Secondary 4 Additional Mathematics Preliminary Examination Paper 1

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

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Section A [40 marks]

1. The curve $C$ has equation $y = 2x^3 - 9x^2 + 12x - 1$.

(a) Find $\frac{dy}{dx}$. [2 marks]

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

Marking: 2 marks for correct differentiation

(b) Find the coordinates of the stationary points of $C$. [5 marks]

Working: $6x^2 - 18x + 12 = 0$ $6(x^2 - 3x + 2) = 0$ $6(x - 1)(x - 2) = 0$ $x = 1$ or $x = 2$

When $x = 1$: $y = 2(1) - 9(1) + 12(1) - 1 = 4$ When $x = 2$: $y = 2(8) - 9(4) + 12(2) - 1 = 16 - 36 + 24 - 1 = 3$

Answer: $(1, 4)$ and $(2, 3)$

Marking: 1 mark for setting derivative = 0, 2 marks for solving quadratic, 2 marks for y-coordinates

(c) Determine the nature of each stationary point. [4 marks]

Working: $\frac{d^2y}{dx^2} = 12x - 18$

At $x = 1$: $\frac{d^2y}{dx^2} = 12 - 18 = -6 < 0$ → maximum At $x = 2$: $\frac{d^2y}{dx^2} = 24 - 18 = 6 > 0$ → minimum

Answer: $(1, 4)$ is a local maximum, $(2, 3)$ is a local minimum

Marking: 1 mark for second derivative, 1 mark for each evaluation, 1 mark for conclusions

(d) Sketch the curve $C$. [3 marks]

Answer: Sketch showing curve with maximum at $(1, 4)$, minimum at $(2, 3)$, y-intercept at $(0, -1)$

Marking: 1 mark for general cubic shape, 1 mark for stationary points, 1 mark for y-intercept

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2. The circle $C_1$ has equation $x^2 + y^2 - 6x + 4y - 12 = 0$.

(a) Find the centre and radius of $C_1$. [3 marks]

Working: Complete the square: $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$ $(x - 3)^2 + (y + 2)^2 = 25$

Answer: Centre $(3, -2)$, radius $5$

Marking: 2 marks for completing the square, 1 mark for centre and radius

(b) Find the coordinates of the points where $C_1$ intersects the x-axis. [4 marks]

Working: On x-axis, $y = 0$: $x^2 - 6x - 12 = 0$ Using quadratic formula: $x = \frac{6 \pm \sqrt{36 + 48}}{2} = \frac{6 \pm \sqrt{84}}{2} = \frac{6 \pm 2\sqrt{21}}{2} = 3 \pm \sqrt{21}$

Answer: $(3 + \sqrt{21}, 0)$ and $(3 - \sqrt{21}, 0)$

Marking: 1 mark for setting $y = 0$, 2 marks for solving quadratic, 1 mark for both coordinates

(c) Find the equation of $C_2$. [3 marks]

Working: Centre $(1, -2)$, passes through origin Radius = $\sqrt{1^2 + (-2)^2} = \sqrt{5}$ $(x - 1)^2 + (y + 2)^2 = 5$ $x^2 - 2x + 1 + y^2 + 4y + 4 = 5$ $x^2 + y^2 - 2x + 4y = 0$

Answer: $x^2 + y^2 - 2x + 4y = 0$

Marking: 1 mark for radius, 2 marks for expanding to general form

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3. Quadrilateral problem.

(a) Find the equation of line $AB$. [2 marks]

Working: $A(6, 0)$, $B(8, 4)$ Gradient = $\frac{4 - 0}{8 - 6} = 2$ $y - 0 = 2(x - 6)$ $y = 2x - 12$

Answer: $y = 2x - 12$

Marking: 1 mark for gradient, 1 mark for equation

(b) Find the equation of line $OC$. [2 marks]

Working: $OC \perp AB$, so gradient of $OC = -\frac{1}{2}$ Through origin: $y = -\frac{1}{2}x$

Answer: $y = -\frac{1}{2}x$

Marking: 1 mark for perpendicular gradient, 1 mark for equation

(c) Find coordinates of point $C$. [3 marks]

Working: $C$ is intersection of $OC$ and line through $B$ perpendicular to $AB$ Line through $B(8, 4)$ with gradient $-\frac{1}{2}$: $y - 4 = -\frac{1}{2}(x - 8)$ $y = -\frac{1}{2}x + 8$

Intersection with $y = -\frac{1}{2}x$: $-\frac{1}{2}x = -\frac{1}{2}x + 8$ This gives $0 = 8$, which is incorrect.

Correct approach: $C$ lies on $OC: y = -\frac{1}{2}x$ and $OC \perp AB$ Need additional constraint from diagram.

Answer: Coordinates depend on diagram constraints

Marking: Method marks for approach

(d) Show that $OABC$ is a trapezium. [2 marks]

Working: Need to show one pair of opposite sides are parallel

Marking: 2 marks for showing parallel sides

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Section B [40 marks]

4. Function $f(x) = x^3 - 3x^2 + 4$.

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

Working: $f'(x) = 3x^2 - 6x = 3x(x - 2) = 0$ $x = 0$ or $x = 2$

When $x = 0$: $f(0) = 4$ When $x = 2$: $f(2) = 8 - 12 + 4 = 0$

Answer: $(0, 4)$ and $(2, 0)$

Marking: 2 marks for derivative and solving, 2 marks for coordinates

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

Working: $f''(x) = 6x - 6$

At $x = 0$: $f''(0) = -6 < 0$ → maximum At $x = 2$: $f''(2) = 6 > 0$ → minimum

Answer: $(0, 4)$ maximum, $(2, 0)$ minimum

Marking: 1 mark for second derivative, 2 marks for nature

(c) Find range where $f(x)$ is decreasing. [2 marks]

Working: $f'(x) < 0$ when $3x(x - 2) < 0$, so $0 < x < 2$

Answer: $0 < x < 2$

Marking: 2 marks for correct interval

(d) Find tangent line at $x = 1$. [4 marks]

Working: At $x = 1$: $f(1) = 1 - 3 + 4 = 2$, $f'(1) = 3 - 6 = -3$ Tangent: $y - 2 = -3(x - 1)$ $y = -3x + 5$

Answer: $m = -3$, $c = 5$

Marking: 2 marks for point and gradient, 2 marks for equation

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5. Curve $y = \frac{x^2 - 4}{x - 1}$.

(a) Express in partial fraction form. [3 marks]

Working: $\frac{x^2 - 4}{x - 1} = \frac{(x-1)(x+1) + 3}{x-1} = x + 1 + \frac{3}{x-1}$

Answer: $y = x + 1 + \frac{3}{x - 1}$

Marking: 3 marks for polynomial division

(b) Find asymptotes. [3 marks]

Working: Vertical asymptote: $x = 1$ Oblique asymptote: $y = x + 1$

Answer: $x = 1$ and $y = x + 1$

Marking: 1 mark for vertical, 2 marks for oblique

(c) Find $\frac{dy}{dx}$. [3 marks]

Working: $\frac{dy}{dx} = 1 + \frac{d}{dx}\left[\frac{3}{x-1}\right] = 1 - \frac{3}{(x-1)^2}$

Answer: $\frac{dy}{dx} = 1 - \frac{3}{(x-1)^2}$

Marking: 3 marks for correct differentiation

(d) Find stationary points. [4 marks]

Working: $1 - \frac{3}{(x-1)^2} = 0$ $(x-1)^2 = 3$ $x - 1 = \pm\sqrt{3}$ $x = 1 \pm \sqrt{3}$

When $x = 1 + \sqrt{3}$: $y = (1 + \sqrt{3}) + 1 + \frac{3}{\sqrt{3}} = 2 + \sqrt{3} + \sqrt{3} = 2 + 2\sqrt{3}$ When $x = 1 - \sqrt{3}$: $y = 2 - 2\sqrt{3}$

Answer: $(1 + \sqrt{3}, 2 + 2\sqrt{3})$ and $(1 - \sqrt{3}, 2 - 2\sqrt{3})$

Marking: 2 marks for solving equation, 2 marks for y-coordinates

(e) Sketch curve. [3 marks]

Answer: Sketch showing asymptotes and stationary points

Marking: 1 mark for asymptotes, 1 mark for stationary points, 1 mark for general shape

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6. Circle problems.

(a) State centre and radius of $S$. [2 marks]

Answer: Centre $(2, -1)$, radius $5$

Marking: 1 mark each

(b) Find tangent at $(6, 2)$. [4 marks]

Working: Gradient of radius = $\frac{2-(-1)}{6-2} = \frac{3}{4}$ Gradient of tangent = $-\frac{4}{3}$ Tangent: $y - 2 = -\frac{4}{3}(x - 6)$ $y = -\frac{4}{3}x + 10$

Answer: $y = -\frac{4}{3}x + 10$

Marking: 2 marks for gradient, 2 marks for equation

(c) Find radius of circle $T$. [3 marks]

Working: Distance between centres = $\sqrt{(-1-2)^2 + (3-(-1))^2} = \sqrt{9 + 16} = 5$ External tangency: $r_T + 5 = 5$, so $r_T = 0$ (impossible) Actually: $r_T + 5 = 5$, so we need distance = sum of radii Let $r_T$ be radius of $T$. Distance = $5$, so $r_T + 5 = 5$ gives $r_T = 0$.

Recalculating: Distance = $5$, for external tangency: $r_T = 5 - 5 = 0$ (impossible) Actually distance between centres should equal sum of radii for external tangency. Distance = $5$, so $r_T + 5 = 5$ is impossible.

Let me recalculate distance: $\sqrt{(-1-2)^2 + (3-(-1))^2} = \sqrt{9 + 16} = 5$

For external tangency: distance = $r_S + r_T = 5 + r_T$ So $5 = 5 + r_T$, giving $r_T = 0$ (impossible)

The question likely has an error. Assuming different interpretation...

Answer: Need clarification on problem setup

(d) Find equation of circle $T$. [2 marks]

Answer: Depends on part (c)

(e) Find line joining centres. [2 marks]

Working: Centres: $(2, -1)$ and $(-1, 3)$ Gradient = $\frac{3-(-1)}{-1-2} = \frac{4}{-3} = -\frac{4}{3}$ $y - (-1) = -\frac{4}{3}(x - 2)$ $y = -\frac{4}{3}x + \frac{5}{3}$

Answer: $y = -\frac{4}{3}x + \frac{5}{3}$

Marking: 1 mark for gradient, 1 mark for equation

Total: 80 marks