Functions

QUESTION 1 3 marks

Easy

Consider the function $f(x) = \sqrt{7-2x}$. (a) Find the largest possible domain of $f$, for $f(x) \in \mathbb{R}$. (b) State the range of $f$ for this domain.

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(a) Using the rule that the expression under a square root must be non-negative for a real output: $$7 - 2x \geq 0$$ $$7 \geq 2x$$ $$x \leq 3.5$$ $$\boxed{\text{Domain: } x \leq 3.5, \; x \in \mathbb{R}}$$ (b) Since a square root is always non-negative, $f(x) \geq 0$ for every $x$ in the domain. As $x \to -\infty$ (moving further left within the domain), the expression $7-2x \to \infty$, so $f(x) = \sqrt{7-2x} \to \infty$ as well. Since $f$ is continuous and takes both the value $0$ (at $x=3.5$) and arbitrarily large values, it takes every value from $0$ upward. $$\boxed{\text{Range: } f(x) \geq 0}$$

QUESTION 2 5 marks

Medium

Consider the function $f(x) = \dfrac{3x+1}{x-2}$, $x \neq 2$. Find $f^{-1}(x)$, and state the domain of $f^{-1}$.

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Using the method of interchanging variables: let $y = f(x)$, then solve for $x$ in terms of $y$. $$y = \frac{3x+1}{x-2}$$ Multiply both sides by $(x-2)$: $$y(x-2) = 3x+1$$ $$yx - 2y = 3x + 1$$ Collect the terms containing $x$ on one side: $$yx - 3x = 1 + 2y$$ $$x(y-3) = 1+2y$$ $$x = \frac{1+2y}{y-3}$$ Replacing $y$ with $x$ (the standard variable for the inverse function): $$\boxed{f^{-1}(x) = \dfrac{2x+1}{x-3}}$$ The domain of $f^{-1}$ equals the range of $f$. Since $f$ has a horizontal asymptote at $y=3$ (the ratio of the leading coefficients, $3/1$), the value $y=3$ is never attained by $f$, so it is excluded from the range of $f$: $$\boxed{\text{Domain of } f^{-1}: x \neq 3}$$ (Consistent check: $f^{-1}(x)=\dfrac{2x+1}{x-3}$ itself has vertical asymptote $x=3$, confirming this.)

QUESTION 3 5 marks

Easy

A function is defined by $f(x) = x^2 - 6x + 5$ for $x \in [4,9]$. (a) Find $f(4)$ and $f(9)$. (b) By finding the axis of symmetry of $f$, explain why $f$ is one-to-one on the domain $[4,9]$. (c) Find the range of $f$ on this domain.

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(a) Substituting the endpoints into $f(x) = x^2-6x+5$: $$f(4) = 4^2 - 6(4) + 5 = 16 - 24 + 5 = -3$$ $$f(9) = 9^2 - 6(9) + 5 = 81 - 54 + 5 = 32$$ $$\boxed{f(4)=-3, \; f(9)=32}$$ (b) Completing the square: $$f(x) = x^2-6x+5 = (x-3)^2 - 9 + 5 = (x-3)^2 - 4$$ The axis of symmetry is $x=3$. Since the domain $[4,9]$ lies entirely to the right of $x=3$, the function is strictly increasing throughout this interval, so no two different $x$-values in $[4,9]$ give the same output. Hence $f$ is one-to-one on $[4,9]$. (c) Since $f$ is increasing on $[4,9]$, the minimum output occurs at the smallest input $x=4$ and the maximum output at the largest input $x=9$. Using the values found in part (a): $$\boxed{\text{Range: } -3 \leq f(x) \leq 32}$$

QUESTION 4 7 marks

Medium

The function $g$ is defined by $g(x) = 2 + \sqrt{x+3}$ for $x \geq -3$. (a) Find $g^{-1}(x)$, stating its domain and range. (b) Solve the equation $g(x) = g^{-1}(x)$, giving your answer in exact form.

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(a) Let $y = 2 + \sqrt{x+3}$, where $y \geq 2$ (since $\sqrt{\phantom{x}} \geq 0$). $$y - 2 = \sqrt{x+3}$$ Squaring both sides: $$(y-2)^2 = x+3$$ $$x = (y-2)^2 - 3$$ Replacing $y$ with $x$: $$\boxed{g^{-1}(x) = (x-2)^2 - 3}$$ The domain of $g^{-1}$ equals the range of $g$: $\boxed{x \geq 2}$ The range of $g^{-1}$ equals the domain of $g$: $\boxed{g^{-1}(x) \geq -3}$ (b) Since $g$ is strictly increasing, its graph can meet the graph of $g^{-1}$ only on the line $y=x$ (a standard property of increasing functions and their inverses). So we solve $g(x) = x$ instead: $$2 + \sqrt{x+3} = x$$ $$\sqrt{x+3} = x - 2 \quad (\text{requires } x \geq 2, \text{ since the left side is non-negative})$$ Squaring both sides: $$x+3 = x^2 - 4x + 4$$ $$0 = x^2 - 5x + 1$$ Using the quadratic formula: $$x = \frac{5 \pm \sqrt{25-4}}{2} = \frac{5\pm\sqrt{21}}{2}$$ Checking against the requirement $x \geq 2$: $\dfrac{5-\sqrt{21}}{2} \approx 0.209$ is rejected, while $\dfrac{5+\sqrt{21}}{2}\approx 4.79$ is valid. $$\boxed{x = \dfrac{5+\sqrt{21}}{2}}$$

QUESTION 5 14 marks

Hard

Consider the function $f(x) = x^2 - 4x + 7$, $x \in \mathbb{R}$. (a) Express $f(x)$ in the form $(x-h)^2+k$, and hence state the range of $f$. (b) By finding two different values of $x$ that give the same output, explain why $f$ does not have an inverse function on its natural domain $\mathbb{R}$. The domain of $f$ is now restricted to $x \geq 2$ to form a new function $g$, where $g(x) = f(x)$ for $x \geq 2$. (c) Find $g^{-1}(x)$, stating its domain and range. (d) Solve the equation $g^{-1}(x) = 3$.

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(a) Completing the square: $$f(x) = x^2-4x+7 = (x-2)^2 - 4 + 7 = (x-2)^2+3$$ $$\boxed{f(x) = (x-2)^2+3}$$ Since $(x-2)^2 \geq 0$ for all $x$, the minimum value of $f$ is $3$, attained at $x=2$, and $f(x) \to \infty$ as $x\to\pm\infty$. $$\boxed{\text{Range: } f(x) \geq 3}$$ (b) Evaluating $f$ at two different inputs: $$f(0) = 0-0+7=7, \qquad f(4) = 16-16+7=7$$ Since $f(0)=f(4)=7$ but $0 \neq 4$, two different inputs give the same output, so $f$ fails the horizontal line test and is not one-to-one on $\mathbb{R}$. A function must be one-to-one to have an inverse function, so $f$ cannot have an inverse function on this domain. (c) For $x \geq 2$, $g(x) = (x-2)^2+3$ is strictly increasing (moving away from the vertex), hence one-to-one, so $g^{-1}$ exists. Let $y = (x-2)^2+3$: $$y - 3 = (x-2)^2$$ $$x - 2 = \pm\sqrt{y-3}$$ Since $x \geq 2$, we take the positive root: $$x = 2 + \sqrt{y-3}$$ $$\boxed{g^{-1}(x) = 2+\sqrt{x-3}}$$ Domain of $g^{-1}$ equals range of $g$: $\boxed{x \geq 3}$ Range of $g^{-1}$ equals domain of $g$: $\boxed{g^{-1}(x) \geq 2}$ (d) Substituting into $g^{-1}(x)=3$: $$2 + \sqrt{x-3} = 3$$ $$\sqrt{x-3} = 1$$ $$x - 3 = 1$$ $$\boxed{x = 4}$$

QUESTION 6 4 marks

Easy

Let $f(x) = 2x-5$ and $g(x) = x^2+1$. (a) Find $(f \circ g)(x)$. (b) Find $(g \circ f)(2)$.

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(a) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = f(g(x)) = f(x^2+1) = 2(x^2+1)-5$$ $$= 2x^2+2-5$$ $$\boxed{(f\circ g)(x) = 2x^2-3}$$ (b) First evaluate the inner function at $x=2$: $$f(2) = 2(2)-5 = -1$$ Then apply $g$ to this result: $$g(f(2)) = g(-1) = (-1)^2+1 = 1+1=2$$ $$\boxed{(g\circ f)(2) = 2}$$

QUESTION 7 5 marks

Medium

Let $f(x) = 3x+2$ and $h(x) = \dfrac{x-2}{3}$. (a) Find $(f\circ h)(x)$ and $(h\circ f)(x)$, simplifying fully. (b) State, with reference to your answers in part (a), the relationship between $f$ and $h$.

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(a) Substituting $h(x)$ into $f$: $$(f\circ h)(x) = f\left(\frac{x-2}{3}\right) = 3\left(\frac{x-2}{3}\right)+2$$ $$= (x-2)+2 = x$$ $$\boxed{(f\circ h)(x) = x}$$ Substituting $f(x)$ into $h$: $$(h\circ f)(x) = h(3x+2) = \frac{(3x+2)-2}{3} = \frac{3x}{3} = x$$ $$\boxed{(h\circ f)(x) = x}$$ (b) Since $(f\circ h)(x) = x$ and $(h\circ f)(x)=x$ for all $x$, the functions undo one another when composed in either order. $$\boxed{h = f^{-1} \text{ (equivalently, } f = h^{-1}\text{)}}$$

QUESTION 8 5 marks

Easy

Let $f(x) = \sqrt{x+5}$, $x \geq -5$, and $g(x) = x^2-3$, $x \in \mathbb{R}$. (a) Find $(g\circ f)(x)$ in simplified form, and state its domain. (b) Find $(g\circ f)(11)$.

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(a) Substituting $f(x)$ into $g$: $$(g\circ f)(x) = g(\sqrt{x+5}) = (\sqrt{x+5})^2 - 3$$ $$= (x+5)-3 = x+2$$ Although the simplified expression $x+2$ is defined for all real $x$, the composite is only defined where the inner function $f$ is defined, i.e. $x\geq-5$. $$\boxed{(g\circ f)(x) = x+2, \; x\geq -5}$$ (b) Using the simplified form found in (a): $$(g\circ f)(11) = 11+2 = 13$$ (Check directly: $f(11) = \sqrt{16}=4$, then $g(4) = 16-3=13$.) $$\boxed{(g\circ f)(11) = 13}$$

QUESTION 9 6 marks

Medium

Let $f(x) = \dfrac{x+1}{x-3}$, $x \neq 3$. (a) Find $f^{-1}(x)$. (b) Find $(f\circ f)(x)$ in simplified form, stating any values of $x$ that must be excluded.

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(a) Let $y = \dfrac{x+1}{x-3}$. Solving for $x$ in terms of $y$: $$y(x-3) = x+1$$ $$yx-3y=x+1$$ $$yx-x=1+3y$$ $$x(y-1)=1+3y$$ $$x = \frac{3y+1}{y-1}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x) = \dfrac{3x+1}{x-1}}$$ (b) Substituting $f(x)$ into $f$ itself: $$(f\circ f)(x) = f(f(x)) = \frac{f(x)+1}{f(x)-3}$$ Simplify the numerator: $$f(x)+1 = \frac{x+1}{x-3}+1 = \frac{x+1+x-3}{x-3} = \frac{2x-2}{x-3}$$ Simplify the denominator: $$f(x)-3 = \frac{x+1}{x-3}-3 = \frac{x+1-3(x-3)}{x-3} = \frac{x+1-3x+9}{x-3} = \frac{-2x+10}{x-3}$$ Combining: $$(f\circ f)(x) = \frac{2x-2}{-2x+10} = \frac{2(x-1)}{-2(x-5)} = -\frac{x-1}{x-5} = \frac{x-1}{5-x}$$ $$\boxed{(f\circ f)(x) = \dfrac{x-1}{5-x}}$$ Excluded values: $x\neq3$ (so that the inner $f(x)$ is defined), and $f(x) \neq 3$ (so that the outer $f$ is defined). Solving $f(x)=3$: $\dfrac{x+1}{x-3}=3 \Rightarrow x+1=3x-9 \Rightarrow x=5$. $$\boxed{x\neq3, \; x\neq5}$$

QUESTION 10 15 marks

Hard

Let $f(x) = \sqrt{2x-4}+1$, $x \geq 2$, and $g(x) = 3x-5$, $x \in \mathbb{R}$. (a) Find $(g\circ f)(x)$ in simplified form, and state its domain. (b) Find $(f\circ g)(x)$ in simplified form, and state its domain. (c) Find $f^{-1}(x)$, stating its domain and range. (d) Using your graphic display calculator, solve the equation $(g\circ f)(x) = (f\circ g)(x)$, giving your answer correct to 3 significant figures.

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(a) Substituting $f(x)$ into $g$: $$(g\circ f)(x) = g(\sqrt{2x-4}+1) = 3(\sqrt{2x-4}+1)-5$$ $$= 3\sqrt{2x-4}+3-5 = 3\sqrt{2x-4}-2$$ Since $g$ accepts every real input, the domain of the composite is simply the domain of $f$: $$\boxed{(g\circ f)(x) = 3\sqrt{2x-4}-2, \; x\geq2}$$ (b) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = f(3x-5) = \sqrt{2(3x-5)-4}+1$$ $$= \sqrt{6x-10-4}+1 = \sqrt{6x-14}+1$$ For this to be defined we need $g(x)$ to lie in the domain of $f$, i.e. $3x-5\geq2$: $$x\geq\frac{7}{3}$$ $$\boxed{(f\circ g)(x) = \sqrt{6x-14}+1, \; x\geq\dfrac{7}{3}}$$ (c) Let $y = \sqrt{2x-4}+1$, where $y\geq1$ (since $\sqrt{\phantom{x}}\geq0$). $$y-1 = \sqrt{2x-4}$$ $$(y-1)^2 = 2x-4$$ $$x = \frac{(y-1)^2+4}{2}$$ $$\boxed{f^{-1}(x) = \dfrac{(x-1)^2+4}{2}}$$ Domain of $f^{-1}$ equals range of $f$: $\boxed{x\geq1}$ Range of $f^{-1}$ equals domain of $f$: $\boxed{f^{-1}(x)\geq2}$ (d) We require $x$ in the domain of both composites, so $x \geq \dfrac{7}{3}$. Using a GDC, graph $Y_1 = 3\sqrt{2x-4}-2$ and $Y_2 = \sqrt{6x-14}+1$ for $x\geq\dfrac{7}{3}$ and find their point of intersection: $$\boxed{x \approx 4.30}$$ (Verification: at $x=4.30$, $Y_1 = 3\sqrt{4.6}-2 \approx 6.434-2=4.434$ and $Y_2=\sqrt{11.8}+1\approx3.435+1=4.435$, which agree to 3 s.f.)

QUESTION 11 6 marks

Easy

Determine, with justification, whether each of the following functions is odd, even, or neither. (a) $f(x) = x^3-4x$ [2] (b) $g(x) = x^4+3x^2-1$ [2] (c) $h(x) = x^2+x$ [2]

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Method: a function is even if $f(-x)=f(x)$ for all $x$ in the domain, and odd if $f(-x)=-f(x)$ for all $x$ in the domain. (a) Substituting $-x$ into $f(x)=x^3-4x$: $$f(-x) = (-x)^3-4(-x) = -x^3+4x = -(x^3-4x) = -f(x)$$ Since $f(-x)=-f(x)$, $\boxed{f \text{ is odd}}$. (b) Substituting $-x$ into $g(x)=x^4+3x^2-1$: $$g(-x) = (-x)^4+3(-x)^2-1 = x^4+3x^2-1 = g(x)$$ Since $g(-x)=g(x)$, $\boxed{g \text{ is even}}$. (c) Substituting $-x$ into $h(x)=x^2+x$: $$h(-x) = (-x)^2+(-x) = x^2-x$$ This is not equal to $h(x)=x^2+x$ (unless $x=0$), and not equal to $-h(x)=-x^2-x$ (unless $x=0$), so neither condition holds for all $x$. $$\boxed{h \text{ is neither odd nor even}}$$

QUESTION 12 5 marks

Medium

Consider the function $f(x) = \dfrac{2x+3}{x-2}$, $x \ne 2$. By finding $f^{-1}(x)$ algebraically, show that $f$ is self-inverse, i.e. that $f^{-1}(x)=f(x)$.

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Method: find $f^{-1}(x)$ algebraically and show it is identical to $f(x)$. Let $y = \dfrac{2x+3}{x-2}$. To find $f^{-1}$, solve for $x$ in terms of $y$. $$y(x-2) = 2x+3$$ $$yx - 2y = 2x+3$$ Collecting terms containing $x$: $$yx - 2x = 3+2y$$ $$x(y-2) = 2y+3$$ $$x = \frac{2y+3}{y-2}$$ So $f^{-1}(y) = \dfrac{2y+3}{y-2}$, which is identical in form to $f(y) = \dfrac{2y+3}{y-2}$. Renaming the variable, $f^{-1}(x) = \dfrac{2x+3}{x-2} = f(x)$. $$\boxed{f^{-1}(x)=f(x), \text{ so } f \text{ is self-inverse.}}$$

QUESTION 13 6 marks

Easy

The function $f(x) = ax^3-4x^2+(b+1)x+c$, where $a,b,c\in\mathbb{R}$, is even for all $x\in\mathbb{R}$. It is also given that $f(1)=10$. Find the values of $a$, $b$ and $c$.

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Method: for $f$ to be even, every odd-power term in $f(x)$ must have coefficient zero, since requiring $f(-x)=f(x)$ for all $x$ forces the odd-degree coefficients to vanish. Coefficient of $x^3$: setting this to zero gives $a=0$. Coefficient of $x^1$: setting this to zero gives $b+1=0 \implies b=-1$. With $a=0,\ b=-1$: the function simplifies to $$f(x) = -4x^2+c$$ Using $f(1)=10$: $$-4(1)^2+c=10$$ $$-4+c=10$$ $$c=14$$ $$\boxed{a=0,\ b=-1,\ c=14}$$ (Check: $f(x)=-4x^2+14$ is indeed even, since $f(-x)=-4x^2+14=f(x)$, and $f(1)=-4+14=10$.)

QUESTION 14 7 marks

Medium

Consider the function $f(x) = \dfrac{ax+b}{cx-a}$, $x\ne \dfrac{a}{c}$, where $a,b,c$ are non-zero real constants with $a^2+bc\ne0$. (a) Show that $f$ is self-inverse, i.e. $f^{-1}(x)=f(x)$, for all valid values of $a$, $b$ and $c$. [5] (b) Hence, given specifically that $a=1,\ b=6,\ c=1$ (so $f(x)=\dfrac{x+6}{x-1}$), evaluate $f(f(3))$ without further algebraic manipulation, and confirm your answer by direct calculation. [2]

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(a) Let $y=\dfrac{ax+b}{cx-a}$. Solve for $x$ in terms of $y$: $$y(cx-a) = ax+b$$ $$cxy - ay = ax+b$$ Collecting terms containing $x$: $$cxy - ax = b+ay$$ $$x(cy-a) = ay+b$$ $$x = \frac{ay+b}{cy-a}$$ So $f^{-1}(y) = \dfrac{ay+b}{cy-a}$, which has exactly the same form as $f(y)=\dfrac{ay+b}{cy-a}$. Hence $f^{-1}(x) = \dfrac{ax+b}{cx-a} = f(x)$ for all valid $a,b,c$. $\blacksquare$ (b) Since $f$ is self-inverse (by part (a), which applies here since $a=1,b=6,c=1$ are all non-zero and $a^2+bc = 1+6=7\neq0$), we know $f(f(x))=x$ for all $x$ in the domain. Therefore $\boxed{f(f(3)) = 3}$ directly, with no further calculation needed. Direct check: $f(3) = \dfrac{3+6}{3-1} = \dfrac{9}{2}$. $$f\left(\frac{9}{2}\right) = \frac{\frac{9}{2}+6}{\frac{9}{2}-1} = \frac{\frac{21}{2}}{\frac{7}{2}} = \frac{21}{7}=3$$

QUESTION 15 18 marks

Hard

(a) Let $f$ be any function defined on all of $\mathbb{R}$. Define $$g(x) = \frac{f(x)+f(-x)}{2}, \qquad h(x) = \frac{f(x)-f(-x)}{2}.$$ Prove that $g$ is an even function, that $h$ is an odd function, and that $f(x)=g(x)+h(x)$ for all $x\in\mathbb{R}$. [6] (b) Let $f(x)=e^x$. Using the definitions in part (a), find $g(x)$ and $h(x)$ in terms of $e^x$ and $e^{-x}$. [4] (c) Now consider the separate function $p(x) = \dfrac{3x-2}{x-3}$, $x\ne3$. (i) By finding $p(p(x))$ directly (using composition, not by finding $p^{-1}$ algebraically), verify that $p(p(x))=x$ for all $x$ in the domain, confirming that $p$ is self-inverse. [4] (ii) Solve the equation $p(x)=x$, and interpret your solutions geometrically in relation to the graph of $y=p(x)$ and the line $y=x$. [4]

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(a) **$g$ is even:** substituting $-x$ into the definition of $g$: $$g(-x) = \frac{f(-x)+f(-(-x))}{2} = \frac{f(-x)+f(x)}{2} = \frac{f(x)+f(-x)}{2} = g(x)$$ Since $g(-x)=g(x)$ for all $x$, $g$ is even. $\blacksquare$ **$h$ is odd:** substituting $-x$ into the definition of $h$: $$h(-x) = \frac{f(-x)-f(-(-x))}{2} = \frac{f(-x)-f(x)}{2} = -\frac{f(x)-f(-x)}{2} = -h(x)$$ Since $h(-x)=-h(x)$ for all $x$, $h$ is odd. $\blacksquare$ **Decomposition:** adding the two definitions: $$g(x)+h(x) = \frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2} = \frac{2f(x)}{2} = f(x)$$ So $f(x)=g(x)+h(x)$ for all $x\in\mathbb{R}$. $\blacksquare$ (b) With $f(x)=e^x$: $f(-x)=e^{-x}$. Substituting into the definitions from (a): $$\boxed{g(x) = \frac{e^x+e^{-x}}{2}}, \qquad \boxed{h(x) = \frac{e^x-e^{-x}}{2}}$$ (These are, respectively, the hyperbolic cosine and hyperbolic sine functions, though recognising this is not required.) (c)(i) Let $u=p(x)=\dfrac{3x-2}{x-3}$. Compute $p(u) = \dfrac{3u-2}{u-3}$ by substituting and simplifying over a common denominator of $(x-3)$. Numerator: $$3u-2 = \frac{3(3x-2)}{x-3}-2 = \frac{3(3x-2)-2(x-3)}{x-3} = \frac{9x-6-2x+6}{x-3} = \frac{7x}{x-3}$$ Denominator: $$u-3 = \frac{3x-2}{x-3}-3 = \frac{3x-2-3(x-3)}{x-3} = \frac{3x-2-3x+9}{x-3} = \frac{7}{x-3}$$ Combining: $$p(u) = \frac{7x/(x-3)}{7/(x-3)} = \frac{7x}{7} = x$$ $$\boxed{p(p(x))=x \text{ for all } x \text{ in the domain, so } p \text{ is self-inverse.}}$$ $\blacksquare$ (ii) Setting $p(x)=x$: $$\frac{3x-2}{x-3} = x$$ $$3x-2 = x(x-3) = x^2-3x$$ $$x^2-3x-3x+2=0$$ $$x^2-6x+2=0$$ By the quadratic formula: $$x = \frac{6\pm\sqrt{36-8}}{2} = \frac{6\pm\sqrt{28}}{2} = \frac{6\pm2\sqrt7}{2} = 3\pm\sqrt7$$ $$\boxed{x = 3+\sqrt7 \text{ or } x=3-\sqrt7}$$ **Geometric interpretation:** Since $p$ is self-inverse, its graph is symmetric about the line $y=x$ (reflecting the graph of a self-inverse function in $y=x$ maps the graph to itself). The solutions of $p(x)=x$ are exactly the $x$-coordinates of the points where the graph of $y=p(x)$ meets the line $y=x$, i.e. the fixed points of $p$, which lie on the axis of symmetry of the curve.

QUESTION 16 4 marks

Easy

Let $f(x) = 3x-1$ and $g(x) = 2x+5$. (a) Find $(f \circ g)(x)$. [2] (b) Find $(g \circ f)(-1)$. [2]

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(a) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = f(2x+5) = 3(2x+5)-1$$ $$=6x+15-1$$ $$\boxed{(f\circ g)(x) = 6x+14}$$ (b) First evaluate the inner function at $x=-1$: $$f(-1) = 3(-1)-1 = -4$$ Then apply $g$ to this result: $$g(f(-1)) = g(-4) = 2(-4)+5 = -8+5=-3$$ $$\boxed{(g\circ f)(-1) = -3}$$

QUESTION 17 5 marks

Easy

Let $f(x) = \dfrac{x+6}{4}$. (a) Find $f^{-1}(x)$. [3] (b) Hence verify that $f(f^{-1}(10)) = 10$. [2]

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(a) Using the method of interchanging variables: let $y = f(x)$, then solve for $x$ in terms of $y$. $$y = \dfrac{x+6}{4}$$ Multiply both sides by $4$: $$4y = x+6$$ $$x = 4y-6$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x) = 4x-6}$$ (b) First find $f^{-1}(10)$ using the result from (a): $$f^{-1}(10) = 4(10)-6 = 34$$ Then substitute into $f$: $$f(34) = \dfrac{34+6}{4} = \dfrac{40}{4}=10$$ Since this equals $10$, the result is verified. $$\boxed{f(f^{-1}(10)) = 10}$$

QUESTION 18 5 marks

Easy

Let $f(x) = x^2-5$ and $g(x) = x+3$. (a) Find $(g\circ f)(x)$. [2] (b) Find $(f\circ g)(x)$, giving your answer in expanded form. [3]

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(a) Substituting $f(x)$ into $g$: $$(g\circ f)(x) = g(x^2-5) = (x^2-5)+3$$ $$\boxed{(g\circ f)(x) = x^2-2}$$ (b) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = f(x+3) = (x+3)^2-5$$ Expanding $(x+3)^2 = x^2+6x+9$: $$(f\circ g)(x) = x^2+6x+9-5$$ $$\boxed{(f\circ g)(x) = x^2+6x+4}$$

QUESTION 19 4 marks

Easy

Let $f(x) = \dfrac{1}{x}$, $x\neq0$, and $g(x)=x-4$. (a) Find $(f\circ g)(x)$ in simplified form. [2] (b) State the value of $x$ that must be excluded from the domain of $f\circ g$, and evaluate $(f\circ g)(6)$. [2]

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(a) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = f(x-4) = \dfrac{1}{x-4}$$ $$\boxed{(f\circ g)(x) = \dfrac{1}{x-4}}$$ (b) The expression $\dfrac{1}{x-4}$ is undefined when its denominator is zero, i.e. when $x-4=0$. $$\boxed{x\neq4 \text{ must be excluded}}$$ Evaluating at $x=6$: $$(f\circ g)(6) = \dfrac{1}{6-4} = \dfrac{1}{2}$$ $$\boxed{(f\circ g)(6) = \dfrac12}$$

QUESTION 20 5 marks

Easy

Let $f(x) = 4x-9$ and $g(x) = \dfrac{x+9}{4}$. Show, by finding both $f(g(x))$ and $g(f(x))$, that $f$ and $g$ are inverse functions of each other.

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Method: two functions are inverses of each other if $f(g(x))=x$ and $g(f(x))=x$ for all $x$ in the relevant domain. Substituting $g(x)$ into $f$: $$f(g(x)) = 4\left(\dfrac{x+9}{4}\right)-9 = (x+9)-9 = x$$ Substituting $f(x)$ into $g$: $$g(f(x)) = \dfrac{(4x-9)+9}{4} = \dfrac{4x}{4}=x$$ Since $f(g(x))=x$ and $g(f(x))=x$ for all $x$, $f$ and $g$ undo one another when composed in either order. $$\boxed{f \text{ and } g \text{ are inverse functions of each other.}}$$

QUESTION 21 4 marks

Easy

Let $f(x)=2x+1$ and $g(x)=x^2$. Solve the equation $f(g(x))=19$, giving both solutions.

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Substituting $g(x)$ into $f$: $$f(g(x)) = 2x^2+1$$ Setting this equal to $19$: $$2x^2+1=19$$ $$2x^2=18$$ $$x^2=9$$ $$x=\pm3$$ $$\boxed{x=3 \text{ or } x=-3}$$

QUESTION 22 4 marks

Easy

Let $f(x) = 6-x$. Show algebraically that $f$ is self-inverse, i.e. that $f^{-1}(x)=f(x)$.

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Let $y=6-x$. Solving for $x$ in terms of $y$: $$y=6-x$$ $$x=6-y$$ Replacing $y$ with $x$: $$f^{-1}(x)=6-x$$ This is identical to $f(x)=6-x$. $$\boxed{f^{-1}(x)=f(x), \text{ so } f \text{ is self-inverse.}}$$ (Check: $f(f(x)) = 6-(6-x)=x$ for all $x$, confirming $f$ is its own inverse.)

QUESTION 23 4 marks

Easy

Let $f(x) = x^3+2$. (a) Find $f^{-1}(x)$. [3] (b) Evaluate $f^{-1}(-6)$. [1]

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(a) Let $y=x^3+2$. Solving for $x$ in terms of $y$: $$y-2=x^3$$ $$x=\sqrt[3]{y-2}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=\sqrt[3]{x-2}}$$ (b) Substituting $x=-6$: $$f^{-1}(-6)=\sqrt[3]{-6-2}=\sqrt[3]{-8}$$ $$\boxed{f^{-1}(-6)=-2}$$

QUESTION 24 6 marks

Easy

A function is defined by $f(x) = x^2-8x+10$ for $x\in[5,10]$. The graph of $y=f(x)$ on this domain is shown below. (a) Find $f(5)$ and $f(10)$. [2] (b) By finding the axis of symmetry of $f$, explain why $f$ is one-to-one on the domain $[5,10]$. [2] (c) Find the range of $f$ on this domain. [2]

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(a) Substituting the endpoints into $f(x)=x^2-8x+10$: $$f(5)=5^2-8(5)+10=25-40+10=-5$$ $$f(10)=10^2-8(10)+10=100-80+10=30$$ $$\boxed{f(5)=-5, \; f(10)=30}$$ (b) Completing the square: $$f(x)=x^2-8x+10=(x-4)^2-16+10=(x-4)^2-6$$ The axis of symmetry is $x=4$. Since the domain $[5,10]$ lies entirely to the right of $x=4$, the function is strictly increasing throughout this interval, so no two different $x$-values in $[5,10]$ give the same output. Hence $f$ is one-to-one on $[5,10]$. (c) Since $f$ is increasing on $[5,10]$, the minimum output occurs at the smallest input $x=5$ and the maximum output at the largest input $x=10$. Using the values found in part (a): $$\boxed{\text{Range: } -5\leq f(x)\leq30}$$

QUESTION 25 4 marks

Easy

Let $f(x)=2x-3$. (a) Find $f(f(x))$ in simplified form. [3] (b) Hence evaluate $f(f(2))$. [1]

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(a) Substituting $f(x)$ into $f$ itself: $$f(f(x)) = 2(2x-3)-3$$ $$=4x-6-3$$ $$\boxed{f(f(x)) = 4x-9}$$ (b) Using the simplified form from (a): $$f(f(2)) = 4(2)-9=8-9=-1$$ (Check directly: $f(2)=2(2)-3=1$, then $f(1)=2(1)-3=-1$.) $$\boxed{f(f(2))=-1}$$

QUESTION 26 5 marks

Easy

The graph of $f(x)=\sqrt{2x+9}$, $x\geq-4.5$, is shown below. (a) Using the graph, state the range of $f$, and hence write down the domain of $f^{-1}$. [2] (b) Find $f^{-1}(x)$ algebraically, stating its domain and range. [3]

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(a) From the graph, the curve starts at the point $(-4.5,0)$ and increases without bound as $x$ increases, so the range of $f$ is $f(x)\geq0$. Since the domain of $f^{-1}$ equals the range of $f$: $$\boxed{\text{Domain of } f^{-1}: x\geq0}$$ (b) Let $y=\sqrt{2x+9}$, where $y\geq0$. $$y^2=2x+9$$ $$x=\dfrac{y^2-9}{2}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=\dfrac{x^2-9}{2}}$$ Domain of $f^{-1}$ equals range of $f$: $\boxed{x\geq0}$ Range of $f^{-1}$ equals domain of $f$: $\boxed{f^{-1}(x)\geq-4.5}$

QUESTION 27 4 marks

Easy

Let $f(x)=\sqrt{x}$, $x\geq0$, and $g(x)=x-6$. Find $(f\circ g)(x)$, and state its domain.

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Substituting $g(x)$ into $f$: $$(f\circ g)(x) = f(x-6) = \sqrt{x-6}$$ For this to be defined, the expression under the square root must be non-negative: $$x-6\geq0$$ $$\boxed{(f\circ g)(x) = \sqrt{x-6}, \; x\geq6}$$

QUESTION 28 4 marks

Easy

Let $f(x)=x+5$ and $g(x)=3x-2$. (a) Find $(f\circ g)(x)$. [2] (b) Solve the equation $(f\circ g)(x)=0$. [2]

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(a) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = f(3x-2) = (3x-2)+5$$ $$\boxed{(f\circ g)(x) = 3x+3}$$ (b) Setting the result equal to $0$: $$3x+3=0$$ $$3x=-3$$ $$\boxed{x=-1}$$

QUESTION 29 4 marks

Easy

Let $f(x)=7x-2$. It is given that $f^{-1}(x)=\dfrac{x+2}{7}$. (a) Using the property of inverse functions, write down the value of $f(f^{-1}(15))$ without further calculation, stating the property used. [2] (b) Verify your answer to (a) by direct substitution. [2]

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(a) By the definition of an inverse function, $f(f^{-1}(x))=x$ for every $x$ in the domain of $f^{-1}$. $$\boxed{f(f^{-1}(15))=15}$$ (b) First find $f^{-1}(15)$: $$f^{-1}(15) = \dfrac{15+2}{7}=\dfrac{17}{7}$$ Then substitute into $f$: $$f\left(\dfrac{17}{7}\right) = 7\left(\dfrac{17}{7}\right)-2 = 17-2=15$$ Since this equals $15$, the result is verified. $$\boxed{f(f^{-1}(15))=15}$$

QUESTION 30 6 marks

Medium

Consider the function $f(x)=\dfrac{2x-1}{x+3}$, $x\neq-3$. The graph of $y=f(x)$ is shown below, with its asymptotes indicated. (a) Using the graph, state the domain of $f^{-1}$. [2] (b) Find $f^{-1}(x)$ algebraically, and confirm your answer to (a). [4]

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(a) The graph shows a horizontal asymptote at $y=2$ (and a vertical asymptote at $x=-3$), so $f$ never attains the value $y=2$. Since the domain of $f^{-1}$ equals the range of $f$: $$\boxed{\text{Domain of } f^{-1}: x\neq2}$$ (b) Using the method of interchanging variables: let $y=\dfrac{2x-1}{x+3}$. $$y(x+3)=2x-1$$ $$yx+3y=2x-1$$ Collecting the terms containing $x$: $$yx-2x=-1-3y$$ $$x(y-2)=-1-3y$$ $$x=\dfrac{-1-3y}{y-2}=\dfrac{3y+1}{2-y}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=\dfrac{3x+1}{2-x}}$$ This is undefined when $2-x=0$, i.e. $x=2$, confirming the domain found in (a). $$\boxed{x\neq2}$$

QUESTION 31 8 marks

Medium

Let $f(x)=x^2+1$, $x\geq0$, and $g(x)=x-4$, $x\in\mathbb{R}$. (a) Find $(f\circ g)(x)$, and state its domain. [3] (b) Find $(g\circ f)(x)$, and state its range. [3] (c) Solve the equation $(f\circ g)(x)=17$. [2]

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(a) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = f(x-4) = (x-4)^2+1$$ For this composite to be defined, the input to $f$ must lie in the domain of $f$, i.e. $g(x)\geq0$: $$x-4\geq0 \implies x\geq4$$ $$\boxed{(f\circ g)(x)=(x-4)^2+1, \; x\geq4}$$ (b) Substituting $f(x)$ into $g$: $$(g\circ f)(x) = g(x^2+1) = (x^2+1)-4 = x^2-3$$ Since $x\geq0$, $x^2$ ranges over all values $\geq0$, so $x^2-3$ ranges over all values $\geq-3$. $$\boxed{(g\circ f)(x)=x^2-3, \; \text{range: } (g\circ f)(x)\geq-3}$$ (c) Using the simplified form from (a), valid for $x\geq4$: $$(x-4)^2+1=17$$ $$(x-4)^2=16$$ $$x-4=\pm4$$ $$x=8 \text{ or } x=0$$ Since the domain from (a) requires $x\geq4$, the solution $x=0$ is rejected. $$\boxed{x=8}$$

QUESTION 32 7 marks

Medium

Let $f(x)=3x+1$ and $g(x)=x^2-2x$. (a) Find $(g\circ f)(x)$ in simplified form. [3] (b) Solve the equation $(g\circ f)(x)=8$. [4]

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(a) Substituting $f(x)$ into $g$: $$(g\circ f)(x) = (3x+1)^2-2(3x+1)$$ Expanding $(3x+1)^2 = 9x^2+6x+1$: $$(g\circ f)(x) = 9x^2+6x+1-6x-2$$ $$\boxed{(g\circ f)(x)=9x^2-1}$$ (b) Setting the result equal to $8$: $$9x^2-1=8$$ $$9x^2=9$$ $$x^2=1$$ $$\boxed{x=1 \text{ or } x=-1}$$

QUESTION 33 6 marks

Medium

The graph of $f(x)=(x-3)^2+2$, $x\geq3$, is shown below. (a) Using the graph, state the range of $f$. [1] (b) Find $f^{-1}(x)$ algebraically, stating its domain and range. [5]

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(a) The graph shows the curve starting at the vertex $(3,2)$ and increasing without bound, so: $$\boxed{\text{Range: } f(x)\geq2}$$ (b) Let $y=(x-3)^2+2$, where $y\geq2$. $$y-2=(x-3)^2$$ $$x-3=\pm\sqrt{y-2}$$ Since $x\geq3$, we take the positive root: $$x=3+\sqrt{y-2}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=3+\sqrt{x-2}}$$ Domain of $f^{-1}$ equals range of $f$: $\boxed{x\geq2}$ Range of $f^{-1}$ equals domain of $f$: $\boxed{f^{-1}(x)\geq3}$

QUESTION 34 8 marks

Medium

Let $f(x)=\dfrac{4x+3}{x-1}$, $x\neq1$, and $g(x)=\dfrac{x+3}{x-4}$, $x\neq4$. Show, by finding both $f(g(x))$ and $g(f(x))$ and simplifying each fully, that $f$ and $g$ are inverse functions of each other.

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Method: two functions are inverses of each other if $f(g(x))=x$ and $g(f(x))=x$ for all $x$ in the relevant domain. Substituting $g(x)$ into $f$: $$f(g(x)) = \dfrac{4g(x)+3}{g(x)-1} = \dfrac{4\left(\dfrac{x+3}{x-4}\right)+3}{\left(\dfrac{x+3}{x-4}\right)-1}$$ Multiplying numerator and denominator by $(x-4)$ to clear the fractions within the fraction: $$f(g(x)) = \dfrac{4(x+3)+3(x-4)}{(x+3)-(x-4)}$$ Simplifying the numerator: $4(x+3)+3(x-4)=4x+12+3x-12=7x$. Simplifying the denominator: $(x+3)-(x-4)=x+3-x+4=7$. $$f(g(x)) = \dfrac{7x}{7}=x$$ Substituting $f(x)$ into $g$: $$g(f(x)) = \dfrac{f(x)+3}{f(x)-4} = \dfrac{\left(\dfrac{4x+3}{x-1}\right)+3}{\left(\dfrac{4x+3}{x-1}\right)-4}$$ Multiplying numerator and denominator by $(x-1)$: $$g(f(x)) = \dfrac{(4x+3)+3(x-1)}{(4x+3)-4(x-1)}$$ Simplifying the numerator: $(4x+3)+3(x-1)=4x+3+3x-3=7x$. Simplifying the denominator: $(4x+3)-4(x-1)=4x+3-4x+4=7$. $$g(f(x)) = \dfrac{7x}{7}=x$$ Since $f(g(x))=x$ and $g(f(x))=x$ for all $x$ in the relevant domain, $f$ and $g$ undo one another when composed in either order. $$\boxed{f \text{ and } g \text{ are inverse functions of each other.}}$$

QUESTION 35 7 marks

Medium

Consider the function $f(x)=\dfrac{x+3}{2x-k}$, $x\neq\dfrac{k}{2}$, where $k$ is a real constant. (a) Find $f^{-1}(x)$ in terms of $k$. [4] (b) Find the value of $k$ for which $f$ is self-inverse, and verify your answer. [3]

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(a) Let $y=\dfrac{x+3}{2x-k}$. Solving for $x$ in terms of $y$: $$y(2x-k)=x+3$$ $$2xy-ky=x+3$$ Collecting the terms containing $x$: $$2xy-x=3+ky$$ $$x(2y-1)=3+ky$$ $$x=\dfrac{3+ky}{2y-1}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=\dfrac{kx+3}{2x-1}}$$ (b) For $f$ to be self-inverse, $f^{-1}(x)$ must be identical to $f(x)=\dfrac{x+3}{2x-k}$ for every $x$. Comparing $f^{-1}(x)=\dfrac{kx+3}{2x-1}$ with $f(x)=\dfrac{x+3}{2x-k}$: the coefficient of $x$ in the numerator gives $k=1$, and the constant term in the denominator gives $-1=-k$, i.e. $k=1$ also. Both comparisons agree: $$\boxed{k=1}$$ Verification: with $k=1$, $f(x)=\dfrac{x+3}{2x-1}$ and, from (a), $f^{-1}(x)=\dfrac{(1)x+3}{2x-1}=\dfrac{x+3}{2x-1}$, which is identical to $f(x)$, confirming $f$ is self-inverse.

QUESTION 36 7 marks

Medium

Let $f(x)=\dfrac{1}{x-1}$, $x\neq1$, and $g(x)=2x+1$. (a) Find $(g\circ f)(x)$ in simplified form. [3] (b) State the value of $x$ that must be excluded from the domain of $g\circ f$. [1] (c) Evaluate $(g\circ f)(3)$. [2]

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(a) Substituting $f(x)$ into $g$: $$(g\circ f)(x) = 2\left(\dfrac{1}{x-1}\right)+1 = \dfrac{2}{x-1}+1$$ Writing over a common denominator: $$(g\circ f)(x) = \dfrac{2+(x-1)}{x-1} = \dfrac{x+1}{x-1}$$ $$\boxed{(g\circ f)(x)=\dfrac{x+1}{x-1}}$$ (b) The inner function $f$ requires $x\neq1$ (its own domain restriction), and this remains the only restriction since $g$ accepts every real input. $$\boxed{x\neq1}$$ (c) Substituting $x=3$ into the simplified form from (a): $$(g\circ f)(3) = \dfrac{3+1}{3-1}=\dfrac{4}{2}=2$$ (Check directly: $f(3)=\dfrac{1}{3-1}=\dfrac12$, then $g\left(\dfrac12\right)=2\left(\dfrac12\right)+1=2$.) $$\boxed{(g\circ f)(3)=2}$$

QUESTION 37 6 marks

Medium

Let $f(x)=2x^3-5$. (a) Find $f^{-1}(x)$. [3] (b) Solve the equation $f^{-1}(x)=3$. [3]

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(a) Let $y=2x^3-5$. Solving for $x$ in terms of $y$: $$y+5=2x^3$$ $$x^3=\dfrac{y+5}{2}$$ $$x=\sqrt[3]{\dfrac{y+5}{2}}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=\sqrt[3]{\dfrac{x+5}{2}}}$$ (b) Method: since $f^{-1}(x)=3$ means $x=f(3)$ (applying $f$ to both sides), it is quickest to evaluate $f(3)$ directly rather than substituting into the formula for $f^{-1}$. $$x=f(3)=2(3)^3-5=2(27)-5=54-5=49$$ $$\boxed{x=49}$$ (Check using the formula from (a): $f^{-1}(49)=\sqrt[3]{\dfrac{49+5}{2}}=\sqrt[3]{27}=3$. $\checkmark$)

QUESTION 38 7 marks

Medium

Let $f(x)=x^2-4$, $x\geq0$, and $g(x)=\sqrt{x}$, $x\geq0$. (a) Find $(f\circ g)(x)$ in simplified form, and state its domain. [3] (b) Find $(g\circ f)(x)$ in simplified form, and state its domain. [4]

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(a) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = (\sqrt{x})^2-4 = x-4$$ The composite is defined wherever $g$ is defined, i.e. $x\geq0$ (and since $g(x)=\sqrt{x}\geq0$ automatically satisfies the domain requirement $g(x)\geq0$ of $f$, no further restriction arises). $$\boxed{(f\circ g)(x)=x-4, \; x\geq0}$$ (b) Substituting $f(x)$ into $g$: $$(g\circ f)(x) = \sqrt{f(x)} = \sqrt{x^2-4}$$ For this composite to be defined, we require both that $x$ lies in the domain of $f$ (so $x\geq0$) and that the input to the square root, $f(x)=x^2-4$, is non-negative: $$x^2-4\geq0 \implies x^2\geq4 \implies x\geq2 \text{ or } x\leq-2$$ Intersecting this with the domain restriction $x\geq0$ from $f$, only $x\geq2$ survives: $$\boxed{(g\circ f)(x)=\sqrt{x^2-4}, \; x\geq2}$$

QUESTION 39 6 marks

Medium

Let $f(x)=\dfrac{5-3x}{2}$. (a) Find $f^{-1}(x)$. [3] (b) Verify algebraically that $(f\circ f^{-1})(x)=x$. [3]

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(a) Let $y=\dfrac{5-3x}{2}$. Solving for $x$ in terms of $y$: $$2y=5-3x$$ $$3x=5-2y$$ $$x=\dfrac{5-2y}{3}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=\dfrac{5-2x}{3}}$$ (b) Substituting $f^{-1}(x)$ into $f$: $$(f\circ f^{-1})(x) = f\left(\dfrac{5-2x}{3}\right) = \dfrac{5-3\left(\dfrac{5-2x}{3}\right)}{2}$$ The factor of $3$ cancels in the numerator: $$= \dfrac{5-(5-2x)}{2} = \dfrac{2x}{2}=x$$ $$\boxed{(f\circ f^{-1})(x)=x}$$

QUESTION 40 6 marks

Medium

Let $f(x)=x^2-3x$ and $g(x)=x+4$. (a) Find $(f\circ g)(x)$ in simplified form. [3] (b) Solve the equation $(f\circ g)(x)=10$. [3]

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(a) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = (x+4)^2-3(x+4)$$ Expanding $(x+4)^2=x^2+8x+16$: $$(f\circ g)(x) = x^2+8x+16-3x-12$$ $$\boxed{(f\circ g)(x)=x^2+5x+4}$$ (b) Setting the result equal to $10$: $$x^2+5x+4=10$$ $$x^2+5x-6=0$$ Factorising: $$(x+6)(x-1)=0$$ $$\boxed{x=-6 \text{ or } x=1}$$

QUESTION 41 7 marks

Medium

Let $f(x)=\dfrac{2x+1}{x-3}$, $x\neq3$. (a) Show that $f(4)=9$. [1] (b) Hence write down the value of $f^{-1}(9)$, justifying your answer using the property of inverse functions. [2] (c) Find $f^{-1}(x)$ algebraically, and use it to verify your answer to (b). [4]

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(a) Substituting $x=4$: $$f(4) = \dfrac{2(4)+1}{4-3} = \dfrac{9}{1}=9$$ $$\boxed{f(4)=9}$$ (b) Since $f(4)=9$, and an inverse function reverses the action of the original function, applying $f^{-1}$ to the output $9$ must return the original input $4$. $$\boxed{f^{-1}(9)=4}$$ (c) Let $y=\dfrac{2x+1}{x-3}$. Solving for $x$ in terms of $y$: $$y(x-3)=2x+1$$ $$yx-3y=2x+1$$ Collecting the terms containing $x$: $$yx-2x=1+3y$$ $$x(y-2)=3y+1$$ $$x=\dfrac{3y+1}{y-2}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=\dfrac{3x+1}{x-2}}$$ Substituting $x=9$: $$f^{-1}(9) = \dfrac{3(9)+1}{9-2} = \dfrac{28}{7}=4$$ This matches the answer found in (b), as required.

QUESTION 42 8 marks

Medium

Let $f(x)=\dfrac{1}{x+2}$, $x\neq-2$, and $g(x)=\dfrac{1}{x-1}$, $x\neq1$. (a) Find $(f\circ g)(x)$ in simplified form. [4] (b) State ALL the values of $x$ that must be excluded from the domain of $f\circ g$. [4]

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(a) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = \dfrac{1}{g(x)+2} = \dfrac{1}{\dfrac{1}{x-1}+2}$$ Writing the denominator over a common denominator of $(x-1)$: $$\dfrac{1}{x-1}+2 = \dfrac{1+2(x-1)}{x-1} = \dfrac{2x-1}{x-1}$$ So: $$(f\circ g)(x) = \dfrac{1}{\dfrac{2x-1}{x-1}} = \dfrac{x-1}{2x-1}$$ $$\boxed{(f\circ g)(x)=\dfrac{x-1}{2x-1}}$$ (b) Two conditions must be satisfied for $(f\circ g)(x)$ to be defined. First, $g$ itself must be defined, requiring $x\neq1$. Second, the output $g(x)$ must lie in the domain of $f$, i.e. $g(x)\neq-2$: $$\dfrac{1}{x-1}\neq-2$$ $$1 \neq -2(x-1)$$ $$1\neq-2x+2$$ $$2x\neq1$$ $$x\neq\dfrac12$$ (Consistency check: the simplified form $\dfrac{x-1}{2x-1}$ found in (a) is itself undefined exactly when $2x-1=0$, i.e. $x=\dfrac12$, confirming this second restriction.) $$\boxed{x\neq1 \text{ and } x\neq\dfrac12}$$

QUESTION 43 6 marks

Medium

Let $f(x)=(x+2)^3-1$. (a) Find $f^{-1}(x)$. [3] (b) Solve the equation $f^{-1}(x)=2$. [3]

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(a) Let $y=(x+2)^3-1$. Solving for $x$ in terms of $y$: $$y+1=(x+2)^3$$ $$x+2=\sqrt[3]{y+1}$$ $$x=\sqrt[3]{y+1}-2$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=\sqrt[3]{x+1}-2}$$ (b) Setting the result equal to $2$: $$\sqrt[3]{x+1}-2=2$$ $$\sqrt[3]{x+1}=4$$ Cubing both sides: $$x+1=64$$ $$\boxed{x=63}$$

QUESTION 44 11 marks

Hard

Let $f(x)=\sqrt{x-1}$, $x\geq1$, and $g(x)=2x-1$, $x\in\mathbb{R}$. (a) Find $(g\circ f)(x)$ in simplified form, and state its domain. [3] (b) Find $(f\circ g)(x)$ in simplified form, and state its domain. [4] (c) Using your graphic display calculator, solve the equation $(g\circ f)(x)=(f\circ g)(x)$ for $x$ in the common domain of the two composites, giving your answer correct to 3 significant figures. [4]

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(a) Substituting $f(x)$ into $g$: $$(g\circ f)(x) = 2\sqrt{x-1}-1$$ The composite is defined wherever $f$ is defined, since $g$ accepts every real input: $$\boxed{(g\circ f)(x)=2\sqrt{x-1}-1, \; x\geq1}$$ (b) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = \sqrt{(2x-1)-1} = \sqrt{2x-2}$$ For this to be defined, the expression under the square root must be non-negative: $$2x-2\geq0 \implies x\geq1$$ $$\boxed{(f\circ g)(x)=\sqrt{2x-2}, \; x\geq1}$$ (c) Both composites share the domain $x\geq1$, so we solve $$2\sqrt{x-1}-1=\sqrt{2x-2}$$ for $x\geq1$. Using a GDC, graph $Y_1=2\sqrt{x-1}-1$ and $Y_2=\sqrt{2x-2}$ for $x\geq1$ and find their point of intersection: $$\boxed{x\approx3.91}$$ (Verification: at $x=3.914$, $Y_1=2\sqrt{2.914}-1\approx2(1.707)-1=2.414$ and $Y_2=\sqrt{5.828}\approx2.414$, which agree to 3 s.f.)

QUESTION 45 11 marks

Hard

Let $f(x)=(x-2)^2-1$, $x\geq2$, and $g(x)=\dfrac{1}{x+3}$, $x\neq-3$. (a) Find the set of values of $x$ for which $(f\circ g)(x)$ is defined, i.e. for which $g(x)$ lies in the domain of $f$. [4] (b) Find $(f\circ g)(x)$ in simplified form. [3] (c) Solve the equation $(f\circ g)(x)=3$, given the domain restriction found in (a). [4]

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(a) For $(f\circ g)(x)$ to be defined, the input to $f$, namely $g(x)$, must satisfy $g(x)\geq2$ (the domain of $f$): $$\dfrac{1}{x+3}\geq2$$ Since the sign of $(x+3)$ affects the direction of the inequality when multiplying through, consider two cases. Case 1: $x+3>0$ (i.e. $x>-3$). Multiplying both sides by the positive quantity $(x+3)$ preserves the inequality: $$1\geq2(x+3)$$ $$1\geq2x+6$$ $$-5\geq2x$$ $$x\leq-2.5$$ Combined with $x>-3$, this gives $-30$ so $|x+3|=x+3$, and $2x+5\leq0$ (since $x\leq-2.5$) so $|2x+5|=-(2x+5)$. Substituting: $$-(2x+5)=2(x+3)$$ $$-2x-5=2x+6$$ $$-11=4x$$ $$x=-\dfrac{11}{4}$$ Checking this lies in the domain $-3

QUESTION 46 9 marks

Hard

Let $f(x)=\dfrac{x+1}{x-2}$, $x\neq2$. (a) Find $f^{-1}(x)$. [4] (b) Solve the equation $f^{-1}(x)=f(x)$ algebraically, giving your answers in exact (surd) form. [5]

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(a) Let $y=\dfrac{x+1}{x-2}$. Solving for $x$ in terms of $y$: $$y(x-2)=x+1$$ $$yx-2y=x+1$$ Collecting the terms containing $x$: $$yx-x=1+2y$$ $$x(y-1)=2y+1$$ $$x=\dfrac{2y+1}{y-1}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=\dfrac{2x+1}{x-1}}$$ (b) Setting $f^{-1}(x)=f(x)$: $$\dfrac{2x+1}{x-1}=\dfrac{x+1}{x-2}$$ Cross-multiplying: $$(2x+1)(x-2)=(x+1)(x-1)$$ Expanding the left side: $$2x^2-4x+x-2=2x^2-3x-2$$ Expanding the right side (difference of squares): $$x^2-1$$ So: $$2x^2-3x-2=x^2-1$$ $$x^2-3x-1=0$$ Using the quadratic formula with $a=1,\;b=-3,\;c=-1$: $$x=\dfrac{3\pm\sqrt{9+4}}{2}=\dfrac{3\pm\sqrt{13}}{2}$$ Both values are distinct from the excluded values $x=1$ (where $f^{-1}$ is undefined) and $x=2$ (where $f$ is undefined), since $\sqrt{13}\approx3.61$. $$\boxed{x=\dfrac{3+\sqrt{13}}{2} \text{ or } x=\dfrac{3-\sqrt{13}}{2}}$$

QUESTION 47 11 marks

Hard

The function $f$ is defined by $f(x)=x^2+6x+5$ for $x\leq-3$. The graph of $y=f(x)$ on this restricted domain is shown below. (a) Express $f(x)$ in the form $(x+p)^2+q$, and hence explain, with reference to the graph, why $f$ is one-to-one on the domain $x\leq-3$. [4] (b) Find $f^{-1}(x)$, stating its domain and range. [5] (c) Solve the equation $f^{-1}(x)=-5$. [2]

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(a) Completing the square: $$f(x)=x^2+6x+5=(x+3)^2-9+5=(x+3)^2-4$$ $$\boxed{f(x)=(x+3)^2-4}$$ The vertex of the parabola is at $x=-3$. As shown in the graph, on the domain $x\leq-3$ the curve is the left-hand branch of the parabola, which is strictly decreasing as $x$ increases towards $-3$ (equivalently, strictly increasing as $x$ decreases away from $-3$). Since the function is strictly monotonic on this interval, no two different $x$-values in the domain give the same output, so $f$ is one-to-one on $x\leq-3$. (b) Let $y=(x+3)^2-4$, where $y\geq-4$ (since $(x+3)^2\geq0$). $$y+4=(x+3)^2$$ $$x+3=\pm\sqrt{y+4}$$ Since $x\leq-3$, we have $x+3\leq0$, so we take the negative root: $$x+3=-\sqrt{y+4}$$ $$x=-3-\sqrt{y+4}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=-3-\sqrt{x+4}}$$ Domain of $f^{-1}$ equals range of $f$: $\boxed{x\geq-4}$ Range of $f^{-1}$ equals domain of $f$: $\boxed{f^{-1}(x)\leq-3}$ (c) Setting the result equal to $-5$: $$-3-\sqrt{x+4}=-5$$ $$-\sqrt{x+4}=-2$$ $$\sqrt{x+4}=2$$ $$x+4=4$$ $$\boxed{x=0}$$

QUESTION 48 12 marks

Hard

Let $f(x)=\dfrac{1}{1-x}$, $x\neq1$. (a) Find $f(f(x))$ in simplified form, stating the additional value of $x$ (beyond $x\neq1$) that must be excluded from its domain. [5] (b) By finding $f(f(f(x)))$, show that $(f\circ f\circ f)(x)=x$ for all $x$ in the domain. [4] (c) Hence, without further calculation, state the value of $f$ applied $100$ times to $x$, justifying your answer using the result of part (b). [3]

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(a) Substituting $f(x)$ into $f$ itself: $$f(f(x)) = \dfrac{1}{1-f(x)} = \dfrac{1}{1-\dfrac{1}{1-x}}$$ Writing the denominator over a common denominator of $(1-x)$: $$1-\dfrac{1}{1-x} = \dfrac{(1-x)-1}{1-x} = \dfrac{-x}{1-x}$$ So: $$f(f(x)) = \dfrac{1}{\dfrac{-x}{1-x}} = \dfrac{1-x}{-x} = \dfrac{x-1}{x}$$ $$\boxed{f(f(x))=\dfrac{x-1}{x}}$$ This expression is undefined when $x=0$, so in addition to the original restriction $x\neq1$ (needed for $f(x)$ itself to be defined), we must also exclude: $$\boxed{x\neq0}$$ (b) Substituting the result of (a) into $f$ once more: $$f(f(f(x))) = \dfrac{1}{1-f(f(x))} = \dfrac{1}{1-\dfrac{x-1}{x}}$$ Writing the denominator over a common denominator of $x$: $$1-\dfrac{x-1}{x} = \dfrac{x-(x-1)}{x} = \dfrac{1}{x}$$ So: $$f(f(f(x))) = \dfrac{1}{\dfrac{1}{x}} = x$$ $$\boxed{(f\circ f\circ f)(x)=x \text{ for all } x \text{ in the domain } (x\neq0,1).}$$ (c) Since applying $f$ three times returns the original input $x$, the function $f$ has period $3$ under repeated composition: applying $f$ any multiple of $3$ times gives back $x$ itself. Dividing $100$ by $3$: $$100 = 3(33)+1$$ So applying $f$ $100$ times is equivalent to applying the period-$3$ cycle $33$ complete times (which returns to $x$) and then applying $f$ one further time: $$\boxed{f \text{ applied } 100 \text{ times to } x \; = \; f(x) = \dfrac{1}{1-x}}$$

QUESTION 49 11 marks

Hard

Let $f(x)=\dfrac{1}{\sqrt{x-4}}$, $x>4$, and $g(x)=x^2-5$, $x\in\mathbb{R}$. (a) Explain why, for $(f\circ g)(x)$ to be defined, we require $x^2-5>4$, and hence find the domain of $f\circ g$. [4] (b) Find $(f\circ g)(x)$ in simplified form. [3] (c) Solve the equation $(f\circ g)(x)=\dfrac12$. [4]

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(a) For $(f\circ g)(x)=f(g(x))$ to be defined, the input to $f$, namely $g(x)$, must lie in the domain of $f$, which is $x>4$. That is, we require: $$g(x)>4$$ $$x^2-5>4$$ $$x^2>9$$ Taking square roots (remembering both branches for a quadratic inequality): $$\boxed{x<-3 \text{ or } x>3}$$ (b) Substituting $g(x)$ into $f$: $$(f\circ g)(x) = \dfrac{1}{\sqrt{g(x)-4}} = \dfrac{1}{\sqrt{(x^2-5)-4}}$$ $$\boxed{(f\circ g)(x)=\dfrac{1}{\sqrt{x^2-9}}, \; x<-3 \text{ or } x>3}$$ (c) Setting the result equal to $\dfrac12$: $$\dfrac{1}{\sqrt{x^2-9}}=\dfrac12$$ $$\sqrt{x^2-9}=2$$ $$x^2-9=4$$ $$x^2=13$$ $$x=\pm\sqrt{13}$$ Checking against the domain from (a): $\sqrt{13}\approx3.61>3$, so $x=\sqrt{13}$ is valid; similarly $-\sqrt{13}\approx-3.61<-3$, so $x=-\sqrt{13}$ is also valid. $$\boxed{x=\sqrt{13} \text{ or } x=-\sqrt{13}}$$

QUESTION 50 11 marks

Hard

Let $f(x)=\dfrac{5x-2}{3-x}$, $x\neq3$. (a) Find $f^{-1}(x)$ algebraically, stating its domain. [5] (b) Solve the equation $f^{-1}(x)=4$, and verify your answer using $f$. [4] (c) Without finding $f(f^{-1}(x))$ explicitly, state its value for every $x$ in the domain of $f^{-1}$, justifying your answer using the definition of an inverse function. [2]

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(a) Let $y=\dfrac{5x-2}{3-x}$. Solving for $x$ in terms of $y$: $$y(3-x)=5x-2$$ $$3y-xy=5x-2$$ Collecting the terms containing $x$ on one side: $$3y+2=5x+xy$$ $$3y+2=x(5+y)$$ $$x=\dfrac{3y+2}{y+5}$$ Replacing $y$ with $x$: $$\boxed{f^{-1}(x)=\dfrac{3x+2}{x+5}}$$ The domain of $f^{-1}$ equals the range of $f$. Since $f$ has a horizontal asymptote at $y=-5$ (the ratio of the leading coefficients, $5/(-1)=-5$), this value is never attained, and this is confirmed since $f^{-1}$ itself has a vertical asymptote at $x=-5$: $$\boxed{\text{Domain of } f^{-1}: x\neq-5}$$ (b) Setting $f^{-1}(x)=4$: $$\dfrac{3x+2}{x+5}=4$$ $$3x+2=4(x+5)$$ $$3x+2=4x+20$$ $$-x=18$$ $$\boxed{x=-18}$$ Verification: since $f^{-1}(-18)=4$, we should have $f(4)=-18$. $$f(4) = \dfrac{5(4)-2}{3-4} = \dfrac{18}{-1}=-18 \; \checkmark$$ (c) By the definition of an inverse function, applying $f$ after $f^{-1}$ returns the original input, for every $x$ for which $f^{-1}(x)$ is defined. $$\boxed{f(f^{-1}(x))=x \text{ for all } x \text{ in the domain of } f^{-1}}$$

QUESTION 1 4 marks

Easy

A line passes through the points $A(-2,5)$ and $B(4,-3)$. (a) Find the gradient of line $AB$. (b) Find the equation of line $AB$, giving your answer in the form $ax+by+d=0$ where $a,b,d \in \mathbb{Z}$.

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(a) The gradient of a line through two points is the change in $y$ divided by the change in $x$. Using $m = \dfrac{y_B - y_A}{x_B - x_A}$ with $A(-2,5)$ and $B(4,-3)$: $$m = \frac{-3-5}{4-(-2)} = \frac{-8}{6} = -\frac{4}{3}$$ $$\boxed{m = -\dfrac{4}{3}}$$ (b) Using the point-gradient form $y - y_1 = m(x-x_1)$ with point $A(-2,5)$ and $m = -\dfrac{4}{3}$: $$y - 5 = -\frac{4}{3}\big(x-(-2)\big)$$ $$y - 5 = -\frac{4}{3}(x+2)$$ Multiply both sides by 3 to clear the fraction: $$3y - 15 = -4(x+2)$$ $$3y - 15 = -4x - 8$$ Rearranging into the form $ax+by+d=0$: $$4x + 3y - 7 = 0$$ $$\boxed{4x+3y-7=0}$$

QUESTION 2 5 marks

Medium

Line $L_1$ has equation $2x - 5y + 10 = 0$. Line $L_2$ is perpendicular to $L_1$ and passes through the point where $L_1$ crosses the $x$-axis. Find the equation of $L_2$, giving your answer in the form $ax+by+d=0$ where $a,b,d \in \mathbb{Z}$.

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**Finding where $L_1$ crosses the $x$-axis:** Setting $y=0$ in $2x-5y+10=0$: $$2x + 10 = 0 \implies x = -5$$ So $L_1$ meets the $x$-axis at $(-5,0)$. **Finding the gradient of $L_1$:** Rearranging $2x-5y+10=0$ into $y=mx+c$ form: $$5y = 2x+10 \implies y = \frac{2}{5}x + 2$$ So $m_1 = \dfrac{2}{5}$. **Finding the gradient of $L_2$:** Since $L_2 \perp L_1$, the gradients satisfy $m_2 = -\dfrac{1}{m_1}$: $$m_2 = -\frac{1}{2/5} = -\frac{5}{2}$$ **Finding the equation of $L_2$:** Using point-gradient form through $(-5,0)$ with $m_2 = -\dfrac{5}{2}$: $$y - 0 = -\frac{5}{2}\big(x-(-5)\big) = -\frac{5}{2}(x+5)$$ $$y = -\frac{5}{2}x - \frac{25}{2}$$ Multiply both sides by 2: $$2y = -5x - 25$$ $$\boxed{5x + 2y + 25 = 0}$$

QUESTION 3 5 marks

Easy

A small business has a linear cost model for manufacturing bicycle helmets. The fixed cost is $\$350$, and each helmet produced costs an additional $\$12.50$. (a) Write down a linear function $C(x)$ for the total cost, in dollars, of producing $x$ helmets. (b) Find the cost of producing 84 helmets. (c) Find the number of helmets produced if the total cost is $\$1200$.

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(a) The total cost is the fixed cost plus the cost per helmet multiplied by the number of helmets: $$\boxed{C(x) = 12.5x + 350}$$ (b) Substituting $x=84$ into $C(x)$: $$C(84) = 12.5(84) + 350$$ $$C(84) = 1050 + 350 = 1400$$ $$\boxed{C(84) = \$1400}$$ (c) Setting $C(x) = 1200$ and solving for $x$: $$12.5x + 350 = 1200$$ $$12.5x = 850$$ $$x = 68$$ $$\boxed{x = 68 \text{ helmets}}$$

QUESTION 4 6 marks

Medium

Line $L_1$ passes through the points $A(1,2)$ and $B(5,14)$. Line $L_2$ passes through the point $C(0,6)$ and has gradient $-\dfrac{1}{2}$. (a) Find the equation of $L_1$ in the form $y = mx+c$. (b) Find the coordinates of the point of intersection of $L_1$ and $L_2$. (c) Find the equation of the line that passes through this point of intersection and is perpendicular to $L_2$.

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(a) Using the gradient formula on points $A(1,2)$ and $B(5,14)$: $$m = \frac{14-2}{5-1} = \frac{12}{4} = 3$$ Using point-gradient form through $A(1,2)$: $$y - 2 = 3(x-1)$$ $$y = 3x - 3 + 2 = 3x-1$$ $$\boxed{L_1: y = 3x - 1}$$ (b) The equation of $L_2$ through $C(0,6)$ with gradient $-\dfrac{1}{2}$ is $y = -\dfrac{1}{2}x + 6$. At the point of intersection, both lines have the same $y$-value, so setting the two expressions equal: $$3x - 1 = -\frac{1}{2}x + 6$$ $$3x + \frac{1}{2}x = 7$$ $$\frac{7}{2}x = 7$$ $$x = 2$$ Substituting back into $L_1$: $$y = 3(2) - 1 = 5$$ $$\boxed{(2,5)}$$ (c) $L_2$ has gradient $-\dfrac{1}{2}$, so a line perpendicular to $L_2$ has gradient $2$ (using $m_2 = -1/m_1$). Using point-gradient form through $(2,5)$ with gradient $2$: $$y - 5 = 2(x-2)$$ $$y = 2x - 4 + 5$$ $$\boxed{y = 2x+1}$$

QUESTION 5 14 marks

Hard

Triangle $ABC$ has vertices $A(7,3)$, $B(2,8)$ and $C(-1,7)$. (a) Find the equation of the perpendicular bisector of $[AB]$. (b) Find the equation of the perpendicular bisector of $[BC]$. (c) Hence find the coordinates of the circumcentre of triangle $ABC$ (the point equidistant from $A$, $B$ and $C$), and find the radius of the circumscribed circle.

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**(a) Perpendicular bisector of $[AB]$** The perpendicular bisector passes through the midpoint of $AB$ and has gradient equal to the negative reciprocal of the gradient of $AB$. Midpoint of $AB$: $$\left(\frac{7+2}{2}, \frac{3+8}{2}\right) = (4.5, 5.5)$$ Gradient of $AB$: $$\frac{8-3}{2-7} = \frac{5}{-5} = -1$$ Perpendicular gradient: $1$. Using point-gradient form through $(4.5,5.5)$: $$y - 5.5 = 1(x - 4.5)$$ $$y = x + 1$$ $$\boxed{y = x+1}$$ **(b) Perpendicular bisector of $[BC]$** Midpoint of $BC$: $$\left(\frac{2+(-1)}{2}, \frac{8+7}{2}\right) = (0.5, 7.5)$$ Gradient of $BC$: $$\frac{7-8}{-1-2} = \frac{-1}{-3} = \frac{1}{3}$$ Perpendicular gradient: $-3$. Using point-gradient form through $(0.5,7.5)$: $$y - 7.5 = -3(x - 0.5)$$ $$y = -3x + 1.5 + 7.5 = -3x+9$$ $$\boxed{y = -3x+9}$$ **(c) Circumcentre** The circumcentre is the point equidistant from all three vertices, found where the two perpendicular bisectors intersect. Setting the two expressions for $y$ equal: $$x + 1 = -3x + 9$$ $$4x = 8$$ $$x = 2$$ $$y = 2+1 = 3$$ $$\boxed{\text{Circumcentre} = (2,3)}$$ The radius is the distance from the circumcentre to any vertex, using the distance formula. Taking $A(7,3)$: $$r = \sqrt{(7-2)^2 + (3-3)^2} = \sqrt{25} = 5$$ (Check with $B(2,8)$: $\sqrt{(2-2)^2+(8-3)^2}=\sqrt{25}=5$; and with $C(-1,7)$: $\sqrt{(-1-2)^2+(7-3)^2}=\sqrt{9+16}=\sqrt{25}=5$.) $$\boxed{r = 5}$$

QUESTION 6 4 marks

Easy

Consider $f(x) = x^2+6x+5$. (a) Write $f(x)$ in factorised form, and hence state the $x$-intercepts of the graph of $f$. (b) Write $f(x)$ in vertex form, and hence state the coordinates of the vertex.

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(a) Factorising $f(x)=x^2+6x+5$, we need two numbers that multiply to $5$ and add to $6$: these are $1$ and $5$. $$f(x) = (x+1)(x+5)$$ $$\boxed{f(x)=(x+1)(x+5)}$$ Setting $f(x)=0$ gives the $x$-intercepts: $$\boxed{x=-1, \; x=-5}$$ (b) Completing the square: $$f(x) = x^2+6x+5 = (x+3)^2 - 9+5 = (x+3)^2-4$$ $$\boxed{f(x)=(x+3)^2-4}$$ $$\boxed{\text{Vertex: } (-3,-4)}$$

QUESTION 7 5 marks

Medium

A quadratic function has zeros at $x=2$ and $x=-6$, and its graph passes through the point $(0,-24)$. (a) Write $f(x)$ in factorised form $f(x) = a(x-p)(x-q)$. (b) Hence write $f(x)$ in general form $f(x) = ax^2+bx+c$.

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(a) Since the zeros are $x=2$ and $x=-6$: $$f(x) = a(x-2)(x+6)$$ Using the given point $(0,-24)$: $$f(0) = a(0-2)(0+6) = a(-2)(6) = -12a$$ $$-12a = -24 \implies a=2$$ $$\boxed{f(x) = 2(x-2)(x+6)}$$ (b) Expanding $(x-2)(x+6)$: $$(x-2)(x+6) = x^2+6x-2x-12 = x^2+4x-12$$ $$f(x) = 2(x^2+4x-12)$$ $$\boxed{f(x) = 2x^2+8x-24}$$

QUESTION 8 5 marks

Easy

The graph of a quadratic function $f(x) = ax^2+bx+c$ passes through the points $(0,3)$, $(1,0)$ and $(3,0)$. Find the values of $a$, $b$ and $c$.

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Since $f(1)=0$ and $f(3)=0$, $x=1$ and $x=3$ are roots of $f$, so $f$ can be written in factorised form: $$f(x)=a(x-1)(x-3)$$ Using the point $(0,3)$: $$f(0) = a(0-1)(0-3) = a(-1)(-3) = 3a$$ $$3a = 3 \implies a=1$$ So: $$f(x) = (x-1)(x-3)$$ Expanding: $$f(x) = x^2-4x+3$$ $$\boxed{a=1, \; b=-4, \; c=3}$$

QUESTION 9 6 marks

Medium

Consider $f(x) = -3x^2+12x-5$. (a) Write $f(x)$ in vertex form. (b) State the coordinates of the vertex, and determine whether it is a maximum or a minimum point. (c) Find the $x$-intercepts of the graph of $f$, correct to 3 significant figures.

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(a) Completing the square on $f(x) = -3x^2+12x-5$: $$f(x) = -3(x^2-4x)-5$$ $$f(x) = -3\big[(x-2)^2-4\big]-5$$ $$f(x) = -3(x-2)^2+12-5$$ $$\boxed{f(x) = -3(x-2)^2+7}$$ (b) From the vertex form, the vertex is: $$\boxed{(2,7)}$$ Since the coefficient of $x^2$ is $a=-3<0$, the parabola opens downward, so this is a $\boxed{\text{maximum}}$ point. (c) Setting $f(x)=0$: $$-3x^2+12x-5=0$$ Multiply by $-1$: $3x^2-12x+5=0$. Using the quadratic formula: $$x = \frac{12\pm\sqrt{144-60}}{6} = \frac{12\pm\sqrt{84}}{6}$$ Since $\sqrt{84}\approx9.1652$: $$x = \frac{12+9.1652}{6} \approx 3.5275 \quad \text{or} \quad x=\frac{12-9.1652}{6}\approx0.4725$$ $$\boxed{x\approx3.53 \text{ or } x\approx0.472}$$

QUESTION 10 14 marks

Hard

A gardener wants to build a rectangular flower bed against a long, straight wall, using $40$ metres of fencing for the other three sides (two widths, perpendicular to the wall, and one length, parallel to the wall). Let $x$ metres be the width of the flower bed (the length of each side perpendicular to the wall). (a) Show that the area of the flower bed, $A(x)$ square metres, is given by $A(x) = 40x-2x^2$. (b) Write $A(x)$ in vertex form, and hence find the maximum possible area of the flower bed and the corresponding value of $x$. (c) Find the values of $x$ for which $A(x) = 168$. (d) State a suitable domain for $x$ in this context, and hence state the range of $A(x)$ over this domain.

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(a) Let the width (perpendicular to the wall) be $x$ and the length (parallel to the wall) be $L$. The fencing covers two widths and one length, since the wall itself forms the fourth side: $$2x+L=40 \implies L=40-2x$$ The area is width times length: $$A = x \times L = x(40-2x)$$ $$\boxed{A(x) = 40x-2x^2}$$ (shown) (b) Completing the square on $A(x)=-2x^2+40x$: $$A(x) = -2(x^2-20x)$$ $$A(x) = -2\big[(x-10)^2-100\big]$$ $$\boxed{A(x) = -2(x-10)^2+200}$$ Since $-2(x-10)^2\leq0$ for all $x$, the maximum occurs at $x=10$: $$\boxed{\text{Maximum area} = 200 \text{ m}^2, \text{ at } x=10 \text{ m}}$$ (c) Setting $A(x) = 168$: $$40x-2x^2=168$$ $$-2x^2+40x-168=0$$ Dividing by $-2$: $$x^2-20x+84=0$$ Factorising: $$(x-6)(x-14)=0$$ $$\boxed{x=6 \text{ or } x=14}$$ (d) For the flower bed to physically exist, we need $x>0$ and $L=40-2x>0$, i.e. $x<20$: $$\boxed{0

QUESTION 11 4 marks

Easy

Solve the equation $3x^{2}-11x-4=0$.

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Method: factorise the quadratic $3x^2-11x-4$. We require two numbers that multiply to give $3\times(-4)=-12$ and combine to give the middle term coefficient $-11$. Using $-12$ and $+1$ (since $-12+1=-11$): $$3x^2-12x+x-4=0$$ Grouping and factorising: $$3x(x-4)+1(x-4)=0$$ $$(3x+1)(x-4)=0$$ So either $3x+1=0$ or $x-4=0$. $$\boxed{x=-\dfrac{1}{3}\ \text{or}\ x=4}$$

QUESTION 12 6 marks

Medium

Find the set of values of $k$ for which the line $y=kx+4$ intersects the curve $y=x^{2}+2x+7$ at two distinct points.

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Method: at points of intersection the $y$-values of the line and curve are equal, giving a quadratic in $x$; two distinct points of intersection require the discriminant of this quadratic to be positive. Setting $kx+4=x^{2}+2x+7$: $$x^{2}+2x+7-kx-4=0$$ $$x^{2}+(2-k)x+3=0$$ For two distinct real solutions, the discriminant $\Delta>0$: $$\Delta=(2-k)^{2}-4(1)(3)$$ $$(2-k)^{2}-12>0$$ $$(2-k)^{2}>12$$ Taking square roots requires a case split, since this is an inequality on a squared expression: $$2-k>2\sqrt{3}\ \ \text{or}\ \ 2-k<-2\sqrt{3}$$ $$-k>2\sqrt{3}-2\ \ \text{or}\ \ -k<-2\sqrt{3}-2$$ Multiplying both inequalities by $-1$ (reversing each): $$k<2-2\sqrt{3}\ \ \text{or}\ \ k>2+2\sqrt{3}$$ $$\boxed{k<2-2\sqrt{3}\ \text{or}\ k>2+2\sqrt{3}}$$

QUESTION 13 5 marks

Easy

A small company's monthly profit, in thousands of dollars, is modelled by $P(x)=-x^{2}+4x+21$, where $x$ is the number of units produced, in hundreds, and $x\ge0$. Find the range of values of $x$ for which the company makes a profit, i.e. $P(x)>0$.

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Method: solve the quadratic inequality by first finding the zeros of $P(x)$. Setting $P(x)=0$: $$-x^{2}+4x+21=0$$ Multiply by $-1$: $x^{2}-4x-21=0$. Factorising: $$(x-7)(x+3)=0$$ $$x=7\ \text{or}\ x=-3$$ Since the coefficient of $x^{2}$ in $P(x)=-x^{2}+4x+21$ is negative, the graph is a downward-opening parabola, so $P(x)>0$ between the roots: $$-3

QUESTION 14 6 marks

Medium

A ball is thrown from a platform. Its height above the ground, in metres, after $t$ seconds is modelled by $h(t)=-4.9t^{2}+14t+1.2$ for $t\ge0$. Find the interval of time during which the ball is more than $8$ metres above the ground, giving your answers correct to three significant figures.

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Method: form and solve the quadratic inequality $h(t)>8$, using the quadratic formula since the numbers do not factorise nicely. $$-4.9t^{2}+14t+1.2>8$$ $$-4.9t^{2}+14t-6.8>0$$ Multiply both sides by $-1$ (reversing the inequality): $$4.9t^{2}-14t+6.8<0$$ Finding the roots of $4.9t^{2}-14t+6.8=0$ using the quadratic formula: $$t=\frac{14\pm\sqrt{(-14)^{2}-4(4.9)(6.8)}}{2(4.9)}=\frac{14\pm\sqrt{196-133.28}}{9.8}=\frac{14\pm\sqrt{62.72}}{9.8}$$ Since $\sqrt{62.72}=7.9198\ldots$: $$t=\frac{14-7.9198}{9.8}=0.6204\ldots\quad\text{or}\quad t=\frac{14+7.9198}{9.8}=2.2367\ldots$$ Since the coefficient of $t^{2}$ in $4.9t^{2}-14t+6.8$ is positive, this expression is negative between its roots, so $h(t)>8$ for $t$ between these values. $$\boxed{0.620\ \text{s}

QUESTION 15 15 marks

Hard

Consider the quadratic equation $2x^{2}-(k+3)x+(k+1)=0$, where $k\in\mathbb{R}$. (a) Show that the discriminant of the equation is $\Delta=(k-1)^{2}$. [3] (b) Hence explain why the equation has real roots for every value of $k$, and find the value of $k$ for which the equation has two equal roots. [3] (c) Solve the equation for the value of $k$ found in part (b). [3] (d) Let $\alpha$ and $\beta$ be the roots of the original equation (in terms of $k$). Given that $\alpha+\beta=\dfrac{k+3}{2}$ and $\alpha\beta=\dfrac{k+1}{2}$, find the value(s) of $k$ for which $\alpha^{2}+\beta^{2}=4$. [6]

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(a) For $2x^{2}-(k+3)x+(k+1)=0$, with $a=2$, $b=-(k+3)$, $c=k+1$, use the discriminant formula $\Delta=b^2-4ac$: $$\Delta=b^{2}-4ac=(k+3)^{2}-4(2)(k+1)$$ Expanding: $$=(k^{2}+6k+9)-(8k+8)$$ $$=k^{2}-2k+1$$ $$=(k-1)^{2}$$ $$\boxed{\Delta=(k-1)^{2}}$$, as required. (b) Since $(k-1)^{2}\ge0$ for all real $k$, we have $\Delta\ge0$ for all $k$, so the equation always has at least two real roots (equal when $\Delta=0$, distinct when $\Delta>0$). Equal roots occur when $\Delta=0$: $$(k-1)^{2}=0\implies k=1$$ $$\boxed{k=1}$$ (c) Substituting $k=1$ into the original equation: $$2x^{2}-(1+3)x+(1+1)=0$$ $$2x^{2}-4x+2=0$$ Dividing by 2: $$x^{2}-2x+1=0$$ $$(x-1)^{2}=0$$ $$\boxed{x=1\ \text{(repeated root)}}$$ (d) Using the identity $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta$: $$\alpha^{2}+\beta^{2}=\left(\frac{k+3}{2}\right)^{2}-2\left(\frac{k+1}{2}\right)$$ $$=\frac{(k+3)^{2}}{4}-(k+1)$$ Setting this equal to 4: $$\frac{(k+3)^{2}}{4}-(k+1)=4$$ Multiplying through by 4: $$(k+3)^{2}-4(k+1)=16$$ $$k^{2}+6k+9-4k-4=16$$ $$k^{2}+2k+5=16$$ $$k^{2}+2k-11=0$$ Applying the quadratic formula: $$k=\frac{-2\pm\sqrt{4-4(1)(-11)}}{2}=\frac{-2\pm\sqrt{48}}{2}=\frac{-2\pm4\sqrt{3}}{2}=-1\pm2\sqrt{3}$$ Since $\Delta=(k-1)^{2}\ge0$ for every real $k$ (from part (a)), the original equation has real roots for both values, so both are valid. $$\boxed{k=-1+2\sqrt{3}\ \text{or}\ k=-1-2\sqrt{3}}$$

QUESTION 16 4 marks

Easy

A line passes through the points $A(-4,1)$ and $B(2,13)$. (a) Find the gradient of line $AB$. (b) Find the equation of line $AB$, giving your answer in the form $y=mx+c$.

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(a) The gradient of a line through two points is the change in $y$ divided by the change in $x$. Using $m=\dfrac{y_B-y_A}{x_B-x_A}$ with $A(-4,1)$ and $B(2,13)$: $$m=\frac{13-1}{2-(-4)}=\frac{12}{6}=2$$ $$\boxed{m=2}$$ (b) Using point-gradient form $y-y_1=m(x-x_1)$ with point $A(-4,1)$ and $m=2$: $$y-1=2\big(x-(-4)\big)$$ $$y-1=2x+8$$ $$\boxed{y=2x+9}$$

QUESTION 17 4 marks

Easy

Line $L$ has equation $y=3x-2$. Find the equation of the line that is parallel to $L$ and passes through the point $P(4,-1)$, giving your answer in the form $y=mx+c$.

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Method: parallel lines have equal gradients. The gradient of $L$ is read directly from $y=3x-2$: the coefficient of $x$ is $m=3$. Since the required line is parallel to $L$, it also has gradient $3$. Using point-gradient form through $P(4,-1)$: $$y-(-1)=3(x-4)$$ $$y+1=3x-12$$ $$\boxed{y=3x-13}$$

QUESTION 18 5 marks

Easy

A ride-share company charges a fixed booking fee of $\$4$, plus $\$2.50$ for every kilometre travelled. (a) Write down a linear function $F(d)$ for the total fare, in dollars, of a trip of $d$ kilometres. (b) Find the fare for a $9$ km trip. (c) Find the length of a trip that costs $\$54$.

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(a) The total fare is the fixed booking fee plus the per-kilometre charge multiplied by the distance: $$\boxed{F(d)=2.5d+4}$$ (b) Substituting $d=9$ into $F(d)$: $$F(9)=2.5(9)+4=22.5+4$$ $$\boxed{F(9)=\$26.50}$$ (c) Setting $F(d)=54$ and solving for $d$: $$2.5d+4=54$$ $$2.5d=50$$ $$d=20$$ $$\boxed{d=20\text{ km}}$$

QUESTION 19 4 marks

Easy

A line has equation $5x+2y-20=0$. (a) Find the $x$-intercept of the line. (b) Find the $y$-intercept of the line.

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(a) The $x$-intercept occurs where $y=0$. Substituting into $5x+2y-20=0$: $$5x+2(0)-20=0$$ $$5x=20$$ $$x=4$$ $$\boxed{x\text{-intercept}=(4,0)}$$ (b) The $y$-intercept occurs where $x=0$. Substituting into $5x+2y-20=0$: $$5(0)+2y-20=0$$ $$2y=20$$ $$y=10$$ $$\boxed{y\text{-intercept}=(0,10)}$$

QUESTION 20 5 marks

Easy

Consider $f(x)=x^2-2x-15$. (a) Write $f(x)$ in factorised form, and hence state the $x$-intercepts of the graph of $f$. (b) Write $f(x)$ in vertex form, and hence state the coordinates of the vertex.

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(a) Factorising $f(x)=x^2-2x-15$, we need two numbers that multiply to give $-15$ and add to give $-2$: these are $-5$ and $3$. $$f(x)=(x-5)(x+3)$$ $$\boxed{f(x)=(x-5)(x+3)}$$ Setting $f(x)=0$ gives the $x$-intercepts: $$\boxed{x=5,\;x=-3}$$ (b) Completing the square: $$f(x)=x^2-2x-15=(x-1)^2-1-15=(x-1)^2-16$$ $$\boxed{f(x)=(x-1)^2-16}$$ $$\boxed{\text{Vertex: }(1,-16)}$$

QUESTION 21 5 marks

Easy

The table below shows three values of a quadratic function $f(x)=ax^2+bx+c$.

$x$	$-1$	$0$	$3$
$f(x)$	$7$	$1$	$7$

Find the values of $a$, $b$ and $c$.

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Method: substitute each pair of values from the table into $f(x)=ax^2+bx+c$ to form a system of equations. Substituting $x=0,\;f(0)=1$: $$a(0)^2+b(0)+c=1 \implies c=1$$ Substituting $x=-1,\;f(-1)=7$: $$a(-1)^2+b(-1)+c=7$$ $$a-b+c=7$$ Since $c=1$: $$a-b=6 \quad \text{(1)}$$ Substituting $x=3,\;f(3)=7$: $$a(3)^2+b(3)+c=7$$ $$9a+3b+c=7$$ Since $c=1$: $$9a+3b=6 \implies 3a+b=2 \quad \text{(2)}$$ Adding equations (1) and (2) to eliminate $b$: $$(a-b)+(3a+b)=6+2$$ $$4a=8 \implies a=2$$ Substituting $a=2$ into equation (1): $$2-b=6 \implies b=-4$$ $$\boxed{a=2,\;b=-4,\;c=1}$$ (Check: $f(x)=2x^2-4x+1$, so $f(-1)=2+4+1=7$, $f(0)=1$, $f(3)=18-12+1=7$, consistent with the table.)

QUESTION 22 3 marks

Easy

Line $L_1$ has equation $2x+3y-6=0$. Line $L_2$ has equation $6x-4y+8=0$. Show that $L_1$ and $L_2$ are perpendicular.

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Method: find the gradient of each line and check that their product is $-1$. Rearranging $L_1: 2x+3y-6=0$ into $y=mx+c$ form: $$3y=-2x+6 \implies y=-\frac{2}{3}x+2$$ So the gradient of $L_1$ is $m_1=-\dfrac{2}{3}$. Rearranging $L_2: 6x-4y+8=0$ into $y=mx+c$ form: $$-4y=-6x-8 \implies y=\frac{3}{2}x+2$$ So the gradient of $L_2$ is $m_2=\dfrac{3}{2}$. Multiplying the gradients: $$m_1\times m_2=-\frac{2}{3}\times\frac{3}{2}=-1$$ Since $m_1 m_2=-1$, the lines $L_1$ and $L_2$ are perpendicular. $\boxed{m_1m_2=-1}$, as required.

QUESTION 23 4 marks

Easy

Points $A(-1,4)$ and $B(5,-2)$ are given. (a) Find the coordinates of the midpoint of $[AB]$. (b) Find the gradient of a line that is perpendicular to $AB$.

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(a) The midpoint formula is $M=\left(\dfrac{x_A+x_B}{2},\dfrac{y_A+y_B}{2}\right)$. Substituting $A(-1,4)$ and $B(5,-2)$: $$M=\left(\frac{-1+5}{2},\frac{4+(-2)}{2}\right)=\left(\frac{4}{2},\frac{2}{2}\right)$$ $$\boxed{M=(2,1)}$$ (b) The gradient of $AB$ is: $$m_{AB}=\frac{-2-4}{5-(-1)}=\frac{-6}{6}=-1$$ Since a line perpendicular to $AB$ has gradient $-\dfrac{1}{m_{AB}}$: $$m_{\perp}=-\frac{1}{-1}=1$$ $$\boxed{m_{\perp}=1}$$

QUESTION 24 3 marks

Easy

Solve the equation $x^2+3x-18=0$.

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Method: factorise the quadratic. We require two numbers that multiply to give $-18$ and add to give $3$: these are $6$ and $-3$. $$x^2+3x-18=(x+6)(x-3)$$ So $(x+6)(x-3)=0$, giving either $x+6=0$ or $x-3=0$. $$\boxed{x=-6\ \text{or}\ x=3}$$

QUESTION 25 5 marks

Easy

Consider $f(x)=2(x-3)^2-8$. (a) Write $f(x)$ in the general form $f(x)=ax^2+bx+c$. (b) Find the $x$-intercepts of the graph of $f$.

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(a) Expanding $(x-3)^2=x^2-6x+9$: $$f(x)=2(x^2-6x+9)-8$$ $$f(x)=2x^2-12x+18-8$$ $$\boxed{f(x)=2x^2-12x+10}$$ (b) Setting $f(x)=0$, it is simplest to use the vertex form given in the question: $$2(x-3)^2-8=0$$ $$(x-3)^2=4$$ $$x-3=\pm2$$ $$x=3+2=5 \quad \text{or} \quad x=3-2=1$$ $$\boxed{x=1\ \text{or}\ x=5}$$

QUESTION 26 5 marks

Easy

A water tank is draining at a constant rate. The volume of water remaining, in litres, after $t$ minutes is modelled by $V(t)=800-20t$, for $0\le t\le 40$. (a) Find the volume of water remaining after $10$ minutes. (b) Find the time taken for the tank to empty completely.

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(a) Substituting $t=10$ into $V(t)=800-20t$: $$V(10)=800-20(10)=800-200$$ $$\boxed{V(10)=600\text{ litres}}$$ (b) The tank is empty when $V(t)=0$: $$800-20t=0$$ $$20t=800$$ $$t=40$$ $$\boxed{t=40\text{ minutes}}$$

QUESTION 27 5 marks

Easy

(a) Find the discriminant of $2x^2-5x+4=0$, and hence state the number of real roots the equation has. (b) Find the discriminant of $x^2-6x+9=0$, and hence state the nature of its roots.

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(a) The discriminant of $ax^2+bx+c=0$ is $\Delta=b^2-4ac$. For $2x^2-5x+4=0$, $a=2$, $b=-5$, $c=4$: $$\Delta=(-5)^2-4(2)(4)=25-32=-7$$ $$\boxed{\Delta=-7}$$ Since $\Delta<0$, the equation has $\boxed{\text{no real roots}}$. (b) For $x^2-6x+9=0$, $a=1$, $b=-6$, $c=9$: $$\Delta=(-6)^2-4(1)(9)=36-36=0$$ $$\boxed{\Delta=0}$$ Since $\Delta=0$, the equation has $\boxed{\text{one repeated (equal) real root}}$.

QUESTION 28 4 marks

Easy

The equation $3x^2-12x+5=0$ has roots $\alpha$ and $\beta$. Without solving the equation, find the value of $\alpha+\beta$ and the value of $\alpha\beta$.

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Method: for a quadratic $ax^2+bx+c=0$ with roots $\alpha,\beta$, the sum and product of the roots are given by $\alpha+\beta=-\dfrac{b}{a}$ and $\alpha\beta=\dfrac{c}{a}$. For $3x^2-12x+5=0$: $a=3$, $b=-12$, $c=5$. $$\alpha+\beta=-\frac{b}{a}=-\frac{-12}{3}=4$$ $$\boxed{\alpha+\beta=4}$$ $$\alpha\beta=\frac{c}{a}=\frac{5}{3}$$ $$\boxed{\alpha\beta=\dfrac{5}{3}}$$

QUESTION 29 5 marks

Easy

(a) Solve the inequality $x^2-5x+6<0$. (b) Solve the inequality $x^2-5x+6\ge0$.

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(a) Method: first find the zeros of $x^2-5x+6$ by factorising. $$x^2-5x+6=(x-2)(x-3)$$ Setting $(x-2)(x-3)=0$ gives zeros at $x=2$ and $x=3$. Since the coefficient of $x^2$ is positive, the graph of $y=x^2-5x+6$ is an upward-opening parabola, so the expression is negative between its zeros: $$\boxed{2

QUESTION 30 7 marks

Medium

Line $L_1$ passes through the points $A(1,-2)$ and $B(4,7)$. Line $L_2$ has equation $3x+y-1=0$. (a) Find the equation of $L_1$, giving your answer in the form $y=mx+c$. [4] (b) Determine whether $L_1$ and $L_2$ are parallel, perpendicular, or neither, justifying your answer. [3]

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(a) Using the gradient formula on points $A(1,-2)$ and $B(4,7)$: $$m_1=\frac{7-(-2)}{4-1}=\frac{9}{3}=3$$ Using point-gradient form through $A(1,-2)$: $$y-(-2)=3(x-1)$$ $$y+2=3x-3$$ $$\boxed{L_1:\;y=3x-5}$$ (b) Rearranging $L_2:3x+y-1=0$ into $y=mx+c$ form: $$y=-3x+1$$ So the gradient of $L_2$ is $m_2=-3$. Comparing gradients: $m_1=3$ and $m_2=-3$. Since $m_1\ne m_2$, the lines are not parallel. Checking the product of gradients: $$m_1\times m_2=3\times(-3)=-9$$ Since $m_1m_2=-9\ne-1$, the lines are not perpendicular either. $$\boxed{L_1\text{ and }L_2\text{ are neither parallel nor perpendicular}}$$

QUESTION 31 7 marks

Medium

Line $L_1$ passes through the points $A(-2,-3)$ and $B(2,5)$. Line $L_2$ has equation $x+2y-8=0$. (a) Find the equation of $L_1$, giving your answer in the form $y=mx+c$. [3] (b) Show that $L_1$ is perpendicular to $L_2$. [2] (c) Find the coordinates of the point of intersection of $L_1$ and $L_2$. [2]

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(a) Using the gradient formula on points $A(-2,-3)$ and $B(2,5)$: $$m_1=\frac{5-(-3)}{2-(-2)}=\frac{8}{4}=2$$ Using point-gradient form through $A(-2,-3)$: $$y-(-3)=2\big(x-(-2)\big)$$ $$y+3=2x+4$$ $$\boxed{L_1:\;y=2x+1}$$ (b) Rearranging $L_2:x+2y-8=0$ into $y=mx+c$ form: $$2y=-x+8 \implies y=-\frac{1}{2}x+4$$ So the gradient of $L_2$ is $m_2=-\dfrac{1}{2}$. Multiplying the gradients of $L_1$ and $L_2$: $$m_1\times m_2=2\times\left(-\frac{1}{2}\right)=-1$$ Since the product of the gradients is $-1$, $\boxed{L_1\text{ is perpendicular to }L_2}$, as required. (c) At the point of intersection, both lines have the same $y$-value, so substitute $y=2x+1$ into $x+2y-8=0$: $$x+2(2x+1)-8=0$$ $$x+4x+2-8=0$$ $$5x-6=0$$ $$x=\frac{6}{5}$$ Substituting back into $L_1$: $$y=2\left(\frac{6}{5}\right)+1=\frac{12}{5}+\frac{5}{5}=\frac{17}{5}$$ $$\boxed{\left(\frac{6}{5},\frac{17}{5}\right)}$$

QUESTION 32 8 marks

Medium

A company manufacturing garden chairs has weekly costs modelled by $C(x)=350+8x$ dollars, and weekly revenue modelled by $R(x)=15x$ dollars, where $x$ is the number of chairs sold per week. (a) Find the break-even number of chairs, i.e. the value of $x$ for which $C(x)=R(x)$. [3] (b) Find an expression for the weekly profit $P(x)=R(x)-C(x)$, simplifying your answer. [2] (c) Find the least number of chairs that must be sold per week for the profit to be at least $\$400$. [3]

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(a) Setting $C(x)=R(x)$: $$350+8x=15x$$ $$350=15x-8x$$ $$350=7x$$ $$x=50$$ $$\boxed{x=50\text{ chairs}}$$ (b) Subtracting the cost function from the revenue function: $$P(x)=R(x)-C(x)=15x-(350+8x)$$ $$P(x)=15x-350-8x$$ $$\boxed{P(x)=7x-350}$$ (c) Setting $P(x)\ge400$: $$7x-350\ge400$$ $$7x\ge750$$ $$x\ge\frac{750}{7}=107.142857\ldots$$ Since $x$ must be a whole number of chairs, the least integer value satisfying this inequality is: $$\boxed{x=108\text{ chairs}}$$

QUESTION 33 6 marks

Medium

Find the value of $c$ for which the line $y=4x+c$ is tangent to the curve $y=x^2+2x+9$ (i.e. meets the curve at exactly one point).

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Method: at points of intersection the $y$-values of the line and curve are equal; tangency means the resulting quadratic in $x$ has exactly one (repeated) solution, so its discriminant is zero. Setting $4x+c=x^2+2x+9$: $$x^2+2x+9-4x-c=0$$ $$x^2-2x+(9-c)=0$$ For a repeated root, the discriminant $\Delta=0$: $$\Delta=(-2)^2-4(1)(9-c)$$ $$4-36+4c=0$$ $$4c-32=0$$ $$c=8$$ $$\boxed{c=8}$$

QUESTION 34 7 marks

Medium

The equation $2x^2-7x+k=0$ has roots $\alpha$ and $\beta$, where $\alpha^2+\beta^2=\dfrac{45}{4}$. Find the value of $k$.

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Method: use the sum and product of roots formulas together with the identity $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$. For $2x^2-7x+k=0$: $a=2$, $b=-7$, $c=k$. $$\alpha+\beta=-\frac{b}{a}=\frac{7}{2}, \qquad \alpha\beta=\frac{c}{a}=\frac{k}{2}$$ Using the identity: $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$ $$\frac{45}{4}=\left(\frac{7}{2}\right)^2-2\left(\frac{k}{2}\right)$$ $$\frac{45}{4}=\frac{49}{4}-k$$ $$k=\frac{49}{4}-\frac{45}{4}=\frac{4}{4}=1$$ $$\boxed{k=1}$$ (Check: with $k=1$, solving $2x^2-7x+1=0$ using the quadratic formula gives $\alpha,\beta=\dfrac{7\pm\sqrt{41}}{4}$; numerically $\alpha\approx3.4753$, $\beta\approx0.1436$, so $\alpha^2+\beta^2\approx12.08+0.0206\approx12.25=\dfrac{45}{4}$, confirming the result.)

QUESTION 35 8 marks

Medium

A farmer wants to build a rectangular enclosure and then divide it into two equal pens using an internal fence parallel to one pair of sides, as shown in the diagram. The farmer has $120$ m of fencing in total, for the three vertical sections (each of length $x$ metres) and the two horizontal sections (each of length $y$ metres). (a) Show that $y=60-1.5x$, and hence show that the total enclosed area is $A(x)=60x-1.5x^2$. [3] (b) Find the value of $x$ that maximises the area, and state the maximum area, using the method of completing the square. [4] (c) State the corresponding value of $y$ at the maximum area. [1]

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(a) The total fencing consists of three vertical sections of length $x$ and two horizontal sections of length $y$: $$3x+2y=120$$ Solving for $y$: $$2y=120-3x$$ $$\boxed{y=60-1.5x}$$, as required. The total enclosed area is the product of the outer width and height, $A=x\times y$: $$A(x)=x(60-1.5x)$$ $$\boxed{A(x)=60x-1.5x^2}$$, as required. (b) Completing the square on $A(x)=-1.5x^2+60x$: $$A(x)=-1.5(x^2-40x)$$ $$A(x)=-1.5\big[(x-20)^2-400\big]$$ $$A(x)=-1.5(x-20)^2+600$$ Since $-1.5(x-20)^2\le0$ for all $x$, with equality when $x=20$, the maximum occurs at $x=20$: $$\boxed{x=20\text{ m, giving a maximum area of }600\text{ m}^2}$$ (c) Substituting $x=20$ into $y=60-1.5x$: $$y=60-1.5(20)=60-30$$ $$\boxed{y=30\text{ m}}$$

QUESTION 36 7 marks

Medium

A stone is thrown into the air. Its height above the ground, in metres, after $t$ seconds is modelled by $h(t)=-5t^2+20t+2$, for $t\ge0$. Find the interval of time during which the stone is more than $15$ metres above the ground, giving your answers correct to three significant figures.

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Method: form and solve the quadratic inequality $h(t)>15$, using the quadratic formula since the numbers do not factorise nicely. $$-5t^2+20t+2>15$$ $$-5t^2+20t-13>0$$ Multiplying both sides by $-1$ (reversing the inequality): $$5t^2-20t+13<0$$ Finding the roots of $5t^2-20t+13=0$ using the quadratic formula: $$t=\frac{20\pm\sqrt{(-20)^2-4(5)(13)}}{2(5)}=\frac{20\pm\sqrt{400-260}}{10}=\frac{20\pm\sqrt{140}}{10}$$ Since $\sqrt{140}=11.8322\ldots$: $$t=\frac{20-11.8322}{10}=0.8168\ldots \quad\text{or}\quad t=\frac{20+11.8322}{10}=3.1832\ldots$$ Since the coefficient of $t^2$ in $5t^2-20t+13$ is positive, this expression is negative between its roots, so $h(t)>15$ for $t$ between these values. $$\boxed{0.817\ \text{s}

QUESTION 37 7 marks

Medium

A parabola has vertex $(2,-5)$ and passes through the point $(5,4)$. (a) Find the equation of the parabola in the form $y=a(x-h)^2+k$. [4] (b) Hence write the equation of the parabola in the general form $y=ax^2+bx+c$. [3]

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(a) Since the vertex is $(2,-5)$, the equation has the form: $$y=a(x-2)^2-5$$ Substituting the point $(5,4)$: $$4=a(5-2)^2-5$$ $$4=9a-5$$ $$9a=9$$ $$a=1$$ $$\boxed{y=(x-2)^2-5}$$ (b) Expanding $(x-2)^2=x^2-4x+4$: $$y=x^2-4x+4-5$$ $$\boxed{y=x^2-4x-1}$$

QUESTION 38 7 marks

Medium

Quadrilateral $ABCD$ has vertices $A(0,0)$, $B(6,2)$, $C(8,8)$ and $D(2,6)$. (a) Show that $[AB]$ is parallel to $[DC]$. [3] (b) Show that $[AD]$ is parallel to $[BC]$. [2] (c) State, with a reason, whether $ABCD$ could be a rectangle. [2]

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(a) Finding the gradient of $[AB]$: $$m_{AB}=\frac{2-0}{6-0}=\frac{2}{6}=\frac{1}{3}$$ Finding the gradient of $[DC]$: $$m_{DC}=\frac{8-6}{8-2}=\frac{2}{6}=\frac{1}{3}$$ Since $m_{AB}=m_{DC}=\dfrac{1}{3}$, $\boxed{[AB]\parallel[DC]}$, as required. (b) Finding the gradient of $[AD]$: $$m_{AD}=\frac{6-0}{2-0}=3$$ Finding the gradient of $[BC]$: $$m_{BC}=\frac{8-2}{8-6}=\frac{6}{2}=3$$ Since $m_{AD}=m_{BC}=3$, $\boxed{[AD]\parallel[BC]}$, as required. (c) Since both pairs of opposite sides are parallel, $ABCD$ is a parallelogram. To check whether it could be a rectangle, test whether adjacent sides $AB$ and $AD$ are perpendicular: $$m_{AB}\times m_{AD}=\frac{1}{3}\times3=1$$ Since the product of the gradients of the adjacent sides is $1$, not $-1$, the adjacent sides are not perpendicular. $$\boxed{ABCD\text{ is a parallelogram but not a rectangle, since its adjacent sides are not perpendicular}}$$

QUESTION 39 8 marks

Medium

A ball is thrown vertically upward from a platform. Its height above the ground, in metres, after $t$ seconds is modelled by $h(t)=-4.9t^2+24.5t+1.5$, for $t\ge0$. (a) Find the time at which the ball reaches its maximum height, and find the maximum height. [4] (b) Find the time at which the ball hits the ground, correct to three significant figures. [4]

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(a) The maximum of a downward-opening parabola $h(t)=at^2+bt+c$ occurs at $t=-\dfrac{b}{2a}$. Here $a=-4.9$, $b=24.5$: $$t=-\frac{24.5}{2(-4.9)}=-\frac{24.5}{-9.8}=2.5$$ Substituting $t=2.5$ into $h(t)$: $$h(2.5)=-4.9(2.5)^2+24.5(2.5)+1.5$$ $$h(2.5)=-4.9(6.25)+61.25+1.5$$ $$h(2.5)=-30.625+61.25+1.5=32.125$$ $$\boxed{t=2.5\text{ s, maximum height }=32.125\text{ m}}$$ (b) The ball hits the ground when $h(t)=0$: $$-4.9t^2+24.5t+1.5=0$$ Using the quadratic formula with $a=-4.9$, $b=24.5$, $c=1.5$: $$t=\frac{-24.5\pm\sqrt{24.5^2-4(-4.9)(1.5)}}{2(-4.9)}=\frac{-24.5\pm\sqrt{600.25+29.4}}{-9.8}=\frac{-24.5\pm\sqrt{629.65}}{-9.8}$$ Since $\sqrt{629.65}=25.0928\ldots$: $$t=\frac{-24.5+25.0928}{-9.8}=-0.0605\ldots \quad\text{or}\quad t=\frac{-24.5-25.0928}{-9.8}=5.0605\ldots$$ Since $t\ge0$ in context, the negative solution is rejected. $$\boxed{t\approx5.06\text{ s (3 s.f.)}}$$

QUESTION 40 7 marks

Medium

The equation $3x^2-5x-2=0$ has roots $\alpha$ and $\beta$. Without solving for $\alpha$ and $\beta$ individually, find a quadratic equation with integer coefficients whose roots are $2\alpha$ and $2\beta$.

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Method: use the sum and product of roots of the original equation to find the sum and product of the new roots, then reconstruct the equation using $x^2-(\text{sum})x+(\text{product})=0$. For $3x^2-5x-2=0$: $a=3$, $b=-5$, $c=-2$. $$\alpha+\beta=-\frac{b}{a}=\frac{5}{3}, \qquad \alpha\beta=\frac{c}{a}=-\frac{2}{3}$$ The new roots are $2\alpha$ and $2\beta$, so: $$(2\alpha)+(2\beta)=2(\alpha+\beta)=2\times\frac{5}{3}=\frac{10}{3}$$ $$(2\alpha)(2\beta)=4\alpha\beta=4\times\left(-\frac{2}{3}\right)=-\frac{8}{3}$$ A quadratic with these roots is: $$x^2-\frac{10}{3}x-\frac{8}{3}=0$$ Multiplying through by $3$ to obtain integer coefficients: $$\boxed{3x^2-10x-8=0}$$ (Check: solving $3x^2-5x-2=0$ numerically gives $\alpha\approx1.8685$, $\beta\approx-0.3568$, so $2\alpha\approx3.7370$, $2\beta\approx-0.7137$; substituting $x=3.7370$ into $3x^2-10x-8$ gives approximately $0$, confirming the result.)

QUESTION 41 7 marks

Medium

Find the set of values of $m$ for which the line $y=mx-3$ does not intersect the curve $y=2x^2-4x+5$.

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Method: at any point of intersection the $y$-values are equal, giving a quadratic in $x$; no intersection requires this quadratic to have no real solutions, i.e. its discriminant is negative. Setting $mx-3=2x^2-4x+5$: $$2x^2-4x+5-mx+3=0$$ $$2x^2-(4+m)x+8=0$$ For no real solutions, the discriminant $\Delta<0$: $$\Delta=(4+m)^2-4(2)(8)$$ $$(4+m)^2-64<0$$ $$(4+m)^2<64$$ Taking square roots requires a case split, since this is an inequality on a squared expression: $$-8<4+m<8$$ Subtracting $4$ throughout: $$\boxed{-12

QUESTION 42 7 marks

Medium

A candle burns at a constant rate. Its height, in cm, decreases linearly over time. When lit ($t=0$), the candle is $20$ cm tall, and after $5$ hours its height is $12.5$ cm. (a) Find a linear function $H(t)$ for the height of the candle, in cm, after $t$ hours. [4] (b) Find the time taken for the candle to burn out completely, correct to three significant figures. [2] (c) State a suitable domain for $t$ in this context. [1]

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(a) The gradient (rate of change of height) is found using the two given points $(0,20)$ and $(5,12.5)$: $$m=\frac{12.5-20}{5-0}=\frac{-7.5}{5}=-1.5$$ Since the initial height (the $y$-intercept) is $20$ cm: $$\boxed{H(t)=20-1.5t}$$ (b) The candle burns out when $H(t)=0$: $$20-1.5t=0$$ $$1.5t=20$$ $$t=\frac{20}{1.5}=13.333\ldots$$ $$\boxed{t\approx13.3\text{ hours (3 s.f.)}}$$ (c) Since $t$ represents time after lighting, and the candle cannot have negative height or burn for longer than the time found in part (b): $$\boxed{0\le t\le\frac{40}{3}}$$

QUESTION 43 8 marks

Medium

Find the value(s) of $p$ (where $p\ne0$) for which the equation $px^2+8x+p=0$ has equal roots, and solve the equation for each value of $p$ found.

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Method: equal roots occur when the discriminant is zero. For $px^2+8x+p=0$: $a=p$, $b=8$, $c=p$. $$\Delta=8^2-4(p)(p)=64-4p^2$$ Setting $\Delta=0$: $$64-4p^2=0$$ $$p^2=16$$ $$\boxed{p=4\ \text{or}\ p=-4}$$ Case $p=4$: the equation becomes: $$4x^2+8x+4=0$$ Dividing through by $4$: $$x^2+2x+1=0$$ $$(x+1)^2=0$$ $$\boxed{x=-1\ \text{(repeated root)}}$$ Case $p=-4$: the equation becomes: $$-4x^2+8x-4=0$$ Dividing through by $-4$: $$x^2-2x+1=0$$ $$(x-1)^2=0$$ $$\boxed{x=1\ \text{(repeated root)}}$$

QUESTION 44 14 marks

Hard

Triangle $PQR$ has vertices $P(-5,2)$, $Q(3,6)$ and $R(1,-4)$. (a) Find the equation of line $PQ$, giving your answer in the form $ax+by+d=0$ where $a,b,d\in\mathbb{Z}$. [3] (b) Find the equation of the line through $R$ that is perpendicular to $PQ$ (the altitude from $R$). [4] (c) Find the coordinates of the point $F$ where this altitude meets line $PQ$. [3] (d) Hence find the area of triangle $PQR$, using $PQ$ as the base and $RF$ as the corresponding height. Confirm your answer using the formula $\text{Area}=\dfrac{1}{2}\left|x_P(y_Q-y_R)+x_Q(y_R-y_P)+x_R(y_P-y_Q)\right|$. [4]

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(a) Finding the gradient of $PQ$ using $P(-5,2)$ and $Q(3,6)$: $$m_{PQ}=\frac{6-2}{3-(-5)}=\frac{4}{8}=\frac{1}{2}$$ Using point-gradient form through $P(-5,2)$: $$y-2=\frac{1}{2}\big(x-(-5)\big)$$ $$2y-4=x+5$$ $$\boxed{x-2y+9=0}$$ (b) Since the altitude from $R$ is perpendicular to $PQ$, its gradient is the negative reciprocal of $m_{PQ}=\dfrac{1}{2}$: $$m_{\perp}=-2$$ Using point-gradient form through $R(1,-4)$: $$y-(-4)=-2(x-1)$$ $$y+4=-2x+2$$ $$\boxed{y=-2x-2}$$ (c) Substituting $y=-2x-2$ into the equation of $PQ$ in the form $y=\dfrac{1}{2}x+\dfrac{9}{2}$ (rearranged from part (a)): $$-2x-2=\frac{1}{2}x+\frac{9}{2}$$ Multiplying every term by $2$ to clear fractions: $$-4x-4=x+9$$ $$-5x=13$$ $$x=-\frac{13}{5}$$ Substituting back into $y=-2x-2$: $$y=-2\left(-\frac{13}{5}\right)-2=\frac{26}{5}-\frac{10}{5}=\frac{16}{5}$$ $$\boxed{F=\left(-\frac{13}{5},\frac{16}{5}\right)}$$ (d) The length of $PQ$, using the distance formula: $$PQ=\sqrt{(3-(-5))^2+(6-2)^2}=\sqrt{8^2+4^2}=\sqrt{64+16}=\sqrt{80}=4\sqrt5$$ The height $RF$, using the distance formula between $R(1,-4)$ and $F\left(-\frac{13}{5},\frac{16}{5}\right)$: $$RF=\sqrt{\left(1-\left(-\frac{13}{5}\right)\right)^2+\left(-4-\frac{16}{5}\right)^2}=\sqrt{\left(\frac{18}{5}\right)^2+\left(-\frac{36}{5}\right)^2}$$ $$RF=\sqrt{\frac{324}{25}+\frac{1296}{25}}=\sqrt{\frac{1620}{25}}=\frac{\sqrt{1620}}{5}=\frac{18\sqrt5}{5}$$ So the area is: $$\text{Area}=\frac{1}{2}\times PQ\times RF=\frac{1}{2}\times4\sqrt5\times\frac{18\sqrt5}{5}=\frac{1}{2}\times\frac{72\times5}{5}=\frac{1}{2}\times72=36$$ $$\boxed{\text{Area}=36}$$ Check using the coordinate (shoelace) formula: $$\text{Area}=\frac{1}{2}\left|(-5)(6-(-4))+3((-4)-2)+1(2-6)\right|$$ $$=\frac{1}{2}\left|(-5)(10)+3(-6)+1(-4)\right|=\frac{1}{2}\left|-50-18-4\right|=\frac{1}{2}(72)=36$$ This matches the area found above, confirming $\boxed{\text{Area}=36}$.

QUESTION 45 14 marks

Hard

Consider the quadratic equation $kx^2+(k+6)x+4=0$, where $k\in\mathbb{R}$, $k\ne0$, with roots $\alpha$ and $\beta$. (a) Show that the discriminant of the equation is $\Delta=k^2-4k+36$. [3] (b) By completing the square on $\Delta$, show that the equation has two distinct real roots for every value of $k\ne0$. [3] (c) Given that $\alpha+\beta=-\dfrac{k+6}{k}$ and $\alpha\beta=\dfrac{4}{k}$, find the value of $k$ for which $\alpha+\beta=3\alpha\beta$. [4] (d) Hence solve the original equation for the value of $k$ found in part (c), giving your answers in exact (surd) form. [4]

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(a) For $kx^2+(k+6)x+4=0$, with $a=k$, $b=k+6$, $c=4$, use the discriminant formula $\Delta=b^2-4ac$: $$\Delta=(k+6)^2-4(k)(4)$$ Expanding: $$=(k^2+12k+36)-16k$$ $$=k^2-4k+36$$ $$\boxed{\Delta=k^2-4k+36}$$, as required. (b) Completing the square on $\Delta=k^2-4k+36$: $$\Delta=(k-2)^2-4+36$$ $$\Delta=(k-2)^2+32$$ Since $(k-2)^2\ge0$ for all real $k$, we have $\Delta\ge32>0$ for every value of $k$. Since $\Delta$ is always strictly positive, $\boxed{\text{the equation always has two distinct real roots, for every }k\ne0}$, as required. (c) Setting $\alpha+\beta=3\alpha\beta$: $$-\frac{k+6}{k}=3\left(\frac{4}{k}\right)$$ $$-\frac{k+6}{k}=\frac{12}{k}$$ Since $k\ne0$, multiply both sides by $k$: $$-(k+6)=12$$ $$-k-6=12$$ $$-k=18$$ $$\boxed{k=-18}$$ (d) Substituting $k=-18$ into the original equation: $$-18x^2+(-18+6)x+4=0$$ $$-18x^2-12x+4=0$$ Dividing through by $-2$: $$9x^2+6x-2=0$$ Using the quadratic formula with $a=9$, $b=6$, $c=-2$: $$x=\frac{-6\pm\sqrt{6^2-4(9)(-2)}}{2(9)}=\frac{-6\pm\sqrt{36+72}}{18}=\frac{-6\pm\sqrt{108}}{18}$$ Since $\sqrt{108}=\sqrt{36\times3}=6\sqrt3$: $$x=\frac{-6\pm6\sqrt3}{18}=\frac{-1\pm\sqrt3}{3}$$ $$\boxed{x=\dfrac{-1+\sqrt3}{3}\ \text{or}\ x=\dfrac{-1-\sqrt3}{3}}$$

QUESTION 46 18 marks

Hard

A ball is kicked from ground level. Its height above the ground, in metres, after $t$ seconds is modelled by $h(t)=-5t^2+18t$, for $t\ge0$. (a) Factorise $h(t)$, and hence find the time at which the ball returns to the ground. [3] (b) Find the maximum height reached by the ball, and the time at which it occurs, using the method of completing the square. [4] (c) Find the time(s) at which the ball is at a height of $10$ m, giving your answers correct to three significant figures. [4] (d) Find the interval of time during which the ball is more than $12$ m above the ground, giving your answers correct to three significant figures. [4] (e) Hence find the total length of time (correct to three significant figures) that the ball spends above $12$ m, expressed as a percentage of its total flight time (correct to three significant figures). [3]

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(a) Factorising $h(t)=-5t^2+18t$ by taking out a common factor of $t$: $$h(t)=t(-5t+18)=t(18-5t)$$ $$\boxed{h(t)=t(18-5t)}$$ The ball is at ground level when $h(t)=0$, i.e. when $t=0$ (the moment it is kicked) or: $$18-5t=0 \implies t=\frac{18}{5}=3.6$$ $$\boxed{t=3.6\text{ s}}$$ (b) Completing the square on $h(t)=-5t^2+18t$: $$h(t)=-5\left(t^2-\frac{18}{5}t\right)$$ $$h(t)=-5\left[\left(t-\frac{9}{5}\right)^2-\frac{81}{25}\right]$$ $$h(t)=-5\left(t-\frac{9}{5}\right)^2+\frac{81}{5}$$ Since $-5\left(t-\frac{9}{5}\right)^2\le0$ for all $t$, the maximum occurs at $t=\dfrac{9}{5}=1.8$: $$\boxed{\text{Maximum height}=\dfrac{81}{5}=16.2\text{ m, at }t=1.8\text{ s}}$$ (c) Setting $h(t)=10$: $$-5t^2+18t=10$$ $$-5t^2+18t-10=0$$ Multiplying by $-1$: $5t^2-18t+10=0$. Using the quadratic formula: $$t=\frac{18\pm\sqrt{(-18)^2-4(5)(10)}}{2(5)}=\frac{18\pm\sqrt{324-200}}{10}=\frac{18\pm\sqrt{124}}{10}$$ Since $\sqrt{124}=11.1355\ldots$: $$t=\frac{18-11.1355}{10}=0.6864\ldots \quad\text{or}\quad t=\frac{18+11.1355}{10}=2.9136\ldots$$ $$\boxed{t\approx0.686\text{ s}\ \text{or}\ t\approx2.91\text{ s (3 s.f.)}}$$ (d) Setting $h(t)>12$: $$-5t^2+18t>12$$ $$-5t^2+18t-12>0$$ Multiplying by $-1$ (reversing the inequality): $5t^2-18t+12<0$. Using the quadratic formula: $$t=\frac{18\pm\sqrt{(-18)^2-4(5)(12)}}{2(5)}=\frac{18\pm\sqrt{324-240}}{10}=\frac{18\pm\sqrt{84}}{10}$$ Since $\sqrt{84}=9.1652\ldots$: $$t=\frac{18-9.1652}{10}=0.8835\ldots \quad\text{or}\quad t=\frac{18+9.1652}{10}=2.7165\ldots$$ Since the coefficient of $t^2$ in $5t^2-18t+12$ is positive, this expression is negative between its roots, so $h(t)>12$ between these values. $$\boxed{0.883\ \text{s}

QUESTION 47 12 marks

Hard

The parabola $C$ has equation $y=x^2-4x+7$. The line $L$ has equation $y=mx+1$, where $m\in\mathbb{R}$. (a) Show that the $x$-coordinates of any points of intersection of $L$ and $C$ satisfy $x^2-(4+m)x+6=0$. [2] (b) Find the value(s) of $m$ for which $L$ is tangent to $C$. [4] (c) For each value of $m$ found in part (b), find the coordinates of the corresponding point of tangency, giving your answers in exact (surd) form. [4] (d) State the coordinates of the vertex of $C$, and hence state the minimum value of $y$ on the graph of $C$. [2]

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(a) At any point of intersection, the $y$-values of $L$ and $C$ are equal: $$mx+1=x^2-4x+7$$ $$0=x^2-4x+7-mx-1$$ $$\boxed{x^2-(4+m)x+6=0}$$, as required. (b) $L$ is tangent to $C$ when the quadratic in part (a) has exactly one (repeated) solution, i.e. its discriminant is zero. With $a=1$, $b=-(4+m)$, $c=6$: $$\Delta=(4+m)^2-4(1)(6)=0$$ $$(4+m)^2=24$$ $$4+m=\pm\sqrt{24}=\pm2\sqrt6$$ $$\boxed{m=-4+2\sqrt6\ \text{or}\ m=-4-2\sqrt6}$$ (c) Since the quadratic in part (a) has a repeated root at tangency, the $x$-coordinate of the point of tangency is $x=\dfrac{4+m}{2}$ (the vertex/repeated-root formula $x=-\dfrac{b}{2a}$ with $b=-(4+m)$, $a=1$). For $m=-4+2\sqrt6$: $4+m=2\sqrt6$, so: $$x=\frac{2\sqrt6}{2}=\sqrt6$$ Substituting into $y=mx+1$: $$y=\left(-4+2\sqrt6\right)\sqrt6+1=-4\sqrt6+2(6)+1=-4\sqrt6+13$$ $$\boxed{\left(\sqrt6,\;13-4\sqrt6\right)}$$ For $m=-4-2\sqrt6$: $4+m=-2\sqrt6$, so: $$x=\frac{-2\sqrt6}{2}=-\sqrt6$$ Substituting into $y=mx+1$: $$y=\left(-4-2\sqrt6\right)\left(-\sqrt6\right)+1=4\sqrt6+2(6)+1=4\sqrt6+13$$ $$\boxed{\left(-\sqrt6,\;13+4\sqrt6\right)}$$ (d) Completing the square on $y=x^2-4x+7$: $$y=(x-2)^2-4+7=(x-2)^2+3$$ The vertex is at $\boxed{(2,3)}$, and since the parabola opens upward ($a=1>0$), this is a minimum point, so the minimum value of $y$ on the graph of $C$ is $\boxed{y=3}$.

QUESTION 48 15 marks

Hard

A company's weekly revenue from selling $x$ units of a product is modelled by $R(x)=-2x^2+240x$ dollars, and its weekly cost is modelled by $C(x)=60x+1000$ dollars, for $0\le x\le100$. (a) Form an expression for the weekly profit $P(x)=R(x)-C(x)$, simplifying your answer. [2] (b) Write $P(x)$ in vertex form, and hence find the maximum weekly profit and the value of $x$ at which it occurs. [4] (c) Find the values of $x$, correct to three significant figures, for which the company breaks even or makes a loss (i.e. $P(x)\le0$), within the given domain. [5] (d) Find the range of values of $x$, correct to three significant figures, for which the weekly profit is at least $\$2000$. [4]

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(a) Subtracting the cost function from the revenue function: $$P(x)=R(x)-C(x)=(-2x^2+240x)-(60x+1000)$$ $$\boxed{P(x)=-2x^2+180x-1000}$$ (b) Completing the square on $P(x)=-2x^2+180x-1000$: $$P(x)=-2(x^2-90x)-1000$$ $$P(x)=-2\big[(x-45)^2-2025\big]-1000$$ $$P(x)=-2(x-45)^2+4050-1000$$ $$\boxed{P(x)=-2(x-45)^2+3050}$$ Since $-2(x-45)^2\le0$ for all $x$, the maximum occurs at $x=45$: $$\boxed{\text{Maximum weekly profit}=\$3050,\text{ at }x=45\text{ units}}$$ (c) Setting $P(x)=0$ using the vertex form: $$-2(x-45)^2+3050=0$$ $$(x-45)^2=1525$$ $$x-45=\pm\sqrt{1525}$$ Since $\sqrt{1525}=39.0512\ldots$: $$x=45-39.0512=5.9488\ldots \quad\text{or}\quad x=45+39.0512=84.0512\ldots$$ Since $P(x)$ is a downward-opening parabola (coefficient of $x^2$ is negative), $P(x)\le0$ outside the interval between these roots. Restricting to the given domain $0\le x\le100$: $$\boxed{0\le x\le5.95\ \text{or}\ 84.1\le x\le100\ \text{(3 s.f.)}}$$ (d) Setting $P(x)\ge2000$: $$-2(x-45)^2+3050\ge2000$$ $$-2(x-45)^2\ge-1050$$ Dividing by $-2$ (reversing the inequality): $$(x-45)^2\le525$$ $$-\sqrt{525}\le x-45\le\sqrt{525}$$ Since $\sqrt{525}=22.9129\ldots$: $$45-22.9129\le x\le45+22.9129$$ $$\boxed{22.1\le x\le67.9\ \text{(3 s.f.)}}$$

QUESTION 49 14 marks

Hard

The quadratic equation $x^2-(m+2)x+(3m-1)=0$, where $m\in\mathbb{R}$, has roots $\alpha$ and $\beta$ such that $\alpha-\beta=4$ (where $\alpha>\beta$). (a) Write down expressions for $\alpha+\beta$ and $\alpha\beta$ in terms of $m$. [2] (b) By using the identity $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$, form an equation in $m$ and solve it, giving your answers in exact (surd) form. [6] (c) For each value of $m$ found in part (b), find the values of $\alpha$ and $\beta$, using $\alpha+\beta$ and the given condition $\alpha-\beta=4$ (without solving the original quadratic directly). [6]

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(a) Comparing $x^2-(m+2)x+(3m-1)=0$ with $x^2-(\text{sum})x+(\text{product})=0$: $$\boxed{\alpha+\beta=m+2, \qquad \alpha\beta=3m-1}$$ (b) Substituting into $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$, using $\alpha-\beta=4$: $$4^2=(m+2)^2-4(3m-1)$$ $$16=(m^2+4m+4)-(12m-4)$$ $$16=m^2+4m+4-12m+4$$ $$16=m^2-8m+8$$ $$0=m^2-8m-8$$ Using the quadratic formula with $a=1$, $b=-8$, $c=-8$: $$m=\frac{8\pm\sqrt{(-8)^2-4(1)(-8)}}{2(1)}=\frac{8\pm\sqrt{64+32}}{2}=\frac{8\pm\sqrt{96}}{2}$$ Since $\sqrt{96}=\sqrt{16\times6}=4\sqrt6$: $$m=\frac{8\pm4\sqrt6}{2}=4\pm2\sqrt6$$ $$\boxed{m=4+2\sqrt6\ \text{or}\ m=4-2\sqrt6}$$ (c) Since $\alpha+\beta=m+2$ and $\alpha-\beta=4$, adding and subtracting these two equations gives $\alpha=\dfrac{(\alpha+\beta)+4}{2}$ and $\beta=\dfrac{(\alpha+\beta)-4}{2}$. For $m=4+2\sqrt6$: $\alpha+\beta=m+2=6+2\sqrt6$, so: $$\alpha=\frac{(6+2\sqrt6)+4}{2}=\frac{10+2\sqrt6}{2}=5+\sqrt6$$ $$\beta=\frac{(6+2\sqrt6)-4}{2}=\frac{2+2\sqrt6}{2}=1+\sqrt6$$ Checking against $\alpha\beta=3m-1=3(4+2\sqrt6)-1=11+6\sqrt6$: $$\alpha\beta=(5+\sqrt6)(1+\sqrt6)=5+5\sqrt6+\sqrt6+6=11+6\sqrt6 \checkmark$$ $$\boxed{\alpha=5+\sqrt6,\ \beta=1+\sqrt6}$$ For $m=4-2\sqrt6$: $\alpha+\beta=m+2=6-2\sqrt6$, so: $$\alpha=\frac{(6-2\sqrt6)+4}{2}=\frac{10-2\sqrt6}{2}=5-\sqrt6$$ $$\beta=\frac{(6-2\sqrt6)-4}{2}=\frac{2-2\sqrt6}{2}=1-\sqrt6$$ Checking against $\alpha\beta=3m-1=3(4-2\sqrt6)-1=11-6\sqrt6$: $$\alpha\beta=(5-\sqrt6)(1-\sqrt6)=5-5\sqrt6-\sqrt6+6=11-6\sqrt6 \checkmark$$ $$\boxed{\alpha=5-\sqrt6,\ \beta=1-\sqrt6}$$

QUESTION 50 10 marks

Hard

A rectangular vegetable plot is to be built in the corner formed by two existing perpendicular walls, so that only the two remaining sides need to be fenced. A gardener has $60$ m of fencing available for these two sides. Let $x$ metres be the length of one fenced side. (a) Show that the enclosed area is $A(x)=x(60-x)$. [2] (b) Find the value of $x$ that maximises the enclosed area, and state this maximum area, using the method of completing the square. [4] (c) The gardener also wants the enclosed area to be at least $800\text{ m}^2$. Find the range of values of $x$ for which this is satisfied. [4]

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(a) Since only two sides are fenced (the other two sides are the existing walls), if one fenced side has length $x$, the other fenced side has length $60-x$ (since the total fencing is $60$ m). The enclosed area is the product of these two side lengths: $$\boxed{A(x)=x(60-x)}$$, as required. (b) Expanding and completing the square on $A(x)=60x-x^2$: $$A(x)=-(x^2-60x)$$ $$A(x)=-\big[(x-30)^2-900\big]$$ $$A(x)=-(x-30)^2+900$$ Since $-(x-30)^2\le0$ for all $x$, the maximum occurs at $x=30$: $$\boxed{x=30\text{ m, giving a maximum area of }900\text{ m}^2}$$ (c) Setting $A(x)\ge800$: $$60x-x^2\ge800$$ $$-x^2+60x-800\ge0$$ Multiplying by $-1$ (reversing the inequality): $$x^2-60x+800\le0$$ Using the quadratic formula with $a=1$, $b=-60$, $c=800$: $$x=\frac{60\pm\sqrt{(-60)^2-4(1)(800)}}{2(1)}=\frac{60\pm\sqrt{3600-3200}}{2}=\frac{60\pm\sqrt{400}}{2}=\frac{60\pm20}{2}$$ $$x=\frac{60-20}{2}=20 \quad\text{or}\quad x=\frac{60+20}{2}=40$$ Since the coefficient of $x^2$ in $x^2-60x+800$ is positive, this expression is $\le0$ between (and including) its roots: $$\boxed{20\le x\le40}$$

QUESTION 1 4 marks

Easy

Consider the function $f(x)=\dfrac{3x-2}{x+5}$, $x\neq-5$. Find (a) the equation of the vertical asymptote, (b) the equation of the horizontal asymptote, and (c) the $y$-intercept of the graph of $f$.

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(a) The vertical asymptote occurs where the denominator is zero (and the numerator is non-zero there): $$x+5=0\implies x=-5$$ At $x=-5$: numerator $=3(-5)-2=-17\neq0$, so this is a genuine vertical asymptote. $$\boxed{x=-5}$$ (b) Since the numerator and denominator both have degree 1, the horizontal asymptote is the ratio of the leading coefficients: $$y=\frac{3}{1}=3$$ $$\boxed{y=3}$$ (c) The $y$-intercept is $f(0)$: $$f(0)=\frac{3(0)-2}{0+5}=\frac{-2}{5}$$ $$\boxed{y\text{-intercept}=-\dfrac{2}{5}}$$

QUESTION 2 6 marks

Medium

Consider $f(x)=\dfrac{x^{2}-4}{x-1}$, $x\neq1$. (a) State the equation of the vertical asymptote of the graph of $f$. (b) By performing polynomial division, express $f(x)$ in the form $f(x)=ax+b+\dfrac{c}{x-1}$, where $a,b,c\in\mathbb{Z}$, and hence state the equation of the oblique (slant) asymptote of the graph of $f$.

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(a) The vertical asymptote occurs where the denominator is zero and the numerator is not: $$x-1=0\implies x=1$$ At $x=1$: numerator $=1^{2}-4=-3\neq0$, so this is a genuine vertical asymptote. $$\boxed{x=1}$$ (b) Method: divide $x^{2}-4$ by $x-1$ using polynomial long division. $$x^{2}\div x=x, \qquad x(x-1)=x^{2}-x$$ Subtract: $(x^{2}-4)-(x^{2}-x)=x-4$ $$x\div x=1, \qquad 1(x-1)=x-1$$ Subtract: $(x-4)-(x-1)=-3$ So $x^{2}-4=(x-1)(x+1)-3$, giving $$f(x)=x+1-\frac{3}{x-1}$$ i.e. $a=1,\ b=1,\ c=-3$. As $x\to\pm\infty$, $\dfrac{-3}{x-1}\to0$, so $f(x)\to x+1$. $$\boxed{f(x)=x+1-\dfrac{3}{x-1};\ \text{oblique asymptote: } y=x+1}$$

QUESTION 3 5 marks

Easy

Consider the function $f(x)=2+\dfrac{3}{x-1}$, $x\neq1$. Find (a) the equations of the vertical and horizontal asymptotes of the graph of $f$, (b) the $y$-intercept, and (c) the $x$-intercept.

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(a) This is a transformation of $y=\dfrac{1}{x}$. The vertical asymptote occurs where $x-1=0$, and the horizontal asymptote is the value $f(x)$ approaches as $x\to\pm\infty$, since $\dfrac{3}{x-1}\to0$: $$\boxed{x=1\ \text{and}\ y=2}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=2+\frac{3}{0-1}=2-3=-1$$ $$\boxed{y\text{-intercept}=-1}$$ (c) The $x$-intercept occurs where $f(x)=0$: $$2+\frac{3}{x-1}=0$$ $$\frac{3}{x-1}=-2$$ Cross-multiplying: $$3=-2(x-1)$$ $$3=-2x+2$$ $$2x=-1$$ $$x=-\frac{1}{2}$$ $$\boxed{x\text{-intercept}=-\dfrac{1}{2}}$$

QUESTION 4 7 marks

Medium

Consider the function $f(x)=\dfrac{2x-1}{x^{2}-5x+6}$. (a) Factorise the denominator, and hence write down the equations of the two vertical asymptotes of the graph of $f$. (b) State the equation of the horizontal asymptote of the graph of $f$, justifying your answer. (c) Find the coordinates of the point(s) where the graph of $f$ crosses the coordinate axes.

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(a) Factorising the denominator, we need two numbers multiplying to $6$ and adding to $-5$: these are $-2$ and $-3$. $$x^{2}-5x+6=(x-2)(x-3)$$ The vertical asymptotes occur where the denominator is zero (checking the numerator, $2x-1$, is non-zero there: at $x=2$, $2x-1=3\neq0$; at $x=3$, $2x-1=5\neq0$, so both are genuine asymptotes). $$\boxed{x=2\ \text{and}\ x=3}$$ (b) The degree of the numerator (1) is less than the degree of the denominator (2). As $x\to\pm\infty$, the denominator grows much faster than the numerator, so $f(x)\to0$. $$\boxed{y=0}$$ (c) The $y$-intercept is $f(0)$: $$f(0)=\frac{2(0)-1}{(0)^{2}-5(0)+6}=\frac{-1}{6}$$ so the point is $\left(0,-\dfrac{1}{6}\right)$. The $x$-intercept occurs where the numerator is zero (since the denominator is non-zero there): $$2x-1=0\implies x=\frac{1}{2}$$ so the point is $\left(\dfrac{1}{2},0\right)$. $$\boxed{\left(0,-\dfrac{1}{6}\right)\ \text{and}\ \left(\dfrac{1}{2},0\right)}$$

QUESTION 5 15 marks

Hard

Consider the function $f(x)=\dfrac{3x+7}{x+2}$, $x\neq-2$. (a) Write down the equations of the vertical and horizontal asymptotes of the graph of $f$. [3] (b) Express $f(x)$ in the form $A+\dfrac{B}{x+2}$, where $A,B\in\mathbb{Z}$, and hence state the range of $f$. [4] (c) The graph of $f$ is transformed to obtain the graph of $g$, where $g(x)=f(x-3)+1$. State the equations of the asymptotes of the graph of $g$, and express $g(x)$ as a single fraction in the form $\dfrac{px+q}{x-1}$, where $p,q\in\mathbb{Z}$. [5] (d) Solve the equation $g(x)=5$ algebraically. [3]

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(a) The vertical asymptote occurs where $x+2=0$, i.e. $x=-2$ (numerator at $x=-2$ is $3(-2)+7=1\neq0$, so genuine). Since numerator and denominator both have degree 1, the horizontal asymptote is the ratio of leading coefficients: $y=\dfrac{3}{1}=3$. $$\boxed{x=-2\ \text{and}\ y=3}$$ (b) Dividing the numerator: $3x+7=3(x+2)+1$, so $$f(x)=\frac{3(x+2)+1}{x+2}=3+\frac{1}{x+2}$$ So $A=3,\ B=1$. Since $\dfrac{1}{x+2}\neq0$ for any $x$ in the domain, $f(x)\neq3$ for all $x$. As $x$ ranges over all values except $-2$, $\dfrac{1}{x+2}$ takes every non-zero real value, so $f$ takes every real value except 3. $$\boxed{f(x)=3+\dfrac{1}{x+2};\ \text{range: } f(x)\in\mathbb{R},\ f(x)\neq3}$$ (c) $g(x)=f(x-3)+1$ shifts the graph of $f$ right by 3 (moving the vertical asymptote from $x=-2$ to $x=-2+3=1$) and up by 1 (moving the horizontal asymptote from $y=3$ to $y=3+1=4$). $$\boxed{\text{vertical asymptote: } x=1;\ \text{horizontal asymptote: } y=4}$$ Using the form from (b): $f(x-3)=3+\dfrac{1}{(x-3)+2}=3+\dfrac{1}{x-1}$, so $$g(x)=3+\frac{1}{x-1}+1=4+\frac{1}{x-1}$$ Writing as a single fraction: $$g(x)=\frac{4(x-1)+1}{x-1}=\frac{4x-4+1}{x-1}=\frac{4x-3}{x-1}$$ $$\boxed{g(x)=\dfrac{4x-3}{x-1}}$$ (so $p=4,\ q=-3$) (d) Using $g(x)=4+\dfrac{1}{x-1}$, solve $g(x)=5$: $$4+\frac{1}{x-1}=5$$ $$\frac{1}{x-1}=1$$ $$x-1=1$$ $$x=2$$ (This is in the domain, since $x\neq1$.) $$\boxed{x=2}$$

QUESTION 6 5 marks

Easy

Consider the function $f(x) = \dfrac{x^2+2x-1}{x+1}$, $x \ne -1$. (a) By performing polynomial division, express $f(x)$ in the form $f(x) = ax+b+\dfrac{c}{x+1}$, where $a,b,c \in \mathbb{Z}$. [3] (b) Hence write down the equations of the vertical asymptote and the oblique asymptote of the graph of $f$. [2]

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(a) Method: divide $x^2+2x-1$ by $x+1$ using polynomial long division. $$x^2+2x-1 = (x+1)(x+1) - 2$$ since $(x+1)(x+1) = x^2+2x+1$, and $x^2+2x-1 = (x^2+2x+1) - 2$. So: $$f(x) = \frac{(x+1)(x+1)-2}{x+1} = (x+1) - \frac{2}{x+1}$$ $$\boxed{f(x) = x+1-\dfrac{2}{x+1}}$$ (i.e. $a=1,\ b=1,\ c=-2$) (b) As $x \to -1$, the denominator $\to 0$ while the numerator remains non-zero there, so there is a vertical asymptote at $\boxed{x=-1}$. As $x \to \pm\infty$, $\dfrac{-2}{x+1} \to 0$, so $f(x) \to x+1$. Hence the oblique asymptote is $\boxed{y = x+1}$.

QUESTION 7 6 marks

Medium

Let $f(x) = \dfrac{2x^2-3x-5}{x-2}$, $x \ne 2$. (a) Find the equations of the vertical and oblique asymptotes of the graph of $f$. [4] (b) Determine whether the graph of $f$ intersects its oblique asymptote, justifying your answer algebraically. [2]

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(a) Method: perform polynomial division of $2x^2-3x-5$ by $x-2$. $$2x^2-3x-5 = (x-2)(2x+1) - 3$$ since $(x-2)(2x+1) = 2x^2+x-4x-2 = 2x^2-3x-2$, and $2x^2-3x-5 = (2x^2-3x-2)-3$. So $f(x) = 2x+1 - \dfrac{3}{x-2}$. Vertical asymptote: $\boxed{x=2}$ (denominator zero here, and numerator at $x=2$ is $8-6-5=-3\neq0$, so this is a genuine asymptote, not a hole). Oblique asymptote: as $x\to\pm\infty$, $\dfrac{3}{x-2}\to 0$, so $f(x)\to 2x+1$. $\boxed{y = 2x+1}$ (b) The graph intersects its oblique asymptote where $f(x) = 2x+1$, i.e. where $-\dfrac{3}{x-2}=0$. This equation has no solution, since a fraction with a non-zero numerator ($-3$) can never equal zero. $$\boxed{\text{The graph of } f \text{ never intersects its oblique asymptote.}}$$

QUESTION 8 6 marks

Easy

Consider the function $f(x) = \dfrac{x^3+2}{x}$, $x \ne 0$. (a) Show that $f(x) = x^2 + \dfrac{2}{x}$. [2] (b) State the equation of the vertical asymptote of the graph of $f$. [1] (c) Describe the behaviour of the graph of $f$ as $x \to \pm\infty$, identifying the non-linear curve that the graph approaches. [3]

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(a) Splitting the fraction into two terms: $$f(x) = \frac{x^3+2}{x} = \frac{x^3}{x}+\frac{2}{x} = x^2+\frac{2}{x}$$ $\blacksquare$ (b) As $x\to 0$, $\dfrac{2}{x}\to\pm\infty$, so there is a vertical asymptote at $\boxed{x=0}$. (c) As $x\to\pm\infty$, the term $\dfrac{2}{x}\to 0$, so $f(x) \to x^2$. $$\boxed{\text{The graph of } f \text{ approaches the parabola } y=x^2 \text{ as } x\to\pm\infty}$$ This is a non-linear (curved) asymptote, since the "distance" between the curve and $y=x^2$, namely $|2/x|$, tends to $0$ as $|x|\to\infty$, even though $y=x^2$ is not itself a straight line.

QUESTION 9 9 marks

Medium

Let $f(x) = \dfrac{x^3+x^2-4x+1}{x^2-1}$, $x \ne \pm1$. (a) Show that $f(x) = x+1+\dfrac{-3x+2}{x^2-1}$, and hence state the equation of the oblique asymptote of the graph of $f$. [4] (b) Write down the equations of the vertical asymptotes of the graph of $f$. [2] (c) Find the coordinates of the point at which the graph of $f$ intersects its oblique asymptote. [3]

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(a) Method: divide $x^3+x^2-4x+1$ by $x^2-1$, trying $f(x) = (x+1) + \dfrac{R(x)}{x^2-1}$. $$(x+1)(x^2-1) = x^3+x^2-x-1$$ $$R(x) = (x^3+x^2-4x+1) - (x^3+x^2-x-1) = -4x+1+x+1 = -3x+2$$ So $f(x) = x+1+\dfrac{-3x+2}{x^2-1}$. $\blacksquare$ As $x\to\pm\infty$, $\dfrac{-3x+2}{x^2-1}\to 0$ (degree of denominator exceeds degree of numerator), so $f(x)\to x+1$. $$\boxed{\text{Oblique asymptote: } y=x+1}$$ (b) The denominator $x^2-1=(x-1)(x+1)$ is zero at $x=1,-1$. Checking the numerator is non-zero there: at $x=1$, $1+1-4+1=-1\neq0$; at $x=-1$, $-1+1+4+1=5\neq0$. Both are genuine vertical asymptotes. $$\boxed{x=1 \text{ and } x=-1}$$ (c) The graph meets its oblique asymptote where $\dfrac{-3x+2}{x^2-1}=0$, i.e. where the numerator $-3x+2=0$. $$-3x+2=0 \implies x=\frac{2}{3}$$ At $x=\dfrac23$: $y = x+1 = \dfrac23+1=\dfrac53$. $$\boxed{\left(\dfrac{2}{3},\ \dfrac{5}{3}\right)}$$

QUESTION 10 14 marks

Hard

Let $f(x) = \dfrac{2x^2+bx+c}{x-1}$, $x\ne1$, where $b,c\in\mathbb{R}$. (a) Show that $f(x) = 2x+(b+2)+\dfrac{b+c+2}{x-1}$. [4] (b) Given that the oblique asymptote of the graph of $f$ is $y=2x+5$, find the value of $b$. [2] (c) Given further that $f(0)=-7$, find the value of $c$. [2] (d) Using your values of $b$ and $c$: state the equation of the vertical asymptote of the graph of $f$, and determine whether the graph intersects its oblique asymptote. [3] (e) Describe the behaviour of $f(x)$ as $x\to1^{-}$ and as $x\to1^{+}$ (state whether $f(x)\to+\infty$ or $f(x)\to-\infty$ in each case). [3]

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(a) Method: divide $2x^2+bx+c$ by $x-1$ using polynomial long division. First term $2x$, since $2x(x-1)=2x^2-2x$. Subtract: $(2x^2+bx+c)-(2x^2-2x) = (b+2)x+c$. Next term $(b+2)$, since $(b+2)(x-1)=(b+2)x-(b+2)$. Subtract: $(b+2)x+c - [(b+2)x-(b+2)] = c+b+2$. So $f(x) = 2x+(b+2)+\dfrac{b+c+2}{x-1}$. $\blacksquare$ (b) As $x\to\pm\infty$, the remainder term $\to0$, so $f(x)\to 2x+(b+2)$. This is the oblique asymptote, so: $$2x+(b+2) = 2x+5 \implies b+2=5 \implies \boxed{b=3}$$ (c) Substituting $x=0$ into $f(x)$: $$f(0) = \frac{2(0)^2+b(0)+c}{0-1} = \frac{c}{-1} = -c$$ Given $f(0)=-7$: $$-c=-7 \implies \boxed{c=7}$$ (d) With $b=3,\ c=7$: $f(x) = \dfrac{2x^2+3x+7}{x-1} = 2x+5+\dfrac{12}{x-1}$ (since $b+c+2=3+7+2=12$). Checking $x=1$ is not also a root of the numerator: $2(1)^2+3(1)+7=12\neq0$, so it is a genuine vertical asymptote. $$\boxed{\text{Vertical asymptote: } x=1}$$ The graph meets its oblique asymptote where $\dfrac{12}{x-1}=0$. Since the numerator $12\neq0$, this equation has no solution. $$\boxed{\text{The graph never intersects its oblique asymptote.}}$$ (e) $f(x) = 2x+5+\dfrac{12}{x-1}$. As $x\to1^{-}$: $(x-1)\to0^{-}$ (a small negative number), so $\dfrac{12}{x-1}\to-\infty$. Hence $\boxed{f(x)\to-\infty}$. As $x\to1^{+}$: $(x-1)\to0^{+}$ (a small positive number), so $\dfrac{12}{x-1}\to+\infty$. Hence $\boxed{f(x)\to+\infty}$.

QUESTION 11 5 marks

Easy

Consider the function $f(x)=\dfrac{5x+1}{x-3}$, $x\neq3$. Find (a) the equation of the vertical asymptote of the graph of $f$, (b) the equation of the horizontal asymptote of the graph of $f$, and (c) the $y$-intercept of the graph of $f$.

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(a) The vertical asymptote occurs where the denominator is zero, provided the numerator is not also zero there: $$x-3=0 \implies x=3$$ Checking the numerator at $x=3$: $5(3)+1=16\neq0$, so this is a genuine vertical asymptote (not a hole). $$\boxed{x=3}$$ (b) The numerator and denominator are both degree 1, so as $x\to\pm\infty$ the function approaches the ratio of the leading coefficients: $$y=\frac{5}{1}=5$$ $$\boxed{y=5}$$ (c) The $y$-intercept is $f(0)$: $$f(0)=\frac{5(0)+1}{0-3}=\frac{1}{-3}=-\frac{1}{3}$$ $$\boxed{y\text{-intercept}=-\dfrac{1}{3}}$$

QUESTION 12 5 marks

Easy

Consider the function $f(x)=\dfrac{4-2x}{x+6}$, $x\neq-6$. Find (a) the equation of the vertical asymptote of the graph of $f$, (b) the equation of the horizontal asymptote of the graph of $f$, and (c) the $x$-intercept of the graph of $f$.

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(a) The vertical asymptote occurs where the denominator is zero: $$x+6=0 \implies x=-6$$ At $x=-6$, the numerator is $4-2(-6)=16\neq0$, so this is a genuine vertical asymptote. $$\boxed{x=-6}$$ (b) Writing the numerator as $-2x+4$, both the numerator and denominator have degree 1. As $x\to\pm\infty$, $f(x)$ approaches the ratio of the leading coefficients: $$y=\frac{-2}{1}=-2$$ $$\boxed{y=-2}$$ (c) The $x$-intercept occurs where the numerator is zero (and the point is in the domain): $$4-2x=0 \implies x=2$$ Since $x=2\neq-6$, this point is in the domain. $$\boxed{x\text{-intercept}=2}$$

QUESTION 13 6 marks

Easy

Consider the function $f(x)=3-\dfrac{4}{x+2}$, $x\neq-2$. Find (a) the equations of the vertical and horizontal asymptotes of the graph of $f$, (b) the $y$-intercept, and (c) the $x$-intercept.

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(a) This is a transformation of $y=\dfrac{1}{x}$: the vertical asymptote occurs where $x+2=0$, and since $\dfrac{4}{x+2}\to0$ as $x\to\pm\infty$, $f(x)\to3$. $$\boxed{x=-2\ \text{and}\ y=3}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=3-\frac{4}{0+2}=3-2=1$$ $$\boxed{y\text{-intercept}=1}$$ (c) Set $f(x)=0$: $$3-\frac{4}{x+2}=0$$ $$\frac{4}{x+2}=3$$ $$x+2=\frac{4}{3}$$ $$x=\frac{4}{3}-2=-\frac{2}{3}$$ $$\boxed{x\text{-intercept}=-\dfrac{2}{3}}$$

QUESTION 14 6 marks

Easy

Consider the function $f(x)=\dfrac{x^{2}-9}{x-3}$, $x\neq3$. (a) Factorise the numerator, and hence show that $f(x)=x+3$ for all $x$ in the domain of $f$. [3] (b) Explain why the graph of $f$ does not have a vertical asymptote at $x=3$, and state the coordinates of the point that is missing from the graph. [3]

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(a) Factorising the numerator as a difference of two squares: $$x^{2}-9=(x-3)(x+3)$$ So for $x\neq3$: $$f(x)=\frac{(x-3)(x+3)}{x-3}=x+3$$ $$\boxed{f(x)=x+3}$$ (b) A vertical asymptote occurs only where the denominator is zero and the numerator is not zero at that same point. Here, at $x=3$ the numerator $x^{2}-9$ is also zero (since $3^{2}-9=0$), so the factor $(x-3)$ cancels. This means the discontinuity at $x=3$ is removable (a "hole"), not an asymptote - the graph is simply the line $y=x+3$ with a single point removed. The missing point has $x$-coordinate $3$, and its $y$-coordinate is found from the simplified form: $y=3+3=6$. $$\boxed{\text{no vertical asymptote; missing point } (3,6)}$$

QUESTION 15 5 marks

Easy

Consider the function $f(x)=\dfrac{2x+5}{3x-6}$, $x\neq2$. (a) State the equation of the vertical asymptote of the graph of $f$. [2] (b) State the equation of the horizontal asymptote of the graph of $f$. [2] (c) Write down the domain of $f$. [1]

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(a) The denominator is zero when $3x-6=0$, i.e. $x=2$. At $x=2$ the numerator is $2(2)+5=9\neq0$, so this is a genuine vertical asymptote. $$\boxed{x=2}$$ (b) The numerator and denominator both have degree 1, so the horizontal asymptote is the ratio of leading coefficients: $$y=\frac{2}{3}$$ $$\boxed{y=\dfrac{2}{3}}$$ (c) The function is undefined only where the denominator is zero, i.e. at $x=2$. $$\boxed{\text{domain: } x\in\mathbb{R},\ x\neq2}$$

QUESTION 16 6 marks

Easy

Consider the function $f(x)=\dfrac{1}{x^{2}-4}$, $x\neq\pm2$. (a) Factorise the denominator, and hence write down the equations of the two vertical asymptotes of the graph of $f$. [3] (b) State the equation of the horizontal asymptote of the graph of $f$. [2] (c) Find the $y$-intercept of the graph of $f$. [1]

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(a) Factorising the denominator as a difference of two squares: $$x^{2}-4=(x-2)(x+2)$$ This is zero when $x=2$ or $x=-2$. The numerator is the constant $1$, which is never zero, so both values give genuine vertical asymptotes. $$\boxed{x=2\ \text{and}\ x=-2}$$ (b) The numerator has degree $0$ and the denominator has degree $2$. Since the degree of the numerator is less than the degree of the denominator, $f(x)\to0$ as $x\to\pm\infty$. $$\boxed{y=0}$$ (c) The $y$-intercept is $f(0)$: $$f(0)=\frac{1}{0^{2}-4}=\frac{1}{-4}=-\frac{1}{4}$$ $$\boxed{y\text{-intercept}=-\dfrac{1}{4}}$$

QUESTION 17 6 marks

Easy

Consider the function $f(x)=\dfrac{x-4}{x^{2}-16}$, $x\neq\pm4$. (a) Factorise the denominator, and hence show that $f(x)=\dfrac{1}{x+4}$ for $x\neq4$. [3] (b) Explain why $x=4$ does not give a vertical asymptote, and state the equation of the vertical asymptote that the graph of $f$ does have. [3]

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(a) Factorising the denominator as a difference of two squares: $$x^{2}-16=(x-4)(x+4)$$ So for $x\neq4$ (and $x\neq-4$): $$f(x)=\frac{x-4}{(x-4)(x+4)}=\frac{1}{x+4}$$ $$\boxed{f(x)=\dfrac{1}{x+4}}$$ (b) At $x=4$, both the original numerator $x-4$ and the original denominator $x^{2}-16$ are zero, so the factor $(x-4)$ cancels - this is a removable discontinuity (a hole), not a vertical asymptote. At $x=-4$, the simplified denominator $x+4$ is zero while the numerator $1$ is not, so this point genuinely is a vertical asymptote. $$\boxed{\text{vertical asymptote: } x=-4\ \text{(no asymptote at } x=4\text{)}}$$

QUESTION 18 5 marks

Easy

Consider the function $f(x)=\dfrac{6x-3}{2x-1}$, $x\neq\dfrac{1}{2}$. (a) Factorise the numerator, and hence show that $f(x)=3$ for all $x$ in the domain of $f$. [3] (b) Describe the graph of $f$, including any point that must be excluded. [2]

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(a) Factorising the numerator: $$6x-3=3(2x-1)$$ So for $x\neq\dfrac{1}{2}$: $$f(x)=\frac{3(2x-1)}{2x-1}=3$$ $$\boxed{f(x)=3}$$ (b) Since $f(x)=3$ for every $x$ in the domain, the graph is the horizontal line $y=3$, with a single point removed where the function is undefined, at $x=\dfrac{1}{2}$. There is no vertical asymptote, because the factor causing the denominator to vanish cancels with an identical factor in the numerator. $$\boxed{\text{the line } y=3 \text{ with a hole at } \left(\tfrac{1}{2},3\right)}$$

QUESTION 19 5 marks

Easy

Solve the equation $\dfrac{x+1}{x-2}=3$ algebraically, giving your reasoning and stating why your solution is valid.

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The equation is defined provided $x\neq2$. Multiplying both sides by $(x-2)$: $$x+1=3(x-2)$$ $$x+1=3x-6$$ $$1+6=3x-x$$ $$7=2x$$ $$x=\frac{7}{2}$$ Checking validity: $x=\dfrac{7}{2}\neq2$, so this value is in the domain of the original equation. Substituting back: $f\left(\tfrac{7}{2}\right)=\dfrac{\frac{7}{2}+1}{\frac{7}{2}-2}=\dfrac{\frac{9}{2}}{\frac{3}{2}}=3$, confirming the solution. $$\boxed{x=\dfrac{7}{2}}$$

QUESTION 20 5 marks

Easy

Solve the equation $\dfrac{2}{x}=\dfrac{3}{x+1}$ algebraically.

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The equation requires $x\neq0$ and $x\neq-1$. Cross-multiplying: $$2(x+1)=3x$$ $$2x+2=3x$$ $$2=3x-2x$$ $$x=2$$ Checking validity: $x=2$ satisfies $x\neq0$ and $x\neq-1$, so it lies in the domain. Substituting back: LHS $=\dfrac{2}{2}=1$, RHS $=\dfrac{3}{3}=1$. The two sides agree, confirming the solution. $$\boxed{x=2}$$

QUESTION 21 6 marks

Easy

Consider the function $f(x)=\dfrac{x+7}{x+1}$, $x\neq-1$. (a) Write down the domain of $f$. [1] (b) State the equations of the vertical and horizontal asymptotes of the graph of $f$. [3] (c) Solve $f(x)=0$ to find the $x$-intercept of the graph of $f$. [2]

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(a) The function is undefined only where the denominator is zero, at $x=-1$. $$\boxed{\text{domain: } x\in\mathbb{R},\ x\neq-1}$$ (b) The vertical asymptote occurs at $x=-1$ (the numerator there is $-1+7=6\neq0$, so it is genuine). Since the numerator and denominator both have degree 1, the horizontal asymptote is the ratio of leading coefficients, $y=\dfrac{1}{1}=1$. $$\boxed{x=-1\ \text{and}\ y=1}$$ (c) Setting $f(x)=0$ means the numerator must be zero (with the point remaining in the domain): $$x+7=0 \implies x=-7$$ Since $-7\neq-1$, this is valid. $$\boxed{x\text{-intercept}=-7}$$

QUESTION 22 5 marks

Easy

Consider the function $g(x)=\dfrac{x-2}{x^{2}-5x+4}$. Factorise the denominator, and hence state the domain of $g$.

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Factorising the denominator by finding two numbers that multiply to $4$ and add to $-5$ (namely $-1$ and $-4$): $$x^{2}-5x+4=(x-1)(x-4)$$ So $g$ is undefined when $(x-1)(x-4)=0$, i.e. when $x=1$ or $x=4$. (The numerator $x-2$ is not zero at either of these values, so both are genuinely excluded - though this is not needed to state the domain, only to confirm each is a true vertical asymptote.) $$\boxed{\text{domain: } x\in\mathbb{R},\ x\neq1,\ x\neq4}$$

QUESTION 23 6 marks

Easy

Consider the function $f(x)=1-\dfrac{2}{x-5}$, $x\neq5$. (a) State the equations of the vertical and horizontal asymptotes of the graph of $f$. [2] (b) Find the $y$-intercept and the $x$-intercept of the graph of $f$. [4]

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(a) This is a transformation of $y=\dfrac{1}{x}$: the vertical asymptote occurs where $x-5=0$, and since $\dfrac{2}{x-5}\to0$ as $x\to\pm\infty$, $f(x)\to1$. $$\boxed{x=5\ \text{and}\ y=1}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=1-\frac{2}{0-5}=1+\frac{2}{5}=\frac{7}{5}$$ For the $x$-intercept, set $f(x)=0$: $$1-\frac{2}{x-5}=0$$ $$\frac{2}{x-5}=1$$ $$x-5=2$$ $$x=7$$ $$\boxed{y\text{-intercept}=\dfrac{7}{5};\ x\text{-intercept}=7}$$

QUESTION 24 6 marks

Easy

Solve the inequality $\dfrac{x-3}{x+1}>0$ algebraically, using a sign diagram.

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The expression $\dfrac{x-3}{x+1}$ changes sign only where the numerator or denominator is zero, i.e. at the critical values $x=3$ and $x=-1$. These split the number line into three intervals: $x<-1$, $-13$. Testing a point in each interval: $x=-2$ (in $x<-1$): $\dfrac{-2-3}{-2+1}=\dfrac{-5}{-1}=5>0$ $x=0$ (in $-13$): $\dfrac{4-3}{4+1}=\dfrac{1}{5}>0$ So the expression is positive for $x<-1$ or $x>3$. Both critical values are excluded: $x=-1$ makes the expression undefined, and $x=3$ makes it equal to zero (not strictly greater than zero). $$\boxed{x<-1\ \text{or}\ x>3}$$

QUESTION 25 7 marks

Easy

Solve the inequality $\dfrac{1}{x-2}\leq3$ algebraically.

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It is tempting to multiply both sides by $(x-2)$, but this is unsafe because $(x-2)$ could be negative, which would reverse the inequality. Instead, rearrange so everything is on one side and combine into a single fraction: $$\frac{1}{x-2}-3\leq0$$ $$\frac{1-3(x-2)}{x-2}\leq0$$ $$\frac{1-3x+6}{x-2}\leq0$$ $$\frac{7-3x}{x-2}\leq0$$ The critical values are where the numerator or denominator is zero: $7-3x=0 \implies x=\dfrac{7}{3}$, and $x-2=0 \implies x=2$. These split the number line into three intervals. Testing a point in each interval: $x=0$ (in $x<2$): $\dfrac{7-0}{0-2}=\dfrac{7}{-2}<0$ (satisfies) $x=2.2$ (in $20$ (fails) $x=3$ (in $x>\frac{7}{3}$): $\dfrac{7-9}{3-2}=\dfrac{-2}{1}<0$ (satisfies) So the expression is $\leq0$ for $x<2$ or $x\geq\dfrac{7}{3}$. Note $x=2$ is excluded (undefined), while $x=\dfrac{7}{3}$ is included since it gives equality. $$\boxed{x<2\ \text{or}\ x\geq\dfrac{7}{3}}$$

QUESTION 26 6 marks

Easy

Consider the function $f(x)=\dfrac{3x}{x^{2}+1}$. (a) Explain why the graph of $f$ has no vertical asymptote. [2] (b) State the domain of $f$ and the equation of its horizontal asymptote. [3] (c) Find the $y$-intercept of the graph of $f$. [1]

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(a) A vertical asymptote can only occur where the denominator is zero. Here $x^{2}+1\geq1>0$ for every real $x$ (a square plus $1$ can never be zero or negative), so the denominator never equals zero and there is no vertical asymptote. $$\boxed{\text{no vertical asymptote, since } x^{2}+1\neq0\ \forall x\in\mathbb{R}}$$ (b) Since the denominator is never zero, $f$ is defined for every real number: $$\boxed{\text{domain: } x\in\mathbb{R}}$$ The numerator has degree 1 and the denominator has degree 2. Since the degree of the numerator is less than that of the denominator, $f(x)\to0$ as $x\to\pm\infty$. $$\boxed{y=0}$$ (c) The $y$-intercept is $f(0)$: $$f(0)=\frac{3(0)}{0^{2}+1}=0$$ $$\boxed{y\text{-intercept}=0}$$

QUESTION 27 8 marks

Medium

Consider the function $f(x)=\dfrac{x^{2}+x-6}{x-2}$, $x\neq2$. (a) Factorise the numerator completely. [2] (b) Hence show that the graph of $f$ has no vertical asymptote, and simplify $f(x)$ to a linear expression. [3] (c) State the coordinates of the point excluded from the graph of $f$. [3]

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(a) We require two numbers that multiply to $-6$ and add to $1$: these are $3$ and $-2$. $$x^{2}+x-6=(x+3)(x-2)$$ $$\boxed{(x+3)(x-2)}$$ (b) A vertical asymptote requires the denominator to be zero while the numerator is non-zero at that point. Here the denominator is zero at $x=2$, but the factorised numerator shows the numerator is also zero at $x=2$ (since $(2+3)(2-2)=0$). The factor $(x-2)$ therefore cancels: $$f(x)=\frac{(x+3)(x-2)}{x-2}=x+3,\quad x\neq2$$ Since the point $x=2$ is only a removable discontinuity, not a genuine asymptote, the graph of $f$ has no vertical asymptote. $$\boxed{f(x)=x+3;\ \text{no vertical asymptote}}$$ (c) The excluded point occurs at $x=2$. Using the simplified form, its $y$-coordinate is $2+3=5$. $$\boxed{(2,5)}$$

QUESTION 28 9 marks

Medium

Consider the function $f(x)=\dfrac{2x^{2}-x-3}{x-1}$, $x\neq1$. (a) Show that $x=1$ gives a genuine vertical asymptote of the graph of $f$ (that is, show the numerator is not also zero at $x=1$). [2] (b) By performing polynomial division, express $f(x)$ in the form $f(x)=ax+b+\dfrac{c}{x-1}$, where $a,b,c\in\mathbb{Z}$. [5] (c) Hence state the equation of the oblique asymptote of the graph of $f$. [2]

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(a) Substituting $x=1$ into the numerator: $$2(1)^{2}-1-3=2-1-3=-2\neq0$$ Since the numerator is non-zero while the denominator is zero at $x=1$, this is a genuine vertical asymptote. $$\boxed{x=1\ \text{is a genuine vertical asymptote}}$$ (b) Dividing $2x^{2}-x-3$ by $x-1$: $$2x^{2}\div x=2x,\qquad 2x(x-1)=2x^{2}-2x$$ Subtract: $(2x^{2}-x-3)-(2x^{2}-2x)=x-3$ $$x\div x=1,\qquad 1(x-1)=x-1$$ Subtract: $(x-3)-(x-1)=-2$ So $2x^{2}-x-3=(x-1)(2x+1)-2$, giving $$f(x)=2x+1-\frac{2}{x-1}$$ i.e. $a=2,\ b=1,\ c=-2$. $$\boxed{f(x)=2x+1-\dfrac{2}{x-1}}$$ (c) As $x\to\pm\infty$, $\dfrac{-2}{x-1}\to0$, so $f(x)\to2x+1$. $$\boxed{y=2x+1}$$

QUESTION 29 9 marks

Medium

Consider the function $f(x)=\dfrac{x^{2}-2x-8}{x^{2}-x-2}$. (a) Factorise the numerator and the denominator completely. [3] (b) Hence state the equations of the two vertical asymptotes of the graph of $f$, explaining why neither is removable. [3] (c) State the equation of the horizontal asymptote, and find the $x$- and $y$-intercepts of the graph of $f$. [3]

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(a) Numerator: two numbers multiplying to $-8$ and adding to $-2$ are $-4$ and $2$: $$x^{2}-2x-8=(x-4)(x+2)$$ Denominator: two numbers multiplying to $-2$ and adding to $-1$ are $-2$ and $1$: $$x^{2}-x-2=(x-2)(x+1)$$ $$\boxed{f(x)=\dfrac{(x-4)(x+2)}{(x-2)(x+1)}}$$ (b) The numerator and denominator share no common factor, so no cancellation occurs. The denominator is zero at $x=2$ and $x=-1$; substituting into the numerator, $(2-4)(2+2)=-8\neq0$ and $(-1-4)(-1+2)=-5\neq0$, so both give genuine vertical asymptotes. $$\boxed{x=2\ \text{and}\ x=-1}$$ (c) The numerator and denominator both have degree 2 with leading coefficient $1$, so the horizontal asymptote is $y=\dfrac{1}{1}=1$. The $x$-intercepts occur where the numerator is zero: $x=4$ and $x=-2$ (both in the domain). The $y$-intercept is $f(0)=\dfrac{(0-4)(0+2)}{(0-2)(0+1)}=\dfrac{-8}{-2}=4$. $$\boxed{y=1;\ x\text{-intercepts: } x=4,\ x=-2;\ y\text{-intercept}=4}$$

QUESTION 30 9 marks

Medium

Consider the function $f(x)=\dfrac{x^{2}-5x+6}{x^{2}-9}$. (a) Factorise the numerator and the denominator completely. [3] (b) Show that the graph of $f$ has only one vertical asymptote (rather than two), and state its equation. [3] (c) State the equation of the horizontal asymptote, and the coordinates of the point excluded from the graph. [3]

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(a) Numerator: two numbers multiplying to $6$ and adding to $-5$ are $-2$ and $-3$: $$x^{2}-5x+6=(x-2)(x-3)$$ Denominator, as a difference of two squares: $$x^{2}-9=(x-3)(x+3)$$ $$\boxed{f(x)=\dfrac{(x-2)(x-3)}{(x-3)(x+3)}}$$ (b) The factor $(x-3)$ appears in both the numerator and denominator, so it cancels: $$f(x)=\frac{x-2}{x+3},\quad x\neq3$$ This means $x=3$ is only a removable discontinuity, not a vertical asymptote. The only genuine vertical asymptote is where the simplified denominator is zero, i.e. $x+3=0$; the numerator there is $-3-2=-5\neq0$, confirming it is genuine. $$\boxed{x=-3}$$ (c) Using the simplified form $\dfrac{x-2}{x+3}$, both numerator and denominator have degree 1, so the horizontal asymptote is $y=\dfrac{1}{1}=1$. The excluded point is at $x=3$; its $y$-value (from the simplified form) is $\dfrac{3-2}{3+3}=\dfrac{1}{6}$. $$\boxed{y=1;\ \text{excluded point} \left(3,\dfrac{1}{6}\right)}$$

QUESTION 31 8 marks

Medium

Consider the function $f(x)=\dfrac{x^{2}+3x-1}{x-2}$, $x\neq2$. (a) State the equation of the vertical asymptote of the graph of $f$. [2] (b) By performing polynomial division, express $f(x)$ in the form $f(x)=x+a+\dfrac{b}{x-2}$, where $a,b\in\mathbb{Z}$, and hence state the equation of the oblique asymptote. [6]

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(a) The denominator is zero at $x=2$. The numerator there is $2^{2}+3(2)-1=4+6-1=9\neq0$, so this is a genuine vertical asymptote. $$\boxed{x=2}$$ (b) Dividing $x^{2}+3x-1$ by $x-2$: $$x^{2}\div x=x,\qquad x(x-2)=x^{2}-2x$$ Subtract: $(x^{2}+3x-1)-(x^{2}-2x)=5x-1$ $$5x\div x=5,\qquad 5(x-2)=5x-10$$ Subtract: $(5x-1)-(5x-10)=9$ So $x^{2}+3x-1=(x-2)(x+5)+9$, giving $$f(x)=x+5+\frac{9}{x-2}$$ i.e. $a=5,\ b=9$. As $x\to\pm\infty$, $\dfrac{9}{x-2}\to0$, so $f(x)\to x+5$. $$\boxed{f(x)=x+5+\dfrac{9}{x-2};\ \text{oblique asymptote: } y=x+5}$$

QUESTION 32 8 marks

Medium

Solve the equation $\dfrac{3}{x-2}-\dfrac{2}{x+1}=1$ algebraically, checking that any solutions are valid.

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The equation requires $x\neq2$ and $x\neq-1$. Combining the left-hand side over a common denominator $(x-2)(x+1)$: $$\frac{3(x+1)-2(x-2)}{(x-2)(x+1)}=1$$ $$\frac{3x+3-2x+4}{(x-2)(x+1)}=1$$ $$\frac{x+7}{(x-2)(x+1)}=1$$ Multiplying both sides by $(x-2)(x+1)$: $$x+7=(x-2)(x+1)$$ $$x+7=x^{2}-x-2$$ $$0=x^{2}-x-2-x-7$$ $$0=x^{2}-2x-9$$ Using the quadratic formula with $a=1,\ b=-2,\ c=-9$: $$x=\frac{2\pm\sqrt{4+36}}{2}=\frac{2\pm\sqrt{40}}{2}=1\pm\sqrt{10}$$ Checking validity: $1+\sqrt{10}\approx4.16$ and $1-\sqrt{10}\approx-2.16$; neither equals $2$ or $-1$, so both solutions lie in the domain. $$\boxed{x=1+\sqrt{10}\ \text{or}\ x=1-\sqrt{10}}$$

QUESTION 33 9 marks

Medium

Solve the inequality $\dfrac{x^{2}-4}{x-1}\geq0$ algebraically, using a sign diagram.

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Factorising the numerator: $x^{2}-4=(x-2)(x+2)$, so the expression is $\dfrac{(x-2)(x+2)}{x-1}$. The critical values are where the numerator or denominator is zero: $x=-2,\ x=1,\ x=2$. These split the number line into four intervals. Testing a point in each interval: $x=-3$ (in $x<-2$): $\dfrac{(-5)(-1)}{-4}=\dfrac{5}{-4}<0$ $x=0$ (in $-20$ $x=1.5$ (in $12$): $\dfrac{(1)(5)}{2}=2.5>0$ So the expression is $\geq0$ on $-2\leq x<1$ or $x\geq2$. The values $x=-2$ and $x=2$ are included (they give equality, and are in the domain), while $x=1$ is excluded (undefined). $$\boxed{-2\leq x<1\ \text{or}\ x\geq2}$$

QUESTION 34 9 marks

Medium

Solve the inequality $\dfrac{2x-1}{x+3}<1$ algebraically.

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Rearranging so all terms are on one side, then combining into a single fraction: $$\frac{2x-1}{x+3}-1<0$$ $$\frac{2x-1-(x+3)}{x+3}<0$$ $$\frac{x-4}{x+3}<0$$ The critical values are $x=4$ and $x=-3$, splitting the number line into three intervals. Testing a point in each interval: $x=-4$ (in $x<-3$): $\dfrac{-8}{-1}=8>0$ $x=0$ (in $-34$): $\dfrac{1}{8}>0$ So the expression is negative for $-3

QUESTION 35 8 marks

Medium

Consider $f(x)=\dfrac{5x-1}{(x-1)(x+2)}$, $x\neq1,-2$. (a) Write down the equations of the two vertical asymptotes of the graph of $f$. [2] (b) Express $f(x)$ in the form $\dfrac{A}{x-1}+\dfrac{B}{x+2}$, where $A,B\in\mathbb{Q}$. [6]

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(a) The denominator is zero when $x=1$ or $x=-2$. At these values the numerator is $5(1)-1=4\neq0$ and $5(-2)-1=-11\neq0$ respectively, so both are genuine vertical asymptotes. $$\boxed{x=1\ \text{and}\ x=-2}$$ (b) Write $$\frac{5x-1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}$$ Multiplying both sides by $(x-1)(x+2)$: $$5x-1=A(x+2)+B(x-1)$$ Substituting $x=1$: $\ 5(1)-1=A(1+2) \implies 4=3A \implies A=\dfrac{4}{3}$ Substituting $x=-2$: $\ 5(-2)-1=B(-2-1) \implies -11=-3B \implies B=\dfrac{11}{3}$ Checking: $\dfrac{4}{3}(x+2)+\dfrac{11}{3}(x-1)=\dfrac{4x+8+11x-11}{3}=\dfrac{15x-3}{3}=5x-1$ (satisfies), which matches the original numerator. $$\boxed{f(x)=\dfrac{4/3}{x-1}+\dfrac{11/3}{x+2}}$$

QUESTION 36 9 marks

Medium

Let $f(x)=\dfrac{7x+6}{x^{2}+x-2}$, $x\neq1,-2$. (a) Factorise the denominator, and hence express $f(x)$ in the form $\dfrac{A}{x-1}+\dfrac{B}{x+2}$, where $A,B\in\mathbb{Q}$. [6] (b) Hence solve the equation $f(x)=0$ algebraically, using your partial-fraction form. [3]

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(a) Factorising: $x^{2}+x-2=(x-1)(x+2)$. Write $$\frac{7x+6}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}$$ Multiplying both sides by $(x-1)(x+2)$: $$7x+6=A(x+2)+B(x-1)$$ Substituting $x=1$: $\ 7(1)+6=13=A(1+2) \implies A=\dfrac{13}{3}$ Substituting $x=-2$: $\ 7(-2)+6=-8=B(-2-1) \implies B=\dfrac{8}{3}$ Checking: $\dfrac{13}{3}(x+2)+\dfrac{8}{3}(x-1)=\dfrac{13x+26+8x-8}{3}=\dfrac{21x+18}{3}=7x+6$ (satisfies). $$\boxed{f(x)=\dfrac{13/3}{x-1}+\dfrac{8/3}{x+2}}$$ (b) Setting $f(x)=0$ using the decomposed form: $$\frac{13/3}{x-1}+\frac{8/3}{x+2}=0$$ Multiplying through by $3$: $$\frac{13}{x-1}=-\frac{8}{x+2}$$ Cross-multiplying: $$13(x+2)=-8(x-1)$$ $$13x+26=-8x+8$$ $$21x=-18$$ $$x=-\frac{18}{21}=-\frac{6}{7}$$ Checking: $-\dfrac{6}{7}\neq1$ and $\neq-2$, so this is valid. (This agrees with setting the original numerator $7x+6=0$ directly, which also gives $x=-\dfrac{6}{7}$.) $$\boxed{x=-\dfrac{6}{7}}$$

QUESTION 37 8 marks

Medium

The function $f(x)=\dfrac{ax+3}{x-b}$, where $a,b\in\mathbb{R}$, has a vertical asymptote at $x=4$ and a horizontal asymptote at $y=\dfrac{3}{2}$. (a) Write down the value of $b$. [1] (b) Find the value of $a$. [3] (c) Hence find the $x$-intercept of the graph of $f$. [3]

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(a) The vertical asymptote occurs where the denominator $x-b$ is zero, i.e. at $x=b$. Since this must equal $4$: $$\boxed{b=4}$$ (b) Since the numerator and denominator both have degree 1, the horizontal asymptote is the ratio of leading coefficients, $y=\dfrac{a}{1}=a$. Setting this equal to $\dfrac{3}{2}$: $$\boxed{a=\dfrac{3}{2}}$$ (c) So $f(x)=\dfrac{\frac{3}{2}x+3}{x-4}$. The $x$-intercept occurs where the numerator is zero: $$\frac{3}{2}x+3=0$$ $$\frac{3}{2}x=-3$$ $$x=-2$$ Since $-2\neq4$, this is a valid intercept. $$\boxed{x\text{-intercept}=-2}$$

QUESTION 38 8 marks

Medium

The function $f(x)=\dfrac{x+p}{x^{2}-9}$, where $p\in\mathbb{R}$, has a removable discontinuity (a hole, not a vertical asymptote) at $x=3$. (a) Find the value of $p$. [3] (b) Using this value of $p$, simplify $f(x)$ and state the equation of the (single) vertical asymptote of the graph of $f$. [5]

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(a) The denominator factorises as $x^{2}-9=(x-3)(x+3)$. For the discontinuity at $x=3$ to be removable, the numerator must also be zero at $x=3$ (so that the factor $(x-3)$ cancels): $$3+p=0$$ $$\boxed{p=-3}$$ (b) With $p=-3$, the numerator is $x-3$, so $$f(x)=\frac{x-3}{(x-3)(x+3)}=\frac{1}{x+3},\quad x\neq3$$ The remaining vertical asymptote occurs where $x+3=0$; the numerator ($1$) is never zero, so this is genuine. $$\boxed{f(x)=\dfrac{1}{x+3};\ \text{vertical asymptote: } x=-3}$$

QUESTION 39 9 marks

Medium

The concentration, in mg/L, of a drug in a patient's bloodstream $t$ hours after an intravenous dose begins is modelled by $$C(t)=\frac{50t}{t+5},\qquad t\geq0$$ (a) Find $C(0)$, and interpret its meaning in context. [2] (b) State the equation of the horizontal asymptote of the graph of $C$, and interpret its meaning in context. [3] (c) Find the time $t$ at which the concentration reaches $40$ mg/L. [4]

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(a) $$C(0)=\frac{50(0)}{0+5}=0$$ This means that at the moment the dose begins ($t=0$), the concentration in the bloodstream is $0$ mg/L. $$\boxed{C(0)=0\ \text{mg/L}}$$ (b) The numerator and denominator both have degree 1 (in $t$), so as $t\to\infty$, $C(t)$ approaches the ratio of leading coefficients: $$y=\frac{50}{1}=50$$ This means that as time goes on, the concentration in the bloodstream levels off, approaching but never quite reaching $50$ mg/L. $$\boxed{y=50\ \text{mg/L (long-term concentration)}}$$ (c) Setting $C(t)=40$: $$\frac{50t}{t+5}=40$$ $$50t=40(t+5)$$ $$50t=40t+200$$ $$10t=200$$ $$t=20$$ Since $t=20\geq0$, this is a valid time. $$\boxed{t=20\ \text{hours}}$$

QUESTION 40 9 marks

Medium

A company manufactures phone cases. The average cost, in dollars, of producing $x$ cases is modelled by $$A(x)=\frac{2000+15x}{x},\qquad x>0$$ where $2000$ is the fixed setup cost and $15$ is the cost of materials per case. (a) State the equation of the vertical asymptote of the graph of $A$, and explain why it is not relevant to the physical context (given $x>0$). [2] (b) State the equation of the horizontal asymptote of the graph of $A$, and interpret its meaning in context. [3] (c) Find the number of cases that must be produced for the average cost to be $\$25$ per case. [4]

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(a) The vertical asymptote occurs where the denominator $x=0$. Although this is a genuine vertical asymptote of the algebraic function, it is not physically relevant here because the model only applies for $x>0$ (a positive number of cases must be produced). $$\boxed{x=0}$$ (b) The numerator and denominator both have degree 1, so as $x\to\infty$, $A(x)\to\dfrac{15}{1}=15$. This means that as more cases are produced, the fixed setup cost is spread across more units, so the average cost per case approaches, but never falls below, $\$15$. $$\boxed{y=15}$$ (c) Setting $A(x)=25$: $$\frac{2000+15x}{x}=25$$ $$2000+15x=25x$$ $$2000=25x-15x$$ $$2000=10x$$ $$x=200$$ Since $x=200>0$, this is valid in context. $$\boxed{200\ \text{cases}}$$

QUESTION 41 10 marks

Medium

Consider the function $f(x)=\dfrac{2x-4}{x+1}$, $x\neq-1$. (a) State the equations of the vertical and horizontal asymptotes of the graph of $f$. [2] (b) Find the $x$- and $y$-intercepts of the graph of $f$. [3] (c) Determine algebraically, using a sign diagram, the set of values of $x$ for which $f(x)<0$. [5]

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(a) The vertical asymptote is where $x+1=0$, i.e. $x=-1$ (numerator there is $2(-1)-4=-6\neq0$, so genuine). Since numerator and denominator both have degree 1, the horizontal asymptote is $y=\dfrac{2}{1}=2$. $$\boxed{x=-1\ \text{and}\ y=2}$$ (b) The $x$-intercept occurs where the numerator is zero: $2x-4=0 \implies x=2$. The $y$-intercept is $f(0)=\dfrac{2(0)-4}{0+1}=-4$. $$\boxed{x\text{-intercept}=2;\ y\text{-intercept}=-4}$$ (c) The critical values of $\dfrac{2x-4}{x+1}$ are where the numerator or denominator is zero: $x=2$ and $x=-1$. These split the number line into three intervals. Testing a point in each: $x=-2$ (in $x<-1$): $\dfrac{-8}{-1}=8>0$ $x=0$ (in $-12$): $\dfrac{2}{4}=0.5>0$ So $f(x)<0$ precisely on the interval $-1

QUESTION 42 9 marks

Medium

Solve the equation $\dfrac{x}{x+2}=\dfrac{8}{x^{2}-4}$ algebraically, checking that any solutions found are valid.

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Factorising: $x^{2}-4=(x-2)(x+2)$, so the equation requires $x\neq2$ and $x\neq-2$. Multiplying both sides by $(x-2)(x+2)$: $$x(x-2)=8$$ $$x^{2}-2x-8=0$$ Factorising: $$(x-4)(x+2)=0$$ $$x=4\ \text{or}\ x=-2$$ Checking validity: $x=4$ is not equal to $2$ or $-2$, so it lies in the domain. However $x=-2$ is excluded from the domain (it makes both the original left-hand side and $x^{2}-4$ zero), so it is an extraneous solution introduced by multiplying through and must be rejected. $$\boxed{x=4\ \text{only}}$$

QUESTION 43 13 marks

Hard

Consider the function $f(x)=\dfrac{2x^{2}+x-1}{x-3}$, $x\neq3$. (a) Show that $x=3$ is a genuine vertical asymptote of the graph of $f$, and by performing polynomial division, express $f(x)$ in the form $f(x)=2x+a+\dfrac{b}{x-3}$, where $a,b\in\mathbb{Z}$. Hence state the equation of the oblique asymptote. [5] (b) Write down the domain of $f$, and find the $y$-intercept of the graph of $f$. [4] (c) By considering the sign of $f(x)-(2x+a)$, determine algebraically the values of $x$ for which the graph of $f$ lies above its oblique asymptote. [4]

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(a) Substituting $x=3$ into the numerator: $2(3)^{2}+3-1=18+3-1=20\neq0$, so $x=3$ is a genuine vertical asymptote (the denominator is zero there but the numerator is not). Dividing $2x^{2}+x-1$ by $x-3$: $$2x^{2}\div x=2x,\qquad 2x(x-3)=2x^{2}-6x$$ Subtract: $(2x^{2}+x-1)-(2x^{2}-6x)=7x-1$ $$7x\div x=7,\qquad 7(x-3)=7x-21$$ Subtract: $(7x-1)-(7x-21)=20$ So $2x^{2}+x-1=(x-3)(2x+7)+20$, giving $$f(x)=2x+7+\frac{20}{x-3}$$ i.e. $a=7,\ b=20$. As $x\to\pm\infty$, $\dfrac{20}{x-3}\to0$, so $f(x)\to2x+7$. $$\boxed{f(x)=2x+7+\dfrac{20}{x-3};\ \text{oblique asymptote: } y=2x+7}$$ (b) The domain excludes only $x=3$: $$\boxed{\text{domain: } x\in\mathbb{R},\ x\neq3}$$ The $y$-intercept is $f(0)=\dfrac{2(0)^{2}+0-1}{0-3}=\dfrac{-1}{-3}=\dfrac{1}{3}$. $$\boxed{y\text{-intercept}=\dfrac{1}{3}}$$ (c) Using the form from (a), the vertical distance between the graph and the oblique asymptote $y=2x+7$ is $$f(x)-(2x+7)=\frac{20}{x-3}$$ The graph lies above its oblique asymptote exactly when this quantity is positive: $$\frac{20}{x-3}>0$$ Since $20>0$, this requires $x-3>0$, i.e. $x>3$. $$\boxed{x>3}$$

QUESTION 44 13 marks

Hard

Consider the function $f(x)=\dfrac{x^{2}+ax+b}{x-2}$, $x\neq2$, where $a,b\in\mathbb{R}$. It is known that the oblique asymptote of the graph of $f$ is $y=x+5$, and that $f(0)=-7$. (a) By performing polynomial division, express $f(x)$ in the form $x+(a+2)+\dfrac{c}{x-2}$ for some constant $c$ in terms of $a$ and $b$. Hence use the given oblique asymptote to find the value of $a$. [6] (b) Use $f(0)=-7$ to find the value of $b$. [3] (c) Confirm that $x=2$ is a genuine vertical asymptote, and show that the graph of $f$ has no $x$-intercepts. [4]

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(a) Dividing $x^{2}+ax+b$ by $x-2$: $$x^{2}\div x=x,\qquad x(x-2)=x^{2}-2x$$ Subtract: $(x^{2}+ax+b)-(x^{2}-2x)=(a+2)x+b$ $$(a+2)x\div x=(a+2),\qquad (a+2)(x-2)=(a+2)x-2(a+2)$$ Subtract: $[(a+2)x+b]-[(a+2)x-2(a+2)]=b+2(a+2)=b+2a+4$ So $$f(x)=x+(a+2)+\frac{b+2a+4}{x-2}$$ As $x\to\pm\infty$, the fractional term vanishes, so the oblique asymptote is $y=x+(a+2)$. Setting this equal to the given $y=x+5$: $$a+2=5$$ $$\boxed{a=3}$$ (b) With $a=3$, $f(x)=\dfrac{x^{2}+3x+b}{x-2}$. Using $f(0)=-7$: $$f(0)=\frac{0+0+b}{0-2}=\frac{b}{-2}=-7$$ $$b=14$$ $$\boxed{b=14}$$ (c) So $f(x)=\dfrac{x^{2}+3x+14}{x-2}$. Substituting $x=2$ into the numerator: $2^{2}+3(2)+14=4+6+14=24\neq0$, so $x=2$ is confirmed as a genuine vertical asymptote (not removable). For $x$-intercepts, we need $x^{2}+3x+14=0$. Using the discriminant: $$\Delta=3^{2}-4(1)(14)=9-56=-47<0$$ Since the discriminant is negative, this quadratic has no real roots, so the numerator is never zero for real $x$. Therefore the graph of $f$ has no $x$-intercepts. $$\boxed{f(x)=\dfrac{x^{2}+3x+14}{x-2};\ \text{no real } x\text{-intercepts since } \Delta=-47<0}$$

QUESTION 45 12 marks

Hard

Consider the inequality $\dfrac{x}{x-1}\geq\dfrac{2}{x+2}$, $x\neq1,-2$. (a) Show that the inequality can be rewritten as $\dfrac{x^{2}+2}{(x-1)(x+2)}\geq0$. [5] (b) Explain why $x^{2}+2$ is positive for all real $x$, and hence show that the inequality reduces to $(x-1)(x+2)>0$. [3] (c) Hence solve the original inequality. [4]

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(a) Moving all terms to one side: $$\frac{x}{x-1}-\frac{2}{x+2}\geq0$$ Combining over the common denominator $(x-1)(x+2)$: $$\frac{x(x+2)-2(x-1)}{(x-1)(x+2)}\geq0$$ Expanding the numerator: $$x(x+2)-2(x-1)=x^{2}+2x-2x+2=x^{2}+2$$ $$\boxed{\frac{x^{2}+2}{(x-1)(x+2)}\geq0}$$ (b) Since $x^{2}\geq0$ for every real $x$, we have $x^{2}+2\geq2>0$ always - the numerator can never be zero or negative. Because the numerator is always strictly positive, the sign of the whole fraction is determined entirely by the sign of the denominator: the fraction is positive exactly when $(x-1)(x+2)>0$, and it can never equal zero (since the numerator is never zero). So the inequality $\dfrac{x^{2}+2}{(x-1)(x+2)}\geq0$ holds precisely when $$\boxed{(x-1)(x+2)>0}$$ (c) The critical values of $(x-1)(x+2)$ are $x=1$ and $x=-2$, splitting the number line into three intervals. Testing a point in each: $x=-3$ (in $x<-2$): $(-4)(-1)=4>0$ (satisfies) $x=0$ (in $-21$): $(1)(4)=4>0$ (satisfies) So $(x-1)(x+2)>0$ for $x<-2$ or $x>1$. Both endpoints are excluded from the domain of the original inequality in any case. $$\boxed{x<-2\ \text{or}\ x>1}$$

QUESTION 46 12 marks

Hard

Consider the function $f(x)=\dfrac{2x^{2}+3x-5}{x^{2}-x-2}$. (a) Factorise the denominator, and hence write down the equations of the two vertical asymptotes of the graph of $f$. [3] (b) By performing polynomial division, show that $f(x)=2+\dfrac{5x-1}{x^{2}-x-2}$. [3] (c) Express $\dfrac{5x-1}{x^{2}-x-2}$ in partial fractions of the form $\dfrac{A}{x-2}+\dfrac{B}{x+1}$, where $A,B\in\mathbb{Z}$, and hence write $f(x)$ as the sum of an integer and two fractions. [4] (d) State the equation of the horizontal asymptote of the graph of $f$, and hence solve the equation $f(x)=2$ algebraically. [3]

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(a) Factorising: two numbers multiplying to $-2$ and adding to $-1$ are $-2$ and $1$: $$x^{2}-x-2=(x-2)(x+1)$$ At $x=2$: numerator $=2(4)+6-5=9\neq0$. At $x=-1$: numerator $=2(1)-3-5=-6\neq0$. Both are genuine. $$\boxed{x=2\ \text{and}\ x=-1}$$ (b) Dividing $2x^{2}+3x-5$ by $x^{2}-x-2$: $$2x^{2}\div x^{2}=2,\qquad 2(x^{2}-x-2)=2x^{2}-2x-4$$ Subtract: $(2x^{2}+3x-5)-(2x^{2}-2x-4)=5x-1$ So $2x^{2}+3x-5=2(x^{2}-x-2)+(5x-1)$, giving $$\boxed{f(x)=2+\dfrac{5x-1}{x^{2}-x-2}}$$ (c) Write $\dfrac{5x-1}{(x-2)(x+1)}=\dfrac{A}{x-2}+\dfrac{B}{x+1}$. Multiplying by $(x-2)(x+1)$: $$5x-1=A(x+1)+B(x-2)$$ Substituting $x=2$: $\ 5(2)-1=9=A(3) \implies A=3$ Substituting $x=-1$: $\ 5(-1)-1=-6=B(-3) \implies B=2$ Checking: $3(x+1)+2(x-2)=3x+3+2x-4=5x-1$ (satisfies) $$\boxed{f(x)=2+\dfrac{3}{x-2}+\dfrac{2}{x+1}}$$ (d) As $x\to\pm\infty$, both $\dfrac{3}{x-2}\to0$ and $\dfrac{2}{x+1}\to0$, so $f(x)\to2$. $$\boxed{y=2}$$ Solving $f(x)=2$ using the decomposed form: $$2+\frac{3}{x-2}+\frac{2}{x+1}=2$$ $$\frac{3}{x-2}+\frac{2}{x+1}=0$$ $$\frac{3}{x-2}=-\frac{2}{x+1}$$ $$3(x+1)=-2(x-2)$$ $$3x+3=-2x+4$$ $$5x=1$$ $$x=\frac{1}{5}$$ Since $\dfrac{1}{5}\neq2$ and $\neq-1$, this is valid. $$\boxed{x=\dfrac{1}{5}}$$

QUESTION 47 12 marks

Hard

The number of bacteria, in thousands, in a laboratory culture $t$ hours after observation begins is modelled by $$N(t)=\frac{200t+50}{t+2},\qquad t\geq0$$ (a) By performing polynomial division, express $N(t)$ in the form $A+\dfrac{B}{t+2}$, where $A,B\in\mathbb{Z}$. [4] (b) State the equation of the horizontal asymptote of the graph of $N$, and interpret its meaning in context. [3] (c) Using your answer to (a), explain why $N(t)$ is an increasing function for $t\geq0$, and hence state the range of $N(t)$ for $t\geq0$. [3] (d) Find the time at which the population reaches $150$ thousand. [2]

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(a) Dividing $200t+50$ by $t+2$: $$200t\div t=200,\qquad 200(t+2)=200t+400$$ Subtract: $(200t+50)-(200t+400)=-350$ So $200t+50=200(t+2)-350$, giving $$N(t)=200-\frac{350}{t+2}$$ i.e. $A=200,\ B=-350$. $$\boxed{N(t)=200-\dfrac{350}{t+2}}$$ (b) As $t\to\infty$, $\dfrac{350}{t+2}\to0$, so $N(t)\to200$. This means that as time passes, the bacteria population grows towards, but never reaches, $200$ thousand. $$\boxed{y=200}$$ (c) As $t$ increases (for $t\geq0$), $t+2$ increases, so $\dfrac{350}{t+2}$ decreases (getting smaller). Since $N(t)=200-\dfrac{350}{t+2}$, subtracting a smaller and smaller positive quantity from $200$ means $N(t)$ increases as $t$ increases. Hence $N$ is increasing on $t\geq0$. At $t=0$: $N(0)=200-\dfrac{350}{2}=200-175=25$. As $t\to\infty$, $N(t)\to200$ (never reaching it). $$\boxed{\text{range: } 25\leq N(t)<200}$$ (d) Setting $N(t)=150$: $$200-\frac{350}{t+2}=150$$ $$\frac{350}{t+2}=50$$ $$t+2=7$$ $$t=5$$ $$\boxed{t=5\ \text{hours}}$$

QUESTION 48 13 marks

Hard

Consider the inequality $\dfrac{(x-1)^{2}(x+3)}{x-2}\geq0$. (a) State the values of $x$ that must be excluded from the domain, and identify all critical values of the expression. [3] (b) Using a sign diagram, determine the sign of the expression on each interval formed by the critical values. Explain carefully how the repeated factor $(x-1)^{2}$ affects the sign. [6] (c) Hence write down the full solution set of the inequality, using set or interval notation. [4]

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(a) The expression is undefined where the denominator is zero, i.e. $x=2$ must be excluded from the domain. The critical values (where the expression is zero or undefined) are where any factor is zero: $x=1$ (from $(x-1)^{2}$), $x=-3$ (from $x+3$), and $x=2$ (from the denominator). $$\boxed{\text{excluded: } x=2;\ \text{critical values: } x=-3,1,2}$$ (b) These three critical values split the number line into four intervals: $x<-3$, $-32$. The factor $(x-1)^{2}$ is a perfect square, so it is always $\geq0$ and never negative - it does not change sign as $x$ crosses $x=1$; it only touches zero there. So the sign of the whole expression across $x=1$ is governed entirely by the remaining factors, $\dfrac{x+3}{x-2}$. Testing a point in each interval: $x=-4$ (in $x<-3$): $(x-1)^{2}=25>0$, $(x+3)=-1<0$, $(x-2)=-6<0$. Overall: $\dfrac{(+)(-)}{(-)}=(+)>0$ $x=0$ (in $-30$, $(x+3)=3>0$, $(x-2)=-2<0$. Overall: $\dfrac{(+)(+)}{(-)}=(-)<0$ $x=1.5$ (in $10$, $(x+3)=4.5>0$, $(x-2)=-0.5<0$. Overall: $\dfrac{(+)(+)}{(-)}=(-)<0$ $x=3$ (in $x>2$): $(x-1)^{2}=4>0$, $(x+3)=6>0$, $(x-2)=1>0$. Overall: $\dfrac{(+)(+)}{(+)}=(+)>0$ Note that at $x=1$ itself, the expression equals exactly $0$ (since $(1-1)^{2}=0$), which satisfies "$\geq0$"; the sign does not flip there because the factor is squared. $$\boxed{\text{positive on } x<-3\ \text{and}\ x>2;\ \text{negative on } -3(c) The inequality is satisfied where the expression is positive, together with the points where it equals exactly zero. It equals zero at $x=-3$ (numerator zero) and at $x=1$ (numerator zero, denominator non-zero). It is undefined at $x=2$, so that point can never be included. Combining: the expression is $\geq0$ for $x\leq-3$, together with the single point $x=1$, together with $x>2$. $$\boxed{x\leq-3\ \text{or}\ x=1\ \text{or}\ x>2}$$

QUESTION 49 19 marks

Hard

Consider the function $f(x)=\dfrac{x^{2}-2x-3}{x-4}$, $x\neq4$. (a) Factorise the numerator, and hence write down the $x$-intercepts of the graph of $f$. Confirm that $x=4$ is not a removable discontinuity. [3] (b) State the equation of the vertical asymptote of the graph of $f$. [1] (c) By performing polynomial division, express $f(x)$ in the form $x+a+\dfrac{b}{x-4}$, where $a,b\in\mathbb{Z}$, and hence state the equation of the oblique asymptote of the graph of $f$. [4] (d) Find the $y$-intercept of the graph of $f$. [2] (e) Using a sign diagram, determine algebraically the set of values of $x$ for which $f(x)>0$. [5] (f) Sketch the graph of $f$ for $-6\leq x\leq12$, showing both asymptotes and all intercepts clearly labelled. [4]

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(a) Factorising: two numbers multiplying to $-3$ and adding to $-2$ are $-3$ and $1$: $$x^{2}-2x-3=(x-3)(x+1)$$ The $x$-intercepts occur where the numerator is zero: $x=3$ and $x=-1$ (both in the domain, since neither equals $4$). At $x=4$: numerator $=(4-3)(4+1)=5\neq0$, so $x=4$ is not a removable discontinuity - it is a genuine vertical asymptote. $$\boxed{x\text{-intercepts}: x=3,\ x=-1}$$ (b) $$\boxed{x=4}$$ (c) Dividing $x^{2}-2x-3$ by $x-4$: $$x^{2}\div x=x,\qquad x(x-4)=x^{2}-4x$$ Subtract: $(x^{2}-2x-3)-(x^{2}-4x)=2x-3$ $$2x\div x=2,\qquad 2(x-4)=2x-8$$ Subtract: $(2x-3)-(2x-8)=5$ So $x^{2}-2x-3=(x-4)(x+2)+5$, giving $$f(x)=x+2+\frac{5}{x-4}$$ i.e. $a=2,\ b=5$. As $x\to\pm\infty$, $\dfrac{5}{x-4}\to0$, so $f(x)\to x+2$. $$\boxed{f(x)=x+2+\dfrac{5}{x-4};\ \text{oblique asymptote: } y=x+2}$$ (d) The $y$-intercept is $f(0)=\dfrac{0-0-3}{0-4}=\dfrac{-3}{-4}=\dfrac{3}{4}$. $$\boxed{y\text{-intercept}=\dfrac{3}{4}}$$ (e) Using the factorised form $f(x)=\dfrac{(x-3)(x+1)}{x-4}$, the critical values are $x=-1,\ 3,\ 4$, splitting the number line into four intervals. Testing a point in each: $x=-2$ (in $x<-1$): $\dfrac{(-5)(-1)}{-6}=\dfrac{5}{-6}<0$ $x=0$ (in $-10$ (satisfies) $x=3.5$ (in $34$): $\dfrac{(2)(6)}{1}>0$ (satisfies) So $f(x)>0$ for $-14$. $$\boxed{-14}$$ (f) The sketch shows: a vertical asymptote (dashed) at $x=4$; an oblique asymptote (dashed) $y=x+2$, which the curve approaches on both branches; the left branch passing through $x$-intercepts $(-1,0)$ and $(3,0)$ and the $y$-intercept $\left(0,\tfrac{3}{4}\right)$, dipping down towards $-\infty$ as $x\to4^{-}$; and the right branch rising from $+\infty$ near $x=4^{+}$ and merging into the line $y=x+2$ for large $x$, consistent with part (e): the curve lies above the $x$-axis exactly on $-14$.

QUESTION 50 14 marks

Hard

(a) Show that $\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4x}{x^{2}-1}$, $x\neq\pm1$. [3] (b) Hence solve the equation $\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{3}{2}$ algebraically. [5] (c) Hence, or otherwise, solve the inequality $\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}>\dfrac{3}{2}$, giving your answer using exact values. [6]

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(a) Combining the left-hand side over the common denominator $(x-1)(x+1)=x^{2}-1$: $$\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{(x+1)^{2}-(x-1)^{2}}{x^{2}-1}$$ Expanding the numerator: $$(x+1)^{2}-(x-1)^{2}=(x^{2}+2x+1)-(x^{2}-2x+1)=4x$$ $$\boxed{\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4x}{x^{2}-1}}$$ (b) Using the result of (a), the equation becomes $$\frac{4x}{x^{2}-1}=\frac{3}{2}$$ Cross-multiplying: $$8x=3(x^{2}-1)$$ $$8x=3x^{2}-3$$ $$0=3x^{2}-8x-3$$ Using the quadratic formula with $a=3,\ b=-8,\ c=-3$: $$x=\frac{8\pm\sqrt{64+36}}{6}=\frac{8\pm\sqrt{100}}{6}=\frac{8\pm10}{6}$$ $$x=\frac{18}{6}=3\quad\text{or}\quad x=\frac{-2}{6}=-\frac{1}{3}$$ Checking validity: neither $3$ nor $-\dfrac{1}{3}$ equals $\pm1$, so both solutions lie in the domain. $$\boxed{x=3\ \text{or}\ x=-\dfrac{1}{3}}$$ (c) Using (a), the inequality is $\dfrac{4x}{x^{2}-1}>\dfrac{3}{2}$. Rearranging so everything is on one side and combining into a single fraction: $$\frac{4x}{x^{2}-1}-\frac{3}{2}>0$$ $$\frac{8x-3(x^{2}-1)}{2(x^{2}-1)}>0$$ $$\frac{8x-3x^{2}+3}{2(x^{2}-1)}>0$$ $$\frac{-(3x^{2}-8x-3)}{2(x^{2}-1)}>0$$ Multiplying both sides by $-1$ (reversing the inequality): $$\frac{3x^{2}-8x-3}{2(x^{2}-1)}<0$$ From part (b), $3x^{2}-8x-3=(3x+1)(x-3)$ (check: $(3x+1)(x-3)=3x^{2}-9x+x-3=3x^{2}-8x-3$ (satisfies)). Also $x^{2}-1=(x-1)(x+1)$. So the inequality is $$\frac{(3x+1)(x-3)}{(x-1)(x+1)}<0$$ The critical values, in order, are $x=-1,\ -\dfrac{1}{3},\ 1,\ 3$, splitting the number line into five intervals. Testing a point in each: $x=-2$: $\dfrac{(-5)(-5)}{(-3)(-1)}=\dfrac{25}{3}>0$ $x=-0.5$: $\dfrac{(-0.5)(-3.5)}{(-1.5)(0.5)}=\dfrac{1.75}{-0.75}<0$ (satisfies) $x=0$: $\dfrac{(1)(-3)}{(-1)(1)}=\dfrac{-3}{-1}=3>0$ $x=2$: $\dfrac{(7)(-1)}{(1)(3)}=\dfrac{-7}{3}<0$ (satisfies) $x=4$: $\dfrac{(13)(1)}{(3)(5)}>0$ So the inequality holds on $-1

QUESTION 1 5 marks

Easy

The graph of $y=f(x)$ is transformed by a horizontal stretch with scale factor $\dfrac{1}{2}$, followed by a translation of 3 units in the positive $x$-direction, to give the graph of $y=g(x)$. (a) Write $g(x)$ in terms of $f$. (b) Given that $f(x)=x^{2}-1$, find $g(x)$ explicitly, and evaluate $g(4)$.

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(a) A horizontal stretch with scale factor $\dfrac{1}{2}$ maps $y=f(x)$ to $y=f(2x)$ (replace $x$ with $\dfrac{x}{1/2}=2x$). A translation of 3 units in the positive $x$-direction then replaces $x$ with $x-3$: $$g(x)=f(2(x-3))=f(2x-6)$$ $$\boxed{g(x)=f(2x-6)}$$ (b) With $f(x)=x^{2}-1$: $$g(x)=f(2x-6)=(2x-6)^{2}-1$$ $$\boxed{g(x)=(2x-6)^{2}-1}$$ Substituting $x=4$: $$g(4)=(2(4)-6)^{2}-1=(8-6)^{2}-1$$ $$=2^{2}-1=4-1=3$$ $$\boxed{g(4)=3}$$

QUESTION 2 6 marks

Medium

The point $(2,-3)$ lies on the graph of $y=f(x)$. (a) State the coordinates of the corresponding point on the graph of $y=-2f(x-1)+5$. (b) Describe, in order, the sequence of transformations that maps the graph of $y=f(x)$ onto the graph of $y=-2f(x-1)+5$.

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(a) Since $(2,-3)$ lies on $y=f(x)$, we have $f(2)=-3$. We want the corresponding point $(X,Y)$ on $y=-2f(x-1)+5$, so we need the same value to be fed into $f$: $X-1=2$, giving $X=3$. $$Y=-2f(X-1)+5=-2f(2)+5$$ $$Y=-2(-3)+5=6+5=11$$ $$\boxed{(3,11)}$$ (b) Writing $y=-2f(x-1)+5$, the transformations applied to the graph of $f$, in the order they act, are: 1. A horizontal translation of 1 unit in the positive $x$-direction (from $x\to x-1$). 2. A vertical stretch with scale factor 2. 3. A reflection in the $x$-axis (from the negative sign). 4. A vertical translation of 5 units in the positive $y$-direction. $$\boxed{\text{translate 1 unit right, stretch vertically (factor 2), reflect in the }x\text{-axis, translate 5 units up}}$$

QUESTION 3 6 marks

Easy

Consider $f(x)=(x-2)^{2}+5$. The graph of $g$ is obtained from the graph of $f$ by a reflection in the $x$-axis, followed by a vertical translation of 3 units in the positive $y$-direction. (a) Write down an expression for $g(x)$. (b) Find the coordinates of the maximum point of the graph of $g$.

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(a) A reflection in the $x$-axis maps $y=f(x)$ to $y=-f(x)$. A subsequent translation of 3 units up gives: $$g(x)=-f(x)+3=-\left[(x-2)^{2}+5\right]+3$$ $$=-(x-2)^{2}-5+3=-(x-2)^{2}-2$$ $$\boxed{g(x)=-(x-2)^{2}-2}$$ (b) This is in vertex form $g(x)=-(x-h)^{2}+k$ with $h=2,\ k=-2$. Since the coefficient of $(x-2)^{2}$ is negative, the parabola opens downward, so the vertex is a maximum point. $$\boxed{\text{maximum point } =(2,-2)}$$ (Check using transformations directly: $f(x)=(x-2)^{2}+5$ has minimum point $(2,5)$. Reflecting in the $x$-axis sends $(2,5)\to(2,-5)$, now a maximum. Translating up 3 sends $(2,-5)\to(2,-2)$, confirming the answer.)

QUESTION 4 7 marks

Medium

The graph of $y=2^{x}$ is transformed to the graph of $y=3\cdot2^{x+1}-4$ by a sequence of transformations. (a) Describe a sequence of transformations that maps $y=2^{x}$ onto $y=3\cdot2^{x+1}-4$. (b) State the equation of the horizontal asymptote of $y=3\cdot2^{x+1}-4$. (c) The point $(0,1)$ lies on the graph of $y=2^{x}$. Find the coordinates of the corresponding point on the graph of $y=3\cdot2^{x+1}-4$.

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(a) Let $f(x)=2^{x}$. Then $3\cdot2^{x+1}-4=3f(x+1)-4$. Reading the transformations in the order they are applied to the graph: 1. A horizontal translation of 1 unit in the negative $x$-direction (from $x\to x+1$). 2. A vertical stretch with scale factor 3. 3. A vertical translation of 4 units in the negative $y$-direction. $$\boxed{\text{translate 1 unit left, stretch vertically (factor 3), translate 4 units down}}$$ (b) The graph of $y=2^{x}$ has horizontal asymptote $y=0$. A vertical stretch (factor 3) keeps the asymptote at $y=3(0)=0$. A translation of 4 units down then moves it to $y=0-4=-4$. $$\boxed{y=-4}$$ (c) $(0,1)$ on $y=f(x)=2^x$ means $f(0)=1$. We want the point $(X,Y)$ on $y=3f(x+1)-4$ such that the same input is fed into $f$, i.e. $X+1=0\implies X=-1$. $$Y=3f(X+1)-4=3f(0)-4=3(1)-4=-1$$ $$\boxed{(-1,-1)}$$ (Check directly: at $x=-1$, $3\cdot2^{-1+1}-4=3\cdot2^{0}-4=3(1)-4=-1$, confirming $y=-1$.)

QUESTION 5 15 marks

Hard

The function $f$ is defined by $f(x)=\dfrac{1}{x-3}+2$, for $x\neq3$. (a) State the equations of the two asymptotes of the graph of $y=f(x)$. [2] (b) The graph of $g$ is obtained from the graph of $f$ by a horizontal stretch with scale factor $\dfrac{1}{3}$, followed by a translation by the vector $\begin{pmatrix}-1\\4\end{pmatrix}$. Find an expression for $g(x)$ in the form $\dfrac{a}{x-h}+k$, stating the values of $a$, $h$ and $k$. [6] (c) Write down the equations of the two asymptotes of the graph of $y=g(x)$. [2] (d) Find the coordinates of the point(s) of intersection of the graph of $y=g(x)$ with the line $y=x$, giving your answers in exact form. [5]

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(a) The vertical asymptote of $f(x)=\dfrac{1}{x-3}+2$ occurs where $x-3=0$, i.e. $x=3$. The horizontal asymptote is the value $f(x)$ approaches as $x\to\pm\infty$: since $\dfrac{1}{x-3}\to0$, $f(x)\to2$. $$\boxed{x=3\ \text{and}\ y=2}$$ (b) A horizontal stretch with scale factor $\dfrac{1}{3}$ maps $y=f(x)$ to $y=f(3x)$ (replace $x$ with $\dfrac{x}{1/3}=3x$). Call this $h(x)=f(3x)$: $$h(x)=\frac{1}{3x-3}+2=\frac{1}{3(x-1)}+2$$ The translation by $\begin{pmatrix}-1\\4\end{pmatrix}$ replaces $x$ with $x-(-1)=x+1$ and adds 4: $$g(x)=h(x+1)+4=\frac{1}{3\big((x+1)-1\big)}+2+4=\frac{1}{3x}+6$$ Writing $\dfrac{1}{3x}=\dfrac{1/3}{x-0}$: $$g(x)=\frac{1/3}{x-0}+6$$ $$\boxed{g(x)=\dfrac{1/3}{x}+6;\quad a=\dfrac{1}{3},\ h=0,\ k=6}$$ (c) From the form in (b), the vertical asymptote is $x=h=0$ and the horizontal asymptote is $y=k=6$. (This is consistent with transforming the asymptotes of $f$ from (a): $x=3\to x=\dfrac{3}{3}=1\to x=1-1=0$; $y=2\to y=2+4=6$.) $$\boxed{x=0\ \text{and}\ y=6}$$ (d) Solve $g(x)=x$: $$\frac{1}{3x}+6=x$$ Multiply both sides by $3x$ (noting $x\neq0$): $$1+18x=3x^{2}$$ $$3x^{2}-18x-1=0$$ Applying the quadratic formula: $$x=\frac{18\pm\sqrt{(-18)^{2}-4(3)(-1)}}{2(3)}=\frac{18\pm\sqrt{324+12}}{6}=\frac{18\pm\sqrt{336}}{6}$$ Simplifying $\sqrt{336}=\sqrt{16\times21}=4\sqrt{21}$: $$x=\frac{18\pm4\sqrt{21}}{6}=\frac{9\pm2\sqrt{21}}{3}=3\pm\frac{2\sqrt{21}}{3}$$ Since the intersection is with the line $y=x$, the $y$-coordinate equals the $x$-coordinate at each point. $$\boxed{\left(3+\dfrac{2\sqrt{21}}{3},\ 3+\dfrac{2\sqrt{21}}{3}\right)\ \text{and}\ \left(3-\dfrac{2\sqrt{21}}{3},\ 3-\dfrac{2\sqrt{21}}{3}\right)}$$

QUESTION 6 5 marks

Easy

Let $f(x)=x^{2}$. The graph of $g$ is obtained from the graph of $f$ by a vertical translation of 6 units in the negative $y$-direction. (a) Write down an expression for $g(x)$. [2] (b) The point $(3,9)$ lies on the graph of $y=f(x)$. State the coordinates of the corresponding point on the graph of $y=g(x)$. [2] (c) Write down the coordinates of the vertex of the graph of $g$. [1]

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(a) A vertical translation of 6 units in the negative $y$-direction maps $y=f(x)$ to $y=f(x)-6$: $$g(x)=f(x)-6=x^{2}-6$$ $$\boxed{g(x)=x^{2}-6}$$ (b) A vertical translation only affects the $y$-coordinate, adding $-6$ to it: $$(3,9)\ \longrightarrow\ (3,\,9-6)$$ $$\boxed{(3,3)}$$ (c) The vertex of $y=f(x)=x^{2}$ is at $(0,0)$. Translating 6 units down moves this to: $$\boxed{(0,-6)}$$

QUESTION 7 4 marks

Easy

Consider $f(x)=x^{3}$ and $g(x)=(x+2)^{3}$. (a) Describe the single transformation that maps the graph of $y=f(x)$ onto the graph of $y=g(x)$. [2] (b) Hence, or otherwise, find the coordinates of the image of the point $(1,1)$ on the graph of $y=f(x)$ under this transformation. [2]

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(a) Since $g(x)=(x+2)^{3}=f(x+2)=f\big(x-(-2)\big)$, this has the form $f(x-c)$ with $c=-2$. $$\boxed{\text{a horizontal translation of 2 units in the negative }x\text{-direction}}$$ (b) A horizontal translation of 2 units in the negative $x$-direction subtracts 2 from the $x$-coordinate only: $$(1,1)\ \longrightarrow\ (1-2,\,1)$$ $$\boxed{(-1,1)}$$ (Check: $g(-1)=(-1+2)^{3}=1^{3}=1$, confirming the image point lies on $y=g(x)$.)

QUESTION 8 6 marks

Easy

Let $f(x)=\cos x$. The graph of $g$ is obtained from the graph of $f$ by a horizontal stretch with scale factor $\dfrac{1}{2}$. (a) Write down an expression for $g(x)$. [2] (b) State the period of the graph of $g$. [2] (c) Write down the maximum value of $g(x)$, and one value of $x$, with $0\le x\le \pi$, at which it occurs. [2]

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(a) A horizontal stretch with scale factor $\dfrac{1}{2}$ maps $y=f(x)$ to $y=f(2x)$ (replace $x$ with $\dfrac{x}{1/2}=2x$): $$\boxed{g(x)=\cos(2x)}$$ (b) The graph of $f(x)=\cos x$ has period $2\pi$. A horizontal stretch with scale factor $\dfrac{1}{2}$ halves all horizontal distances, so the period is also halved: $$2\pi\times\dfrac{1}{2}=\pi$$ $$\boxed{\text{period}=\pi}$$ (c) A horizontal stretch does not change $y$-values, so the maximum value of $g$ is the same as the maximum value of $f$, namely 1. This occurs where $2x=0$, i.e. $x=0$ (within $0\le x\le\pi$). $$\boxed{\text{maximum value }1\text{ at }x=0}$$

QUESTION 9 5 marks

Easy

Let $f(x)=\sqrt{x}$, for $x\ge0$. The graph of $g$ is the reflection of the graph of $f$ in the $y$-axis. (a) Write down an expression for $g(x)$, and state its domain. [3] (b) The point $(9,3)$ lies on the graph of $y=f(x)$. State the coordinates of the corresponding point on the graph of $y=g(x)$. [2]

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(a) A reflection in the $y$-axis maps $y=f(x)$ to $y=f(-x)$: $$g(x)=f(-x)=\sqrt{-x}$$ For $\sqrt{-x}$ to be defined we need $-x\ge0$, i.e. $x\le0$. $$\boxed{g(x)=\sqrt{-x},\ \ \text{domain } x\le0}$$ (b) A reflection in the $y$-axis negates the $x$-coordinate and leaves the $y$-coordinate unchanged: $$(9,3)\ \longrightarrow\ (-9,3)$$ $$\boxed{(-9,3)}$$ (Check: $g(-9)=\sqrt{-(-9)}=\sqrt{9}=3$, confirming the image point lies on $y=g(x)$.)

QUESTION 10 5 marks

Easy

Let $f(x)=e^{x}$. The graph of $g$ is obtained from the graph of $f$ by a vertical stretch with scale factor 2. (a) Write down an expression for $g(x)$. [1] (b) Write down the $y$-intercept of the graph of $g$. [2] (c) State the equation of the horizontal asymptote of the graph of $g$. [2]

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(a) A vertical stretch with scale factor 2 maps $y=f(x)$ to $y=2f(x)$: $$\boxed{g(x)=2e^{x}}$$ (b) The $y$-intercept occurs at $x=0$: $$g(0)=2e^{0}=2(1)=2$$ $$\boxed{(0,2)}$$ (c) The graph of $f(x)=e^{x}$ has horizontal asymptote $y=0$. A vertical stretch multiplies $y$-values by 2, and $2\times0=0$, so the asymptote is unchanged. $$\boxed{y=0}$$

QUESTION 11 6 marks

Easy

Let $f(x)=\ln x$, for $x>0$. The graph of $g$ is obtained from the graph of $f$ by a horizontal translation of 3 units in the positive $x$-direction. (a) Write down an expression for $g(x)$, and state its domain. [2] (b) Write down the equation of the vertical asymptote of the graph of $g$. [2] (c) The point $(1,0)$ lies on the graph of $y=f(x)$. Find the corresponding point on the graph of $y=g(x)$. [2]

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(a) A translation of 3 units in the positive $x$-direction maps $y=f(x)$ to $y=f(x-3)$: $$g(x)=f(x-3)=\ln(x-3)$$ For $\ln(x-3)$ to be defined we need $x-3>0$, i.e. $x>3$. $$\boxed{g(x)=\ln(x-3),\ \ \text{domain } x>3}$$ (b) The graph of $f(x)=\ln x$ has vertical asymptote $x=0$. Translating 3 units right moves this to: $$\boxed{x=3}$$ (c) We require the point $(X,Y)$ on $y=g(x)$ such that the same value is fed into $f$: $X-3=1\implies X=4$. $$Y=f(X-3)=f(1)=0$$ $$\boxed{(4,0)}$$

QUESTION 12 4 marks

Easy

Let $f(x)=\dfrac{1}{x}$, for $x\neq0$. The graph of $g$ is the reflection of the graph of $f$ in the $x$-axis. (a) Write down an expression for $g(x)$. [1] (b) Show that $g(x)=f(-x)$ for all $x\neq0$, and hence describe a second single transformation, different from a reflection in the $x$-axis, that also maps the graph of $f$ onto the graph of $g$. [3]

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(a) A reflection in the $x$-axis maps $y=f(x)$ to $y=-f(x)$: $$\boxed{g(x)=-f(x)=-\dfrac{1}{x}}$$ (b) Computing $f(-x)$: $$f(-x)=\frac{1}{-x}=-\frac{1}{x}$$ This is identical to $g(x)=-\dfrac{1}{x}$ found in (a), so: $$\boxed{g(x)=f(-x)}$$ Since $g(x)=f(-x)$ also has the form of a reflection in the $y$-axis (replacing $x$ with $-x$), a second transformation that maps the graph of $f$ onto the graph of $g$ is: $$\boxed{\text{a reflection in the }y\text{-axis}}$$ (This works because $f(x)=\dfrac{1}{x}$ is an odd function: $f(-x)=-f(x)$, so reflecting in the $x$-axis and reflecting in the $y$-axis produce the same image graph.)

QUESTION 13 4 marks

Easy

The point $(-2,5)$ lies on the graph of $y=g(x)$. The graph of $y=g(x)$ is translated by the vector $\begin{pmatrix}4\\-3\end{pmatrix}$ to give the graph of $y=h(x)$. (a) Write down an expression for $h(x)$ in terms of $g$. [2] (b) Find the coordinates of the point on the graph of $y=h(x)$ that corresponds to $(-2,5)$. [2]

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(a) A translation by the vector $\begin{pmatrix}4\\-3\end{pmatrix}$ replaces $x$ with $x-4$ and subtracts 3 from the output: $$\boxed{h(x)=g(x-4)-3}$$ (b) A translation by $\begin{pmatrix}4\\-3\end{pmatrix}$ adds 4 to the $x$-coordinate and subtracts 3 from the $y$-coordinate of every point: $$(-2,5)\ \longrightarrow\ (-2+4,\,5-3)$$ $$\boxed{(2,2)}$$

QUESTION 14 4 marks

Easy

Let $f(x)=|x|$. The graph of $g$ is obtained from the graph of $f$ by a vertical stretch with scale factor 3. (a) Write down an expression for $g(x)$. [1] (b) State the coordinates of the vertex of the graph of $g$. [1] (c) Find the value of $g(-5)$. [2]

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(a) A vertical stretch with scale factor 3 maps $y=f(x)$ to $y=3f(x)$: $$\boxed{g(x)=3|x|}$$ (b) A vertical stretch leaves points on the $x$-axis fixed, and the vertex of $f$ is at $(0,0)$: $$\boxed{(0,0)}$$ (c) Substituting $x=-5$: $$g(-5)=3|-5|=3(5)=15$$ $$\boxed{g(-5)=15}$$

QUESTION 15 6 marks

Easy

Let $f(x)=x^{2}$ and let $g(x)=f(2x)$. (a) Describe the transformation of the graph of $f$ that produces the graph of $g$. [2] (b) Show that $g(x)=4f(x)$, and hence describe a second, different, single transformation of the graph of $f$ that also produces the graph of $g$. [4]

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(a) $g(x)=f(2x)$ has the form $f(bx)$ with $b=2$, which is a horizontal stretch with scale factor $\dfrac{1}{b}$: $$\boxed{\text{a horizontal stretch with scale factor }\dfrac{1}{2}}$$ (b) Since $f(x)=x^{2}$: $$g(x)=f(2x)=(2x)^{2}=4x^{2}=4\big(x^{2}\big)=4f(x)$$ $$\boxed{g(x)=4f(x)}$$ The equation $g(x)=4f(x)$ has the form $af(x)$ with $a=4$, which is a vertical stretch with scale factor 4: $$\boxed{\text{a vertical stretch with scale factor }4}$$ (Both transformations produce the identical graph because $f(x)=x^2$ is a pure power function: stretching it horizontally by scale factor $\dfrac12$ multiplies every $y$-value by $2^2=4$, the same effect as stretching vertically by scale factor 4.)

QUESTION 16 4 marks

Easy

Let $f(x)=\sin x$. The graph of $g$ is obtained from the graph of $f$ by a vertical translation of 2 units in the positive $y$-direction. (a) Write down an expression for $g(x)$. [1] (b) State the range of $g$. [3]

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(a) A translation of 2 units in the positive $y$-direction maps $y=f(x)$ to $y=f(x)+2$: $$\boxed{g(x)=\sin x+2}$$ (b) The range of $f(x)=\sin x$ is $-1\le f(x)\le1$. Adding 2 to every value of $f$ shifts this interval up by 2: $$-1+2\le g(x)\le1+2$$ $$\boxed{1\le g(x)\le3}$$

QUESTION 17 5 marks

Easy

Let $f(x)=x^{3}+x$. (a) Show algebraically that $f$ is an odd function. [3] (b) Given that the point $(2,10)$ lies on the graph of $y=f(x)$, use the result from (a) to write down the coordinates of another point that must lie on the graph of $y=f(x)$. [2]

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(a) Computing $f(-x)$: $$f(-x)=(-x)^{3}+(-x)=-x^{3}-x=-(x^{3}+x)=-f(x)$$ Since $f(-x)=-f(x)$ for all $x$, $f$ is an odd function. $$\boxed{f(-x)=-f(x)\ \Rightarrow\ f\text{ is odd}}$$ (b) Since $(2,10)$ lies on the graph, $f(2)=10$. As $f$ is odd, $f(-2)=-f(2)=-10$. $$\boxed{(-2,-10)}$$ (Check directly: $f(-2)=(-2)^{3}+(-2)=-8-2=-10$, confirming the answer.)

QUESTION 18 6 marks

Easy

Let $f(x)=\dfrac{1}{x}$, for $x\neq0$. The graph of $g$ is obtained from the graph of $f$ by a vertical translation of 3 units in the negative $y$-direction. (a) Write down an expression for $g(x)$. [1] (b) Write down the equations of the two asymptotes of the graph of $g$. [3] (c) Find the $x$-intercept of the graph of $g$. [2]

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(a) A vertical translation of 3 units in the negative $y$-direction maps $y=f(x)$ to $y=f(x)-3$: $$\boxed{g(x)=\dfrac{1}{x}-3}$$ (b) The graph of $f(x)=\dfrac1x$ has asymptotes $x=0$ and $y=0$. A vertical translation does not affect a vertical asymptote, so $x=0$ is unchanged; it shifts a horizontal asymptote by the same amount as the graph, so $y=0$ becomes $y=0-3=-3$. $$\boxed{x=0\ \text{and}\ y=-3}$$ (c) Setting $g(x)=0$: $$\frac{1}{x}-3=0\ \Rightarrow\ \frac{1}{x}=3\ \Rightarrow\ x=\frac{1}{3}$$ $$\boxed{x=\dfrac{1}{3}}$$

QUESTION 19 5 marks

Easy

Let $f(x)=\cos x$. The graph of $g$ is obtained from the graph of $f$ by a horizontal translation of $\dfrac{\pi}{3}$ units in the positive $x$-direction. (a) Write down an expression for $g(x)$. [1] (b) Find the exact value of $g\left(\dfrac{\pi}{3}\right)$. [2] (c) Write down the smallest positive value of $x$ for which $g(x)=1$. [2]

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(a) A translation of $\dfrac{\pi}{3}$ units in the positive $x$-direction maps $y=f(x)$ to $y=f\left(x-\dfrac{\pi}{3}\right)$: $$\boxed{g(x)=\cos\left(x-\dfrac{\pi}{3}\right)}$$ (b) Substituting $x=\dfrac{\pi}{3}$: $$g\left(\frac{\pi}{3}\right)=\cos\left(\frac{\pi}{3}-\frac{\pi}{3}\right)=\cos(0)=1$$ $$\boxed{g\left(\dfrac{\pi}{3}\right)=1}$$ (c) $f(x)=\cos x$ first reaches its maximum value of 1 (for $x>0$) at $x=0$ and then next at $x=2\pi$. Since $g$ is $f$ shifted $\dfrac\pi3$ to the right, $g$ reaches its maximum value of 1 at $x=0+\dfrac{\pi}{3}=\dfrac{\pi}{3}$, which is the point found in (b). $$\boxed{x=\dfrac{\pi}{3}}$$

QUESTION 20 4 marks

Easy

The point $(-4,-7)$ lies on the graph of $y=f(x)$. The graph of $y=h(x)$ is the reflection of the graph of $y=f(x)$ in the $x$-axis. (a) Write down an expression for $h(x)$ in terms of $f$. [1] (b) State the coordinates of the point on the graph of $y=h(x)$ that corresponds to $(-4,-7)$. [2] (c) Given that $f$ is an odd function, find the value of $f(4)$. [1]

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(a) A reflection in the $x$-axis maps $y=f(x)$ to $y=-f(x)$: $$\boxed{h(x)=-f(x)}$$ (b) A reflection in the $x$-axis negates the $y$-coordinate and leaves the $x$-coordinate unchanged: $$(-4,-7)\ \longrightarrow\ (-4,\,7)$$ $$\boxed{(-4,7)}$$ (c) Since $(-4,-7)$ lies on $y=f(x)$, $f(-4)=-7$. As $f$ is odd, $f(-x)=-f(x)$, so: $$f(4)=f\big(-(-4)\big)=-f(-4)=-(-7)$$ $$\boxed{f(4)=7}$$

QUESTION 21 4 marks

Easy

Let $f(x)=x^{2}$. The graph of $g$ is obtained from the graph of $f$ by a vertical stretch with scale factor $\dfrac{1}{2}$. (a) Write down an expression for $g(x)$. [1] (b) The point $(2,4)$ lies on the graph of $y=f(x)$. State the corresponding point on the graph of $y=g(x)$. [2] (c) Compared to the graph of $f$, state whether the graph of $g$ is narrower or wider, giving a reason. [1]

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(a) A vertical stretch with scale factor $\dfrac12$ maps $y=f(x)$ to $y=\dfrac12f(x)$: $$\boxed{g(x)=\dfrac12x^{2}}$$ (b) A vertical stretch multiplies the $y$-coordinate by $\dfrac12$ and leaves the $x$-coordinate unchanged: $$(2,4)\ \longrightarrow\ \left(2,\,4\times\dfrac12\right)$$ $$\boxed{(2,2)}$$ (c) Since the scale factor $\dfrac12<1$, every $y$-value of $f$ is compressed towards the $x$-axis, so: $$\boxed{\text{the graph of }g\text{ is wider than the graph of }f}$$

QUESTION 22 5 marks

Easy

Let $f(x)=\sqrt{x}$, for $x\ge0$. The graph of $g$ is obtained from the graph of $f$ by a horizontal translation of 4 units in the negative $x$-direction. (a) Write down an expression for $g(x)$, and state its domain. [3] (b) The point $(4,2)$ lies on the graph of $y=f(x)$. Find the coordinates of the corresponding point on the graph of $y=g(x)$. [2]

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(a) A translation of 4 units in the negative $x$-direction maps $y=f(x)$ to $y=f(x+4)$: $$g(x)=f(x+4)=\sqrt{x+4}$$ For $\sqrt{x+4}$ to be defined we need $x+4\ge0$, i.e. $x\ge-4$. $$\boxed{g(x)=\sqrt{x+4},\ \ \text{domain } x\ge-4}$$ (b) A translation of 4 units in the negative $x$-direction subtracts 4 from the $x$-coordinate and leaves the $y$-coordinate unchanged: $$(4,2)\ \longrightarrow\ (4-4,\,2)$$ $$\boxed{(0,2)}$$ (Check: $g(0)=\sqrt{0+4}=\sqrt4=2$, confirming the image point lies on $y=g(x)$.)

QUESTION 23 4 marks

Easy

Let $f(x)=e^{x}$. The graph of $g$ is the reflection of the graph of $f$ in the $y$-axis. (a) Write down an expression for $g(x)$. [1] (b) Write down the $y$-intercept of the graph of $g$. [1] (c) Find the value of $g(-3)$, giving your answer in the form $e^{k}$. [2]

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(a) A reflection in the $y$-axis maps $y=f(x)$ to $y=f(-x)$: $$\boxed{g(x)=f(-x)=e^{-x}}$$ (b) The $y$-intercept occurs at $x=0$: $$g(0)=e^{-0}=e^{0}=1$$ $$\boxed{(0,1)}$$ (c) Substituting $x=-3$: $$g(-3)=e^{-(-3)}=e^{3}$$ $$\boxed{g(-3)=e^{3}}$$ (Reflecting $y=e^x$, which has horizontal asymptote $y=0$ as $x\to-\infty$, in the $y$-axis gives $y=e^{-x}$, whose horizontal asymptote is $y=0$ as $x\to+\infty$ instead; the asymptote's equation $y=0$ is unchanged, only the direction of approach reverses.)

QUESTION 24 8 marks

Medium

Let $f(x)=(x+1)^{2}-4$. The graph of $g$ is obtained from the graph of $f$ by a vertical stretch with scale factor 2, followed by a reflection in the $x$-axis, followed by a vertical translation of 5 units in the positive $y$-direction. (a) Write $g(x)$ in terms of $f$, and hence find $g(x)$ in expanded vertex form $a(x-h)^{2}+k$. [5] (b) State the coordinates of the vertex of the graph of $g$, and determine whether it is a maximum or a minimum point. [3]

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(a) Applying the transformations to $f$ in the order given: a vertical stretch with scale factor 2 gives $2f(x)$; a reflection in the $x$-axis then gives $-2f(x)$; a vertical translation of 5 units up then gives $-2f(x)+5$: $$g(x)=-2f(x)+5$$ Substituting $f(x)=(x+1)^{2}-4$: $$g(x)=-2\left[(x+1)^{2}-4\right]+5=-2(x+1)^{2}+8+5$$ $$\boxed{g(x)=-2(x+1)^{2}+13}$$ (b) The graph of $f(x)=(x+1)^{2}-4$ is a parabola opening upward with vertex (minimum point) at $(-1,-4)$. Tracking the vertex through each transformation: the vertical stretch (scale factor 2) keeps the $x$-coordinate fixed and doubles the $y$-coordinate: $(-1,-4)\to(-1,-8)$. The reflection in the $x$-axis then negates the $y$-coordinate: $(-1,-8)\to(-1,8)$. The vertical translation up 5 then adds 5: $(-1,8)\to(-1,13)$. Since the graph of $f$ opened upward (vertex a minimum) and was reflected in the $x$-axis, the graph of $g$ opens downward, so the vertex is now a maximum. $$\boxed{(-1,13)\ \text{is a maximum point}}$$

QUESTION 25 7 marks

Medium

Let $f(x)=e^{x}$. The graph of $g$ is obtained from the graph of $f$ by a horizontal translation of 2 units in the negative $x$-direction, followed by a reflection in the $x$-axis, followed by a vertical translation of 1 unit in the positive $y$-direction. (a) Show that $g(x)=-e^{x+2}+1$. [3] (b) Write down the equation of the horizontal asymptote of the graph of $g$. [2] (c) Find the exact $y$-intercept of the graph of $g$, giving your answer in terms of $e$. [2]

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(a) Applying the transformations to $f$ in the order given: a horizontal translation 2 units left gives $f(x+2)$; a reflection in the $x$-axis then gives $-f(x+2)$; a vertical translation 1 unit up then gives $-f(x+2)+1$: $$g(x)=-f(x+2)+1$$ Substituting $f(x)=e^{x}$: $$\boxed{g(x)=-e^{x+2}+1}$$ (b) The graph of $f(x)=e^{x}$ has horizontal asymptote $y=0$. A horizontal translation does not change a horizontal asymptote, so it remains $y=0$. Reflecting in the $x$-axis sends $y=0$ to $y=-0=0$. Translating up 1 then sends $y=0$ to $y=0+1=1$. $$\boxed{y=1}$$ (c) Substituting $x=0$: $$g(0)=-e^{0+2}+1=-e^{2}+1$$ $$\boxed{g(0)=1-e^{2}}$$

QUESTION 26 7 marks

Medium

Let $f(x)=\sin x$ and $g(x)=3\sin(2x)$. (a) Describe fully the sequence of two transformations that maps the graph of $f$ onto the graph of $g$. [4] (b) Write down the amplitude and the period of $g$. [3]

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(a) We can write $g(x)=3f(2x)$. Reading the transformations applied to $x$ first, then to the output: Step 1: replacing $x$ with $2x$ gives $f(2x)$, a horizontal stretch with scale factor $\dfrac12$. Step 2: multiplying the result by 3 gives $3f(2x)$, a vertical stretch with scale factor 3. $$\boxed{\text{a horizontal stretch, scale factor }\dfrac12\text{, followed by a vertical stretch, scale factor }3}$$ (b) The amplitude of $f(x)=\sin x$ is 1; the vertical stretch (scale factor 3) multiplies this by 3, giving amplitude 3. The period of $f$ is $2\pi$; the horizontal stretch (scale factor $\dfrac12$) halves this, giving period $2\pi\times\dfrac12=\pi$. $$\boxed{\text{amplitude}=3,\ \ \text{period}=\pi}$$

QUESTION 27 8 marks

Medium

Let $f(x)=\dfrac1x$, for $x\neq0$. The graph of $g$ is obtained from the graph of $f$ by a vertical stretch with scale factor 2, followed by a translation by the vector $\begin{pmatrix}3\\1\end{pmatrix}$. (a) Find an expression for $g(x)$. [4] (b) Write down the equations of the two asymptotes of the graph of $g$. [2] (c) Verify that the point $(4,3)$ lies on the graph of $y=g(x)$. [2]

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(a) A vertical stretch with scale factor 2 gives $2f(x)=\dfrac2x$. The translation by $\begin{pmatrix}3\\1\end{pmatrix}$ then replaces $x$ with $x-3$ and adds 1: $$g(x)=2f(x-3)+1$$ $$\boxed{g(x)=\dfrac{2}{x-3}+1}$$ (b) The graph of $f$ has asymptotes $x=0$ and $y=0$. The vertical stretch leaves $x=0$ unchanged and sends $y=0$ to $y=2(0)=0$. The translation by $\begin{pmatrix}3\\1\end{pmatrix}$ then sends $x=0\to x=3$ and $y=0\to y=0+1=1$. $$\boxed{x=3\ \text{and}\ y=1}$$ (c) Substituting $x=4$ into $g(x)=\dfrac{2}{x-3}+1$: $$g(4)=\frac{2}{4-3}+1=\frac{2}{1}+1=2+1=3$$ Since $g(4)=3$, the point $(4,3)$ does lie on the graph of $y=g(x)$. $\boxed{\checkmark}$

QUESTION 28 7 marks

Medium

The graph of $y=g(x)$ is obtained from the graph of $y=f(x)$ by the transformation $g(x)=f(3x-6)$. (a) By writing $3x-6$ in the form $3(x-c)$, describe fully, in the correct order, the sequence of two transformations that maps the graph of $f$ onto the graph of $g$. [4] (b) Given that $f(9)=7$, find the value of $x$ for which $g(x)=7$. [3]

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(a) Factorising: $$3x-6=3(x-2)$$ so $g(x)=f\big(3(x-2)\big)$. To build this from $f$, we must apply the horizontal stretch to $x$ before the translation, since the translation acts on the already-stretched variable: Step 1: replacing $x$ with $3x$ gives $f(3x)$, a horizontal stretch with scale factor $\dfrac13$. Step 2: replacing $x$ with $x-2$ in $f(3x)$ gives $f\big(3(x-2)\big)=f(3x-6)$, a horizontal translation of 2 units in the positive $x$-direction. $$\boxed{\text{a horizontal stretch (scale factor }\tfrac13\text{), then a horizontal translation of 2 units right}}$$ (Applying the translation first would give $f(3(x-2))$, a different, incorrect route unless carefully re-derived; the safe method is always to factor out the coefficient of $x$ first, as done above, so the stretch is applied to the original variable before the shift.) (b) We require $g(x)=f(3(x-2))=7$. Since $f(9)=7$, we need the input of $f$ to equal 9: $$3(x-2)=9$$ $$x-2=3$$ $$\boxed{x=5}$$

QUESTION 29 6 marks

Medium

Let $f(x)=\cos x$. The graph of $g$ is obtained from the graph of $f$ by a horizontal translation of $\dfrac{\pi}{2}$ units in the negative $x$-direction, followed by a vertical stretch with scale factor 4. (a) Write down an expression for $g(x)$. [2] (b) Find the exact value of $g(0)$. [2] (c) Show that $g(x)=-4\sin x$ for all $x$. [2]

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(a) The horizontal translation $\dfrac\pi2$ units left gives $f\left(x+\dfrac\pi2\right)$. The vertical stretch with scale factor 4 then gives: $$\boxed{g(x)=4\cos\left(x+\dfrac{\pi}{2}\right)}$$ (b) Substituting $x=0$: $$g(0)=4\cos\left(0+\frac{\pi}{2}\right)=4\cos\left(\frac{\pi}{2}\right)=4(0)$$ $$\boxed{g(0)=0}$$ (c) Using the identity $\cos\left(x+\dfrac{\pi}{2}\right)=-\sin x$ (a standard shift identity for cosine): $$g(x)=4\cos\left(x+\frac{\pi}{2}\right)=4\big(-\sin x\big)$$ $$\boxed{g(x)=-4\sin x}$$ (As a check, at $x=0$ this gives $g(0)=-4\sin(0)=-4(0)=0$, consistent with part (b).)

QUESTION 30 7 marks

Medium

Let $f(x)=x^{3}$. The graph of $g$ is obtained from the graph of $f$ by a horizontal stretch with scale factor $\dfrac12$, followed by a reflection in the $x$-axis. (a) Write $g(x)$ in terms of $f$, and hence show that $g(x)=-8x^{3}$. [4] (b) Hence write down the value of the single vertical stretch scale factor that, applied directly to $f(x)=x^3$ together with a reflection in the $x$-axis, would produce the same graph as $g$. [1] (c) State whether $g$ is even, odd, or neither, justifying your answer. [2]

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(a) The horizontal stretch with scale factor $\dfrac12$ maps $f(x)$ to $f(2x)$. The reflection in the $x$-axis then gives: $$g(x)=-f(2x)$$ Substituting $f(x)=x^{3}$: $$g(x)=-f(2x)=-(2x)^{3}=-8x^{3}$$ $$\boxed{g(x)=-8x^{3}}$$ (b) Since $g(x)=-8x^3=-8f(x)$, applying a vertical stretch with scale factor 8 to $f$ (together with the reflection) produces the identical graph. $$\boxed{8}$$ (This mirrors part (a) of a horizontal-vs-vertical stretch equivalence: because $f(x)=x^3$ is a pure power function of odd degree 3, a horizontal stretch scale factor $\dfrac12$ multiplies every $y$-value by $2^3=8$, the same as a vertical stretch scale factor 8.) (c) Computing $g(-x)$: $$g(-x)=-8(-x)^{3}=-8(-x^{3})=8x^{3}=-g(x)$$ Since $g(-x)=-g(x)$ for all $x$, $g$ is an odd function. $$\boxed{g\text{ is odd}}$$

QUESTION 31 7 marks

Medium

Let $f(x)=\ln x$, for $x>0$. The graph of $g$ is obtained from the graph of $f$ by a horizontal stretch with scale factor $\dfrac12$, followed by a vertical translation of 3 units in the positive $y$-direction. (a) Write down an expression for $g(x)$, and state its domain. [3] (b) Write down the equation of the vertical asymptote of the graph of $g$. [1] (c) Find the value of $x$ for which $g(x)=3$. [3]

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(a) The horizontal stretch with scale factor $\dfrac12$ maps $f(x)$ to $f(2x)$. The vertical translation of 3 units up then gives: $$g(x)=f(2x)+3=\ln(2x)+3$$ For $\ln(2x)$ to be defined we need $2x>0$, i.e. $x>0$. $$\boxed{g(x)=\ln(2x)+3,\ \ \text{domain } x>0}$$ (b) The graph of $f(x)=\ln x$ has vertical asymptote $x=0$. A horizontal stretch about the $y$-axis leaves $x=0$ fixed (since $2(0)=0$), and a vertical translation does not affect a vertical asymptote. $$\boxed{x=0}$$ (c) Setting $g(x)=3$: $$\ln(2x)+3=3\ \Rightarrow\ \ln(2x)=0\ \Rightarrow\ 2x=e^{0}=1$$ $$\boxed{x=\dfrac12}$$

QUESTION 32 9 marks

Medium

Let $f(x)=\sqrt{x}$, for $x\ge0$. The graph of $g$ is obtained from the graph of $f$ by a horizontal translation of 1 unit in the positive $x$-direction, a reflection in the $x$-axis, and a vertical translation of 2 units in the positive $y$-direction, applied in that order. (a) Find an expression for $g(x)$. [3] (b) State the domain and range of $g$. [3] (c) Find the $x$-intercept of the graph of $g$. [3]

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(a) Applying the transformations in order: the horizontal translation gives $f(x-1)$; the reflection in the $x$-axis gives $-f(x-1)$; the vertical translation up 2 then gives: $$g(x)=-f(x-1)+2$$ $$\boxed{g(x)=-\sqrt{x-1}+2}$$ (b) For $\sqrt{x-1}$ to be defined we need $x-1\ge0$, so the domain is $x\ge1$ (translating the domain $x\ge0$ of $f$ right by 1 does not change with reflection/vertical shift). The range of $f$ is $y\ge0$. Reflecting in the $x$-axis gives $y\le0$; translating up 2 then gives $y\le2$. $$\boxed{\text{domain: } x\ge1,\ \ \text{range: } y\le2}$$ (c) Setting $g(x)=0$: $$-\sqrt{x-1}+2=0\ \Rightarrow\ \sqrt{x-1}=2\ \Rightarrow\ x-1=4$$ $$\boxed{x=5}$$

QUESTION 33 8 marks

Medium

Let $f(x)=|x|$. The graph of $g$ is obtained from the graph of $f$ by a horizontal translation of 3 units in the negative $x$-direction, followed by a vertical stretch with scale factor 2, followed by a vertical translation of 5 units in the negative $y$-direction. (a) Find an expression for $g(x)$. [3] (b) State the coordinates of the vertex of the graph of $g$. [2] (c) Solve $g(x)=0$. [3]

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(a) The horizontal translation gives $f(x+3)$. The vertical stretch with scale factor 2 gives $2f(x+3)$. The vertical translation down 5 then gives: $$g(x)=2f(x+3)-5$$ $$\boxed{g(x)=2|x+3|-5}$$ (b) The vertex of $f(x)=|x|$ is at $(0,0)$. Tracking it through the transformations: the horizontal translation sends $(0,0)\to(-3,0)$; the vertical stretch keeps the $x$-coordinate and doubles the $y$-coordinate: $(-3,0)\to(-3,0)$; the vertical translation down 5 sends $(-3,0)\to(-3,-5)$. $$\boxed{(-3,-5)}$$ (c) Setting $g(x)=0$: $$2|x+3|-5=0\ \Rightarrow\ |x+3|=\frac{5}{2}$$ $$x+3=\frac52\ \ \text{or}\ \ x+3=-\frac52$$ $$\boxed{x=-\dfrac12\ \ \text{or}\ \ x=-\dfrac{11}{2}}$$

QUESTION 34 7 marks

Medium

The point $(6,-2)$ lies on the graph of $y=f(x)$. The graph of $y=h(x)$ is obtained from the graph of $y=f(x)$ by a horizontal stretch with scale factor $\dfrac12$, followed by a reflection in the $x$-axis, followed by a vertical translation of 4 units in the positive $y$-direction. (a) Write $h(x)$ in terms of $f$. [3] (b) Find the coordinates of the point on the graph of $y=h(x)$ that corresponds to $(6,-2)$. [4]

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(a) The horizontal stretch with scale factor $\dfrac12$ maps $f(x)$ to $f(2x)$. The reflection in the $x$-axis gives $-f(2x)$. The vertical translation up 4 then gives: $$\boxed{h(x)=-f(2x)+4}$$ (b) We need the point $(X,Y)$ on $y=h(x)$ that uses the same input into $f$ as the given point, i.e. $2X=6\implies X=3$. $$Y=-f(2X)+4=-f(6)+4$$ Since $(6,-2)$ lies on $y=f(x)$, $f(6)=-2$: $$Y=-(-2)+4=2+4=6$$ $$\boxed{(3,6)}$$

QUESTION 35 6 marks

Medium

Let $f(x)=x^{2}$. The graph of $g$ is obtained from the graph of $f$ by a horizontal stretch with scale factor 2, followed by a reflection in the $x$-axis. (a) Write $g(x)$ in terms of $f$, and hence show that $g(x)=-\dfrac{x^{2}}{4}$. [4] (b) State the coordinates of the vertex of the graph of $g$, and determine whether it is a maximum or minimum point. [2]

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(a) A horizontal stretch with scale factor 2 maps $f(x)$ to $f\left(\dfrac{x}{2}\right)$ (replace $x$ with $\dfrac x2$, since scale factor $2=\dfrac{1}{1/2}$). The reflection in the $x$-axis then gives: $$g(x)=-f\left(\frac{x}{2}\right)$$ Substituting $f(x)=x^{2}$: $$g(x)=-\left(\frac{x}{2}\right)^{2}=-\frac{x^{2}}{4}$$ $$\boxed{g(x)=-\dfrac{x^{2}}{4}}$$ (b) The vertex of $f(x)=x^2$ is at $(0,0)$, a minimum. Both the horizontal stretch and the reflection in the $x$-axis fix the origin, so the vertex of $g$ remains at: $$\boxed{(0,0)}$$ Since $g(x)=-\dfrac{x^2}{4}$ has a negative leading coefficient, the parabola opens downward, so $(0,0)$ is a $\boxed{\text{maximum}}$ point.

QUESTION 36 7 marks

Medium

Let $f(x)=e^{x}$. The graph of $g$ is obtained from the graph of $f$ by a horizontal stretch with scale factor 2, followed by a vertical translation of 4 units in the negative $y$-direction. (a) Write down an expression for $g(x)$. [2] (b) Write down the equation of the horizontal asymptote of the graph of $g$. [2] (c) Find the exact $y$-intercept of the graph of $g$. [3]

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(a) A horizontal stretch with scale factor 2 maps $f(x)$ to $f\left(\dfrac x2\right)$. The vertical translation 4 units down then gives: $$\boxed{g(x)=e^{x/2}-4}$$ (b) The graph of $f(x)=e^x$ has horizontal asymptote $y=0$. A horizontal stretch does not change a horizontal asymptote, so it remains $y=0$. The vertical translation down 4 then sends $y=0$ to $y=0-4=-4$. $$\boxed{y=-4}$$ (c) Substituting $x=0$: $$g(0)=e^{0/2}-4=e^{0}-4=1-4$$ $$\boxed{g(0)=-3}$$

QUESTION 37 6 marks

Medium

Let $f(x)=\sin x$. The graph of $g$ is the reflection of the graph of $f$ in the $y$-axis, followed by a vertical translation of 1 unit in the positive $y$-direction. (a) Write down an expression for $g(x)$ in terms of $f$. [2] (b) Using the fact that $\sin x$ is an odd function, show that $g(x)=1-\sin x$. [2] (c) Hence describe a different single transformation of the graph of $f$ (not a reflection in the $y$-axis) that, together with the same vertical translation, would also produce the graph of $g$. [2]

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(a) The reflection in the $y$-axis gives $f(-x)$. The vertical translation 1 unit up then gives: $$\boxed{g(x)=f(-x)+1}$$ (b) Since $f(x)=\sin x$ is odd, $f(-x)=-\sin x$ for all $x$. Substituting: $$g(x)=f(-x)+1=-\sin x+1$$ $$\boxed{g(x)=1-\sin x}$$ (c) Since $g(x)=1-\sin x=-\sin x+1=-f(x)+1$, the same graph $g$ can also be obtained by applying a reflection in the $x$-axis to $f$ (giving $-f(x)$) and then the same vertical translation of 1 unit up. $$\boxed{\text{a reflection in the }x\text{-axis}}$$ (This works because $\sin x$ is odd: reflecting an odd function in the $y$-axis gives exactly the same graph as reflecting it in the $x$-axis, since $f(-x)=-f(x)$.)

QUESTION 38 7 marks

Medium

Let $f(x)=\dfrac1x$, for $x\neq0$. The graph of $g$ is obtained from the graph of $f$ by a reflection in the $x$-axis, followed by a horizontal translation of 2 units in the negative $x$-direction. (a) Find an expression for $g(x)$. [3] (b) Write down the equations of the two asymptotes of the graph of $g$. [2] (c) Explain why the order in which the two transformations are applied does not affect the final equation of $g$ in this particular case. [2]

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(a) The reflection in the $x$-axis gives $-f(x)$. The horizontal translation 2 units left then gives: $$g(x)=-f(x+2)$$ $$\boxed{g(x)=-\dfrac{1}{x+2}}$$ (b) The graph of $f$ has asymptotes $x=0$ and $y=0$. The reflection in the $x$-axis leaves $x=0$ unchanged and sends $y=0\to y=-0=0$. The translation 2 units left then sends $x=0\to x=-2$, while $y=0$ is unaffected by a horizontal translation. $$\boxed{x=-2\ \text{and}\ y=0}$$ (c) A reflection in the $x$-axis acts only on the output (the $y$-value), while a horizontal translation acts only on the input ($x$ is replaced by $x+2$ before $f$ is evaluated). Since these two transformations act on independent parts of the function (one on $x$ before evaluating $f$, one on the result afterwards), applying them in either order gives the same composite function: $$-f(x+2)=-\left(\frac{1}{x+2}\right)\quad\text{and}\quad \big(-f\big)(x+2)=-\frac{1}{x+2}$$ $$\boxed{\text{both orders give } g(x)=-\dfrac{1}{x+2}\text{, since the reflection (on the output) and the horizontal translation (on the input) do not interfere with each other}}$$

QUESTION 39 6 marks

Medium

The graph of $y=g(x)$ is obtained from the graph of $y=f(x)$ by applying, in the following order: a vertical stretch with scale factor 3, then a reflection in the $x$-axis, then a horizontal translation of 1 unit in the negative $x$-direction. (a) Write down an expression for $g(x)$ in terms of $f$. [3] (b) Given that $f(2)=-5$, find the value of $x$ such that $g(x)=15$. [3]

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(a) Applying the transformations in the given order: the vertical stretch (scale factor 3) gives $3f(x)$. The reflection in the $x$-axis gives $-3f(x)$. The horizontal translation of 1 unit in the negative $x$-direction replaces $x$ with $x-(-1)=x+1$, giving: $$\boxed{g(x)=-3f(x+1)}$$ (b) We require $g(x)=-3f(x+1)=15$, i.e. $f(x+1)=-5$. Since $f(2)=-5$, we need: $$x+1=2$$ $$\boxed{x=1}$$ (Check: $g(1)=-3f(1+1)=-3f(2)=-3(-5)=15$. $\checkmark$)

QUESTION 40 9 marks

Medium

The graph of $y=f(x)=x^{2}$ and the graph of $y=g(x)$, obtained from it by a sequence of transformations, are shown below. The vertex of $g$ is at $(-1,-3)$, and the point $(1,5)$ lies on the graph of $g$. (a) Explaining your reasoning, find the values of $a$, $h$ and $k$ such that $g(x)=a(x-h)^{2}+k$. [5] (b) Hence describe, in order, a sequence of transformations that maps the graph of $f$ onto the graph of $g$. [4]

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(a) The vertex of $y=a(x-h)^2+k$ is at $(h,k)$. From the graph, the vertex of $g$ is at $(-1,-3)$, so $h=-1$ and $k=-3$: $$g(x)=a(x+1)^{2}-3$$ Using the given point $(1,5)$ on the graph of $g$: $$5=a(1+1)^{2}-3$$ $$5=4a-3$$ $$4a=8\ \Rightarrow\ a=2$$ $$\boxed{a=2,\ h=-1,\ k=-3\ \Rightarrow\ g(x)=2(x+1)^{2}-3}$$ (b) Comparing $g(x)=2(x+1)^{2}-3=2f(x+1)-3$ with the vertex of $f$ at $(0,0)$ moving to $(-1,-3)$: Step 1: a horizontal translation of 1 unit in the negative $x$-direction (giving $f(x+1)$, vertex $\to(-1,0)$). Step 2: a vertical stretch with scale factor 2 (giving $2f(x+1)$, vertex stays at $(-1,0)$ since it is on the $x$-axis). Step 3: a vertical translation of 3 units in the negative $y$-direction (giving $2f(x+1)-3$, vertex $\to(-1,-3)$, matching the graph). $$\boxed{\text{translate 1 unit left, stretch vertically (factor 2), translate 3 units down}}$$

QUESTION 41 7 marks

Medium

Let $f(x)=\cos x$. The graph of $g$ is obtained from the graph of $f$ by a horizontal translation of $\pi$ units in the positive $x$-direction, a vertical stretch with scale factor 2, and a vertical translation of 1 unit in the negative $y$-direction, applied in that order. (a) Find an expression for $g(x)$. [3] (b) Find the maximum and minimum values of $g$, and one value of $x\in[0,2\pi]$ at which each occurs. [4]

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(a) The horizontal translation gives $f(x-\pi)$. The vertical stretch (scale factor 2) gives $2f(x-\pi)$. The vertical translation down 1 then gives: $$g(x)=2f(x-\pi)-1$$ $$\boxed{g(x)=2\cos(x-\pi)-1}$$ (b) The maximum value of $f(x)=\cos x$ is 1. The vertical stretch (scale factor 2) doubles this to 2, and the vertical translation down 1 gives a maximum of $2-1=1$ for $g$. This occurs where $\cos(x-\pi)=1$, i.e. $x-\pi=0$, so $x=\pi$ (within $[0,2\pi]$). The minimum value of $f$ is $-1$. Scaling by 2 gives $-2$, and translating down 1 gives a minimum of $-2-1=-3$ for $g$. This occurs where $\cos(x-\pi)=-1$, i.e. $x-\pi=\pi$, so $x=2\pi$; equivalently $x-\pi=-\pi$, so $x=0$. $$\boxed{\text{maximum}=1\text{ at }x=\pi;\quad \text{minimum}=-3\text{ at }x=0\ (\text{and }x=2\pi)}$$

QUESTION 42 12 marks

Hard

The graph of $y=f(x)=x^{3}-3x$ has a local maximum point at $(-1,2)$ and a local minimum point at $(1,-2)$. The graph of $y=f(x)$ is transformed to the graph of $y=g(x)=af\big(b(x-c)\big)+d$, where $a,b>0$. The image of the local maximum point of $f$ is the local maximum point $(0.5,5)$ of $g$, and the image of the local minimum point of $f$ is the local minimum point $(1.5,1)$ of $g$. (a) By comparing the local maximum and minimum points of $f$ and $g$, form and solve a system of equations to find the values of $a$, $b$, $c$ and $d$. [8] (b) Given also that $f(0)=0$, find the coordinates of the point on the graph of $g$ that corresponds to the point $(0,0)$ on the graph of $f$. [2] (c) State, giving a reason, whether $g$ is an odd function. [2]

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(a) Under $g(x)=af(b(x-c))+d$, the input to $f$ is $b(x-c)$ and the output is scaled by $a$ and shifted by $d$. Since a local max/min of $f$ maps to the corresponding local max/min of $g$ at the same relative position, the inputs must match: Local maximum: $f$'s local max input is $x=-1$; $g$'s local max is at $x=0.5$, so: $$b(0.5-c)=-1 \quad (1)$$ Local minimum: $f$'s local min input is $x=1$; $g$'s local min is at $x=1.5$, so: $$b(1.5-c)=1 \quad (2)$$ Subtracting (1) from (2): $$b(1.5-c)-b(0.5-c)=1-(-1)$$ $$b(1.5-c-0.5+c)=2$$ $$b(1)=2\ \Rightarrow\ b=2$$ Substituting $b=2$ into (1): $2(0.5-c)=-1\Rightarrow 1-2c=-1\Rightarrow c=1$. Now use the $y$-values. $g$'s local max value is $5=a\,f(-1)+d=a(2)+d$, and $g$'s local min value is $1=a\,f(1)+d=a(-2)+d$: $$2a+d=5 \quad (3)$$ $$-2a+d=1 \quad (4)$$ Adding (3) and (4): $2d=6\Rightarrow d=3$. Substituting into (3): $2a+3=5\Rightarrow a=1$. $$\boxed{a=1,\ b=2,\ c=1,\ d=3\ \ \Rightarrow\ \ g(x)=f\big(2(x-1)\big)+3}$$ (b) We require $2(X-1)=0\Rightarrow X=1$. $$g(1)=f(0)+3=0+3=3$$ $$\boxed{(1,3)}$$ (c) A function $g$ is odd only if $g(-x)=-g(x)$ for all $x$, which in particular requires $g(0)=0$ (since $g(0)=-g(0)\Rightarrow 2g(0)=0$). From part (b), $g(1)=3$, and $g$'s local max/min points $(0.5,5)$ and $(1.5,1)$ are not symmetric about the origin (nor is $g(0)$ equal to 0, since $2(0-1)=-2$ and $g(0)=f(-2)+3=(-8+6)+3=1\neq0$). $$\boxed{g\text{ is not odd (}g(0)=1\neq0\text{)}}$$

QUESTION 43 10 marks

Hard

The function $f$ is even. It is known that $f(0)=4$, $f(1)=5$ and $f(3)=-2$. The function $g$ is defined by $g(x)=2f(x+2)-1$. (a) Explain why the value of $f(-1)$ can be determined even though it is not one of the values given directly, and state its value. [2] (b) Find $g(-2)$, $g(-3)$ and $g(-5)$, showing your method clearly in each case. [8]

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(a) Since $f$ is even, $f(-x)=f(x)$ for all $x$ in the domain of $f$. Taking $x=1$: $$f(-1)=f(1)$$ Since $f(1)=5$ is given: $$\boxed{f(-1)=5}$$ (b) $g(x)=2f(x+2)-1$; we evaluate $f$ at $x+2$ for each required $x$. $g(-2)$: here $x+2=-2+2=0$, and $f(0)=4$ is given directly: $$g(-2)=2f(0)-1=2(4)-1=8-1$$ $$\boxed{g(-2)=7}$$ $g(-3)$: here $x+2=-3+2=-1$, which is not given directly, but by evenness $f(-1)=f(1)=5$ as in part (a): $$g(-3)=2f(-1)-1=2(5)-1=10-1$$ $$\boxed{g(-3)=9}$$ $g(-5)$: here $x+2=-5+2=-3$, which is not given directly; by evenness $f(-3)=f(3)=-2$: $$g(-5)=2f(-3)-1=2(-2)-1=-4-1$$ $$\boxed{g(-5)=-5}$$

QUESTION 44 10 marks

Hard

Let $f(x)=|x|$. The graph of $g$ shown below is obtained from the graph of $f$ by a sequence of transformations, and can be written as $g(x)=a|x-h|+k$. The vertex of the graph of $g$ is $(3,4)$, and the point $(5,-4)$ lies on the graph. (a) Find the values of $h$, $k$ and $a$. [6] (b) State whether $g$ represents a reflection of $f$ in the $x$-axis in addition to a stretch, and justify your answer using the sign of $a$. [2] (c) A function is even if its graph is symmetric about the $y$-axis. Explain why $g$, despite being built from the even function $f$, is not itself an even function. [2]

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(a) Since the vertex of $a|x-h|+k$ is at $(h,k)$, reading the vertex from the graph gives: $$h=3,\ k=4$$ $$g(x)=a|x-3|+4$$ Using the point $(5,-4)$: $$-4=a|5-3|+4$$ $$-4=2a+4$$ $$2a=-8\ \Rightarrow\ a=-4$$ $$\boxed{h=3,\ k=4,\ a=-4\ \Rightarrow\ g(x)=-4|x-3|+4}$$ (b) Since $a=-4<0$, the graph has been reflected in the $x$-axis (in addition to a vertical stretch with scale factor $|a|=4$): the vertex $(3,4)$ is a maximum point of a "$\vee$"-shaped graph turned upside-down into a "$\wedge$"-shape, confirming the reflection. $$\boxed{\text{yes; }a<0\text{ indicates a reflection in the }x\text{-axis}}$$ (c) $f(x)=|x|$ is even because its graph is symmetric about the $y$-axis (the line $x=0$). The transformation that produces $g$ includes a horizontal translation of 3 units (moving the vertex from $x=0$ to $x=3$), so the graph of $g$ is symmetric about the vertical line $x=3$, not about $x=0$. Since $g(-x)\neq g(x)$ in general (for example $g(0)=-4|{-3}|+4=-8$ while $g(-0)=g(0)=-8$ trivially, but comparing e.g. $g(1)=-4(2)+4=-4$ and $g(-1)=-4(4)+4=-12$, so $g(1)\neq g(-1)$): $$\boxed{g\text{ is not even, since its axis of symmetry } x=3\text{ has been shifted away from the }y\text{-axis by the horizontal translation}}$$

QUESTION 45 13 marks

Hard

The graph of $y=g(x)$ shown below is obtained from the graph of $y=\sin x$ by a sequence of transformations, and can be written as $g(x)=a\sin\big(b(x-c)\big)+d$, where $a,b>0$ and $0(a) Use the maximum and minimum values of $g$ to find $a$ and $d$. [3] (b) Use the horizontal distance between the maximum and minimum points to find the period of $g$, and hence find $b$. [3] (c) Given that a maximum of $\sin x$ itself occurs at $x=\dfrac{\pi}{2}$, use the location of $g$'s maximum point to find $c$. [3] (d) Write down the sequence of transformations, in the order they are applied to $y=\sin x$, that produces $y=g(x)$. [4]

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(a) The maximum value of $g$ is 4 and the minimum is $-2$. For $a\sin(\cdot)+d$ with $a>0$, the maximum is $a+d$ and the minimum is $-a+d$: $$a+d=4,\qquad -a+d=-2$$ Adding: $2d=2\Rightarrow d=1$. Substituting: $a=4-1=3$. $$\boxed{a=3,\ d=1}$$ (b) The horizontal distance from a maximum to the next minimum is half a period: $$\frac{11\pi}{12}-\frac{5\pi}{12}=\frac{6\pi}{12}=\frac{\pi}{2}$$ So the period is $2\times\dfrac{\pi}{2}=\pi$. Since the period of $a\sin(b(x-c))+d$ is $\dfrac{2\pi}{b}$: $$\frac{2\pi}{b}=\pi\ \Rightarrow\ b=2$$ $$\boxed{b=2}$$ (c) The graph of $\sin x$ has a maximum at $x=\dfrac{\pi}{2}$; under the horizontal translation (replacing $x$ with $x-c$, combined with the horizontal stretch already accounted for by $b$), the maximum of $b(x-c)=\dfrac{\pi}{2}$ occurs at $g$'s maximum $x=\dfrac{5\pi}{12}$: $$2\left(\frac{5\pi}{12}-c\right)=\frac{\pi}{2}$$ $$\frac{5\pi}{12}-c=\frac{\pi}{4}=\frac{3\pi}{12}$$ $$c=\frac{5\pi}{12}-\frac{3\pi}{12}=\frac{2\pi}{12}=\frac{\pi}{6}$$ $$\boxed{c=\dfrac{\pi}{6}}$$ So $g(x)=3\sin\left(2\left(x-\dfrac{\pi}{6}\right)\right)+1$. (d) Reading $g(x)=3f\big(2(x-\tfrac{\pi}{6})\big)+1$ with $f(x)=\sin x$, and applying the horizontal operations to $x$ before the vertical ones, in the correct order: $$\boxed{\text{horizontal stretch (scale factor }\tfrac12\text{), then horizontal translation (}\tfrac{\pi}{6}\text{ right), then vertical stretch (scale factor 3), then vertical translation (1 up)}}$$

QUESTION 46 12 marks

Hard

The function $f$ is odd. The graph of $y=f(x)$ is shown below for $x\ge0$ only, where it passes through the points $(0,0)$, $(2,3)$ and $(5,-1)$; no information about $f$ for $x<0$ is given directly. The function $g$ is defined by $g(x)=f(x+1)-2$. (a) Explain how the values of $f(-2)$ and $f(-5)$ can be found using the given information, and state their values. [3] (b) Find $g(-3)$, $g(4)$ and $g(-6)$, showing your method clearly. [7] (c) State the value of $f(0)$, and hence explain why the graph of $y=f(x)$ must pass through the origin for any odd function $f$ that is defined at $x=0$. [2]

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(a) Since $f$ is odd, $f(-x)=-f(x)$ for all $x$ in the domain. Taking $x=2$ and $x=5$: $$f(-2)=-f(2)=-3,\qquad f(-5)=-f(5)=-(-1)=1$$ $$\boxed{f(-2)=-3,\ \ f(-5)=1}$$ (b) $g(x)=f(x+1)-2$; we evaluate $f$ at $x+1$ in each case. $g(-3)$: here $x+1=-3+1=-2$. From (a), $f(-2)=-3$: $$g(-3)=f(-2)-2=-3-2$$ $$\boxed{g(-3)=-5}$$ $g(4)$: here $x+1=4+1=5$, given directly: $f(5)=-1$: $$g(4)=f(5)-2=-1-2$$ $$\boxed{g(4)=-3}$$ $g(-6)$: here $x+1=-6+1=-5$. From (a), $f(-5)=1$: $$g(-6)=f(-5)-2=1-2$$ $$\boxed{g(-6)=-1}$$ (c) Since $(0,0)$ lies on the given graph, $f(0)=0$. For any odd function defined at $x=0$: setting $x=0$ in $f(-x)=-f(x)$ gives $f(0)=-f(0)$, so $2f(0)=0$, forcing $f(0)=0$. Hence the graph of every odd function that is defined at $x=0$ must pass through the origin. $$\boxed{f(0)=0;\ \text{this follows for every odd function since }f(0)=-f(0)\Rightarrow f(0)=0}$$

QUESTION 47 11 marks

Hard

The graph of $y=g(x)$ shown below is obtained from the graph of $y=\dfrac1x$ by a sequence of transformations, and can be written as $g(x)=\dfrac{a}{x-h}+k$. The graph has asymptotes $x=-1$ and $y=3$, and passes through the point $(1,4)$. (a) Write down the values of $h$ and $k$, explaining how they follow from the asymptotes. [3] (b) Use the point $(1,4)$ to find the value of $a$. [3] (c) Describe, in order, a sequence of three transformations that maps the graph of $y=\dfrac1x$ onto the graph of $y=g(x)$. [5]

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(a) For $g(x)=\dfrac{a}{x-h}+k$, the vertical asymptote occurs where the denominator is zero, i.e. $x=h$; the horizontal asymptote is $y=k$ (the value the fraction term tends to 0 towards). Matching to the given asymptotes $x=-1$ and $y=3$: $$\boxed{h=-1,\ \ k=3}$$ (b) With $h=-1,k=3$: $g(x)=\dfrac{a}{x+1}+3$. Substituting the point $(1,4)$: $$4=\frac{a}{1+1}+3$$ $$1=\frac{a}{2}$$ $$\boxed{a=2}$$ So $g(x)=\dfrac{2}{x+1}+3$. (c) Writing $g(x)=2\left(\dfrac{1}{x+1}\right)+3$ in terms of $f(x)=\dfrac1x$: $g(x)=2f(x+1)+3$. Applying the horizontal operation to $x$ first, then the vertical ones, in order: $$\boxed{\text{a horizontal translation of 1 unit in the negative }x\text{-direction, then a vertical stretch (scale factor 2), then a vertical translation of 3 units in the positive }y\text{-direction}}$$ (As a check on the asymptotes: $f$'s asymptotes $x=0,y=0$ translate to $x=-1,y=0$, then the vertical stretch keeps $y=2(0)=0$, then the final translation gives $y=0+3=3$, matching $x=-1,y=3$ as given.)

QUESTION 48 12 marks

Hard

Let $f(x)=x^{2}$. The graph of $g$ shown below is obtained from the graph of $f$ by a sequence of transformations that includes a reflection, and can be written as $g(x)=a(x-h)^{2}+k$. The vertex of the graph of $g$ is $(3,4)$, and the point $(5,-4)$ lies on the graph. (a) Write down the values of $h$ and $k$. [2] (b) Use the point $(5,-4)$ to find the value of $a$, and explain how its sign confirms that a reflection in the $x$-axis was applied. [4] (c) Describe, in order, a sequence of three transformations that maps the graph of $f$ onto the graph of $g$. [4] (d) Explain why $g$ is neither an even nor an odd function, even though $f(x)=x^2$ is even. [2]

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(a) The vertex of $a(x-h)^2+k$ is $(h,k)$; reading the vertex of $g$ from the graph: $$\boxed{h=3,\ k=4}$$ (b) With $h=3,k=4$: $g(x)=a(x-3)^{2}+4$. Substituting $(5,-4)$: $$-4=a(5-3)^{2}+4$$ $$-4=4a+4$$ $$4a=-8\ \Rightarrow\ a=-2$$ $$\boxed{a=-2}$$ Since $a=-2<0$, the parabola opens downward rather than upward as $f(x)=x^2$ does; the vertex $(3,4)$ is a maximum point rather than a minimum. This sign change in the leading coefficient is exactly the effect of a reflection in the $x$-axis, confirming one occurred. (c) With $g(x)=-2(x-3)^2+4=-2f(x-3)+4$, applying the horizontal translation to $x$ first (before the vertical operations), then the vertical stretch and reflection (combined, since both act on the output of $f$), then the final vertical translation: $$\boxed{\text{translate 3 units right, stretch vertically by factor 2 and reflect in the }x\text{-axis, translate 4 units up}}$$ (d) $f(x)=x^2$ is even because its graph is symmetric about $x=0$. The transformation producing $g$ includes a horizontal translation of 3 units, shifting the axis of symmetry from $x=0$ to $x=3$; since $x=3\neq0$, $g(-x)\neq g(x)$ in general (for instance $g(0)=-2(9)+4=-14$ while $g(6)=-2(9)+4=-14\neq g(-6)$, or more simply $g(3)=4\neq g(-3)=-2(36)+4=-68$). $g$ is also not odd, since $g(0)=-14\neq0$ (an odd function must satisfy $g(0)=0$). $$\boxed{g\text{ is neither even nor odd: its axis of symmetry has moved off }x=0\text{ (not even), and }g(0)=-14\neq0\text{ (not odd)}}$$

QUESTION 49 12 marks

Hard

The function $f$ is odd, and it is known that $f(3)=-6$. The function $g$ is defined by $g(x)=f(2x-4)+1$. (a) By writing $2x-4$ in the form $2(x-c)$, state the value of $c$, and describe, in order, the sequence of three transformations that maps the graph of $f$ onto the graph of $g$. [5] (b) Find the value of $x$ for which $g(x)=f(3)+1$ directly (i.e. for which the input to $f$ equals 3). Hence state this value of $g(x)$. [3] (c) Find the value of $x$ for which the input to $f$ in $g(x)=f(2x-4)+1$ equals $-3$, and hence, using the fact that $f$ is odd, find $g(x)$ at this value of $x$. [4]

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(a) Factorising: $2x-4=2(x-2)$, so $c=2$. $$\boxed{c=2}$$ Building $f(2(x-2))+1$ from $f$, applying the horizontal stretch to $x$ before the horizontal translation: $$\boxed{\text{a horizontal stretch (scale factor }\tfrac12\text{), then a horizontal translation of 2 units in the positive }x\text{-direction, then a vertical translation of 1 unit up}}$$ (b) We require $2x-4=3$: $$2x=7\ \Rightarrow\ x=3.5$$ At this $x$: $g(3.5)=f(3)+1=-6+1$ $$\boxed{x=3.5,\ \ g(3.5)=-5}$$ (c) We require $2x-4=-3$: $$2x=1\ \Rightarrow\ x=0.5$$ At this $x$, $g(0.5)=f(-3)+1$. Since $f$ is odd, $f(-3)=-f(3)=-(-6)=6$: $$g(0.5)=6+1$$ $$\boxed{x=0.5,\ \ g(0.5)=7}$$

QUESTION 50 11 marks

Hard

Let $f(x)=x^{3}$. The graph of $y=g(x)$ shown below is obtained from the graph of $f$ by a sequence of transformations of the form $g(x)=af\big(b(x-c)\big)+d$, where $a,b>0$. The point of inflection of the graph of $g$, corresponding to the point of inflection $(0,0)$ of $f$, is at $(2,1)$, and the graph of $g$ also passes through the point $(3,6)$. (a) Explain why the point of inflection of $g$ gives the values of $c$ and $d$ directly, and state these values. [2] (b) Using the point $(3,6)$, show that $g(x)=5(x-2)^{3}+1$. [4] (c) Verify that $a=5,\ b=1$ and $a=40,\ b=\dfrac12$ both satisfy $g(x)=af\big(b(x-2)\big)+1$, and hence explain why the individual values of $a$ and $b$ cannot be uniquely determined from the graph alone. [5]

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(a) The point of inflection $(0,0)$ of $f$ is the unique point about which the graph of $f(x)=x^3$ is symmetric; whatever values $a$ and $b$ take, the horizontal translation $c$ and vertical translation $d$ move this point to $(c,d)$ (the scale factors $a,b$ do not move a point already at the "center" of the transformation, $x=c$, $y=d$). Reading the point of inflection of $g$ from the graph: $$\boxed{c=2,\ d=1}$$ (b) With $c=2,d=1$: $g(x)=af(b(x-2))+1=a\big(b(x-2)\big)^{3}+1=ab^{3}(x-2)^{3}+1$. Let $A=ab^3$ denote the combined coefficient, so $g(x)=A(x-2)^3+1$. Substituting the point $(3,6)$: $$6=A(3-2)^{3}+1$$ $$6=A(1)+1$$ $$A=5$$ $$\boxed{g(x)=5(x-2)^{3}+1}$$ (c) Checking $a=5,b=1$: $ab^{3}=5(1)^{3}=5$. Checking $a=40,b=\dfrac12$: $ab^{3}=40\left(\dfrac12\right)^{3}=40\times\dfrac18=5$. Both pairs give $ab^3=5=A$, so both satisfy $g(x)=A(x-2)^3+1=5(x-2)^3+1$ exactly, for every value of $x$: e.g. at $x=3$, $5(1)^3=40(\tfrac12)^3=5$ in each case. $$\boxed{\text{both pairs give } ab^3=5, \text{ so both produce the identical graph}}$$ Since $f(x)=x^3$ is a pure power function, $f(bx)=b^3f(x)$, so a horizontal stretch (factor $\tfrac1b$) and a vertical stretch (factor $a$) always combine into the single effective coefficient $A=ab^3$ in front of $(x-c)^3$. Any two data points on the graph of $g$ (beyond the point of inflection) can only ever determine this single product $A$, never $a$ and $b$ individually: infinitely many $(a,b)$ pairs with the same $ab^3$ produce exactly the same graph. $$\boxed{a\text{ and }b\text{ are not individually identifiable; only the combined coefficient }ab^{3}=5\text{ is determined by the graph}}$$

QUESTION 1 5 marks

Easy

Let $f(x) = 2x^3 - 5x^2 - 4x + 3$. (a) Find the remainder when $f(x)$ is divided by $(x-2)$. [2] (b) Show that $(x+1)$ is a factor of $f(x)$. [3]

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(a) By the remainder theorem, the remainder when $f(x)$ is divided by $(x-2)$ is $f(2)$. $$f(2) = 2(2)^3 - 5(2)^2 - 4(2) + 3$$ $$= 16 - 20 - 8 + 3 = -9$$ $$\boxed{\text{Remainder} = -9}$$ (b) By the factor theorem, $(x+1)$ is a factor of $f(x)$ if and only if $f(-1) = 0$. $$f(-1) = 2(-1)^3 - 5(-1)^2 - 4(-1) + 3$$ $$= -2 - 5 + 4 + 3 = 0$$ Since $f(-1) = 0$, by the factor theorem $(x+1)$ is a factor of $f(x)$. $\blacksquare$

QUESTION 2 6 marks

Medium

Let $f(x) = x^3 + kx^2 - 5x - 6$, where $k \in \mathbb{R}$. It is given that $(x-2)$ is a factor of $f(x)$. (a) Find the value of $k$. [3] (b) Hence express $f(x)$ as a product of three linear factors. [3]

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(a) By the factor theorem, since $(x-2)$ is a factor, $f(2) = 0$. $$f(2) = (2)^3 + k(2)^2 - 5(2) - 6$$ $$= 8 + 4k - 10 - 6 = 4k - 8$$ Setting $4k - 8 = 0$: $$\boxed{k = 2}$$ (b) With $k=2$: $f(x) = x^3 + 2x^2 - 5x - 6$. Dividing $f(x)$ by the known factor $(x-2)$ using polynomial long division gives: $$x^3 + 2x^2 - 5x - 6 = (x-2)(x^2 + 4x + 3)$$ Factorising the quadratic: $$x^2 + 4x + 3 = (x+1)(x+3)$$ $$\boxed{f(x) = (x-2)(x+1)(x+3)}$$, with roots $x = 2, -1, -3$.

QUESTION 3 6 marks

Easy

The cubic equation $2x^3 - 3x^2 + 4x - 5 = 0$ has roots $\alpha, \beta, \gamma$ (not necessarily real). (a) Write down the values of $\alpha+\beta+\gamma$, $\alpha\beta+\beta\gamma+\gamma\alpha$, and $\alpha\beta\gamma$. [3] (b) Hence find the exact value of $\alpha^2+\beta^2+\gamma^2$. [3]

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(a) For $ax^3+bx^2+cx+d=0$ with roots $\alpha,\beta,\gamma$, Vieta's formulas give: $$\alpha+\beta+\gamma = -\frac{b}{a}, \qquad \alpha\beta+\beta\gamma+\gamma\alpha = \frac{c}{a}, \qquad \alpha\beta\gamma = -\frac{d}{a}$$ Here $a=2, b=-3, c=4, d=-5$. Substituting these values: $$\boxed{\alpha+\beta+\gamma = \dfrac{3}{2}}$$, $$\boxed{\alpha\beta+\beta\gamma+\gamma\alpha = 2}$$, $$\boxed{\alpha\beta\gamma = \dfrac{5}{2}}$$ (b) Using the identity $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)$: $$\alpha^2+\beta^2+\gamma^2 = \left(\frac{3}{2}\right)^2 - 2(2)$$ $$= \frac{9}{4} - 4 = -\frac{7}{4}$$ $$\boxed{\alpha^2+\beta^2+\gamma^2 = -\dfrac{7}{4}}$$ (Note: this value being negative is not a contradiction, since this cubic has one real root and a complex conjugate pair, so $\alpha^2+\beta^2+\gamma^2$ need not be non-negative. The identity holds algebraically regardless of whether the roots are real.)

QUESTION 4 8 marks

Medium

The cubic equation $x^3 - 7x^2 + kx - 8 = 0$ has three roots that form a geometric sequence. Find the value of $k$ and solve the equation.

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Let the roots in geometric sequence be $\dfrac{a}{r}, a, ar$. By Vieta's formulas for $x^3 - 7x^2 + kx - 8=0$: Sum of roots: $\dfrac{a}{r} + a + ar = 7$ Product of roots: $\dfrac{a}{r}\cdot a \cdot ar = a^3 = 8$ From the product: $$a^3 = 8 \implies a = 2$$ Substituting into the sum: $$\frac{2}{r} + 2 + 2r = 7 \implies \frac{2}{r} + 2r = 5$$ Multiplying through by $r$: $$2 + 2r^2 = 5r$$ $$2r^2 - 5r + 2 = 0$$ Solving by the quadratic formula: $$r = \frac{5 \pm \sqrt{25-16}}{4} = \frac{5\pm 3}{4}$$ so $r = 2$ or $r = \dfrac12$. Taking $r=2$ (the case $r=\tfrac12$ gives the same set of roots in reverse order): the roots are $$\frac{a}{r} = 1, \quad a = 2, \quad ar = 4$$ Check: sum $=1+2+4=7$, product $=1\times2\times4=8$. By Vieta's formulas, $k$ equals the sum of pairwise products of the roots: $$k = (1)(2)+(2)(4)+(4)(1) = 2+8+4=14$$ $$\boxed{k = 14, \text{ roots } x = 1, 2, 4}$$ (Verification: $x^3-7x^2+14x-8 = (x-1)(x-2)(x-4)$, which expands correctly to the given cubic.)

QUESTION 5 18 marks

Hard

Let $f(x) = x^4 - 2x^3 - 7x^2 + 8x + 12$. (a) (i) Use the factor theorem to show that $(x-3)$ is a factor of $f(x)$. [2] (ii) Find the remainder when $f(x)$ is divided by $(x-1)$. [2] (b) Given that $(x+1)$ is also a factor of $f(x)$, express $f(x)$ as a product of four linear factors. [6] (c) Write down the sum and the product of the four roots of $f(x)=0$ directly from the coefficients of $f(x)$, and verify that these are consistent with the roots found in part (b). [4] (d) The roots of $f(x) = 0$ are $\alpha, \beta, \gamma, \delta$. Using Vieta's formulas for a quartic (i.e. without simply substituting the numerical roots found in (b)), find the exact value of $\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}+\dfrac{1}{\delta}$, giving your answer as a single fraction. [4]

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(a)(i) By the factor theorem, $(x-3)$ is a factor if $f(3)=0$. $$f(3) = 3^4 - 2(3)^3 - 7(3)^2 + 8(3) + 12$$ $$= 81 - 54 - 63 + 24 + 12 = 0$$ Since $f(3)=0$, $(x-3)$ is a factor. $\blacksquare$ (a)(ii) By the remainder theorem, the remainder is $f(1)$. $$f(1) = 1 - 2 - 7 + 8 + 12 = 12$$ $$\boxed{\text{Remainder} = 12}$$ (b) Since $(x-3)$ and $(x+1)$ are both factors, their product $(x-3)(x+1) = x^2-2x-3$ is a factor of $f(x)$. Dividing $f(x)$ by $x^2-2x-3$: write $f(x) = (x^2-2x-3)(x^2+ax+b)$. Expanding: $(x^2-2x-3)(x^2+ax+b) = x^4+(a-2)x^3+(b-2a-3)x^2+(-2b-3a)x-3b$. Comparing coefficients with $f(x)=x^4-2x^3-7x^2+8x+12$: $x^3$ coefficient: $a-2=-2 \implies a=0$ constant term: $-3b=12 \implies b=-4$ (checks: $x^2$ coefficient: $b-2a-3 = -4-0-3=-7$; $x^1$ coefficient: $-2b-3a = 8-0=8$) So the quadratic factor is $x^2+0x-4 = x^2-4 = (x-2)(x+2)$. $$\boxed{f(x) = (x-3)(x+1)(x-2)(x+2)}$$, with roots $x = 3, -1, 2, -2$. (c) For a quartic $x^4+px^3+qx^2+rx+s=0$ with roots $\alpha,\beta,\gamma,\delta$, Vieta's formulas give: sum of roots $=-p$ and product of roots $=s$ (since the leading coefficient is $1$ and the quartic has an even number of roots, $\alpha\beta\gamma\delta = s$). Here $p=-2, s=12$: $$\alpha+\beta+\gamma+\delta = -(-2) = 2, \qquad \alpha\beta\gamma\delta = 12$$ Check against part (b): $3+(-1)+2+(-2) = 2$, and $3\times(-1)\times2\times(-2)=12$. Consistent. (d) For a quartic $x^4+px^3+qx^2+rx+s$, using the elementary symmetric sums $e_1=-p,\, e_2=q,\, e_3=-r,\, e_4=s$: $$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta} = \frac{\beta\gamma\delta+\alpha\gamma\delta+\alpha\beta\delta+\alpha\beta\gamma}{\alpha\beta\gamma\delta} = \frac{e_3}{e_4}$$ (the sum of products of roots taken three at a time, divided by the product of all four roots). Here $p=-2,\ q=-7,\ r=8,\ s=12$, so $e_3 = -r = -8$ and $e_4 = s = 12$. $$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta} = \frac{-8}{12} = -\frac{2}{3}$$ $$\boxed{\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}+\dfrac{1}{\delta} = -\dfrac{2}{3}}$$ (Check using the actual roots: $\tfrac{1}{3}+\tfrac{1}{-1}+\tfrac{1}{2}+\tfrac{1}{-2} = \tfrac{1}{3}-1+\tfrac12-\tfrac12 = \tfrac13-1=-\tfrac23$.)

QUESTION 6 5 marks

Easy

Let $f(x) = x^3 - 3x^2 - 6x + 8$. (a) Find the remainder when $f(x)$ is divided by $(x-3)$. [2] (b) Show that $(x-1)$ is a factor of $f(x)$. [3]

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(a) By the remainder theorem, the remainder when $f(x)$ is divided by $(x-3)$ is $f(3)$. $$f(3) = (3)^3 - 3(3)^2 - 6(3) + 8$$ $$= 27 - 27 - 18 + 8 = -10$$ $$\boxed{\text{Remainder} = -10}$$ (b) By the factor theorem, $(x-1)$ is a factor of $f(x)$ if and only if $f(1) = 0$. $$f(1) = (1)^3 - 3(1)^2 - 6(1) + 8$$ $$= 1 - 3 - 6 + 8 = 0$$ Since $f(1) = 0$, by the factor theorem $(x-1)$ is a factor of $f(x)$. $\blacksquare$

QUESTION 7 5 marks

Easy

Let $f(x) = x^3 + kx^2 + 7x - 4$, where $k \in \mathbb{R}$. It is given that $(x+4)$ is a factor of $f(x)$. Find the value of $k$.

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By the factor theorem, since $(x+4)$ is a factor, $f(-4) = 0$. $$f(-4) = (-4)^3 + k(-4)^2 + 7(-4) - 4$$ $$= -64 + 16k - 28 - 4 = 16k - 96$$ Setting $16k - 96 = 0$: $$16k = 96$$ $$\boxed{k = 6}$$

QUESTION 8 4 marks

Easy

Let $p(x) = 3x^3 - 4x^2 - 5x + 7$. Find the remainder when $p(x)$ is divided by $(x+1)$.

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By the remainder theorem, the remainder when $p(x)$ is divided by $(x+1)$ is $p(-1)$. $$p(-1) = 3(-1)^3 - 4(-1)^2 - 5(-1) + 7$$ $$= -3 - 4 + 5 + 7 = 5$$ $$\boxed{\text{Remainder} = 5}$$

QUESTION 9 6 marks

Easy

Let $f(x) = x^3 - 5x^2 + 2x + 8$. (a) Show that $(x+1)$ is a factor of $f(x)$. [3] (b) Find the remainder when $f(x)$ is divided by $(x-1)$. [3]

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(a) By the factor theorem, $(x+1)$ is a factor of $f(x)$ if and only if $f(-1) = 0$. $$f(-1) = (-1)^3 - 5(-1)^2 + 2(-1) + 8$$ $$= -1 - 5 - 2 + 8 = 0$$ Since $f(-1) = 0$, $(x+1)$ is a factor of $f(x)$. $\blacksquare$ (b) By the remainder theorem, the remainder when $f(x)$ is divided by $(x-1)$ is $f(1)$. $$f(1) = (1)^3 - 5(1)^2 + 2(1) + 8$$ $$= 1 - 5 + 2 + 8 = 6$$ $$\boxed{\text{Remainder} = 6}$$

QUESTION 10 4 marks

Easy

The cubic equation $3x^3 + 6x^2 - 9x + 2 = 0$ has roots $\alpha, \beta, \gamma$ (not necessarily real). (a) Write down the values of $\alpha+\beta+\gamma$ and $\alpha\beta+\beta\gamma+\gamma\alpha$. [2] (b) Write down the value of $\alpha\beta\gamma$. [2]

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(a) For $ax^3+bx^2+cx+d=0$ with roots $\alpha,\beta,\gamma$, Vieta's formulas give: $$\alpha+\beta+\gamma = -\frac{b}{a}, \qquad \alpha\beta+\beta\gamma+\gamma\alpha = \frac{c}{a}$$ Here $a=3, b=6, c=-9$. Substituting: $$\boxed{\alpha+\beta+\gamma = -2}, \qquad \boxed{\alpha\beta+\beta\gamma+\gamma\alpha = -3}$$ (b) $$\alpha\beta\gamma = -\frac{d}{a} = -\frac{2}{3}$$ $$\boxed{\alpha\beta\gamma = -\dfrac{2}{3}}$$

QUESTION 11 7 marks

Easy

Let $f(x) = x^3 - 2x^2 - 5x + 6$. (a) Show that $(x-1)$ is a factor of $f(x)$. [2] (b) Hence express $f(x)$ as a product of three linear factors, and state the roots of $f(x)=0$. [3] (c) Sketch the graph of $y=f(x)$, clearly marking the $x$- and $y$-intercepts. [2]

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(a) By the factor theorem, $(x-1)$ is a factor of $f(x)$ if and only if $f(1)=0$. $$f(1) = 1 - 2 - 5 + 6 = 0$$ Since $f(1)=0$, $(x-1)$ is a factor. $\blacksquare$ (b) Dividing $f(x)$ by $(x-1)$ using synthetic (or long) division: $$x^3 - 2x^2 - 5x + 6 = (x-1)(x^2 - x - 6)$$ Factorising the quadratic: $$x^2 - x - 6 = (x-3)(x+2)$$ $$\boxed{f(x) = (x-1)(x-3)(x+2)}$$, with roots $x = 1, 3, -2$. (c) Since the leading coefficient is positive and $f(x)$ is a cubic, $f(x)\to -\infty$ as $x\to-\infty$ and $f(x)\to +\infty$ as $x\to+\infty$. The curve crosses the $x$-axis at $x=-2, 1, 3$ (each a simple root) and crosses the $y$-axis at $f(0)=6$.

QUESTION 12 6 marks

Easy

Let $f(x) = x^4 - 3x^3 + 2x^2 + x - 5$. When $f(x)$ is divided by $x^2-1$, the remainder has the form $ax+b$. Find the values of $a$ and $b$.

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Since $x^2-1=(x-1)(x+1)$, we can write $$f(x) = (x-1)(x+1)Q(x) + ax+b$$ for some quotient $Q(x)$. Substituting $x=1$ and $x=-1$ makes the $(x-1)(x+1)Q(x)$ term vanish, so: $$f(1) = a+b, \qquad f(-1) = -a+b$$ Evaluating $f$ at these points: $$f(1) = 1 - 3 + 2 + 1 - 5 = -4$$ $$f(-1) = 1 + 3 + 2 - 1 - 5 = 0$$ So $a+b=-4$ and $-a+b=0$, giving $a=b$. Substituting into the first equation: $2a = -4$, so $a=-2$ and $b=-2$. $$\boxed{a = -2, \ b = -2}$$ (Check: dividing directly gives $f(x) = (x^2-1)(x^2-3x+3) + (-2x-2)$, confirming the remainder is $-2x-2$.)

QUESTION 13 5 marks

Easy

Let $f(x) = 2x^3 - 3x^2 - kx + 10$, where $k \in \mathbb{R}$. It is given that $(x+2)$ is a factor of $f(x)$. Find the value of $k$.

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By the factor theorem, since $(x+2)$ is a factor, $f(-2) = 0$. $$f(-2) = 2(-2)^3 - 3(-2)^2 - k(-2) + 10$$ $$= -16 - 12 + 2k + 10 = 2k - 18$$ Setting $2k-18=0$: $$\boxed{k = 9}$$

QUESTION 14 6 marks

Easy

Let $h(x) = x^3 - x^2 - 10x - 8$. (a) Show that $x=-1$ is a root of $h(x)=0$, using the factor theorem. [2] (b) Hence find the other two roots of $h(x)=0$. [4]

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(a) By the factor theorem, $x=-1$ is a root of $h(x)=0$ if and only if $h(-1)=0$. $$h(-1) = (-1)^3 - (-1)^2 - 10(-1) - 8$$ $$= -1 - 1 + 10 - 8 = 0$$ Since $h(-1)=0$, $x=-1$ is a root and $(x+1)$ is a factor of $h(x)$. $\blacksquare$ (b) Dividing $h(x)$ by $(x+1)$: $$x^3 - x^2 - 10x - 8 = (x+1)(x^2 - 2x - 8)$$ Factorising the quadratic: $$x^2 - 2x - 8 = (x-4)(x+2)$$ So $h(x) = (x+1)(x-4)(x+2)$, giving roots $x=-1$, and: $$\boxed{x = 4, \ x = -2}$$

QUESTION 15 5 marks

Easy

Consider the cubic function $f(x) = (x-1)^2(x+2)$. (a) State the roots of $f(x)=0$ and their multiplicities. [2] (b) Sketch the graph of $y=f(x)$, showing the $x$- and $y$-intercepts and describing the behaviour of the curve at $x=1$. [3]

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(a) Reading directly from the factorised form $f(x)=(x-1)^2(x+2)$: $x=1$ is a root of multiplicity $2$ (a repeated/double root), and $x=-2$ is a root of multiplicity $1$ (a simple root). $$\boxed{x=1 \text{ (multiplicity 2)}, \quad x=-2 \text{ (multiplicity 1)}}$$ (b) The $y$-intercept is found by setting $x=0$: $$f(0) = (0-1)^2(0+2) = (1)(2) = 2$$ So the curve crosses the $x$-axis at $x=-2$ (simple root - the curve passes straight through, changing sign), and touches the $x$-axis at $x=1$ without crossing it (since the multiplicity there is even) - the curve turns back on the same side of the axis. The $y$-intercept is at $(0,2)$. Since $f(x)$ is a cubic with positive leading coefficient, $f(x)\to-\infty$ as $x\to-\infty$ and $f(x)\to+\infty$ as $x\to+\infty$.

QUESTION 16 6 marks

Easy

The cubic equation $x^3 - 4x^2 - 7x + 10 = 0$ has roots $\alpha, \beta, \gamma$. It is given that $\alpha = 1$. (a) Using Vieta's formulas (without dividing the cubic), find $\beta+\gamma$ and $\beta\gamma$. [3] (b) Hence find the values of $\beta$ and $\gamma$. [3]

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(a) By Vieta's formulas, for $x^3-4x^2-7x+10=0$: $$\alpha+\beta+\gamma = 4, \qquad \alpha\beta\gamma = -10$$ Since $\alpha=1$: $$1+\beta+\gamma = 4 \implies \boxed{\beta+\gamma = 3}$$ $$1\cdot\beta\gamma = -10 \implies \boxed{\beta\gamma = -10}$$ (b) Since $\beta$ and $\gamma$ have sum $3$ and product $-10$, they are the roots of $$t^2 - 3t - 10 = 0$$ $$(t-5)(t+2) = 0$$ $$\boxed{\beta = 5, \ \gamma = -2} \text{ (in either order)}$$

QUESTION 17 5 marks

Easy

Let $f(x) = x^3 + 2x^2 + ax - 3$, where $a \in \mathbb{R}$. When $f(x)$ is divided by $(x-2)$, the remainder is $11$. Find the value of $a$.

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By the remainder theorem, the remainder when $f(x)$ is divided by $(x-2)$ is $f(2)$. $$f(2) = (2)^3 + 2(2)^2 + a(2) - 3$$ $$= 8 + 8 + 2a - 3 = 2a + 13$$ Setting this equal to the given remainder: $$2a + 13 = 11$$ $$2a = -2$$ $$\boxed{a = -1}$$

QUESTION 18 4 marks

Easy

Let $g(x) = x^3 - 3x^2 - 4x + 10$. Determine, showing your working, whether $(x-2)$ is a factor of $g(x)$.

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By the factor theorem, $(x-2)$ is a factor of $g(x)$ if and only if $g(2)=0$. $$g(2) = (2)^3 - 3(2)^2 - 4(2) + 10$$ $$= 8 - 12 - 8 + 10 = -2$$ Since $g(2) = -2 \neq 0$, $(x-2)$ is $\boxed{\text{not}}$ a factor of $g(x)$ (by the remainder theorem, dividing $g(x)$ by $(x-2)$ leaves a remainder of $-2$).

QUESTION 19 6 marks

Easy

Let $p(x) = x^3 + 3x^2 - 10x - 24$. (a) Show that $(x+2)$ is a factor of $p(x)$. [2] (b) Hence fully factorise $p(x)$ and state all the roots of $p(x)=0$. [4]

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(a) By the factor theorem, $(x+2)$ is a factor of $p(x)$ if and only if $p(-2)=0$. $$p(-2) = (-2)^3 + 3(-2)^2 - 10(-2) - 24$$ $$= -8 + 12 + 20 - 24 = 0$$ Since $p(-2)=0$, $(x+2)$ is a factor. $\blacksquare$ (b) Dividing $p(x)$ by $(x+2)$: $$x^3+3x^2-10x-24 = (x+2)(x^2+x-12)$$ Factorising the quadratic: $$x^2+x-12 = (x+4)(x-3)$$ $$\boxed{p(x) = (x+2)(x+4)(x-3)}$$, with roots $x=-2, -4, 3$.

QUESTION 20 7 marks

Easy

Let $f(x) = x^3 - 4x^2 - x + 4$. (a) Show that $(x-1)$ and $(x+1)$ are both factors of $f(x)$. [3] (b) Hence write $f(x)$ as a product of three linear factors and sketch the graph of $y=f(x)$, labelling all intercepts. [4]

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(a) By the factor theorem: $$f(1) = 1 - 4 - 1 + 4 = 0 \implies (x-1) \text{ is a factor}$$ $$f(-1) = -1 - 4 + 1 + 4 = 0 \implies (x+1) \text{ is a factor}$$ $\blacksquare$ (b) Since $(x-1)$ and $(x+1)$ are both factors, their product $(x-1)(x+1) = x^2-1$ is a factor of $f(x)$. Dividing: $$f(x) = (x^2-1)(x-4)$$ $$\boxed{f(x) = (x-1)(x+1)(x-4)}$$ The graph crosses the $x$-axis at $x=-1, 1, 4$ and the $y$-axis at $f(0)=4$. As a positive-leading-coefficient cubic, $f(x)\to-\infty$ as $x\to-\infty$ and $f(x)\to+\infty$ as $x\to+\infty$.

QUESTION 21 5 marks

Easy

Let $f(x) = 4x^3 - 2x^2 + 5x - 1$. Find the remainder when $f(x)$ is divided by $(2x-1)$.

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The divisor $2x-1$ is zero when $x=\dfrac12$. By the remainder theorem, the remainder when $f(x)$ is divided by $(2x-1)$ is $f\left(\dfrac12\right)$. $$f\left(\tfrac12\right) = 4\left(\tfrac12\right)^3 - 2\left(\tfrac12\right)^2 + 5\left(\tfrac12\right) - 1$$ $$= 4\left(\tfrac18\right) - 2\left(\tfrac14\right) + \tfrac52 - 1 = \tfrac12 - \tfrac12 + \tfrac52 - 1$$ $$= \tfrac32$$ $$\boxed{\text{Remainder} = \dfrac{3}{2}}$$

QUESTION 22 4 marks

Easy

The quartic equation $2x^4 - 3x^3 + x^2 + 5x - 6 = 0$ has roots $\alpha,\beta,\gamma,\delta$ (not necessarily real). (a) Write down the value of $\alpha+\beta+\gamma+\delta$. [2] (b) Write down the value of $\alpha\beta\gamma\delta$. [2]

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For a quartic $ax^4+bx^3+cx^2+dx+e=0$ with roots $\alpha,\beta,\gamma,\delta$, Vieta's formulas give: $$\alpha+\beta+\gamma+\delta = -\frac{b}{a}, \qquad \alpha\beta\gamma\delta = \frac{e}{a}$$ Here $a=2,\ b=-3,\ e=-6$. (a) $$\alpha+\beta+\gamma+\delta = -\frac{-3}{2} = \boxed{\dfrac{3}{2}}$$ (b) $$\alpha\beta\gamma\delta = \frac{-6}{2} = \boxed{-3}$$

QUESTION 23 6 marks

Easy

Let $h(x) = x^3 + x^2 - 8x - 12$. (a) Show that $x=-2$ is a repeated (double) root of $h(x)=0$. [3] (b) Hence find all the roots of $h(x)=0$. [3]

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(a) By the factor theorem, $x=-2$ is a root since: $$h(-2) = (-2)^3+(-2)^2-8(-2)-12 = -8+4+16-12 = 0$$ Dividing $h(x)$ by $(x+2)$: $$h(x) = (x+2)(x^2 - x - 6)$$ Testing whether $x=-2$ is also a root of the quotient $x^2-x-6$: $$(-2)^2 - (-2) - 6 = 4+2-6 = 0$$ So $(x+2)$ divides the quotient again, confirming $x=-2$ is a repeated root. $\blacksquare$ (b) Factorising the quotient $x^2-x-6 = (x-3)(x+2)$, so $$h(x) = (x+2)(x+2)(x-3) = (x+2)^2(x-3)$$ $$\boxed{x = -2 \text{ (double root)}, \quad x = 3}$$

QUESTION 24 8 marks

Medium

Let $p(x) = 3x^3 - 10x^2 - 9x + 4$. (a) Show that $(x+1)$ is a factor of $p(x)$. [2] (b) Hence express $p(x)$ as a product of three linear factors, and state all the roots of $p(x)=0$. [4] (c) Verify that the sum of the roots found in (b) agrees with the value given by Vieta's formula $-\dfrac{b}{a}$. [2]

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(a) By the factor theorem, $(x+1)$ is a factor if $p(-1)=0$. $$p(-1) = 3(-1)^3 - 10(-1)^2 - 9(-1) + 4$$ $$= -3 - 10 + 9 + 4 = 0$$ Since $p(-1)=0$, $(x+1)$ is a factor. $\blacksquare$ (b) Dividing $p(x)$ by $(x+1)$ using synthetic division with coefficients $3, -10, -9, 4$ and root $-1$: Bring down $3$; multiply by $-1$ to get $-3$, add to $-10$ to get $-13$; multiply by $-1$ to get $13$, add to $-9$ to get $4$; multiply by $-1$ to get $-4$, add to $4$ to get remainder $0$. $$p(x) = (x+1)(3x^2 - 13x + 4)$$ Factorising the quadratic (using the quadratic formula or by inspection, since $3\times4=12$ and $-1\times-12=12$ with $-1-12=-13$): $$3x^2-13x+4 = (3x-1)(x-4)$$ $$\boxed{p(x) = (x+1)(3x-1)(x-4)}$$, with roots $x = -1, \ \dfrac13, \ 4$. (c) Sum of roots found: $-1+\dfrac13+4 = \dfrac{10}{3}$. By Vieta's formula, $-\dfrac{b}{a} = -\dfrac{-10}{3} = \dfrac{10}{3}$. $$\boxed{\text{Both equal } \dfrac{10}{3}, \text{ confirming consistency.}}$$

QUESTION 25 8 marks

Medium

Let $f(x) = x^3 + ax^2 + bx - 6$, where $a, b \in \mathbb{R}$. It is given that $(x-1)$ and $(x+2)$ are both factors of $f(x)$. (a) Form two equations in $a$ and $b$, and solve them to find the values of $a$ and $b$. [4] (b) Hence express $f(x)$ as a product of three linear factors and state all the roots of $f(x)=0$. [4]

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(a) By the factor theorem, $f(1)=0$ and $f(-2)=0$. $$f(1) = 1+a+b-6 = a+b-5 = 0 \implies a+b=5 \quad (1)$$ $$f(-2) = -8+4a-2b-6 = 4a-2b-14=0 \implies 4a-2b=14 \implies 2a-b=7 \quad (2)$$ Adding $(1)$ and $(2)$: $$(a+b)+(2a-b) = 5+7$$ $$3a = 12 \implies a = 4$$ Substituting into $(1)$: $4+b=5 \implies b=1$. $$\boxed{a=4, \ b=1}$$ (b) With $a=4,\ b=1$: $f(x) = x^3+4x^2+x-6$. Since $(x-1)$ and $(x+2)$ are factors, so is their product $(x-1)(x+2)=x^2+x-2$. Dividing: $$f(x) = (x^2+x-2)(x+3)$$ $$\boxed{f(x) = (x-1)(x+2)(x+3)}$$, with roots $x = 1, -2, -3$. (Check: product of roots $= 1\times(-2)\times(-3) = 6 = -\dfrac{-6}{1}$, consistent with Vieta's formula.)

QUESTION 26 9 marks

Medium

Let $f(x) = x^3 - 3x^2 + 4x - 2$. (a) Show that $x=1$ is a root of $f(x)=0$. [2] (b) Hence find the remaining quadratic factor of $f(x)$. [2] (c) Solve $f(x)=0$ completely, giving the complex roots in the form $p+qi$. [3] (d) Verify that the sum of all three roots (real and complex) agrees with Vieta's formula. [2]

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(a) By the factor theorem: $$f(1) = 1 - 3 + 4 - 2 = 0$$ so $x=1$ is a root and $(x-1)$ is a factor. $\blacksquare$ (b) Dividing $f(x)$ by $(x-1)$: $$f(x) = (x-1)(x^2-2x+2)$$ $$\boxed{x^2-2x+2}$$ (c) Solving $x^2-2x+2=0$ using the quadratic formula: $$x = \frac{2 \pm \sqrt{(-2)^2-4(1)(2)}}{2} = \frac{2\pm\sqrt{4-8}}{2} = \frac{2\pm\sqrt{-4}}{2} = \frac{2\pm2i}{2}$$ $$\boxed{x = 1+i, \ x=1-i}$$, together with $x=1$ from part (a). The complex roots occur as a conjugate pair, as expected for a cubic with real coefficients. (d) Sum of all three roots: $1+(1+i)+(1-i) = 3$. By Vieta's formula, sum of roots $= -\dfrac{b}{a} = -\dfrac{-3}{1} = 3$. $$\boxed{\text{Both equal } 3, \text{ confirming consistency.}}$$

QUESTION 27 8 marks

Medium

Let $f(x) = x^4 - 3x^3 - 15x^2 + 19x + 30$. It is given that $(x+3)$ and $(x-2)$ are both factors of $f(x)$. (a) Show that the product $(x+3)(x-2)$ divides $f(x)$ exactly, by dividing $f(x)$ by $x^2+x-6$. [4] (b) Hence find all four roots of $f(x)=0$. [4]

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(a) Since $(x+3)$ and $(x-2)$ are both factors of $f(x)$, their product $$(x+3)(x-2) = x^2+x-6$$ must also divide $f(x)$ exactly. Dividing $f(x) = x^4-3x^3-15x^2+19x+30$ by $x^2+x-6$ using polynomial long division: $x^4-3x^3-15x^2+19x+30 \div (x^2+x-6)$: the first term of the quotient is $x^2$, since $x^2\cdot(x^2+x-6) = x^4+x^3-6x^2$. Subtracting: $$(x^4-3x^3-15x^2) - (x^4+x^3-6x^2) = -4x^3-9x^2$$ Bringing down the remaining terms gives $-4x^3-9x^2+19x+30$. The next quotient term is $-4x$, since $-4x\cdot(x^2+x-6)=-4x^3-4x^2+24x$. Subtracting: $$(-4x^3-9x^2+19x)-(-4x^3-4x^2+24x) = -5x^2-5x$$ Bringing down the constant gives $-5x^2-5x+30$. The next quotient term is $-5$, since $-5\cdot(x^2+x-6)=-5x^2-5x+30$. Subtracting gives remainder $0$. $$f(x) = (x^2+x-6)(x^2-4x-5)$$ Since the remainder is $0$, $(x+3)(x-2)$ divides $f(x)$ exactly. $\blacksquare$ (b) Factorising the remaining quadratic: $$x^2-4x-5 = (x-5)(x+1)$$ $$\boxed{f(x) = (x+3)(x-2)(x-5)(x+1)}$$, with roots $x = -3, 2, 5, -1$.

QUESTION 28 8 marks

Medium

The cubic equation $x^3 - 12x^2 + kx - 28 = 0$ has three roots that form an arithmetic sequence. Find the value of $k$ and solve the equation.

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Let the roots in arithmetic progression be $a-d,\ a,\ a+d$. By Vieta's formulas for $x^3-12x^2+kx-28=0$: Sum of roots: $(a-d)+a+(a+d) = 3a = 12 \implies a=4$ Product of roots: $(a-d)(a)(a+d) = a(a^2-d^2) = 28$ Substituting $a=4$: $$4(16-d^2) = 28$$ $$16-d^2 = 7$$ $$d^2 = 9 \implies d = 3$$ So the roots are $4-3=1,\ 4,\ 4+3=7$. Check: sum $=1+4+7=12$ (correct), product $=1\times4\times7=28$ (correct). By Vieta's formulas, $k$ equals the sum of pairwise products of the roots: $$k = (1)(4)+(4)(7)+(7)(1) = 4+28+7 = 39$$ $$\boxed{k=39, \text{ roots } x=1,4,7}$$ (Verification: $x^3-12x^2+39x-28 = (x-1)(x-4)(x-7)$, which expands correctly to the given cubic.)

QUESTION 29 8 marks

Medium

The cubic equation $x^3 + px^2 + qx + r = 0$ has roots $\alpha,\beta,\gamma$ satisfying $$\alpha+\beta+\gamma = 6, \qquad \alpha\beta+\beta\gamma+\gamma\alpha = -1, \qquad \alpha\beta\gamma = -30$$ (a) Find the values of $p$, $q$ and $r$. [3] (b) Hence solve the cubic completely, given that all three roots are integers. [5]

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(a) By Vieta's formulas for $x^3+px^2+qx+r=0$: $$p = -(\alpha+\beta+\gamma), \qquad q = \alpha\beta+\beta\gamma+\gamma\alpha, \qquad r = -\alpha\beta\gamma$$ Substituting the given values: $$\boxed{p=-6, \ q=-1, \ r=30}$$ (b) The cubic is $x^3-6x^2-x+30=0$. Since the roots are integers, by the rational root theorem any integer root must divide $30$ (the constant term). Testing $x=-2$: $$(-2)^3-6(-2)^2-(-2)+30 = -8-24+2+30 = 0$$ So $(x+2)$ is a factor. Dividing: $$x^3-6x^2-x+30 = (x+2)(x^2-8x+15)$$ Factorising the quadratic: $$x^2-8x+15 = (x-3)(x-5)$$ $$\boxed{x=-2, \ x=3, \ x=5}$$ (Check: sum $=-2+3+5=6$ (correct), pairwise sum $=(-2)(3)+(3)(5)+(5)(-2)=-6+15-10=-1$ (correct), product $=(-2)(3)(5)=-30$ (correct) - all consistent with the given data.)

QUESTION 30 8 marks

Medium

Let $f(x) = x^4 + ax^3 + bx^2 + 5x + 6$, where $a,b\in\mathbb{R}$. When $f(x)$ is divided by $x^2-4$, the remainder is $13x+10$. (a) Using the fact that $x^2-4=(x-2)(x+2)$, form two equations in $a$ and $b$ by evaluating $f(2)$ and $f(-2)$. [4] (b) Solve these equations to find $a$ and $b$. [4]

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(a) Since $x^2-4=(x-2)(x+2)$, we may write $f(x) = (x-2)(x+2)Q(x) + 13x+10$ for some quotient $Q(x)$. Substituting $x=2$ and $x=-2$ makes the $(x-2)(x+2)Q(x)$ term vanish, so $f(2)$ and $f(-2)$ must equal the remainder evaluated at those points. $$f(2) = 16+8a+4b+10+6 = 8a+4b+32$$ This must equal $13(2)+10=36$: $$8a+4b+32=36 \implies 8a+4b=4 \implies 2a+b=1 \quad (1)$$ $$f(-2) = 16-8a+4b-10+6 = -8a+4b+12$$ This must equal $13(-2)+10=-16$: $$-8a+4b+12=-16 \implies -8a+4b=-28 \implies -2a+b=-7 \quad (2)$$ (b) Adding $(1)$ and $(2)$: $$(2a+b)+(-2a+b) = 1+(-7)$$ $$2b = -6 \implies b=-3$$ Subtracting $(2)$ from $(1)$: $$(2a+b)-(-2a+b) = 1-(-7)$$ $$4a = 8 \implies a=2$$ $$\boxed{a=2, \ b=-3}$$ (Check: with $a=2,b=-3$, $f(x)=x^4+2x^3-3x^2+5x+6$. Dividing by $x^2-4$ by long division gives quotient $x^2+2x+1$ and remainder $13x+10$, confirming the given remainder.)

QUESTION 31 8 marks

Medium

Let $f(x) = (x+1)(x-2)^2(x-4)$. (a) State the roots of $f(x)=0$ and their multiplicities. [2] (b) Expand $f(x)$ into the form $x^4+px^3+qx^2+rx+s$. [3] (c) Sketch the graph of $y=f(x)$, describing the behaviour of the curve at $x=2$, and use your expanded form together with Vieta's formulas to state the sum of the four roots (counted with multiplicity). [3]

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(a) Reading directly from the factorised form: $x=-1$ (multiplicity $1$), $x=2$ (multiplicity $2$), $x=4$ (multiplicity $1$). $$\boxed{x=-1,\ x=2\ (\text{double}),\ x=4}$$ (b) First expand $(x-2)^2 = x^2-4x+4$. Then: $$(x+1)(x-4) = x^2-3x-4$$ Now multiply $(x^2-4x+4)(x^2-3x-4)$: $$= x^2(x^2-3x-4) -4x(x^2-3x-4) + 4(x^2-3x-4)$$ $$= (x^4-3x^3-4x^2) + (-4x^3+12x^2+16x) + (4x^2-12x-16)$$ $$= x^4 + (-3-4)x^3 + (-4+12+4)x^2 + (16-12)x - 16$$ $$\boxed{f(x) = x^4-7x^3+12x^2+4x-16}$$ (c) The curve crosses the $x$-axis at $x=-1$ and $x=4$ (simple roots), and touches the $x$-axis at $x=2$ without crossing (even multiplicity) - the curve turns back on the same side. The $y$-intercept is $f(0)=-16$. Since the leading coefficient is positive and the degree is even, $f(x)\to+\infty$ as $x\to\pm\infty$. By Vieta's formula, the sum of the roots (with multiplicity) is $-p = -(-7) = 7$. $$\boxed{\text{Sum of roots} = 7}$$ (Check: $-1+2+2+4=7$.)

QUESTION 32 9 marks

Medium

Let $f(x) = x^4 + 2x^3 - 13x^2 - 14x + 24$. Find all four roots of $f(x)=0$, showing your method clearly.

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By the rational root theorem, any rational root of $f(x)=0$ must be an integer dividing the constant term $24$ (since the leading coefficient is $1$). Candidates include $\pm1,\pm2,\pm3,\pm4,\pm6,\ldots$ Testing $x=1$: $$f(1) = 1+2-13-14+24 = 0$$ So $(x-1)$ is a factor. Dividing by synthetic division (coefficients $1,2,-13,-14,24$, root $1$): $$f(x) = (x-1)(x^3+3x^2-10x-24)$$ Testing the cubic factor at $x=-2$: $$(-2)^3+3(-2)^2-10(-2)-24 = -8+12+20-24=0$$ So $(x+2)$ is a factor of the cubic. Dividing $x^3+3x^2-10x-24$ by $(x+2)$: $$x^3+3x^2-10x-24 = (x+2)(x^2+x-12)$$ Factorising the remaining quadratic: $$x^2+x-12 = (x+4)(x-3)$$ Putting it all together: $$f(x) = (x-1)(x+2)(x+4)(x-3)$$ $$\boxed{x = 1,\ -2,\ -4,\ 3}$$

QUESTION 33 9 marks

Medium

The cubic equation $x^3 + px^2 - 13x + q = 0$ has $x=3$ as one root. The product of the other two roots is $-4$. (a) Using Vieta's formula for the sum of pairwise products of the roots, find $p$. [4] (b) Using Vieta's formula for the product of all three roots, find $q$. [2] (c) Hence find the other two roots. [3]

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Let the other two roots be $r_1$ and $r_2$, so $r_1 r_2 = -4$ (given). (a) By Vieta's formula, the sum of pairwise products of all three roots equals $\dfrac{-13}{1}=-13$ (the coefficient of $x$): $$3r_1+3r_2+r_1r_2 = -13$$ $$3(r_1+r_2) + (-4) = -13$$ $$3(r_1+r_2) = -9 \implies r_1+r_2 = -3$$ By Vieta's formula, the sum of all three roots equals $-p$: $$3+r_1+r_2 = -p \implies 3+(-3) = -p \implies 0=-p$$ $$\boxed{p = 0}$$ (b) By Vieta's formula, the product of all three roots equals $-q$: $$3\cdot r_1 r_2 = -q \implies 3(-4) = -q \implies -12=-q$$ $$\boxed{q=12}$$ (c) Since $r_1+r_2=-3$ and $r_1r_2=-4$, $r_1$ and $r_2$ are roots of $$t^2+3t-4=0$$ $$(t+4)(t-1)=0$$ $$\boxed{r_1=1, \ r_2=-4}$$ (in either order) (Check: the cubic is $x^3-13x+12=0$; indeed $(x-3)(x-1)(x+4) = x^3-13x+12$.)

QUESTION 34 7 marks

Medium

Let $f(x) = x^4 + 2x^3 - x^2 + 3x - 5$. Find the quotient and remainder when $f(x)$ is divided by $x^2+x+1$.

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Using polynomial long division: The first term of the quotient is $x^2$, since $x^2(x^2+x+1) = x^4+x^3+x^2$. Subtracting from $f(x)$: $$(x^4+2x^3-x^2) - (x^4+x^3+x^2) = x^3-2x^2$$ Bringing down the next term: $x^3-2x^2+3x$. The next quotient term is $x$, since $x(x^2+x+1)=x^3+x^2+x$. Subtracting: $$(x^3-2x^2+3x)-(x^3+x^2+x) = -3x^2+2x$$ Bringing down the constant: $-3x^2+2x-5$. The next quotient term is $-3$, since $-3(x^2+x+1)=-3x^2-3x-3$. Subtracting: $$(-3x^2+2x-5)-(-3x^2-3x-3) = 5x-2$$ Since $5x-2$ has lower degree than $x^2+x+1$, this is the remainder. $$\boxed{\text{Quotient} = x^2+x-3, \quad \text{Remainder} = 5x-2}$$ (Check: $(x^2+x+1)(x^2+x-3)+(5x-2) = x^4+2x^3-x^2+3x-5$, which matches $f(x)$.)

QUESTION 35 8 marks

Medium

Let $f(x) = -x^3 + 6x^2 - 3x - 10$. (a) Show that $(x+1)$ is a factor of $f(x)$. [2] (b) Hence express $f(x)$ as a product of three linear factors (including any constant factor) and state the roots of $f(x)=0$. [4] (c) State the sign of $f(x)$ as $x\to\infty$ and as $x\to-\infty$, justifying your answer using the leading term. [2]

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(a) By the factor theorem, $(x+1)$ is a factor if $f(-1)=0$. $$f(-1) = -(-1)^3+6(-1)^2-3(-1)-10$$ $$= 1+6+3-10 = 0$$ Since $f(-1)=0$, $(x+1)$ is a factor. $\blacksquare$ (b) Dividing $f(x)$ by $(x+1)$ using synthetic division (coefficients $-1,6,-3,-10$, root $-1$): Bring down $-1$; multiply by $-1$ to get $1$, add to $6$ to get $7$; multiply by $-1$ to get $-7$, add to $-3$ to get $-10$; multiply by $-1$ to get $10$, add to $-10$ to get remainder $0$. $$f(x) = (x+1)(-x^2+7x-10)$$ Factorising $-x^2+7x-10 = -(x^2-7x+10) = -(x-2)(x-5)$: $$\boxed{f(x) = -(x+1)(x-2)(x-5)}$$, with roots $x=-1, 2, 5$. (c) The leading term of $f(x)$ is $-x^3$. As $x\to\infty$, $-x^3\to-\infty$, so $f(x)\to-\infty$. As $x\to-\infty$, $-x^3\to+\infty$, so $f(x)\to+\infty$. $$\boxed{f(x)\to-\infty \text{ as } x\to\infty; \quad f(x)\to+\infty \text{ as } x\to-\infty}$$

QUESTION 36 8 marks

Medium

The quartic equation $x^4 - 5x^3 + 3x^2 + 7x - 2 = 0$ has roots $\alpha,\beta,\gamma,\delta$ (not necessarily real). Without finding the individual roots, find the exact value of $\alpha^2+\beta^2+\gamma^2+\delta^2$.

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By Vieta's formulas for $x^4+px^3+qx^2+rx+s=0$ with roots $\alpha,\beta,\gamma,\delta$: $$\alpha+\beta+\gamma+\delta = -p, \qquad \alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta = q$$ Here $p=-5,\ q=3$, so: $$\alpha+\beta+\gamma+\delta = 5, \qquad \sum_{i

QUESTION 37 7 marks

Medium

Let $f(x) = x^3 - x^2 + kx + 8$, where $k \in \mathbb{R}$. The remainder when $f(x)$ is divided by $(x-2)$ exceeds the remainder when $f(x)$ is divided by $(x+1)$ by $15$. Find the value of $k$.

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By the remainder theorem, the two remainders are $f(2)$ and $f(-1)$. $$f(2) = 8-4+2k+8 = 2k+12$$ $$f(-1) = -1-1-k+8 = 6-k$$ The condition states $f(2) - f(-1) = 15$: $$(2k+12)-(6-k) = 15$$ $$3k+6 = 15$$ $$3k=9$$ $$\boxed{k=3}$$ (Check: with $k=3$, $f(2)=2(3)+12=18$ and $f(-1)=6-3=3$; indeed $18-3=15$.)

QUESTION 38 7 marks

Medium

Given that two of the roots of $x^3-2x^2-9x+18=0$ are negatives of each other, find all three roots. (Hint: group the terms of the cubic in pairs.)

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Group the terms of $x^3-2x^2-9x+18$ in pairs: $$x^3-2x^2-9x+18 = (x^3-2x^2) + (-9x+18)$$ $$= x^2(x-2) - 9(x-2)$$ $$= (x-2)(x^2-9)$$ Factorising the difference of squares $x^2-9=(x-3)(x+3)$: $$x^3-2x^2-9x+18 = (x-2)(x-3)(x+3)$$ Indeed the roots $x=3$ and $x=-3$ are negatives of each other, consistent with the given information. $$\boxed{x = 2, \ x=3, \ x=-3}$$

QUESTION 39 8 marks

Medium

Let $f(x) = x^3 + ax^2 - 7x + b$, where $a,b\in\mathbb{R}$. It is given that $(x+2)$ is a factor of $f(x)$, and that $f(x)$ leaves a remainder of $-18$ when divided by $(x-1)$. (a) Form two equations in $a$ and $b$, and solve them. [4] (b) Hence express $f(x)$ as a product of a linear factor and a quadratic factor, and solve $f(x)=0$, giving any irrational roots in exact (surd) form. [4]

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(a) By the factor theorem, $f(-2)=0$: $$f(-2) = -8+4a+14+b = 4a+b+6 = 0 \implies 4a+b=-6 \quad (1)$$ By the remainder theorem, $f(1)=-18$: $$f(1) = 1+a-7+b = a+b-6 = -18 \implies a+b=-12 \quad (2)$$ Subtracting $(2)$ from $(1)$: $$(4a+b)-(a+b) = -6-(-12)$$ $$3a = 6 \implies a=2$$ Substituting into $(2)$: $2+b=-12 \implies b=-14$. $$\boxed{a=2, \ b=-14}$$ (b) With $a=2,\ b=-14$: $f(x) = x^3+2x^2-7x-14$. Dividing by the known factor $(x+2)$: $$f(x) = (x+2)(x^2-7)$$ Setting each factor to zero: $x+2=0 \implies x=-2$, and $x^2-7=0 \implies x=\pm\sqrt7$. $$\boxed{f(x)=(x+2)(x^2-7), \quad x=-2, \ x=\sqrt7, \ x=-\sqrt7}$$

QUESTION 40 7 marks

Medium

The quartic equation $2x^4 + px^3 - x^2 + qx - 8 = 0$ has roots $\alpha,\beta,\gamma,\delta$ satisfying $$\alpha+\beta+\gamma+\delta = 3, \qquad \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta = -7$$ (a) State Vieta's formulas for the sum of the roots and the sum of the products of the roots taken three at a time, for a general quartic $ax^4+bx^3+cx^2+dx+e=0$. [2] (b) Find the value of $p$. [2] (c) Find the value of $q$. [3]

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(a) For $ax^4+bx^3+cx^2+dx+e=0$ with roots $\alpha,\beta,\gamma,\delta$: $$\alpha+\beta+\gamma+\delta = -\frac{b}{a}, \qquad \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta = -\frac{d}{a}$$ (b) Here $a=2,\ b=p$. Using the sum-of-roots formula: $$3 = -\frac{p}{2}$$ $$\boxed{p=-6}$$ (c) Here $d=q$. Using the sum-of-triple-products formula: $$-7 = -\frac{q}{2}$$ $$\boxed{q=14}$$

QUESTION 41 8 marks

Medium

A cubic function has the form $f(x) = a(x+2)(x-1)(x-3)$ and passes through the point $(0,-12)$. (a) Find the value of $a$. [2] (b) Expand $f(x)$ into the form $Ax^3+Bx^2+Cx+D$. [3] (c) Using the factorised form from part (a), find the remainder when $f(x)$ is divided by $(x-5)$. [3]

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(a) Substituting $x=0$ and $f(0)=-12$: $$f(0) = a(0+2)(0-1)(0-3) = a(2)(-1)(-3) = 6a$$ $$6a=-12 \implies \boxed{a=-2}$$ (b) $f(x) = -2(x+2)(x-1)(x-3)$. First expand $(x+2)(x-1) = x^2+x-2$, then multiply by $(x-3)$: $$(x^2+x-2)(x-3) = x^3-3x^2+x^2-3x-2x+6 = x^3-2x^2-5x+6$$ Multiplying by $-2$: $$\boxed{f(x) = -2x^3+4x^2+10x-12}$$ (c) By the remainder theorem, the remainder when $f(x)$ is divided by $(x-5)$ is $f(5)$. Using the factorised form (avoiding re-expanding): $$f(5) = -2(5+2)(5-1)(5-3) = -2(7)(4)(2) = -2(56) = -112$$ $$\boxed{\text{Remainder} = -112}$$ (Check using the expanded form: $-2(125)+4(25)+10(5)-12 = -250+100+50-12=-112$, which agrees.)

QUESTION 42 13 marks

Hard

Let $f(x) = x^4 - x^3 - 3x^2 + 17x - 30$. (a) (i) Use the factor theorem to show that $(x-2)$ is a factor of $f(x)$. [2] (ii) Use the factor theorem to show that $(x+3)$ is a factor of $f(x)$. [2] (b) Hence find the quadratic factor of $f(x)$ that remains after dividing out $(x-2)(x+3)$. [4] (c) Given that this quadratic factor has no real roots, solve $f(x)=0$ completely, expressing the complex roots in the form $p+qi$. [3] (d) Let the four roots (two real, two complex) be $\alpha,\beta,\gamma,\delta$. Using Vieta's formula for a quartic (i.e. without simply adding the four numerical roots), find $\alpha+\beta+\gamma+\delta$, and verify this is consistent with the roots found in part (c). [2]

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(a)(i) By the factor theorem, $(x-2)$ is a factor if $f(2)=0$. $$f(2) = 2^4-2^3-3(2)^2+17(2)-30 = 16-8-12+34-30 = 0$$ Since $f(2)=0$, $(x-2)$ is a factor. $\blacksquare$ (a)(ii) By the factor theorem, $(x+3)$ is a factor if $f(-3)=0$. $$f(-3) = (-3)^4-(-3)^3-3(-3)^2+17(-3)-30 = 81+27-27-51-30 = 0$$ Since $f(-3)=0$, $(x+3)$ is a factor. $\blacksquare$ (b) Since $(x-2)$ and $(x+3)$ are both factors, their product $(x-2)(x+3)=x^2+x-6$ is a factor of $f(x)$. Write $f(x) = (x^2+x-6)(x^2+ax+b)$. Expanding: $$(x^2+x-6)(x^2+ax+b) = x^4+(a+1)x^3+(b+a-6)x^2+(b-6a)x-6b$$ Comparing coefficients with $f(x)=x^4-x^3-3x^2+17x-30$: $x^3$ coefficient: $a+1=-1 \implies a=-2$ constant term: $-6b=-30 \implies b=5$ (check: $x^2$ coefficient: $b+a-6 = 5-2-6=-3$ (correct); $x^1$ coefficient: $b-6a = 5+12=17$ correct) $$\boxed{x^2-2x+5}$$ (c) Solving $x^2-2x+5=0$ by the quadratic formula: $$x = \frac{2\pm\sqrt{(-2)^2-4(1)(5)}}{2} = \frac{2\pm\sqrt{4-20}}{2} = \frac{2\pm\sqrt{-16}}{2} = \frac{2\pm4i}{2}$$ $$\boxed{x=2,\ x=-3,\ x=1+2i,\ x=1-2i}$$ (d) For a quartic $x^4+px^3+qx^2+rx+s=0$, Vieta's formula gives the sum of roots as $-p$. Here $p=-1$, so: $$\alpha+\beta+\gamma+\delta = -(-1) = 1$$ Check against part (c): $2+(-3)+(1+2i)+(1-2i) = 2-3+1+1 = 1$. Consistent (the imaginary parts of the conjugate pair cancel, as expected). $\blacksquare$

QUESTION 43 11 marks

Hard

Let $f(x) = x^3 + ax^2 + bx - 24$, where $a,b\in\mathbb{R}$. It is given that $(x-4)$ and $(x+1)$ are both factors of $f(x)$. (a) Form two equations in $a$ and $b$, and solve them. [4] (b) Using Vieta's formula for the product of the roots of $f(x)=0$ (i.e. without dividing $f(x)$ by its known factors), find the third root. [3] (c) Hence write $f(x)$ as a product of three linear factors, and sketch the graph of $y=f(x)$, labelling all axis intercepts. [4]

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(a) By the factor theorem, $f(4)=0$ and $f(-1)=0$. $$f(4) = 64+16a+4b-24 = 16a+4b+40=0 \implies 4a+b=-10 \quad (1)$$ $$f(-1) = -1+a-b-24 = a-b-25=0 \implies a-b=25 \quad (2)$$ From $(2)$: $a=b+25$. Substituting into $(1)$: $$4(b+25)+b=-10$$ $$4b+100+b=-10$$ $$5b=-110 \implies b=-22$$ Then $a=b+25=-22+25=3$. $$\boxed{a=3, \ b=-22}$$ (b) With $a=3,\ b=-22$: $f(x)=x^3+3x^2-22x-24$. Let the third root be $r$. By Vieta's formula, the product of all three roots is $-\dfrac{d}{a}$ where $d=-24,\ a=1$: $$4\times(-1)\times r = -\frac{-24}{1} = 24$$ $$-4r = 24 \implies \boxed{r=-6}$$ (c) $$\boxed{f(x) = (x-4)(x+1)(x+6)}$$ The curve crosses the $x$-axis at $x=-6,-1,4$ and the $y$-axis at $f(0)=-24$. As a positive-leading-coefficient cubic, $f(x)\to-\infty$ as $x\to-\infty$ and $f(x)\to+\infty$ as $x\to+\infty$. (Check: $(x-4)(x+1)(x+6)$ expands to $x^3+3x^2-22x-24$, matching $f(x)$.)

QUESTION 44 12 marks

Hard

The quartic equation $x^4 - 15x^3 + 70x^2 - 120x + 64 = 0$ has four roots that form a geometric sequence, all of which are real and positive. Find the four roots.

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Let the roots in geometric progression be $a,\ ar,\ ar^2,\ ar^3$ (all positive, so $a>0,\ r>0$). By Vieta's formulas for $x^4-15x^3+70x^2-120x+64=0$: Product of roots: $a\cdot ar\cdot ar^2\cdot ar^3 = a^4r^6 = \dfrac{64}{1}=64$ Sum of roots: $a+ar+ar^2+ar^3 = a(1+r+r^2+r^3) = 15$ From the product equation, $a^4r^6=64=2^6$, so $(a^{2}r^{3})^2 = 2^6$, giving $a^2r^3 = 2^3=8$ (taking the positive root since $a,r>0$). This suggests trying integer values of $a$ and $r$ that satisfy $a^4r^6=64$; testing $a=1,\ r=2$: $$a^4r^6 = 1^4\cdot2^6 = 64 \ \checkmark$$ Checking the sum: $1(1+2+4+8) = 1(15)=15$ (correct), which matches the required sum exactly. So $a=1,\ r=2$, giving roots: $$a=1,\quad ar=2,\quad ar^2=4,\quad ar^3=8$$ $$\boxed{x=1,\ 2,\ 4,\ 8}$$ (Check: the product $1\times2\times4\times8=64$ (correct), and expanding $(x-1)(x-2)(x-4)(x-8)$ gives exactly $x^4-15x^3+70x^2-120x+64$, confirming the roots. Also, the coefficient of $x^2$, $70$, equals the sum of all pairwise products of the roots: $1(2)+1(4)+1(8)+2(4)+2(8)+4(8) = 2+4+8+8+16+32=70$ (correct).)

QUESTION 45 11 marks

Hard

A quartic equation with real coefficients, $x^4+px^3+qx^2+rx+s=0$, has $x=-1$ and $x=4$ as two of its roots. Its other two roots are a complex conjugate pair $x=m\pm ni$ where $m,n\in\mathbb{R},\ n>0$. It is given that the sum of all four roots is $7$ and the sum of the products of the roots taken three at a time is $-1$. (a) Find $m$. [3] (b) Given additionally that the product of all four roots is $-20$, find $n$. [4] (c) Write down the quartic equation in expanded form $x^4+px^3+qx^2+rx+s=0$. [4]

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(a) By Vieta's formula, the sum of all four roots is $7$: $$(-1)+4+(m+ni)+(m-ni) = 7$$ The imaginary parts cancel (since $ni$ and $-ni$ sum to zero), leaving: $$3+2m = 7$$ $$\boxed{m=2}$$ (b) By Vieta's formula, the product of all four roots is $-20$: $$(-1)(4)(m+ni)(m-ni) = -20$$ Since $(m+ni)(m-ni)=m^2+n^2$: $$-4(m^2+n^2) = -20$$ $$m^2+n^2 = 5$$ With $m=2$: $4+n^2=5 \implies n^2=1 \implies n=1$ (taking $n>0$). $$\boxed{n=1}$$ So the complex roots are $2+i$ and $2-i$. (c) The four roots are $-1,\ 4,\ 2+i,\ 2-i$. Since $-1$ and $4$ are roots, $(x+1)(x-4)=x^2-3x-4$ is a factor; since $2\pm i$ are roots, $(x-(2+i))(x-(2-i)) = (x-2)^2-(i)^2 = x^2-4x+4+1=x^2-4x+5$ is the other quadratic factor. Multiplying: $$(x^2-3x-4)(x^2-4x+5)$$ $$= x^2(x^2-4x+5) -3x(x^2-4x+5) -4(x^2-4x+5)$$ $$= (x^4-4x^3+5x^2) + (-3x^3+12x^2-15x) + (-4x^2+16x-20)$$ $$= x^4 + (-4-3)x^3 + (5+12-4)x^2 + (-15+16)x - 20$$ $$\boxed{x^4-7x^3+13x^2+x-20=0}$$

QUESTION 46 10 marks

Hard

Let $f(x) = x^3 - 4x^2 - 2x + k$, where $k\in\mathbb{R}$. It is given that $(x-2)$ is a factor of $f(x)$. (a) Find the value of $k$. [3] (b) Hence find the quadratic factor that remains after dividing $f(x)$ by $(x-2)$. [3] (c) Solve $f(x)=0$ completely, giving the irrational roots in exact (surd) form. [4]

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(a) By the factor theorem, $(x-2)$ is a factor if $f(2)=0$. $$f(2) = 8-16-4+k = k-12$$ Setting $k-12=0$: $$\boxed{k=12}$$ (b) With $k=12$: $f(x)=x^3-4x^2-2x+12$. Dividing by $(x-2)$ using synthetic division (coefficients $1,-4,-2,12$, root $2$): Bring down $1$; multiply by $2$ to get $2$, add to $-4$ to get $-2$; multiply by $2$ to get $-4$, add to $-2$ to get $-6$; multiply by $2$ to get $-12$, add to $12$ to get remainder $0$. $$f(x) = (x-2)(x^2-2x-6)$$ $$\boxed{x^2-2x-6}$$ (c) The factor $(x-2)=0$ gives $x=2$. For $x^2-2x-6=0$, using the quadratic formula: $$x = \frac{2\pm\sqrt{(-2)^2-4(1)(-6)}}{2} = \frac{2\pm\sqrt{4+24}}{2} = \frac{2\pm\sqrt{28}}{2}$$ Since $\sqrt{28}=2\sqrt7$: $$x = \frac{2\pm2\sqrt7}{2} = 1\pm\sqrt7$$ $$\boxed{x=2, \quad x=1+\sqrt7, \quad x=1-\sqrt7}$$

QUESTION 47 12 marks

Hard

Let $f(x) = x^4 + 3x^3 - 15x^2 + kx + 30$, where $k\in\mathbb{R}$. It is given that $(x-3)$ is a factor of $f(x)$. (a) Find the value of $k$. [3] (b) Given that $(x+2)$ is also a factor of $f(x)$, find the quadratic factor that remains after dividing out $(x-3)(x+2)$. [5] (c) Hence find all four roots of $f(x)=0$. [4]

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(a) By the factor theorem, $(x-3)$ is a factor if $f(3)=0$. $$f(3) = 81+81-135+3k+30 = 3k+57$$ Setting $3k+57=0$: $$\boxed{k=-19}$$ (b) With $k=-19$: $f(x) = x^4+3x^3-15x^2-19x+30$. Since $(x-3)$ and $(x+2)$ are both factors, their product $$(x-3)(x+2) = x^2-x-6$$ is a factor of $f(x)$. Dividing $f(x)$ by $x^2-x-6$: the first quotient term is $x^2$, since $x^2(x^2-x-6)=x^4-x^3-6x^2$. Subtracting: $$(x^4+3x^3-15x^2)-(x^4-x^3-6x^2) = 4x^3-9x^2$$ Bringing down the remaining terms: $4x^3-9x^2-19x+30$. The next quotient term is $4x$, since $4x(x^2-x-6)=4x^3-4x^2-24x$. Subtracting: $$(4x^3-9x^2-19x)-(4x^3-4x^2-24x) = -5x^2+5x$$ Bringing down the constant: $-5x^2+5x+30$. The next quotient term is $-5$, since $-5(x^2-x-6)=-5x^2+5x+30$. Subtracting gives remainder $0$. $$\boxed{x^2+4x-5}$$ (c) Factorising the remaining quadratic: $$x^2+4x-5 = (x+5)(x-1)$$ $$f(x) = (x-3)(x+2)(x+5)(x-1)$$ $$\boxed{x=3,\ -2,\ -5,\ 1}$$

QUESTION 48 12 marks

Hard

Let $f(x) = 2x^4 + ax^3 + 12x^2 + bx - 30$, where $a,b\in\mathbb{R}$. It is given that $(x-1)$ and $(x+3)$ are both factors of $f(x)$. (a) Find the values of $a$ and $b$. [5] (b) Hence factorise $f(x)$ completely over the reals, given that the remaining quadratic factor has no real roots, and find its complex roots. [5] (c) Verify Vieta's product-of-roots relation using all four roots (two real, two complex). [2]

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(a) By the factor theorem, $f(1)=0$ and $f(-3)=0$. $$f(1) = 2+a+12+b-30 = a+b-16=0 \implies a+b=16 \quad (1)$$ $$f(-3) = 2(81)+a(-27)+12(9)+b(-3)-30 = 162-27a+108-3b-30 = 240-27a-3b=0$$ $$\implies 27a+3b=240 \implies 9a+b=80 \quad (2)$$ Subtracting $(1)$ from $(2)$: $$(9a+b)-(a+b) = 80-16$$ $$8a=64 \implies a=8$$ Substituting into $(1)$: $8+b=16 \implies b=8$. $$\boxed{a=8, \ b=8}$$ (b) With $a=8,\ b=8$: $f(x) = 2x^4+8x^3+12x^2+8x-30$. Since $(x-1)$ and $(x+3)$ are factors, $(x-1)(x+3)=x^2+2x-3$ divides $f(x)$. Write $f(x) = 2(x^2+2x-3)(x^2+cx+d)$ and expand: $$2(x^2+2x-3)(x^2+cx+d) = 2\left[x^4+(c+2)x^3+(d+2c-3)x^2+(2d-3c)x-3d\right]$$ Comparing the $x^3$ coefficient: $2(c+2)=8 \implies c+2=4 \implies c=2$. Comparing the constant term: $2(-3d)=-30 \implies -6d=-30 \implies d=5$. (check: $x^2$ coefficient $2(d+2c-3)=2(5+4-3)=12$ (correct); $x^1$ coefficient $2(2d-3c)=2(10-6)=8$ correct) So the remaining quadratic factor is $x^2+2x+5$. Solving $x^2+2x+5=0$: $$x = \frac{-2\pm\sqrt{4-20}}{2} = \frac{-2\pm4i}{2} = -1\pm2i$$ $$\boxed{f(x) = 2(x-1)(x+3)(x^2+2x+5), \quad x=1,\ -3,\ -1+2i,\ -1-2i}$$ (c) The product of all four roots: $1\times(-3)\times(-1+2i)\times(-1-2i)$. Since $(-1+2i)(-1-2i) = 1+4=5$: $$1\times(-3)\times5 = -15$$ By Vieta's formula for $2x^4+8x^3+12x^2+8x-30=0$, the product of the roots is $\dfrac{e}{a}=\dfrac{-30}{2}=-15$. $$\boxed{\text{Both equal } -15, \text{ confirming consistency.}}$$

QUESTION 49 11 marks

Hard

Let $f(x) = x^4 - 6x^3 + 15x^2 - 20x + 12$. (a) Show that $x=2$ is a repeated (double) root of $f(x)=0$. [4] (b) Find the remaining quadratic factor and solve it to find the two complex roots of $f(x)=0$, giving your answers in the form $p+qi$. [4] (c) State the sum of all four roots using Vieta's formula, and verify it against the roots found in (a) and (b). [3]

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(a) By the factor theorem, $x=2$ is a root since: $$f(2) = 16-48+60-40+12 = 0$$ Dividing $f(x)$ by $(x-2)$ using synthetic division (coefficients $1,-6,15,-20,12$, root $2$): Bring down $1$; $\times2=2$, $-6+2=-4$; $\times2=-8$, $15-8=7$; $\times2=14$, $-20+14=-6$; $\times2=-12$, $12-12=0$. $$f(x) = (x-2)(x^3-4x^2+7x-6)$$ Testing whether $x=2$ is also a root of the cubic factor: $$(2)^3-4(2)^2+7(2)-6 = 8-16+14-6=0$$ So $(x-2)$ divides the cubic factor again, confirming $x=2$ is (at least) a double root. $\blacksquare$ (b) Dividing $x^3-4x^2+7x-6$ by $(x-2)$ (coefficients $1,-4,7,-6$, root $2$): Bring down $1$; $\times2=2$, $-4+2=-2$; $\times2=-4$, $7-4=3$; $\times2=6$, $-6+6=0$. $$x^3-4x^2+7x-6 = (x-2)(x^2-2x+3)$$ So $f(x) = (x-2)^2(x^2-2x+3)$, confirming $x=2$ is exactly a double root (the quadratic factor $x^2-2x+3$ does not have $x=2$ as a root, since $4-4+3=3\neq0$). Solving $x^2-2x+3=0$: $$x = \frac{2\pm\sqrt{4-12}}{2} = \frac{2\pm\sqrt{-8}}{2} = \frac{2\pm2i\sqrt2}{2}$$ $$\boxed{x = 1+i\sqrt2, \quad x=1-i\sqrt2}$$ (c) For $f(x)=x^4-6x^3+15x^2-20x+12=0$, Vieta's formula gives the sum of the four roots (with multiplicity) as $-p$ where $p=-6$: $$\text{Sum} = -(-6) = 6$$ Check using the roots found: $2+2+(1+i\sqrt2)+(1-i\sqrt2) = 4+2 = 6$. Consistent. $\blacksquare$ $$\boxed{\text{Sum of roots} = 6}$$

QUESTION 50 10 marks

Hard

The cubic equation $x^3 + px^2 - 16x + q = 0$ has $x=4$ as one root. The sum of the other two roots is $-1$. (a) Using Vieta's formula for the sum of all three roots, find $p$. [2] (b) Using Vieta's formula for the sum of the pairwise products of the roots (the coefficient of $x$), find the product of the other two roots. [5] (c) Hence find $q$, and find the other two roots. [3]

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Let the other two roots be $r_1,r_2$, so $r_1+r_2=-1$ (given). (a) By Vieta's formula, the sum of all three roots is $-p$: $$4+r_1+r_2 = -p$$ $$4+(-1) = -p$$ $$3=-p \implies \boxed{p=-3}$$ (b) By Vieta's formula, the sum of the pairwise products of the roots equals the coefficient of $x$, which is $-16$: $$4r_1+4r_2+r_1r_2 = -16$$ $$4(r_1+r_2)+r_1r_2=-16$$ $$4(-1)+r_1r_2 = -16$$ $$-4+r_1r_2=-16$$ $$\boxed{r_1r_2 = -12}$$ (c) By Vieta's formula, the product of all three roots is $-q$: $$4\times r_1r_2 = -q$$ $$4(-12)=-q \implies -48=-q \implies \boxed{q=48}$$ Since $r_1+r_2=-1$ and $r_1r_2=-12$, $r_1$ and $r_2$ are roots of $$t^2+t-12=0$$ $$(t+4)(t-3)=0$$ $$\boxed{r_1=-4, \ r_2=3}$$ (in either order) (Check: the cubic is $x^3-3x^2-16x+48=0$; indeed $(x-4)(x+4)(x-3) = (x^2-16)(x-3) = x^3-3x^2-16x+48$, confirming all three roots $4,-4,3$.)

QUESTION 1 4 marks

Easy

Solve the equation $|2x-3| = 7$.

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By the definition of the modulus function, $|2x-3|=7$ means $2x-3=7$ or $2x-3=-7$. **Case 1:** $2x-3=7 \Rightarrow 2x=10 \Rightarrow x=5$. **Case 2:** $2x-3=-7 \Rightarrow 2x=-4 \Rightarrow x=-2$. $$\boxed{x=5 \text{ or } x=-2}$$ Check: $|2(5)-3|=|7|=7$, and $|2(-2)-3|=|-7|=7$.

QUESTION 2 6 marks

Medium

Solve the inequality $|2x-1| \ge |x+4|$.

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Method: since both sides are non-negative moduli, and squaring preserves order for non-negative quantities, we may square both sides directly. $$(2x-1)^2 \ge (x+4)^2$$ $$(2x-1)^2-(x+4)^2 \ge 0$$ Using the difference of two squares, $A^2-B^2=(A-B)(A+B)$ with $A=2x-1,\ B=x+4$: $$[(2x-1)-(x+4)][(2x-1)+(x+4)] \ge 0$$ $$(x-5)(3x+3) \ge 0$$ $$3(x-5)(x+1) \ge 0$$ $$(x-5)(x+1) \ge 0$$ This is an upward-opening parabola in factored form, non-negative outside its roots $x=-1$ and $x=5$: $$\boxed{x \le -1 \ \text{ or } \ x \ge 5}$$ Check: $x=0$: $|-1|=1 \ge |4|=4$? False, correctly excluded. $x=6$: $|11|=11\ge|10|=10$, correctly included.

QUESTION 3 5 marks

Easy

Solve the equation $|x^2-4| = 5$.

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By definition of the modulus, $|x^2-4|=5$ means $x^2-4=5$ or $x^2-4=-5$. **Case 1:** $x^2-4=5 \Rightarrow x^2=9 \Rightarrow x=\pm3$. **Case 2:** $x^2-4=-5 \Rightarrow x^2=-1$. There is no real solution, since $x^2\ge0$ for all real $x$. $$\boxed{x=3 \text{ or } x=-3}$$ Check: $|3^2-4|=|5|=5$, and $|(-3)^2-4|=|5|=5$.

QUESTION 4 8 marks

Medium

Solve the equation $|x^2-4| = 2x+1$, giving exact answers. (Note that any solution must satisfy $2x+1\ge0$, since the left-hand side is non-negative.)

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Method: split into cases based on the sign of $x^2-4$, since the definition of $|x^2-4|$ depends on it. **Case A: $x^2-4\ge0$, i.e. $x\le-2$ or $x\ge2$.** Here $|x^2-4|=x^2-4$, so the equation becomes: $$x^2-4 = 2x+1$$ $$x^2-2x-5=0$$ $$x = \frac{2\pm\sqrt{4+20}}{2} = \frac{2\pm\sqrt{24}}{2} = 1\pm\sqrt6$$ Checking against the case condition ($x\le-2$ or $x\ge2$): - $x=1+\sqrt6 \approx 3.449$: satisfies $x\ge2$. Also $2x+1\approx7.90\ge0$. **Valid.** - $x=1-\sqrt6 \approx -1.449$: does **not** satisfy $x\le-2$ (since $-1.449>-2$). **Rejected**, extraneous to this case. **Case B: $x^2-4<0$, i.e. $-2

QUESTION 5 15 marks

Hard

Let $f(x) = |3x-6| - x$. (a) Solve the equation $f(x)=0$. [4] (b) Solve the inequality $f(x) < 4$. [5] (c) Solve the inequality $f(x) \ge 2x-1$. [6]

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(a) $f(x)=0$ means $|3x-6|-x=0$, i.e. $|3x-6|=x$. (For a solution to exist, we need $x\ge0$, since the left side is non-negative.) **Case 1: $3x-6\ge0$, i.e. $x\ge2$.** Then $3x-6=x \Rightarrow 2x=6 \Rightarrow x=3$. This satisfies $x\ge2$ and $x\ge0$. **Valid.** **Case 2: $3x-6<0$, i.e. $x<2$.** Then $-(3x-6)=x \Rightarrow -3x+6=x \Rightarrow 6=4x \Rightarrow x=1.5$. This satisfies $x<2$ and $x\ge0$. **Valid.** $$\boxed{x=1.5 \ \text{ or } \ x=3}$$ Check: $x=1.5$: $|4.5-6|-1.5 = 1.5-1.5=0$. $x=3$: $|9-6|-3=3-3=0$. (b) $f(x)<4$ means $|3x-6|-x<4$, i.e. $|3x-6| < x+4$. For $|A|0.5$ **Right part:** $3x-67/6$): $|4.5-6|=1.5$; RHS $=3.5$; $1.5\ge3.5$? False, correctly excluded. $x=-5$: $|-21|=21 \ge -16$ (included, very negative $x$).

QUESTION 6 3 marks

Easy

Solve the equation $|x+5| = 9$.

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By the definition of the modulus function, $|x+5|=9$ means $x+5=9$ or $x+5=-9$. **Case 1:** $x+5=9 \Rightarrow x=4$. **Case 2:** $x+5=-9 \Rightarrow x=-14$. $$\boxed{x=4 \text{ or } x=-14}$$ Check: $|4+5|=|9|=9$, and $|-14+5|=|-9|=9$.

QUESTION 7 4 marks

Easy

Solve the equation $|3x+2| = 11$.

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By the definition of the modulus function, $|3x+2|=11$ means $3x+2=11$ or $3x+2=-11$. **Case 1:** $3x+2=11 \Rightarrow 3x=9 \Rightarrow x=3$. **Case 2:** $3x+2=-11 \Rightarrow 3x=-13 \Rightarrow x=-\dfrac{13}{3}$. $$\boxed{x=3 \text{ or } x=-\dfrac{13}{3}}$$ Check: $|3(3)+2|=|11|=11$, and $\left|3\left(-\dfrac{13}{3}\right)+2\right|=|-13+2|=|-11|=11$.

QUESTION 8 4 marks

Easy

Solve the equation $|4-2x| = 6$.

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By the definition of the modulus function, $|4-2x|=6$ means $4-2x=6$ or $4-2x=-6$. **Case 1:** $4-2x=6 \Rightarrow -2x=2 \Rightarrow x=-1$. **Case 2:** $4-2x=-6 \Rightarrow -2x=-10 \Rightarrow x=5$. $$\boxed{x=-1 \text{ or } x=5}$$ Check: $|4-2(-1)|=|4+2|=|6|=6$, and $|4-2(5)|=|4-10|=|-6|=6$.

QUESTION 9 5 marks

Easy

Solve the equation $|x-7| = |2x+1|$.

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Two quantities have equal modulus exactly when they are equal or when they are negatives of each other. **Case 1:** $x-7 = 2x+1 \Rightarrow -x = 8 \Rightarrow x=-8$. **Case 2:** $x-7 = -(2x+1) \Rightarrow x-7=-2x-1 \Rightarrow 3x=6 \Rightarrow x=2$. $$\boxed{x=-8 \text{ or } x=2}$$ Check: $x=-8$: $|-8-7|=|-15|=15$ and $|2(-8)+1|=|-15|=15$. $x=2$: $|2-7|=|-5|=5$ and $|2(2)+1|=|5|=5$. Both check out.

QUESTION 10 5 marks

Easy

Solve the equation $|5-3x| = |x+9|$.

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Two quantities have equal modulus exactly when they are equal or when they are negatives of each other. **Case 1:** $5-3x = x+9 \Rightarrow -4 = 4x \Rightarrow x=-1$. **Case 2:** $5-3x = -(x+9) \Rightarrow 5-3x=-x-9 \Rightarrow 14=2x \Rightarrow x=7$. $$\boxed{x=-1 \text{ or } x=7}$$ Check: $x=-1$: $|5-3(-1)|=|8|=8$ and $|-1+9|=|8|=8$. $x=7$: $|5-21|=|-16|=16$ and $|7+9|=16$. Both check out.

QUESTION 11 4 marks

Easy

Solve the inequality $|x-4| < 6$.

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The inequality $|x-4|<6$ means the distance from $x$ to $4$ on the number line is less than $6$, which is equivalent to the double inequality $$-6 < x-4 < 6$$ Adding $4$ throughout: $$-2 < x < 10$$ $$\boxed{x \in (-2,\ 10)}$$ Check: $x=0$ (inside): $|0-4|=4<6$, true. $x=-2$ (boundary): $|-2-4|=6<6$? False, correctly excluded. $x=11$ (outside): $|11-4|=7<6$? False, correctly excluded.

QUESTION 12 4 marks

Easy

Solve the inequality $|2x+5| \le 9$.

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The inequality $|2x+5|\le9$ is equivalent to the double inequality $$-9 \le 2x+5 \le 9$$ Subtracting $5$ throughout: $$-14 \le 2x \le 4$$ Dividing by $2$: $$-7 \le x \le 2$$ $$\boxed{x \in [-7,\ 2]}$$ Check: $x=0$: $|5|=5\le9$, true. $x=2$ (boundary): $|9|=9\le9$, true. $x=3$ (outside): $|11|=11\le9$? False, correctly excluded.

QUESTION 13 4 marks

Easy

Solve the inequality $|x+1| > 3$.

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For $|A|>k$ with $k>0$, the solution is $A>k$ or $A<-k$ (the point is more than $k$ away from $0$ in either direction). Here $A=x+1$, $k=3$. **Branch 1:** $x+1>3 \Rightarrow x>2$. **Branch 2:** $x+1<-3 \Rightarrow x<-4$. $$\boxed{x < -4 \ \text{ or } \ x>2}$$ Check: $x=3$: $|4|=4>3$, true. $x=-5$: $|-4|=4>3$, true. $x=0$ (excluded region): $|1|=1>3$? False, correctly excluded.

QUESTION 14 4 marks

Easy

Solve the inequality $|3x-2| \ge 10$.

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For $|A|\ge k$ with $k>0$, the solution is $A\ge k$ or $A\le-k$. Here $A=3x-2$, $k=10$. **Branch 1:** $3x-2 \ge 10 \Rightarrow 3x \ge 12 \Rightarrow x \ge 4$. **Branch 2:** $3x-2 \le -10 \Rightarrow 3x \le -8 \Rightarrow x \le -\dfrac{8}{3}$. $$\boxed{x \le -\dfrac{8}{3} \ \text{ or } \ x \ge 4}$$ Check: $x=5$: $|13|=13\ge10$, true. $x=-3$: $|-11|=11\ge10$, true. $x=0$ (excluded region): $|-2|=2\ge10$? False, correctly excluded.

QUESTION 15 5 marks

Easy

Solve the equation $|x^2-9| = 16$.

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By definition of the modulus, $|x^2-9|=16$ means $x^2-9=16$ or $x^2-9=-16$. **Case 1:** $x^2-9=16 \Rightarrow x^2=25 \Rightarrow x=\pm5$. **Case 2:** $x^2-9=-16 \Rightarrow x^2=-7$. There is no real solution, since $x^2\ge0$ for all real $x$. $$\boxed{x=5 \text{ or } x=-5}$$ Check: $|5^2-9|=|16|=16$, and $|(-5)^2-9|=|16|=16$.

QUESTION 16 6 marks

Easy

Solve the equation $|x^2-2x| = 3$.

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By definition of the modulus, $|x^2-2x|=3$ means $x^2-2x=3$ or $x^2-2x=-3$. **Case 1:** $x^2-2x=3 \Rightarrow x^2-2x-3=0 \Rightarrow (x-3)(x+1)=0 \Rightarrow x=3 \text{ or } x=-1$. **Case 2:** $x^2-2x=-3 \Rightarrow x^2-2x+3=0$. The discriminant is $(-2)^2-4(1)(3)=4-12=-8<0$, so there is no real solution. $$\boxed{x=3 \text{ or } x=-1}$$ Check: $|3^2-2(3)|=|9-6|=3$, and $|(-1)^2-2(-1)|=|1+2|=3$.

QUESTION 17 4 marks

Easy

The diagram shows the graph of a linear function $y=f(x)$, where $f(x)=2x-4$ for $-1 \le x \le 5$. Sketch the graph of $y=|f(x)|$ for $-1 \le x \le 5$ on the same axes, clearly labelling the coordinates of the points where the graph meets or turns at the $x$-axis and the two endpoints.

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The graph of $y=|f(x)|$ is obtained from $y=f(x)$ by reflecting in the $x$-axis any part of the graph that lies below it, and leaving unchanged any part that lies on or above it. First find where $f(x)=0$: $2x-4=0 \Rightarrow x=2$. Also $f(-1)=2(-1)-4=-6$ and $f(5)=2(5)-4=6$. So on $-1\le x\le2$, $f(x)\le0$ (it runs from $-6$ up to $0$), and this part of the graph must be reflected: the reflected segment runs from $(-1,6)$ down to $(2,0)$. On $2\le x\le5$, $f(x)\ge0$ (it runs from $0$ up to $6$), so this part is unchanged: it runs from $(2,0)$ up to $(5,6)$. The resulting graph of $y=|f(x)|$ is a "V" shape with vertex at $(2,0)$, endpoints at $(-1,6)$ and $(5,6)$, entirely on or above the $x$-axis, as shown (amber dashed = original $y=f(x)$, green solid = $y=|f(x)|$).

QUESTION 18 4 marks

Easy

Let $f(x) = x-3$. Sketch the graph of $y=f(|x|)$ for $-5 \le x \le 5$, clearly labelling the coordinates of the vertex and the two endpoints.

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To sketch $y=f(|x|)$, note that for $x\ge0$, $|x|=x$, so the graph is simply $y=f(x)=x-3$ on the right-hand side (for $0\le x\le5$, running from $(0,-3)$ to $(5,2)$). For $x<0$, $|x|=-x>0$, so $f(|x|)=f(-x)$: the value at a negative $x$ equals the value the right-hand branch takes at the corresponding positive point $-x$. This means the left-hand half of the graph ($-5\le x\le0$) is the mirror image of the right-hand half in the $y$-axis. Indeed, algebraically $f(|x|) = |x|-3$, a "V" shape with vertex at $(0,-3)$ and slopes $\pm1$, so it passes through $(5,2)$ and $(-5,2)$. The graph of $y=f(|x|)$ is therefore symmetric about the $y$-axis (even), with vertex $(0,-3)$ and endpoints $(-5,2)$ and $(5,2)$, as shown (amber dashed = original right-hand branch $y=f(x)$ for $x\ge0$, green solid = full graph of $y=f(|x|)$).

QUESTION 19 3 marks

Easy

A function is defined by $f(x) = |x-2| + 3$ for all real $x$. (a) State the minimum value of $f(x)$ and the value of $x$ at which it occurs. [1] (b) Hence state the range of $f$. [2]

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(a) Since $|x-2|\ge0$ for all real $x$, with equality exactly when $x=2$, the smallest possible value of $f(x)=|x-2|+3$ occurs at $x=2$, giving $f(2)=0+3=3$. $$\boxed{\text{Minimum value } 3 \text{, occurring at } x=2}$$ (b) Since $|x-2|$ can take any value in $[0,\infty)$ as $x$ ranges over the reals, $f(x)=|x-2|+3$ can take any value in $[3,\infty)$. $$\boxed{\text{Range: } f(x) \ge 3}$$

QUESTION 20 4 marks

Easy

Solve the equation $\left|\dfrac{x}{2} - 1\right| = 4$.

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By the definition of the modulus function, $\left|\dfrac{x}{2}-1\right|=4$ means $\dfrac{x}{2}-1=4$ or $\dfrac{x}{2}-1=-4$. **Case 1:** $\dfrac{x}{2}-1=4 \Rightarrow \dfrac{x}{2}=5 \Rightarrow x=10$. **Case 2:** $\dfrac{x}{2}-1=-4 \Rightarrow \dfrac{x}{2}=-3 \Rightarrow x=-6$. $$\boxed{x=10 \text{ or } x=-6}$$ Check: $\left|\dfrac{10}{2}-1\right|=|5-1|=4$, and $\left|\dfrac{-6}{2}-1\right|=|-3-1|=|-4|=4$.

QUESTION 21 5 marks

Easy

Solve the equation $|x+3| = 2x-1$. (Note that any solution must satisfy $2x-1 \ge 0$, since the left-hand side is non-negative.)

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Method: split into cases based on the sign of $x+3$, since the definition of $|x+3|$ depends on it. **Case 1: $x+3\ge0$, i.e. $x\ge-3$.** Here $|x+3|=x+3$, so the equation becomes: $$x+3 = 2x-1 \Rightarrow 4 = x \Rightarrow x=4$$ This satisfies $x\ge-3$. Also $2x-1=7\ge0$. **Valid.** **Case 2: $x+3<0$, i.e. $x<-3$.** Here $|x+3|=-(x+3)$, so the equation becomes: $$-(x+3) = 2x-1 \Rightarrow -x-3=2x-1 \Rightarrow -2=3x \Rightarrow x=-\dfrac{2}{3}$$ This does **not** satisfy $x<-3$ (since $-\dfrac23>-3$). **Rejected**, extraneous to this case. $$\boxed{x=4}$$ Check: $|4+3|=|7|=7$, and $2(4)-1=7$. They match.

QUESTION 22 3 marks

Easy

Solve the inequality $(x-2)(x+5) \le 0$.

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The expression $(x-2)(x+5)$ is a quadratic in factored form with roots $x=2$ and $x=-5$, and it opens upward (positive leading coefficient), so it is negative *between* its roots and positive *outside* them. Sign diagram: for $x<-5$, both factors are negative, so the product is positive. For $-50$ and $(x-2)<0$, so the product is negative. For $x>2$, both factors are positive, so the product is positive. Since we require the product to be $\le0$ (negative or zero), the solution is the closed interval between the roots: $$\boxed{-5 \le x \le 2}$$ Check: $x=0$: $(0-2)(0+5)=(-2)(5)=-10\le0$, true. $x=2$ (boundary): $(0)(7)=0\le0$, true. $x=3$: $(1)(8)=8\le0$? False, correctly excluded.

QUESTION 23 4 marks

Easy

Solve the inequality $\dfrac{x-3}{x+2} < 0$.

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A quotient is negative exactly when the numerator and denominator have opposite signs. The critical values are $x=3$ (numerator zero) and $x=-2$ (denominator zero, so the expression is undefined there). Sign diagram: - For $x<-2$: numerator $x-3<0$, denominator $x+2<0$, quotient $=\dfrac{-}{-}$ is positive. - For $-20$, quotient $=\dfrac{-}{+}$ is negative. - For $x>3$: numerator $x-3>0$, denominator $x+2>0$, quotient $=\dfrac{+}{+}$ is positive. We need the quotient to be strictly negative, which happens on the middle interval (and $x=-2$ must be excluded as it is undefined; $x=3$ is excluded since equality is not permitted): $$\boxed{-2 < x < 3}$$ Check: $x=0$: $\dfrac{-3}{2}=-1.5<0$, true. $x=-3$: $\dfrac{-6}{-1}=6<0$? False, correctly excluded. $x=4$: $\dfrac{1}{6}<0$? False, correctly excluded.

QUESTION 24 6 marks

Medium

Solve the equation $|2x-5| = x+1$. (Note that any solution must satisfy $x+1 \ge 0$, since the left-hand side is non-negative.)

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Method: split into cases based on the sign of $2x-5$. **Case 1: $2x-5\ge0$, i.e. $x\ge2.5$.** Then: $$2x-5 = x+1 \Rightarrow x=6$$ This satisfies $x\ge2.5$, and $x+1=7\ge0$. **Valid.** **Case 2: $2x-5<0$, i.e. $x<2.5$.** Then: $$-(2x-5) = x+1 \Rightarrow -2x+5=x+1 \Rightarrow 4=3x \Rightarrow x=\dfrac{4}{3}$$ This satisfies $x<2.5$, and $x+1=\dfrac73\ge0$. **Valid.** $$\boxed{x=\dfrac{4}{3} \ \text{ or } \ x=6}$$ Check: $x=6$: $|12-5|=7=6+1$. $x=\dfrac43$: $\left|\dfrac83-5\right|=\left|-\dfrac73\right|=\dfrac73$, and $\dfrac43+1=\dfrac73$. Both match exactly.

QUESTION 25 8 marks

Medium

Solve the equation $|x+2| + |x-5| = 9$.

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The expression has two critical points, $x=-2$ and $x=5$, which split the real line into three regions. **Case A: $x<-2$.** Here $x+2<0$ and $x-5<0$, so $|x+2|=-(x+2)$ and $|x-5|=-(x-5)$: $$-(x+2)-(x-5)=9 \Rightarrow -2x+3=9 \Rightarrow x=-3$$ This satisfies $x<-2$. **Valid.** **Case B: $-2\le x\le5$.** Here $x+2\ge0$ and $x-5\le0$, so $|x+2|=x+2$ and $|x-5|=5-x$: $$(x+2)+(5-x)=9 \Rightarrow 7=9$$ This is never true, so there are **no solutions** in this region (the sum is constant, equal to $7$, throughout). **Case C: $x>5$.** Here $x+2>0$ and $x-5>0$, so both bars simply drop: $$(x+2)+(x-5)=9 \Rightarrow 2x-3=9 \Rightarrow x=6$$ This satisfies $x>5$. **Valid.** $$\boxed{x=-3 \ \text{ or } \ x=6}$$ Check: $x=-3$: $|-1|+|-8|=1+8=9$. $x=6$: $|8|+|1|=8+1=9$. Both check out.

QUESTION 26 7 marks

Medium

Solve the equation $|x^2-1| = |x+1|$.

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Factorise the expression inside the first modulus: $x^2-1=(x-1)(x+1)$, so $$|x^2-1| = |(x-1)(x+1)| = |x-1|\,|x+1|$$ The equation becomes $|x-1|\,|x+1| = |x+1|$, i.e. $$|x+1|\big(|x-1|-1\big) = 0$$ so either $|x+1|=0$ or $|x-1|=1$. **Branch 1:** $|x+1|=0 \Rightarrow x=-1$. **Branch 2:** $|x-1|=1 \Rightarrow x-1=1 \text{ or } x-1=-1 \Rightarrow x=2 \text{ or } x=0$. $$\boxed{x=-1,\ 0,\ \text{or } 2}$$ Check: $x=-1$: $|1-1|=0=|0|$. $x=0$: $|-1|=1=|1|$. $x=2$: $|4-1|=3=|3|$. All three check out exactly.

QUESTION 27 6 marks

Medium

Solve the inequality $|2x-3| < x+2$.

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Since the left side is non-negative, a solution can only exist where the right side is also non-negative: $x+2\ge0$, i.e. $x\ge-2$ is a necessary condition. For $x\ge-2$ (so $x+2\ge0$), $|2x-3|\dfrac13$ **Right part:** $2x-3

QUESTION 28 7 marks

Medium

Solve the inequality $|x+4| \ge 3x-2$.

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**If $3x-2<0$, i.e. $x<\dfrac23$:** the inequality holds automatically, since the left side is $\ge0$ while the right side is negative. So **every** $x<\dfrac23$ is a solution. **If $3x-2\ge0$, i.e. $x\ge\dfrac23$:** for non-negative $B$, $|A|\ge B \iff A\ge B \text{ or } A\le-B$. With $A=x+4$, $B=3x-2$: Branch (i): $x+4\ge3x-2 \Rightarrow 6\ge2x \Rightarrow x\le3$. Combined with $x\ge\dfrac23$: $\dfrac23\le x\le3$. Branch (ii): $x+4\le-(3x-2) \Rightarrow x+4\le-3x+2 \Rightarrow 4x\le-2 \Rightarrow x\le-\dfrac12$. Combined with $x\ge\dfrac23$: no solution (impossible to have $x\ge\dfrac23$ and $x\le-\dfrac12$ simultaneously). So from this case we get $\dfrac23\le x\le3$. **Combining both cases:** $x<\dfrac23$ together with $\dfrac23\le x\le3$ merges into a single interval: $$\boxed{x \le 3}$$ Check: $x=0$ ($<\frac23$): $|4|=4\ge-2$, true. $x=3$ (boundary): $|7|=7\ge7$, true. $x=4$: $|8|=8\ge10$? False, correctly excluded.

QUESTION 29 9 marks

Medium

Solve the inequality $|x^2-5x+4| \le 4$.

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The condition $|x^2-5x+4|\le4$ is equivalent to the double inequality $$-4 \le x^2-5x+4 \le 4$$ **Right part:** $x^2-5x+4\le4 \Rightarrow x^2-5x\le0 \Rightarrow x(x-5)\le0$. This is a sign-diagram condition on an upward parabola with roots $0$ and $5$, so it holds for $0\le x\le5$. **Left part:** $x^2-5x+4\ge-4 \Rightarrow x^2-5x+8\ge0$. The discriminant of $x^2-5x+8$ is $(-5)^2-4(1)(8)=25-32=-7<0$, and the leading coefficient is positive, so $x^2-5x+8>0$ for **all** real $x$ -- this part is automatically satisfied everywhere. The binding constraint is therefore the right part alone: $$\boxed{0 \le x \le 5}$$ Check: $x=0$ (boundary): $|0-0+4|=4\le4$, true. $x=5$ (boundary): $|25-25+4|=4\le4$, true. $x=2.5$: $|6.25-12.5+4|=|-2.25|=2.25\le4$, true. $x=-1$: $|1+5+4|=10\le4$? False, correctly excluded.

QUESTION 30 8 marks

Medium

Let $f(x) = x^2-4x+3$. (a) Sketch the graph of $y=|f(x)|$, indicating the coordinates of the points where the graph meets the $x$-axis and the coordinates of any local maximum or minimum point introduced by the reflection. [5] (b) Describe how the graph of $y=f(|x|)$ can be obtained from the graph of $y=f(x)$ for $x\ge0$, and state the value of $f(|x|)$ at $x=-3$. [3]

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(a) Factorising, $f(x)=x^2-4x+3=(x-1)(x-3)$, so $f$ has $x$-intercepts at $x=1$ and $x=3$, and vertex at $x=\dfrac{1+3}{2}=2$, where $f(2)=4-8+3=-1$. Since $f(x)<0$ exactly between the roots (for $10$, so $f(|x|)$ takes the value that $f$ takes at the positive number $-x$; this means the graph for $x<0$ is the mirror image, in the $y$-axis, of the graph for $x\ge0$. So $y=f(|x|)$ is an **even function**, symmetric about the $y$-axis. At $x=-3$: $f(|-3|)=f(3)=(3-1)(3-3)=0$. $$\boxed{f(|-3|)=0}$$

QUESTION 31 7 marks

Medium

Solve the equation $|x^2-4| = 3x$. (Note that any solution must satisfy $x \ge 0$, since the left-hand side is non-negative.)

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Method: split into cases based on the sign of $x^2-4$. **Case A: $x^2-4\ge0$, i.e. $x\le-2$ or $x\ge2$.** Then: $$x^2-4 = 3x \Rightarrow x^2-3x-4=0 \Rightarrow (x-4)(x+1)=0 \Rightarrow x=4 \text{ or } x=-1$$ - $x=4$: satisfies $x\ge2$, and $x\ge0$. **Valid.** - $x=-1$: does **not** satisfy $x\le-2$ or $x\ge2$ (since $-2<-1<2$), and also fails $x\ge0$. **Rejected.** **Case B: $x^2-4<0$, i.e. $-2

QUESTION 32 7 marks

Medium

Solve the inequality $\dfrac{2x-1}{x+3} \le 1$.

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Rearrange so that one side is zero. Note we must **not** simply multiply through by $(x+3)$, since its sign is unknown; instead combine into a single fraction: $$\dfrac{2x-1}{x+3} - 1 \le 0 \ \Longrightarrow\ \dfrac{(2x-1)-(x+3)}{x+3} \le 0 \ \Longrightarrow\ \dfrac{x-4}{x+3} \le 0$$ The critical values are $x=4$ (numerator zero) and $x=-3$ (denominator zero, undefined). Sign diagram: - $x<-3$: $\dfrac{(-)}{(-)}$ is positive. - $-34$: $\dfrac{(+)}{(+)}$ is positive. We need the quotient $\le0$: this holds on the middle interval, together with $x=4$ where the quotient equals zero, but $x=-3$ must be excluded (undefined): $$\boxed{-3 < x \le 4}$$ Check: $x=0$: $\dfrac{-1}{3}\le1$, true. $x=4$ (boundary): $\dfrac{7}{7}=1\le1$, true. $x=5$: $\dfrac{9}{8}=1.125\le1$? False, correctly excluded. $x=-4$: $\dfrac{-9}{-1}=9\le1$? False, correctly excluded.

QUESTION 33 6 marks

Medium

Solve the inequality $2x^2-5x-3 \ge 0$.

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Factorise: we need two numbers multiplying to $2\times(-3)=-6$ and summing to $-5$, namely $-6$ and $1$: $$2x^2-5x-3 = 2x^2-6x+x-3 = 2x(x-3)+(x-3) = (2x+1)(x-3)$$ This is an upward-opening parabola with roots $x=-\dfrac12$ and $x=3$, so it is non-negative *outside* the roots. Sign diagram: for $x<-\dfrac12$, both factors negative, product positive. For $-\dfrac123$, both positive, product positive. $$\boxed{x \le -\dfrac12 \ \text{ or } \ x \ge 3}$$ Check: $x=-1$: $2+5-3=4\ge0$, true. $x=0$: $-3\ge0$? False, correctly excluded. $x=4$: $32-20-3=9\ge0$, true.

QUESTION 34 7 marks

Medium

Solve the inequality $\dfrac{x+2}{x-1} > 2$.

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Rearrange so one side is zero, combining into a single fraction (do not multiply by $(x-1)$ directly, as its sign is unknown): $$\dfrac{x+2}{x-1} - 2 > 0 \ \Longrightarrow\ \dfrac{(x+2)-2(x-1)}{x-1} > 0 \ \Longrightarrow\ \dfrac{-x+4}{x-1} > 0 \ \Longrightarrow\ \dfrac{4-x}{x-1} > 0$$ Critical values: $x=4$ (numerator zero) and $x=1$ (denominator zero, undefined). Sign diagram: - $x<1$: numerator $4-x>0$, denominator $x-1<0$, quotient negative. - $10$, denominator $x-1>0$, quotient positive. - $x>4$: numerator $4-x<0$, denominator $x-1>0$, quotient negative. We need the quotient strictly positive, which holds only on the middle interval: $$\boxed{1 < x < 4}$$ Check: $x=2$: $\dfrac{4}{1}=4>2$, true. $x=0$: $\dfrac{2}{-1}=-2>2$? False, correctly excluded. $x=5$: $\dfrac{7}{4}=1.75>2$? False, correctly excluded.

QUESTION 35 7 marks

Medium

Solve the equation $|x^2-3x| = 2x$. (Note that any solution must satisfy $x \ge 0$, since the left-hand side is non-negative.)

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Method: split into cases based on the sign of $x^2-3x=x(x-3)$. **Case A: $x^2-3x\ge0$, i.e. $x\le0$ or $x\ge3$.** Then: $$x^2-3x = 2x \Rightarrow x^2-5x=0 \Rightarrow x(x-5)=0 \Rightarrow x=0 \text{ or } x=5$$ - $x=0$: satisfies $x\le0$ (boundary) and $x\ge0$. **Valid.** - $x=5$: satisfies $x\ge3$ and $x\ge0$. **Valid.** **Case B: $x^2-3x<0$, i.e. $0

QUESTION 36 8 marks

Medium

Solve the inequality $\left|\dfrac{x-1}{x+2}\right| < 3$.

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Since $x\ne-2$ (undefined there), multiply both sides by $|x+2|>0$: $$|x-1| < 3|x+2|$$ Both sides are non-negative, so we may square: $$(x-1)^2 < 9(x+2)^2$$ $$(x-1)^2 - 9(x+2)^2 < 0$$ Using the difference of two squares with $A=x-1$, $B=3(x+2)$: $$\big[(x-1)-3(x+2)\big]\big[(x-1)+3(x+2)\big] < 0$$ $$(-2x-7)(4x+5) < 0$$ Multiplying both sides by $-1$ (which reverses the inequality): $$(2x+7)(4x+5) > 0$$ This is an upward-opening parabola in $x$ with roots $x=-\dfrac72$ and $x=-\dfrac54$, so it is positive *outside* its roots: $$\boxed{x < -\dfrac{7}{2} \ \text{ or } \ x > -\dfrac{5}{4}}$$ (The excluded point $x=-2$ lies strictly between $-\dfrac72$ and $-\dfrac54$, inside the rejected region, so no special exception is needed.) Check: $x=0$: $\left|\dfrac{-1}{2}\right|=0.5<3$, true, and $0>-\dfrac54$. $x=-3$: $\left|\dfrac{-4}{-1}\right|=4<3$? False, correctly excluded ($-3$ is between $-\dfrac72$ and $-\dfrac54$). $x=-4$: $\left|\dfrac{-5}{-2}\right|=2.5<3$, true, and $-4<-\dfrac72$.

QUESTION 37 8 marks

Medium

A function is defined by $g(x) = |x+1| - |x-3|$. (a) Express $g(x)$ as a piecewise linear function, giving the three separate expressions and the intervals on which each applies. [4] (b) Hence solve the equation $g(x) = 2$. [4]

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(a) The critical points are $x=-1$ and $x=3$. **For $x<-1$:** $x+1<0$ and $x-3<0$, so $g(x)=-(x+1)-[-(x-3)]=-(x+1)+(x-3)=-4$. **For $-1\le x<3$:** $x+1\ge0$ and $x-3<0$, so $g(x)=(x+1)-[-(x-3)]=(x+1)+(x-3)=2x-2$. **For $x\ge3$:** $x+1>0$ and $x-3\ge0$, so $g(x)=(x+1)-(x-3)=4$. $$\boxed{g(x)=\begin{cases}-4 & x<-1\\ 2x-2 & -1\le x<3\\ 4 & x\ge3\end{cases}}$$ (b) Solve $g(x)=2$ on each piece: - $x<-1$: $g(x)=-4$, never equal to $2$. No solution here. - $-1\le x<3$: $2x-2=2 \Rightarrow x=2$. This satisfies $-1\le2<3$. **Valid.** - $x\ge3$: $g(x)=4$, never equal to $2$. No solution here. $$\boxed{x=2}$$ Check: $g(2)=|3|-|-1|=3-1=2$, as required.

QUESTION 38 6 marks

Medium

Solve the inequality $-x^2+2x+8 \ge 0$.

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Multiply both sides by $-1$, which reverses the inequality: $$x^2-2x-8 \le 0$$ Factorise: we need two numbers multiplying to $-8$ and summing to $-2$, namely $-4$ and $2$: $$(x-4)(x+2) \le 0$$ This is an upward-opening parabola with roots $x=-2$ and $x=4$, so it is non-positive *between* the roots: $$\boxed{-2 \le x \le 4}$$ Check: $x=0$: $0+0+8=8\ge0$, true. $x=5$: $-25+10+8=-7\ge0$? False, correctly excluded. $x=-3$: $-9-6+8=-7\ge0$? False, correctly excluded.

QUESTION 39 7 marks

Medium

Solve the inequality $\dfrac{3x}{x+1} < 2$.

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Rearrange so one side is zero, combining into a single fraction: $$\dfrac{3x}{x+1} - 2 < 0 \ \Longrightarrow\ \dfrac{3x-2(x+1)}{x+1} < 0 \ \Longrightarrow\ \dfrac{x-2}{x+1} < 0$$ Critical values: $x=2$ (numerator zero) and $x=-1$ (denominator zero, undefined). Sign diagram: - $x<-1$: $\dfrac{(-)}{(-)}$ is positive. - $-12$: $\dfrac{(+)}{(+)}$ is positive. We need the quotient strictly negative, which holds only on the middle interval: $$\boxed{-1 < x < 2}$$ Check: $x=0$: $\dfrac{0}{1}=0<2$, true. $x=-2$: $\dfrac{-6}{-1}=6<2$? False, correctly excluded. $x=3$: $\dfrac{9}{4}=2.25<2$? False, correctly excluded.

QUESTION 40 6 marks

Medium

The diagram shows the graph of $y=f(x)$, where $f(x)=-2x+6$ for $0 \le x \le 5$. Sketch the graph of $y=|f(x)|$ for $0 \le x \le 5$ on the same axes, clearly labelling the coordinates of the endpoints and the point where the graph meets the $x$-axis, and state the range of $y=|f(x)|$ on this domain.

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First find where $f(x)=0$: $-2x+6=0 \Rightarrow x=3$. Also $f(0)=6$ and $f(5)=-2(5)+6=-4$. On $0\le x\le3$, $f(x)\ge0$ (it runs from $6$ down to $0$), so this part of $y=|f(x)|$ is unchanged: it runs from $(0,6)$ down to $(3,0)$. On $3\le x\le5$, $f(x)<0$ (it runs from $0$ down to $-4$), so this part must be reflected in the $x$-axis: $|f(x)|=-f(x)=2x-6$, which runs from $(3,0)$ up to $(5,4)$. The graph of $y=|f(x)|$ is therefore a "V" shape with vertex at $(3,0)$, endpoints at $(0,6)$ and $(5,4)$ (amber dashed = original $y=f(x)$, green solid = $y=|f(x)|$). Since the graph's lowest value is $0$ (at the vertex) and its highest value is $6$ (at $x=0$): $$\boxed{\text{Range: } 0 \le y \le 6}$$

QUESTION 41 7 marks

Medium

Solve the inequality $|x-1| \le |x+5|$.

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Both sides are non-negative, so we may square both sides (squaring preserves order for non-negative quantities): $$(x-1)^2 \le (x+5)^2$$ $$(x-1)^2-(x+5)^2 \le 0$$ Using the difference of two squares, $A^2-B^2=(A-B)(A+B)$ with $A=x-1,\ B=x+5$: $$\big[(x-1)-(x+5)\big]\big[(x-1)+(x+5)\big] \le 0$$ $$(-6)(2x+4) \le 0$$ Dividing both sides by $-6$ (which reverses the inequality): $$2x+4 \ge 0 \ \Longrightarrow\ x \ge -2$$ $$\boxed{x \ge -2}$$ Check: $x=0$: $|-1|=1\le|5|=5$, true. $x=-2$ (boundary): $|-3|=3\le|3|=3$, true. $x=-3$: $|-4|=4\le|2|=2$? False, correctly excluded.

QUESTION 42 10 marks

Hard

Solve the inequality $|x-2| + |x+3| \le 7$.

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The critical points are $x=2$ and $x=-3$, giving three regions. **Case A: $x<-3$.** Here $x-2<0$ and $x+3<0$: $$-(x-2)-(x+3) \le 7 \Rightarrow -2x-1 \le 7 \Rightarrow -2x \le 8 \Rightarrow x \ge -4$$ Combined with $x<-3$: $-4\le x<-3$. **Case B: $-3\le x\le2$.** Here $x-2\le0$ and $x+3\ge0$, so $|x-2|=2-x$ and $|x+3|=x+3$: $$(2-x)+(x+3) \le 7 \Rightarrow 5 \le 7$$ This is always true, so **every** $x$ in $[-3,2]$ is a solution. **Case C: $x>2$.** Here both bars drop: $$(x-2)+(x+3) \le 7 \Rightarrow 2x+1 \le 7 \Rightarrow x \le 3$$ Combined with $x>2$: $2

QUESTION 43 11 marks

Hard

Solve the inequality $\left|\dfrac{2x-1}{x+3}\right| \ge 1$.

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Since $x\ne-3$ (undefined there), multiply both sides by $|x+3|>0$: $$|2x-1| \ge |x+3|$$ Both sides are non-negative, so we may square: $$(2x-1)^2 \ge (x+3)^2$$ $$(2x-1)^2-(x+3)^2 \ge 0$$ Using the difference of two squares with $A=2x-1,\ B=x+3$: $$\big[(2x-1)-(x+3)\big]\big[(2x-1)+(x+3)\big] \ge 0$$ $$(x-4)(3x+2) \ge 0$$ This is an upward-opening parabola in $x$ with roots $x=4$ and $x=-\dfrac23$, so it is non-negative *outside* the roots: $$x \le -\dfrac{2}{3} \ \text{ or } \ x \ge 4$$ Finally, we must exclude $x=-3$ from the domain, since the original expression is undefined there (and $-3\le-\dfrac23$, so it lies inside the region just found): $$\boxed{x \in (-\infty,\,-3)\cup\left(-3,\,-\dfrac23\right] \cup [4,\,\infty)}$$ Check: $x=-1$: $\left|\dfrac{-3}{2}\right|=1.5\ge1$, true, and $-1\le-\dfrac23$. $x=0$: $\left|\dfrac{-1}{3}\right|=\dfrac13\ge1$? False, correctly excluded (between $-\dfrac23$ and $4$). $x=5$: $\left|\dfrac{9}{8}\right|=1.125\ge1$, true, and $5\ge4$. $x=-3.5$: $\left|\dfrac{-8}{-0.5}\right|=16\ge1$, true, consistent with $x<-\dfrac23$ (excluding $x=-3$).

QUESTION 44 15 marks

Hard

Let $f(x) = |x^2-4x|$. (a) Sketch the graph of $y=f(x)$ for $-1 \le x \le 5$, marking the coordinates of the points where the graph meets the $x$-axis and any local maximum or minimum point. [4] (b) Solve the equation $f(x) = 3$, giving exact answers. [5] (c) Solve the inequality $f(x) < 3x$. [6]

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(a) Let $g(x)=x^2-4x=x(x-4)$, so $g$ has roots at $x=0$ and $x=4$, and vertex at $x=2$, $g(2)=4-8=-4$. Since $g(x)<0$ exactly between the roots ($00$, i.e. $x>0$ (if $x\le0$, the right side is $\le0\le$ left side, so the strict inequality fails). For $x>0$, this is equivalent to the double inequality $$-3x < x^2-4x < 3x$$ **Left part:** $-3x0 \Rightarrow x<0 \text{ or } x>1$. Combined with $x>0$: $x>1$. **Right part:** $x^2-4x<3x \Rightarrow x^2-7x<0 \Rightarrow x(x-7)<0 \Rightarrow 01$ and $07$).

QUESTION 45 10 marks

Hard

Solve the inequality $\dfrac{x^2-x-6}{x+1} \ge 0$.

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Factorise the numerator: $x^2-x-6=(x-3)(x+2)$, so the inequality is $$\dfrac{(x-3)(x+2)}{x+1} \ge 0$$ The critical values, in order, are $x=-2$, $x=-1$ (undefined), and $x=3$. Sign diagram: - $x<-2$, e.g. $x=-3$: $(x-3)<0$, $(x+2)<0$, $(x+1)<0$ $\Rightarrow \dfrac{(-)(-)}{(-)}$ is negative. - $-20$, $(x+1)<0$ $\Rightarrow \dfrac{(-)(+)}{(-)}$ is positive. - $-10$, $(x+1)>0$ $\Rightarrow \dfrac{(-)(+)}{(+)}$ is negative. - $x>3$, e.g. $x=4$: all three factors positive $\Rightarrow$ positive. We need the expression $\ge0$: this holds on $-23$, together with the zeros of the numerator $x=-2$ and $x=3$ (where the expression equals $0$), but $x=-1$ must be excluded (undefined): $$\boxed{x \in [-2,\,-1) \cup [3,\,\infty)}$$ Check: $x=-2$ (boundary): $\dfrac{(-5)(0)}{-1}=0\ge0$, true. $x=-1.5$: $\dfrac{(-4.5)(0.5)}{-0.5}=4.5\ge0$, true. $x=0$: $\dfrac{(-3)(2)}{1}=-6\ge0$? False, correctly excluded. $x=3$ (boundary): $\dfrac{(0)(5)}{4}=0\ge0$, true. $x=4$: $\dfrac{(1)(6)}{5}=1.2\ge0$, true.

QUESTION 46 11 marks

Hard

Solve the equation $|x^2-5x+6| = |x-4|$.

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Two quantities have equal modulus exactly when their squares are equal, so we may square both sides directly: $$(x^2-5x+6)^2 = (x-4)^2$$ $$(x^2-5x+6)^2 - (x-4)^2 = 0$$ Using the difference of two squares with $A=x^2-5x+6,\ B=x-4$: $$\big[(x^2-5x+6)-(x-4)\big]\big[(x^2-5x+6)+(x-4)\big] = 0$$ $$(x^2-6x+10)(x^2-4x+2) = 0$$ **First factor:** $x^2-6x+10=0$ has discriminant $36-40=-4<0$, so no real roots. **Second factor:** $x^2-4x+2=0 \Rightarrow x=\dfrac{4\pm\sqrt{16-8}}{2}=\dfrac{4\pm2\sqrt2}{2}=2\pm\sqrt2$. $$\boxed{x = 2-\sqrt2 \ \text{ or } \ x=2+\sqrt2}$$ Check: at $x=2+\sqrt2$: $x^2-5x+6=(2+\sqrt2)^2-5(2+\sqrt2)+6=(6+4\sqrt2)-(10+5\sqrt2)+6=2-\sqrt2$, and since $\sqrt2\approx1.41<2$, $|2-\sqrt2|=2-\sqrt2$. Meanwhile $|x-4|=|(2+\sqrt2)-4|=|\sqrt2-2|=2-\sqrt2$. Equal. At $x=2-\sqrt2$: $x^2-5x+6=(2-\sqrt2)^2-5(2-\sqrt2)+6=(6-4\sqrt2)-(10-5\sqrt2)+6=2+\sqrt2$, and $|x-4|=|(2-\sqrt2)-4|=|-2-\sqrt2|=2+\sqrt2$. Equal. Both solutions confirmed exactly.

QUESTION 47 12 marks

Hard

Solve the inequality $|3x+2| - |x-1| \ge 4$.

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The critical points are $x=-\dfrac23$ and $x=1$, giving three regions. **Case A: $x<-\dfrac23$.** Here $3x+2<0$ and $x-1<0$, so $|3x+2|=-3x-2$ and $|x-1|=1-x$: $$(-3x-2)-(1-x) \ge 4 \Rightarrow -2x-3 \ge 4 \Rightarrow -2x \ge 7 \Rightarrow x \le -\dfrac72$$ Combined with $x<-\dfrac23$: $x\le-\dfrac72$. **Case B: $-\dfrac23\le x<1$.** Here $3x+2\ge0$ and $x-1<0$, so $|3x+2|=3x+2$ and $|x-1|=1-x$: $$(3x+2)-(1-x) \ge 4 \Rightarrow 4x+1 \ge 4 \Rightarrow x \ge \dfrac34$$ Combined with $-\dfrac23\le x<1$: $\dfrac34\le x<1$. **Case C: $x\ge1$.** Here both bars drop directly: $$(3x+2)-(x-1) \ge 4 \Rightarrow 2x+3 \ge 4 \Rightarrow x \ge \dfrac12$$ Combined with $x\ge1$: all $x\ge1$. **Combining all three cases:** $x\le-\dfrac72$, together with $\dfrac34\le x<1$ and $x\ge1$ (which merge into $x\ge\dfrac34$): $$\boxed{x \le -\dfrac{7}{2} \ \text{ or } \ x \ge \dfrac{3}{4}}$$ Check: $x=-4$: $|-10|-|-5|=10-5=5\ge4$, true, $-4\le-3.5$. $x=-2$: $|-4|-|-3|=4-3=1\ge4$? False, correctly excluded ($-2$ between $-3.5$ and $0.75$). $x=1$: $|5|-|0|=5\ge4$, true. $x=0.75$ (boundary): $|4.25|-|-0.25|=4.25-0.25=4\ge4$, true. $x=-3.5$ (boundary): $|-8.5|-|-4.5|=8.5-4.5=4\ge4$, true.

QUESTION 48 9 marks

Hard

Solve the inequality $(x-1)(x+2)(3-x) > 0$.

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Rewrite the last factor to standard form: $(3-x)=-(x-3)$, so the inequality becomes $$-(x-1)(x+2)(x-3) > 0 \ \Longrightarrow\ (x-1)(x+2)(x-3) < 0$$ The critical values, in order, are $x=-2,\ 1,\ 3$. This is a cubic with positive leading coefficient, so its sign alternates across the four intervals; testing one point in each: - $x<-2$, e.g. $x=-3$: $(-4)(-1)(-6)=-24$, negative. - $-23$, e.g. $x=4$: $(3)(6)(1)=18$, positive. We need $(x-1)(x+2)(x-3)<0$, which holds on the first and third intervals: $$\boxed{x < -2 \ \text{ or } \ 1 < x < 3}$$ Check (in the original form): $x=-3$: $(-4)(-1)(6)=24>0$, true. $x=2$: $(1)(4)(1)=4>0$, true. $x=0$: $(-1)(2)(3)=-6>0$? False, correctly excluded. $x=4$: $(3)(6)(-1)=-18>0$? False, correctly excluded.

QUESTION 49 11 marks

Hard

Let $f(x) = |x+1|$ and $g(x) = |2x-5|$. (a) Solve the equation $f(x) = g(x)$. [5] (b) Solve the inequality $f(x) < g(x)$. [6]

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(a) $|x+1|=|2x-5|$. Two quantities have equal modulus exactly when they are equal or negatives of each other. **Case 1:** $x+1=2x-5 \Rightarrow 6=x \Rightarrow x=6$. **Case 2:** $x+1=-(2x-5) \Rightarrow x+1=-2x+5 \Rightarrow 3x=4 \Rightarrow x=\dfrac43$. $$\boxed{x=\dfrac43 \ \text{ or } \ x=6}$$ Check: $x=6$: $|7|=7$ and $|12-5|=7$. $x=\dfrac43$: $\left|\dfrac73\right|=\dfrac73$ and $\left|\dfrac83-5\right|=\left|-\dfrac73\right|=\dfrac73$. Both match. (b) $|x+1|<|2x-5|$. Both sides are non-negative, so we may square: $$(x+1)^2 < (2x-5)^2$$ $$(x+1)^2-(2x-5)^2 < 0$$ Using the difference of two squares with $A=x+1,\ B=2x-5$: $$\big[(x+1)-(2x-5)\big]\big[(x+1)+(2x-5)\big] < 0$$ $$(6-x)(3x-4) < 0$$ This is a downward-opening parabola in $x$ (the coefficient of $x^2$, from expanding, is $-3$), with roots $x=6$ and $x=\dfrac43$, so it is negative *outside* the roots: $$\boxed{x < \dfrac{4}{3} \ \text{ or } \ x > 6}$$ Check: $x=0$: $|1|=1<|-5|=5$, true, and $0<\dfrac43$. $x=2$: $|3|=3<|-1|=1$? False, correctly excluded (between $\dfrac43$ and $6$). $x=7$: $|8|=8<|9|=9$, true, and $7>6$.

QUESTION 50 10 marks

Hard

Solve the inequality $\dfrac{1}{x-2} < \dfrac{1}{x+1}$.

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Do **not** multiply directly by $(x-2)(x+1)$, since its sign is unknown. Instead move everything to one side and combine into a single fraction: $$\dfrac{1}{x-2} - \dfrac{1}{x+1} < 0$$ $$\dfrac{(x+1)-(x-2)}{(x-2)(x+1)} < 0$$ $$\dfrac{3}{(x-2)(x+1)} < 0$$ Since the numerator $3$ is always positive, this is negative exactly when the denominator is negative: $$(x-2)(x+1) < 0$$ This is an upward-opening parabola with roots $x=-1$ and $x=2$, negative *between* the roots: $$\boxed{-1 < x < 2}$$ (Both $x=-1$ and $x=2$ are automatically excluded, as required, since the original expression is undefined there.) Check: $x=0$: $\dfrac{1}{-2}=-0.5$ and $\dfrac{1}{1}=1$; is $-0.5<1$? True, and $0$ is in $(-1,2)$. $x=3$: $\dfrac{1}{1}=1$ and $\dfrac{1}{4}=0.25$; is $1<0.25$? False, correctly excluded. $x=-2$: $\dfrac{1}{-4}=-0.25$ and $\dfrac{1}{-1}=-1$; is $-0.25<-1$? False, correctly excluded.

QUESTION 1 4 marks

Easy

Consider the linear function $f(x) = -2x+6$. (a) Find the coordinates of the $x$-intercept and the $y$-intercept of the graph of $f$. (b) State whether $f$ is increasing or decreasing, justifying your answer.

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(a) The $x$-intercept occurs where $f(x)=0$: $$-2x+6=0$$ $$x=3$$ giving the point $(3,0)$. The $y$-intercept is $f(0)$: $$f(0) = -2(0)+6 = 6$$ giving the point $(0,6)$. $$\boxed{(3,0) \text{ and } (0,6)}$$ (b) The gradient of $f(x)=-2x+6$ is $-2$. Since this is negative, as $x$ increases, $f(x)$ decreases. $$\boxed{f \text{ is decreasing}}$$

QUESTION 2 5 marks

Medium

Let $f(x) = x^2$. The graph of $y=f(x)$ is transformed to give the graph of $y = f(x-3)+2$. (a) Write the equation of the transformed function in expanded (general) form. (b) State the coordinates of the vertex of the transformed graph, and describe the geometric transformation of the graph of $f$ that produces it.

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(a) Substituting into the transformed function: $$g(x) = f(x-3)+2 = (x-3)^2 + 2$$ Expanding $(x-3)^2$: $$(x-3)^2 = x^2 - 6x+9$$ $$g(x) = x^2 - 6x + 9 + 2$$ $$\boxed{g(x) = x^2-6x+11}$$ (b) The vertex of $y=f(x)=x^2$ is $(0,0)$. In the transformation $x \to x-3$, replacing $x$ with $x-3$ translates the graph $3$ units in the positive $x$-direction, and adding $2$ translates it $2$ units in the positive $y$-direction. $$\boxed{\text{Vertex: } (3,2); \text{ translation by } \begin{pmatrix}3\\2\end{pmatrix}}$$

QUESTION 3 6 marks

Easy

The height of a ball above the ground, $t$ seconds after being thrown, is modelled by $$h(t) = -5t^2+20t+2, \quad 0 \leq t \leq 4.2$$ where $h$ is measured in metres. (a) Find the height of the ball at $t=0$. (b) By writing $h(t)$ in the form $-5(t-p)^2+q$, find the maximum height reached by the ball and the time at which this occurs. (c) Find the time at which the ball hits the ground, correct to 3 significant figures.

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(a) Substituting $t=0$: $$h(0) = -5(0)^2+20(0)+2 = 2$$ $$\boxed{h(0) = 2 \text{ m (the initial height of the ball)}}$$ (b) Completing the square on $h(t) = -5t^2+20t+2$: $$h(t) = -5(t^2-4t)+2$$ $$h(t) = -5\big[(t-2)^2-4\big]+2$$ $$h(t) = -5(t-2)^2+20+2 = -5(t-2)^2+22$$ Since $-5(t-2)^2 \leq 0$ for all $t$, the maximum occurs when $(t-2)^2=0$, i.e. $t=2$, giving $h=22$. $$\boxed{\text{Maximum height } = 22 \text{ m, at } t=2 \text{ s}}$$ (c) The ball hits the ground when $h(t)=0$: $$-5t^2+20t+2=0$$ Dividing by $-1$: $5t^2-20t-2=0$. Using the quadratic formula: $$t = \frac{20 \pm \sqrt{(-20)^2 - 4(5)(-2)}}{2(5)} = \frac{20\pm\sqrt{400+40}}{10} = \frac{20\pm\sqrt{440}}{10}$$ Since $\sqrt{440} \approx 20.976$: $$t = \frac{20+20.976}{10} \approx 4.0976$$ (taking the positive root, since $t\geq0$ in context). $$\boxed{t \approx 4.10 \text{ s}}$$

QUESTION 4 6 marks

Medium

Consider $f(x) = \dfrac{x^2-9}{x-3}$, $x \neq 3$. (a) Simplify $f(x)$, and explain why $x=3$ must still be excluded from the domain even after simplifying. (b) State the coordinates of the point of discontinuity ("hole") on the graph of $f$. (c) State the value that $f(3)$ would need to be defined as in order to make the function continuous at $x=3$.

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(a) Factorising the numerator as a difference of two squares: $$f(x) = \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3}$$ For $x \neq 3$, the factor $(x-3)$ cancels: $$f(x) = x+3, \quad x \neq 3$$ $x=3$ must still be excluded because the original expression involves division by $x-3$, which equals $0$ when $x=3$: the function is undefined there regardless of the algebraic simplification, since cancelling $(x-3)$ is only a valid step when $x-3 \neq 0$. $$\boxed{f(x) = x+3, \; x\neq 3}$$ (b) The simplified expression gives $x+3 = 3+3 = 6$ at $x=3$, so the graph is the line $y=x+3$ with an open circle (a "hole") at: $$\boxed{(3,6)}$$ (c) For continuity at $x=3$, $f(3)$ must be defined to equal the value the simplified function approaches there: $$\boxed{f(3) = 6}$$

QUESTION 5 14 marks

Hard

The number of bacteria in a culture, $N(t)$ (measured in thousands), $t$ hours after the start of an experiment, is modelled by $$N(t) = 3+12t-2t^2, \quad 0 \leq t \leq 6.$$ (a) Find $N(0)$ and interpret this value in context. (b) By writing $N(t)$ in the form $-2(t-p)^2+q$, find the maximum number of bacteria in the culture and the value of $t$ at which it occurs. (c) Find the value(s) of $t>0$ for which $N(t)=3$, and interpret your answer in context. (d) Hence, state the range of $N$ over the given domain $0\leq t \leq 6$.

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(a) Substituting $t=0$: $$N(0) = 3+12(0)-2(0)^2 = 3$$ $$\boxed{N(0)=3}$$: there are $3$ thousand ($3000$) bacteria at the start of the experiment. (b) Writing $N(t)=-2t^2+12t+3$ and completing the square: $$N(t) = -2(t^2-6t)+3$$ $$N(t) = -2\big[(t-3)^2-9\big]+3$$ $$N(t) = -2(t-3)^2+18+3 = -2(t-3)^2+21$$ Since $-2(t-3)^2 \leq 0$ for all $t$, the maximum occurs at $t=3$: $$\boxed{\text{Maximum} = 21 \text{ thousand bacteria, at } t=3 \text{ hours}}$$ (c) Setting $N(t) = 3$: $$-2t^2+12t+3 = 3$$ $$-2t^2+12t = 0$$ Factorising: $$-2t(t-6) = 0$$ $$t = 0 \text{ or } t=6$$ Since we require $t>0$: $$\boxed{t=6 \text{ hours}}$$ - the population returns to its initial size ($3000$ bacteria) after 6 hours. (d) Since $N$ is a downward-opening parabola with vertex at $t=3$, which lies within the domain $[0,6]$, the maximum on $[0,6]$ is $21$ (from part (b)). Because the domain $[0,6]$ is symmetric about $t=3$, the two endpoints give the minimum value on this interval: $$N(0)=3, \qquad N(6) = -2(36)+72+3 = -72+72+3=3$$ $$\boxed{\text{Range: } 3 \leq N(t) \leq 21}$$

QUESTION 6 4 marks

Easy

Consider $f(x) = x^2-8x+15$. (a) Find the $x$-intercepts of the graph of $f$. (b) State the equation of the axis of symmetry of the graph of $f$.

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(a) Factorising $f(x)=x^2-8x+15$, we need two numbers that multiply to $15$ and add to $-8$: these are $-3$ and $-5$. $$f(x) = (x-3)(x-5)$$ Setting $f(x)=0$: $$x=3 \text{ or } x=5$$ $$\boxed{x=3, \; x=5}$$ (b) For a quadratic, the axis of symmetry passes through the midpoint of the two roots: $$x = \frac{3+5}{2} = 4$$ $$\boxed{x=4}$$

QUESTION 7 5 marks

Medium

Consider the function $f(x) = \dfrac{2x+1}{x-4}$, $x \neq 4$. (a) State the equation of the vertical asymptote of the graph of $f$. (b) State the equation of the horizontal asymptote of the graph of $f$. (c) Find the coordinates of the points where the graph of $f$ crosses the $x$-axis and the $y$-axis.

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(a) The vertical asymptote occurs where the denominator is zero: $$x-4=0$$ $$\boxed{x=4}$$ (b) Since the numerator and denominator of $f(x)=\dfrac{2x+1}{x-4}$ both have degree $1$, the horizontal asymptote is the ratio of the leading coefficients: as $x \to \pm\infty$, $f(x) \to \dfrac{2x}{x} = 2$. $$\boxed{y=2}$$ (c) The $x$-intercept occurs where $f(x)=0$, i.e. where the numerator is zero: $$2x+1=0 \implies x=-\frac{1}{2}$$ giving the point $\left(-\dfrac{1}{2},0\right)$. The $y$-intercept is $f(0)$: $$f(0) = \frac{1}{-4} = -\frac{1}{4}$$ giving the point $\left(0,-\dfrac{1}{4}\right)$. $$\boxed{\left(-\dfrac{1}{2},0\right) \text{ and } \left(0,-\dfrac{1}{4}\right)}$$

QUESTION 8 6 marks

Easy

Consider $f(x) = x^3-6x^2+9x+1$, $x \in \mathbb{R}$. (a) Find the $y$-intercept of the graph of $f$. (b) Using your graphic display calculator, find the coordinates of the local maximum point and the local minimum point of the graph of $f$. (c) State the interval(s) of $x$ for which $f$ is increasing.

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(a) Substituting $x=0$: $$f(0) = 0-0+0+1 = 1$$ $$\boxed{(0,1)}$$ (b) Using a graphic display calculator to graph $f(x) = x^3-6x^2+9x+1$ and locate the turning points: $$\boxed{\text{Local maximum at } (1,5)}$$ $$\boxed{\text{Local minimum at } (3,1)}$$ (For reference, these values may be verified algebraically: $f'(x)=3x^2-12x+9=3(x-1)(x-3)$ gives critical points $x=1,3$; substituting, $f(1)=1-6+9+1=5$ and $f(3)=27-54+27+1=1$.) (c) The graph rises to the local maximum at $x=1$, falls to the local minimum at $x=3$, then rises again. So $f$ is increasing before the local maximum and after the local minimum: $$\boxed{x<1 \text{ or } x>3}$$

QUESTION 9 6 marks

Medium

Consider $g(x) = \dfrac{x-1}{x^2-4}$. (a) State the equations of the vertical asymptotes of the graph of $g$. (b) State the equation of the horizontal asymptote of the graph of $g$. (c) Determine whether $g$ is an odd function, an even function, or neither, showing full reasoning.

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(a) Vertical asymptotes occur where the denominator is zero, provided the numerator is not also zero there: $$x^2-4=0 \implies x=2 \text{ or } x=-2$$ At $x=2$: numerator $=2-1=1\neq0$. At $x=-2$: numerator $=-2-1=-3\neq0$. So both are genuine vertical asymptotes. $$\boxed{x=2 \text{ and } x=-2}$$ (b) The degree of the numerator ($1$) is less than the degree of the denominator ($2$), so as $x\to\pm\infty$, $g(x)\to 0$. $$\boxed{y=0}$$ (c) Testing for even/odd symmetry, substitute $-x$ for $x$: $$g(-x) = \frac{-x-1}{(-x)^2-4} = \frac{-x-1}{x^2-4} = -\frac{x+1}{x^2-4}$$ Compare this to $g(x) = \dfrac{x-1}{x^2-4}$ and to $-g(x) = -\dfrac{x-1}{x^2-4} = \dfrac{1-x}{x^2-4}$. Since $-(x+1) \neq x-1$ in general, $g(-x) \neq g(x)$, so $g$ is not even. Since $-(x+1) \neq -(x-1)$ in general (this would require $-1=1$), $g(-x) \neq -g(x)$, so $g$ is not odd. $$\boxed{g \text{ is neither odd nor even}}$$

QUESTION 10 13 marks

Hard

Consider the function $h(x) = \dfrac{2x^2-8}{x^2-1}$, $x \neq \pm 1$. (a) By carrying out polynomial division, express $h(x)$ in the form $A + \dfrac{B}{x^2-1}$, where $A,B \in \mathbb{Z}$. (b) Write down the equations of all asymptotes of the graph of $h$. (c) Find the coordinates of the $x$-intercepts and the $y$-intercept of the graph of $h$. (d) By considering $h(-x)$, determine whether the graph of $h$ has any symmetry, stating its type.

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(a) Performing polynomial division on $2x^2-8$ by $x^2-1$: since $2(x^2-1) = 2x^2-2$, $$2x^2-8 = 2(x^2-1) - 6$$ So: $$h(x) = \frac{2x^2-8}{x^2-1} = \frac{2(x^2-1) - 6}{x^2-1} = 2 - \frac{6}{x^2-1}$$ $$\boxed{A=2, \; B=-6}$$ (b) Vertical asymptotes occur where $x^2-1=0$, i.e. $x=1$ or $x=-1$ (the numerator, $-6$, is never zero, so both are genuine asymptotes). From part (a), as $x\to\pm\infty$, $\dfrac{-6}{x^2-1}\to 0$, so $h(x)\to 2$. $$\boxed{x=1, \; x=-1, \; y=2}$$ (c) The $x$-intercepts occur where the numerator is zero: $$2x^2-8=0 \implies x^2=4 \implies x=\pm2$$ giving points $(2,0)$ and $(-2,0)$. The $y$-intercept is $h(0)$: $$h(0) = \frac{-8}{-1} = 8$$ giving the point $(0,8)$. $$\boxed{(2,0), \; (-2,0), \; (0,8)}$$ (d) Substituting $-x$ for $x$: $$h(-x) = \frac{2(-x)^2-8}{(-x)^2-1} = \frac{2x^2-8}{x^2-1} = h(x)$$ Since $h(-x)=h(x)$ for all $x$ in the domain, $h$ is an even function. $$\boxed{\text{The graph of } h \text{ is symmetric about the } y\text{-axis}}$$

QUESTION 11 5 marks

Easy

Solve the equation $e^{2x}-5e^{x}+6=0$, giving your answers in exact form.

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Method: this equation is quadratic in $e^{x}$. Let $u=e^{x}$, noting $u>0$. Since $e^{2x}=(e^{x})^{2}=u^{2}$, the equation becomes: $$u^{2}-5u+6=0$$ Factorising: $$(u-2)(u-3)=0$$ $$u=2\ \text{or}\ u=3$$ Both are positive, so both are valid values of $e^{x}$. Substituting back $u=e^{x}$: $$e^{x}=2\implies x=\ln2$$ $$e^{x}=3\implies x=\ln3$$ $$\boxed{x=\ln2\ \text{or}\ x=\ln3}$$

QUESTION 12 6 marks

Medium

Solve the equation $\dfrac{2}{x-1}+\dfrac{3}{x+2}=1$ algebraically, giving your answers in exact form.

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The equation is undefined at $x=1$ and $x=-2$, so these values are excluded from the solution set. Multiplying both sides by $(x-1)(x+2)$ to clear the fractions: $$2(x+2)+3(x-1)=(x-1)(x+2)$$ Expanding the left-hand side: $$2x+4+3x-3=5x+1$$ Expanding the right-hand side: $$(x-1)(x+2)=x^{2}+x-2$$ So the equation becomes: $$5x+1=x^{2}+x-2$$ $$0=x^{2}+x-2-5x-1$$ $$x^{2}-4x-3=0$$ Applying the quadratic formula: $$x=\frac{4\pm\sqrt{(-4)^{2}-4(1)(-3)}}{2}=\frac{4\pm\sqrt{16+12}}{2}=\frac{4\pm\sqrt{28}}{2}=\frac{4\pm2\sqrt{7}}{2}=2\pm\sqrt{7}$$ Checking against the excluded values $x=1,-2$: since $\sqrt{7}\approx2.6458$, $2+\sqrt7\approx4.646$ and $2-\sqrt7\approx-0.646$, neither of which equals 1 or $-2$, so both solutions are valid. $$\boxed{x=2+\sqrt{7}\ \text{or}\ x=2-\sqrt{7}}$$

QUESTION 13 5 marks

Easy

The equation $2^{x}=3x$ cannot be solved algebraically using elementary methods. Use your graphic display calculator to find all solutions of $2^{x}=3x$, giving your answers correct to three significant figures.

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Method: graph $Y_{1}=2^{x}$ and $Y_{2}=3x$ on a GDC and find the $x$-coordinates of their points of intersection (equivalently, graph $Y_{1}=2^{x}-3x$ and find its zeros). The two curves intersect at two points. Reading the intersection points from the GDC: $$x=0.458\ \text{(3 s.f.)}$$ $$x=3.31\ \text{(3 s.f.)}$$ (Check: at $x=0.458$, $2^{0.458}=1.3736\ldots\approx3(0.458)=1.374$. At $x=3.31$, $2^{3.31}=9.92\ldots\approx3(3.31)=9.93$, confirming both solutions.) $$\boxed{x\approx0.458\ \text{or}\ x\approx3.31}$$

QUESTION 14 7 marks

Medium

(a) Use your graphic display calculator to solve $\ln x=4-x$, giving your answer correct to three significant figures. [4] (b) Hence, or otherwise, solve $\ln(x^{2})=8-2x$ for $x>0$. [3]

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(a) Method: graph $Y_{1}=\ln x$ and $Y_{2}=4-x$ on a GDC and find the $x$-coordinate of their point of intersection (equivalently, find the zero of $Y_{1}=\ln x+x-4$). Reading from the GDC: $$\boxed{x\approx2.93\ \text{(3 s.f.)}}$$ (b) For $x>0$, use the log law $\ln(x^{2})=2\ln x$: $$\ln(x^{2})=8-2x$$ $$2\ln x=8-2x$$ Dividing both sides by 2: $$\ln x=4-x$$ This is exactly the equation solved in part (a), so the solution is the same. $$\boxed{x\approx2.93\ \text{(3 s.f.)}}$$

QUESTION 15 16 marks

Hard

Consider the equation $xe^{-x}=k$, where $k\in\mathbb{R}$ and $x\ge0$. Let $f(x)=xe^{-x}$. (a) Find $f'(x)$, and hence show that $f$ has a maximum value at $x=1$. State this maximum value in exact form. [5] (b) State the equation of the horizontal asymptote of the graph of $y=f(x)$ as $x\to\infty$, and find $f(0)$. [3] (c) Using your results from (a) and (b), determine the values of $k$ for which the equation $xe^{-x}=k$ has, for $x\ge0$: (i) no solutions; (ii) exactly one solution; (iii) exactly two solutions. [5] (d) Solve the equation $xe^{-x}=0.2$ for $0\le x<1$, giving your answer correct to three significant figures. [3]

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(a) Using the product rule with $u=x,\ v=e^{-x}$ (so $u'=1,\ v'=-e^{-x}$): $$f'(x)=u'v+uv'=e^{-x}+x(-e^{-x})=e^{-x}(1-x)$$ Since $e^{-x}>0$ for all $x$, the sign of $f'(x)$ is determined by $(1-x)$: For $x<1$: $1-x>0\implies f'(x)>0$ (f is increasing). For $x>1$: $1-x<0\implies f'(x)<0$ (f is decreasing). Since $f'$ changes sign from positive to negative at $x=1$, $f$ has a (local, and since it's the only turning point on $x\ge0$, global) maximum at $x=1$. Maximum value: $f(1)=1\cdot e^{-1}=\dfrac{1}{e}$ $$\boxed{f'(x)=e^{-x}(1-x);\ \text{maximum value } =\dfrac{1}{e}\ \text{at } x=1}$$ (b) As $x\to\infty$, the exponential decay $e^{-x}\to0$ dominates the linear factor $x$, so $f(x)\to0$. $$\boxed{\text{horizontal asymptote: } y=0}$$ $$f(0)=0\cdot e^{0}=0$$ $$\boxed{f(0)=0}$$ (c) On $x\ge0$: $f(0)=0$, $f$ increases to a maximum of $\dfrac{1}{e}$ at $x=1$, then decreases and approaches (but never reaches) $y=0$ as $x\to\infty$. Also $f(x)>0$ for all $x>0$, and $f(x)\ge0$ throughout. (i) $xe^{-x}=k$ has no solutions when $k$ is outside the range of $f$ on $[0,\infty)$, i.e. when $k<0$ or $k>\dfrac{1}{e}$. $$\boxed{k<0\ \text{or}\ k>\dfrac{1}{e}}$$ (ii) Exactly one solution occurs at the endpoints of the achievable range: $k=0$ (only at $x=0$, since $f(x)>0$ for all $x>0$) or $k=\dfrac{1}{e}$ (only at the turning point $x=1$). $$\boxed{k=0\ \text{or}\ k=\dfrac{1}{e}}$$ (iii) Exactly two solutions occur for every value strictly between the endpoint values (one on the increasing branch $01$): $$\boxed{0

QUESTION 16 5 marks

Easy

Solve the inequality $x^2+1 \ge 4x$, giving your answer in exact (surd) form.

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Method: rearrange so that one side is zero, then use the quadratic formula and the shape of the parabola. Rearranging $x^2+1 \ge 4x$: $$x^2-4x+1 \ge 0$$ Solving the corresponding equation $x^2-4x+1=0$ using the quadratic formula: $$x = \frac{4\pm\sqrt{16-4}}{2} = \frac{4\pm\sqrt{12}}{2} = \frac{4\pm2\sqrt3}{2} = 2\pm\sqrt3$$ Since the coefficient of $x^2$ is positive, the graph of $y=x^2-4x+1$ is an upward-opening parabola, which is $\ge0$ outside its roots (i.e. for $x$ less than or equal to the smaller root, or greater than or equal to the larger root). $$\boxed{x \le 2-\sqrt3 \ \text{ or } \ x \ge 2+\sqrt3}$$

QUESTION 17 6 marks

Medium

Solve the inequality $\dfrac{x+3}{x-2} \ge 1$. Give your answer in interval notation.

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Method: do not multiply both sides by $(x-2)$ directly, since its sign is unknown. Instead, move everything to one side first and combine into a single fraction. $$\frac{x+3}{x-2} - 1 \ge 0$$ $$\frac{(x+3)-(x-2)}{x-2} \ge 0$$ $$\frac{5}{x-2} \ge 0$$ Since the numerator $5$ is a positive constant, the sign of the fraction is determined entirely by the sign of the denominator: the fraction is positive exactly when $x-2>0$, i.e. $x>2$. (Note $x=2$ must be excluded, as it makes the expression undefined, and the fraction can never equal $0$ since the numerator is never $0$.) $$\boxed{x \in (2,\infty)}$$ Check: at $x=3$, $\dfrac{3+3}{3-2}=6\ge1$. At $x=0$, $\dfrac{0+3}{0-2}=-1.5$, not $\ge1$, correctly excluded.

QUESTION 18 7 marks

Easy

Solve the inequality $\sqrt{2x+3} \ge x$, stating clearly the domain restriction required for the square root to be defined.

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Method: this involves squaring an inequality, which is only valid when both sides are non-negative, so we split into cases based on the sign of $x$. **Domain:** for $\sqrt{2x+3}$ to be defined, we require $2x+3\ge0$, i.e. $x\ge-\dfrac{3}{2}$. **Case 1: $x<0$.** Since $\sqrt{2x+3}\ge0$ always, and here the right-hand side $x$ is negative, the inequality $\sqrt{2x+3}\ge x$ holds automatically for every $x$ in the domain with $x<0$, i.e. for all $-\dfrac32\le x<0$. **Case 2: $x\ge0$.** Both sides are now non-negative, so it is valid to square both sides (this preserves the inequality direction since both sides are $\ge0$): $$2x+3 \ge x^2$$ $$x^2-2x-3 \le 0$$ $$(x-3)(x+1) \le 0$$ $$-1 \le x \le 3$$ Combined with the case condition $x\ge0$: $0\le x\le3$. **Combining both cases:** $-\dfrac32 \le x < 0$ together with $0\le x\le3$ gives: $$\boxed{-\dfrac{3}{2} \le x \le 3}$$ Check: at $x=3$: $\sqrt{9}=3\ge3$ (equality holds). At $x=4$ (outside): $\sqrt{11}\approx3.317 \ge 4$? False, correctly excluded.

QUESTION 19 8 marks

Medium

Solve the inequality $\dfrac{x-1}{x+2} \ge x-3$, giving your answer in exact form using interval notation.

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Method: move everything to one side and combine into a single fraction (do not cross-multiply by $x+2$, whose sign is unknown). $$\frac{x-1}{x+2} - (x-3) \ge 0$$ $$\frac{(x-1)-(x-3)(x+2)}{x+2} \ge 0$$ Expanding $(x-3)(x+2) = x^2-x-6$: $$\frac{(x-1)-(x^2-x-6)}{x+2} \ge 0 \ \Rightarrow\ \frac{-x^2+2x+5}{x+2} \ge 0$$ Multiplying both sides by $-1$ (reversing the inequality): $$\frac{x^2-2x-5}{x+2} \le 0$$ Finding the critical values. Denominator zero: $x=-2$ (excluded from domain). Numerator zero: $x^2-2x-5=0 \Rightarrow x = \dfrac{2\pm\sqrt{4+20}}{2} = \dfrac{2\pm\sqrt{24}}{2} = 1\pm\sqrt6$. So the critical points, in order, are: $x=-2,\quad x=1-\sqrt6\ (\approx-1.449), \quad x=1+\sqrt6\ (\approx3.449)$. Testing the sign of $\dfrac{x^2-2x-5}{x+2}$ in each interval: - $x<-2$ (e.g. $x=-3$): numerator $=9+6-5=10>0$; denominator $=-1<0$ $\Rightarrow$ negative. Satisfies $\le0$. - $-20$; denominator $=0.3>0$ $\Rightarrow$ positive. Does not satisfy. - $1-\sqrt60$ $\Rightarrow$ negative. Satisfies $\le0$. - $x>1+\sqrt6$ (e.g. $x=4$): numerator $=3>0$; denominator $=6>0$ $\Rightarrow$ positive. Does not satisfy. Including the roots $1\pm\sqrt6$ (where the expression equals $0$, satisfying $\le0$) but excluding $x=-2$ (undefined): $$\boxed{x \in (-\infty,-2) \cup [\,1-\sqrt6,\ 1+\sqrt6\,]}$$

QUESTION 20 16 marks

Hard

Consider the functions $f(x) = \dfrac{2x+1}{x-1}$ and $g(x) = x+2$, where $x\ne1$. (a) Show that solving $f(x) \ge g(x)$ is equivalent to solving $\dfrac{x^2-x-3}{x-1} \le 0$. [5] (b) Find the roots of $x^2-x-3=0$ in exact (surd) form. [3] (c) Hence solve the inequality $f(x)\ge g(x)$, giving your answer in interval notation using exact values. [5] (d) State the number of integer values of $x$ that satisfy the inequality within the interval $-5\le x\le5$. [3]

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(a) Method: subtract $g$ from $f$ and combine over a common denominator. $$f(x)-g(x) = \frac{2x+1}{x-1} - (x+2) = \frac{(2x+1)-(x+2)(x-1)}{x-1}$$ Expanding $(x+2)(x-1) = x^2+x-2$: Numerator: $(2x+1)-(x^2+x-2) = 2x+1-x^2-x+2 = -x^2+x+3$. So $f(x)-g(x) = \dfrac{-x^2+x+3}{x-1}$. The inequality $f(x)\ge g(x)$ becomes $\dfrac{-x^2+x+3}{x-1}\ge0$. Multiplying by $-1$ (reversing the inequality direction): $\dfrac{x^2-x-3}{x-1}\le0$. $\blacksquare$ (b) Using the quadratic formula on $x^2-x-3=0$: $$x = \frac{1\pm\sqrt{1+12}}{2} = \frac{1\pm\sqrt{13}}{2}$$ $$\boxed{x = \dfrac{1+\sqrt{13}}{2}\ (\approx2.303) \quad \text{or} \quad x=\dfrac{1-\sqrt{13}}{2}\ (\approx-1.303)}$$ (c) Critical values, in order: $x=\dfrac{1-\sqrt{13}}{2}\ (\approx-1.303)$, $x=1$ (excluded, denominator zero), $x=\dfrac{1+\sqrt{13}}{2}\ (\approx2.303)$. Testing the sign of $\dfrac{x^2-x-3}{x-1}$: - $x<\dfrac{1-\sqrt{13}}{2}$ (e.g. $x=-2$): numerator $=4+2-3=3>0$; denominator $=-3<0$ $\Rightarrow$ negative. Satisfies $\le0$. - $\dfrac{1-\sqrt{13}}{2}0$ $\Rightarrow$ negative. Satisfies $\le0$. - $x>\dfrac{1+\sqrt{13}}{2}$ (e.g. $x=3$): numerator $=9-3-3=3>0$; denominator $=2>0$ $\Rightarrow$ positive. Does not satisfy. Including roots (equality, satisfies $\le0$) but excluding $x=1$: $$\boxed{x \in \left(-\infty,\ \dfrac{1-\sqrt{13}}{2}\right] \cup \left(1,\ \dfrac{1+\sqrt{13}}{2}\right]}$$ (d) Numerically, $\dfrac{1-\sqrt{13}}{2}\approx-1.303$ and $\dfrac{1+\sqrt{13}}{2}\approx2.303$. From the first interval $\left(-\infty,-1.303\right]$: integers satisfying $x\le-1.303$ within $-5\le x\le5$ are $x=-5,-4,-3,-2$ (note $x=-1$ is excluded since $-1>-1.303$). That is $4$ integers. From the second interval $(1,\ 2.303]$: integers satisfying $1

QUESTION 21 4 marks

Easy

The graph shows $y=f(x)$, where $f$ is a quadratic function. (a) Write down the coordinates of the $x$-intercepts of the graph. [2] (b) Write down the coordinates of the vertex of the graph, and state whether it is a maximum or minimum point. [2]

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(a) The graph crosses the $x$-axis at two points. Reading off the grid, the curve passes through the axis at $x=-1$ and at $x=3$. $$\boxed{(-1,0) \text{ and } (3,0)}$$ (b) The turning point of the graph is the highest point shown, since the parabola opens downward. Reading its coordinates from the grid, it is located directly above the midpoint of the two $x$-intercepts (at $x=\frac{-1+3}{2}=1$), at height $4$. $$\boxed{(1,4), \text{ a maximum point}}$$

QUESTION 22 7 marks

Medium

The graph shows $y=f(x)=x^4-4x^2$ for $-2.5 \le x \le 2.5$. (a) Write down the $x$-intercepts shown on the graph. [2] (b) State the coordinates of the three turning points shown, classifying each as a local maximum or a local minimum. [3] (c) Write down the interval(s) of $x$ for which $f$ is decreasing. [2]

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(a) Reading from the graph, the curve crosses or touches the $x$-axis at three places: it touches the axis in the middle and crosses it once on each side. $$\boxed{x=-2, \; x=0, \; x=2}$$ (b) Between the two outer roots, the graph dips down to a low point on either side of the central touching point, and rises to a local peak exactly at the centre. Reading their coordinates from the grid: $$\boxed{\text{Local minima: } (-\sqrt2,-4) \text{ and } (\sqrt2,-4); \quad \text{Local maximum: } (0,0)}$$ (c) The graph falls from the top-left of the picture down to the first local minimum at $x=-\sqrt2$: this is the first decreasing stretch. It then rises to the local maximum at $x=0$ (increasing), then falls again from $x=0$ down to the second local minimum at $x=\sqrt2$ (decreasing again), before rising to the top-right of the picture. $$\boxed{x<-\sqrt2 \; \text{ or } \; 0

QUESTION 23 5 marks

Easy

The graph shows $y=f(x)$, where $f$ is a cubic function. (a) Write down the $x$-intercepts of the graph, stating which one corresponds to a repeated root of $f(x)=0$. Justify your choice by describing the shape of the curve at that point. [3] (b) Write down the coordinates of the local maximum point shown on the graph. [2]

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(a) The curve crosses the $x$-axis at $x=-2$, passing straight through from below to above the axis, so this is a root of odd (here, simple) multiplicity. The curve also meets the axis at $x=1$, but there it only touches the axis and turns back upward without crossing to negative values - this "flattening" contact is the signature of a repeated (even-multiplicity) root. $$\boxed{x=-2 \text{ (simple root)}, \quad x=1 \text{ (repeated root, curve touches the axis)}}$$ (b) Between the two $x$-intercepts the curve rises to a peak before turning back down towards $x=1$. Reading the coordinates of this peak from the grid: $$\boxed{(-1,4)}$$

QUESTION 24 7 marks

Medium

Consider the function $f(x) = \dfrac{2x-3}{x+1}$, $x\neq-1$. (a) Write down the equations of the vertical and horizontal asymptotes of the graph of $f$. [2] (b) Find the coordinates of the $x$-intercept and the $y$-intercept of the graph of $f$. [2] (c) By writing $f(x)$ in the form $2+\dfrac{c}{x+1}$, determine whether the graph of $f$ approaches its horizontal asymptote from above or from below as $x\to\infty$, and as $x\to-\infty$. [3]

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(a) The vertical asymptote occurs where the denominator is zero: $$x+1=0 \implies x=-1$$ Since the numerator and denominator both have degree $1$, the horizontal asymptote is the ratio of the leading coefficients: as $x\to\pm\infty$, $f(x)\to\dfrac{2x}{x}=2$. $$\boxed{x=-1 \text{ and } y=2}$$ (b) The $x$-intercept occurs where the numerator is zero: $$2x-3=0 \implies x=\frac32$$ giving the point $\left(\dfrac32,0\right)$. The $y$-intercept is $f(0)$: $$f(0) = \frac{-3}{1} = -3$$ giving the point $(0,-3)$. $$\boxed{\left(\dfrac32,0\right) \text{ and } (0,-3)}$$ (c) Since $2(x+1)=2x+2$, we can write: $$2x-3 = 2(x+1) - 5$$ So: $$f(x) = \frac{2(x+1)-5}{x+1} = 2 - \frac{5}{x+1}$$ As $x\to\infty$, $x+1\to\infty$ (positive), so $\dfrac{5}{x+1}\to0^+$, meaning $-\dfrac{5}{x+1}\to0^-$. Hence $f(x)\to2$ from below. As $x\to-\infty$, $x+1\to-\infty$ (negative), so $\dfrac{5}{x+1}\to0^-$, meaning $-\dfrac{5}{x+1}\to0^+$. Hence $f(x)\to2$ from above. $$\boxed{\text{As } x\to\infty, \; f(x)\to2^- \text{ (from below)}; \quad \text{as } x\to-\infty, \; f(x)\to2^+ \text{ (from above)}}$$

QUESTION 25 12 marks

Hard

Consider the function $f(x) = \dfrac{x^2-4}{x-1}$, $x\neq1$. (a) By performing polynomial division, express $f(x)$ in the form $x+1+\dfrac{c}{x-1}$, where $c\in\mathbb{Z}$. [3] (b) Write down the equation of the oblique (slant) asymptote and the equation of the vertical asymptote of the graph of $f$. [2] (c) Find the coordinates of the $x$-intercepts and the $y$-intercept of the graph of $f$. [3] (d) Find $f'(x)$, and hence show that the graph of $f$ has no turning points. [3] (e) State whether the graph of $f$ approaches its oblique asymptote from above or from below as $x\to\infty$. [1]

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(a) Performing polynomial division on $x^2-4$ by $x-1$: since $(x-1)(x+1)=x^2-1$, $$x^2-4 = (x^2-1)-3 = (x-1)(x+1)-3$$ So: $$f(x) = \frac{x^2-4}{x-1} = \frac{(x-1)(x+1)-3}{x-1} = x+1-\frac{3}{x-1}$$ $$\boxed{f(x) = x+1+\dfrac{-3}{x-1}, \text{ i.e. } c=-3}$$ (b) As $x\to\pm\infty$, the term $-\dfrac{3}{x-1}\to0$, so $f(x)$ approaches the line $y=x+1$: this is the oblique asymptote. The vertical asymptote occurs where the denominator of the original expression is zero. $$\boxed{\text{Oblique asymptote: } y=x+1; \quad \text{Vertical asymptote: } x=1}$$ (c) The $x$-intercepts occur where the numerator is zero: $$x^2-4=0 \implies x=\pm2$$ giving the points $(2,0)$ and $(-2,0)$. The $y$-intercept is $f(0)$: $$f(0) = \frac{0-4}{0-1} = \frac{-4}{-1}=4$$ giving the point $(0,4)$. $$\boxed{(2,0), \; (-2,0), \; (0,4)}$$ (d) Differentiating the form found in part (a), $f(x)=x+1-3(x-1)^{-1}$: $$f'(x) = 1 - 3\cdot(-1)(x-1)^{-2} = 1+\frac{3}{(x-1)^2}$$ For all $x\neq1$ in the domain, $(x-1)^2>0$, so $\dfrac{3}{(x-1)^2}>0$. Therefore: $$f'(x) = 1+\frac{3}{(x-1)^2} > 1 > 0 \quad \text{for all } x\neq1$$ Since $f'(x)$ is strictly positive everywhere it is defined, $f$ is increasing on each branch of its domain and $f'(x)$ is never zero - so the graph of $f$ has no turning points. $\blacksquare$ (e) From part (a), $f(x)-(x+1) = -\dfrac{3}{x-1}$. As $x\to\infty$, $x-1\to\infty$ (positive), so $-\dfrac{3}{x-1}\to0^-$ (a small negative number). Since $f(x)$ is slightly less than $x+1$, the graph approaches the oblique asymptote from below. $$\boxed{\text{From below}}$$

QUESTION 26 5 marks

Easy

The graph shows $y=f(x)$, where $f$ is a rational function. The dashed lines indicate the asymptotes of the graph. (a) Write down the equation of the vertical asymptote. [1] (b) Write down the equation of the horizontal asymptote. [1] (c) Write down the coordinates of the $x$-intercept of the graph. [1] (d) Hence state the domain and range of $f$. [2]

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(a) The branches of the curve both run alongside the vertical dashed line without ever crossing it. Reading its position on the horizontal grid: $$\boxed{x=2}$$ (b) Similarly, both branches flatten out and hug the horizontal dashed line as $x\to\pm\infty$. Reading its height on the vertical grid: $$\boxed{y=1}$$ (c) The right-hand branch crosses the $x$-axis once, at the point where the curve meets $y=0$. Reading this off the grid: $$\boxed{(1,0)}$$ (d) Since the graph never touches the vertical asymptote $x=2$, this value is excluded from the domain. Since the graph never touches the horizontal asymptote $y=1$ (both branches approach it but never reach it), this value is excluded from the range. $$\boxed{\text{Domain: } x\neq2; \quad \text{Range: } y\neq1}$$

QUESTION 27 8 marks

Medium

The graph shows $y=f(x)=2\sin(x)+1$ for $0 \le x \le 2\pi$. The dashed line shows the principal axis of the graph. (a) Write down the amplitude of $f$ and the equation of the principal axis. [2] (b) Write down the coordinates of the maximum and minimum points shown on the graph. [2] (c) Using the graph, or otherwise, solve the equation $2\sin(x)+1=2$ for $0\le x\le2\pi$, giving your answers in exact form. [4]

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(a) The dashed principal axis lies at height $y=1$ on the vertical scale, and the curve rises $2$ units above this line at its peak. So the amplitude is $2$ and the principal axis is $y=1$. $$\boxed{\text{Amplitude} = 2; \quad \text{Principal axis: } y=1}$$ (b) Reading from the graph, the curve reaches its highest point a quarter of the way along the domain, and its lowest point three-quarters of the way along. $$\boxed{\text{Maximum: } \left(\dfrac{\pi}{2},3\right); \quad \text{Minimum: } \left(\dfrac{3\pi}{2},-1\right)}$$ (c) Setting $f(x)=2$: $$2\sin(x)+1=2$$ $$2\sin(x)=1$$ $$\sin(x)=\frac12$$ This corresponds to the height $y=2$, which lies between the principal axis ($y=1$) and the maximum ($y=3$); on the graph, the horizontal line $y=2$ crosses the curve exactly twice within $[0,2\pi]$ - once while the curve is rising towards its first maximum, and once while it is falling away from that maximum. The reference angle for $\sin(x)=\dfrac12$ is $\dfrac{\pi}{6}$, and since sine is positive in the first and second quadrants: $$x=\frac{\pi}{6} \quad \text{or} \quad x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$$ Both lie within $[0,2\pi]$, consistent with the two crossings seen on the graph. $$\boxed{x=\dfrac{\pi}{6} \text{ or } x=\dfrac{5\pi}{6}}$$

QUESTION 28 4 marks

Easy

The graph shows $y=f(x)$, where $f$ is an exponential function. The dashed line indicates the horizontal asymptote of the graph. (a) Write down the equation of the horizontal asymptote. [1] (b) Write down the coordinates of the $y$-intercept. [1] (c) State the range of $f$. [2]

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(a) As $x\to-\infty$, the curve flattens out and approaches the dashed horizontal line without ever reaching it. Reading its height from the grid: $$\boxed{y=-3}$$ (b) The curve crosses the vertical axis at the point where $x=0$. Reading its height from the grid: $$\boxed{(0,-2)}$$ (c) Since the curve rises without bound as $x\to\infty$ and approaches, but never reaches, the horizontal asymptote $y=-3$ as $x\to-\infty$, every output value is greater than $-3$. $$\boxed{f(x)>-3}$$

QUESTION 29 7 marks

Medium

Consider the function $f(x) = 3-2^{-x}$. (a) State the equation of the horizontal asymptote of the graph of $f$. [1] (b) Find the $y$-intercept of the graph of $f$. [1] (c) Find the exact $x$-intercept of the graph of $f$, giving your answer in the form $-\log_2 k$ for some integer $k$. [2] (d) State the range of $f$. [1] (e) Describe fully the sequence of transformations that maps the graph of $y=2^x$ onto the graph of $f$. [2]

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(a) As $x\to\infty$, $2^{-x}\to0$, so $f(x)\to3$. $$\boxed{y=3}$$ (b) Substituting $x=0$: $$f(0)=3-2^{0}=3-1=2$$ $$\boxed{(0,2)}$$ (c) Setting $f(x)=0$: $$3-2^{-x}=0$$ $$2^{-x}=3$$ Taking $\log_2$ of both sides: $$-x=\log_23$$ $$x=-\log_23$$ $$\boxed{x=-\log_23}$$ (d) As $x\to-\infty$, $2^{-x}\to\infty$, so $f(x)\to-\infty$; as $x\to\infty$, $f(x)\to3$ but never reaches it (part (a)). So $f$ takes every value strictly below $3$. $$\boxed{f(x)<3}$$ (e) Write $f(x)=3-2^{-x}=-2^{-x}+3=-\left(2^{x}\right)\Big|_{x\to-x}+3$. Starting from $y=2^x$: reflecting in the $y$-axis gives $y=2^{-x}$; reflecting this in the $x$-axis gives $y=-2^{-x}$; translating this $3$ units in the positive $y$-direction gives $y=-2^{-x}+3=f(x)$. $$\boxed{\text{Reflect in the } y\text{-axis, then reflect in the } x\text{-axis, then translate } 3 \text{ units vertically upward}}$$

QUESTION 30 12 marks

Hard

Consider the function $f(x) = (x-2)e^x$, $x\in\mathbb{R}$. (a) Find $f(0)$, and find the exact $x$-intercept of the graph of $f$. [2] (b) Find $f'(x)$ using the product rule, and hence find the exact coordinates of the local minimum point of $f$. [4] (c) Find $f''(x)$, and hence find the exact coordinates of the point of inflection of $f$, justifying that the concavity changes there. [4] (d) State the equation of the horizontal asymptote of the graph of $f$ as $x\to-\infty$, briefly justifying why $(x-2)e^x\to0$ in this limit. [2]

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(a) Substituting $x=0$: $$f(0) = (0-2)e^0 = -2(1) = -2$$ Setting $f(x)=0$: since $e^x\neq0$ for all $x$, we require $x-2=0$. $$\boxed{f(0)=-2; \quad x\text{-intercept: } (2,0)}$$ (b) Using the product rule with $u=x-2$, $v=e^x$ (so $u'=1$, $v'=e^x$): $$f'(x) = u'v+uv' = e^x + (x-2)e^x = e^x\big[1+(x-2)\big] = e^x(x-1)$$ Since $e^x>0$ for all $x$, the sign of $f'(x)$ is determined by $(x-1)$: For $x<1$: $x-1<0 \implies f'(x)<0$ ($f$ decreasing). For $x>1$: $x-1>0 \implies f'(x)>0$ ($f$ increasing). Since $f'$ changes from negative to positive at $x=1$, this is a local minimum. Substituting: $$f(1) = (1-2)e^1 = -e$$ $$\boxed{\text{Local minimum: } (1,-e)}$$ (c) Differentiating $f'(x)=e^x(x-1)$ using the product rule again, with $u=x-1$, $v=e^x$: $$f''(x) = 1\cdot e^x + (x-1)e^x = e^x\big[1+(x-1)\big] = e^x\cdot x$$ Setting $f''(x)=0$: since $e^x\neq0$, we require $x=0$. $$f(0) = -2 \quad \text{(from part (a))}$$ Checking the sign of $f''$ either side of $x=0$: for $x<0$ (e.g. $x=-1$), $f''(-1)=-e^{-1}<0$ (concave down); for $x>0$ (e.g. $x=1$), $f''(1)=e>0$ (concave up). Since the sign changes, the concavity genuinely changes at $x=0$. $$\boxed{\text{Point of inflection: } (0,-2)}$$ (d) As $x\to-\infty$, $e^x\to0$ extremely rapidly, while the factor $(x-2)$ only grows in magnitude at a linear (much slower) rate. The exponential decay dominates the linear growth, so their product $(x-2)e^x\to0$. $$\boxed{y=0}$$

QUESTION 31 5 marks

Easy

The graph shows $y=f(x)$, where $f$ is a logarithmic function. The dashed line indicates the vertical asymptote of the graph. (a) Write down the equation of the vertical asymptote. [1] (b) Write down the coordinates of the $x$-intercept. [1] (c) State the domain of $f$. [1] (d) Use the graph to estimate the $y$-intercept, correct to 1 decimal place. [2]

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(a) As $x$ decreases towards a certain value, the curve plunges downward, running alongside the dashed vertical line without ever crossing it. Reading its position from the grid: $$\boxed{x=-3}$$ (b) The curve crosses the $x$-axis at the point where $y=0$. Reading this off the grid, where the curve intersects a labelled grid intersection: $$\boxed{(1,0)}$$ (c) The graph exists only to the right of the vertical asymptote identified in part (a). $$\boxed{x>-3}$$ (d) Reading the height of the curve where it crosses the $y$-axis ($x=0$) from the grid, the curve passes just below the $x$-axis at that point, at approximately $-0.4$ on the vertical scale. $$\boxed{(0,-0.4) \text{ (1 d.p.)}}$$

QUESTION 32 9 marks

Medium

Consider the function $f(x) = x^3-3x^2$. (a) By factorising $f(x)$, find the $x$-intercepts of the graph of $f$. [2] (b) Find $f'(x)$, and hence find the coordinates of the local maximum and local minimum points of $f$. [3] (c) Find $f''(x)$, and hence find the coordinates of the point of inflection of $f$, confirming that the concavity changes there. [3] (d) State the interval(s) of $x$ for which the graph of $f$ is concave up. [1]

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(a) Factorising: $$f(x) = x^3-3x^2 = x^2(x-3)$$ Setting $f(x)=0$: $x^2=0$ or $x-3=0$. $$\boxed{x=0 \text{ (repeated root)}, \; x=3}$$ (b) Differentiating using the power rule: $$f'(x) = 3x^2-6x = 3x(x-2)$$ Setting $f'(x)=0$: $x=0$ or $x=2$. For $x<0$ (e.g. $x=-1$): $f'(-1)=3(-1)(-3)=9>0$ (increasing). For $02$ (e.g. $x=3$): $f'(3)=3(3)(1)=9>0$ (increasing). Since $f'$ changes from $+$ to $-$ at $x=0$, this is a local maximum; since $f'$ changes from $-$ to $+$ at $x=2$, this is a local minimum. $$f(0)=0, \qquad f(2)=8-12=-4$$ $$\boxed{\text{Local maximum: } (0,0); \quad \text{Local minimum: } (2,-4)}$$ (c) Differentiating $f'(x)=3x^2-6x$: $$f''(x) = 6x-6$$ Setting $f''(x)=0$: $$6x-6=0 \implies x=1$$ $$f(1)=1-3=-2$$ Checking the sign of $f''$ either side of $x=1$: $f''(0)=6(0)-6=-6<0$ (concave down), while $f''(2)=6(2)-6=6>0$ (concave up). Since the sign of $f''$ changes from negative to positive at $x=1$, the concavity genuinely changes there, confirming a point of inflection. $$\boxed{(1,-2)}$$ (d) From part (c), $f''(x)>0$ (concave up) exactly when $x>1$. $$\boxed{x>1}$$

QUESTION 33 5 marks

Easy

Consider the quadratic function $f(x) = -x^2+4x+5$. (a) By factorising $f(x)$, find the $x$-intercepts of the graph of $f$. [2] (b) By writing $f(x)$ in the form $-(x-p)^2+q$, find the coordinates of the vertex of the graph of $f$. [2] (c) State the equation of the axis of symmetry. [1]

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(a) Factorising $f(x)=-x^2+4x+5$, first take out a factor of $-1$: $$f(x) = -(x^2-4x-5)$$ We need two numbers that multiply to $-5$ and add to $-4$: these are $-5$ and $1$. $$f(x) = -(x-5)(x+1)$$ Setting $f(x)=0$: $$x=5 \text{ or } x=-1$$ $$\boxed{(-1,0) \text{ and } (5,0)}$$ (b) Completing the square on $f(x)=-x^2+4x+5$: $$f(x) = -(x^2-4x)+5$$ $$f(x) = -\big[(x-2)^2-4\big]+5$$ $$f(x) = -(x-2)^2+4+5 = -(x-2)^2+9$$ $$\boxed{\text{Vertex: } (2,9)}$$ (c) The axis of symmetry of a parabola passes vertically through its vertex, so it has equation $x=p$ where $p$ is the $x$-coordinate of the vertex found in part (b). $$\boxed{x=2}$$

QUESTION 34 7 marks

Medium

Consider the function $f(x) = \ln(x-2)+1$. (a) State the domain of $f$. [1] (b) State the equation of the vertical asymptote of the graph of $f$. [1] (c) Find the exact $x$-intercept of the graph of $f$, giving your answer in terms of $e$. [2] (d) Explain why the graph of $f$ has no $y$-intercept. [1] (e) Briefly explain why the graph of $f$ has no horizontal asymptote as $x\to\infty$. [2]

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(a) The natural logarithm requires its argument to be strictly positive: $$x-2>0 \implies x>2$$ $$\boxed{x>2}$$ (b) As $x\to2^+$, $x-2\to0^+$, so $\ln(x-2)\to-\infty$, meaning $f(x)\to-\infty$: the graph plunges downward alongside the line $x=2$. $$\boxed{x=2}$$ (c) Setting $f(x)=0$: $$\ln(x-2)+1=0$$ $$\ln(x-2)=-1$$ Rewriting in exponential form: $$x-2=e^{-1}$$ $$x=2+e^{-1}$$ $$\boxed{x=2+\dfrac1e}$$ (d) A $y$-intercept would require evaluating $f(0)$. But $x=0$ does not satisfy the domain restriction $x>2$ found in part (a) - substituting $x=0$ into $\ln(x-2)$ gives $\ln(-2)$, which is undefined, since the logarithm of a negative number does not exist. $$\boxed{f(0) \text{ is undefined, since } x=0 \text{ is not in the domain of } f}$$ (e) As $x\to\infty$, $x-2\to\infty$ also, and $\ln(x-2)$ increases without bound (though very slowly). So $f(x)=\ln(x-2)+1\to\infty$ rather than settling towards a fixed finite value - this unbounded growth means there is no horizontal asymptote as $x\to\infty$. $$\boxed{f(x)\to\infty \text{ as } x\to\infty, \text{ so no horizontal asymptote exists on that side}}$$

QUESTION 35 10 marks

Hard

The graph shows $y=k(x)$, where $k(x)=\dfrac{x^2}{x-2}$, $x\neq2$. The dashed lines show the vertical asymptote and the oblique (slant) asymptote of the graph. (a) State the domain of $k$. [1] (b) Using the graph, write down the coordinates of the local maximum point and the local minimum point. [2] (c) Write down the interval(s) of $x$ for which $k$ is increasing. [2] (d) Write down the equation of the vertical asymptote, and describe the behaviour of $k(x)$ as $x\to2^-$ and as $x\to2^+$. [3] (e) Hence, using the graph, write down the solution set of $k(x) \le 0$. [2]

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(a) The function is undefined where the denominator is zero. $$\boxed{x\in\mathbb{R}, \; x\neq2}$$ (b) The left-hand branch of the curve rises to a peak that just touches the $x$-axis before falling away towards the asymptote; the right-hand branch dips to a low point before rising again. Reading their coordinates from the grid: $$\boxed{\text{Local maximum: } (0,0); \quad \text{Local minimum: } (4,8)}$$ (c) Tracing the curve from left to right: it rises up to the local maximum at $x=0$ (increasing), then falls all the way through the vertical asymptote down to the local minimum at $x=4$ (decreasing throughout this whole middle stretch, on both sides of the asymptote), then rises again beyond $x=4$ (increasing). $$\boxed{x<0 \; \text{ or } \; x>4}$$ (d) The dashed vertical line on the graph is positioned at: $$\boxed{x=2}$$ Looking at the graph immediately to the left of this line, the curve plunges steeply downward; immediately to the right, it shoots steeply upward. $$\boxed{\text{As } x\to2^-, \; k(x)\to-\infty; \quad \text{as } x\to2^+, \; k(x)\to+\infty}$$ (e) Reading the sign of the curve across the picture: for $x<0$ the curve lies below the $x$-axis, touching zero only at the local maximum $x=0$; for $02$, the entire right-hand branch (including its local minimum at height $8$) lies above the $x$-axis and never returns to zero or below. So $k(x)\le0$ everywhere to the left of the vertical asymptote, with equality only at the touching point $x=0$. $$\boxed{x<2}$$

QUESTION 36 6 marks

Easy

The graph shows $y=f(x)=3\cos(2x)$ for $0 \le x \le \pi$. (a) Write down the amplitude of $f$. [1] (b) Write down the period of $f$. [1] (c) Write down the coordinates of the maximum and minimum points shown on the graph. [2] (d) Write down the $x$-intercepts shown on the graph, giving your answers in terms of $\pi$. [2]

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(a) The curve oscillates between a highest value of $3$ and a lowest value of $-3$ on the vertical scale, so the amplitude (the distance from the centre line $y=0$ to a peak) is: $$\boxed{3}$$ (b) The graph completes exactly one full oscillation (from a maximum, down to a minimum, and back up to a maximum) over the domain shown, which has length $\pi$. $$\boxed{\text{Period} = \pi}$$ (c) Reading from the graph, the curve begins at its maximum height at $x=0$, and returns to this maximum height again at $x=\pi$; between these it dips to its minimum height at the point midway along, $x=\dfrac{\pi}{2}$. $$\boxed{\text{Maxima: } (0,3) \text{ and } (\pi,3); \quad \text{Minimum: } \left(\dfrac{\pi}{2},-3\right)}$$ (d) The curve crosses the $x$-axis twice within the domain shown, at the quarter and three-quarter points of the interval $[0,\pi]$ marked on the horizontal scale. $$\boxed{x=\dfrac{\pi}{4} \text{ and } x=\dfrac{3\pi}{4}}$$

QUESTION 37 7 marks

Medium

Consider the function $g(x) = \dfrac{x^2-4}{(x-2)(x+1)}$, $x\neq2,\,x\neq-1$. (a) Show that $g(x)$ simplifies to $\dfrac{x+2}{x+1}$, and explain why $x=2$ must still be excluded from the domain even after simplifying. [2] (b) State the coordinates of the point of discontinuity ("hole") on the graph of $g$. [2] (c) State the equations of the vertical and horizontal asymptotes of the graph of $g$. [2] (d) State the domain of $g$. [1]

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(a) Factorising the numerator as a difference of two squares: $$g(x) = \frac{x^2-4}{(x-2)(x+1)} = \frac{(x-2)(x+2)}{(x-2)(x+1)}$$ For $x\neq2$, the factor $(x-2)$ cancels: $$g(x) = \frac{x+2}{x+1}, \quad x\neq2$$ $x=2$ must still be excluded because the original expression involves the factor $(x-2)$ in the denominator, which equals $0$ when $x=2$: the function is undefined there regardless of the algebraic simplification, since cancelling $(x-2)$ is only valid when $x-2\neq0$. $\blacksquare$ (b) The simplified expression gives $\dfrac{2+2}{2+1}=\dfrac43$ at $x=2$, so the graph is the curve $y=\dfrac{x+2}{x+1}$ with an open circle (a "hole") at: $$\boxed{\left(2,\dfrac43\right)}$$ (c) After simplification, the vertical asymptote occurs where the remaining denominator is zero: $$x+1=0 \implies x=-1$$ Since the numerator and denominator of $\dfrac{x+2}{x+1}$ both have degree $1$, the horizontal asymptote is the ratio of leading coefficients: as $x\to\pm\infty$, $g(x)\to\dfrac{x}{x}=1$. $$\boxed{x=-1 \text{ and } y=1}$$ (d) The domain excludes the value that makes the original denominator zero at $x=-1$ (a true vertical asymptote) and the value $x=2$ (the hole). $$\boxed{x\in\mathbb{R}, \; x\neq2, \; x\neq-1}$$

QUESTION 38 5 marks

Easy

The table below shows values of a function for five consecutive integer values of $x$. $$\begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline y & 5 & 10 & 20 & 40 & 80 \end{array}$$ (a) By considering the first differences between consecutive $y$-values, explain why this data cannot be modelled by a linear function. [2] (b) Show that the data can be modelled by an exponential function of the form $y=ab^x$, and find the values of $a$ and $b$. [3]

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(a) For a linear function, the first differences between consecutive $y$-values (for equally spaced $x$-values) must be constant. Here the first differences are: $$10-5=5, \quad 20-10=10, \quad 40-20=20, \quad 80-40=40$$ These differences ($5,10,20,40$) are not constant - each is double the previous one - so the data cannot be modelled by a linear function. $\blacksquare$ (b) For an exponential model $y=ab^x$, the ratio between consecutive $y$-values should be constant and equal to $b$. Checking the ratios: $$\frac{10}{5}=2, \quad \frac{20}{10}=2, \quad \frac{40}{20}=2, \quad \frac{80}{40}=2$$ Since the ratio is constant ($=2$) for equally spaced $x$-values, the data can indeed be modelled by an exponential function with $b=2$. $\blacksquare$ Since $y=ab^x$ and at $x=0$, $y=a b^0 = a$, and the table gives $y=5$ when $x=0$: $$\boxed{a=5, \; b=2 \quad \text{i.e. } y=5(2)^x}$$

QUESTION 39 8 marks

Medium

Consider the quartic function $f(x) = -(x+2)(x-1)^2(x-3)$. (a) Write down the $x$-intercepts of the graph of $f$, and state which one corresponds to a repeated (touching) root. [3] (b) Find the $y$-intercept of the graph of $f$. [1] (c) State whether the leading coefficient of $f$, in expanded form, is positive or negative, and describe the end behaviour of $f$ as $x\to\pm\infty$. [2] (d) Using your graphic display calculator, find the coordinates of the three turning points of $f$, correct to 3 significant figures where necessary. [2]

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(a) Since $f(x)$ is given in factorised form, setting each factor to zero: $$x+2=0 \implies x=-2, \qquad x-1=0 \implies x=1 \text{ (from the squared factor)}, \qquad x-3=0 \implies x=3$$ $$\boxed{x=-2,\; x=3 \text{ (simple roots)}; \quad x=1 \text{ (repeated root, since it comes from } (x-1)^2\text{)}}$$ (b) Substituting $x=0$: $$f(0) = -(0+2)(0-1)^2(0-3) = -(2)(1)(-3) = 6$$ $$\boxed{(0,6)}$$ (c) Expanding the leading term: the product of the four linear factors $(x+2)(x-1)^2(x-3)$ has leading term $x\cdot x\cdot x\cdot x=x^4$, and the overall expression is $-1$ times this, so the leading coefficient of $f$ is $-1$, which is negative. For a quartic with negative leading coefficient, both ends of the graph point downward. $$\boxed{\text{Leading coefficient negative; } f(x)\to-\infty \text{ as } x\to\infty \text{ and as } x\to-\infty}$$ (d) Using a GDC to graph $f(x)=-(x+2)(x-1)^2(x-3)$ and locate its turning points: $$\boxed{\text{Local maximum } \approx(-1.15,16.3); \quad \text{Local minimum} = (1,0); \quad \text{Local maximum} \approx(2.40,5.17)}$$ (The local minimum at $(1,0)$ is exact, since it coincides with the repeated root found in part (a): the curve touches the $x$-axis there.)

QUESTION 40 11 marks

Hard

Consider the function $f(x)=2\cos(2x)-1$ for $0 \le x \le 2\pi$. (a) Write down the amplitude and the period of $f$. [2] (b) Find $f'(x)$, and hence find the exact coordinates of all local minimum and local maximum points of $f$ on the given domain (including at the endpoints, if applicable). [4] (c) Find the exact roots of $f(x)=0$ on $0\le x\le2\pi$. [3] (d) Hence, using a sign diagram or your graph, solve the inequality $f(x)\ge0$ for $0\le x\le2\pi$. [2]

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(a) Comparing $f(x)=2\cos(2x)-1$ to the general form $a\cos(bx)+d$, the amplitude is $|a|=2$ and the period is $\dfrac{2\pi}{b}=\dfrac{2\pi}{2}=\pi$. $$\boxed{\text{Amplitude} = 2; \quad \text{Period} = \pi}$$ (b) Differentiating using the chain rule: $$f'(x) = 2\cdot(-\sin(2x))\cdot2 = -4\sin(2x)$$ Setting $f'(x)=0$: $\sin(2x)=0$, so $2x=0,\pi,2\pi,3\pi,4\pi$, giving $x=0,\dfrac{\pi}{2},\pi,\dfrac{3\pi}{2},2\pi$ within the given domain. Evaluating $f$ at each: $\cos(0)=1$, $\cos(\pi)=-1$, $\cos(2\pi)=1$, $\cos(3\pi)=-1$, $\cos(4\pi)=1$, so: $$f(0)=2(1)-1=1, \; f\left(\frac{\pi}{2}\right)=2(-1)-1=-3, \; f(\pi)=1, \; f\left(\frac{3\pi}{2}\right)=-3, \; f(2\pi)=1$$ $$\boxed{\text{Local minima: } \left(\dfrac{\pi}{2},-3\right) \text{ and } \left(\dfrac{3\pi}{2},-3\right); \quad \text{Local/endpoint maxima: } (0,1), \; (\pi,1), \; (2\pi,1)}$$ (c) Setting $f(x)=0$: $$2\cos(2x)-1=0 \implies \cos(2x)=\frac12$$ The general solutions of $\cos(\theta)=\dfrac12$ are $\theta=\dfrac{\pi}{3}+2k\pi$ or $\theta=-\dfrac{\pi}{3}+2k\pi$ (equivalently $\dfrac{5\pi}{3}+2k\pi$), for integer $k$. With $\theta=2x$ and $0\le2x\le4\pi$: $$2x=\frac{\pi}{3}, \; \frac{5\pi}{3}, \; \frac{\pi}{3}+2\pi=\frac{7\pi}{3}, \; \frac{5\pi}{3}+2\pi=\frac{11\pi}{3}$$ Dividing by $2$: $$\boxed{x=\dfrac{\pi}{6}, \; \dfrac{5\pi}{6}, \; \dfrac{7\pi}{6}, \; \dfrac{11\pi}{6}}$$ (d) Using the local maxima/minima from (b) and the roots from (c) to build a sign diagram: at $x=0$, $f(0)=1>0$; between the roots $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ the graph dips to the minimum $-3$ (negative); between $\frac{5\pi}{6}$ and $\frac{7\pi}{6}$ it rises to the maximum $1$ at $x=\pi$ (positive); between $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ it dips to $-3$ again (negative); and from $\frac{11\pi}{6}$ to $2\pi$ it rises back to $1$ (positive). So $f(x)\ge0$ on: $$\boxed{0\leq x\leq\dfrac{\pi}{6} \;\cup\; \dfrac{5\pi}{6}\leq x\leq\dfrac{7\pi}{6} \;\cup\; \dfrac{11\pi}{6}\leq x\leq2\pi}$$

QUESTION 41 5 marks

Easy

Consider the function $f(x) = |x-2|+1$. (a) State the coordinates of the vertex of the graph of $f$, and state whether it is a maximum or a minimum point. [2] (b) Find the $y$-intercept of the graph of $f$. [1] (c) Explain why the graph of $f$ has no $x$-intercepts. [2]

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(a) The graph of $f(x)=|x-2|+1$ is the graph of $y=|x|$ translated $2$ units in the positive $x$-direction and $1$ unit in the positive $y$-direction. The vertex of $y=|x|$ is at the origin, so the vertex of $f$ is at: $$\boxed{(2,1), \text{ a minimum point}}$$ (It is a minimum because $|x-2|\ge0$ for all $x$, so $f(x)=|x-2|+1\ge1$ for all $x$, with equality exactly at the vertex.) (b) Substituting $x=0$: $$f(0) = |0-2|+1 = |-2|+1 = 2+1=3$$ $$\boxed{(0,3)}$$ (c) An $x$-intercept would require $f(x)=0$, i.e. $|x-2|+1=0$, i.e. $|x-2|=-1$. But an absolute value can never be negative, so $|x-2|=-1$ has no solution. Equivalently, from part (a), the minimum value of $f$ is $1$, which is greater than $0$, so the graph never reaches the $x$-axis. $$\boxed{\text{No } x\text{-intercepts exist, since } f(x)\geq1>0 \text{ for all } x}$$

QUESTION 42 6 marks

Medium

The graph shows $y=h(x)$, where $h(x)=\dfrac{x+1}{(x-2)(x+4)}$, together with its vertical asymptotes (dashed). (a) Write down the root of $h$ shown on the graph. [1] (b) Write down the equations of the vertical asymptotes. [2] (c) Using the graph, write down the solution set of $h(x) \le 0$. [3]

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(a) The curve crosses the $x$-axis exactly once. Reading this off the grid: $$\boxed{x=-1}$$ (b) The two dashed vertical lines mark where the curve runs off to $\pm\infty$. Reading their positions from the grid: $$\boxed{x=-4 \text{ and } x=2}$$ (c) Looking at the graph region by region: to the left of $x=-4$, the curve lies below the $x$-axis (negative). Between $x=-4$ and the root at $x=-1$, the curve lies above the axis (positive). Between the root $x=-1$ and the asymptote $x=2$, the curve lies below the axis (negative) - and it touches zero exactly at $x=-1$ itself. To the right of $x=2$, the curve lies above the axis (positive). So $h(x)\le0$ on the leftmost branch, and from the root up to (but not including) the middle vertical asymptote: $$\boxed{x<-4 \; \text{ or } \; -1\leq x<2}$$

QUESTION 43 5 marks

Easy

Consider the function $f(x) = \sqrt{x+3}-1$. (a) State the domain of $f$. [1] (b) State the range of $f$. [1] (c) Find the exact $x$-intercept of the graph of $f$. [2] (d) Find the $y$-intercept of the graph of $f$, correct to 3 significant figures. [1]

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(a) The expression under the square root must be non-negative: $$x+3\geq0 \implies x\geq-3$$ $$\boxed{x\geq-3}$$ (b) Since $\sqrt{x+3}\geq0$ for all $x$ in the domain, $f(x)=\sqrt{x+3}-1\geq-1$, with the minimum value $-1$ attained at $x=-3$. As $x$ increases without bound, $\sqrt{x+3}$ (and hence $f(x)$) increases without bound. $$\boxed{f(x)\geq-1}$$ (c) Setting $f(x)=0$: $$\sqrt{x+3}-1=0 \implies \sqrt{x+3}=1$$ Squaring both sides: $$x+3=1 \implies x=-2$$ (Check: this is in the domain $x\geq-3$, and $\sqrt{-2+3}-1=\sqrt1-1=0$, confirming the solution is valid.) $$\boxed{(-2,0)}$$ (d) Substituting $x=0$: $$f(0) = \sqrt{0+3}-1 = \sqrt3-1$$ Since $\sqrt3 \approx 1.7320508$: $$f(0) \approx 1.7320508-1 = 0.7320508$$ $$\boxed{(0,0.732) \text{ (3 s.f.)}}$$

QUESTION 44 7 marks

Medium

The graph shows a function $f$ defined piecewise on $\mathbb{R}$. Filled circles indicate points included on the graph; open circles indicate points excluded. (a) State the domain and the range of $f$, as shown on the graph. [2] (b) Write down the value of $f(-1)$, and state whether $f$ is continuous at $x=-1$, justifying your answer using the circles shown at that point. [2] (c) Write down the value of $f(2)$, and state whether $f$ is continuous at $x=2$, justifying your answer using the circles shown at that point. [2] (d) Write down the value of $f(0)$. [1]

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(a) The graph consists of pieces that together cover every real value of $x$ (a line for $x<-1$, a curve for $-1\le x\le2$, and a horizontal segment for $x>2$), so the domain is all real numbers. Vertically, the lowest point reached on the graph is $y=0$ (the bottom of the curved piece), and the graph extends upward without bound on the left-hand linear piece, so every $y$-value from $0$ upward is attained. $$\boxed{\text{Domain: } x\in\mathbb{R}; \quad \text{Range: } f(x)\ge0}$$ (b) At $x=-1$, the graph shows a filled circle at height $1$ and an open circle at height $3$ directly above it. The filled circle is the one that is actually part of the graph, so $f(-1)$ takes that value: $$\boxed{f(-1)=1}$$ Since the curve approaches height $3$ (the open circle) as $x\to-1$ from the left, but the actual value $f(-1)=1$ (the filled circle) is different, the two do not match - there is a jump in the graph at $x=-1$, so $f$ is **not** continuous there. (c) At $x=2$, the graph shows a filled circle at height $4$ (the end of the curved piece) and an open circle at height $5$ (where the horizontal piece for $x>2$ would begin). The filled circle gives the actual function value: $$\boxed{f(2)=4}$$ Since the horizontal piece approaches height $5$ (open circle) as $x\to2$ from the right, but $f(2)=4$ (filled circle) is different, there is again a jump - $f$ is **not** continuous at $x=2$. (d) Reading the height of the curved middle piece at $x=0$ directly from the grid: $$\boxed{f(0)=0}$$

QUESTION 45 10 marks

Hard

Consider the function $g(x) = -\ln(x+3)+2$. (a) State the domain of $g$ and the equation of the vertical asymptote of its graph. [2] (b) Find $g'(x)$, and hence explain why $g$ has no turning points and is decreasing throughout its domain. [3] (c) Find the exact $x$-intercept of the graph of $g$, giving your answer in terms of $e$. [2] (d) Find the $y$-intercept of the graph of $g$, correct to 3 significant figures. [1] (e) Describe fully the sequence of transformations that maps the graph of $y=\ln x$ onto the graph of $g$. [2]

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(a) The natural logarithm requires its argument to be strictly positive: $$x+3>0 \implies x>-3$$ As $x\to-3^+$, $\ln(x+3)\to-\infty$, so $g(x)=-\ln(x+3)+2\to+\infty$. $$\boxed{\text{Domain: } x>-3; \quad \text{Vertical asymptote: } x=-3}$$ (b) Differentiating using the chain rule, with the derivative of $\ln(x+3)$ being $\dfrac{1}{x+3}$: $$g'(x) = -\frac{1}{x+3}$$ For every $x$ in the domain ($x>-3$), $x+3>0$, so $\dfrac{1}{x+3}>0$, and therefore: $$g'(x) = -\frac{1}{x+3} < 0 \quad \text{for all } x>-3$$ Since $g'(x)$ is never zero, $g$ has no turning points; since $g'(x)$ is always negative, $g$ is strictly decreasing throughout its entire domain. $\blacksquare$ (c) Setting $g(x)=0$: $$-\ln(x+3)+2=0$$ $$\ln(x+3)=2$$ Rewriting in exponential form: $$x+3=e^2$$ $$x=e^2-3$$ $$\boxed{x=e^2-3}$$ (d) Substituting $x=0$: $$g(0) = -\ln(0+3)+2 = -\ln3+2$$ Since $\ln3\approx1.0986$: $$g(0)\approx-1.0986+2=0.9014$$ $$\boxed{(0,0.901) \text{ (3 s.f.)}}$$ (e) Write $g(x)=-\ln(x+3)+2$. Starting from $y=\ln x$: reflecting in the $x$-axis gives $y=-\ln x$; translating this $3$ units in the negative $x$-direction gives $y=-\ln(x+3)$; translating this $2$ units in the positive $y$-direction gives $y=-\ln(x+3)+2=g(x)$. $$\boxed{\text{Reflect in the } x\text{-axis, then translate } 3 \text{ units left and } 2 \text{ units up}}$$

QUESTION 46 4 marks

Easy

The graph shows $y=f(x)$. (a) State, with reference to the symmetry of the graph, whether $f$ is an even function, an odd function, or neither. [2] (b) The graph passes through the point $\left(\sqrt2,-1\right)$. Using the symmetry identified in part (a), write down another point that must lie on the graph. [1] (c) Write down the number of $x$-intercepts shown on the graph. [1]

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(a) The graph is symmetric about the $y$-axis: for every point on the curve, its mirror image in the vertical axis is also on the curve (the left- and right-hand halves of the picture are reflections of one another). This is exactly the defining property of an even function, $f(-x)=f(x)$. $$\boxed{f \text{ is an even function}}$$ (b) Since $f$ is even, $f(-x)=f(x)$ for every $x$, so the point $(-\sqrt2,-1)$, the reflection of $(\sqrt2,-1)$ in the $y$-axis, must also lie on the graph. $$\boxed{(-\sqrt2,-1)}$$ (c) Counting the points where the curve crosses the horizontal axis in the picture (two on the left of the $y$-axis, two on the right, by the symmetry of part (a)): $$\boxed{4}$$

QUESTION 47 8 marks

Medium

The graph shows $y=f(x)=x^2-2x$, with $x$-intercepts, $y$-intercept and vertex marked. (a) Write down the coordinates of the vertex of the graph of $f$. [1] (b) Write down the coordinates of the vertex of the graph of $y=f(x)+3$. [2] (c) Write down the coordinates of the $x$-intercepts of the graph of $y=f(x-2)$. [2] (d) Describe the transformation that maps the graph of $f$ onto the graph of $y=-f(x)$, and state the coordinates of the vertex of $y=-f(x)$. [2] (e) Given that the range of $f$ is $f(x)\ge-1$, state the range of $y=|f(x)|$. [1]

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(a) Reading the lowest point of the parabola from the graph (or by completing the square: $f(x)=x^2-2x=(x-1)^2-1$): $$\boxed{(1,-1)}$$ (b) The transformation $y=f(x)+3$ translates every point on the graph of $f$ vertically upward by $3$ units, so the vertex moves from $(1,-1)$ to: $$\boxed{(1,2)}$$ (c) The $x$-intercepts of $f$ are found by factorising: $f(x)=x^2-2x=x(x-2)$, so $f(x)=0$ at $x=0$ and $x=2$. The transformation $y=f(x-2)$ translates the graph $2$ units in the positive $x$-direction, so each $x$-intercept also shifts $2$ units to the right: $$\boxed{(2,0) \text{ and } (4,0)}$$ (d) The transformation $y=-f(x)$ reflects the graph of $f$ in the $x$-axis. Under this reflection, a point $(x,y)$ maps to $(x,-y)$, so the vertex $(1,-1)$ maps to: $$\boxed{\text{Reflection in the } x\text{-axis}; \quad \text{vertex of } y=-f(x) \text{ is } (1,1)}$$ (e) Since the range of $f$ is $f(x)\ge-1$, $f$ takes all values from $-1$ up to $0$ (exclusive of positive values only between the roots) and all non-negative values outside the roots - in short, $f$ attains every value in $[-1,\infty)$. Taking the absolute value reflects the negative portion of this range (the values in $[-1,0)$) up to become $(0,1]$, while the already non-negative portion $[0,\infty)$ is unchanged. The union of $(0,1]$ and $[0,\infty)$ is $[0,\infty)$. $$\boxed{|f(x)|\ge0}$$

QUESTION 48 6 marks

Easy

Consider the function $g(x) = (x+3)(x-1)(2-x)$. (a) Write down the roots of $g(x)=0$. [1] (b) Construct a sign diagram for $g(x)$. [3] (c) Hence solve the inequality $g(x) \ge 0$. [2]

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(a) Since $g(x)$ is already given in factorised form, setting each factor to zero: $$x+3=0 \implies x=-3, \qquad x-1=0 \implies x=1, \qquad 2-x=0 \implies x=2$$ $$\boxed{x=-3, \; x=1, \; x=2}$$ (b) These three roots divide the number line into four intervals. Testing a convenient value of $x$ in each interval: For $x<-3$, try $x=-4$: $g(-4)=(-1)(-5)(6)=30>0$. For $-30$. For $x>2$, try $x=3$: $g(3)=(6)(2)(-1)=-12<0$. $$\boxed{\begin{array}{c|ccccccc} x & & -3 & & 1 & & 2 & \\ \hline g(x) & + & 0 & - & 0 & + & 0 & - \end{array}}$$ (c) From the sign diagram, $g(x)$ is positive on $x<-3$ and on $1

QUESTION 49 8 marks

Medium

The graph shows $y=f(x)=x^3-6x^2+9x$. (a) Using the graph, write down the coordinates of the local maximum and local minimum points. [2] (b) Find $f''(x)$, and hence find the exact coordinates of the point of inflection of the graph of $f$. [3] (c) By evaluating $f''(x)$ at a value of $x$ on either side of the point of inflection, confirm that the concavity changes there, stating the interval(s) on which the graph of $f$ is concave down. [3]

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(a) Reading the coordinates of the two turning points marked on the graph: $$\boxed{\text{Local maximum: } (1,4); \quad \text{Local minimum: } (3,0)}$$ (b) Differentiating $f(x)=x^3-6x^2+9x$: $$f'(x) = 3x^2-12x+9$$ Differentiating again: $$f''(x) = 6x-12$$ Setting $f''(x)=0$: $$6x-12=0 \implies x=2$$ Substituting back into $f$: $$f(2) = 8-24+18=2$$ $$\boxed{(2,2)}$$ (c) Evaluating $f''(x)$ just below and just above $x=2$: $$f''(1) = 6(1)-12=-6<0 \quad \text{(concave down)}$$ $$f''(3) = 6(3)-12=6>0 \quad \text{(concave up)}$$ Since $f''$ changes sign from negative to positive at $x=2$, the concavity genuinely changes there, confirming that $(2,2)$ is a point of inflection. Since $f''(x)=6x-12<0$ whenever $x<2$, the graph is concave down on this interval. $$\boxed{\text{Concave down for } x<2}$$

QUESTION 50 14 marks

Hard

Consider the function $f(x)=x^4-4x^3$, which can also be written as $f(x)=x^3(x-4)$. (a) Write down the $x$-intercepts of the graph of $f$, and state the multiplicity of each root. [2] (b) Find $f'(x)$, and use it to find the exact coordinates of the local minimum point of $f$. [4] (c) By evaluating $f'(x)$ at values of $x$ on either side of $x=0$, show that $x=0$ is a stationary point of $f$, but that it is neither a local maximum nor a local minimum. [3] (d) Find $f''(x)$, and hence find the $x$-coordinates of all points of inflection of $f$. [3] (e) Describe the end behaviour of $f$ as $x\to\infty$ and as $x\to-\infty$. [2]

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(a) From the factorised form $f(x)=x^3(x-4)$, setting each factor to zero: $$x^3=0 \implies x=0, \qquad x-4=0 \implies x=4$$ The factor $x^3$ contributes three copies of the root $x=0$, so this root has multiplicity $3$; the factor $(x-4)$ contributes a single copy of the root $x=4$. $$\boxed{x=0 \text{ (multiplicity 3)}, \quad x=4 \text{ (multiplicity 1)}}$$ (b) Differentiating $f(x)=x^4-4x^3$ using the power rule: $$f'(x) = 4x^3-12x^2 = 4x^2(x-3)$$ Setting $f'(x)=0$: $4x^2=0$ or $x-3=0$, giving $x=0$ (repeated) or $x=3$. Checking the sign of $f'$ either side of $x=3$: for $x<3$ (e.g. $x=1$), $f'(1)=4(1)(1-3)=-8<0$ (decreasing); for $x>3$ (e.g. $x=4$), $f'(4)=4(16)(1)=64>0$ (increasing). Since $f'$ changes from negative to positive at $x=3$, this is a local minimum. $$f(3) = 3^4-4(3)^3 = 81-108=-27$$ $$\boxed{\text{Local minimum: } (3,-27)}$$ (c) Since $f'(x)=4x^2(x-3)$ and the factor $4x^2\ge0$ always, the sign of $f'(x)$ is controlled entirely by $(x-3)$, which is negative for all $x<3$ - this includes both sides of $x=0$. Checking directly: for $x=-1$ (just left of $0$), $f'(-1)=4(1)(-1-3)=4(1)(-4)=-16<0$. For $x=1$ (just right of $0$), $f'(1)=4(1)(1-3)=-8<0$. Since $f'(0)=4(0)^2(0-3)=0$, $x=0$ is a stationary point. But since $f'(x)<0$ on both sides of $x=0$ (the sign does not change), $f$ is decreasing throughout a neighbourhood of $x=0$ - it is neither a local maximum nor a local minimum, but a **stationary point of inflection**. $\blacksquare$ (d) Differentiating $f'(x)=4x^3-12x^2$: $$f''(x) = 12x^2-24x = 12x(x-2)$$ Setting $f''(x)=0$: $x=0$ or $x=2$. Checking the sign of $f''$ across each: for $x<0$ (e.g. $x=-1$), $f''(-1)=12(-1)(-3)=36>0$ (concave up); for $02$ (e.g. $x=3$), $f''(3)=12(3)(1)=36>0$ (concave up). The sign of $f''$ changes at both $x=0$ and $x=2$, confirming genuine points of inflection at each. $$\boxed{x=0 \text{ and } x=2}$$ (The point of inflection at $x=0$ coincides with the stationary point found in part (c), since $f'(0)=0$ there too - this is the stationary point of inflection.) (e) The leading term of $f(x)=x^4-4x^3$ is $x^4$, with a positive coefficient. For a quartic with positive leading coefficient, both ends of the graph rise without bound. $$\boxed{f(x)\to+\infty \text{ as } x\to\infty \text{ and as } x\to-\infty}$$

QUESTION 1 5 marks

Easy

Consider the function $f(x)=3^{x+1}-2$. Find (a) the equation of the horizontal asymptote of the graph of $f$, (b) the $y$-intercept, and (c) the range of $f$.

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(a) As $x\to-\infty$, $3^{x+1}\to0$, so $f(x)\to-2$. The graph never actually reaches this value. $$\boxed{y=-2}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=3^{0+1}-2=3^{1}-2=3-2=1$$ $$\boxed{y\text{-intercept}=1}$$ (c) Since $3^{x+1}>0$ for all real $x$, we have $f(x)=3^{x+1}-2>-2$ for all $x$, and $f(x)$ can be made arbitrarily large (as $x\to\infty$) or arbitrarily close to $-2$ (as $x\to-\infty$). $$\boxed{f(x)>-2,\ \text{i.e. range } =\left]-2,\infty\right[}$$

QUESTION 2 6 marks

Medium

Solve the equation $\log_{2}(x+3)+\log_{2}(x-1)=5$.

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Method: combine the logarithms using $\log_{a}m+\log_{a}n=\log_{a}(mn)$, being careful to state the domain restrictions first. For the original logarithms to be defined, we require $x+3>0$ and $x-1>0$, i.e. $x>1$. Combining the left-hand side: $$\log_{2}\big[(x+3)(x-1)\big]=5$$ Converting to exponential form: $$(x+3)(x-1)=2^{5}=32$$ Expanding: $$x^{2}+2x-3=32$$ $$x^{2}+2x-35=0$$ Factorising: $$(x+7)(x-5)=0$$ $$x=-7\ \text{or}\ x=5$$ Applying the domain restriction $x>1$: reject $x=-7$ (since $x=-7$ makes $x-1<0$, undefined for the original equation). $$\boxed{x=5}$$

QUESTION 3 6 marks

Easy

The number of bacteria in a culture is modelled by $N(t)=200e^{0.15t}$, where $t$ is the time in hours after the culture is set up. (a) Find the initial number of bacteria. (b) Find the number of bacteria after 10 hours, correct to the nearest integer. (c) Find the time taken for the number of bacteria to reach 1000, correct to three significant figures.

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(a) The initial number of bacteria is $N(0)$: $$N(0)=200e^{0.15(0)}=200e^{0}=200$$ $$\boxed{N(0)=200\ \text{bacteria}}$$ (b) Substituting $t=10$: $$N(10)=200e^{0.15(10)}=200e^{1.5}$$ Using a calculator, $e^{1.5}=4.4817\ldots$: $$N(10)=200\times4.4817\ldots=896.3\ldots$$ $$\boxed{N(10)\approx896\ \text{bacteria}}$$ (c) Setting $N(t)=1000$: $$200e^{0.15t}=1000$$ $$e^{0.15t}=5$$ Taking the natural logarithm of both sides: $$0.15t=\ln5$$ $$t=\frac{\ln5}{0.15}=\frac{1.60944\ldots}{0.15}=10.7295\ldots$$ $$\boxed{t\approx10.7\ \text{hours (3 s.f.)}}$$

QUESTION 4 7 marks

Medium

Consider the function $f(x)=\ln(2x-4)+1$. (a) State the domain of $f$. (b) State the equation of the vertical asymptote of the graph of $f$. (c) Find $f^{-1}(x)$, and state its range.

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(a) The argument of the logarithm must be strictly positive: $$2x-4>0\implies x>2$$ $$\boxed{\text{domain: } x>2}$$ (b) The graph has a vertical asymptote where the argument of the logarithm approaches zero, i.e. at $x=2$ (the boundary of the domain found in (a)). $$\boxed{x=2}$$ (c) Let $y=\ln(2x-4)+1$. Solve for $x$ in terms of $y$ first, then swap variables. $$y-1=\ln(2x-4)$$ Exponentiating both sides: $$e^{y-1}=2x-4$$ $$2x=e^{y-1}+4$$ $$x=\frac{e^{y-1}+4}{2}$$ Swapping $x\leftrightarrow y$: $$f^{-1}(x)=\frac{e^{x-1}+4}{2}$$ The range of $f^{-1}$ equals the domain of $f$, which is $x>2$ from part (a). $$\boxed{f^{-1}(x)=\dfrac{e^{x-1}+4}{2};\ \text{range: } f^{-1}(x)>2}$$

QUESTION 5 15 marks

Hard

Consider the function $f(x)=\ln(x^{2}-4)$, for $x>2$. (a) Explain why the range of $f$ is $\mathbb{R}$ (all real numbers). [3] (b) Show that $f^{-1}(x)=\sqrt{e^{x}+4}$. [4] (c) Solve the equation $f(x)=3$, giving your answer correct to three significant figures. [4] (d) The graph of $y=f(x)$ is transformed by a vertical stretch with scale factor $\dfrac{1}{2}$, followed by a translation of 2 units in the positive $y$-direction, to give the graph of $y=g(x)$. Write down an expression for $g(x)$, and state the equation of its vertical asymptote. [4]

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(a) For $x>2$, $x^{2}-4$ is continuous and strictly increasing (since $x>0$ throughout this interval), taking values from $0^{+}$ (as $x\to2^{+}$) up to $\infty$ (as $x\to\infty$); that is, $x^{2}-4$ takes every value in $(0,\infty)$. Since $\ln$ is a continuous, strictly increasing bijection from $(0,\infty)$ onto $\mathbb{R}$, the composite function $f(x)=\ln(x^{2}-4)$ takes every real value exactly once as $x$ ranges over $(2,\infty)$. $$\boxed{\text{range of } f=\mathbb{R}}$$ (b) Let $y=\ln(x^{2}-4)$, with $x>2$. Solve for $x$: $$e^{y}=x^{2}-4$$ $$x^{2}=e^{y}+4$$ $$x=\pm\sqrt{e^{y}+4}$$ Since $x>2>0$, we take the positive root: $$x=\sqrt{e^{y}+4}$$ Swapping $x\leftrightarrow y$: $$\boxed{f^{-1}(x)=\sqrt{e^{x}+4}}$$, as required. (c) Setting $f(x)=3$: $$\ln(x^{2}-4)=3$$ $$x^{2}-4=e^{3}$$ $$x^{2}=e^{3}+4$$ Using a calculator, $e^{3}=20.0855\ldots$, so $x^{2}=24.0855\ldots$: $$x=\pm\sqrt{24.0855\ldots}=\pm4.9077\ldots$$ Since the domain of $f$ is $x>2$, reject the negative root. $$\boxed{x\approx4.91\ \text{(3 s.f.)}}$$ (d) A vertical stretch with scale factor $\dfrac{1}{2}$ gives $\dfrac{1}{2}f(x)=\dfrac{1}{2}\ln(x^{2}-4)$. Translating this 2 units up gives $$g(x)=\frac{1}{2}\ln(x^{2}-4)+2$$ Both transformations act only in the $y$-direction (a vertical stretch and a vertical translation), so they do not affect the $x$-value at which the function is undefined. The vertical asymptote is therefore unchanged from that of $f$. $$\boxed{g(x)=\dfrac{1}{2}\ln(x^{2}-4)+2;\ \text{vertical asymptote: } x=2}$$

QUESTION 6 6 marks

Easy

Consider the function $f(x)=2^{x}+3$. (a) Find the equation of the horizontal asymptote of the graph of $f$. (b) Find the $y$-intercept of the graph of $f$. (c) State the range of $f$.

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(a) As $x\to-\infty$, $2^{x}\to0$ (a positive quantity shrinking towards zero, never reaching it), so $f(x)\to0+3=3$. The graph approaches this height but never touches it. $$\boxed{y=3}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=2^{0}+3=1+3=4$$ $$\boxed{y\text{-intercept}=4}$$ (c) Since $2^{x}>0$ for every real $x$, adding 3 gives $f(x)=2^{x}+3>3$ for all $x$. As $x$ increases without bound $f(x)$ grows without bound, and as $x$ decreases without bound $f(x)$ gets arbitrarily close to (but never reaches) 3. $$\boxed{f(x)>3,\ \text{i.e. range}=\left]3,\infty\right[}$$

QUESTION 7 6 marks

Easy

Consider the function $f(x)=5^{x-2}-1$. (a) Find the equation of the horizontal asymptote of the graph of $f$. (b) Find the $y$-intercept of the graph of $f$, giving your answer as an exact fraction. (c) State the range of $f$.

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(a) As $x\to-\infty$, the exponent $x-2\to-\infty$ as well, so $5^{x-2}\to0$ (it shrinks towards zero but stays positive). Hence $f(x)\to0-1=-1$. $$\boxed{y=-1}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=5^{0-2}-1=5^{-2}-1=\frac{1}{25}-1=\frac{1-25}{25}=-\frac{24}{25}$$ $$\boxed{y\text{-intercept}=-\frac{24}{25}}$$ (c) Since $5^{x-2}>0$ for all real $x$, we get $f(x)=5^{x-2}-1>-1$ for every $x$. The graph can be pushed arbitrarily high (as $x\to\infty$) but only ever approaches $-1$ from above (as $x\to-\infty$). $$\boxed{f(x)>-1,\ \text{i.e. range}=\left]-1,\infty\right[}$$

QUESTION 8 6 marks

Easy

Consider the function $f(x)=-2^{x}+4$. (a) Find the equation of the horizontal asymptote of the graph of $f$. (b) Find the $y$-intercept of the graph of $f$. (c) State the range of $f$.

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(a) As $x\to-\infty$, $2^{x}\to0$, so $-2^{x}\to0$ too, and $f(x)\to0+4=4$. This limiting height is approached but never reached. $$\boxed{y=4}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=-2^{0}+4=-1+4=3$$ $$\boxed{y\text{-intercept}=3}$$ (c) Since $2^{x}>0$ for all real $x$, we have $-2^{x}<0$, so $f(x)=-2^{x}+4<4$ for every $x$. Because of the negative sign in front of $2^{x}$, the graph is a reflection of $y=2^{x}$ in the $x$-axis, then shifted up 4 units - it falls without bound as $x$ increases, and approaches $4$ from below as $x\to-\infty$. $$\boxed{f(x)<4,\ \text{i.e. range}=\left]-\infty,4\right[}$$

QUESTION 9 5 marks

Easy

Consider the function $f(x)=4^{x+3}$. (a) Find the equation of the horizontal asymptote of the graph of $f$. (b) Find the $y$-intercept of the graph of $f$. (c) State the range of $f$.

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(a) As $x\to-\infty$, the exponent $x+3\to-\infty$, so $4^{x+3}\to0$ - the curve flattens towards the $x$-axis but never crosses or touches it. $$\boxed{y=0}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=4^{0+3}=4^{3}=64$$ $$\boxed{y\text{-intercept}=64}$$ (c) Since $4^{x+3}>0$ for every real $x$ (a positive base raised to any real power is positive), $f(x)$ takes every positive value and only positive values. $$\boxed{f(x)>0,\ \text{i.e. range}=\left]0,\infty\right[}$$

QUESTION 10 6 marks

Easy

Consider the function $f(x)=3^{-x}+2$. (a) Find the equation of the horizontal asymptote of the graph of $f$. (b) Find the $y$-intercept of the graph of $f$. (c) State the range of $f$, and state whether $f$ is increasing or decreasing.

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(a) As $x\to+\infty$, the exponent $-x\to-\infty$, so $3^{-x}\to0$, and $f(x)\to0+2=2$. (Note the limiting behaviour happens as $x\to+\infty$ here, because of the minus sign in the exponent - the opposite direction to a standard exponential.) $$\boxed{y=2}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=3^{-0}+2=3^{0}+2=1+2=3$$ $$\boxed{y\text{-intercept}=3}$$ (c) Since $3^{-x}>0$ for all real $x$, $f(x)=3^{-x}+2>2$ for every $x$, so the range is $f(x)>2$. Replacing $x$ with $-x$ reflects the graph of $y=3^{x}$ in the $y$-axis, so as $x$ increases, $-x$ decreases, and $3^{-x}$ decreases; hence $f$ is a decreasing function. $$\boxed{\text{range}=\left]2,\infty\right[;\ f\text{ is decreasing}}$$

QUESTION 11 5 marks

Easy

Consider the function $f(x)=\log_{2}(x-3)$. (a) State the domain of $f$. (b) Find the equation of the vertical asymptote of the graph of $f$. (c) Find the $x$-intercept of the graph of $f$.

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(a) A logarithm is only defined when its argument is strictly positive, so we require $$x-3>0\implies x>3$$ $$\boxed{\text{domain: } x>3}$$ (b) As $x\to3^{+}$, the argument $x-3\to0^{+}$, and $\log_{2}(\text{a small positive number})\to-\infty$. The graph plunges downward as $x$ approaches 3 from the right, so the vertical asymptote sits at the domain boundary found in (a). $$\boxed{x=3}$$ (c) The $x$-intercept occurs where $f(x)=0$: $$\log_{2}(x-3)=0$$ $$x-3=2^{0}=1$$ $$x=4$$ $$\boxed{x\text{-intercept}=4}$$

QUESTION 12 6 marks

Easy

Consider the function $f(x)=\log_{5}(x+4)-2$. (a) State the domain of $f$. (b) Find the equation of the vertical asymptote of the graph of $f$. (c) Find the $x$-intercept of the graph of $f$.

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(a) The argument of the logarithm must be strictly positive: $$x+4>0\implies x>-4$$ $$\boxed{\text{domain: } x>-4}$$ (b) As $x\to-4^{+}$, $x+4\to0^{+}$, so $\log_{5}(x+4)\to-\infty$, and hence $f(x)\to-\infty$ as well (subtracting 2 does not change this). The graph has a vertical asymptote at the left edge of the domain. $$\boxed{x=-4}$$ (c) Setting $f(x)=0$: $$\log_{5}(x+4)-2=0$$ $$\log_{5}(x+4)=2$$ $$x+4=5^{2}=25$$ $$x=21$$ $$\boxed{x\text{-intercept}=21}$$

QUESTION 13 5 marks

Easy

Consider the function $f(x)=\ln(x)+3$. (a) State the domain of $f$. (b) Find the equation of the vertical asymptote of the graph of $f$. (c) Find the $x$-intercept of the graph of $f$, giving your answer in exact form.

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(a) The natural logarithm requires a strictly positive argument: $$x>0$$ $$\boxed{\text{domain: } x>0}$$ (b) As $x\to0^{+}$, $\ln(x)\to-\infty$, so $f(x)=\ln(x)+3\to-\infty$. The graph drops without bound as $x$ approaches 0 from the right, giving a vertical asymptote at the domain boundary. $$\boxed{x=0}$$ (c) Setting $f(x)=0$: $$\ln(x)+3=0$$ $$\ln(x)=-3$$ $$x=e^{-3}$$ $$\boxed{x\text{-intercept}=e^{-3}\ (\approx0.0498)}$$

QUESTION 14 5 marks

Easy

Consider the function $f(x)=-\log_{3}(x)$. (a) State the domain of $f$. (b) Find the equation of the vertical asymptote of the graph of $f$. (c) State the range of $f$, and describe how the graph of $f$ relates to the graph of $y=\log_{3}(x)$.

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(a) The argument $x$ must be strictly positive: $$\boxed{\text{domain: } x>0}$$ (b) As $x\to0^{+}$, $\log_{3}(x)\to-\infty$, so $f(x)=-\log_{3}(x)\to+\infty$. Either way the function is unbounded as $x$ approaches the domain boundary, so there is a vertical asymptote there. $$\boxed{x=0}$$ (c) Since $\log_{3}(x)$ takes every real value as $x$ ranges over $(0,\infty)$, so does $-\log_{3}(x)$ (multiplying by $-1$ still covers all of $\mathbb{R}$, just in reverse order). $$\boxed{\text{range}=\mathbb{R}}$$ The graph of $f$ is the reflection of the graph of $y=\log_{3}(x)$ in the $x$-axis: every point $(x,y)$ on $y=\log_3(x)$ maps to $(x,-y)$ on $f$.

QUESTION 15 5 marks

Easy

Consider the function $f(x)=2\ln(x-1)$. (a) State the domain of $f$. (b) Find the equation of the vertical asymptote of the graph of $f$. (c) Find the $x$-intercept of the graph of $f$.

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(a) We require the argument of the logarithm to be strictly positive: $$x-1>0\implies x>1$$ $$\boxed{\text{domain: } x>1}$$ (b) As $x\to1^{+}$, $x-1\to0^{+}$, so $\ln(x-1)\to-\infty$, and multiplying by 2 (a positive constant) preserves this: $f(x)\to-\infty$. $$\boxed{x=1}$$ (c) Setting $f(x)=0$: $$2\ln(x-1)=0$$ $$\ln(x-1)=0$$ $$x-1=e^{0}=1$$ $$x=2$$ $$\boxed{x\text{-intercept}=2}$$

QUESTION 16 6 marks

Easy

The value of an investment, in dollars, after $t$ years is modelled by $$P(t)=500(1.08)^{t},\quad t\ge0.$$ (a) Find $P(0)$, and explain what this value represents. (b) Find $P(5)$, correct to the nearest dollar. (c) State the range of $P$ for $t\ge0$.

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(a) $$P(0)=500(1.08)^{0}=500(1)=500$$ $$\boxed{P(0)=\$500}$$ This represents the initial value of the investment, before any growth has occurred (at $t=0$ years). (b) Substituting $t=5$: $$P(5)=500(1.08)^{5}$$ Using a calculator, $(1.08)^{5}=1.46932\ldots$: $$P(5)=500\times1.46932\ldots=734.66\ldots$$ $$\boxed{P(5)\approx\$735}$$ (c) Since the base $1.08>1$, the function $(1.08)^{t}$ is increasing, and at $t=0$ it equals 1, giving $P(0)=500$. For all $t\ge0$, $(1.08)^t\ge1$, so $P(t)\ge500$, and $P(t)$ increases without bound as $t\to\infty$. $$\boxed{P(t)\ge500,\ \text{i.e. range}=[500,\infty)}$$

QUESTION 17 5 marks

Easy

The mass, in grams, of a decaying radioactive sample after $t$ days is modelled by $$M(t)=80e^{-0.03t},\quad t\ge0.$$ (a) Find the initial mass of the sample. (b) Find the equation of the horizontal asymptote of the graph of $M$, and explain its meaning in this context. (c) State the range of $M$ for $t\ge0$.

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(a) The initial mass is $M(0)$: $$M(0)=80e^{-0.03(0)}=80e^{0}=80$$ $$\boxed{M(0)=80\ \text{g}}$$ (b) As $t\to\infty$, the exponent $-0.03t\to-\infty$, so $e^{-0.03t}\to0$, giving $M(t)\to0$. $$\boxed{M=0}$$ This means that, according to the model, the mass of the sample decreases towards zero as time passes but never actually reaches zero. (c) Since $e^{-0.03t}>0$ for all $t$, $M(t)=80e^{-0.03t}>0$. On the domain $t\ge0$, $M$ is decreasing (as $-0.03t$ becomes more negative), starting at $M(0)=80$ and approaching (but never reaching) 0. $$\boxed{0

QUESTION 18 5 marks

Easy

Consider the function $f(x)=10^{x}-5$. (a) Find the equation of the horizontal asymptote of the graph of $f$. (b) Find the $y$-intercept of the graph of $f$. (c) State the range of $f$.

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(a) As $x\to-\infty$, $10^{x}\to0$, so $f(x)\to0-5=-5$. The graph approaches this height from above but never reaches it. $$\boxed{y=-5}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=10^{0}-5=1-5=-4$$ $$\boxed{y\text{-intercept}=-4}$$ (c) Since $10^{x}>0$ for all real $x$, $f(x)=10^{x}-5>-5$ for every $x$, and $f(x)$ can be made arbitrarily large as $x\to\infty$. $$\boxed{f(x)>-5,\ \text{i.e. range}=\left]-5,\infty\right[}$$

QUESTION 19 6 marks

Easy

Consider the function $f(x)=\left(\dfrac{1}{2}\right)^{x}+1$. (a) Find the equation of the horizontal asymptote of the graph of $f$. (b) Find the $y$-intercept of the graph of $f$. (c) State the range of $f$, and state whether $f$ is increasing or decreasing.

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(a) Since the base $\dfrac12$ lies between 0 and 1, as $x\to+\infty$ the value $\left(\dfrac12\right)^{x}\to0$ (each increase in $x$ halves the previous value). Hence $f(x)\to0+1=1$. $$\boxed{y=1}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=\left(\frac12\right)^{0}+1=1+1=2$$ $$\boxed{y\text{-intercept}=2}$$ (c) Since $\left(\dfrac12\right)^{x}>0$ for all real $x$, $f(x)>1$ for every $x$. Because the base is less than 1, larger values of $x$ produce smaller values of $\left(\dfrac12\right)^x$, so $f$ is a decreasing function. $$\boxed{\text{range}=\left]1,\infty\right[;\ f\text{ is decreasing}}$$

QUESTION 20 6 marks

Easy

Consider the function $f(x)=\log_{4}(2x)-1$. (a) State the domain of $f$. (b) Find the equation of the vertical asymptote of the graph of $f$. (c) Find the $x$-intercept of the graph of $f$.

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(a) The argument of the logarithm must be strictly positive: $$2x>0\implies x>0$$ $$\boxed{\text{domain: } x>0}$$ (b) As $x\to0^{+}$, $2x\to0^{+}$, so $\log_{4}(2x)\to-\infty$, and hence $f(x)\to-\infty$. $$\boxed{x=0}$$ (c) Setting $f(x)=0$: $$\log_{4}(2x)-1=0$$ $$\log_{4}(2x)=1$$ $$2x=4^{1}=4$$ $$x=2$$ $$\boxed{x\text{-intercept}=2}$$

QUESTION 21 5 marks

Easy

Consider the function $f(x)=e^{x}-4$. (a) Find the equation of the horizontal asymptote of the graph of $f$. (b) Find the $y$-intercept of the graph of $f$. (c) State the range of $f$.

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(a) As $x\to-\infty$, $e^{x}\to0$, so $f(x)\to0-4=-4$. $$\boxed{y=-4}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=e^{0}-4=1-4=-3$$ $$\boxed{y\text{-intercept}=-3}$$ (c) Since $e^{x}>0$ for all real $x$, $f(x)=e^{x}-4>-4$ for every $x$, and $f(x)$ increases without bound as $x\to\infty$. $$\boxed{f(x)>-4,\ \text{i.e. range}=\left]-4,\infty\right[}$$

QUESTION 22 5 marks

Easy

Consider the function $f(x)=\ln(x+5)$. (a) State the domain of $f$. (b) Find the equation of the vertical asymptote of the graph of $f$. (c) Find the $x$-intercept of the graph of $f$.

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(a) The argument of the natural logarithm must be strictly positive: $$x+5>0\implies x>-5$$ $$\boxed{\text{domain: } x>-5}$$ (b) As $x\to-5^{+}$, $x+5\to0^{+}$, so $\ln(x+5)\to-\infty$. $$\boxed{x=-5}$$ (c) Setting $f(x)=0$: $$\ln(x+5)=0$$ $$x+5=e^{0}=1$$ $$x=-4$$ $$\boxed{x\text{-intercept}=-4}$$

QUESTION 23 6 marks

Easy

Consider the function $f(x)=6-3^{x}$. (a) Find the equation of the horizontal asymptote of the graph of $f$. (b) Find the $y$-intercept of the graph of $f$. (c) State the range of $f$.

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(a) As $x\to-\infty$, $3^{x}\to0$, so $f(x)=6-3^{x}\to6-0=6$. $$\boxed{y=6}$$ (b) The $y$-intercept is $f(0)$: $$f(0)=6-3^{0}=6-1=5$$ $$\boxed{y\text{-intercept}=5}$$ (c) Since $3^{x}>0$ for all real $x$, we have $-3^{x}<0$, so $f(x)=6-3^{x}<6$ for every $x$. As $x\to\infty$, $3^x\to\infty$, so $f(x)\to-\infty$; as $x\to-\infty$, $f(x)\to6$ from below. $$\boxed{f(x)<6,\ \text{i.e. range}=\left]-\infty,6\right[}$$

QUESTION 24 8 marks

Medium

Consider the function $f(x)=e^{2x-1}+3$, for $x\in\mathbb{R}$. (a) Find the equation of the horizontal asymptote of the graph of $f$. (b) State the range of $f$. (c) Find $f^{-1}(x)$, and state its domain.

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(a) As $x\to-\infty$, the exponent $2x-1\to-\infty$, so $e^{2x-1}\to0$, and $f(x)\to0+3=3$. $$\boxed{y=3}$$ (b) Since $e^{2x-1}>0$ for all real $x$, $f(x)=e^{2x-1}+3>3$ for every $x$, and $f(x)$ increases without bound as $x\to\infty$. $$\boxed{\text{range}=\left]3,\infty\right[}$$ (c) Let $y=e^{2x-1}+3$. Isolate the exponential term first: $$y-3=e^{2x-1}$$ Take the natural logarithm of both sides: $$\ln(y-3)=2x-1$$ $$2x=\ln(y-3)+1$$ $$x=\frac{\ln(y-3)+1}{2}$$ Swap $x\leftrightarrow y$: $$f^{-1}(x)=\frac{\ln(x-3)+1}{2}$$ The domain of $f^{-1}$ must equal the range of $f$ found in (b), so we require $x-3>0$. $$\boxed{f^{-1}(x)=\dfrac{\ln(x-3)+1}{2};\ \text{domain: } x>3}$$

QUESTION 25 8 marks

Medium

Consider the function $f(x)=\ln(3-x)$. (a) State the domain of $f$. (b) Find the equation of the vertical asymptote of the graph of $f$. (c) Find $f^{-1}(x)$, and state its range.

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(a) The argument of the logarithm must be strictly positive: $$3-x>0\implies x<3$$ $$\boxed{\text{domain: } x<3}$$ (b) As $x\to3^{-}$, $3-x\to0^{+}$, so $\ln(3-x)\to-\infty$. The graph plunges downward as $x$ approaches 3 from the left. $$\boxed{x=3}$$ (c) Let $y=\ln(3-x)$. Solve for $x$ first: $$e^{y}=3-x$$ $$x=3-e^{y}$$ Swap $x\leftrightarrow y$: $$f^{-1}(x)=3-e^{x}$$ The range of $f^{-1}$ equals the domain of $f$ found in (a), namely $x<3$. $$\boxed{f^{-1}(x)=3-e^{x};\ \text{range: } f^{-1}(x)<3}$$

QUESTION 26 7 marks

Medium

Consider the function $f(x)=4^{x}-3$. (a) Find the equation of the horizontal asymptote of the graph of $f$, and state the range of $f$. (b) Find $f^{-1}(x)$, and state its domain and range.

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(a) As $x\to-\infty$, $4^{x}\to0$, so $f(x)\to0-3=-3$, giving asymptote $y=-3$. Since $4^x>0$ for all real $x$, $f(x)=4^x-3>-3$ for every $x$. $$\boxed{\text{asymptote } y=-3;\ \text{range}=\left]-3,\infty\right[}$$ (b) Let $y=4^{x}-3$. Isolate the exponential term: $$y+3=4^{x}$$ Taking $\log_4$ of both sides: $$x=\log_{4}(y+3)$$ Swap $x\leftrightarrow y$: $$f^{-1}(x)=\log_{4}(x+3)$$ The domain of $f^{-1}$ equals the range of $f$, i.e. $x>-3$; the range of $f^{-1}$ equals the domain of $f$, i.e. all real numbers. $$\boxed{f^{-1}(x)=\log_{4}(x+3);\ \text{domain: } x>-3;\ \text{range}=\mathbb{R}}$$

QUESTION 27 9 marks

Medium

The value, in dollars, of an investment under continuous compounding after $t$ years is modelled by $$A(t)=2000e^{0.045t},\quad t\ge0.$$ (a) Find $A(0)$, and explain its meaning. (b) State the range of $A$ for $t\ge0$. (c) Show that the inverse function is $A^{-1}(v)=\dfrac{\ln(v/2000)}{0.045}$, and use it to find the time taken for the investment to reach $\$3000$, correct to two decimal places.

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(a) $$A(0)=2000e^{0.045(0)}=2000e^{0}=2000$$ $$\boxed{A(0)=\$2000}$$ This is the initial value of the investment, before any interest has accrued. (b) On $t\ge0$, $0.045t\ge0$, so $e^{0.045t}\ge1$, giving $A(t)\ge2000$. Since $A$ is increasing (a positive exponent grows), $A(t)$ increases without bound as $t\to\infty$. $$\boxed{\text{range}=[2000,\infty)}$$ (c) Let $v=2000e^{0.045t}$. Isolate the exponential term: $$\frac{v}{2000}=e^{0.045t}$$ Take the natural logarithm of both sides: $$\ln\!\left(\frac{v}{2000}\right)=0.045t$$ $$t=\frac{\ln(v/2000)}{0.045}$$ which, after swapping variable names, gives $A^{-1}(v)=\dfrac{\ln(v/2000)}{0.045}$, as required. Substituting $v=3000$: $$A^{-1}(3000)=\frac{\ln(3000/2000)}{0.045}=\frac{\ln(1.5)}{0.045}=\frac{0.405465\ldots}{0.045}=9.0103\ldots$$ $$\boxed{t\approx9.01\ \text{years}}$$

QUESTION 28 9 marks

Medium

Consider the function $f(x)=-\ln(x+2)+3$. (a) State the domain of $f$, and find the equation of the vertical asymptote of the graph of $f$. (b) Find the $x$-intercept of the graph of $f$, giving your answer in exact form. (c) Find $f^{-1}(x)$.

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(a) The argument of the logarithm must be strictly positive: $$x+2>0\implies x>-2$$ $$\boxed{\text{domain: } x>-2}$$ As $x\to-2^{+}$, $x+2\to0^{+}$, so $\ln(x+2)\to-\infty$, and hence $-\ln(x+2)\to+\infty$, so $f(x)\to+\infty$. $$\boxed{\text{vertical asymptote: } x=-2}$$ (b) Setting $f(x)=0$: $$-\ln(x+2)+3=0$$ $$\ln(x+2)=3$$ $$x+2=e^{3}$$ $$x=e^{3}-2$$ $$\boxed{x\text{-intercept}=e^{3}-2\ (\approx18.09)}$$ (c) Let $y=-\ln(x+2)+3$. Isolate the logarithm: $$3-y=\ln(x+2)$$ Exponentiate both sides: $$e^{3-y}=x+2$$ $$x=e^{3-y}-2$$ Swap $x\leftrightarrow y$: $$\boxed{f^{-1}(x)=e^{3-x}-2}$$

QUESTION 29 7 marks

Medium

Consider the function $f(x)=2^{x+1}-8$. (a) Find the equation of the horizontal asymptote of the graph of $f$. (b) Find the $x$-intercept and the $y$-intercept of the graph of $f$. (c) State the range of $f$.

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(a) As $x\to-\infty$, $2^{x+1}\to0$, so $f(x)\to0-8=-8$. $$\boxed{y=-8}$$ (b) $x$-intercept: set $f(x)=0$: $$2^{x+1}-8=0$$ $$2^{x+1}=8=2^{3}$$ $$x+1=3\implies x=2$$ $$\boxed{x\text{-intercept}=2}$$ $y$-intercept: $f(0)=2^{0+1}-8=2-8=-6$. $$\boxed{y\text{-intercept}=-6}$$ (c) Since $2^{x+1}>0$ for all real $x$, $f(x)=2^{x+1}-8>-8$ for every $x$, and $f(x)$ grows without bound as $x\to\infty$. $$\boxed{f(x)>-8,\ \text{i.e. range}=\left]-8,\infty\right[}$$

QUESTION 30 8 marks

Medium

Consider the function $f(x)=\log_{2}(4-x)$. (a) State the domain of $f$, and find the equation of the vertical asymptote of the graph of $f$. (b) Explain why $f$ is a decreasing function. (c) Find $f^{-1}(x)$, and state its range.

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(a) The argument must be strictly positive: $$4-x>0\implies x<4$$ $$\boxed{\text{domain: } x<4}$$ As $x\to4^{-}$, $4-x\to0^{+}$, so $\log_2(4-x)\to-\infty$. $$\boxed{\text{vertical asymptote: } x=4}$$ (b) As $x$ increases (towards 4), the argument $4-x$ decreases towards $0^+$. Since $\log_2$ is an increasing function of its argument, a decreasing argument produces a decreasing output. Hence $f(x)=\log_2(4-x)$ decreases as $x$ increases. (c) Let $y=\log_{2}(4-x)$. Exponentiate both sides: $$2^{y}=4-x$$ $$x=4-2^{y}$$ Swap $x\leftrightarrow y$: $$f^{-1}(x)=4-2^{x}$$ The range of $f^{-1}$ equals the domain of $f$ found in (a), namely $x<4$. $$\boxed{f^{-1}(x)=4-2^{x};\ \text{range: } f^{-1}(x)<4}$$

QUESTION 31 9 marks

Medium

The temperature, in $^{\circ}\text{C}$, of a cooling liquid $t$ minutes after being placed in a room is modelled by $$T(t)=20+75e^{-0.2t},\quad t\ge0.$$ (a) Find the initial temperature of the liquid. (b) Find the equation of the horizontal asymptote of the graph of $T$, and explain its meaning in this context. (c) State the range of $T$ for $t\ge0$. (d) Find the time taken for the temperature to fall to $50^{\circ}\text{C}$, correct to three significant figures.

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(a) $$T(0)=20+75e^{-0.2(0)}=20+75(1)=95$$ $$\boxed{T(0)=95^{\circ}\text{C}}$$ (b) As $t\to\infty$, $e^{-0.2t}\to0$, so $T(t)\to20+0=20$. $$\boxed{T=20}$$ This represents the temperature of the surrounding room: the liquid cools towards room temperature but, according to this model, never quite reaches it. (c) On $t\ge0$, $T$ decreases from $T(0)=95$ towards (but never reaching) 20. $$\boxed{20

QUESTION 32 8 marks

Medium

Consider the function $f(x)=3e^{-x}-1$. (a) Find the equation of the horizontal asymptote of the graph of $f$, and state the range of $f$. (b) Find the $y$-intercept of the graph of $f$. (c) Find $f^{-1}(x)$, and state its domain.

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(a) As $x\to+\infty$, $e^{-x}\to0$, so $f(x)\to0-1=-1$. Since $e^{-x}>0$ for all real $x$, $f(x)=3e^{-x}-1>-1$ for every $x$. $$\boxed{\text{asymptote } y=-1;\ \text{range}=\left]-1,\infty\right[}$$ (b) $$f(0)=3e^{0}-1=3-1=2$$ $$\boxed{y\text{-intercept}=2}$$ (c) Let $y=3e^{-x}-1$. Isolate the exponential term: $$y+1=3e^{-x}$$ $$e^{-x}=\frac{y+1}{3}$$ Take the natural logarithm of both sides: $$-x=\ln\!\left(\frac{y+1}{3}\right)$$ $$x=-\ln\!\left(\frac{y+1}{3}\right)$$ Swap $x\leftrightarrow y$: $$f^{-1}(x)=-\ln\!\left(\frac{x+1}{3}\right)$$ The domain of $f^{-1}$ equals the range of $f$ from (a). $$\boxed{f^{-1}(x)=-\ln\!\left(\dfrac{x+1}{3}\right);\ \text{domain: } x>-1}$$

QUESTION 33 8 marks

Medium

Consider the function $f(x)=\log_{10}(4x)$. (a) State the domain of $f$, and find the equation of the vertical asymptote of the graph of $f$. (b) Find the $x$-intercept of the graph of $f$. (c) Find $f^{-1}(x)$, and state its domain and range.

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(a) The argument must be strictly positive: $$4x>0\implies x>0$$ $$\boxed{\text{domain: } x>0}$$ As $x\to0^{+}$, $4x\to0^{+}$, so $\log_{10}(4x)\to-\infty$. $$\boxed{\text{vertical asymptote: } x=0}$$ (b) Setting $f(x)=0$: $$\log_{10}(4x)=0$$ $$4x=10^{0}=1$$ $$x=\frac14$$ $$\boxed{x\text{-intercept}=\frac14}$$ (c) Let $y=\log_{10}(4x)$. Exponentiate both sides: $$10^{y}=4x$$ $$x=\frac{10^{y}}{4}$$ Swap $x\leftrightarrow y$: $$f^{-1}(x)=\frac{10^{x}}{4}$$ The domain of $f^{-1}$ equals the range of $f$, which is all real numbers (since $\log_{10}$ of a positive quantity that ranges over all of $(0,\infty)$ takes every real value); the range of $f^{-1}$ equals the domain of $f$, i.e. $x>0$. $$\boxed{f^{-1}(x)=\dfrac{10^{x}}{4};\ \text{domain}=\mathbb{R};\ \text{range: } f^{-1}(x)>0}$$

QUESTION 34 9 marks

Medium

Consider the function $f(x)=5-2\ln(x-3)$. (a) State the domain of $f$, and find the equation of the vertical asymptote of the graph of $f$. (b) Find the $x$-intercept of the graph of $f$, giving your answer in exact form. (c) Find $f^{-1}(x)$.

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(a) The argument of the logarithm must be strictly positive: $$x-3>0\implies x>3$$ $$\boxed{\text{domain: } x>3}$$ As $x\to3^{+}$, $x-3\to0^{+}$, so $\ln(x-3)\to-\infty$; multiplying by $-2$ flips the sign, so $-2\ln(x-3)\to+\infty$, and hence $f(x)\to+\infty$. $$\boxed{\text{vertical asymptote: } x=3}$$ (b) Setting $f(x)=0$: $$5-2\ln(x-3)=0$$ $$\ln(x-3)=\frac{5}{2}$$ $$x-3=e^{5/2}$$ $$x=3+e^{5/2}$$ $$\boxed{x\text{-intercept}=3+e^{5/2}\ (\approx15.18)}$$ (c) Let $y=5-2\ln(x-3)$. Isolate the logarithm: $$\frac{5-y}{2}=\ln(x-3)$$ Exponentiate both sides: $$e^{(5-y)/2}=x-3$$ $$x=3+e^{(5-y)/2}$$ Swap $x\leftrightarrow y$: $$\boxed{f^{-1}(x)=3+e^{(5-x)/2}}$$

QUESTION 35 9 marks

Medium

Consider the function $f(x)=\left(\dfrac13\right)^{x-2}+1$. (a) Find the equation of the horizontal asymptote of the graph of $f$, and state the range of $f$. (b) Find the $y$-intercept of the graph of $f$. (c) Find $f^{-1}(x)$, giving your answer in terms of natural logarithms, and state its domain.

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(a) As $x\to+\infty$, the exponent $x-2\to+\infty$, and since the base $\dfrac13<1$, $\left(\dfrac13\right)^{x-2}\to0$, so $f(x)\to0+1=1$. Since $\left(\dfrac13\right)^{x-2}>0$ for all real $x$, $f(x)>1$ for every $x$. $$\boxed{\text{asymptote } y=1;\ \text{range}=\left]1,\infty\right[}$$ (b) $$f(0)=\left(\frac13\right)^{0-2}+1=\left(\frac13\right)^{-2}+1=9+1=10$$ $$\boxed{y\text{-intercept}=10}$$ (c) Let $y=\left(\dfrac13\right)^{x-2}+1$. Isolate the exponential term: $$y-1=\left(\frac13\right)^{x-2}$$ Write $\left(\dfrac13\right)^{x-2}=e^{(x-2)\ln(1/3)}$ and take $\ln$ of both sides: $$\ln(y-1)=(x-2)\ln\!\left(\frac13\right)$$ $$x-2=\frac{\ln(y-1)}{\ln(1/3)}$$ $$x=2+\frac{\ln(y-1)}{\ln(1/3)}$$ Swap $x\leftrightarrow y$: $$f^{-1}(x)=2+\frac{\ln(x-1)}{\ln(1/3)}$$ The domain of $f^{-1}$ equals the range of $f$ found in (a). $$\boxed{f^{-1}(x)=2+\dfrac{\ln(x-1)}{\ln(1/3)};\ \text{domain: } x>1}$$

QUESTION 36 9 marks

Medium

Consider the function $f(x)=\ln(x^{2}+1)$, for $x\in\mathbb{R}$. (a) State the domain and range of $f$. (b) Explain why $f$ does not have an inverse function on its full domain $\mathbb{R}$. (c) The domain of $f$ is now restricted to $x\ge0$. Find $f^{-1}(x)$ on this restricted domain, and state its domain and range.

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(a) Since $x^{2}+1\ge1>0$ for every real $x$, $\ln(x^{2}+1)$ is defined for all $x$. $$\boxed{\text{domain}=\mathbb{R}}$$ Because $x^{2}+1\ge1$, and $\ln$ is increasing, $\ln(x^{2}+1)\ge\ln(1)=0$. As $|x|\to\infty$, $x^2+1\to\infty$, so $f(x)\to\infty$ too. Every value in between is attained since $f$ is continuous. $$\boxed{\text{range}=[0,\infty)}$$ (b) Since $f(-x)=\ln((-x)^{2}+1)=\ln(x^{2}+1)=f(x)$, the function is even: every output value (other than $f(0)$) is produced by two different inputs, $x$ and $-x$. A function is only invertible if it is one-to-one (each output comes from exactly one input), so $f$ has no inverse on all of $\mathbb{R}$. (c) On $x\ge0$, $f$ is one-to-one (it is strictly increasing there, since $x^2+1$ is strictly increasing for $x\ge0$), so an inverse exists. Let $y=\ln(x^{2}+1)$ with $x\ge0$. Solve for $x$: $$e^{y}=x^{2}+1$$ $$x^{2}=e^{y}-1$$ $$x=\pm\sqrt{e^{y}-1}$$ Since $x\ge0$, take the positive root: $$x=\sqrt{e^{y}-1}$$ Swap $x\leftrightarrow y$: $$f^{-1}(x)=\sqrt{e^{x}-1}$$ The domain of $f^{-1}$ equals the range of the restricted $f$, i.e. $x\ge0$; the range of $f^{-1}$ equals the restricted domain, i.e. $x\ge0$. $$\boxed{f^{-1}(x)=\sqrt{e^{x}-1};\ \text{domain: } x\ge0;\ \text{range: } f^{-1}(x)\ge0}$$

QUESTION 37 7 marks

Medium

Let $f(x)=2^{x}$ and $g(x)=\log_{2}(x)$. (a) Show algebraically that $f$ and $g$ are inverse functions of each other. (b) State the domain and range of $f$, and the domain and range of $g$. (c) Describe how the graph of $g$ can be obtained from the graph of $f$.

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(a) Compute the composite $f(g(x))$ for $x>0$: $$f(g(x))=f(\log_{2}x)=2^{\log_{2}x}=x$$ Compute the composite $g(f(x))$ for $x\in\mathbb{R}$: $$g(f(x))=g(2^{x})=\log_{2}(2^{x})=x$$ Since $f(g(x))=x$ for all $x$ in the domain of $g$, and $g(f(x))=x$ for all $x$ in the domain of $f$, the functions $f$ and $g$ are inverses of each other. (b) $f(x)=2^{x}$: domain $=\mathbb{R}$, range $=\left]0,\infty\right[$ (since $2^x>0$ for all $x$ and takes every positive value). $g(x)=\log_{2}(x)$: domain $=\left]0,\infty\right[$, range $=\mathbb{R}$. $$\boxed{f:\ \text{domain}=\mathbb{R},\ \text{range}=\left]0,\infty\right[;\quad g:\ \text{domain}=\left]0,\infty\right[,\ \text{range}=\mathbb{R}}$$ (c) Since $g=f^{-1}$, the graph of $g$ is the reflection of the graph of $f$ in the line $y=x$. (This is a general property of any function and its inverse: swapping the roles of $x$ and $y$ swaps the domain and range, which geometrically corresponds to reflecting the graph in $y=x$.)

QUESTION 38 8 marks

Medium

The number of bacteria in a culture $t$ hours after the start of an experiment is modelled by $$P(t)=150\times2^{t/3},\quad t\ge0.$$ (a) Find $P(0)$, and interpret this value. (b) Find the time taken for the population to reach $4800$ bacteria. (c) Show that the inverse function is $P^{-1}(n)=3\log_{2}\!\left(\dfrac{n}{150}\right)$, and state its domain in the context of this model.

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(a) $$P(0)=150\times2^{0/3}=150\times1=150$$ $$\boxed{P(0)=150\ \text{bacteria}}$$ This is the size of the culture at the start of the experiment. (b) Setting $P(t)=4800$: $$150\times2^{t/3}=4800$$ $$2^{t/3}=32$$ Since $32=2^{5}$: $$\frac{t}{3}=5\implies t=15$$ $$\boxed{t=15\ \text{hours}}$$ (c) Let $n=150\times2^{t/3}$. Isolate the exponential term: $$\frac{n}{150}=2^{t/3}$$ Take $\log_{2}$ of both sides: $$\log_{2}\!\left(\frac{n}{150}\right)=\frac{t}{3}$$ $$t=3\log_{2}\!\left(\frac{n}{150}\right)$$ which, after renaming the variable, gives $P^{-1}(n)=3\log_{2}\!\left(\dfrac{n}{150}\right)$, as required. Since $P(t)\ge150$ for $t\ge0$ (the population never falls below its starting size in this growth model), the domain of $P^{-1}$ in context is $$\boxed{n\ge150}$$

QUESTION 39 8 marks

Medium

Consider the function $f(x)=\log_{3}(4x-8)$. (a) State the domain of $f$, and find the equation of the vertical asymptote of the graph of $f$. (b) Find the $x$-intercept of the graph of $f$. (c) Find $f^{-1}(x)$.

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(a) The argument of the logarithm must be strictly positive: $$4x-8>0\implies x>2$$ $$\boxed{\text{domain: } x>2}$$ As $x\to2^{+}$, $4x-8\to0^{+}$, so $\log_{3}(4x-8)\to-\infty$. $$\boxed{\text{vertical asymptote: } x=2}$$ (b) Setting $f(x)=0$: $$\log_{3}(4x-8)=0$$ $$4x-8=3^{0}=1$$ $$4x=9$$ $$x=\frac94$$ $$\boxed{x\text{-intercept}=\frac94}$$ (c) Let $y=\log_{3}(4x-8)$. Exponentiate both sides: $$3^{y}=4x-8$$ $$4x=3^{y}+8$$ $$x=\frac{3^{y}+8}{4}$$ Swap $x\leftrightarrow y$: $$\boxed{f^{-1}(x)=\dfrac{3^{x}+8}{4}}$$

QUESTION 40 9 marks

Medium

Consider the function $f(x)=7-e^{-x/2}$. (a) Find the equation of the horizontal asymptote of the graph of $f$, and state the range of $f$. (b) Find the $y$-intercept of the graph of $f$. (c) Find $f^{-1}(x)$, and state its domain.

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(a) As $x\to+\infty$, the exponent $-x/2\to-\infty$, so $e^{-x/2}\to0$, and $f(x)\to7-0=7$. Since $e^{-x/2}>0$ for all real $x$, $-e^{-x/2}<0$, so $f(x)=7-e^{-x/2}<7$ for every $x$. $$\boxed{\text{asymptote } y=7;\ \text{range}=\left]-\infty,7\right[}$$ (b) $$f(0)=7-e^{0}=7-1=6$$ $$\boxed{y\text{-intercept}=6}$$ (c) Let $y=7-e^{-x/2}$. Isolate the exponential term: $$e^{-x/2}=7-y$$ Take the natural logarithm of both sides: $$-\frac{x}{2}=\ln(7-y)$$ $$x=-2\ln(7-y)$$ Swap $x\leftrightarrow y$: $$f^{-1}(x)=-2\ln(7-x)$$ The domain of $f^{-1}$ equals the range of $f$ found in (a). $$\boxed{f^{-1}(x)=-2\ln(7-x);\ \text{domain: } x<7}$$

QUESTION 41 7 marks

Medium

Let $f(x)=2^{x}$ and $g(x)=2^{-x}$. (a) Find the coordinates of the point of intersection of the graphs of $f$ and $g$. (b) Describe the geometric relationship between the graphs of $f$ and $g$. (c) State the range of $f$ and the range of $g$, and the equation of the horizontal asymptote common to both graphs.

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(a) Setting $f(x)=g(x)$: $$2^{x}=2^{-x}$$ Since the bases are equal, the exponents must be equal: $$x=-x\implies2x=0\implies x=0$$ Then $f(0)=2^{0}=1$. $$\boxed{(0,1)}$$ (b) Since $g(x)=2^{-x}=f(-x)$, the graph of $g$ is the reflection of the graph of $f$ in the $y$-axis (replacing $x$ with $-x$ reflects a graph in the $y$-axis). (c) Since $2^{x}>0$ for all real $x$, and likewise $2^{-x}>0$ for all real $x$: $$\boxed{\text{range of }f=\left]0,\infty\right[;\ \text{range of }g=\left]0,\infty\right[}$$ Both graphs approach the $x$-axis (in opposite directions - $f$ as $x\to-\infty$, $g$ as $x\to+\infty$) without reaching it. $$\boxed{\text{common horizontal asymptote: } y=0}$$

QUESTION 42 13 marks

Hard

Consider the function $f(x)=\ln(9-x^{2})$, for $-3(a) Show that the range of $f$ is $\left]-\infty,\ln9\right[$. [4] (b) Find $f^{-1}(x)$, and state its domain and range. [5] (c) Solve the equation $f(x)=1$, giving your answer correct to three significant figures. [4]

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(a) On $-3

QUESTION 43 12 marks

Hard

Consider the function $f(x)=2^{x^{2}-4}$, for $x\ge2$. (a) Explain why $f$ is an increasing function on this domain, and hence show that the range of $f$ is $f(x)\ge1$. [4] (b) Find $f^{-1}(x)$, and state its domain and range. [5] (c) Hence solve the equation $f(x)=64$. [3]

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(a) For $x\ge2$, the exponent $u(x)=x^{2}-4$ is increasing (since $x\ge2>0$, larger $x$ gives larger $x^{2}$). Since the base $2>1$, $2^{u}$ is an increasing function of $u$; combining these, $f(x)=2^{x^{2}-4}$ is an increasing function of $x$ on $x\ge2$. At $x=2$: $f(2)=2^{4-4}=2^{0}=1$. Since $f$ is increasing on $x\ge2$ and continuous, and $f(x)\to\infty$ as $x\to\infty$, $f$ takes every value from $1$ (attained at $x=2$) up to $\infty$. $$\boxed{\text{range: } f(x)\ge1}$$, as required. (b) Let $y=2^{x^{2}-4}$, with $x\ge2$. Take $\log_{2}$ of both sides: $$\log_{2}y=x^{2}-4$$ $$x^{2}=\log_{2}y+4$$ $$x=\pm\sqrt{\log_{2}y+4}$$ Since $x\ge2>0$, take the positive root: $$x=\sqrt{\log_{2}y+4}$$ Swap $x\leftrightarrow y$: $$f^{-1}(x)=\sqrt{\log_{2}x+4}$$ The domain of $f^{-1}$ equals the range of $f$ found in (a); the range of $f^{-1}$ equals the domain of $f$. $$\boxed{f^{-1}(x)=\sqrt{\log_{2}x+4};\ \text{domain: } x\ge1;\ \text{range: } f^{-1}(x)\ge2}$$ (c) Since $f$ is one-to-one on $x\ge2$, $f(x)=64$ has at most one solution, given by $x=f^{-1}(64)$: $$x=\sqrt{\log_{2}64+4}=\sqrt{6+4}=\sqrt{10}$$ $$\boxed{x=\sqrt{10}\ (\approx3.16)}$$

QUESTION 44 14 marks

Hard

Consider the function $f(x)=\log_{2}\!\left(\dfrac{x-1}{x+1}\right)$, for $x>1$. (a) Find the equation of the vertical asymptote of the graph of $f$. [2] (b) Show that, as $x\to\infty$, $f(x)\to0$ from below, and hence state the range of $f$. [4] (c) Find $f^{-1}(x)$, and state its domain and range. [5] (d) Hence, or otherwise, solve the equation $f(x)=-1$. [3]

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(a) As $x\to1^{+}$, the numerator $x-1\to0^{+}$ while the denominator $x+1\to2$, so $\dfrac{x-1}{x+1}\to0^{+}$, and $\log_2$ of a small positive number tends to $-\infty$. $$\boxed{x=1}$$ (b) Write $\dfrac{x-1}{x+1}=1-\dfrac{2}{x+1}$. As $x\to\infty$, $\dfrac{2}{x+1}\to0^{+}$, so $\dfrac{x-1}{x+1}\to1^{-}$ (approaching 1 from below, always slightly less than 1 for finite $x$). Since $\log_2$ is continuous and $\log_2(1)=0$, and $\log_2$ is increasing, a value slightly less than 1 gives a value of $f$ slightly less than 0. $$\boxed{f(x)\to0^{-}\text{ as }x\to\infty}$$ The derivative of $\dfrac{x-1}{x+1}$ with respect to $x$ is $\dfrac{(x+1)-(x-1)}{(x+1)^2}=\dfrac{2}{(x+1)^2}>0$, so the ratio is strictly increasing on $x>1$; hence $f$ is strictly increasing too, rising from $-\infty$ (at $x=1$) towards, but never reaching, $0$ (as $x\to\infty$). $$\boxed{\text{range: } f(x)<0}$$ (c) Let $y=\log_{2}\!\left(\dfrac{x-1}{x+1}\right)$. Exponentiate both sides: $$2^{y}=\frac{x-1}{x+1}$$ $$2^{y}(x+1)=x-1$$ $$2^{y}x+2^{y}=x-1$$ $$x\left(2^{y}-1\right)=-1-2^{y}$$ $$x=\frac{-1-2^{y}}{2^{y}-1}=\frac{1+2^{y}}{1-2^{y}}$$ Swap $x\leftrightarrow y$: $$f^{-1}(x)=\frac{1+2^{x}}{1-2^{x}}$$ The domain of $f^{-1}$ equals the range of $f$ from (b), and the range of $f^{-1}$ equals the domain of $f$. $$\boxed{f^{-1}(x)=\dfrac{1+2^{x}}{1-2^{x}};\ \text{domain: } x<0;\ \text{range: } f^{-1}(x)>1}$$ (d) Since $f$ is one-to-one, $f(x)=-1$ has exactly one solution, given by $x=f^{-1}(-1)$: $$x=\frac{1+2^{-1}}{1-2^{-1}}=\frac{1+0.5}{1-0.5}=\frac{1.5}{0.5}=3$$ Check: $f(3)=\log_{2}\!\left(\dfrac{3-1}{3+1}\right)=\log_{2}\!\left(\dfrac12\right)=-1$ (correct) $$\boxed{x=3}$$

QUESTION 45 15 marks

Hard

Consider the function $f(x)=e^{2x}-4e^{x}+3$, for $x\in\mathbb{R}$. (a) By writing $u=e^{x}$, show that $f(x)=(e^{x}-2)^{2}-1$. [3] (b) Show that $f(0)=f(\ln3)=0$, and hence explain why $f$ does not have an inverse function on its full domain $\mathbb{R}$. [4] (c) The domain of $f$ is now restricted to $x\ge\ln2$. Show that, on this restricted domain, $f$ is increasing, and find $f^{-1}(x)$ stating its domain and range. [6] (d) Using the restricted domain from (c), solve $f(x)=0$, giving an exact answer. [2]

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(a) Let $u=e^{x}$, so $u^{2}=e^{2x}$. Then $$f(x)=u^{2}-4u+3$$ Completing the square: $$u^{2}-4u+3=(u-2)^{2}-4+3=(u-2)^{2}-1$$ Substituting back $u=e^{x}$: $$\boxed{f(x)=(e^{x}-2)^{2}-1}$$, as required. (b) $$f(0)=e^{0}-4e^{0}+3=1-4+3=0$$ $$f(\ln3)=e^{2\ln3}-4e^{\ln3}+3=3^{2}-4(3)+3=9-12+3=0$$ $$\boxed{f(0)=f(\ln3)=0}$$ Since two different inputs, $x=0$ and $x=\ln3$, produce the same output, $0$, the function $f$ is not one-to-one on $\mathbb{R}$. A function must be one-to-one to have an inverse function, so $f$ has no inverse on its full domain. (c) From (a), $f(x)=(e^{x}-2)^{2}-1$. For $x\ge\ln2$, $e^{x}\ge2$ (since $e^{x}$ is increasing), so $e^{x}-2\ge0$. On this range, squaring $(e^{x}-2)$ is an increasing operation (squaring a non-negative, increasing quantity gives an increasing result), and $e^x-2$ itself is increasing in $x$; hence $f$ is increasing on $x\ge\ln2$. At $x=\ln2$, $f(\ln2)=(2-2)^{2}-1=-1$, and $f(x)\to\infty$ as $x\to\infty$, so the range is $f(x)\ge-1$. Let $y=(e^{x}-2)^{2}-1$. Solve for $x$: $$y+1=(e^{x}-2)^{2}$$ $$e^{x}-2=\pm\sqrt{y+1}$$ Since $x\ge\ln2$ gives $e^x-2\ge0$, take the positive root: $$e^{x}-2=\sqrt{y+1}$$ $$e^{x}=2+\sqrt{y+1}$$ $$x=\ln\!\left(2+\sqrt{y+1}\right)$$ Swap $x\leftrightarrow y$: $$\boxed{f^{-1}(x)=\ln\!\left(2+\sqrt{x+1}\right);\ \text{domain: } x\ge-1;\ \text{range: } f^{-1}(x)\ge\ln2}$$ (d) From (b), the solutions to $f(x)=0$ over all of $\mathbb{R}$ are $x=0$ and $x=\ln3$. On the restricted domain $x\ge\ln2$ (note $\ln2\approx0.693$), $x=0$ is excluded, but $x=\ln3\approx1.099$ satisfies $\ln3\ge\ln2$. $$\boxed{x=\ln3}$$

QUESTION 46 14 marks

Hard

A cup of coffee is placed in a room where the air temperature is a constant $18^{\circ}\text{C}$. The temperature of the coffee, in $^{\circ}\text{C}$, $t$ minutes after being placed in the room is modelled by $$T(t)=18+Ae^{-kt},\quad t\ge0,$$ where $A,k$ are positive constants. When the coffee is first placed in the room ($t=0$), its temperature is $90^{\circ}\text{C}$. After 5 minutes, its temperature has fallen to $70^{\circ}\text{C}$. (a) Show that $A=72$. [2] (b) Show that $k=\dfrac{1}{5}\ln\!\left(\dfrac{18}{13}\right)$, and find its value correct to four significant figures. [5] (c) Write down the equation of the horizontal asymptote of the graph of $T$, and state the range of $T$ for $t\ge0$. [3] (d) Find $T^{-1}(T_0)$ in terms of $T_0$, and use it to find how long it takes for the coffee to cool to $40^{\circ}\text{C}$, correct to three significant figures. [4]

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(a) Substituting $t=0$, $T=90$: $$90=18+Ae^{-k(0)}=18+A(1)=18+A$$ $$\boxed{A=72}$$, as required. (b) Substituting $t=5$, $T=70$, and $A=72$: $$70=18+72e^{-5k}$$ $$72e^{-5k}=52$$ $$e^{-5k}=\frac{52}{72}=\frac{13}{18}$$ Taking the natural logarithm of both sides: $$-5k=\ln\!\left(\frac{13}{18}\right)$$ $$k=-\frac{1}{5}\ln\!\left(\frac{13}{18}\right)=\frac{1}{5}\ln\!\left(\frac{18}{13}\right)$$ (using $-\ln(p)=\ln(1/p)$), as required. $$\boxed{k=\frac15\ln\!\left(\frac{18}{13}\right)\approx0.06508\ \text{(4 s.f.)}}$$ (c) As $t\to\infty$, $e^{-kt}\to0$ (since $k>0$), so $T(t)\to18+0=18$. $$\boxed{\text{asymptote: } T=18}$$ On $t\ge0$, $T$ decreases from $T(0)=90$ towards, but never reaching, $18$. $$\boxed{\text{range: } 18

QUESTION 47 13 marks

Hard

Consider the function $g(x)=2\log_{5}(3x-6)-1$. (a) State the domain of $g$, and find the equation of the vertical asymptote of the graph of $g$. [3] (b) Find the $x$-intercept of the graph of $g$, giving your answer in exact form. [4] (c) Find $g^{-1}(x)$, and state its domain and range. [5] (d) Hence find the value of $x$ for which $g(x)=5$, giving your answer correct to three significant figures. [3]

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(a) The argument of the logarithm must be strictly positive: $$3x-6>0\implies x>2$$ $$\boxed{\text{domain: } x>2}$$ As $x\to2^{+}$, $3x-6\to0^{+}$, so $\log_{5}(3x-6)\to-\infty$, and hence $g(x)\to-\infty$. $$\boxed{\text{vertical asymptote: } x=2}$$ (b) Setting $g(x)=0$: $$2\log_{5}(3x-6)-1=0$$ $$\log_{5}(3x-6)=\frac12$$ $$3x-6=5^{1/2}=\sqrt5$$ $$3x=6+\sqrt5$$ $$x=\frac{6+\sqrt5}{3}$$ $$\boxed{x\text{-intercept}=\dfrac{6+\sqrt5}{3}\ (\approx2.75)}$$ (c) Let $y=2\log_{5}(3x-6)-1$. Isolate the logarithm: $$\frac{y+1}{2}=\log_{5}(3x-6)$$ Exponentiate both sides: $$5^{(y+1)/2}=3x-6$$ $$3x=5^{(y+1)/2}+6$$ $$x=\frac{5^{(y+1)/2}+6}{3}$$ Swap $x\leftrightarrow y$: $$g^{-1}(x)=\frac{5^{(x+1)/2}+6}{3}$$ Since $g$ is a logarithm scaled and shifted (with positive coefficient $2$), it is continuous and strictly increasing on its domain, taking every real value as $x$ ranges over $x>2$; hence the range of $g$ is $\mathbb{R}$, so the domain of $g^{-1}$ is $\mathbb{R}$, and the range of $g^{-1}$ equals the domain of $g$. $$\boxed{g^{-1}(x)=\dfrac{5^{(x+1)/2}+6}{3};\ \text{domain}=\mathbb{R};\ \text{range: } g^{-1}(x)>2}$$ (d) Since $g$ is one-to-one, $g(x)=5$ has exactly one solution, given by $x=g^{-1}(5)$: $$x=\frac{5^{(5+1)/2}+6}{3}=\frac{5^{3}+6}{3}=\frac{125+6}{3}=\frac{131}{3}=43.6\overline{6}$$ $$\boxed{x\approx43.7\ \text{(3 s.f.)}}$$

QUESTION 48 16 marks

Hard

A fossil sample initially contains 12 grams of carbon-14. The mass, in grams, remaining after $t$ years is modelled by $$N(t)=12\left(\frac12\right)^{t/5730},\quad t\ge0,$$ since carbon-14 has a half-life of 5730 years. (a) Find $N(1000)$, correct to three significant figures. [2] (b) State the domain and range of $N$ in the context of this model. [3] (c) Show that the inverse function is $N^{-1}(m)=5730\cdot\dfrac{\ln(12/m)}{\ln2}$, and explain what $N^{-1}(m)$ represents. [6] (d) A different fossil is found to contain $30\%$ of its original carbon-14. Use $N^{-1}$ to estimate the age of this fossil, correct to the nearest 10 years. [5]

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(a) $$N(1000)=12\left(\frac12\right)^{1000/5730}=12(0.5)^{0.174520\ldots}$$ Using a calculator, $(0.5)^{0.174520\ldots}=0.886028\ldots$: $$N(1000)=12\times0.886028\ldots=10.6323\ldots$$ $$\boxed{N(1000)\approx10.6\ \text{g (3 s.f.)}}$$ (b) In context, time cannot be negative, so $t\ge0$. Since $\left(\frac12\right)^{t/5730}$ decreases from $1$ (at $t=0$) towards, but never reaching, $0$ as $t\to\infty$, $N(t)$ decreases from $N(0)=12$ towards $0$. $$\boxed{\text{domain: } t\ge0;\ \text{range: } 0

QUESTION 49 12 marks

Hard

Consider the function $f(x)=\sqrt{\ln(x)}$. (a) State the domain of $f$, explaining why $x\ge1$ is required. [3] (b) State the range of $f$. [2] (c) Show that $f^{-1}(x)=e^{x^{2}}$, and verify your answer by showing that $f\big(f^{-1}(x)\big)=x$ for $x\ge0$. [5] (d) Find the coordinates of the point(s) where the graphs of $f$ and $f^{-1}$ intersect. [3]

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(a) Two conditions must both hold: the argument of $\ln$ must be positive ($x>0$), and the quantity under the square root must be non-negative ($\ln x\ge0$, i.e. $x\ge1$). Since $x\ge1$ automatically satisfies $x>0$, the binding condition is $x\ge1$. $$\boxed{\text{domain: } x\ge1}$$ (b) For $x\ge1$, $\ln(x)\ge0$, and $\ln(x)$ is unbounded above as $x\to\infty$; since the square root of a quantity ranging over $[0,\infty)$ also ranges over $[0,\infty)$, and $f(1)=\sqrt{\ln1}=\sqrt0=0$: $$\boxed{\text{range: } f(x)\ge0}$$ (c) Let $y=\sqrt{\ln x}$, with $x\ge1$ (so $y\ge0$). Square both sides: $$y^{2}=\ln x$$ Exponentiate: $$x=e^{y^{2}}$$ Swap $x\leftrightarrow y$: $$\boxed{f^{-1}(x)=e^{x^{2}}}$$ Verification: for $x\ge0$, $$f\big(f^{-1}(x)\big)=f\!\left(e^{x^{2}}\right)=\sqrt{\ln\!\left(e^{x^{2}}\right)}=\sqrt{x^{2}}=|x|=x$$ (the last step uses $x\ge0$, so $|x|=x$), confirming $f^{-1}(x)=e^{x^2}$ is correct. (d) Since $f^{-1}$ is the reflection of $f$ in the line $y=x$, any intersection points of $f$ and $f^{-1}$ that lie off the line $y=x$ would occur in mirrored pairs; here it is simplest to search directly on $y=x$, since $f$ is increasing (as $\ln x$ and $\sqrt{\cdot}$ are both increasing) and so can cross its own reflection only where $f(x)=x$: $$\sqrt{\ln x}=x$$ Testing $x=1$: $f(1)=\sqrt{\ln1}=\sqrt0=0\neq1$, so $(1,1)$ is not a solution. Testing values numerically (e.g. using a graphical or numerical method, as this equation cannot be solved algebraically in closed form) shows $f(x)1$ and $f(x)$ is undefined for $x<1$, with equality only in the limit - a careful check (e.g. $g(x)=x-\sqrt{\ln x}$, $g(1)=1>0$, $g'(x)=1-\frac{1}{2x\sqrt{\ln x}}>0$ for $x>1$) shows $g(x)>0$ throughout, so $f(x)

QUESTION 50 12 marks

Hard

Let $f(x)=3^{x}$ and $g(x)=\log_{3}(x-4)$. (a) State the domain and range of $f$, and the domain and range of $g$. [4] (b) Find $(f\circ g)(x)$ in simplified form, and state its domain. [4] (c) Find $(g\circ f)(x)$ in simplified form, and state its domain and range. [4]

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(a) $f(x)=3^{x}$: since $3^x>0$ for every real $x$, and $3^x$ takes every positive value, $$\boxed{\text{domain of }f=\mathbb{R};\ \text{range of }f=\left]0,\infty\right[}$$ $g(x)=\log_{3}(x-4)$ requires $x-4>0$, i.e. $x>4$; a logarithm takes every real value, $$\boxed{\text{domain of }g=\left]4,\infty\right[;\ \text{range of }g=\mathbb{R}}$$ (b) $(f\circ g)(x)=f(g(x))=3^{\log_{3}(x-4)}$. Since $3^{\log_3 A}=A$ for $A>0$ (raising the base to the log base-3 of a quantity recovers that quantity, because exponentiation and $\log_3$ are inverse operations), $$\boxed{(f\circ g)(x)=x-4}$$ For the composite to be defined, $x$ must lie in the domain of $g$ (so that $g(x)$ exists), i.e. $x>4$; every such $g(x)$ is a real number, which lies in the domain of $f$ ($\mathbb{R}$) automatically, so no further restriction is added. $$\boxed{\text{domain of }(f\circ g)=\left]4,\infty\right[}$$ (Note this is narrower than the domain of the simplified expression $x-4$ on its own, which would be all of $\mathbb{R}$ - the domain of a composite function is inherited from the *original* functions, not just the simplified formula.) (c) $(g\circ f)(x)=g(f(x))=\log_{3}\!\left(3^{x}-4\right)$. $$\boxed{(g\circ f)(x)=\log_{3}\!\left(3^{x}-4\right)}$$ For this to be defined, we need $3^{x}-4>0$, i.e. $3^{x}>4$, i.e. $x>\log_{3}4$ (taking $\log_3$ of both sides, valid since $\log_3$ is increasing). $$\boxed{\text{domain: } x>\log_{3}4\ (\approx1.26)}$$ As $x\to\left(\log_{3}4\right)^{+}$, $3^{x}-4\to0^{+}$, so $\log_3(3^x-4)\to-\infty$; as $x\to\infty$, $3^x-4\to\infty$, so $\log_3(3^x-4)\to\infty$. Since the expression is continuous and increasing on this domain, it takes every real value. $$\boxed{\text{range}=\mathbb{R}}$$

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