Secondary 3 Additional Mathematics Semestral Assessment 2 (End of Year) Paper 5

TuitionGoWhere Practice Paper - Additional Mathematics Secondary 3 (SA2 Version 5) - Answer Key

Total Marks: 60

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Section A [30 marks]

1

(a) $f(x) = 2x^2 - 8x + 5$

Complete the square: $f(x) = 2(x^2 - 4x) + 5$ $= 2[(x - 2)^2 - 4] + 5$ $= 2(x - 2)^2 - 8 + 5$ $= 2(x - 2)^2 - 3$

So $a = 2$, $h = 2$, $k = -3$.

Answer: $f(x) = 2(x - 2)^2 - 3$ [2 marks]

(b) From the completed square form $2(x - 2)^2 - 3$, the minimum value is $-3$ (since the square term is always $\ge 0$ and multiplied by positive 2). This occurs when $x - 2 = 0$, i.e., $x = 2$.

Answer: Minimum value $= -3$ at $x = 2$ [1 mark]

(c)

• Vertex: $(2, -3)$
• $y$-intercept: $x = 0 \Rightarrow f(0) = 5$, so $(0, 5)$
• Additional points: $f(1) = -1$, $f(3) = -1$, $f(4) = 5$, $f(5) = 15$

Answer: Graph sketched with vertex $(2, -3)$, $y$-intercept $(0, 5)$, parabola opening upwards. [2 marks]

Marking notes: 1 mark for correct vertex and intercept labelled, 1 mark for correct shape and symmetry.

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2

(a) For equal roots, discriminant $\Delta = 0$: $k^2 - 4(3)(12) = 0$ $k^2 - 144 = 0$ $k^2 = 144$ $k = \pm 12$

Answer: $k = 12$ or $k = -12$ [2 marks]

(b) When $k = 12$: $3x^2 + 12x + 12 = 0 \Rightarrow 3(x + 2)^2 = 0 \Rightarrow x = -2$ When $k = -12$: $3x^2 - 12x + 12 = 0 \Rightarrow 3(x - 2)^2 = 0 \Rightarrow x = 2$

Answer: For $k = 12$, root $= -2$; for $k = -12$, root $= 2$ [1 mark]

Common trap: Forgetting $\pm$ when taking square root of $k^2 = 144$.

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3

Given $2x^2 - 5x + 3 = 0$, roots $\alpha, \beta$.

(a) Sum of roots: $\alpha + \beta = -\frac{-5}{2} = \frac{5}{2}$ Product of roots: $\alpha\beta = \frac{3}{2}$

Answer: $\alpha + \beta = \frac{5}{2}$, $\alpha\beta = \frac{3}{2}$ [1 mark]

(b) $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ $= \left(\frac{5}{2}\right)^2 - 2\left(\frac{3}{2}\right)$ $= \frac{25}{4} - 3$ $= \frac{25}{4} - \frac{12}{4}$ $= \frac{13}{4}$

Answer: $\frac{13}{4}$ [2 marks]

(c) $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$ $= \frac{5/2}{3/2}$ $= \frac{5}{3}$

Answer: $\frac{5}{3}$ [2 marks]

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4

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

Answer: $g^{-1}(x) = \frac{x + 2}{3 - x}$ [3 marks]

Marking: 1 mark for swapping, 1 mark for algebraic manipulation, 1 mark for final simplified form.

(b) Domain of $g^{-1}$ = Range of $g$. $g(x) = \frac{3x - 2}{x + 1} = 3 - \frac{5}{x + 1}$. As $x \to \pm\infty$, $g(x) \to 3$ but never equals 3. So range of $g$ is $\mathbb{R} \setminus \{3\}$. Thus domain of $g^{-1}$ is $x \neq 3$.

Range of $g^{-1}$ = Domain of $g$ = $x \neq -1$.

Answer: Domain: $x \in \mathbb{R}, x \neq 3$; Range: $y \in \mathbb{R}, y \neq -1$ [2 marks]

(c) Solve $g(x) = g^{-1}(x)$: $\frac{3x - 2}{x + 1} = \frac{x + 2}{3 - x}$ $(3x - 2)(3 - x) = (x + 2)(x + 1)$ $9x - 3x^2 - 6 + 2x = x^2 + 3x + 2$ $-3x^2 + 11x - 6 = x^2 + 3x + 2$ $0 = 4x^2 - 8x + 8$ $0 = x^2 - 2x + 2$

Discriminant: $(-2)^2 - 4(1)(2) = 4 - 8 = -4 < 0$. No real solutions.

Answer: No real solutions [2 marks]

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5

(a) By Remainder Theorem: $P(1) = 10$: $2(1)^3 + a(1)^2 + b(1) - 6 = 10 \Rightarrow 2 + a + b - 6 = 10 \Rightarrow a + b = 14$ ... (1)

$P(-2) = -20$: $2(-2)^3 + a(-2)^2 + b(-2) - 6 = -20 \Rightarrow -16 + 4a - 2b - 6 = -20 \Rightarrow 4a - 2b = 2 \Rightarrow 2a - b = 1$ ... (2)

Add (1) and (2): $3a = 15 \Rightarrow a = 5$ Substitute into (1): $5 + b = 14 \Rightarrow b = 9$

Answer: $a = 5$, $b = 9$ [4 marks]

(b) $P(x) = 2x^3 + 5x^2 + 9x - 6$ Try $x = \frac{1}{2}$: $P(\frac{1}{2}) = 2(\frac{1}{8}) + 5(\frac{1}{4}) + 9(\frac{1}{2}) - 6 = \frac{1}{4} + \frac{5}{4} + \frac{9}{2} - 6 = \frac{6}{4} + \frac{18}{4} - \frac{24}{4} = 0$ So $(2x - 1)$ is a factor.

Divide: $(2x^3 + 5x^2 + 9x - 6) \div (2x - 1) = x^2 + 3x + 6$

Check discriminant of $x^2 + 3x + 6$: $9 - 24 = -15 < 0$, so no further real factors.

Answer: $P(x) = (2x - 1)(x^2 + 3x + 6)$ [3 marks]

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6

$x^2 - 4x - 5 > 0$ Factorise: $(x - 5)(x + 1) > 0$

Roots: $x = 5$, $x = -1$. Parabola opens upwards. Inequality $> 0$ holds outside the roots.

Answer: $x < -1$ or $x > 5$ [3 marks]

Marking: 1 mark for factorisation/roots, 1 mark for correct region identification, 1 mark for final answer in correct notation.

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7

(a) $y = \sqrt{4 - x}$, $x \le 4$, $y \ge 0$. Square: $y^2 = 4 - x \Rightarrow x = 4 - y^2$. Swap: $h^{-1}(x) = 4 - x^2$. Domain of $h^{-1}$ = Range of $h$ = $y \ge 0$, so $x \ge 0$.

Answer: $h^{-1}(x) = 4 - x^2$, domain $x \ge 0$ [3 marks]

(b)

• $h(x)$: $x$-intercept at $4 - x = 0 \Rightarrow x = 4$, so $(4, 0)$; $y$-intercept at $x = 0 \Rightarrow y = 2$, so $(0, 2)$.
• $h^{-1}(x)$: $y$-intercept at $x = 0 \Rightarrow y = 4$, so $(0, 4)$; $x$-intercept at $4 - x^2 = 0 \Rightarrow x = 2$ (since $x \ge 0$), so $(2, 0)$.
• Line $y = x$ as line of symmetry.

Answer: Graphs sketched with correct intercepts and reflection across $y = x$. [3 marks]

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8

Line: $y = kx + 3$ Curve: $y = x^2 - 2x + 5$

No intersection $\Rightarrow$ equation $kx + 3 = x^2 - 2x + 5$ has no real roots. $x^2 - 2x + 5 - kx - 3 = 0$ $x^2 - (k + 2)x + 2 = 0$

Discriminant $< 0$: $[-(k + 2)]^2 - 4(1)(2) < 0$ $(k + 2)^2 - 8 < 0$ $(k + 2)^2 < 8$ $-\sqrt{8} < k + 2 < \sqrt{8}$ $-2\sqrt{2} - 2 < k < 2\sqrt{2} - 2$

Answer: $-2 - 2\sqrt{2} < k < 2\sqrt{2} - 2$ [4 marks]

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9

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

$f(2) = 8 - 12 + 4 = 0$, so $x - 2$ is a factor.

Divide: $(x^3 - 3x^2 + 4) \div (x - 2) = x^2 - x - 2$ Factorise: $x^2 - x - 2 = (x - 2)(x + 1)$

So $f(x) = (x - 2)(x - 2)(x + 1) = (x - 2)^2(x + 1)$

Answer: $f(x) = (x - 2)^2(x + 1)$ [4 marks]

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10

(a) $fg(x) = f(g(x)) = f(x^2 + 1) = 5 - 2(x^2 + 1) = 5 - 2x^2 - 2 = 3 - 2x^2$

$gf(x) = g(f(x)) = g(5 - 2x) = (5 - 2x)^2 + 1 = 25 - 20x + 4x^2 + 1 = 4x^2 - 20x + 26$

Answer: $fg(x) = 3 - 2x^2$, $gf(x) = 4x^2 - 20x + 26$ [2 marks]

(b) $fg(x) = gf(x)$ $3 - 2x^2 = 4x^2 - 20x + 26$ $0 = 6x^2 - 20x + 23$

Discriminant: $(-20)^2 - 4(6)(23) = 400 - 552 = -152 < 0$

No real solutions.

Answer: No real solutions [3 marks]

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Section B [30 marks]

11

(a) Perimeter $= 2(x + \text{width}) = 60 \Rightarrow \text{width} = 30 - x$ Area $A = x(30 - x) = 30x - x^2$

Answer: $A = 30x - x^2$ [1 mark]

(b) $A = -x^2 + 30x = -(x^2 - 30x) = -[(x - 15)^2 - 225] = -(x - 15)^2 + 225$

Answer: $A = -(x - 15)^2 + 225$ [2 marks]

(c) Maximum area $= 225$ m² when $x = 15$ m. Then width $= 30 - 15 = 15$ m. So the garden is a square of side 15 m.

Answer: Maximum area $= 225$ m², dimensions $15$ m by $15$ m [2 marks]

(d) $A \ge 200 \Rightarrow 30x - x^2 \ge 200$ $x^2 - 30x + 200 \le 0$ $(x - 10)(x - 20) \le 0$ $10 \le x \le 20$

Since length must be positive and less than 30, this is valid.

Answer: $10 \le x \le 20$ [3 marks]

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12

(a) $y = x^3 - 6x^2 + 9x + 1$ $\frac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$

Stationary points when $\frac{dy}{dx} = 0$: $x = 1$ or $x = 3$.

When $x = 1$: $y = 1 - 6 + 9 + 1 = 5$. Point $(1, 5)$. When $x = 3$: $y = 27 - 54 + 27 + 1 = 1$. Point $(3, 1)$.

Second derivative: $\frac{d^2y}{dx^2} = 6x - 12$. At $x = 1$: $\frac{d^2y}{dx^2} = -6 < 0$ $\Rightarrow$ local maximum. At $x = 3$: $\frac{d^2y}{dx^2} = 6 > 0$ $\Rightarrow$ local minimum.

Answer: Local maximum at $(1, 5)$; local minimum at $(3, 1)$ [5 marks]

Marking: 1 mark for derivative, 1 mark for finding x-coordinates, 1 mark for y-coordinates, 1 mark for second derivative test, 1 mark for nature conclusion.

(b) $y$-intercept: $x = 0 \Rightarrow y = 1$, so $(0, 1)$. Graph: cubic with positive leading coefficient, local max at $(1, 5)$, local min at $(3, 1)$, passing through $(0, 1)$ and $(4, 5)$.

Answer: Graph sketched with correct features. [3 marks]

(c) Curve increasing when $\frac{dy}{dx} > 0$: $3(x - 1)(x - 3) > 0$ $(x - 1)(x - 3) > 0$ $x < 1$ or $x > 3$

Answer: $x < 1$ or $x > 3$ [2 marks]

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13

(a) $f(x) = \frac{2x + 3}{x - 1}$ $f^2(x) = f(f(x)) = f\left(\frac{2x + 3}{x - 1}\right)$ $= \frac{2\left(\frac{2x + 3}{x - 1}\right) + 3}{\left(\frac{2x + 3}{x - 1}\right) - 1}$ $= \frac{\frac{4x + 6 + 3(x - 1)}{x - 1}}{\frac{2x + 3 - (x - 1)}{x - 1}}$ $= \frac{4x + 6 + 3x - 3}{2x + 3 - x + 1}$ $= \frac{7x + 3}{x + 4}$

Answer: $f^2(x) = \frac{7x + 3}{x + 4}$ [3 marks]

(b) Find $f^{-1}(x)$: $y = \frac{2x + 3}{x - 1}$ $y(x - 1) = 2x + 3$ $xy - y = 2x + 3$ $xy - 2x = y + 3$ $x(y - 2) = y + 3$ $x = \frac{y + 3}{y - 2}$

So $f^{-1}(x) = \frac{x + 3}{x - 2}$. Domain of $f^{-1}$ = Range of $f$. $f(x) = 2 + \frac{5}{x - 1}$, so $f(x) \neq 2$. Domain: $x \neq 2$.

Answer: $f^{-1}(x) = \frac{x + 3}{x - 2}$, domain $x \neq 2$ [3 marks]

(c) $f^2(x) = x$ $\frac{7x + 3}{x + 4} = x$ $7x + 3 = x^2 + 4x$ $x^2 - 3x - 3 = 0$ $x = \frac{3 \pm \sqrt{9 + 12}}{2} = \frac{3 \pm \sqrt{21}}{2}$

Check: $x \neq -4$ (domain of $f^2$), both solutions valid.

Answer: $x = \frac{3 + \sqrt{21}}{2}$ or $x = \frac{3 - \sqrt{21}}{2}$ [3 marks]

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14

(a) $P(1) = 1 - 5 + 5 + 5 - 6 = 0$. So $x = 1$ is a root.

Answer: Shown. [1 mark]

(b) Divide by $(x - 1)$: $P(x) = (x - 1)(x^3 - 4x^2 + x + 6)$

Try $x = -1$ for cubic: $-1 - 4 - 1 + 6 = 0$. So $(x + 1)$ is a factor. Divide: $(x^3 - 4x^2 + x + 6) \div (x + 1) = x^2 - 5x + 6 = (x - 2)(x - 3)$

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

Answer: $P(x) = (x - 1)(x + 1)(x - 2)(x - 3)$ [4 marks]

(c) Roots: $x = 1, -1, 2, 3$

Answer: $x = -1, 1, 2, 3$ [1 mark]

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15

(a) $f(x) = 3x^2 - 12x + 11 = 3(x^2 - 4x) + 11 = 3[(x - 2)^2 - 4] + 11 = 3(x - 2)^2 - 12 + 11 = 3(x - 2)^2 - 1$

Answer: $f(x) = 3(x - 2)^2 - 1$ [2 marks]

(b) $g(x) = f(x) + 5$. This is a translation of the graph of $y = f(x)$ by 5 units in the positive $y$-direction (upwards).

Answer: Translation by $\begin{pmatrix} 0 \\ 5 \end{pmatrix}$ [2 marks]

(c) $h(x) = f(2x) = 3(2x)^2 - 12(2x) + 11 = 12x^2 - 24x + 11$ Vertex of $h$: $x = -\frac{-24}{2(12)} = 1$, $h(1) = 12 - 24 + 11 = -1$. Alternatively, $f$ has vertex $(2, -1)$. $h(x) = f(2x)$ is a horizontal stretch by factor $\frac{1}{2}$, so vertex moves to $(1, -1)$.

Answer: Vertex at $(1, -1)$ [2 marks]

(d)

• $f(x)$: vertex $(2, -1)$, $y$-intercept $(0, 11)$
• $h(x)$: vertex $(1, -1)$, $y$-intercept $(0, 11)$ Both parabolas open upwards.

Answer: Graphs sketched with correct vertices and intercepts. [3 marks]

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16

$|2x - 5| = x + 1$

Case 1: $2x - 5 \ge 0 \Rightarrow x \ge 2.5$ $2x - 5 = x + 1 \Rightarrow x = 6$. Check: $6 \ge 2.5$, valid.

Case 2: $2x - 5 < 0 \Rightarrow x < 2.5$ $-(2x - 5) = x + 1 \Rightarrow -2x + 5 = x + 1 \Rightarrow 3x = 4 \Rightarrow x = \frac{4}{3}$. Check: $\frac{4}{3} < 2.5$, valid.

Also need $x + 1 \ge 0 \Rightarrow x \ge -1$. Both solutions satisfy this.

Answer: $x = \frac{4}{3}$ or $x = 6$ [4 marks]

Marking: 1 mark for setting up cases, 1 mark each for correct solutions with checks, 1 mark for final answer.

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17

(a) $f(x) = x^2 - 4x + 7 = (x - 2)^2 + 3$ for $x \ge 2$. On $x \ge 2$, $(x - 2)^2$ is strictly increasing (since $x - 2 \ge 0$ and squaring preserves order for non-negative numbers). So $f$ is strictly increasing on its domain, hence one-to-one, so an inverse exists.

Answer: $f$ is strictly increasing on $x \ge 2$, so it is one-to-one and has an inverse. [1 mark]

(b) $y = (x - 2)^2 + 3$, $x \ge 2$, $y \ge 3$. $(x - 2)^2 = y - 3$ $x - 2 = \sqrt{y - 3}$ (positive root since $x \ge 2$) $x = 2 + \sqrt{y - 3}$

$f^{-1}(x) = 2 + \sqrt{x - 3}$ Domain of $f^{-1}$ = Range of $f$ = $x \ge 3$ Range of $f^{-1}$ = Domain of $f$ = $y \ge 2$

Answer: $f^{-1}(x) = 2 + \sqrt{x - 3}$, domain $x \ge 3$, range $y \ge 2$ [3 marks]

(c)

• $f(x)$: vertex $(2, 3)$, domain $x \ge 2$
• $f^{-1}(x)$: vertex $(3, 2)$, domain $x \ge 3$
• Reflection across $y = x$

Answer: Graphs sketched with correct vertices and reflection. [3 marks]

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18

$kx^2 + (k - 3)x + 1 = 0$ has real and distinct roots $\Rightarrow \Delta > 0$.

$(k - 3)^2 - 4k(1) > 0$ $k^2 - 6k + 9 - 4k >

<stage3_exam_answers_md> = -6 < 0 \Rightarrow$local maximum. At$x = 3$:$\frac{d^2y}{dx^2} = 6 > 0 \Rightarrow$ local minimum.

Answer: Local maximum at $(1, 5)$, local minimum at $(3, 1)$ [5 marks]

(b)

• $y$-intercept: $x = 0 \Rightarrow y = 1$, so $(0, 1)$.
• Stationary points: $(1, 5)$ (max), $(3, 1)$ (min).
• Additional point: $x = 4 \Rightarrow y = 64 - 96 + 36 + 1 = 5$, so $(4, 5)$.

Answer: Graph sketched with correct intercepts and stationary points. [3 marks]

(c) Curve increasing when $\frac{dy}{dx} > 0$: $3(x - 1)(x - 3) > 0 \Rightarrow x < 1$ or $x > 3$.

Answer: $x < 1$ or $x > 3$ [2 marks]

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13

(a) $f(x) = \frac{2x + 3}{x - 1}$ $f^2(x) = f(f(x)) = \frac{2\left(\frac{2x + 3}{x - 1}\right) + 3}{\left(\frac{2x + 3}{x - 1}\right) - 1}$ $= \frac{\frac{4x + 6 + 3(x - 1)}{x - 1}}{\frac{2x + 3 - (x - 1)}{x - 1}}$ $= \frac{4x + 6 + 3x - 3}{2x + 3 - x + 1}$ $= \frac{7x + 3}{x + 4}$

Answer: $f^2(x) = \frac{7x + 3}{x + 4}$ [3 marks]

(b) Let $y = \frac{2x + 3}{x - 1}$. Swap $x$ and $y$: $x = \frac{2y + 3}{y - 1}$ $x(y - 1) = 2y + 3$ $xy - x = 2y + 3$ $xy - 2y = x + 3$ $y(x - 2) = x + 3$ $y = \frac{x + 3}{x - 2}$

So $f^{-1}(x) = \frac{x + 3}{x - 2}$. Domain of $f^{-1}$ = Range of $f$. $f(x) = 2 + \frac{5}{x - 1}$, so $f(x) \neq 2$. Domain: $x \neq 2$.

Answer: $f^{-1}(x) = \frac{x + 3}{x - 2}$, domain $x \neq 2$ [3 marks]

(c) $f^2(x) = x$ $\frac{7x + 3}{x + 4} = x$ $7x + 3 = x^2 + 4x$ $x^2 - 3x - 3 = 0$ $x = \frac{3 \pm \sqrt{9 + 12}}{2} = \frac{3 \pm \sqrt{21}}{2}$

Check: neither root is $-4$ (excluded from domain of $f^2$), so both valid.

Answer: $x = \frac{3 + \sqrt{21}}{2}$ or $x = \frac{3 - \sqrt{21}}{2}$ [3 marks]

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14

(a) $P(1) = 1 - 5 + 5 + 5 - 6 = 0$. So $x = 1$ is a root.

Answer: Shown [1 mark]

(b) Divide by $(x - 1)$: $P(x) = (x - 1)(x^3 - 4x^2 + x + 6)$

Test $x = -1$ for cubic: $-1 - 4 - 1 + 6 = 0$. So $(x + 1)$ is a factor. Divide: $x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6) = (x + 1)(x - 2)(x - 3)$

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

Answer: $P(x) = (x - 1)(x + 1)(x - 2)(x - 3)$ [4 marks]

(c) $P(x) = 0 \Rightarrow x = 1, -1, 2, 3$

Answer: $x = -1, 1, 2, 3$ [1 mark]

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15

(a) $f(x) = 3x^2 - 12x + 11 = 3(x^2 - 4x) + 11 = 3[(x - 2)^2 - 4] + 11 = 3(x - 2)^2 - 12 + 11 = 3(x - 2)^2 - 1$

Answer: $f(x) = 3(x - 2)^2 - 1$ [2 marks]

(b) $g(x) = f(x) + 5 = 3(x - 2)^2 + 4$. This is a translation of $y = f(x)$ by 5 units in the positive $y$-direction (upwards).

Answer: Translation by $\begin{pmatrix} 0 \\ 5 \end{pmatrix}$ [2 marks]

(c) $h(x) = f(2x) = 3(2x)^2 - 12(2x) + 11 = 12x^2 - 24x + 11 = 12(x^2 - 2x) + 11 = 12[(x - 1)^2 - 1] + 11 = 12(x - 1)^2 - 1$. Vertex at $(1, -1)$.

Answer: Vertex at $(1, -1)$ [2 marks]

(d)

• $f(x)$: vertex $(2, -1)$, $y$-intercept $(0, 11)$.
• $h(x)$: vertex $(1, -1)$, $y$-intercept $(0, 11)$.
• Both parabolas open upwards.

Answer: Graphs sketched with correct vertices and intercepts. [3 marks]

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16

$|2x - 5| = x + 1$

Case 1: $2x - 5 \ge 0 \Rightarrow x \ge 2.5$ $2x - 5 = x + 1 \Rightarrow x = 6$. Valid since $6 \ge 2.5$.

Case 2: $2x - 5 < 0 \Rightarrow x < 2.5$ $-(2x - 5) = x + 1 \Rightarrow -2x + 5 = x + 1 \Rightarrow 3x = 4 \Rightarrow x = \frac{4}{3}$. Valid since $\frac{4}{3} < 2.5$.

Check: $x = 6 \Rightarrow |7| = 7$ ✓; $x = \frac{4}{3} \Rightarrow |-\frac{7}{3}| = \frac{7}{3}$ ✓.

Answer: $x = 6$ or $x = \frac{4}{3}$ [4 marks]

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17

(a) $f(x) = x^2 - 4x + 7 = (x - 2)^2 + 3$ for $x \ge 2$. On this domain, $f$ is strictly increasing (derivative $2x - 4 \ge 0$ for $x \ge 2$), hence one-to-one, so an inverse exists.

Answer: $f$ is one-to-one on $x \ge 2$ (strictly increasing) [1 mark]

(b) $y = (x - 2 + \sqrt{y - 3}$ (positive root since $x \ge 2$). Swap: $f^{-1}(x) = 2 + \sqrt{x - 3}$. Domain of $f^{-1}$ = Range of $f$ = $[3, \infty)$. Range of $f^{-1}$ = Domain of $f$ = $[2, \infty)$.

Answer: $f^{-1}(x) = 2 + \sqrt{x - 3}$, domain $x \ge 3$, range $y \ge 2$ [3 marks]

(c)

• $f(x)$: vertex $(2, 3)$, domain $x \ge 2$.
• $f^{-1}(x)$: vertex $(3, 2)$, domain $x \ge 3$.
• Reflection across $y = x$.

Answer: Graphs sketched with correct vertices and reflection. [3 marks]

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18

$kx^2 + (k - 3)x + 1 = 0$ has real and distinct roots $\Rightarrow \Delta > 0$. $(k - 3)^2 - 4k(1) > 0$ $k^2 - 6k + 9 - 4k > 0$ $k^2 - 10k + 9 > 0$ $(k - 1)(k - 9) > 0$ $k < 1$ or $k > 9$.

Also need $k \neq 0$ for quadratic (if $k = 0$, equation is $-3x + 1 = 0$, only one root). But $k = 0$ is in $k < 1$, so exclude $k = 0$.

Answer: $k < 0$ or $0 < k < 1$ or $k > 9$ [4 marks]

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19

(a) $f(x) = 2x + 1$, $g(x) = \frac{x - 1}{2}$. $fg(x) = f(g(x)) = 2\left(\frac{x - 1}{2}\right) + 1 = x - 1 + 1 = x$. $gf(x) = g(f(x)) = \frac{(2x + 1) - 1}{2} = \frac{2x}{2} = x$. Since $fg(x) = gf(x) = x$, $g$ is the inverse of $f$.

Answer: Shown [2 marks]

(b) $h(x) = f(x^2) = 2x^2 + 1$. For $h$ to have an inverse, we need to restrict domain. Since $x^2$ is not one-to-one on $\mathbb{R}$, typically we take $x \ge 0$ (implied by context of finding inverse). Let $y = 2x^2 + 1$, $x \ge 0$. $2x^2 = y - 1 \Rightarrow x^2 = \frac{y - 1}{2} \Rightarrow x = \sqrt{\frac{y - 1}{2}}$ (positive root). Swap: $h^{-1}(x) = \sqrt{\frac{x - 1}{2}}$. Domain of $h^{-1}$ = Range of $h$ = $[1, \infty)$.

Answer: $h^{-1}(x) = \sqrt{\frac{x - 1}{2}}$, domain $x \ge 1$ [4 marks]

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20

(a) Substitute points: $(1, 6)$: $a + b + c = 6$ ... (1) $(2, 11)$: $4a + 2b + c = 11$ ... (2) $(3, 18)$: $9a + 3b + c = 18$ ... (3)

Answer: Three equations formed [1 mark]

(b) (2) - (1): $3a + b = 5$ ... (4) (3) - (2): $5a + b = 7$ ... (5) (5) - (4): $2a = 2 \Rightarrow a = 1$. Sub into (4): $3 + b = 5 \Rightarrow b = 2$. Sub into (1): $1 + 2 + c = 6 \Rightarrow c = 3$.

Answer: $a = 1$, $b = 2$, $c = 3$ [3 marks]

(c) $y = x^2 + 2x + 3 = (x + 1)^2 + 2$. Minimum value $= 2$ at $x = -1$.

Answer: Minimum value $= 2$ at $x = -1$ [2 marks]

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END OF ANSWER KEY