Page 52 - Spotlight A+ SPM Additional Mathematics Form 4 & 5

ANSWERS Complete answers
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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