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Secondary 4 Additional Mathematics Preliminary Examination Paper 5

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TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

TuitionGoWhere Exam Practice (AI)
PRELIMINARY EXAMINATION 2024
Version 5 of 5

Subject: Additional Mathematics (4049)
Level: Secondary 4
Paper: 1
Duration: 1 hour 30 minutes
Total Marks: 80

Name: __________________________
Class: __________________________
Date: __________________________

INSTRUCTIONS TO CANDIDATES

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided in the question paper.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
The use of an approved scientific calculator is expected, where appropriate.
If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to 3 significant figures.
Solutions by accurate drawing will not be accepted.

Section A [40 Marks]

Answer all questions in this section.

1. The line $L_1$ has equation $y = 2x + 5$ . The line $L_2$ is perpendicular to $L_1$ and passes through the point $A(4, -1)$ . Find the coordinates of the point of intersection of $L_1$ and $L_2$ .

Answer space

2. The curve $C$ has equation $y = x^2 - 6x + 10$ . (a) Express $x^2 - 6x + 10$ in the form $(x - a)^2 + b$ .
[1]

(b) Hence, state the coordinates of the minimum point of the curve $C$ .
[2]

Answer space

3. A circle has centre $C(3, -2)$ and radius 5 units. (a) Write down the equation of the circle in the form $(x - a)^2 + (y - b)^2 = r^2$ .
[1]

(b) Show that the circle intersects the y-axis at two distinct points and find the coordinates of these points.
[3]

Answer space

4. The points $A(-2, 3)$ and $B(4, 7)$ are endpoints of a diameter of a circle. Find the equation of the circle in the form $x^2 + y^2 + 2gx + 2fy + c = 0$ .

Answer space

5. The line $y = mx + 3$ is a tangent to the curve $y = x^2 - 4x + 7$ . Find the possible values of $m$ .

Answer space

6. Find the coordinates of the stationary points on the curve $y = x^3 - 12x + 5$ and determine the nature of each stationary point.

Answer space

7. The diagram shows a triangle $ABC$ with vertices $A(1, 2)$ , $B(5, 6)$ , and $C(7, 0)$ . (Note: Diagram not to scale. Solutions by accurate drawing will not be accepted.)

Find the equation of the perpendicular bisector of the side $AB$ .

Answer space

8. The curve $y = \frac{4}{x}$ and the line $y = x + 3$ intersect at points $P$ and $Q$ . Find the coordinates of $P$ and $Q$ .

Answer space

9. A circle passes through the origin $O(0,0)$ and the points $A(6,0)$ and $B(0,8)$ . Find the coordinates of the centre of this circle.

Answer space

10. The line $L$ has equation $3x - 4y + 12 = 0$ . (a) Find the gradient of $L$ .
[1]

(b) Find the distance from the origin to the line $L$ .
[2]

Answer space

Section B [40 Marks]

Answer all questions in this section.

11. The curve $C_1$ has equation $y = x^2 + 2x - 3$ . (a) Find the coordinates of the points where $C_1$ crosses the x-axis.
[2]

(b) Find the coordinates of the vertex of $C_1$ .
[2]

(c) The curve $C_2$ is a translation of $C_1$ by the vector $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ . Write down the equation of $C_2$ and determine whether $C_2$ intersects the x-axis. Justify your answer.
[3]

Answer space

12. The points $A(-1, 4)$ , $B(3, 6)$ , and $C(5, 2)$ are vertices of a triangle. (a) Show that triangle $ABC$ is right-angled at $B$ .
[3]

(b) Find the area of triangle $ABC$ .
[2]

Answer space

13. A circle $C$ has equation $x^2 + y^2 - 6x + 8y - 11 = 0$ . (a) Find the coordinates of the centre and the length of the radius of $C$ .
[3]

(b) The line $y = k$ is a tangent to the circle $C$ . Find the possible values of $k$ .
[3]

Answer space

14. The curve $y = x^3 - 3x^2 - 9x + 10$ has stationary points at $A$ and $B$ . (a) Find the x-coordinates of $A$ and $B$ .
[3]

(b) Determine the nature of the stationary point at $A$ where $x > 0$ .
[2]

Answer space

15. The line $L_1$ passes through the points $P(2, 5)$ and $Q(6, 1)$ . (a) Find the equation of $L_1$ in the form $ax + by + c = 0$ .
[2]

(b) The line $L_2$ is parallel to $L_1$ and passes through the point $R(0, -3)$ . Find the equation of $L_2$ .
[2]

(c) The line $L_3$ is perpendicular to $L_1$ and passes through the midpoint of $PQ$ . Find the coordinates of the intersection of $L_2$ and $L_3$ .
[4]

Answer space

16. A circle $C_1$ has centre $(2, 3)$ and radius 4. A second circle $C_2$ has centre $(8, 3)$ and radius $r$ . (a) Given that the two circles touch externally, find the value of $r$ .
[2]

(b) Given instead that the two circles touch internally, find the possible values of $r$ .
[2]

(c) For the case where $r = 2$ and the circles do not touch, find the range of distances between the centres for which the circles intersect at two distinct points.
[2]

Answer space

17. The curve $y = 2x^2 - 8x + 5$ is reflected in the y-axis to form curve $C'$ . (a) Find the equation of $C'$ .
[2]

(b) Find the coordinates of the minimum point of $C'$ .
[2]

Answer space

18. The points $A(1, 1)$ , $B(5, 3)$ , and $C(3, 7)$ form a triangle. (a) Find the gradient of $AC$ .
[1]

(b) Find the equation of the altitude from $B$ to $AC$ .
[3]

Answer space

19. Consider the curve $y = x^2 + kx + 9$ . (a) Find the set of values of $k$ for which the curve lies entirely above the x-axis.
[3]

(b) Find the set of values of $k$ for which the line $y = 2x$ intersects the curve at two distinct points.
[3]

Answer space

20. The diagram shows a rectangle $OABC$ where $O$ is the origin, $A$ lies on the x-axis, and $C$ lies on the y-axis. The point $B$ has coordinates $(6, 4)$ . (Note: Diagram not to scale. Solutions by accurate drawing will not be accepted.)

(a) Find the equation of the diagonal $OB$ .
[1]

(b) Find the equation of the diagonal $AC$ .
[2]

(d) A circle is drawn with $OB$ as diameter. Find the equation of this circle.
[3]

Answer space

END OF PAPER

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Answers

TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

Answer Key & Marking Scheme

Version 5 of 5

Section A

1. Gradient of $L_1$ , $m_1 = 2$ . Since $L_2 \perp L_1$ , gradient of $L_2$ , $m_2 = -\frac{1}{2}$ . Equation of $L_2$ : $y - (-1) = -\frac{1}{2}(x - 4) \Rightarrow y + 1 = -\frac{1}{2}x + 2 \Rightarrow y = -\frac{1}{2}x + 1$ . Intersection: $2x + 5 = -\frac{1}{2}x + 1$ $2.5x = -4 \Rightarrow x = -1.6$ . $y = 2(-1.6) + 5 = 1.8$ . Coordinates: $(-1.6, 1.8)$ or $(-\frac{8}{5}, \frac{9}{5})$ . [3 marks: 1 for grad L2, 1 for eq L2, 1 for coords]

2. (a) $x^2 - 6x + 10 = (x - 3)^2 - 9 + 10 = **(x - 3)^2 + 1**$ . [1 mark] (b) Minimum point at vertex. Coordinates: $(3, 1)$ . [2 marks]

3. (a) Equation: $(x - 3)^2 + (y + 2)^2 = 25$ . [1 mark] (b) At y-axis, $x = 0$ . $(0 - 3)^2 + (y + 2)^2 = 25$ $9 + (y + 2)^2 = 25$ $(y + 2)^2 = 16$ $y + 2 = \pm 4$ $y = 2$ or $y = -6$ . Coordinates: $(0, 2)$ and $(0, -6)$ . Since there are two distinct real solutions, it intersects at two points. [3 marks: 1 for sub x=0, 1 for solving, 1 for coords]

4. Midpoint (Centre) $M = (\frac{-2+4}{2}, \frac{3+7}{2}) = (1, 5)$ . Radius squared $r^2 = (4 - 1)^2 + (7 - 5)^2 = 3^2 + 2^2 = 9 + 4 = 13$ . Equation: $(x - 1)^2 + (y - 5)^2 = 13$ . $x^2 - 2x + 1 + y^2 - 10y + 25 = 13$ . $x^2 + y^2 - 2x - 10y + 13 = 0$ . [4 marks: 1 for centre, 1 for r^2, 1 for expansion, 1 for final form]

5. Intersection: $x^2 - 4x + 7 = mx + 3$ . $x^2 - (4 + m)x + 4 = 0$ . For tangent, discriminant $\Delta = 0$ . $(-(4 + m))^2 - 4(1)(4) = 0$ . $(4 + m)^2 - 16 = 0$ . $(4 + m)^2 = 16$ . $4 + m = \pm 4$ . $m = 0$ or $m = -8$ . [4 marks: 1 for quadratic, 1 for Delta condition, 1 for solving, 1 for both values]

6. $\frac{dy}{dx} = 3x^2 - 12$ . Stationary points when $\frac{dy}{dx} = 0 \Rightarrow 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$ . When $x = 2, y = 8 - 24 + 5 = -11$ . Point $(2, -11)$ . When $x = -2, y = -8 + 24 + 5 = 21$ . Point $(-2, 21)$ . $\frac{d^2y}{dx^2} = 6x$ . At $x = 2, \frac{d^2y}{dx^2} = 12 > 0 \Rightarrow$ Minimum. At $x = -2, \frac{d^2y}{dx^2} = -12 < 0 \Rightarrow$ Maximum. Coords: $(2, -11)$ [Min], $(-2, 21)$ [Max]. [4 marks: 1 for dy/dx, 1 for x values, 1 for coords, 1 for nature]

7. Midpoint of $AB = (\frac{1+5}{2}, \frac{2+6}{2}) = (3, 4)$ . Gradient of $AB = \frac{6-2}{5-1} = \frac{4}{4} = 1$ . Gradient of perpendicular bisector = $-1$ . Equation: $y - 4 = -1(x - 3) \Rightarrow y = -x + 7$ or $x + y - 7 = 0$ . [3 marks: 1 for midpt, 1 for grad, 1 for eq]

8. $\frac{4}{x} = x + 3 \Rightarrow 4 = x^2 + 3x \Rightarrow x^2 + 3x - 4 = 0$ . $(x + 4)(x - 1) = 0$ . $x = -4$ or $x = 1$ . If $x = -4, y = -1$ . Point $P(-4, -1)$ . If $x = 1, y = 4$ . Point $Q(1, 4)$ . Coordinates: $(-4, -1)$ and $(1, 4)$ . [3 marks: 1 for quadratic, 1 for x values, 1 for coords]

9. Since $\angle AOB = 90^\circ$ (axes are perpendicular), $AB$ is the diameter. Centre is midpoint of $AB$ . $A(6,0), B(0,8)$ . Midpoint = $(\frac{6+0}{2}, \frac{0+8}{2}) = **(3, 4)**$ . [2 marks: 1 for identifying diameter/midpoint logic, 1 for coords]

10. (a) $3x + 12 = 4y \Rightarrow y = \frac{3}{4}x + 3$ . Gradient = $\frac{3}{4}$ . [1 mark] (b) Distance from $(0,0)$ to $3x - 4y + 12 = 0$ . $d = \frac{|3(0) - 4(0) + 12|}{\sqrt{3^2 + (-4)^2}} = \frac{12}{\sqrt{25}} = \frac{12}{5} = **2.4$ **. [2 marks: 1 for formula/sub, 1 for answer]

Section B

11. (a) $x^2 + 2x - 3 = 0 \Rightarrow (x+3)(x-1)=0$ . $x = -3, 1$ . Coords: $(-3, 0)$ and $(1, 0)$ . [2 marks] (b) $y = (x+1)^2 - 1 - 3 = (x+1)^2 - 4$ . Vertex: $(-1, -4)$ . [2 marks] (c) Translation up 4 units: $y = (x^2 + 2x - 3) + 4 = x^2 + 2x + 1 = (x+1)^2$ . Equation: $y = (x+1)^2$ . Vertex is $(-1, 0)$ . Since min value is 0, it touches x-axis at one point. Does it intersect at two distinct points? No. [3 marks: 1 for eq, 1 for reasoning, 1 for conclusion]

12. (a) $m_{AB} = \frac{6-4}{3-(-1)} = \frac{2}{4} = \frac{1}{2}$ . $m_{BC} = \frac{2-6}{5-3} = \frac{-4}{2} = -2$ . Product $m_{AB} \times m_{BC} = \frac{1}{2}(-2) = -1$ . Therefore $AB \perp BC$ , so $\angle B = 90^\circ$ . [3 marks: 1 for m1, 1 for m2, 1 for product] (b) $AB = \sqrt{4^2 + 2^2} = \sqrt{20}$ . $BC = \sqrt{2^2 + (-4)^2} = \sqrt{20}$ . Area = $\frac{1}{2} \times \sqrt{20} \times \sqrt{20} = **10$ **. [2 marks] (c) Since right-angled at B, AC is diameter. Midpoint of AC = Centre = $(\frac{-1+5}{2}, \frac{4+2}{2}) = (2, 3)$ . Radius squared $r^2 = (2 - (-1))^2 + (3 - 4)^2 = 3^2 + (-1)^2 = 10$ . Eq: $(x-2)^2 + (y-3)^2 = 10$ . $x^2 - 4x + 4 + y^2 - 6y + 9 = 10$ . $x^2 + y^2 - 4x - 6y + 3 = 0$ . [3 marks: 1 for centre, 1 for r^2, 1 for eq]

13. (a) $x^2 - 6x + y^2 + 8y = 11$ . $(x-3)^2 - 9 + (y+4)^2 - 16 = 11$ . $(x-3)^2 + (y+4)^2 = 36$ . Centre: $(3, -4)$ . Radius: $\sqrt{36} = **6$ **. [3 marks: 1 for completing square, 1 for centre, 1 for radius] (b) Tangent is horizontal line $y=k$ . Distance from centre y-coord to line equals radius. $|k - (-4)| = 6$ . $k + 4 = 6 \Rightarrow k = 2$ . $k + 4 = -6 \Rightarrow k = -10$ . Values: $2, -10$ . [3 marks: 1 for logic, 1 for each value]

14. (a) $\frac{dy}{dx} = 3x^2 - 6x - 9$ . $3(x^2 - 2x - 3) = 0 \Rightarrow 3(x-3)(x+1) = 0$ . $x = 3, x = -1$ . [3 marks] (b) $A$ has $x > 0$ , so $x = 3$ . $\frac{d^2y}{dx^2} = 6x - 6$ . At $x = 3, \frac{d^2y}{dx^2} = 18 - 6 = 12 > 0$ . Nature: Minimum. [2 marks] (c) At $x = 1, y = 1 - 3 - 9 + 10 = -1$ . Point $(1, -1)$ . Gradient $m = 3(1)^2 - 6(1) - 9 = -12$ . Eq: $y - (-1) = -12(x - 1) \Rightarrow y + 1 = -12x + 12$ . $y = -12x + 11$ . [3 marks: 1 for pt, 1 for grad, 1 for eq]

15. (a) $m = \frac{1-5}{6-2} = \frac{-4}{4} = -1$ . $y - 5 = -1(x - 2) \Rightarrow y = -x + 7 \Rightarrow **x + y - 7 = 0**$ . [2 marks] (b) $L_2$ parallel $\Rightarrow m = -1$ . Passes $(0, -3)$ . $y = -x - 3 \Rightarrow **x + y + 3 = 0**$ . [2 marks] (c) Midpoint $PQ = (\frac{2+6}{2}, \frac{5+1}{2}) = (4, 3)$ . $L_3 \perp L_1 \Rightarrow m = 1$ . Eq $L_3: y - 3 = 1(x - 4) \Rightarrow y = x - 1$ . Intersection $L_2$ and $L_3$ : $-x - 3 = x - 1 \Rightarrow 2x = -2 \Rightarrow x = -1$ . $y = -1 - 1 = -2$ . Coords: $(-1, -2)$ . [4 marks: 1 for midpt, 1 for L3 eq, 1 for solving, 1 for coords]

16. Distance between centres $d = \sqrt{(8-2)^2 + (3-3)^2} = 6$ . (a) External touch: $d = r_1 + r_2 \Rightarrow 6 = 4 + r \Rightarrow **r = 2$ . [2 marks] (b) Internal touch: $d = |r_1 - r_2| \Rightarrow 6 = |4 - r|$ . $4 - r = 6 \Rightarrow r = -2$ (impossible). $r - 4 = 6 \Rightarrow **r = 10$ . [2 marks] (c) Intersect at 2 points: $|r_1 - r_2| < d < r_1 + r_2$ . $|4 - 2| < d < 4 + 2 \Rightarrow 2 < d < 6$ . Range: $2 < d < 6$ . [2 marks]

17. (a) Reflection in y-axis: replace $x$ with $-x$ . $y = 2(-x)^2 - 8(-x) + 5 \Rightarrow **y = 2x^2 + 8x + 5**$ . [2 marks] (b) $x_{vertex} = -\frac{b}{2a} = -\frac{8}{4} = -2$ . $y = 2(4) - 16 + 5 = -3$ . Coords: $(-2, -3)$ . [2 marks] (c) Min value is -3. For 2 intersections, line must be above minimum. $c > -3$ . [2 marks]

18. (a) $m_{AC} = \frac{7-1}{3-1} = \frac{6}{2} = **3$ **. [1 mark] (b) Altitude from B is $\perp AC$ . Gradient $= -\frac{1}{3}$ . Passes $B(5,3)$ . $y - 3 = -\frac{1}{3}(x - 5) \Rightarrow 3y - 9 = -x + 5 \Rightarrow **x + 3y - 14 = 0**$ . [3 marks] (c) Eq of AC: $y - 1 = 3(x - 1) \Rightarrow y = 3x - 2$ . Sub into altitude eq: $x + 3(3x - 2) - 14 = 0$ . $x + 9x - 6 - 14 = 0 \Rightarrow 10x = 20 \Rightarrow x = 2$ . $y = 3(2) - 2 = 4$ . Coords: $(2, 4)$ . [3 marks]

19. (a) Above x-axis $\Rightarrow$ No real roots AND $a > 0$ . $\Delta < 0 \Rightarrow k^2 - 4(1)(9) < 0 \Rightarrow k^2 < 36$ . $-6 < k < 6$ . [3 marks] (b) Intersection: $x^2 + kx + 9 = 2x \Rightarrow x^2 + (k-2)x + 9 = 0$ . Two distinct points $\Rightarrow \Delta > 0$ . $(k-2)^2 - 36 > 0$ . $(k-2)^2 > 36$ . $k - 2 > 6$ or $k - 2 < -6$ . $k > 8$ or $k < -4$ . [3 marks]

20. (a) $O(0,0), B(6,4)$ . $m = \frac{4}{6} = \frac{2}{3}$ . Eq: $y = \frac{2}{3}x$ or $2x - 3y = 0$ . [1 mark] (b) $A(6,0), C(0,4)$ . $m = \frac{4-0}{0-6} = -\frac{2}{3}$ . Eq: $y = -\frac{2}{3}x + 4 \Rightarrow **2x + 3y - 12 = 0**$ . [2 marks] (c) Diagonals of rectangle bisect each other. Midpoint of OB. $(\frac{6}{2}, \frac{4}{2}) = **(3, 2)$ **. [1 mark] (d) Centre $(3,2)$ . Radius = dist from $(3,2)$ to $(0,0) = \sqrt{9+4} = \sqrt{13}$ . Eq: $(x-3)^2 + (y-2)^2 = 13$ . $x^2 - 6x + 9 + y^2 - 4y + 4 = 13$ . $x^2 + y^2 - 6x - 4y = 0$ . [3 marks]