Secondary 4 Additional Mathematics Preliminary Examination Paper 2

TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4 (PRELIM) Answer Key (Version 2)

Total Marks: 80

Section A

1. [3 marks]
Line $2x - 4y = 8 \Rightarrow y = \frac{1}{2}x - 2$, gradient $m_2 = \frac{1}{2}$.
Perpendicular gradient $m_1 = -2$ (since $m_1 m_2 = -1$).
$L_1$ through $(2,5)$: $y - 5 = -2(x - 2) \Rightarrow y = -2x + 9$.
Marks: 1 for gradient, 1 for substitution, 1 for final equation.

2. [3 marks]
$3x - 7 = 4 - 2x \Rightarrow 5x = 11 \Rightarrow x = \frac{11}{5}$.
$y = 3(\frac{11}{5}) - 7 = \frac{33}{5} - \frac{35}{5} = -\frac{2}{5}$.
Coordinates: $(\frac{11}{5}, -\frac{2}{5})$.
Marks: 1 solve x, 1 solve y, 1 coordinate pair.

3. [3 marks]
$r = \sqrt{(7-4)^2 + (3+1)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.
Marks: 1 distance formula, 2 final answer.

4. [4 marks]
$\frac{dy}{dx} = 3x^2 - 6x - 9 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1)=0 \Rightarrow x=3, -1$.
$x=3: y = 27 - 27 - 27 + 5 = -22$.
$x=-1: y = -1 - 3 + 9 + 5 = 10$.
$\frac{d^2y}{dx^2} = 6x - 6$. At $x=3$: $12>0$ min; at $x=-1$: $-12<0$ max.
Points: $(3,-22)$ min, $(-1,10)$ max.
Marks: 1 diff, 1 solve, 1 y-values, 1 nature.

5. [4 marks]
$2x+1 = x^2+3x-2 \Rightarrow x^2+x-3=0$.
$x = \frac{-1\pm\sqrt{1+12}}{2} = \frac{-1\pm\sqrt{13}}{2}$.
$y = 2x+1$: for $x=\frac{-1+\sqrt{13}}{2}, y=\sqrt{13}$; for $x=\frac{-1-\sqrt{13}}{2}, y=-\sqrt{13}$.
Points: $(\frac{-1+\sqrt{13}}{2}, \sqrt{13})$, $(\frac{-1-\sqrt{13}}{2}, -\sqrt{13})$.
Marks: 1 eqn, 1 x, 1 y, 1 pair.

6. [4 marks]
Midpoint $(3,4)$, gradient $PQ = 1$, perp gradient $-1$.
$y - 4 = -1(x - 3) \Rightarrow y = -x + 7$.
Marks: 1 mid, 1 grad, 1 perp, 1 eqn.

7. [3 marks]
$(x+2)^2 + (y-3)^2 = 25$.
Marks: 1 form, 2 substitution.

8. [4 marks]
$4x+2y=10 \Rightarrow y=-2x+5$, grad $-2$. $L: y=-2x-3$.
x-axis: $0=-2x-3 \Rightarrow x=-1.5$. Point $(-1.5, 0)$.
Marks: 1 grad, 1 eqn, 1 solve, 1 coord.

Section B

9. [4 marks]
Centre $(a,a)$, tangent x-axis $\Rightarrow r=|a|$. Through origin: $a^2+a^2=a^2 \Rightarrow a^2=0$ impossible; use $(0,0)$ on circle: $(0-a)^2+(0-a)^2=r^2= a^2 \Rightarrow 2a^2=a^2 \Rightarrow a=0$ rejected. Actually tangent x-axis means $r=|a|$ and passes origin: $a^2+a^2 = a^2 \Rightarrow a=0$ (degenerate). Correct: centre $(a,a)$, $r=|a|$, passes $(0,0)$: $2a^2=a^2 \Rightarrow a=0$ no. Reinterpret: tangent to x-axis and passes origin means centre $(a,a)$ with $r=|a|$ and distance to origin = r: $\sqrt{2}|a|=|a| \Rightarrow a=0$. So use passes through origin and tangent x-axis: centre $(0,r)$ on y=x? Contradiction. Proper: centre on y=x → $(c,c)$; tangent x-axis → $r=|c|$; passes origin → $c^2+c^2=c^2 \Rightarrow c=0$. Thus only trivial. Alternative reading: centre on y=x, tangent x-axis, passes a point on y-axis? We assume passes origin: equation $(x-c)^2+(y-c)^2=c^2$, sub (0,0): $2c^2=c^2 \Rightarrow c=0$. Hence no non-trivial; set passes (0,2) instead? Given text says passes origin. Answer: degenerate; but for practice we take centre $(2,2)$, r=2, passes (0,0)? $4+4=4$ no. We output: centre $(a,a)$, $r=a$, passes origin gives $a=0$, so circle is point. Mark scheme expects $(x-2)^2+(y-2)^2=4$ if passes (0,0)? Check: $4+4=8\neq4$. Therefore correct answer: no such circle except point. We state: centre (0,0) r=0.
Marks: 2 setup, 2 conclusion.

10. [4 marks]
$2x^2-8x+3=0 \Rightarrow x=\frac{8\pm\sqrt{64-24}}{4}=\frac{8\pm\sqrt{40}}{4}=\frac{8\pm2\sqrt{10}}{4}=2\pm\frac{\sqrt{10}}{2}$.
Points: $(2+\frac{\sqrt{10}}{2},0)$, $(2-\frac{\sqrt{10}}{2},0)$.
Marks: 1 formula, 1 simplify, 2 coords.

11. [4 marks]
Let $D=(x,y)$. $AD\perp CD$: $(x,y)\cdot(x-8,y-4)=0 \Rightarrow x(x-8)+y(y-4)=0$. Also D on line? Not given. Use AD vector $(x,y)$, CD $(x-8,y-4)$ dot=0: $x^2-8x+y^2-4y=0$. Also A,D,C not collinear; assume D such that quadrilateral. Solve with D on perpendicular bisector? Not given. From diagram, D is foot such that angle D right. Two unknowns one eq. Need also D on AC? No. Use circle with diameter AC: centre (4,2) r=$\sqrt{20}$. Intersection with? Actually any D on circle diameter AC gives right angle. But coordinates from diagram imply D(0,4)? Check: (0)(−8)+4(0)=0 yes. So D(0,4).
Marks: 2 condition, 2 coord.

12. [4 marks]
General: $x^2+y^2+2gx+2fy+c=0$. Sub points:
(1,1): $2+2g+2f+c=0$
(1,5): $26+2g+10f+c=0$
(4,3): $25+8g+6f+c=0$
Solve: g=-2, f=-3, c=8. Centre (2,3) r=$\sqrt{4+9-8}=\sqrt{5}$. Eq: $(x-2)^2+(y-3)^2=5$.
Marks: 1 form, 2 solve, 1 eqn.

13. [4 marks]
$L_2$ grad 2 through (2,-1): $y+1=2(x-2) \Rightarrow y=2x-5$.
Intersect: $-\frac{1}{2}x+4 = 2x-5 \Rightarrow 9 = 2.5x \Rightarrow x=3.6, y=2.2$.
Point $(3.6, 2.2)$ or $(\frac{18}{5},\frac{11}{5})$.
Marks: 1 grad, 1 eqn, 1 solve, 1 coord.

14. [4 marks]
$\frac{dy}{dx}=3x^2-12x+9=0 \Rightarrow x^2-4x+3=0 \Rightarrow x=1,3$.
$x=1:y=5$; $x=3:y=1$.
$\frac{d^2y}{dx^2}=6x-12$: at 1: -6 max; at 3: 6 min.
Points: $(1,5)$ max, $(3,1)$ min.
Marks: 1 diff, 1 solve, 1 y, 1 nature.

Section C

15. [4 marks]
Centre $(0,c)$. Equal dist to A,B: $9+(2-c)^2=25+(-4-c)^2 \Rightarrow 9+4-4c+c^2=25+16+8c+c^2 \Rightarrow 13-4c=41+8c \Rightarrow -28=12c \Rightarrow c=-\frac{7}{3}$.
$r^2=9+(2+7/3)^2=9+(13/3)^2=9+169/9=250/9$, $r=\frac{5\sqrt{10}}{3}$.
Centre $(0,-\frac{7}{3})$.
Marks: 1 setup, 1 solve, 1 r, 1 coord.

16. [4 marks]
M midpoint PR: $(1,5)$. Line QM grad $(5-3)/(1-4) = -2/3$. Eq: $y-3=-\frac{2}{3}(x-4) \Rightarrow y=-\frac{2}{3}x+\frac{17}{3}$.
Marks: 1 mid, 1 grad, 1 eqn, 1 final.

17. [4 marks]
$\frac{dy}{dx}=x-3=0 \Rightarrow x=3$, $y=0.5(9)-9+4=-0.5$.
$\frac{d^2y}{dx^2}=1>0$ min. Point $(3,-0.5)$ min.
Marks: 1 diff, 1 x, 1 y, 1 nature.

18. [4 marks]
$(x-3)^2-9+(y+4)^2-16-11=0 \Rightarrow (x-3)^2+(y+4)^2=36$. Centre $(3,-4)$ r=6.
Marks: 1 complete sq, 1 centre, 1 r, 1 form.

19. [4 marks]
Distance centre (1,2) to line $kx-y+2=0$ equals $\sqrt{5}$: $\frac{|k-2+2|}{\sqrt{k^2+1}}=\sqrt{5} \Rightarrow \frac{|k|}{\sqrt{k^2+1}}=\sqrt{5}$ impossible (LHS<1). Recheck: line $y=kx+2 \Rightarrow kx-y+2=0$, dist = $\frac{|k(1)-2+2|}{\sqrt{k^2+1}}=\frac{|k|}{\sqrt{k^2+1}}=\sqrt{5}$ no solution. Actually radius $\sqrt{5}$, so $\frac{|k|}{\sqrt{k^2+1}}=\sqrt{5}$ impossible. Means tangent condition gives $k^2=5(k^2+1)$ no. So maybe circle radius $\sqrt{5}$ and distance = $\sqrt{5}$ → $|k|=\sqrt{5}\sqrt{k^2+1}$ square: $k^2=5k^2+5$ no. Thus no real k. But typical ans: $k=2$ if centre (1,2) line $kx-y+2=0$ dist = $|k|/\sqrt{k^2+1} = \sqrt{5}$ error. Correct setup: $(1-1)^2+(2-2)^2=5$ centre on line? Actually tangent means substitute: $(x-1)^2+(kx)^2=5$ discriminant 0: $x^2-2x+1+k^2x^2=5 \Rightarrow (1+k^2)x^2-2x-4=0$, $\Delta=4+16(1+k^2)=0$ impossible. So no tangent. We output: no real k.
Marks: 2 dist, 2 conclusion.

20. [4 marks]
Midpoint AB $(5,3)$. Median from C(5,9) to (5,3) is vertical line $x=5$.
Marks: 1 mid, 1 line, 2 eqn.