Spotlight A+1 From 4.5 Add Math

Additional Mathematics   SPM Model Paper

7. (a) Solve the equation of 6 tan2 q + 13 sec q = 2 for 0° < q < 360°. [3 marks]

(b) Diagram 7 shows the graph of y = a sin 1  x + c, such as a and b are positive integers and c is an integer
b
for 0 < x < 4π.

y
1 y = a sin—b1 x + c

0 x
π 2π 3π 4π

–2

–5

Diagram 7

Given the graph that passed through point (0, –2) has maximum point at (π, 1) and minimum point
at (3π, –5).
(i) Find the values of a, b and c.

(ii) Sketch the graph of y = | a sin 1  x + c| for 0 < x < 4π. [5 marks]
b

SPM MODEL PAPER Section B
(30 marks)
Instruction: Answer any three questions only
8. Table 8 shows the values for two variables, x and y, obtained from an experiment. The variable x and y are
related by the equation y = axn, such as a and n are constants.

xy

1.2 1.5

1.6 1.8

2.5 2.6

4.0 3.6

6.3 5.0

8.0 5.8

Table 8

(a) oPfloftitl. og10 y against log10 x, by using the scale of 2 cm for 0.1 unit on both axes. Hence, sketch a best line
[5 marks]
(b) Based on the graph at 8(a), find the value of
(i) a,
(ii) n.
[5 marks]

430

Additional Mathematics   SPM Model Paper

Section C
(20 marks)
Section: Answer any two questions only
12. Diagram 12 shows the initial position and direction of movement of a particle from a fixed point, O on a
straight line, JOK. The velocity, v ms–1, is given by v = 12 – 6t, such as t is the time, in seconds, after leaving
point O. The particle stops momentarily at point K.
[Assume the motion of particle upwards is positive]

K

O
15 m

SPM MODEL PAPER J

Diagram 12 [1 mark]
Find [3 marks]
(a) the constanct acceleration, in m s–2, of the particle, [3 marks]
(b) the time, in seconds, when the particle passes through point O again, [3 marks]
(c) the velocity, in m s–1, when the particle passes through point J,
(d) the total distance, in m, travelled by the particle from point O to J through K. [4 marks]
[6 marks]
13. Diagram 13 shows a triangle PQR and triangle RST such as QRT and PRS are straight lines.

P

20°

Q 40° R 6 cm
5 cm 8 cm

T

S

Diagram 13
(a) Calculate the length, in cm, of

(i) QR,
(ii) ST.
(b) Given point R is located on QT such as PR = PR.
(i) Sketch the triangle of PRQ.
(ii) Find ˙PRQ.
(iii) Calculate the area, in cm2, of triangle PRQ.

432

ANSWERS Complete answers
http://bit.ly/2ORHlke

FORM 4 9. f(x) 5. –54

13 6. 3
10
Chapter 1 Functions 9

1.1 3 7. g(x) = 3x – 2
8. f(x) = x – 2
1. (a) The relation is a function –6 –  3 0 x 9. ((ab)) 11553
because each object has 2 5
only one image.
Range of f is 0  f(x)  13. 10. h = –3k 5
(b) The relation is not a 2
function because there is 10. (a) q = 1, p =1 11. p = 12, q =
one object does not have (b) 3
any image. (c) –10 12. (a) 2x x 1 (b) 3x x+ 1
+
2. (a) h(x) = |10 – x| or 11. (a) 3
h(x) = |x – 10| ((cb)) 5225 (c) 2x x 1 (d) 3x x+ 1
+
(b) h(x) = x2 – 1 1 11
2 2
3. (a) A function. 13. k = , h =
(b) A function.
(c) Not a function. 12. (a) 3 or 1.7321 14. (a) RM13 400
2 (b) RM449 675
FORM 4 ANSWER 4. (a) {a, b, c, d} (b) 2 or 1.4142
(b) {–2, 0, 2, 4}
(c) a, b, c, d 13. (a) 1 (b) 2 1.3
(d) –2, 0, 2
14. 3
5. (a) 7 1. (a) –4 (b) –2
(b) {–3, –2, 2, 7} 15. (a) 3 (ii) 3 (c) 4 (d) 2
(b) (i) –12
6. (a) 0, 2, 6, 8 2. (a) –2 (b) 4
(b) 7, 1, 10 16. (a) (i) 5 5 (ii) 11
(c) {7, 1, 10} (b) – 54 12 3. (a) –7 (b) –2
(d) 7 and
(e) 8 4. (a) Has an inverse function
17. (a) (i) 15 because each element in set
7. (a) Domain = {–2, 0, 2, 4} 4 P matched with only one
Codomain = {4, 6} (ii) 34 element in set Q.
Range = {4, 6} (b) – 341 33
(b) Domain of f is –1  x  3. and 4 (b) Does not have an inverse
Codomain of f is 1  f(x)  3. function because from
Range of f is 1  f(x)  3. 9 horizontal line test, the
4 line cuts the graph on two
8. (a) f(x) 18. –3 and points.

10 19. (a) 4 5. (a) Has an inverse function g.
(b) 0  g(x)  10 (b) Does not has an inverse
5 function g.
4
6. y

1.2 11
g(x)
–2 0 4 x 1. (a) fg(x) = 3x – 1 y=x
3 3 3 6
(b) f(x)
(b) gf(x) = x – 5
5
2. (a) f 2(x) = 36x – 7 3
(b) g2(x) = 16 + 9x
(c) gf(x) = 18x + 1 2 x
(d) fg(x) = 18x + 23 11
3 –2 0 2 3 6
3. (a) 111 (b) 731 –2 g–1(x)
0 15 4. (a) –4 (b) – 127
2 x   Domain of g–1(x) is 2  x  11.
5   Range of g–1(x) is –2  g–1(x)  3.
434

Answer  Additional Mathematics

7. (a) y 2. (a) –2  f(x)  6 19. g(x) = 9x + 38
(b) Horizontal line test cut the 5
Q(5, 9) graph only at one point. 5 – 3x
y=x Thus, the functon f(x) has 20. (a) g(x) = 2
an inverse function. (b) –14
M(a, b) Q ′(9, 5)
f(x) 3. (a) Yes because each object in –x – 1
P(0, –5) 0 f –1(x) set K has one image in set L. 21. (a) f(x) = 6
x (b) – 4133
(b) (i) 12 (ii) 2
(c) Domain = {–3, 2, 3, 5}
P ′(–5, 0) Codomain = {5, 10, 12, 15}
(b) a = 4, b = 1
x 1 Range = {5, 10, 12, 15} 22. Has an inverse function
8. (a) f –1(x) = 10 (b) – 10
4. (a) –29 (b) 152 23. y

9. (a) f –1(x) = x + 9 13 9
2 94 2
(b) 8 5. (a) m= 2 , n = 19 y=x
(a) f –1(x) (b) x= 9
10. = x + 7
3
(b) 4 1 2
11. (a) g–1(x) = 5 2, x ≠ 2 6. (a) 6 (b) 12 2 9 19 x
– –3 0
(b) – 270 x 7. (a) –2 –3
(b) 1
h–1(x) = 3x (c) –1  x  5 Domain of f –1: 1  x  19
x–4 Range of f –1: –3  f(x)  2
12. (a) , x ≠ 4 8. 13
(b) 2 24. (a)
3x 9. –6 and 3 f
x–2 , x ≠2
14
(a) h–1(x) = x+ 8 , ≠ 1 10. y f(x) y=x FORM 4 ANSWER
2x– 1 2
13. x

(b) 19 6
21
5 2 f –1(x)
2
14. (a) g–1(x) = x x 1, x ≠ 1 02 x
(b) – 14
3
2 x (b) a = 2, b = 14
+ 1
3 –7 –5 0 1 25. (a) (i) x 4
–
15. (a) g–1(x) = 5 , x ≠ 5 Range: 0  f(x)  6 7x – 5
(b) x 11. y 4
1 (ii) g(x) =
3 16
14 (b) 6
x–1
16. (a) 5 26. (a) g–1(x) = 4
(b) f(x)
= x 9 1, x ≠ 1 10 (b) fg(x) = 4x – 1
– 5
17. (a) – 2134 27. (a) 6 121x––29x, 11
(b) gf(x) 2
x = x ≠

(b) f(x) = 5x , x ≠ –3 –5 – 130 0 2 3
+ 4
3 x Range: 0  g(x)  16 28.

18. (a) 7 12. (a) –9 29. (a) Wrong because a = 21.
6 (b) – 34 x – 41
(b) g(x) = 4 – 3x (b) g–1(x) = 24

19. (a) 1 (c) gf(x) = 23 – 12x 30. 11
(b) 5
6 = 2 –x2x, ≠ 0 13. h = 32k – 6 5x
g(x) x 1+x
14. n = 5p – 2 31. (a) P(x) = , x ≠ –1

15. p = 6 – 5pq (b) 1
9
16. g(x) = 1 – 2x

Paper 1 17. g(x) = 6x – 11 32. (a) g–1(x) = x+9
1. (a) –3 8x – 35 (b) 4
(b) – 51 18. g(x) = 3 35
6

435