O Level Additional Mathematics Practice Paper 1

TuitionGoWhere Practice Paper - Additional Mathematics O-Level

Answer Key and Marking Scheme

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Section A [30 marks]

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

(a) Find the coordinates of the centre and the radius of circle C. [4 marks]

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

Centre: $(4, -3)$ [2 marks] Radius: $\sqrt{36} = 6$ [2 marks]

(b) The line $y = x + k$ is tangent to circle C. Find the possible values of $k$. [4 marks]

Answer: Substitute $y = x + k$ into circle equation: $x^2 + (x + k)^2 - 8x + 6(x + k) - 11 = 0$ $x^2 + x^2 + 2kx + k^2 - 8x + 6x + 6k - 11 = 0$ $2x^2 + (2k - 2)x + (k^2 + 6k - 11) = 0$ [2 marks]

For tangency, discriminant = 0: $(2k - 2)^2 - 4(2)(k^2 + 6k - 11) = 0$ $4k^2 - 8k + 4 - 8k^2 - 48k + 88 = 0$ $-4k^2 - 56k + 92 = 0$ $k^2 + 14k - 23 = 0$ [1 mark]

$k = \frac{-14 \pm \sqrt{196 + 92}}{2} = \frac{-14 \pm \sqrt{288}}{2} = \frac{-14 \pm 12\sqrt{2}}{2} = -7 \pm 6\sqrt{2}$ [1 mark]

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2. The curve $y = 2x^3 - 9x^2 + 12x - 1$ has two stationary points.

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

Answer: $\frac{dy}{dx} = 6x^2 - 18x + 12$ [1 mark]

For stationary points: $6x^2 - 18x + 12 = 0$ $x^2 - 3x + 2 = 0$ $(x - 1)(x - 2) = 0$ $x = 1$ or $x = 2$ [2 marks]

When $x = 1$: $y = 2(1) - 9(1) + 12(1) - 1 = 4$ When $x = 2$: $y = 2(8) - 9(4) + 12(2) - 1 = 3$

Stationary points: $(1, 4)$ and $(2, 3)$ [1 mark]

(b) Determine the nature of each stationary point. [3 marks]

Answer: $\frac{d^2y}{dx^2} = 12x - 18$ [1 mark]

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

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3. Express $\frac{5x + 7}{(x + 1)(x - 2)}$ in partial fractions. [4 marks]

Answer: $\frac{5x + 7}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}$ [1 mark]

$5x + 7 = A(x - 2) + B(x + 1)$ [1 mark]

Let $x = -1$: $5(-1) + 7 = A(-3) \Rightarrow 2 = -3A \Rightarrow A = -\frac{2}{3}$ Let $x = 2$: $5(2) + 7 = B(3) \Rightarrow 17 = 3B \Rightarrow B = \frac{17}{3}$ [1 mark]

$\frac{5x + 7}{(x + 1)(x - 2)} = \frac{-2}{3(x + 1)} + \frac{17}{3(x - 2)}$ [1 mark]

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4. The quadratic function $f(x) = x^2 - 4x + k$ is always positive.

(a) Find the range of values of $k$. [3 marks]

Answer: For $f(x) > 0$ for all real $x$, discriminant < 0 [1 mark] $\Delta = (-4)^2 - 4(1)(k) = 16 - 4k < 0$ [1 mark] $k > 4$ [1 mark]

(b) Given that $k = 5$, find the minimum value of $f(x)$ and the value of $x$ at which this occurs. [3 marks]

Answer: $f(x) = x^2 - 4x + 5$ Complete the square: $f(x) = (x - 2)^2 + 1$ [2 marks] Minimum value: 1, occurring at $x = 2$ [1 mark]

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5. Solve the equation $3\cos 2x + 5\sin x = 1$ for $0° \leq x \leq 360°$. [5 marks]

Answer: Using $\cos 2x = 1 - 2\sin^2 x$: $3(1 - 2\sin^2 x) + 5\sin x = 1$ $3 - 6\sin^2 x + 5\sin x = 1$ $6\sin^2 x - 5\sin x - 2 = 0$ [2 marks]

Let $u = \sin x$: $6u^2 - 5u - 2 = 0$ $(6u + 3)(u - \frac{2}{3}) = 0$ or $(3u + 1)(2u - 2) = 0$ $u = -\frac{1}{2}$ or $u = \frac{2}{3}$ [2 marks]

$\sin x = -\frac{1}{2}$: $x = 210°, 330°$ $\sin x = \frac{2}{3}$: $x = 41.8°, 138.2°$ [1 mark]

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Section B [35 marks]

6. The diagram shows the graph of $y = f(x)$ where $f(x) = \frac{x^2 - 4}{x - 1}$.

(a) Find the equations of the asymptotes of the curve. [3 marks]

Answer: Vertical asymptote: $x = 1$ (denominator = 0) [1 mark]

For oblique asymptote, divide: $\frac{x^2 - 4}{x - 1} = x + 1 - \frac{3}{x - 1}$ As $x \to \infty$, $y \to x + 1$ Oblique asymptote: $y = x + 1$ [2 marks]

(b) Find the coordinates of the points where the curve intersects the coordinate axes. [4 marks]

Answer: $y$-intercept (when $x = 0$): $y = \frac{0 - 4}{0 - 1} = 4$ Point: $(0, 4)$ [2 marks]

$x$-intercepts (when $y = 0$): $x^2 - 4 = 0$ $x = \pm 2$ Points: $(-2, 0)$ and $(2, 0)$ [2 marks]

(c) The line $y = mx + c$ passes through the point $(3, 5)$ and is tangent to the curve. Find the values of $m$ and $c$. [6 marks]

Answer: $f'(x) = \frac{(x-1)(2x) - (x^2-4)(1)}{(x-1)^2} = \frac{x^2 - 2x + 4}{(x-1)^2}$ [2 marks]

At tangent point $(a, f(a))$: slope = $f'(a) = m$ Line equation: $y - f(a) = m(x - a)$ Since line passes through $(3, 5)$: $5 - f(a) = m(3 - a)$ [2 marks]

Also: $5 = 3m + c$ and $f(a) = ma + c$ Solving system with tangency condition gives $a = 0$ or $a = 4$

When $a = 0$: $m = 4, c = -7$ When $a = 4$: $m = \frac{4}{9}, c = \frac{31}{9}$ [2 marks]

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7. A particle moves along a straight line such that its displacement $s$ metres from a fixed point O at time $t$ seconds is given by $s = t^3 - 6t^2 + 9t + 2$.

(a) Find expressions for the velocity and acceleration of the particle at time $t$. [2 marks]

Answer: $v = \frac{ds}{dt} = 3t^2 - 12t + 9$ [1 mark] $a = \frac{dv}{dt} = 6t - 12$ [1 mark]

(b) Find the times when the particle is momentarily at rest. [3 marks]

Answer: $v = 0$: $3t^2 - 12t + 9 = 0$ $t^2 - 4t + 3 = 0$ $(t - 1)(t - 3) = 0$ [2 marks] $t = 1$ or $t = 3$ seconds [1 mark]

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

Answer: At $t = 0$: $s = 2$ At $t = 1$: $s = 1 - 6 + 9 + 2 = 6$ At $t = 3$: $s = 27 - 54 + 27 + 2 = 2$ At $t = 4$: $s = 64 - 96 + 36 + 2 = 6$ [2 marks]

Distance = $|6 - 2| + |2 - 6| + |6 - 2| = 4 + 4 + 4 = 12$ metres [2 marks]

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8. The population of a bacterial culture grows according to the model $P = P_0 e^{kt}$.

(a) Given that the population doubles in 3 hours, find the value of $k$ to 3 significant figures. [3 marks]

Answer: $2P_0 = P_0 e^{3k}$ $2 = e^{3k}$ $\ln 2 = 3k$ [2 marks] $k = \frac{\ln 2}{3} = 0.231$ (3 s.f.) [1 mark]

(b) If the initial population is 500 bacteria, find the population after 8 hours. [2 marks]

Answer: $P = 500e^{0.231 \times 8} = 500e^{1.848} = 500 \times 6.35 = 3175$ bacteria [2 marks]

(c) Find the time taken for the population to reach 10,000 bacteria. [3 marks]

Answer: $10000 = 500e^{0.231t}$ $20 = e^{0.231t}$ $\ln 20 = 0.231t$ [2 marks] $t = \frac{\ln 20}{0.231} = 13.0$ hours [1 mark]

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9. Express $4\cos x - 3\sin x$ in the form $R\cos(x + \alpha)$.

(a) Find the values of $R$ and $\alpha$. [3 marks]

Answer: $R = \sqrt{4^2 + (-3)^2} = \sqrt{25} = 5$ [1 mark] $\tan \alpha = \frac{3}{4}$ $\alpha = 36.9°$ [2 marks]

(b) Hence, or otherwise, find the maximum and minimum values of $4\cos x - 3\sin x + 2$. [2 marks]

Answer: $4\cos x - 3\sin x = 5\cos(x + 36.9°)$ Maximum value of $5\cos(x + 36.9°) + 2 = 5 + 2 = 7$ [1 mark] Minimum value = $-5 + 2 = -3$ [1 mark]

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Section C [25 marks]

10. The circle $C_1$ has centre $(2, -1)$ and radius 3. The circle $C_2$ has equation $x^2 + y^2 - 6x + 4y + 9 = 0$.

(a) Find the equation of circle $C_1$ in the form $x^2 + y^2 + 2gx + 2fy + c = 0$. [2 marks]

Answer: $(x - 2)^2 + (y + 1)^2 = 9$ $x^2 - 4x + 4 + y^2 + 2y + 1 = 9$ $x^2 + y^2 - 4x + 2y - 4 = 0$ [2 marks]

(b) Show that the circles $C_1$ and $C_2$ intersect at two points. [4 marks]

Answer: $C_2$: Complete the square: $(x - 3)^2 + (y + 2)^2 = 4$ Centre of $C_2$: $(3, -2)$, radius = 2 [2 marks]

Distance between centres: $\sqrt{(3-2)^2 + (-2-(-1))^2} = \sqrt{2} = 1.41$ Since $|3 - 2| < 1.41 < 3 + 2$, circles intersect at two points [2 marks]

(c) Find the coordinates of the points of intersection. [6 marks]

Answer: Subtract equations: $(x^2 + y^2 - 4x + 2y - 4) - (x^2 + y^2 - 6x + 4y + 9) = 0$ $2x - 2y - 13 = 0$ $y = x - 6.5$ [2 marks]

Substitute into $C_1$: $(x - 2)^2 + (x - 6.5 + 1)^2 = 9$ $(x - 2)^2 + (x - 5.5)^2 = 9$ $x^2 - 4x + 4 + x^2 - 11x + 30.25 = 9$ $2x^2 - 15x + 25.25 = 0$ [2 marks]

$x = \frac{15 \pm \sqrt{225 - 202}}{4} = \frac{15 \pm \sqrt{23}}{4}$ Points: $(\frac{15 + \sqrt{23}}{4}, \frac{3 + \sqrt{23}}{4})$ and $(\frac{15 - \sqrt{23}}{4}, \frac{3 - \sqrt{23}}{4})$ [2 marks]

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11. A rectangular piece of cardboard has dimensions 20 cm by 15 cm.

(a) Show that the volume $V$ cm³ of the box is given by $V = x(20 - 2x)(15 - 2x)$. [2 marks]

Answer: After cutting squares of side $x$ from corners: Length = $20 - 2x$, Width = $15 - 2x$, Height = $x$ $V = x(20 - 2x)(15 - 2x)$ [2 marks]

(b) Find the value of $x$ that maximizes the volume. [5 marks]

Answer: $V = x(300 - 40x - 30x + 4x^2) = x(300 - 70x + 4x^2)$ $V = 4x^3 - 70x^2 + 300x$ [1 mark]

$\frac{dV}{dx} = 12x^2 - 140x + 300$ [1 mark]

For maximum: $12x^2 - 140x + 300 = 0$ $3x^2 - 35x + 75 = 0$ [1 mark]

$x = \frac{35 \pm \sqrt{1225 - 900}}{6} = \frac{35 \pm \sqrt{325}}{6} = \frac{35 \pm 5\sqrt{13}}{6}$

$x = 2.49$ or $x = 10.01$ [1 mark]

Since $x < 7.5$ (constraint), $x = 2.49$ cm [1 mark]

(c) Calculate the maximum volume. [2 marks]

Answer: $V = 2.49(20 - 4.98)(15 - 4.98) = 2.49 \times 15.02 \times 10.02 = 375$ cm³ [2 marks]

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12. The polynomial $P(x) = 2x^3 + ax^2 + bx - 12$ has factors $(x - 2)$ and $(x + 3)$.

(a) Find the values of $a$ and $b$. [4 marks]

Answer: $P(2) = 0$: $2(8) + a(4) + b(2) - 12 = 0$ $16 + 4a + 2b - 12 = 0$ $4a + 2b = -4$ $2a + b = -2$ ... (1) [2 marks]

$P(-3) = 0$: $2(-27) + a(9) + b(-3) - 12 = 0$ $-54 + 9a - 3b - 12 = 0$ $9a - 3b = 66$ $3a - b = 22$ ... (2) [1 mark]

From (1) and (2): $a = 4, b = -10$ [1 mark]

(b) Hence, solve the equation $P(x) = 0$. [2 marks]

Answer: $P(x) = 2x^3 + 4x^2 - 10x - 12 = 2(x - 2)(x + 3)(x + 1)$ [1 mark] Solutions: $x = 2, -3, -1$ [1 mark]