Secondary 4 Additional Mathematics Preliminary Examination Paper 3

Answer Key - Additional Mathematics Secondary 4 (Prelim V3)

Section A

Question 1

• Gradient $m_1 = (6 - (-3)) / (5 - 2) = 9 / 3 = 3$.
• $L_2$ is parallel, so $m_2 = 3$.
• Equation: $y - 4 = 3(x - 0) \Rightarrow y = 3x + 4$.
• Marks: 1 for gradient, 1 for parallel condition, 1 for final equation.

Question 2

• Substitute $y = 2x - 5$ into $x^2 + y^2 = 25$:
• $x^2 + (2x - 5)^2 = 25 \Rightarrow x^2 + 4x^2 - 20x + 25 = 25 \Rightarrow 5x^2 - 20x = 0$.
• $5x(x - 4) = 0 \Rightarrow x = 0$ or $x = 4$.
• If $x = 0, y = -5$. If $x = 4, y = 3$.
• Coordinates: $(0, -5)$ and $(4, 3)$.
• Marks: 1 for substitution, 2 for solving quadratic, 1 for coordinate pairs.

Question 3

• $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$.
• $(x - 3)^2 + (y + 2)^2 = 25$.
• Centre: $(3, -2)$, Radius: $\sqrt{25} = 5$.
• Marks: 1 for completing square, 1 for centre, 1 for radius.

Question 4

• Midpoint (Centre): $((-1+3)/2, (4+2)/2) = (1, 3)$.
• Radius squared: $r^2 = (1 - (-1))^2 + (3 - 4)^2 = 2^2 + (-1)^2 = 5$.
• Equation: $(x - 1)^2 + (y - 3)^2 = 5$.
• Marks: 1 for centre, 1 for $r^2$, 1 for final equation.

Question 5

• $dy/dx = 2x - 8$. Set $2x - 8 = 0 \Rightarrow x = 4$.
• $y = 4^2 - 8(4) + 15 = 16 - 32 + 15 = -1$.
• Point: $(4, -1)$.
• $d^2y/dx^2 = 2 > 0$, therefore it is a minimum point.
• Marks: 1 for $dy/dx$, 1 for coordinates, 1 for nature.

Question 6

• Midpoint $M' = ((1+5)/2, (2+10)/2) = (3, 6)$.
• Gradient $m_{MN} = (10-2)/(5-1) = 8/4 = 2$.
• Perpendicular gradient $m_\perp = -1/2$.
• Equation: $y - 6 = -1/2(x - 3) \Rightarrow 2y - 12 = -x + 3 \Rightarrow x + 2y = 15$.
• Marks: 1 for midpoint, 1 for $m_{MN}$, 1 for $m_\perp$, 1 for final equation.

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Section B

Question 7 (a) $m_{QR} = (4-0)/(2-6) = 4/-4 = -1$. Equation: $y - 0 = -1(x - 6) \Rightarrow y = -x + 6$. [3] (b) Midpoint $S = ((0+2)/2, (0+4)/2) = (1, 2)$. [2] (c) Line $QS$: $m = (2-0)/(1-6) = 2/-5 = -0.4$. Equation: $y - 0 = -0.4(x - 6) \Rightarrow y = -0.4x + 2.4$. [3] (d) Centroid $G = ((0+6+2)/3, (0+0+4)/3) = (8/3, 4/3)$. [2]

Question 8 (a) $dy/dx = 6x^2 - 18x + 12$. Set $6(x^2 - 3x + 2) = 0 \Rightarrow (x-1)(x-2) = 0$.

• $x = 1 \Rightarrow y = 2(1)^3 - 9(1)^2 + 12(1) - 5 = 0$. Point $(1, 0)$.
• $x = 2 \Rightarrow y = 2(8) - 9(4) + 12(2) - 5 = 16 - 36 + 24 - 5 = -1$. Point $(2, -1)$. [5] (b) $d^2y/dx^2 = 12x - 18$.
• At $x = 1, d^2y/dx^2 = 12 - 18 = -6 < 0 \Rightarrow$ Maximum.
• At $x = 2, d^2y/dx^2 = 24 - 18 = 6 > 0 \Rightarrow$ Minimum. [4] (c) Set $x = 0 \Rightarrow y = -5$. Point $(0, -5)$. [2]

Question 9 (a) Set $y = 0$: $(x-2)^2 + (0-3)^2 = 25 \Rightarrow (x-2)^2 + 9 = 25 \Rightarrow (x-2)^2 = 16$.

• $x - 2 = \pm 4 \Rightarrow x = 6$ or $x = -2$. Points $(6, 0)$ and $(-2, 0)$. [4] (b) Centre $C_1(2, 3)$, $T(6, 6)$. Vector $C_1T = (4, 3)$.
• Since $C_2$ touches externally and has radius 5, and $C_1$ has radius 5, the centre $C_2$ is the reflection of $C_1$ across $T$ or simply $T + \text{vector } C_1T$.
• Centre $C_2 = (6+4, 6+3) = (10, 9)$.
• Equation: $(x-10)^2 + (y-9)^2 = 5^2 \Rightarrow x^2 - 20x + 100 + y^2 - 18y + 81 = 25$.
• $x^2 + y^2 - 20x - 18y + 156 = 0$. [6]

Question 10 (a) $\log_{10} y = \log_{10}(Ax^n) = \log_{10} A + \log_{10} x^n = \log_{10} A + n \log_{10} x$. [2] (b) Gradient $n = 2.5$. Y-intercept $\log_{10} A = 0.8 \Rightarrow A = 10^{0.8} \approx 6.31$. [4] (c) $y = 6.31 \times 10^{2.5} = 6.31 \times 316.23 \approx 1995$. [3]