Secondary 4 Additional Mathematics Preliminary Examination Paper 3

TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

Answer Key and Marking Scheme

Paper: Preliminary Examination (Version 3 of 5) Total Marks: 60

Section A (36 marks)

1. A(−2, 5), B(4, −1)

(a) Length AB = (\sqrt{(4 - (-2))^2 + (-1 - 5)^2}) [M1] = (\sqrt{6^2 + (-6)^2}) = (\sqrt{36 + 36}) = (\sqrt{72}) = (6\sqrt{2}) units [A1]

(b) Midpoint of AB = (\left(\frac{-2 + 4}{2}, \frac{5 + (-1)}{2}\right) = (1, 2)) [M1] Gradient of AB = (\frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1) [M1] Gradient of perpendicular bisector = 1 Equation: (y - 2 = 1(x - 1)) (y = x + 1) [A1]

2. (y = 2x + 1), (x^2 + y^2 = 25)

Substitute: (x^2 + (2x + 1)^2 = 25) [M1] (x^2 + 4x^2 + 4x + 1 = 25) (5x^2 + 4x - 24 = 0) [M1] ((5x - 6)(x + 4) = 0) or using quadratic formula [M1] (x = \frac{6}{5}) or (x = -4) When (x = \frac{6}{5}): (y = 2(\frac{6}{5}) + 1 = \frac{17}{5}) When (x = -4): (y = 2(-4) + 1 = -7) Points: (\left(\frac{6}{5}, \frac{17}{5}\right)) and ((-4, -7)) [A1]

3. (y = x^3 - 6x^2 + 9x + 1)

(a) (\frac{dy}{dx} = 3x^2 - 12x + 9) [M1] Set (\frac{dy}{dx} = 0): (3x^2 - 12x + 9 = 0) (x^2 - 4x + 3 = 0) ((x - 1)(x - 3) = 0) [M1] (x = 1) or (x = 3) [M1] When (x = 1): (y = 1 - 6 + 9 + 1 = 5) When (x = 3): (y = 27 - 54 + 27 + 1 = 1) Stationary points: (1, 5) and (3, 1) [A1]

(b) (\frac{d^2y}{dx^2} = 6x - 12) [M1] At (1, 5): (\frac{d^2y}{dx^2} = 6(1) - 12 = -6 < 0) → maximum [M1] At (3, 1): (\frac{d^2y}{dx^2} = 6(3) - 12 = 6 > 0) → minimum [A1]

4. (C_1: x^2 + y^2 - 4x + 6y - 12 = 0)

(a) Complete the square: ((x^2 - 4x) + (y^2 + 6y) = 12) [M1] ((x - 2)^2 - 4 + (y + 3)^2 - 9 = 12) ((x - 2)^2 + (y + 3)^2 = 25) [M1] Centre: (2, −3), Radius: 5 units [A1]

(b) (C_2) centre: (1, −2) Distance between centres = (\sqrt{(2 - 1)^2 + (-3 - (-2))^2} = \sqrt{1 + 1} = \sqrt{2}) [M1] For external tangency: distance = (r_1 + r_2) (\sqrt{2} = 5 + r_2) [M1] (r_2 = \sqrt{2} - 5) (negative, so external tangency impossible with (C_1) containing (C_2)) Check: (r_2 = 5 - \sqrt{2}) (for internal tangency where (C_2) is inside (C_1)) Distance = (r_1 - r_2) → (\sqrt{2} = 5 - r_2) → (r_2 = 5 - \sqrt{2}) [M1] Equation of (C_2): ((x - 1)^2 + (y + 2)^2 = (5 - \sqrt{2})^2) ((x - 1)^2 + (y + 2)^2 = 25 - 10\sqrt{2} + 2) ((x - 1)^2 + (y + 2)^2 = 27 - 10\sqrt{2}) [A1]

5. P(1, 2), Q(5, 6), R(9, 2)

(a) PQ = (\sqrt{(5-1)^2 + (6-2)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}) [M1] QR = (\sqrt{(9-5)^2 + (2-6)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}) Since PQ = QR, triangle PQR is isosceles. [A1]

(b) Area = (\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|) [M1] = (\frac{1}{2}|1(6 - 2) + 5(2 - 2) + 9(2 - 6)|) [M1] = (\frac{1}{2}|4 + 0 - 36|) = (\frac{1}{2}|-32| = 16) square units [A1]

6. (y = ax^n)

(a) (\log y = \log a + n \log x) Plot (\log y) vs (\log x):

Generated graph for this question.

(b) From graph: gradient (n = 3) [M1] y-intercept = (\log a \approx -0.398) [M1] (a = 10^{-0.398} \approx 0.4) [A1]

Section B (24 marks)

7. A(0, 0), B(6, 0), C(8, 4), D(2, 4)

(a) Gradient of AC = (\frac{4 - 0}{8 - 0} = \frac{1}{2}) [M1] Equation: (y = \frac{1}{2}x) [A1]

(b) Gradient of BD = (\frac{4 - 0}{2 - 6} = \frac{4}{-4} = -1) [M1] Equation: (y - 0 = -1(x - 6)) (y = -x + 6) [A1]

(c) Gradient of AC = (\frac{1}{2}), Gradient of BD = (-1) [M1] (\frac{1}{2} \times (-1) = -\frac{1}{2} \neq -1) Wait - recheck: AC gradient = (\frac{4}{8} = \frac{1}{2}), BD gradient = (\frac{4-0}{2-6} = -1) Product = (-\frac{1}{2}) → NOT perpendicular. [Note: Original question intent was to show perpendicularity. With given coordinates, diagonals are NOT perpendicular. Accept correct working showing product ≠ −1.]

(d) Intersection: (\frac{1}{2}x = -x + 6) [M1] (\frac{3}{2}x = 6) [M1] (x = 4), (y = 2) [A1] Intersection point: (4, 2)

8. (y = \frac{4}{x} + x), (x > 0)

(a) (\frac{dy}{dx} = -\frac{4}{x^2} + 1) [M1] Set (\frac{dy}{dx} = 0): (-\frac{4}{x^2} + 1 = 0) (\frac{4}{x^2} = 1) [M1] (x^2 = 4), (x = 2) (since (x > 0)) [M1] When (x = 2): (y = \frac{4}{2} + 2 = 4) Stationary point: (2, 4) [A1]

(b) (\frac{d^2y}{dx^2} = \frac{8}{x^3}) [M1] At (x = 2): (\frac{d^2y}{dx^2} = \frac{8}{8} = 1 > 0) → minimum [A1]

(c) Sketch: Vertical asymptote at (x = 0), curve approaches (y = x) as (x \to \infty), minimum at (2, 4). [3 marks: correct shape, asymptote, stationary point marked]

9. (C: x^2 + y^2 - 2x + 4y - 20 = 0)

(a) ((x^2 - 2x) + (y^2 + 4y) = 20) [M1] ((x - 1)^2 - 1 + (y + 2)^2 - 4 = 20) ((x - 1)^2 + (y + 2)^2 = 25) Centre: (1, −2), Radius: 5 [A1]

(b) Substitute (y = 2x + k) into circle equation: (x^2 + (2x + k)^2 - 2x + 4(2x + k) - 20 = 0) [M1] (x^2 + 4x^2 + 4kx + k^2 - 2x + 8x + 4k - 20 = 0) (5x^2 + (4k + 6)x + (k^2 + 4k - 20) = 0) [M1] For tangency, discriminant = 0: ((4k + 6)^2 - 4(5)(k^2 + 4k - 20) = 0) [M1] (16k^2 + 48k + 36 - 20k^2 - 80k + 400 = 0) (-4k^2 - 32k + 436 = 0) (k^2 + 8k - 109 = 0) [M1] (k = \frac{-8 \pm \sqrt{64 + 436}}{2} = \frac{-8 \pm \sqrt{500}}{2} = \frac{-8 \pm 10\sqrt{5}}{2}) (k = -4 \pm 5\sqrt{5}) [A1]

END OF ANSWER KEY

Marking Notes:

• M1: Method mark for correct approach
• A1: Accuracy mark for correct answer
• Allow follow-through marks where appropriate
• Accept equivalent forms of answers
• Deduct 1 mark for missing units where applicable