SPM Additional Mathematics (Model Test Paper)


SPM Additional Mathematics (Model Test Paper)

Section A
[40 marks]
 Answer all questions.

Question 5
Diagram 2 shows a triangle EFG.

It is given ER:RF=1:2, FG:TG=3:1,  ER =4 x ˜  and  EG =6 y ˜ .
(a) Express in terms of  x ˜  and  y ˜ :      (i)  GR      (ii)  GT
[3 marks]

(b) If line GR is extended to point K such that GK =h GR  and  EK =6 x ˜ 3 y ˜ , find the value of h.   
[3 marks]
Answer and Solution:
(a)(i)
GR = GE + ER =6 y ˜ +4 x ˜

(a)(ii)
GT = 1 3 GF      = 1 3 ( GR + RF )      = 1 3 ( 6 y ˜ +4 x ˜ +8 x ˜ ) RF =2 ER      =2 y ˜ +4 x ˜

(b)
GK =h GR GE + EK =h GR 6 y ˜ +( 6 x ˜ 3 y ˜ )=h( 6 y ˜ +4 x ˜ ) 9 y ˜ +6 x ˜ =6h y ˜ +4h x ˜ Compare, 6h=9 h= 3 2



Question 6
Solution by scale drawing is not accepted.


Diagram 3 shows three points, A, B and D in a Cartesian plane. The straight line AB is perpendicular to straight line BD which intersect the y-axis at D. The equation of the line BD is y = 2x – 5.
(a)    Find the equation of the straight line AB.                                                             [2 Marks]
(b)   The straight line AB is extended to a point C such that AB : BC = 2 : 3. Find the coordinates of C.                                                                                                                           [3 Marks]

(c)    Point P moves such that its distance from A is equal to its distance from B. Find the equation of the locus P.                                                                                                       [2 Marks]

Answer and Solution:
(a)
y = 2x – 5
mBD = 2, mAB = –½
y – 6 = –½ (x – 3)
2y – 12 = –x + 3
2y = –x + 15

(b)
y = 2x – 5 ----- (1)
2y = –x + 15 ----- (2)
Substitute (1) into (2),
2(2x – 5) = –x + 15
4x – 10 = –x + 15
5x = 25
x = 5
When x = 5, From (1)
y = 2(5) – 5 = 5
B = (5, 5)

AB : BC = 2 : 3
Let coordinates of point C = (h, k)
( 3(3)+2h 3+2 , 3(6)+2k 3+2 )=( 5,5 ) 9+2h 5 =5         or         18+2k 5 =5 9+2h=25                   18+2k=25 h=8                                      k=3 1 2 Coordinates of point C=( 8, 3 1 2 )

(c)
Let point P = (x, y)
√(x – 3)2 + (y – 6)2 = √(x – 5)2 + (y – 5)2 
(x – 3)2 + (y – 6)2 = (x – 5)2 + (y – 5)2  ← (Square for the both sides)
x2 – 6x + 9 + y2 – 12y + 36 = x2 – 10x + 25 + y2 – 10y + 25
4x – 2y – 5 = 0