Secondary 4 Additional Mathematics Practice Paper 4

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TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

TuitionGoWhere Practice Paper (AI)
Subject: Additional Mathematics
Level: Secondary 4
Paper: Practice Paper 4 (Version 4 of 5)
Duration: 2 hours 15 minutes
Total Marks: 100

Name: ___________________________
Class: ___________________________
Date: ___________________________

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Instructions

1. Write your name, class, and date in the spaces provided above.
2. Answer all questions.
3. Write your answers in the spaces provided in this question paper.
4. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
5. The use of an approved scientific calculator is expected, where appropriate.
6. You are reminded of the need for clear presentation in your answers.
7. The number of marks is given in brackets [ ] at the end of each question or part question.
8. The total number of marks for this paper is 100.

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Section A (40 marks)

Answer all questions in this section.

Question 1 [4 marks]

The curve $C$ has equation $y = x^3 - 6x^2 + 9x + 2$.

(a) Find the coordinates of the stationary points of $C$. [3]

(b) Determine the nature of each stationary point. [1]

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Question 2 [5 marks]

The line $L$ passes through the points $A(-2, 5)$ and $B(4, -1)$.

(a) Find the equation of the perpendicular bisector of $AB$. [3]

(b) The perpendicular bisector of $AB$ meets the $x$-axis at point $C$. Find the coordinates of $C$. [2]

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Question 3 [6 marks]

A circle has equation $x^2 + y^2 - 8x + 6y - 11 = 0$.

(a) Find the centre and radius of the circle. [3]

(b) The line $y = 2x - 3$ intersects the circle at points $P$ and $Q$. Find the coordinates of $P$ and $Q$. [3]

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Question 4 [5 marks]

The function $f$ is defined by $f(x) = \frac{2x+3}{x-1}$ for $x \neq 1$.

(a) Find $f^{-1}(x)$, the inverse function of $f$. [3]

(b) State the domain and range of $f^{-1}$. [2]

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Question 5 [6 marks]

The diagram shows part of the curve $y = \ln(2x+1)$ for $x \geq 0$.

Image pending generation: graph for Q5.

The region bounded by the curve, the $x$-axis, the $y$-axis, and the line $x = 2$ is rotated completely about the $x$-axis.

Find the exact volume of the solid generated. [6]

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Question 6 [5 marks]

The polynomial $p(x) = 2x^3 + ax^2 + bx - 6$ has a factor $(x-2)$ and leaves a remainder of $12$ when divided by $(x+1)$.

(a) Find the values of $a$ and $b$. [3]

(b) Hence factorise $p(x)$ completely. [2]

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Question 7 [4 marks]

Solve the equation $\log_2(x+3) + \log_2(x-1) = 3$ for $x$. [4]

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Question 8 [5 marks]

The variables $x$ and $y$ are related by the equation $y = \frac{k}{x^2} + c$, where $k$ and $c$ are constants.

When $x = 1$, $y = 5$. When $x = 2$, $y = 2$.

(a) Find the values of $k$ and $c$. [3]

(b) Find the value of $x$ when $y = 3$. [2]

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Section B (60 marks)

Answer all questions in this section.

Question 9 [12 marks]

A particle moves in a straight line such that its velocity $v$ m/s at time $t$ seconds is given by $v = 3t^2 - 12t + 9$ for $t \geq 0$.

(a) Find the times when the particle is at rest. [2]

(b) Find the acceleration of the particle when $t = 1$. [2]

(c) Find the total distance travelled by the particle in the first 4 seconds. [4]

(d) Sketch the velocity-time graph for $0 \leq t \leq 4$, indicating clearly the coordinates of any intercepts and turning points. [4]

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Question 10 [10 marks]

The diagram shows a sector $OAB$ of a circle with centre $O$ and radius $r$ cm. The angle $AOB$ is $\theta$ radians. The perimeter of the sector is 30 cm.

Image pending generation: diagram for Q10.

(a) Show that the area $A$ cm$^2$ of the sector is given by $A = 15r - r^2$. [3]

(b) Given that $r$ can vary, find the maximum area of the sector and the corresponding value of $r$. [4]

(c) Find the value of $\theta$ when the area is maximum. [3]

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Question 11 [12 marks]

The curve $C$ has equation $y = \frac{x^2 - 4}{x - 3}$ for $x \neq 3$.

(a) Find the equations of the asymptotes of $C$. [3]

(b) Find the coordinates of the stationary points of $C$. [4]

(c) Sketch the curve $C$, showing clearly the asymptotes, stationary points, and intercepts with the axes. [5]

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Question 12 [10 marks]

The function $f$ is defined by $f(x) = 3\sin 2x + 4\cos 2x$ for $0 \leq x \leq \pi$.

(a) Express $f(x)$ in the form $R\sin(2x + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. [3]

(b) Hence, or otherwise, find the maximum and minimum values of $f(x)$ and the corresponding values of $x$ in the given interval. [4]

(c) Solve the equation $f(x) = 2$ for $0 \leq x \leq \pi$. [3]

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Question 13 [8 marks]

The equation of a curve is $y = x^2 e^{-x}$ for $x \geq 0$.

(a) Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$. [4]

(b) Show that the curve has a maximum point at $x = 2$ and find the coordinates of this point. [2]

(c) Determine the range of values of $x$ for which the curve is concave downwards. [2]

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Question 14 [8 marks]

In the triangle $ABC$, $AB = 10$ cm, $BC = 7$ cm, and $\angle ABC = 60^\circ$.

(a) Find the length of $AC$. [2]

(b) Find the area of triangle $ABC$. [2]

(c) The point $D$ lies on $AC$ such that $BD$ is perpendicular to $AC$. Find the length of $BD$. [4]

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TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4 (Answer Key)

Subject: Additional Mathematics
Level: Secondary 4
Paper: Practice Paper 4 (Version 4 of 5)
Total Marks: 100

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Section A (40 marks)

Question 1 [4 marks]

Curve: $y = x^3 - 6x^2 + 9x + 2$

(a) Stationary points [3 marks]

$\frac{dy}{dx} = 3x^2 - 12x + 9$

At stationary points, $\frac{dy}{dx} = 0$: $3x^2 - 12x + 9 = 0$ $x^2 - 4x + 3 = 0$ $(x-1)(x-3) = 0$ $x = 1$ or $x = 3$

When $x = 1$: $y = 1 - 6 + 9 + 2 = 6$ → Point $(1, 6)$ When $x = 3$: $y = 27 - 54 + 27 + 2 = 2$ → Point $(3, 2)$

Stationary points: $(1, 6)$ and $(3, 2)$

(b) Nature of stationary points [1 mark]

$\frac{d^2y}{dx^2} = 6x - 12$

At $x = 1$: $\frac{d^2y}{dx^2} = 6 - 12 = -6 < 0$ → Maximum at $(1, 6)$ At $x = 3$: $\frac{d^2y}{dx^2} = 18 - 12 = 6 > 0$ → Minimum at $(3, 2)$

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Question 2 [5 marks]

Points: $A(-2, 5)$, $B(4, -1)$

(a) Perpendicular bisector of $AB$ [3 marks]

Midpoint of $AB$: $\left(\frac{-2+4}{2}, \frac{5+(-1)}{2}\right) = (1, 2)$

Gradient of $AB$: $m_{AB} = \frac{-1-5}{4-(-2)} = \frac{-6}{6} = -1$

Gradient of perpendicular bisector: $m = 1$ (since $m \times (-1) = -1$)

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

Equation: $y = x + 1$

(b) Intersection with $x$-axis [2 marks]

At $x$-axis, $y = 0$: $0 = x + 1 \Rightarrow x = -1$

Coordinates of $C$: $(-1, 0)$

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Question 3 [6 marks]

Circle: $x^2 + y^2 - 8x + 6y - 11 = 0$

(a) Centre and radius [3 marks]

Complete the square: $(x^2 - 8x) + (y^2 + 6y) = 11$ $(x - 4)^2 - 16 + (y + 3)^2 - 9 = 11$ $(x - 4)^2 + (y + 3)^2 = 36$

Centre: $(4, -3)$, Radius: $6$

(b) Intersection with $y = 2x - 3$ [3 marks]

Substitute $y = 2x - 3$ into circle equation: $x^2 + (2x-3)^2 - 8x + 6(2x-3) - 11 = 0$ $x^2 + 4x^2 - 12x + 9 - 8x + 12x - 18 - 11 = 0$ $5x^2 - 8x - 20 = 0$

$x = \frac{8 \pm \sqrt{64 + 400}}{10} = \frac{8 \pm \sqrt{464}}{10} = \frac{8 \pm 4\sqrt{29}}{10} = \frac{4 \pm 2\sqrt{29}}{5}$

When $x = \frac{4 + 2\sqrt{29}}{5}$: $y = 2\left(\frac{4 + 2\sqrt{29}}{5}\right) - 3 = \frac{8 + 4\sqrt{29} - 15}{5} = \frac{4\sqrt{29} - 7}{5}$

When $x = \frac{4 - 2\sqrt{29}}{5}$: $y = 2\left(\frac{4 - 2\sqrt{29}}{5}\right) - 3 = \frac{8 - 4\sqrt{29} - 15}{5} = \frac{-4\sqrt{29} - 7}{5}$

Points: $P\left(\frac{4 + 2\sqrt{29}}{5}, \frac{4\sqrt{29} - 7}{5}\right)$, $Q\left(\frac{4 - 2\sqrt{29}}{5}, \frac{-4\sqrt{29} - 7}{5}\right)$

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Question 4 [5 marks]

Function: $f(x) = \frac{2x+3}{x-1}$, $x \neq 1$

(a) Inverse function [3 marks]

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

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

$x(y-1) = 2y+3$ $xy - x = 2y + 3$ $xy - 2y = x + 3$ $y(x-2) = x+3$ $y = \frac{x+3}{x-2}$

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

(b) Domain and range of $f^{-1}$ [2 marks]

Domain of $f^{-1}$ = $Range of$f$=${x \in \mathbb{R} : x \neq 2}$(since$f(x) \neq 2$) Range of$f^{-1}$= Domain of$f$=${x \in \mathbb{R} : x \neq 1}$

Domain: $x \neq 2$, Range: $y \neq 1$

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Question 5 [6 marks]

Volume of revolution about $x$-axis: $y = \ln(2x+1)$, $0 \leq x \leq 2$

$V = \pi \int_0^2 [\ln(2x+1)]^2 \, dx$

Let $u = 2x+1$, then $du = 2\,dx$, $dx = \frac{du}{2}$ When $x = 0$, $u = 1$; when $x = 2$, $u = 5$

$V = \pi \int_1^5 (\ln u)^2 \cdot \frac{du}{2} = \frac{\pi}{2} \int_1^5 (\ln u)^2 \, du$

Use integration by parts: $\int (\ln u)^2 \, du = u(\ln u)^2 - 2\int \ln u \, du$ $= u(\ln u)^2 - 2(u\ln u - u) + C$ $= u[(\ln u)^2 - 2\ln u + 2] + C$

$V = \frac{\pi}{2} \left[ u((\ln u)^2 - 2\ln u + 2) \right]_1^5$ $= \frac{\pi}{2} \left[ 5((\ln 5)^2 - 2\ln 5 + 2) - 1(0 - 0 + 2) \right]$ $= \frac{\pi}{2} \left[ 5(\ln 5)^2 - 10\ln 5 + 10 - 2 \right]$ $= \frac{\pi}{2} \left[ 5(\ln 5)^2 - 10\ln 5 + 8 \right]$

Exact volume: $\frac{\pi}{2} \left[ 5(\ln 5)^2 - 10\ln 5 + 8 \right]$ cubic units

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Question 6 [5 marks]

Polynomial: $p(x) = 2x^3 + ax^2 + bx - 6$

(a) Find $a$ and $b$ [3 marks]

Factor $(x-2)$: $p(2) = 0$ $2(8) + a(4) + b(2) - 6 = 0$ $16 + 4a + 2b - 6 = 0$ $4a + 2b = -10$ $2a + b = -5$ ... (1)

Remainder when divided by $(x+1)$ is 12: $p(-1) = 12$ $2(-1) + a(1) + b(-1) - 6 = 12$ $-2 + a - b - 6 = 12$ $a - b = 20$ ... (2)

Add (1) and (2): $3a = 15 \Rightarrow a = 5$ From (2): $5 - b = 20 \Rightarrow b = -15$

$a = 5$, $b = -15$

(b) Factorise completely [2 marks]

$p(x) = 2x^3 + 5x^2 - 15x - 6$

Since $(x-2)$ is a factor, divide: $2x^3 + 5x^2 - 15x - 6 = (x-2)(2x^2 + 9x + 3)$

Quadratic $2x^2 + 9x + 3$ has discriminant $81 - 24 = 57$, not a perfect square.

$p(x) = (x-2)(2x^2 + 9x + 3)$

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Question 7 [4 marks]

Equation: $\log_2(x+3) + \log_2(x-1) = 3$

Combine logs: $\log_2[(x+3)(x-1)] = 3$

$(x+3)(x-1) = 2^3 = 8$ $x^2 + 2x - 3 = 8$ $x^2 + 2x - 11 = 0$

$x = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2} = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3}$

Check domain: $x+3 > 0 \Rightarrow x > -3$ and $x-1 > 0 \Rightarrow x > 1$

$-1 + 2\sqrt{3} \approx -1 + 3.46 = 2.46 > 1$ ✓ $-1 - 2\sqrt{3} \approx -4.46 < 1$ ✗

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

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Question 8 [5 marks]

Equation: $y = \frac{k}{x^2} + c$

(a) Find $k$ and $c$ [3 marks]

When $x=1$, $y=5$: $5 = k + c$ ... (1) When $x=2$, $y=2$: $2 = \frac{k}{4} + c$ ... (2)

Subtract (2) from (1): $3 = k - \frac{k}{4} = \frac{3k}{4} \Rightarrow k = 4$

From (1): $5 = 4 + c \Rightarrow c = 1$

$k = 4$, $c = 1$

(b) Find $x$ when $y = 3$ [2 marks]

$3 = \frac{4}{x^2} + 1$ $2 = \frac{4}{x^2}$ $x^2 = 2$ $x = \sqrt{2}$ (since $x > 0$ typically for this context)

$x = \sqrt{2}$

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Section B (60 marks)

Question 9 [12 marks]

Velocity: $v = 3t^2 - 12t + 9$, $t \geq 0$

(a) Times at rest [2 marks]

$v = 0 \Rightarrow 3t^2 - 12t + 9 = 0$ $t^2 - 4t + 3 = 0$ $(t-1)(t-3) = 0$ $t = 1, 3$

At rest at $t = 1$ s and $t = 3$ s

(b) Acceleration at $t = 1$ [2 marks]

$a = \frac{dv}{dt} = 6t - 12$ At $t = 1$: $a = 6 - 12 = -6$

Acceleration = $-6$ m/s$^2$

(c) Total distance in first 4 seconds [4 marks]

$v = 3(t-1)(t-3)$ $v > 0$ for $t < 1$ and $t > 3$; $v < 0$ for $1 < t < 3$

Distance = $\int_0^1 v\,dt + \int_1^3 (-v)\,dt + \int_3^4 v\,dt$

$\int v\,dt = t^3 - 6t^2 + 9t$

$\int_0^1 v\,dt = [1 - 6 + 9] - 0 = 4$ $\int_1^3 v\,dt = [27 - 54 + 27] - [1 - 6 + 9] = 0 - 4 = -4$ $\int_3^4 v\,dt = [64 - 96 + 36] - [27 - 54 + 27] = 4 - 0 = 4$

Total distance = $4 + 4 + 4 = 12$

Total distance = 12 m

(d) Velocity-time graph [4 marks]

• $v$-intercept: $t=0$, $v=9$ → $(0, 9)$
• $t$-intercepts: $t=1$, $t=3$ → $(1, 0)$, $(3, 0)$
• Vertex at $t = 2$: $v = 12 - 24 + 9 = -3$ → $(2, -3)$
• At $t=4$: $v = 48 - 48 + 9 = 9$ → $(4, 9)$

Parabola opening upwards, minimum at $(2, -3)$, symmetric about $t=2$.

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Question 10 [10 marks]

Sector: Perimeter = $2r + r\theta = 30$

(a) Area formula [3 marks]

$2r + r\theta = 30 \Rightarrow r\theta = 30 - 2r \Rightarrow \theta = \frac{30 - 2r}{r}$

Area $A = \frac{1}{2}r^2\theta = \frac{1}{2}r^2\left(\frac{30 - 2r}{r}\right) = \frac{1}{2}r(30 - 2r) = 15r - r^2$

$A = 15r - r^2$ (shown)

(b) Maximum area [4 marks]

$A = 15r - r^2 = -(r^2 - 15r) = -\left[(r - \frac{15}{2})^2 - \frac{225}{4}\right] = \frac{225}{4} - (r - 7.5)^2$

Maximum when $r = 7.5$ cm Maximum area = $\frac{225}{4} = 56.25$ cm$^2$

Maximum area = 56.25 cm$^2$ at $r = 7.5$ cm

(c) Value of $\theta$ at maximum [3 marks]

$\theta = \frac{30 - 2r}{r} = \frac{30 - 15}{7.5} = \frac{15}{7.5} = 2$

$\theta = 2$ radians

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Question 11 [12 marks]

Curve: $y = \frac{x^2 - 4}{x - 3}$, $x \neq 3$

(a) Asymptotes [3 marks]

Vertical asymptote: $x = 3$ (denominator = 0)

Oblique asymptote: Divide $x^2 - 4$ by $x - 3$: $x^2 - 4 = (x-3)(x+3) + 5$ $y = x + 3 + \frac{5}{x-3}$

As $x \to \pm\infty$, $y \to x + 3$

Vertical: $x = 3$, Oblique: $y = x + 3$

(b) Stationary points [4 marks]

$y = \frac{x^2 - 4}{x - 3}$

Quotient rule: $\frac{dy}{dx} = \frac{2x(x-3) - (x^2-4)(1)}{(x-3)^2} = \frac{2x^2 - 6x - x^2 + 4}{(x-3)^2} = \frac{x^2 - 6x + 4}{(x-3)^2}$

Set $\frac{dy}{dx} = 0$: $x^2 - 6x + 4 = 0$ $x = \frac{6 \pm \sqrt{36 - 16}}{2} = \frac{6 \pm \sqrt{20}}{2} = 3 \pm \sqrt{5}$

When $x = 3 + \sqrt{5}$: $y = \frac{(3+\sqrt{5})^2 - 4}{\sqrt{5}} = \frac{9 + 6\sqrt{5} + 5 - 4}{\sqrt{5}} = \frac{10 + 6\sqrt{5}}{\sqrt{5}} = 2\sqrt{5} + 6$

When $x = 3 - \sqrt{5}$: $y = \frac{(3-\sqrt{5})^2 - 4}{-\sqrt{5}} = \frac{9 - 6\sqrt{5} + 5 - 4}{-\sqrt{5}} = \frac{10 - 6\sqrt{5}}{-\sqrt{5}} = -2\sqrt{5} + 6$

Stationary points: $(3+\sqrt{5}, 6+2\sqrt{5})$ and $(3-\sqrt{5}, 6-2\sqrt{5})$

(c) Sketch [5 marks]

Key features:

• Vertical asymptote $x = 3$ (dashed line)
• Oblique asymptote $y = x + 3$ (dashed line)
• $y$-intercept: $x=0$, $y = \frac{-4}{-3} = \frac{4}{3}$ → $(0, \frac{4}{3})$
• $x$-intercepts: $x^2 - 4 = 0 \Rightarrow x = \pm 2$ → $(-2, 0)$, $(2, 0)$
• Stationary points as above
• As $x \to 3^-$, $y \to -\infty$; as $x \to 3^+$, $y \to +\infty$
• Curve approaches $y = x+3$ from above for $x < 3$, from below for $x > 3$

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Question 12 [10 marks]

Function: $f(x) = 3\sin 2x + 4\cos 2x$, $0 \leq x \leq \pi$

(a) Express as $R\sin(2x + \alpha)$ [3 marks]

$R = \sqrt{3^2 + 4^2} = 5$

$\sin\alpha = \frac{4}{5}$, $\cos\alpha = \frac{3}{5} \Rightarrow \alpha = \arcsin(0.8) \approx 0.927$ rad (or $\arctan(4/3)$)

$f(x) = 5\sin(2x + \alpha)$ where $\alpha = \arctan(4/3)$

(b) Maximum and minimum [4 marks]

Maximum of $5\sin(2x+\alpha)$ is $5$, minimum is $-5$

Maximum $= 5$ when $2x + \alpha = \frac{\pi}{2} \Rightarrow 2x = \frac{\pi}{2} - \alpha \Rightarrow x = \frac{\pi}{4} - \frac{\alpha}{2}$

Minimum $= -5$ when $2x + \alpha = \frac{3\pi}{2} \Rightarrow 2x = \frac{3\pi}{2} - \alpha \Rightarrow x = \frac{3\pi}{4} - \frac{\alpha}{2}$

Check interval $0 \leq x \leq \pi$: $\alpha \approx 0.927$, so $\frac{\alpha}{2} \approx 0.464$ $x_{\text{max}} \approx 0.785 - 0.464 = 0.321$ ✓ $x_{\text{min}} \approx 2.356 - 0.464 = 1.892$ ✓

Maximum: 5 at $x = \frac{\pi}{4} - \frac{1}{2}\arctan(4/3)$; Minimum: -5 at $x = \frac{3\pi}{4} - \frac{1}{2}\arctan(4/3)$

(c) Solve $f(x) = 2$ [3 marks]

$5\sin(2x + \alpha) = 2$ $\sin(2x + \alpha) = 0.4$

$2x + \alpha = \arcsin(0.4)$ or $\pi - \arcsin(0.4)$ $2x + \alpha \approx 0.4115$ or $2.730$

$2x \approx 0.4115 - 0.9273 = -0.5158$ (reject, $x < 0$) $2x \approx 2.730 - 0.9273 = 1.8027 \Rightarrow x \approx 0.901$

Also add $2\pi$: $2x + \alpha = 0.4115 + 2\pi \approx 6.695$ $2x \approx 5.767 \Rightarrow x \approx 2.884$ (within $[0, \pi]$)

$x \approx 0.901$ and $x \approx 2.884$ (3 s.f.)

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Question 13 [8 marks]

Curve: $y = x^2 e^{-x}$, $x \geq 0$

(a) Derivatives [4 marks]

$\frac{dy}{dx} = 2x e^{-x} + x^2(-e^{-x}) = e^{-x}(2x - x^2) = xe^{-x}(2 - x)$

$\frac{d^2y}{dx^2} = \frac{d}{dx}[e^{-x}(2x - x^2)]$ $= -e^{-x}(2x - x^2) + e^{-x}(2 - 2x)$ $= e^{-x}(-2x + x^2 + 2 - 2x)$ $= e^{-x}(x^2 - 4x + 2)$

$\frac{dy}{dx} = xe^{-x}(2-x)$, $\frac{d^2y}{dx^2} = e^{-x}(x^2 - 4x + 2)$

(b) Maximum at $x = 2$ [2 marks]

$\frac{dy}{dx} = xe^{-x}(2-x) = 0 \Rightarrow x = 0$ or $x = 2$

At $x = 2$: $\frac{d^2y}{dx^2} = e^{-2}(4 - 8 + 2) = -2e^{-2} < 0$ → Maximum

$y = 4e^{-2}$

Maximum at $(2, 4e^{-2})$

(c) Concave downwards [2 marks]

Concave down when $\frac{d^2y}{dx^2} < 0$: $e^{-x}(x^2 - 4x + 2) < 0$ Since $e^{-x} > 0$: $x^2 - 4x + 2 < 2 < 0$

Roots: $x = \frac{4 \pm \sqrt{16-8}}{2} = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}$

Quadratic opens upwards, so $< 0$ between roots.

$2 - \sqrt{2} < x < 2 + \sqrt{2}$ (approximately $0.586 < x < 3.414$)

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Question 14 [8 marks]

Triangle $ABC$: $AB = 10$, $BC = 7$, $\angle ABC = 60^\circ$

(a) Length $AC$ [2 marks]

Cosine rule: $AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos 60^\circ$ $= 100 + 49 - 2(10)(7)(\frac{1}{2})$ $= 149 - 70 = 79$

$AC = \sqrt{79}$ cm

$AC = \sqrt{79}$ cm

(b) Area [2 marks]

Area $= \frac{1}{2}(AB)(BC)\sin 60^\circ = \frac{1}{2}(10)(7)\frac{\sqrt{3}}{2} = \frac{35\sqrt{3}}{2}$

Area = $\frac{35\sqrt{3}}{2}$ cm$^2$

(c) Length $BD$ (perpendicular from $B$ to $AC$) [4 marks]

Area also $= \frac{1}{2}(AC)(BD)$ $\frac{1}{2}(\sqrt{79})(BD) = \frac{35\sqrt{3}}{2}$ $BD = \frac{35\sqrt{3}}{\sqrt{79}} = \frac{35\sqrt{237}}{79}$

$BD = \frac{35\sqrt{3}}{\sqrt{79}}$ cm (or $\frac{35\sqrt{237}}{79}$ cm)

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