Spotlight A+ SPM Additional Mathematics Form 4 & 5

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 , is given by v = 12 – 6t, such as t is the time, in seconds, after leaving
–1
point O. The particle stops momentarily at point K.
[Assume the motion of particle upwards is positive]

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K





O




15 m




J
SPM MODEL PAPER Find –1 –2 Diagram 12 [3 marks]




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

20°

6 cm
Q 40°

R
8 cm
5 cm

T
S
Diagram 13
(a) Calculate the length, in cm, of
(i) QR,
(ii) ST. [4 marks]
(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 cm , of triangle PRQ. [6 marks]
2

432




Model Paper Spotlight Add Math F5.indd 432 20/04/2021 10:37 AM

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




9. f(x) 4
FORM 4 5. –
5
13 6. 3
Chapter 1 Functions 10
9
7. g(x) = 3x – 2
1.1 8. f(x) = x – 2
3
1. (a) The relation is a function x 9. (a) 15
13
because each object has –6 3 0 5 (b) 5
only one image. – 2
(b) The relation is not a Range of f is 0  f(x)  13. 10. h = –3k
function because there is 11. p = 12, q = 5
one object does not have 10. (a) q = 1, p =1 2
any image. (b) 3 12. (a) x (b) x
2. (a) h(x) = |10 – x| or (c) –10 2x + 1 3x + 1
h(x) = |x – 10| 11. (a) 3 (c) x (d) x
2
(b) h(x) = x – 1 (b) 5 25 2x + 1 3x + 1
1
3. (a) A function. (c) 2 13. k = , h = 11
(b) A function. 2 2
(c) Not a function. 12. (a) 2 3 or 1.7321 14. (a) RM13 400
4. (a) {a, b, c, d}
(b) RM449 675
FORM 4 ANSWER 5. (a) 7 ©PAN ASIA PUBLICATIONS
or 1.4142
(b)
(b) {–2, 0, 2, 4}
2
(c) a, b, c, d
(b) 2
13. (a) 1
1.3
(d) –2, 0, 2
14. 3
(b) –2
1. (a) –4
15. (a) 3
(b) {–3, –2, 2, 7}
(d) 2
(c) 4
(ii) 3
(b) (i) –12
6. (a) 0, 2, 6, 8
(b) 7, 1, 10
5
5
(b) –2
3. (a) –7
(c) {7, 1, 10}
(b) – and
(d) 7 16. (a) (i) 5 12 (ii) 11 2. (a) –2 (b) 4
4
4. (a) Has an inverse function
(e) 8 15 because each element in set
7. (a) Domain = {–2, 0, 2, 4} 17. (a) (i) 4 P matched with only one
Codomain = {4, 6} 3 element in set Q.
Range = {4, 6} (ii) 4 (b) Does not have an inverse
(b) Domain of f is –1  x  3. 31 33 function because from
Codomain of f is 1  f(x)  3. (b) – and horizontal line test, the
Range of f is 1  f(x)  3. 4 4 line cuts the graph on two
8. (a) f(x) 18. –3 and 9 points.
4 5. (a) Has an inverse function g.
10 (b) Does not has an inverse
19. (a) 4
(b) 0  g(x)  10 function g.
5
6. y
4
1.2
11
x 3x – 1
–2 0 4 3 1. (a) fg(x) = y = x
3 3 g(x)
(b) f(x) (b) gf(x) = x – 5 6
2. (a) f (x) = 36x – 7
2
5 2
(b) g (x) = 16 + 9x 3
(c) gf(x) = 18x + 1 2
(d) fg(x) = 18x + 23 x
3 –2 0 2 3 6 11
3. (a) 111 (b) 731 –2 g (x)
–1
17 –1
x 4. (a) –4 (b) – Domain of g (x) is 2  x  11.
–1
–1
0 1 5 5 2 Range of g (x) is –2  g (x)  3.
2
434

Answer Additional Mathematics

7. (a) y 2. (a) –2  f(x)  6 19. g(x) = 9x + 38
(b) Horizontal line test cut the
Q(5, 9) 5
graph only at one point.
y = x 5 – 3x
Thus, the functon f(x) has 20. (a) g(x) =
an inverse function. 2
M(a, b) Q ′(9, 5) (b) –14
f(x) 3. (a) Yes because each object in –x – 1
–1
f (x) 21. (a) f(x) =
x set K has one image in set L. 6
P(0, –5) 0 (b) (i) 12 (ii) 2 43
(c) Domain = {–3, 2, 3, 5} (b) –
P ′(–5, 0) 13
(b) a = 4, b = 1 Codomain = {5, 10, 12, 15} 22. Has an inverse function
x 1 Range = {5, 10, 12, 15}
–1
8. (a) f (x) = (b) – 5 23.
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10 10 4. (a) –29 (b) y
x + 9 12
–1
9. (a) f (x) = 13 9
2 5. (a) m = , n = 19
(b) 8 9 4 2 y = x
–1
10. (a) f (x) = x + 7 (b) x = 9
3 2
(b) 4 1 2
–1
11. (a) g (x) = 5 , x ≠ 2 6. (a) 6 (b) 12 0 x
x – 2 –3 2 9 19
20 7. (a) –2 –3
(b) – (b) 1
7 (c) –1  x  5 Domain of f : 1  x  19
–1
3x Range of f : –3  f(x)  2
–1
–1
12. (a) h (x) = , x ≠ 4 8. 13
x – 4 2 24. (a)
3x 9. –6 and f
(b) , x ≠ 2 3
x – 2 14
x + 8 1 10. y y = x
–1
13. (a) h (x) = , x ≠
2x– 1 2 f(x)
19 6
(b) –1
21 5 2 f (x) FORM 4 ANSWER
–1
14. (a) g (x) = x , x ≠ 1 0 x
x – 1 2 2 14
3
(b) (b) a = 2, b = 14
2 0 1 x x + 1
3 –7 –5 25. (a) (i)
–1
15. (a) g (x) = , x ≠ 5 Range: 0  f(x)  6 4
5 – x 7x – 5
1 11. (ii) g(x) =
(b) y 4
3 (b) 6
14 16 x – 1
16. (a) 26. (a) g (x) =
–1
5 4
9 10
(b) f(x) = , x ≠ 1 (b) fg(x) = 4x – 1
x – 1 27. (a) 6
24 5 2x – 9 11
17. (a) – (b) gf(x) = , x ≠
13 0 x 11 – 2x 2
(b) f(x) = 5x , x ≠ –3 –5 – 10 2 28. 3
3
3 + x Range: 0  g(x)  16 4
7
18. (a) 12. (a) –9 29. (a) Wrong because a = 21.
6 4 –1 x – 41
(b) g(x) = 4 – 3x (b) – 3 (b) g (x) = 24
1 11
19. (a) (c) gf(x) = 23 – 12x 30.
6 5
2 – 2x 13. h = 32k – 6
(b) g(x) = , x ≠ 0 5x
x 14. n = 5p – 2 31. (a) P(x) = , x ≠ –1
15. p = 6 – 5pq 1 1 + x
16. g(x) = 1 – 2x (b) 9
Paper 1 17. g(x) = 6x – 11 32. (a) g (x) = x + 9
–1
8x – 35
4
1. (a) –3 (b) – 1 18. g(x) = 3 35
5 (b)
6
435

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