k=6 \text{ or } k=-6}$$
Check: when $k=6$, $x^2+6x+9=(x+3)^2=0$, a repeated root at $x=-3$. When $k=-6$, $x^2-6x+9=(x-3)^2=0$, a repeated root at $x=3$. Both confirm $\Delta=0$.
QUESTION 30
7 marks
Medium
A parabola has $x$-intercepts at $(-2,0)$ and $(4,0)$, and $y$-intercept at $(0,-16)$. Find the equation of the parabola in standard form.
Show complete worked solution
Since the $x$-intercepts are $-2$ and $4$, write the equation in factored form:
$$y = a(x+2)(x-4)$$
Substitute the $y$-intercept $(0,-16)$:
$$-16 = a(0+2)(0-4) = a(2)(-4) = -8a$$
$$a = \frac{-16}{-8}=2$$
So $y=2(x+2)(x-4)$. Expand:
$$(x+2)(x-4) = x^2-2x-8$$
$$y = 2(x^2-2x-8) = 2x^2-4x-16$$
$$\boxed{y=2x^2-4x-16}$$
Check: $y(0)=2(0)-4(0)-16=-16$. Correct.
QUESTION 31
7 marks
Medium
Solve the inequality $x^2-3x-10\le 0$, giving your answer as an interval.
Show complete worked solution
First solve the corresponding equation $x^2-3x-10=0$ to find the critical values. Factorising ($a=1$, so we seek two numbers multiplying to $-10$ and summing to $-3$: these are $-5$ and $2$):
$$x^2-3x-10 = (x-5)(x+2) = 0$$
$$x=5 \text{ or } x=-2$$
Since the coefficient of $x^2$ is positive ($a=1$), the parabola $y=x^2-3x-10$ opens upward. It lies below or on the $x$-axis (i.e. $y\le 0$) between its two roots.
Testing $x=0$ (between the roots) in the original expression: $0-0-10=-10\le 0$. This confirms the region between the roots satisfies the inequality.
$$\boxed{-2\le x\le 5}$$
QUESTION 32
8 marks
Medium
A company's monthly cost, in dollars, for producing $x$ units is $C(x)=5x+200$. Its monthly revenue from selling all $x$ units is modelled by $R(x)=-0.5x^2+40x$.
(a) Write an equation for the break-even quantity, where revenue equals cost. [2]
(b) Solve your equation to find the break-even quantities, correct to 2 decimal places. [6]
Show complete worked solution
(a) Break-even occurs when $R(x)=C(x)$:
$$-0.5x^2+40x = 5x+200$$
Rearranging so one side is zero:
$$-0.5x^2+40x-5x-200=0$$
$$\boxed{-0.5x^2+35x-200=0}$$
(b) Multiply through by $-2$ to clear the decimal and simplify signs:
$$x^2-70x+400=0$$
Apply the quadratic formula with $a=1$, $b=-70$, $c=400$:
$$\Delta = (-70)^2-4(1)(400) = 4900-1600=3300$$
$$x = \frac{70\pm\sqrt{3300}}{2} = \frac{70\pm 57.4456\ldots}{2}$$
$$x = \frac{70+57.4456\ldots}{2}=63.7228\ldots \qquad \text{or} \qquad x=\frac{70-57.4456\ldots}{2}=6.2772\ldots$$
$$\boxed{x\approx 6.28 \text{ or } x\approx 63.72 \text{ units (2 d.p.)}}$$
QUESTION 33
7 marks
Medium
An arch over a footpath is modelled by $h(x)=-0.05x^2+4$, where $h$ is the height above the ground in metres and $x$ is the horizontal distance in metres from the centre of the arch.
(a) Find the width of the arch at ground level (i.e. where $h(x)=0$). [4]
(b) Find the height of the arch $5$ m from its centre. [3]
Show complete worked solution
(a) Set $h(x)=0$:
$$-0.05x^2+4=0$$
$$0.05x^2=4$$
$$x^2 = \frac{4}{0.05}=80$$
$$x = \pm\sqrt{80} = \pm 4\sqrt{5} \approx \pm 8.944$$
The arch spans from $x=-4\sqrt{5}$ to $x=4\sqrt{5}$, so its total width is
$$2\times 4\sqrt{5} = 8\sqrt{5} \approx 17.9 \text{ m}$$
$$\boxed{\text{Width} \approx 17.9 \text{ m (3 s.f.)}}$$
(b) Substitute $x=5$:
$$h(5) = -0.05(5)^2+4 = -0.05(25)+4 = -1.25+4=2.75$$
$$\boxed{h(5) = 2.75 \text{ m}}$$
QUESTION 34
8 marks
Medium
A ball is thrown from a height of $1.5$ m with height, in metres, above the ground modelled by $h(t)=-5t^2+12t+1.5$, where $t$ is time in seconds.
Find the time(s) at which the ball is at a height of $6$ m. Give your answers correct to 3 significant figures.
Show complete worked solution
Set $h(t)=6$:
$$-5t^2+12t+1.5=6$$
$$-5t^2+12t-4.5=0$$
Multiply through by $-1$ for convenience:
$$5t^2-12t+4.5=0$$
Apply the quadratic formula with $a=5$, $b=-12$, $c=4.5$:
$$\Delta = (-12)^2-4(5)(4.5) = 144-90=54$$
$$t = \frac{12\pm\sqrt{54}}{10} = \frac{12\pm 7.34847\ldots}{10}$$
$$t = \frac{12+7.34847\ldots}{10}=1.93485\ldots \qquad \text{or} \qquad t=\frac{12-7.34847\ldots}{10}=0.46515\ldots$$
Both values are positive and physically valid: the ball first passes $6$ m on the way up, then again on the way back down.
$$\boxed{t\approx 0.465 \text{ s (rising) and } t\approx 1.93 \text{ s (falling), to 3 s.f.}}$$
QUESTION 35
7 marks
Medium
A parabola $f(x)=x^2+bx+c$ has axis of symmetry $x=3$ and $y$-intercept $(0,5)$.
(a) Find the values of $b$ and $c$. [4]
(b) Hence find $f(7)$. [3]
Show complete worked solution
(a) For $f(x)=x^2+bx+c$ (with $a=1$), the axis of symmetry is $x=-\dfrac{b}{2a}=-\dfrac{b}{2}$. Setting this equal to $3$:
$$-\frac{b}{2}=3 \implies b=-6$$
The $y$-intercept gives $f(0)=c$, so $c=5$.
$$\boxed{b=-6,\ c=5, \text{ so } f(x)=x^2-6x+5}$$
(b) Substitute $x=7$:
$$f(7) = (7)^2-6(7)+5 = 49-42+5 = 12$$
$$\boxed{f(7)=12}$$
QUESTION 36
7 marks
Medium
Find the value of $k$ for which the line $y=2x+k$ is tangent to the parabola $y=x^2-4x+7$ (i.e. touches it at exactly one point).
Show complete worked solution
At points of intersection, the $y$-values are equal:
$$x^2-4x+7 = 2x+k$$
$$x^2-6x+(7-k)=0$$
For the line to be tangent to the parabola, this equation must have exactly one (repeated) solution, so its discriminant must equal zero. With $a=1$, $b=-6$, $c=7-k$:
$$\Delta = (-6)^2-4(1)(7-k) = 36-28+4k = 8+4k$$
Set $\Delta=0$:
$$8+4k=0$$
$$k=-2$$
$$\boxed{k=-2}$$
Check: with $k=-2$, $x^2-6x+9=(x-3)^2=0$, giving the single touching point $x=3$, $y=2(3)-2=4$.
QUESTION 37
7 marks
Medium
A ball is thrown upward from the edge of a building. Its height above the ground, in metres, is modelled by $h(t)=-5t^2+10t+45$, where $t$ is time in seconds after release.
Find the time at which the ball hits the ground, correct to 3 significant figures.
Show complete worked solution
The ball hits the ground when $h(t)=0$:
$$-5t^2+10t+45=0$$
Divide through by $-5$:
$$t^2-2t-9=0$$
Apply the quadratic formula with $a=1$, $b=-2$, $c=-9$:
$$\Delta = (-2)^2-4(1)(-9) = 4+36=40$$
$$t = \frac{2\pm\sqrt{40}}{2} = \frac{2\pm 6.32456\ldots}{2}$$
$$t = \frac{2+6.32456\ldots}{2}=4.16228\ldots \qquad \text{or} \qquad t=\frac{2-6.32456\ldots}{2}=-2.16228\ldots$$
Since $t$ represents time after release, the negative solution is not physically meaningful and is rejected.
$$\boxed{t\approx 4.16 \text{ s (3 s.f.)}}$$
QUESTION 38
6 marks
Medium
Two numbers have a sum of $24$. Let one number be $x$, so the other is $(24-x)$. Find the two numbers that maximise their product, and state the maximum product.
Show complete worked solution
The product is
$$P(x) = x(24-x) = 24x-x^2$$
Writing this as $P(x)=-x^2+24x$, we have $a=-1$, $b=24$. Since $a<0$, the vertex is a maximum, occurring at
$$x = -\frac{b}{2a} = -\frac{24}{2(-1)}=12$$
Then the other number is $24-12=12$, and the maximum product is
$$P(12) = 12(24-12) = 12(12) = 144$$
$$\boxed{\text{The numbers are 12 and 12, with maximum product } 144}$$
QUESTION 39
7 marks
Medium
A parabola has vertex $(-1,-8)$ and passes through the point $(2,1)$.
(a) Find the equation of the parabola in vertex form. [4]
(b) Hence state the $y$-intercept of the parabola. [3]
Show complete worked solution
(a) With vertex $(-1,-8)$, write the equation as
$$y = a(x+1)^2-8$$
Substitute the point $(2,1)$:
$$1 = a(2+1)^2-8$$
$$1 = 9a-8$$
$$9 = 9a$$
$$a=1$$
$$\boxed{y=(x+1)^2-8}$$
(b) Expand to standard form to read off the $y$-intercept, or substitute $x=0$ directly:
$$y(0) = (0+1)^2-8 = 1-8=-7$$
$$\boxed{y\text{-intercept} = (0,-7)}$$
QUESTION 40
8 marks
Medium
A company's monthly profit, in dollars, from selling $x$ items is modelled by
$$P(x) = -2x^2+80x-350$$
Find the break-even quantities (the values of $x$ for which $P(x)=0$), correct to 3 significant figures.
Show complete worked solution
Set $P(x)=0$:
$$-2x^2+80x-350=0$$
Divide through by $-2$:
$$x^2-40x+175=0$$
Apply the quadratic formula with $a=1$, $b=-40$, $c=175$:
$$\Delta = (-40)^2-4(1)(175) = 1600-700=900$$
Since $\sqrt{900}=30$:
$$x = \frac{40\pm 30}{2}$$
$$x = \frac{40+30}{2}=35 \qquad \text{or} \qquad x=\frac{40-30}{2}=5$$
$$\boxed{x=5 \text{ or } x=35 \text{ items}}$$
The company breaks even when it sells 5 or 35 items; between these values profit is positive.
QUESTION 41
10 marks
Hard
A parabola has $x$-intercepts at $x=-1$ and $x=4$, and passes through the point $(2,-12)$.
(a) Find the equation of the parabola in factored form. [3]
(b) Expand your answer to give the equation in standard form. [2]
(c) Find the coordinates of the vertex and state the minimum value of the function. [3]
(d) Find the $y$-intercept of the parabola. [2]
Show complete worked solution
(a) With $x$-intercepts $-1$ and $4$, and unknown leading coefficient $a$:
$$y = a(x+1)(x-4)$$
Substitute the point $(2,-12)$:
$$-12 = a(2+1)(2-4) = a(3)(-2) = -6a$$
$$a = \frac{-12}{-6}=2$$
$$\boxed{y=2(x+1)(x-4)}$$
(b) Expand $(x+1)(x-4)$:
$$(x+1)(x-4) = x^2-4x+x-4 = x^2-3x-4$$
Multiply by $2$:
$$y = 2(x^2-3x-4) = 2x^2-6x-8$$
$$\boxed{y=2x^2-6x-8}$$
(c) The $x$-coordinate of the vertex is
$$x = -\frac{b}{2a} = -\frac{-6}{2(2)} = \frac{6}{4}=1.5$$
(This also equals the midpoint of the intercepts: $\dfrac{-1+4}{2}=1.5$.)
$$y(1.5) = 2(1.5)^2-6(1.5)-8 = 2(2.25)-9-8 = 4.5-9-8=-12.5$$
$$\boxed{\text{Vertex} = (1.5,-12.5), \text{ minimum value } -12.5 \text{ (since } a=2>0\text{)}}$$
(d) The $y$-intercept occurs at $x=0$:
$$y(0) = 2(0)^2-6(0)-8=-8$$
$$\boxed{(0,-8)}$$
QUESTION 42
11 marks
Hard
A ball is launched from a platform $2$ m above the ground with initial vertical velocity $16$ m/s. Its height above the ground, in metres, $t$ seconds after launch is modelled by
$$h(t) = -5t^2+16t+2$$
(a) Find the maximum height reached by the ball, and the time at which it occurs. [4]
(b) Find the time at which the ball hits the ground, correct to 3 significant figures, explaining why one solution of your equation is rejected. [4]
(c) Find the time(s) at which the ball is at a height of $10$ m, correct to 3 significant figures. [3]
Show complete worked solution
(a) Since $a=-5<0$, the vertex of $h(t)$ is a maximum, at time
$$t = -\frac{b}{2a} = -\frac{16}{2(-5)} = \frac{16}{10}=1.6$$
$$h(1.6) = -5(1.6)^2+16(1.6)+2 = -5(2.56)+25.6+2 = -12.8+25.6+2=14.8$$
$$\boxed{\text{Maximum height} = 14.8 \text{ m at } t=1.6 \text{ s}}$$
(b) The ball hits the ground when $h(t)=0$:
$$-5t^2+16t+2=0$$
Divide by $-1$:
$$5t^2-16t-2=0$$
$$\Delta = (-16)^2-4(5)(-2) = 256+40=296$$
$$t = \frac{16\pm\sqrt{296}}{10} = \frac{16\pm 17.2047\ldots}{10}$$
$$t = \frac{16+17.2047\ldots}{10}=3.32047\ldots \qquad \text{or} \qquad t=\frac{16-17.2047\ldots}{10}=-0.12047\ldots$$
The negative solution, $t\approx -0.120$, occurs before launch ($t=0$) and has no physical meaning in this context, so it is rejected.
$$\boxed{t\approx 3.32 \text{ s (3 s.f.)}}$$
(c) Set $h(t)=10$:
$$-5t^2+16t+2=10$$
$$-5t^2+16t-8=0$$
$$5t^2-16t+8=0$$
$$\Delta = (-16)^2-4(5)(8) = 256-160=96$$
$$t = \frac{16\pm\sqrt{96}}{10} = \frac{16\pm 9.79796\ldots}{10}$$
$$t = \frac{16+9.79796\ldots}{10}=2.57980\ldots \qquad \text{or} \qquad t=\frac{16-9.79796\ldots}{10}=0.62020\ldots$$
Both solutions occur before the ball lands (before $t\approx 3.32$ s from part (b)), so both are physically valid: the ball passes $10$ m once on the way up and once on the way down.
$$\boxed{t\approx 0.620 \text{ s and } t\approx 2.58 \text{ s (3 s.f.)}}$$
QUESTION 43
10 marks
Hard
A company sells a product at price $\$p$ per unit. Market research shows that the number of units sold per week, $x$, is related to price by the demand function $p(x)=60-0.5x$. The company's weekly cost of producing $x$ units is $C(x)=15x+200$.
(a) Show that the weekly revenue is $R(x)=60x-0.5x^2$. [2]
(b) Find an expression for the weekly profit $P(x)=R(x)-C(x)$, simplified to standard form. [2]
(c) Find the number of units that maximises profit, and state the maximum profit. [3]
(d) Find the break-even quantities, correct to 2 decimal places. [3]
Show complete worked solution
(a) Revenue is price multiplied by quantity sold:
$$R(x) = x\cdot p(x) = x(60-0.5x) = 60x-0.5x^2$$
$$\boxed{R(x)=60x-0.5x^2}$$
(b) Profit is revenue minus cost:
$$P(x) = R(x)-C(x) = (60x-0.5x^2)-(15x+200)$$
$$= -0.5x^2+60x-15x-200$$
$$\boxed{P(x)=-0.5x^2+45x-200}$$
(c) $P(x)=-0.5x^2+45x-200$ has $a=-0.5$, $b=45$. Since $a<0$, the vertex is a maximum, at
$$x = -\frac{b}{2a} = -\frac{45}{2(-0.5)} = \frac{45}{1}=45$$
$$P(45) = -0.5(45)^2+45(45)-200 = -0.5(2025)+2025-200 = -1012.5+2025-200=812.5$$
$$\boxed{\text{Sell 45 units for a maximum profit of } \$812.50}$$
(d) Set $P(x)=0$:
$$-0.5x^2+45x-200=0$$
Multiply through by $-2$:
$$x^2-90x+400=0$$
$$\Delta = (-90)^2-4(1)(400) = 8100-1600=6500$$
$$x = \frac{90\pm\sqrt{6500}}{2} = \frac{90\pm 80.6226\ldots}{2}$$
$$x = \frac{90+80.6226\ldots}{2}=85.3113\ldots \qquad \text{or} \qquad x=\frac{90-80.6226\ldots}{2}=4.6887\ldots$$
$$\boxed{x\approx 4.69 \text{ or } x\approx 85.31 \text{ units (2 d.p.)}}$$
QUESTION 44
10 marks
Hard
A farmer has $120$ m of fencing to build a rectangular enclosure alongside an existing straight wall, which forms one side of the enclosure (so fencing is only needed for the other three sides). Let $x$ metres be the width of each of the two sides perpendicular to the wall.
(a) Show that the area enclosed is $A(x)=120x-2x^2$. [2]
(b) Find the value of $x$ that maximises the area, and state the maximum area and the corresponding dimensions. [4]
(c) Find the range of values of $x$ for which the enclosed area is at least $1750$ m$^2$. [4]
Show complete worked solution
(a) Let $x$ be the width of each of the two sides perpendicular to the wall, and $y$ be the length of the side parallel to the wall. Since only three sides need fencing:
$$2x+y=120 \implies y=120-2x$$
The area is
$$A(x) = x\cdot y = x(120-2x) = 120x-2x^2$$
$$\boxed{A(x)=120x-2x^2}$$
(b) Writing $A(x)=-2x^2+120x$, we have $a=-2$, $b=120$. Since $a<0$, the vertex is a maximum, at
$$x = -\frac{b}{2a} = -\frac{120}{2(-2)} = \frac{120}{4}=30$$
$$A(30) = 120(30)-2(30)^2 = 3600-1800=1800$$
Then $y=120-2(30)=60$.
$$\boxed{x=30 \text{ m gives dimensions } 30\text{ m}\times 60\text{ m and a maximum area of } 1800 \text{ m}^2}$$
(c) Require $A(x)\ge 1750$:
$$120x-2x^2\ge 1750$$
$$-2x^2+120x-1750\ge 0$$
Multiply by $-1$ (reversing the inequality):
$$2x^2-120x+1750\le 0$$
Divide by $2$:
$$x^2-60x+875\le 0$$
Solve the corresponding equation: $\Delta=(-60)^2-4(1)(875)=3600-3500=100$, $\sqrt{100}=10$.
$$x = \frac{60\pm 10}{2}$$
$$x=35 \qquad \text{or} \qquad x=25$$
Since $a=1>0$ for this rearranged quadratic, it is $\le 0$ between its roots.
$$\boxed{25\le x\le 35 \text{ (metres)}}$$
Check: $A(25)=120(25)-2(625)=3000-1250=1750$ and $A(35)=120(35)-2(1225)=4200-2450=1750$, confirming both boundary values give exactly $1750$ m$^2$.
QUESTION 45
9 marks
Hard
An arch bridge is modelled by a downward parabola with vertex at $(0,8)$, where $8$ is the maximum height of the arch (in metres) above the road, and the arch meets the road at $(20,0)$ and $(-20,0)$.
(a) Find the equation of the parabola in the form $h(x)=ax^2+8$. [3]
(b) A truck of height $5.5$ m travels along a path $15$ m horizontally from the centre of the arch. Determine, showing your working, whether the truck passes under the arch without touching it. [3]
(c) Find the width of the arch at a height of $4.5$ m above the road. [3]
Show complete worked solution
(a) Since the vertex is $(0,8)$, the equation has the form $h(x)=ax^2+8$. Substitute the point $(20,0)$:
$$0 = a(20)^2+8$$
$$0 = 400a+8$$
$$a = -\frac{8}{400} = -0.02$$
$$\boxed{h(x)=-0.02x^2+8}$$
(b) Substitute $x=15$ into the model:
$$h(15) = -0.02(15)^2+8 = -0.02(225)+8 = -4.5+8=3.5$$
Since the arch height at $x=15$ m is only $3.5$ m, which is less than the truck's height of $5.5$ m, the truck would not pass under the arch at that position without touching it.
$$\boxed{h(15)=3.5 \text{ m} < 5.5 \text{ m, so the truck does not clear the arch}}$$
(c) Set $h(x)=4.5$:
$$-0.02x^2+8=4.5$$
$$-0.02x^2=-3.5$$
$$x^2 = \frac{3.5}{0.02}=175$$
$$x = \pm\sqrt{175} = \pm 5\sqrt{7} \approx \pm 13.229$$
The width at this height is the distance between the two solutions:
$$2\times 5\sqrt{7} \approx 26.5 \text{ m}$$
$$\boxed{\text{Width} \approx 26.5 \text{ m (3 s.f.)}}$$
QUESTION 46
9 marks
Hard
Consider $f(x)=x^2-2kx+(k+6)$, where $k$ is a real parameter.
(a) Show that the discriminant of $f(x)=0$ can be written as $\Delta=4(k-3)(k+2)$. [3]
(b) Hence find the set of values of $k$ for which $f(x)=0$ has two distinct real roots. [3]
(c) For $k=5$, solve $f(x)=0$ using the quadratic formula, giving answers correct to 3 significant figures. [3]
Show complete worked solution
(a) For $f(x)=x^2-2kx+(k+6)$, $a=1$, $b=-2k$, $c=k+6$:
$$\Delta = b^2-4ac = (-2k)^2-4(1)(k+6) = 4k^2-4k-24$$
Factor out $4$:
$$\Delta = 4(k^2-k-6)$$
Factorise the quadratic in $k$: we need two numbers multiplying to $-6$ and summing to $-1$, namely $-3$ and $2$.
$$k^2-k-6 = (k-3)(k+2)$$
$$\boxed{\Delta = 4(k-3)(k+2)}$$
(b) Two distinct real roots require $\Delta>0$:
$$4(k-3)(k+2)>0 \implies (k-3)(k+2)>0$$
This product is positive when both factors are positive, or both are negative:
Both positive: $k>3$ and $k>-2$, i.e. $k>3$.
Both negative: $k<3$ and $k<-2$, i.e. $k<-2$.
$$\boxed{k<-2 \text{ or } k>3}$$
(c) With $k=5$ (which satisfies $k>3$, consistent with part (b)): $f(x)=x^2-10x+11$.
$$\Delta = (-10)^2-4(1)(11) = 100-44=56$$
$$x = \frac{10\pm\sqrt{56}}{2} = \frac{10\pm 7.48331\ldots}{2}$$
$$x = \frac{10+7.48331\ldots}{2}=8.74166\ldots \qquad \text{or} \qquad x=\frac{10-7.48331\ldots}{2}=1.25834\ldots$$
$$\boxed{x\approx 8.74 \text{ or } x\approx 1.26 \text{ (3 s.f.)}}$$
QUESTION 47
10 marks
Hard
The height of a ball above the ground, in metres, was recorded at three instants: at $t=0$ s the height was $2$ m, at $t=1$ s the height was $15$ m, and at $t=3$ s the height was $11$ m. The height is known to follow a quadratic model $h(t)=at^2+bt+c$.
(a) Use the three data points to form a system of equations, and solve it to find $a$, $b$ and $c$. [4]
(b) Find the maximum height of the ball and the time at which it occurs. [3]
(c) Find the time at which the ball hits the ground, correct to 3 significant figures. [3]
Show complete worked solution
(a) Substitute each data point into $h(t)=at^2+bt+c$.
From $(0,2)$: $\quad c=2$
From $(1,15)$: $\quad a+b+c=15 \implies a+b=13$
From $(3,11)$: $\quad 9a+3b+c=11 \implies 9a+3b=9 \implies 3a+b=3$
Subtract the second reduced equation from the third:
$$(3a+b)-(a+b) = 3-13$$
$$2a=-10$$
$$a=-5$$
Substitute into $a+b=13$: $b=13-(-5)=18$.
$$\boxed{a=-5,\ b=18,\ c=2, \text{ so } h(t)=-5t^2+18t+2}$$
Check: $h(3)=-5(9)+18(3)+2=-45+54+2=11$. Correct.
(b) Since $a=-5<0$, the vertex is a maximum, at
$$t = -\frac{b}{2a} = -\frac{18}{2(-5)} = \frac{18}{10}=1.8$$
$$h(1.8) = -5(1.8)^2+18(1.8)+2 = -5(3.24)+32.4+2 = -16.2+32.4+2=18.2$$
$$\boxed{\text{Maximum height} = 18.2 \text{ m at } t=1.8 \text{ s}}$$
(c) The ball hits the ground when $h(t)=0$:
$$-5t^2+18t+2=0$$
Divide by $-1$:
$$5t^2-18t-2=0$$
$$\Delta = (-18)^2-4(5)(-2) = 324+40=364$$
$$t = \frac{18\pm\sqrt{364}}{10} = \frac{18\pm 19.0788\ldots}{10}$$
$$t = \frac{18+19.0788\ldots}{10}=3.70788\ldots \qquad \text{or} \qquad t=\frac{18-19.0788\ldots}{10}=-0.10788\ldots$$
The negative solution is rejected as it occurs before the ball was thrown.
$$\boxed{t\approx 3.71 \text{ s (3 s.f.)}}$$
QUESTION 48
10 marks
Hard
Consider $f(x)=2x^2-8x+6$.
(a) Write $f(x)$ in factored form. [2]
(b) Write $f(x)$ in vertex form by completing the square. [3]
(c) Find the set of values of $x$ for which $f(x)<0$. [3]
(d) The graph of $y=f(x)$ is reflected in the $x$-axis. Find the equation of the image, in standard form. [2]
Show complete worked solution
(a) Factor out the common factor of $2$ first:
$$f(x) = 2x^2-8x+6 = 2(x^2-4x+3)$$
Factorise the bracket: we need two numbers multiplying to $3$ and summing to $-4$, namely $-1$ and $-3$.
$$x^2-4x+3 = (x-1)(x-3)$$
$$\boxed{f(x)=2(x-1)(x-3)}$$
(b) Complete the square on $x^2-4x$ (half of $-4$ is $-2$):
$$f(x) = 2(x^2-4x)+6 = 2\big[(x-2)^2-4\big]+6 = 2(x-2)^2-8+6$$
$$\boxed{f(x)=2(x-2)^2-2}$$
(c) From part (a), $f(x)=0$ at $x=1$ and $x=3$. Since the leading coefficient $a=2>0$, the parabola opens upward, so $f(x)<0$ strictly between the roots.
Checking $x=2$ (between the roots): $f(2)=2(2)^2-8(2)+6=8-16+6=-2<0$. This confirms the sign.
$$\boxed{1
QUESTION 49
8 marks
Hard
Two real numbers differ by $6$. Let the smaller number be $x$, so the larger is $(x+6)$.
(a) Write an expression, $S(x)$, for the sum of the squares of the two numbers, and simplify it to standard quadratic form. [3]
(b) Find the value of $x$ that minimises $S(x)$, and state the minimum sum of squares. [3]
(c) State the two numbers that give this minimum. [2]
Show complete worked solution
(a) The sum of squares is
$$S(x) = x^2+(x+6)^2$$
Expand $(x+6)^2=x^2+12x+36$:
$$S(x) = x^2+x^2+12x+36 = 2x^2+12x+36$$
$$\boxed{S(x)=2x^2+12x+36}$$
(b) $S(x)=2x^2+12x+36$ has $a=2$, $b=12$. Since $a>0$, the vertex is a minimum, at
$$x = -\frac{b}{2a} = -\frac{12}{2(2)} = -\frac{12}{4}=-3$$
$$S(-3) = 2(-3)^2+12(-3)+36 = 2(9)-36+36 = 18-36+36=18$$
$$\boxed{\text{Minimum sum of squares} = 18, \text{ at } x=-3}$$
(c) With $x=-3$, the two numbers are
$$x=-3 \qquad \text{and} \qquad x+6=-3+6=3$$
$$\boxed{\text{The numbers are } -3 \text{ and } 3}$$
Check: $(-3)^2+3^2=9+9=18$, which matches the minimum found in part (b), and the numbers indeed differ by $6$.
QUESTION 50
10 marks
Hard
A company's monthly profit, in dollars, from selling a product at price $\$p$ per unit is modelled by
$$P(p) = -4p^2+320p-4800$$
(a) Find the break-even prices (the values of $p$ for which $P(p)=0$). [3]
(b) Find the price that maximises profit, and state the maximum monthly profit. [3]
(c) Find the range of prices for which the monthly profit is at least $\$1200$. [4]
Show complete worked solution
(a) Set $P(p)=0$:
$$-4p^2+320p-4800=0$$
Divide through by $-4$:
$$p^2-80p+1200=0$$
$$\Delta = (-80)^2-4(1)(1200) = 6400-4800=1600$$
Since $\sqrt{1600}=40$:
$$p = \frac{80\pm 40}{2}$$
$$p = \frac{80+40}{2}=60 \qquad \text{or} \qquad p=\frac{80-40}{2}=20$$
$$\boxed{\text{Break-even at } p=\$20 \text{ and } p=\$60}$$
(b) $P(p)=-4p^2+320p-4800$ has $a=-4$, $b=320$. Since $a<0$, the vertex is a maximum, at
$$p = -\frac{b}{2a} = -\frac{320}{2(-4)} = \frac{320}{8}=40$$
$$P(40) = -4(40)^2+320(40)-4800 = -4(1600)+12800-4800 = -6400+12800-4800=1600$$
$$\boxed{\text{Maximum profit } = \$1600 \text{ at price } p=\$40}$$
(c) Require $P(p)\ge 1200$:
$$-4p^2+320p-4800\ge 1200$$
$$-4p^2+320p-6000\ge 0$$
Divide by $-4$ (reversing the inequality):
$$p^2-80p+1500\le 0$$
$$\Delta = (-80)^2-4(1)(1500) = 6400-6000=400$$
Since $\sqrt{400}=20$:
$$p = \frac{80\pm 20}{2}$$
$$p=50 \qquad \text{or} \qquad p=30$$
Since $a=1>0$ for this rearranged quadratic, it is $\le 0$ between its roots.
$$\boxed{30\le p\le 50 \text{ (dollars)}}$$
Check: $P(30)=-4(900)+320(30)-4800=-3600+9600-4800=1200$ and $P(50)=-4(2500)+320(50)-4800=-10000+16000-4800=1200$, confirming both endpoints give exactly $\$1200$.
Cubic Functions 50 questions
QUESTION 1
4 marks
Easy
The graph of the cubic function $y=(x-1)(x-2)(x-3)$ is considered.
Write down the $x$-intercepts of the graph of $y=(x-1)(x-2)(x-3)$.
Show complete worked solution
The $x$-intercepts of a factored cubic $y=(x-r_1)(x-r_2)(x-r_3)$ occur where $y=0$, which happens when any one of the linear factors is zero.
Setting each factor equal to zero:
$$x-1=0 \Rightarrow x=1$$
$$x-2=0 \Rightarrow x=2$$
$$x-3=0 \Rightarrow x=3$$
$$\boxed{x=1,\ x=2,\ x=3}$$
QUESTION 2
5 marks
Easy
A cubic function is given in factored form by $y=2(x+1)(x-2)(x-5)$.
(a) Write down the $x$-intercepts of the graph. [2]
(b) Find the $y$-intercept of the graph. [3]
Show complete worked solution
(a) The $x$-intercepts occur where each factor is zero:
$$x+1=0 \Rightarrow x=-1$$
$$x-2=0 \Rightarrow x=2$$
$$x-5=0 \Rightarrow x=5$$
$$\boxed{x=-1,\ x=2,\ x=5}$$
(b) The $y$-intercept occurs where $x=0$. Substituting $x=0$ into $y=2(x+1)(x-2)(x-5)$:
$$y=2(0+1)(0-2)(0-5)=2(1)(-2)(-5)$$
$$y=2\times 10=20$$
$$\boxed{y\text{-intercept}=(0,20)}$$
QUESTION 3
4 marks
Easy
Expand and simplify $y=(x+3)(x-1)(x-4)$ to give the equation of the cubic in the form $y=ax^3+bx^2+cx+d$.
Show complete worked solution
First multiply the last two factors:
$$(x-1)(x-4)=x^2-5x+4$$
Now multiply this result by $(x+3)$:
$$y=(x+3)(x^2-5x+4)$$
$$y=x^3-5x^2+4x+3x^2-15x+12$$
$$y=x^3-2x^2-11x+12$$
$$\boxed{y=x^3-2x^2-11x+12}$$
QUESTION 4
5 marks
Easy
Consider the cubic function $y=x^3-6x^2+11x-6$.
By factorising, or by using your GDC's factor/root-finding feature, find the three $x$-intercepts of the graph of $y=x^3-6x^2+11x-6$.
Show complete worked solution
Testing small integer values using the factor theorem, try $x=1$:
$$1^3-6(1)^2+11(1)-6=1-6+11-6=0$$
So $x=1$ is a root, meaning $(x-1)$ is a factor. Dividing $x^3-6x^2+11x-6$ by $(x-1)$ (by inspection or polynomial division):
$$x^3-6x^2+11x-6=(x-1)(x^2-5x+6)$$
Factorising the quadratic:
$$x^2-5x+6=(x-2)(x-3)$$
So:
$$y=(x-1)(x-2)(x-3)$$
Setting each factor to zero (this can also be confirmed directly using the GDC's graph or "zero"/root-finding feature):
$$\boxed{x=1,\ x=2,\ x=3}$$
QUESTION 5
5 marks
Easy
Consider the cubic function $y=x^3-x^2-4x+4$.
By grouping terms, factorise $y$ completely and hence write down the $x$-intercepts of its graph.
Show complete worked solution
Group the terms in pairs:
$$y=(x^3-x^2)+(-4x+4)$$
Factor each pair:
$$y=x^2(x-1)-4(x-1)$$
Both terms now share the common factor $(x-1)$:
$$y=(x-1)(x^2-4)$$
The quadratic factor is a difference of squares:
$$x^2-4=(x-2)(x+2)$$
So:
$$y=(x-1)(x-2)(x+2)$$
Setting each factor to zero gives the $x$-intercepts:
$$\boxed{x=-2,\ x=1,\ x=2}$$
QUESTION 6
4 marks
Easy
Consider the cubic function $y=x^3-4x$.
(a) Write down the value of the $y$-intercept. [1]
(b) Describe the behaviour of $y$ as $x\to+\infty$ and as $x\to-\infty$. [3]
Show complete worked solution
(a) The $y$-intercept occurs at $x=0$:
$$y=0^3-4(0)=0$$
$$\boxed{y\text{-intercept}=(0,0)}$$
(b) The leading coefficient of $y=x^3-4x$ is $a=1$, which is positive, and the degree of the polynomial is odd (degree 3). For a cubic with a positive leading coefficient, the graph falls to the left and rises to the right:
$$\text{as } x\to+\infty,\ y\to+\infty$$
$$\text{as } x\to-\infty,\ y\to-\infty$$
$$\boxed{y\to+\infty \text{ as } x\to+\infty; \quad y\to-\infty \text{ as } x\to-\infty}$$
This matches the graph shown, which comes up from the bottom-left, crosses the $x$-axis three times, and continues up to the top-right.
QUESTION 7
4 marks
Easy
Consider the cubic function $y=-2x^3+3x-5$.
(a) Write down the value of the $y$-intercept. [1]
(b) Describe the behaviour of $y$ as $x\to+\infty$ and as $x\to-\infty$. [3]
Show complete worked solution
(a) The $y$-intercept occurs at $x=0$:
$$y=-2(0)^3+3(0)-5=-5$$
$$\boxed{y\text{-intercept}=(0,-5)}$$
(b) The leading coefficient of $y=-2x^3+3x-5$ is $a=-2$, which is negative. For a cubic with a negative leading coefficient, the graph rises to the left and falls to the right (the opposite of the standard positive cubic shape):
$$\text{as } x\to+\infty,\ y\to-\infty$$
$$\text{as } x\to-\infty,\ y\to+\infty$$
$$\boxed{y\to-\infty \text{ as } x\to+\infty; \quad y\to+\infty \text{ as } x\to-\infty}$$
This matches the graph shown, which comes down from the top-left and continues down to the bottom-right.
QUESTION 8
4 marks
Easy
Let $f(x)=2x^3-5x^2+x-7$.
Find the value of $f(3)$.
Show complete worked solution
Substitute $x=3$ directly into $f(x)=2x^3-5x^2+x-7$:
$$f(3)=2(3)^3-5(3)^2+(3)-7$$
$$f(3)=2(27)-5(9)+3-7$$
$$f(3)=54-45+3-7$$
$$f(3)=5$$
$$\boxed{f(3)=5}$$
QUESTION 9
5 marks
Easy
Let $f(x)=x^3-2x^2-5x+6$.
(a) By evaluating $f(2)$, determine whether $x=2$ is a root of $f(x)$. [2]
(b) By trial and error with small integers, or by using your GDC, find the three roots of $f(x)$. [3]
Show complete worked solution
(a) Substitute $x=2$:
$$f(2)=(2)^3-2(2)^2-5(2)+6=8-8-10+6=-4$$
Since $f(2)=-4\neq 0$, $x=2$ is not a root of $f(x)$.
$$\boxed{f(2)=-4\neq 0,\text{ so } x=2 \text{ is not a root}}$$
(b) Trying $x=-2$:
$$f(-2)=(-2)^3-2(-2)^2-5(-2)+6=-8-8+10+6=0$$
So $x=-2$ is a root. Dividing $f(x)$ by $(x+2)$ gives $f(x)=(x+2)(x^2-4x+3)=(x+2)(x-1)(x-3)$, which can also be confirmed by graphing $f(x)$ on the GDC and using the root/zero-finding feature.
$$\boxed{x=-2,\ x=1,\ x=3}$$
QUESTION 10
5 marks
Easy
An open box is made by cutting a square of side $x$ cm from each corner of a square sheet of card of side $20$ cm, then folding up the sides. The volume of the resulting box is given by
$$V(x)=x(20-2x)^2 \text{ cm}^3$$
Find the volume of the box when $x=3$.
Show complete worked solution
Substitute $x=3$ into $V(x)=x(20-2x)^2$:
$$V(3)=3(20-2(3))^2$$
$$V(3)=3(20-6)^2$$
$$V(3)=3(14)^2$$
$$V(3)=3(196)$$
$$V(3)=588$$
$$\boxed{V(3)=588 \text{ cm}^3}$$
QUESTION 11
4 marks
Easy
The weekly cost, in dollars, of running a small workshop that produces $x$ hundred units of a product is modelled by
$$C(x)=x^3-9x^2+24x+10$$
Find the value of $C(2)$ and state what this value represents.
Show complete worked solution
Substitute $x=2$ into $C(x)=x^3-9x^2+24x+10$:
$$C(2)=(2)^3-9(2)^2+24(2)+10$$
$$C(2)=8-36+48+10$$
$$C(2)=30$$
$$\boxed{C(2)=30}$$
This means that when the workshop produces $x=2$ hundred (i.e. $200$) units in a week, the running cost is \$30.
QUESTION 12
4 marks
Easy
The number of active users (in hundreds) of a new phone app, $t$ months after its launch, is modelled by
$$N(t)=t^3-6t^2+12t, \quad 0\le t\le 5$$
Find the value of $N(2)$ and state what this value represents in context.
Show complete worked solution
Substitute $t=2$ into $N(t)=t^3-6t^2+12t$:
$$N(2)=(2)^3-6(2)^2+12(2)$$
$$N(2)=8-24+24$$
$$N(2)=8$$
$$\boxed{N(2)=8}$$
Since $N(t)$ is measured in hundreds of users, this means that $2$ months after launch, the app has approximately $800$ active users.
QUESTION 13
5 marks
Easy
Consider the cubic function $y=x^3-3x$.
Using your GDC's graphing (or maximum/minimum-finding) feature, find the coordinates of the local maximum point and the local minimum point on the graph of $y=x^3-3x$.
Show complete worked solution
Enter $y=x^3-3x$ into the GDC and view the graph. The graph shows one "hump" on the left (a local maximum) and one "dip" on the right (a local minimum).
Using the GDC's maximum-finding feature on the left-hand turning point:
$$\text{local maximum at } (-1,\ 2)$$
Using the GDC's minimum-finding feature on the right-hand turning point:
$$\text{local minimum at } (1,\ -2)$$
$$\boxed{\text{local maximum } (-1,2); \quad \text{local minimum } (1,-2)}$$
QUESTION 14
5 marks
Easy
Consider the cubic function $y=-x^3+3x^2$.
Using your GDC's graphing (or maximum/minimum-finding) feature, find the coordinates of the local minimum point and the local maximum point on the graph of $y=-x^3+3x^2$.
Show complete worked solution
Enter $y=-x^3+3x^2$ into the GDC and view the graph. Since the leading coefficient is negative, the curve dips down on the left before rising, giving a local minimum first, followed by a local maximum.
Using the GDC's minimum-finding feature on the left-hand turning point:
$$\text{local minimum at } (0,\ 0)$$
Using the GDC's maximum-finding feature on the right-hand turning point:
$$\text{local maximum at } (2,\ 4)$$
$$\boxed{\text{local minimum } (0,0); \quad \text{local maximum } (2,4)}$$
QUESTION 15
4 marks
Easy
The graph of a cubic function $y=f(x)$ is shown below.
(a) Write down the three $x$-intercepts of the graph. [3]
(b) State whether the leading coefficient of $f(x)$ is positive or negative, giving a reason. [1]
Show complete worked solution
(a) Reading the points where the curve crosses the $x$-axis directly from the graph:
$$\boxed{x=-3,\ x=1,\ x=4}$$
(b) As $x\to+\infty$, the graph shows $y\to+\infty$, and as $x\to-\infty$, the graph shows $y\to-\infty$ (the curve rises to the right and falls to the left). This "bottom-left to top-right" behaviour is characteristic of a cubic with a positive leading coefficient.
$$\boxed{\text{leading coefficient is positive}}$$
QUESTION 16
4 marks
Easy
The graph of a cubic function $y=g(x)$ is shown below, with its two turning points labelled.
(a) Write down the coordinates of the local maximum point. [2]
(b) Write down the coordinates of the local minimum point. [2]
Show complete worked solution
Reading the labelled turning points directly from the graph:
(a) The local maximum point (the higher of the two turning points, on the left) is at
$$\boxed{(0,\ 0)}$$
(b) The local minimum point (the lower of the two turning points, on the right) is at
$$\boxed{(2,\ -4)}$$
QUESTION 17
5 marks
Easy
A cubic function has $x$-intercepts at $x=-2$, $x=0$ and $x=3$, and a leading coefficient of $1$.
(a) Write down the equation of the function in factored form. [2]
(b) Expand your answer to part (a) to write the equation in the form $y=ax^3+bx^2+cx+d$. [3]
Show complete worked solution
(a) Since the roots are $x=-2,\ 0,\ 3$ and the leading coefficient is $1$, the factored form is
$$\boxed{y=(x+2)(x)(x-3)}$$
(b) First multiply $(x+2)(x-3)$:
$$(x+2)(x-3)=x^2-x-6$$
Now multiply by the remaining factor $x$:
$$y=x(x^2-x-6)$$
$$\boxed{y=x^3-x^2-6x}$$
QUESTION 18
4 marks
Easy
Consider the cubic function $f(x)=x^3+2x^2-5x-6$. It is given that $x=-3$ is a root of $f(x)$.
(a) Verify, by substitution, that $x=-3$ is indeed a root of $f(x)$. [2]
(b) Write down the $y$-intercept of the graph of $f(x)$. [2]
Show complete worked solution
(a) Substitute $x=-3$ into $f(x)=x^3+2x^2-5x-6$:
$$f(-3)=(-3)^3+2(-3)^2-5(-3)-6$$
$$f(-3)=-27+2(9)+15-6$$
$$f(-3)=-27+18+15-6$$
$$f(-3)=0$$
Since $f(-3)=0$, this confirms that $x=-3$ is a root of $f(x)$.
$$\boxed{f(-3)=0 \implies x=-3 \text{ is a root}}$$
(b) The $y$-intercept occurs at $x=0$:
$$f(0)=(0)^3+2(0)^2-5(0)-6=-6$$
$$\boxed{y\text{-intercept}=(0,-6)}$$
QUESTION 19
5 marks
Easy
The table below shows values of $f(x)=x^3-5x^2+2x+8$ for integer values of $x$ from $-2$ to $5$.
(a) Using the table, write down the three roots of $f(x)$. [3]
(b) Write down the $y$-intercept of the graph of $f(x)$. [2]
| $x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|---|---|---|
| $f(x)$ | $-24$ | $0$ | $8$ | $6$ | $0$ | $-4$ | $0$ | $18$ |
Show complete worked solution
(a) A root of $f(x)$ occurs at any $x$-value for which $f(x)=0$. Scanning the table for entries equal to $0$:
$$f(-1)=0,\quad f(2)=0,\quad f(4)=0$$
$$\boxed{x=-1,\ x=2,\ x=4}$$
(b) The $y$-intercept occurs at $x=0$. Reading directly from the table, $f(0)=8$.
$$\boxed{y\text{-intercept}=(0,8)}$$
QUESTION 20
4 marks
Easy
A small company's weekly profit, in thousands of dollars, when it produces $x$ hundred units is modelled by
$$P(x)=-x^3+12x^2-20x$$
Find the value of $P(1)$ and state what this value means in context.
Show complete worked solution
Substitute $x=1$ into $P(x)=-x^3+12x^2-20x$:
$$P(1)=-(1)^3+12(1)^2-20(1)$$
$$P(1)=-1+12-20$$
$$P(1)=-9$$
$$\boxed{P(1)=-9}$$
Since $P(x)$ is measured in thousands of dollars, this means that when the company produces $x=1$ hundred (i.e. $100$) units in a week, it makes a loss of \$9000.
QUESTION 21
7 marks
Medium
An open-topped box is to be made from a rectangular sheet of card measuring $30$ cm by $20$ cm, by cutting a square of side $x$ cm from each corner and folding up the four flaps, as shown below.
(a) Show that the volume of the box, in $\text{cm}^3$, is given by
$$V(x)=4x^3-100x^2+600x$$ [3]
(b) Write down the possible values of $x$ for the box to have a physically valid (positive) volume. [2]
(c) Find the volume of the box when $x=4$. [2]
Show complete worked solution
(a) After the corners are removed and the flaps folded up, the base of the box has dimensions $(30-2x)$ cm by $(20-2x)$ cm, and the box has height $x$ cm. So the volume is
$$V(x)=x(30-2x)(20-2x)$$
Expanding $(30-2x)(20-2x)$ first:
$$(30-2x)(20-2x)=600-60x-40x+4x^2=4x^2-100x+600$$
Multiplying by $x$:
$$V(x)=x(4x^2-100x+600)$$
$$\boxed{V(x)=4x^3-100x^2+600x}$$
(b) For the box to exist physically, $x$ must be positive, and each side of the base, $30-2x$ and $20-2x$, must also be positive. The more restrictive condition is $20-2x>0$, i.e. $x<10$.
$$\boxed{0(c) Substitute $x=4$ into $V(x)=4x^3-100x^2+600x$:
$$V(4)=4(4)^3-100(4)^2+600(4)$$
$$V(4)=4(64)-100(16)+2400$$
$$V(4)=256-1600+2400$$
$$V(4)=1056$$
$$\boxed{V(4)=1056 \text{ cm}^3}$$
QUESTION 22
8 marks
Medium
Consider the cubic function $f(x)=x^3-x^2-7x+3$.
(a) Show that $x=3$ is a root of $f(x)$. [2]
(b) Hence write $f(x)$ in the form $(x-3)(x^2+bx+c)$, stating the values of $b$ and $c$. [3]
(c) Use your GDC to find the other two roots of $f(x)$, giving your answers to 3 significant figures. [3]
Show complete worked solution
(a) Substitute $x=3$ into $f(x)=x^3-x^2-7x+3$:
$$f(3)=(3)^3-(3)^2-7(3)+3=27-9-21+3=0$$
Since $f(3)=0$, $x=3$ is a root of $f(x)$.
$$\boxed{f(3)=0 \implies x=3 \text{ is a root}}$$
(b) Since $(x-3)$ is a factor, dividing $f(x)$ by $(x-3)$ (by inspection, matching coefficients with $(x-3)(x^2+bx+c)=x^3+(b-3)x^2+(c-3b)x-3c$):
Comparing the constant terms: $-3c=3 \Rightarrow c=-1$.
Comparing the $x^2$ coefficients: $b-3=-1 \Rightarrow b=2$.
Check with the $x$ coefficient: $c-3b=-1-6=-7$ ✓ (matches the coefficient of $x$ in $f(x)$)
$$\boxed{f(x)=(x-3)(x^2+2x-1)}$$
(c) The other two roots come from solving $x^2+2x-1=0$. Entering this into the GDC's polynomial/equation solver, or the graph's root-finding feature applied to $y=x^2+2x-1$:
$$\boxed{x\approx -2.41, \quad x\approx 0.414}$$
QUESTION 23
8 marks
Medium
A company's monthly profit, in thousands of dollars, when it produces $x$ hundred units ($x\ge 0$) is modelled by
$$P(x)=-2x^3+15x^2-24x-10$$
(a) Find $P(2)$ and interpret your answer. [2]
(b) Use your GDC to find the roots of $P(x)=0$, giving your answers to 3 significant figures. Hence state the production level(s), for $x\ge 0$, at which the company breaks even. [3]
(c) Use your GDC to find the coordinates of the local maximum point of $P(x)$ for $x\ge0$, and state the maximum monthly profit predicted by the model. [3]
Show complete worked solution
(a) Substitute $x=2$ into $P(x)=-2x^3+15x^2-24x-10$:
$$P(2)=-2(2)^3+15(2)^2-24(2)-10$$
$$P(2)=-2(8)+15(4)-48-10$$
$$P(2)=-16+60-48-10=-14$$
$$\boxed{P(2)=-14}$$
This means that when the company produces $200$ units a month, it makes a loss of \$14,000.
(b) Graphing $y=P(x)$ on the GDC and using the root/zero-finding feature, the roots are
$$x\approx -0.341, \quad x\approx 3.09, \quad x\approx 4.76$$
Since $x\ge 0$ in this context, the negative root is not physically valid. The company breaks even at approximately
$$\boxed{x\approx 3.09 \text{ and } x\approx 4.76 \text{ (i.e. about } 309 \text{ and } 476 \text{ units)}}$$
(c) Using the GDC's maximum-finding feature on the graph of $P(x)$ for $x\ge0$:
$$\text{local maximum at } (4,\ 6)$$
$$\boxed{\text{maximum profit} = \$6000 \text{, occurring at } x=4 \text{ (i.e. } 400 \text{ units)}}$$
QUESTION 24
7 marks
Medium
The depth of water, in cm, in a decorative fountain basin, $t$ hours after the fountain is switched on, is modelled by
$$d(t)=-t^3+6t^2-3t+10, \quad 0\le t\le 6$$
(a) Find $d(2)$. [2]
(b) Use your GDC to find the coordinates of the local minimum point and the local maximum point of $d(t)$, and briefly describe what happens to the water depth over the interval $0\le t \le 6$. [3]
(c) Use your GDC to find the value of $t$ at which the basin is empty (i.e. $d(t)=0$), giving your answer to 3 significant figures. [2]
Show complete worked solution
(a) Substitute $t=2$ into $d(t)=-t^3+6t^2-3t+10$:
$$d(2)=-(2)^3+6(2)^2-3(2)+10$$
$$d(2)=-8+24-6+10=20$$
$$\boxed{d(2)=20 \text{ cm}}$$
(b) Graphing $y=d(t)$ on the GDC and using the minimum- and maximum-finding features:
$$\text{local minimum at } (0.268,\ 9.61) \qquad \text{local maximum at } (3.73,\ 30.4)$$
$$\boxed{\text{local min }(0.268, 9.61); \text{ local max }(3.73, 30.4)}$$
This means the depth dips slightly from $10$ cm to about $9.61$ cm shortly after the fountain starts, then rises to a peak depth of about $30.4$ cm at $t\approx 3.73$ hours, before falling away towards zero.
(c) Using the GDC's root-finding feature on $y=d(t)$ for $t>3.73$:
$$\boxed{t\approx 5.78 \text{ hours (the basin empties after about } 5.78 \text{ hours)}}$$
QUESTION 25
7 marks
Medium
An open-topped box is made by cutting a square of side $x$ cm from each corner of a rectangular sheet of card measuring $16$ cm by $10$ cm, and folding up the flaps.
(a) Show that the volume of the box is given by $V(x)=4x^3-52x^2+160x$. [3]
(b) Write down the domain of $x$ for which the box is physically valid. [1]
(c) Use your GDC to find all values of $x$, within the domain from part (b), for which the volume of the box is exactly $100\text{ cm}^3$, giving your answers to 3 significant figures. [3]
Show complete worked solution
(a) The base of the box has dimensions $(16-2x)$ by $(10-2x)$, and the height is $x$, so
$$V(x)=x(16-2x)(10-2x)$$
Expanding $(16-2x)(10-2x)$:
$$(16-2x)(10-2x)=160-32x-20x+4x^2=4x^2-52x+160$$
Multiplying by $x$:
$$\boxed{V(x)=4x^3-52x^2+160x}$$
(b) The height and both base dimensions must be positive, and the tighter restriction is $10-2x>0$:
$$\boxed{0(c) Graph $y=V(x)$ and $y=100$ on the GDC and find their points of intersection (equivalently, find the roots of $V(x)-100=0$) within $0
QUESTION 26
6 marks
Medium
The cubic function $f(x)=x^3+kx^2-5x-6$ has $(x+1)$ as a factor.
(a) Use the factor theorem to find the value of $k$. [3]
(b) Hence find the other two roots of $f(x)$. [3]
Show complete worked solution
(a) Since $(x+1)$ is a factor, $f(-1)=0$ (factor theorem). Substituting $x=-1$ into $f(x)=x^3+kx^2-5x-6$:
$$f(-1)=(-1)^3+k(-1)^2-5(-1)-6$$
$$f(-1)=-1+k+5-6$$
$$f(-1)=k-2$$
Setting this equal to zero:
$$k-2=0$$
$$\boxed{k=2}$$
(b) With $k=2$, $f(x)=x^3+2x^2-5x-6$. Dividing by the known factor $(x+1)$:
$$f(x)=(x+1)(x^2+x-6)$$
Factorising the quadratic:
$$x^2+x-6=(x+3)(x-2)$$
So $f(x)=(x+1)(x+3)(x-2)$, giving roots at $x=-1$, $x=-3$, $x=2$. The remaining two roots (other than the given $x=-1$) are
$$\boxed{x=-3 \text{ and } x=2}$$
QUESTION 27
7 marks
Medium
The number of fish (in a stocked pond), $t$ years after the pond is first stocked, is modelled by
$$P(t)=t^3-9t^2+15t+200, \quad 0\le t\le 8$$
(a) State the initial number of fish in the pond, $P(0)$. [1]
(b) Use your GDC to find the coordinates of the local maximum and local minimum points of $P(t)$, and interpret each in context. [4]
(c) Find $P(8)$ and describe the overall trend in the fish population between $t=5$ and $t=8$. [2]
Show complete worked solution
(a) $$\boxed{P(0)=200 \text{ fish}}$$
(b) Graphing $y=P(t)$ on the GDC and using the maximum- and minimum-finding features on $0\le t\le 8$:
$$\text{local maximum at } (1,\ 207) \qquad \text{local minimum at } (5,\ 175)$$
$$\boxed{\text{local max }(1,207); \text{ local min }(5,175)}$$
This means the fish population rises quickly from $200$ to a peak of $207$ after about $1$ year, then declines to a low of $175$ fish after $5$ years, before beginning to recover.
(c) Substitute $t=8$ into $P(t)=t^3-9t^2+15t+200$:
$$P(8)=(8)^3-9(8)^2+15(8)+200$$
$$P(8)=512-576+120+200=256$$
$$\boxed{P(8)=256 \text{ fish}}$$
Since $P(5)=175$ and $P(8)=256$, the fish population is increasing (recovering) over the interval $5\le t\le 8$, rising by $81$ fish over the $3$ years.
QUESTION 28
7 marks
Medium
The graph below shows a company's weekly revenue, $R(x)$, in thousands of dollars, as a cubic function of the number of items produced, $x$ hundred, for $0\le x\le 10$.
(a) Write down the values of $x$ at which the revenue is zero. [2]
(b) Write down the coordinates of the local maximum point shown on the graph, and state what it represents. [2]
(c) Based on the graph, state the interval of $x$-values, other than the endpoints, for which the company generates a positive revenue. [3]
Show complete worked solution
(a) Reading the $x$-intercepts directly from the graph:
$$\boxed{x=0,\ x=6,\ x=10}$$
(b) Reading the labelled local maximum point directly from the graph:
$$\boxed{(8.24,\ 32.5)}$$
This represents the production level ($824$ items) at which the model predicts the maximum weekly revenue (\$32,500) is generated.
(c) The graph lies above the $x$-axis (i.e. $R(x)>0$) between the second and third $x$-intercepts:
$$\boxed{6
QUESTION 29
6 marks
Medium
Consider the cubic function $f(x)=x^3-3x^2-4x+12$. It is given that $(x-2)$ is a factor of $f(x)$.
(a) Divide $f(x)$ by $(x-2)$ to find the remaining quadratic factor, and hence write down all three roots of $f(x)$. [4]
(b) Describe the behaviour of $f(x)$ as $x\to+\infty$ and as $x\to-\infty$. [2]
Show complete worked solution
(a) Dividing $f(x)=x^3-3x^2-4x+12$ by $(x-2)$: matching coefficients with $(x-2)(x^2+bx+c)=x^3+(b-2)x^2+(c-2b)x-2c$,
Constant term: $-2c=12 \Rightarrow c=-6$.
$x^2$ coefficient: $b-2=-3 \Rightarrow b=-1$.
Check $x$ coefficient: $c-2b=-6+2=-4$ ✓ (matches)
So $f(x)=(x-2)(x^2-x-6)=(x-2)(x-3)(x+2)$.
$$\boxed{x=-2,\ x=2,\ x=3}$$
(b) The leading coefficient of $f(x)$ is $+1$ (positive), and the degree is odd, so:
$$\boxed{y\to+\infty \text{ as } x\to+\infty; \quad y\to-\infty \text{ as } x\to-\infty}$$
QUESTION 30
8 marks
Medium
Consider the cubic function $f(x)=x^3-4x^2-2x+5$.
(a) Show that $x=1$ is a root of $f(x)$. [2]
(b) Hence write $f(x)$ in the form $(x-1)(x^2+bx+c)$, and use your GDC to find the other two roots of $f(x)$, giving your answers to 3 significant figures. [4]
(c) Describe the behaviour of $f(x)$ as $x\to\pm\infty$. [2]
Show complete worked solution
(a) Substitute $x=1$ into $f(x)=x^3-4x^2-2x+5$:
$$f(1)=(1)^3-4(1)^2-2(1)+5=1-4-2+5=0$$
Since $f(1)=0$, $x=1$ is a root of $f(x)$.
$$\boxed{f(1)=0 \implies x=1 \text{ is a root}}$$
(b) Dividing $f(x)$ by $(x-1)$: matching coefficients with $(x-1)(x^2+bx+c)=x^3+(b-1)x^2+(c-b)x-c$,
Constant term: $-c=5 \Rightarrow c=-5$.
$x^2$ coefficient: $b-1=-4 \Rightarrow b=-3$.
Check $x$ coefficient: $c-b=-5+3=-2$ ✓ (matches)
So $f(x)=(x-1)(x^2-3x-5)$.
Solving $x^2-3x-5=0$ using the GDC's polynomial solver (or the quadratic formula $x=\dfrac{3\pm\sqrt{9+20}}{2}$):
$$\boxed{x\approx -1.19 \text{ and } x\approx 4.19}$$
(c) The leading coefficient is $+1$ (positive) and the degree is odd:
$$\boxed{y\to+\infty \text{ as } x\to+\infty; \quad y\to-\infty \text{ as } x\to-\infty}$$
QUESTION 31
8 marks
Medium
A start-up sells a gadget for a fixed price, giving weekly revenue $R(x)=40x$ (in dollars), where $x$ is the number of gadgets sold. Its weekly production cost, in dollars, is modelled by
$$C(x)=x^3-9x^2+40x+20$$
(a) Show that the weekly profit is given by $\text{Profit}(x)=-x^3+9x^2-20$. [3]
(b) Find the profit when $x=4$. [2]
(c) Use your GDC to find the local maximum point of $\text{Profit}(x)$ for $x\ge0$, and state the maximum weekly profit predicted by the model. [3]
Show complete worked solution
(a) Profit is revenue minus cost:
$$\text{Profit}(x)=R(x)-C(x)=40x-(x^3-9x^2+40x+20)$$
$$\text{Profit}(x)=40x-x^3+9x^2-40x-20$$
$$\boxed{\text{Profit}(x)=-x^3+9x^2-20}$$
(b) Substitute $x=4$:
$$\text{Profit}(4)=-(4)^3+9(4)^2-20=-64+144-20=60$$
$$\boxed{\text{Profit}(4)=\$60}$$
(c) Using the GDC's maximum-finding feature on the graph of $\text{Profit}(x)$ for $x\ge0$:
$$\text{local maximum at } (6,\ 88)$$
$$\boxed{\text{maximum weekly profit}=\$88\text{, at } x=6 \text{ gadgets sold}}$$
QUESTION 32
7 marks
Medium
The concentration of a drug in a patient's bloodstream, in mg/L, $t$ hours after it is administered, is modelled by
$$C(t)=-t^3+9t^2, \quad 0\le t\le 9$$
(a) Find $C(3)$. [2]
(b) Use your GDC to find the maximum concentration reached, and the time at which it occurs. [3]
(c) Verify that $C(9)=0$, and state what this means in context. [2]
Show complete worked solution
(a) Substitute $t=3$ into $C(t)=-t^3+9t^2$:
$$C(3)=-(3)^3+9(3)^2=-27+81=54$$
$$\boxed{C(3)=54 \text{ mg/L}}$$
(b) Using the GDC's maximum-finding feature on $y=C(t)$:
$$\text{local maximum at } (6,\ 108)$$
$$\boxed{\text{maximum concentration}=108 \text{ mg/L, occurring at } t=6 \text{ hours}}$$
(c) Substitute $t=9$:
$$C(9)=-(9)^3+9(9)^2=-729+729=0$$
$$\boxed{C(9)=0}$$
This means that, according to the model, the drug has been completely eliminated from the bloodstream $9$ hours after it was administered.
QUESTION 33
7 marks
Medium
The number of bacteria (in millions) in a laboratory culture, $t$ hours after the culture is started, is modelled by
$$N(t)=t^3-6t^2+9t+4, \quad t\ge 0$$
(a) State the initial number of bacteria, $N(0)$. [1]
(b) Use your GDC to find the coordinates of the local maximum and local minimum points of $N(t)$, and interpret each in context. [4]
(c) Find $N(5)$. [2]
Show complete worked solution
(a) $$\boxed{N(0)=4 \text{ million bacteria}}$$
(b) Graphing $y=N(t)$ on the GDC and using the maximum- and minimum-finding features:
$$\text{local maximum at } (1,\ 8) \qquad \text{local minimum at } (3,\ 4)$$
$$\boxed{\text{local max }(1,8); \text{ local min }(3,4)}$$
This means the population grows from $4$ million to a peak of $8$ million after $1$ hour, then falls back to a low of $4$ million after $3$ hours, before growing again.
(c) Substitute $t=5$ into $N(t)=t^3-6t^2+9t+4$:
$$N(5)=(5)^3-6(5)^2+9(5)+4$$
$$N(5)=125-150+45+4=24$$
$$\boxed{N(5)=24 \text{ million bacteria}}$$
QUESTION 34
7 marks
Medium
The graph of the cubic function $y=x^3+x^2-6x$ is shown below.
(a) Write down the three $x$-intercepts shown on the graph. [2]
(b) Using the graph, write down the intervals of $x$ for which $y>0$. [3]
(c) Describe the behaviour of $y$ as $x\to+\infty$ and as $x\to-\infty$. [2]
Show complete worked solution
(a) Reading the $x$-intercepts directly from the graph:
$$\boxed{x=-3,\ x=0,\ x=2}$$
(b) The graph lies above the $x$-axis (i.e. $y>0$) between the first and second intercepts, and again beyond the third intercept:
$$\boxed{-32}$$
(c) The leading coefficient of $y=x^3+x^2-6x$ is $+1$ (positive), so:
$$\boxed{y\to+\infty \text{ as } x\to+\infty; \quad y\to-\infty \text{ as } x\to-\infty}$$
QUESTION 35
8 marks
Medium
The price of a small company's stock, in dollars, $t$ months after it is first listed, is modelled by
$$S(t)=0.5t^3-6t^2+18t+50, \quad 0\le t\le 10$$
(a) Find $S(4)$. [2]
(b) Use your GDC to find the coordinates of the local maximum and local minimum points of $S(t)$, and interpret each in context. [4]
(c) By comparing $S(0)$ and $S(10)$, state whether the stock price is higher at the start or at the end of the $10$-month period, and by how much. [2]
Show complete worked solution
(a) Substitute $t=4$ into $S(t)=0.5t^3-6t^2+18t+50$:
$$S(4)=0.5(4)^3-6(4)^2+18(4)+50$$
$$S(4)=0.5(64)-6(16)+72+50$$
$$S(4)=32-96+72+50=58$$
$$\boxed{S(4)=\$58}$$
(b) Using the GDC's maximum- and minimum-finding features on $y=S(t)$:
$$\text{local maximum at } (2,\ 66) \qquad \text{local minimum at } (6,\ 50)$$
$$\boxed{\text{local max }(2,66); \text{ local min }(6,50)}$$
This means the stock price rises from \$50 to a peak of \$66 after $2$ months, then falls back to \$50 after $6$ months, before rising again.
(c) $$S(0)=50, \qquad S(10)=0.5(1000)-6(100)+180+50=500-600+180+50=130$$
$$\boxed{S(10)-S(0)=130-50=80}$$
The stock price is \$80 higher at the end of the $10$-month period than at the start.
QUESTION 36
7 marks
Medium
A cubic function has roots at $x=-1$, $x=2$ and $x=4$, and its graph passes through the point $(1,12)$.
(a) Write the function in the form $y=k(x+1)(x-2)(x-4)$ and use the given point to find the value of $k$. [4]
(b) Hence find the $y$-intercept of the graph. [3]
Show complete worked solution
(a) Substituting the point $(1,12)$ into $y=k(x+1)(x-2)(x-4)$:
$$12=k(1+1)(1-2)(1-4)$$
$$12=k(2)(-1)(-3)$$
$$12=6k$$
$$\boxed{k=2}$$
So the function is $y=2(x+1)(x-2)(x-4)$.
(b) Substitute $x=0$ into $y=2(x+1)(x-2)(x-4)$:
$$y=2(0+1)(0-2)(0-4)$$
$$y=2(1)(-2)(-4)$$
$$y=2\times 8=16$$
$$\boxed{y\text{-intercept}=(0,16)}$$
QUESTION 37
6 marks
Medium
The graph of $y=x^3-16x$ is shown below, representing the height of the water surface, $y$ cm relative to the calm-water level, in a wave pool at a horizontal distance $x$ m from the wave generator ($-4\le x\le 4$).
(a) Write down the $x$-intercepts and the $y$-intercept shown on the graph. [2]
(b) Write down the coordinates of the two turning points shown on the graph. [2]
(c) Describe the symmetry of the graph about the origin. [2]
Show complete worked solution
(a) Reading directly from the graph, the $x$-intercepts are $x=-4,\ 0,\ 4$, and the $y$-intercept is at $x=0$ (which coincides with one of the $x$-intercepts here):
$$\boxed{x\text{-intercepts: } -4,\ 0,\ 4; \quad y\text{-intercept}: (0,0)}$$
(b) Reading the labelled turning points directly from the graph:
$$\boxed{\text{local maximum }(-2.31,\ 24.6); \quad \text{local minimum }(2.31,\ -24.6)}$$
(c) The graph has point symmetry (rotational symmetry of order $2$) about the origin: rotating the curve $180^{\circ}$ about $(0,0)$ maps it onto itself. This is because $y=x^3-16x$ contains only odd powers of $x$, so $f(-x)=-f(x)$ for every point on the curve.
QUESTION 38
8 marks
Medium
The temperature deviation from a reference level, in degrees Celsius, inside a greenhouse, $x$ hours after midnight, is modelled by
$$T(x)=x^3-3x^2-9x+5, \quad -3\le x\le 5$$
(a) Find $T(2)$. [2]
(b) Use your GDC to find the coordinates of the local maximum and local minimum points of $T(x)$, and interpret each in context. [3]
(c) Use your GDC to find the roots of $T(x)=0$ to 3 significant figures, and interpret what these values represent. [3]
Show complete worked solution
(a) Substitute $x=2$ into $T(x)=x^3-3x^2-9x+5$:
$$T(2)=(2)^3-3(2)^2-9(2)+5$$
$$T(2)=8-12-18+5=-17$$
$$\boxed{T(2)=-17^{\circ}\text{C (relative to the reference level)}}$$
(b) Using the GDC's maximum- and minimum-finding features on $y=T(x)$:
$$\text{local maximum at } (-1,\ 10) \qquad \text{local minimum at } (3,\ -22)$$
$$\boxed{\text{local max }(-1,10); \text{ local min }(3,-22)}$$
This means the temperature deviation reaches a peak of $10^{\circ}$C above the reference level at $x=-1$ (i.e. $11$ hours before midnight, or equivalently at 11pm the previous day), and reaches its lowest point of $22^{\circ}$C below the reference level at $x=3$ (3am).
(c) Using the GDC's root-finding feature on $y=T(x)$:
$$\boxed{x\approx -2.18,\ x\approx 0.489,\ x\approx 4.69}$$
These are the times at which the temperature inside the greenhouse is exactly equal to the reference level (the deviation is zero).
QUESTION 39
8 marks
Medium
Consider the cubic function $f(x)=2x^3-5x^2-4x+3$.
(a) Use your GDC to find the three roots of $f(x)$. [3]
(b) Use your GDC to find the coordinates of the local maximum and local minimum points of $f(x)$. [3]
(c) Describe the behaviour of $f(x)$ as $x\to+\infty$ and as $x\to-\infty$. [2]
Show complete worked solution
(a) Graphing $y=f(x)$ on the GDC and using the root/zero-finding feature:
$$\boxed{x=-1,\ x=0.5,\ x=3}$$
(b) Using the GDC's maximum- and minimum-finding features:
$$\text{local maximum at } (-0.333,\ 3.70) \qquad \text{local minimum at } (2,\ -9)$$
$$\boxed{\text{local max }(-0.333,3.70); \text{ local min }(2,-9)}$$
(c) The leading coefficient of $f(x)=2x^3-5x^2-4x+3$ is $+2$ (positive), so:
$$\boxed{y\to+\infty \text{ as } x\to+\infty; \quad y\to-\infty \text{ as } x\to-\infty}$$
QUESTION 40
8 marks
Medium
Recall from an earlier question the open box formed by cutting squares of side $x$ cm from a $30$ cm by $20$ cm sheet, with volume
$$V(x)=4x^3-100x^2+600x, \quad 0(a) Find $V(2)$ and $V(6)$, and state which of these two values of $x$ gives the larger volume. [3]
(b) Use your GDC to find the value of $x$ that maximises $V(x)$ on $0(c) State whether either of the values of $x$ used in part (a) gives the true maximum volume. [2]
Show complete worked solution
(a) Substituting $x=2$:
$$V(2)=4(2)^3-100(2)^2+600(2)=32-400+1200=832$$
Substituting $x=6$:
$$V(6)=4(6)^3-100(6)^2+600(6)=864-3600+3600=864$$
$$\boxed{V(2)=832 \text{ cm}^3, \quad V(6)=864 \text{ cm}^3 \implies x=6 \text{ gives the larger volume of the two}}$$
(b) Using the GDC's maximum-finding feature on $y=V(x)$ for $0(c) Neither $x=2$ nor $x=6$ gives the true maximum. The true maximum volume of about $1056\text{ cm}^3$, found using the GDC in part (b), occurs at $x\approx 3.92$ cm, which is larger than both $V(2)=832\text{ cm}^3$ and $V(6)=864\text{ cm}^3$ found in part (a).
$$\boxed{\text{No -- the true maximum } (\approx1056\text{ cm}^3 \text{ at } x\approx3.92) \text{ exceeds both } V(2) \text{ and } V(6)}$$
QUESTION 41
11 marks
Hard
An open-topped storage box is to be manufactured from a rectangular sheet of thin metal measuring $24$ cm by $15$ cm, by cutting a square of side $x$ cm from each corner and folding up the four resulting flaps, as shown below.
(a) Show that the volume of the box, in $\text{cm}^3$, is given by
$$V(x)=4x^3-78x^2+360x$$ [3]
(b) Write down the domain of $x$ for which the box is physically valid. [2]
(c) Use your GDC to find the value of $x$, within the domain from part (b), that maximises the volume, and state this maximum volume. [3]
(d) State the dimensions (length, width and height) of the box when its volume is maximised. [3]
Show complete worked solution
(a) After the corner squares are removed, the base has dimensions $(24-2x)$ by $(15-2x)$, and the height is $x$, so
$$V(x)=x(24-2x)(15-2x)$$
Expanding $(24-2x)(15-2x)$:
$$(24-2x)(15-2x)=360-48x-30x+4x^2=4x^2-78x+360$$
Multiplying by $x$:
$$\boxed{V(x)=4x^3-78x^2+360x}$$
(b) Each dimension must be positive; the more restrictive condition is $15-2x>0$, i.e. $x<7.5$:
$$\boxed{0(c) Graphing $y=V(x)$ on the GDC (restricted to $0(d) With $x=3$:
$$\text{length}=24-2(3)=18 \text{ cm}, \quad \text{width}=15-2(3)=9 \text{ cm}, \quad \text{height}=3 \text{ cm}$$
Check: $18\times 9\times 3=486\text{ cm}^3$ ✓ (matches part (c))
$$\boxed{18\text{ cm}\times 9\text{ cm}\times 3\text{ cm}}$$
QUESTION 42
11 marks
Hard
Consider the cubic function $f(x)=x^3-6x^2+kx-8$, where $k$ is a constant. It is given that $x=4$ is a root of $f(x)$.
(a) Find the value of $k$. [3]
(b) Hence write $f(x)$ as a product of $(x-4)$ and a quadratic factor, and find the other two roots of $f(x)$. [3]
(c) Use your GDC to find the coordinates of the local maximum and local minimum points of $f(x)$, giving your answers to 3 significant figures. [3]
(d) Describe the behaviour of $f(x)$ as $x\to+\infty$ and as $x\to-\infty$. [2]
Show complete worked solution
(a) Since $x=4$ is a root, $f(4)=0$. Substitute $x=4$ into $f(x)=x^3-6x^2+kx-8$:
$$f(4)=(4)^3-6(4)^2+k(4)-8$$
$$f(4)=64-96+4k-8$$
$$f(4)=4k-40$$
Setting this equal to zero:
$$4k-40=0$$
$$\boxed{k=10}$$
(b) With $k=10$, $f(x)=x^3-6x^2+10x-8$. Dividing by $(x-4)$: matching coefficients with $(x-4)(x^2+bx+c)=x^3+(b-4)x^2+(c-4b)x-4c$,
Constant term: $-4c=-8 \Rightarrow c=2$.
$x^2$ coefficient: $b-4=-6 \Rightarrow b=-2$.
Check $x$ coefficient: $c-4b=2+8=10$ ✓ (matches)
So $f(x)=(x-4)(x^2-2x+2)$. The discriminant of $x^2-2x+2$ is $(-2)^2-4(1)(2)=4-8=-4<0$, so this quadratic has no real roots.
$$\boxed{x=4 \text{ is the only real root of } f(x)}$$
(c) Graphing $y=f(x)$ on the GDC and using the maximum- and minimum-finding features:
$$\text{local maximum at } (1.18,\ -2.91) \qquad \text{local minimum at } (2.82,\ -5.09)$$
$$\boxed{\text{local max }(1.18,-2.91); \text{ local min }(2.82,-5.09)}$$
(d) The leading coefficient of $f(x)$ is $+1$ (positive), so:
$$\boxed{y\to+\infty \text{ as } x\to+\infty; \quad y\to-\infty \text{ as } x\to-\infty}$$
QUESTION 43
12 marks
Hard
The clearance height beneath a footbridge, relative to a minimum safety threshold, $h(x)$ metres, at a horizontal distance $x$ metres from the left-hand support, is modelled by
$$h(x)=x^3-x^2-8x+12, \quad -3\le x \le 5$$
(a) Show that $x=2$ is a root of $h(x)$, and by evaluating $h(1.9)$ and $h(2.1)$, explain why the graph touches the $x$-axis at $x=2$ rather than crossing it. [4]
(b) Find the other root of $h(x)$. [3]
(c) Use your GDC to find the coordinates of the local maximum and local minimum points of $h(x)$. [3]
(d) Using your answers above, state the values of $x$ (within the given domain) for which the clearance falls below the safety threshold (i.e. $h(x)<0$). [2]
Show complete worked solution
(a) Substitute $x=2$ into $h(x)=x^3-x^2-8x+12$:
$$h(2)=(2)^3-(2)^2-8(2)+12=8-4-16+12=0$$
Since $h(2)=0$, $x=2$ is a root.
Now evaluate $h(x)$ just either side of $x=2$:
$$h(1.9)=(1.9)^3-(1.9)^2-8(1.9)+12=0.049$$
$$h(2.1)=(2.1)^3-(2.1)^2-8(2.1)+12=0.051$$
Both $h(1.9)$ and $h(2.1)$ are positive (very close to, but not below, zero). Since $h(x)$ does not change sign either side of $x=2$ (it is positive on both sides), the graph only touches the $x$-axis at $x=2$ and turns back upward, rather than crossing through it.
$$\boxed{h(2)=0,\ h(1.9)>0,\ h(2.1)>0 \implies \text{the graph touches (does not cross) the axis at } x=2}$$
(b) Since $x=2$ is a repeated root, $(x-2)^2$ is a factor of $h(x)$. Dividing $h(x)=x^3-x^2-8x+12$ by $(x-2)^2=x^2-4x+4$:
$$h(x)=(x^2-4x+4)(x+3)=(x-2)^2(x+3)$$
(This can be checked by expanding: $(x-2)^2(x+3)=(x^2-4x+4)(x+3)=x^3+3x^2-4x^2-12x+4x+12=x^3-x^2-8x+12$ ✓)
The other root is where $x+3=0$:
$$\boxed{x=-3}$$
(c) Graphing $y=h(x)$ on the GDC and using the maximum- and minimum-finding features:
$$\text{local maximum at } (-1.33,\ 18.5) \qquad \text{local minimum at } (2,\ 0)$$
$$\boxed{\text{local max }(-1.33,18.5); \text{ local min }(2,0)}$$
(Note the local minimum coincides exactly with the repeated root found in part (a), confirming that the curve touches the $x$-axis there.)
(d) Since $h(x)$ has roots only at $x=-3$ and $x=2$ (a touching point, not a crossing), and the graph is above the axis everywhere else in the domain (rising to a maximum of $18.5$ then dipping only down to $0$ at $x=2$ before rising again), the clearance $h(x)$ is never negative anywhere in $-3\le x\le 5$.
$$\boxed{h(x)\ge 0 \text{ throughout } -3\le x\le5 \text{ -- the clearance never falls below the safety threshold}}$$
QUESTION 44
11 marks
Hard
The height of a hot air balloon relative to a viewing platform, $h(t)$ metres (positive values are above the platform, negative values below), $t$ hours after launch, is modelled by
$$h(t)=t^3-12t^2+36t-20, \quad 0\le t\le 8$$
(a) Find $h(1)$ and $h(0)$, and interpret both values. [3]
(b) Use your GDC to find the roots of $h(t)=0$ to 3 significant figures, and interpret what these values represent. [3]
(c) Use your GDC to find the coordinates of the local maximum and local minimum points of $h(t)$, and interpret each in context. [3]
(d) Find $h(8)$ and hence describe the overall journey of the balloon between $t=6$ and $t=8$. [2]
Show complete worked solution
(a) Substitute $t=1$ into $h(t)=t^3-12t^2+36t-20$:
$$h(1)=(1)^3-12(1)^2+36(1)-20=1-12+36-20=5$$
Substitute $t=0$:
$$h(0)=(0)^3-12(0)^2+36(0)-20=-20$$
$$\boxed{h(1)=5 \text{ m}, \quad h(0)=-20 \text{ m}}$$
At launch ($t=0$), the balloon starts $20$ m below the platform level. After $1$ hour, it has risen to $5$ m above the platform.
(b) Graphing $y=h(t)$ on the GDC and using the root/zero-finding feature:
$$\boxed{t\approx 0.716,\ t\approx 3.66,\ t\approx 7.62}$$
These are the times (in hours) at which the balloon is exactly level with the viewing platform.
(c) Using the GDC's maximum- and minimum-finding features:
$$\text{local maximum at } (2,\ 12) \qquad \text{local minimum at } (6,\ -20)$$
$$\boxed{\text{local max }(2,12); \text{ local min }(6,-20)}$$
This means the balloon first climbs to a height of $12$ m above the platform after $2$ hours, then descends to $20$ m below the platform after $6$ hours, before climbing again.
(d) Substitute $t=8$ into $h(t)=t^3-12t^2+36t-20$:
$$h(8)=(8)^3-12(8)^2+36(8)-20=512-768+288-20=12$$
$$\boxed{h(8)=12 \text{ m}}$$
Between $t=6$ and $t=8$, the balloon rises from its lowest point of $-20$ m (below the platform) up to $12$ m above the platform, a net climb of $32$ m over $2$ hours, passing through the platform level ($h=0$) at $t\approx 7.62$ hours along the way.
QUESTION 45
11 marks
Hard
A cubic function is of the form $f(x)=x^3+bx^2+cx+6$, where $b$ and $c$ are constants. It is known that $x=-1$ and $x=2$ are both roots of $f(x)$.
(a) Using the two given roots, form a system of two linear equations in $b$ and $c$. [3]
(b) Solve the system (by hand, or using your GDC's simultaneous equation solver) to find the values of $b$ and $c$. [2]
(c) Hence find the third root of $f(x)$. [3]
(d) Use your GDC to find the coordinates of the local maximum and local minimum points of $f(x)$, giving your answers to 3 significant figures. [3]
Show complete worked solution
(a) Since $x=-1$ is a root, $f(-1)=0$:
$$(-1)^3+b(-1)^2+c(-1)+6=0$$
$$-1+b-c+6=0$$
$$b-c=-5 \quad \text{...(1)}$$
Since $x=2$ is a root, $f(2)=0$:
$$(2)^3+b(2)^2+c(2)+6=0$$
$$8+4b+2c+6=0$$
$$4b+2c=-14 \implies 2b+c=-7 \quad \text{...(2)}$$
$$\boxed{\begin{aligned}b-c&=-5\\2b+c&=-7\end{aligned}}$$
(b) Adding equations (1) and (2) eliminates $c$:
$$(b-c)+(2b+c)=-5+(-7)$$
$$3b=-12$$
$$b=-4$$
Substituting into (1): $-4-c=-5 \Rightarrow c=1$.
$$\boxed{b=-4,\ c=1}$$
(c) So $f(x)=x^3-4x^2+x+6$. Since $x=-1$ and $x=2$ are roots, $(x+1)(x-2)=x^2-x-2$ is a factor. Dividing $f(x)$ by $x^2-x-2$:
$$f(x)=(x^2-x-2)(x-3)$$
(Check: expanding $(x^2-x-2)(x-3)=x^3-3x^2-x^2+3x-2x+6=x^3-4x^2+x+6$ ✓)
$$\boxed{x=3}$$
(d) Graphing $y=f(x)=x^3-4x^2+x+6$ on the GDC and using the maximum- and minimum-finding features:
$$\text{local maximum at } (0.131,\ 6.06) \qquad \text{local minimum at } (2.54,\ -0.879)$$
$$\boxed{\text{local max }(0.131,6.06); \text{ local min }(2.54,-0.879)}$$
QUESTION 46
12 marks
Hard
The graph of a cubic function $y=f(x)$ is shown below. It has $x$-intercepts at $x=-3,\ x=1,\ x=4$, and its $y$-intercept is at $(0,-12)$.
(a) Explain why the graph shows that the leading coefficient of $f(x)$ is negative. [2]
(b) Write $f(x)$ in factored form as $f(x)=k(x+3)(x-1)(x-4)$, and use the given $y$-intercept to find the value of $k$. [4]
(c) Expand your answer to part (b) to express $f(x)$ in the form $ax^3+bx^2+cx+d$. [3]
(d) Use your GDC to find the coordinates of the local maximum and local minimum points of $f(x)$. [3]
Show complete worked solution
(a) The graph rises to the left and falls to the right (i.e. $y\to-\infty$ as $x\to+\infty$, and $y\to+\infty$ as $x\to-\infty$). This "top-left to bottom-right" behaviour only occurs for a cubic whose leading coefficient is negative.
$$\boxed{\text{leading coefficient} < 0}$$
(b) Substituting the $y$-intercept $(0,-12)$ into $f(x)=k(x+3)(x-1)(x-4)$:
$$-12=k(0+3)(0-1)(0-4)$$
$$-12=k(3)(-1)(-4)$$
$$-12=12k$$
$$\boxed{k=-1}$$
So $f(x)=-(x+3)(x-1)(x-4)$, which has a negative leading coefficient, consistent with part (a).
(c) First expand $(x+3)(x-1)(x-4)$. Multiply the last two factors:
$$(x-1)(x-4)=x^2-5x+4$$
Multiply by $(x+3)$:
$$(x+3)(x^2-5x+4)=x^3-5x^2+4x+3x^2-15x+12=x^3-2x^2-11x+12$$
Now apply the factor of $k=-1$:
$$\boxed{f(x)=-x^3+2x^2+11x-12}$$
(d) Graphing $y=f(x)$ on the GDC and using the maximum- and minimum-finding features:
$$\text{local maximum at } (2.69,\ 12.6) \qquad \text{local minimum at } (-1.36,\ -20.7)$$
$$\boxed{\text{local max }(2.69,12.6); \text{ local min }(-1.36,-20.7)}$$
QUESTION 47
10 marks
Hard
The weekly cost, in dollars, of running a small production line at $x$ (hundred units of) output is modelled by
$$\text{Cost}(x)=x^3-12x^2+45x+500, \quad 0\le x\le 8$$
(a) Find $\text{Cost}(2)$. [2]
(b) Use your GDC to find the coordinates of the local maximum and local minimum points of $\text{Cost}(x)$. [3]
(c) By also evaluating $\text{Cost}(0)$ and $\text{Cost}(8)$, determine the output level $x$, for $0\le x\le 8$, at which the weekly cost is lowest, and state this minimum cost. [3]
(d) Interpret your result from part (c) in context. [2]
Show complete worked solution
(a) Substitute $x=2$ into $\text{Cost}(x)=x^3-12x^2+45x+500$:
$$\text{Cost}(2)=(2)^3-12(2)^2+45(2)+500$$
$$\text{Cost}(2)=8-48+90+500=550$$
$$\boxed{\text{Cost}(2)=\$550}$$
(b) Using the GDC's maximum- and minimum-finding features on $y=\text{Cost}(x)$:
$$\text{local maximum at } (3,\ 554) \qquad \text{local minimum at } (5,\ 550)$$
$$\boxed{\text{local max }(3,554); \text{ local min }(5,550)}$$
(c) Evaluating the domain endpoints:
$$\text{Cost}(0)=500, \qquad \text{Cost}(8)=(8)^3-12(8)^2+45(8)+500=512-768+360+500=604$$
Comparing all four candidate points -- the two endpoints and the two turning points -- for the lowest cost:
$$\text{Cost}(0)=500,\quad \text{Cost}(3)=554,\quad \text{Cost}(5)=550,\quad \text{Cost}(8)=604$$
The smallest of these is $\text{Cost}(0)=500$.
$$\boxed{\text{minimum weekly cost} = \$500 \text{, occurring at } x=0}$$
(d) Although $x=5$ is a local minimum of the cubic model (a "dip" in the cost curve), the overall lowest weekly cost on the given domain actually occurs at the left-hand endpoint $x=0$ (producing no units), where the fixed costs alone total \$500. This shows that a local minimum found using the GDC is not always the true minimum over a restricted domain -- the domain endpoints must always be checked too.
QUESTION 48
10 marks
Hard
The number of people (in hundreds) inside a festival venue, $x$ hours after midnight, is modelled by
$$N(x)=-x^3+15x^2-36x$$
valid only for $3\le x\le 12$ (the model is not considered reliable outside this window).
(a) Find $N(5)$ and interpret your answer. [2]
(b) Show that $x=3$ and $x=12$ are both roots of $N(x)$, and interpret what this means in context. [3]
(c) Use your GDC to find the coordinates of the local maximum point of $N(x)$ within the domain $3\le x\le 12$, and interpret it in context. [3]
(d) Given that $N(1)=-22$, explain why the domain of the model is restricted to $3\le x\le 12$ rather than $x\ge 0$. [2]
Show complete worked solution
(a) Substitute $x=5$ into $N(x)=-x^3+15x^2-36x$:
$$N(5)=-(5)^3+15(5)^2-36(5)$$
$$N(5)=-125+375-180=70$$
$$\boxed{N(5)=70 \text{ (i.e. } 7000 \text{ people at } 5\text{am)}}$$
(b) Substitute $x=3$:
$$N(3)=-(3)^3+15(3)^2-36(3)=-27+135-108=0$$
Substitute $x=12$:
$$N(12)=-(12)^3+15(12)^2-36(12)=-1728+2160-432=0$$
$$\boxed{N(3)=0 \text{ and } N(12)=0}$$
This means that, according to the model, there are effectively no attendees present at $x=3$ (3am) and again at $x=12$ (noon) -- these mark the boundaries of the meaningful attendance window.
(c) Graphing $y=N(x)$ on the GDC for $3\le x\le 12$ and using the maximum-finding feature:
$$\text{local maximum at } (8.61,\ 164)$$
$$\boxed{\text{peak attendance} \approx 16{,}400 \text{ people, at } x\approx 8.61 \text{ (about 8:37am)}}$$
(d) Substituting $x=1$ gives $N(1)=-1+15-36=-22$, a negative value. Since a number of attendees cannot be negative, the model $N(x)$ does not give a sensible (physically meaningful) result for $x<3$, which is why its domain is restricted to $3\le x\le 12$, where $N(x)\ge 0$ throughout.
$$\boxed{N(1)=-22<0 \text{ is not physically possible, so the model is restricted to } 3\le x\le12}$$
QUESTION 49
11 marks
Hard
The size of a protected owl population, relative to its 2015 baseline count, $t$ years after 2015, is modelled by
$$P(t)=-t^3+12t^2-21t, \quad 0\le t\le 10$$
(positive values of $P(t)$ mean the population is above the 2015 baseline; negative values mean it is below).
(a) Find $P(3)$ and interpret your answer. [2]
(b) Use your GDC to find the non-zero roots of $P(t)=0$ to 3 significant figures, and interpret what they represent. [3]
(c) Use your GDC to find the coordinates of the local minimum and local maximum points of $P(t)$, and interpret each in context. [3]
(d) Find $P(10)$ and $P(0)$, and hence describe the long-term trend in the population over the full $10$-year period. [3]
Show complete worked solution
(a) Substitute $t=3$ into $P(t)=-t^3+12t^2-21t$:
$$P(3)=-(3)^3+12(3)^2-21(3)$$
$$P(3)=-27+108-63=18$$
$$\boxed{P(3)=18}$$
This means that $3$ years after 2015, the owl population was $18$ above its 2015 baseline count.
(b) Graphing $y=P(t)$ on the GDC and using the root/zero-finding feature (excluding the obvious root at $t=0$):
$$\boxed{t\approx 2.13 \text{ and } t\approx 9.87}$$
These are the times (other than $t=0$) at which the owl population returned to exactly its 2015 baseline level.
(c) Using the GDC's minimum- and maximum-finding features:
$$\text{local minimum at } (1,\ -10) \qquad \text{local maximum at } (7,\ 98)$$
$$\boxed{\text{local min }(1,-10); \text{ local max }(7,98)}$$
This means the population first dropped to $10$ below the baseline after $1$ year, then recovered and grew to a peak of $98$ above the baseline after $7$ years.
(d) $$P(0)=0, \qquad P(10)=-(10)^3+12(10)^2-21(10)=-1000+1200-210=-10$$
$$\boxed{P(0)=0, \quad P(10)=-10}$$
Over the full $10$-year period, the population started at the baseline, dipped slightly below it in year $1$, recovered strongly to a substantial peak $98$ above baseline by year $7$, and then declined again, ending the period $10$ below the original 2015 baseline -- an overall long-term decline despite the mid-period peak.
QUESTION 50
12 marks
Hard
Two competing designs for an open-topped storage container are made by cutting squares of side $x$ cm from the corners of a sheet of metal, then folding up the sides.
Design A uses a square sheet measuring $20$ cm by $20$ cm. Design B uses a rectangular sheet measuring $22$ cm by $14$ cm.
(a) Show that the volumes of the two designs are given by
$$V_A(x)=4x^3-80x^2+400x, \qquad V_B(x)=4x^3-72x^2+308x$$ [3]
(b) Write down the valid domain of $x$ for each design. [2]
(c) Use your GDC to find the maximum volume achievable by each design, and the value of $x$ at which each maximum occurs, giving your answers to 3 significant figures where necessary. [4]
(d) Hence determine which design achieves the larger maximum volume, and by how much. [3]
Show complete worked solution
(a) For Design A, the base is $(20-2x)$ by $(20-2x)$ and the height is $x$:
$$V_A(x)=x(20-2x)(20-2x)=x(20-2x)^2$$
Expanding $(20-2x)^2=400-80x+4x^2$, then multiplying by $x$:
$$\boxed{V_A(x)=4x^3-80x^2+400x}$$
For Design B, the base is $(22-2x)$ by $(14-2x)$ and the height is $x$:
$$V_B(x)=x(22-2x)(14-2x)$$
Expanding $(22-2x)(14-2x)=308-44x-28x+4x^2=4x^2-72x+308$, then multiplying by $x$:
$$\boxed{V_B(x)=4x^3-72x^2+308x}$$
(b) For Design A, both base dimensions require $20-2x>0$, i.e. $x<10$:
$$\boxed{00$, i.e. $x<7$:
$$\boxed{0(c) Graphing $y=V_A(x)$ on the GDC for $0(d) Comparing the two maximum volumes found in part (c):
$$593-386=207$$
$$\boxed{\text{Design A achieves the larger maximum volume, exceeding Design B's maximum by approximately } 207\text{ cm}^3}$$
Direct & Inverse Variation 50 questions
QUESTION 1
4 marks
Easy
In a physics experiment, a spring is hung vertically and different weights are attached to its end. The extension $e$ (in cm) of the spring varies directly as the force $F$ (in newtons) applied to it, at least while the spring is not stretched beyond its elastic limit.
When a force of $8$ N is applied, the spring extends by $6$ cm.
(a) Write down an equation connecting $e$ and $F$, and find the value of the constant of variation $k$. [2]
(b) Find the extension of the spring when a force of $20$ N is applied. [2]
Show complete worked solution
(a) Since $e$ varies directly as $F$, the relationship has the form
$$e = kF$$
Substituting the given data point $F=8$, $e=6$:
$$6 = k(8) \Rightarrow k = \frac{6}{8} = 0.75$$
$$\boxed{e = 0.75F}$$
(b) Substituting $F=20$ into the model:
$$e = 0.75(20) = 15$$
$$\boxed{e = 15 \text{ cm}}$$
QUESTION 2
4 marks
Easy
A craft shop sells decorative ribbon by the metre. The total cost $C$ (in dollars) of a length of ribbon varies directly as the length $L$ (in metres) purchased.
A customer buys $5$ m of ribbon and pays \$12.50.
(a) Find the constant of variation $k$ (the price per metre) and write down the equation connecting $C$ and $L$. [2]
(b) Find the cost of $14$ m of ribbon. [2]
Show complete worked solution
(a) Since $C$ varies directly as $L$:
$$C = kL$$
Using $L=5$, $C=12.5$:
$$12.5 = k(5) \Rightarrow k = \frac{12.5}{5} = 2.5$$
$$\boxed{C = 2.5L}$$
(b) Substituting $L=14$:
$$C = 2.5(14) = 35$$
$$\boxed{C = \$35.00}$$
QUESTION 3
5 marks
Easy
A ball is dropped from the top of a tall building. Air resistance is ignored. The distance $d$ (in metres) the ball has fallen varies directly as the square of the time $t$ (in seconds) since it was released.
After $2$ seconds, the ball has fallen $19.6$ m.
(a) Find the constant of variation $k$ and write down the equation connecting $d$ and $t$. [2]
(b) Find the distance the ball has fallen after $5$ seconds. [3]
Show complete worked solution
(a) Since $d$ varies directly as the square of $t$:
$$d = kt^2$$
Substituting $t=2$, $d=19.6$:
$$19.6 = k(2)^2 = 4k \Rightarrow k = \frac{19.6}{4} = 4.9$$
$$\boxed{d = 4.9t^2}$$
(b) Substituting $t=5$:
$$d = 4.9(5)^2 = 4.9 \times 25 = 122.5$$
$$\boxed{d = 122.5 \text{ m}}$$
QUESTION 4
4 marks
Easy
For an object of fixed mass, its kinetic energy $KE$ (in joules) varies directly as the square of its speed $v$ (in m/s).
When the object travels at $10$ m/s, its kinetic energy is $2500$ J.
(a) Find the constant of variation $k$ and write down the equation connecting $KE$ and $v$. [2]
(b) Find the kinetic energy of the object when it travels at $15$ m/s. [2]
Show complete worked solution
(a) Since $KE$ varies directly as the square of $v$:
$$KE = kv^2$$
Substituting $v=10$, $KE=2500$:
$$2500 = k(10)^2 = 100k \Rightarrow k = \frac{2500}{100} = 25$$
$$\boxed{KE = 25v^2}$$
(b) Substituting $v=15$:
$$KE = 25(15)^2 = 25 \times 225 = 5625$$
$$\boxed{KE = 5625 \text{ J}}$$
QUESTION 5
5 marks
Easy
The power output $P$ (in watts) generated by a small wind turbine varies directly as the cube of the wind speed $v$ (in m/s).
When the wind speed is $8$ m/s, the turbine generates $1024$ W.
(a) Find the constant of variation $k$ and write down the equation connecting $P$ and $v$. [2]
(b) Find the power output of the turbine when the wind speed is $12$ m/s. [3]
Show complete worked solution
(a) Since $P$ varies directly as the cube of $v$:
$$P = kv^3$$
Substituting $v=8$, $P=1024$:
$$1024 = k(8)^3 = 512k \Rightarrow k = \frac{1024}{512} = 2$$
$$\boxed{P = 2v^3}$$
(b) Substituting $v=12$:
$$P = 2(12)^3 = 2 \times 1728 = 3456$$
$$\boxed{P = 3456 \text{ W}}$$
QUESTION 6
5 marks
Easy
A factory manufactures spherical steel ball bearings. The mass $m$ (in grams) of a ball bearing varies directly as the cube of its radius $r$ (in cm), since all ball bearings are made of the same steel.
A ball bearing of radius $2$ cm has a mass of $64$ g.
(a) Find the constant of variation $k$ and write down the equation connecting $m$ and $r$. [2]
(b) Find the mass, in kilograms, of a ball bearing of radius $5$ cm. [3]
Show complete worked solution
(a) Since $m$ varies directly as the cube of $r$:
$$m = kr^3$$
Substituting $r=2$, $m=64$:
$$64 = k(2)^3 = 8k \Rightarrow k = \frac{64}{8} = 8$$
$$\boxed{m = 8r^3}$$
(b) Substituting $r=5$:
$$m = 8(5)^3 = 8 \times 125 = 1000 \text{ g}$$
Converting to kilograms:
$$1000 \text{ g} = 1000 \div 1000 = 1$$
$$\boxed{m = 1 \text{ kg}}$$
QUESTION 7
4 marks
Easy
In shallow water, the speed $v$ (in m/s) of an ocean wave such as a tsunami varies directly as the square root of the water depth $d$ (in metres).
In water of depth $16$ m, a wave travels at $12.4$ m/s.
(a) Find the constant of variation $k$ and write down the equation connecting $v$ and $d$. [2]
(b) Find the speed of the wave in water of depth $81$ m. [2]
Show complete worked solution
(a) Since $v$ varies directly as the square root of $d$:
$$v = k\sqrt{d}$$
Substituting $d=16$, $v=12.4$: note $\sqrt{16}=4$.
$$12.4 = k(4) \Rightarrow k = \frac{12.4}{4} = 3.1$$
$$\boxed{v = 3.1\sqrt{d}}$$
(b) Substituting $d=81$: note $\sqrt{81}=9$.
$$v = 3.1(9) = 27.9$$
$$\boxed{v = 27.9 \text{ m/s}}$$
QUESTION 8
5 marks
Easy
A sailor standing on the deck of a ship can see the horizon. The distance $D$ (in km) to the horizon varies directly as the square root of the sailor's height $h$ (in metres) above sea level.
When the sailor's eyes are $4$ m above sea level, the horizon is $7.2$ km away.
(a) Find the constant of variation $k$ and write down the equation connecting $D$ and $h$. [2]
(b) The sailor climbs to the top of the mast, so that their eyes are $25$ m above sea level. Find the new distance to the horizon. [3]
Show complete worked solution
(a) Since $D$ varies directly as the square root of $h$:
$$D = k\sqrt{h}$$
Substituting $h=4$, $D=7.2$: note $\sqrt{4}=2$.
$$7.2 = k(2) \Rightarrow k = \frac{7.2}{2} = 3.6$$
$$\boxed{D = 3.6\sqrt{h}}$$
(b) Substituting $h=25$: note $\sqrt{25}=5$.
$$D = 3.6(5) = 18$$
$$\boxed{D = 18 \text{ km}}$$
QUESTION 9
4 marks
Easy
A fixed mass of gas is held in a cylinder at a constant temperature. Boyle's Law states that the pressure $P$ (in kPa) of the gas varies inversely as its volume $V$ (in litres).
When the volume is $4$ L, the pressure is $150$ kPa.
(a) Find the constant of variation $k$ and write down the equation connecting $P$ and $V$. [2]
(b) Find the pressure of the gas when it is compressed to a volume of $6$ L.
(Note: use technology if it is quicker for you to check the arithmetic.) [2]
Show complete worked solution
(a) Since $P$ varies inversely as $V$:
$$P = \frac{k}{V}$$
Substituting $V=4$, $P=150$:
$$150 = \frac{k}{4} \Rightarrow k = 150 \times 4 = 600$$
$$\boxed{P = \frac{600}{V}}$$
(b) Substituting $V=6$:
$$P = \frac{600}{6} = 100$$
$$\boxed{P = 100 \text{ kPa}}$$
QUESTION 10
4 marks
Easy
A team of bricklayers is building a garden wall. The time $t$ (in hours) needed to complete the wall varies inversely as the number of bricklayers $n$ working on it, assuming they all work at the same steady rate.
With $4$ bricklayers working, the wall takes $15$ hours to complete.
(a) Find the constant of variation $k$ and write down the equation connecting $t$ and $n$. [2]
(b) Find how long the wall would take to complete with $6$ bricklayers working. [2]
Show complete worked solution
(a) Since $t$ varies inversely as $n$:
$$t = \frac{k}{n}$$
Substituting $n=4$, $t=15$:
$$15 = \frac{k}{4} \Rightarrow k = 15 \times 4 = 60$$
$$\boxed{t = \frac{60}{n}}$$
(b) Substituting $n=6$:
$$t = \frac{60}{6} = 10$$
$$\boxed{t = 10 \text{ hours}}$$
QUESTION 11
4 marks
Easy
In an electrical circuit with a fixed voltage supply, the current $I$ (in amps) flowing through a resistor varies inversely as its resistance $R$ (in ohms).
When the resistance is $5$ ohms, the current is $2.4$ A.
(a) Find the constant of variation $k$ and write down the equation connecting $I$ and $R$. [2]
(b) Find the current when the resistance is $8$ ohms. [2]
Show complete worked solution
(a) Since $I$ varies inversely as $R$:
$$I = \frac{k}{R}$$
Substituting $R=5$, $I=2.4$:
$$2.4 = \frac{k}{5} \Rightarrow k = 2.4 \times 5 = 12$$
$$\boxed{I = \frac{12}{R}}$$
(b) Substituting $R=8$:
$$I = \frac{12}{8} = 1.5$$
$$\boxed{I = 1.5 \text{ A}}$$
QUESTION 12
4 marks
Easy
When a guitarist presses a string down onto a fret, they shorten the vibrating length of the string, which raises the pitch. The frequency $f$ (in Hz) of the note produced varies inversely as the vibrating length $L$ (in metres) of the string, all other factors being fixed.
An open string of vibrating length $0.65$ m produces a note of frequency $440$ Hz.
(a) Find the constant of variation $k$ and write down the equation connecting $f$ and $L$. [2]
(b) The guitarist presses the string at a fret so that the vibrating length becomes $0.5$ m. Find the frequency of the new note. [2]
Show complete worked solution
(a) Since $f$ varies inversely as $L$:
$$f = \frac{k}{L}$$
Substituting $L=0.65$, $f=440$:
$$440 = \frac{k}{0.65} \Rightarrow k = 440 \times 0.65 = 286$$
$$\boxed{f = \frac{286}{L}}$$
(b) Substituting $L=0.5$:
$$f = \frac{286}{0.5} = 572$$
$$\boxed{f = 572 \text{ Hz}}$$
QUESTION 13
5 marks
Easy
The gravitational attraction $F$ (in newtons) between two fixed point masses varies inversely as the square of the distance $d$ (in metres) between them.
When the distance between the masses is $2$ m, the force between them is $90$ N.
(a) Find the constant of variation $k$ and write down the equation connecting $F$ and $d$. [2]
(b) Find the force between the masses when the distance between them is $6$ m. [3]
Show complete worked solution
(a) Since $F$ varies inversely as the square of $d$:
$$F = \frac{k}{d^2}$$
Substituting $d=2$, $F=90$:
$$90 = \frac{k}{(2)^2} = \frac{k}{4} \Rightarrow k = 90 \times 4 = 360$$
$$\boxed{F = \frac{360}{d^2}}$$
(b) Substituting $d=6$:
$$F = \frac{360}{(6)^2} = \frac{360}{36} = 10$$
$$\boxed{F = 10 \text{ N}}$$
QUESTION 14
5 marks
Easy
The illumination intensity $I$ (in lux) produced by a lamp varies inversely as the square of the distance $d$ (in metres) from the lamp.
At a distance of $3$ m from the lamp, the intensity is $80$ lux.
(a) Find the constant of variation $k$ and write down the equation connecting $I$ and $d$. [2]
(b) Find the intensity at a distance of $5$ m from the lamp. [3]
Show complete worked solution
(a) Since $I$ varies inversely as the square of $d$:
$$I = \frac{k}{d^2}$$
Substituting $d=3$, $I=80$:
$$80 = \frac{k}{(3)^2} = \frac{k}{9} \Rightarrow k = 80 \times 9 = 720$$
$$\boxed{I = \frac{720}{d^2}}$$
(b) Substituting $d=5$:
$$I = \frac{720}{(5)^2} = \frac{720}{25} = 28.8$$
$$\boxed{I = 28.8 \text{ lux}}$$
QUESTION 15
5 marks
Easy
At an outdoor concert, the sound intensity $I$ (in W/m$^2$) reaching a listener varies inversely as the square of the listener's distance $d$ (in metres) from the speaker.
A listener standing $10$ m from the speaker measures an intensity of $0.02$ W/m$^2$.
(a) Find the constant of variation $k$ and write down the equation connecting $I$ and $d$. [2]
(b) Find the sound intensity at a distance of $25$ m from the speaker. [3]
Show complete worked solution
(a) Since $I$ varies inversely as the square of $d$:
$$I = \frac{k}{d^2}$$
Substituting $d=10$, $I=0.02$:
$$0.02 = \frac{k}{(10)^2} = \frac{k}{100} \Rightarrow k = 0.02 \times 100 = 2$$
$$\boxed{I = \frac{2}{d^2}}$$
(b) Substituting $d=25$:
$$I = \frac{2}{(25)^2} = \frac{2}{625} = 0.0032$$
$$\boxed{I = 0.0032 \text{ W/m}^2}$$
QUESTION 16
4 marks
Easy
The air resistance (drag force) $F$ (in newtons) acting on a moving cyclist varies directly as the square of their speed $v$ (in m/s), for a fixed riding position.
At a speed of $20$ m/s, the drag force is $48$ N.
(a) Find the constant of variation $k$ and write down the equation connecting $F$ and $v$. [2]
(b) Find the drag force at a speed of $35$ m/s. [2]
Show complete worked solution
(a) Since $F$ varies directly as the square of $v$:
$$F = kv^2$$
Substituting $v=20$, $F=48$:
$$48 = k(20)^2 = 400k \Rightarrow k = \frac{48}{400} = 0.12$$
$$\boxed{F = 0.12v^2}$$
(b) Substituting $v=35$:
$$F = 0.12(35)^2 = 0.12 \times 1225 = 147$$
$$\boxed{F = 147 \text{ N}}$$
QUESTION 17
4 marks
Easy
Malia is paid at a fixed hourly rate. Her total wage $W$ (in dollars) for a shift varies directly as the number of hours $h$ she works.
For a $13$-hour shift, Malia earns \$97.50.
(a) Find Malia's hourly rate of pay $k$ and write down the equation connecting $W$ and $h$. [2]
(b) Find how much Malia earns for a $22$-hour shift. [2]
Show complete worked solution
(a) Since $W$ varies directly as $h$:
$$W = kh$$
Substituting $h=13$, $W=97.5$:
$$97.5 = k(13) \Rightarrow k = \frac{97.5}{13} = 7.5$$
$$\boxed{W = 7.5h}$$
(b) Substituting $h=22$:
$$W = 7.5(22) = 165$$
$$\boxed{W = \$165.00}$$
QUESTION 18
4 marks
Easy
A train travels at a constant speed. The distance $d$ (in km) it covers varies directly as the time $t$ (in hours) travelled.
In $3$ hours, the train covers $252$ km.
(a) Find the train's speed $k$ and write down the equation connecting $d$ and $t$. [2]
(b) Find the distance the train covers in $7.5$ hours. [2]
Show complete worked solution
(a) Since $d$ varies directly as $t$:
$$d = kt$$
Substituting $t=3$, $d=252$:
$$252 = k(3) \Rightarrow k = \frac{252}{3} = 84$$
$$\boxed{d = 84t}$$
(b) Substituting $t=7.5$:
$$d = 84(7.5) = 630$$
$$\boxed{d = 630 \text{ km}}$$
QUESTION 19
4 marks
Easy
A research station has a fixed total supply of food. The number of days $D$ the supply lasts varies inversely as the number of people $p$ at the station, since the food is shared equally.
With $8$ people at the station, the food supply lasts $15$ days.
(a) Find the constant of variation $k$ and write down the equation connecting $D$ and $p$. [2]
(b) Four more people arrive, making $12$ people at the station in total. Find how many days the food supply will now last. [2]
Show complete worked solution
(a) Since $D$ varies inversely as $p$:
$$D = \frac{k}{p}$$
Substituting $p=8$, $D=15$:
$$15 = \frac{k}{8} \Rightarrow k = 15 \times 8 = 120$$
$$\boxed{D = \frac{120}{p}}$$
(b) Substituting $p=12$:
$$D = \frac{120}{12} = 10$$
$$\boxed{D = 10 \text{ days}}$$
QUESTION 20
5 marks
Easy
Highway engineers model the maximum safe speed $v$ (in mph) for a car travelling around a banked curve of radius $r$ (in metres). For a fixed banking angle, $v$ varies directly as the square root of $r$.
A curve of radius $64$ m has a maximum safe speed of $28$ mph.
(a) Find the constant of variation $k$ and write down the equation connecting $v$ and $r$. [2]
(b) Find the maximum safe speed for a curve of radius $225$ m. [3]
Show complete worked solution
(a) Since $v$ varies directly as the square root of $r$:
$$v = k\sqrt{r}$$
Substituting $r=64$, $v=28$: note $\sqrt{64}=8$.
$$28 = k(8) \Rightarrow k = \frac{28}{8} = 3.5$$
$$\boxed{v = 3.5\sqrt{r}}$$
(b) Substituting $r=225$: note $\sqrt{225}=15$.
$$v = 3.5(15) = 52.5$$
$$\boxed{v = 52.5 \text{ mph}}$$
QUESTION 21
6 marks
Medium
The elastic potential energy $E$ (in joules) stored in a stretched elastic band varies directly as the square of its extension $x$ (in cm): $E = kx^2$.
(a) Show algebraically that if the extension is tripled, the stored energy increases by a factor of $9$. [3]
(b) An elastic band stores $12$ J of energy when stretched by $4$ cm. Find the constant of variation $k$, and hence find the energy stored when the band is stretched by $12$ cm ($3$ times the original extension), confirming your answer to part (a). [3]
Show complete worked solution
(a) Let the original extension be $x$, so the original energy is $E = kx^2$.
If the extension is tripled, the new extension is $3x$, so the new energy is
$$E_{\text{new}} = k(3x)^2 = k \times 9x^2 = 9(kx^2) = 9E$$
So the stored energy increases by a factor of $\boxed{9}$.
(b) Substituting $x=4$, $E=12$ into $E=kx^2$:
$$12 = k(4)^2 = 16k \Rightarrow k = \frac{12}{16} = 0.75$$
$$E = 0.75x^2$$
When the extension is $12$ cm (three times $4$ cm):
$$E = 0.75(12)^2 = 0.75 \times 144 = 108 \text{ J}$$
Checking against part (a): $9 \times 12 = 108$ J, which matches.
$$\boxed{E = 108 \text{ J}}$$
QUESTION 22
6 marks
Medium
A photographer's light meter measures the illumination intensity $I$ from a studio lamp. The intensity varies inversely as the square of the distance $d$ from the lamp: $I = \dfrac{k}{d^2}$.
(a) Show algebraically that if the distance from the lamp is halved, the intensity increases by a factor of $4$. [3]
(b) At a distance of $4$ m from the lamp, the light meter reads $50$ lux. Find the constant of variation $k$, and hence find the reading at a distance of $2$ m (half the original distance), confirming your answer to part (a). [3]
Show complete worked solution
(a) Let the original distance be $d$, so the original intensity is $I = \dfrac{k}{d^2}$.
If the distance is halved, the new distance is $\dfrac{d}{2}$, so the new intensity is
$$I_{\text{new}} = \frac{k}{\left(\frac{d}{2}\right)^2} = \frac{k}{\frac{d^2}{4}} = \frac{4k}{d^2} = 4\left(\frac{k}{d^2}\right) = 4I$$
So the intensity increases by a factor of $\boxed{4}$.
(b) Substituting $d=4$, $I=50$ into $I=\dfrac{k}{d^2}$:
$$50 = \frac{k}{(4)^2} = \frac{k}{16} \Rightarrow k = 50 \times 16 = 800$$
$$I = \frac{800}{d^2}$$
At $d=2$ m:
$$I = \frac{800}{(2)^2} = \frac{800}{4} = 200 \text{ lux}$$
Checking against part (a): $4 \times 50 = 200$ lux, which matches.
$$\boxed{I = 200 \text{ lux}}$$
QUESTION 23
7 marks
Medium
A painting company charges a fee that varies jointly as the wall area $A$ (in m$^2$) to be painted and the number of coats $n$ applied: $C = kAn$, where $C$ is the total cost in dollars.
Painting an area of $40$ m$^2$ with $2$ coats costs \$60.
(a) Write down the equation connecting $C$, $A$ and $n$. [2]
(b) Find the constant of variation $k$. [2]
(c) Find the cost of painting a wall area of $75$ m$^2$ with $3$ coats. [3]
Show complete worked solution
(a) Since $C$ varies jointly as $A$ and $n$:
$$C = kAn$$
(b) Substituting $A=40$, $n=2$, $C=60$:
$$60 = k(40)(2) = 80k \Rightarrow k = \frac{60}{80} = 0.75$$
$$\boxed{C = 0.75An}$$
(c) Substituting $A=75$, $n=3$:
$$C = 0.75(75)(3) = 0.75 \times 225 = 168.75$$
$$\boxed{C = \$168.75}$$
QUESTION 24
8 marks
Medium
For a fixed amount of an ideal gas, the pressure $P$ (in kPa) varies directly as the absolute temperature $T$ (in kelvin) and inversely as the volume $V$ (in litres): $P = \dfrac{kT}{V}$.
A sample of gas has $P=150$ kPa when $T=300$ K and $V=2$ L.
(a) Write down the equation connecting $P$, $T$ and $V$. [2]
(b) Find the constant of variation $k$. [3]
(c) The gas is heated to $T=350$ K and allowed to expand to $V=2.5$ L. Find the new pressure. [3]
Show complete worked solution
(a) Since $P$ varies directly as $T$ and inversely as $V$:
$$P = \frac{kT}{V}$$
(b) Substituting $T=300$, $V=2$, $P=150$:
$$150 = \frac{k(300)}{2} \Rightarrow 150 = 150k \Rightarrow k = 1$$
$$\boxed{P = \frac{T}{V}}$$
(c) Substituting $T=350$, $V=2.5$:
$$P = \frac{350}{2.5} = 140$$
$$\boxed{P = 140 \text{ kPa}}$$
QUESTION 25
7 marks
Medium
The electrical resistance $R$ (in ohms) of a wire varies directly as its length $L$ (in metres) and inversely as its cross-sectional area $A$ (in mm$^2$): $R = \dfrac{kL}{A}$.
A wire of length $2$ m and cross-sectional area $0.5$ mm$^2$ has a resistance of $14$ ohms.
(a) Write down the equation connecting $R$, $L$ and $A$. [2]
(b) Find the constant of variation $k$. [2]
(c) Find the resistance of a wire made from the same material, of length $5$ m and cross-sectional area $0.8$ mm$^2$. [3]
Show complete worked solution
(a) Since $R$ varies directly as $L$ and inversely as $A$:
$$R = \frac{kL}{A}$$
(b) Substituting $L=2$, $A=0.5$, $R=14$:
$$14 = \frac{k(2)}{0.5} = 4k \Rightarrow k = \frac{14}{4} = 3.5$$
$$\boxed{R = \frac{3.5L}{A}}$$
(c) Substituting $L=5$, $A=0.8$:
$$R = \frac{3.5(5)}{0.8} = \frac{17.5}{0.8} = 21.875$$
$$\boxed{R = 21.875 \text{ ohms}}$$
QUESTION 26
7 marks
Medium
The kinetic energy $KE$ (in joules) of a moving object varies jointly as its mass $m$ (in kg) and the square of its speed $v$ (in m/s): $KE = kmv^2$.
An object of mass $4$ kg travelling at $5$ m/s has kinetic energy $50$ J.
(a) Write down the equation connecting $KE$, $m$ and $v$. [2]
(b) Find the constant of variation $k$. [2]
(c) Find the kinetic energy of an object of mass $6$ kg travelling at $8$ m/s. [3]
Show complete worked solution
(a) Since $KE$ varies jointly as $m$ and $v^2$:
$$KE = kmv^2$$
(b) Substituting $m=4$, $v=5$, $KE=50$:
$$50 = k(4)(5)^2 = k(4)(25) = 100k \Rightarrow k = \frac{50}{100} = 0.5$$
$$\boxed{KE = 0.5mv^2}$$
(This matches the well-known physics formula $KE = \tfrac{1}{2}mv^2$.)
(c) Substituting $m=6$, $v=8$:
$$KE = 0.5(6)(8)^2 = 0.5(6)(64) = 0.5 \times 384 = 192$$
$$\boxed{KE = 192 \text{ J}}$$
QUESTION 27
6 marks
Medium
The volume $V$ (in cm$^3$) of a scaled replica of a car varies directly as the cube of the scale factor $s$ used to build it: $V = ks^3$.
(a) Show algebraically that if the scale factor is halved, the volume of the replica decreases to $\dfrac{1}{8}$ of its original value. [3]
(b) A replica built at scale factor $s=0.2$ has a volume of $40$ cm$^3$. Find the constant of variation $k$, and hence find the volume of a replica built at scale factor $s=0.1$ (half the original scale factor), confirming your answer to part (a). [3]
Show complete worked solution
(a) Let the original scale factor be $s$, so the original volume is $V = ks^3$.
If the scale factor is halved, the new scale factor is $\dfrac{s}{2}$, so the new volume is
$$V_{\text{new}} = k\left(\frac{s}{2}\right)^3 = k \times \frac{s^3}{8} = \frac{1}{8}(ks^3) = \frac{1}{8}V$$
So the volume decreases to $\boxed{\dfrac{1}{8}}$ of its original value.
(b) Substituting $s=0.2$, $V=40$ into $V=ks^3$:
$$40 = k(0.2)^3 = k(0.008) \Rightarrow k = \frac{40}{0.008} = 5000$$
$$V = 5000s^3$$
At $s=0.1$ (half of $0.2$):
$$V = 5000(0.1)^3 = 5000(0.001) = 5 \text{ cm}^3$$
Checking against part (a): $\dfrac{40}{8} = 5$ cm$^3$, which matches.
$$\boxed{V = 5 \text{ cm}^3}$$
QUESTION 28
7 marks
Medium
A construction job takes $18$ days to complete with $10$ workers on site. The time $t$ (in days) to complete the job varies inversely as the number of workers $n$, assuming all workers work at the same steady rate.
(a) Find the constant of variation $k$. [2]
(b) The contractor increases the workforce by $50\%$ (to $15$ workers). Find the new number of days needed to complete the job. [2]
(c) Find the percentage decrease in the time taken, giving your answer to $1$ decimal place. [3]
Show complete worked solution
(a) Since $t$ varies inversely as $n$: $t = \dfrac{k}{n}$. Substituting $n=10$, $t=18$:
$$18 = \frac{k}{10} \Rightarrow k = 18 \times 10 = 180$$
$$\boxed{t = \frac{180}{n}}$$
(b) A $50\%$ increase in workers means $n = 10 \times 1.5 = 15$ workers.
$$t = \frac{180}{15} = 12$$
$$\boxed{t = 12 \text{ days}}$$
(c) The time decreased from $18$ days to $12$ days, a decrease of $18 - 12 = 6$ days.
$$\text{percentage decrease} = \frac{6}{18} \times 100\% = 33.3\%$$
$$\boxed{33.3\% \text{ decrease}}$$
QUESTION 29
6 marks
Medium
In an experiment, a student records the following data, and believes that $y$ varies directly as the square of $x$, i.e. $y = kx^2$.
(a) By computing $\dfrac{y}{x^2}$ for each data pair, show that the data is consistent with a direct square variation law, and state the value of the constant of variation $k$. [3]
(b) Write down the equation connecting $y$ and $x$. [1]
(c) Use the equation to predict the value of $y$ when $x=10$. [2]
| $x$ | 2 | 4 | 6 |
|---|---|---|---|
| $y$ | 8 | 32 | 72 |
Show complete worked solution
(a) Computing $\dfrac{y}{x^2}$ for each data pair:
$$x=2: \quad \frac{8}{(2)^2} = \frac{8}{4} = 2$$
$$x=4: \quad \frac{32}{(4)^2} = \frac{32}{16} = 2$$
$$x=6: \quad \frac{72}{(6)^2} = \frac{72}{36} = 2$$
Since $\dfrac{y}{x^2} = 2$ for every data pair, the data is consistent with $y = kx^2$, with
$$\boxed{k = 2}$$
(b)
$$\boxed{y = 2x^2}$$
(c) Substituting $x=10$:
$$y = 2(10)^2 = 2 \times 100 = 200$$
$$\boxed{y = 200}$$
QUESTION 30
6 marks
Medium
The graph below shows a curve of the form $y = \dfrac{k}{x}$, for $x>0$, passing through the labelled point.
(a) Use the labelled point on the graph to find the constant of variation $k$, and write down the equation of the curve. [2]
(b) Use the equation to find the value of $y$ when $x=6$. [2]
(c) Use the equation to find the value of $x$ for which $y=4$. [2]
Show complete worked solution
(a) The graph passes through $(2, 18)$, so substituting into $y=\dfrac{k}{x}$:
$$18 = \frac{k}{2} \Rightarrow k = 18 \times 2 = 36$$
$$\boxed{y = \frac{36}{x}}$$
(b) Substituting $x=6$:
$$y = \frac{36}{6} = 6$$
$$\boxed{y = 6}$$
(c) Substituting $y=4$:
$$4 = \frac{36}{x} \Rightarrow x = \frac{36}{4} = 9$$
$$\boxed{x = 9}$$
QUESTION 31
8 marks
Medium
The period $T$ (in seconds) of a simple pendulum -- the time taken for one complete swing -- varies directly as the square root of its length $L$ (in metres): $T = k\sqrt{L}$.
(a) Show algebraically that if the length of a pendulum is quadrupled, its period doubles. [3]
(b) A pendulum of length $0.81$ m has a period of $1.8$ s. Find the constant of variation $k$. [2]
(c) Find the length required for a pendulum to have a period of $3.6$ s (double the period in part (b)), and confirm this is consistent with your answer to part (a). [3]
Show complete worked solution
(a) Let the original length be $L$, so the original period is $T = k\sqrt{L}$.
If the length is quadrupled, the new length is $4L$, so the new period is
$$T_{\text{new}} = k\sqrt{4L} = k \times 2\sqrt{L} = 2(k\sqrt{L}) = 2T$$
So the period doubles, as required. $\blacksquare$
(b) Substituting $L=0.81$, $T=1.8$ into $T=k\sqrt{L}$: note $\sqrt{0.81}=0.9$.
$$1.8 = k(0.9) \Rightarrow k = \frac{1.8}{0.9} = 2$$
$$\boxed{T = 2\sqrt{L}}$$
(c) We require $T=3.6$ s. Rearranging $T=2\sqrt{L}$:
$$3.6 = 2\sqrt{L} \Rightarrow \sqrt{L} = 1.8 \Rightarrow L = (1.8)^2 = 3.24$$
$$\boxed{L = 3.24 \text{ m}}$$
Since the period has doubled (from $1.8$ s to $3.6$ s), part (a) predicts the length should have quadrupled: $0.81 \times 4 = 3.24$ m, which matches. $\blacksquare$
QUESTION 32
7 marks
Medium
A hardware shop sells rope by the metre. The shop manager believes the price $P$ (in dollars) varies directly as the length $L$ (in metres) purchased. The following prices were recorded:
(a) Compute the ratio $\dfrac{P}{L}$ for each row of the table. [3]
(b) Identify which row does not fit a direct variation law, and suggest a reason (in the context of a shop) why this row might differ from the other two. [2]
(c) Using the constant of variation suggested by the two consistent rows, predict the price of $12$ m of rope. [2]
| $L$ (m) | 2 | 5 | 9 |
|---|---|---|---|
| $P$ (\$) | 7.00 | 17.50 | 29.70 |
Show complete worked solution
(a) Computing $\dfrac{P}{L}$ for each row:
$$L=2: \quad \frac{7.00}{2} = 3.5$$
$$L=5: \quad \frac{17.50}{5} = 3.5$$
$$L=9: \quad \frac{29.70}{9} = 3.3$$
(b) The first two rows both give a ratio of $3.5$, but the row with $L=9$ gives a ratio of $3.3$, which does not match. This row does not fit the direct variation law $P = 3.5L$.
A likely reason is that the shop offers a bulk discount for larger purchases of rope, so buying $9$ m works out slightly cheaper per metre than the standard rate.
(c) Using $k=3.5$ (the rate suggested by the two consistent rows) and $P=kL$:
$$P = 3.5(12) = 42$$
$$\boxed{P = \$42.00}$$
QUESTION 33
6 marks
Medium
At a concert, the sound intensity $I$ received by a listener varies inversely as the square of their distance $d$ from the speaker stack: $I = \dfrac{k}{d^2}$. A listener moves to a new position at $60\%$ of their original distance from the speakers (i.e. $d_{\text{new}} = 0.6d$).
(a) Show algebraically that the sound intensity at the new position is approximately $2.78$ times the original intensity. [3]
(b) At the original position, the intensity was $15$ W/m$^2$. Find the intensity at the new position, giving your answer to $2$ decimal places. [3]
Show complete worked solution
(a) Let the original distance be $d$, so the original intensity is $I = \dfrac{k}{d^2}$.
At the new distance $d_{\text{new}} = 0.6d$:
$$I_{\text{new}} = \frac{k}{(0.6d)^2} = \frac{k}{0.36d^2} = \frac{1}{0.36}\left(\frac{k}{d^2}\right) = \frac{1}{0.36}I$$
$$\frac{1}{0.36} = 2.7\overline{7} \approx 2.78$$
So $I_{\text{new}} \approx 2.78 \, I$, as required. $\blacksquare$
(b) With $I=15$ W/m$^2$:
$$I_{\text{new}} = \frac{15}{0.36} = 41.6\overline{6} \approx 41.67$$
$$\boxed{I_{\text{new}} \approx 41.67 \text{ W/m}^2}$$
QUESTION 34
6 marks
Medium
In a physics simulation, the attractive force $F$ (in newtons) between two objects of masses $m_1$ and $m_2$ (in kg) varies jointly as $m_1$ and $m_2$, and inversely as the square of the distance $d$ (in metres) between them:
$$F = \frac{km_1m_2}{d^2}$$
Two objects of masses $5$ kg and $8$ kg, placed $2$ m apart, attract each other with a force of $100$ N.
(a) Find the constant of variation $k$. [3]
(b) The first mass is doubled to $10$ kg, and the distance between the objects is doubled to $4$ m (the second mass, $8$ kg, is unchanged). Find the new attractive force. [3]
Show complete worked solution
(a) Substituting $m_1=5$, $m_2=8$, $d=2$, $F=100$:
$$100 = \frac{k(5)(8)}{(2)^2} = \frac{40k}{4} = 10k \Rightarrow k = \frac{100}{10} = 10$$
$$\boxed{F = \frac{10m_1m_2}{d^2}}$$
(b) Substituting $m_1=10$, $m_2=8$, $d=4$:
$$F = \frac{10(10)(8)}{(4)^2} = \frac{800}{16} = 50$$
$$\boxed{F = 50 \text{ N}}$$
QUESTION 35
6 marks
Medium
A cookbook states that the roasting time $t$ (in minutes) for a turkey varies directly as its mass $m$ (in kg): $t = km$.
A $5$ kg turkey needs $175$ minutes of roasting time.
(a) Find the constant of variation $k$. [2]
(b) A larger turkey has a mass $20\%$ greater than $5$ kg. Find its mass, and the roasting time required. [2]
(c) Find the percentage increase in roasting time from part (a) to part (b), and explain why this percentage is equal to the percentage increase in mass. [2]
Show complete worked solution
(a) Substituting $m=5$, $t=175$:
$$175 = k(5) \Rightarrow k = \frac{175}{5} = 35$$
$$\boxed{t = 35m}$$
(b) A $20\%$ increase in mass gives $m = 5 \times 1.2 = 6$ kg.
$$t = 35(6) = 210$$
$$\boxed{t = 210 \text{ minutes}}$$
(c) Percentage increase in time:
$$\frac{210-175}{175} \times 100\% = \frac{35}{175} \times 100\% = 20\%$$
This equals the percentage increase in mass because $t=km$ is a direct linear ($n=1$) variation: multiplying $m$ by any factor $(1+r)$ multiplies $t$ by the exact same factor $(1+r)$, since
$$t_{\text{new}} = k \times m(1+r) = (1+r)(km) = (1+r)t$$
$$\boxed{20\% \text{ increase}}$$
QUESTION 36
7 marks
Medium
For a fixed distance journey, the travel time $t$ (in hours) varies inversely as the average speed $s$ (in km/h): $t = \dfrac{k}{s}$.
(a) Show algebraically that if the average speed decreases by $20\%$, the travel time increases by exactly $25\%$. [3]
(b) A journey at an average speed of $80$ km/h takes $3.5$ hours. Find the constant of variation $k$ (the distance travelled, in km). [2]
(c) Due to roadworks, the average speed drops by $20\%$ to $64$ km/h. Find the new travel time, and confirm it represents a $25\%$ increase on the original time. [2]
Show complete worked solution
(a) Let the original speed be $s$, so the original time is $t = \dfrac{k}{s}$.
A $20\%$ decrease in speed gives a new speed of $0.8s$, so the new time is
$$t_{\text{new}} = \frac{k}{0.8s} = \frac{1}{0.8}\left(\frac{k}{s}\right) = 1.25\left(\frac{k}{s}\right) = 1.25t$$
Since $1.25t = t + 0.25t$, this is a $25\%$ increase on the original time. $\blacksquare$
(b) Substituting $s=80$, $t=3.5$:
$$k = st = 80(3.5) = 280$$
$$\boxed{k = 280 \text{ km}}$$
(c) New speed: $80 \times 0.8 = 64$ km/h.
$$t_{\text{new}} = \frac{280}{64} = 4.375 \text{ hours}$$
Checking the percentage increase: $\dfrac{4.375 - 3.5}{3.5} \times 100\% = \dfrac{0.875}{3.5} \times 100\% = 25\%$, which confirms part (a).
$$\boxed{t_{\text{new}} = 4.375 \text{ hours} \; (25\% \text{ increase})}$$
QUESTION 37
6 marks
Medium
The rate of heat loss $H$ (in watts) from an uninsulated pipe varies jointly as its surface area $A$ (in m$^2$) and the temperature difference $\Delta T$ (in $^{\circ}$C) between the pipe and the surrounding air: $H = kA\,\Delta T$.
A pipe with surface area $2.5$ m$^2$ and a temperature difference of $15^{\circ}$C loses heat at a rate of $225$ W.
(a) Write down the equation connecting $H$, $A$ and $\Delta T$, and find the constant of variation $k$. [3]
(b) Find the rate of heat loss from a pipe with surface area $4$ m$^2$ and a temperature difference of $22^{\circ}$C. [3]
Show complete worked solution
(a) Substituting $A=2.5$, $\Delta T=15$, $H=225$:
$$225 = k(2.5)(15) = 37.5k \Rightarrow k = \frac{225}{37.5} = 6$$
$$\boxed{H = 6A\,\Delta T}$$
(b) Substituting $A=4$, $\Delta T=22$:
$$H = 6(4)(22) = 6 \times 88 = 528$$
$$\boxed{H = 528 \text{ W}}$$
QUESTION 38
8 marks
Medium
The graph below shows two curves, Curve A and Curve B, each passing through the point $(2, 12)$. One curve has the equation $y=kx^2$ (a direct square variation); the other has the equation $y=\dfrac{k}{x}$ (an inverse variation).
(a) By considering the shape of each curve, state which curve (A or B) represents $y=kx^2$ and which represents $y=\dfrac{k}{x}$. Justify your choice with one property of each shape. [2]
(b) Find the value of $k$ for Curve A. [2]
(c) Find the value of $k$ for Curve B. [2]
(d) Use your equations to find the value of $y$ on each curve when $x=6$. [2]
Show complete worked solution
(a) Curve A rises steadily away from the origin and continues to increase without bound as $x$ increases -- this is the shape of a direct square variation $y=kx^2$ (a parabola-type curve through the origin).
Curve B falls steeply and gets closer and closer to the $x$-axis without ever touching it as $x$ increases (it never touches either axis) -- this is the shape of an inverse variation $y=\dfrac{k}{x}$ (a hyperbola-type curve).
$$\boxed{\text{Curve A: } y=kx^2, \qquad \text{Curve B: } y=\frac{k}{x}}$$
(b) Substituting $(2, 12)$ into $y=kx^2$:
$$12 = k(2)^2 = 4k \Rightarrow k = \frac{12}{4} = 3$$
$$\boxed{k_A = 3}$$
(c) Substituting $(2, 12)$ into $y=\dfrac{k}{x}$:
$$12 = \frac{k}{2} \Rightarrow k = 12 \times 2 = 24$$
$$\boxed{k_B = 24}$$
(d) Curve A: $y = 3(6)^2 = 3 \times 36 = 108$.
Curve B: $y = \dfrac{24}{6} = 4$.
$$\boxed{\text{Curve A: } y=108, \qquad \text{Curve B: } y=4}$$
QUESTION 39
7 marks
Medium
A bakery finds that the cost $C$ (in dollars) of ingredients for a circular cake varies directly as the cube of its diameter $d$ (in cm), since the amount of batter needed scales with the volume of the cake: $C = kd^3$.
(a) Show algebraically that if the diameter of a cake increases by $50\%$, the ingredient cost increases by a factor of $3.375$. [3]
(b) A cake of diameter $20$ cm costs \$18.00 in ingredients. Find the constant of variation $k$. [2]
(c) Find the ingredient cost of a cake with diameter $30$ cm ($50\%$ larger), and confirm it is consistent with part (a). [2]
Show complete worked solution
(a) Let the original diameter be $d$, so the original cost is $C=kd^3$.
A $50\%$ increase in diameter gives a new diameter of $1.5d$, so the new cost is
$$C_{\text{new}} = k(1.5d)^3 = k(1.5)^3d^3 = 3.375(kd^3) = 3.375C$$
So the cost increases by a factor of $\boxed{3.375}$ (a $237.5\%$ increase). $\blacksquare$
(b) Substituting $d=20$, $C=18$:
$$18 = k(20)^3 = 8000k \Rightarrow k = \frac{18}{8000} = 0.00225$$
$$\boxed{C = 0.00225d^3}$$
(c) At $d=30$ cm:
$$C = 0.00225(30)^3 = 0.00225 \times 27000 = 60.75$$
Checking against part (a): $18 \times 3.375 = 60.75$, which matches.
$$\boxed{C = \$60.75}$$
QUESTION 40
7 marks
Medium
A mobile phone's signal strength $S$ varies inversely as the square of the distance $d$ from the nearest cell tower: $S = \dfrac{k}{d^2}$. A user moves to a position at $70\%$ of their original distance from the tower (i.e. $d_{\text{new}} = 0.7d$).
(a) Show algebraically that the signal strength at the new position is $\dfrac{1}{0.49}$ times the original signal strength, and evaluate this factor to $3$ significant figures. [3]
(b) At the original position, the signal strength was $25$ units. Find the signal strength at the new position, to $3$ significant figures. [2]
(c) Find the percentage increase in signal strength, to $1$ decimal place. [2]
Show complete worked solution
(a) Let the original distance be $d$, so the original signal strength is $S=\dfrac{k}{d^2}$.
At the new distance $d_{\text{new}}=0.7d$:
$$S_{\text{new}} = \frac{k}{(0.7d)^2} = \frac{k}{0.49d^2} = \frac{1}{0.49}\left(\frac{k}{d^2}\right) = \frac{1}{0.49}S$$
$$\frac{1}{0.49} = 2.0408\ldots \approx 2.04$$
$$\boxed{S_{\text{new}} \approx 2.04\,S}$$
(b) With $S=25$:
$$S_{\text{new}} = \frac{25}{0.49} = 51.0204\ldots \approx 51.0$$
$$\boxed{S_{\text{new}} \approx 51.0 \text{ units}}$$
(c) Percentage increase:
$$\frac{51.0204\ldots - 25}{25} \times 100\% = \frac{26.0204\ldots}{25} \times 100\% \approx 104.1\%$$
$$\boxed{104.1\% \text{ increase}}$$
QUESTION 41
11 marks
Hard
Two point masses attract each other with a force $F$ (in newtons) that varies jointly as their masses $m_1$ and $m_2$ (in kg), and inversely as the square of the distance $d$ (in metres) between them:
$$F = \frac{km_1m_2}{d^2}$$
Two masses, $3$ kg and $6$ kg, placed $1.5$ m apart, attract each other with a force of $48$ N.
(a) Write down the equation connecting $F$, $m_1$, $m_2$ and $d$. [2]
(b) Find the constant of variation $k$. [3]
(c) The first mass is doubled to $6$ kg, the second mass is tripled to $18$ kg, and the distance between them is doubled to $3$ m. Find the new attractive force, both by (i) direct substitution into the model, and (ii) by reasoning about the combined factor of change. Confirm the two methods agree. [3]
(d) With the original masses ($3$ kg and $6$ kg) restored, find the distance $d$ at which the attractive force would be $12$ N. [3]
Show complete worked solution
(a)
$$F = \frac{km_1m_2}{d^2}$$
(b) Substituting $m_1=3$, $m_2=6$, $d=1.5$, $F=48$:
$$48 = \frac{k(3)(6)}{(1.5)^2} = \frac{18k}{2.25} = 8k \Rightarrow k = \frac{48}{8} = 6$$
$$\boxed{F = \frac{6m_1m_2}{d^2}}$$
(c)(i) Substituting $m_1=6$, $m_2=18$, $d=3$:
$$F = \frac{6(6)(18)}{(3)^2} = \frac{648}{9} = 72$$
(c)(ii) $m_1$ is multiplied by $2$, $m_2$ is multiplied by $3$, and $d$ is multiplied by $2$ (so $d^2$ is multiplied by $4$). The overall factor on $F$ is
$$\frac{2 \times 3}{2^2} = \frac{6}{4} = 1.5$$
$$F_{\text{new}} = 1.5 \times 48 = 72$$
Both methods give $\boxed{F = 72 \text{ N}}$, confirming agreement.
(d) With $m_1=3$, $m_2=6$ fixed, and $F=12$:
$$12 = \frac{6(3)(6)}{d^2} = \frac{108}{d^2} \Rightarrow d^2 = \frac{108}{12} = 9 \Rightarrow d = \sqrt{9} = 3$$
$$\boxed{d = 3 \text{ m}}$$
QUESTION 42
10 marks
Hard
For a fixed quantity of an ideal gas, $P = \dfrac{kT}{V}$, where $P$ is pressure (kPa), $T$ is absolute temperature (K), and $V$ is volume (L).
A sample has $P=200$ kPa when $T=400$ K and $V=3$ L.
(a) Write down the equation connecting $P$, $T$ and $V$. [2]
(b) Find the constant of variation $k$. [2]
(c) The gas is cooled to $T=350$ K and its volume expands to $V=4$ L. Find the new pressure, to $2$ decimal places. [3]
(d) Starting again from the original state ($P=200$ kPa, $T=400$ K, $V=3$ L), the gas is heated to $T=500$ K. Find the volume $V$ required to keep the pressure unchanged at $200$ kPa. [3]
Show complete worked solution
(a)
$$P = \frac{kT}{V}$$
(b) Substituting $T=400$, $V=3$, $P=200$:
$$200 = \frac{k(400)}{3} \Rightarrow 600 = 400k \Rightarrow k = \frac{600}{400} = 1.5$$
$$\boxed{P = \frac{1.5T}{V}}$$
(c) Substituting $T=350$, $V=4$:
$$P = \frac{1.5(350)}{4} = \frac{525}{4} = 131.25$$
$$\boxed{P = 131.25 \text{ kPa}}$$
(d) We require $P=200$ when $T=500$. Rearranging $P=\dfrac{1.5T}{V}$ to make $V$ the subject:
$$V = \frac{1.5T}{P}$$
Substituting $T=500$, $P=200$:
$$V = \frac{1.5(500)}{200} = \frac{750}{200} = 3.75$$
$$\boxed{V = 3.75 \text{ L}}$$
QUESTION 43
10 marks
Hard
Kepler's Third Law of planetary motion states that the square of a planet's orbital period $T$ (measured in Earth years) varies directly as the cube of its average orbital radius $r$ (measured in astronomical units, AU, where $1$ AU is the average Earth-Sun distance): $T^2 = kr^3$.
(a) Using the fact that Earth has $r=1$ AU and $T=1$ year (by definition), find the constant of variation $k$. [2]
(b) Write down the equation connecting $T$ and $r$. [2]
(c) Mars has an average orbital radius of $r=1.524$ AU. Use the equation to predict the orbital period of Mars, giving your answer to $3$ significant figures. (The true value is approximately $1.88$ years -- your answer should be close to this.) [3]
(d) A hypothetical planet is discovered with an orbital period of $T=8$ years. Find its orbital radius $r$, in AU. [3]
Show complete worked solution
(a) Substituting $r=1$, $T=1$ into $T^2=kr^3$:
$$(1)^2 = k(1)^3 \Rightarrow 1 = k \Rightarrow k = 1$$
$$\boxed{k=1}$$
(b)
$$\boxed{T^2 = r^3}$$
(c) Substituting $r=1.524$:
$$T^2 = (1.524)^3 = 3.5396\ldots$$
$$T = \sqrt{3.5396\ldots} = 1.8814\ldots \approx 1.88$$
$$\boxed{T \approx 1.88 \text{ years}}$$
This matches the true orbital period of Mars very closely.
(d) Substituting $T=8$ into $T^2=r^3$:
$$(8)^2 = r^3 \Rightarrow 64 = r^3 \Rightarrow r = \sqrt[3]{64} = 4$$
$$\boxed{r = 4 \text{ AU}}$$
QUESTION 44
9 marks
Hard
The intensity $I$ received from a loudspeaker (in W/m$^2$) varies directly as the power $P$ (in watts) of the speaker, and inversely as the square of the distance $d$ (in metres) from it:
$$I = \frac{kP}{d^2}$$
A $50$ W speaker produces an intensity of $25$ W/m$^2$ at a distance of $4$ m.
(a) Write down the equation connecting $I$, $P$ and $d$. [2]
(b) Find the constant of variation $k$. [2]
(c) An $80$ W speaker is used instead. Find the intensity at a distance of $5$ m. [2]
(d) Using the $80$ W speaker, find the distance at which the intensity would be $20$ W/m$^2$, giving your answer to $3$ significant figures. [3]
Show complete worked solution
(a)
$$I = \frac{kP}{d^2}$$
(b) Substituting $P=50$, $d=4$, $I=25$:
$$25 = \frac{k(50)}{(4)^2} = \frac{50k}{16} \Rightarrow k = \frac{25 \times 16}{50} = \frac{400}{50} = 8$$
$$\boxed{I = \frac{8P}{d^2}}$$
(c) Substituting $P=80$, $d=5$:
$$I = \frac{8(80)}{(5)^2} = \frac{640}{25} = 25.6$$
$$\boxed{I = 25.6 \text{ W/m}^2}$$
(d) Substituting $P=80$, $I=20$ and solving for $d$:
$$20 = \frac{8(80)}{d^2} = \frac{640}{d^2} \Rightarrow d^2 = \frac{640}{20} = 32 \Rightarrow d = \sqrt{32} = 5.6568\ldots$$
$$\boxed{d \approx 5.66 \text{ m}}$$
QUESTION 45
12 marks
Hard
An engineer is testing a specialised sensor component. She suspects that its output resistance $R$ (in ohms) varies inversely as the square of the applied voltage $V$ (in volts): $R = \dfrac{k}{V^2}$. The following readings were recorded:
(a) By computing $R \times V^2$ for each of the four data pairs, show that the data is consistent with an inverse square variation law, and state the constant of variation $k$. [3]
(b) Write down the equation connecting $R$ and $V$. [2]
(c) Use the equation to predict the resistance when $V=10$ volts. [2]
(d) Find the voltage required for the resistance to equal exactly $1$ ohm. [2]
(e) Show algebraically that if the applied voltage is tripled (from any starting value), the resistance decreases to $\dfrac{1}{9}$ of its previous value. [3]
| $V$ (volts) | 2 | 4 | 5 | 8 |
|---|---|---|---|---|
| $R$ (ohms) | 100 | 25 | 16 | 6.25 |
Show complete worked solution
(a) Computing $R \times V^2$ for each pair:
$$V=2: \; 100 \times (2)^2 = 100 \times 4 = 400$$
$$V=4: \; 25 \times (4)^2 = 25 \times 16 = 400$$
$$V=5: \; 16 \times (5)^2 = 16 \times 25 = 400$$
$$V=8: \; 6.25 \times (8)^2 = 6.25 \times 64 = 400$$
Since $R \times V^2 = 400$ for every data pair, the data is consistent with $R=\dfrac{k}{V^2}$, with
$$\boxed{k = 400}$$
(b)
$$\boxed{R = \frac{400}{V^2}}$$
(c) Substituting $V=10$:
$$R = \frac{400}{(10)^2} = \frac{400}{100} = 4$$
$$\boxed{R = 4 \text{ ohms}}$$
(d) Substituting $R=1$:
$$1 = \frac{400}{V^2} \Rightarrow V^2 = 400 \Rightarrow V = \sqrt{400} = 20$$
$$\boxed{V = 20 \text{ volts}}$$
(e) Let the original voltage be $V$, so the original resistance is $R=\dfrac{k}{V^2}$.
If the voltage is tripled, the new voltage is $3V$, so the new resistance is
$$R_{\text{new}} = \frac{k}{(3V)^2} = \frac{k}{9V^2} = \frac{1}{9}\left(\frac{k}{V^2}\right) = \frac{1}{9}R$$
So the resistance decreases to $\boxed{\dfrac{1}{9}}$ of its previous value, regardless of the starting voltage. $\blacksquare$
QUESTION 46
11 marks
Hard
The power $P$ (in watts) that a wind turbine can generate varies jointly as the square of its blade radius $r$ (in metres) and the cube of the wind speed $v$ (in m/s), since the swept area is proportional to $r^2$ and the kinetic energy flux is proportional to $v^3$:
$$P = kr^2v^3$$
A turbine with blade radius $20$ m generates $125\,000$ W ($125$ kW) of power in a wind speed of $10$ m/s.
(a) Write down the equation connecting $P$, $r$ and $v$. [2]
(b) Find the constant of variation $k$. [3]
(c) A new turbine design doubles the blade radius to $40$ m, and is tested in a wind speed $50\%$ higher, at $15$ m/s. Find the new power output (i) by direct substitution, and (ii) by finding the overall factor of change and applying it to the original power. Confirm the two methods agree. [3]
(d) Keeping the wind speed fixed at the original $10$ m/s, find the blade radius $r$ required to double the original power output ($125$ kW), giving your answer to $3$ significant figures. [3]
Show complete worked solution
(a)
$$P = kr^2v^3$$
(b) Substituting $r=20$, $v=10$, $P=125000$:
$$125000 = k(20)^2(10)^3 = k(400)(1000) = 400000k$$
$$k = \frac{125000}{400000} = 0.3125$$
$$\boxed{P = 0.3125\,r^2v^3}$$
(c)(i) Substituting $r=40$, $v=15$:
$$P = 0.3125(40)^2(15)^3 = 0.3125(1600)(3375) = 0.3125 \times 5\,400\,000 = 1\,687\,500$$
(c)(ii) $r$ is multiplied by $2$, so $r^2$ is multiplied by $2^2=4$. $v$ is multiplied by $1.5$, so $v^3$ is multiplied by $(1.5)^3=3.375$. The overall factor on $P$ is
$$4 \times 3.375 = 13.5$$
$$P_{\text{new}} = 13.5 \times 125\,000 = 1\,687\,500$$
Both methods agree: $\boxed{P = 1\,687\,500 \text{ W} = 1687.5 \text{ kW}}$.
(d) With $v=10$ fixed, we require $P = 250\,000$ W (double $125\,000$). Since $v$ is unchanged, $v^3$ is unchanged, so the factor of $2$ on $P$ must come entirely from $r^2$:
$$r_{\text{new}}^2 = 2r^2 \Rightarrow r_{\text{new}} = r\sqrt{2} = 20\sqrt{2} = 28.284\ldots$$
$$\boxed{r \approx 28.3 \text{ m}}$$
QUESTION 47
10 marks
Hard
The power $P$ (in watts) generated by a small hydroelectric dam varies jointly as the water flow rate $F$ (in m$^3$/s) and the square of the head height $h$ (in metres, the vertical drop of the water): $P = kFh^2$.
At a flow rate of $50$ m$^3$/s and a head height of $20$ m, the dam generates $4\,000\,000$ W ($4$ MW).
(a) Write down the equation connecting $P$, $F$ and $h$. [2]
(b) Find the constant of variation $k$. [2]
(c) During a drought, the flow rate falls by $10\%$. At the same time, a new spillway increases the effective head height by $25\%$. Find, algebraically, the overall percentage change in power output. [4]
(d) Verify your answer to part (c) by computing the new power output directly using the model, and comparing it to the original $4$ MW. [2]
Show complete worked solution
(a)
$$P = kFh^2$$
(b) Substituting $F=50$, $h=20$, $P=4\,000\,000$:
$$4\,000\,000 = k(50)(20)^2 = k(50)(400) = 20\,000k$$
$$k = \frac{4\,000\,000}{20\,000} = 200$$
$$\boxed{P = 200Fh^2}$$
(c) The flow rate falls by $10\%$, so the new flow rate is $0.9F$. The head height increases by $25\%$, so the new head height is $1.25h$, and $h^2$ is multiplied by $(1.25)^2=1.5625$.
The overall factor on $P$ is
$$0.9 \times 1.5625 = 1.40625$$
This corresponds to an increase of $(1.40625-1)\times 100\% = 40.625\%$.
$$\boxed{\text{Power increases by } 40.625\%}$$
(d) New flow rate: $F=50 \times 0.9=45$ m$^3$/s. New head height: $h = 20 \times 1.25 = 25$ m.
$$P = 200(45)(25)^2 = 200(45)(625) = 200 \times 28\,125 = 5\,625\,000$$
Checking: $\dfrac{5\,625\,000}{4\,000\,000} = 1.40625$, which matches the factor found in part (c).
$$\boxed{P = 5\,625\,000 \text{ W} = 5.625 \text{ MW}}$$
QUESTION 48
10 marks
Hard
A scientist believes that $y$ varies directly as some power $n$ of $x$, so that $y = kx^n$, but does not yet know the values of $k$ or $n$. Two experimental data points are recorded: $x=4, y=16$ and $x=9, y=54$ (shown on the graph below).
(a) By substituting both data points into $y=kx^n$ and dividing one equation by the other, show that
$$\left(\frac{9}{4}\right)^n = \frac{54}{16}$$
[2]
(b) Solve the equation in part (a) for $n$, using logarithms. [3]
(c) Using your value of $n$ and the data point $(4, 16)$, find the value of $k$. [3]
(d) Use your equation to predict the value of $y$ when $x=16$. [2]
Show complete worked solution
(a) Substituting each data point into $y=kx^n$:
$$16 = k(4)^n \qquad \text{and} \qquad 54 = k(9)^n$$
Dividing the second equation by the first (the constant $k$ cancels):
$$\frac{54}{16} = \frac{k(9)^n}{k(4)^n} = \left(\frac{9}{4}\right)^n$$
$$\boxed{\left(\frac{9}{4}\right)^n = \frac{54}{16}}$$
(b) Simplifying both sides: $\dfrac{9}{4}=2.25$ and $\dfrac{54}{16}=3.375$, so
$$(2.25)^n = 3.375$$
Taking $\log$ of both sides:
$$n\log(2.25) = \log(3.375)$$
$$n = \frac{\log(3.375)}{\log(2.25)} = \frac{0.52827\ldots}{0.35218\ldots} = 1.5$$
$$\boxed{n = 1.5}$$
(c) Substituting $n=1.5$ and the point $(4, 16)$ into $y=kx^n$:
$$16 = k(4)^{1.5}$$
Since $4^{1.5} = 4\sqrt{4} = 4 \times 2 = 8$:
$$16 = 8k \Rightarrow k = \frac{16}{8} = 2$$
$$\boxed{k=2, \qquad y = 2x^{1.5}}$$
(d) Substituting $x=16$: note $16^{1.5} = 16\sqrt{16} = 16 \times 4 = 64$.
$$y = 2(64) = 128$$
$$\boxed{y = 128}$$
QUESTION 49
10 marks
Hard
The daily running cost $C$ (in dollars) of a small factory varies jointly as the number of machines $M$ in operation, and the square root of the number of hours $H$ the machines run per day:
$$C = kM\sqrt{H}$$
With $6$ machines running for $9$ hours in a day, the running cost is \$5400.
(a) Write down the equation connecting $C$, $M$ and $H$. [2]
(b) Find the constant of variation $k$. [3]
(c) The factory keeps $6$ machines running, but management wants to find the number of hours $H$ needed to reach a daily running cost of exactly \$7200. Find $H$. [3]
(d) Show algebraically that if the number of hours $H$ is quadrupled (with the number of machines $M$ unchanged), the running cost exactly doubles. [2]
Show complete worked solution
(a)
$$C = kM\sqrt{H}$$
(b) Substituting $M=6$, $H=9$, $C=5400$: note $\sqrt{9}=3$.
$$5400 = k(6)(3) = 18k \Rightarrow k = \frac{5400}{18} = 300$$
$$\boxed{C = 300M\sqrt{H}}$$
(c) Substituting $M=6$, $C=7200$ and solving for $H$:
$$7200 = 300(6)\sqrt{H} = 1800\sqrt{H}$$
$$\sqrt{H} = \frac{7200}{1800} = 4 \Rightarrow H = (4)^2 = 16$$
$$\boxed{H = 16 \text{ hours}}$$
(d) Let the original number of hours be $H$, so the original cost is $C = kM\sqrt{H}$ (with $M$ fixed).
If the hours are quadrupled, the new number of hours is $4H$, so the new cost is
$$C_{\text{new}} = kM\sqrt{4H} = kM \times 2\sqrt{H} = 2\left(kM\sqrt{H}\right) = 2C$$
So the running cost exactly doubles. $\blacksquare$
QUESTION 50
10 marks
Hard
In a simple LC electronic circuit (with fixed capacitance), the resonant frequency $f$ (in Hz) varies inversely as the square root of the inductance $L$ (in Henries): $f = \dfrac{k}{\sqrt{L}}$. The graph below shows this relationship, together with two labelled points.
An inductance of $L=0.25$ Henries gives a resonant frequency of $2000$ Hz.
(a) Write down the equation connecting $f$ and $L$. [2]
(b) Find the constant of variation $k$. [2]
(c) Find the inductance $L$ required to double the resonant frequency to $4000$ Hz, and check this value is consistent with the second labelled point on the graph. [3]
(d) Show algebraically that if the inductance $L$ is quadrupled, the resonant frequency is halved. [2]
(e) With reference to the shape of the graph, describe what happens to the resonant frequency $f$ as the inductance $L$ approaches $0$. [1]
Show complete worked solution
(a)
$$f = \frac{k}{\sqrt{L}}$$
(b) Substituting $L=0.25$, $f=2000$: note $\sqrt{0.25}=0.5$.
$$2000 = \frac{k}{0.5} \Rightarrow k = 2000 \times 0.5 = 1000$$
$$\boxed{f = \frac{1000}{\sqrt{L}}}$$
(c) Substituting $f=4000$ and solving for $L$:
$$4000 = \frac{1000}{\sqrt{L}} \Rightarrow \sqrt{L} = \frac{1000}{4000} = 0.25 \Rightarrow L = (0.25)^2 = 0.0625$$
$$\boxed{L = 0.0625 \text{ Henries}}$$
This matches the second labelled point on the graph, $(0.0625, 4000)$.
(d) Let the original inductance be $L$, so the original frequency is $f=\dfrac{k}{\sqrt{L}}$.
If the inductance is quadrupled, the new inductance is $4L$, so the new frequency is
$$f_{\text{new}} = \frac{k}{\sqrt{4L}} = \frac{k}{2\sqrt{L}} = \frac{1}{2}\left(\frac{k}{\sqrt{L}}\right) = \frac{1}{2}f$$
So the resonant frequency is exactly halved. $\blacksquare$
(e) As $L$ approaches $0$, the curve rises steeply and $f$ increases without bound (the curve approaches the vertical axis but never touches it), so the resonant frequency becomes arbitrarily large.
$$\boxed{f \to \infty \text{ as } L \to 0}$$
Exponential Functions 50 questions
QUESTION 1
5 marks
Easy
A microbiologist models the number of bacteria in a laboratory colony using
$$P(t)=500(1.08)^t$$
where $P(t)$ is the number of bacteria $t$ hours after the colony is first observed.
(a) Write down the number of bacteria in the colony at $t=0$. [1]
(b) Find $P(6)$, giving your answer to the nearest whole bacterium. [2]
(c) State whether $P$ models growth or decay, and write down the percentage rate of change per hour. [2]
Show complete worked solution
(a) At $t=0$:
$$P(0)=500(1.08)^0=500(1)=500$$
$$\boxed{P(0)=500 \text{ bacteria}}$$
(b) Substituting $t=6$ into the model, using the GDC:
$$P(6)=500(1.08)^6=793.4372\ldots$$
$$\boxed{P(6)\approx 793 \text{ bacteria}}$$
(c) Since the base is $b=1.08$ and $1.08>1$, the function $P(t)$ is an increasing (growth) function.
The percentage rate of change per hour is found from $b=1+r$, so $r=0.08$, i.e.
$$\boxed{\text{8 percent growth per hour}}$$
QUESTION 2
5 marks
Easy
The mass, in grams, of a radioactive sample is modeled by
$$m(t)=150(0.92)^t$$
where $t$ is the time in years after the sample is first measured.
(a) Write down the initial mass of the sample. [1]
(b) State the percentage rate of decay per year. [2]
(c) Find $m(10)$, giving your answer correct to 1 decimal place. [2]
Show complete worked solution
(a) At $t=0$, $m(0)=150(0.92)^0=150$.
$$\boxed{m(0)=150 \text{ g}}$$
(b) The base is $b=0.92$. Writing $b=1-r$ gives $r=1-0.92=0.08$, so the mass decreases by
$$\boxed{\text{8 percent per year}}$$
(c) Using the GDC:
$$m(10)=150(0.92)^{10}=65.15827\ldots$$
$$\boxed{m(10)\approx 65.2 \text{ g}}$$
QUESTION 3
5 marks
Easy
The concentration of a drug in a patient's bloodstream, in mg/L, is modeled by
$$C(t)=60e^{-0.25t}$$
where $t$ is the time in hours after the drug is administered.
(a) Write down the concentration of the drug at $t=0$. [1]
(b) Find $C(4)$, giving your answer correct to 2 decimal places. [2]
(c) Explain why $C$ models decay, and find the percentage decrease in concentration per hour, correct to 2 decimal places. [2]
Show complete worked solution
(a) At $t=0$: $C(0)=60e^{0}=60$.
$$\boxed{C(0)=60 \text{ mg/L}}$$
(b) Using the GDC:
$$C(4)=60e^{-0.25(4)}=60e^{-1}=22.072766\ldots$$
$$\boxed{C(4)\approx 22.07 \text{ mg/L}}$$
(c) Since the exponent has coefficient $k=-0.25<0$, the concentration decreases as $t$ increases, so $C$ models decay.
Each hour, the concentration is multiplied by $e^{-0.25}$. The percentage decrease per hour is
$$(1-e^{-0.25})\times 100=22.11992\ldots$$
$$\boxed{\text{decrease} \approx 22.12\text{ percent per hour}}$$
QUESTION 4
5 marks
Easy
The number of downloads of a new app, $t$ weeks after its launch, is modeled by
$$N(t)=1200(1.45)^t$$
(a) Write down the number of downloads in the launch week ($t=0$). [1]
(b) Find $N(3)$, giving your answer to the nearest whole download. [2]
(c) Write down the percentage rate of growth per week. [2]
Show complete worked solution
(a) $N(0)=1200(1.45)^0=1200$
$$\boxed{N(0)=1200 \text{ downloads}}$$
(b) Using the GDC:
$$N(3)=1200(1.45)^3=3658.350\ldots$$
$$\boxed{N(3)\approx 3658 \text{ downloads}}$$
(c) The base is $b=1.45=1+r$, so $r=0.45$.
$$\boxed{\text{45 percent growth per week}}$$
QUESTION 5
4 marks
Easy
The population of an endangered bird species on a protected island is modeled by
$$P(t)=40(1.06)^t$$
where $t$ is the number of years since monitoring began, $P(t)>0$.
(a) Write down the horizontal asymptote of the graph of $P$, and state the $y$-intercept. [2]
(b) Find $P(15)$, giving your answer to the nearest whole bird. [2]
Show complete worked solution
(a) As $t\to -\infty$, $1.06^t\to 0$, so $P(t)\to 0$. The graph of $P$ has horizontal asymptote
$$\boxed{P=0}$$
The $y$-intercept occurs at $t=0$: $P(0)=40(1.06)^0=40$, so the $y$-intercept is $\boxed{(0,40)}$.
(b) Using the GDC:
$$P(15)=40(1.06)^{15}=95.86233\ldots$$
$$\boxed{P(15)\approx 96 \text{ birds}}$$
QUESTION 6
5 marks
Easy
A cup of coffee cools according to the model
$$T(t)=22+68e^{-0.11t}$$
where $T(t)$ is the temperature of the coffee in degrees Celsius, $t$ minutes after it is poured.
(a) Write down the horizontal asymptote of the graph of $T$, and explain what it represents in this context. [2]
(b) Write down the temperature of the coffee when it is poured. [1]
(c) Find the temperature of the coffee after 10 minutes, correct to 1 decimal place. [2]
Show complete worked solution
(a) As $t\to\infty$, $e^{-0.11t}\to 0$, so $T(t)\to 22$. The horizontal asymptote is
$$\boxed{T=22}$$
This represents the temperature of the surrounding room, i.e. the temperature that the coffee approaches (but never quite reaches) as time passes.
(b) At $t=0$:
$$T(0)=22+68e^{0}=22+68=90$$
$$\boxed{T(0)=90^{\circ}\text{C}}$$
(c) Using the GDC:
$$T(10)=22+68e^{-0.11(10)}=22+68e^{-1.1}=44.63523\ldots$$
$$\boxed{T(10)\approx 44.6^{\circ}\text{C}}$$
QUESTION 7
4 marks
Easy
A newly-discovered isotope decays according to
$$m(t)=200(0.5)^{t/8}$$
where $m(t)$ is the mass remaining, in grams, and $t$ is the time in days.
(a) Write down the half-life of the isotope, as given directly by the form of the model. [1]
(b) Find $m(8)$. [1]
(c) Find $m(16)$, and verify that this is consistent with your answer to part (a). [2]
Show complete worked solution
(a) The model has the form $m(t)=m_0(0.5)^{t/h}$ with $h=8$, so the mass halves every 8 days.
$$\boxed{\text{half-life}=8 \text{ days}}$$
(b) $$m(8)=200(0.5)^{8/8}=200(0.5)^1=100$$
$$\boxed{m(8)=100 \text{ g}}$$
(c) $$m(16)=200(0.5)^{16/8}=200(0.5)^2=200(0.25)=50$$
$$\boxed{m(16)=50 \text{ g}}$$
This is consistent with part (a): going from $t=8$ to $t=16$ is exactly one more half-life (8 more days), and the mass has indeed halved again, from 100 g to 50 g.
QUESTION 8
4 marks
Easy
The number of times a video has been shared on social media is modeled by
$$N(t)=10(2)^t$$
where $t$ is the number of days since it was posted.
(a) Write down $N(0)$ and explain its meaning. [1]
(b) Find $N(5)$. [2]
(c) Describe, in words, what the base of this exponential model tells you about how the number of shares changes each day. [1]
Show complete worked solution
(a) $N(0)=10(2)^0=10$. This means the video had already been shared 10 times at the moment it was posted (e.g. by early viewers).
$$\boxed{N(0)=10 \text{ shares}}$$
(b) $$N(5)=10(2)^5=10(32)=320$$
$$\boxed{N(5)=320 \text{ shares}}$$
(c) Since the base is $b=2$, the number of shares doubles (increases by 100 percent) every day.
$$\boxed{\text{the number of shares doubles each day}}$$
QUESTION 9
5 marks
Easy
The number of cells in a yeast culture is modeled by
$$y(t)=250e^{0.18t}$$
where $y(t)$ is the number of cells and $t$ is the time in hours.
(a) Write down the initial number of cells in the culture. [1]
(b) Find $y(6)$, giving your answer to the nearest whole cell. [2]
(c) Find the percentage growth rate of the culture per hour, correct to 2 decimal places. [2]
Show complete worked solution
(a) $y(0)=250e^{0}=250$
$$\boxed{y(0)=250 \text{ cells}}$$
(b) Using the GDC:
$$y(6)=250e^{0.18(6)}=250e^{1.08}=736.1699\ldots$$
$$\boxed{y(6)\approx 736 \text{ cells}}$$
(c) Each hour, the population is multiplied by $e^{0.18}$. The percentage growth rate per hour is
$$(e^{0.18}-1)\times 100=19.72174\ldots$$
$$\boxed{\text{growth rate}\approx 19.72\text{ percent per hour}}$$
QUESTION 10
5 marks
Easy
Atmospheric pressure, in kPa, is modeled as a function of altitude by
$$P(h)=101(0.88)^h$$
where $h$ is the altitude in kilometers above sea level.
(a) Write down the atmospheric pressure at sea level ($h=0$). [1]
(b) Write down the percentage rate of decrease in pressure per kilometer of altitude. [2]
(c) Find $P(5)$, giving your answer correct to 1 decimal place. [2]
Show complete worked solution
(a) $P(0)=101(0.88)^0=101$
$$\boxed{P(0)=101 \text{ kPa}}$$
(b) The base is $b=0.88=1-r$, so $r=0.12$.
$$\boxed{\text{12 percent decrease per km}}$$
(c) Using the GDC:
$$P(5)=101(0.88)^5=53.30092\ldots$$
$$\boxed{P(5)\approx 53.3 \text{ kPa}}$$
QUESTION 11
4 marks
Easy
An infestation of insects in a greenhouse is modeled by
$$N(t)=30(1.6)^t$$
where $N(t)$ is the number of insects and $t$ is the time in days.
(a) Write down the initial number of insects. [1]
(b) Find $N(4)$, giving your answer to the nearest whole insect. [2]
(c) Write down the percentage growth rate per day. [1]
Show complete worked solution
(a) $N(0)=30(1.6)^0=30$
$$\boxed{N(0)=30 \text{ insects}}$$
(b) Using the GDC:
$$N(4)=30(1.6)^4=196.6080\ldots$$
$$\boxed{N(4)\approx 197 \text{ insects}}$$
(c) The base is $b=1.6=1+r$, so $r=0.6$.
$$\boxed{\text{60 percent growth per day}}$$
QUESTION 12
4 marks
Easy
A patient is given a dose of medication. The amount remaining in their body is modeled by
$$A(t)=500(0.85)^t$$
where $A(t)$ is measured in mg and $t$ is the time in hours since the dose was taken.
(a) Write down the initial dose. [1]
(b) Write down the percentage rate at which the medication is eliminated from the body each hour. [1]
(c) Find $A(6)$, giving your answer to the nearest mg. [2]
Show complete worked solution
(a) $A(0)=500(0.85)^0=500$
$$\boxed{A(0)=500 \text{ mg}}$$
(b) The base is $b=0.85=1-r$, so $r=0.15$.
$$\boxed{\text{15 percent eliminated per hour}}$$
(c) Using the GDC:
$$A(6)=500(0.85)^6=188.5748\ldots$$
$$\boxed{A(6)\approx 189 \text{ mg}}$$
QUESTION 13
5 marks
Easy
Due to rising ocean temperatures, the living area of a coral reef, in $\text{m}^2$, is modeled by
$$A(t)=1000e^{-0.04t}$$
where $t$ is the time in years since monitoring began.
(a) Write down $A(0)$ and explain its meaning in context. [1]
(b) Find $A(20)$, giving your answer to the nearest $\text{m}^2$. [2]
(c) Write down the horizontal asymptote of $A$, and briefly explain what it suggests (unrealistically) about the long-term behavior of the reef area under this model. [2]
Show complete worked solution
(a) $A(0)=1000e^{0}=1000$. This is the living area of the coral reef, in $\text{m}^2$, at the start of monitoring.
$$\boxed{A(0)=1000 \text{ m}^2}$$
(b) Using the GDC:
$$A(20)=1000e^{-0.04(20)}=1000e^{-0.8}=449.3290\ldots$$
$$\boxed{A(20)\approx 449 \text{ m}^2}$$
(c) As $t\to\infty$, $e^{-0.04t}\to 0$, so $A(t)\to 0$.
$$\boxed{A=0}$$
This suggests that the living area of the reef shrinks towards, but never reaches, zero -- i.e. the model predicts the reef never fully disappears but becomes arbitrarily small, which is only a reasonable approximation and would eventually break down in reality (e.g. once the reef area is too small to sustain a living colony at all).
QUESTION 14
4 marks
Easy
The number of active users of a mobile app is modeled by
$$N(t)=5000(0.9)^t$$
where $t$ is the number of months after launch.
(a) Write down the number of active users at launch. [1]
(b) Find $N(12)$, giving your answer to the nearest whole user. [2]
(c) Write down the percentage rate of decrease in active users per month. [1]
Show complete worked solution
(a) $N(0)=5000(0.9)^0=5000$
$$\boxed{N(0)=5000 \text{ users}}$$
(b) Using the GDC:
$$N(12)=5000(0.9)^{12}=1412.1477\ldots$$
$$\boxed{N(12)\approx 1412 \text{ users}}$$
(c) The base is $b=0.9=1-r$, so $r=0.1$.
$$\boxed{\text{10 percent decrease per month}}$$
QUESTION 15
5 marks
Easy
The volume of a glacier, in $\text{km}^3$, is decreasing according to the model
$$V(t)=2.5e^{-0.02t}$$
where $t$ is the number of years since measurements began.
(a) Write down the initial volume of the glacier. [1]
(b) Find $V(30)$, giving your answer correct to 3 significant figures. [2]
(c) Find the percentage rate of decrease in volume per year, correct to 2 decimal places. [2]
Show complete worked solution
(a) $V(0)=2.5e^{0}=2.5$
$$\boxed{V(0)=2.5 \text{ km}^3}$$
(b) Using the GDC:
$$V(30)=2.5e^{-0.02(30)}=2.5e^{-0.6}=1.372029\ldots$$
$$\boxed{V(30)\approx 1.37 \text{ km}^3}$$
(c) The percentage rate of decrease per year is
$$(1-e^{-0.02})\times 100=1.98013\ldots$$
$$\boxed{\text{decrease}\approx 1.98\text{ percent per year}}$$
QUESTION 16
4 marks
Easy
The intensity of light passing through seawater, in lumens, is modeled by
$$I(d)=100(0.8)^d$$
where $d$ is the depth in meters below the surface.
(a) Write down the light intensity at the surface ($d=0$). [1]
(b) Write down the percentage rate of decrease in light intensity per meter of depth. [1]
(c) Find $I(3)$, correct to 1 decimal place. [2]
Show complete worked solution
(a) $I(0)=100(0.8)^0=100$
$$\boxed{I(0)=100 \text{ lumens}}$$
(b) The base is $b=0.8=1-r$, so $r=0.2$.
$$\boxed{\text{20 percent decrease per meter}}$$
(c) Using the GDC:
$$I(3)=100(0.8)^3=51.2000\ldots$$
$$\boxed{I(3)\approx 51.2 \text{ lumens}}$$
QUESTION 17
4 marks
Easy
A population of rabbits on an island, unchecked by predators, is modeled by
$$P(t)=60(1.22)^t$$
where $t$ is the time in months.
(a) Write down the initial population. [1]
(b) Write down the percentage growth rate per month. [1]
(c) Find $P(5)$, giving your answer to the nearest whole rabbit. [2]
Show complete worked solution
(a) $P(0)=60(1.22)^0=60$
$$\boxed{P(0)=60 \text{ rabbits}}$$
(b) The base is $b=1.22=1+r$, so $r=0.22$.
$$\boxed{\text{22 percent growth per month}}$$
(c) Using the GDC:
$$P(5)=60(1.22)^5=162.1625\ldots$$
$$\boxed{P(5)\approx 162 \text{ rabbits}}$$
QUESTION 18
5 marks
Easy
The number of bacteria (in thousands) in a pond is modeled by
$$N(t)=80e^{0.3t}$$
where $t$ is the time in days.
(a) Write down the initial number of bacteria (in thousands). [1]
(b) Find $N(4)$, giving your answer correct to 1 decimal place. [2]
(c) Find the percentage growth rate per day, correct to 2 decimal places. [2]
Show complete worked solution
(a) $N(0)=80e^{0}=80$
$$\boxed{N(0)=80 \text{ thousand bacteria}}$$
(b) Using the GDC:
$$N(4)=80e^{0.3(4)}=80e^{1.2}=265.60935\ldots$$
$$\boxed{N(4)\approx 265.6 \text{ thousand bacteria}}$$
(c) $$(e^{0.3}-1)\times 100=34.98588\ldots$$
$$\boxed{\text{growth rate}\approx 34.99\text{ percent per day}}$$
QUESTION 19
4 marks
Easy
The number of subscribers to a streaming service is modeled by
$$N(t)=4000(1.12)^t$$
where $t$ is the number of months since the service launched.
(a) Write down the number of subscribers at launch. [1]
(b) Find $N(6)$, giving your answer to the nearest whole subscriber. [2]
(c) Write down the percentage growth rate per month. [1]
Show complete worked solution
(a) $N(0)=4000(1.12)^0=4000$
$$\boxed{N(0)=4000 \text{ subscribers}}$$
(b) Using the GDC:
$$N(6)=4000(1.12)^6=7895.2907\ldots$$
$$\boxed{N(6)\approx 7895 \text{ subscribers}}$$
(c) The base is $b=1.12=1+r$, so $r=0.12$.
$$\boxed{\text{12 percent growth per month}}$$
QUESTION 20
5 marks
Easy
The concentration of a pollutant in a lake, in mg/L, is modeled by
$$C(t)=45(0.93)^t$$
where $t$ is the number of weeks since a clean-up program began.
(a) Write down the horizontal asymptote of the graph of $C$, and write down the $y$-intercept. [2]
(b) Find $C(10)$, giving your answer correct to 2 decimal places. [2]
(c) Write down the percentage rate of decrease in concentration per week. [1]
Show complete worked solution
(a) As $t\to\infty$, $0.93^t\to 0$, so $C(t)\to 0$. The horizontal asymptote is
$$\boxed{C=0}$$
The $y$-intercept is at $t=0$: $C(0)=45(0.93)^0=45$, so the $y$-intercept is $\boxed{(0,45)}$.
(b) Using the GDC:
$$C(10)=45(0.93)^{10}=21.77920\ldots$$
$$\boxed{C(10)\approx 21.78 \text{ mg/L}}$$
(c) The base is $b=0.93=1-r$, so $r=0.07$.
$$\boxed{\text{7 percent decrease per week}}$$
QUESTION 21
6 marks
Medium
The population of a small town is modeled by
$$P(t)=800(1.045)^t$$
where $P(t)$ is the population and $t$ is the number of years after the year 2020.
(a) Find the annual percentage growth rate of the population. [2]
(b) Find the smallest integer value of $t$ for which $P(t)>1500$. [4]
Show complete worked solution
(a) Writing $b=1.045=1+r$, the annual growth rate is $r=0.045$.
$$\boxed{\text{4.5 percent per year}}$$
(b) We require $800(1.045)^t>1500$.
Dividing both sides by 800:
$$1.045^t>\frac{1500}{800}=1.875$$
Taking $\ln$ of both sides (valid since both sides are positive):
$$t\ln(1.045)>\ln(1.875)$$
Since $\ln(1.045)>0$, dividing does not reverse the inequality:
$$t>\frac{\ln(1.875)}{\ln(1.045)}=14.28108\ldots$$
Since $t$ must be a whole number of years and $t>14.281\ldots$, the smallest integer satisfying this is $t=15$.
Check using the GDC: $P(15)=800(1.045)^{15}=1548.23>1500$, while $P(14)=800(1.045)^{14}=1481.56<1500$, confirming the result.
$$\boxed{t=15}$$
QUESTION 22
6 marks
Medium
A radioactive substance decays according to
$$m(t)=250e^{-0.083t}$$
where $m(t)$ is the mass in grams and $t$ is the time in years.
(a) Find the percentage rate of decay per year, correct to 2 decimal places. [2]
(b) Find the half-life of the substance algebraically, correct to 2 decimal places. [4]
Show complete worked solution
(a) Each year, the mass is multiplied by $e^{-0.083}$. The percentage decrease per year is
$$(1-e^{-0.083})\times 100=7.96489\ldots$$
$$\boxed{\text{decay rate}\approx 7.96\text{ percent per year}}$$
(b) The half-life is the value of $t$ for which $m(t)=\dfrac{1}{2}m(0)=125$.
$$250e^{-0.083t}=125$$
$$e^{-0.083t}=0.5$$
Taking the natural logarithm of both sides:
$$-0.083t=\ln(0.5)$$
$$t=\frac{\ln(0.5)}{-0.083}=\frac{\ln 2}{0.083}=8.35117\ldots$$
$$\boxed{\text{half-life}\approx 8.35 \text{ years}}$$
QUESTION 23
6 marks
Medium
The concentration of a drug in a patient's bloodstream, in mg/L, is modeled by
$$C(t)=80(0.75)^t$$
where $t$ is the time in hours after the drug is administered.
(a) Find the percentage decrease in concentration per hour. [2]
(b) Find the smallest integer number of hours after which the concentration first drops below 10 mg/L. [4]
Show complete worked solution
(a) The base is $b=0.75=1-r$, so $r=0.25$.
$$\boxed{\text{25 percent decrease per hour}}$$
(b) We require $80(0.75)^t<10$.
$$0.75^t<\frac{10}{80}=0.125$$
Taking $\ln$ of both sides. Since $\ln(0.75)<0$, the inequality reverses when dividing by it:
$$t\ln(0.75)<\ln(0.125)$$
$$t>\frac{\ln(0.125)}{\ln(0.75)}=7.22826\ldots$$
The smallest integer satisfying this is $t=8$.
Check using the GDC: $C(8)=80(0.75)^{8}=8.0090<10$, confirming the drug concentration first drops below 10 mg/L during the 8th hour.
$$\boxed{t=8 \text{ hours}}$$
QUESTION 24
6 marks
Medium
A bacteria population is modeled by
$$P(t)=100e^{0.25t}$$
where $P(t)$ is the number of bacteria and $t$ is the time in hours.
(a) Find the doubling time of the population algebraically, correct to 3 significant figures. [4]
(b) Evaluate $P(t)$ at the doubling time found in part (a), and confirm this is consistent with your answer. [2]
Show complete worked solution
(a) The doubling time $T$ satisfies $P(T)=2P(0)$, i.e.
$$100e^{0.25T}=200$$
$$e^{0.25T}=2$$
Taking the natural logarithm of both sides:
$$0.25T=\ln 2$$
$$T=\frac{\ln 2}{0.25}=2.772589\ldots$$
$$\boxed{T\approx 2.77 \text{ hours}}$$
(b) Using the GDC:
$$P(2.772589)=100e^{0.25(2.772589)}=100e^{\ln 2}=100(2)=200$$
Since $P(0)=100$, this confirms $P(T)=200=2\times100=2P(0)$, so the population has indeed doubled after $T\approx 2.77$ hours.
$$\boxed{P(T)=200}$$
QUESTION 25
7 marks
Medium
Two towns' populations are modeled by
$$P_A(t)=5000(1.02)^t \qquad P_B(t)=4000(1.035)^t$$
where $t$ is the number of years from now.
(a) Find the annual percentage growth rate of each town's population. [2]
(b) Using technology, find the smallest integer value of $t$ for which the population of Town B exceeds the population of Town A. [5]
Show complete worked solution
(a) For Town A, $b_A=1.02=1+r_A$, so $r_A=0.02$: growth rate $\boxed{2\text{ percent per year}}$.
For Town B, $b_B=1.035=1+r_B$, so $r_B=0.035$: growth rate $\boxed{3.5\text{ percent per year}}$.
(b) Although Town A starts larger (5000 vs 4000), Town B grows at a faster percentage rate (3.5 percent vs 2 percent), so its population will eventually overtake Town A's.
Using the GDC table or graph feature to evaluate $P_A(t)$ and $P_B(t)$ for consecutive integer values of $t$:
At $t=15$, $P_B(t)P_A(t)$ for the first time.
$$\boxed{t=16 \text{ years}}$$
| $t$ | $P_A(t)$ | $P_B(t)$ |
|---|---|---|
| 15 | 6729.3 | 6701.4 |
| 16 | 6863.9 | 6935.9 |
QUESTION 26
6 marks
Medium
The number of views of a viral video is modeled by
$$V(t)=500(1.85)^t$$
where $V(t)$ is the number of views and $t$ is the number of days since it was posted.
(a) Write down the daily percentage growth rate of the number of views. [1]
(b) Find the smallest integer number of days after which the number of views first exceeds 1,000,000. [5]
Show complete worked solution
(a) The base is $b=1.85=1+r$, so $r=0.85$.
$$\boxed{\text{85 percent growth per day}}$$
(b) We require $500(1.85)^t>1\,000\,000$.
$$1.85^t>\frac{1\,000\,000}{500}=2000$$
Taking $\ln$ of both sides:
$$t\ln(1.85)>\ln(2000)$$
$$t>\frac{\ln(2000)}{\ln(1.85)}=12.35546\ldots$$
The smallest integer satisfying this is $t=13$.
Check using the GDC: $V(12)=500(1.85)^{12}=803583.0<1\,000\,000$, but $V(13)=500(1.85)^{13}=1486628.6>1\,000\,000$.
$$\boxed{t=13 \text{ days}}$$
QUESTION 27
7 marks
Medium
A newly-discovered isotope decays according to
$$m(t)=40(0.5)^{t/18}$$
where $m(t)$ is the mass remaining, in grams, and $t$ is the time in days.
(a) This decay can equivalently be written in the form $m(t)=40e^{-kt}$. Find the value of $k$, correct to 4 significant figures. [4]
(b) Use either form of the model to find the mass remaining after 50 days, correct to 3 significant figures. [3]
Show complete worked solution
(a) Since both expressions represent the same function, $40e^{-kt}=40(0.5)^{t/18}$, so $e^{-kt}=(0.5)^{t/18}$ for all $t$.
Taking natural logarithms:
$$-kt=\frac{t}{18}\ln(0.5)$$
Dividing both sides by $t$ (this holds for all $t\neq 0$, so it determines $k$):
$$-k=\frac{\ln(0.5)}{18}=\frac{-\ln 2}{18}$$
$$k=\frac{\ln 2}{18}=0.038508\ldots$$
$$\boxed{k\approx 0.0385}$$
(b) Using the original half-life form on the GDC:
$$m(50)=40(0.5)^{50/18}=5.83265\ldots$$
$$\boxed{m(50)\approx 5.83 \text{ g}}$$
(As a check, using the equivalent form: $m(50)=40e^{-0.0385(50)}=5.83265\ldots$, which agrees.)
QUESTION 28
7 marks
Medium
A metal rod, heated in a furnace, cools according to
$$T(t)=18+82e^{-0.09t}$$
where $T(t)$ is the temperature in degrees Celsius and $t$ is the time in minutes after it is removed from the furnace.
(a) Write down the horizontal asymptote of $T$, and interpret it in context. [2]
(b) Find the time taken for the rod to cool to 30 degrees Celsius, correct to 2 decimal places. [5]
Show complete worked solution
(a) As $t\to\infty$, $e^{-0.09t}\to 0$, so $T(t)\to 18$.
$$\boxed{T=18}$$
This represents the temperature of the surrounding air, which the rod's temperature approaches as it cools.
(b) We require $T(t)=30$:
$$18+82e^{-0.09t}=30$$
$$82e^{-0.09t}=12$$
$$e^{-0.09t}=\frac{12}{82}=0.146341\ldots$$
Taking the natural logarithm of both sides:
$$-0.09t=\ln\left(\frac{12}{82}\right)$$
$$t=\frac{\ln(12/82)}{-0.09}=21.353473\ldots$$
$$\boxed{t\approx 21.35 \text{ minutes}}$$
QUESTION 29
7 marks
Medium
A population is modeled by
$$P(t)=P_0(1.07)^t$$
where $P_0$ is the initial population and $t$ is measured in years.
(a) By solving $2=1.07^t$, show that the doubling time of the population is approximately 10.2 years. [4]
(b) Find the time taken for the population to triple, correct to 3 significant figures. [3]
Show complete worked solution
(a) The doubling time $T$ satisfies $P(T)=2P_0$, i.e. $P_0(1.07)^T=2P_0$, so
$$1.07^T=2$$
Taking the natural logarithm of both sides:
$$T\ln(1.07)=\ln 2$$
$$T=\frac{\ln 2}{\ln(1.07)}=10.244768\ldots$$
$$\boxed{T\approx 10.2 \text{ years} \approx 10.2 \text{ years}}$$
(b) Similarly, the tripling time $T_3$ satisfies $1.07^{T_3}=3$:
$$T_3=\frac{\ln 3}{\ln(1.07)}=16.237574\ldots$$
$$\boxed{T_3\approx 16.2 \text{ years}}$$
QUESTION 30
6 marks
Medium
The amount of a medication remaining in a patient's body is modeled by
$$A(t)=300(0.5)^{t/6}$$
where $A(t)$ is measured in mg and $t$ is the time in hours after the dose is taken.
(a) By evaluating $A(6)$, verify that the half-life of the medication is 6 hours. [2]
(b) Find the time taken for the amount remaining to fall to 20 mg, correct to 2 decimal places. [4]
Show complete worked solution
(a) $$A(6)=300(0.5)^{6/6}=300(0.5)^1=150$$
Since $A(0)=300(0.5)^0=300$ and $A(6)=150=\dfrac{1}{2}(300)$, the amount has halved after 6 hours, confirming the half-life is $\boxed{6 \text{ hours}}$.
(b) We require $A(t)=20$:
$$300(0.5)^{t/6}=20$$
$$(0.5)^{t/6}=\frac{20}{300}=0.066667\ldots$$
Taking the natural logarithm of both sides:
$$\frac{t}{6}\ln(0.5)=\ln\left(\frac{1}{15}\right)$$
$$t=\frac{6\ln(1/15)}{\ln(0.5)}=23.441344\ldots$$
$$\boxed{t\approx 23.44 \text{ hours}}$$
QUESTION 31
6 marks
Medium
A population is modeled by
$$N(t)=1000e^{kt}$$
where $t$ is the time in years. It is known that the population grows by 4 percent each year.
(a) Find the value of $k$, correct to 4 decimal places. [3]
(b) Find $N(10)$, giving your answer to the nearest whole number. [3]
Show complete worked solution
(a) A 4 percent annual growth rate means that each year, the population is multiplied by $1.04$, i.e. $e^{k}=1.04$.
Taking the natural logarithm of both sides:
$$k=\ln(1.04)=0.039221\ldots$$
$$\boxed{k\approx 0.0392}$$
(b) Using the GDC:
$$N(10)=1000e^{k(10)}=1000e^{0.0392\times10}=1480.2443\ldots$$
(equivalently, $N(10)=1000(1.04)^{10}=1480.2443\ldots$, since $e^{10k}=(e^k)^{10}=1.04^{10}$)
$$\boxed{N(10)\approx 1480}$$
QUESTION 32
7 marks
Medium
The concentration of a pollutant in a reservoir, in mg/L, is modeled by
$$C(t)=200(0.88)^t$$
where $t$ is the time in weeks since a filtration system was installed.
(a) Write down the percentage rate of decrease in concentration per week. [2]
(b) Find the smallest integer number of weeks after which the concentration first falls below 5 percent of its initial value. [5]
Show complete worked solution
(a) The base is $b=0.88=1-r$, so $r=0.12$.
$$\boxed{\text{12 percent decrease per week}}$$
(b) 5 percent of the initial concentration $C(0)=200$ is $0.05(200)=10$ mg/L.
We require $200(0.88)^t<10$:
$$0.88^t<\frac{10}{200}=0.05$$
Taking $\ln$ of both sides. Since $\ln(0.88)<0$, the inequality reverses:
$$t>\frac{\ln(0.05)}{\ln(0.88)}=23.43467\ldots$$
The smallest integer satisfying this is $t=24$.
Check using the GDC: $C(24)=200(0.88)^{24}=9.3028<10$, confirming this is the first week in which the concentration is below 5 percent of the initial value.
$$\boxed{t=24 \text{ weeks}}$$
QUESTION 33
6 marks
Medium
Two radioactive isotopes, X and Y, each start with a mass of 100 g, and decay according to
$$m_X(t)=100e^{-0.05t} \qquad m_Y(t)=100e^{-0.08t}$$
where $t$ is measured in years.
(a) Find the half-life of each isotope, correct to 2 decimal places. [4]
(b) State, with a reason, which isotope reaches half of its initial mass first. [2]
Show complete worked solution
(a) For a general model $m(t)=m_0e^{-kt}$, the half-life $T$ satisfies $e^{-kT}=0.5$, so $T=\dfrac{\ln 2}{k}$.
For isotope X, $k_X=0.05$:
$$T_X=\frac{\ln 2}{0.05}=13.86294\ldots$$
$$\boxed{T_X\approx 13.86 \text{ years}}$$
For isotope Y, $k_Y=0.08$:
$$T_Y=\frac{\ln 2}{0.08}=8.66434\ldots$$
$$\boxed{T_Y\approx 8.66 \text{ years}}$$
(b) Since $T_Y\approx8.66$ years is less than $T_X\approx13.86$ years, isotope Y reaches half its initial mass first. This makes sense because Y has the larger decay constant ($k_Y=0.08>k_X=0.05$), meaning it decays more rapidly.
$$\boxed{\text{isotope Y}}$$
QUESTION 34
6 marks
Medium
An insect population is modeled by
$$N(t)=25(1.3)^t$$
where $t$ is the time in weeks.
(a) Find the doubling time of the population algebraically, correct to 3 significant figures. [4]
(b) Write down the percentage growth rate of the population per week. [2]
Show complete worked solution
(a) The doubling time $T$ satisfies $25(1.3)^T=50$, i.e.
$$1.3^T=2$$
Taking the natural logarithm of both sides:
$$T\ln(1.3)=\ln 2$$
$$T=\frac{\ln 2}{\ln(1.3)}=2.641927\ldots$$
$$\boxed{T\approx 2.64 \text{ weeks}}$$
(b) The base is $b=1.3=1+r$, so $r=0.3$.
$$\boxed{\text{30 percent growth per week}}$$
QUESTION 35
7 marks
Medium
A yeast culture doubles in size every 4 hours. The population is modeled by
$$N(t)=50e^{kt}$$
where $N(t)$ is the number of cells and $t$ is the time in hours.
(a) Find the value of $k$, correct to 4 decimal places. [4]
(b) Find $N(10)$, giving your answer to the nearest whole cell. [3]
Show complete worked solution
(a) Since the population doubles every 4 hours, $N(4)=2N(0)$:
$$50e^{4k}=100$$
$$e^{4k}=2$$
Taking the natural logarithm of both sides:
$$4k=\ln 2$$
$$k=\frac{\ln 2}{4}=0.173287\ldots$$
$$\boxed{k\approx 0.1733}$$
(b) Using the GDC:
$$N(10)=50e^{0.1733(10)}=282.842712\ldots$$
$$\boxed{N(10)\approx 283 \text{ cells}}$$
QUESTION 36
7 marks
Medium
The volume of a glacier, in $\text{km}^3$, is modeled by
$$V(t)=5e^{-0.015t}$$
where $t$ is the time in years.
(a) Find the time taken for the volume of the glacier to halve, correct to 1 decimal place. [5]
(b) Find the percentage rate of decrease in volume per year, correct to 2 decimal places. [2]
Show complete worked solution
(a) We require $V(t)=\dfrac{1}{2}V(0)=2.5$:
$$5e^{-0.015t}=2.5$$
$$e^{-0.015t}=0.5$$
Taking the natural logarithm of both sides:
$$-0.015t=\ln(0.5)$$
$$t=\frac{\ln 2}{0.015}=46.209812\ldots$$
$$\boxed{t\approx 46.2 \text{ years}}$$
(b) $$(1-e^{-0.015})\times 100=1.48881\ldots$$
$$\boxed{\text{decrease}\approx 1.49\text{ percent per year}}$$
QUESTION 37
7 marks
Medium
The number of active users of an app is declining according to
$$N(t)=8000(0.85)^t$$
where $t$ is the number of months since a rival app launched.
(a) Find the smallest integer number of months after which the number of active users first falls below 1000. [5]
(b) Write down the percentage rate of decrease in active users per month. [2]
Show complete worked solution
(a) We require $8000(0.85)^t<1000$:
$$0.85^t<\frac{1000}{8000}=0.125$$
Taking $\ln$ of both sides. Since $\ln(0.85)<0$, the inequality reverses:
$$t>\frac{\ln(0.125)}{\ln(0.85)}=12.79507\ldots$$
The smallest integer satisfying this is $t=13$.
Check: $N(13)=8000(0.85)^{13}=967.244<1000$.
$$\boxed{t=13 \text{ months}}$$
(b) The base is $b=0.85=1-r$, so $r=0.15$.
$$\boxed{\text{15 percent decrease per month}}$$
QUESTION 38
6 marks
Medium
Two quantities are modeled by
$$A(t)=100(1.06)^t \qquad B(t)=100(0.94)^t$$
where $A$ grows at 6 percent per period and $B$ decays at 6 percent per period, both starting from an initial value of 100.
(a) Find $A(10)$ and $B(10)$, each correct to 2 decimal places. [4]
(b) Explain, with reference to your answers in part (a), why the increase in $A$ is not "cancelled out" by the decrease in $B$, i.e. why $A(10)$ is more than 100 more above the starting value than $B(10)$ is below it. [2]
Show complete worked solution
(a) Using the GDC:
$$A(10)=100(1.06)^{10}=179.084770\ldots$$
$$\boxed{A(10)\approx 179.08}$$
$$B(10)=100(0.94)^{10}=53.861511\ldots$$
$$\boxed{B(10)\approx 53.86}$$
(b) $A(10)-100\approx79.08$, while $100-B(10)\approx46.14$. The increase in $A$ (about 79.1) is considerably larger than the decrease in $B$ (about 46.1).
This is because percentage growth compounds on an increasing base: each year, $A$ grows by 6 percent of an ever-larger amount, so the actual (absolute) increases get bigger and bigger over time. In contrast, $B$ decays by 6 percent of an ever-shrinking amount, so its absolute decreases get smaller and smaller, meaning $B$ can never fall below 0 and its total drop is bounded, whereas $A$'s total rise is unbounded. Growth and decay at the "same" percentage rate are therefore not symmetric.
$$\boxed{\text{compounding growth outpaces compounding decay at the same percentage rate}}$$
QUESTION 39
7 marks
Medium
The number of trees in a rapidly-growing plantation increases by a factor of 1.5 every 3 years. The initial number of trees is 200, so the population can be modeled by
$$P(t)=200(1.5)^{t/3}$$
where $t$ is the time in years.
(a) Find the equivalent annual percentage growth rate of the plantation, correct to 2 decimal places. [4]
(b) Find $P(10)$, giving your answer to the nearest whole tree. [3]
Show complete worked solution
(a) We want to write $P(t)=200(1+r)^t$ for some annual growth rate $r$, where $(1+r)^3=1.5$ (since 3 years of annual growth at rate $r$ must match one factor of 1.5).
$$1+r=1.5^{1/3}=1.144714\ldots$$
$$r=0.144714\ldots$$
$$\boxed{r\approx 14.47\text{ percent per year}}$$
(b) Using the original model directly on the GDC:
$$P(10)=200(1.5)^{10/3}=772.68211\ldots$$
$$\boxed{P(10)\approx 773 \text{ trees}}$$
QUESTION 40
7 marks
Medium
A radioactive sample decays according to
$$m(t)=m_0e^{-0.12t}$$
where $m(t)$ is the mass in grams, $t$ is the time in years, and $m_0$ is the initial mass. It is known that $m(5)=40$.
(a) Find the value of $m_0$, correct to 3 significant figures. [3]
(b) Find the half-life of the sample, correct to 2 decimal places. [4]
Show complete worked solution
(a) Substituting $t=5$ and $m(5)=40$:
$$m_0e^{-0.12(5)}=40$$
$$m_0e^{-0.6}=40$$
$$m_0=\frac{40}{e^{-0.6}}=40e^{0.6}=72.88475\ldots$$
$$\boxed{m_0\approx 72.9 \text{ g}}$$
(b) The half-life $T$ satisfies $e^{-0.12T}=0.5$, independent of $m_0$:
$$-0.12T=\ln(0.5)$$
$$T=\frac{\ln 2}{0.12}=5.776227\ldots$$
$$\boxed{T\approx 5.78 \text{ years}}$$
QUESTION 41
11 marks
Hard
A student places 80 bacteria in a petri dish. The population grows at a constant rate of 15 percent per hour.
(a) Write down a model $N(t)=ab^t$ for the number of bacteria present after $t$ hours. [2]
(b) Find $N(8)$, giving your answer to the nearest whole bacterium. [2]
(c) Find the doubling time of the population algebraically, correct to 3 significant figures. [3]
(d) Find the smallest integer number of hours after which the population first exceeds 5000 bacteria. [3]
(e) State one reason why this exponential model is unlikely to remain realistic for very large values of $t$. [1]
Show complete worked solution
(a) The initial number of bacteria is $a=80$. A growth rate of 15 percent per hour means the population is multiplied by $1+0.15=1.15$ each hour, so $b=1.15$.
$$\boxed{N(t)=80(1.15)^t}$$
(b) Using the GDC:
$$N(8)=80(1.15)^8=244.72183\ldots$$
$$\boxed{N(8)\approx 245 \text{ bacteria}}$$
(c) The doubling time $T$ satisfies $80(1.15)^T=160$, i.e. $1.15^T=2$.
Taking the natural logarithm of both sides:
$$T\ln(1.15)=\ln 2$$
$$T=\frac{\ln 2}{\ln(1.15)}=4.959484\ldots$$
$$\boxed{T\approx 4.96 \text{ hours}}$$
(d) We require $80(1.15)^t>5000$:
$$1.15^t>\frac{5000}{80}=62.5$$
Taking $\ln$ of both sides:
$$t>\frac{\ln(62.5)}{\ln(1.15)}=29.58721\ldots$$
The smallest integer satisfying this is $t=30$.
Check: $N(29)=80(1.15)^{29}=4606.04<5000$, while $N(30)=80(1.15)^{30}=5296.94>5000$.
$$\boxed{t=30 \text{ hours}}$$
(e) In reality, exponential growth cannot continue indefinitely: as the bacteria population grows, it will eventually be limited by factors such as the finite size of the petri dish, depletion of nutrients, or build-up of waste products, causing the actual growth rate to slow down (unlike the unbounded model $N(t)=80(1.15)^t$).
$$\boxed{\text{limited resources/space eventually slow real growth, but the exponential model has no such limit}}$$
QUESTION 42
10 marks
Hard
Two radioactive substances, X and Y, are being studied. Substance X has an initial mass of 80 g and a half-life of 25 years. Substance Y has an initial mass of 50 g and a half-life of 40 years. Both may be modeled in the form $m(t)=m_0e^{-kt}$, with $t$ in years.
(a) Find the decay constant $k$ for each substance, correct to 5 significant figures. [4]
(b) Find the mass of each substance remaining after 60 years, correct to 2 decimal places. [2]
(c) Find the time, in years, at which the two substances have equal mass. [4]
Show complete worked solution
(a) For a substance with half-life $H$, the decay constant satisfies $e^{-kH}=0.5$, so $k=\dfrac{\ln 2}{H}$.
For substance X, $H=25$:
$$k_X=\frac{\ln 2}{25}=0.027726\ldots$$
$$\boxed{k_X\approx 0.02773}$$
For substance Y, $H=40$:
$$k_Y=\frac{\ln 2}{40}=0.017329\ldots$$
$$\boxed{k_Y\approx 0.01733}$$
(b) $$m_X(60)=80e^{-0.02773(60)}=15.157166\ldots$$
$$\boxed{m_X(60)\approx 15.16 \text{ g}}$$
$$m_Y(60)=50e^{-0.01733(60)}=17.677670\ldots$$
$$\boxed{m_Y(60)\approx 17.68 \text{ g}}$$
(c) We require $m_X(t)=m_Y(t)$:
$$80e^{-k_Xt}=50e^{-k_Yt}$$
Dividing both sides by $50e^{-k_Xt}$:
$$\frac{80}{50}=e^{-k_Yt+k_Xt}=e^{(k_X-k_Y)t}$$
$$1.6=e^{(k_X-k_Y)t}$$
Taking the natural logarithm of both sides:
$$(k_X-k_Y)t=\ln(1.6)$$
$$t=\frac{\ln(1.6)}{k_X-k_Y}=\frac{\ln(1.6)}{0.02773-0.01733}=45.204794\ldots$$
$$\boxed{t\approx 45.20 \text{ years}}$$
Check using the GDC: at this value of $t$, $m_X(t)\approx22.844$ g and $m_Y(t)\approx22.844$ g, which agree (small difference due to rounding), confirming the masses are equal at this time.
QUESTION 43
10 marks
Hard
After a patient takes a dose of medication, the concentration of the drug in their bloodstream, in mg/L, is modeled by
$$C(t)=100e^{-0.2t}$$
where $t$ is the time in hours after the dose is taken. The medication is considered to be within its therapeutic window when its concentration is between 10 mg/L and 50 mg/L, inclusive.
(a) Find the time at which the concentration first falls to 50 mg/L, correct to 2 decimal places. [3]
(b) Find the time at which the concentration falls to 10 mg/L, correct to 2 decimal places. [3]
(c) Hence find the total length of time, in hours, during which the drug concentration lies within the therapeutic window. [2]
(d) Find the percentage of the initial dose remaining in the bloodstream after 10 hours, correct to 2 decimal places. [2]
Show complete worked solution
(a) We require $C(t)=50$:
$$100e^{-0.2t}=50$$
$$e^{-0.2t}=0.5$$
Taking the natural logarithm of both sides:
$$-0.2t=\ln(0.5)$$
$$t=\frac{\ln 2}{0.2}=3.465736\ldots$$
$$\boxed{t\approx 3.47 \text{ hours}}$$
(b) We require $C(t)=10$:
$$100e^{-0.2t}=10$$
$$e^{-0.2t}=0.1$$
$$-0.2t=\ln(0.1)$$
$$t=\frac{-\ln(0.1)}{0.2}=11.512925\ldots$$
$$\boxed{t\approx 11.51 \text{ hours}}$$
(c) The concentration starts at $C(0)=100$ mg/L, which is above the therapeutic window. It first enters the window (falling to 50 mg/L) at $t\approx3.47$ hours, and leaves the window (falling to 10 mg/L) at $t\approx11.51$ hours. The length of time within the window is
$$11.5129-3.4657=8.0472\ldots$$
$$\boxed{\text{duration}\approx 8.05 \text{ hours}}$$
(d) Using the GDC:
$$C(10)=100e^{-0.2(10)}=100e^{-2}=13.53353\ldots$$
Since $C(0)=100$, the percentage of the initial dose remaining is
$$\frac{C(10)}{100}\times 100=13.53353\ldots$$
$$\boxed{\approx 13.53\text{ percent}}$$
QUESTION 44
9 marks
Hard
The number of users of a new social media app is modeled by
$$U(t)=300(1.4)^t$$
where $U(t)$ is the number of users and $t$ is the number of days since launch.
(a) Write down the daily percentage growth rate of the number of users. [1]
(b) Find the doubling time of the number of users algebraically, correct to 3 significant figures. [3]
(c) Find the smallest integer number of days after which the number of users first exceeds 100,000. [3]
(d) Explain why this exponential model is unlikely to remain a good description of the number of users indefinitely, and identify a feature the true growth curve would need that this model does not have. [2]
Show complete worked solution
(a) The base is $b=1.4=1+r$, so $r=0.4$.
$$\boxed{\text{40 percent growth per day}}$$
(b) The doubling time $T$ satisfies $1.4^T=2$:
$$T\ln(1.4)=\ln 2$$
$$T=\frac{\ln 2}{\ln(1.4)}=2.060043\ldots$$
$$\boxed{T\approx 2.06 \text{ days}}$$
(c) We require $300(1.4)^t>100\,000$:
$$1.4^t>\frac{100\,000}{300}=333.\overline{3}$$
$$t>\frac{\ln(333.\overline{3})}{\ln(1.4)}=17.26485\ldots$$
The smallest integer satisfying this is $t=18$.
Check: $U(17)=300(1.4)^{17}=91474.0<100\,000$, while $U(18)=300(1.4)^{18}=128063.7>100\,000$.
$$\boxed{t=18 \text{ days}}$$
(d) The total number of users of any real app is bounded above by the size of the population that could plausibly use it (its potential market), whereas $U(t)=300(1.4)^t$ grows without bound as $t\to\infty$. In reality, growth would need to slow down and level off as the app approaches this maximum, i.e. the true growth curve would need a horizontal asymptote (a saturation/carrying-capacity level) that this simple exponential model does not have.
$$\boxed{\text{a purely exponential model has no upper bound, but real user numbers must saturate near a maximum market size}}$$
QUESTION 45
9 marks
Hard
Two bacteria colonies are being cultured. Colony A starts with 400 bacteria and doubles in size every 5 hours. Colony B starts with only 20 bacteria, but triples in size every 3 hours. Let $t$ denote time in hours.
(a) Show that colony A can be modeled by $N_A(t)=400b_A^t$ and colony B by $N_B(t)=20b_B^t$, finding the values of $b_A$ and $b_B$ correct to 4 significant figures. [4]
(b) Find the number of bacteria in each colony after 15 hours, correct to the nearest whole bacterium. [2]
(c) Using technology, find the smallest integer number of hours after which colony B's population first exceeds colony A's population. [3]
Show complete worked solution
(a) Colony A doubles every 5 hours, so $N_A(5)=2N_A(0)$:
$$400b_A^5=800 \implies b_A^5=2 \implies b_A=2^{1/5}=1.148698\ldots$$
$$\boxed{b_A\approx 1.1487}$$
Colony B triples every 3 hours, so $N_B(3)=3N_B(0)$:
$$20b_B^3=60 \implies b_B^3=3 \implies b_B=3^{1/3}=1.442250\ldots$$
$$\boxed{b_B\approx 1.4422}$$
So $N_A(t)=400(1.1487)^t$ and $N_B(t)=20(1.4422)^t$.
(b) Using the GDC:
$$N_A(15)=400(1.1487)^{15}=3200.0000\ldots$$
$$\boxed{N_A(15)\approx 3200 \text{ bacteria}}$$
$$N_B(15)=20(1.4422)^{15}=4860.0000\ldots$$
$$\boxed{N_B(15)\approx 4860 \text{ bacteria}}$$
(This makes sense: $15$ hours is exactly $3$ doubling-periods for A ($400\times2^3=3200$) and exactly $5$ tripling-periods for B ($20\times3^5=4860$).)
(c) Colony A starts far ahead (400 vs 20), but colony B grows at a much faster proportional rate per hour ($b_B\approx1.4422$ implies roughly 44.2 percent growth per hour, versus $b_A\approx1.1487$, roughly 14.9 percent growth per hour for A), so colony B is expected to catch up eventually. Using the GDC table feature to compare $N_A(t)$ and $N_B(t)$ for consecutive integer hours:
At $t=13$, $N_B(t)N_A(t)$ for the first time.
$$\boxed{t=14 \text{ hours}}$$
| $t$ | $N_A(t)$ | $N_B(t)$ |
|---|---|---|
| 13 | 2425.1 | 2336.4 |
| 14 | 2785.8 | 3369.7 |
QUESTION 46
10 marks
Hard
Conservationists are monitoring a population of an endangered species. The population is modeled by
$$P(t)=2000(0.96)^t$$
where $P(t)$ is the number of individuals and $t$ is the number of years since monitoring began.
(a) Write down the annual percentage rate of decline of the population. [1]
(b) Complete the table below, giving each value of $P(t)$ correct to the nearest whole individual. [3]
(c) Find the smallest integer number of years after which the population is first predicted to fall below 800 individuals (the level classified as "critically endangered" for this species). [3]
(d) Find the number of years it takes for the population to halve, correct to 2 decimal places. [3]
| $t$ (years) | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| $P(t)$ | ? | ? | ? | ? | ? |
Show complete worked solution
(a) The base is $b=0.96=1-r$, so $r=0.04$.
$$\boxed{\text{4 percent decline per year}}$$
(b) Substituting each value of $t$ into $P(t)=2000(0.96)^t$ using the GDC:
(c) We require $2000(0.96)^t<800$:
$$0.96^t<\frac{800}{2000}=0.4$$
Taking $\ln$ of both sides. Since $\ln(0.96)<0$, the inequality reverses:
$$t>\frac{\ln(0.4)}{\ln(0.96)}=22.44601\ldots$$
The smallest integer satisfying this is $t=23$.
Check: $P(22)=2000(0.96)^{22}=814.699>800$, but $P(23)=2000(0.96)^{23}=782.111<800$.
$$\boxed{t=23 \text{ years}}$$
(d) The halving time $T$ satisfies $0.96^T=0.5$:
$$T\ln(0.96)=\ln(0.5)$$
$$T=\frac{\ln(0.5)}{\ln(0.96)}=16.979748\ldots$$
$$\boxed{T\approx 16.98 \text{ years}}$$
| $t$ | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| $P(t)$ | 2000 | 1631 | 1330 | 1084 | 884 |
QUESTION 47
10 marks
Hard
A liquid is heated and then left to cool in a room. Its temperature is modeled by
$$T(t)=15+80e^{-kt}$$
where $T(t)$ is the temperature in degrees Celsius, $t$ is the time in minutes after cooling begins, and $k$ is a positive constant. It is known that $T(5)=55$.
(a) Find the value of $k$, correct to 4 decimal places. [4]
(b) Write down the horizontal asymptote of $T$, and interpret it in context. [1]
(c) Find the time taken for the liquid to cool to 25 degrees Celsius, correct to 1 decimal place. [4]
(d) Write down the temperature of the liquid at the moment cooling begins ($t=0$). [1]
Show complete worked solution
(a) Substituting $t=5$ and $T(5)=55$:
$$15+80e^{-5k}=55$$
$$80e^{-5k}=40$$
$$e^{-5k}=0.5$$
Taking the natural logarithm of both sides:
$$-5k=\ln(0.5)$$
$$k=\frac{\ln 2}{5}=0.138629\ldots$$
$$\boxed{k\approx 0.1386}$$
(b) As $t\to\infty$, $e^{-kt}\to 0$, so $T(t)\to 15$.
$$\boxed{T=15}$$
This represents the (constant) temperature of the room, which the liquid's temperature approaches as $t$ increases.
(c) Using $k\approx0.1386$, we require $T(t)=25$:
$$15+80e^{-0.1386t}=25$$
$$80e^{-0.1386t}=10$$
$$e^{-0.1386t}=\frac{10}{80}=0.125$$
Taking the natural logarithm of both sides:
$$-0.1386t=\ln(0.125)$$
$$t=\frac{\ln(0.125)}{-0.1386}=15.000000\ldots$$
$$\boxed{t\approx 15.0 \text{ minutes}}$$
(d) $T(0)=15+80e^{0}=15+80=95$
$$\boxed{T(0)=95^{\circ}\text{C}}$$
QUESTION 48
10 marks
Hard
The number of trees in a forest plantation, $A(t)$, is being reduced by logging, where $t$ is the number of years since records began. $A$ is modeled by a function of the form $A(t)=ab^t$. It is known that $A(4)=5000$ and $A(10)=2000$.
(a) By dividing the equation for $A(10)$ by the equation for $A(4)$, show that $b^6=0.4$, and hence find the value of $b$ correct to 4 significant figures. [4]
(b) Find the value of $a$, correct to the nearest whole number. [2]
(c) Hence write down the number of trees in the plantation when records began ($t=0$). [1]
(d) Find the "half-life" of the tree population, i.e. the time taken for the number of trees to halve, correct to 2 decimal places. [3]
Show complete worked solution
(a) From the model, $A(4)=ab^4=5000$ and $A(10)=ab^{10}=2000$. Dividing the second equation by the first:
$$\frac{ab^{10}}{ab^4}=\frac{2000}{5000}$$
$$b^6=0.4$$
Taking the sixth root of both sides (or raising both sides to the power $\tfrac{1}{6}$):
$$b=0.4^{1/6}=0.858374\ldots$$
$$\boxed{b\approx 0.8584}$$
(b) Substituting $b\approx0.8584$ into $ab^4=5000$:
$$a=\frac{5000}{b^4}=\frac{5000}{(0.8584)^4}=9210.07875\ldots$$
$$\boxed{a\approx 9210}$$
(c) Since $a=A(0)$ is the value of the model at $t=0$:
$$\boxed{A(0)\approx 9210 \text{ trees}}$$
(d) The half-life $T$ satisfies $b^T=0.5$:
$$T\ln(b)=\ln(0.5)$$
$$T=\frac{\ln(0.5)}{\ln(0.8584)}=4.538825\ldots$$
$$\boxed{T\approx 4.54 \text{ years}}$$
(As a check: since $A(4)=5000$ and $A(10)=2000$ is a drop of more than half over 6 years, a half-life somewhat less than 6 years, as found here, is consistent.)
QUESTION 49
10 marks
Hard
An invasive plant species is spreading across the surface of a lake. The area it covers, in $\text{m}^2$, is modeled by
$$A(t)=5e^{0.09t}$$
where $t$ is the number of days since the plant was first observed. The total surface area of the lake is $5000 \text{ m}^2$.
(a) Write down the daily percentage growth rate of the area covered, correct to 2 decimal places. [1]
(b) Find the doubling time of the covered area algebraically, correct to 3 significant figures. [3]
(c) Find the smallest integer number of days after which the covered area first exceeds half of the lake's surface area. [3]
(d) Find the smallest integer number of days after which the entire lake is first predicted to be covered (i.e. $A(t)\ge 5000$). [3]
Show complete worked solution
(a) $$(e^{0.09}-1)\times 100=9.41743\ldots$$
$$\boxed{\approx 9.42\text{ percent per day}}$$
(b) The doubling time $T$ satisfies $e^{0.09T}=2$:
$$0.09T=\ln 2$$
$$T=\frac{\ln 2}{0.09}=7.701635\ldots$$
$$\boxed{T\approx 7.7 \text{ days}}$$
(c) Half of the lake's area is $\dfrac{5000}{2}=2500\ \text{m}^2$. We require $5e^{0.09t}>2500$:
$$e^{0.09t}>\frac{2500}{5}=500$$
Taking the natural logarithm of both sides:
$$0.09t>\ln(500)$$
$$t>\frac{\ln(500)}{0.09}=69.05120\ldots$$
The smallest integer satisfying this is $t=70$.
$$\boxed{t=70 \text{ days}}$$
(d) We require $5e^{0.09t}\ge 5000$:
$$e^{0.09t}\ge 1000$$
$$0.09t\ge\ln(1000)$$
$$t\ge\frac{\ln(1000)}{0.09}=76.75284\ldots$$
The smallest integer satisfying this is $t=77$.
Check: $A(76)=5e^{0.09(76)}=4672.45<5000$, while $A(77)=5e^{0.09(77)}=5112.47\ge5000$.
$$\boxed{t=77 \text{ days}}$$
(As expected, the lake is half-covered ($t=70$) well before it is fully covered ($t=77$), consistent with the area more than doubling between these two events.)
QUESTION 50
9 marks
Hard
Two formulations of a drug are being compared. Formulation A has concentration modeled by $C_A(t)=80e^{-0.3t}$ and formulation B by $C_B(t)=50e^{-0.15t}$, both in mg/L, where $t$ is the time in hours after administration.
(a) Find the half-life of each formulation, correct to 2 decimal places. [4]
(b) Find the time, in hours, at which the two formulations have equal concentration, correct to 2 decimal places. [4]
(c) State which formulation has the higher concentration for $0
Show complete worked solution
(a) For each model $C(t)=C_0e^{-kt}$, the half-life is $T=\dfrac{\ln 2}{k}$.
For formulation A, $k=0.3$:
$$T_A=\frac{\ln 2}{0.3}=2.310491\ldots$$
$$\boxed{T_A\approx 2.31 \text{ hours}}$$
For formulation B, $k=0.15$:
$$T_B=\frac{\ln 2}{0.15}=4.620981\ldots$$
$$\boxed{T_B\approx 4.62 \text{ hours}}$$
(b) We require $C_A(t)=C_B(t)$:
$$80e^{-0.3t}=50e^{-0.15t}$$
Dividing both sides by $50e^{-0.3t}$:
$$\frac{80}{50}=e^{-0.15t+0.3t}=e^{0.15t}$$
$$1.6=e^{0.15t}$$
Taking the natural logarithm of both sides:
$$0.15t=\ln(1.6)$$
$$t=\frac{\ln(1.6)}{0.15}=3.133358\ldots$$
$$\boxed{t\approx 3.13 \text{ hours}}$$
Check using the GDC: $C_A(3.13)\approx31.250$ mg/L and $C_B(3.13)\approx31.250$ mg/L, which agree, confirming the concentrations are equal at this time.
(c) Since $C_A(0)=80$ mg/L is greater than $C_B(0)=50$ mg/L, and both concentrations decrease continuously and only cross once (at $t\approx3.13$), formulation A has the higher concentration for $0
Logarithmic Functions 50 questions
QUESTION 1
5 marks
Easy
The following data is believed to follow an exponential model y = ab^x.
x: 0, 1, 2, 3, 4
y: 8.0, 12.2, 18.7, 28.6, 43.8
(a) Explain why, if y = ab^x, plotting ln y against x should produce an approximately straight line. [1]
(b) Use your GDC to find the values of a and b, giving each to 3 significant figures. State the value of R$^{2}$ for this model. [3]
(c) Use your model to estimate the value of y when x = 6, giving your answer to 3 significant figures. [1]
Show complete worked solution
(a) If $y=ab^x$, taking the natural logarithm of both sides:
$$\ln y = \ln(ab^x) = \ln a + \ln(b^x)$$
$$\ln y = \ln a + x\ln b$$
This has the form $Y=mX+c$ with $Y=\ln y$, $X=x$, gradient $m=\ln b$ and intercept $c=\ln a$. Since $\ln y$ is a linear function of $x$, plotting $\ln y$ against $x$ (a semi-log plot) should produce an approximately straight line if the exponential model is appropriate.
(b) Using exponential regression on the GDC (equivalent to linear regression of $\ln y$ on $x$) on the five data points:
$$\boxed{a \approx 7.99,\quad b \approx 1.53,\quad R^2 \approx 1.00\text{ (3 s.f.)}}$$
giving the model $y = 7.99(1.53)^x$, which is an excellent fit to the data.
(c) Substituting $x=6$ into the model (using the calculator's stored, unrounded regression values for accuracy):
$$y = 7.99(1.53)^6$$
$$y \approx 102.49\ldots$$
$$\boxed{y \approx 102\text{ (3 s.f.)}}$$
QUESTION 2
5 marks
Easy
The following data is believed to follow a power model y = ax^n.
x: 1, 2, 4, 8, 16
y: 3.0, 8.5, 24.0, 68.0, 192.0
(a) Explain why, if y = ax^n, plotting log y against log x should produce an approximately straight line, and state how n can be found from this graph. [2]
(b) Use your GDC to find the values of a and n, giving each to 3 significant figures. [2]
(c) Use your model to estimate the value of y when x = 10, giving your answer to 3 significant figures. [1]
Show complete worked solution
(a) If $y=ax^n$, taking the logarithm (base 10) of both sides:
$$\log y = \log(ax^n) = \log a + \log(x^n)$$
$$\log y = \log a + n\log x$$
This has the form $Y=mX+c$ with $Y=\log y$, $X=\log x$, gradient $m=n$ and intercept $c=\log a$. Since $\log y$ is a linear function of $\log x$, plotting $\log y$ against $\log x$ (a log-log plot) should produce an approximately straight line, and the gradient of that line gives an estimate of the power $n$.
(b) Using power regression on the GDC (equivalent to linear regression of $\log y$ on $\log x$) on the five data points:
$$\boxed{n \approx 1.50,\quad a \approx 3.00}$$
giving the model $y = 3.00x^{1.50}$, with $R^2 \approx 1.00$, an excellent fit.
(c) Substituting $x=10$ into the model:
$$y = 3.00(10)^{1.50}$$
$$y = 3.00\times 31.62\ldots$$
$$\boxed{y \approx 94.9\text{ (3 s.f.)}}$$
QUESTION 3
5 marks
Medium
The magnitude, M, of an earthquake on a particular logarithmic scale is related to the intensity, I, of its shaking by
M = log10(I / I0),
where I0 is a fixed reference intensity.
Earthquake A has magnitude M1 = 5.2. Earthquake B has magnitude M2 = 6.7.
(a) Show that the intensity of earthquake B is approximately 31.6 times the intensity of earthquake A. [3]
(b) A third earthquake, C, has an intensity 500 times that of earthquake A. Find the magnitude of earthquake C, giving your answer to 3 significant figures. [2]
Show complete worked solution
(a) From the definition $M=\log_{10}(I/I_0)$, rearranging for intensity in each case:
$$I_1 = I_0 \times 10^{M_1}, \qquad I_2 = I_0 \times 10^{M_2}$$
Forming the ratio of intensities:
$$\dfrac{I_2}{I_1} = \dfrac{I_0\times10^{M_2}}{I_0\times10^{M_1}} = 10^{M_2-M_1}$$
Substituting $M_1=5.2$ and $M_2=6.7$:
$$\dfrac{I_2}{I_1} = 10^{6.7-5.2} = 10^{1.5}$$
$$10^{1.5} = 31.6227\ldots \approx 31.6\text{ (3 s.f.)}$$
$$\boxed{I_2 \approx 31.6\, I_1}$$
(b) Since $I_C = 500 \times I_A$:
$$\dfrac{I_C}{I_A} = 500$$
Using the same relationship as in part (a):
$$M_C-M_A = \log_{10}\left(\dfrac{I_C}{I_A}\right) = \log_{10}(500) = 2.69897\ldots$$
Solving for $M_C$, using $M_A=5.2$:
$$M_C = M_A+2.699 = 5.2+2.699$$
$$M_C = 7.899\ldots$$
$$\boxed{M_C \approx 7.90\text{ (3 s.f.)}}$$
QUESTION 4
7 marks
Medium
A biologist investigates how resting metabolic rate, R (in watts), depends on body mass, m (in kg), for a group of mammal species, and collects the following data.
m (kg): 2, 10, 50, 200, 800
R (W): 1.2, 4.5, 16.8, 52.0, 155.0
It is believed that R and m are related by a power model R = am^n.
(a) The biologist calculates log10(m) and log10(R) for each data pair and finds the equation of the regression line of log10(R) on log10(m) to be
log10(R) = 0.813 log10(m) - 0.161.
Use this equation to find the values of a and n in the power model R = am^n, giving each to 3 significant figures. [3]
(b) Use the model to estimate the resting metabolic rate of a mammal with body mass 400 kg, giving your answer to 3 significant figures. [2]
(c) Comment on the reliability of your estimate in part (b), with reference to the given data. [2]
Show complete worked solution
(a) Taking $\log_{10}$ of both sides of $R=am^n$:
$$\log_{10}(R) = \log_{10}(a) + n\log_{10}(m)$$
Comparing this with the given regression equation:
$$\log_{10}(R) = 0.813\log_{10}(m) - 0.161$$
the gradient gives $n$ and the intercept gives $\log_{10}(a)$:
$$n = 0.813$$
$$\log_{10}(a) = -0.161$$
Solving for $a$:
$$a = 10^{-0.161} = 0.690\ldots \approx 0.691\text{ (3 s.f.)}$$
$$\boxed{a \approx 0.691,\quad n \approx 0.813}$$
giving the model $R = 0.691m^{0.813}$.
(b) Substituting $m=400$ into the model (using the unrounded values $a=0.6905$, $n=0.8127$ for accuracy):
$$R = 0.6905 \times 400^{0.8127}$$
$$400^{0.8127} \approx 130.2$$
$$R \approx 0.6905 \times 130.2$$
$$\boxed{R \approx 89.9\text{ W (3 s.f.)}}$$
(c) Since $m=400$ kg lies between two data values already collected ($m=200$ kg and $m=800$ kg), this estimate is an interpolation rather than an extrapolation, so it can reasonably be trusted - provided the power-law relationship continues to hold for mammals of this size, and the (very good) linear fit of the log-log data supports the model's reliability in this range.
QUESTION 5
16 marks
Hard
Astronomers measure the apparent brightness (flux), F, of a distant star, in units of 10$^{-9}$ W/m$^{2}$, at several distances, d, in light-years, from an observing satellite. Theory predicts that flux should follow an inverse-square law with distance, of the form F = kd^n where n is expected to be close to -2.
d (light-years): 5, 10, 20, 40, 80
F (10$^{-9}$ W/m$^{2}$): 20.3, 4.9, 1.27, 0.304, 0.080
(a) Explain why plotting log10(F) against log10(d) is an appropriate way to test whether F and d follow a power-law relationship, and state what feature of the resulting graph would give the value of n. [2]
(b) Use your GDC to find the equation of the regression line of log10(F) on log10(d), and hence find the values of k and n in the model F = kd^n, giving each to 3 significant figures. [5]
(c) State, with a reason, whether the data supports the theoretical inverse-square law (n = -2). [2]
(d) Use the model to estimate the flux at a distance of d = 60 light-years, giving your answer to 3 significant figures. [3]
(e) A separate team wants to use this model to estimate the flux at d = 5000 light-years. Explain why this would be much less reliable than the estimate found in part (d). [4]
Show complete worked solution
(a) If $F=kd^n$, taking $\log_{10}$ of both sides:
$$\log_{10}(F) = \log_{10}(k) + n\log_{10}(d)$$
This is a linear equation in $\log_{10}(d)$, with gradient $n$ and vertical intercept $\log_{10}(k)$. Plotting $\log_{10}(F)$ against $\log_{10}(d)$ therefore tests the power-law hypothesis: if the data lies close to a straight line, $F=kd^n$ is a good model, and the gradient of that line gives the value of the power $n$ directly.
(b) Using power regression on the GDC (equivalent to linear regression of $\log_{10}(F)$ on $\log_{10}(d)$) on the five data points:
$$\boxed{n \approx -2.00,\quad k \approx 498\text{ (3 s.f.)}}$$
with $R^2 \approx 1.00$, an excellent fit, giving the model $F = 498d^{-2.00}$.
(c) The regression gives $n=-2.00$ (3 s.f.), which matches the theoretical value $n=-2$ predicted by the inverse-square law almost exactly, together with an $R^2$ value extremely close to $1$.
$$\boxed{\text{Yes - the data strongly supports the inverse-square law, since } n\approx-2.00 \text{ with an excellent fit.}}$$
(d) Substituting $d=60$ into the model (using the unrounded regression values $k=498.45$, $n=-1.99852$ for accuracy):
$$F = 498.45 \times 60^{-1.99852}$$
$$F \approx 0.139$$
$$\boxed{F \approx 0.139\times10^{-9}\text{ W/m}^2\text{ (3 s.f.)}}$$
(e) The value $d=60$ lies within the range of the collected data ($5$ to $80$ light-years), so it is an interpolation and can be estimated with reasonable confidence, as in part (d). By contrast, $d=5000$ light-years is far outside this range - over 60 times the largest distance measured - so using the model there would be a substantial extrapolation. While the inverse-square law is a well-established physical law for light spreading through empty space, over such an enormous additional distance other real-world effects not present in the original small-scale data - such as interstellar dust or gas absorbing or scattering some of the light, or gravitational effects - could cause the actual flux to differ from what the fitted power model predicts. The estimate at $d=5000$ would therefore be far less reliable than the estimate at $d=60$.
QUESTION 6
5 marks
Easy
Without using a calculator, evaluate the following.
(a) $\log_3 81$ [2]
(b) $\log_2\left(\dfrac{1}{16}\right)$ [2]
(c) $\log_7 1$ [1]
Show complete worked solution
(a) We require the power to which $3$ must be raised to give $81$. Since $3^4=81$:
$$\boxed{\log_3 81 = 4}$$
(b) We require the power to which $2$ must be raised to give $\dfrac{1}{16}$. Since $2^{-4}=\dfrac{1}{16}$:
$$\boxed{\log_2\left(\dfrac{1}{16}\right) = -4}$$
(c) Since $\log_b 1 = 0$ for any valid base $b$ (because $b^0=1$):
$$\boxed{\log_7 1 = 0}$$
QUESTION 7
5 marks
Easy
(a) Use the laws of logarithms to write $\log_4 8 + \log_4 32$ as a single logarithm. [2]
(b) Hence evaluate $\log_4 8 + \log_4 32$ exactly, without using a calculator. [2]
(c) State the name of the log law used in part (a). [1]
Show complete worked solution
(a) Using the product law of logarithms, $\log_b m + \log_b n = \log_b(mn)$:
$$\log_4 8 + \log_4 32 = \log_4(8\times32) = \log_4 256$$
$$\boxed{\log_4 8+\log_4 32 = \log_4 256}$$
(b) Since $4^4 = 256$:
$$\boxed{\log_4 8+\log_4 32 = 4}$$
(c) $\boxed{\text{The product law (addition law) of logarithms.}}$
QUESTION 8
4 marks
Easy
(a) Use the laws of logarithms to write $2\log_3 6 - \log_3 4$ as a single logarithm. [2]
(b) Hence evaluate $2\log_3 6 - \log_3 4$ exactly. [2]
Show complete worked solution
(a) Using the power law, $2\log_3 6 = \log_3(6^2) = \log_3 36$. Then, using the quotient law, $\log_b m - \log_b n = \log_b\left(\dfrac{m}{n}\right)$:
$$2\log_3 6 - \log_3 4 = \log_3 36 - \log_3 4 = \log_3\left(\dfrac{36}{4}\right) = \log_3 9$$
$$\boxed{2\log_3 6-\log_3 4 = \log_3 9}$$
(b) Since $3^2=9$:
$$\boxed{2\log_3 6-\log_3 4 = 2}$$
QUESTION 9
4 marks
Easy
The diagram shows the graph of $y=\log_2 x$.
(a) State the domain of $y=\log_2 x$. [1]
(b) State the equation of the vertical asymptote of $y=\log_2 x$. [1]
(c) State the coordinates of the $x$-intercept of $y=\log_2 x$. [1]
(d) Find the value of $y$ when $x=4$. [1]
Show complete worked solution
(a) A logarithm is only defined for a positive argument, so:
$$\boxed{\text{Domain: } x>0 \ (x\in\mathbb{R}, x>0)}$$
(b) As $x\to0^{+}$, $\log_2 x\to-\infty$, so the curve approaches but never reaches the $y$-axis:
$$\boxed{x=0}$$
(c) The $x$-intercept occurs where $y=0$, i.e. where $\log_2 x=0$, so $x=2^0=1$:
$$\boxed{(1,\ 0)}$$
(d) Substituting $x=4$: $y=\log_2 4$. Since $2^2=4$:
$$\boxed{y=2}$$
QUESTION 10
4 marks
Easy
The graph of $y=\log_5 x$ is transformed to give the graph of $y=\log_5(x-3)$.
(a) State the equation of the vertical asymptote of $y=\log_5(x-3)$. [2]
(b) State the domain of $y=\log_5(x-3)$. [2]
Show complete worked solution
(a) The transformation $y=\log_5(x-3)$ is a horizontal translation of $y=\log_5 x$ by $3$ units in the positive $x$-direction. Since $y=\log_5 x$ has vertical asymptote $x=0$, translating by $3$ gives:
$$\boxed{x=3}$$
(b) The expression $x-3$ inside the logarithm must be strictly positive, so $x-3>0$, i.e. $x>3$:
$$\boxed{\text{Domain: } x>3}$$
QUESTION 11
4 marks
Easy
(a) Solve the equation $\log_6 x = 2$, giving your answer as an exact value. [2]
(b) Verify your answer to part (a) by substituting it back into the original equation. [2]
Show complete worked solution
(a) Writing the equation in exponential form, $\log_6 x = 2$ means $x = 6^2$:
$$\boxed{x=36}$$
(b) Substituting $x=36$ into $\log_6 x$: we require the power to which $6$ must be raised to give $36$. Since $6^2=36$, $\log_6 36 = 2$, which matches the right-hand side of the original equation.
$$\boxed{\text{Verified: } \log_6 36 = 2 \ \checkmark}$$
QUESTION 12
4 marks
Easy
(a) Solve the equation $3^x = 50$, giving your answer to 3 significant figures. [3]
(b) Verify that your answer is reasonable by evaluating $3^{3.5}$ and $3^{3.6}$. [1]
Show complete worked solution
(a) Taking $\log_{10}$ (or $\ln$) of both sides of $3^x=50$:
$$\log(3^x) = \log(50)$$
Using the power law of logarithms, $\log(3^x)=x\log 3$:
$$x\log 3 = \log 50$$
$$x = \dfrac{\log 50}{\log 3} = \dfrac{1.69897\ldots}{0.47712\ldots}$$
$$x = 3.5608\ldots$$
$$\boxed{x \approx 3.56\text{ (3 s.f.)}}$$
(b) Evaluating: $3^{3.5}\approx 46.8$ and $3^{3.6}\approx 52.5$. Since $46.8<50<52.5$, the solution $x\approx3.56$ (which lies between $3.5$ and $3.6$) is consistent with these bracketing values, confirming the answer is reasonable.
$$\boxed{\text{Reasonable, since } 3^{3.5}<50<3^{3.6}}$$
QUESTION 13
4 marks
Easy
(a) Use the change of base formula to evaluate $\log_8 20$, giving your answer to 3 significant figures. [2]
(b) Verify your answer to part (a) by evaluating $8^{1.44}$ and comparing it to $20$. [2]
Show complete worked solution
(a) By the change of base formula, $\log_8 20 = \dfrac{\log 20}{\log 8}$ (using logarithms in any consistent base, e.g. base 10 on the GDC):
$$\log_8 20 = \dfrac{\log_{10} 20}{\log_{10} 8} = \dfrac{1.30103\ldots}{0.90309\ldots}$$
$$\log_8 20 = 1.44064\ldots$$
$$\boxed{\log_8 20 \approx 1.44\text{ (3 s.f.)}}$$
(b) Evaluating $8^{1.44} = 19.86\ldots \approx 19.9$, which is very close to $20$ (the small difference is due to rounding the exponent to 3 s.f.), confirming the answer.
$$\boxed{8^{1.44}\approx19.9\approx20\ \checkmark}$$
QUESTION 14
4 marks
Easy
(a) Solve the equation $\ln x = 1.8$, giving your answer to 3 significant figures. [2]
(b) State the value of $\ln(e^{1.8})$, giving a reason for your answer. [2]
Show complete worked solution
(a) Writing $\ln x = 1.8$ in exponential form (base $e$):
$$x = e^{1.8}$$
$$x = 6.0496\ldots$$
$$\boxed{x\approx6.05\text{ (3 s.f.)}}$$
(b) Since $\ln$ and the exponential function $e^{(\cdot)}$ are inverse functions, $\ln(e^{k})=k$ for any real $k$:
$$\boxed{\ln(e^{1.8}) = 1.8, \text{ because } \ln \text{ and } e^{(\cdot)} \text{ are inverse functions.}}$$
QUESTION 15
4 marks
Easy
The pH of an aqueous solution is defined by pH $= -\log_{10}[\text{H}^+]$, where $[\text{H}^+]$ is the hydrogen ion concentration in mol/L.
(a) A solution has $[\text{H}^+] = 4.7\times10^{-3}$ mol/L. Find its pH, giving your answer to 2 decimal places. [3]
(b) State whether this solution is acidic, neutral, or basic (given that pure water has pH $7$). [1]
Show complete worked solution
(a) Substituting $[\text{H}^+]=4.7\times10^{-3}$ into $\text{pH}=-\log_{10}[\text{H}^+]$:
$$\text{pH} = -\log_{10}(4.7\times10^{-3})$$
$$\text{pH} = -(-2.32790\ldots)$$
$$\text{pH} = 2.32790\ldots$$
$$\boxed{\text{pH} \approx 2.33\text{ (2 d.p.)}}$$
(b) Since $2.33 < 7$, the solution has a pH below that of pure water.
$$\boxed{\text{Acidic}}$$
QUESTION 16
4 marks
Easy
Using pH $= -\log_{10}[\text{H}^+]$, find the hydrogen ion concentration $[\text{H}^+]$, in mol/L, of a solution with pH $5.6$, giving your answer in the form $a\times10^{k}$ where $1\leq a <10$, to 3 significant figures.
Show complete worked solution
Rearranging $\text{pH}=-\log_{10}[\text{H}^+]$ for $[\text{H}^+]$:
$$-\text{pH} = \log_{10}[\text{H}^+]$$
$$[\text{H}^+] = 10^{-\text{pH}}$$
Substituting $\text{pH}=5.6$:
$$[\text{H}^+] = 10^{-5.6}$$
$$[\text{H}^+] = 2.51188\ldots\times10^{-6}$$
$$\boxed{[\text{H}^+] \approx 2.51\times10^{-6}\text{ mol/L (3 s.f.)}}$$
QUESTION 17
4 marks
Easy
The loudness of a sound, in decibels (dB), is given by $L = 10\log_{10}\left(\dfrac{I}{I_0}\right)$, where $I$ is the sound intensity and $I_0$ is a fixed reference intensity.
(a) A sound has intensity $I$ such that $\dfrac{I}{I_0} = 2.5\times10^{6}$. Find its loudness $L$, giving your answer to 3 significant figures. [3]
(b) Comment on whether this loudness is closer to that of a quiet whisper (approximately $30$ dB) or a jet engine at close range (approximately $140$ dB). [1]
Show complete worked solution
(a) Substituting $\dfrac{I}{I_0}=2.5\times10^6$ into $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$:
$$L = 10\log_{10}(2.5\times10^6)$$
$$L = 10\times6.39794\ldots$$
$$L = 63.9794\ldots$$
$$\boxed{L\approx64.0\text{ dB (3 s.f.)}}$$
(b) Since $64.0$ dB is much closer in value to $30$ dB than to $140$ dB:
$$\boxed{\text{Closer to the quiet whisper level of about 30 dB.}}$$
QUESTION 18
4 marks
Easy
The magnitude of an earthquake on a logarithmic scale is given by $M = \log_{10}\left(\dfrac{I}{I_0}\right)$, where $I_0$ is a fixed reference intensity.
(a) An earthquake has intensity $I$ such that $\dfrac{I}{I_0} = 6300$. Find its magnitude $M$, giving your answer to 3 significant figures. [3]
(b) Would this be classed as a small earthquake (magnitude below 4) or a major earthquake (magnitude above 7)? [1]
Show complete worked solution
(a) Substituting $\dfrac{I}{I_0}=6300$ into $M=\log_{10}\left(\dfrac{I}{I_0}\right)$:
$$M = \log_{10}(6300)$$
$$M = 3.79934\ldots$$
$$\boxed{M\approx3.80\text{ (3 s.f.)}}$$
(b) Since $3.80<4$:
$$\boxed{\text{Small earthquake (magnitude below 4).}}$$
QUESTION 19
4 marks
Easy
A set of data believed to follow the model $y=ab^x$ is entered into a GDC, which performs a linear regression of $\ln y$ on $x$ and returns the equation
$$\ln y = 0.847x + 1.02$$
(a) State the values of $\ln b$ and $\ln a$ from this regression equation. [1]
(b) Hence find the values of $a$ and $b$ in the model $y=ab^x$, giving each to 3 significant figures. [3]
Show complete worked solution
(a) Comparing $\ln y = 0.847x+1.02$ with the linearized form $\ln y = (\ln b)x + \ln a$:
$$\boxed{\ln b = 0.847, \quad \ln a = 1.02}$$
(b) Exponentiating both sides:
$$b = e^{0.847} = 2.33263\ldots$$
$$a = e^{1.02} = 2.77319\ldots$$
$$\boxed{a\approx2.77,\quad b\approx2.33\text{ (3 s.f.)}}$$
QUESTION 20
4 marks
Easy
A set of data believed to follow the model $y=ax^n$ is entered into a GDC, which performs a linear regression of $\log_{10} y$ on $\log_{10} x$ and returns the equation
$$\log_{10} y = 1.35\log_{10} x - 0.602$$
(a) State the value of $n$ in the model $y=ax^n$. [1]
(b) Hence find the value of $a$ in the model $y=ax^n$, giving your answer to 3 significant figures. [3]
Show complete worked solution
(a) Comparing $\log_{10}y = 1.35\log_{10}x - 0.602$ with the linearized form $\log_{10}y = n\log_{10}x + \log_{10}a$:
$$\boxed{n=1.35}$$
(b) Reading off the intercept, $\log_{10}a = -0.602$, so:
$$a = 10^{-0.602}$$
$$a = 0.250034\ldots$$
$$\boxed{a\approx0.250\text{ (3 s.f.)}}$$
QUESTION 21
4 marks
Easy
The following data is believed to follow an exponential model $y=ab^x$.
x: 1, 2, 3, 4, 5
y: 6.1, 9.3, 14.2, 21.6, 33.0
Use your GDC to find the values of $a$ and $b$, giving each to 3 significant figures.
Show complete worked solution
Using exponential regression on the GDC (equivalent to linear regression of $\ln y$ on $x$) on the five data points:
$$\boxed{a\approx4.00,\quad b\approx1.52\text{ (3 s.f.)}}$$
giving the model $y=4.00(1.52)^x$, with $R^2\approx1.00$, an excellent fit.
QUESTION 22
4 marks
Easy
(a) Solve the equation $\log_2 x + \log_2 3 = 5$ for $x$, giving an exact value. [3]
(b) Verify your solution satisfies the original equation. [1]
Show complete worked solution
(a) Using the product law, $\log_2 x + \log_2 3 = \log_2(3x)$, so the equation becomes:
$$\log_2(3x) = 5$$
Writing this in exponential form:
$$3x = 2^5 = 32$$
$$x = \dfrac{32}{3}$$
$$\boxed{x = \dfrac{32}{3}\approx10.7}$$
(b) Substituting $x=\dfrac{32}{3}$: $\log_2\left(\dfrac{32}{3}\right)+\log_2 3 = \log_2\left(\dfrac{32}{3}\times3\right) = \log_2 32 = 5$, which matches the right-hand side.
$$\boxed{\text{Verified} \ \checkmark}$$
QUESTION 23
4 marks
Easy
Use the laws of logarithms to write $\log x + 2\log y - \log z$ as a single logarithm.
Show complete worked solution
Using the power law, $2\log y = \log(y^2)$:
$$\log x + 2\log y - \log z = \log x + \log(y^2) - \log z$$
Using the product law, $\log x + \log(y^2) = \log(xy^2)$:
$$= \log(xy^2) - \log z$$
Using the quotient law:
$$\boxed{\log x+2\log y-\log z = \log\left(\dfrac{xy^2}{z}\right)}$$
QUESTION 24
6 marks
Medium
(a) Solve the equation $\log_2 x + \log_2(x-2) = 3$. [4]
(b) Explain why any negative root of the quadratic equation formed in part (a) must be rejected. [2]
Show complete worked solution
(a) Using the product law, $\log_2 x + \log_2(x-2) = \log_2(x(x-2))$, so the equation becomes:
$$\log_2(x(x-2)) = 3$$
Writing this in exponential form:
$$x(x-2) = 2^3 = 8$$
$$x^2 - 2x - 8 = 0$$
Factorising:
$$(x-4)(x+2) = 0$$
$$x = 4 \quad \text{or} \quad x = -2$$
Checking the domain (both $x>0$ and $x-2>0$ are required, i.e. $x>2$): $x=4$ satisfies this, but $x=-2$ does not.
$$\boxed{x=4}$$
(b) Both $\log_2 x$ and $\log_2(x-2)$ in the original equation require positive arguments, so we need $x>0$ and $x>2$ simultaneously, i.e. $x>2$. Substituting $x=-2$ would require evaluating $\log_2(-2)$, which is undefined (logarithms of negative numbers do not exist), so:
$$\boxed{x=-2 \text{ is rejected as it lies outside the domain } x>2.}$$
QUESTION 25
5 marks
Medium
Solve the equation $4^x = 3^{x+1}$, giving your answer to 3 significant figures.
Show complete worked solution
Taking $\ln$ of both sides of $4^x = 3^{x+1}$:
$$\ln(4^x) = \ln(3^{x+1})$$
Using the power law of logarithms:
$$x\ln 4 = (x+1)\ln 3$$
Expanding the right-hand side:
$$x\ln 4 = x\ln 3 + \ln 3$$
Collecting the $x$ terms on the left-hand side:
$$x\ln 4 - x\ln 3 = \ln 3$$
$$x(\ln 4 - \ln 3) = \ln 3$$
Solving for $x$:
$$x = \dfrac{\ln 3}{\ln 4-\ln 3} = \dfrac{1.09861\ldots}{1.38629\ldots-1.09861\ldots}$$
$$x = 3.81884\ldots$$
$$\boxed{x\approx3.82\text{ (3 s.f.)}}$$
QUESTION 26
6 marks
Medium
(a) Solve the equation $\log_3(x+5) - \log_3 x = 1$. [4]
(b) State the restriction on $x$ required for the original equation to be defined, and confirm your answer satisfies it. [2]
Show complete worked solution
(a) Using the quotient law, $\log_3(x+5)-\log_3 x = \log_3\left(\dfrac{x+5}{x}\right)$, so the equation becomes:
$$\log_3\left(\dfrac{x+5}{x}\right) = 1$$
Writing this in exponential form:
$$\dfrac{x+5}{x} = 3^1 = 3$$
$$x+5 = 3x$$
$$5 = 2x$$
$$\boxed{x=2.5}$$
(b) The original equation requires both $x+5>0$ and $x>0$; since $x>0$ is the stricter condition, we need $x>0$. As $x=2.5>0$, the solution is valid.
$$\boxed{x=2.5>0 \ \checkmark}$$
QUESTION 27
5 marks
Medium
Solve the equation $2(5^x) = 150$, giving your answer to 3 significant figures.
Show complete worked solution
Dividing both sides of $2(5^x)=150$ by $2$:
$$5^x = 75$$
Taking $\log_{10}$ of both sides:
$$\log(5^x) = \log 75$$
Using the power law:
$$x\log 5 = \log 75$$
$$x = \dfrac{\log 75}{\log 5} = \dfrac{1.87506\ldots}{0.69897\ldots}$$
$$x = 2.68261\ldots$$
$$\boxed{x\approx2.68\text{ (3 s.f.)}}$$
QUESTION 28
6 marks
Medium
The following data is believed to follow an exponential model $y=ab^x$.
x: 0, 2, 4, 6, 8
y: 5.0, 9.8, 19.4, 38.0, 75.0
(a) Explain why plotting $\ln y$ against $x$ is an appropriate way to test this model. [1]
(b) Use your GDC to find the values of $a$, $b$ and $R^2$, giving $a$ and $b$ to 3 significant figures. [3]
(c) Use your model to estimate the value of $y$ when $x=5$, giving your answer to 3 significant figures. [2]
Show complete worked solution
(a) If $y=ab^x$, taking $\ln$ of both sides gives $\ln y = \ln a + x\ln b$, which is linear in $x$ (gradient $\ln b$, intercept $\ln a$). So if the model is appropriate, a plot of $\ln y$ against $x$ should be approximately a straight line.
$$\boxed{\ln y = (\ln b)x+\ln a \text{ is linear in } x, \text{ so a semi-log plot tests the exponential model.}}$$
(b) Using exponential regression on the GDC (equivalent to linear regression of $\ln y$ on $x$) on the five data points:
$$\boxed{a\approx4.99,\quad b\approx1.40,\quad R^2\approx1.00\text{ (3 s.f.)}}$$
giving the model $y=4.99(1.40)^x$, an excellent fit.
(c) Substituting $x=5$ into the model (using the unrounded regression values $a=4.9937$, $b=1.40293$ for accuracy):
$$y = 4.9937\times1.40293^5$$
$$y\approx27.1\ldots$$
$$\boxed{y\approx27.1\text{ (3 s.f.)}}$$
QUESTION 29
6 marks
Medium
The following data is believed to follow a power model $y=ax^n$.
x: 2, 4, 6, 8, 10
y: 6.0, 18.0, 31.5, 51.5, 68.0
(a) Use your GDC to find the values of $a$, $n$ and $R^2$, giving $a$ and $n$ to 3 significant figures. [3]
(b) Use your model to estimate the value of $y$ when $x=7$, giving your answer to 3 significant figures. [2]
(c) Comment on what the value of $R^2$ found in part (a) indicates about the fit of the model. [1]
Show complete worked solution
(a) Using power regression on the GDC (equivalent to linear regression of $\log y$ on $\log x$) on the five data points:
$$\boxed{a\approx2.13,\quad n\approx1.52,\quad R^2\approx0.999\text{ (3 s.f.)}}$$
giving the model $y=2.13x^{1.52}$.
(b) Substituting $x=7$ into the model (using the unrounded regression values $a=2.1323$, $n=1.51543$ for accuracy):
$$y = 2.1323\times7^{1.51543}$$
$$y\approx40.7\ldots$$
$$\boxed{y\approx40.7\text{ (3 s.f.)}}$$
(c) $R^2\approx0.999$ is very close to $1$, indicating that the power model explains almost all of the variation in $y$, so it is an excellent fit to the data.
$$\boxed{\text{Excellent fit -- } R^2\approx0.999 \text{ is very close to } 1.}$$
QUESTION 30
6 marks
Medium
The pH of a solution is given by pH $=-\log_{10}[\text{H}^+]$.
(a) A solution has $[\text{H}^+]=8.9\times10^{-4}$ mol/L. Find its pH, giving your answer to 3 significant figures. [2]
(b) A second solution has a pH exactly $2$ units higher than the solution in part (a). Find the ratio $\dfrac{[\text{H}^+]_1}{[\text{H}^+]_2}$ of the hydrogen ion concentration of the first solution to that of the second solution. [4]
Show complete worked solution
(a) Substituting $[\text{H}^+]=8.9\times10^{-4}$ into $\text{pH}=-\log_{10}[\text{H}^+]$:
$$\text{pH} = -\log_{10}(8.9\times10^{-4}) = 3.05060\ldots$$
$$\boxed{\text{pH}\approx3.05\text{ (3 s.f.)}}$$
(b) Let the first solution have $[\text{H}^+]_1$ and pH $=P_1$, and the second have $[\text{H}^+]_2$ and pH $=P_2=P_1+2$. From the definition, $[\text{H}^+]=10^{-\text{pH}}$, so:
$$\dfrac{[\text{H}^+]_1}{[\text{H}^+]_2} = \dfrac{10^{-P_1}}{10^{-P_2}} = 10^{P_2-P_1} = 10^{2}$$
$$\boxed{\dfrac{[\text{H}^+]_1}{[\text{H}^+]_2} = 100}$$
So the first (lower-pH) solution has a hydrogen ion concentration $100$ times greater than the second.
QUESTION 31
6 marks
Medium
The loudness of a sound is given by $L = 10\log_{10}\left(\dfrac{I}{I_0}\right)$ decibels.
(a) A sound has $\dfrac{I}{I_0}=4.0\times10^{4}$. Find its loudness $L$, giving your answer to 3 significant figures. [2]
(b) Show that an increase in loudness of exactly $10$ dB corresponds to the sound intensity $I$ increasing by a factor of $10$. [4]
Show complete worked solution
(a) Substituting $\dfrac{I}{I_0}=4.0\times10^4$ into $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$:
$$L = 10\log_{10}(4.0\times10^4) = 10\times4.60206\ldots$$
$$\boxed{L\approx46.0\text{ dB (3 s.f.)}}$$
(b) Let two sounds have intensities $I_1$ and $I_2$, with loudness $L_1=10\log_{10}\left(\dfrac{I_1}{I_0}\right)$ and $L_2=10\log_{10}\left(\dfrac{I_2}{I_0}\right)$. Suppose $L_2-L_1=10$:
$$10\log_{10}\left(\dfrac{I_2}{I_0}\right) - 10\log_{10}\left(\dfrac{I_1}{I_0}\right) = 10$$
Dividing through by $10$ and using the quotient law:
$$\log_{10}\left(\dfrac{I_2}{I_0}\right) - \log_{10}\left(\dfrac{I_1}{I_0}\right) = 1$$
$$\log_{10}\left(\dfrac{I_2/I_0}{I_1/I_0}\right) = 1$$
$$\log_{10}\left(\dfrac{I_2}{I_1}\right) = 1$$
Writing in exponential form:
$$\dfrac{I_2}{I_1} = 10^1 = 10$$
$$\boxed{\text{A 10 dB increase in loudness corresponds to a 10-fold increase in intensity.}}$$
QUESTION 32
5 marks
Medium
Solve the equation $\log_x 81 = 4$ for the base $x$, where $x>0$ and $x\neq1$.
Show complete worked solution
Writing the equation $\log_x 81 = 4$ in exponential form:
$$x^4 = 81$$
Taking the fourth root of both sides (and requiring $x>0$):
$$x = 81^{1/4}$$
Since $81 = 3^4$:
$$x = (3^4)^{1/4} = 3$$
$$\boxed{x=3}$$
Checking: $\log_3 81 = 4$ since $3^4=81$. $\checkmark$
QUESTION 33
5 marks
Medium
Solve the equation $e^{2x-1} = 25$, giving your answer to 3 significant figures.
Show complete worked solution
Taking $\ln$ of both sides of $e^{2x-1}=25$:
$$\ln\left(e^{2x-1}\right) = \ln 25$$
Since $\ln$ and $e^{(\cdot)}$ are inverse functions, $\ln(e^{2x-1})=2x-1$:
$$2x-1 = \ln 25$$
$$2x-1 = 3.21887\ldots$$
$$2x = 4.21887\ldots$$
$$x = 2.10943\ldots$$
$$\boxed{x\approx2.11\text{ (3 s.f.)}}$$
QUESTION 34
7 marks
Medium
(a) Solve the equation $\ln x + \ln(x-3) = \ln 10$. [5]
(b) Explain why one of the two roots of the quadratic equation formed in part (a) must be rejected. [2]
Show complete worked solution
(a) Using the product law, $\ln x + \ln(x-3) = \ln(x(x-3))$, so the equation becomes:
$$\ln(x(x-3)) = \ln 10$$
Since $\ln$ is a one-to-one function, the arguments must be equal:
$$x(x-3) = 10$$
$$x^2-3x-10 = 0$$
Factorising:
$$(x-5)(x+2) = 0$$
$$x = 5 \quad \text{or} \quad x = -2$$
Checking the domain (both $x>0$ and $x-3>0$ are required, i.e. $x>3$): $x=5$ satisfies this, but $x=-2$ does not.
$$\boxed{x=5}$$
(b) The terms $\ln x$ and $\ln(x-3)$ in the original equation require $x>0$ and $x>3$ respectively, so overall $x>3$ is required. Substituting $x=-2$ would require evaluating $\ln(-2)$ and $\ln(-5)$, both of which are undefined, since the natural logarithm of a negative number does not exist.
$$\boxed{x=-2 \text{ is rejected as it lies outside the domain } x>3.}$$
QUESTION 35
6 marks
Medium
A biologist finds that a population, $P$, of bacteria (in thousands) grows with time $t$ (in hours). A GDC performs a linear regression of $\ln P$ on $t$ and returns the equation
$$\ln P = 0.223t + 4.605$$
(a) Find the values of $P_0$ and $b$ in the model $P = P_0 b^t$, giving each to 3 significant figures. [3]
(b) Find the doubling time of the population, that is, the time taken for $P$ to double, giving your answer to 3 significant figures. [3]
Show complete worked solution
(a) Comparing $\ln P = 0.223t + 4.605$ with the linearized form $\ln P = (\ln b)t + \ln P_0$:
$$\ln P_0 = 4.605 \quad\Rightarrow\quad P_0 = e^{4.605} = 99.98\ldots$$
$$\ln b = 0.223 \quad\Rightarrow\quad b = e^{0.223} = 1.24982\ldots$$
$$\boxed{P_0\approx100,\quad b\approx1.25\text{ (3 s.f.)}}$$
(b) The population doubles when $b^t=2$, i.e. when $t\ln b = \ln 2$:
$$t = \dfrac{\ln 2}{\ln b} = \dfrac{\ln 2}{0.223}$$
$$t = \dfrac{0.69315\ldots}{0.223}$$
$$t = 3.10828\ldots$$
$$\boxed{t\approx3.11\text{ hours (3 s.f.)}}$$
QUESTION 36
6 marks
Medium
The magnitude of an earthquake on a logarithmic scale is given by $M=\log_{10}\left(\dfrac{I}{I_0}\right)$, where $I_0$ is a fixed reference intensity.
(a) Earthquake P has magnitude $M_1=4.8$ and earthquake Q has magnitude $M_2=6.1$. Show that the intensity of Q is approximately $20.0$ times the intensity of P. [4]
(b) A third earthquake, R, has an intensity $250$ times that of a reference earthquake with $\dfrac{I}{I_0}=1$. Find the magnitude of earthquake R, giving your answer to 3 significant figures. [2]
Show complete worked solution
(a) From $M=\log_{10}\left(\dfrac{I}{I_0}\right)$, rearranging for intensity: $I = I_0\times10^M$. Forming the ratio:
$$\dfrac{I_2}{I_1} = \dfrac{I_0\times10^{M_2}}{I_0\times10^{M_1}} = 10^{M_2-M_1}$$
Substituting $M_1=4.8$ and $M_2=6.1$:
$$\dfrac{I_2}{I_1} = 10^{6.1-4.8} = 10^{1.3}$$
$$10^{1.3} = 19.9526\ldots \approx20.0\text{ (3 s.f.)}$$
$$\boxed{I_2\approx20.0\,I_1}$$
(b) With $\dfrac{I}{I_0}=250$:
$$M = \log_{10}(250) = 2.39794\ldots$$
$$\boxed{M\approx2.40\text{ (3 s.f.)}}$$
QUESTION 37
7 marks
Medium
The number of subscribers, $N$, to a streaming service $t$ years after launch is modelled by $N = 250(1.18)^t$.
(a) Find the doubling time of the number of subscribers, that is, the time taken for $N$ to double, giving your answer to 3 significant figures. [3]
(b) Find the time taken for the number of subscribers to reach $1000$, giving your answer to 3 significant figures. [4]
Show complete worked solution
(a) The number of subscribers doubles when $1.18^t=2$. Taking $\ln$ of both sides:
$$t\ln 1.18 = \ln 2$$
$$t = \dfrac{\ln 2}{\ln 1.18} = \dfrac{0.69315\ldots}{0.16551\ldots}$$
$$t = 4.18784\ldots$$
$$\boxed{t\approx4.19\text{ years (3 s.f.)}}$$
(b) Setting $N=1000$:
$$250(1.18)^t = 1000$$
$$1.18^t = 4$$
Taking $\ln$ of both sides:
$$t\ln 1.18 = \ln 4$$
$$t = \dfrac{\ln 4}{\ln 1.18} = \dfrac{1.38629\ldots}{0.16551\ldots}$$
$$t = 8.37567\ldots$$
$$\boxed{t\approx8.38\text{ years (3 s.f.)}}$$
QUESTION 38
5 marks
Medium
Solve the equation $\log_5(x^2) = 4$, giving all possible exact values of $x$.
Show complete worked solution
Writing the equation $\log_5(x^2)=4$ in exponential form:
$$x^2 = 5^4 = 625$$
Taking the square root of both sides:
$$x = \pm\sqrt{625} = \pm25$$
Since $x^2>0$ for both $x=25$ and $x=-25$ (the original expression $\log_5(x^2)$ only requires $x^2>0$, i.e. $x\neq0$, not $x>0$), both solutions are valid.
$$\boxed{x=25 \text{ or } x=-25}$$
QUESTION 39
7 marks
Medium
A dataset believed to satisfy $y=ab^x$ is linearized, giving the following table of $x$ and $\ln y$ values.
x: 1, 3, 5, 7, 9
ln y: 2.708, 3.198, 3.597, 4.086, 4.477
(a) Use your GDC to perform a linear regression of $\ln y$ on $x$, and state the equation of the regression line, giving the gradient and intercept to 3 significant figures. [2]
(b) Hence find the values of $a$ and $b$ in the model $y=ab^x$, and state the value of $R^2$, giving each to 3 significant figures. [3]
(c) Use your model to predict the value of $y$ when $x=12$, giving your answer to 3 significant figures. [2]
Show complete worked solution
(a) Using linear regression on the GDC with $x$ as the independent variable and $\ln y$ as the dependent variable, on the five data points:
$$\boxed{\ln y \approx 0.221x + 2.51 \ \text{(3 s.f.)}}$$
(b) Comparing with $\ln y = (\ln b)x + \ln a$: $\ln b\approx0.221$ and $\ln a\approx2.51$, so:
$$b = e^{0.221294\ldots} = 1.24769\ldots, \qquad a = e^{2.507001\ldots} = 12.2681\ldots$$
$$\boxed{a\approx12.3,\quad b\approx1.25,\quad R^2\approx0.999\text{ (3 s.f.)}}$$
(c) Substituting $x=12$ into the model (using the unrounded values $a=12.2681$, $b=1.24769$ for accuracy):
$$y = 12.2681\times1.24769^{12}$$
$$y\approx174.6\ldots$$
$$\boxed{y\approx175\text{ (3 s.f.)}}$$
QUESTION 40
6 marks
Medium
A semi-log plot of $\ln y$ against $x$, for data believed to satisfy $y=ab^x$, passes through the points $(2,\ 3.10)$ and $(10,\ 5.98)$.
(a) Find the gradient and the vertical intercept of this line. [3]
(b) Hence find the values of $a$ and $b$, giving each to 3 significant figures. [3]
Show complete worked solution
(a) The gradient of the line through $(2,3.10)$ and $(10,5.98)$ is:
$$m = \dfrac{5.98-3.10}{10-2} = \dfrac{2.88}{8} = 0.36$$
Using $y-y_1=m(x-x_1)$ with $(2,3.10)$: $\ln y - 3.10 = 0.36(x-2)$, so the vertical intercept ($x=0$) is:
$$c = 3.10 - 0.36\times2 = 3.10-0.72 = 2.38$$
$$\boxed{\text{gradient}=0.36,\quad \text{intercept}=2.38}$$
(b) Comparing $\ln y = 0.36x+2.38$ with $\ln y = (\ln b)x+\ln a$:
$$b = e^{0.36} = 1.43333\ldots, \qquad a = e^{2.38} = 10.8049\ldots$$
$$\boxed{a\approx10.8,\quad b\approx1.43\text{ (3 s.f.)}}$$
QUESTION 41
5 marks
Medium
Two sounds have loudness $L_1=62$ dB and $L_2=85$ dB, where $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$.
Find the ratio $\dfrac{I_2}{I_1}$ of their intensities, giving your answer to 3 significant figures.
Show complete worked solution
From $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$, rearranging for intensity: $I = I_0\times10^{L/10}$. Forming the ratio:
$$\dfrac{I_2}{I_1} = \dfrac{I_0\times10^{L_2/10}}{I_0\times10^{L_1/10}} = 10^{(L_2-L_1)/10}$$
Substituting $L_1=62$ and $L_2=85$:
$$\dfrac{I_2}{I_1} = 10^{(85-62)/10} = 10^{2.3}$$
$$10^{2.3} = 199.526\ldots$$
$$\boxed{\dfrac{I_2}{I_1}\approx200\text{ (3 s.f.)}}$$
QUESTION 42
13 marks
Hard
An acoustics engineer measures the intensity, $I$ (in $10^{-6}$ W/m$^2$), of the sound from a loudspeaker at various distances, $d$ (in metres), from the speaker. Theory predicts that, in an open space, intensity should follow an inverse-square law with distance, of the form $I=kd^n$ where $n$ is expected to be close to $-2$.
d (m): 2, 4, 8, 16, 32
I (10$^{-6}$ W/m$^2$): 318.0, 79.6, 19.9, 4.97, 1.24
(a) Explain why plotting $\log_{10}I$ against $\log_{10}d$ is an appropriate way to test whether $I$ and $d$ follow a power-law relationship, and state what feature of the resulting graph would give the value of $n$. [2]
(b) Use your GDC to find the equation of the regression line of $\log_{10}I$ on $\log_{10}d$, and hence find the values of $k$ and $n$ in the model $I=kd^n$, giving each to 3 significant figures. State the value of $R^2$. [5]
(c) State, with a reason, whether the data supports the theoretical inverse-square law ($n=-2$). [2]
(d) Use the model to estimate the intensity at a distance of $d=20$ m, giving your answer to 3 significant figures. [2]
(e) Explain why using this model to estimate the intensity at $d=500$ m would be unreliable. [2]
Show complete worked solution
(a) If $I=kd^n$, taking $\log_{10}$ of both sides:
$$\log_{10}I = \log_{10}k + n\log_{10}d$$
This is linear in $\log_{10}d$, with gradient $n$ and vertical intercept $\log_{10}k$. Plotting $\log_{10}I$ against $\log_{10}d$ therefore tests the power-law hypothesis: if the data lies close to a straight line, $I=kd^n$ is a good model, and the gradient of that line gives the value of the power $n$ directly.
$$\boxed{\log_{10}I=\log_{10}k+n\log_{10}d \text{ is linear in } \log_{10}d; \text{ the gradient gives } n.}$$
(b) Using power regression on the GDC (equivalent to linear regression of $\log_{10}I$ on $\log_{10}d$) on the five data points:
$$\boxed{k\approx1270,\quad n\approx-2.00,\quad R^2\approx1.00\text{ (3 s.f.)}}$$
giving the model $I=1270\,d^{-2.00}$, an excellent fit.
(c) The regression gives $n\approx-2.00$ (3 s.f.), which matches the theoretical value $n=-2$ predicted by the inverse-square law almost exactly, with $R^2$ extremely close to $1$.
$$\boxed{\text{Yes -- the data strongly supports the inverse-square law, since } n\approx-2.00 \text{ with an excellent fit.}}$$
(d) Substituting $d=20$ into the model (using the unrounded regression values $k=1273.99$, $n=-2.00065$ for accuracy):
$$I = 1273.99\times20^{-2.00065}$$
$$I\approx3.18\ldots$$
$$\boxed{I\approx3.18\times10^{-6}\text{ W/m}^2\text{ (3 s.f.)}}$$
(e) The value $d=20$ m lies within the range of the collected data ($2$ to $32$ m), so it is an interpolation, as in part (d). By contrast, $d=500$ m is far outside this range -- more than $15$ times the largest distance measured -- so using the model there would be a substantial extrapolation. Over such a large additional distance, real-world effects not captured by the original small-scale data (such as reflections from walls or the ground, background noise, or air absorption) could cause the actual intensity to differ significantly from what the fitted power model predicts.
$$\boxed{d=500\text{ m is far outside the data range, so this would be an unreliable extrapolation.}}$$
QUESTION 43
12 marks
Hard
A microbiologist measures the population, $P$ (in thousands of cells), of a bacterial culture at various times, $t$ (in hours), after the start of an experiment. It is believed that $P$ and $t$ are related by an exponential model $P=P_0b^t$.
t (hours): 0, 1, 2, 3, 4
P (thousands): 200, 320, 512, 819, 1311
(a) Explain why plotting $\ln P$ against $t$ is an appropriate way to test this exponential model. [2]
(b) Use your GDC to find the values of $P_0$, $b$ and $R^2$, giving $P_0$ and $b$ to 3 significant figures. [4]
(c) Find the doubling time of the population, giving your answer to 3 significant figures. [2]
(d) Use the model to estimate the population at $t=8$ hours, giving your answer to 3 significant figures. [2]
(e) Comment on the reliability of the estimate in part (d). [2]
Show complete worked solution
(a) If $P=P_0b^t$, taking $\ln$ of both sides:
$$\ln P = \ln P_0 + t\ln b$$
This is linear in $t$, with gradient $\ln b$ and intercept $\ln P_0$. So plotting $\ln P$ against $t$ should give an approximately straight line if the exponential model is appropriate, allowing the model to be tested and its parameters found.
$$\boxed{\ln P=(\ln b)t+\ln P_0 \text{ is linear in } t.}$$
(b) Using exponential regression on the GDC (equivalent to linear regression of $\ln P$ on $t$) on the five data points:
$$\boxed{P_0\approx200,\quad b\approx1.60,\quad R^2\approx1.00\text{ (3 s.f.)}}$$
giving the model $P=200(1.60)^t$, an excellent fit.
(c) The population doubles when $b^t=2$, i.e. when $t\ln b=\ln 2$:
$$t = \dfrac{\ln 2}{\ln 1.60003\ldots} = \dfrac{0.69315\ldots}{0.47003\ldots}$$
$$t = 1.47471\ldots$$
$$\boxed{t\approx1.47\text{ hours (3 s.f.)}}$$
(d) Substituting $t=8$ into the model (using the unrounded values $P_0=199.991$, $b=1.60003$ for accuracy):
$$P = 199.991\times1.60003^8$$
$$P\approx8590.8\ldots$$
$$\boxed{P\approx8590\text{ thousand cells (3 s.f.)}}$$
(e) The value $t=8$ hours is well outside the range of the collected data ($t=0$ to $t=4$ hours), so this is a substantial extrapolation -- double the largest time measured. Exponential growth models for real populations typically break down over longer timescales, as limited food, space, or the build-up of waste products cause the growth rate to slow (a carrying capacity is eventually reached), so the true population at $t=8$ is likely to be considerably lower than this model predicts.
$$\boxed{\text{Unreliable: } t=8\text{ is a large extrapolation, and real bacterial growth cannot remain exponential indefinitely.}}$$
QUESTION 44
13 marks
Hard
A physicist measures the mass, $M$ (in grams), of a radioactive sample remaining at various times, $t$ (in days). It is believed that $M$ and $t$ are related by an exponential decay model $M=M_0b^t$, where $0(a) Explain why plotting $\ln M$ against $t$ is an appropriate way to test this exponential decay model, and explain why you would expect the gradient of the resulting line to be negative. [2]
(b) Use your GDC to find the values of $M_0$, $b$ and $R^2$, giving $M_0$ and $b$ to 3 significant figures. [4]
(c) Find the half-life of the sample, that is, the time taken for the mass to halve, giving your answer to 3 significant figures. Comment on how this compares with the data given in the table. [4]
(d) Use the model to estimate the remaining mass at $t=50$ days, giving your answer to 3 significant figures. [3]
Show complete worked solution
(a) If $M=M_0b^t$, taking $\ln$ of both sides:
$$\ln M = \ln M_0 + t\ln b$$
This is linear in $t$, with gradient $\ln b$. Since the sample is decaying, $0
QUESTION 45
12 marks
Hard
A zoologist investigates how brain mass, $B$ (in grams), depends on body mass, $m$ (in kg), across a range of mammal species, and collects the following data. It is believed that $B$ and $m$ are related by a power (allometric) model $B=am^n$.
m (kg): 10, 50, 100, 500, 1000
B (g): 4.2, 11.0, 19.5, 52.0, 92.0
(a) Explain why plotting $\log_{10}B$ against $\log_{10}m$ is an appropriate way to test this power model. [2]
(b) Use your GDC to find the values of $a$, $n$ and $R^2$, giving $a$ and $n$ to 3 significant figures. [4]
(c) Use your model to estimate the brain mass of a mammal with body mass $300$ kg, giving your answer to 3 significant figures. [2]
(d) A separate researcher wants to use the model to estimate the brain mass of a mammal with body mass $5000$ kg. Find this estimate, and comment on its reliability. [4]
Show complete worked solution
(a) If $B=am^n$, taking $\log_{10}$ of both sides:
$$\log_{10}B = \log_{10}a + n\log_{10}m$$
This is linear in $\log_{10}m$, with gradient $n$ and intercept $\log_{10}a$. Plotting $\log_{10}B$ against $\log_{10}m$ should therefore give an approximately straight line if the power model is appropriate.
$$\boxed{\log_{10}B=n\log_{10}m+\log_{10}a \text{ is linear in } \log_{10}m.}$$
(b) Using power regression on the GDC (equivalent to linear regression of $\log_{10}B$ on $\log_{10}m$) on the five data points:
$$\boxed{a\approx0.872,\quad n\approx0.667,\quad R^2\approx0.998\text{ (3 s.f.)}}$$
giving the model $B=0.872\,m^{0.667}$, an excellent fit.
(c) Substituting $m=300$ into the model (using the unrounded values $a=0.871772$, $n=0.666714$ for accuracy):
$$B = 0.871772\times300^{0.666714}$$
$$B\approx39.1\ldots$$
$$\boxed{B\approx39.1\text{ g (3 s.f.)}}$$
(d) Substituting $m=5000$ into the model:
$$B = 0.871772\times5000^{0.666714}$$
$$B\approx255\ldots$$
$$\boxed{B\approx255\text{ g (3 s.f.)}}$$
Since $m=300$ kg lies within the range of the collected data ($10$ kg to $1000$ kg), the estimate in part (c) is an interpolation and can be trusted reasonably well. However, $m=5000$ kg is well outside this range -- five times the largest body mass measured -- so this estimate is a substantial extrapolation and is far less reliable. Interestingly, the fitted exponent $n\approx0.667$ is close to the value $\dfrac{2}{3}$ predicted by some biological scaling theories, which lends some support to using the model beyond the data range, but the estimate at $m=5000$ kg should still be treated with caution.
$$\boxed{B(5000)\approx255\text{ g, but this is an unreliable extrapolation well beyond the data range.}}$$
QUESTION 46
11 marks
Hard
The energy, $E$ (in joules), released by an earthquake is related to its magnitude, $M$, on the Richter-type scale by
$$\log_{10}E = 1.5M + 4.8$$
(a) Show that an increase in magnitude of exactly $1$ corresponds to the energy released increasing by a factor of approximately $31.6$. [4]
(b) An earthquake of magnitude $M=3.0$ releases energy $E_1$. A second earthquake releases $1\,000\,000$ times as much energy as the first. Find the magnitude of the second earthquake. [4]
(c) Earthquake X has magnitude $5.5$ and earthquake Y has magnitude $7.5$. Find how many times more energy earthquake Y releases compared with earthquake X. [3]
Show complete worked solution
(a) Let earthquake $1$ have magnitude $M_1$ and energy $E_1$, and earthquake $2$ have magnitude $M_2=M_1+1$ and energy $E_2$. From $\log_{10}E=1.5M+4.8$, rearranging for $E$:
$$E = 10^{1.5M+4.8}$$
Forming the ratio:
$$\dfrac{E_2}{E_1} = \dfrac{10^{1.5M_2+4.8}}{10^{1.5M_1+4.8}} = 10^{1.5(M_2-M_1)}$$
Since $M_2-M_1=1$:
$$\dfrac{E_2}{E_1} = 10^{1.5\times1} = 10^{1.5}$$
$$10^{1.5} = 31.6227\ldots \approx31.6\text{ (3 s.f.)}$$
$$\boxed{\dfrac{E_2}{E_1}\approx31.6, \text{ so each unit increase in magnitude multiplies energy by about } 31.6.}$$
(b) Using the general result from part (a), $\dfrac{E_2}{E_1}=10^{1.5(M_2-M_1)}$. Setting this equal to $1\,000\,000=10^6$, with $M_1=3.0$:
$$10^{1.5(M_2-3.0)} = 10^6$$
Since the bases are equal, the exponents must be equal:
$$1.5(M_2-3.0) = 6$$
$$M_2-3.0 = 4$$
$$M_2 = 7.0$$
$$\boxed{M_2 = 7.0}$$
(c) Using $\dfrac{E_Y}{E_X}=10^{1.5(M_Y-M_X)}$ with $M_X=5.5$, $M_Y=7.5$:
$$\dfrac{E_Y}{E_X} = 10^{1.5\times(7.5-5.5)} = 10^{1.5\times2} = 10^3$$
$$\boxed{\text{Earthquake Y releases } 1000 \text{ times more energy than earthquake X.}}$$
QUESTION 47
12 marks
Hard
A planetary geologist studies impact craters and records the crater diameter, $D$ (in metres), produced by impacts of different energy, $E$ (in kilotons of TNT equivalent). It is believed that $D$ and $E$ are related by a power model $D=kE^n$.
E (kilotons): 1, 10, 100, 1000, 10000
D (m): 50.0, 95.0, 178.0, 340.0, 640.0
(a) Explain why plotting $\log_{10}D$ against $\log_{10}E$ is an appropriate way to test this power model, and state how the value of $n$ can be found from the resulting graph. [2]
(b) Use your GDC to find the values of $k$, $n$ and $R^2$, giving $k$ and $n$ to 3 significant figures. [4]
(c) Use your model to estimate the crater diameter produced by an impact of energy $5000$ kilotons, giving your answer to 3 significant figures. [3]
(d) Explain why using this model to estimate the crater diameter for an impact of energy $50\,000\,000$ kilotons (roughly the scale of the impact linked to the extinction of the dinosaurs) would be unreliable. [3]
Show complete worked solution
(a) If $D=kE^n$, taking $\log_{10}$ of both sides:
$$\log_{10}D = \log_{10}k + n\log_{10}E$$
This is linear in $\log_{10}E$, with gradient $n$ and intercept $\log_{10}k$. Plotting $\log_{10}D$ against $\log_{10}E$ should therefore give an approximately straight line if the power model is appropriate, and the gradient of that line gives the value of $n$ directly.
$$\boxed{\log_{10}D=n\log_{10}E+\log_{10}k \text{ is linear in } \log_{10}E; \text{ the gradient gives } n.}$$
(b) Using power regression on the GDC (equivalent to linear regression of $\log_{10}D$ on $\log_{10}E$) on the five data points:
$$\boxed{k\approx50.0,\quad n\approx0.277,\quad R^2\approx1.00\text{ (3 s.f.)}}$$
giving the model $D=50.0\,E^{0.277}$, an excellent fit.
(c) Substituting $E=5000$ into the model (using the unrounded values $k=50.0402$, $n=0.276818$ for accuracy):
$$D = 50.0402\times5000^{0.276818}$$
$$D\approx528.8\ldots$$
$$\boxed{D\approx529\text{ m (3 s.f.)}}$$
(d) The value $E=5000$ kilotons lies within the range of the collected data ($1$ to $10\,000$ kilotons), so the estimate in part (c) is an interpolation and can be reasonably trusted. In contrast, $E=50\,000\,000$ kilotons is many thousands of times larger than the largest energy in the dataset, so using the model there would be an extreme extrapolation, far beyond the scale on which the power law was tested. At such enormous energies, additional physical effects not present in the small-scale cratering data used here (such as changes in target rock behaviour, atmospheric effects, or the impact forming a crater so large that gravitational collapse and rim slumping significantly alter its final shape) could cause the true crater diameter to differ substantially from the model's prediction.
$$\boxed{E=50\,000\,000\text{ kilotons is far outside the data range, so this extrapolation would be highly unreliable.}}$$
QUESTION 48
12 marks
Hard
A cup of coffee is left to cool in a room with a constant ambient temperature of $20\,^{\circ}\text{C}$. The temperature difference, $D$ (in $^{\circ}\text{C}$), between the coffee and the room is recorded at various times, $t$ (in minutes), where $D=T-20$ and $T$ is the coffee's temperature. It is believed that $D$ and $t$ are related by an exponential decay model $D=D_0b^t$.
t (min): 0, 4, 8, 12, 16
D ($^{\circ}$C): 70.0, 52.0, 37.0, 28.5, 19.5
(a) Explain why plotting $\ln D$ against $t$ is an appropriate way to test this exponential decay model. [2]
(b) Use your GDC to find the values of $D_0$, $b$ and $R^2$, giving $D_0$ and $b$ to 3 significant figures, and hence write down a model for the coffee's temperature $T$ in terms of $t$. [4]
(c) Use your model to estimate the coffee's temperature at $t=25$ minutes, giving your answer to 3 significant figures. [3]
(d) Explain why this exponential model cannot realistically be used to predict the coffee's temperature for very large values of $t$. [3]
Show complete worked solution
(a) If $D=D_0b^t$, taking $\ln$ of both sides:
$$\ln D = \ln D_0 + t\ln b$$
This is linear in $t$, with gradient $\ln b$ and intercept $\ln D_0$. Plotting $\ln D$ against $t$ should therefore give an approximately straight line if the exponential decay model is appropriate.
$$\boxed{\ln D=(\ln b)t+\ln D_0 \text{ is linear in } t.}$$
(b) Using exponential regression on the GDC (equivalent to linear regression of $\ln D$ on $t$) on the five data points:
$$\boxed{D_0\approx70.6,\quad b\approx0.924,\quad R^2\approx0.997\text{ (3 s.f.)}}$$
giving $D=70.6(0.924)^t$. Since $D=T-20$, the model for temperature is:
$$\boxed{T = 20 + 70.6(0.924)^t}$$
(c) Substituting $t=25$ into the model (using the unrounded values $D_0=70.647$, $b=0.924098$ for accuracy):
$$D = 70.647\times0.924098^{25} \approx 9.82\ldots$$
$$T = 20 + 9.82 = 29.82\ldots$$
$$\boxed{T\approx29.8\,^{\circ}\text{C (3 s.f.)}}$$
(d) As $t\to\infty$, the model $D=D_0b^t$ with $0
QUESTION 49
13 marks
Hard
An engineer records the number of transistors, $N$ (in millions), on a typical microprocessor chip, $t$ years after a reference year. It is believed that $N$ and $t$ are related by an exponential model $N=N_0b^t$.
t (years): 0, 2, 4, 6, 8
N (millions): 1000, 2100, 4000, 8900, 17000
(a) Explain why plotting $\ln N$ against $t$ is an appropriate way to test this exponential model. [2]
(b) Use your GDC to find the values of $N_0$, $b$ and $R^2$, giving $N_0$ and $b$ to 3 significant figures. [4]
(c) Find the doubling time of the number of transistors, giving your answer to 3 significant figures. [3]
(d) A colleague wants to use the model to predict the number of transistors at $t=40$ years. Find this prediction, and explain why it would be unreliable even though $R^2$ is very close to $1$. [4]
Show complete worked solution
(a) If $N=N_0b^t$, taking $\ln$ of both sides:
$$\ln N = \ln N_0 + t\ln b$$
This is linear in $t$, with gradient $\ln b$ and intercept $\ln N_0$. So plotting $\ln N$ against $t$ should give an approximately straight line if the exponential growth model is appropriate.
$$\boxed{\ln N=(\ln b)t+\ln N_0 \text{ is linear in } t.}$$
(b) Using exponential regression on the GDC (equivalent to linear regression of $\ln N$ on $t$) on the five data points:
$$\boxed{N_0\approx1010,\quad b\approx1.43,\quad R^2\approx0.999\text{ (3 s.f.)}}$$
giving the model $N=1010(1.43)^t$, an excellent fit.
(c) The value doubles when $b^t=2$, i.e. when $t\ln b=\ln 2$:
$$t = \dfrac{\ln 2}{\ln 1.42693\ldots} = \dfrac{0.69315\ldots}{0.35553\ldots}$$
$$t = 1.94964\ldots$$
$$\boxed{t\approx1.95\text{ years (3 s.f.)}}$$
(d) Substituting $t=40$ into the model (using the unrounded values $N_0=1007.42$, $b=1.42693$ for accuracy):
$$N = 1007.42\times1.42693^{40}$$
$$N\approx4.37\times10^{8}$$
$$\boxed{N\approx4.37\times10^8\text{ million transistors (3 s.f.)}}$$
Although $R^2\approx0.999$ shows the model fits the given data ($t=0$ to $t=8$ years) extremely well, $t=40$ is five times beyond the largest value of $t$ in the dataset, so this is a huge extrapolation. A high $R^2$ only measures how well the model fits the data it was built from -- it says nothing about whether the same trend continues far outside that range. In reality, exponential growth in transistor counts is limited by physical constraints (the size of atoms, manufacturing precision, heat dissipation) and economic factors, so growth is expected to slow down well before $t=40$, making this prediction highly unreliable despite the excellent $R^2$ value.
$$\boxed{\text{Unreliable: } t=40 \text{ is a huge extrapolation, and physical/economic limits mean growth cannot stay exponential forever, whatever } R^2 \text{ says about the original data.}}$$
QUESTION 50
11 marks
Hard
The apparent magnitude, $m$, of a star (a measure of how bright it appears) is related to its flux (brightness), $F$, by the Pogson formula
$$m = -2.5\log_{10}\left(\dfrac{F}{F_0}\right)$$
where $F_0$ is a fixed reference flux. Note that a smaller (or more negative) value of $m$ corresponds to a brighter star.
(a) Two stars have fluxes $F_1$ and $F_2$, with magnitudes $m_1$ and $m_2$ respectively, where $m_2-m_1=5$. Show that $\dfrac{F_1}{F_2}=100$ exactly. [4]
(b) Two other stars differ in magnitude by exactly $2.3$ (the brighter star having the smaller magnitude). Find the ratio of their fluxes, giving your answer to 3 significant figures. [3]
(c) Star A has magnitude $m_A=1.2$ and has a flux $15$ times that of Star B. Find the magnitude of Star B, giving your answer to 3 significant figures. [4]
Show complete worked solution
(a) From $m=-2.5\log_{10}\left(\dfrac{F}{F_0}\right)$, rearranging for flux:
$$-\dfrac{m}{2.5} = \log_{10}\left(\dfrac{F}{F_0}\right) \quad\Rightarrow\quad F = F_0\times10^{-m/2.5}$$
Forming the ratio for the two stars:
$$\dfrac{F_1}{F_2} = \dfrac{F_0\times10^{-m_1/2.5}}{F_0\times10^{-m_2/2.5}} = 10^{(m_2-m_1)/2.5}$$
Substituting $m_2-m_1=5$:
$$\dfrac{F_1}{F_2} = 10^{5/2.5} = 10^2$$
$$\boxed{\dfrac{F_1}{F_2} = 100}$$
as required.
(b) Using the general result from part (a), $\dfrac{F_{\text{bright}}}{F_{\text{dim}}}=10^{\Delta m/2.5}$, with $\Delta m=2.3$:
$$\dfrac{F_{\text{bright}}}{F_{\text{dim}}} = 10^{2.3/2.5} = 10^{0.92}$$
$$10^{0.92} = 8.31763\ldots$$
$$\boxed{\dfrac{F_{\text{bright}}}{F_{\text{dim}}}\approx8.32\text{ (3 s.f.)}}$$
(c) Since Star A has flux $15$ times that of Star B, $\dfrac{F_A}{F_B}=15$, so Star A is brighter and therefore $m_A