Secondary 3 Additional Mathematics Practice Paper 3

TuitionGoWhere Practice Paper - Additional Mathematics Secondary 3

Answer Key and Marking Scheme (Version 3)

Total Marks: 80

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Section A: Short Answer Questions (40 marks)

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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.

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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.

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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 $b$.

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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.

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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.

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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.

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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).

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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.

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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°).

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Section B: Structured Questions (40 marks)

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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]

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

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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]

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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]

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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.

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END OF ANSWER KEY