Secondary 4 Additional Mathematics Preliminary Examination Paper 5

TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

PRELIMINARY EXAMINATION — Version 5 — ANSWER KEY & MARKING SCHEME

TuitionGoWhere Secondary School (AI)

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Section A (36 marks)

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Question 1

(a) Gradient of AB: $m = \dfrac{10 - 2}{7 - 1} = \dfrac{8}{6} = \dfrac{4}{3}$ [M1]

Equation: $y - 2 = \dfrac{4}{3}(x - 1)$ → $y = \dfrac{4}{3}x + \dfrac{2}{3}$ [A1]

(b) Midpoint of AB: $\left(\dfrac{1+7}{2}, \dfrac{2+10}{2}\right) = (4, 6)$ [M1]

Gradient of L: $m_L = -\dfrac{3}{4}$ (perpendicular to AB) [M1]

Equation of L: $y - 6 = -\dfrac{3}{4}(x - 4)$ → $y = -\dfrac{3}{4}x + 9$ [A1]

(c) At x-axis, $y = 0$: $0 = -\dfrac{3}{4}x + 9$ → $x = 12$ [M1]

Coordinates: (12, 0) [A1]

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Question 2

(a) $y = x^3 - 6x^2 + 9x + 4$

$\dfrac{dy}{dx} = 3x^2 - 12x + 9$ [M1]

Set $\dfrac{dy}{dx} = 0$: $3x^2 - 12x + 9 = 0$ → $x^2 - 4x + 3 = 0$ → $(x-1)(x-3) = 0$ [M1]

$x = 1$ or $x = 3$ [A1]

When $x = 1$: $y = 1 - 6 + 9 + 4 = 8$ → (1, 8)

When $x = 3$: $y = 27 - 54 + 27 + 4 = 4$ → (3, 4) [A1]

(b) $\dfrac{d^2y}{dx^2} = 6x - 12$ [M1]

At $x = 1$: $\dfrac{d^2y}{dx^2} = 6(1) - 12 = -6 < 0$ → maximum point [A1]

At $x = 3$: $\dfrac{d^2y}{dx^2} = 6(3) - 12 = 6 > 0$ → minimum point [A1]

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Question 3

(a) $x^2 + y^2 - 8x + 6y + 9 = 0$

Complete the square: $(x^2 - 8x) + (y^2 + 6y) = -9$

$(x - 4)^2 - 16 + (y + 3)^2 - 9 = -9$ [M1]

$(x - 4)^2 + (y + 3)^2 = 16$ [M1]

Centre: (4, -3), Radius: 4 [A1]

(b) Centre C(4, -3), point P(5, -2)

Gradient of CP: $m_{CP} = \dfrac{-2 - (-3)}{5 - 4} = \dfrac{1}{1} = 1$ [M1]

Gradient of tangent at P: $m_T = -1$ (perpendicular to radius) [M1]

Equation of tangent: $y - (-2) = -1(x - 5)$ → $y = -x + 3$ [A1]

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Question 4

(a) Gradient of AB: $m_{AB} = \dfrac{3 - 1}{8 - 2} = \dfrac{2}{6} = \dfrac{1}{3}$ [M1]

Gradient of DC: $m_{DC} = \dfrac{7 - 9}{4 - 10} = \dfrac{-2}{-6} = \dfrac{1}{3}$

Since $m_{AB} = m_{DC}$, AB ∥ DC. [A1]

(b) Gradient of AD: $m_{AD} = \dfrac{7 - 1}{4 - 2} = \dfrac{6}{2} = 3$ [M1]

$m_{AB} \times m_{AD} = \dfrac{1}{3} \times 3 = 1$... Wait, check: $\dfrac{1}{3} \times 3 = 1 \neq -1$.

Recheck: AB gradient = $\dfrac{3-1}{8-2} = \dfrac{2}{6} = \dfrac{1}{3}$. AD gradient = $\dfrac{7-1}{4-2} = \dfrac{6}{2} = 3$.

$m_{AB} \times m_{AD} = \dfrac{1}{3} \times 3 = 1$. This is NOT perpendicular.

Correction: The question should be checked. For AB ⟂ AD, we need $m_{AB} \times m_{AD} = -1$. With A(2,1), B(8,3), D(4,7): $m_{AB} = \frac{1}{3}$, $m_{AD} = 3$. Product = 1, not -1. The quadrilateral as given is a parallelogram but not necessarily a rectangle.

Revised marking based on actual coordinates: $m_{AB} = \frac{1}{3}$, $m_{AD} = 3$ [M1] $m_{AB} \times m_{AD} = 1 \neq -1$, so AB is NOT perpendicular to AD. [A1 — accept correct conclusion]

Note: If the question intended perpendicularity, coordinates would need adjustment. Mark according to correct mathematical working.

(c) Since AB ∥ DC and AD ∥ BC (parallelogram), area = base × height.

Length of AB: $\sqrt{(8-2)^2 + (3-1)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$ [M1]

Perpendicular distance from D to line AB:

Line AB: $y - 1 = \frac{1}{3}(x - 2)$ → $x - 3y + 1 = 0$ [M1]

Distance from D(4,7): $\dfrac{|4 - 3(7) + 1|}{\sqrt{1^2 + (-3)^2}} = \dfrac{|4 - 21 + 1|}{\sqrt{10}} = \dfrac{16}{\sqrt{10}}$ [M1]

Area = $2\sqrt{10} \times \dfrac{16}{\sqrt{10}} = 32$ square units [A1]

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Question 5

(a) $y = \dfrac{2x + 1}{x - 3}$

Crosses y-axis ($x = 0$): $y = \dfrac{1}{-3} = -\dfrac{1}{3}$ → $(0, -\frac{1}{3})$ [M1]

Crosses x-axis ($y = 0$): $2x + 1 = 0$ → $x = -\dfrac{1}{2}$ → $(-\frac{1}{2}, 0)$ [A1]

(b) $\dfrac{dy}{dx} = \dfrac{(x-3)(2) - (2x+1)(1)}{(x-3)^2} = \dfrac{2x - 6 - 2x - 1}{(x-3)^2} = \dfrac{-7}{(x-3)^2}$ [M1]

For stationary points, $\dfrac{dy}{dx} = 0$: $\dfrac{-7}{(x-3)^2} = 0$ [M1]

Since numerator is -7 (never zero) and denominator is always positive (for $x \neq 3$), $\dfrac{dy}{dx}$ is never zero. Therefore, the curve has no stationary points. [A1]

(c) Vertical asymptote: $x = 3$ (denominator = 0) [A1]

Horizontal asymptote: As $x \to \pm\infty$, $y \to \dfrac{2x}{x} = 2$, so $y = 2$ [A1]

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Section B (24 marks)

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Question 6

(a) Intersection: $2x + k = x^2 + 3x - 1$

$x^2 + 3x - 1 - 2x - k = 0$ → $x^2 + x - (k + 1) = 0$ [M1]

Discriminant: $\Delta = 1^2 - 4(1)(-(k+1)) = 1 + 4k + 4 = 4k + 5$ [M1]

Wait — recheck: $x^2 + x - k - 1 = 0$, so $a=1, b=1, c=-(k+1)$.

$\Delta = 1^2 - 4(1)(-(k+1)) = 1 + 4k + 4 = 4k + 5$.

The question states "show that its discriminant is $4k + 17$". This does not match.

Correction to question: For discriminant to be $4k + 17$, the curve should be $y = x^2 + 3x + 4$ or similar. With $y = x^2 + 3x - 1$ and line $y = 2x + k$:

$x^2 + 3x - 1 = 2x + k$ → $x^2 + x - (k+1) = 0$. Discriminant = $1 + 4(k+1) = 4k + 5$.

Revised marking: Accept correct discriminant $4k + 5$ with full working. [A1 for correct discriminant]

(b) For two distinct points: $\Delta > 0$ → $4k + 5 > 0$ → $k > -\dfrac{5}{4}$ [M1, A1]

(c) For tangent: $\Delta = 0$ → $4k + 5 = 0$ → $k = -\dfrac{5}{4}$ [M1]

When $k = -\frac{5}{4}$: $x^2 + x - (-\frac{5}{4} + 1) = 0$ → $x^2 + x + \frac{1}{4} = 0$ → $(x + \frac{1}{2})^2 = 0$ → $x = -\frac{1}{2}$ [M1]

$y = 2(-\frac{1}{2}) - \frac{5}{4} = -1 - \frac{5}{4} = -\frac{9}{4}$

Point of contact: $(-\frac{1}{2}, -\frac{9}{4})$ [A1]

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Question 7

(a) Centre (4, -1), passes through (0, 0).

$r^2 = (0-4)^2 + (0-(-1))^2 = 16 + 1 = 17$ [M1]

Equation: $(x - 4)^2 + (y + 1)^2 = 17$ [A1]

(b) Radius of $C_2$: $r_2 = \sqrt{17}$

Distance between centres: $d = \sqrt{(10-4)^2 + (7-(-1))^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ [M1]

For external tangency: $d = r_2 + r_3$ → $10 = \sqrt{17} + r_3$ [M1]

$r_3 = 10 - \sqrt{17}$ [M1]

Equation of $C_3$: $(x - 10)^2 + (y - 7)^2 = (10 - \sqrt{17})^2$ [A1]

(c) Point of contact lies on line joining centres. Parametric: from (4,-1) toward (10,7), ratio $\sqrt{17} : (10 - \sqrt{17})$.

Point of contact: $\left(4 + \dfrac{\sqrt{17}}{10}(6), -1 + \dfrac{\sqrt{17}}{10}(8)\right)$ [M1]

Gradient from origin to point of contact = gradient of line joining centres = $\dfrac{8}{6} = \dfrac{4}{3}$.

Line from origin with gradient $\frac{4}{3}$: $y = \frac{4}{3}x$. The point of contact lies on this line (since centres and contact point are collinear, and origin lies on the same line as centres? Check: (0,0), (4,-1), (10,7) — gradient (0,0) to (4,-1) is $-\frac{1}{4}$, not $\frac{4}{3}$. So origin is NOT collinear with centres.)

Revised approach: The common tangent at the point of contact is perpendicular to the line of centres. Gradient of line of centres = $\frac{8}{6} = \frac{4}{3}$. Gradient of common tangent = $-\frac{3}{4}$.

Equation of common tangent: passes through point of contact. Does it pass through origin? Check if origin satisfies. This requires the specific point of contact.

Point of contact: $\left(\dfrac{4(10-\sqrt{17}) + 10\sqrt{17}}{10}, \dfrac{-1(10-\sqrt{17}) + 7\sqrt{17}}{10}\right)$ = $\left(\dfrac{40 + 6\sqrt{17}}{10}, \dfrac{-10 + 8\sqrt{17}}{10}\right)$ = $\left(4 + 0.6\sqrt{17}, -1 + 0.8\sqrt{17}\right)$

Tangent gradient = $-\frac{3}{4}$. Equation: $y - (-1 + 0.8\sqrt{17}) = -\frac{3}{4}(x - (4 + 0.6\sqrt{17}))$.

Check if (0,0) satisfies: LHS = $1 - 0.8\sqrt{17}$; RHS = $-\frac{3}{4}(-4 - 0.6\sqrt{17}) = 3 + 0.45\sqrt{17}$. These are not equal for general $\sqrt{17}$.

Conclusion: The statement in part (c) is not generally true with the given numbers. Mark according to correct mathematical reasoning — accept valid demonstration that it does or does not pass through the origin. [M1 for method, A1 for correct conclusion with working]

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Question 8

(a) $y = ax^n$ → $\log y = \log a + n \log x$ [M1]

Plot $\log y$ on vertical axis against $\log x$ on horizontal axis. The graph will be a straight line with gradient $n$ and vertical intercept $\log a$. [A1]

(b) Using points (2, 5.66) and (10, 63.2):

$\log 2 \approx 0.3010$, $\log 5.66 \approx 0.7528$

$\log 10 = 1$, $\log 63.2 \approx 1.8007$ [M1]

Gradient $n = \dfrac{1.8007 - 0.7528}{1 - 0.3010} = \dfrac{1.0479}{0.6990} \approx 1.50$ [M1]

$\log a = \log y - n \log x$. Using (2, 5.66): $\log a = 0.7528 - 1.50(0.3010) = 0.7528 - 0.4515 = 0.3013$ [M1]

$a = 10^{0.3013} \approx 2.00$ [A1]

$n \approx 1.5$, $a \approx 2.0$

(c) $y = 2.0 \times 12^{1.5} = 2.0 \times \sqrt{12^3} = 2.0 \times \sqrt{1728}$ [M1]

$\sqrt{1728} \approx 41.57$ → $y \approx 83.1$ [A1]

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Question 9

(a) $s = t^3 - 9t^2 + 24t + 5$

$v = \dfrac{ds}{dt} = 3t^2 - 18t + 24$ [M1]

$a = \dfrac{dv}{dt} = 6t - 18$ [A1]

(b) At rest: $v = 0$ → $3t^2 - 18t + 24 = 0$ → $t^2 - 6t + 8 = 0$ [M1]

$(t - 2)(t - 4) = 0$ → $t = 2$ or $t = 4$ [A1]

(c) At $t = 2$: $a = 6(2) - 18 = -6$ m/s² [A1]

At $t = 4$: $a = 6(4) - 18 = 6$ m/s² [A1]

(d) Displacement at key times:

$t = 0$: $s = 5$

$t = 2$: $s = 8 - 36 + 48 + 5 = 25$

$t = 4$: $s = 64 - 144 + 96 + 5 = 21$

$t = 5$: $s = 125 - 225 + 120 + 5 = 25$ [M1]

Motion: 0→2: from 5 to 25 (forward 20 m)

2→4: from 25 to 21 (backward 4 m)

4→5: from 21 to 25 (forward 4 m) [M1]

Total distance = $20 + 4 + 4 = 28$ m [A1]

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Question 10

(a) Crosses y-axis ($x = 0$): $y = (0-2)^2(0+1) = 4 \times 1 = 4$ → (0, 4) [A1]

(b) $y = (x-2)^2(x+1) = (x^2 - 4x + 4)(x+1) = x^3 - 3x^2 + 0x + 4$ [M1]

$\dfrac{dy}{dx} = 3x^2 - 6x$ [M1]

Set $\dfrac{dy}{dx} = 0$: $3x(x - 2) = 0$ → $x = 0$ or $x = 2$ [M1]

When $x = 0$: $y = 4$ → (0, 4)

When $x = 2$: $y = 0$ → (2, 0) [A1]

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

At $x = 0$: $\dfrac{d^2y}{dx^2} = -6 < 0$ → maximum at (0, 4)

At $x = 2$: $\dfrac{d^2y}{dx^2} = 6 > 0$ → minimum at (2, 0) [A1]

(c) Sketch: [4 marks awarded for:]

• Correct intercepts: (-1, 0), (0, 4), (2, 0) [1]
• Stationary points correctly plotted: max at (0, 4), min at (2, 0) [1]
• Correct shape: cubic with positive leading coefficient (rises to right, falls to left) [1]
• Behaviour: as $x \to \infty$, $y \to \infty$; as $x \to -\infty$, $y \to -\infty$ [1]

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— END OF ANSWER KEY —