A Level H2 Mathematics Practice Paper 3

TuitionGoWhere Practice Paper — Maths H2 A-Level

Answer Key — Algebra & Functions (Version 3 of 5)

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Question 1

(a) Let $y = \dfrac{3x - 1}{x + 2}$.

Swap $x$ and $y$: $x = \dfrac{3y - 1}{y + 2}$

$x(y + 2) = 3y - 1$

$xy + 2x = 3y - 1$

$xy - 3y = -1 - 2x$

$y(x - 3) = -1 - 2x$

$f^{-1}(x) = \frac{-1 - 2x}{x - 3} = \frac{2x + 1}{3 - x}$

Domain of $f^{-1}$ = range of $f$. Since $f(x) = \dfrac{3x-1}{x+2}$, the horizontal asymptote is $y = 3$, so $f(x) \neq 3$.

Domain of $f^{-1}$: $x \neq 3$, i.e. $x \in \mathbb{R}, x \neq 3$. [3]

(b) Range of $f$: Since the horizontal asymptote is $y = 3$ and the function never equals 3 (solving $\frac{3x-1}{x+2} = 3$ gives $3x - 1 = 3x + 6$, i.e. $-1 = 6$, impossible), the range is $\{y \in \mathbb{R} : y \neq 3\}$.

Domain of $f^{-1}$ = range of $f$: $x \neq 3$. [2]

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Question 2

(a) For $gf$ to exist, we need $\text{range}(f) \subseteq \text{domain}(g)$.

$f(x) = x^2 - 4x + 6 = (x-2)^2 + 2$, with domain $x \geq 2$.

Since $(x-2)^2 \geq 0$ for all $x$, the minimum value of $f$ is $2$ (at $x = 2$), and $f(x)$ increases without bound.

So $\text{range}(f) = [2, \infty)$.

Domain of $g$ is $x > 1$. Since $[2, \infty) \subset (1, \infty)$, we have $\text{range}(f) \subseteq \text{domain}(g)$.

Therefore $gf$ exists. [2]

(b) $gf(x) = g(f(x)) = g(x^2 - 4x + 6) = \dfrac{1}{(x^2 - 4x + 6) - 1} = \dfrac{1}{x^2 - 4x + 5}$

Now $x^2 - 4x + 5 = (x-2)^2 + 1 \geq 1$ for $x \geq 2$, with minimum $1$ at $x = 2$.

So $0 < gf(x) \leq 1$.

Range of $gf$: $(0, 1]$ [3]

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Question 3

(a) Let $y = \ln(2x - 5)$. Then $e^y = 2x - 5$, so $x = \dfrac{e^y + 5}{2}$.

$f^{-1}(x) = \frac{e^x + 5}{2}$ [2]

(b) Domain of $f^{-1}$ = range of $f$: Since $2x - 5$ can take any positive value, $\ln(2x-5)$ can take any real value. Domain of $f^{-1}$ is $\mathbb{R}$.

Range of $f^{-1}$ = domain of $f$: $x > \dfrac{5}{2}$, so range of $f^{-1}$ is $\left(\dfrac{5}{2}, \infty\right)$. [2]

(c) Graph of $y = f(x) = \ln(2x - 5)$:

• Vertical asymptote: $2x - 5 = 0 \Rightarrow x = \dfrac{5}{2}$
• $x$-intercept: $\ln(2x-5) = 0 \Rightarrow 2x - 5 = 1 \Rightarrow x = 3$, so $(3, 0)$
• Passes through $\left(\dfrac{7}{2}, \ln 2\right)$

Graph of $y = f^{-1}(x) = \dfrac{e^x + 5}{2}$:

• Horizontal asymptote: as $x \to -\infty$, $e^x \to 0$, so $y \to \dfrac{5}{2}$
• $y$-intercept: $f^{-1}(0) = \dfrac{1+5}{2} = 3$, so $(0, 3)$

The two graphs are reflections of each other in the line $y = x$. [2]

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Question 4

(a) $f(1) = 4 - 1^2 = 3$. [1]

$\displaystyle\lim_{x \to 1^+} f(x) = 2(1) + 1 = 3$. [1]

(b) For $x \leq 1$: $f(x) = 4 - x^2$, which is increasing on $(-\infty, 1]$ (since $f'(x) = -2x > 0$ for $x < 0$... actually, $f(x) = 4 - x^2$ is a downward parabola with vertex at $x = 0$, so it increases on $(-\infty, 0]$ and decreases on $[0, 1]$).

Since $f(x) = 4 - x^2$ is not monotonic on $(-\infty, 1]$ (e.g., $f(-1) = 3 = f(1)$), $f$ is not one-one. [2]

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Question 5

We need $f^{-1}(3)$, i.e. the value of $x$ such that $f(x) = 3$.

$e^{3x} + 2 = 3$

$e^{3x} = 1$

$3x = 0$

$x = 0$

Therefore $f^{-1}(3) = 0$. [3]

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Question 6

(a) Since $f(0) = f(4) = 7$ and $f(x) = (x-a)^2 + b$ is a parabola, the axis of symmetry is at $x = \dfrac{0+4}{2} = 2$. Therefore $a = 2$. [2]

(b) $f(0) = (0-2)^2 + b = 4 + b = 7$, so $b = 3$. [2]

(c) $f(x) = (x-2)^2 + 3 \geq 3$. Range of $f$: $[3, \infty)$. [1]

(d) The line $y = k$ intersects the graph at exactly one point when $k$ equals the minimum value, i.e. $k = 3$. [2]

(e) We minimise the distance from $(5, 0)$ to a point $(x, (x-2)^2 + 3)$ on the curve.

Distance squared: $D^2 = (x - 5)^2 + ((x-2)^2 + 3)^2$

Let $u = x - 2$, so $x = u + 2$:

$D^2 = (u + 2 - 5)^2 + (u^2 + 3)^2 = (u - 3)^2 + (u^2 + 3)^2$

$= u^2 - 6u + 9 + u^4 + 6u^2 + 9 = u^4 + 7u^2 - 6u + 18$

Differentiate: $\dfrac{d(D^2)}{du} = 4u^3 + 14u - 6 = 0$

Try $u = \dfrac{1}{2}$: $4\left(\dfrac{1}{8}\right) + 7 - 6 = \dfrac{1}{2} + 1 = \dfrac{3}{2} \neq 0$

Try factoring: $4u^3 + 14u - 6 = 0 \Rightarrow 2u^3 + 7u - 3 = 0$

Try $u = \dfrac{1}{2}$: $2\left(\dfrac{1}{8}\right) + \dfrac{7}{2} - 3 = \dfrac{1}{4} + \dfrac{7}{2} - 3 = \dfrac{1}{4} + \dfrac{1}{2} = \dfrac{3}{4} \neq 0$

Try $u = \dfrac{3 - \sqrt{57}}{8}$... Let me reconsider. Actually, let me try a cleaner approach.

Setting $\dfrac{d(D^2)}{du} = 4u^3 + 14u - 6 = 0$.

By inspection or numerical methods, $u \approx 0.413$. But for an exact answer, let me check if there's a rational root.

By the Rational Root Theorem, possible rational roots are $\pm\dfrac{1}{4}, \pm\dfrac{1}{2}, \pm\dfrac{3}{4}, \pm\dfrac{3}{2}, \pm 1, \pm 3, \pm\dfrac{3}{4}$.

$u = \dfrac{1}{2}$: $4\left(\dfrac{1}{8}\right) + 7 - 6 = \dfrac{1}{2} + 1 = 1.5 \neq 0$

Actually, let me reconsider the problem. The closest point should satisfy that the line from $(5,0)$ to the curve is perpendicular to the tangent at that point.

Tangent slope at $(x, f(x))$: $f'(x) = 2(x-2)$

Slope of line from $(5, 0)$ to $(x, (x-2)^2+3)$: $\dfrac{(x-2)^2 + 3}{x - 5}$

For perpendicularity: $2(x-2) \cdot \dfrac{(x-2)^2 + 3}{x - 5} = -1$

This is equivalent to the derivative condition above. Let me solve $4u^3 + 14u - 6 = 0$ where $u = x - 2$.

Using the cubic formula or numerical approximation: $u \approx 0.4131$, so $x \approx 2.4131$.

$f(2.4131) \approx (0.4131)^2 + 3 \approx 3.1707$.

Hmm, this doesn't give a clean answer. Let me re-examine — perhaps the question expects a specific nice value.

Actually, let me try $u = \dfrac{\sqrt{7} - 1}{2}$... this is getting complex. For a well-designed question, let me note that the answer is:

$x = 2 + u$ where $4u^3 + 14u - 6 = 0$.

Using numerical methods: $u \approx 0.413$, $x \approx 2.41$, $f(x) \approx 3.17$.

The closest point is approximately $(2.41, 3.17)$. [3]

Note: This requires solving a cubic. In an exam, the cubic would factor nicely. For this practice paper, students should set up the derivative equation and solve numerically or verify their answer.

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Question 7

(a) $f(x) = \dfrac{ax + b}{cx + d}$

$f(0) = \dfrac{b}{d} = -1 \Rightarrow b = -d$ ... (i)

$f(1) = \dfrac{a + b}{c + d} = 0 \Rightarrow a + b = 0 \Rightarrow a = -b = d$ ... (ii)

$f^{-1}(2) = 3 \Rightarrow f(3) = 2$

$f(3) = \dfrac{3a + b}{3c + d} = 2$

$3a + b = 2(3c + d) = 6c + 2d$ ... (iii)

From (i): $b = -d$. From (ii): $a = d$.

Substitute into (iii): $3d + (-d) = 6c + 2d$

$2d = 6c + 2d$

$0 = 6c \Rightarrow c = 0$

Wait, but the problem states $c \neq 0$. Let me re-read... The problem says $c \neq 0$. So there's an inconsistency. Let me re-examine.

Actually, if $c = 0$, then $f(x) = \dfrac{ax+b}{d}$ which is linear, and $f^{-1}(2) = 3$ gives $f(3) = 2$, which is fine. But the problem states $c \neq 0$.

Let me reconsider. Perhaps I made an error. From $f(1) = 0$: $a + b = 0$, so $a = -b$. From $f(0) = -1$: $b/d = -1$, so $b = -d$, hence $a = d$.

From $f(3) = 2$: $\dfrac{3a + b}{3c + d} = 2$, so $3d + (-d) = 2(3c + d)$, giving $2d = 6c + 2d$, so $c = 0$.

This is a contradiction with $c \neq 0$. The question as stated has no solution with $c \neq 0$.

This is a design issue. For a corrected version, let me adjust: if $f^{-1}(2) = 3$ is changed to $f^{-1}(3) = 2$:

$f(2) = 3$: $\dfrac{2a + b}{2c + d} = 3$, so $2d + (-d) = 3(2c + d)$, giving $d = 6c + 3d$, so $-2d = 6c$, $d = -3c$.

Then $a = d = -3c$ and $b = -d = 3c$.

So $f(x) = \dfrac{-3cx + 3c}{cx - 3c} = \dfrac{-3x + 3}{x - 3} = \dfrac{3(1-x)}{x-3}$ (taking $c = -1$ for simplicity, or $c = 1$: $f(x) = \dfrac{-3x+3}{x-3}$).

Let me verify: $f(0) = \dfrac{3}{-3} = -1$ ✓, $f(1) = \dfrac{0}{-2} = 0$ ✓, $f(2) = \dfrac{-3}{-1} = 3$ ✓, so $f^{-1}(3) = 2$ ✓.

Corrected answer: $a = -3$, $b = 3$, $c = 1$, $d = -3$ (or any scalar multiple). [5]

(b) $f(x) = \dfrac{-3x + 3}{x - 3}$. Horizontal asymptote: $y = \dfrac{-3}{1} = -3$. So range is $\{y \in \mathbb{R} : y \neq -3\}$. [2]

(c) Vertical asymptote: $x = 3$. Horizontal asymptote: $y = -3$.

$x$-intercept: $-3x + 3 = 0 \Rightarrow x = 1$, so $(1, 0)$.

$y$-intercept: $f(0) = -1$, so $(0, -1)$.

The graph is a rectangular hyperbola with asymptotes $x = 3$ and $y = -3$. [3]

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Question 8

(a) $g(x) = f(x + 3) - 1$ represents a translation of $f(x)$ by 3 units left and 1 unit down.

Vertical asymptote of $f$: $x = -1$. After shifting left by 3: $x = -4$.

Horizontal asymptote of $f$: $y = 2$. After shifting down by 1: $y = 1$.

Asymptotes of $g$: $x = -4$ and $y = 1$. [2]

(b) The point $(0, 0)$ on $y = f(x)$: shifting left by 3 gives $x = -3$, shifting down by 1 gives $y = -1$.

Corresponding point on $y = g(x)$: $(-3, -1)$. [2]

(c) $y = 2f(x) + 4$: the horizontal asymptote $y = 2$ becomes $y = 2(2) + 4 = 8$.

New horizontal asymptote: $y = 8$. [2]

(d) $y = f(2x)$ represents a horizontal stretch with scale factor $\dfrac{1}{2}$ (i.e., compression towards the $y$-axis by factor 2). [2]

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Question 9

(a) $f(x) = \sqrt{x + 3}$, domain $x \geq -3$.

Minimum value: $f(-3) = 0$. As $x \to \infty$, $f(x) \to \infty$.

Range of $f$: $[0, \infty)$. [1]

(b) $fg(x) = f(g(x)) = f(x^2 - 4) = \sqrt{x^2 - 4 + 3} = \sqrt{x^2 - 1}$.

For this to be defined, we need $x^2 - 1 \geq 0$, i.e., $x^2 \geq 1$, i.e., $x \leq -1$ or $x \geq 1$.

But the range of $g$ is $[-4, \infty)$ (since $g(x) = x^2 - 4 \geq -4$), and the domain of $f$ is $[-3, \infty)$.

For $fg$ to exist, we need $\text{range}(g) \subseteq \text{domain}(f)$, i.e., $[-4, \infty) \subseteq [-3, \infty)$.

This is not true since $g$ can output values in $[-4, -3)$ which are not in the domain of $f$.

Therefore $fg$ does not exist (with the given unrestricted domain of $g$). [2]

(c) We need $\text{range}(g) \subseteq [-3, \infty)$, i.e., $x^2 - 4 \geq -3$, so $x^2 \geq 1$.

The largest possible domain of $g$ is $(-\infty, -1] \cup [1, \infty)$.

For this domain: $fg(x) = \sqrt{x^2 - 4 + 3} = \sqrt{x^2 - 1}$.

Domain of $fg$: $(-\infty, -1] \cup [1, \infty)$. [4]

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Question 10

(a) Let $y = \dfrac{2x + 3}{x - 1}$. Swap: $x = \dfrac{2y + 3}{y - 1}$.

$x(y - 1) = 2y + 3$

$xy - x = 2y + 3$

$xy - 2y = x + 3$

$y(x - 2) = x + 3$

$f^{-1}(x) = \frac{x + 3}{x - 2}$ [2]

(b) $f(f(x)) = f\left(\dfrac{2x+3}{x-1}\right) = \dfrac{2\cdot\dfrac{2x+3}{x-1} + 3}{\dfrac{2x+3}{x-1} - 1}$

Numerator: $\dfrac{2(2x+3) + 3(x-1)}{x-1} = \dfrac{4x + 6 + 3x - 3}{x-1} = \dfrac{7x + 3}{x-1}$

Denominator: $\dfrac{2x+3 - (x-1)}{x-1} = \dfrac{x + 4}{x-1}$

$f(f(x)) = \dfrac{7x+3}{x+4}$

Hmm, this doesn't equal $x$. Let me recheck... Actually, $f(f(x)) = x$ would mean $f = f^{-1}$, which requires the function to be self-inverse.

Let me check: $f(x) = \dfrac{2x+3}{x-1}$ and $f^{-1}(x) = \dfrac{x+3}{x-2}$. These are not the same, so $f(f(x)) \neq x$.

Let me recalculate $f(f(x))$:

$f(f(x)) = \dfrac{2\cdot\dfrac{2x+3}{x-1}+3}{\dfrac{2x+3}{x-1}-1} = \dfrac{\dfrac{4x+6+3x-3}{x-1}}{\dfrac{2x+3-x+1}{x-1}} = \dfrac{7x+3}{x+4}$

This is not equal to $x$. So the question as stated is incorrect.

Design note: For $f(f(x)) = x$, we need $f = f^{-1}$. A function of the form $f(x) = \dfrac{ax+b}{cx+d}$ is self-inverse when $a + d = 0$. Here $a = 2, d = -1$, so $a + d = 1 \neq 0$.

For a corrected version, let $f(x) = \dfrac{2x+3}{x-2}$ (so $a = 2, d = -2$, and $a + d = 0$).

Then $f^{-1}(x) = \dfrac{-2x+3}{-x+1} = \dfrac{2x-3}{x-1}$... let me redo this.

Actually, let me use $f(x) = \dfrac{x+3}{x+1}$ where $a = 1, d = 1$, so $a + d = 2 \neq 0$. Still not self-inverse.

For self-inverse: $a + d = 0$. Let $f(x) = \dfrac{2x+3}{x-2}$. Then $a = 2, d = -2$, so $a + d = 0$.

$f^{-1}$: $y = \dfrac{2x+3}{x-2}$, swap: $x = \dfrac{2y+3}{y-2}$, $x(y-2) = 2y+3$, $xy - 2x = 2y + 3$, $xy - 2y = 2x + 3$, $y(x-2) = 2x+3$, $f^{-1}(x) = \dfrac{2x+3}{x-2} = f(x)$. ✓

So the corrected function is $f(x) = \dfrac{2x+3}{x-2}$.

Let me redo parts (a)-(e) with $f(x) = \dfrac{2x+3}{x-2}$:

(a) $f^{-1}(x) = \dfrac{2x+3}{x-2}$ (since $f = f^{-1}$). [2]

(b) $f(f(x)) = f\left(\dfrac{2x+3}{x-2}\right) = \dfrac{2\cdot\dfrac{2x+3}{x-2}+3}{\dfrac{2x+3}{x-2}-2}$

Numerator: $\dfrac{4x+6+3x-6}{x-2} = \dfrac{7x}{x-2}$

Denominator: $\dfrac{2x+3-2x+4}{x-2} = \dfrac{7}{x-2}$

$f(f(x)) = \dfrac{7x}{7} = x$ ✓

Geometric interpretation: Since $f = f^{-1}$, the graph of $y = f(x)$ is symmetric about the line $y = x$. Applying $f$ twice returns the original input, so the function is its own inverse. [4]

(c) $f(x) = f^{-1}(x)$: Since $f = f^{-1}$, this holds for all $x$ in the domain. But the question asks for $f(x) = f^{-1}(x)$ which means $f(x) = x$ (the intersection with $y = x$).

$\dfrac{2x+3}{x-2} = x$

$2x + 3 = x^2 - 2x$

$x^2 - 4x - 3 = 0$

$x = \dfrac{4 \pm \sqrt{16+12}}{2} = \dfrac{4 \pm \sqrt{28}}{2} = \dfrac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}$

Solutions: $x = 2 + \sqrt{7}$ or $x = 2 - \sqrt{7}$. [4]

(d) Asymptotes of $y = f(x)$: vertical $x = 2$, horizontal $y = 2$.

Asymptotes of $y = f^{-1}(x)$: Since $f = f^{-1}$, the same: vertical $x = 2$, horizontal $y = 2$.

(For a general function, the vertical asymptote of $f^{-1}$ is $y =$ horizontal asymptote of $f$, and vice versa.) [3]

(e) Graph: Rectangular hyperbola with asymptotes $x = 2$ and $y = 2$, symmetric about $y = x$. Passes through the fixed points $(2+\sqrt{7}, 2+\sqrt{7})$ and $(2-\sqrt{7}, 2-\sqrt{7})$. [2]

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Question 11

(a) $P(2) = P_0 e^{2k} = 1500$ ... (i)

$P(5) = P_0 e^{5k} = 6000$ ... (ii)

Divide (ii) by (i): $\dfrac{P_0 e^{5k}}{P_0 e^{2k}} = \dfrac{6000}{1500} = 4$

$e^{3k} = 4$

$3k = \ln 4$

$k = \frac{\ln 4}{3}$ [3]

(b) From (i): $P_0 \cdot e^{2\ln 4 / 3} = 1500$

$P_0 \cdot 4^{2/3} = 1500$

$P_0 = \dfrac{1500}{4^{2/3}} = \dfrac{1500}{(4^{1/3})^2} = \dfrac{1500}{2^{4/3}}$

Alternatively: $P_0 \cdot e^{\ln(4^{2/3})} = 1500$, so $P_0 = 1500 \cdot 4^{-2/3} = 1500 \cdot 2^{-4/3}$.

$P_0 = 1500 \times 4^{-2/3} = 1500 \times \dfrac{1}{4^{2/3}}$

$4^{2/3} = (2^2)^{2/3} = 2^{4/3} \approx 2.5198$

$P_0 \approx \dfrac{1500}{2.5198} \approx 595.3$

$P_0 = 1500 \cdot 4^{-2/3} \approx 595$ (or exactly $\dfrac{1500}{2^{4/3}}$). [2]

(c) $P = P_0 e^{kt}$, so $\dfrac{P}{P_0} = e^{kt}$, $kt = \ln\dfrac{P}{P_0}$, $t = \dfrac{1}{k}\ln\dfrac{P}{P_0}$.

$T(P) = \frac{1}{k}\ln\frac{P}{P_0} = \frac{3}{\ln 4}\ln\frac{P}{P_0}$ [3]

(d) $T(20000) = \dfrac{3}{\ln 4}\ln\dfrac{20000}{1500 \cdot 4^{-2/3}} = \dfrac{3}{\ln 4}\ln\dfrac{20000 \cdot 4^{2/3}}{1500}$

$= \dfrac{3}{\ln 4}\ln\dfrac{40 \cdot 4^{2/3}}{3} = \dfrac{3}{\ln 4}\ln\dfrac{40}{3} + \dfrac{3}{\ln 4}\cdot\dfrac{2}{3}\ln 4$

$= \dfrac{3}{\ln 4}\ln\dfrac{40}{3} + 2$

$\dfrac{40}{3} \approx 13.333$, $\ln(13.333) \approx 2.590$

$T(20000) \approx \dfrac{3 \times 2.590}{1.386} + 2 \approx 5.605 + 2 = 7.605$

Time ≈ 7.61 hours (to 2 d.p.). [2]

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Question 12

(a) $f(x) = x^2 - 6x + 13 = (x - 3)^2 + 4$

Minimum value is 4, occurring at $x = 3$. [3]

(b) $f$ does not have an inverse function because it is not one-one. The parabola $y = (x-3)^2 + 4$ fails the horizontal line test — for any $y > 4$, there are two distinct $x$-values giving the same output. [2]

(c) The smallest value of $c$ is $3$ (the $x$-coordinate of the vertex). For $x \geq 3$, $f$ is strictly increasing and hence one-one.

Let $y = (x-3)^2 + 4$, with $x \geq 3$.

$y - 4 = (x-3)^2$

$x - 3 = \sqrt{y - 4}$ (taking positive root since $x \geq 3$)

$x = 3 + \sqrt{y - 4}$

$f^{-1}(x) = 3 + \sqrt{x - 4}$

Domain of $f^{-1}$ = range of $f$ (restricted): $[4, \infty)$. [4]

(d) Graph of $y = f(x)$ for $x \geq 3$: right half of parabola with vertex at $(3, 4)$, opening upward.

Graph of $y = f^{-1}(x) = 3 + \sqrt{x-4}$: starts at $(4, 3)$ and increases, concave down.

Line of symmetry: $y = x$. [3]

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Question 13

(a) For $fg$ to exist: $\text{range}(g) \subseteq \text{domain}(f)$.

$g(x) = \dfrac{4}{x+1}$, domain $x > -1, x \neq 0$.

For $x > -1$ (excluding 0): as $x \to -1^+$, $g(x) \to +\infty$; as $x \to \infty$, $g(x) \to 0^+$; as $x \to 0^-$, $g(x) \to -\infty$... wait, $x > -1$ and $x \neq 0$.

For $-1 < x < 0$: $0 < x + 1 < 1$, so $g(x) = \dfrac{4}{x+1} > 4$.

For $x > 0$: $x + 1 > 1$, so $0 < g(x) < 4$.

So $\text{range}(g) = (0, 4) \cup (4, \infty)$.

Domain of $f$ is $x \geq 0$. Since $(0, 4) \cup (4, \infty) \subseteq [0, \infty)$, we have $\text{range}(g) \subseteq \text{domain}(f)$.

Therefore $fg$ exists. ✓

$fg(x) = f(g(x)) = f\left(\dfrac{4}{x+1}\right) = \dfrac{1}{2}\left(\dfrac{4}{x+1}\right)^2 - 2 = \dfrac{8}{(x+1)^2} - 2$

Domain of $fg$: $x > -1, x \neq 0$ (same as domain of $g$). [4]

(b) For $x > -1, x \neq 0$: $(x+1)^2 \in (0, 1) \cup (1, \infty)$.

So $\dfrac{8}{(x+1)^2} \in (0, 8) \cup (8, \infty)$.

$fg(x) = \dfrac{8}{(x+1)^2} - 2 \in (-2, 6) \cup (6, \infty)$.

Range of $fg$: $(-2, 6) \cup (6, \infty)$. [3]

(c) For $gf$ to exist: $\text{range}(f) \subseteq \text{domain}(g)$.

$f(x) = \dfrac{1}{2}x^2 - 2$, domain $x \geq 0$. Range: $[-2, \infty)$.

Domain of $g$: $x > -1, x \neq 0$.

Since $[-2, \infty) \not\subseteq (-1, 0) \cup (0, \infty)$ (e.g., $f(0) = -2 \notin \text{domain}(g)$), $gf$ does not exist. **[3]

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Question 14

(a) From the graph:

• Line segment from $(-3, 0)$ to $(0, 3)$: range $[0, 3]$
• Upper semicircle centre $(2, 0)$ radius 2: from $x = 0$ to $x = 4$, the $y$-values go from $0$ up to $2$ and back to $0$. Range $[0, 2]$.
• Horizontal line $y = 0$ for $x \geq 4$: value $0$.

Combined range: $[0, 3]$. [2]

(b) $f(2)$: At $x = 2$, we're on the semicircle. The semicircle is the upper half of $(x-2)^2 + y^2 = 4$, so $y = \sqrt{4 - (x-2)^2}$.

$f(2) = \sqrt{4 - 0} = 2$. [1]

(c) $h(x) = f(x - 2)$: horizontal translation 2 units right.

• Line segment: from $(-3, 0)$ to $(0, 3)$ becomes from $(-1, 0)$ to $(2, 3)$.
• Semicircle: centre moves from $(2, 0)$ to $(4, 0)$, still radius 2, from $x = 2$ to $x = 6$.
• Horizontal line: from $x \geq 4$ to $x \geq 6$. [3]

(d) $f(x) = 1$:

• On the line segment from $(-3, 0)$ to $(0, 3)$: slope is $1$, so $y = x + 3$. Setting $x + 3 = 1$: $x = -2$. ✓ (in range)
• On the semicircle: $\sqrt{4-(x-2)^2} = 1$, so $4-(x-2)^2 = 1$, $(x-2)^2 = 3$, $x = 2\pm\sqrt{3}$. Both in $[0, 4]$. ✓ (2 solutions)
• On the horizontal line $y = 0$: no solution.

Total: 3 solutions. [2]

---

Question 15

To show $g = f^{-1}$, we show $f(g(x)) = x$ and $g(f(x)) = x$.

$f(g(x)) = f\left(\dfrac{x+3}{1-x}\right) = \dfrac{\dfrac{x+3}{1-x} - 3}{\dfrac{x+3}{1-x} + 1}$

Numerator: $\dfrac{x+3 - 3(1-x)}{1-x} = \dfrac{x+3-3+3x}{1-x} = \dfrac{4x}{1-x}$

Denominator: $\dfrac{x+3 + (1-x)}{1-x} = \dfrac{4}{1-x}$

$f(g(x)) = \dfrac{4x}{4} = x$ ✓

$g(f(x)) = g\left(\dfrac{x-3}{x+1}\right) = \dfrac{\dfrac{x-3}{x+1}+3}{1-\dfrac{x-3}{x+1}}$

Numerator: $\dfrac{x-3+3(x+1)}{x+1} = \dfrac{x-3+3x+3}{x+1} = \dfrac{4x}{x+1}$

Denominator: $\dfrac{x+1-(x-3)}{x+1} = \dfrac{4}{x+1}$

$g(f(x)) = \dfrac{4x}{4} = x$ ✓

Therefore $g = f^{-1}$. [4]

---

Question 16

(a) Let $y = 2 + \ln(x - 1)$. Then $y - 2 = \ln(x - 1)$, so $x - 1 = e^{y-2}$, $x = e^{y-2} + 1$.

$f^{-1}(x) = e^{x-2} + 1$

Domain of $f^{-1}$ = range of $f$: $\mathbb{R}$.

Range of $f^{-1}$ = domain of $f$: $(1, \infty)$. [3]

(b) $f(x) = f^{-1}(x)$:

$2 + \ln(x - 1) = e^{x-2} + 1$

$\ln(x - 1) = e^{x-2} - 1$

This is a transcendental equation. We look for solutions where $f(x) = f^{-1}(x)$, which occur on the line $y = x$:

$f(x) = x$: $2 + \ln(x-1) = x$, so $\ln(x-1) = x - 2$, $x - 1 = e^{x-2}$.

Let $u = x - 1$: $u = e^{u-1}$, so $u e^{1-u} = 1$, $u e^{-u} = e^{-1}$, $-u e^{-u} = -e^{-1}$.

So $-u = W(-e^{-1})$ where $W$ is the Lambert W function. Since $-e^{-1}$ is the branch point, $W(-e^{-1}) = -1$.

So $-u = -1$, $u = 1$, $x = 2$.

Check: $f(2) = 2 + \ln(1) = 2$. ✓

Solution: $x = 2$. [4]

(c) Graph of $y = f(x) = 2 + \ln(x-1)$:

• Vertical asymptote: $x = 1$
• $y$-intercept: none (domain is $x > 1$)
• Passes through $(2, 2)$ since $f(2) = 2$
• Passes through $(1+e, 3)$ since $f(1+e) = 2 + \ln(e) = 3$

Graph of $y = f^{-1}(x) = e^{x-2} + 1$:

• Horizontal asymptote: $y = 1$ (as $x \to -\infty$)
• Passes through $(2, 2)$ since $f^{-1}(2) = e^0 + 1 = 2$
• Passes through $(3, 1+e)$ since $f^{-1}(3) = e^1 + 1 = e + 1$

The graphs are reflections of each other in $y = x$. [3]

---

Question 17

(a) $f(x) = \dfrac{x^2 + 1}{x} = x + \dfrac{1}{x}$, for $x > 0$.

By AM-GM inequality: $x + \dfrac{1}{x} \geq 2\sqrt{x \cdot \dfrac{1}{x}} = 2$.

Equality when $x = \dfrac{1}{x}$, i.e., $x = 1$.

Therefore $f(x) \geq 2$ for all $x > 0$. [3]

(b) Let $y = x + \dfrac{1}{x}$. Then $yx = x^2 + 1$, so $x^2 - yx + 1 = 0$.

$x = \dfrac{y \pm \sqrt{y^2 - 4}}{2}$

For $x > 0$ and $y \geq 2$: both roots are positive (since $y > \sqrt{y^2-4}$), so we need to restrict to one branch.

Since $f$ is not one-one on $x > 0$ (it decreases on $(0, 1]$ and increases on $[1, \infty)$), we need to restrict the domain.

For the inverse to exist, restrict to $x \geq 1$ (where $f$ is increasing). On this domain, $x = \dfrac{y + \sqrt{y^2 - 4}}{2}$ (taking the larger root since $x \geq 1$).

$f^{-1}(x) = \frac{x + \sqrt{x^2 - 4}}{2}$

Domain of $f^{-1}$ = range of $f$ (restricted to $x \geq 1$): $[2, \infty)$. [4]

(c) $f(x) = 3$: $x + \dfrac{1}{x} = 3$, so $x^2 - 3x + 1 = 0$.

$x = \dfrac{3 \pm \sqrt{9-4}}{2} = \dfrac{3 \pm \sqrt{5}}{2}$

Both are positive: $\dfrac{3+\sqrt{5}}{2} \approx 2.618$ and $\dfrac{3-\sqrt{5}}{2} \approx 0.382$.

Solutions: $x = \dfrac{3 + \sqrt{5}}{2}$ or $x = \dfrac{3 - \sqrt{5}}{2}$. [3]

---

Question 18

(a) $f(x) = x^2 - 2x = (x-1)^2 - 1$. Minimum value is $-1$ at $x = 1$.

Range of $f$: $[-1, \infty)$. [1]

(b) $fg(x) = f(g(x)) = f(2x + k) = (2x+k)^2 - 2(2x+k) = 4x^2 + 4kx + k^2 - 4x - 2k$

$= 4x^2 + (4k - 4)x + (k^2 - 2k)$ [2]

(c) $fg(x) = 4x^2 + (4k-4)x + (k^2-2k)$.

This is a quadratic in $x$ with leading coefficient $4 > 0$, so it has a minimum.

The minimum occurs at $x = -\dfrac{4k-4}{8} = \dfrac{1-k}{2}$.

Minimum value: $fg\left(\dfrac{1-k}{2}\right) = 4\left(\dfrac{1-k}{2}\right)^2 + (4k-4)\left(\dfrac{1-k}{2}\right) + k^2 - 2k$

$= (1-k)^2 + (2k-2)(1-k) + k^2 - 2k$

$= (1-k)^2 - 2(1-k)^2 + k^2 - 2k$

$= -(1-k)^2 + k^2 - 2k$

$= -(1 - 2k + k^2) + k^2 - 2k$

$= -1 + 2k - k^2 + k^2 - 2k$

$= -1$

Wait, the minimum is always $-1$ regardless of $k$. So the range is always $[-1, \infty)$, not $[-4, \infty)$.

This means the question as stated has an issue — the range of $fg$ is always $[-1, \infty)$ for any $k$.

Design note: The range of $fg$ is always $[-1, \infty)$ because $f$ has range $[-1, \infty)$ and $g$ is surjective onto $\mathbb{R}$ (for appropriate $k$). To get range $[-4, \infty)$, we'd need a different setup.

Let me reconsider. If the question asks for range $[-4, \infty)$, perhaps $f(x) = x^2 - 2x - 3 = (x-1)^2 - 4$, giving range $[-4, \infty)$.

With $f(x) = x^2 - 2x - 3$:

$fg(x) = (2x+k)^2 - 2(2x+k) - 3 = 4x^2 + 4kx + k^2 - 4x - 2k - 3$

Minimum at $x = \dfrac{1-k}{2}$:

Minimum value $= -(1-k)^2 + k^2 - 2k - 3 = -1 + 2k - k^2 + k^2 - 2k - 3 = -4$.

So the range is always $[-4, \infty)$ regardless of $k$. This means any value of $k$ works, which also isn't a well-designed question.

For a properly designed question, let me adjust: suppose $f(x) = x^2 - 2x$ and we want the range of $fg$ to be $[m, \infty)$ where $m$ depends on $k$. But as shown, the minimum is always $-1$.

Revised interpretation: The question is correct as stated, and the answer is that any value of $k$ gives range $[-1, \infty)$. But since the question asks for range $[-4, \infty)$, there may be a typo in the original.

For the purposes of this answer key, I'll note that with the given functions, the range of $fg$ is $[-1, \infty)$ for all $k$, so no value of $k$ gives range $[-4, \infty)$.

Answer: No such value of $k$ exists. (The range of $fg$ is always $[-1, \infty)$.) [3]

(d) Since no valid $k$ exists for part (c), this part is moot. However, if we proceed with any $k$:

$fg(x)

<stage5_exam_answers_md>
# TuitionGoWhere Practice Paper — Maths H2 A-Level  
## Answer Key — Algebra & Functions (Version 3 of 5)

---

### Question 1

**(a)** Let $y = \dfrac{3x - 1}{x + 2}$.

Swap $x$ and $y$: $x = \dfrac{3y - 1}{y + 2}$

$x(y + 2) = 3y - 1$

$xy + 2x = 3y - 1$

$xy - 3y = -1 - 2x$

$y(x - 3) = -1 - 2x$

$$f^{-1}(x) = \frac{-1 - 2x}{x - 3} = \frac{2x + 1}{3 - x}$$

Domain of $f^{-1}$ = range of $f$. Since $f(x) = \dfrac{3x-1}{x+2}$, the horizontal asymptote is $y = 3$, so $f(x) \neq 3$.

**Domain of $f^{-1}$:** $x \neq 3$, i.e. $x \in \mathbb{R}, x \neq 3$. **[3]**

**(b)** Range of $f$: Since the horizontal asymptote is $y = 3$ and the function never equals 3 (solving $\frac{3x-1}{x+2} = 3$ gives $3x - 1 = 3x + 6$, i.e. $-1 = 6$, impossible), the range is $\{y \in \mathbb{R} : y \neq 3\}$.

Domain of $f^{-1}$ = range of $f$: $x \neq 3$. **[2]**

---

### Question 2

**(a)** For $gf$ to exist, we need $\text{range}(f) \subseteq \text{domain}(g)$.

$f(x) = x^2 - 4x + 6 = (x-2)^2 + 2$, with domain $x \geq 2$.

Since $(x-2)^2 \geq 0$ for all $x$, the minimum value of $f$ is $2$ (at $x = 2$), and $f(x)$ increases without bound.

So $\text{range}(f) = [2, \infty)$.

Domain of $g$ is $x > 1$. Since $[2, \infty) \subset (1, \infty)$, we have $\text{range}(f) \subseteq \text{domain}(g)$.

Therefore $gf$ exists. **[2]**

**(b)** $gf(x) = g(f(x)) = g(x^2 - 4x + 6) = \dfrac{1}{(x^2 - 4x + 6) - 1} = \dfrac{1}{x^2 - 4x + 5}$

Now $x^2 - 4x + 5 = (x-2)^2 + 1 \geq 1$ for $x \geq 2$, with minimum $1$ at $x = 2$.

So $0 < gf(x) \leq 1$.

**Range of $gf$:** $(0, 1]$ **[3]**

---

### Question 3

**(a)** Let $y = \ln(2x - 5)$. Then $e^y = 2x - 5$, so $x = \dfrac{e^y + 5}{2}$.

$$f^{-1}(x) = \frac{e^x + 5}{2}$$ **[2]**

**(b)** Domain of $f^{-1}$ = range of $f$: Since $2x - 5$ can take any positive value, $\ln(2x-5)$ can take any real value. Domain of $f^{-1}$ is $\mathbb{R}$.

Range of $f^{-1}$ = domain of $f$: $x > \dfrac{5}{2}$, so range of $f^{-1}$ is $\left(\dfrac{5}{2}, \infty\right)$. **[2]**

**(c)** Graph of $y = f(x) = \ln(2x - 5)$:
- Vertical asymptote: $2x - 5 = 0 \Rightarrow x = \dfrac{5}{2}$
- $x$-intercept: $\ln(2x-5) = 0 \Rightarrow 2x - 5 = 1 \Rightarrow x = 3$, so $(3, 0)$
- Passes through $\left(\dfrac{7}{2}, \ln 2\right)$

Graph of $y = f^{-1}(x) = \dfrac{e^x + 5}{2}$:
- Horizontal asymptote: as $x \to -\infty$, $e^x \to 0$, so $y \to \dfrac{5}{2}$
- $y$-intercept: $f^{-1}(0) = \dfrac{1+5}{2} = 3$, so $(0, 3)$

The two graphs are reflections of each other in the line $y = x$. **[2]**

---

### Question 4

**(a)** $f(1) = 4 - 1^2 = 3$. **[1]**

$\displaystyle\lim_{x \to 1^+} f(x) = 2(1) + 1 = 3$. **[1]**

**(b)** For $x \leq 1$: $f(x) = 4 - x^2$, which is a downward parabola with vertex at $x = 0$, so it increases on $(-\infty, 0]$ and decreases on $[0, 1]$.

Since $f(x) = 4 - x^2$ is not monotonic on $(-\infty, 1]$ (e.g., $f(-1) = 3 = f(1)$), $f$ is **not one-one**. **[2]**

---

### Question 5

We need $f^{-1}(3)$, i.e. the value of $x$ such that $f(x) = 3$.

$e^{3x} + 2 = 3$

$e^{3x} = 1$

$3x = 0$

$$x = 0$$

Therefore $f^{-1}(3) = 0$. **[3]**

---

### Question 6

**(a)** Since $f(0) = f(4) = 7$ and $f(x) = (x-a)^2 + b$ is a parabola, the axis of symmetry is at $x = \dfrac{0+4}{2} = 2$. Therefore $a = 2$. **[2]**

**(b)** $f(0) = (0-2)^2 + b = 4 + b = 7$, so $b = 3$. **[2]**

**(c)** $f(x) = (x-2)^2 + 3 \geq 3$. **Range of $f$:** $[3, \infty)$. **[1]**

**(d)** The line $y = k$ intersects the graph at exactly one point when $k$ equals the minimum value, i.e. $k = 3$. **[2]**

**(e)** We minimise the distance from $(5, 0)$ to a point $(x, (x-2)^2 + 3)$ on the curve.

The squared distance is:

$$D^2 = (x - 5)^2 + ((x-2)^2 + 3 - 0)^2 = (x-5)^2 + ((x-2)^2 + 3)^2$$

Let $u = x - 2$, so $x = u + 2$:

$$D^2 = (u - 3)^2 + (u^2 + 3)^2 = u^2 - 6u + 9 + u^4 + 6u^2 + 9 = u^4 + 7u^2 - 6u + 18$$

Differentiating:

$$\frac{d(D^2)}{du} = 4u^3 + 14u - 6$$

Setting equal to zero:

$$4u^3 + 14u - 6 = 0$$
$$2u^3 + 7u - 3 = 0$$

By the Rational Root Theorem, testing $u = \dfrac{1}{2}$:

$$2\left(\frac{1}{8}\right) + \frac{7}{2} - 3 = \frac{1}{4} + \frac{7}{2} - 3 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \neq 0$$

Testing values numerically, $2u^3 + 7u - 3 = 0$ has a root near $u \approx 0.413$. However, for an A-Level problem, let us check whether the problem has been set up with a nice answer.

Alternatively, the closest point occurs where the normal to the curve at that point passes through $(5, 0)$.

The gradient of the tangent to $y = (x-2)^2 + 3$ is $\dfrac{dy}{dx} = 2(x-2)$.

The gradient of the normal is $-\dfrac{1}{2(x-2)}$.

The normal at point $(x, (x-2)^2+3)$ passing through $(5, 0)$ gives:

$$\frac{(x-2)^2 + 3 - 0}{x - 5} = -\frac{1}{2(x-2)}$$

$$2(x-2)((x-2)^2 + 3) = -(x-5)$$

$$2(x-2)^3 + 6(x-2) + (x-5) = 0$$

Let $u = x - 2$:

$$2u^3 + 6u + u + 2 - 5 = 0$$
$$2u^3 + 7u - 3 = 0$$

This cubic has one real root. Using numerical methods or the cubic formula:

$$u = \sqrt[3]{\frac{3}{4} + \sqrt{\frac{9}{16} + \frac{343}{216}}} + \sqrt[3]{\frac{3}{4} - \sqrt{\frac{9}{16} + \frac{343}{216}}}$$

This does not simplify nicely. For the purposes of this question, the exact answer is the solution to $2u^3 + 7u - 3 = 0$, giving $x = u + 2$ and $y = u^2 + 3$.

Numerically, $u \approx 0.4131$, so $x \approx 2.413$ and $y \approx (0.4131)^2 + 3 \approx 3.171$.

**Coordinates:** approximately $(2.41, 3.17)$ (to 3 s.f.) **[3]**

---

### Question 7

**(a)** $f(x) = \dfrac{ax + b}{cx + d}$

From $f(0) = -1$: $\dfrac{b}{d} = -1$, so $b = -d$. — (i)

From $f(1) = 0$: $\dfrac{a + b}{c + d} = 0$, so $a + b = 0$, i.e. $a = -b = d$. — (ii)

From $f^{-1}(2) = 3$: this means $f(3) = 2$.

$f(3) = \dfrac{3a + b}{3c + d} = 2$

$3a + b = 2(3c + d) = 6c + 2d$

From (ii): $a = d$, and from (i): $b = -d$.

$3d + (-d) = 6c + 2d$

$2d = 6c + 2d$

$0 = 6c$, so $c = 0$.

But the problem states $c \neq 0$. Let me re-read: the problem says $c \neq 0$. This means I need to reconsider.

Actually, let me re-examine. If $c = 0$, then $f(x) = \dfrac{ax+b}{d}$, which is linear, and $f^{-1}(2) = 3$ would still work. But the problem states $c \neq 0$.

Let me recheck: $f(1) = 0$ means $a(1) + b = 0$, so $a + b = 0$, giving $b = -a$.

From $f(0) = -1$: $\dfrac{b}{d} = -1$, so $b = -d$, hence $a = d$.

From $f(3) = 2$: $\dfrac{3a + b}{3c + d} = 2$

$3a + b = 6c + 2d$

Since $b = -a$ and $a = d$:

$3a - a = 6c + 2a$

$2a = 6c + 2a$

$0 = 6c$, so $c = 0$.

This contradicts $c \neq 0$. The problem as stated has an inconsistency. However, if we proceed with $c = 0$ (a linear fractional function that is actually linear), we get:

Let $d = 1$, then $a = 1$, $b = -1$, $c = 0$.

$f(x) = \dfrac{x - 1}{1} = x - 1$.

Check: $f(0) = -1$ ✓, $f(1) = 0$ ✓, $f(3) = 2$ ✓.

So $a = 1$, $b = -1$, $c = 0$, $d = 1$.

But since the problem states $c \neq 0$, let me re-interpret: perhaps the condition is that the function is genuinely a rational function (not linear), and there may be a typo in the problem. Proceeding with the values found:

**$a = 1$, $b = -1$, $c = 0$, $d = 1$** (noting that $c = 0$ makes $f$ linear) **[5]**

**(b)** $f(x) = x - 1$, so the range is $\mathbb{R}$. **[2]**

**(c)** The graph of $y = x - 1$ is a straight line with gradient 1, passing through $(0, -1)$ and $(1, 0)$. There are no asymptotes. **[3]**

---

### Question 8

**(a)** $g(x) = f(x + 3) - 1$ represents a translation of $f$ by 3 units left and 1 unit down.

Vertical asymptote of $f$: $x = -1$ → vertical asymptote of $g$: $x = -1 - 3 = -4$.

Horizontal asymptote of $f$: $y = 2$ → horizontal asymptote of $g$: $y = 2 - 1 = 1$.

**Asymptotes of $g$:** $x = -4$ and $y = 1$. **[2]**

**(b)** The point $(0, 0)$ on $y = f(x)$: shifting left by 3 gives $x = 0 - 3 = -3$, and shifting down by 1 gives $y = 0 - 1 = -1$.

**Corresponding point on $y = g(x)$:** $(-3, -1)$. **[2]**

**(c)** $y = 2f(x) + 4$: the horizontal asymptote $y = 2$ becomes $y = 2(2) + 4 = 8$.

**New horizontal asymptote:** $y = 8$. **[2]**

**(d)** $y = f(2x)$ represents a **horizontal stretch** with scale factor $\dfrac{1}{2}$ (i.e., compression towards the $y$-axis by factor $\dfrac{1}{2}$). **[2]**

---

### Question 9

**(a)** $f(x) = \sqrt{x + 3}$, domain $x \geq -3$.

As $x$ ranges from $-3$ to $\infty$, $\sqrt{x+3}$ ranges from $0$ to $\infty$.

**Range of $f$:** $[0, \infty)$. **[1]**

**(b)** $fg(x) = f(g(x)) = f(x^2 - 4) = \sqrt{x^2 - 4 + 3} = \sqrt{x^2 - 1}$.

For this to be defined, we need $x^2 - 1 \geq 0$, i.e. $x^2 \geq 1$, i.e. $x \leq -1$ or $x \geq 1$.

But the domain of $g$ is $\mathbb{R}$, and for $x \in (-1, 1)$, $g(x) = x^2 - 4 \in (-4, -1)$, which is outside the domain of $f$ (which requires input $\geq -3$... actually, $x^2 - 4 \geq -4$, and $f$ requires input $\geq -3$, so we need $x^2 - 4 \geq -3$, i.e. $x^2 \geq 1$).

Since $g(x)$ can take values less than $-3$ (e.g., $g(0) = -4 < -3$), and $f$ is only defined for inputs $\geq -3$, the composite $fg$ does not exist for the full domain of $g$.

More precisely: range of $g$ is $[-4, \infty)$ (since $x^2 - 4 \geq -4$), and domain of $f$ is $[-3, \infty)$. Since $[-4, \infty) \not\subseteq [-3, \infty)$ (e.g., $g(0) = -4 \notin [-3, \infty)$), the composite $fg$ does not exist. **[2]**

**(c)** We need $\text{range}(g) \subseteq \text{domain}(f) = [-3, \infty)$.

$g(x) = x^2 - 4 \geq -3$

$x^2 \geq 1$

$x \leq -1$ or $x \geq 1$.

**Largest possible domain of $g$:** $(-\infty, -1] \cup [1, \infty)$.

For this domain:

$fg(x) = f(g(x)) = \sqrt{x^2 - 4 + 3} = \sqrt{x^2 - 1}$. **[4]**

---

### Question 10

**(a)** Let $y = \dfrac{2x + 3}{x - 1}$.

Swap: $x = \dfrac{2y + 3}{y - 1}$

$x(y - 1) = 2y + 3$

$xy - x = 2y + 3$

$xy - 2y = x + 3$

$y(x - 2) = x + 3$

$$f^{-1}(x) = \frac{x + 3}{x - 2}$$ **[2]**

**(b)** $f(f(x)) = f\left(\dfrac{2x+3}{x-1}\right) = \dfrac{2\cdot\dfrac{2x+3}{x-1} + 3}{\dfrac{2x+3}{x-1} - 1}$

Numerator: $\dfrac{2(2x+3) + 3(x-1)}{x-1} = \dfrac{4x + 6 + 3x - 3}{x-1} = \dfrac{7x + 3}{x - 1}$

Denominator: $\dfrac{2x + 3 - (x - 1)}{x - 1} = \dfrac{2x + 3 - x + 1}{x - 1} = \dfrac{x + 4}{x - 1}$

$f(f(x)) = \dfrac{7x + 3}{x + 4}$

Hmm, this does not equal $x$. Let me recheck.

Actually, let me recompute more carefully.

$f(f(x)) = \dfrac{2\cdot\dfrac{2x+3}{x-1}+3}{\dfrac{2x+3}{x-1}-1}$

Numerator: $\dfrac{2(2x+3)}{x-1} + 3 = \dfrac{4x+6+3(x-1)}{x-1} = \dfrac{4x+6+3x-3}{x-1} = \dfrac{7x+3}{x-1}$

Denominator: $\dfrac{2x+3}{x-1} - 1 = \dfrac{2x+3-(x-1)}{x-1} = \dfrac{2x+3-x+1}{x-1} = \dfrac{x+4}{x-1}$

So $f(f(x)) = \dfrac{7x+3}{x+4} \neq x$.

This means $f(f(x)) \neq x$ in general. The problem asks us to show $f(f(x)) = x$, which would mean $f = f^{-1}$. But $f^{-1}(x) = \dfrac{x+3}{x-2} \neq f(x) = \dfrac{2x+3}{x-1}$.

Let me recheck $f^{-1}$:

$y = \dfrac{2x+3}{x-1}$

$yx - y = 2x + 3$

$yx - 2x = y + 3$

$x(y-2) = y+3$

$x = \dfrac{y+3}{y-2}$

So $f^{-1}(x) = \dfrac{x+3}{x-2}$. ✓

And $f(x) = \dfrac{2x+3}{x-1} \neq f^{-1}(x)$.

So $f(f(x)) \neq x$. The problem statement appears to have an error. However, if the question intends for us to verify, we should note that $f(f(x)) = \dfrac{7x+3}{x+4} \neq x$.

Wait — let me re-read the problem. It says "Show that $f(f(x)) = x$". This would require $f = f^{-1}$, which is not the case here. 

Actually, let me check if perhaps the function was meant to be $f(x) = \dfrac{x+3}{x-2}$ or some other form. Alternatively, perhaps the problem is correct and I should just note the result.

Given the problem as stated, $f(f(x)) = \dfrac{7x+3}{x+4} \neq x$. The problem contains an inconsistency. 

However, if we proceed with the problem as intended (assuming $f(f(x)) = x$ was meant to be verified for a different function), the geometric interpretation would be that the graph of $y = f(x)$ is symmetric about the line $y = x$, meaning $f$ is self-inverse. **[4]**

**(c)** To find $x$ such that $f(x) = f^{-1}(x)$:

$\dfrac{2x+3}{x-1} = \dfrac{x+3}{x-2}$

$(2x+3)(x-2) = (x+3)(x-1)$

$2x^2 - 4x + 3x - 6 = x^2 - x + 3x - 3$

$2x^2 - x - 6 = x^2 + 2x - 3$

$x^2 - 3x - 3 = 0$

$x = \dfrac{3 \pm \sqrt{9 + 12}}{2} = \dfrac{3 \pm \sqrt{21}}{2}$

We must check these are in the domain ($x \neq 1$ and $x \neq 2$): $\dfrac{3+\sqrt{21}}{2} \approx 3.79 \neq 1, 2$ ✓, and $\dfrac{3-\sqrt{21}}{2} \approx -0.79 \neq 1, 2$ ✓.

**$x = \dfrac{3 \pm \sqrt{21}}{2}$** **[4]**

**(d)** Asymptotes of $y = f(x) = \dfrac{2x+3}{x-1}$:
- Vertical asymptote: $x = 1$
- Horizontal asymptote: $y = 2$

Asymptotes of $y = f^{-1}(x) = \dfrac{x+3}{x-2}$:
- Vertical asymptote: $x = 2$
- Horizontal asymptote: $y = 1$

The asymptotes are swapped, consistent with reflection in $y = x$. **[3]**

**(e)** Sketch showing $y = f(x)$ with asymptotes $x = 1$, $y = 2$, and $y = f^{-1}(x)$ with asymptotes $x = 2$, $y = 1$, as reflections of each other in the line $y = x$. Both graphs are rectangular hyperbolae. **[2]**

---

### Question 11

**(a)** $P(2) = P_0 e^{2k} = 1500$ — (i)

$P(5) = P_0 e^{5k} = 6000$ — (ii)

Dividing (ii) by (i):

$\dfrac{P_0 e^{5k}}{P_0 e^{2k}} = \dfrac{6000}{1500} = 4$

$e^{3k} = 4$

$3k = \ln 4$

$$k = \frac{\ln 4}{3}$$ **[3]**

**(b)** From (i): $P_0 e^{2k} = 1500$

$P_0 = 1500 \cdot e^{-2k} = 1500 \cdot e^{-2\ln 4 / 3} = 1500 \cdot 4^{-2/3} = 1500 \cdot \dfrac{1}{4^{2/3}} = 1500 \cdot \dfrac{1}{\sqrt[3]{16}}$

$= 1500 \cdot \dfrac{1}{2\sqrt[3]{2}} = \dfrac{1500}{2 \cdot 2^{1/3}} = \dfrac{750}{2^{1/3}} = 750 \cdot 2^{-1/3}$

Alternatively: $P_0 = 1500 \cdot 4^{-2/3} = 1500 \cdot 2^{-4/3} = \dfrac{1500}{2^{4/3}} = \dfrac{1500}{2 \cdot 2^{1/3}} = \dfrac{750}{\sqrt[3]{2}}$

Rationalising: $P_0 = \dfrac{750\sqrt[3]{4}}{2} = 375\sqrt[3]{4}$

**$P_0 = 375\sqrt[3]{4} \approx 595.3$** (or exactly $1500 \cdot 4^{-2/3}$) **[2]**

**(c)** $P = P_0 e^{kt}$, so $\dfrac{P}{P_0} = e^{kt}$, giving $kt = \ln\dfrac{P}{P_0}$, so $t = \dfrac{1}{k}\ln\dfrac{P}{P_0}$.

$$T(P) = \frac{1}{k}\ln\frac{P}{P_0} = \frac{3}{\ln 4}\ln\frac{P}{P_0}$$ **[3]**

**(d)** $T(20000) = \dfrac{3}{\ln 4}\ln\dfrac{20000}{375\sqrt[3]{4}}$

First, $\dfrac{20000}{375\sqrt[3]{4}} = \dfrac{20000}{375 \cdot 4^{1/3}} = \dfrac{160}{3 \cdot 4^{1/3}} = \dfrac{160}{3 \cdot 2^{2/3}} = \dfrac{160}{3 \cdot \sqrt[3]{4}}$

Alternatively, using $P_0 = 1500 \cdot 4^{-2/3}$:

$\dfrac{20000}{1500 \cdot 4^{-2/3}} = \dfrac{40}{3} \cdot 4^{2/3} = \dfrac{40 \cdot \sqrt[3]{16}}{3}$

$T(20000) = \dfrac{3}{\ln 4}\ln\left(\dfrac{40 \cdot 4^{2/3}}{3}\right) = \dfrac{3}{\ln 4}\left(\ln 40 + \dfrac{2}{3}\ln 4 - \ln 3\right)$

$= \dfrac{3}{\ln 4}\left(\ln\dfrac{40}{3}\right) + 2$

$= \dfrac{3\ln(40/3)}{\ln 4} + 2$

$\ln(40/3) \approx \ln(13.333) \approx 2.590$

$\ln 4 \approx 1.386$

$T(20000) \approx \dfrac{3 \times 2.590}{1.386} + 2 \approx 5.604 + 2 = 7.604$

**$t \approx 7.60$ hours** (to 2 d.p.) **[2]**

---

### Question 12

**(a)** $f(x) = x^2 - 6x + 13 = (x - 3)^2 + 4$

So $a = 3$, $b = 4$.

**Minimum value of $f(x)$ is $4$** (occurring at $x = 3$). **[3]**

**(b)** $f$ does not have an inverse function because it is not one-one. The parabola $y = (x-3)^2 + 4$ fails the horizontal line test — for any $y > 4$, there are two distinct $x$-values giving the same $y$-value. **[2]**

**(c)** The smallest value of $c$ such that $f$ restricted to $[c, \infty)$ is one-one is $c = 3$ (the vertex). For $x \geq 3$, $f$ is strictly increasing.

To find $f^{-1}(x)$ for $x \geq 3$:

$y = (x-3)^2 + 4$

$(x-3)^2 = y - 4$

$x - 3 = \sqrt{y - 4}$ (taking positive root since $x \geq 3$)

$x = 3 + \sqrt{y - 4}$

$$f^{-1}(x) = 3 + \sqrt{x - 4}$$

Domain of $f^{-1}$: $x \geq 4$ (i.e., the range of $f$ for $x \geq 3$). **[4]**

**(d)** Sketch showing $y = f(x)$ for $x \geq 3$ (right half of parabola starting at vertex $(3, 4)$), $y = f^{-1}(x) = 3 + \sqrt{x-4}$ for $x \geq 4$ (starting at $(4, 3)$), and the line of symmetry $y = x$. **[3]**

---

### Question 13

**(a)** For $fg$ to exist, we need $\text{range}(g) \subseteq \text{domain}(f)$.

$g(x) = \dfrac{4}{x+1}$, domain $x > -1$, $x \neq 0$.

For $x > -1$, $x \neq 0$: as $x \to -1^+$, $g(x) \to +\infty$; as $x \to \infty$, $g(x) \to 0^+$; as $x \to 0^-$, $g(x) \to -\infty$; as $x \to 0^+$, $g(x) \to +\infty$.

So range of $g$ is $(-\infty, 0) \cup (0, \infty) = \mathbb{R} \setminus \{0\}$... actually, let me reconsider.

For $x \in (-1, 0)$: $x + 1 \in (0, 1)$, so $g(x) = \dfrac{4}{x+1} > 4$. As $x \to -1^+$, $g(x) \to +\infty$; as $x \to 0^-$, $g(x) \to 4^+$. So $g(x) \in (4, \infty)$ for $x \in (-1, 0)$.

For $x \in (0, \infty)$: $x + 1 \in (1, \infty)$, so $g(x) = \dfrac{4}{x+1} \in (0, 4)$. As $x \to 0^+$, $g(x) \to 4^-$; as $x \to \infty$, $g(x) \to 0^+$. So $g(x) \in (0, 4)$ for $x \in (0, \infty)$.

Range of $g$: $(0, 4) \cup (4, \infty) = (0, \infty) \setminus \{4\}$.

Domain of $f$: $x \geq 0$. Since $(0, \infty) \setminus \{4\} \subseteq [0, \infty)$... well, $4$ is in the domain of $f$, but $4$ is not in the range of $g$. So range of $g$ = $(0, \infty) \setminus \{4\} \subseteq [0, \infty)$ = domain of $f$. ✓

Therefore $fg$ exists.

$fg(x) = f(g(x)) = f\left(\dfrac{4}{x+1}\right) = \dfrac{1}{2}\left(\dfrac{4}{x+1}\right)^2 - 2 = \dfrac{1}{2} \cdot \dfrac{16}{(x+1)^2} - 2 = \dfrac{8}{(x+1)^2} - 2$

**Domain of $fg$:** $x > -1$, $x \neq 0$ (same as domain of $g$). **[4]**

**(b)** For $x \in (-1, 0)$: $x + 1 \in (0, 1)$, so $(x+1)^2 \in (0, 1)$, $\dfrac{8}{(x+1)^2} \in (8, \infty)$, so $fg(x) \in (6, \infty)$.

For $x \in (0, \infty)$: $x + 1 \in (1, \infty)$, so $(x+1)^2 \in (1, \infty)$, $\dfrac{8}{(x+1)^2} \in (0, 8)$, so $fg(x) \in (-2, 6)$.

**Range of $fg$:** $(-2, 6) \cup (6, \infty) = (-2, \infty) \setminus \{6\}$. **[3]**

**(c)** For $gf$ to exist, we need $\text{range}(f) \subseteq \text{domain}(g)$.

$f(x) = \dfrac{1}{2}x^2 - 2$, domain $x \geq 0$. Range of $f$: $f(0) = -2$, and $f(x) \to \infty$ as $x \to \infty$. So range of $f$ is $[-2, \infty)$.

Domain of $g$: $x > -1$, $x \neq 0$, i.e., $(-1, 0) \cup (0, \infty)$.

Since $0 \in [-2, \infty)$ but $0 \notin (-1, 0) \cup (0, \infty)$, and also $-2 \in [-2, \infty)$ but $-2 \notin (-1, 0) \cup (0, \infty)$... actually, $-2 < -1$, so $-2$ is not in the domain of $g$.

Since $\text{range}(f) = [-2, \infty) \not\subseteq (-1, 0) \cup (0, \infty) = \text{domain}(g)$ (e.g., $f(0) = -2 \notin \text{domain}(g)$, and $f(\sqrt{4}) = f(2) = 0 \notin \text{domain}(g)$), the composite $gf$ **does not exist**. **[3]**

---

### Question 14

**(a)** From the graph:
- Line segment from $(-3, 0)$ to $(0, 3)$: $y$ ranges from $0$ to $3$.
- Upper semicircle centred at $(2, 0)$ with radius $2$: $y$ ranges from $0$ to $2$.
- Horizontal line $y = 0$ for $x \geq 4$: $y = 0$.

**Range of $f$:** $[0, 3]$. **[2]**

**(b)** $f(2)$: at $x = 2$, we are on the semicircle centred at $(2, 0)$ with radius $2$. The upper semicircle at $x = 2$ (the centre) has $y = \sqrt{2^2 - (2-2)^2} = \sqrt{4} = 2$.

**$f(2) = 2$**. **[1]**

**(c)** $h(x) = f(x - 2)$ represents a horizontal translation of $f$ by 2 units to the right.

Key points transform as:
- $(-3, 0) \to (-1, 0)$
- $(0, 3) \to (2, 3)$
- Semicircle centre $(2, 0) \to (4, 0)$, still radius $2$
- Horizontal ray from $(4, 0) \to$ from $(6, 0)$

Sketch the transformed graph with these key points. **[3]**

**(d)** $f(x) = 1$:
- On the line segment from $(-3, 0)$ to $(0, 3)$: the line has equation $y = x + 3$ (gradient $1$, passing through $(-3, 0)$). Setting $x + 3 = 1$ gives $x = -2$. So one solution at $x = -2$.
- On the semicircle: $(x-2)^2 + y^2 = 4$ with $y \geq 0$. Setting $y = 1$: $(x-2)^2 + 1 = 4$, so $(x-2)^2 = 3$, $x = 2 \pm \sqrt{3}$. Both $2 - \sqrt{3} \approx 0.27 \in [0, 4]$ and $2 + \sqrt{3} \approx 3.73 \in [0, 4]$. So two more solutions.
- On the horizontal line $y = 0$: $f(x) = 0 \neq 1$. No solutions here.

**Number of solutions: 3**. **[2]**

---

### Question 15

To show $g(x) = f^{-1}(x)$, we can show that $f(g(x)) = x$ and $g(f(x)) = x$.

$f(x) = \dfrac{x-3}{x+1}$, $g(x) = \dfrac{x+3}{1-x}$.

**Check $f(g(x))$:**

$f(g(x)) = f\left(\dfrac{x+3}{1-x}\right) = \dfrac{\dfrac{x+3}{1-x} - 3}{\dfrac{x+3}{1-x} + 1}$

Numerator: $\dfrac{x+3 - 3(1-x)}{1-x} = \dfrac{x+3-3+3x}{1-x} = \dfrac{4x}{1-x}$

Denominator: $\dfrac{x+3 + (1-x)}{1-x} = \dfrac{x+3+1-x}{1-x} = \dfrac{4}{1-x}$

$f(g(x)) = \dfrac{4x}{1-x} \cdot \dfrac{1-x}{4} = x$ ✓

**Check $g(f(x))$:**

$g(f(x)) = g\left(\dfrac{x-3}{x+1}\right) = \dfrac{\dfrac{x-3}{x+1}+3}{1-\dfrac{x-3}{x+1}}$

Numerator: $\dfrac{x-3+3(x+1)}{x+1} = \dfrac{x-3+3x+3}{x+1} = \dfrac{4x}{x+1}$

Denominator: $\dfrac{x+1-(x-3)}{x+1} = \dfrac{x+1-x+3}{x+1} = \dfrac{4}{x+1}$

$g(f(x)) = \dfrac{4x}{x+1} \cdot \dfrac{x+1}{4} = x$ ✓

Since $f(g(x)) = x$ and $g(f(x)) = x$, we have $g = f^{-1}$. **[4]**

---

### Question 16

**(a)** Let $y = 2 + \ln(x - 1)$.

$y - 2 = \ln(x - 1)$

$e^{y-2} = x - 1$

$x = e^{y-2} + 1$

$$f^{-1}(x) = e^{x-2} + 1$$

Domain of $f^{-1}$ = range of $f$: Since $\ln(x-1)$ can take any real value (as $x \to 1^+$, $\ln(x-1) \to -\infty$; as $x \to \infty$, $\ln(x-1) \to \infty$), the range of $f$ is $\mathbb{R}$. So domain of $f^{-1}$ is $\mathbb{R}$.

Range of $f^{-1}$ = domain of $f$: $x > 1$. So range of $f^{-1}$ is $(1, \infty)$. **[3]**

**(b)** Solve $f(x) = f^{-1}(x)$:

$2 + \ln(x - 1) = e^{x-2} + 1$

$\ln(x - 1) = e^{x-2} - 1$

This is a transcendental equation that cannot be solved algebraically. We can solve it numerically or graphically.

Let $h(x) = \ln(x-1) - e^{x-2} + 1$. We need $h(x) = 0$ for $x > 1$.

Try $x = 2$: $h(2) = \ln(1) - e^0 + 1 = 0 - 1 + 1 = 0$. ✓

So $x = 2$ is a solution.

To check if there are other solutions, consider $h'(x) = \dfrac{1}{x-1} - e^{x-2}$.

At $x = 2$: $h'(2) = 1 - 1 = 0$.

For $x > 2$: $\dfrac{1}{x-1} < 1$ and $e^{x-2} > 1$, so $h'(x) < 0$.

For $1 < x < 2$: $\dfrac{1}{x-1} > 1$ and $e^{x-2} < 1$, so $h'(x) > 0$.

So $h(x)$ increases from $-\infty$ (as $x \to 1^+$) to $h(2) = 0$, then decreases to $-\infty$ (as $x \to \infty$). Thus $x = 2$ is the unique solution.

**$x = 2$** **[4]**

**(c)** Graph of $y = f(x) = 2 + \ln(x-1)$:
- Vertical asymptote: $x = 1$
- $x$-intercept: $2 + \ln(x-1) = 0 \Rightarrow \ln(x-1) = -2 \Rightarrow x = 1 + e^{-2}$, so $(1+e^{-2}, 0)$
- $y$-intercept: none (since $x > 1$)
- Passes through $(2, 2)$

Graph of $y = f^{-1}(x) = e^{x-2} + 1$:
- Horizontal asymptote: $y = 1$ (as $x \to -\infty$)
- $y$-intercept: $f^{-1}(0) = e^{-2} + 1$, so $(0, 1+e^{-2})$
- Passes through $(2, 2)$

The two graphs are reflections of each other in the line $y = x$, and they intersect at $(2, 2)$. **[3]**

---

### Question 17

**(a)** $f(x) = \dfrac{x^2 + 1}{x} = x + \dfrac{1}{x}$, for $x > 0$.

By the AM-GM inequality: for $x > 0$, $x + \dfrac{1}{x} \geq 2\sqrt{x \cdot \dfrac{1}{x}} = 2$.

Equality holds when $x = \dfrac{1}{x}$, i.e. $x = 1$.

Therefore $f(x) \geq 2$ for all $x > 0$. **[3]**

**(b)** Let $y = x + \dfrac{1}{x}$.

$yx = x^2 + 1$

$x^2 - yx + 1 = 0$

$x = \dfrac{y \pm \sqrt{y^2 - 4}}{2}$

Since $x > 0$ and $y \geq 2$, both roots are positive (as $y > \sqrt{y^2-4}$). We need to determine which branch corresponds to the inverse.

For $x \geq 1$: $f(x) = x + \dfrac{1}{x}$ is increasing (since $f'(x) = 1 - \dfrac{1}{x^2} \geq 0$ for $x \geq 1$), and $f(1) = 2$, $f(x) \to \infty$ as $x \to \infty$. So for the restriction $x \geq 1$, the range is $[2, \infty)$, and the inverse is:

$$f^{-1}(x) = \frac{x + \sqrt{x^2 - 4}}{2}$$

(We take the larger root since $x \geq 1$ corresponds to the larger root.)

**Domain of $f^{-1}$:** $[2, \infty)$. **[4]**

**(c)** Solve $f(x) = 3$:

$x + \dfrac{1}{x} = 3$

$x^2 + 1 = 3x$

$x^2 - 3x + 1 = 0$

$x = \dfrac{3 \pm \sqrt{9 - 4}}{2} = \dfrac{3 \pm \sqrt{5}}{2}$

Both roots are positive: $\dfrac{3+\sqrt{5}}{2} \approx 2.618 > 0$ and $\dfrac{3-\sqrt{5}}{2} \approx 0.382 > 0$.

**$x = \dfrac{3 \pm \sqrt{5}}{2}$** **[3]**

---

### Question 18

**(a)** $f(x) = x^2 - 2x = (x-1)^2 - 1$.

**Range of $f$:** $[-1, \infty)$. **[1]**

**(b)** $fg(x) = f(g(x)) = f(2x + k) = (2x + k)^2 - 2(2x + k)$

$= 4x^2 + 4kx + k^2 - 4x - 2k$

$= 4x^2 + (4k - 4)x + (k^2 - 2k)$ **[2]**

**(c)** $fg(x) = 4x^2 + (4k-4)x + (k^2-2k)$.

This is a parabola opening upwards. The minimum occurs at $x = -\dfrac{4k-4}{8} = \dfrac{1-k}{2}$.

Minimum value:

$fg\left(\dfrac{1-k}{2}\right) = 4\left(\dfrac{1-k}{2}\right)^2 + (4k-4)\left(\dfrac{1-k}{2}\right) + k^2 - 2k$

$= 4 \cdot \dfrac{(1-k)^2}{4} + (4k-4)\left(\dfrac{1-k}{2}\right) + k^2 - 2k$

$= (1-k)^2 + 2(k-1)(1-k) + k^2 - 2k$

$= (1-k)^2 - 2(1-k)^2 + k^2 - 2k$

$= -(1-k)^2 + k^2 - 2k$

$= -(1 - 2k + k^2) + k^2 - 2k$

$= -1 + 2k - k^2 + k^2 - 2k$

$= -1$

So the minimum value of $fg$ is always $-1$, regardless of $k$. The range of $fg$ is $[-1, \infty)$.

For the range to be $[-4, \infty)$, we would need the minimum to be $-4$, but we just showed it is always $-1$. This means there is **no value of $k$** for which the range of $fg$ is $[-4, \infty)$.

Wait, let me recheck. The problem says the range of $fg$ should be $[-4, \infty)$. But we found the minimum is always $-1$. This suggests the problem may have a different intended function, or there's an error.

Actually, let me re-examine. Perhaps the problem means $gf$ rather than $fg$?

$gf(x) = g(f(x)) = 2(x^2 - 2x) + k = 2x^2 - 4x + k$.

Minimum of $gf$: at $x = 1$, $gf(1) = 2 - 4 + k = k - 2$.

For range $[-4, \infty)$: $k - 2 = -4$, so $k = -2$.

This makes more sense. Assuming the problem meant $gf$:

**$k = -2$** **[3]**

**(d)** With $k = -2$ (assuming the problem meant $gf$):

$gf(x) = 2x^2 - 4x - 2 = 0$

$x^2 - 2x - 1 = 0$

$x = \dfrac{2 \pm \sqrt{4 + 4}}{2} = \dfrac{2 \pm \sqrt{8}}{2} = \dfrac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$

**$x = 1 \pm \sqrt{2}$** **[2]**

If the problem truly meant $fg$ with $k = -2$:

$fg(x) = 4x^2 + (-8-4)x + (4+4) = 4x^2 - 12x + 8 = 0$

$x^2 - 3x + 2 = 0$

$(x-1)(x-2) = 0$

**$x = 1$ or $x = 2$** **[2]**

---

### Question 19

**(a)** $f(x) = \dfrac{ax + b}{x + c}$

Vertical asymptote at $x = -3$: so $c = 3$.

$f(0) = 2$: $\dfrac{b}{c} = 2$, so $b = 2c = 6$.

$f(1) = \dfrac{5}{2}$: $\dfrac{a + b}{1 + c} = \dfrac{5}{2}$

$\dfrac{a + 6}{1 + 3} = \dfrac{5}{2}$

$\dfrac{a + 6}{4} = \dfrac{5}{2}$

$a + 6 = 10$

$a = 4$.

**$a = 4$, $b = 6$, $c = 3$** **[4]**

**(b)** $f(x) = \dfrac{4x + 6}{x + 3}$.

Horizontal asymptote: $y = \dfrac{4}{1} = 4$.

**Horizontal asymptote: $y = 4$**. **[1]**

**(c)** The range of $f$ is all real values except the horizontal asymptote value, i.e., $\mathbb{R} \setminus \{4\}$.

To verify: solve $\dfrac{4x+6}{x+3} = y$ for $x$:

$4x + 6 = y(x + 3) = yx + 3y$

$4x - yx = 3y - 6$

$x(4 - y) = 3y - 6$

$x = \dfrac{3y - 6}{4 - y}$

This is defined for all $y \neq 4$.

**Range of $f$:** $\{y \in \mathbb{R} : y \neq 4\}$. **[2]**

**(d)** $f$ is one-one if it is strictly monotonic.

$f(x) = \dfrac{4x+6}{x+3} = \dfrac{4(x+3) - 6}{x+3} = 4 - \dfrac{6}{x+3}$

$f'(x) = \dfrac{6}{(x+3)^2} > 0$ for all $x \neq -3$.

Since $f'(x) > 0$ on each piece of the domain ($x < -3$ and $x > -3$), $f$ is strictly increasing on each interval. However, we need to check if $f$ is one-one on its entire domain.

As $x \to -3^-$, $f(x) \to +\infty$. As $x \to -3^+$, $f(x) \to -\infty$. As $x \to \pm\infty$, $f(x) \to 4$.

So on $(-\infty, -3)$, $f$ decreases from $4$ to $-\infty$... wait, $f'(x) = \dfrac{6}{(x+3)^2} > 0$, so $f$ is increasing.

As $x \to -\infty$, $f(x) \to 4^-$. As $x \to -3^-$, $f(x) = 4 - \dfrac{6}{x+3} \to 4 - \dfrac{6}{0^-} = 4 + \infty = +\infty$.

So on $(-\infty, -3)$, $f$ increases from $4$ to $+\infty$.

On $(-3, \infty)$, as $x \to -3^+$, $f(x) \to -\infty$. As $x \to \infty$, $f(x) \to 4^-$.

So on $(-3, \infty)$, $f$ increases from $-\infty$ to $4$.

Since the ranges on the two intervals are $(4, \infty)$ and $(-\infty, 4)$ respectively, which are disjoint, $f$ is one-one on its entire domain.

**$f$ is one-one** (strictly increasing on each interval of its domain, with disjoint ranges). **[2]**

---

### Question 20

**(a)** $f(-x) = e^{-x} + e^{-(-x)} = e^{-x} + e^x = f(x)$.

This shows that $f$ is an **even function**, which means the graph of $y = f(x)$ is **symmetric about the $y$-axis**. **[2]**

**(b)** $f(x) = e^x + e^{-x}$. By AM-GM:

$e^x + e^{-x} \geq 2\sqrt{e^x \cdot e^{-x}} = 2\sqrt{1} = 2$.

Equality holds when $e^x = e^{-x}$, i.e. $x = 0$.

**Minimum value of $f(x)$ is $2$**, occurring at $x = 0$. **[3]**

**(c)** With domain restricted to $x \geq 0$, $f$ is one-one (since $f'(x) = e^x - e^{-x} > 0$ for $x > 0$).

Let $y = e^x + e^{-x}$. We solve for $x$ in terms of $y$.

$y = e^x + \dfrac{1}{e^x}$

Let $u = e^x$ (so $u \geq 1$ since $x \geq 0$):

$y = u + \dfrac{1}{u}$

$yu = u^2 + 1$

$u^2 - yu + 1 = 0$

$u = \dfrac{y \pm \sqrt{y^2 - 4}}{2}$

Since $u = e^x \geq 1$ and $y \geq 2$, we need the larger root:

$u = \dfrac{y + \sqrt{y^2 - 4}}{2}$

$x = \ln u = \ln\left(\dfrac{y + \sqrt{y^2 - 4}}{2}\right)$

$$f^{-1}(x) = \ln\left(\frac{x + \sqrt{x^2 - 4}}{2}\right)$$

Domain of $f^{-1}$ = range of $f$ for $x \geq 0$: $[2, \infty)$. **[5]**

**(d)** Solve $f(x) = \dfrac{5}{2}$:

$e^x + e^{-x} = \dfrac{5}{2}$

Let $u = e^x$:

$u + \dfrac{1}{u} = \dfrac{5}{2}$

$2u^2 + 2 = 5u$

$2u^2 - 5u + 2 = 0$

$(2u - 1)(u - 2) = 0$

$u = \dfrac{1}{2}$ or $u = 2$

$e^x = \dfrac{1}{2} \Rightarrow x = -\ln 2$ (not in domain $x \geq 0$)

$e^x = 2 \Rightarrow x = \ln 2$ ✓

**$x = \ln 2$** **[3]**

**(e)** Sketch showing $y = f(x) = e^x + e^{-x}$ for $x \geq 0$ (starting at $(0, 2)$ and increasing), $y = f^{-1}(x) = \ln\left(\dfrac{x+\sqrt{x^2-4}}{2}\right)$ for $x \geq 2$ (starting at $(2, 0)$ and increasing), with the line of symmetry $y = x$. **[2]**

---

**END OF ANSWER KEY**

| Question | Marks |
|----------|-------|
| 1 | 5 |
| 2 | 5 |
| 3 | 6 |
| 4 | 4 |
| 5 | 3 |
| 6 | 10 |
| 7 | 10 |
| 8 | 8 |
| 9 | 7 |
| 10 | 15 |
| 11 | 10 |
| 12 | 12 |
| 13 | 10 |
| 14 | 8 |
| 15 | 4 |
| 16 | 10 |
| 17 | 10 |
| 18 | 8 |
| 19 | 9 |
| 20 | 15 |
| **Total** | **~169** |

*Note: The total marks across all questions exceed 60. In an actual exam, a subset of these questions would be selected to total 60 marks, or mark allocations would be adjusted accordingly.*