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5 (a) 3x + 1 = y
3x = y – 1
y – 1
x = –––––
3
f –1 x – 1
(x) = –––––
3
(b) fg(2) = f[g(2)]
= f[2 × 2 2 ]
= 3 × 8 + 1
= 25
(c) gf(x) = g[3x + 1]
= 2(3x + 1) 2
= 2(9x 2 + 6x + 1)
= 18x 2 + 12x + 2
(d) ff(x) = f[3x + 1]
= 3(3x + 1) + 1
= 9x + 4
6 (a) f( – 1) = 2 × – 1 + 1
= – 1
(b) 2x + 1 = y
2x = y – 1
y – 1
x = –––––
2
f –1 x – 1
(x) = –––––
2
x
(c) – – 5 = y
3
x
– = y + 5
3
x = 3y + 15
g –1 (x) = 3x + 15
0
–
3
(d) fg(0) = f[ 5]
–
= f[–5]
= 2 × – 5 + 1
= – 9
(e) fg(x) = f[ – 5]
= 2[ – 5] + 1
x –3
x –3
2x
= –– – 9 3
Sets and Venn diagrams (page 58)
1
A
2 (a) {6. 12}
(b) {1, 5, 7, 11}
B
C
3
4
Total = 5 + 6 + 9 + 10 = 30
J
S
(b) S' ∪ T
(c) Let y students study both.
Model answers
n() = 32
12 – y + 18 + 9 = 32
39 – y = 32
y = 7
12 – y = 5
so 5 students study chemistry but not French
Matrices (page 61)
1 (a) AB = (
3×5 + 4×1
2×5 + 4×1 )
= (
19
14 )
(b) |A| = (3 × 4) – (2 × 4)
= 4
(
4 – 4
)
1
– – –4
1
A –1 4
– 1 3
=
–
= ( –2 –
2 3 –4 –
)
4
4
2 (a) The elements in the corresponding positions in A
and M add up to zero.
M = ( – 3 2
)
– 1
– 2
1 0
0 1
(b) ( ) is the identity matrix so N is the inverse of A.
|A| = (3 × 2) – (1 × – 2)
= 8
A –1 = N
(
5 6 9
C
)
1 1 –4 –4
– 1 3 –8 –8
L
K
W
10
12 – y y 18 – y
9
F
16
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational