Page 33 - Focus SPM KSSM Tg 4.5 - Add Maths

Additional Mathematics SPM Answers
Try This! 1.2 16. (a) f (x) = x 2 2 (b) f (x) = x 2 30
2
30
f (x) = x (–2) 3 f (x) = x (–2) 31
31
3
1. (a) 4 2 4
h g f (x) = x
17. (a) gf(x)
x gh(x) (b) h[gf(x)] = 3 600 + 0.06x, x . 15 000
h(x) (c) RM5 280
Try This! 1.3
gh 1
1. (a) –3 (b) p = 10
2
(b) 2. (a) 0 (b) 2 (c) 0 (d) –2
g h
3. (a) f has an inverse function. f is a one-to-one function.
(b) g does not have an inverse function. The inverse of g
x hg(x) maps two elements in the codomain to one element a
g(x) in the domain.
(c) h has an inverse function. h is a one-to-one function.
4. (a) The graph of function f has an inverse function. f is a
hg one-to-one function.
(b) The graph of function g does not have an inverse
(c) g g function. g is not a one-to-one function.
(c) The graph of function h has an inverse function. h is a
one-to-one function.
x gg(x)
g(x) 5. (a) The graph of function f has an inverse function. f is a
one-to-one function.
(b) The graph of function g does not have an inverse
function. g is not a one-to-one function.
gg
2
6. f [g(x)] = f 1 5 x + 2 , x ≠ 0
(d) 5
h h =
5
x + 2 – 2
x h (x)
2
h(x) = 5
5
x
= x, x ≠ 0
h 2
and g[f(x)] = g 1 5 2 , x ≠ 2
2. (a) 3x + 2 (b) 12x + 5 x – 2
2
3. (a) (i) 6x – 1 (ii) 18x – 12x + 2 = 5 + 2
2
(iii) 9x – 4 (iv) 8x 4 5
1 1 x – 2
(b) (i) , x ≠ –5 (ii) + 5, x ≠ 0
x + 5 x = x – 2 + 2
(iii) x, x ≠ 0 (iv) x + 10 = x, x ≠ 2
x + 1
4. (a) (i) x + 1 (ii)  Since fg(x) = x, where x is in the domain of g, and
2
(b) (i) |x – 6| (ii) |x| – 6 gf(x) = x, where x is in the domain of f, therefore g is an
5. (a) p = 10, k = 35 (b) gf(x) = x – 8x + 15 inverse function of f.
2
2
6. (a) 8 (b) 12 7. f [h(x)] = f 1 1 x + 2 , x ≠ 0
7. 61 1
=
8. 2x + 3 1 + 2 – 2
x
2
9. 2x + 3x + 4
1
2
10. 2x – 3 = = x, x ≠ 0
1
11. x + 7 x
12. x + 1 1
2
h[f(x)] = h 1 2 , x ≠ 2
13. x + x + 1 x – 2
2
1 1
14. (a) f (x) = x, f (x) = – , x ≠ 0, f (x) = x = + 2
3
2
4
x 1
1 x – 2
8
25
(b) f (x) = x, f (x) = – , x ≠ 0
x = x – 2 + 2 = x, x ≠ 2
1
15. (a) – , x ≠ 0 (b) x Since fh(x) = x, where x is in the domain of h, and
x hf(x) = x, where x is in the domain of f, therefore h(x) is
1 + x
(c) , x ≠ 1 the inverse of f(x).
1 – x
400