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Secondary 3 Additional Mathematics Practice Paper 3

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TuitionGoWhere Practice Paper - Additional Mathematics Secondary 3

TuitionGoWhere Practice Paper (AI)

Subject: Additional Mathematics Level: Secondary 3 Paper: Practice Paper Version 3 Duration: 1 hour 30 minutes Total Marks: 80

Name: _________________________ Class: _________________________ Date: _________________________

Instructions to Candidates

This paper consists of two sections. Answer all questions.
Write your answers in the spaces provided.
Show all working clearly. Marks are awarded for method, not just the final answer.
Non-exact numerical answers should be given correct to three significant figures, unless otherwise stated.
You are reminded of the need for clear presentation in your answers.
The number of marks is given in brackets [ ] at the end of each question or part question.
The total mark for this paper is 80.

Section A: Short Answer Questions (40 marks)

Answer all questions in this section.

1. Express $2x^2 - 12x + 23$ in the form $a(x - h)^2 + k$ .

Hence state the minimum value of the expression and the value of $x$ at which it occurs.

[4 marks]

2. Find the range of values of $k$ for which the equation $x^2 + (k - 3)x + 4 = 0$ has two distinct real roots.

[4 marks]

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

[5 marks]

4. Express $\frac{4x + 1}{(x - 1)(x + 2)}$ in partial fractions.

[5 marks]

5. Find the equation of the circle with centre $(3, -2)$ that passes through the point $(7, 1)$ .

Express your answer in the form $(x - a)^2 + (y - b)^2 = r^2$ .

[4 marks]

6. Find the coordinates of the points of intersection of the line $y = 2x - 1$ and the curve $y = x^2 - x - 3$ .

[5 marks]

7. The variables $x$ and $y$ are related by the equation $y = ax^n$ . The table below shows experimental values of $x$ and $y$ .

$x$	2	4	6	8	10
$y$	5.6	22.6	50.9	90.5	141.4

By plotting $\lg y$ against $\lg x$ on graph paper, it is found that the points lie approximately on a straight line with gradient $1.5$ and vertical intercept $0.15$ .

Estimate the values of $a$ and $n$ .

[4 marks]

8. Prove the identity $\frac{\cos 2\theta}{1 - \sin 2\theta} = \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta}$ .

[4 marks]

9. Solve the equation $3\sin^2 x - 2\cos x - 3 = 0$ for $0^\circ \le x \le 360^\circ$ .

[5 marks]

Section B: Structured Questions (40 marks)

Answer all questions in this section.

10. A curve has equation $y = x^3 - 6x^2 + 9x + 5$ .

(a) Find the coordinates of the stationary points of the curve. [4 marks]

(b) Determine the nature of each stationary point using the second derivative test. [3 marks]

(c) Find the equation of the tangent to the curve at the point where $x = 2$ . [3 marks]

11.

(a) Express $7\sin \theta + 24\cos \theta$ in the form $R\sin(\theta + \alpha)$ , where $R > 0$ and $0^\circ < \alpha < 90^\circ$ . [3 marks]

(b) Hence find the maximum value of $7\sin \theta + 24\cos \theta + 10$ , and the smallest positive value of $\theta$ at which it occurs. [3 marks]

(c) Solve the equation $7\sin \theta + 24\cos \theta = 12$ for $0^\circ \le \theta \le 360^\circ$ . [4 marks]

12. A rectangular box with an open top is to be constructed from a rectangular sheet of cardboard measuring 30 cm by 20 cm. Squares of side $x$ cm are cut from each corner, and the sides are folded up to form the box.

(a) Show that the volume $V \text{ cm}^3$ of the box is given by $V = 4x^3 - 100x^2 + 600x$ . [3 marks]

(b) Find the value of $x$ that maximises the volume of the box. [4 marks]

(c) Calculate the maximum volume, giving your answer correct to the nearest $\text{cm}^3$ . [2 marks]

13.

(a) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $2x^2 - 5x + 1 = 0$ , find the values of:

(i) $\alpha + \beta$ [1 mark]
(ii) $\alpha\beta$ [1 mark]

(b) Find the quadratic equation whose roots are $\alpha^2$ and $\beta^2$ , giving your answer in the form $ax^2 + bx + c = 0$ where $a$ , $b$ , and $c$ are integers. [5 marks]

14.

(a) Simplify $\frac{\sqrt{48} - \sqrt{27}}{\sqrt{3}}$ . [2 marks]

(b) Solve the equation $\sqrt{2x + 5} - \sqrt{x - 1} = 2$ . [5 marks]

END OF PAPER

Answers

TuitionGoWhere Practice Paper - Additional Mathematics Secondary 3

Answer Key and Marking Scheme (Version 3)

Total Marks: 80

Section A: Short Answer Questions (40 marks)

1. Express $2x^2 - 12x + 23$ in the form $a(x - h)^2 + k$ . Hence state the minimum value and the value of $x$ at which it occurs.

Answer: $2x^2 - 12x + 23$ $= 2(x^2 - 6x) + 23$ $= 2[(x - 3)^2 - 9] + 23$ [M1] $= 2(x - 3)^2 - 18 + 23$ $= 2(x - 3)^2 + 5$ [A1]

Minimum value is $5$ [A1], occurring at $x = 3$ [A1].

Marking notes: Award M1 for correctly factoring out 2 and completing the square inside the bracket. A1 for correct completed square form. A1 for minimum value. A1 for correct $x$ -value.

2. Find the range of values of $k$ for which $x^2 + (k - 3)x + 4 = 0$ has two distinct real roots.

Answer: For two distinct real roots, discriminant $> 0$ . $a = 1$ , $b = k - 3$ , $c = 4$ $\Delta = (k - 3)^2 - 4(1)(4)$ [M1] $= k^2 - 6k + 9 - 16$ $= k^2 - 6k - 7$ [A1]

$k^2 - 6k - 7 > 0$ $(k - 7)(k + 1) > 0$ [M1] $k < -1$ or $k > 7$ [A1]

Marking notes: M1 for setting up discriminant correctly. A1 for simplified quadratic inequality. M1 for factorising and solving. A1 for correct range.

3. $P(x) = x^3 + ax^2 + bx - 6$ has factor $(x - 2)$ and remainder $-12$ when divided by $(x + 1)$ . Find $a$ and $b$ .

Answer: Factor Theorem: $P(2) = 0$ $8 + 4a + 2b - 6 = 0$ $4a + 2b + 2 = 0$ $2a + b = -1$ ... (1) [M1, A1]

Remainder Theorem: $P(-1) = -12$ $-1 + a - b - 6 = -12$ $a - b - 7 = -12$ $a - b = -5$ ... (2) [M1, A1]

Solving (1) and (2): From (2): $a = b - 5$ Sub into (1): $2(b - 5) + b = -1$ $2b - 10 + b = -1$ $3b = 9$ $b = 3$ [A1] $a = 3 - 5 = -2$ [A1]

Marking notes: M1 for applying Factor Theorem correctly. A1 for equation (1). M1 for applying Remainder Theorem correctly. A1 for equation (2). A1 each for correct $a$ and $b$ .

4. Express $\frac{4x + 1}{(x - 1)(x + 2)}$ in partial fractions.

Answer: Let $\frac{4x + 1}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}$ [M1] $4x + 1 = A(x + 2) + B(x - 1)$ [M1]

When $x = 1$ : $4(1) + 1 = A(3) + B(0)$ $5 = 3A \implies A = \frac{5}{3}$ [A1]

When $x = -2$ : $4(-2) + 1 = A(0) + B(-3)$ $-7 = -3B \implies B = \frac{7}{3}$ [A1]

$\frac{4x + 1}{(x - 1)(x + 2)} = \frac{5}{3(x - 1)} + \frac{7}{3(x + 2)}$ [A1]

Marking notes: M1 for correct partial fraction form. M1 for multiplying by denominator. A1 for correct $A$ . A1 for correct $B$ . A1 for final expression.

5. Find the equation of the circle with centre $(3, -2)$ passing through $(7, 1)$ .

Answer: Radius $r = \sqrt{(7 - 3)^2 + (1 - (-2))^2}$ [M1] $= \sqrt{4^2 + 3^2}$ $= \sqrt{16 + 9}$ $= \sqrt{25} = 5$ [A1]

Equation: $(x - 3)^2 + (y - (-2))^2 = 5^2$ [M1] $(x - 3)^2 + (y + 2)^2 = 25$ [A1]

Marking notes: M1 for using distance formula. A1 for correct radius. M1 for substituting into standard form. A1 for correct equation.

6. Find the intersection of $y = 2x - 1$ and $y = x^2 - x - 3$ .

Answer: $2x - 1 = x^2 - x - 3$ [M1] $x^2 - 3x - 2 = 0$ [A1] $(x - ?)(x - ?) = 0$ — use quadratic formula: $x = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}$ [M1]

When $x = \frac{3 + \sqrt{17}}{2}$ : $y = 2\left(\frac{3 + \sqrt{17}}{2}\right) - 1 = 3 + \sqrt{17} - 1 = 2 + \sqrt{17}$ [A1]

When $x = \frac{3 - \sqrt{17}}{2}$ : $y = 2\left(\frac{3 - \sqrt{17}}{2}\right) - 1 = 3 - \sqrt{17} - 1 = 2 - \sqrt{17}$ [A1]

Points: $\left(\frac{3 + \sqrt{17}}{2}, 2 + \sqrt{17}\right)$ and $\left(\frac{3 - \sqrt{17}}{2}, 2 - \sqrt{17}\right)$ .

Marking notes: M1 for equating. A1 for quadratic. M1 for solving. A1 each for correct coordinates.

7. Estimate $a$ and $n$ from the linearised graph.

Answer: $y = ax^n \implies \lg y = \lg a + n \lg x$ [M1]

Gradient $= n = 1.5$ [A1] Vertical intercept $= \lg a = 0.15$ [M1] $a = 10^{0.15} \approx 1.41$ [A1]

Marking notes: M1 for stating linear form. A1 for $n$ . M1 for interpreting intercept. A1 for $a$ (accept 1.41 or equivalent).

8. Prove $\frac{\cos 2\theta}{1 - \sin 2\theta} = \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta}$ .

Answer: LHS $= \frac{\cos^2\theta - \sin^2\theta}{1 - 2\sin\theta\cos\theta}$ [M1] $= \frac{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)}{\cos^2\theta + \sin^2\theta - 2\sin\theta\cos\theta}$ [M1] $= \frac{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)}{(\cos\theta - \sin\theta)^2}$ [M1] $= \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} =$ RHS [A1]

Marking notes: M1 for double angle expansions. M1 for factorising numerator. M1 for recognising denominator as perfect square. A1 for complete proof.

9. Solve $3\sin^2 x - 2\cos x - 3 = 0$ for $0^\circ \le x \le 360^\circ$ .

Answer: $3(1 - \cos^2 x) - 2\cos x - 3 = 0$ [M1] $3 - 3\cos^2 x - 2\cos x - 3 = 0$ $-3\cos^2 x - 2\cos x = 0$ $3\cos^2 x + 2\cos x = 0$ [A1] $\cos x(3\cos x + 2) = 0$ [M1]

$\cos x = 0 \implies x = 90^\circ, 270^\circ$ [A1] $\cos x = -\frac{2}{3} \implies x = 180^\circ - 48.2^\circ = 131.8^\circ$ and $x = 180^\circ + 48.2^\circ = 228.2^\circ$ [A1]

Solutions: $x = 90^\circ, 131.8^\circ, 228.2^\circ, 270^\circ$ .

Marking notes: M1 for using $\sin^2 x = 1 - \cos^2 x$ . A1 for simplified equation. M1 for factorising. A1 for $\cos x = 0$ solutions. A1 for $\cos x = -2/3$ solutions (accept 132°, 228°).

Section B: Structured Questions (40 marks)

10. $y = x^3 - 6x^2 + 9x + 5$

(a) Find stationary points.

$\frac{dy}{dx} = 3x^2 - 12x + 9$ [M1] $= 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$ [M1]

Stationary points when $\frac{dy}{dx} = 0$ : $x = 1$ or $x = 3$ [A1]

When $x = 1$ : $y = 1 - 6 + 9 + 5 = 9$ → $(1, 9)$ When $x = 3$ : $y = 27 - 54 + 27 + 5 = 5$ → $(3, 5)$ [A1]

(b) Nature using second derivative.

$\frac{d^2y}{dx^2} = 6x - 12$ [M1]

At $x = 1$ : $\frac{d^2y}{dx^2} = 6 - 12 = -6 < 0$ → maximum [A1] At $x = 3$ : $\frac{d^2y}{dx^2} = 18 - 12 = 6 > 0$ → minimum [A1]

(c) Tangent at $x = 2$ .

At $x = 2$ : $y = 8 - 24 + 18 + 5 = 7$ [M1] Gradient $= 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3$ [M1] Tangent: $y - 7 = -3(x - 2)$ $y = -3x + 13$ [A1]

11.

(a) $7\sin\theta + 24\cos\theta = R\sin(\theta + \alpha)$

$R = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ [M1, A1] $\tan\alpha = \frac{24}{7} \implies \alpha = \tan^{-1}\left(\frac{24}{7}\right) \approx 73.7^\circ$ [A1]

$7\sin\theta + 24\cos\theta = 25\sin(\theta + 73.7^\circ)$

(b) Maximum value of $25\sin(\theta + 73.7^\circ) + 10$ is $25 + 10 = 35$ [A1] Occurs when $\sin(\theta + 73.7^\circ) = 1$ $\theta + 73.7^\circ = 90^\circ \implies \theta = 16.3^\circ$ [A1, A1]

(c) $25\sin(\theta + 73.7^\circ) = 12$ $\sin(\theta + 73.7^\circ) = 0.48$ [M1] $\theta + 73.7^\circ = 28.7^\circ, 180^\circ - 28.7^\circ = 151.3^\circ$ [M1] $\theta = 28.7^\circ - 73.7^\circ = -45^\circ$ (reject) or $\theta = -45^\circ + 360^\circ = 315^\circ$ [A1] $\theta = 151.3^\circ - 73.7^\circ = 77.6^\circ$ [A1]

Solutions: $\theta = 77.6^\circ, 315^\circ$ .

12.

(a) Dimensions of box: length $= 30 - 2x$ , width $= 20 - 2x$ , height $= x$ [M1] $V = x(30 - 2x)(20 - 2x)$ [M1] $= x(600 - 100x + 4x^2)$ $= 4x^3 - 100x^2 + 600x$ [A1]

(b) $\frac{dV}{dx} = 12x^2 - 200x + 600$ [M1] $= 4(3x^2 - 50x + 150) = 0$ [M1] $x = \frac{50 \pm \sqrt{2500 - 1800}}{6} = \frac{50 \pm \sqrt{700}}{6} = \frac{50 \pm 10\sqrt{7}}{6}$ [M1] $x \approx \frac{50 \pm 26.46}{6}$ $x \approx 12.74$ (reject, $> 10$ ) or $x \approx 3.92$ [A1]

$\frac{d^2V}{dx^2} = 24x - 200$ . At $x \approx 3.92$ : $24(3.92) - 200 = -105.9 < 0$ → maximum. [A1]

(c) $V = 4(3.92)^3 - 100(3.92)^2 + 600(3.92)$ $\approx 4(60.24) - 100(15.37) + 2352$ $\approx 240.96 - 1537 + 2352 = 1056 \text{ cm}^3$ [A1, A1]

13. $2x^2 - 5x + 1 = 0$

(a) (i) $\alpha + \beta = \frac{5}{2}$ [A1] (ii) $\alpha\beta = \frac{1}{2}$ [A1]

(b) Sum of new roots $= \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ [M1] $= \left(\frac{5}{2}\right)^2 - 2\left(\frac{1}{2}\right) = \frac{25}{4} - 1 = \frac{21}{4}$ [A1]

Product of new roots $= \alpha^2\beta^2 = (\alpha\beta)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$ [A1]

New equation: $x^2 - \frac{21}{4}x + \frac{1}{4} = 0$ [M1] Multiply by 4: $4x^2 - 21x + 1 = 0$ [A1]

14.

(a) $\frac{\sqrt{48} - \sqrt{27}}{\sqrt{3}} = \frac{4\sqrt{3} - 3\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} = 1$ [M1, A1]

(b) $\sqrt{2x + 5} - \sqrt{x - 1} = 2$ $\sqrt{2x + 5} = 2 + \sqrt{x - 1}$ [M1] Square both sides: $2x + 5 = 4 + 4\sqrt{x - 1} + (x - 1)$ [M1] $2x + 5 = x + 3 + 4\sqrt{x - 1}$ $x + 2 = 4\sqrt{x - 1}$ [A1] Square again: $(x + 2)^2 = 16(x - 1)$ $x^2 + 4x + 4 = 16x - 16$ $x^2 - 12x + 20 = 0$ [M1] $(x - 2)(x - 10) = 0$ $x = 2$ or $x = 10$ [A1]

Check: For $x = 2$ : $\sqrt{9} - \sqrt{1} = 3 - 1 = 2$ ✓ For $x = 10$ : $\sqrt{25} - \sqrt{9} = 5 - 3 = 2$ ✓ [A1]

Both solutions valid.

END OF ANSWER KEY