Secondary 4 Additional Mathematics Preliminary Examination Paper 3

TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4 (Version 3) Answer Key

Total Marks: 60

Section A Answers

1. Gradient $= \frac{7-3}{4-2} = \frac{4}{2} = 2$. [1]
Teaching note: Gradient formula $m = \frac{y_2-y_1}{x_2-x_1}$. Student must subtract y's over x's in same order.

2. Given line gradient $= 2$. Perpendicular gradient $= -\frac{1}{2}$. Through $(0,0)$: $y = -\frac{1}{2}x$. [2]
Marks: 1 for perpendicular gradient, 1 for equation.
Common mistake: Using same gradient (parallel) instead of negative reciprocal.

3. Midpoint $= \left(\frac{1+5}{2}, \frac{2+6}{2}\right) = (3, 4)$. [1]

4. $(x-3)^2 + (y+2)^2 = 16$. [2]
Standard form $(x-a)^2+(y-b)^2=r^2$ with $a=3,b=-2,r=4$.

5. Set $y=0$: $0 = 3x-6 \Rightarrow x=2$. Point $(2,0)$. [1]

6. Parallel lines have equal gradient, so $m = -1$. [1]

7. Distance $= \sqrt{(3-0)^2+(4-0)^2} = \sqrt{9+16} = \sqrt{25} = 5$. [1]

8. Centre $(1, -3)$. [1]
From $(x-1)^2+(y+3)^2=25$, centre is $(1,-3)$.

Section B Answers

9.
(a) Gradient $= \frac{11-3}{5-1} = 2$. Equation: $y-3 = 2(x-1) \Rightarrow y = 2x+1$. [2]
(b) Solve $2x+1 = x+1 \Rightarrow x=0, y=1$. Point $(0,1)$. [2]

10.
(a) Centre $= \left(\frac{2+8}{2}, \frac{4+10}{2}\right) = (5,7)$. [1]
(b) Radius $= \sqrt{(8-5)^2+(10-7)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$. [2]
(c) $(x-5)^2+(y-7)^2 = 18$. [1]

11. $A(4,0), B(6,4) \Rightarrow m_{AB} = \frac{4-0}{6-4}=2$. Perpendicular gradient $= -\frac{1}{2}$. Line $BC$: through $B(6,4)$, $y-4 = -\frac{1}{2}(x-6)$. At $x=0$: $y-4 = 3 \Rightarrow y=7$. So $C(0,7)$. [4]
Marks: 1 gradient AB, 1 perp gradient, 1 eqn BC, 1 coord C.
Image must show C on y-axis at (0,7) and right angle at B.

12.
(a) $\frac{dy}{dx} = 3x^2 - 6x - 9$. [1]
(b) Set $=0$: $3(x^2-2x-3)=0 \Rightarrow (x-3)(x+1)=0 \Rightarrow x=3,-1$.
$x=3: y=27-27-27+5=-22$; $x=-1: y=-1-3+9+5=10$. Points $(3,-22), (-1,10)$. [3]
(c) $\frac{d^2y}{dx^2}=6x-6$. At $x=3$: $12>0$ min; at $x=-1$: $-12<0$ max. [2]

13.
(a) Centre on $x=3$, passes through $(0,0)$ and $(6,0)$ symmetric → centre $(3, k)$. Using distance to $(0,0)$: $9+k^2 = 25 \Rightarrow k=\pm4$. Centre $(3,4)$ or $(3,-4)$. [2]
(b) With $r=5$: $(x-3)^2+(y-4)^2=25$ or $(x-3)^2+(y+4)^2=25$. [2]

14.
(a) $2y=x-3 \Rightarrow y=\frac{1}{2}x-\frac{3}{2}$, gradient $\frac{1}{2}$. Perp gradient $= -2$. [1]
(b) Through $(1,2)$: $y-2 = -2(x-1) \Rightarrow y = -2x+4$. [2]
(c) At $x=0$, $y=4$. Point $(0,4)$. [1]

Section C Answers

15.
(a) $DE$ midpoint $(4,1)$, gradient $0$ → perp bisector vertical $x=4$. [3]
(b) $EF$ midpoint $(5.5,3)$, $m_{EF}=\frac{5-1}{4-7}=-\frac{4}{3}$ → perp grad $\frac{3}{4}$. Eq: $y-3 = \frac{3}{4}(x-5.5)$. [3]
(c) At $x=4$: $y-3 = \frac{3}{4}(-1.5) = -1.125 \Rightarrow y=1.875 = \frac{15}{8}$. Circumcentre $(4, \frac{15}{8})$. [2]

16. Substitute $y=2x+k$ into $x^2+y^2=20$: $x^2+(2x+k)^2=20 \Rightarrow 5x^2+4kx+k^2-20=0$. Tangent ⇒ discriminant $0$: $(4k)^2-4(5)(k^2-20)=0 \Rightarrow 16k^2-20k^2+400=0 \Rightarrow -4k^2=-400 \Rightarrow k^2=100 \Rightarrow k=\pm10$. [4]

17. $M = (6,3)$. Perp bisector $x=6$. $PA=5$: $(6-2)^2+(y-3)^2=25 \Rightarrow 16+(y-3)^2=25 \Rightarrow (y-3)^2=9 \Rightarrow y=6$ or $0$. So $P(6,6)$ or $(6,0)$. [4]
Image must show both possible P positions on vertical line x=6.

18.
(a) $x^2-4x+3=0 \Rightarrow (x-1)(x-3)=0 \Rightarrow A(1,0), B(3,0)$. [2]
(b) $x=0 \Rightarrow y=3$, $C(0,3)$. [1]
(c) Base $AB=2$, height from C to x-axis $=3$, area $=\frac{1}{2}\times2\times3=3$. [2]

19.
(a) $m=\frac{7-1}{4+2}=1$. Eq: $y-1 = 1(x+2) \Rightarrow y=x+3$. [2]
(b) $S$ midpoint of $RT$: $(4,7) = (\frac{-2+x_T}{2}, \frac{1+y_T}{2}) \Rightarrow x_T=10, y_T=13$. $T(10,13)$. [2]

20.
(a) Centre equidistant from $(1,2)$ and $(5,2)$ ⇒ on perp bisector $x=3$, so $h=3$. [2]
(b) $y=x-1 \Rightarrow k=2$. Radius to $(1,2)$: $\sqrt{(1-3)^2+(2-2)^2}=2$. [2]
(c) $(x-3)^2+(y-2)^2=4$. [1]