A Level H1 Mathematics Algebra Functions Quiz

A-Level Maths H1 Quiz - Algebra Functions

Answer Key and Teaching Notes

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Question 1 [4 marks]

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

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

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

$xy + x = 3y - 2$

$xy - 3y = -2 - x$

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

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

Teaching note: To find an inverse, swap $x$ and $y$, then rearrange to make $y$ the subject. The last step factors out $-1$ from numerator and denominator to give a cleaner form.

(b) The domain of $f^{-1}$ is the range of $f$. Since $f(x) = \dfrac{3x-2}{x+1}$, the horizontal asymptote is $y = 3$ (ratio of leading coefficients). So the range of $f$ is all real $y \neq 3$.

$\text{Domain of } f^{-1}: x \in \mathbb{R}, x \neq 3$

[Marking: 2 marks for correct inverse, 2 marks for correct domain]

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Question 2 [6 marks]

(a) $f(x) = x^2 - 4x + 7 = (x - 2)^2 + 3$

Since $(x - 2)^2 \geq 0$ for all real $x$, the minimum value is $3$.

$\text{Range of } f: f(x) \geq 3$

(b) $g(x) = \sqrt{x - 3}$, domain $x \geq 3$.

When $x = 3$, $g(3) = 0$. As $x$ increases, $g(x)$ increases without bound.

$\text{Range of } g: g(x) \geq 0$

(c) For $fg$ to exist, the range of $g$ must be a subset of the domain of $f$.

Range of $g$ is $[0, \infty)$ and domain of $f$ is $\mathbb{R}$. Since $[0, \infty) \subset \mathbb{R}$, the composite $fg$ exists.

$fg(x) = f(g(x)) = f(\sqrt{x - 3}) = (\sqrt{x - 3})^2 - 4\sqrt{x - 3} + 7$

$fg(x) = x - 3 - 4\sqrt{x - 3} + 7 = x + 4 - 4\sqrt{x - 3}$

Domain of $fg$: $x \geq 3$ (same as domain of $g$).

[Marking: 1 mark for range of f, 1 mark for range of g, 1 mark for explaining existence, 2 marks for fg(x), 1 mark for domain]

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Question 3 [5 marks]

(a) Let $y = \dfrac{5}{2x - 3}$.

Swap: $x = \dfrac{5}{2y - 3}$

$x(2y - 3) = 5$

$2xy - 3x = 5$

$2xy = 5 + 3x$

$h^{-1}(x) = \dfrac{5 + 3x}{2x}$

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

$\dfrac{5}{2x - 3} = \dfrac{5 + 3x}{2x}$

Cross-multiply: $5 \cdot 2x = (5 + 3x)(2x - 3)$

$10x = 10x - 15 + 6x^2 - 9x$

$10x = 6x^2 + x - 15$

$0 = 6x^2 - 9x - 15$

$0 = 2x^2 - 3x - 5$

$(2x - 5)(x + 1) = 0$

$x = \dfrac{5}{2} \text{ or } x = -1$

Check: $x = \dfrac{5}{2}$ makes $h(x)$ undefined (denominator $2(\frac{5}{2}) - 3 = 2 \neq 0$, so it's valid). Actually $2(\frac{5}{2}) - 3 = 5 - 3 = 2 \neq 0$, so $x = \frac{5}{2}$ is valid.

$x = \dfrac{5}{2} \text{ or } x = -1$

[Marking: 3 marks for inverse, 2 marks for solving h(x) = h⁻¹(x)]

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Question 4 [6 marks]

(a) $f(x) = x^2 - 6x + 5 = (x - 3)^2 - 4$

The vertex is at $x = 3$. For $f^{-1}$ to exist, $f$ must be one-to-one, so we restrict to one side of the vertex.

$k = 3$

(b) With $k = 3$: $f(x) = (x - 3)^2 - 4$, domain $x \geq 3$.

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

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

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

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

Domain of $f^{-1}$: Since range of $f$ is $[-4, \infty)$, domain of $f^{-1}$ is $x \geq -4$.

Range of $f^{-1}$: Since domain of $f$ is $[3, \infty)$, range of $f^{-1}$ is $y \geq 3$.

[Marking: 1 mark for k, 2 marks for f⁻¹(x), 2 marks for domain, 1 mark for range]

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Question 5 [6 marks]

(a) $f$ is undefined at $x = -2$, so the denominator $x + c = 0$ when $x = -2$, giving $c = 2$.

$f(1) = 2$: $\dfrac{a(1) + b}{1 + 2} = 2 \Rightarrow a + b = 6$ ... (i)

$f(4) = 1$: $\dfrac{4a + b}{4 + 2} = 1 \Rightarrow 4a + b = 6$ ... (ii)

Subtract (i) from (ii): $3a = 0 \Rightarrow a = 0$

From (i): $0 + b = 6 \Rightarrow b = 6$

$a = 0, \quad b = 6, \quad c = 2$

So $f(x) = \dfrac{6}{x + 2}$.

(b) Let $y = \dfrac{6}{x + 2}$.

Swap: $x = \dfrac{6}{y + 2}$

$x(y + 2) = 6$

$xy + 2x = 6$

$xy = 6 - 2x$

$f^{-1}(x) = \dfrac{6 - 2x}{x}$

[Marking: 3 marks for a, b, c, 3 marks for f⁻¹(x)]

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Question 6 [6 marks]

(a) $y = f(x + 2)$: Translation of $y = f(x)$ by 2 units in the negative $x$-direction.

Points become: $(-3, 0)$, $(-2, -3)$, $(0, 5)$, minimum at $(-1, -4)$.

(b) $y = 2f(x)$: Stretch parallel to the $y$-direction, scale factor 2.

Points become: $(-1, 0)$, $(0, -6)$, $(2, 10)$, minimum at $(1, -8)$.

(c) $y = |f(x)|$: Reflect any negative parts of the graph in the $x$-axis.

The point $(0, -3)$ becomes $(0, 3)$ and the minimum $(1, -4)$ becomes $(1, 4)$. Points $(-1, 0)$ and $(2, 5)$ remain unchanged. The graph touches the $x$-axis at $x = -1$ and has a V-shaped turning point at $(1, 4)$.

[Marking: 2 marks each — 1 for correct shape, 1 for correct coordinates of key points]

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Question 7 [5 marks]

(a) $f(-1) = (-1)^2 - 1 = 1 - 1 = 0$ (using $x < 2$ branch)

$f(2) = 3(2) - 1 = 5$ (using $x \geq 2$ branch)

$f(3) = 3(3) - 1 = 8$ (using $x \geq 2$ branch)

(b) For $x < 2$: parabola $y = x^2 - 1$, vertex at $(0, -1)$, passing through $(-1, 0)$, approaching $(2, 3)$ from the left (open circle at $(2, 3)$).

For $x \geq 2$: line $y = 3x - 1$, starting at closed circle $(2, 5)$, passing through $(3, 8)$, $(4, 11)$.

(c) For $x < 2$: $x^2 - 1 \geq -1$, so range is $[-1, 3)$ (approaching but not reaching 3).

For $x \geq 2$: $3x - 1 \geq 5$, so range is $[5, \infty)$.

$\text{Range of } f: [-1, 3) \cup [5, \infty)$

[Marking: 1 mark for values, 2 marks for sketch, 2 marks for range]

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Question 8 [5 marks]

(a) Vertical asymptote: $x = 1$; Horizontal asymptote: $y = 2$.

(b) Domain: $x \in \mathbb{R}, x \neq 1$; Range: $y \in \mathbb{R}, y \neq 2$.

(c) The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$.

Asymptotes swap: vertical asymptote becomes $y = 2$ (horizontal), horizontal asymptote becomes $x = 1$ (vertical).

The curve passes through $(0, 0)$ (unchanged since it lies on $y = x$).

The inverse graph has a horizontal asymptote $y = 2$ and vertical asymptote $x = 1$.

[Marking: 1 mark for asymptotes, 2 marks for domain/range, 2 marks for inverse sketch with correct asymptotes]

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Question 9 [6 marks]

(a) Vertical asymptote $x = -3 \Rightarrow h = -3$.

Horizontal asymptote $y = 4 \Rightarrow k = 4$.

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

Passes through $(0, 6)$: $\dfrac{a}{0 + 3} + 4 = 6 \Rightarrow \dfrac{a}{3} = 2 \Rightarrow a = 6$.

$a = 6, \quad h = -3, \quad k = 4$

(b) Graph: vertical asymptote $x = -3$, horizontal asymptote $y = 4$, passing through $(0, 6)$. Since $a = 6 > 0$, the left branch (below $y = 4$) is in $x < -3$ and the right branch (above $y = 4$) is in $x > -3$.

$y$-intercept: $(0, 6)$. No $x$-intercept since $\dfrac{6}{x+3} + 4 = 0 \Rightarrow x = -\dfrac{9}{2}$, so $x$-intercept at $\left(-\dfrac{9}{2}, 0\right)$.

(c) Set $\dfrac{6}{x + 3} + 4 = x$:

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

$6 + 4x + 12 = x^2 + 3x$

$x^2 + 3x - 4x - 18 = 0$

$x^2 - x - 18 = 0$

$x = \dfrac{1 \pm \sqrt{1 + 72}}{2} = \dfrac{1 \pm \sqrt{73}}{2}$

Points of intersection: $\left(\dfrac{1 + \sqrt{73}}{2}, \dfrac{1 + \sqrt{73}}{2}\right)$ and $\left(\dfrac{1 - \sqrt{73}}{2}, \dfrac{1 - \sqrt{73}}{2}\right)$.

[Marking: 3 marks for a, h, k; 1 mark for sketch; 2 marks for intersection points]

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Question 10 [6 marks]

(a) Vertical asymptote: $x = 1$.

For oblique asymptote, perform polynomial division:

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

Oblique asymptote: $y = x + 1$.

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

$\dfrac{dy}{dx} = 1 - \dfrac{2}{(x-1)^2}$

Set $\dfrac{dy}{dx} = 0$: $(x-1)^2 = 2 \Rightarrow x = 1 \pm \sqrt{2}$

When $x = 1 + \sqrt{2}$: $y = 1 + \sqrt{2} + 1 + \dfrac{2}{\sqrt{2}} = 2 + \sqrt{2} + \sqrt{2} = 2 + 2\sqrt{2}$

When $x = 1 - \sqrt{2}$: $y = 1 - \sqrt{2} + 1 + \dfrac{2}{-\sqrt{2}} = 2 - \sqrt{2} - \sqrt{2} = 2 - 2\sqrt{2}$

Stationary points: $(1 + \sqrt{2}, 2 + 2\sqrt{2})$ and $(1 - \sqrt{2}, 2 - 2\sqrt{2})$.

Second derivative: $\dfrac{d^2y}{dx^2} = \dfrac{4}{(x-1)^3}$

At $x = 1 + \sqrt{2}$: $\dfrac{d^2y}{dx^2} > 0$ → minimum

At $x = 1 - \sqrt{2}$: $\dfrac{d^2y}{dx^2} < 0$ → maximum

(c) Sketch showing vertical asymptote $x = 1$, oblique asymptote $y = x + 1$, maximum at $(1 - \sqrt{2}, 2 - 2\sqrt{2})$, minimum at $(1 + \sqrt{2}, 2 + 2\sqrt{2})$, passing through $(0, -1)$.

[Marking: 2 marks for asymptotes, 3 marks for stationary points with nature, 1 mark for sketch]

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Question 11 [3 marks]

$\dfrac{2x - 3}{x + 4} \leq 1$

$\dfrac{2x - 3}{x + 4} - 1 \leq 0$

$\dfrac{2x - 3 - (x + 4)}{x + 4} \leq 0$

$\dfrac{x - 7}{x + 4} \leq 0$

Critical values: $x = 7$ and $x = -4$.

Sign chart:

Interval	$x - 7$	$x + 4$	$\dfrac{x-7}{x+4}$
$x < -4$	$-$	$-$	$+$
$-4 < x < 7$	$-$	$+$	$-$
$x > 7$	$+$	$+$	$+$

We need $\leq 0$, so $-4 < x \leq 7$.

$x \in (-4, 7]$

Common mistake: Multiplying both sides by $x + 4$ without considering the sign. Always bring everything to one side and use a sign chart.

[Marking: 1 mark for combining fractions, 1 mark for critical values/sign chart, 1 mark for final answer]

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Question 12 [3 marks]

$\dfrac{3x + 5}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$

$3x + 5 = A(x + 2) + B(x - 1)$

Let $x = 1$: $3(1) + 5 = A(3) \Rightarrow A = \dfrac{8}{3}$

Let $x = -2$: $3(-2) + 5 = B(-3) \Rightarrow -1 = -3B \Rightarrow B = \dfrac{1}{3}$

$\dfrac{3x + 5}{(x - 1)(x + 2)} = \dfrac{8/3}{x - 1} + \dfrac{1/3}{x + 2}$

Or equivalently: $\dfrac{1}{3}\left(\dfrac{8}{x-1} + \dfrac{1}{x+2}\right)$

[Marking: 1 mark for setup, 1 mark for each value of A and B]

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Question 13 [3 marks]

$f(x) = x^2 + px + q$ has a minimum at $x = 3$.

$f'(x) = 2x + p$

At $x = 3$: $f'(3) = 6 + p = 0 \Rightarrow p = -6$

Minimum value is $-7$: $f(3) = 9 + 3p + q = -7$

$9 + 3(-6) + q = -7$

$9 - 18 + q = -7$

$q = 2$

$p = -6, \quad q = 2$

Alternative method (completing the square):

$f(x) = (x - 3)^2 - 7 = x^2 - 6x + 9 - 7 = x^2 - 6x + 2$

So $p = -6$, $q = 2$.

[Marking: 1 mark for p, 1 mark for q, 1 mark for method]

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Question 14 [4 marks]

(a) Perform polynomial division of $2x^2 + 3x + 4$ by $x + 1$:

$2x^2 + 3x + 4 = (x+1)(2x + 1) + 3$

Check: $(x+1)(2x+1) = 2x^2 + x + 2x + 1 = 2x^2 + 3x + 1$, remainder $4 - 1 = 3$. ✓

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

So $a = 2$, $b = 1$, $c = 3$.

(b) As $x \to \pm\infty$, $\dfrac{3}{x+1} \to 0$, so the graph approaches the line $y = 2x + 1$.

$\text{Oblique asymptote: } y = 2x + 1$

[Marking: 3 marks for division, 1 mark for asymptote]

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Question 15 [5 marks]

(a) $f(x) = e^{2x}$. Let $y = e^{2x}$.

$\ln y = 2x \Rightarrow x = \dfrac{\ln y}{2}$

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

Domain of $f^{-1}$: $x > 0$ (since $\ln x$ requires $x > 0$).

(b) $g(x) = \ln(x + 1)$. Let $y = \ln(x + 1)$.

$e^y = x + 1 \Rightarrow x = e^y - 1$

$g^{-1}(x) = e^x - 1$

Domain of $g^{-1}$: $x \in \mathbb{R}$ (since $e^x$ is defined for all real $x$).

(c) $fg(x) = f(g(x)) = f(\ln(x+1)) = e^{2\ln(x+1)} = e^{\ln(x+1)^2} = (x+1)^2$

Set $(x + 1)^2 = e^4$:

$x + 1 = \pm e^2$

$x = -1 \pm e^2$

Since $g(x)$ requires $x > -1$, we need $x = -1 + e^2$ (since $-1 - e^2 < -1$).

$x = e^2 - 1$

[Marking: 2 marks for f⁻¹, 2 marks for g⁻¹, 1 mark for solving]

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Question 16 [8 marks]

(a) $y = \dfrac{x^2 - 4x + 6}{x - 2}$

Using the quotient rule: $\dfrac{dy}{dx} = \dfrac{(2x - 4)(x - 2) - (x^2 - 4x + 6)(1)}{(x - 2)^2}$

Numerator: $(2x - 4)(x - 2) - (x^2 - 4x + 6)$

$= 2x^2 - 4x - 4x + 8 - x^2 + 4x - 6$

$= x^2 - 4x + 2$

$\dfrac{dy}{dx} = \dfrac{x^2 - 4x + 2}{(x - 2)^2}$

(b) Set $\dfrac{dy}{dx} = 0$: $x^2 - 4x + 2 = 0$

$x = \dfrac{4 \pm \sqrt{16 - 8}}{2} = \dfrac{4 \pm \sqrt{8}}{2} = \dfrac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}$

When $x = 2 + \sqrt{2}$: $y = \dfrac{(2+\sqrt{2})^2 - 4(2+\sqrt{2}) + 6}{\sqrt{2}} = \dfrac{4 + 4\sqrt{2} + 2 - 8 - 4\sqrt{2} + 6}{\sqrt{2}} = \dfrac{4}{\sqrt{2}} = 2\sqrt{2}$

When $x = 2 - \sqrt{2}$: $y = \dfrac{(2-\sqrt{2})^2 - 4(2-\sqrt{2}) + 6}{-\sqrt{2}} = \dfrac{4 - 4\sqrt{2} + 2 - 8 + 4\sqrt{2} + 6}{-\sqrt{2}} = \dfrac{4}{-\sqrt{2}} = -2\sqrt{2}$

Stationary points: $(2 + \sqrt{2}, 2\sqrt{2})$ and $(2 - \sqrt{2}, -2\sqrt{2})$.

Using the first derivative test or second derivative:

$\dfrac{d^2y}{dx^2}$ evaluated at $x = 2 + \sqrt{2}$ gives a positive value → minimum

At $x = 2 - \sqrt{2}$ gives a negative value → maximum

(c) Vertical asymptote: $x = 2$

Oblique asymptote: $\dfrac{x^2 - 4x + 6}{x - 2} = x - 2 + \dfrac{2}{x - 2}$

Oblique asymptote: $y = x - 2$

[Marking: 2 marks for derivative, 2 marks for stationary points, 2 marks for nature, 2 marks for asymptotes]

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Question 17 [5 marks]

(a) $f(x) = \sqrt{2x + 5}$, domain $x \geq -\dfrac{5}{2}$.

Let $y = \sqrt{2x + 5}$

$y^2 = 2x + 5$

$x = \dfrac{y^2 - 5}{2}$

$f^{-1}(x) = \dfrac{x^2 - 5}{2}$

Domain of $f^{-1}$: Since range of $f$ is $[0, \infty)$, domain of $f^{-1}$ is $x \geq 0$.

Range of $f^{-1}$: Since domain of $f$ is $[-\frac{5}{2}, \infty)$, range of $f^{-1}$ is $y \geq -\frac{5}{2}$.

(b) $y = f(x)$ is the upper half of a sideways parabola starting at $(-\frac{5}{2}, 0)$ and increasing. $y = f^{-1}(x)$ is a rightward-opening parabola with vertex at $(0, -\frac{5}{2})$. They are reflections of each other in the line $y = x$.

(c) The graphs intersect on the line $y = x$, so solve $f(x) = x$:

$\sqrt{2x + 5} = x$

$2x + 5 = x^2$

$x^2 - 2x - 5 = 0$

$x = \dfrac{2 \pm \sqrt{4 + 20}}{2} = \dfrac{2 \pm \sqrt{24}}{2} = 1 \pm \sqrt{6}$

Since $\sqrt{2x+5} = x$ requires $x \geq 0$, we take $x = 1 + \sqrt{6}$.

Point of intersection: $(1 + \sqrt{6}, 1 + \sqrt{6})$.

[Marking: 2 marks for f⁻¹ with domain/range, 1 mark for sketch, 2 marks for intersection]

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Question 18 [4 marks]

$\dfrac{4}{x + 1} - \dfrac{3}{x - 2} = 1$

Multiply through by $(x+1)(x-2)$:

$4(x - 2) - 3(x + 1) = (x + 1)(x - 2)$

$4x - 8 - 3x - 3 = x^2 - 2x + x - 2$

$x - 11 = x^2 - x - 2$

$0 = x^2 - 2x - 9$

$x = \dfrac{2 \pm \sqrt{4 + 36}}{2} = \dfrac{2 \pm \sqrt{40}}{2} = \dfrac{2 \pm 2\sqrt{10}}{2} = 1 \pm \sqrt{10}$

$x = 1 + \sqrt{10} \approx 4.16$

$x = 1 - \sqrt{10} \approx -2.16$

$x \approx 4.16 \text{ or } x \approx -2.16 \text{ (2 d.p.)}$

Check: Neither value makes the original denominators zero. ✓

[Marking: 2 marks for forming quadratic, 2 marks for solutions to 2 d.p.]

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Question 19 [7 marks]

(a) Vertical asymptote: $x = 2$; Horizontal asymptote: $y = 1$.

(b) Domain: $x \in \mathbb{R}, x \neq 2$; Range: $y \in \mathbb{R}, y \neq 1$.

(c) $f(x) = \dfrac{ax + b}{x - 2} + 1$

Using $(0, -1)$: $\dfrac{b}{-2} + 1 = -1 \Rightarrow \dfrac{b}{-2} = -2 \Rightarrow b = 4$

Using $(4, 3)$: $\dfrac{4a + b}{2} + 1 = 3 \Rightarrow \dfrac{4a + 4}{2} = 2 \Rightarrow 2a + 2 = 2 \Rightarrow a = 0$

Wait, let me recheck: $\dfrac{4a + 4}{2} + 1 = 3 \Rightarrow 2a + 2 + 1 = 3 \Rightarrow 2a = 0 \Rightarrow a = 0$.

So $f(x) = \dfrac{4}{x - 2} + 1$.

Check with $(0, -1)$: $\dfrac{4}{-2} + 1 = -2 + 1 = -1$ ✓

Check with $(4, 3)$: $\dfrac{4}{2} + 1 = 2 + 1 = 3$ ✓

$a = 0, \quad b = 4$

(d) $f(x) = \dfrac{4}{x - 2} + 1 = \dfrac{4 + x - 2}{x - 2} = \dfrac{x + 2}{x - 2}$

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

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

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

$xy - 2x = y + 2$

$xy - y = 2x + 2$

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

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

Domain of $f^{-1}$: $x \neq 1$ (since range of $f$ is $y \neq 1$).

[Marking: 1 mark for asymptotes, 1 mark for domain/range, 2 marks for a and b, 3 marks for f⁻¹ with domain]

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Question 20 [10 marks]

(a) Polynomial division of $x^2 - 2x + 3$ by $x - 1$:

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

Check: $(x-1)^2 = x^2 - 2x + 1$, so remainder is $3 - 1 = 2$. ✓

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

So $a = 1$, $b = -1$, $c = 2$.

(b) Vertical asymptote: $x = 1$

Oblique asymptote: $y = x - 1$ (as $x \to \pm\infty$, the fraction term vanishes).

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

$\dfrac{dy}{dx} = 1 - \dfrac{2}{(x-1)^2}$

Set $\dfrac{dy}{dx} = 0$: $(x-1)^2 = 2 \Rightarrow x = 1 \pm \sqrt{2}$

When $x = 1 + \sqrt{2}$: $y = \sqrt{2} + \dfrac{2}{\sqrt{2}} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$

When $x = 1 - \sqrt{2}$: $y = -\sqrt{2} + \dfrac{2}{-\sqrt{2}} = -\sqrt{2} - \sqrt{2} = -2\sqrt{2}$

Stationary points: $(1 + \sqrt{2}, 2\sqrt{2})$ and $(1 - \sqrt{2}, -2\sqrt{2})$.

$\dfrac{d^2y}{dx^2} = \dfrac{4}{(x-1)^3}$

At $x = 1 + \sqrt{2}$: $\dfrac{d^2y}{dx^2} > 0$ → minimum at $(1 + \sqrt{2}, 2\sqrt{2})$

At $x = 1 - \sqrt{2}$: $\dfrac{d^2y}{dx^2} < 0$ → maximum at $(1 - \sqrt{2}, -2\sqrt{2})$

(d) Sketch should show:

• Vertical asymptote $x = 1$ (dashed)
• Oblique asymptote $y = x - 1$ (dashed)
• Maximum at $(1 - \sqrt{2}, -2\sqrt{2}) \approx (-0.41, -2.83)$
• Minimum at $(1 + \sqrt{2}, 2\sqrt{2}) \approx (2.41, 2.83)$
• $y$-intercept at $(0, -1 - 2) = (0, -3)$
• No $x$-intercepts (since $x - 1 + \frac{2}{x-1} = 0 \Rightarrow (x-1)^2 + 2 = 0$ has no real solutions)

(e) From the graph, the range is:

$f(x) \leq -2\sqrt{2} \quad \text{or} \quad f(x) \geq 2\sqrt{2}$

i.e., $\text{Range}: (-\infty, -2\sqrt{2}] \cup [2\sqrt{2}, \infty)$

[Marking: 2 marks for part (a), 2 marks for part (b), 3 marks for part (c) including nature, 2 marks for part (d) sketch, 1 mark for part (e) range]

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Mark Summary

Q	Marks		Q	Marks
1	4		11	3
2	6		12	3
3	5		13	3
4	6		14	4
5	6		15	5
6	6		16	8
7	5		17	5
8	5		18	4
9	6		19	7
10	6		20	10
			Total	60