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Model answers
(b)
d (m)
50
40
30
20
10
0
F
1 2 3 4 5
6 7
t (s)
(c) F on graph (d must be increasing because the fish
is swimming away from Dimitra.)
t = 3·6 seconds
(d) 3·35 – 0·8 = 2.55 s
(e) Tangent at t = 2·5 drawn on graph.
18·5 – 0
Gradient = speed = ––––––– = 3·7 m/s.
6 – 1
4 (a) Tangent at t = 30.
33·5 – 12
Gradient = acceleration = –––––––– = 1·08 m/s
35 – 15
2
(b) Taking strips of width = 10 s
1 – × 5 × 10
2
= 25
1 – (5 + 14) × 10
2
= 95
1 – (14 + 28) × 10
2
= 210
1 – (28 + 32) × 10
2
= 300
10 × 32 = 320
10 × 32 = 320
1 – (32 + 25) × 10
2
= 285
1 – × 25 × 10
2
= 125
Total = 1680
Approximate distance = 1680 m
1
5 (a) Using strips of width –2 hour
1 1 – × –2 × 2 = 0·5
2
1 1 – (2 + 8) × –2
2
= 2.5
Approximate distance = 19.85 km or 20 km
(b) The acceleration is greatest when the graph is
steepest. This is approximately at 9.45 a.m.
Functions (page 55)
1 (a) f(x) = 2x – 1
y = 2x – 1
2x = y + 1
y + 1
x = –––––
2
(b)
f –1 x + 1
(x) = –––––
2
gf(x) = g[f(x)]
= g[2x – 1]
= (2x – 1) 2 – 1
= 4x 2 – 4x
2 (a) (i) g(4) = 2 × 4 2 – 5
= 27
1 –
3
(ii) fg(4) = 27
= 3
(b) (i) gf(x)= g[f(x)]
1 –
3
= g[x ]
1 –3 1 –
3
= 2 × x × x – 5
1 –
3
= 2x – 5
(ii) x = y
x = y 3
f –1 (x)= x 3
3 (a) f( – 1) = 3 × – 1 – 5
= – 8
(b) 3x – 5 = y
y + 5
x = –––––
3
f –1 x + 5
(x)=–––––
3
(c) fg(x)= f[x + 1]
= 3(x + 1) – 5
= 3x – 2
(d) 3f(x) = 5g(x)
3(3x – 5) = 5(x + 1)
9x – 15 = 5x + 5
4x = 20
x = 5
4 (a) f( – 1
4) = – (2 × – 4 + 5)
3
2 –
3
1 1 – (8 + 9.1) × –2
2
= 4.275
1 1 – (9.1 + 8.7) × –2
2
= 4.45
1 1 – (8.7 + 8.2) × –2
2
= 4.225
1 1 – (8.2 + 7.4) × –
2
2
= 3.9
Total = 19.85
(b)
= – 1
1 – (2x + 5) = y
3
2x + 5 = 3y
2x = 3y – 5
3y – 5
x= ––––––
2
f –1 (x) = ––––––
3x – 5
2
IGCSE Revision Guide for Mathematics Model Answers © 2004, Hodder & Stoughton Educational
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