Cambridge International
AS & A Level Mathematics
Pure Mathematics 2 & 3
STUDENT’S BOOK: Worked solutions
Tom Andrews, Helen Ball,
Michael Kent, Chris Pearce
Series Editor: Dr Adam Boddison
Pure Mathematics 1 International Students Book Title page.indd 1
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1 Mathematics 2 & 3.indd 1
WS TITLE PAGE_Pure
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6/18/18
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PM
8/1/18 4:57
1
WORKED SOLUTIONS
Worked solutions
1 Algebra
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge
1
2 a (x + 1)(x + 2) = x2 + 3x + 2
b (x − 3)(x + 4) = x2 + x − 12
a y = 2x + 5
c (x − 2)(x − 3) = x2 − 5x + 6
y
d (x − 3)2 = x2 − 6x + 9
6
e (2x + 1)(x + 2) = 2x2 + 5x + 2
5
f (x + 1)(x + 2)(x + 3) = x3 + 6x2 + 11x + 6
4
3
3
2
 n
 n  n −1
 n
a b +   a n − 2b 2 +   a n − 3b 3 + ...

3
1 
 2
( a + b )n = a n + 
1
–5
–4
–3
–2
0
–1
–1
 n  n −1
 n
 n
a b +   a n − 2b 2 +   a n − 3b 3 + ...

1 
3
 2
( a + b )n = a n + 
1
2
3
4
5
6
x
x3 term = 6 ! 23 x 3 = 160x 3 so the coefficient is 160.
3! 3!
Exercise 1.1A
b y = 3x − 2
1
y
a y = 3x + 2
y
3
5
2
4
1
3
–5
–4
–3
–2
0
–1
–1
1
2
3
4
5
6
x
2
1
–2
–3
3
–7
–6
–5
–4
–3
–2
0
–1
–1
c y=3−x
–4
2
3
4
5
x
3
4
5
6
7
8
–2
b y = x−2
y
–5
1
–3
–2
4
y
3
5
2
4
1
3
0
–1
–1
–2
1
2
3
4
5
6
2
x
1
–4
–3
–2
0
–1
–1
1
2
x
–2
1
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1 ALGEBRA
3
y =3x +2
c
a y = 6−x
y
y
7
6
6
5
5
4
4
3
3
2
2
1
–6
–5
–4
–3
–2
1
0
–1
–1
1
2
3
4
5
6
x
–3
–2
–2
0
–1
–1
1
2
3
9
x
4
2
3
1
–2
0
–1
–1
2
1
2
3
4
5
6
x
–2
1
–6
–5
–4
–3
–2
–3
2
8
5
3
–3
7
6
4
–4
6
y
y
–5
5
b y = −x
d y = x −2
–6
4
0
–1
–1
1
2
3
4
5
6
x
1
2
3
4
5
6
x
1
2
3
4
5
6
c y =6− x
Not A as y = 2x − 5 has all positive y values.
Not B as y = 2x − 5 has all positive y values.
y
C is the correct graph.
7
y
6
5
8
4
6
3
4
2
2
1
–12 –10 –8
–6
–4
0
–2
–2
2
4
6
8
10 12
x
–6
–5
–4
–3
–2
0
–1
–4
–6
d y=−x
Not D as y = 2x − 5 has a positive x-intercept.
y
2
1
–6
–5
–4
–3
–2
0
–1
–1
x
–2
–3
–4
–5
2
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1
WORKED SOLUTIONS
4
7
a y = f (x )
y
y
5
5
4
4
3
3
2
2
1
1
–6
–5
–4
b y = f( x
a f (x) = 3 – 4 (|x + 2|)
–3
–2
0
–1
–1
–6
1
2
3
4
5
6
–5
–4
–3
–2
x
)
8
–2
0
–1
–1
f(x) = 5 − 2x, y = f ( x
3
2
1
2
3
4
5
6
x
)
1
–5
–4
–3
6
–2
25
3
20
2
15
1
10
0
–1
–1
1
2
3
4
5
x
y
4
1
2
3
4
5
6
x
5
0
–25 –20 –15 –10 –5
–5
–2
6
0
–1
–1
b y = |f(x – 2)|
5
–3
–2
–2
y
–4
x
4
1
–5
6
y
2
–6
5
5
3
5
4
x
a y = f .
 2
4
–3
3
b It is the modulus of a horizontal translation
to the left of two units.
5
–4
2
–3
6
–5
1
–2
y
–6
0
–1
–1
f(x) = 3 – |4x|
5
10 15 20
x
–10
y
4
3
2
1
–6
–5
–4
–3
–2
0
–1
–1
1
2
3
4
5
6
x
–2
–3
–4
3
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1 ALGEBRA
c y = |f(x)| + 3
2
a y = x − 5 and y = 2x
y
y
14
5
12
4
10
3
8
2
6
1
4
2
–10 –8
–6
–4
–4
0
–2
–2
2
4
6
8
10
–2
x
0
–1
–1
1
b x<
f(x) = – |2x – 7|
3
–3
–2
4
3x2 − 2x = 0
3
x(3x − 2) = 0
0
–1
–1
x = 0 or x =
1
2
3
4
5
6
5
–3
4
–4
3
–4
1
–2
b x=
2
or x = 2
3
0
–1
–1
5
1 − 3x = 4x + 5
2
1
–2
2 − 5x < 4x
−4x < 2 − 5x < 4x
2 – 5x < 4x
2
x>
9
2 − 5x > −4x
x<2
2
<x<2
9
3
0
–1
–1
–3
4
y
–2
4
5
6
7
x
2
3
1
2
3
–2
a y = 3x − 4 and y = 2.
–3
x
1
Exercise 1.1B
–4
7
2
( 12 )
23
b y = |f(x – 4)|, A at ( 0, )
2
c y = |f(x)| + 5, A at ( 0, 11 )
2
–5
6
y
–2
10 a y = f(|2x|), A at 0,
–6
5
b y = 2x − 1 and y = 1 − x
x
The transformation is –(f(|2x|) + 4)
1
4
5
3
1
–4
3
4x2 − 4x + 1 = 1 − 2x + x2
2
–5
2
a (2x − 1)2 = (1 − x)2
y
–6
1
–2
–4
9
–3
2
3
4
5
6
x
(1 − 3x)2 = (4x + 5)2
7x2 + 46x + 24 = 0
x=
−b ± b 2 − 4ac
2a
x=
−46 ± 46 2 − 4 × 7 × 24
14
x = −0.571 or x = −6
4
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1
WORKED SOLUTIONS
3
6
4 − x < 2x − 7
2x − 7 > 4 − x
11
x>
3
2x − 7 < −(4 − x)
x<3
7
(2x – 7)2 = (6 – x)2
4x2 – 28x + 49 = 36 – 12x + x2
3x2 – 16x + 13 = 0
8
So (x +1) is not a factor of 3x3 + 2x2 − 7x + 2.
16 ± 16 2 − 4 × 3 × 13
or ( x − 1)( 3x − 13) = 0
6
13
x = 1 or
3
9
b
a
Use the relation |x – a |  b ⇔ x  a – b or x  a + b.
b 3 – 7x > 2x or 3 – 7x < –2x
1
3
x < or x >
3
5
9x2 – 30x + 25 = 16 – 16x + 4x2
5x2 – 14x + 9 = 0
14 ± 14 2 − 4 × 5 × 9
or ( x − 1)( 5x − 9) = 0
10
9
x = 1 or 5
6x + 6
6x + 6
0
So (x + 1) is a factor of 2x4 + 3x3 − 12x2 − 7x + 6.
4
a
10 f(x) = 5 – 4x and g(x) = 3x
|5 – 4x| > 3|x|
25 – 40x + 16x2 > 9x2
7x2 – 40x + 25 > 0
61
8x3 − 6x2 − 47x + 84
x + 1 8x4 + 2x3 − 53x2 + 37x − 6
8x4 + 8x3
−6x3 − 53x2
−6x3 − 6x2
−47x2 + 37x
−47x2 − 47x
6x – 6
6x – 6
84x − 6
84x + 84
0
2
−10x + 81
−10x + 20
b
x2 + 5x + 6
x − 1 x3 + 4x2 + x − 6
x3 − x2
5x2 + x
5x2 − 5x
x3 + 2x2 + 4x − 10
x − 2 x4 + 0x3 + 0x2 − 18x + 81
x4 − 2x3
2x3 + 0x2
2x3 − 4x2
4x2 − 18x
4x2 − 8x
40 ± 40 2 − 4 × 7 × 25
14
5
x < or x > 5
7
x=
1
2x3 + x2 − 13x + 6
x + 1 2x4 + 3x3 − 12x2 − 7x + 6
2x4 + 2x3
x3 − 12x2
x3 + x2
−13x2 − 7x
−13x2 − 13x
x=
Exercise 1.2A
3x2 − x − 6
x + 1 3x3 + 2x2 − 7x + 2
3x3 + 3x2
−x2 − 7x
−x2 − x
−6x + 2
−6x − 6
x=
8
a
2x2 + 17x + 48
x − 1 2x3 + 15x2 + 31x + 12
2x3 − 2x2
17x2 + 31x
17x2 − 17x
−90
48x + 12
48x − 48
60
So the remainder is 60.
5
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1 Algebra
5
3 2 11
11
x − x+
2
4
8
8
2x + 1 3x3 − 4x2 + 0x + 1
3
3x3 + x 2
2
11 2
− x + 0x
2
11 2 11
− x − x
2
4
11
x +1
4
11
11
x+
4
8
3
−
8
6
(12 – 5a)x + b
(12 – 5a)x + 5(12 – 5a)
b – 5(12 – 5a) = 0 1
x2
+ (1 + b)x + (a + 1 + b)
x – 1 x3 + bx2 + ax + 10
x3 – x2
(1 + b)x2 + ax
(1 + b)x2 – (1 + b)x
(a + 1 + b)x + 10
(a + 1 + b)x – (a + 1 + b)
2x2 + 9x + (p + 18)
x – 2 2x3 + 5x2 + px – q
2x3 – 4x2
9x2 + px
9x2 – 18x
(p + 18)x – q
(p + 18)x – 2(p + 18)
– q + 2(p + 18) = 0 1
11 + a + b = –1 2
25a + b = 60 1
a + b = –12
a = 3 and b = –15
9
2x2 – 3x + (p + 12)
x + 4 2x3 + 5x2 + px – q
2x3 + 8x2
– 3x2 + px
– 3x2 – 12x
mx2 + 13x + 6
x – 2 mx3 + x2 – 10x – 6
mx3 – 2mx2
(1 + 2m)x2 – 10x
(1 + 2m)x2 – 2(1 + 2m)x
(4m – 8) x – 6
(4m – 8)x – 2(4m – 8)
(p + 12)x – q
(p + 12)x + 4(p + 12)
8m – 22
2x2 + m
x – 2 2x3 – 4x2 + mx + 8
2x3 – 4x2
0 + mx + 8
mx – 2m
– q – 4(p + 12) 2
2p – q = – 36 1
4p + q = – 48 2
1 + 2
7
x2 + (a – 5)x + (12 – 5a)
x + 5 x3 + ax2 – 13x + b
x3 + 5x2
(a – 5)x2 – 13x
(a – 5)x2 + 5(a – 5)x
8 + 2m
6p = – 84
8m – 22 = 8 + 2m
p = – 14 and q = 8
m=5
x2
– 2x + (10 – 2p)
x + 5 x3 + 3x2 – 2px + p
x3 + 5x2
–2x2 – 2px
–2x2 – 10x
(10 – 2p)x + p
(10 – 2p)x + 5(10 – 2p)
p – 5(10 – 2p) = –33
p = 17
11
10
5x – 7
x2 + 3x – 4 5x3 + 8x2 – 41x + 28
5x3 + 15x2 – 20x
–7x2 – 21x + 28
–7x2 – 21x + 28
0
Exercise 1.3A
1
f(1) = 13 + 4(12) + 1 − 6 = 0 so (x − 1) is a factor of f(x).
2
f(1) = 2(13) + 15(12) + 31 + 12 = 60 so (x − 1) is not a
factor of f(x) and the remainder is 60.
6
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1
WORKED SOLUTIONS
3
a f(x) = 3x3 + 2x2 − 7x + 2
f(1) = 3 + 2 − 7 + 2 = 0 so (x − 1) is a factor.
8
f(–3) = –81 – 45 + 141 – 15 = 0
1
1 5 47
f − =− − +
− 15 = 0
3
9 9 3
3
2
3x – 5x – 47x – 15 = (x + 3)(3x + 1)(x – 5)
( )
Using algebraic division,
3x3 + 2x2 – 7x + 2 = (x – 1)(3x2 + 5x – 2)
and 3x2 + 5x – 2 = (3x – 1)(x + 2)
b 5x3 – 15x2 – 47x – 15 = 2x3 – 10x2
3x3 + 2x2 − 7x + 2 = (x − 1)(3x − 1)(x + 2)
3x3 – 5x2 – 47x – 15 = 0
b f(x) = 2x4 + 3x3 − 12x2 − 7x + 6
(x + 3)(3x + 1)(x – 5) = 0
1
So x = −3, − or 5.
3
f(−1) = 2 − 3 − 12 + 7 + 6 = 0 so (x + 1) is a factor.
By algebraic division,
2x4 + 3x3 − 12x2 − 7x + 6
= (x + 1)(2x3 + x2 − 13x + 6)
9
( 12 ) = − 14 + 94 + 10 − 12 = 0
g(−3) = −54 + 9 + 39 + 6 = 0 so (x + 3) is a factor.
f −
By algebraic division,
2x3 + x2 – 13x + 6 = (x + 3)(2x2 – 5x + 2) and
2x2 – 5x + 2 = (2x – 1)(x − 2)
2x3 + 9x2 – 20x – 12 = (x + 6)(2x + 1)(x – 2)
2x4 + 3x3 − 12x2 − 7x + 6
= (x + 1)(x + 3)(2x − 1)(x − 2)
a f(x) = x4 − 18x + 81
f(2) = 16 − 36 + 81 = 61
b 2x3 + 9x2 = 20x + 12
2x3 + 9x2 – 20x – 12 = 0
(x + 6)(2x + 1)(x – 2) = 0
1
x = −6, − or 2.
2
10 f(x) = 12x2 + 5x + 7
There are no values of x such that f(x) = 0.
b f(x) = 8x4 + 2x3 − 53x2 + 37x − 6
f(−1) = 8 − 2 − 53 − 37 − 6 = −90
5
6
a f(2) = 16 + 36 – 40 – 12 = 0
f(–6) = – 432 + 324 + 120 – 12 = 0
g(x) = 2x3 + x2 − 13x + 6
4
a f(5) = 375 – 125 – 235 – 15 = 0
f(x) = 3x3 − 4x2 + 1
3
3
1
f − = − − 1 + 1 = − so (2x + 1) is not a factor.
2
8
8
( )
a f(x) = x4 − 7x3 + 13x2 + 3x − 18
f(2) = 16 − 56 + 52 + 6 − 18 = 0 so (x − 2) is a factor.
x4 − 7x3 + 13x2 + 3x − 18 = (x − 2)(x3 − 5x2 + 3x + 9)
g(x) = x3 − 5x2 + 3x + 9
g(3) = 27 − 45 + 9 + 9 = 0 so (x − 3) is a factor.
g(x) = (x − 3)(x2 − 2x − 3)
f(x) = (x + 1)(x − 2)(x − 3)2
Exercise 1.4A
1
a i
2x − 5 ≡ A(x + 3) + B(x + 2)
Substitute x = −3
−11 = −B
B = 11
Substitute x = −2
−9 = A
A = −9
b (x + 1)(x − 2)(x − 3)2 = 0
So x = −1, 2 or 3
7
a f(–1) = –2 – 13 + 10 + 21 = 16 so not a factor.
b f(1) = 2 – 13 – 10 + 21 = 0
f(7) = 686 – 637 – 70 + 21 = 0
( 32 ) = − 274 − 1174 + 302 + 21 = 0
f −
2x3 – 13x2 – 10x + 21 = (x – 1)(2x + 3)(x – 7)
c 2x3 – 13x2 – 10x + 21 = 0
(x – 1)(2x + 3)(x – 7) = 0
3
So x = 1, − or 7.
2
2x − 5
A
B
=
+
(x + 2)(x + 3) x + 2 x + 3
−9
11
2x − 5
=
+
(x + 2)(x + 3) x + 2 x + 3
2x − 5
A
B
=
+
(x + 2)(x + 3) x + 2 x + 3
2x – 5 = A(x + 3) + B(x + 2)
ii
2 = A + B
1
−5 = 3A + 2B
2
1 ×2
4 = 2A + 2B
3
2 – 3
A = −9
Substitute into 1 .
B = 11
−9
2x − 5
11
=
+
(x + 2)(x + 3) x + 2 x + 3
7
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1 Algebra
6x − 12
A
B
b i (x − 1)(x + 5) = x − 1 + x + 5
2
6x − 12 = A(x + 5) + B(x − 1)
Substitute x = 1
−6 = 6A
A = −1
Substitute x = −5
−42 = −6B
B=7
11 − 2x
1
3
=
−
(x − 2)(x + 5) x − 2 x + 5
−1
6x − 12
7
=
+
(x − 1)(x + 5) x − 1 x + 5
ii
11 − 2x
A
B
=
+
(x − 2)(x + 5) x − 2 x + 5
11 − 2x = A(x + 5) + B(x − 2)
Substitute x = −5
21 = −7B
B = −3
Substitute x = 2
7 = 7A
A=1
3
6x − 12
A
B
=
+
(x − 1)(x + 5) x − 1 x + 5
a
2 − 3x
A
B
=
+
(2x + 1)(3 − x) 2x + 1 3 − x
2 − 3x = A(3 − x) + B(2x + 1)
6x − 12 = A(x + 5) + B(x − 1)
6 = A + B
1
−12 = 5A − B
2
Substitute x = 3
−7 = 7B
B = −1
1 + 2
Substitute x = −
−6 = 6A
A = −1
7 7
= A
2 2
A=1
Substitute into 1 B = 7
−1
6x − 12
7
=
+
(x − 1)(x + 5) x − 1 x + 5
2 − 3x
( 2x + 1)(3 − x )
3 − 5x
A
B
=
+
(x − 3)(x − 7) x − 3 x − 7
c i
b
3 − 5x = A(x − 7) + B(x − 3)
A = 11
Substitute x = 7
Substitute x = −2
−32 = 4B
18 = −2B
B = −8
B = −9
3 − 5x
= 3 − 8
(x − 3)(x − 7) x − 3 x − 7
2x + 22 11
9
=
−
x x+2
x 2 + 2x
3 − 5x
A
B
=
+
(x − 3)(x − 7) x − 3 x − 7
c
3 − 5x = A(x − 7) + B(x − 3)
1
2
2 + 3
−12 = −4A
A=3
Substitute into 1 B = −8
3 − 5x
3
8
=
−
(x − 3)(x − 7) x − 3 x − 7
4x − 30 = A + B
x 2 − 8x + 15 x − 5 x − 3
4x – 30 = A(x − 3) + B (x − 5)
Substitute x = 3
−18 = −2B
1 ×3
−15 = 3A + 3B
2x + 22 A
B
= +
x 2 + 2x x x + 2
22 = 2A
A=3
3 = −7A − 3B
1
1
−
2x + 1 3 − x
Substitute x = 0
−12 = −4A
−5 = A + B
=
2x + 22 = A (x + 2) + B (x)
Substitute x = 3
ii
1
2
3
B=9
Substitute x = 5
−10 = 2A
A = −5
4x − 30
−5
9
+
=
x 2 − 8x + 15 x − 5 x − 3
8
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1
WORKED SOLUTIONS
4
2 = A(−1)(1)
A
1
B
=
+
x −9 x−3 x+3
1 = A(x + 3) + B(x − 3)
Substitute x = 3
1 = 6A
1
A=
6
Substitute x = −3
1 = −6B
1
B=−
6
1
1
1
=
−
x 2 − 9 6 ( x − 3) 6 ( x + 3)
2
A = −2
Substitute x = 1
2 − 3 − 4 = B(1)(−1)
B=5
Substitute x =
5
a
C=2
2 − 3x − 4x
c
6 − 6x − 5x 2
( x − 1)( x − 2)( x + 4 )
B=−
A=1
Substitute x = −4
6 + 24 − 80 = C(−5)(−6)
C=−
6
8 − 8 + 8 = B(1)(−1)
B = −8
Substitute x = 3
18 − 12 + 8 = C(2)(1)
C=7
2 − 3x − 4x 2
=
3
8
7
−
+
x −1 x − 2 x − 3
A
B
C
+
+
x x − 1 1 − 2x
2 − 3x − 4x2 = A(x − 1)(1 − 2x)
+ B(x)(1 − 2x)
+ C(x)(x − 1)
Substitute x = 0
5
3
6 − 6x − 5x 2
1
13
5
=
−
−
( x − 1)( x − 2)( x + 4 ) x − 1 3 ( x − 2) 3 ( x + 4 )
Substitute x = 2
( x )( x − 1)(1 − 2x )
13
3
6 − 6 − 5 = A(−1)(5)
A=3
b
A
B
C
+
+
x −1 x − 2 x + 4
Substitute x = 1
Substitute x = 1
=
=
6 − 12 − 20 = B(1)(6)
(x − 2)(x − 3)
2x2 − 4x + 8 = A
+ B(x − 1)(x − 3)
+ C(x − 1)(x − 2)
2x − 4x + 8
−2
5
2
+
+
x
x − 1 1 − 2x
Substitute x = 2
2x 2 − 4x + 8
A
B
C
=
+
+
( x − 1)( x − 2)( x − 3) x − 1 x − 2 x − 3
( x − 1)( x − 2)( x − 3)
=
6 − 6x − 5x2 = A(x − 2)(x + 4)
+ B(x − 1)(x + 4)
+ C(x − 1)(x − 2)
1
1
1
=
−
.
x − a 2 2a ( x − a ) 2a ( x + a )
2
2
( x )( x − 1)(1 − 2x )
2
2 − 4 + 8 = A(−1)(−2)
( )( )
4 3 2
1
1
− − =C
−
2 2 2
2
2
A
1
B
=
+
x 2 − 16 x − 4 x + 4
1 = A(x + 4) + B(x − 4)
Substitute x = 4
1 = 8A
1
A=
8
Substitute x = −4
1 = −8B
1
B=−
8
1
1
1
=
−
x 2 − 16 8 ( x − 4 ) 8 ( x + 4 )
Notice that
1
2
7
A
1
B
=
+
x 2 − a2 x − a x + a
1 = A(x + a) + B(x − a)
Substitute x = a
1 = 2aA
1
A=
2a
Substitute x = −a
1 = −2aB
1
B=−
2a
1
1
1
=
−
x 2 − a 2 2a ( x − a ) 2a ( x + a )
5 + 3x − x 2
−x + 3x 2 + 4x − 12
3
Using the factor theorem:
Substitute x = 2 into denominator
−8 + 12 + 8 − 12 = 0
So (x − 2) is a factor.
9
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1 Algebra
Using the factor theorem:
Substitute x = − 2 into denominator
8 + 12 − 8 − 12 = 0
So (x + 2) is a factor.
10
Using the factor theorem:
Substitute x = 3 into denominator
−27 + 27 + 12 − 12 = 0
So (3 − x) is a factor as coefficient of x3 is negative.
Express the expression as partial fractions.
1 = A(2x + 1)(x – 5) + B (x + 3)
(x – 5) + C (x + 3)(2x + 1)
1
x=−
2
5
11
1= B
−
2
2
5 + 3x − x 2
A
B
C
=
+
+
−x 3 + 3x 2 + 4x − 12 x + 2 x − 2 3 − x
5 + 3x − x2 = A(x − 2)(3 − x) + B(x + 2)(3 − x)
+ C(x + 2)(x − 2)
Substitute x = 2
5 + 6 − 4 = B(4)(1)
7
B=
4
Substitute x = 3
5 + 9 − 9 = C(5)(1)
C=1
Substitute x = −2
5 − 6 − 4 = A(−4)(5)
1
A=
4
B=−
( )( )
The coefficient of
1
1
7
1
5 + 3x − x 2
=
+
+
−x + 3x 2 + 4x − 12 4 ( x + 2) 4 ( x − 2) 3 − x
5
cannot be split into partial fractions
x 2 − x + 10
because x2 – x + 10 does not factorise.
( x + 3)( x − 2)
x2 + x − 6
=
3
2
x
−
1)( x + 2)( x + 4 )
(
x + 5x + 2x − 8
9
x2 + x − 6
A
B
C
=
+
+
( x − 1)( x + 2)( x + 4 ) x − 1 x + 2 x + 4
2
1 + 1 – 6 = A(3)(5)
B=
2
3
x=–4
16 – 4 – 6 = C (–5)(–2)
C=
3
5
x2 + x − 6
2
3
4
=
+
−
x + 5x 2 + 2x − 8 3 ( x + 2) 5 ( x + 4 ) 15 ( x − 1)
3
p
3p
5x
.
2 ≡ x +3 −
( x + 3)
( x + 3 )2
5x
( x + 3 )2
4
A=−
15
4 – 2 – 6 = B(–3)(2)
x 2 + 8x + 4 A B
C
= + 2+
x x
x−2
x 2 ( x − 2)
x2 + 8x + 4 = A(x)(x − 2) + B(x − 2) + C(x2)
Substitute x = 0
4 = −2B
B = −2
Substitute x = 2
4 + 16 + 4 = 4C
C=6
Substitute x = 1
1 + 8 + 4 = A(1)(−1) + (−2)(−1) + (6)(1)
13 = −A + 2 + 6
A = −5
5x = p(x + 3) −3p
Substitute x = −3
−15 = −3p
p=5
x=1
x = –2
1
4
is −
2x + 1
55
x 2 + 8x + 4 −5 2
6
− 2+
=
x
x−2
x 2 ( x − 2)
x
x2 + x – 6 = A
(x + 2)(x + 4) + B(x – 1)
(x + 4) + C(x – 1)(x + 2)
4
55
Exercise 1.4B
3
8
1
1
=
2x 3 − 3x 2 − 32x − 15 ( x + 3)( 2x + 1)( x − 5)
1
A
B
C
=
+
+
( x + 3)( 2x + 1)( x − 5) x + 3 2x + 1 x − 5
3
≡
5
15
−
.
x + 3 ( x + 3 )2
7x − 3
7x − 3
=
x 2 − 8x + 16 ( x − 4 )2
7x − 3
( x − 4 )2
=
A
B
+
x − 4 ( x − 4 )2
7x − 3 = A(x − 4) + B
Substitute x = 4
B = 25
Substitute x = 0
−3 = −4A + 25
A=7
7x − 3
7
25
=
+
x 2 − 8x + 16 x − 4 ( x − 4 )2
10
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1
WORKED SOLUTIONS
4
Substitute x = 0
−20 = 4A – 2B
Substitute x = 1
−12 = A − B
2 ×2
−24 = 2A − 2B
3 – 1
−4 = −2A
A=2
The errors are shown in red.
2x 2 − x − 6 ≡ 2x 2 − x − 6
x 3 + 4x 2 + 4x
x(x + 2)2
2x 2 − x − 6
A
B
C
≡
+
+
x 3 + 4x 2 + 4x (x) (x + 2) (x + 2)2
2x 2 − x − 6
A(x + 2)2 + B(x)(x + 2) + C(x)
≡
3
2
x + 4x + 4x
x(x + 2)2
2x2 − x − 6 ≡ A(x + 2)2 + B(x)(x + 2) + C(x)
Substitute x = 0.
−6 = 4A
3
A=−
2
Substitute x = −2.
8 + 2 − 6 = −2C
C = −2
Substitute x = 1.
3
2 − 1 − 6 = − (9) + B(1)(3) + (−2)(1)
2
( )
5
7
A
1
B
C
=
+
+
(x + 1)(x − 2)2 x + 1 x − 2 (x − 2)2
1 = A(x − 2)2 + B(x + 1)(x − 2)
+ C(x + 1)
Substitute x = 2
1 = 3C
1
C=
3
()
B=−
()
1
9
1
1
1
1
=
−
+
(x + 1)(x − 2)2 9(x + 1) 9(x − 2) 3(x − 2)2
Yes,
6
1
can be split into partial fractions.
(x + 1)(x − 2)2
2x 2 + 6x + 5
A
B
C
=
+
+
x − 2 (x − 2)2 (x − 2)3
(x − 2)3
2x2 + 6x + 5 = A(x − 2)2 +B(x − 2) + C
Substitute x = 2
8 + 12 + 5 = C
C = 25
3
2x 2 + 6x + 5
2
14
25
=
+
+
(x − 2) (x − 2)2 (x − 2)3
(x − 2)3
2x 2 − x − 6
3
7
2
≡−
+
−
2(x) 2(x + 2) (x + 2)2
x 3 + 4x 2 + 4x
Substitute x = −1
1 = 9A
1
A=
9
Substitute x = 0
1
1
(1)
1=
(−2)2 + B(1)(−2) +
9
3
2
Substitute into 2 :
−12 = 2 − B
B = 14
−5 = − 27 + 3B − 2
2
21
3B =
2
∴
1
8
The two different methods to split a rational
function with linear factors in its denominator
into partial fractions are substitution and equating
coefficients.
For partial fractions without repeated terms,
the substitution method relies only on basic
mathematical operations without the need for
any algebraic manipulation, whereas the equating
coefficients method results in the solving of
simultaneous equations.
For partial fractions with repeated terms, the
substitution method relies on a combination of
basic mathematical operations and the solving
of simultaneous equations, whereas the equating
coefficients method results in the solving of
simultaneous equations with three unknowns.
1
A
B
C
= +
+
x x − 3 ( x − 3 )2
x 3 − 6x 2 + 9x
1 = A(x – 3)2 + Bx (x – 3) + Cx
x=3
1 = 3C
1
C=
3
x=0
1 = 9A
1
A=
9
x=1
1
1=
( 4 ) + B (1)( −2) + 13 (1)
9
()
B=−
()
1
9
1
1
1
1
=
−
+
9x 9(x − 3) 3 ( x − 3)2
x 3 − 6x 2 + 9x
11
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1 Algebra
2x − 1
A
B
C
=
+
+
( 2x + 1)3 2x + 1 ( 2x + 1)2 ( 2x + 1)3
9
2
2x – 1 = A(2x + 1)2 + B(2x + 1) + C
x=−
A(x 2 + 1) + (Bx + C )(x + 3)
(x + 3)(x 2 + 1)
2
2
x − 7x + 6 ≡ A(x + 1) + (Bx + C)(x + 3)
Let x = −3.
9 + 21 + 6 = 10A
18
So A = .
5
Let x = 0.
18
6 = 5 + 3C
≡
1
2
–2 = C
x=0
–1 = A + B – 2
x=1
( )
1 = 9A + 3B – 2
A = 0 and B = 1
2x − 1
1
2
=
−
( 2x + 1)3 ( 2x + 1)2 ( 2x + 1)3
4
.
5
Let x = 1.
So C =
1−7+6 =
x 2 + 7x − 1
A
B
C
=
+
+
x + 5 ( x + 5)2 ( x + 5)3
( x + 5)3
10
36
16
+ 4B +
5
5
−13
So B =
.
5
All of the above working is correct. The denominator
of A, B and C has been missed from the final
statement:
State the original fraction as partial fractions.
x = –5
–11 = C
x=0
–1 = 25A + 5B – 11
x=1
7 = 36A + 6B – 11
∴
A = 1 and B = –3
1
( x + 5 )2
is –3.
x 2 − 5x + 6
Ax + B
C
≡ 2
+
(x + 2)(x 2 + 1)
x +1 x + 2
( Ax + B)(x + 2) + C(x 2 + 1)
(x 2 + 1)(x + 2)
2
x − 5x + 6 ≡ (Ax + B)(x + 2) + C(x2 + 1)
Let x = −2.
4 + 10 + 6 = 5C
So C = 4.
Let x = 0.
6 = 2B + 4
So B = 1.
Let x = 1.
1 − 5 + 6 = (A + 1)(3) + 8
So A = −3.
≡
∴
3
x 2 − 7x + 6
18
4 − 13x
≡
+
(x + 3)(x 2 + 1) 5(x + 3) 5(x 2 + 1)
(x − 3)2
A Bx + C
≡ + 2
x(x 2 − 6) x
x −6
A(x 2 − 6) + (Bx + C )x
x(x 2 − 6)
(x − 3)2 ≡ A(x2 − 6) + (Bx + C)x
Let x = 0.
9 = −6A.
−3
So A =
2
Let x = 1.
15
4=
+B+C
2
−7
B+C=
2
Let x = −1.
15
16 =
+B−C
2
17
B–C=
2
5
B=
2
C = −6
≡
Exercise 1.4C
1
(185 )(2) + ( B + 45 )(4)
0=
x2 + 7x – 1 = A(x + 5)2 + B(x + 5) + C
The coefficient of
x 2 − 7x + 6
A
Bx + C
≡
+
(x + 3)(x 2 + 1) x + 3 x 2 + 1
x 2 − 5x + 6
1 − 3x
4
≡
+
(x + 2)(x 2 + 1) x 2 + 1 x + 2
∴
5x − 12
3
(x − 3)2
≡
−
x(x 2 − 6) 2(x 2 − 6) 2x
12
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1
WORKED SOLUTIONS
4
Let x = 2
8p
p
x 2 + 8x + 7 ≡ px − 37 −
+
.
2 2
2
9
(
−
1
)
x
3(x − 1)2
(x − 1) (x + 2) 9(x + 2)
(
5 = 2A +
Put the right-hand side over a common
denominator.
)
17
(1) − 14 (7 )
4
5
A=
4
2
(px − 37)(x − 1) − p(x − 1)(x + 2) + 8p × 3(x + 2)
x + 8x + 7
=
x 2 + 3x − 5
5x + 17
1
(x − 1)2 (x 2 + 2)
9(x − 1)2 (x 2 + 2)
≡
−
∴
2
2
− 1)
4
x
(
x
−
x
+
x
+
1
3
4
3
(
)
(px − 37)(x − 1)2 − p(x − 1)(x 2 + 2) + 8p × 3(x 2 + 2)
x 2 + 8x + 7
=
(x − 1)2 (x 2 + 2)
9(x − 1)2 (x 2 + 2)
3x − 1
A
B
C
≡
+
+
7
2
x + 5 x −1 x +1
x
+
5
x
−
1
(
)
9(x2 + 8x + 7) = (px − 37)(x − 1)2 − p(x − 1)(x2 + 2)
+ 24p(x2 + 2)
3x – 1 ≡ A(x – 1)(x + 1) + B(x + 5)(x + 1)
Let x = 1.
+ C(x + 5)(x – 1)
9(1 + 8 + 7) = 72p
Let x = 1
So p = 2.
2 = B(6)(2)
2
2
2
(
)
(
5
So
6
x2 − 5
can be split into partial fractions.
x(x 2 − 3)
x 2 + 3x − 5
Ax + B
C
≡ 2
+
x + 3 x −1
( x − 1) x 2 + 3
(
)
x2 + 3x – 5 ≡ (Ax + B)(x – 1) + C(x2 + 3)
Let x = 1
B=
17
4
So B =
– 4 = C(4)(– 2)
1
2
Let x = – 5
So C =
– 16 = A(– 6)(– 4)
2
3
So A =
∴
8
3x − 1
( x + 5 ) ( x 2 − 1)
−2
1
1
+
+
3(x + 5) 6(x − 1) 2(x + 1)
3x + 2 ≡ (Ax + B)(2x – 5) + C(x2 + 5)
Let x =
5
2
(
38
45
= C( )
4
4
15
25
+ 2 =C
+5
2
4
)
38
45
Let x = 0
C=
2 = (B)(– 5) +
4
9
Let x = 1
38
( 5)
45
B=
A =−
1
( 3)
4
≡
3x + 2
Ax + B
C
≡
+
(x 2 + 5)(2x − 5) x 2 + 5 2x − 5
5= A+
1
C=−
4
Let x = 0
– 5 = (B)(– 1) −
)
(
– 1 = C(4)
)
1
6
Let x = –1
x2 − 5
A Bx + C
≡ + 2
x(x 2 − 3) x
x −3
x2 − 5
A(x 2 − 3) + (Bx + C )x
≡
2
x(x 2 − 3)
x(x − 3)
2
2
x − 5 ≡ A(x − 3) + (Bx + C)x
Let x = 0.
−5 = −3A
5
So A =
3
Let x = 1.
−10
−4 =
+B+C
3
−2
B+C=
3
Let x = −1.
−10
−4 =
+B−C
3
−2
B–C=
3
−2
B=
3
C=0
2
2x
∴ x 2− 5 ≡ 5 −
x(x − 3) 3x 3(x 2 − 3)
(
∴
)
20
( −3) + 38
(6 )
45
45
19
45
3x + 2
20 − 19x
38
≡
+
(x 2 + 5)( 2x − 5) 45(x 2 + 5) 45 ( 2x − 5)
13
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1 Algebra
2x + 1
9
( 2x + 3)( x 2 + 1)
≡
Ax + B
C
+
x 2 + 1 2x + 3
2
( )
2x + 1 ≡ (Ax + B)(2x + 3) + C(x2 + 1)
Let x = −
−2 = C
3
2
( 94 + 1)
3
8
13
Let x = 0
C=−
7
13
Let x = 1
8
(1)
13
4
B=
(
∴
10
(
)
7
8
(2)
(5) − 13
13
2x + 1
( 2x + 3)( x 2 + 1)
≡
4x + 7
8
−
13(x 2 + 1) 13(2x + 3)
1
Ax + B
C
≡ 2
+
x 2 + 2 ( 2x − 1) x + 2 2x − 1
)
5
1 ≡ (Ax + B)(2x – 1) + C(x2 + 2)
Let x =
1 =C
1
2
( 14 + 2)
4
9
Let x = 0
1
B=−
9
Let x = 1
(
1= A−
4
( 2)
9
(1 + x )
1
2
0.1 0.01
−
= 1.03 to 3 s.f.
3
9
1  − 1  − 3
1 −1
2  2  2 2  2   2  3
1
≈1+ x +
x +
x
2
2
6
= 1 + 1 x − 1 x2 + 1 x3
2
8
16
1 = + x −1 ≈ + − x + ( −1)( −2) x 2
1
) 1 ( 1)
1+ x (
2
−1)( −2)( −3) 3
(
x
+
3!
= 1 – x + x2 – x3
1
b
Replace x with x2:
= (1 + x2)−1
1 + x2
a
6
1 3
27 2 27 3
− x+
x −
x
4 4
16
8
3x < 1 so |x| < 2
3
2
7
The term in x3 in the expansion of (1 − 2x )
)
−2x − 1
1
∴ coefficient of 2
is
.
9
x +2
Exercise 1.5A
(−2)(−3) 2
x = 1 − 2x + 3x2
2
b Replace x with –x to get (1 − x)−2
a (1 + x)−2 ≈ 1 + (−2)x +
( )( )( )
−1
2
is
−1 −3 −5
2
2
2
3
5
5
× ( −2x ) = −
× −8 x 3 = x 3 so
3× 2×1
16 ( )
2
5
the coefficient of x3 is
2
1
(1) + 49 (3)
9
2
A =−
9
1
1 + 0.1 ≈ 1 +
≈ 1 – x2 + x4 – x6.
C=
1 = (B)(– 1) +
a
3
b
Replace x with 2x in part a to get
2
3
1
1
1
1 + 2x ≈ 1 + ( 2x ) − ( 2x ) + ( 2x )
2
8
16
1
1
= 1 + x − x2 + x3
2
2
c
The expansion is valid if −1 < 2x < 1
so –0.5 < x < 0.5 or x < 0.5.
4
13
A=
( )
c
1 = (B)(3) −
3= A+
5×3×1
a The coefficient of x3 is 2 2 2 = 5 .
3!
16
5 × 3 × 1 × −1
2 2 2
2
5
.
b The coefficient of x4 is
=−
4!
128
1 −2
1
3 3 2
1
1
1
3
x = 1 + x − x2
a (1 + x ) ≈ 1 + x +
3
2
3
9
b −1 < x < 1 or x < 1
8
3
3
3
3
4.01 2 = ( 4 × 1.0025) 2 = 4 2 × (1.0025) 2
3
= 8 × (1 + 0.0025) 2
so you need an expansion of
(1 + x )
3
2
( )
3 × 1 × −1
3×1
2 2
2 3
3
2
2
2
=1+ x +
x +
x + ...
2
2
6
3
3
1 x 3 ...
= 1 + x + x2 −
+
2
8
16
= (1 + (−x))−2 ≈ 1 − 2(−x) + 3(−x)2 = 1 + 2x +3x2
14
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1
WORKED SOLUTIONS
y
So when x = 0.0025 then
3
1
4.01 2 = 8 + 12x + 3x 2 − x 3 + ...
2
= 8 + 0.03 + 0.000 018 75 – ....
which is 8.030 02 to 5 d.p. Three terms are enough
for this accuracy.
9
30
25
20
15
10
−1
1
= (1 − 3x ) 2
1 − 3x
1
(1 − 3x )− 2
( ) ( )( )
(− 12 )(− 32 )(− 52 )(3x) +
3!
1
(− 2 )(− 32 )(− 52 )(− 72 )(3x) +
4!
1
3
(− 2 )(− 2 )(− 52 )(− 72 )(− 92 )(3x)
5
−1 −3
2
2
1
= 1 + − (3x) +
(3x)2 +
2
2
–8
–6
–4
0
–2
–5
2
4
6
8
x
–10
3
4
x < –5 or x > –1
2
a f(x) = 2x3 + x2 + px + 12
f(3) = 0
2(33) + 32 + 3p + 12 = 0
5
3p = –75
5!
3
27x
135x 3 2835 4 15309 5
=1− x +
−
+
x −
x
2
8
16
128
256
p = –25
2
1
1
b
1
10 3.99 2 = ( 4 − 0.01) 2 = 2 (1 − 0.0025) 2
x x2 x3
−
−
2
8 16
1

0.0025 0.00252 0.00253 
∴ 3.99 2 2  1 −
−
−
= 1.997498
2
8
16 

1.99750 to 5 d.p.
[this gives approximation correct to 9 d.p]
1
(1 − x)2 = 1 −
−4x + 12
−4x + 12
0
2x2 + 7x – 4 = (2x – 1)(x + 4)
Exam-style questions
1
a |3x + 5| = 4x
2x2 + 7x – 4
x − 3 2x3 + x2 – 25x + 12
2x3 − 6x2
7x2 − 25x
7x2 − 21x
3
Means 3x + 5 = 4x so x = 5
Or – (3x + 5) = 4x
7x = –5
5
x =−
7
However this is not a valid solution since it
makes 4x a negative value, and the modulus
cannot be negative.
b 2|3x + 5| > |4x|
|6x + 10| > |4x|
2x3 + x2 – 25x + 12 = (x– 3)(2x – 1)(x + 4)
a p(x) = x3 + 2ax2 + 3ax + 14
p(7) = 0
73 + 2a(72) + 3a(7) + 14 = 0
119a = –357
a = –3
b i p(x) = x3 – 6x2 – 9x + 14 = 0
x2 + x − 2
x − 7 x3 − 6x2 − 9x + 14
x3 − 7x2
x2 − 9x
x2 − 7x
−2x + 14
−2x + 14
(6x + 10)2 > 16x2
36x2 + 120x + 100 > 16x2
0
20x2 + 120x + 100 > 0
x2
x2 + 6x + 5 > 0
+ x − 2 = (x + 2)(x − 1)
x3 − 6x2 − 9x + 14 = (x − 7)(x + 2)(x − 1) = 0
So x = −2, 1 or 7
ii Remainder is −20
4
|3 – 2x| < |4x + 3|
(3 – 2x)2 < (4x + 3)2
15
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1 Algebra
9 – 12x + 4x2 < 16x2 + 24x + 9
12x2 + 36x > 0
x(x + 3) > 0
y
7
30
25
20
15
a f(x) = 4x4 + 8x3 − 21x2 − 18x + 27
f(1) = 4 + 8 − 21 − 18 + 27 = 0 so (x − 1) is a factor
of f(x).
4x3 + 12x2 − 9x − 27
(x − 1) 4x4 + 8x3 − 21x2 − 18x + 27
4x4 − 4x3
12x3 − 21x2
12x3 − 12x2
−9x2 − 18x
−9x2 + 9x
10
5
–8
–6
0
–2
–5
–4
2
4
6
8
−27x + 27
−27x + 27
x
0
x4 − 9x2 − 4x + 12 = (x − 1)(4x3 + 12x2 − 9x − 27)
g(x) = 4x3 + 12x2 − 9x − 27
g(−3) = −108 + 108 + 27 − 27 = 0 so (x + 3) is a
factor of g(x).
–10
x < –3 or x > 0
5
a p(x) = x4 – 9x2 – ax + 12
p(3) = 0
34 – 9(32) – 3a + 12 = 0
3a = 12
a=4
b p(x) = x4 – 9x2 – 4x + 12
p(1) = 1 – 9 – 4 + 12 = 0
Hence (x – 1)is a factor of f(x).
4x2 − 9
(x + 3) 4x3 + 12x2 − 9x − 27
4x3 + 12x2
0 −9x − 27
−9x − 27
0
f(x) = 4x4 + 8x3 − 21x2 − 18x + 27
= (x − 1)(x + 3) (2x − 3)(2x + 3)
b 6x4 + 2x3 – 10x = 2x4 – 6x3 + 21x2 + 8x – 27
c f(x) = x4 − 9x2 − 4x + 12 = 0
4x4 + 8x3 – 21x2 – 18x + 27 = 0
(x – 3)(x –1)(x + 2)2 = 0
(x – 1)(x + 3)(2x – 3)(2x + 3) = 0
3
3
so x = 1, – 3, or −
2
2
x = –2, 1 or 3
6
a 7 – 3x < |11x – 5|
either 11x – 5 > 7 – 3x
8
Using remainder theorem
m(3)3 – 3(3)2 + 5(3) – 4m = 3(3)3 – 5(3)2 – m(3) + 2m
14x > 12
6
x>
7
or – (11x – 5) > 7 – 3x
⇒ 27m – 27 + 15 – 4m = 81 – 45 – 3m + 2m
– 11x + 5 > 7 – 3x
⇒ 24m = 48
8x < – 2
x<– 1
4
b 7 – 3(y – 1)2 < |11(y – 1)2 – 5|
7 – 3(y2 – 2y + 1) < |11(y2 – 2y + 1) – 5|
7 – 3y2 + 6y – 3 < |11y2 – 22y + 11 – 5|
4 + 6y – 3y2 < |11y2 – 22y + 6|
c x = (y –
1)2
either (y –
y >1±
1)2
6
>
7
6
7
or (y – 1)2 < −
no solutions
1
4
⇒ 23m – 12 = 36 – m
⇒m=2
9
2x – 1
4x2 – 2x + 3 8x3 – 8x2 + 8x – 3
8x3 – 4x2 + 6x
–4x2 + 2x – 3
–4x2 + 2x – 3
No remainder.
10 a
x2
0
x2 – 3x + 5
+ 2x + 3 x4 – x3 + 2x2 + px + q
x4 + 2x3 + 3x2
–3x3 – x2 + px
–3x3 – 6x2 – 9x
5x2 + (p + q)x + q
5x2 + 10x + 15
(p – 1) + (q – 15) = 0
16
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1
WORKED SOLUTIONS
(x2 + 2x + 3)(x2 – 3x + 5) = x4 – x3 + 2x2 + x + 15
So p = 1 and q = 15
14
b (x2 + 2x + 3)(x2 – 3x + 5) = x4 – x3 + 2x2 + x + 15
x2 + 2x + 3
b2 – 4ac = 4 – 4.1.3 < 0 so no solutions
x2 – 3x + 5
A
6x + 7
B
=
+
(x + 1)2 (x + 1) (x + 1)2
6x + 7 = A(x + 1) + B
Substitute x = −1
B=1
Substitute x = 0
A=6
6x + 7
6
1
=
+
(x + 1)2 (x + 1) (x + 1)2
b2 – 4ac = 9 – 4.1.5 < 0 so no solutions
11 a
2x – 3
2x2 + 13x + 20 4x3 + 20x2 + px – 60
4x3 + 26x2 + 40x
– 6x2 + (p – 40)x – 60
– 6x2 – 39x – 60
(p – 1)x
b p=1
c 4x3 + 20x2 + x – 60 = (2x2 + 13x + 20)(2x – 3) = 0
(2x + 5)(x + 4)(2x – 3) = 0
3 5
So x = , − or − 4.
2 2
y
12 a
25
20
15
10
5
–8
–6
–4
0
–2
–5
2
4
6
8
x
15
9
A
Bx + C
≡
+
(x + 1)(x 2 + 2) x + 1 x 2 + 2
9
A(x 2 + 2) + (Bx + C )(x + 1)
≡
2
(x + 1)(x + 2)
(x + 1)(x 2 + 2)
2
9 ≡ A(x + 2) + (Bx + C) (x + 1)
Let x = −1.
9 = 3A
So A = 3.
Let x = 0.
9=6+C
So C = 3.
Let x = 1.
9 = 9 + (B + 3)(2)
0 = 2B + 6
So B = −3
9
3
−3x + 3
∴
≡
+
(x + 1)(x 2 + 2) x + 1 x 2 + 2
16 a (1 − x)−1 = 1 + (−1)(−x) +
(−1)(−2)(−3)
(−x)3 + ….
3!
= 1 + x + x2 + x3…
–10
b
2x + 4
0 can never be negative so
2 − 12 x
2x + 4
=0
2 − 12 x
⇒ 2x + 4 = 0 ⇒ x = −2 (NB x ≠ 4 or
denominator = 0)
13
11x − 5
A
B
=
+
(x − 3)(3x − 2) (x − 3) (3x − 2)
11x − 5 = A(3x − 2) + B(x − 3)
Substitute x = 3
7A = 28
A=4
Substitute x = 0
−5 = (4)(−2) − 3B
B = −1
11x − 5
4
1
=
−
(x − 3)(3x − 2) (x − 3) (3x − 2)
(−1)(−2) (−x)2
2!
+
b
1
=
2−x
1
( )
2 1− x
2
=
( ) + ( x2 ) + ...
x
x
1
1+ +
2 
2
2
2
3
1 x x2 x3
+ +
+
+ ...
2 4
8 16
c The expansion is valid if x < 1 which means
2
x < 2.
=
17 a
16x 2 + 29x + 7
A
B
C
=
+
+
( x + 4 )( 2x + 1)2 x + 4 2x + 1 ( 2x + 1)2
16x2 + 29x + 7 = A(2x + 1)2 + B(x + 4)(2x + 1) +
C(x + 4)
Let x = –4
256 – 116 + 7 = 49A
A=3
Let x = −
1
2
17
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1 Algebra
x=0
29
7
+7= C
2
2
C = –1
4−
1 = 4A – 2B + 5
2A – B = –2 (1)
Let x = 0
x=3
7 = 3 + 4B – 4
10 = A + B + 5
B=2
A + B = 5(2)
16x 2 + 29x + 7
3
2
1
=
+
−
( x + 4 )( 2x + 1)2 x + 4 2x + 1 ( 2x + 1)2
b
3A = 3
−1
3
= 3( x + 4)
x+4
A=1
( )
3
x
x
3  4 (1 + )  = (1 + )
4 
4
4

3( x + 4)
x
= 3  4 1 + 
4 

−1
−1
−1
( )
3
x
1+
4
4
−1
B=4
x2 + 1
1
4
5
=
+
+
( x − 2)3 x − 2 ( x − 2)2 ( x − 2)3
−1
( ) 

( −1)( −2) x4
3
x
= 1 + ( −1)
+
4
4
2


()
2


3
x x2 
= 1 − +
4 16 
4
3 3x 3x 2
= −
+
−…
4 16 64
−1
2
= 2 ( 2x + 1)
2x + 1

( −1)( −2)( 2x )2 
2 ( 2x + 1) = 2 1 + ( −1)( 2x ) +

2


2
= 2[1 – 2x + 4x ]
−1
= 2 – 4x + 8x2 – …
−
−2
1
2 = −1( 2x + 1)
2
x
+
1
(
)

( −2)( −3)( 2x )2 
−1( 2x + 1) = −1 1 + ( −2)( 2x ) +

2


= – 1[1 – 4x + 12x2]
−2
= – 1 + 4x – 12x2
3 + 2 −
1
x + 4 2x + 1 ( 2x + 1)2
2
= 3 − 3x + 3x + 2 − 4x + 8x 2 − 1 + 4x − 12x 2
4 16 64
=
18 a
(1) + (2)
7 3
253 2...
−
x−
x …
4 16
64
x2 + 1
A
B
C
=
+
+
( x − 2)3 x − 2 ( x − 2)2 ( x − 2)3
b
1 = x − 2 −1
)
x−2 (
( x2 )
( x − 2)−1 =  −2 1 −

( )
 −2 1 − x 

2 
( )
x
1
− 1−
2
2
−1
−1
= ( −2)
−1
−1
(1 − x2 )
−1
( ) 

( −1)( −2) − x2
x
1
= − 1 + ( −1) − +
2
2
2


( )
2


1
x x2 
= − 1 + +
2
4 
2
=−
1 x x2
− −
…
2 4
8
( )
4
x
2 = 1− 2
( x − 2)
( x2 ) +
1 + ( −2) −
=1+ x +
−2
( −2)( −3)
2
(− x2 )
2
3x 2
…
4
( )
5
5
x
=− 1−
8
2
( x − 2 )2
−3
( ) 

( −3)( −4 ) − x2
5
x
− 1 + ( −3) − +
8
2
2


( )
2


5
3x 3x 2 
= − 1 +
+
8
2
2 
=−
5 15x 15x 2
−
−
…
8 16
16
x2 + 1 = A(x – 2)2 + B(x – 2) + C
x=2
5=C
18
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1
WORKED SOLUTIONS
1
4
5
+
+
x − 2 ( x − 2)2 ( x − 2)3
=−
∴
3x 2 5 15x 15x 2
1 x x2
− −
+1+ x +
− −
−
2 4
8
4
8 16
16
b
1 3
5 2
=− −
x−
x
8 16
16
x
c Expansion valid if 2 < 1 which means |x| < 2.
19 a
4 − 8x = 4(1 − 2x) = 2(1 −
( )
1
2x)2
( )( )
1 −1 −3
1 −1




2 2
2
2 2
1
(−2x)3 
(−2x)2 +
≈ 2 1 + (−2x) +
6
2
2




2
3
= 2 − 2x − x − x
1
b The expansion is valid if 2x < 1 so x <
2
7x − 5
9
b
20
≡−
+
x−a x−3
( x − a )( x − 3)
7x – 5 = –9(x – 3) + b(x – a)
x=3
16 = b(3 – a)
x=0
– 5 = 27 – ab
– 32 = – ab
32
b=
a
32
16 =
(3 − a )
a
96
16 =
− 32
a
a=2
b = 16
3x 2 + 4x + 1
A
Bx + C
≡
+
21 a
(x − 1)(x 2 + 2) x − 1 x 2 + 2
3x 2 + 4x + 1
8
x + 13
=
+
( x − 1) x 2 + 2 3 ( x − 1) 3 x 2 + 2
−1
1
= − (1 − x ) valid for x < 1
( x − 1)
(
(
)
(

1
x2 
= 21 + 
2
2

x +2
−1
)
valid for
)
x2
<1
2
Mathematics in life and work
The distance travelled by vehicle A at time t s is given
by d = 7 − t. When t = 4 vehicle B starts its journey. At
time t s the distance travelled by vehicle B is given by
d = 2t − 8.
1
d
7
B
A
6
5
4
3
2
1
0
1
2
3
4
5
6
7
8
9
10
t
2
The vehicles A and B meet 1 second after B
starts (t = 5).
3
7 − t = 2t − 8
(7 − t)2 = (2t − 8)2
49 − 14t + t2 = 4t2 − 32t + 64
3t2 − 18t + 15 = 0
t2 − 6t + 5 = 0
(t − 1)(t − 5) = 0
t = 1 or 5
4
The vehicles only meet once despite there being
two solutions to part 3. Vehicle B doesn’t start
moving until t = 4.
3x 2 + 4x + 1
A(x 2 + 2) + (Bx + C )(x − 1)
≡
2
(x − 1)(x + 2)
(x − 1)(x 2 + 2)
3x2 + 4x + 1 ≡ A(x2 + 2) + (Bx + C)(x − 1)
Let x = 1.
3 + 4 + 1 = 3A
So A = 8 .
3
Let x = 0.
16
−C
1=
3
13
So C = .
3
Let x = −1.
26
3 − 4 + 1 = 8 + 2B −
3
24 26
−
+
= 2B
3
3
1
So B = .
3
3x 2 + 4x + 1
8
x + 13
≡
+
(x − 1)(x 2 + 2) 3(x − 1) 3(x 2 + 2)
19
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LOGARITHMS AND EXPONENTIAL FUNCTIONS
2 Logarithms and exponential functions
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge
1
a 24
d
2
3
b 2−5
1
(23)2
=
3
22
c
4
1
22
5
e 20
a n5
Exercise 2.1A
b n−1
c n6
a
1
n2
d
×
1
n3
=
5
n6
y
20
1
a log 103 = 3
c log 10−3 = −3
2
a log10 2 =
15
10
3
5
0
–30 –25 –20 –15 –10 –5
–5
5
10 15 20 25 30
10
4
8
7
5
3
2
1
c
–1
0
1
2
3
4
5
x
6
5
−1
2
=−
1
2
a log (2 × 10) = log 2 + log 10 = 0.3010 + 1
= 1.3010
−1
2
=−
1
× log 2 = −0.1505
2
a log (8 × 10) = log 8 + log 10 = 0.9031 + 1
= 1.9031
a 2k
b log 100 + log x = 2 + k
c log 1 − log x = 0 − k = −k
y
1
1
k
2
a log x + log y2 = k + 2h
d log10 − log x 2 = 1 −
10
8
6
6
1
1
k −h
2
c log 100 + log x2 − log y3 = 2 + 2k − 3h
b log x 2 − log y =
4
2
0
d log10
c log (12 × 8 ÷ 10) = log 12 + log 8 − log 10
= 1.0792 + 0.9031 − 1 = 0.9823
d log 8 = log (8 ÷ 12) = log 8 − log 12
12
= 0.9031 − 1.0792 = −0.1761
4
–2
b log 10 = 1
b log (12 ÷ 8) = log 12 − log 8 = 1.0792 − 0.9031
= 0.1761
6
–3
1
2
3
3
c log10 2 =
2
d log 2
9
–4
1
c log (10 ÷ 2) = log 10 − log 2 = 1 − 0.3010
= 0.6990
y
–5
b log 106 = 6
d log 10−6 = −6
b log (2 ÷ 10) = log 2 − log 10 = 0.3010 − 1
= −0.6990
x
b
–6
20 − 18
= −0.4
0−5
Equation is y = −0.4x + 20
New price = 24 000 × 1.04 × 1.04 = $25 958.40
Gradient =
2
4
6
8
10 12 14 16 18 20
x
d log 1 − log x − log y = −k – h
20
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WORKED SOLUTIONS
7
a x = 102 = 100
b 2y =
b T
he curve is y = 10x for the reflection in the
line y = x, swap x and y in the equation.
102
x = 10y and this can be written as log x = y
y = 50
8
c
c log z = 102 = 100
z = 10100
a log 4 = log 22 = 2 log 2
y
y = 10x
y=x
b log 5 is not possible
d log 7 is not possible
e log 8 =
log 23
–1
= 3 log 2
f log 9 = log 32 = 2 log 3
h log 12 = log (22 × 3) = 2 log 2 + log 3
1
10 aIf y = log a then 10 y = 1 ; a = 10–y; log a = –y;
a
hence y = log 1 = − log a
a
n
blog an = n log a and a = n a so
n
log a = n log
( a)
n
n
( )
The point on the line y = log x is the
reflection of this, it is (2.174, 0.337)
Exercise 2.2A
1
2
2
= n × n log n a = n log n a
( )
12 aSome points on the curve are (0, 1), (1, 10)
and (−1, 0.1)
y
1
a log 9 9 2 = 1
2
1
b 27 = 9 × 3 = 9 × 9 2
3
2
c log9 9−2 = −2
1
d 3 = 92
1
1
log 9 9 4 =
3
1
4
2 loga 4 = loga 16
loga 8 − loga 2 = loga 4
log a
1
1
− log a = loga 3
2
6
3 loga 2 = loga 8
4
a 7
c
b
1
log2 512 = 4.5
2
1
= log3 1 − log3 5 = −c
5
c log3 53 = 3c
1
x
d
1
= − log2 32 = −5
32
1
8
log2 256 =
3
3
log 2
a log3 (3 × 5) = log3 3 + log3 5 = 1 + c
b log 3
–1
1
3 = 92 = 94
5
1
b log5 53 = 3
d log4 44 = 4
3
The gradient is log ( a + 1) − log a
( a + 1) − a
= log (a + 1) – log a = log a + 1 = log 1 + 1
a
a
The gradient is positive for all values of a so the
function is increasing.
1
1
cAs a increases, a → 0 , 1 + a → 1 and so the
gradient log 1 + 1 → log1 = 0
a
a log2 26 = 6
c log3 35 = 5
log 9 9 2 =
11 aThe coordinates of P and Q are (a, log a) and
((a + 1), log (a + 1))
( )
1
1
1
bSince a > 0, 1 + a > 1 so log (1 + a ) > 0
x
1
( )
aIf y = log a then 10y = a; hence ak = (10y)k = 10yk
using the law of indices.
Hence log ak = yk and from the definition of y,
log ak = k log a
log 216 log 63 3log 6 3
=
=
= = 1.5
b
log 36
log 6 2 2log 6 2
0
–1
If the gradient on y = log x is 0.2, this is the
reflection of the point on y = 10x where the
1
5
gradient is 0.2 = 5. That is 2.3 × 10x = 5; 10 x = 2.3 ;
x = log 5 = 0.337 and y = 100.337 = 2.174
2.3
g log 11 is not possible
9
y = log x
1
c log 6 = log (2 × 3) = log 2 + log 3
c
1c
 1
d 5 = 3c =  9 2  = 9 2
 
log9 5 =
1
c
2
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LOGARITHMS AND EXPONENTIAL FUNCTIONS
6
Exercise 2.3A
a n3 = 216 so n = 6
b
1
n2
= 4 so n = 4² = 16
1
a 7.39
d 6.57
2
a To 5 d.p. the values are:
i 2.704 81
ii 2.718 15
iii 2.718 28
b The answers increase and approach the
value of e as n increases.
c n0 = 1 so n can be any positive integer.
d n1.5 = 27
n × n = 27
n=9
7
a = n3.5
b=
8
n4.5
n1 + 3.5 =
=
a i b=
ax
b In a =
by
n×
n3.5
3
= na
b 1.65
e 0.0498
c The limit is e
a
y
ii a = by
7
6
from ii above
5
substitute for b using i: a = (ax ) y = a xy
This means that xy = 1 and the result follows.
9
a 16 =
24
so log2 16 = 4; 8 =
4
y = e–x
2
4
1
b If x = logc a then c x = a; if y = logc b then c y = b
(c y)z =
–4
10 a (x – 1)5 = 1024 = 45; hence x – 1 = 4 and x = 5
–3
–2
–1
4
5
b logx 9x = 3
= 9x(x ≠ 0)
y = ex+1 is A.
= 9 so x = 3
y = (e + 1)x is B.
11 Add the two equations: 2 log4 x = 9; log 4 x = 9 ;
2
9
9
2
x = 4 = 2 = 512
3
4
x
 −1
b y = ex + 1 is A and a translation of   .
 0
3
Subtract the two equations: 2 log4 y = 3; log 4 y = 2 ;
 0
y = ex + 1is C and a translation of   .
1 
3
y = 4 2 = 23 = 8
x
a2
6
c A is y = ex + 1 = e × ex so it is a stretch of y = ex
by a factor of e.
a If x = 0, y = ex + 2 = e2.
The intercept is (0, e2).
2
b A translation of   will map y = ex + 4 onto
0
 
log2 x + log4 x = 12; hence
log 2 x + 1 log 2 x = 12 ; 3 log 2 x = 12;
2
2
log2 x = 8; x = 28 = 256
1
c log 8 x = log 3 x = 3 log 2 x;
2
log 2 x + 1 log 2 x + 1 log 2 x = 22 ; 11 log 2 x = 22;
2
3
6
log 2 x = 6 × 22 = 12 ; x = 212 = 4096
11
2
y = ex + 1 is C.
x2
then (a2)y = x; a2y = x; hence
1
2y = loga x and y = log a x
2
b From part a, log 4 x = log 2 x = 1 log 2 x ;
2
2
1
Gradient = ex = y
a e0 = 1
b e2
c 2
a One method is to find where each curve crosses
the y-axis.
x3
12 a If y = log
0
b Reflection in the y-axis
cy z =
If z = logb a then
a;
a;
a;
log c a log c a
=
which is the
yz = logc a; z =
y
log c b
required result.
y = ex
3
23 so log2 8 = 3
4
 1
log 216
16 = 24 =  8 3  = 8 3 so log 8 16 = 4 =
3
log 2 8
 
b z=
c 0.368
y = e(x − 2) + 4 which is y = ex + 2.
7
a Correct graphs sketched
()
2 x
8
9
b e0.5x = e 4
An x-direction stretch of stretch factor 4.
x = −e, e
2
a Using the chain rule, if y = e0.5x + a then
dy
dy
= e0.5x + a × 0.5 = 0.5y ; hence 2
=y
dx
dx
b At A x = 0 and y = ea so A has coordinates (0, ea)
22
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WORKED SOLUTIONS
dy
= 0.5y = 1 e a and this is the gradient of
2
dx
the tangent.
1
The equation of the tangent is y − e a = 2 e ax ;
if y = 0 then −e a = 1 e ax ; x = –2
2
The point is (−2, 0)
c At A,
Exercise 2.4A
1
11 a Using the chain rule f'(x) =
4e2x +1 ×
2
x = 3.51
c 2x − 5 = ln 125
x=
d −
5 + ln125
= 4.91
2
1 2
x = ln 0.5
2
x = ± −2 ln 0.5 = ±1.18
3
a 0.5x = e4
x = 109
b 4x + 2 = e3.5
b At P, x = −1 and f(–1) = 4e–2 +1 = 4e–1; the
point is (–1, 4e–1)
e3.5 − 2
= 7.78
4
c x = 1 e−4 = 0.009 16
2
120
d x = 5 = 0.809
e
1
ln = ln 1 − ln a = −ln a because ln 1 = 0.
a
a y = ln 4x = ln 4 + ln x
x=
c f'(–1) = 2f (–1) = 2 × 4e–1 = 8e–1 The gradient of
the tangent is 8e–1 and the equation is
y – 4e–1 = 8e–1 (x + 1)
– 4 = 8x + 8; 8x = –12; x = –1.5 and the point is
(–1.5, 0)
a x = ln 2000 = 7.60
b −x = ln 0.03
2 = 2f (x)
Where this meets the x-axis, y = 0 and
– 4e–1 = 8e–1 (x + 1); – 4 = 8 (x + 1);
b 4 × ln a = 12
d 1 (ln a + ln b) = 3.75
2
c ln b − 2 × ln a = −1.5
10 a f'(x) = ex = y and this is the gradient of the
tangent at (x, y)
At P the gradient of the tangent is y so y = y
x
x
and hence x = 1; P is (1, e)
b At P the gradient of the tangent is e so the
gradient of the normal at P is − 1
e
1
The equation of the normal is y − e = − e ( x − 1);
ey – e2 = – x + 1; x + ey = 1 + e2
a ln a + ln b = 7.5
4
5
12
 0 
The graph is a translation of y = ln x by 
.
 ln 4
b
y
y = ln 4x
3
A
2
ln 4
1
T
–3
N
–2
y = ln x
0
–1
–1
1
2
3
4
5
6
7
8
x
–2
dy
= y so the gradient of the tangent at A is ea
dx
The equation of the tangent is y – ea = ea(x – a);
where it crosses the x-axis y = 0 and so
– ea = ea(x – a); –1= x – a; x = a –1
The gradient of the normal is − 1a = −e −a and the
e
equation is y – ea = – e–a(x – a)
where it crosses the x-axis y = 0 and so
– ea = – e–a(x – a); e2a = x – a; x = a + e2a
The base of the triangle TN = a + e2a – (a – 1) = e2a + 1;
the height is ea
The area = 1 × e 2a + 1 × e a = 1 e a(e 2a + 1)
2
2
(
)
–3
6
a Either ln 20 + t = ln 100 or t = ln
100
= ln 5.
20
t = 1.609
b −t = ln
35
40
t = 0.134
c 3t = ln
8000
250
t = 1.155
d −0.85t = ln
14.8
32.5
t = ln 0.4554 = 0.925
−0.85
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LOGARITHMS AND EXPONENTIAL FUNCTIONS
7
a
12 a f'(x) = 2ex – e–x. At a stationary point, f'(x) = 0 so
2ex – e–x = 0; 2ex = e–x;
1
1
2e2x = 1; e 2x = 2 ; 2x = ln 2 = − ln 2 ; so
x = − 1 ln 2 at a stationary point.
2
1 ln 2
− 1 ln 2
x
=
− 1 ln 2, y = 2e 2 + e 2
If
21
1
−
= 2 e ln 2 2 + e ln 2 2 = 2 × 1 + 2
2
= 2+ 2=2 2
y
y = 6e–x
7
6
5
y = 2ex
4
3
( )
2
1
–4
–3
–2
–1
0
1
2
3
4
5
6
x
b f"(x) = 2ex + e–x, which is positive for all
values of x and in particular if x = − 1 ln 2 so
2
the value is a minimum.
b Where they cross, 2ex = 6e−x.
e2x = 3
2x = ln 3
x = ln 3 = 0.5493;
2
y = 2e0.5493 = 3.464
Coordinates, to 3 s.f., are (0.549, 3.46).
8
a B is (ln 2, 2)
b The gradient is the y-coordinate = 2.
c The gradient is
9
1
.
2
1
d The gradient of y = ln x at (x, ln x) is or you
x
dy 1
=
could say
dx x
a y2 + y = 6 and this is a quadratic equation in y.
13 a If 10x ≡ ecx then ln 10x ≡ ln ecx; x ln 10 ≡ cx so
c = ln 10 or 2.30 to 3 s.f.
dy
(ln10)x
b If y = 10x = e (ln 10) x then dx = ( ln10 ) e
= ( ln10 ) × 10 x
c If (ln 10) × 10x = 10 then 10 x = 10 and
ln10
10
x = log
= log10 − log ( ln10 )
ln10
( )
= 1 – log(ln 10) or 0.638 to 3 d.p.
14 a At Q, 0.2e0.5x = a; e0.5x = 5a; 0.5x = ln 5a; x = 2 ln 5a
b
(y + 3)(y − 2) = 0
y – a = 0.5a(x – 2 ln 5a)
y = −3 or 2
Where it crosses the x-axis, y = 0 and
– a = 0.5a(x – 2 ln 5a); – 2 = x – 2 ln 5a;
c Either ex = −3 or ex = 2.
ex = −3 has no solution because ex is always
positive.
x = 2 ln 5a – 2; hence 2 ln 5a – 2 = 6;
4
2 ln 5a = 8; ln 5a = 4; 5a = e4; a = e
5
If ex = 2, then x = ln 2 = 0.693 to 3 s.f.
Exercise 2.5A
The logarithms in these solutions can be to any base.
1
a x log 3 = log 11
x=
log11
= 2.183 to 3 d.p.
log 3
b x log 4 = log 175
11 a ln (e + e2) = ln (e(1 + e)) = ln e + ln (1 + e)
= 1 + ln (1 + e)
x=
log175
= 3.726 to 3 d.p.
log 4
c x log 12 < log 6
b ln (e2 − e4) = ln (e2(1 − e2)) = ln e2(1 + e)(1 − e)
x<
= ln e2 + ln (1 + e) + ln (1 − e)
= 2 + ln (1 + e) + ln (1 − e)
dy
= 0.2e0.5x × 0.5 = 0.5y ; At Q, y = a and so
dx
dy
= 0.5a = gradient of tangent.
dx
The equation of the tangent is
b Rearrange and factorise: y2 + y − 6 = 0.
10 Write y = ex.
2y2 − 9y + 4 = 0
(2y − 1)(y − 4) = 0
1
y = or 4
2
1
1
x
If e = then x = ln = −0.693 to 3 s.f.
2
2
If ex = 4 then x = ln 4 = 1.39 to 3 s.f.
( )
2
log 6
hence x < 0.721 to 3 d.p.
log12
Answers are rounded to 3 d.p.
a x log 0.5 = log 0.4 hence x =
log 0.4
= 1.322.
log 0.5
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2
WORKED SOLUTIONS
d The multiplier is now 1.2.
b log 0.7x ⩾ log 0.25 hence x log 0.7 ⩾ log 0.25.
log 0.25
Hence x log 0.7 .
When 10 × 1.2t = 80
1.2t = 8
The inequality reverses because log 0.7 is
negative.
Hence x ⩽ 3.887.
c x log 0.9 = log 2.55 hence x =
3
a 1.8t =
t=
750
= 3.75
200
log 2.55
= −8.885.
log 0.9
c
Growth will not continue when the pond is
fully covered.
7
1.04t
b 1.65 × 1.05t million
c When 1.65 × 1.05t = 2.5
> 123 = 1.2947
95
1.05t =
log1.2947
= 6.586
log1.04
t=
a If it takes t years to increase by 50%, 1.09t = 1.5.
Changes in travel patterns could make the
model incorrect.
b If it takes t years to double, 1.09t = 2.
8
Rearrange: 3 log 5 + log 7 = x log 7 − x log 5.
a (x + 2) log 4 = log 90
3 log 5 + log 7
= 20.1 to 3 s.f.
log 7 − log 5
b log 3x + 2 > log 4x
Hence x =
log 90
x=
− 2 =1.246
log 4
b (2x + 1) log 6 > log 35
Hence (x + 2) log 3 > x log 4.
log 35
2x + 1 >
=1.9843
log 6
1.9843 − 1
x>
and hence x > 0.492.
2
c (4x − 3) log 15 = log 8
4x − 3 =
log 8
log15
a y = 10 ×
Hence x <
9
2 log 3
; x < 7.64 to 3 s.f.
log 4 − log 3
Take logarithms base e: (x +1) ln a = x ln (a + 1).
Hence x ln a + ln a = x ln (a + 1).
But ln (a + 1) − ln a = ln
1.5t
b 10 × 1.53 = 33.75 m²
c When 10 × 1.5t = 80
1.5t = 8
t log 1.5 = log 8
t=
5.1 weeks
Rearrange: 2 log 3 > x (log 4 − log 3).
Rearrange: x{ln (a + 1) − ln a} = ln a.
x = 0.942
6
a log 5x + 3 = log 7x − 1
Hence (x + 3) log 5 = (x − 1) log 7.
Just over 8 years
5
log1.51515
= 8.52
log1.05
d The number cannot increase beyond the
capacity of the airport.
4.7 years
log 2
= 8.04
log1.09
2.5
= 1.51515
1.65
8.5 years
log1.5
t=
= 4.70
log1.09
t=
a 1.65 × 1.05 = 1.7325 million
1.7325 × 1.05 = 1.82 million (to 3 s.f.)
log 0.6429
= 3.173
log 0.87
t>
4
e Changes in temperature and light conditions
can change the growth rate.
log 3.75
= 2.249
log1.8
b 0.87t  4500 = 0.6429
7000
t⩾
t log 1.2 = log 8
log 8
t=
=11.4
log1.2
11.4 weeks
log 8
= 5.1
log1.5
Hence x =
ln a
( )
ln 1 + 1
a
( )
a +1
1
= ln 1 + .
a
a
.
10 a If the annual multiplier is m, then m² = 2.
m = 2 =1.4142
If 1.4142t = 100 then t =
log100
= 13.3.
log1.4142
Just over 13 years
25
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Logarithms and exponential functions
b This time m1.5 = 2
m=
2
23
=1.5874
log100
If 1.5874t = 100 then t =
= 9.97.
log1.5874
10 years
16 aThe nth term is 7 × 1.3n – 1 ≈ 1000; 1.3n – 1 ≈ 142.9;
(n – 1) log 1.3 = log 142.9;
n – 1 ≈ 18.9 so n must be 19 + 1 = 20
7 1.3k − 1
b The sum of k terms is
≈ 10 000;
1.3 − 1
1.3k – 1 ≈ 428.6; 1.3k ≈ 429.6;
(
c The size of an atom puts a limit on how
small components can be made.
log 3
11 aTake logs:
;
x =
log 2
log 3
= 1.259 to 3 d.p.
x=±
log 2
x2 log 2 = log 3;
b Take logarithms: x2 log 2 = (x + 5) log 3
Rearrange:
x2
k≈
2
)
log 429.6
= 23.1 so k = 23
log1.3
Exercise 2.6A
1
log 2 − x log 3 − 5 log 3 = 0.
a log y = log 250x2
log y = log 250 + 2 log x
This is a quadratic equation in x. Use the
quadratic formula to solve it.
b 2
c (0, log 250)
4
2 log V = log 4 πr 3 = log π + log r 3
−b ± b − 4ac log 3 ± (log 3) + 20 log 2 log 3
x=
=
= 3.717
3
3
2a
2 log 2
4 π + 3 log r
log
V
=
log
= 3.717 or –2.132
3
The graph is a straight line with a gradient of 3.
51
12 aThe first term, a = 60 and r = 60 = 0.85;
Gm1m2
3 a ln F = ln
= ln Gm1m2 − ln r2
a
60
r2
S∞ =
=
= 400
1 − r 1 − 0.85
= ln Gm1m2 − 2 ln r
n
60(1
−
0.85
)
So
the
graph is a straight line.
b The sum of n terms is
1 − 0.85
b Gradient is −2
= 400 1 − 0.85n
c Intercept is (0, ln Gm1m2).
If 400(1 – 0.85n) = 390 then 1 – 0.85n = 0.975;
4 a log P = log Act = log A + t log c
0.85n = 0.025;
log 0.025
This is a straight line graph with a gradient
n log 0.85 = log 0.025; n =
= 22.7;
log 0.85
of log c.
need at least 23 terms.
2
2
(
)
b log c = 0.0128 so c = 100.0128 = 1.03
13 a The value is £120 000 × 1.043 = £135 000 to 3 s.f.
log A = 1.97
b If the value is £200 000 after n years, 120 000 ×
A = 101.97 = 93.3
1.04n = 200 000; 1.04n = 1.667
log1.6667
n log 1.04 = log 1.667; =
= 13.0 ; it
log1.04
will be in 13 years.
c 1.03 is the multiplier for an annual
percentage increase of 3%.
d Changes in birth rates, life expectancy and
immigration rates or emigration rates will
change the annual rate of growth.
14 aAfter 2 years the value is £22 000 × 0.85 × 0.95
= £17 765
b If the value is £11 000 after n years then
22 000 × 0.85 × 0.95n–1 = 11 000
5
0.95n – 1 = 0.5882; (n – 1)log 0.95 = log 0.5882;
n – 1 = 10.34; n = 11.34; after 11.3 years.
15 aThe probability that Sam wins each month is 2
100
= 0.02; the probability that Sam does not win each
c The gradient is n so n = 0.5.
log a = 1.5 so a = 101.5 = 31.6.
month is 0.98; The probability that Sam does not
win in 3 months is 0.983 = 0.9412 to 4 d.p.
b If the probability Sam does not win for n
months is 0.5 then 0.98n = 0.5;
n log 0.98 = log 0.5; n = 34.3; he must play for
at least 35 months.
a Gradient = 6.5 − 1.5 = 0.5.
10
b log v = log arn = log a + n log r. This is the
equation of a straight line and the intercept
on the y-axis is log a . From the graph,
log a = 1.5.
d v = 31.6 × 1000.5 = 31.6 × 10 = 316
6
a log x2y = log k
2 log x + log y = log k
log y = −2 log x + log k
26
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2
WORKED SOLUTIONS
10 a 227 = ka2
165 = ka5
165
⇒
= a3 ⇒ a =
227
A graph of log y against log x is a straight line
with a gradient of −2.
b log k = 4.6
104.6
7
k=
= 40 000 to 2 s.f.
a log y = log A + t log c
b 227 = ka2
2
4
Value ($y)
18 490
log y
4.267
6
8
10
13 675 10 114
7480
5533
4.136
3.874
3.743
4.005
c x = 281 × 0.9t; if x = 100 then
log 0.3559
0.899t = 100 = 0.3559 ; t =
log 0.899 = 9.70
281
log y = log a + n log x so the graph will be a
straight line with a gradient of n
5
4
b The gradient is log11.05 − log 8.98 = 0.3
log 20 − log10
3
2
c n = 0.3 and the equation of the line is
log y – log 8.98 = 0.3 (log x – log 10)
1
0
1
2
3
4
5
6
7
8
9
10
If x = 0, y = a and log a – log 8.98 = – 0.3;
log a = 0.6533; a = 4.5
t
b From the graph, log A = 4.4.
A = 104.4 = 25 000 to 2 s.f.
The gradient of the graph is −0.066 = log c.
c = 10−0.066 = 0.86 to 2 s.f.
The equation is y = 4.5x0.3
12 a If p = ak t then ln p = ln a + t ln k; if t = 0 then
from the graph ln a = 2; a = e2 = 7.39
b The gradient of the graph is
ln k = 3.5 − 2 = 0.06; k = e0.06 = 1.062
25 − 0
c Since p = 7.4 × 1.062t the initial investment,
when t = 0, is $7390.
(Note: values from a graph could vary
slightly.)
The gradient is − 2 and the intercept on the
3
y-axis is 2.
If log y = 2 then y = 100.
2
The equation is log y = − log x + 2.
3
2
log y = − 2 log x + log 100 = log 100 − log x 3
3
2
100
y = 2 or x 3 y = 100
x3
9
2
11 a If y α xn then y = axn for some constant a. Hence
log y = log axn;
log y
8
165
= 0.899
227


⇒ 227 = k  3 165  ⇒ k = 281
 227 
A graph of log y against t is a straight line.
Age (t years)
3
This is multiplied by 1.062 each year so the
rate of interest is 6.2%
Exam-style questions
1
a If y = 4, x ≈ 3.4.
b
y
a log pv c = log k; log p + c log v = log k;
log p = – c log v + log k so a graph of log p against
log v will be a straight line with a gradient of –c
and an intercept of log k
y = 1.5−x
log 80 − log 41
= −1.50 to 3 s.f;
log16 − log 25
−c = −1.50 so c = 1.50
y = 1.5x
b The gradient is
1
c The equation of the line is log p – log 41
= – 1.5 (log v – log 25)
If log v = 0 then log p – log 41 = – 1.5 ×
( – log 25 ) = 2.097;
so log k = 2.097 + log 41 = 3.710; k = 103.710
= 5120 to 3 s.f.
0
x
c (0, 1)
2
3t + 8 = 52.1
t=
52.1 − 8
= 7.12
3
27
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Logarithms and exponential functions
3
a If x = 0, y = e3
9
Crosses at (0, e3)
b If x = −2 , k = e0.5 × (−2) + 3 = e2 or 7.39.
c 100 = e0.5h+3
10 aln (15 – 2x) = ln(31x – 12) – ln x2;
15 − 2x = 31x −2 12 ; 15x2 – 2x3 = 31x – 12
x
2x3 – 15x2 + 31x – 12 = 0
0.5h + 3 = ln 100
h=
4
ln100 − 3
= 3.21
0.5
a Let y = log
Then
(a2)y
a2
a.
b Write f(x) = 2x3 – 15x2 + 31x – 12; f(1) = 6;
f(2) = 6; f(3) = 0
= a.
Hence a2y = a.
So x – 3 is a factor;
1
Hence 2y = 1 and y = log 2 a = .
2
a
b Let y = logb a.
1
Then by = a and hence b = a y .
Hence log a b =
5
6
7
1
1
=
.
y log b a
2x < 32x – 1; x ln 2 < (2x – 1) ln 3; x ln 2 < 2x ln 3 – ln 3;
ln 3 < x ln 9 – x ln 2
x (ln 9 – ln 2) > ln 3; x ln 4.5 > ln 3; x > ln 3
ln 4.5
42x – 1 = 53x + 1; (2x – 1) ln 4 = (3x + 1) ln 5;
2x ln 4 – ln 4 = 3x ln 5 + ln 5
3x ln 5 – 2x ln 4 = – ln 4 – ln 5;
x (ln 125 – ln 16) = – ln 20; x ln 125 = − ln 20
16
x = − ln 20 = –1.46 to 3 s.f.
ln 7.8125
35 + 8x
a
= ln 3;
ln(35 + 8x) − 2 ln x = ln 3; ln
x2
35 + 8x = 3 2
; 3x – 8x – 35 = 0
x2
(3x + 7) (x – 5) = 0; x = − 7 or 5
3
e 2x +1 = − 7 or 5; it cannot be negative and so
b
3
e2x+1 = 5; 2x + 1 = ln 5
x = ln 5 − 1 = 0.305 to 3 s.f.
2
8
a When x = 0, y = 375
375 =
Ae0
=A
b When x = 400, y = 945
945 = 375e400k
e400k =
k=
590
n
y = 20xn; 590 = 20 × 15n; 15 = 20 = 29.5;
n ln 15 = ln 29.5;
n = ln 29.5 = 1.25 to 3 s.f.
ln15
945
= 2.52
375
ln 2.52
= 0.002 31
400
cIf y = 540 then 375e0.002 31x = 540;
ln 375 + 0.002 31 x = ln 540
ln 540 − ln 375
x=
= 158 to 3 s.f.
0.002 31
2x3 – 15x2 + 31x – 12 = (x – 3) (2x2 – 9x + 4)
= (x – 3) (x – 4) (2x – 1)
c If f(x) = 0 then x = 3, 4 or 1
2
11 6 2x − 3 = 3 × 2x +1 ; divide by 6 to get 2x − 3 = 2x
Hence 2x – 3 = ± 2x; the positive sign has no
solution hence 2x – 3 = – 2x
2 × 2x = 3; 2x + 1 = 3; (x + 1) ln 2 = ln 3; x = ln 3 − 1
ln 2
12 log y = log k + t log a
A graph of log y against t crosses the log y-axis at
log k so log k = 3.30
k = 103.3 = 2000 to 3 s.f.
The gradient is log a.
4.65 − 3.30
log a =
= 0.45
3
a = 100.45 = 2.82 to 3 s.f.
13 a Multiply both sides by ex: e2x = 4
2x = ln 4
x = ln 4 = 0.693 to 3 s.f.
2
b Multiply both sides by ex.
e2x − 3ex − 4 = 0
This is a quadratic in ex.
Factorise.
(ex − 4)(ex + 1) = 0
ex = 4 or ex = −1.
The second has no solution.
If ex = 4 then x = ln 4 or 1.39 to 3 s.f.
14 22x + 8 = 8 × 2x
Write y = 2x and then y2 − 8y + 8 = 0.
Solve with the quadratic formula:
y = 8 ± 64 − 4 × 8 = 8 ± 32 .
2
2
y = 6.828 or 1.172
Either 2x = 6.828 or 2x = 1.172.
28
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2
WORKED SOLUTIONS
Take logs: either x =
x=
1
b ln s = 2 ln a + 0.2ln t ; in particular,
log 6.828
log 2 = 2.77 or
2.364 = 1 ln a + 0.2 × 2.303
2
log1.172
= 0.228.
log 2
2.364 − 0.2 × 2.303 = 1 ln a ; ln a = 3.807;
2
15 Multiply by 3x: 32x + 4 = 5 × 3x; 32x – 5 × 3x + 4 = 0;
this is a quadratic in 3x;
a = e3.807 = 45
132
0.4
c s2 = 45 × t0.4; if s = 13 then t = 45 = 3.7556 ;
(3x – 1)(3x – 4) = 0; either 3x = 1 or 3x = 4; if 3x = 1
then x = 0;
log 4
if 3x = 4 then x log 3 = log 4; x =
= 1.26 to 3 s.f.
log 3
16 3x + 3x + 2 = 3x + 1 + 20; 3x + 9 × 3x = 3 × 3x + 20;
20
;
7 × 3x = 20; 3x =
7
log 20
20
7 = 0.956 to 3 s.f.
x log 3 = log ; x =
7
log 3
0.4 ln t = ln 3.7556;
ln t = 3.3081; t = e3.3081 = 27.3
20 a2 log2 (x + 4) – log2 x = 5; log 2
(x + 4)2 = 25 ; x2 + 8x + 16 = 32x;
x
x2 – 24x + 16 = 0
39
17 a
The common ratio is 40 = 0.975; the third term is
39 × 0.975 = 38.025
b S∞ =
2
b x = 24 ± 24 − 64 = 24 ± 22.627 = 23.3 or
2
2
0.686 to 3 s.f.
a =
40
= 1600
1 − r 1 − 0.975
c If 1600 – Sn < 1 then Sn > 1599; hence
(
)
40 1 − 0.975n
> 1599;
1 − 0.975
1 − 0.975n > 1599 × 0.025 = 0.999 375
40
(x + 4)2 = 5
;
x
Mathematics in life and work
1
0.975n < 0.000 625; n log 0.975 < log 0.000 625
The population is multiplied by 1.015 every year.
If it increases by 20% in n years then 1.015n = 1.2
Take logarithms: log 1.015n = log 1.2.
n log 1.015 = log 1.2
log0.000 625
= 291.4;
log0.975
if S∞ – Sn < 1 then n ⩾ 292
log1.2
n = log1.015 =12.25 so it will be just over 12
years.
n>
18 aAt P, 1 e 2x = 4; e2x = 16; 2x = ln 16;
4
x = 1 ln16 = 1 ln 4 2 = ln 4 ;
2
2
The coordinates are (ln 4, 4)
b T
he gradient of the tangent is 8; the
equation is y – 4 = 8(x – ln 4)
or y = 8x + 4 – 8 ln 4
c Where the tangent meets the y-axis, x = 0
and y = 4 – 8 ln 4 (this is a negative number)
Where the tangent meets the x-axis, y = 0
1
and 0 = 8x + 4 – 8 ln 4 so x = ln 4 –
2
1
The area of the triangle = × ( 8 ln 4 − 4 ) ×
2
1
2
ln 4 −
2 = (2 ln 4 – 1)
2
16
= (2 ln 4 – ln e)2 = ln
e
(
)
( )
2
If the annual rate of growth is r % then you want
20
r
1+
< 1.1.
100
(
)
(
Take logarithms: 20 log 1 +
(
Hence log 1 +
)
)
r
< log 1.1.
100
log1.1
r
<
= 0.002 070.
20
100
r
< 100.002 070 = 1.004 78
100
Hence r < 100(1.004 78 − 1) = 0.478.
The annual growth rate must be less than 0.478%.
If ln is used the numbers in the calculation are
r
ln1.1
<
= 0.004 766 and
ln 1 +
20
100
1+
(
)
1 + r < e0.004 766 = 1.004 78 which gives the
100
same answer.
1
1
19 a 2 ln s = ln a + n ln t so ln s = 2 ln a + 2 n ln t
1
A graph of ln t against ln s has a gradient of 2 n
1 n = 2.722 − 2.364 = 0.2;
n = 0.4
2
4.094 − 2.303
29
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TRIGONOMETRY
3 Trigonometry
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge
1
sin 600° = sin (360 + 240)° = sin 240° = – sin 60°
2
=− 3
2
The amplitude is 3 and the period is 360 ÷ 2 = 180°.
6

1
1 
+ cos A ×
= 2  sin A ×


2
2
= sin A + cos A = left-hand side.
π + 2 sin x + π = sin x cos π − cos x sin π
a sin x −
6
6
6
6
( )
( )
+ 2 sin x cos π + 2 cos x sin π
6
6
π
π
3
1
= 3sin x cos + cos x sin = 3sin x ×
+ cos x ×
6
6
2
2
y
3
2
1
=
0
x
1
3 3
sin x + cos x
2
2
π then x − π = π and x + π = π + π = 5π
6 12
6 4 6 12
4
–2
1 +1× 1 =3 3
π
5
3
3
π
1
π 3 3
Hence sin 12 + 2 sin 12 = 2 sin 4 + 2 cos 4 = 2 ×
2
2
2 2
–3
+
π
5
3
3
π
1
π
3
3
1
1
1
3
3
1
=
=
sin + cos =
×
+ ×
3 cos (x + 0.3) = 0.4 hence x + 0.3 = ±1.159sin + 2 sin
12
12
2
4 2
4
2
2 2
2
2 2
Hence x = 0.859 or – 1.46 to 3 s.f.
sin A sin B sin A cos B + cos A sin B
7 a Left-hand side = tan A + tan B = cos A + cos B =
cos A cos B
2
2
sin x sin x = 1 − cos x
sin A sin B sin A cos B + cos A sin B sin( A + B)
4 sin x tan x = sin x × cos x = cos x tan
x B=
Acos
+ tan
+
=
=
cos A cos B
cos A cos B
cos A cos B
(
1
+
cos
x
)(
1
−
cos
x
)
=
= right-hand side.
cos x
b tan A − tan B = tan A + tan (−B) because
tan (−B) = −tan B
Exercise 3.1A
90
–1
1
270
360
b If x =
cos 75° = cos (45° + 30°)
= cos 45° cos 30° – sin 45° sin 30°
=
2
180
3 −1
1
3
1
1
×
−
× =
2
2 2
2
2 2
8
π radians = 15° and hence
12
π = tan 15°
tan
12
= 2 sin A cos B = right-hand side.
b Let A + B = x and A − B = y.
tan 45 − tan 30
= tan (45° – 30°) =
1 + tan 45 tan 30
=
1−
1+
1
3
1
3
3
The expression is sin (2A + 3A) = sin 5A.
4
3
The expression is cos (10° + 20°) = cos 30° =
2
π
Right-hand side = 2 sin A + 4
π
π
= 2 sin A cos + cos A sin
4
4
5
(
30
Add the equations to get 2A = x + y.
x+y
Hence A =
.
2
Subtract the equations to get 2B = x − y.
x−y
Hence B =
.
2
Substitute these expressions into the
identity in part a and the result follows.
3 −1
3 +1
=
(
Hence, from part a, tan A + tan (−B)
= sin( A − B) = sin( A − B) because
cos A cos(−B) cos A cos B
cos (−B) = cos B
a Left-hand side = sin (A + B) + sin (A − B) =
(sin A cos B + cos A sin B) + (sin A cos B − cos A sin B)
)
)
9
a tan (A + B) =
tan A + tan B
= tan π = 1
1 − tan A tan B
4
Hence tan A + tan B = 1− tan A tan B.
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3
WORKED SOLUTIONS
Rearrange: tan A + tan A tan B = 1 − tan B.
tan x (1 – tan x) + tan x + 1 = 2(1 – tan x);
writing t = tan x, t – t2 + t + 1 = 2 – 2t
Factorise: tan A (1 + tan B) = 1 − tan B.
Hence tan A =
Rearrange as t2 – 4t + 1 = 0;
4 ± 16 − 4 4 ± 12
t=
=
=2± 3
2
2
(or completing the square: (t – 2)2 – 4 + 1 = 0;
(t – 2)2 = 3; this gives the same result)
1 − tan B
.
1 + tan B
1 − tan A
.
1 + tan A
10 acos (A + B) – cos (A – B) = cos A cos B – sin A sin B
– (cos A cos B + sin A sin B )
= cos A cos B – sin A sin B – cos A cos B
– sin A sin B = – 2 sin A sin B
b Simply swap A and B to get tan B =
b Write A + B = α and A – B = β so that
cos (A + B) – cos (A – B) ≡ cos α – cos β;
If tan x = 2 + 3 then x = 75° or 255°; if
tan x = 2 − 3 then x = 15° or 195°
Exercise 3.2A
then 2A = α + β and 2B = α – β
α +β
α −β
Hence A =
and B =
; substitute to
2
2
α +β
α −β
get cosα − cos β = −2sin  2  sin  2 




11 a sin ( A − B ) = sin A cos B − cos A sin B
cos A cos B
cos A cos B
sin A sin B
=
−
= tan A − tan B
cos A cos B
sin 2A ≡ 2 sin A cos A
x
x
x
Let x = 2A so = A and sin x ≡ 2sin cos .
2
2
2
1
So
a cos 2 ≡ 2cos2  − 1
2
Rearrange: 2cos2  ≡ 1 + cos 2.
1
(1 + cos 2).
2
b If  = 15° then the identity becomes
cos2 15° = 1 (1 + cos 30°).
2
1
3 2 + 3
2
.
=
So cos 15° =  1 +
4
2
2 
Hence cos2  ≡
sin ( A − B ) sin ( B − C ) sin (C − A )
+
+
cos A cos B cos B cos C cos C cos A
= tan A – tan B + tan B – tan C + tan C – tan A = 0
b
(
12 a
sin x +
)
π
π
π
= sin x cos + cos x sin =
6
6
6
cos15° is positive so take the positive square root:
1
3
1
3
sin x + cos x so a =
and b =
2
2
2
2
(
b sin x +
cos15° =
)
π
π
π
= sin x cos + cos x sin
3
3
3
2 sin2  = 1 − cos 2
sin 2 θ =
(
)
Hence tan x =
(
)
3 −1
=
3 +1
3 −1
×
3 +1
3 −1
3 −1
tan x + tan 45°
= 2;
1 − tan x tan 45°
tan x + 1
= 2;
tan x +
1 − tan x
b tan x +
(
(
)




sin 2 π = 1 1 − cos π = 1  1 − 1  = 1  1 − 2  = 2 − 2
4
8 2
4
2
2 
2 2
)




sin 2 π = 1 1 − cos π = 1  1 − 1  = 1  1 − 2  = 2 − 2 .
8 2
4
2
2
2
4



2
3− 2 3 +1
= 2− 3
=
3−1
13 asin θ cos 40° – cosθ sin 40° = cosθ cos 20° + sinθ
sin 20°
(cos 40° – sin 20°) sin θ = (cos 20 °+ sin 40°)
sin θ cos20° + sin 40°
cos θ;
=
= tan θ
cosθ cos40° − sin 20°
Hence tan θ = 3.732 and θ = 75° or 255°
1
(1 − cos 2θ )
2
b If θ = π radians then
8

 3 1
3 1
 1 + 2 − 2  sin x =  2 − 2  cos x or
3 − 1 cos x
2+ 3 .
2
a cos 2 = 1 − 2 sin2 
3
= 1 sin x + 3 cos x
2
2
3
1
1
3
Hence sin x + 2 sin x + 2 cos x = 2 sin x + 2 cos x ;
3 + 1 sin x =
1
x
x
sin x ≡ sin cos .
2
2
2
4
2− 2
.
Hence, taking the positive root, sin π =
8
2
Use the double angle formula:
2 sin  cos  = 1.5 sin .
Either sin  = 0 or 2 cos  = 1.5.
If sin  = 0 then  = 180°.
If 2 cos  = 1.5 then cos  = 0.75 and  = 41.4° or
360° – 41.4° = 318.6°.
The three possible solutions in the given domain
are 41.4°, 180° and 318.6°.
31
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TRIGONOMETRY
5
2 cos2 x = 1 + cos 2x so 2 cos3 x = cos x + cos x cos 2x
Hence right-hand side =
6
7
1
(sin x sin 2x + 2 cos3 x)
2
1
= (sin x sin 2x + cos x + cos x cos 2x)
2
1
= (cos x + cos (2x − x)) = cos x = left-hand side.
2
Left-hand side = cos 2 + 2 cos  + 1
= 2 cos2  − 1 + 2 cos  + 1 = 2 cos2  + 2 cos 
= 2 cos  (cos  + 1) = right-hand side.
1
2× 1
2tan θ
4 = 2 = 1 × 16 = 8
a tan 2θ ≡
2 =
15 2 15 15
1 − tan θ 1 − 1
16 16
b If tan  = t then 2 =
2t
.
1 − t2
Rearrange: 2(1 − t2) = 2t and hence t2 + t − 1 = 0.
−1 ± 1 + 4
.
2
t is positive so take the positive root:
−1 + 5
tan θ =
.
2
The quadratic formula gives t =
8
2t
= 4t .
1 − t2
Write tan  = t and then
Rearrange: 2t = 4t(1 − t2).
t cannot be 0 for the given range of  and so you
can divide by it: 2 = 4(1 − t2).
9
1 and so  = 0.615 to 3 d.p.
2
Hence t 2 = 1 and t =
2
a
y
4
3
2
1
–1
0
–1
1
45
90
135
180
x
–2
–3
–4
b 3(2 cos2  − 1) = cos  − 2
Write c = cos  and then 6c2 − 3 = c − 2.
Rearrange: 6c2 − c – 1 = 0.
Factorise: (3c + 1)(2c − 1) = 0.
Hence either 3c + 1 = 0 or 2c − 1 = 0.
If 3c + 1 = 0 then c = −
If 2c − 1 = 0 then c =
1
and  = 109.5°.
3
1
and  = 60°.
2
10 a If x = π then 2x = π so tan 2x = 1
4
8
tan 2x =
2tan x
=1
1 − tan 2 x
Hence 2 tan x = 1 − tan2 x.
b tan2 x + 2 tan x − 1 = 0
Use the quadratic formula:
−2 ± 22 + 4 −2 ± 8
=
.
2
2
Since tan x is positive, take the positive root:
tan x =
tan π = −2 + 8 = −2 + 2 2 = 2 − 1
8
2
2
11 left-hand side = sin (2A + A)
= sin 2A cos A + cos 2A sin A
= 2 sin A cos2 A + (1 − 2 sin2 A) sin A
= 2 sin A (1 − sin2 A) + sin A − 2 sin3 A
= 2 sin A − 2 sin3 A + sin A − 2 sin3 A
= 3 sin A − 4 sin3 A = right-hand side.
12 a sin 2x ≡ 2 sin x cos x so 4 sin x cos x = cos x;
cos x (4 sin x – 1) = 0;
1
either cos x = 0 or sin x = ; hence x = 90° or
4
14.5° or 165.5°
b 2(1 – 2 sin2 x) = sin x + 1
4 sin2 x + sin x – 1 = 0
x = 22.9˚ or 157.0˚
13 a cos 4A = cos 2(2A) = 2 cos2 2A – 1
= 2(2 cos2 A – 1)2 – 1
= 2(4 cos4 A – 4 cos2 A + 1) – 1
= 8 cos4 A – 8 cos2 A + 1
π 3π 5π
b If x = cos A then cos 4A = 0; 4 A = ,
,
2 2 2
7π
are solutions;
and
2
π 3π 5π
7π
A= ,
,
or
; x = cos A = 0.924, 0.383,
8 8 8
8
−0.383 or −0.924
(In degrees, 4A = 90°, 270°, 450° or 630° give
the same solutions)
tan A + tan 2A
14 a tan 3A = tan( A + 2A) =
1 − tan A tan 2A
2
tan A + 2tan A
1 − tan 2 A = tan A 1 − tan A + 2tan A
=
2
2
1 − tan A − 2tan 2 A
1 − 2tan 2A
1 − tan A
(
=
)
3tan A − tan 3 A
1 − 3tan 2 A
b Rearrange as 1 – 3 tan2 A = 3 tan A – tan3 A;
3tan A − tan 3 A
= 1 ; tan 3A = 1
1 − 3tan 2 A
3A = 45° or 225° or 405°
A = 15°, 75°, 135° are the three solutions
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3
WORKED SOLUTIONS
15 a tan ( A + 45° ) = tan A + tan 45° ; tan 45° = 1 so
1 − tan A tan 45°
1 + tan A
tan ( A + 45° ) ≡
1 − tan A
b 5 sin( + 53.1°) = 5 hence sin( + 53.1°) = 1
Hence  + 53.1° = 90° and so  = 36.9°.
3
a cos  + sin  ≡ r sin  cos α + r cos  sin α
2tan x
1 + tan x
− 2;
b
=
1 − tan 2 x 1 − tan x
2 tan x = (1 + tan x)2 – 2(1 – tan2 x );
Therefore r cos α = 1 and r sin α = 1
r = 2 and tan α = 1 so α = π .
4
(
2 tan x = 1 + 2 tan x + tan2 x – 2 + 2 tan2 x;
1
3 tan2 x = 1; tan x = ±
;
3
x = 30° or 150°
2tan A
A
2 = 2t
16 a tan A = tan 2 ×
=
2
2
1 − tan 2 A 1 − t
2
(
b 2t =
=
sin  + cos  = 2 sin θ + π
4
b cos  − 7 sin  ≡ r cos  cos α − r sin  sin α
Therefore r cos α = 1 and r sin α = 7.
)
 sin A 
2sin A
2 and 1 + t 2 = 1 + 
2
 cos A 
cos A
2
2
r = 12 + 7 2 = 50 and tan α = 7 so α = 81.9°.
cos  − 7 sin  = 50 cos ( + 81.9°).
2
cos2 A + sin 2 A
1
2
2 =
2 A
2 A
cos
cos
2
2
c
4
Therefore r cos α = 8 and r sin α = 6.
6
r = 8 2 + 6 2 = 10 and tan α = 8 so α = 0.644
radians or 36.9°
8 sin x – 6 cos x = 10 sin ( − 0.644) using
radians, or 10 sin ( − 36.9°).
b The smallest possible value is when the sine
is –1 to give a minimum of –10.
sin A
2t
2t
1 − t2
=
÷
=
2
2
tan A 1 + t
1−t
1 + t2
5
Exercise 3.3A
1
a 12 sin  + 5 cos  ≡ r sin  cos α + r cos  sin α
so r cos α = 12 and r sin α = 5
a cos  + 3 sin  ≡ r sin  cos α + r cos  sin α
r = 122 + 52 = 13 and tan α =
Therefore r sin α = 1 and r cos α = 3 .
α = 0.395.
1
r = 1 + 3 = 2 and tan α =
so α = π
6
3
(
π
cos  + 3 sin  = 2 sin θ + 6
b
5
so
12
12 sin  + 5 cos  = 13 sin ( + 0.395)
)
b 13 sin ( + 0.395) = 8 so
sin ( + 0.395) = 8 = 0.615
13
 + 0.395 = 0.663
y
 = 0.268
2
6
1
0
π
2π
a 10 cos  − 12 sin  ≡ r cos  cos α − r sin  sin α
Therefore r cos α = 10 and r sin α = 12.
r = 10 2 + 122 = 244 and tan α = 12 = 1.2 so
10
α = 0.876.
–1
10 cos  − 12 sin  = 244 cos( + 0.876)
b
–2
2
50 cos ( + 81.9°) = −5 and hence
5
1
=−
.
cos ( + 81.9°) = −
50
2
Hence  + 81.9° = 135° and  = 53.1°.
a 8 sin x − 6 cos x ≡ r sin x cos α − r cos x sin α
2sin A
2t
1
2 ÷
=
A
1 + t2
cos
cos2 A
2
2
A
A
= 2sin cos = sin A
2
2
c cos A =
)
a 3 sin  + 4 cos  ≡ r sin  cos α + r cos  sin α
Therefore r cos α = 3 and r sin α = 4.
4
r = 3 + 4 = 5 and tan α = so α = 53.1°.
3
2
244 cos( + 0.876) = 5 so
cos( + 0.876) = 0.320
 + 0.876 = 1.245 or 2π – 1.245 = 5.038
 = 0.369 or 4.162
2
3 sin  + 4 cos  = 5 sin( + 53.1°)
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Trigonometry
7
(
a
Vertical distance from R to OB is 6 sin  and
vertical distance from P to R is 4 cos  so the
sum of these is the height of P.
r cosα = 1 + 2 and r sin α = 2
b Write 6 sin  + 4 cos  ≡ r sin ( + α)
= r sin  cos α + r cos  sin α.
2 and α = 0.530;
1+ 2
r 2 = 2 + (1 + 2)2 so r = 2.798
Hence tan α =
Therefore r cos α = 6 and r sin α = 4.
The equation is 2.798 cos (θ + 0.530) = 2;
cos (θ + 0.530 ) = 2 = 0.715;
2.798
θ + 0.530 = 0.774 or 5.509; θ = 0.244 or 4.98
r = 4 2 + 6 2 = 52
The maximum height is 52 = 7.21 m.
4
c tan α = 6 so α = 0.588.
52 sin ( + 0.588) = 7
sin ( + 0.588) = 0.9707
 + 0.588 = 1.328 or 1.813
Exercise 3.4A
1
a
 = 0.740 or 1.23
8
c
a
If 0.5 sin  + 0.4 cos  ≡ r sin ( + α) ≡ r sin  cos α
+ r cos  sin α then r cos α = 0.5 and r sin α = 0.4.
r = 0.52 + 0.4 2 = 0.41 = 0.640
tan α =
sin 
π
6
1
2
3
2
π
3
3
2
1
2
3
The two values in the given range are
 = 2.15 or 5.93.
9
b If 0.640 sin ( + 0.675) = 0.7 then
sin ( + 0.675) = 1.094
4
This has no solution because
–1  sin ( + 0.675)  1.
5
asin x + sin (x + 10°) = sin x + sin x cos 10° + cos x
sin 10°
tan α =
tan 
sec 
cosec 
cot 
1
3
2
3
2
3
3
2
2
3
1
3
The two solutions in the given domain are
38.2° and 218.2°.
b cos (x − 25°) = 0.5
Hence x − 25° = 60° or 300° or …
Hence x = 85° or 325°.
3
2
=
and so 3 cos x = 2 sin x which
sin x cos x
rearranges to sin x = tan x = 1.5.
cos x
Hence x = 56.3° or 180° + 56.3° = 236.3°.
d sin (2x + 20°) = 1 = 0.625
1.6
Hence 2x + 20° = 38.7° or 180° – 38.7° or 360°
+ 38.7° or 540° – 38.7° or …
c
The equation is 1.993 sin (x + 5°) = 1; sin (x + 5°)
= 0.502; x + 5° = 30.1° or 149.9°;
x = 25.1° or 144.9°
π
π
π
10 a
cos θ +
= cosθ cos − sin θ sin
4
4
4
1
1
1
cosθ −
sin θ ; so a =
and b = − 1
=
2
2
2
2
(
)
(
)
b cosθ + 2cos θ + π = cosθ + 2 cosθ − 2 sin θ
4
(
cos 
Hence x = 38.2° or 218.2° or 398.2° or …
0.174
so α = 5°; r2 = 1.9852 + 0.1742 so
1.985
r = 1.993
3
2
3
a
sin x = 0.25 therefore x = 0.253 or π – 0.253
= 2.89.
b cos x = − 1 therefore x = 1.91 or 2π – 1.91
3
= 4.37.
2
3
c
a 2
b
3
a tan (x + 30°) = 1 = 2.5
0.4
Hence x + 30° = 68.2° or 180° + 68.2° or 360° +
68.2° or …
= sin x + 0.985 sin x + 0.174 cos x = 1.985 sin x
+ 0.174 cos x
b If 1.985 sin x + 0.174 cos x = r sin (x + α) then
r cos α = 1.985 and r sin α = 0.174
d

The equation is 0.640 sin ( + 0.675) = 0.2 so
sin ( + 0.675) = 0.312.
 = –0.357 or 2.149 or 5.926 or...
b
2
2
3
2
0.4
= 0.8 so α = 0.675
0.5
 + 0.675 = 0.318 or π – 0.318 or 2π + 0.318 or...
)
If 1 + 2 cosθ − 2 sin θ ≡ r cos (θ + α ) then
)
2x + 20° = 38.7° or 141.3° or 398.7 or 501.3° or …
Hence x = 9.3° or 60.7° or 189.3° or 240.7°.
= 1 + 2 cosθ − 2 sin θ
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3
WORKED SOLUTIONS
6
a
=
y
2
(
=4
1
1 sin 2x
2
cosec2 2x
10 a cotA − cosec 2A =
1
)
2
=
1
1 sin 2 2x
4
= right-hand side.
cos A
1
−
sin A sin 2A
cos A
1
2cos2 A − 1 cos2A
−
=
=
sin A 2sin A cos A 2sin A cos A sin 2A
= cot 2A
=
0
2π
x
b cot x – cot 2x = 4; and cot x – cot 2x = cosec 2x;
hence cosec 2x = 4; sin 2x = 0.25;
–1
–2
7
x = 0.126, 1.44, 3.27, 4.59
b f(x)  1 and f(x)  –1
a
y
Exercise 3.5A
1
sec2  ≡ 1 + tan2  so 9 = 1 + tan2 
Hence tan2  = 8 and tan θ = ± 8 .
6
2
4
2
cosec2  ≡ 1 + cot2  and so 1 + cot2  + cot2  = 9
Hence 2 cot2  = 8
cot2  = 4
cot  = ±2.
Hence  = 26.6° or 153.4°.
0
90
180
270
360
x
3
Hence sec2  – tan2  = 1
–2
2
(sec  – tan )(sec  + tan ) = 1.
(sec  − tan )k = 1 and so sec  – tan  = 1 .
k
b Add the simultaneous equations
sec  + tan  = k and sec  – tan  = 1 to get
k
1
1
1
2 sec  = k + and hence secθ = k + .
2
k
k
–4
4
b
2
cos x
=
cos x sin x
(
Rearrange: 2 sin x = cos2 x
Hence 2 sin x = 1 − sin2x.
Write s = sin x and rearrange: s2 + 2s − 1 = 0
Use the quadratic formula: s =
= –2.4142 or 0.4142.
4
−2 ± 4 + 4
2
If sin x = 0.4142 then x = 24.5° or 155.5°.
2
If x = 24.5 then y =
= 2.20.
cos 24.5°
5
The coordinates are (24.5, 2.20) and
(155.5, –2.20).
a f(x)  0.5 and f(x)  –0.5
b f(x)  1.5 and f(x)  –0.5
6
c f(x)  1 and f(x)  –1
9
Left-hand side = sec2 x + cosec2 x =
)
1 + tan2  + 2 tan  = 4
tan2  + 2 tan  − 3 = 0
Factorise: (tan  − 1)(tan  + 3) = 0.
Either tan  = 1 or tan  = −3.
 = π = 0.785 or  = π – 1.249 = 1.89.
4
sin x = –2.4142 has no solution.
8
a sec2  ≡ 1 + tan2 
1
1
+
cos2 x sin 2 x
2
2
1
= sin x2 + cos2 x =
cos x sin x
(cos x sin x )2
sec2x − 1 = 6 sec x − 10
sec2 x − 6 sec x + 9 = 0
(sec x − 3)2 = 0
sec x = 3
1
cos x =
3
x = 70.5°
1
sin x 1 + sin x
a sec x + tan x =
+
=
cos x cos x
cos x
2
2
2
b sec x = 1 + tan x so sec x – tan2 x = 1;
(sec x + tan x) (sec x – tan x) = 1;
sec x − tan x =
1
cos x
=
sec x + tan x 1 + sin x
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TRIGONOMETRY
θ
θ
θ
1
sin θ cosθ sin 2 θ + cos2 θ
b θIn
the θsame way, sin = 2cos sin
cosec
+
=
=
= sec
2
4
4
cosθ sin θ
cosθ sin θ
cosθ sin θ
θ
θ
θ
1
sin θ cosθ sin 2 θ + cos2 θ
So sin θ = 2cos 2 × 2cos 4 sin 4
=
+
=
=
= secθ cosecθ
cosθ sin θ
cosθ sin θ
cosθ sin θ
θ
θ
θ
1
1
1
= 4cos cos sin .
8 cosec 2x =
=
= cosec x sec x
2
4
4
sin 2x 2sin x cos x 2
2  − sin  = 0
4
2
sin
1 − tan 2 x
1
9 cot 2x =
Factorise: sin  (2 sin  − 1) = 0.
=
tan 2x
2tan x
Either sin  = 0 or 2 sin  − 1 = 0.
Divide the numerator and the denominator by
If sin  = 0 then  = 0 or π.
1 −1
2
1
2
cot
x
1
−
.
If 2 sin  − 1 = 0 then sin θ = so θ = π or 5π .
tan2 x to get tan x
=
2
6
6
1
2cot
x
2×
tan x
There are 4 values for .
5° =
7
tan θ + cot θ =
sin x cos x
cos 2θ
cos2 θ − sin 2 θ
sin 2 θ
2
2
+
= 3 ; Multiply by sin x cos x to get
+ tan 2 θ = 1 −
5
cos x sin x
2 + tan θ =
2
2 + tan θ
cos
θ
cos
θ
cos
θ
sin2 x + cos2 x = 3 sin x cos x;
2
2
2
3
2 cos 22θ + tan 2 θ = cos θ −2sin θ + tan 2 θ = 1 − sin 2θ + tan 2 θ
× 2sin x cos x = 1 ; sin 2x = ; 2xcos
= 0.730
or
θ
cos θ
cos θ
2
3
2.412; x = 0.365 or 1.21
= 1 − tan2  + tan2  = 1
7
5
; multiply by tan x;
b tan x +
=
tan x sin x
6 cos4  = 2 sin2  cos2 
2
2
tan x + 7 = 5 sec x; sec x – 1 + 7 = 5 sec x
Either cos2  = 0 or cos2  = 2 sin2 .
2
sec x – 5 sec x + 6 = 0; (sec x – 2) (sec x – 3) = 0;
If cos2  = 0 then cos  = 0 and  = 90°.
1
1
If cos2  = 2 sin2  then tan2  = 0.5 and tan θ = 0.5
sec x = 2 or 3; cos x = or ;
2
3
or tan θ = − 0.5.
π
Then  = 35.3° or 144.7°.
x = (1.05) or 1.23
3
π
π
π
7 cos θ + 4 = cos θ cos 4 − sin θ sin 4
10 a
(
Exam-style questions
1
a i
3
2
)
(
b cos 105° = cos (135° − 30°)
= cos 135° cos 30° + sin 135° sin 30°
8
)
a
1+
2 a tan 75° = tan (45° + 30°) = tan 45° + tan 30° =
1 − tan 45° tan 30° 1 −
1
1+
tan 45° + tan 30°
3 = 3 +1
=
1 − tan 45° tan 30° 1 − 1
3 −1
3
3 −1
=
3 +1
2
1
3 =
1
3
1
3 +1
3 −1
–180
θ and then
2
θ
θ
sin  = 2 cos sin .
2
2
90
180
x
–1
3 −1
=
3 +1
Let  = 2x so x =
0
–90
3 −1
3 −1 3− 2 3 +1 4 − 2 3
=
=
= 2− 3
×
3−1
2
3 +1
3 −1
1
cos x
3 −1
3 −1 3− 2 3 +1 4 − 2 3
b
=
×
=
=
= 2− 3
cos x sin x
3
1
2
−
3 +1
3 −1
sin x = cos2 x
3 a sin 2x = 2 sin x cos x
b cot 75° =
2
y
=− 1 × 3 + 1 ×1
2
2
2 2
1− 3
=
2 2
()
4
3
cos π = sin π = 1 and sin θ = 1 −
=
5
5
4
4
2
So cos θ + π = 4 × 1 − 3 × 1 = 1 = 2 .
4
5
2 5
2 5 2 10
1
ii cos 135° = − cos 45° = −
2
––2
sin x = 1 − sin2 x
sin2 x + sin x − 1 = 0
sin x =
−1 ± 1 + 4
2
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3
WORKED SOLUTIONS
sin x =
−1 − 5
−1 + 5
or sin x =
2
2
sin x =
−1 − 5
= −1.62 has no solution
2
−1 + 5
= 0.618 has solutions
2
x = 38.2° or 180° − 38.2° = 141.8
These are the two solutions in the interval
−180°  x  180°.
sin x =
9
a 10 sin  + 14 cos  = r sin  cos α + r cos  sin α
r cos α = 10 and r sin α = 14
r = 10 2 + 14 2 = 296 = 17.2
14
tan α =
= 1.4 so α = 0.95.
10
b
296 sin ( + 0.951) = 15
sin ( + 0.951) = 0.872
 + 0.951 = 1.059 or π – 1.059 = 2.083
 = 0.108 or 1.13
10 a cos 2A = 2 cos2 A − 1 so
cos2 2A = (2 cos2 A − 1)2 = 4 cos4 A − 4 cos2 A + 1
b cos 4A = 2 cos2 2A − 1
= 2(4 cos4 A − 4 cos2 A + 1) − 1
= 8 cos4 A − 8 cos2 A + 1
1
sin x
2 sin2 x + sin x = 1
2 sin2 x + sin x – 1 = 0
Factorise this quadratic: (2 sin x − 1)(sin x + 1) = 0.
1
sin x = or –1
2
π
x = or 5π or 3π
2
6
6
12 a tan x + cot x = 5
11 2 sin x + 1 =
Multiply by tan x : tan2 x + 1 = 5 tan x
Rearrange: tan2 x − 5 tan x + 1 = 0.
Use the quadratic formula.
tan x = 5 ± 25 − 4 = 4.791 or 0.2087
2
Therefore x = 1.37 or 0.206.
b tan x + cot x = k; tan2 x − k tan x + 1 = 0
This only has a solution if k2 − 4  0.
Hence k2  4 so that k  2 or k  −2.
13 6(1 − cos2 x) + cos x = 5
6 − 6 cos2 x + cos x = 5
6 cos2 x − cos x − 1 = 0
Factorise: (2 cos x − 1)(3 cos x + 1) = 0
1
1
so cos x = or − .
2
3
If cos x =
1
π
5π
then x = or
.
2
3
3
1
If cos x = − 3 then x = 4.37 or 1.91.
14 a5 sin θ + 12 cos θ ≡ R cos (θ – α) = R cos θ cos α +
R sin θ sin α
So 5 = R sin α and 12 = R cos α;
R = 52 + 122 = 13 and tan α = 5 so
12
α = 0.39 to 2 d.p. So
5 sin θ + 12 cos θ = 13 cos (θ − 0.39)
10
;
b 13 cos (θ – 0.39) = 10; cos (θ − 0.39) =
13
θ – 0.39 = 0.693 or 2π – 0.693 = 5.590; hence
θ = 1.09 or 5.98
15 If 3 sin θ + 4 cos θ ≡ r sin (θ + a)
= r sin θ cos α + r cos θ sin α then
r cos α = 3 and r sin α = 4; hence r = 32 + 4 2 = 5
4
and tan α = so α = 53.1°
3
Hence 5 sin (θ + 53.1°) = 2; sin (θ + 53.1°) = 0.4;
θ + 53.1° = 23.6° or 156.4° or 383.6˚ …; values of θ in
the interval when 0°  θ  360° are
θ = 103.3° or 330.4˚ to 1 d.p.
16 3 sin θ + 1 – 2 sin2 θ = 2; 2 sin2 θ – 3 sin θ + 1 = 0;
(2 sin θ – 1)(sin θ – 1) = 0
sin θ = 1 or 1; θ = 30°, 150° or 90°
2
17 asin A cos B + cos A sin B = 2(sin A cos B –
cos A sin B);
sin A cos B + cos A sin B = 2 sin A cos B –
2 cos A sin B; 3 cos A sin B = sin A cos B;
3cos A sin B = sin A cos B ; 3 tan B = tan A
cos A cos B
cos A cos B
b Replacing A with θ and B with 0.5 you get
tan θ = 3 tan 0.5 = 1.639; θ = 1.02 or 4.16
π
π
18 a cos + A + cos − A
4
4
(
)
(
)
π
π
= cos cos A − sin sin A
4
4
π
π
+ cos cos A + sin sin A
4
4
π
= 2cos cos A = 2 × 1 cos A = 2 cos A
4
2
π
5π
5π π π
+A=
− = and also
then A =
4
12
12 4 6
π
π π
π
−A= − =
4
4 6 12
b If
37
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Trigonometry
Hence cos 5π + cos π = 2 cos π
12
12
6
c cos
(
3 1
= 2×
=
6
2
2
)
(
5π
π
π π
π π
− cos = cos +
− cos −
12
4 6
4 6
12
)
= cos π cos π − sin π sin π − cos π cos π
4
6
4
6
4
6
π
π
− sin sin
4
6
π
π
= −2sin sin
4
6
= −2 × 1 × 1 = − 1 or − 1 2
2
2 2
2
Mathematics in life and work
1
r sin ( + α) = r sin  cos α + r cos  sin α ≡ 5 sin  +
k cos 
Hence r cos α = 5 and r sin α = k.
Square and add: r2 cos2 α + r2 sin2 α = 52 + k2.
Hence r2 = 25 + k2 and r = 25 + k 2 .
2
Dividing:
r sin α k
=
r cos α 5
So k = 5 tan α = 5 tan 0.3 = 1.55.
3
r = 25 + 1.5472 = 5.234
5 sin  + 1.547 cos  = −4 so 5.234 sin ( + 0.3) = −4
4
sin ( + 0.3) = −
= −0.7643
5.234
Hence ( + 0.3) = −0.8699 or π + 0.8699 or …
The smallest positive solution is
 = π + 0.8699 − 0.3 = 3.71 to 3 s.f.
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4
WORKED SOLUTIONS
4 Differentiation
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering
the question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge
1
4
a y = 2x(x2 − 1) = 2x3 − 2x
b 1.5 = eln 1.5 so 1.5x = (eln 1.5)x = e(ln 1.5)x.
dy
= 6x 2 − 2
dx
b y=
dy
= (ln1.5)e(ln1.5)x
dx
= (ln 1.5)1.5x.
c y = e(ln 1.5)x so
1
1
x − x −1
2
2
dy 1 1 −2
1
1
x2 + 1
= + x or + 2 or
2 2x
dx 2 2
2x 2
c y=x
−1
2
dy
1
=−
2
dx
2
5
6
−3
x 2
t = 3,
c If
7
1
dy
= 0.2N then 0.1 N e0.1k = 0.2N and
dt
ln 2
= 6.93 hours.
0.1
a f(x) = 2e2x − 2e−2x and
f(x) = 4e2x − 2e−2x × −2 = 4e2x + 4e−2x
= 4(e2x + e−2x) = 4f(x)
b A
t a stationary point f(x) = 0; 2e2x – 2e–2x = 0;
e2x = e–2x; e4x = 1; x = 0
−−11
ff′(′(xx))== 11((11++ xx22))22 ××22xx == xx
22
11++xx22 .
0.2y = e2x − 1 and hence ln 0.2y = 2x − 1 and
1
x = (1 + ln 0.2y).
2
1
An equivalent answer is x = (1 + ln y − ln 5).
2
Exercise 4.1A
1
a 2e2x
c 0.4e0.4x
2
a 4 × 0.5e0.5x = 2e0.5x
b −e−x
d e4x + 2 × 4 = 4e4x + 2
b 100e−0.1x × −0.1 = −10e−0.1x
3
dy
= 0.1 N e0.3 = 0.135N h−1.
dt
Hence 0.1k = ln 2 and k =
f(x) = (1 + x 2)2
50e2x−10
dy
= 0.1 N e0.1t and when
dt
hence e0.1k = 2.
Using the chain rule,
4
2
dy
−0.5x 2
= 4e −0.5x × −0.5 × 2x = −4xe
dx
dy
b When x = –2,
= −4 × −2 × e−2 = 8e−2.
dx
a
If t = 3 then the number of bacteria is
N e0.1 × 3 = e0.3 N = 1.35N which is an increase
of 35%.
a
b If y = N e0.1t then
dy
= 3x 2 − 12x
dx
dy
If x = 6 then
= 108 − 72 = 36.
dx
The gradient of the normal is − 1 and the
36
1
equation is y − 0 = − (x − 6) which simplifies
36
to 36y + x = 6.
3
aBy definition, if y = ln 1.5 then ey = 1.5 so
eln 1.5 = 1.5.
100e2x−10
c
×2=
a f(x) = 4e2x × 2 = 8e2x
b f(x) = 8e2x × 2 = 16e2x
If x = 0, f(x) = 4f(x) = 4 × 2 = 8 > 0 showing
this is a minimum point.
8
dy
= ex − 4e−x.
dx
At a stationary point ex − 4e−x = 0.
Multiply by ex:
e2x − 4 = 0 so e2x = 4.
Take logarithms: 2x = ln 4 and
 1
1
x = ln 4 = ln  4 2  = ln 2 .
2
 
d 2y
= ex + 4e−x which equals y.
d 2x
When x = ln 2 then y = eln 2 + 4e−ln 2 = 2 + 4 ×
1
= 4.
2
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DIFFERENTIATION
This means
minimum.
d2y
= 4 > 0 so the stationary point is a
dx 2
b The gradient of the normal at P is −
1
the equation is y − k = − x .
k
The coordinates are (ln 2, 4).
9
a
dy
= ae ax ; if the gradient is 1 then aea x= 1;
dx
1
1
e ax = ; ax = ln = − ln a.
a
a
ln a
; then
a
1
1
= e − ln a = ln a = ; the point P is
a
e
Hence x = −
y = e ax
This meets the x-axis at (100, 0) so
1
0 − k = − × 100; k2 = 100; k = 10.
k
Exercise 4.2A
1
(
)
1
ln a or x + y = 1 − ln a .
y− =− x+
a
a
a
10 a
a y = ln 3 + ln x and hence
b y = 3 ln x and hence
(− lnaa , a1 ).
b The gradient of the tangent is 1 so the gradient
of the normal is −1 and the equation is
dy 1
= .
dx x
dy 3
= .
dx x
dy
1
3x 2
= 3
× 3x 2 = 3
dx x + 2
x +2
dy 4
1
4
1
a
b
×4=
c y = 4 ln x so
= .
x
4x
x
dx x
c
2
3
a
y
y = ln 2x
dy
= e x − a; at a stationary point ex – a = 0;
dx
ex
1
and
k
y = ln x
= a; x = ln a;
then = eln a – a ln a = a – a ln a;
coordinates are (ln a, a – a ln a).
0
d 2y
= e x > 0 for all x so the point is a minimum.
dx 2
dy
c If x = 0 then y = 1 and
= 1 − a; The
dx
equation of the tangent is
1
x
b
 0 
b ln 2x = ln 2 + ln x so the translation is 
.
 ln 2
c f(x) = ln kx = ln k + ln x and since ln k is a
1
constant, f ′(x) = .
x
y – 1 = (1 – a)x or y = 1 + (1 – a)x.
11 a f(x) = ex + 3e–x; f(x) = ex – 3e–x. If the gradient is 4
then f(x) = ex – 3e–x = 4; e2x – 4ex – 3 = 0.
ex = 4.646 (ignore the negative solution).
x = ln 4.646 = 1.536
At x = 1.536, f(x) = 5.292.
Coordinates are (1.54, 5.29).
4
b At a stationary point, f(x) = 0 so ex – 3e–x = 0;
ex = 3e–x; e2x = 3.
It crosses the x-axis at (–4, 0) and there
dy
1
=
= 1.
dx −4 + 5
1
ln 3 = ln 3 ; f (x) = ex + 3e–x
2
which is positive for all values of x and in
particular if x = ln 3 so this is a minimum
point.
2x = ln 3; x =
The equation of the tangent is
y − 0 = 1 (x − −4) or y = x + 4.
b It crosses the y-axis at (0, ln 5) and there
dy
1
1
=
= .
dx 0 + 5 5
The minimum value is f(ln 3).
3
= 3+ 3=2 3
3
dy
dy
12 a
= ke x = y . At P, x = 0, y = k and
= k.
dx
dx
= e ln
3
+ 3e − ln
3
= 3+
The equation of the tangent is y – k = kx.
Where the tangent meets the x-axis, y = 0
and so –k = kx; x = –1 and the point is (−1, 0).
d The translation is parallel to the y-axis so
the gradient for any particular x value is
unchanged.
dy
1
=
.
a If y = ln(x + 5) then
dx x + 5
5
The equation of the tangent is y − ln 5 = 1 x
5
1
or y = x + ln 5.
5
a If x = e, then y = ln e = 1 so (e, 1) is on the graph.
dy 1
dy 1
b dx = x and if x = e then
= .
dx e
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4
WORKED SOLUTIONS
The equation of the tangent is
1
1
y − 1 = (x − e) which simplifies to y = x .
e
e
This line passes through the origin.
1
(x − a); if this passes through (0, 0)
a
1
then 0 − b = (0 − a); b = 1.
a
So ln a = 1; a = e; P is (e, 1).
y−b=
c The gradient of the normal is –e and the
equation is y −1 = −e(x − e)
b At Q, gradient
Where it crosses the x-axis, y = 0 so
−1 = −e(x − e)
1
1
= x − e and so x = e + .
e
e
10
1
= ln 10 − ln x and so f ′(x) = − .
a f(x) = ln
x
x
10
b
f(x) = ln 2 = ln 10 − ln x2 = ln 10 − 2 ln x and
x
2
so f ′(x) = − .
x
y = ln x = ln e–1 = –1; Q is (e–1, –1).
c The equation of the tangent at Q is
Which rearranges to
6
y + 1 = e(x – e–1);
if x = 0 then y + 1 = e × (–e –1) = –1; y = −2 and
the point is (0, −2).
dy
1
a
×a =
; if the
=
dx ax + b
ax + b
a
= 1;
gradient is 1 then
ax + b
10 a
y = ln (ax + b) so
c f(x) = ln 10n = ln 10 − ln xn = ln 10 − n ln x and
x
n
so f ′(x) = − .
x
7
a = ax + b; ax = a – b; x = 1 −
and b are positive, so is
a y = log10 x therefore x = 10y = (eln 10)y = e(ln 10)y.
Take logs and then (ln 10)y = ln x.
1
y=
ln x
ln10
b y = xn – ln x; if x = 1 then y = 1 – 0 = 1 so S is
on the curve.
c If y = xn – ln x then
dy
1
= nx n −1 − ; if x = 1
x
dx
dy
a
1
1
=
=
and if the gradient is
2
dx ax + a x + 1
1
1
= so x = 1.
then
x +1 2
Then y = ln (ax + a) = y = ln(e2 + e2) = ln 2 e2
= ln 2 + ln e2 = 2 + ln 2.
The point is ( 1, 2 + ln 2).
Exercise 4.3A
1
d If y =
b −sin (5x − 2) × 5 = −5 sin (5x − 2)
2
dy
1
1
1
1
= 2x − = 0 then 2x = ; x 2 = ; x =
.
x
x
2
dx
2
9
c −sin (x2 − 3x − 4) × (2x − 3)
= −(2x − 3) sin (x2 − 3x − 4)
3
aSuppose P is the point (a, b).
gradient is
Hence f"(x) + 16f(x) = 0.
4
dy 1
= ; at P the
dx x
1
and the tangent is
a
f'(x) = 4 cos 4x − 4 sin 4x
f"(x) = −16 sin 4x − 16 cos 4x
= −16(sin 4x + cos 4x) = −16f(x)
1
1
1
1 1
− ln
= + ln 2 = + ln 2
2
2 2
2 2
= 1 (1 + ln 2).
2
a 10 cos 0.5x × 0.5 = 5 cos 0.5x
b 3 cos 3x − 6 sin 6x
dy
1
– ln x then
= 2x − ; if
x
dx
Then y =
a 2 cos 2x
c cos (x2 + 1) × 2x = 2x cos (x2 + 1)
dy
then
= n − 1.
dx
x2
a
a
= 0; = 1; a = b
b
b
y = ln (ax + a) and 2 = ln (0 + a) = ln a so a = e2.
b y = log10 ax2 = log10 a + log10 x2
8
a
. Because a
b
a
hence x < 1.
b
b On the y-axis x = 0 so 1 −
dy
1
=
dx x ln10
= log10 a + 2 log10 x;
dy
1
2 .
so
= 2×
dx
x ln10 = x ln10
dy
1
1
a = 1 − ; at a stationary point, 1 − = 0;
x
x
dx
x = 1 and y = 1 – ln 1 = 1 – 0 = 1; S is (1, 1).
1
1
= e; x = = e −1 ; then
x
e
a y = (sin x)2
Use the chain rule:
dy
= 2 sin x × cos x = 2 sin x cos x = sin 2x
dx
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Differentiation
dy
= 2 cos x × −sin x.
dx
= −2sin x cos x = −sin 2x.
b If y = cos2 x then
1
1
sin = 0 .
x
x2
Since x > 0, this means sin 1 = 0 and hence
x
1 = π, 2π, 3π, 4π, …
x
1 1 1 1
Hence x = , , , ,…
π 2π 3π 4π
The turning points are
1
1
1
1
,1 ,
, −1 ,
,1 , …
, −1 ,
π
2π
3π
4π
b At a stationary point
c sin2 x + cos2 x = 1 so the derivative of
sin2 x + cos2 x is 0.
Hence the two derivatives are identical apart
from the sign.
5
dy
= esin x × cos x = cos x esin x
dx
dy
If x = 0 then
= e0 × 1 = 1. The gradient is 1.
dx
b At a stationary point, cos x esin x = 0.
( ) ( )(
a
10 a
Since esin x is always positive, cos x = 0.
π
sin
One solution is x = π and then y = e 2 = e
2
π
so 2 ,e is a stationary point.
( )
6
a f(x) =
3 cos x − 3 sin2 x × cos x = 3 cos x(1 − sin2 x)
= 3 cos x × cos2 x = 3 cos3 x
b f"(x) = 3 × 3 cos2 x × −sin x = −9 cos2 x sin x
7
a f(x) = (cos x)−1
Using the chain rule,
f(x) = ( −1)(cos x)−2 × (−sin x) =
=
sin x
cos2 x
1
sin x
×
= sec x tan x .
cos x cos x
a
The velocity is
dy
= 0.1 cos 2.4t × 2.4
dt
= 0.24 cos 2.4t.
dy
= 0.24 cos 2.4 = −0.177.
When t = 1,
dt
The speed is 0.177 m s−1.
b The speed is zero when cos 2.4t = 0
Then sin 2.4t = ±1 and the displacement is
0.1 m, the largest possible value.
d 2y
= −0.576 sin 2.4t.
dt 2
This is zero when sin 2.4t = 0 and then y is
also zero.
c The acceleration is
The bob is in the central position.
9
1
1
is − 2
x
x
Using the chain rule,
a The derivative of
f ′(x) = − sin
dy
= 2cos x + 4sin x ; if x = 0 then
dx
dy
= 2×1+ 4 × 0 = 2
dx
dy
π
then y = 2 – 0 = 2 and
= 0 + 4 = 4 ; the
2
dx
π
tangent is y − 2 = 4 x − ;
2
bIf x =
(
)
y – 2 = 4x – 2π; y = 4x + 2 – 2π.
dy
= 0 then 2 cos x + 4 sin x = 0;
dx
4 sin x = –2 cos x; tan x = –0.5; x = 2.68.
c If
11 a f(x) = –2 sin 2x + 2 sin x
b y = (sin x)−1 and
dy
cos x
= (−1)(sin x)−2 × cos x = −
dx
sin 2 x
1
cos x
=−
×
= −cosec x cot x
sin x sin x
8
)( )
1
1
1
1
× − 2 = 2 sin .
x
x
x
x
b If f(x) = 0 then –2 sin 2 x + 2 sin x = 0;
–4 sin x cos x + 2 sin x = 0;
–2 sin x cos x + sin x = 0; sin x (–2 cos x + 1) = 0;
sin x = 0 or cos x = 1 ;
2
1
π
if sin x = 0 then x = 0; if cos x = then x =
3
2
f(0) = 1 – 2 = – 1;
f π = cos 2π − 2cos π = −0.5 − 2 = −1.5.
3
3
3
π
Stationary points are (0, −1) and , −1.5 .
3
()
(
)
dy
dy
12 a = 1 − 2sin 2x ; if x = 0, y = 1 and
= 1 ; the
dx
dx
gradient of the normal is –1; the equation is
y – 1 = – 1(x – 0) or y + x = 1
b At a stationary point, 1 – 2 sin 2x = 0;
1
π
5π
π
5π
sin 2x = ; 2x = or
;x=
or
;
12
2
6
6
12
two points in the given interval.
c
2
d 2y
π d y = −4cos2x
–3.46 < 0 so
2 = −4cos2x ; if x = 12 ,
dx
dx 2
this is the maximum point.
The y-coordinate is
π
π
π
3 π+6 3
+ cos =
+
=
or 1.13 to
12
6 12
2
12
3 s.f. This is the maximum value.
13 a y = sin 2θ + 2 sin θ so
dy
= 2cos2θ + 2cosθ
dθ
b At a maximum point, 2 cos 2θ + 2 cos θ = 0;
cos 2θ + cos θ = 0;
42
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4
WORKED SOLUTIONS
2cos2 θ – 1 + cos θ = 0; (2 cos θ – 1)(cos θ + 1) = 0;
1
cosθ = or −1.
2
θ = π or π are two solutions.
3
d 2y
π
= – 4 sin 2θ – 2 sin θ; if θ = then
3
dθ 2
d 2y
3
3
= −4 ×
−2×
= −3 3 < 0 so this
2
2
dθ 2
π
will be a maximum value. If θ = Then
3
3 3
2
π
π
3
y = sin
+ 2sin = 3 ×
=
3
3
2
2 and this
Then f(t) = sin 10t × 0.2e0.2t + 10 cos 10t × e0.2t
= (0.2 sin 10t + 10 cos 10t)e0.2t.
b f(t) = cos t2 e−1.5t
If u =cos t2 and v = e−1.5t then
and
= −(1.5 cos t2 + 2t sin t2)e−1.5t.
c f(t) = sin (2t + 1)e −t
)
( )
a
Then f(x)
2
2
= {−2t sin (2t + 1) + 2 cos (2t + 1)}e −t .
6
a
dy
= sin x + x cos x
dx
dy
= 2 then 1 + ln x = 2.
dx
dy
= cos x − x sin x
dx
c dy = 2x sin 2x + x2 × 2 cos 2x
dx
= 2x sin 2x + 2x2 cos 2x
c If
a ex + xex
d At a stationary point 1 + ln x = 0 therefore
ln x = −1.
Therefore ln x = 1.
Therefore x = e and y = e ln e = e so the
coordinates are (e, e).
So x = e−1 and y = e−1 × −1 = −e−1. The
coordinates are (e−1, −e−1).
a
f(x) = cos x × cos x + sin x × (−sin x)
= cos2 x − sin2 x
1
b sin 2x = 2 sin x cos x so y = sin 2x and
2
dy 1
= × 2 cos 2x = cos 2x.
dx 2
7
a
a
dy
= ex sin x + ex cos x
dx
d 2y
b
= ex sin x + ex cos x + ex cos x − ex sin x
dx 2
= 2ex cos x
a f(t) = sin 10t e0.2t
du
If u = sin 10t and v = e0.2t then dt =10 cos 10t
dv = 0.2e0.2t .
and
dt
dy
= e−x − xe−x
dx
At a stationary point, e−x − xe−x = 0 so
e−x(1 − x) = 0.
e−x cannot be 0 so the only solution is x = 1.
c The identity cos 2x ≡ cos2 x − sin2 x means
that the two answers are identical.
5
dy
= 1 + 0 = 1.
dx
The gradient is 1.
c 2xe2x + x2 × 2e2x = 2xe2x + 2x2 e2x = 2x(x + 1)e2x
4
dy
1
= ln x + x × = 1 + ln x
x
dx
b If x = 1,
b e−x − (x + 1)e−x = e−x − xe−x − e−x = −xe−x
3
2
−t
= sin (2t + 1) × −2te −t + e × 2 cos (2t + 1)
b
2
2
2
dv
then du = 2 cos (2t + 1) and
= −2te −t .
dt
dt
Exercise 4.4A
1
2
If u = sin (2t + 1) and v = e −t
=
(
dv
= −1.5e −1.5t .
dt
Then f(t) = cos t2 × −1.5e−1.5t + (−2t sin t2) × e−1.5t
is the maximum value.
dy
π
= cos x + 3cos3x; if x = then y = sin π + sin 3π
14
4
4
4
dx
1
1
2
+
=
= 2
=
2
2
2
dy
3
and
= cos π + 3cos π
4
4
dx
1
3
2
−
=−
= − 2 ; the tangent is
2
2
2
π
π
y− 2=− 2 x−
or y + 2x = 2 1 +
4
4
du
= −2t sin t2
dt
Then y = 1 × e−1 and the coordinates are
(1, e−1).
b
dy
= 2xe−x − x2 e−x
dx
At a stationary point, 2xe−x − x2 e−x = 0 so
x(2 − x)e−x = 0.
Either x = 0 or x = 2.
If x = 0, y = 0.
If x = 2, y = 4e−2.
The points are (0, 0) and (2, 4e−2).
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Differentiation
8
2
a If u = x2 and v = e x then
2
2
dv
= e −x × ( −2x ) = −2xe −x
dx
(
2
2
So dy = 10 × 2xe −x + 10x 2 × −2xe −x
dx
2
= 20xe −x − 20x 3e −x
c At a stationary point ex sin x + ex cos x = 0;
3π
sin x = – cos x; tan x = –1; x =
4
du
= 2x and
dx
)
Then y =
2
b At a stationary point, 20xe −x − 20x 3e −x = 0;
2
hence 20x (1 − x )e
−x 2
d 2y
dy
1 34π
0
−
2
×
e
y
=
2
−
2
=
dx
2
dx 2
which is negative so stationary point is a
maximum.
Also
=0
2
e −x cannot be 0 so x(1 – x2) = 0; Hence
x(1 – x)(1 + x) = 0.
x = 0, 1 or −1; if x = 0, y = 0; if x = 1, y = 10e–1;
if x = −1, y = 10e–1
The stationary points are (0, 0), (1, 10e–1)
and (−1, 10e–1).
9
dy
= −0.2e−0.2x sin 2x + 2e−0.2x cos 2x
dx
At stationary point,
−0.2e−0.2x sin 2x + 2e−0.2x cos 2x = 0.
2e−0.2x cos 2x = 0.2e−0.2x sin 2x
2 cos 2x = 0.2 sin 2x
sin 2x
2
=
cos 2x 0.2
tan 2x = 10
2x = arctan 10 = 1.4711 or 1.4711 + π or
1.4711 + 2π or...
2x = 1.4711 + nπ where n = 0, 1, 2, 3, ...
So x = nπ + 0.736 where n is 0, 1, 2, 3, ...
2
2
2
10 dy = 15e −0.05t + 15te −0.05t × (−0.1t )
dt
2
2
= 15e −0.05t − 1.5t 2e −0.05t
When the speed is maximum the graph has a
stationary point so
−0.05t 2
2 −0.05t 2
15e
− 1.5t e
= 0.
Therefore 15 − 1.5t2 = 0.
t2 = 10
t = 10
Then y = 15 10e −0.5 = 28.77.
The maximum speed is 28.8 m s−1
11 a Using the product rule,
b
dy
= e x sin x + e x cos x
dx
d 2y
= derivative of ex sin x + derivative of ex cos x
dx 2
= ex sin x + ex cos x + ex cos x – ex sin x = 2ex cos x
dy
2
− 2y = 2(ex sin x + ex cos x) – 2ex sin x
dx
d 2y
= 2ex cos x =
dx 2
e4
=
4
2
3π


4
The stationary point is  3π , e 
2
 4

2
2
3π
3π
3π
e 4 sin
12 a
dy
= −e −x sin x + e−x cos x;
dx
d 2y
= e −x sin x − e −x cos x − e −x cos x − e −x sin x
dx 2
= −2e −x cos x
b At a stationary point – e–x sin x + e–x cos x = 0;
sin x = cos x; tan x = 1; x =
π 5π 9π
,
,
…
4 4 4
−π
−π
If x = π , y = e 4 sin π = 1 e 4 ;
4
4
2
−π
−π
−π
d 2y
π
1
4
4
cos = −2e ×
= − 2e 4 < 0
2 = −2e
4
2
dx
so the value is a maximum.
c When x =
2
−π
5π d y
,
= 2e 4 > 0 a minimum
2
4 dx
point.
When x =
2
−π
9π d y
,
= − 2e 4 < 0 a maximum
4 dx 2
 9π 1 − 94π 
point at  ,
e 
 4

2
d The maximum points occur 2π apart when
π 9π 17π
x= ,
,
,…
4 4
4
The values of y are
1 − π4
e ,
2
1 − 94π
1 − π4
=
e × e −2π ,
e
2
2
1 − 174π
1 − π4
=
e × e −4π , … and these form
e
2
2
a geometric progression with r = e–2π
1
13 a If u = 1 + x = (1 + x)2 and v = sin πx
2
−1
1
and
then du = 1 (1 + x) 2 =
dx 2
2 1+ x
dv π
= cos π x
2
dx 2
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4
WORKED SOLUTIONS
dy
πx
1
× sin
Hence dx =
2
2 1+ x
3
πx
π x = sin 2
π
π 1+ x
+ 1 + x × cos
cos π x
+
2
2
2
2
2 1+ x
3π = −2 and
b If x = 3, y = 2sin
2
3
π
dy sin 2
2π
3π = − 1 + 0 = − 1
cos
=
+
4
4
4
2
2
dx
4
14 If u = x2 and v = 20 − x = (20 −
−1
du
dv 1
= 2x and
= × (20 − x) 2 × −1
dx
dx 2
1
=−
2 20 − x
5
Then
b
dy
= sec2 (4x + 3) × 4 = 4 sec2 (4x + 3)
dx
c
dy
= 3 sec2 (x2) × 2x = 6x sec2 (x2)
dx
dv
2
a If u = x and v = tan x then du = 1 and dx = sec x .
dx
f(x) = x sec2 x + tan x
c f(x) = e−x × sec2 ax × a − e−x tan ax
= e−x (a sec2 ax − tan ax)
dy
= −2e−2x sin x + e−2x cos x
a
dx
b
cos x − 2 sin x = cos x − 2 sin x which is the
e 2x
e 2x
e 2x
same answer.
x2
= 0;
2 20 − x
x2
; 4x (20 – x) = x2; x ≠ 0 so
2 20 − x
4 (20 – x) = x; 80 – 4x = x;
5x = 80; x = 16. Then y = 16 2 20 − 16 = 512; the
stationary point is (16, 512).
2x 20 − x =
6
a
dy (x + 1) × 1 − x × 1 x + 1 − x
1
=
=
=
dx
(x + 1)2
(x + 1)2
(x + 1)2
b
dy (x 2 + 1) × 1 − (x + 1) × 2x x 2 + 1 − 2x 2 − 2x
=
=
dx
(x 2 + 1)2
(x 2 + 1)2
=
c
a f ′(x) =
1 − 2x − x 2
(x 2 + 1)2
The coordinates of the stationary point are
e, 1 .
e
d2y
When x = e then 2 = 2 ln e3 − 3 = − 13 which
dx
e
e
is negative so the stationary point is a
maximum point.
( )
dy (2x − 1) × 3x 2 − x 3 × 2 6x 3 − 3x 2 − 2x 3
=
=
dx
(2x − 1)2
(2x − 1)2
3
2
= 4x − 3x2
(2x − 1)
2
a If f(x) =
b If f(x) =
c If f(x) =
x cos x − sin x
sin x
then f ′(x) =
.
x
x2
x × 1 − ln x × 1 1 − ln x
x
=
x2
x2
x 2 ×  − 1  − (1 − ln x) × 2x
 x
b f ′′(x) =
x4
= −x − 2x +4 2x ln x = 2x ln x4 − 3x
x
x
ln
−
2
3
x
=
x3
1 − ln x
c At a stationary point
= 0.
x2
ln e 1
= .
Therefore ln x = 1 and x = e and y =
e
e
Exercise 4.5A
1
dy e 2x cos x − sin x × 2e 2x
=
dx
e 4x
Divide numerator and denominator by e2x to get
dy
x2
Hence
= 2x 20 − x −
; at a stationary
dx
2 20 − x
point 2x 20 − x −
7
a
dy (1 + e −x ) × e −x − (1 − e −x ) × (−e −x )
=
dx
(1 + e −x )2
x +1
× (− sin x)
then f ′(x) = cos x − (x + 1)
cos x
cos2 x
cos x + (x + 1)sin x
=
.
cos2 x
b When x = 0,
2
x2
then f ′(x) = 2x × sin 2x −2x × 2cos2x
sin 2x
sin 2x
c At a stationary point,
=
2x sin 2x − 2x 2 cos2x
.
sin 2 2x
dy
= sec2 ax × a = a sec2 ax.
dx
b f(x) = 2x tan 2x + x2 × sec2 2x × 2
= 2x tan 2x + 2x2 sec2 2x
The equation is y + 2 = − 1 (x − 3); 4y + 8 = – x + 3;
4
x + 4y + 5 = 0.
1
x)2
a Using the chain rule,
−2x
−x
−2x
−x
−x
= e + e +−ex 2 − e
= 2e −x 2
(1 + e )
(1 + e )
dy 2 × 1 1
1
= 2 = so the gradient is .
2
2
dx
2
2e −x
= 0 but since
(1 + e −x )2
e−x is always positive, this can never happen.
Therefore there is no turning point.
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Differentiation
8
a f(x) = cos x
sin x
d
sin x × (− sin x) − cos x × cos x
So f ′(x) =
sin 2 x
2
(sin x + cos2 x)
=−
= −12 = − cosec2 x .
sin 2 x
sin x
b Using the product rule, if y = x cot x then
dy
= cot x − x cosec2 x.
dx
9
a If v =
dy
=
dx
x
2
x +1
du
= 2x and
12 aIf u = x2 and v = cos x then
dx
dv
= − sin x
dx
2x cos x − x 2 × (− sin x)
f′(x ) =
cos2 x
x
then
x2 + 1
x2 + 1 − x ×
x2 + 1
x
2
x +1 .
=
The numerator is
x2 + 1 − x2
x
x2 + 1 − x ×
=
=
x2 + 1
x2 + 1
So
2
−(x − 1)2
1
1
2 
−4
=
=− 
=
2
2
2  y − 1
dy
2 ( y − 1)
dx
−2
dx
=
=
( y − 1)2 dy
1
So if y =
dy
1
=
=
dx (x 2 + 1) x 2 + 1
y +1− y +1
2
=
y −1
y −1
=
So then
x 2 + 1 = (x 2 + 1)2 then
−1
dv 1 2
= (x + 1) 2 × 2x =
dx 2
1
−(x − 1)2
from part a. From part b,
=
2
dy
dx
y + 1 − ( y − 1)
y +1
−1=
x −1 =
y −1
y −1
1
2
x +1
1
3
x2 + 1
x2 + 1
then
x


x
2

 × x − x +1
2
dy
x + 1
x 2 − (x 2 + 1)
=
=
2
dx
x
x2 x2 + 1
1
=−
.
2
x x2 + 1
b If y =
10 tan x = sin x
cos x
Use the quotient rule.
dy cos x × cos x − sin x × (− sin x)
=
dx
cos2 x
2
2
= cos x +2sin x = 12
cos x
cos x
dy ( x − 1) × 1 − (x + 1) × 1 x − 1 − x − 1
=
11 a dx =
( x − 1 )2
( x − 1)2
−
2
=
2
x − 1)
(
b y(x – 1) = x + 1; so xy – y = x + 1; hence
xy – x = y + 1; x (y – 1) = y + 1;
y +1
x=
y −1
c The structure of the formula is the same so
dx
−2
=
dy ( y − 1)2
.
2x cos x + x 2 sin x
cos2 x
2cos1 + sin1
= 6.584
cos2 1
b If x = 1, f ′ ( x ) =
12
= 1.851; the equation is
cos1
y – 1.851 = 6.584(x – 1) or y = 6.584x – 4.733
c At P, y =
13 a If u = x – 1 and v =
then
(
dv 1 2
du
= 1 and
= x +1
dx 2
dx
(x
+1
)
−1
2
3
2
=
b At a stationary point
and x = −1; y =
(x
+1
1+ x
(x
2
+1
)
)
1
2
× 2x =
x2 + 1 − x2 + x
2
)
x
x2 + 1
x
x2 + 1
x2 + 1
x 2 + 1 − x(x − 1)
2
)
x 2 + 1 − (x − 1) ×
dy
=
Hence
dx
=
(
x2 + 1 = x2 + 1
3
2
3
2
=
1+ x
(x
2
+1
)
3
2
= 0 so 1 + x = 0
−1 − 1 −2
=
=− 2
1+1
2
The stationary point is (−1, − 2)
c Where the curve crosses the y-axis, x = 0 and
1+0
the gradient is
3 =1
( 0 + 1) 2
Where the curve crosses the x-axis, y = 0 and
x −1
so
= 0;
x2 + 1
x = 1 and the gradient is
or
1
2
2
1+1
(1 + 1)
3
2
=
2
3
22
=
1
2
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4
WORKED SOLUTIONS
14 aIf u = x and v = (x + 1)2 then
16 aLet u = sin x + cos x and v = ex; then
du
dv
= cos x − sin x and
= ex
dx
dx
du
= 1 and
dx
dv
= 2(x + 1)
dx
dy (x + 1) − x × 2(x + 1)
=
dx
(x + 1)4
Hence
=
Hence
dy e x ( cos x − sin x ) − e x (sin x + cos x )
=
dx
e 2x
2
x + 1 − 2x
1− x
=
(x + 1)3
(x + 1)3
At a stationary point,
y=
=
ex is always positive and sin x is positive if
−2sin x
0 < x < π which means that
is
ex
negative in that interval; the gradient is
dy
= 0 so x = 1; then
dx
( )
1
1
= ; The point is 1, 1
4
(1 + 1)2 4
negative.
1− x
= 1;
(x + 1)3
1 – x = (x + 1)3; 1 – x = x3 + 3x2 + 3x + 1;
b If the gradient is 1 then
dy
−2sin x
= 0 then
= 0 and so sin x = 0;
dx
ex
x = 0, π, 2π, 3π, … Stationary points occur
every π units along the axis.
b If
x3 + 3x2 + 4x = 0; x(x2 + 3x + 4) = 0; either x = 0
or x2 + 3x + 4 = 0
The determinant of the quadratic is
9 – 16 = −7 so it has no solution. The only
0
point is where x = 0 and then y = = 0 ; it is
1
the origin (0, 0).
du
15 aLet u = 1 – x2 and v = 1 + x2; then
= −2x and
dx
dv
= 2x
dx
(
)
(
2
2
dy −2x 1 + x − 2x 1 − x
Hence
=
2
dx
1 + x2
=
(
−2x − 2x 3 − 2x + 2x 2
(1 + x )
2 2
)
=
c To find
v = (1 + x2)2
Then
−4x
(1 + x )
2 2
Hence
=
(1 + x )
becomes
2 2
So
=
d y
=
dx 2
(
(
)
−4 1 + x 2 + 16x 2
(1 + x )
2 3
−2cos x
e 2x
When x = 0, π, 2π, 3π, … cos x is 1, –1, 1, –1 …
d 2y
alternates in sign +, −, +, …
This means
dx 2
and the stationary points are alternately
maximum and minimum.
let u = –4x and
(
)
2
1
(
− (−4x) × 4x 1 + x 2
(1 + x )
=
)
2 4
a If f(x) = tan−1 ax then f ′(x) =
1
a
.
×a =
1 + (ax)2
1 + a 2x 2
b If f(x) = tan−1 (0.5x − 1) then
1
0.5
× 0.5 =
.
f ′(x) =
1 + (0.5x − 1)2
1 + (0.5x − 1)2
)
c If f(x) = 3tan−1 (x2) then
3
6x
f ′(x) =
× 2x =
.
1 + (x 2)2
1 + x4
12x 2 − 4
(1 + x )
2 3
d 2y
= −4 < 0 so the stationary point
dx 2
is a maximum.
c If x = 0,
d 2y
dx 2
Exercise 4.6A
)
−4 1 + x 2
)
−2( cos x − sin x )
ex
At a stationary point sin x = 0 and
−4x
dv
= 2 1 + x 2 × 2x = 4x 1 + x 2
dx
2
(
du
dv
= cos x and
= ex
dx
dx
x
x
d 2 y −2 e cos x − e sin x
2 =
2x
dx
e
)
du
= −4 and
dx
(
d 2y
dy −2sin x
, differentiate
=
.
dx
dx 2
ex
If u = sin x and v = ex then
dy
There is a stationary point where dx = 0;
–4x = 0; x = 0; the stationary point is (0, 1).
b To differentiate
cos x − sin x − sin x − cos x −2sin x
=
ex
ex
2
a
dy
1
10
= 10 ×
=
dx
1 + x2 1 + x2
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Differentiation
b Using the chain rule,
dy
1
10
=
× 10 =
dx 1 + (10x)2
1 + 100x 2
c
3
dy
=
dx
dy
a
=
dx
1
1+ x
10
( )
1
1+ x
a
()
2
2
×
1
=
10
Hence if f(x) = tan−1 x − 1 then
1
1
1
f ′(x) =
×
=
1 + (x − 1) 2 x − 1 2x x − 1 .
0.1
10
2 =
+ x2
x
100
1+
100
b If f(x) = tan−1
f(x) =
1
a
× = 2
a a + x2
7
If the gradient is
a
1
1
then 2
=
and so
2a
a + x 2 2a
x2 = a2 and so x = ±a.
But a > 0, so x = a then y = tan−1 1 = π .
4
π
The coordinates are a, .
4
( )
π
b At a,
the gradient of the normal is −2a.
4
The derivative of x tan−1 x is tan−1 x +
The derivative of 0.5 ln (1 + x2) is
1
x
× 2x =
.
0.5 ×
1 + x2
1 + x2
8
=
x .
1 + x2
dy
1
2
=
×2=
.
dx 1 + (2x)2
1 + 4x 2
If the gradient is 1 then
2
= 1.
1 + 4x 2
Therefore 2 = 1 + 4x2.
4x2 = 1
1
x2 =
4
1
x=±
2
1
If x = then y = tan−1 1 = π .
2
4
1
−1
If x = − then y = tan (−1) = − π .
2
4
The points where the gradient is 1 are  1 , π 
 2 4
and  − 1 , − π  .
 2 4
6
a If y =
1
x − 1 = (x − 1)2 then
−1
dy 1
1
= (x − 1) 2 =
.
dx 2
2 x −1
(
(
du
= 2x and
dx
)
)
)
(
aThe derivative of tan–1 x is
b f ′(x ) =
)
1
so
1 + x2
1
×
a
1
a2 + x 2
1
()
1+ x
a
2
×
1
1
×
=
a a2
1
()
1+ x
a
2
1
1
c ⌠
dx = ⌠
dx ; using the


⌡ 25 + x 2
⌡ 52 + x 2
answer to part b with a = 5 you get
x
x
Hence f ′(x) = tan x +
−
= tan −1 x .
1 + x2 1 + x2
If y = tan−1 2x then
aLet u = 1 + x2 and v = tan–1 x; then
1
⌠
dx = tan −1 x + c

⌡ 1 + x2
−1
5
×  − 12  = − 21 .
 x 
x +1
(
The equation is y − π = −2a(x − a) or
4
y + 2ax = 2a2 + π .
4
4
()
2
1
dv
=
dx 1 + x 2
dy
1
= 2x tan −1 x + 1 + x 2 ×
= 2x tan −1 x + 1
dx
1 + x2
2x tan −1 x − 1 + x 2 × 1 2
−1
1 + x = 2x tan x − 1
b dy =
2
2
dx
tan −1 x
tan −1 x
2a2 = a2 + x2.
( )
1
1+ 1
x
( 1x ) then
1
1
x
⌠
dx = tan −1 + c

5
5
⌡ 25 + x 2
9
dy
1
a
a =
×a=
; at the origin x = 0
dx 1 + (ax)2
1 + a 2x 2
and the gradient is a
a
a
a
then
= ; hence
2
1 + a 2x 2 2
1
1 + a2x2 = 2; a2x2 = 1; x = ±
a
b If the gradient is
( a1 ) = tan ( ±1) = π4 or − π4 ;
1 π
the points are ( , ) and ( − 1 , − π )
a
4
a 4
Then y = tan −1 ±a ×
10 aIf y = tan–1(x3) then
b If y =
(tan–1 x)3
(
dy
= 3 tan −1 x
dx
then
)
2
–1
dy
=
dx
1
( )
1+ x
(
3 2
× 3x 2 =
3 tan −1 x
1
×
2 =
1+ x
1 + x2
)
3x 2
1 + x6
2
c = x3 tan–1 x; let u = x3 and v = tan–1 x; then
du
dv
1
= 3x 2 and
=
dx
dx 1 + x 2
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4
WORKED SOLUTIONS
So dy = 3x 2 tan −1 x + x 3 × 1 2
dx
1+ x
= 3x 2 tan −1 x +
4
x3
1 + x2
When t = 1, x = 3, y = 0 and
a If t = π then x = 4 cos π = 2.828 and
4
4
π
y = 3 sin = 2.121 and the coordinates are
4

3 2
(2.83, 2.12) to 3 s.f. or  2 2,
2 

5
( )
( )
dy dy dx 2t
=
÷
=
= 4t .
1
dx dt dt
2
1
When t = −1, x = − and y = −3 and the gradient of
2
1 1
the normal is − = .
4t 4
1
1
The equation of the normal is y + 3 = x +
4
2
( )
3π
3π
b If t = 2 then x = 4 cos 2 = 0 and
y = 3 sin 3π = −3 and the point is (0, –3).
2
c
therefore 4y + 12 = x +
3
6
2
1
–3
0
–1
–1
–2
1
2
3
4
a
dy dy dx −5sin t
5
=
÷
=
= − tan t
4cos t
4
dx dt dt
b
dy dy dx −6sin 2t 3sin 2t
=
÷
=
=
sin t
−2sin t
dx dt dt
x
This simplifies to
–2
–3
3
2
a
b
t
c
−3 −2 −1
0
1
2
3
x
9
4
1
0
1
4
9
y
−1
0
1
2
3
4
5
7
y
8
5
3
2
1
0
c
1
2
3
4
5
6
7
dy
dy
dx
1
= 2t and
= 1 so
= .
dt
dx 2t
dt
d At (4, 0) t = −2 and the gradient
dy
1
1
1
=
=
=− .
4
dx 2t 2 × −2
3
a
dy dy dx 2t
=
÷
=
= 2t
1
dx dt dt
dy dy dx 4t 3
b
=
÷
=
= 4t 3
1
dx dt dt
c
8
dy dy dx 2t − 3
=
÷
=
or 1 − 3
2t
2t
dx dt dt
9
3 × 2sin t cos t
= 6 cos t.
sin t
dy dy dx
2cos 2t
=
÷
=
= cos 2t × cos2 2t
dx dt dt 2sec2 2t
= cos3 2t
dy dy dx
7cos t
7
=
÷
=
= − cot t
3
dx dt dt −3sin t
7
π
7
7
π
When t = the gradient is − 3 cot 4 = − 3 × 1 = − 3 .
4
dy dy dx −3sin t
=
÷
=
dx dt dt 4cos 2t
π
The coordinates are (0, 0) when t = or 3π
2
2
π
dy −3 sin 2
=
= −3 = 3 .
When t = π the gradient
dx
4cos π
4 × −1 4
2
3
π
When t =
the gradient
2
3π
dy −3 sin 2
=
= −3 × −1 = − 3 .
dx
4cos3π
4 × −1
4
4
–1
1
.
2
So 8y + 24 = 2x + 1.
So 8y − 2x + 23 = 0.
y
–4
dy
= 2.
dx
The equation of the tangent is y − 0 = 2(x − 3) or
y = 2x − 6 or 2x − y − 6 = 0.
Exercise 4.7A
1
dy dy dx 2t
=
÷
=
= 2t
1
dx dt dt
x
9
dy dy dx
2sin t
=
÷
=
.
dx dt dt 2 − 2cos t
2× 1
2sin π
dy
π
2
2 =
4
When t = ,
=
=
4 dx 2 − 2cos π 2 − 2 × 1
2− 2
4
2
=
2 × 2 + 2 = 2 2 + 2 = 2 + 1.
4−2
2− 2 2+ 2
So the gradient of the tangent is 2 + 1.
49
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Differentiation
The gradient of the normal is
1
2 −1
2 −1
1
−
=−
×
=−
= 1 − 2.
2−1
2 +1
2 +1
2 −1
10
dy dy dx
b cos t
b
=
÷
=
= − cot t
a
dx dt dt −a sin t
d
y
= − b cot π = − b
When t = π ,
4
a
a
4 dx
a
b
π
and x = a cos =
and y = b sin π =
.
4
4
2
2
b = −bx − a 
The equation of the tangent is y −
.
a 
2
2
At the y-intercept x = 0 and so
b
b
a
y−
=− ×−
a
2
2
b
b
y−
=
2
2
2
b
y=
= 2b
2
The y-intercept is (0, 2b ).
11 a
dy dy dx −4sin 2t −0.4sin 2t
=
÷
=
=
10cos t
cos t
dx dt dt
b At the highest point the gradient is 0 so
−0.4sin 2t
= 0.
cos t
Therefore sin 2t = 0 and t = 0.
c Where it meets the x-axis, y = 0 and x = 20t;
the point is (20t, 0).
Where it meets the y-axis, x = 0 and t2 y = 20t
20
20
so y =
.
; the point is 0,
t
t
( )
1
20
× 20t ×
= 200
2
t
which is a constant and does not depend on t.
d The area of the triangle =
14 a
dy 
1
t2 − 1 1 t2 − 1
= 1 − 2  ÷ 2t =
×
=
2t
dx 
t 
t2
2t 3
b At a stationary point, t2 – 1 = 0 so t = ±1
If t = 1 the point is (1, 2) and if t = −1 the
point is (1, −2)
c Consider points near (1, 2) where t = 1;
if t = 0.9 then point is (0.81, 2.01) and
dy 0.92 − 1
=
< 0; if t = 1.1 then the point
dx 2 × 0.93
dy 1.12 − 1
is (1.21, 2.01) then
=
> 0; this
dx 2 × 1.13
means (1, 2) is a minimum point.
Consider points near (1, −2) where t = −1;
if t = −0.9 then point is (0.81, −2.01) and
dy
0.92 − 1
=
> 0; if t = 1.1 then the point
dx −2 × 0.93
dy
1.12 − 1
is (1.21, −2.01) then
=
< 0; this
dx −2 × 1.13
c If x = 9, 10 sin t = 9, sin t = 0.9 and t = 1.120.
Then
dy −0.4sin 2.24
=
= −0.720.
cos1.12
dx
This means that the tangent of the angle of
slope is 0.720.
So the angle is arctan 0.720 = 0.624 radians.
That is 0.624 × 180 = 36° to the nearest degree.
π
dy
dx
12 a
= 2sin θ cosθ and
= −2sin θ ;
dθ
dθ
dy
−2sin θ
−1
=
=
dx 2sin θ cosθ cosθ
b When the gradient is 1, cos θ = – 1 and the
point is (sin2 θ, 2cos θ) = (0, – 2) and the
equation of the tangent is y = x – 2
c When the tangent is parallel to the y-axis the
gradient is infinite so cos θ = 0.
Then sin θ = ±1; For θ > 0, the point on the
curve (sin2 θ, 2 cos θ) is (1, 0) where the tangent
is parallel to the y-axis.
dy
dy
dx
10
10
1
= 10 and
= − 2 so
= − 2 ÷ 10 = − 2
dt
dt
dx
t
t
t
10
1
b The equation is y −
= − 2 (x − 10t );
t
t
multiply by t2 : t2y – 10t = –x +10t;
hence x + t2y = 20t.
13 a
dy
1
dx
= 2t and
= 1 − 2 so
dt
dt
t
means (1, −2) is a maximum point.
15 a
dy
dy
dx
2 1
= 2t and
= 2 so
=
=
dt
dt
dx 2t t
b The gradient of the normal is –t. The
equation is y – 2t = –t(x – t2) or
y + tx = 2t + t3
c Where the normal meets the y-axis, x = 0 and
y = 2t + t3
Where the normal meets the x-axis, y = 0 and
tx = 2t + t3; so x = 2 + t2
The area of the triangle is
1
1
× 2t + t 3 2 + t 2 = t 2 + t 2 2 + t 2 =
2
2
1
2 2
t 2+t
2
(
(
)(
)
(
)(
)
)
dy
dy 3cos3t
dx
= 3cos3t ;
=
16 a = 2cos2t and
dt
dt
dx 2cos2t
At A the gradient is 0 so cos 3t = 0; one
π
π
solution is 3t = ; t = .
2
6
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4
WORKED SOLUTIONS
π
3
π
=
and y = sin = 1; the
3
2
2
 3 
coordinates of A are  2 , 1 .


b Differentiate: 2xy + x2
Then x = sin
b At B the gradient is infinite so cos 2t = 0; one
π
π
solution is 2t = ; t =
2
4
Then x = sin
3π
1
π
=
; the
= 1 and y = sin
4
2
2
If x = 2 and y = 1 then 4 + 4
So 8
4
dy
dy
+ 2y
=0
dx
dx
dy
If x = 5 and y = 0, 10 + 0 + 20
+0=0
dx
dy
dy
= −10 and
= −0.5
Therefore 20
dx
dx
b Differentiate: 2x + 4y + 4x
dy 3cos0 3
=
=
dx 2cos0 2
Another solution is t = π because
sin 2π = sin 3π = 0; then dy = 3cos3π = −3
2
dx 2cos2π
c 2x + 4y + (4x + 2y)
dy
= −(2x + 4y)
dx
dy −(2x + 4y)
x + 2y
=−
=
4x + 2y
2x + y
dx
Exercise 4.8A
5
a If x = 3 and y = 5 then y2 − x2 = 25 − 9 = 16
b Differentiate: 2y
b Using the product rule on the left-hand side
( x + y ) + x  1 + ddxy  = 4y ddxy ;
x+y+x+x
c 0=−
dy 3
= and the equation of the
dx 5
tangent is y − 5 = 3 (x − 3)
5
Or 5y − 25 = 3x − 9 or 3x − 5y + 16 = 0
6
y3
2
2 dy 2
2 dy dy
3 − 3 dx ; 3 = − 3 dx ; dx = − 3
x
y
x
y
x
sin x sin y
dy
= 0 and so
dx
2
2
2(x + y ) × (2x) = 50(2x).
Therefore 2(x2 + y2) = 50
x2 + y2 = 25
This is a circle of radius 5.
so
2
dy
= 0;
dx
dy
= cos x cos y ;
dx
dy cos x cos y
=
or cot x cot y
dx sin x sin y
a 4x + 2y
dy
dy
= 0 therefore 2x = −y
and
dx
dx
dy
2x
=− .
y
dx
dy
6
3
=− =− .
4
2
dx
dy
6
3
c If x = 3 and y = −4 then dx = − −4 = 2 .
b If x = 3 and y = 4 then
3
a
If x = 2 and y = 1 then x2y = 4 and 4(2 − y) = 4 so
the point is on the line.
Differentiate:
dy
dy
2(x 2 + y 2) ×  2x + 2y  = 50  2x − 2y 
dx 
dx 


At a stationary point,
d Using the produce rule,
cos x cos y + sin x (− sin y)
dy
− 2x = 0 so dy = x .
dx
dx y
At (3, 5)
dy
dy
dy
= 4y
; 2x + y = (4y − x) ;
dx
dx
dx
dy 2x + y
=
dx 4y − x
dy
=0
dx
(4x + 2y)
3
3
The gradients are and −
2
2
1
dy
dy
1
= −4 and so
=− .
2
dx
dx
a If x = 5 and y = 0 then x2 + 4xy + y2 = 25 + 0 + 0 = 25
c At the origin x = y = 0 so sin 2t = sin 3t = 0
dy
dy
x
ax
= 0;
2 − 4y
=
dx
dx 2y
dy
dy
= −4 .
dx
dx
If x = 0 and y = 5 then x2 + 4xy + y2 = 0 + 0 + 25 = 25

 1 
2
coordinates of B are  1,
 or  1, 2 



2
One solution is t = 0; then
dy
dy
= −4 .
dx
dx
7
a Differentiate: 2x − 2y
dy
dy x
= 0 and so
= .
dx
dx y
dy dy dx
sec2 t
sec t
=
÷
=
=
dx dt dt sec t tan t tan t
x sec t
and
c If x = sec t and y = tan t then =
y tan t
the expressions are equivalent.
b
sec t
d If the gradient is 2,
= 2 therefore
tan t
sec t = 2 tan t.
1 = 2sin t
cos t
cos t
So 2 sin t = 1
sin t =
1
so t = π and 5π .
2
6
6
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Differentiation
π
If t = 6 then x = sec π = 2 and
6
3
π
1
 2 1 
y = tan =
,
.
and the point is 
6
 3 3 
3
1
The gradient of the normal is − and the
4
1
equation is y + 2 = − (x − 3)
4
or 4y + 8 = –x + 3 or x + 4y + 5 = 0
If t = 5π then x = sec 5π = − 2 and
6
6
3
5
1
π
y = tan
=−
and the point is
6
3
 2
1 
,−
−
.

3
3
8
c Where the normal and the curve cross,
x2 + xy + y2 = 7 and x = –4y – 5
Substitute: (–4y – 5)2 + (–4y – 5)y + y2 = 7;
16y2 + 40y + 25 – 4y2 – 5y + y2 = 7
13y2 + 35y + 18 = 0; (y + 2)(13y + 9) = 0; y = −2
9
as given or − ; there are two solutions which
13
shows that the normal meets the curve again.
aIf x = −4 and y = 3 then x2 + xy + y 2 = 16 – 12 + 9 = 13
so the point is on the curve.
b Differentiate: 2x + y + x
dy
dy
+ 2y
= 0;
dx
dx
If x = −4 and y = 3 then
dy
dy
dy
−8 + 3 − 4
+6
= 0 ; −5 + 2
= 0;
dx
dx
dx
dy 5
=
dx 2
The equation of the tangent is y − 3 =
Exam-style questions
1
1
2
a y = (1 + x 2) and
=
5
(x + 4);
2
2y – 6 = 5x + 20; 2y = 5x + 26
9
dy
dy
aDifferentiate: 2x + 2y
−4+6
= 0;
dx
dx
( 2y + 6 ) ddxy = 4 − 2x ; ddxy = 42y−+2x6
dy
= 0 so 4 – 2x = 0 and
At a stationary point
dx
hence x = 2
b
2
3
10 aIf x = 3 and y = −2 then x2 + xy + y2 = 9 – 6 + 4 = 7
so (3, −2) is on the curve.
dy
dy
+ 2y
= 0 ; at (3, −2)
dx
dx
dy
dy
dy
6−2+3
−4
= 0;
=4
dx
dx
dx
b 2x + y + x
a f(x) = −2e−2x. At x = 0, gradient = –2
a
y = x2 cos x;
dy
π
= 2x cos x − x 2 sin x ; if x = then
2
dx
2
cos x dy x ( − sin x ) − 2x cos x
; if x = π
2 ; dx =
x
x4
dy π 2 ( − sin π ) − 2π cos π 2π
2
then dx =
= 4 = 3
π4
π
π
= 1 + 1 + 4 + 6 = 12 so (−1, 1) is on the curve.
The midpoint of AB is (2, −3) and if x = 2 and
y = −3 then 3y + 4x + 1 = –9 + 8 + 1 = 0 which
shows the midpoint is on the normal.
dy
x
x
x
=
=
1 =
y
1 + x 2 (1 + x 2)2 dx
b y=
b If x = −1 and y = 1 then x 2 + y 2 − 4x + 6y
4
y − 1 = − (x + 1); 3y – 3 = –4x – 4; 3y + 4x + 1 = 0
3
x
1 + x2
()
A and B are (2, 2) and (2, −8)
4
and the equation is
3
1
(1 + x 2)2
2
2
dy
= π cos π − π sin π = − π
2
2
2
dx
4
y2 + 6y – 16 = 0; (y – 2)(y + 8) = 0 y = 2 or −8
of the normal is −
x
b f(x) = e−x + x × (−e−x) = (1 − x)e−x. At x = 0,
gradient = 1
Substitute x = 2 into the equation of the
curve: 4 + y2 – 8 + 6y = 12
c If x = −1 and y = 1 then
dy 4 − 2x 4 + 2 6 3
=
=
= = ; the gradient
dx 2y + 6 2 + 6 8 4
1
−
dy 1
= (1 + x 2) 2 × 2x =
dx 2
4
a
Using the quotient rule, the derivative is
2x(x 2 + 1) − 2x(x 2 − 1) 2x 3 + 2x − 2x 3 + 2x
=
(x 2 + 1)2
(x 2 + 1)2
= 24x 2 .
(x + 1)
n −1 n
n −1 n
b The derivative is nx (x + 1n) − nx2 (x − 1)
(x + 1)
2n −1
+ nx n −1 − nx 2n −1 + nx n −1 = 2nx n −1 .
= nx
(x n + 1)2
(x n + 1)2
5 y = 2 ln x so
dy 2
= .
dx x
If the gradient is 0.5 then 2 = 0.5 so x = 4 and
x
y = ln 16.
The coordinates are (4, ln 16).
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4
WORKED SOLUTIONS
6
y = e2x – 11ex + 12x;
dy
= 2e 2x − 11e x + 12
dx
11 y = x2 tan 2x;
= 2x tan 2x + 2x2 sec2 2x
dy π
π
2 2
= tan π + π sec2 π =
If x = then
3 36
3
6
dx 3
At a stationary point, 2e2x – 11ex + 12 = 0;
(2ex – 3)(ex – 4) = 0; ex = 1.5 or 4;
x = ln 1.5 or ln 4
d 2y
= 4e 2x − 11e x
dx 2
d 2y
If x = ln 1.5 then
= 4e 2 ln1.5 − 11e ln1.5
dx 2
= 4 × 1.52 – 11 × 1.5 = −7.5 which is negative. This is
7
8
a maximum point.
d 2y
= 4e 2 ln 4 − 11e ln 4
If x = In 4 then
dx 2
= 4 × 16 – 11 × 4 = 20 which is positive. This is a
minimum point.
dy
a = 0.3 × 2 cos 2x − 0.4 × (−2sin 2x)
dx
= 0.6 cos 2x + 0.8 sin 2x
d 2y
= −1.2 sin 2x +1.6 cos 2x
b
dx 2
d 2y
d 2y
+ 4y = 0
4y = 1.2 sin 2x − 1.6 cos 2x = − 2 so
dx
dx 2
dy
a
= 0.1ex − 0.5e− 0.5x
dx
b At a minimum point, 0.1ex − 0.5e−0.5x = 0.
Multiply by
e0.5x:
0.1e1.5x
= 0.5 and so
e1.5x
Take logs: 1.5x = ln 5.
ln 5
x=
= 1.073
1.5
Then y = 0.1e1.073 + e−0.536 = 0.877.
The coordinates are (1.073, 0.877).
9
dy dy dx
sin θ
a dθ = dθ ÷ dθ = 1 − cosθ
sin θ
= 0.5 so sin θ = 0.5(1 − cos θ).
b
1 − cosθ
2 sin θ = 1 − cos θ
2sin θ + cos θ = 1
dy
10 a
= −e−x sin 10x + 10e−x cos 10x
dx
b At a stationary point
−e−x sin 10x + 10e−x cos 10x = 0.
Therefore −sin 10x + 10 cos 10x = 0.
sin 10x = 10 cos 10x
tan 10x = 10
10x = 1.471 or 1.471 + π or...
x = 0.147 is the smallest value and is the
x-coordinate of the first maximum point.
c The second value is given by 10x = 4.613.
So x = 0.461 at the first minimum point.
dy
= 2x tan 2x + x 2 sec2 2x × 2
dx
π
π2
π
2
2
× 3+
×4=
+ π 2 ; a = and b = 3
3
18
9
3 9
12 (10, 20) is on the curve so 20 = ae10k.
dy
= kaekx The gradient of the line from (10, 20) to
dx
25
= 2.5.
(0, −5) is
10
Therefore kae10k = 2.5 From these two equations,
20k = 2.5 and so k = 0.125.
Hence 20 = ae10 ×0.125 = ae1.25
a = 20 ÷ e1.25 = 5.73
dy
= 3 sin2 x × cos x
13 a y = sin3 x so
dx
= 3 cos x sin2 x = 3 cos x (1 − cos2 x)
= 3 cos x − 3 cos3 x
b
d 2y
− 3 sin x − 9 cos2 x × (−sin x)
=
dx 2
= −3 sin x + 9 sin x cos2 x
= −3 sin x + 9 sin x (1 − sin2 x)
= 5.
= −3 sin x + 9 sin x − 9sin3 x
= 6 sin x − 9 sin3 x
14 a If y = e
− 1 x2
2
then
−1 x2
e 2
−1 x2
dy
× (−x) = −xe 2 … = −xy which is the
=
dx
result required.
b When x = a, y = e
dy
=
dx
− 1 a2
2
and
− 1 a2
−ae 2 .
The equation of the tangent at P is
y−e
− 1 a2
2
= −ae
− 1 a2
2 (x
− a).
Where this crosses the x-axis, y = 0
and −e
− 1 a2
2
= −ae
− 1 a2
2 (x
− a) .
Hence 1 = a(x − a),
which you can rearrange to x = a +
15 a Differentiate implicitly: 2x − 2y − 2x
At a maximum point,
dy
= 0.
dx
1
a
dy
dy
+ 4y
= 0.
dx
dx
So 2x − 2y = 0 and y = x.
Substitute in the initial equation.
x2 − 2x2 + 2x2 = 9.
x2 = 9
x = 3 or −3.
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Differentiation
From the sketch it is clear that the maximum
point is where x = 3.
Then 9 − 6y + 2y2 = 9.
c At a stationary point,
dy
= 5e −2x cos ( x + 1.107 ) = 0 ;
dx
cos(x + 1.107) = 0
y2 − 3y = 0
y(y − 3) = 0
y = 0 or 3.
From the diagram, the maximum point must
be (3, 3).
b Where the graph crosses the x-axis, y = 0,
and from part a the point is (3, 0).
Substitute in 2x − 2y − 2x
Hence 6 − 6
dy
dy
+ 4y
= 0.
dx
dx
dy
dy
= 0 and
= 1.
dx
dx
The gradient of the normal is –1.
The equation of the normal is y = −1(x − 3) or
x + y = 3.
16 a Let u = x and v = x + 1 then du = dv = 1
dx dx
Hence, using the quotient rule,
dy ( x + 1) − x
1
=
=
dx
(x + 1)2
( x + 1 )2
b Differentiate
to x:
y
x
+
= xy 3 with respect
x +1 y +1
dy
dy
1
1
+
= y 3 + x × 3y 2
dx
(x + 1)2 ( y + 1)2 dx
dy
dy
1
1
+
= y 3 + 3xy 2
;
dx
(x + 1)2 ( y + 1)2 dx
dy
1 1 dy
+
=1+3
4 4 dx
dx
11 dy −3 dy −3
=
=
;
4 dx
4 dx 11
if x = y = 1 then
11
and the
c The gradient of the normal is−
3
11
equation is y − 1 = (x − 1) which is
3
3y – 3 = 11x – 11 or 11x – 3y = 8
17 a
y = e–2x sin x;
=
e–2x
dy
= −2e −2x sin x + e −2x cos x
dx
(cos x – 2 sin x)
If cos x – 2 sin x ≡ R cos(x + α) = R cos x cos α
– R sin x sin α
Hence R cos α = 1 and R sin α = 2;
2
2
R = 1 + 2 = 5 ; tan α = 2;
a = tan–1 2 = 1.107 to 3 d.p.
Hence
b Where the curve crosses the y-axis, x = 0 and
dy
= 5 cos1.107 = 1.00
dx
dy
= 5e −2x cos ( x + 1.107 )
dx
π
; x = 0.464 to 3 d.p.
2
and this is the closest to the y-axis.
A solution is x + 1.107 =
Then y = e–2x sin x = 0.177 to 3 d.p. The
coordinates are (0.464, 0.177)
dy
dx
18 a = −5sin t and
= 5cos t ; hence
dt
dt
dy
5cost
=
= − cot t
dx −5sin t
b At a stationary point, − cot t = −
So cos t = 0 and then t = ±
cost
=0
sin t
π
and sin t = ±1
2
Stationary points are when t =
π
at (1, 7) or
2
π
at (1, −3)
2
c At the point with parameter t the gradient of
the normal is tan t and the equation is
when t = −
y – 2 – 5 sin t = tan t (x – 1 – 5 cos t ); so
sin t
y − 2 − 5 sin t =
( x − 1 − 5cost )
cost
y cos t – 2 cos t – 5 sin t cos t
= x sin t – sin t – 5 sin t cos t
y cos t – 2 cos t = x sin t – sin t; or
(y – 2) cos t = (x – 1) sin t
d x = 1 and y = 2 satisfies this equation so (1, 2)
is on the normal.
19 aIf y = 2x then ln y = x ln 2; differentiate both sides
1 dy
= ln 2
with respect to x;
y dx
so dy = ( ln 2) y = ( ln 2) 2x
dx
2
b If y = a x then ln y = x2 ln a; differentiate
both sides with respect to x:
dy
1 dy
= 2x ( ln a ) so
= 2 ( ln a ) xy
y dx
dx
c If y = xx then ln y = x ln x; differentiate using
1 dy
1
= ln x + x × ;
the product rule:
y dx
x
dy
so 1
= ln x + 1 ;
y dx
dy
= ( ln x + 1) y = ( ln x + 1) x x
dx
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4
WORKED SOLUTIONS
20 a
x sec x =
and
dv
= − sin x
dx
22 ady = 2tan −1 x + 2x 2
dx
1+x
If f(x) = 1 + x2, then f(x) = 2x
dy cos x + x sin x
x sin x
1
=
=
+
and
cos x cos2 x
dx
cos2 x
x tan x
= sec x +
= sec x + x tan x sec x
cos x
= sec x (1 + xtan x)
b If u = x + x 2 + 1 then
du = 1 + 1 ×
1
× 2x = 1 +
2
dx
x2 + 1
(
)
y 2x tan −1 x
=
= 2tan −1x
x
x
f(x) + y = 2x + 2tan −1 x = dy
f(x) x 1 + x 2
dx
x
x2 + 1

dy
1
=
× 1+
dx x + x 2 + 1 


x + 1
x
2
1
x2 + 1 + x =
x2 + 1
x2 + 1
dy
dx
21 ax = t2 + 2 and y = 2t + 3 so
= 2t and
=2
dt
dt
dy
2 1
and
=
=
dx 2t t
1
1
1
If the gradient is − then = − and t = −2;
t
2
2
A is (6, −1)
1
×
x + x2 + 1
1
b A
t B, = 1 so t = 1 and B is (3, 5)
t
The tangent at A is y + 1 = −
dy π + 2
π
=
and y =
2
2
dx
−2
Gradient of normal =
π+2
Equation of normal: y − π = −2(x − 1)
2
π+2
x
dy
23 = e 2x
dx 1 + e
dy 1
π
At x = 0: y = and
=
4
dx 2
b At x = 1,
So if y = ln x + x 2 + 1 then
=
The base of the triangle is 6 and the area is
1
×6×2=6
2
x
du
; if u = x and v = cos x then
=1
cos x
dx
1
( x − 6 ) so
2
Equation of tangent at x = 0:
π x
x π
y − = or y = +
4 2
2 4
Mathematics in life and work
1
dy e x × (e x + 1) − e x × e x
ex
=
= x
x
2
dx
(e + 1)
(e + 1)2
2
1−y=1−
3
dy kekx × (ekx + 1) − kekx × ekx
kekx
=
= kx
2
kx
dx
(e + 1)
(e + 1)2
2y + 2 = –x + 6 or x + 2y = 4
The tangent at B is y – 5 = x – 3 or y = x + 2
Where these lines cross, x + 2 (x + 2) = 4;
x + 2x + 4 = 4; x = 0 and y = 2 so they meet on
the y-axis at (0, 2)
c
y
ex = ex + 1 − ex = 1
e +1
ex + 1
ex + 1
x
x
dy
e
× x1 = xe 2 =
Hence y(1 − y) = x
dx .
e + 1 e + 1 (e + 1)
=k×
x
1
ekx
×
= ky(1 − y)
e + 1 ekx + 1
kx
6
B
4
2
–2
0
2
4
6
x4
A
–2
The height of the triangle is 2.
x + 2y = 4 meets the x-axis at 4;
y = x + 2 meets the x-axis at −2.
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INTEGRATION
5 Integration
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge
1
a If f(x) = sin (x2 + 1) then by the chain rule
f ′(x) = cos(x 2 + 1) × 2x = 2x cos (x2 + 1)
= 11.3889.
The area of the ellipse ≈ 4 × 11.3889 = 45.56
to 4 s.f.
b y = uv where u = e–x and v = cos 2x
Then
du
dv
= −e −x and
= −2sin 2x.
dx
dx
dy
dv
du
=u
+v
dx
dx
dx
= e–x × (–2 sin 2x ) – e–x × cos 2x
= –e–x(2 sin 2x + cos 2 x).
Hence
4
a
∫ 4x
−1
2dx
3
1
= 4 × 2x 2 + c = = 8 x + c
2
b The integral is a multiple of (2x – 1)6.
If f(x) = (2x – 1)6 then
1
f ′(x) = 6 (2x − 1)5 × 2 = 12 (2x − 1)5.
Hence ∫ (2x − 1)5 dx =
c
3
1
(2x − 1)6 + c
12
0
4x 3 + 3
dx = ∫ 4x + 3x −2 dx = 2x2 – 3x–1 + c
x2
3
or 2x 2 − + c
x
∫
1
∫0 10e
−2x
x
2
Area =
−x 2
0.5
1
1.5
2
4 3.1152 1.4715 0.4216 0.0733
2
∫0 y dx = 4e
−x 2
dx
≈ 1 × 0.5 {4 + 2(3.1152 + 1.4715 + 0.4216) + 0.0733}
2
1
In this case h = 2.
× 0.5 {4 + 2(3.1152 + 1.4715 + 0.4216) + 0.0733}
4
2
1
∫−2 f(x)dx ≈ 2 × 2 {12.25 + 2(17.89 + 21.04) + 23.38}
= 3.5225
= 113.49
h = 0.1 and
0.6
∫0.2
1 − x 2 dx ≈
0.9798 + 2(0.9539

1
× 0.1 

2
+
0.9165
+
0.8660)
+
0.8


= 0.3626.
3
x
2
0
y = 4e
1
Exercise 5.1A
1
1
b The width of each strip is 2 ÷ 4 = 0.5.
dx =  −5e −2x  = [–5e–2] – [–5] =

0
5 – 5e–2 = 4.323 to 3 d.p.
The area is
b Every chord is below the curve and so
every trapezium is smaller than the area it
approximates.
a y
4
c If f(x) = ln (3x2 + 2) then using the chain rule,
6x
1
f ′(x) = 2
× 6x = 2
.
3x + 2
3x + 2
2
1
× 1 {3 + 2(2.9394 + 2.7495 + 2.4 + 1.8) + 0}
2
≈
a h = 1.
5
c Statement C is correct.
a
21
1
1
1
1
1
1
∫1 x dx ≈ 2 × 0.1 1 + 2 1.1 + 1.2 + 1.3 + ... + 1.9 + 2
= 0.6938 to 4 d.p.
{ (
) }
b It is an overestimate.
The area of a quarter of the ellipse
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5
WORKED SOLUTIONS
6
b
a
15
8
10
5
–1
0
1
2
3
4
5
–5
0
–10
4
c The area between 0 and 4 under y = 4 x is
(
)
b Area = ∫ 12x − 3x 2 dx = 6x 2 −
0
4
x 3 
0
≈
= [96 – 64] – [0] = 32
c Each chord will be below the curve so each
trapezium will have an area less than the area
between the curve and the x-axis.
Percentage error =
a h = 1; area ≈
{
(
) }
(
1 2
x is
2
1
0 + 2 ( 0.5 + 2 + 4.5) + 8} = 11
2{
The estimate of the area between them is
20.585 – 11 = 9.585
≈
4
41
0
0
d The exact area = ∫ 4 x dx −∫
32 − 30
× 100 = 6.25%
32
1
5+2
2
{
1
0 + 2 4 + 4 2 + 4 3 + 8 = 20.585
2
The corresponding area for y =
d Trapezium rule estimate is
1
× 1 × {0 + 2 ( 9 + 12 + 9) + 0} = 30
2
7
4
2
x 2 dx
4
4

2 3
1
= 4 × x 2  −  x 3 


3
6
0

0 
) }
24 + 21 + 4 + 3 + 0
2
64 64
1
2
−
= 21 − 10 = 10
3
3
6
3
3
a Using the product rule, f(x) = e–x – xe–x . At a
stationary point, e–x – xe–x = 0.
=
= 18.98
b y = 25 − x 2 → y2 = 25 – x2 → x2 + y2 = 25
This is the equation of a circle of radius 5.
9
e–x (1 – x) = 0; e–x is always positive
so 1 – x = 0 and x = 1
f (x) = –e–x – e–x + xe–x; f (1) = –e–1 which is
negative. The stationary point is a maximum
and the maximum value is f(1) = e–1
5
b
y
0.5
–5
0
5
The area is a quarter of the circle.
1
Area = 4 × π × 52 = 6.25π
8
1 2
x = 4 x ; so x 2 = 8 x ;
2
so x4 = 64x; so x (x3 – 64) = 0
a Where they meet,
–2
0
1
x
c h = 0.5; area
0.5
≈
0 + 2 0.5e −0.5 + e −1 + 1.5e −1.5 + 2e −2
2
{
(
)
}
= 0.5706 to 4 d.p.
Either x = 0 or x3 = 64 and x = 4
The graphs cross at (0, 0) and (4, 8)
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INTEGRATION
c If f(x) = ln (5x + 2) then
1
5
.
f ′(x) =
×5=
5x + 2
5x + 2
10 a
1
Hence
0.5
5
0
0.5
1
1.5
3
b h = 1 and ∫ 0.5x dx ≈
0
{
100
∫0
{
100
∫0
2.5
(
0.5x dx ≈
(
{
)
(
b
3
1
1 + 2 0.5 + 0.52 + 0.53
2
= 1.3125
c h = 1 and
2
100
Hence ∫
0
{
}
1
∫
1
Hence ∫ sin(2x − 3)dx = − 2 cos(2x − 3) + c.
a –2e–2x + c
b 2e0.5x + 2e–0.5x + c
3
c x 2 + e 4x + 5 + c
4
1
1
a − cos 2x − sin 2x + c
2
2
b
d
sin(0.1x + 1.3) = 0.1cos(0.1x + 1.3) so the
dx
integral is
4
sin(0.1x + 1.3) + c = 40sin(0.1x + 1.3) + c.
0.1
4
4
5
= − 5 cos5x − 4 sin 4x + c
c
∫ 4sin 5x − 5cos4x dx
a
∫ x dx = 5 ∫ x dx = 5 ln |x| + c
5
a
4
dx = [ 4 ln x ]1 = 4 ln a
x
Therefore 4 ln a = 10.
ln a = 2.5
a = e2.5 or 12.18
a
a 2 ln | x + 1| + c
1
2
d
b dx ln(2x − 1) = 2x − 1 × 2 = 2x − 1
8
1
2
2 1
2
dx = ∫ dx = ln x + c
b ∫
5x
5 x
5
∫
1
x2 − 6
3
1
dx =∫ x − dx = x 2 − 3 ln x + c
2x
2
x
4
Area =
4
4 0.5x − 2
∫2 e
(
1 2x − 3
e
+ c.
2
c If f(x) = cos(2x – 3) then f ′(x) = − sin(2x − 3) × 2
= –2sin (2x – 3).
x
∫1
c
1
dx = 2x − 3 + c.
2x − 3
Hence ∫ e 2x − 3 dx =
π
3
Therefore integral = 2 ln | 2x − 1| + c.
1
.
2x − 3
b If f(x) = e2x – 3 then f ′(x) = e 2x − 3 × 2 = 2e 2x − 3.
3
6
a If f(x) = (2x − 3)2 then
Hence
}
0
7
−1
1
f ′(x) = (2x − 3) 2 × 2 =
2
2
)
1
1 + 2 × 1 + 0.5100 = 1.5
2
Exercise 5.2A
1
π
2
}
)
0.5x dx ≈
2cos0.5x dx = [ 4sin 0.5x ]0 = [ 4 ] − [ 0 ] = 4
1
1
1 + 2 0.5 + 0.52 + 0.53 + ... + 0.599 + 0.5100
2
)
1
y
}
1
0.5x dx ≈ 1 + 2 0.5 + 0.52 + 0.53 + ... + 0.599 + 0.5100
2
Now 0.5 + 0.52 + 0.53 + ... + 0.599 is the sum of a
geometric progression with a = 0.5, r = 0.5 and
0.5 1 − 0.599
n = 99; S99
=1
1 − 0.5
(
π
∫0
a
1
∫ 5x + 2 dx = 5 ln 5x + 2 + c.
dx =  2e0.5x − 2  =  2  −  2e −1 
2
)
= 2 1 − e −1 = 1.264.
d
sin(x 2) = cos(x 2) × 2x = 2x cos(x 2)
dx
1
b
sin(x 2) + c
2
d
c
cos(x 2) = − sin(x 2) × 2x = −2x sin(x 2)
dx
1
Hence ∫ x sin(x 2)dx = − cos(x 2) + c.
2
10 Both are correct as they have different constants.
ln 2x = ln 2 + ln x so Bhaskar’s answer is
1
1
ln 2 + ln x + c which differs from Alice’s only by
2
2
1
the constant ln 2.
2
11 a Using the product rule,
f ′(x) = e x(sin x + cos x) + e x(cos x − sin x)
9
a
= 2ex cos x.
b
∫e
x
cos x dx =
1 x
e (sin x + cos x) + c
2
c Try g(x) = ex (sin x – cos x).
58
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5
WORKED SOLUTIONS
x
1+ x −1
=
= 1+ x − 1
1+ x
1+ x
1+ x
1+ x
1
= 1+ x −
;
1+ x
Then g′(x) = e x(sin x − cos x) − e x(cos x − sin x)
= 2ex sin x.
ii 1 x
Hence ∫ e x sin x dx = 2 e (sin x − cos x ) + c.
x
dx =
Hence ⌠

⌡ 1+ x
12 aUsing the product rule with u = x and v = ln x;
du
dv 1
= 1 and
=
dx
dx x
Hence f ′ ( x ) = ln x + x ×
1
= ln x + 1
x
3
= 2 (1 + x)2 − 2 1 + x + c
3
1
x ln x + c = ∫ ln x dx + x ; rearrange as
a tan x + c
b
∫ ln x dx = x ln x − x + c
c ∫ ln x dx = [ x ln x − x ]1 = [ a ln a − a ] − [ 0 − 1]
a
= a ln a – a + 1
If a ln a – a + 1 = 1 then a ln a = a; hence ln a = 1
and a = e
)
2
3
2
13 aLet f(x) = x 2 + 1 ; using the chain rule,
f ′(x ) =
(
3 2
x +1
2
)
1
2
(
× 2x = 3x x 2 + 1
)
3
(
x
2
x +1
f ′(x ) =
(
= x x2 + 1
(
2 2
x +1
3
)
−1
3
)
−1
3
(
)
)
3
2
× 2x =
(
4
x x2 + 1
3
)
)
2
=
3
2
+c
4
1
= (x − 1)−2 then
(x − 1)2
f(x) = –2(x – 1)–3 × 1 = –2(x – 1)–3
⌠ 1
1
Hence 
3 dx = −
2 +c
⌡ (x − 1)
2 ( x − 1)
x
x −1+1
ii =
(x − 1)3
(x − 1)3
x −1
1
1
1
+
=
+
(x − 1)3 (x − 1)3 (x − 1)2 (x − 1)3
x
1
⌠ 1 dx +⌠
Hence ⌠
dx =
dx


⌡ (x − 1)3
⌡ (x − 1)2
⌡ (x − 1)3
1
1
−
1
b i ⌠
dx =∫ (1 + x) 2 = 2(1 + x)2 + c

⌡ 1+ x
=2 1+ x +c
)dx = 1 tan(x 2) + c.
2
2
−(sin 2 x + cos2 x)
1
=−
= −cosec2x.
sin 2 x
sin 2 x
b From part a it follows that
2
∫ cosec x dx = − cot x + c.
14 a i If f ( x ) =
1
=− 1 −
+c
x − 1 2 ( x − 1)2
2
cos x
then using the quotient rule,
sin x
sin x × (− sin x) − cos x × cos x
f ′(x) =
sin 2 x
+c
−1
3
(
∫ x sec (x
aIf f(x) =
so try f ( x ) = x 2 + 1 3 ;
⌠
x
3
3
Hence  3
dx = f(x) = x 2 + 1
4
4
⌡ x2 + 1
=
d
tan(2x − 1) = 2sec2(2x − 1) so the integral
dx
1
is tan(2x − 1) + c.
2
aUsing the chain rule, the derivative of tan (x2) is
sec2 (x2) × 2x = 2x sec2 (x2)
b Hence
1
2
1
1 2
2
Hence ⌠
 x x + 1 dx = 3 f ( x ) + c = 3 x + 1
⌡
b3
d
tan 2x = 2sec2(2x) so the integral is
dx
1
tan 2x + c.
2
c
1
(
1
1 + x dx −⌠
dx

⌡ 1+ x
Exercise 5.3A
b
∫f ′(x)dx = ∫(ln x + 1)dx; hence
a
∫
acos 2x = 2 cos2 x – 1 so 2 cos2 x = 1 + cos 2x and
1 1
therefore cos2 x = + cos 2x
2 2
1 1
2
Integrate: ∫ cos x dx = ∫ + cos 2x dx
2 2
1
1
1
1
= x + × sin 2x + c = x + 1 sin 2x + c
2
2 2
2
4
b Similarly to a, from the double angle formula
1 1
cos2 2x = + cos4x
2 2
1 1
2
so ∫ cos 2 x dx =∫ + cos4x dx
2 2
1 1
1
1
1
= x + × sin 4x + c = x + sin 4x + c
2 4
2
8
2
5
a
1
d
ln cos x dx =
× (− sin x) = − tan x
cos x
dx
b
∫ tan x dx = − ln cos x
c
1
d
ln sin x dx =
× cos x = cot x ;
sin x
dx
+c
therefore ∫ cot x dx = ln sin x + c.
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INTEGRATION
6
b If f(x) = cot x then f (x) = –cosec2 x (this was
shown in question 3)
Try f(x) = cot(3x + 2) then
f (x) = – cosec2 (3x + 2) × 3 = – 3 cosec2 (3x + 2)
Hence ∫ cosec2(3x + 2)dx = − 1 cot(3x + 2)
3
a sin 2x = 2 sin x cos x hence
π
⌠2
π
∫02 sin x cos x dx =⌡
0
1
1
1
=   −  −  = or
 4   4  2
π
2
sin x cos x dx
0
∫
π
2
1
1
sin 2x dx =  − cos2x
 4
2
0
π
2
=  1 sin 2 x  =  1 × 12  −  1 × 0 2  = 1
 2
0  2
  2
 2
π
b sin 4x = 2 sin 2x cos 2x hence ∫ 8 sin 2x cos2x dx
π
8
0
=∫
Exercise 5.4A
1
0
1
sin 4x dx
2
 2 50 dx = 50∫ 2 1 2 dx
b ⌠
⌡ x + 100
x + 10
π
x
1
tan −1
+ c = 5tan −1 x + c
10
10
10
6
1
c ∫ 2
dx = 6∫
2 dx
x +6
x2 + 6
8
1
1
π
1
=  − cos4x  =  − cos  −  − cos0 
2   8

 8
0  8
=0+
7
= 50 ×
1 1
=
8 8
1 1
a cos 10x = 2
– 1 so cos 5x = + cos10x
2 2
1
1
2
and ∫ cos 5x dx = x +
sin10x + c.
2
20
1 1
b cos x = 1 – 2 sin2 0.5x so sin 2 0.5x = − cos x
2 2
1
1
2
and ∫ sin 0.5x dx = 2 x − 2 sin x + c.
( )
2
cos2 5x
The area is
π
∫0 2 cos
2
2
10
1
a ⌠
dx = 10⌠
dx
 2
 2
⌡ x + 100
⌡ x + 10 2
= 10 ×
1
x
x
tan −1
+ c = tan −1
+c
10
10
10
⌠
10
10
⌠
b 
dx = 
dx
2

1
⌡ 100x + 1
2
⌡ 100 x + 100
(
= 0.1 ×
0.5x dx.
π
∫0 1 + cos x dx = [ x + sin x ]
π
0
= [π + 0]–
1
x
tan −1
+c
0.1
0.1
= tan −1 10x + c
3
a
y
[0 + 0] = π.
9
)
1
dx
= 0.1⌠
 2
⌡ x + 0.12
Now cos x = 2 cos2 0.5x – 1 so 2cos2 0.5x = 1 + cos x.
The area is
1
x
+ c = 6 tan −1 x + c
tan −1
6
6
6
= 6×
c cos 2ax = 1 – 2 sin2 ax so
1 1
sin 2 ax = − cos 2ax
2 2
1
1
2
and ∫ sin ax dx = x −
sin 2ax + c.
2
4a
8
⌠ 1
⌠ 1 dx = 1 tan −1 x + c
dx =
a 
5
5
⌡ x 2 + 25
⌡ x 2 + 52
4
cos x cos 3x – sin x sin 3x = cos (x + 3x) = cos 4x
π
4
cos4x
0
∫
π
2
4
1
dx =  sin 4x 
 4
0
1
1
=  sin π  −  sin 0  = 0 − 0 = 0
 4
  4

π
Hence ∫ 4 ( cos x cos3x − sin x sin 3x ) dx = 0 ;
b Area =
0
Hence
π
π
π
π
0
Hence ∫ sec2(2x + 1)dx =
1
tan(2x + 1) + c
2
16
4
x
1
∫−4 x 2 + 4 dx = 16∫−4 x 2 + 22 dx =
16 × 1 tan −1 x 
2
2  −4

= [8 tan–1 2] – [8 tan–1 (–2)]
= [8.8572] – [–8.8572] = 17.71 to 4 s.f.
Hence ∫ 4 cos x cos3x dx =∫ 4 sin x sin 3x dx
10 a The derivative of tan x is sec2 x so try f(x) =
tan (2x + 1); then f’(x) = sec2 (2x + 1) × 2 =
2 sec2 (2x + 1)
4
5
4
∫04 cos x cos3x dx − ∫04 sin x sin 3x dx = 0;
0
0
–5
4
a
a
a dx 
1
x
⌠
Area = 
= a × tan −1 
⌡0 x 2 + a 2

a
a 0
a
x
= tan −1  = [tan–1 1] – [tan–1 0] = π .
a 0
4

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5
WORKED SOLUTIONS
5
1
−1 x
1
⌠
a
=
 21 dx = ⌠
dx = 5 5 tan
1
⌡ 5x + 1
2
1

5 x + 5
5
⌡
1
−1
tan
5x + c
5
(
b
)
10 a
2.5
–5
12
12
∫ 5x 2 + 4 dx = ⌠ 5 x 2 + 4 dx =

5
⌡
(
)
2
−2
= 5tan −1 1 − 5tan −1(−1)
= 5 × π + 5 × π = 5π
4
4
2
= 6 tan −1 5x + c
2
5
1
∞
1
dx =
q

2
p x + 
p

1 p
−1 p
x+c =
= p q tan
2 dx
q
∞
10
x
c
Area = ⌠
dx = 5tan −1  =
 2

2 0
⌡0 x + 22
5 × π  − [ 0 ] = 5π
2 

2
∫ px 2 + q dx = ∫
1
p∫
1
 q
x2 + 
 p 
Exercise 5.5A
1
p
1
tan −1
x+c
q
pq
6
∫
b
∫ x 2 + 1 dx = ∫
x2
8
Hence
x4 + 1 = (x2)2 + 1 so try f(x) = tan–1 (x2).
2x
1
Then f ′(x) = 4
× 2x = 4
.
x +1
x +1
x
1
Hence ∫ 4
dx = tan −1(x 2) + c.
2
x +1
= 2tan −1
x
+c
5
(
Hence
2
)
=⌠

⌡
Hence ∫
⌠
x + a − 2a
2a 
dx =   1 − 2
dx

x
x 2 + a2
+ a 2 
⌡
2
2
= x − 2a 2 ×
2
2
1
3x 2
× 3x 2 = 3
.
x −3
x −3
3
x2
∫ x 3 − 3 dx = 13 ln x 3 − 3 + c.
x +1
∫ x 2 + 2x + 5 dx =
1
ln x 2 + 2x + 5 + c.
2
b If f(x) = ln |2x3 – 3x2 + 12| then
2
1
f ′(x) = 3
× (6x 2 − 6x)= 36(x −2 x) .
2
2x − 3x + 12
2x − 3x + 12
x
+c
5
(x + a)(x − a)
x 2 − a2
⌠
dx = ⌠
dx

 2
2
2
⌡
x +a
⌡ x + a2
x
aIf f(x) = ln |x2 + 2x + 5| then
1
2(x + 1)
f ′(x) = 2
.
× ( 2x + 2) = 2
x + 2x + 5
x + 2x + 5
Hence
250 
1
x

=⌠
dx = 10x − 250 × tan −1 + c
  10 − 2
5
5
x + 25 
⌡
9
1
× 2x =
x2 + 1
∫ x 2 + 1 dx = 12 ln ( x 2 + 1) + c.
c If f(x) = ln |x3 – 3| then f ′(x) =
2
⌠ 10x 2 dx = ⌠ 10 x + 25 − 250 dx
b
 2

x 2 + 25
⌡ x + 25
⌡
= 10x − 50tan −1
1
∫ x + 1 dx = ln |x + 1| + c.
2x
.
x2 + 1
1
x2 + 1 − 1
dx =
dx = ∫ 1 − 2
x +1
x2 + 1
10
1
1
x
⌠
⌠
a
dx = 10 ×
dx = 10 × tan −1
 2
5
5
⌡ x + 25
⌡ x 2 + 25
1
x +1.
b If f(x) = ln (x2 + 1) then f ′(x) =
x – tan–1 x + c
7
a If f(x) = ln |x + 1| then f ′(x) =
Hence
x2 + 1
1
dx = ∫ 1 + 2 dx = x − 1 + c
x
x2
x
a
5
2
10
b
Area = ⌠
dx = 10 × 1 tan −1 x 
 2
2

2
2  −2
⌡x +2
12
5
5x
12 ⌠
1
×
tan −1
+c
dx =
5  2  2 2
5
2
2
x
+



5
⌡
c
0
3
a
∫
1
x2 − x
dx = ln 2x 3 − 3x 2 + 12 + c.
6
2x − 3x 2 + 12
3
1
ex + 1
d x = ∫ 1 + x d x = ∫ 1 + e −x d x
e
ex
= x – e–x + c
1
x
x
tan −1 + c = x − 2a tan −1 + c
a
a
a
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Integration
b To find
ex
Then f ′(x) =
Hence
4
5
Area =
1
ex + 1 − ex
ex
cx
=
=1− x
; hence
x
e +1
e +1
e +1
∫ e x + 4 dx try f(x) = ln (ex + 4).
1
ex
1
× ex = x
.
e +4
e +4
1
⌠ 
1
ex 
⌠
dx =   1 − x
dx
 x
⌡0 e + 1
e + 1 
⌡0 
x
ex
∫ e x + 4 dx = ln (ex + 4) + c.
π
4
tan x dx
0
∫
=
π
4
0
∫
sin x
dx
cos x
= 1 − ln
π

1 
= [ − ln cos x ]04 =  − ln
− [ − ln1]
2 

= ln 2 or 0.347 to 3 s.f.
sin 2x
a tan 2x =
cos 2x
9
sec x × sec x
sec2 x
sec x
=
=
b
sin x − cos x sec x (sin x − cos x ) tan x − 1
1
cos x
If f(x) = tan x – 1 then f(x) = sec2x
sec x

dx = ln |tan x – 1| + c
Hence ⌠
⌡ sin x − cos x
since sec x =
cos0.5x
sin 0.5x
Hence ∫ cot 0.5x dx = 2 ln sin 0.5x + c.
6
10 af(x) = (cos x )–1; hence using the chain rule
f(x)= –1 × (cos x)–2 × (–sin x)
1
× 2x
If f(x) = ln (x2 + a2) then f ′(x) = 2
x + a2
2x
.
= 2
x + a2
Hence
x
1
∫0 x 2 + a 2 dx =  2 ln(x
a
2
=
+ a 2) =
0
if f(x) = 2 sec x + 1 then f(x) = 2 sec x tan x
Hence ∫
 1 ln 2a 2  −  1 ln a 2  = 1 ln 2a = 1 ln 2 = ln 2.
2
a2
  2
 2
 2
If f(x) = ln |ln x| then f ′(x) =
Hence
1
1
1
× =
.
ln x x x ln x
1
∫ x ln x dx = ln ( ln x ) + c.
Since x > 1, ln x > 0 and the modulus sign is
unnecessary.
1
8
(
=
1
a
If x = –5 then –5B = –15 and B = 3.
So
ex;
1
b
1
ex
dx =  ln e x + 1 
hence ⌠
 x
0
⌡0 e + 1
= [ ln e + 1 ] − [ ln 1 + 1 ] = ln (e + 1) – ln 2
( )
A
B
A(x + 5) + Bx
x − 10
≡ +
=
x(x + 5) x x + 5
x(x + 5)
If x = 0 then 5A = –10 and A = –2.
= 1 − e −1  − [ 0 − 1] = 2 − e −1
= ln e + 1
2
1
ln 2sec x + 1 + c
2
So A(x + 5) + Bx ≡ x – 10.
)
b
If f ( x ) = e + 1 then f’(x) =
tan x
sec x tan x
dx = ∫
dx
2 + cos x
2sec x + 1
Exercise 5.6A
⌠ e x + 1 dx = 1 1 + e −x dx =  x − e −x 1
a

∫0

0
⌡0 e x
x
sin x
1
sin x
=
×
= sec x tan x
cos2 x cos x cos x
tan x
sec x tan x
sec x tan x
=
=
b
;
2 + cos x sec x ( 2 + cos x ) 2sec x + 1
a
2
7
cos2x
cos2x
a
=
; if f(x) = 0.5 sin 2x
sin x cos x + 4
0.5sin 2x + 4
+ 4 then f (x) = cos 2x
cos2x
dx = ln|0.5 sin 2x + 4| + c
sin x cos x + 4
or ln (0.5 sin 2x + 4) + c since the expression in
brackets is always positive.
1
Hence ∫ tan 2x dx = − ln cos 2x + c.
2
If f(x) = ln |sin 0.5x| then
1
f ′(x) =
× 0.5cos0.5x ) = 0.5 cot 0.5x.
sin 0.5x (
( e 2+ 1 )
Hence ∫
If f(x) = ln |cos 2x | then
1
f ′(x) =
× (−2sin 2x) = –2 tan 2x.
cos 2x
b cot 0.5x =
( )
1
1
1
ex
e+1
dx = [ x ]0 − ln
= ∫ 1 dx −⌠

x
2
0
⌡0 e + 1
x − 10
3
2
=
− and
x(x + 5) x + 5 x
x − 10
3
2
∫ x(x + 5) dx =∫ x + 5 dx −∫ x dx
= 3 ln |x + 5| – 2 ln |x| + c or ln
2
a Write
(x + 5)3
+ c.
x2
4
A
B
A(x + 4) + Bx
.
≡ +
=
x(x + 4) x x + 4
x(x + 4)
Then A(x + 4) + Bx ≡ 4.
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5
WORKED SOLUTIONS
If x = 0 then 4A = 4 and A = 1.
If x = 2 then 4A = 4 and A = 1.
If x = –4 then –4B = 4 and B = –1.
If x = –2 then –4B = 4 and B = –1.
So
4
1
1
and
≡ −
x(x + 4) x x + 4
1
4
1
1
=
−
and
x2 − 4 x − 2 x + 2
4
1
1
∫ x 2 − 4 dx = ∫ x − 2 dx − ∫ x + 2 dx
4
∫ x(x + 4) dx =
Then
1
∫ x dx − ∫ x + 4
= ln |x| – ln|x + 4| + c or ln
x
+ c.
x+4
= ln |x – 2| – ln |x + 2| + c = ln
1
A
B
b Write (x − 2)(x − 4) ≡ x − 2 + x − 4 .
b If f(x) = ln(x2 + 4)
then f ′(x) = 21 × 2x = 22x
x +4
x +4
4
2
x
Hence ∫ 2
dx = 2∫ 2 x dx
x +4
x +4
Then A(x – 4) + B(x – 2) = 1.
If x = 2 then –2A = 1 and A = −
1
2.
(
−
c
1
∫ (x − 2)(x − 4) dx =
1
1
1
1
dx + ∫
dx
2∫x −2
2 x−4
=x−∫
1
2
1
x−4
x−4
ln
+ c or ln
x−2
2
x−2
4
5
If x = 3 then 6A = 6 and A = 1.
If x = –3 then –6B = 6 and B = –1.
1
1 and
6
So
≡
−
(x − 3)(x + 3) x − 3 x + 3
1
1
∫ x 2 − 9 dx = ∫ x − 3 dx − ∫ x + 3 dx
= ln |x – 3| – ln |x + 3| + c = ln
3
Let
x−3
+ c.
x+3
3x + 4
A
B
A(x + 1) + Bx
.
≡ +
=
x(x + 1) x x + 1
x(x + 1)
Then A(x + 1) + Bx = 3x + 4.
If x = 0 then A = 4.
If x = –1 then –B = –3 + 4 and B = –1.
3x + 4
4
1
and
≡ −
x(x + 1) x x + 1
3x + 4
4
1
∫ x(x + 1) dx =∫ x dx −∫ x + 1 dx
= 4 ln |x| – ln |x + 1| + c = ln
4
a Write
x4
+c
x +1
4
4
A
B
=
≡
+
.
x 2 − 4 (x − 2)(x + 2) x − 2 x + 2
Hence A(x + 2) + B(x – 2) = 4.
2
+c
4
x2
d x =∫ 1 − 2
dx
x +4
x +4
2
4
dx.
x2 + 4
1
x
1
= 4 × tan −1 + c
2
2
2
x
so ∫ 2
dx = x − 2tan −1 x + c
2
x +4
+c
Then A(x + 3) + B(x – 3) = 6.
6
)
∫ x 2 + 4 d x = 4∫ x 2 + 22 d x
6
6
A
B
=
≡
+
x 2 − 9 (x − 3)(x + 3) x − 3 x + 3
c
(
x2
(x 2 + 4) − 4
4
=
=1− 2
x2 + 4
x2 + 4
x +4
Hence ∫
1
1
= − ln x − 2 + ln x − 4 + c
2
2
=
)
= 2 ln x 2 + 4 + c = ln x 2 + 4
1
.
2
1
1
1
1
1
≡− ×
+ ×
(x − 2)(x − 4)
2 x−2 2 x−4
If x = 4 then 2B = 1 and B =
and
x−2
+ c.
x+2
6
3x − 6
A
B
≡
+
(x + 2)(x − 1) x + 2 x − 1
A(x − 1) + B(x + 2)
=
(x + 2)(x − 1)
Equate the numerators.
3x – 6 = A(x – 1) + B(x + 2)
Let x = –2: –12 = –3A so A = 4.
Let x = 1: –3 = 3B so B = –1.
3x − 6
4
1
So
≡
−
(x + 2)(x − 1) x + 2 x − 1
3x − 6
4
1
and ∫ (x + 2)(x − 1) dx = ∫ x + 2 − x − 1 dx
= 4 ln |x + 2| – ln |x – 1| + c
(x + 2)4
+ c.
or ln |x + 2|4 – ln |x –1| + c or ln
x −1
Write
4x + 10
A
B
≡
+
(2x − 3)(2x + 1) 2x − 3 2x + 1
A(2x + 1) + B(2x − 3)
.
=
(2x − 3)(2x + 1)
Then A(2x + 1) + B(2x – 3) = 4x + 10.
3
If x = then 4A = 16 and A = 4.
2
1
If x = − then –4B = 8 and B = –2.
2
4
2
4x + 10
.
≡
−
Hence
(2x − 3)(2x + 1) 2x − 3 2x + 1
Write
63
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Integration
Then
4x + 10
4
= 2 ln |2x – 3| – ln |2x + 1| + c = ln
7
(2x − 3)
+ c.
2x + 1
A(x − 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x − 2)
.
(x + 1)(x − 2)(x + 3)
Then A(x – 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x – 2)
= 2x2 + 9x – 11.
If x = –1 then –6A = 2 – 9 – 11 = –18 and A = 3.
If x = 2 then 15B = 8 + 18 – 11 = 15 and B = 1.
If x = –3 then 10C = 18 – 27 – 11 = –20 and C = –2.
2x 2 + 9x − 11
3
1
2
So
≡
+
−
(x + 1)(x − 2)(x + 3) x + 1 x − 2 x + 3
and
2
2x + 9x − 11
∫ (x + 1)(x − 2)(x + 3) dx = 3 ln x + 1
(x + 1)3(x − 2)
+ ln x − 2 − 2 ln x + 3 + c = ln
+ c.
(x + 3)2
8
a Write
(x + 1)(x + 3) A Bx + C
≡ + 2
x
x(x 2 + 1)
x +1
=
Then
A(x 2 + 1) + x(Bx + C )
.
x(x 2 + 1)
A(x2
+ 1) + x(Bx + C) = (x + 1)(x + 3).
If x = 0 then A = 3.
So
3(x2
+ 1) + x(Bx + C) = (x + 1)(x + 3).
If x = 1 then 6 + B + C = 8 and hence B + C = 2.
If x = –1 then 6 – (–B + C) = 0 and hence –B + C = 6.
Add the last two equations: 2C = 8 so C = 4
and B = –2.
Hence
b
∫
(x + 1)(x + 3) 3 4 − 2x
≡ + 2
.
x x +1
x(x 2 + 1)
(x + 1)(x + 3)
3
4
2x
dx = ∫ dx + ∫ 2
dx − ∫ 2
dx
x
x(x 2 + 1)
x +1
x +1
= 3 ln |x| + 4 tan–1 x – ln (x2 + 1) + c
= ln
9
x3
x2 + 1
a
Write
Hence
2
2x 2 + 9x − 11
A
B
C
Write
≡
+
+
(x + 1)(x − 2)(x + 3) x + 1 x − 2 x + 3
=
If x = 0 then 2 – 2B + 4 = 8 and B = –1.
2
∫ (2x − 3)(2x + 1) dx =∫ 2x − 3 dx −∫ 2x + 1 dx
+ 4tan −1 x + c
x 2 − 3x + 8 ≡ A + B + C
(x + 2)(x − 1)2 x + 2 x − 1 (x − 1)2
2
) + C(x + 2).
= A(x − 1) + B(x + 2)(x − 1
(x + 2)(x − 1)2
Hence A(x – 1)2 + B(x + 2)(x – 1) + C(x + 2)
= x2 – 3x + 8.
If x = –2 then 9A = 4 + 6 + 8 = 18 and A = 2.
If x = 1 then 3C = 1 – 3 + 8 = 6 and C = 2.
So 2(x – 1)2 + B(x + 2)(x – 1) + 2(x + 2)
= x2 – 3x + 8.
b
x 2 − 3x + 8
2
1
2
≡
−
+
.
(x + 2)(x − 1)2 x + 2 x − 1 (x − 1)2
x 2 − 3x + 8
∫ (x + 2)(x − 1)2 dx
=
2
1
2
∫ x + 2 dx − ∫ x − 1 dx + ∫ (x − 1)2 dx
2
= 2 ln x + 2 − ln x − 1 − x − 1 + c
= ln
(x + 2)2
2
−
+c
x −1
x −1
10 a If f(x) = x2 – 2x – 8 then f(x) = 2x – 2
x −1
1
Hence ⌠
dx = ln x 2 − 2x − 8 + c
 2
2
⌡ x − 2x − 8
x − 10
x − 10
b 2
=
; suppose
x − 2x − 8 ( x + 2)( x − 4 )
x − 10
≡
A
( x + 2)( x − 4 ) x + 2
= A ( x − 4 ) + B(x + 2)
+
B
x−4
(x + 2)(x − 4)
If x = −2, then −12 = −6 A and A = 2; If x = 4 ,
then –6 = 6B and B = –1
2
x − 10
1
≡
−
;
x 2 − 2x − 8 x + 2 x − 4
x − 10
2
1
⌠
dx
dx =⌠
−
 2

⌡ x − 2x − 8
⌡ x+2 x−4
Hence
(
)
= 2 ln x + 2 − ln x − 4 + c
= ln
(x + 2)2
+c
x−4
x2 + 3 x2 + 4 − 1
1
11 a 2
=
=1− 2
; hence
x +4
x2 + 4
x +4
⌠ x 2 + 3 dx = ⌠  1 − 1  dx
 2

x2 + 4
⌡x +4
⌡
1
−1 x
= x − 2 tan 2 + c
x2 − 3 x2 − 4 + 1
x2 − 3
1
b 2
=
=1+ 2
so ⌠
dx
 2
2
x −4
x −4
x −4
⌡x −4
1 
1

=⌠
dx = x + ⌠
dx
 2
 1 + 2
⌡x −4
x − 4
⌡
1
Write 2
in partial fractions:
x −4
1
1
A
B
=
≡
+
x 2 − 4 (x − 2)(x + 2) x − 2 x + 2
A ( x + 2) + B(x − 2)
x2 − 4
1 ≡ A (x + 2) + B(x – 2); let x = 2 then 1= 4A and
1
A= ;
2
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5
WORKED SOLUTIONS
let x = −2 then 1 = –4B and B = − 1 ; hence
4
1
1⌠ 1
1
⌠
dx
dx = 
−
 2
4⌡ x − 2 x + 2
⌡x −4
(
=
)
1
1
x−2
ln x − 2 − ln x + 2 } + c = ln
+c
4
x+2
4{
x2 − 3
1
x−2
Hence ⌠
dx = x + ln
+c
 2
4
x+2
⌡x −4
12
3
4
x2
x2 − 1 + 1
1
=
=1+ 2
x −1
x2 − 1
x −1
2
If
1
1
B
A
=
≡
+
;
x 2 − 1 (x − 1)(x + 1) x − 1 x + 1
π
b
Using the answer to a, ∫ 2 x sin x 2 dx
so 1 ≡ A(x + 1) + B(x – 1)
0
If x = 1 then 1 = 2A and A =
5
{
π
22
 1
π2   1 
1
=  − 2 cos 4  −  − 2 
=  − cos x
 2
0 

1
;
2
if x = – 1 then 1 = – 2B and B = −
5
1
2
1
1
⌠ x 2 dx = ⌠
 1 + 2(x − 1) − 2(x + 1)
 2
⌡
2
⌡2 x − 1
}
5
5
1
x −1 
1 2
1 1
=  x + ln
= 5 + ln  −  2 + ln 
2 x + 1  2 
2 3  
2 3 

c As in part b, let u = x + 2 then x = u – 2 and
2
1
1 3
ln = 3 + ln 2
2
2 1
3
Therefore
x2
∫ x + 2 dx = ∫
Exercise 5.7A
dx
= 1 so the integral =
du
1 5 1 4
3
4
3
∫ (u − 2)u du =∫ u − 2u du = 5 u − 2 u + c
a u = x2 + 4 and
Then ∫
=
∫2
x
2
x +4
dx = ∫
1
(x + 2)2 − 4(x + 2) + 4 ln x + 2 + c
2
du
dx
1
a Let u = x2 + 3 then
= 2x and
=
dx
du 2x
6
1
x 2 + 3 dx = ∫ xu 2 ×
=
x
1
dx
du = ∫
du
×
2
x
d
u
u
x +4
x
2
1 d = 1 − 12 d = 12 + =
u ∫ u
u u
c
2
u
(u − 2)2 du = u 2 − 4u + 4 du
∫ u
u
=
∫x
du
dx
1
.
= 2x so
=
dx
du 2x
x 2 + 4 + c.
x
du
dx
dx =
b u = x + 4 so
=1=
;
dx
du ∫ x + 4
1
1
u − 4 dx
∫ u du du = ∫ u 2 − 4u − 2 du
1
3
1
2 3
2
= u 2 − 8u 2 + c = (x + 4)2 − 8(x + 4)2 + c
3
3
dx
= 1.
du
= ∫ u − 4 + 4 du = 1 u 2 − 4u + 4 ln u + c
u
2
x = u – 2 and
1
1
= (x + 2)5 − (x + 2)4 + c.
2
5
2
dx
= 1.
du
Therefore
x
u−2
2
∫ x + 2 dx = ∫ u du = ∫ 1 − u du
= u − 2 ln u + c = x + 2 − 2 ln x + 2 + c
5
1
= 0.891 to 3 s.f.
a ln |x + 2| + c
b Let u = x + 2 then x = u – 2 and
1
1
dx =  x + ln x − 1 − ln x + 1 
 2

2
2
=3+
du
= 6x 2 and dx = 1 2 .
dx
du 6x
4
dx
1 4
2
3
2 4 1
∫ x (2x + 5) du du = ∫ x u 6x 2 du = 6 ∫ u du
= 1 × 1 u 5 + c = 1 (2x 3 + 5)5 + c
6 5
30
du
2
a Let u = x then
= 2x
dx
dx
1
;
=
du 2x
1
1
2 dx
∫ x sin x du du = ∫ x sin u 2x du = ∫ 2 sin u du
1
1
= − cos u + c = − cos x 2 + c
2
2
Let u = 2x3 + 5 then
1
1 1
du = ∫ u 2 du
2x
2
3
1 2 32
1
× u + c = (x 2 + 3)2 + c
2 3
3
b Let u = x + 3 then x = u – 3 and dx = 1.
du
∫x
1
x + 3 dx = ∫ (u − 3)u 2 du
3
1
5
3
= ∫ u 2 − 3u 2 du = 2 u 2 − 3 × 2 u 2 + c
5
3
=
5
3
2
(x + 3)2 − 2(x + 3)2 + c
5
c Let u = x + 3 then x = u – 3 and
dx
= 1.
du
1
Then ∫ x 2 x + 3 dx = ∫ (u − 3)2u 2 du
1
5
3
1
= ∫ (u 2 − 6u + 9) u 2du = ∫ u 2 − 6u 2 + 9u 2 du.
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Integration
7
5
3
10 Where the graph meets the x-axis, x x + 4 = 0 so
x = 0 or −4
0
The integral required is ∫ x x + 4 dx
= 2u2 − 6 × 2u2 + 9 × 2u2 + c
7
5
3
5
7
3
= 2 (x + 3)2 − 12 (x + 3)2 + 6(x + 3)2 + c
7
5
7
a If u = cos x then
−4
du
dx
1
= − sin x and
=−
.
sin x
dx
du
sin x
sin x
dx
∫ tan x dx = ∫ cos x × du du =∫ cos x ×
(− sin1 x )du
1
= −∫ du = − ln u + c
u
b
To find
∫
π
2
π
4
=∫
= ln sin x + c
π
8
π

∫π42 cot x dx = [ ln sin x ]π42 = [ ln1] −  ln
1 
2 
= − ln 1 = ln 2 = 1 ln 2
2
2
dx
Let x = a sin u and then
= a cos u and
du
1
dx du =
2
2 du
a −x
1
∫ a 2 − a 2 sin 2 u × a cosu du
1
=∫
× a cos u du
a 1 − sin 2 u
∫
= ∫ 1 du
=u+c
x
x
so u = sin −1 and the integral
a
a
x
+ c.
is u = sin −1
a
Now sin u =
9
aIf y = 0 then x (x – 2)4 = 0 so x = 0 or 2; the point
is (2, 0)
2
b
Area = ∫ x ( x − 2) dx ; let u = x – 2 and then
4
0
∫0 x ( x − 2)
∫u = − 2 (u
u=0
4
x=2
∫x = 0 (u + 2)u
dx =
)
4
du
0
1
2
+ 2u 4 du =  u 6 + u 5 
5  −2
 6
2
64 64 
= [0 ] −  −
=2
 6
15
5 
=
5
4
1
⌠  3
2 5 8 3

=   u 2 − 4u 2  du =  u 2 − u 2 
3 


⌡0
5
0
64 64 
128
8
= −
= −8
− 0 =−
3  [ ]
15
15
 5
8
Hence the area is 8
15
du
dθ
1
11 aLet u = 1 − cos θ then
= sin θ and
=
dθ
du sin θ
1
1
dθ
∫ sin θ 1 − cosθ dθ = ⌠⌡ sin θ × u 2 du du = ∫ u 2 du
3
2
2 3
= u 2 + c = (1 − cosθ ) 2 + c
3
3
An alternative method is to recognise that
3
the integral is a multiple of (1 − cosθ ) 2 and
differentiate that function and find the multiple.
b
1 − cos2 2θ = sin 2 2θ so ∫ sin θ 1 − cos2 2θ dθ
= ∫ sin θ sin 2θ dx
= ∫ 2sin 2 θ cosθ dθ
Try f ( x ) = sin 3 x then f(x) = 3 sin2 x cos x and so
∫ sin θ
2
1 − cos2 2θ dθ = sin 3 x + c
3
An alternative method is to find ∫ 2sin 2 θ cosθ dθ
by using the substitution u = sin θ or by spotting
that sin3 x differentiates to 3 sin2 xcos x
(
= ∫ ( 2sin θ cos θ − sin θ ) dθ
)
2
csin
∫ θ cos2θ dθ = ∫ sin θ 2cos θ − 1 dθ
2
2
= − cos3 θ + cosθ + c
3
dx
=1
du
If x = 0 u = −2 and if x = 2, u = 0
x = u + 2 and
2
Using the values of u for the limits, the integral is
4
cos x dx , let u = sin x then
sin x
du
= cos x and dx = 1 ;
dx
du cos x
cos x dx
cos x
1
∫ cot x dx =⌠⌡ sin x × du du = ∫ sin x × cosu du
= ∫ 1 du = ln u + c
u
Hence,
When x = −4, u = 0 and when x = 0, u = 4
4
⌠ (u − 4) u dx du
du
⌡0
= – ln |cos x| + c
π
2
π cot x d x
4
Make the substitution u = x + 4; then x = u − 4 and
dx
=1
du
1
−1
1
12 aIf f ( x ) = (1 + x)2 then f( x ) = (1 + x) 2 × 1
2
1
1
1
= 2 × 1 + x so ∫ 1 + x dx
=2 1+ x +c
b
Use the substitution u = 1 + x; then x = u − 1 and
dx
=1
du
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5
WORKED SOLUTIONS
⌠ 1
−1 
x
u −1
⌠
dx = ⌠
du =   u 2 − u 2  du


⌡ 1+ x
⌡ u
⌡
3
1
3
1
= 2 u 2 − 2u 2 + c = 2 (1 + x ) 2 − 2 (1 + x ) 2 + c
3
3
1
2
This can be written as (1 + x ) 2 {(1 + x ) − 3} + c
3
1
2
= (1 + x ) 2 (x − 2) + c
3
du
= 2x and
c
Try the substitution u = 1 + x2 ;
dx
dx
1
=
du 2x
1
3
∫ xe
2
2
du
= 1 and v = –e–x.
dx
−x
−x
−x
−x
−x
∫ x e d x = −x e + ∫ e d x = −x e − e + c
c Let u = x and
∫ xe
2
2
= 2 − 2 = 0.1953
3
4
0
To find ∫ xe xdx let u = x and
Then
dv
= e x then
dx
dv
= ex .
dx
du
= 1 and v = e x and
dx
∫ xe dx = xe
x
x
− ∫e xdx = xe x − e x + c
Therefore ∫ x 2e xdx = x 2e x − 2∫ xe xdx
b To find ∫ (x + 1)cos x dx let u = x + 1 and
(
)
= x 2e x − 2 xe x − e x + c
dv
= cos x .
dx
= x2 ex – 2xex + 2ex + c
= (x + 1)sin x + cos x + c
a To find ∫ x sin 2x dx let u = x and
dv
= sin 2x .
dx
du
1
= 1 and v = − cos 2x.
Then
2
dx
1
1
∫ x sin 2x dx = − 2 x cos 2x + ∫ 2 cos 2x dx
= − 1 x cos 2x + 1 sin 2x + c
2
4
b To find ∫ x cos 4x dx let u = x and
1
1
0
0
Hence ∫ x 2e x dx =  x 2e x − 2xe x + 2e x 
du
= 1 and v = sin x.
dx
∫ (x + 1)cos x dx = (x + 1)sin x − ∫ sin x dx
dv
= cos 4x .
dx
du
1
= 1 and v = sin 4x.
4
dx
1
1
∫ x cos 4x dx = 4 x sin 4x − ∫ 4 sin 4x dx
= 1 x sin 4x + 1 cos 4x + c
4
16
Then
1
To find ∫ x 2e x dx let u = x2 and
So ∫ x 2e xdx = x 2e x − 2∫ xe xdx
= x sin x + cos x + c
2
dx = 2xe0.5x − ∫ 2e0.5x dx
du
= 2x and v = e x
dx
dv
a To find ∫ x cos x dx let u = x and dx = cos x .
du
= 1 and v = sin x.
Then
dx
∫ x cos x dx = x sin x − ∫ sin x dx
Then
0.5x
= 2xe0.5x − 4e0.5x + c
Exercise 5.8A
1
dv
= e0.5x .
dx
du
= 1 and v = 2e0.5x.
dx
Then
2
1
1  2 32
1⌠  1
−1 
2
 u 2 − u 2  du = 2  3 u − 2u 
2

1
⌡1
dv
= e −x .
dx
Then
1⌠ x
⌠ x
1
u −1
du = 1 ⌠
×
du = 
= 
du
2 ⌡1 u
2
⌡1 u
⌡1 u 2x
=
dx = xe x − ∫ e x dx = xe x − e x + c
b Let u = x and
x =1
3
x
dv
= ex .
dx
du
= 1 and v = ex.
dx
Then
⌠
⌠
x3
x 3 dx
du
dx = 
Then 
2
⌡x = 0 u du
⌡0 1 + x
2
a Let u = x and
= [ e − 2e + 2e ] − [ 0 − 0 + 2] = e – 2
5
To find ∫ (2x 2 − 1)e −x dx let u = 2x2 – 1 and
Then
dv
= e −x .
dx
du
= 4x and v = –e–x.
dx
Therefore ∫ (2x 2 − 1)e −x dx
= −(2x 2 − 1)e −x + 4 ∫ xe −x dx.
To find ∫ x e −x dx let u = x and
dv
= e −x .
dx
Then
du
= 1 and v = –e–x and
dx
∫xe
dx = −xe −x + ∫ e −x dx = −xe −x − e −x + c.
−x
Therefore
2
−x
2
−x
−x
∫ (2x − 1)e dx = −(2x − 1)e + 4 ∫ xe dx
= – (2x2 – 1)e–x + 4(–xe–x – e–x) + c
= (–2x2 + 1 – 4x – 4)e–x + c
= – (2x2 + 4x + 3)e–x + c
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Integration
6
dv
To find I = ∫e x cos x dx let u = cos x and
= ex
dx
du
then dx = − sin x and v = e x
8
dv
du
= e −x then
= 2cos 2x
dx
dx
and v = –e–x.
x
x
x
So I = ∫e cos x dx = e cos x + ∫e sin x dx
To find ∫e x sin x dx let u = sin x and
then
∫e
x
So I = −e −x sin 2x + 2∫ e −x cos 2x dx.
dv
= ex
dx
To find ∫ e −x cos 2x dx let u = cos 2x and
du
= cos x and v = e x
dx
dv
du
= e −x then
= −2sin 2x and v = –e–x.
dx
dx
sin x dx = e x sin x − ∫e x cos x dx = e x sin x − I
∫e
Hence ∫
= − e −x sin 2x − 2e −x cos2x − 4I .
1 x
e ( cos x + sin x ) + c
2
Rearrange to get 5I = –e–x sin 2x – 2e–x cos 2x
so I = − 1 e −x(sin 2x + 2cos 2x) + c.
5
π
2
1
dx =  e x ( cos x + sin x ) 
 2
0

1 π
1 π 
1
=  e 2  −   = 2  e 2 − 1


2
2



  
7
 1 −π

2
=  − e 2 × (−2) −  − 
 5
  5 
e
1
dv
=1
dx
=
du 1
= and v = x; hence
dx x
Then
∫ ln x dx
= x ln x − ∫1 dx = x ln x + x + c
e
Hence ∫ ln x dx = [ x ln x − x ]1 = [e – e] – [0 – 1] = 1
e
9
2 − π2 2
e + = 0.483 to 3 s.f.
5
5
a Let u = x + 1 so x = u – 1 and dx = 1.
du
∫x
1
e
dx = ∫ 2 ln x dx = 2[ x ln x − x ]1 = 2
b
∫1 ln x
c
x d2x+=ln∫ (xln
) d2x + ln x ) dx
∫ ln∫ 2( ln
2
1
5
3
= 2u2 − 2u2 + c
5
3
e
1
3
x + 1 dx = ∫ (u − 1)u 2 du = ∫ u 2 − u 2 du
1
e
π
π
2
1
b Area = ∫ 2 e −x sin 2x dx =  − e −x(sin 2x + 2cos2x)
0
0
 5
a
To find ∫ ln x dx use integration by parts with
u = ln x and
cos2x dx = −e −x cos2x − 2∫ e −x sin 2x dx
So I = − e −x sin 2x + 2∫ e −x cos2x dx
Rearrange to get
2I = e x cos x + e x sin x so I =
−x
= −e −x cos2x − 2I
So I = e x cos x + ∫e x sin x dx = e x cos x + e x sin x − I
π
2 x
e cos x
0
aTo find I = ∫ e −x sin 2x dx let u = sin 2x and
=
5
3
2
2
(x + 1)2 − (x + 1)2 + c
5
3
1
b Let u = x and dv = (x + 1)2.
dx
x (xln
= ( ln 2) x =
+ (xln
ln2x) x− +x x+ln
c x= −x (xln+2c+=ln
) −2 x+ +lncx ) − x + c
3
So du = 1 and v = 2 (x + 1)2 .
3
dx
= x ln 2x − x + c
e
3
Hence ∫ ln 2x dx = [ x ln 2x − x ]1
e
1
= [ e ln 2e − e ] − [ ln 2 − 1] = e ln 2e – e – ln 2 + 1
d
To find ∫ ln(x + 1)dx let u = x + 1 and
e
e +1
Then ∫ ln(x + 1)dx = ⌠
⌡2
1
=∫
e +1
2
ln(x + 1)
du
dx
=1=
dx
du
dx
du
du
ln u du = [u ln u − u ]2
e +1
= [(e + 1)ln e + 1 − (e + 1)] − [ 2 ln 2 − 2]
= ( e + 1) ln e + 1 − (e + 1) − 2 ln 2 + 2
3
So ∫ x x + 1 dx = x × 2 (x + 1)2 − ∫ 2(x + 1)2 dx
3
3
3
5
= 2 x(x + 1)2 − 2 × 2 (x + 1)2 + c
3 5
3
=
3
5
2
4
x(x + 1)2 − (x + 1)2 + c.
3
15
c The answer to part a can be written
5
3
3
2
2
2
2
(x + 1)2 − (x + 1)2 = (x + 1)2 (x + 1) −
5
3
5
3
= (x
3
+ 1)2
( 52 x − 154 ).
{
}
The answer to part b can be written
3
5
3
2
4
2
4
x(x + 1)2 − (x + 1)2 = (x + 1)2 x − (x + 1)
3
15
3
15
{
}
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5
WORKED SOLUTIONS
3
= (x + 1)2
{ 32 x − 154 x − 154 } = (x + 1) ( 52 x − 154 )
Exam-style questions
3
2
1
so the two are the same.
10
∫ cos
3
a =∫
xdx = ∫ cos2 x × cos x dx ; let u = cos2 x and
b
dv
= cos x
dx
du
= −2cos x sin x and v = sin x
Then
dx
2
Hence ∫ cos3 x dx
e
1
1.2
1.4
2 3
sin x + c
3
dv
du
11 Let u = x2 and
= sin x ; then
= 2x and
dx
dx
v = – cos x
≈
1.8
2
0.2
0.3333 + 0.2 + 2 ( 0.3219 + 0.2951 + 0.2625 + 0.2298 )}
2 {
= 0.275 to 3 s.f.
2
2
Then ∫ x sin x dx = −x cos x + 2∫ x cos x dx
3
a If f(x) = cos (4x + 1) then f ′(x) = −4sin(4x + 1)
Hence ∫ sin(4x + 1)dx = − 1 cos(4x + 1) + c
4
dv
= cos x; then
dx
b If f(x) = ln |4x + 1| then f(x) =
Hence ∫ x cos x dx = x sin x − ∫ sin x dx
Hence ∫
= x sin x + cos x + c
3
2
Hence ∫ x 2 sin x dx = −x 2 cos x + 2x sin x + 2cos x + c
and ∫ 2 x 2 sin x dx =  −x 2 cos x + 2x sin x + 2cos x  2
0
0
1 ×4= 4
4x + 1
4x + 1
3
1
dx =  1 ln 4x + 1 
4x + 1
 2
 4
1
1
1 13
=  ln13  −  ln 9  = ln
9
  4
 4
 4
π
π
1.6
2
x
⌠
dx
 3
⌡1 x + 2
= cos2 x sin x +
4
a If f(x) = tan 0.5x then f ′(x) = sec2 0.5x × 0.5
= 0.5sec2 0.5x.
=π−2
12 af(x) = e–x – xe–x; at a stationary point,
e–x – xe–x = 0; e–x(1 – x) = 0
Hence ∫ sec2 0.5x dx = 2tan 0.5x + c
b 1 + tan2 0.5 x = sec2 0.5x or
tan2 0.5x = sec2 0.5x – 1
e −x is always positive so the only solution is x = 1.
f (x) = –e–x – e–x + xe–x and f (1) = –e–1 < 0 so the
stationary point is a maximum.
Hence ∫ tan 2 0.5x dx = ∫ (sec2 0.5x − 1) dx
= 2tan 0.5x − x + c
b
5
0
1.8
dx =  1 e 2x + 1  = 0.5e 4.6  − 0.5e3 
1
 2
= 39.699 to =
3 d.p.
0.5(e4.6 – e3)
1.8 2x + 1
∫1
x
x 3 + 2 0.3333 0.3219 0.2951 0.2625 0.2298 0.2
= cos2 x sin x + 2∫ sin 2 x cos x dx
du
= 1 and v = sin x
dx
e
x
= cos2 x sin x − ∫ ( −2cos x sin x ) × sin x dx
To find ∫ x cos x dx let u = x and
dx ≈ 1 × 0.2{20.09 + 2(29.96 + 44.70
2
+ 66.69) + 99.48} = 40.227
1.8 2x + 1
1
3
3
Area = ∫ e0.5x − e −0.5xdx =  2e0.5x + 2e −0.5x 
0
0
=  2e1.5 + 2e −1.5  − [2 + 2]
1
= 9.410 – 4 = 5.41 to 3 s.f.
6
1
The area is ∫ xe −x dx
0
Let u = x and
1
dv
du
= e −x then
= 1 and v = −e −x
dx
dx
1
1
1
x
−x
−x
−x
−x
∫0 xe dx =  −xe 0 + ∫0 e dx =  −xe − e 0
=  −2e −1  − [ −1] = 1 − 2e −1


a cos 4x = 1 – 2 sin2 2x and so sin 2 2x =
1 1
− cos4x .
2 2
Hence ∫ sin 2 2x dx = ∫ 1 − 1 cos4x dx
2 2
1
1
= x − sin 4x + c
2
8
1
1
1
1
b Area = ∫ sin 2 2x dx =  x − sin 4x 
8
0
0
2
1
= 0.5 − sin 4  − [0 − 0]= 0.595 to 3 s.f.

8

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Integration
7
3
11 a
a If f(x) = (2x + 1)2 then
f ′(x) =
1
1
3
(2x + 1)2 × 2 = 3(2x + 1)2.
2
1
2
Hence ∫ (2x + 1) dx =
b Let u = 2x + 1 then
∫ 2x
3
+ 1)2
+ c.
×
Hence
1
du
2
c
3
1
5
3
= ∫ 1 u 2 − 1 u 2 du = 1 u 2 − 1 u 2 + c
2
2
5
3
x2
c If 2x 2x + 1 = 0 then x = 0 or –0.5.
0
3
5
1
1
2 − (2x + 1) 2
2
x
2
x
+
1
d
x
=
2
x
+
1
(
)


∫−0.5
3
5
 −0.5
12 a
=  1 − 1  − [0] = − 2
15
 5 3 
8
)
(
)
= sin 3x – tan x + x + c
2.5
2.5
1 2
1
4x − 8
1
b
∫2 2e dx =  2 e4x −8 2 =  2 e  −  2 
9
(
)
1 2
e −1
2
2
acos 2x – 4 sin2 x cos2 x = cos2 2x – (2 sin x cos x)2
= cos2 2x – sin2 2x
= cos (2 × 2x) = cos 4x
b
Integral =
π
16
cos4x dx
0
∫
b9 cos2 θ – sin2 θ – 6 sin θ cos θ
= 9(1 – sin2 θ) – sin2 θ – 3 × 2 sin θ cos θ
= 9 – 10 sin2 θ – 3 sin 2θ
= 4 + 5(1 – 2 sin2 θ) – 3 sin 2θ = 4 + 5 cos 2θ – 3 sin 2θ
c
∫ (9cos θ − sin
2
=
2
)
θ − 6sin θ cosθ dθ
∫ (5cos2θ − 3sin 2θ + 4 ) dθ
5
3
= sin 2θ + cos2θ + 4θ + c
2
2
2
+ 9) + c
x 2 + 9 − 9 dx = 1 − 9 dx
∫ x2 + 9
x2 + 9
= x − 9∫ 21 dx
x +9
= x − 9 × 1 tan −1 x + c
3
3
x
= x − 3tan −1 + c
3
)
sin 0.5x
π
b
π
∫02 tan 0.5x dx =  −2 ln cos 0.5x 02
=  −2 ln cos π  − [−2 ln 1]

4 
= −2 ln 1 − 0 = −2 ln 1 − ln 2
2
(
π
2 1
2
2+x
dx = ∫  2 +  dx = − + ln x + c
x
x
x
x2
∫
1
∫ tan 0.5x dx = ∫ cos0.5x dx .
16
=  1 sin 4x 
0
 4
1
1
2
=  1 sin π  − [ 0 ] = ×
=
4 
4
8
 4
2
10 a
1
× 2x
x2 + 9
du
= −0.5sin 0.5x .
dx
sin 0.5x
sin 0.5x dx
∫ cos 0.5x dx = ∫ u du du
1
= ∫ sin 0.5x ×
du
− 0.5 sin 0.5x
u
= ∫ −2 du = −2 ln u + c
u
= – 2 ln |cos 0.5x| + c
2
2
a
∫ 3cos3x − tan x dx = sin 3x − ∫ sec x − 1 dx
=
x
+c
3
If u = cos 0.5x then
2
.
15
(
x
∫ x 2 + 9 dx = 2 ln(x
(
0
−1
∫ x 2 + 9 dx = ∫
5
3
= 1 (2x + 1)2 − 1 (2x + 1)2 + c
5
3
The area is
1
2x
.
x2 + 9
=
du
dx 1
= 2 and
= ;
dx
du 2
2x + 1 dx = ∫ (u
1
b If f(x) = ln(x 2 + 9) then f ′ ( x ) =
1
1
f(x) = (2x
3
3
1
− 1)u 2
1
∫ x 2 + 9 dx = ∫ x 2 + 32 dx = 3 tan
)
1
= 2 ln 2 2 = 2 × 1 ln 2 = ln 2
2
13 aUsing the product rule,
f(x) = sin 0.5x + 0.5x cos 0.5x
At stationary point, sin 0.5x + 0.5x cos 0.5x = 0;
sin 0.5x = – 0.5x cos 0.5x
tan 0.5x = – 0.5x; tan 0.5x + 0.5x = 0
b
4
2
0
2π
c
To find ∫ x sin 0.5x dx, let u = x and
then
dv
= sin 0.5x ;
dx
du
= 1 and v = – 2 cos 0.5x
dx
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5
WORKED SOLUTIONS
So 4x ≡ A(x + 1) + B(x – 3)
Hence ∫ x sin 0.5x dx = −2x cos 0.5x
If x = 3 then 12 = 4A and A = 3; if x = – 1
then –4 = – 4 B and B = 1
+ ∫ 2cos 0.5x dx
= −2x cos0.5x + 4sin 0.5x + c
So
2π
d
Area = ∫ x sin 0.5x dx = [ – 2x cos 0.5x + 4 sin 0.5x]02π
0
= [4π] – [0] = 4π
π
2
sin x cos x dx
0
π
2 1 
π
1
1
∫02 2 sin 2x dx =  − 4 cos2x 0
Alternative method
π
2
sin x cos x dx
0
∫
b
=
 4 
( )
π
1
∫ x sin x cos x dx = 2 ∫ x sin 2x dx
}
x+7 =
x+7
≡ A + B
x 2 − x − 6 (x − 3)(x + 2) x − 3 x + 2
2
x − 3)
= A(x + ) + B(x
(x − 3)(x + 2)
x+7
2
1
∫ x 2 − x − 6 dx = ∫ x − 3 − x + 2 dx
= 2 ln x − 3 − ln x + 2 + c
= ln
(x − 3)2
+c
x+2
16 aIf f(x) = x2 – 2x – 3 then f(x) = 2x – 2 and
x −1
= 0.5f(x)
f(x)
x 2 − 2x − 3




x − 1 dx = 0.5ln f(x) + c
⌠
Hence 
⌡ x 2 − 2x − 3
= 0.5ln x 2 − 2x − 3 + c
b
4x
4x
A
B
=
≡
+
x 2 − 2x − 3 (x − 3)(x + 1) (x − 3) ( x + 1)
=
A ( x + 1) + B(x − 3)
(x − 3)(x + 1)
then f(x) = 2xe–x – x2 e–x
At a stationary point 2xe–x –x2 e–x = 0;
x (2 – x) e–x = 0; x = 0 or 2; if x > 0 then the only
solution is x = 2
So the maximum value is f(2) = 4e–2
Equate the numerators: A(x + 2) + B(x – 3) = x + 7.
Put x = 3: 5A = 10 so A = 2.
Put x = –2: –5B = 5 so B = –1.
Therefore 2 x + 7 = 2 − 1 .
x −x−6 x−3 x+2
b
17 a If f(x) =
x2 e–x
f (x) = 2e–x – 2xe–x – 2xe–x + x2 e–x
= (2 – 4x + x2)e–x so f  (2) = – 2e–2 < 0
= − 1 x cos 2x + ∫ 1 cos 2x dx
4
4
= − 1 x cos 2x + 1 sin 2x + c
4
8
15 a
)
= [3 ln 2 + ln 2] – [3 ln 3] = 4 ln 2 – 3 ln 3
16
= ln 24 – ln 33 = ln
27
2
=  1 sin 2 x  =  1 × 12  −  1 × 0 2  = 1
 2
0  2
  2
 2
{
(
1
1
1
− −  =
 4  2
Write u = x and dv = sin 2x.
dx
du
1
= 1 and v = − cos 2x
Then
2
dx
1 1
1
∫ x sin x cos x dx = 2 − 2 x cos 2x − ∫ − 2 cos 2x dx
1
1
4x
3
1
⌠
dx =⌠
c
 2
 x − 3 + x + 1 dx
⌡0 x − 2x − 3
⌡0
= [ 3 ln x − 3 + ln x + 1 ]0
14 a sin 2x = 2 sin x cos x so ∫
=
4x
3
1
≡
+
x 2 − 2x − 3 x − 3 x + 1
b
I = ∫ x 2e −xdx ; let u = x2 and
dv
= e −x then
dx
du
= 2x and v = –e–x
dx
Then I = −x 2e −x + 2∫ xe −x dx
dv
To find ∫ xe −x dx let u = x and
= e −x then
dx
du
= 1 and v = –e–x
dx
Hence ∫ xe −x dx = −xe −x + ∫e −x dx = – xe– x – e– x + c
Hence I = –x2e–x + 2(–xe–x – e–x) + c
= – (x2 + 2x + 2)e–x + c
3
18 aIf u = (1 − x ) 2 then
= −3 1− x
2
1
du 3
= (1 − x)2 × −1
dx 2
1
3
1 − x dx =  − 2 (1 − x ) 2 
 3

0
0
2 2

=
= [0 ] − −
 3  3
1
Hence ∫
1
b
To find ∫ x 1 − x dx let u = 1 – x; then x = 1 – u
0
dx
= −1
du
When x = 0, u = 1 and when x = 1, u = 0
and
u=0
1
dx
Hence ∫ x 1 − x dx = ⌠ (1 − u) u
du
du
⌡u =1
0
0
⌠  1
3
=  −u 2 + u 2  du


⌡1
0
 2 3 2 5
=  − u 2 + u 2  = [0 ] −  − 2 + 2  = 4
5 
 3 5  15
 3
1
(This could also be done using integration by
parts)
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Integration
c Use the same substitution as part b:
π
1 2
dx
2
∫0 x 1 − x dx = ⌠⌡u =1 (1 − u) u du du
0
(
)
1
 1 − 2u + u 2 u 2 du
= −⌠
⌡1
b
If f (x) = sin3 x then f(x) = 3 sin2 x cos x; hence
1
π
2
sin 2 x cos x
0
⌠  1
3
5
3
5
=   u 2 − 2u 2 + u 2  du =  2 u 2 − 4 u 2 +
3

5
7 
⌡0 

0
2 4 2
16
=  − +  − [0 ] =
 3 5 7 
105
19 a If f ( x ) = ln 1 + 4x then f( x ) = 1 × 4 = 4
1 + 4x
1 + 4x
1
2 72 
u
∫
c h=
8
dx = 2 ln 1 + 4x + c
1 + 4x
This can also be solved with the substitution
dx 1
1
π
π
π
π
=
u = 1 + 4x;
0 + sin 2 + sin 2 + sin 2
≈
du 4
24
2
6
3
2
2
8
8
1
Then ∫
dx = ∫ × du = ∫ du = 2 ln u + c
1 + 4x
u 4
u
= 2 ln 1 + 4x + c as before.
d
2
8
8
⌠
⌠
⌠
d
x
b
d
x
=
=

d
x
2
 2

⌡ 1 + 4x 2
 4 x2 + 1
x + 1
4
⌡
2
⌡
= 2 × 2 tan–1 (2x) + c = 4 tan–1 (2x) + c
{ (
)
()
A (1 + 2x ) + B(1 − 2x)
(1 − 2x)(1 + 2x)
8
4
4
8
2 ≡
+
so ⌠
dx

1 − 2x 1 + 2x
⌡ 1 − 4x 2
1 − 4x
4
4
=⌠
 1 − 2x + 1 + 2x dx
⌡
1
× −2
Now if f ( x ) = ln 1 − 2x then f ′ ( x ) =
1 − 2x
−2
;
=
1 − 2x
(
(
)
4
4
So ⌠
 1 − 2x + 1 + 2x dx
⌡
= −2 ln 1 − 2x + 2 ln 1 + 2x + c
{ (
) }
π
π 1 3
3 1
+ +1+ + }≈
=
48 { 4 4
16
4 4
+ sin 2
2π
5π
+ sin 2
+0
3
6
cos 4x = 1 – 2 sin2 2x so sin 2 2x =
π
2
0
∫
) }
1
2π
5π
π
π
π
π
0 + sin 2 + sin 2 + sin 2 + sin 2
+ sin 2
+0
24
2
6
3
2
3
6
π
(
1 1
− cos4x and
2 2
)
2 1
1
1
sin 2 2x dx = ⌠
 8 − 8 cos4x dx
4
⌡0
π
1
× 5{5.80 + 2(2.52 + 1.55
2
+1.10) + 0.84} = 42.45
25
1
h = 5 and ∫ p dv ≈
2
If pv1.2 = 40 then p = 40v–1.2.
5
25
Then ∫ p dv =
)
2
Similarly if f ( x ) = ln 1 + 2x then f ′ ( x ) =
1 + 2x
π
Mathematics in life and work
So 8 ≡ A (1 + 2x) + B (1 – 2x)
1
1
Let x = then 8 = 2A and A = 4; let x = − then
2
2
8 = 2B and B = 4
≈
π
1
π
and ∫ 2 sin 2 x cos2 x dx =∫ 2 sin 2 2x dx
12
0
0 4
2
=  1 x − 1 sin 4x  =  π  − [ 0 ] = π
0
 8
 16 
8
16
8
8
A
B
c
=
≡
+
1 − 4x 2 (1 − 2x)(1 + 2x) 1 − 2x 1 + 2x
=
π
2
1
dx =  sin 3 x 
 3
0
1
1 
=  3  − [ 0 ] = 3
Hence ∫
(
π
π
2
1
1
20 a∫ 2 sin x cos x dx =∫ 2 sin 2x dx =  − cos2x 
 4
0
0
0 2
1
1
1
=   − −  =
 4   4  2
u=0
5
25
∫5
25
40v −1.2dv =  40 v −0.2 
 −0.2
5
25
=  −200v −0.2 
5
= [–105.06] – [–144.96] = 39.9 to 3 s.f.
3
If pv = 24 then p = 24 and
v
25
25 24
p
d
v
=
d
v
=
[ 24 ln v ]525 = 24 ln 25 − 24 ln 5
∫5
∫5 v
25
= 24 ln 5 = 24 ln 5 = 38.6 to 3 s.f.
The work done is less by 39.9 – 38.6 = 1.3.
= ln(1 + 2x)2 − ln(1 − 2x)2 + c
= ln
(11 +− 22xx ) + c
2
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6
WORKED SOLUTIONS
6 Numerical solution of equations
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility
for the example answers to questions taken from its past question papers, which are contained in this publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question. When using numerical methods, it is
generally acceptable for intermediate working to be accurate to 6 decimal places. For completeness, some of the
intermediate working in this chapter is accurate to 9 decimal places. When using numerical methods, it is generally
acceptable for intermediate working to be accurate to 6 decimal places. For completeness, some of the intermediate
working in this chapter is accurate to 9 decimal places.
Prerequisite knowledge
Exercise 6.1A
1
1
y
a f(x) = 3x2 + 4x – 11
f(1.2) = 4.32 + 4.8 – 11 = –1.88
f(1.4) = 5.88 + 5.6 – 11 = 0.48
0
–1
There is a sign change between f(1.2) and
f(1.4).
x
5
Therefore there is a root of f(x) in the interval
1.2 < x < 1.4.
–5
b f(x) = x3 + 6x2 + 11x + 6
2
f(–1.4) = –2.744 + 11.76 – 15.4 + 6 = –0.384
y
f(–0.8)= –0.512 + 3.84 – 8.8 + 6 = 0.528
There is a sign change between f(–1.4) and
f(–0.8), therefore there is a root of f(x) in the
interval –1.4 < x < –0.8.
–5
–1
0
2
c f(x) = 8x4 + 2x3 – 53x2 + 37x – 6
x
f(1) = 8 + 2 – 53 + 37 – 6 = –12
f(3) = 648 + 54 – 477 + 111 – 6 = 330
3
There is a sign change between f(1) and f(3),
therefore there is a root of f(x) in the interval
1 < x < 3.
y
–10
2
f(x) = 1 – ex – ln x
f(0.2) = 1.388
f(0.7) = –0.657
There is a sign change between f(0.2) and f(0.7).
Therefore there is a root of f(x) in the interval
0.2 < x < 0.7.
3
f(x) = sinx x ; −0.8 < x < −0.7
e
f(–0.8) = –1.597
f(–0.7) = –1.297
There is not a sign change between f(–0.8) and
f(–0.7).
Therefore there is not a root of f(x) in the interval
–0.8 < x < –0.7.
x
–3
Asymptotes at x = −1 and y = 0.
4
y
0 1
2
–2
x
Asymptotes at x = 1 and y = 0.
2
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NUMERICAL SOLUTION OF EQUATIONS
4
a
y
y=
1
x2
b 7+
y = ex
5 13
=
x x2
f(x) = 7 +
5 13
−
x x2
f(0.95) = –2.141
f(1.1) = 0.802
There is a sign change between f(0.95) and f(1.1).
0
Therefore there is a root of f(x) in the interval
0.95 < x < 1.1.
x
c Rearranging 7 +
One point of intersection of the two graphs
1
therefore 2 = e x has one root.
x
Consequently, by solving the quadratic, the
roots to the original equation are also found.
1
b f(x) = 2 − e x
x
a = 7, b = 5, c = −13
f(0.6) = 0.956
x=
f(0.8) = –0.663
There is a sign change between f(0.6) and
f(0.8).
Therefore there is a root of f(x) in the interval
0.6 < x < 0.8.
sin2 x – cos2 x = 0
f(x) = sin2 x – cos2 x
5
−5 ± 52 − (4)(7)(−13)
14
x = 1.05 or –1.77
7
There is a sign change between f
( π8 ) and f ( 38π ).
π
π
3π
is in the interval < x <
.
4
8
8
Sign change so x = 3 − 1 − 22 has a root in the
x x
interval –1 < x < 0.
f(x) = 3 − 1 − 22 − x
x x
f(1.4) = –0.134 693 877 6
y
f(1.8) = 0.027 160 493 83
5
y=7+ x
y=
13
x2
0
–12
8
Sign change so x = 3x 2 − 1 has a root in the
x
interval 0 < x < 1.
f(x) = 3 − 1 − 22 − x
x x
f(–1) = 3
f(–0.5) = –2.5
Therefore there is a root near x = π
4
a
f( x ) = 3x 2 − 1 − x
x
f(10–6) = –1 000 000
f(1) = 1
( π8 ) = −0.707
3π
f ( ) = 0.707
8
6
−b ± b 2 − 4ac
2a
x=
f
x=
5 13
gives 7x2 + 5x – 13 = 0.
=
x x2
x
There are two points of intersection.
Therefore the equation 7 +
5 13
=
has two roots.
x x2
Sign change so x = 3 − 1 − 22 has a root in the
x x
interval 1.4 < x < 1.8.
f(x) = 3 − 1 − 22 − x
x x
f(1.8) = 0.027 160 493 83
f(2.2) = –0.067 768 595 04
Sign change so x = 3 − 1 − 22 has a root in the
x x
interval 1.8 < x < 2.2.
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6
WORKED SOLUTIONS
9
f ( x ) = 1 + 7 − ex
2x
y
f(10–6) = 500 006
f 3 = 2.851 644 263
2
()
No sign change so either ex = 1 + 7 has no roots
2x
3
in the interval 0 < x < or has more than one root
2
in this interval.
–2
0
x
3
10 f(x) = 3x2 – 1 – 2x
f(–5) = 73.968 75
f(0) = –2
There is a discontinuity in the interval
2.4 < x < 3.3, so although there is a sign change
between f(2.4) and f(3.3) there isn’t a root, as
shown on the graph sketch.
Sign change so 2x = 3x2 – 1 has a root in the interval
–5 < x < 0.
f(x) = 3x2 – 1 – 2x
f(0) = –2
3
Functions in the form f(x) = (x ± a)2 do not
intersect the x-axis. Instead the curve ‘sits on’
the x-axis (the discriminant equals zero and so
the function has one distinct, repeated root).
Consequently in the case of f(x) = (x ± a)2, f(a – 1)
and f(a + 1) are both positive.
4
f(x) =
f(2) = 7
Sign change so 2x = 3x2 – 1 has a root in the interval
0 < x < 2.
f(x) = 3x2 – 1 – 2x
f(2) = 7
f(10) = –725
1
+2
x
Sign change so 2x = 3x2 – 1 has a root in the interval
2 < x < 10.
y
Exercise 6.2A
1
2
f(x) = 6x2 – x – 2
Root in the interval –2 < x < 0.
f(–2) = 24
f(0) = –2
There is a sign change between f(–2) and f(0).
Therefore there is a root of f(x) in the interval
–2 < x < 0.
Root in the interval 0 < x < 2.
f(0) = –2
f(2) = 20
There is a sign change between f(0) and f(2).
Therefore there is a root of f(x) in the interval
0 < x < 2.
f(x) =
1
; 2.4 < x < 3.3
(x − 3)
−5
3
10
f(3.3) =
3
f(2.4) =
2
0
–6
a f(x) =
x
1
+2
x
Root in the interval –1 < x < –0.2.
f(–1) = 1
f(–0.2) = –3
There is a sign change between f(–1) and
f(–0.2).
Therefore there is a root of f(x) in the interval
–1 < x < –0.2.
b f(x) =
1
+2
x
Asymptote in the interval –0.2 < x < 0.2.
f(–0.2) = –3
f(0.2) = 7
75
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Numerical solution of equations
5
There is a sign change between f(–0.2) and
f(0.2).
b & c f( x ) = 12 − 2
x
Therefore there is an asymptote of f(x) in the
interval –0.2 < x < 0.2.
( )
f (1 ) = 2
2
f(x) = tan x
Asymptote in the interval π < x < 3π .
4
4
π
f
=1
4
()
3π
f ( ) = −1
4
There is a sign change between f
No sign change in the interval − 1 < x < 1
2
2
however there is a discontinuity.
f(–1) = –1
f −1 = 2
2
( )
( π4 ) and f ( 34π ).
Sign change so there is a root in the interval
–1 < x < − 1 .
2
1
f
=2
2
f(1) = –1
Therefore there is an asymptote of f(x) in the
π
3π
interval < x <
.
4
4
Root in the interval 3π < x < 5π .
4
4
3π
f
= −1
4
( )
5π
f( ) =1
4
There is a sign change between f
()
Sign change so there is a root in the interval
1 < x < 1.
2
( 34π ) and f ( 54π ).
8 f( x ) =
Therefore there is a root of f(x) in the interval
3π
5π
<x<
.
4
4
6
7
1
(3x + 2)(2x − 1)
Discontinuity in the interval –1 < x < –0.1.
f(–1) = 0.333
f(–0.1) = –0.490
There is a sign change between f(–1) and f(–0.1).
Therefore there is a discontinuity of f(x) in the
interval –1 < x < –0.1.
Discontinuity in the interval 0.1 < x < 1.
f(0.1) = –0.543
f(1) = 0.2
There is a sign change between f(0.1) and f(1).
Therefore there is a discontinuity of f(x) in the
interval 0.1 < x < 1.
f( −6 ) = − 11
2
f(–3.9999) = 9995
4
2
–12 –10 –8
–2 0
–2
9
6
8
10
ex+1 – 7 = x3
4
f(x) = x3 – ex+1 + 7
2
f(–2) = –1.367 879 441
–4
4
Using the graph – the sign change in the interval
–3.9999 < x < –2 is caused by a root. The sign
change in the interval –6 < x < –3.9999 is caused by
a discontinuity.
6
0
–2
–2
2
–8
y
–4
–4
–6
8
–6
–6
–4
10
–14 –12 –10 –8
1 −5
x+4
f(–3.9999) = 9995
f( −2) = − 9
2
f(x) =
a f( x ) = 12 − 2
x
f −1 = 2
2
2
4
6
8
10 12
x
f(0) = 4.281 718 172
Sign change so there is a root in the interval
–2 < x < 0.
76
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6
WORKED SOLUTIONS
b x2 + 4x – 7 = 0
f(0) = 4.281 718 172
x2 = 7 – 4x
xn + 1 = 7 − 4
xn
f(2) = –5.085 536 923
Sign change so there is a root in the interval
0 < x < 2.
c x0 = –5
10 f(x) = x4 – 5x2 + 1
x1 = –5.4
x2 = –5.2963
x3 = –5.321 68
x4 = –5.315 37
x5 = –5.316 93
x = –5.32 to 2 d.p.
f(–3) = 37
f(–2) = –3
Sign change so there is a root in the interval
–3 < x < –2.
f(–2) = –3
f(0) = 1
d
Sign change so there is a root in the interval
–2 < x < 0.
0.5
y
–7.5 –7 –6.5 –6 –5.5 –5 –4.5 –4 –3.5 –3 –2.5 –2 –1.5
–3.5
–4
y=x
f(0) = 1
–4.5
–5
f(2) = –3
–5.5
Sign change so there is a root in the interval
0<x<2
–6
–6.5
f(2) = –3
f(3) = 37
4
Sign change so there is a root in the interval 2 < x < 3.
Exercise 6.3A
1
Taking the square root of both sides gives
x = 11 − 5x .
b x2 = 11 – 5x
Dividing both sides by x gives x =
11
− 5.
x
c x2 = 11 – 5x
x2 – 11 = –5x
2
3
a f(x) = 2x3 – 5x + 1
f(–2) = –5
f(–1) = 4
a x2 = 11 – 5x
x=
x
x 2 − 11
−5
There is a sign change between f(–2) and
f(–1).
Therefore there is a root of f(x) in the interval
–2 < x < –1.
b 2x 3 − 5x + 1 = 0
x 2 = 5x − 1 = 5 − 1
2x
2 2x
x = − 5 − 1 (for a negative root)
2 2x
xn+1 = − 5 − 1
2 2xn
 6 
x n +1 = 
 xn − 1 
x 0 = –1
x0 = 2
x1 = 2.449 49
x2 = 2.034 549
x3 = 2.408 242
x3 = 2.408 to 3 d.p.
x 2 = –1.669 932 673
a f(x) = x2 + 4x – 7
x1 = –1.732 050 808
x 3 = –1.673144 724
x 4 = –1.672 972 942
x 5 = –1.672 982113
So the root is −1.673 to 3 d.p.
f(–6) = 5
f(–5) = –2
There is a sign change between f(–6) and
f(–5).
Therefore there is a root of f(x) in the interval
–6 < x < –5.
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NUMERICAL SOLUTION OF EQUATIONS
c
1
y
xn + 1 =
0.2
–2.4 –2.2 –2 –1.8 –1.6 –1.4 –1.2 –1 –0.8 –0.6 –0.4 –0.20
–0.2
0.2
x0 = 0.5
x1 = 0.281 25
x2 = 0.255 561 829
x3 = 0.254 172 804
x4 = 0.254 105 133
x = 0.25 to 2 d.p.
Interval 1 < x < 2.
x
0.2 0.4 0.6 2
–0.4
y=x
–0.6
–0.8
–1
–1.2
–1.4
–1.6
1
xn +1 = ( 4xn − 1) 3
x0 = 1
x1 = 1.442 249 57
x2 = 1.683 225 836
x3 = 1.789 746 575
x4 = 1.833 030 731
x5 = 1.850 048 494
x6 = 1.856 654 291
x7 = 1.859 205 847
x = 1.86 to 2 d.p.
–1.8
–2.0
–2.2
–2.4
5
(
2
a xn + 1 = 7 − 3xn
x0 = 1
)
1
5
x1 = 1.319 507 911
x2 = 1.121 818 649
x3 = 1.263 846 495
x4 = 1.171 663 249
7
x5 = 1.235 740 306
x7 = 1.222 334 153
x8 = 1.202 820 206
x9 = 1.216 089 595
x10 = 1.207 153 618
x = 1.21 to 2 d.p.
y
1.2
1
y=x
0.8
0.6
0.2 0.4 0.6 0.8
6
1.2 1.4 1.6 x
1
f(x) = x4 – 4x2 + x
y
8
1
–6
–5
–4
–3
–2
0
–1
–1
1
x n +1 =
xn3 − xn2 + 3
3
x0 = 1.5
x1 = 1.375
x2 = 1.236 328
x3 = 1.120 41
x4 = 1.050 384
x5 = 1.018 53
x6 = 1.006 408
x7 = 1.002 163
x8 = 1.000 724
x9 = 1.000 242
x10 = 1.000 081
x11 = 1.000 027
x12 = 1.000 009
x13 = 1.000 003
x14 = 1.000 001
x15 = 1
x16 = 1
x6 = 1.193 222 257
b
xn3 + 1
4
a f(x) = x2 – 3x – 13
f(–3) = 9 + 9 – 13 = 5
1
2
3
4
5
6
x
–2
––3
–4
–5
Root in the interval 0 < x < 0.5 and in the interval
1 < x < 2.
Interval 0 < x < 1.
f(–2) = 4 + 6 – 13 = – 3
Sign change so there is a root in the interval
–3 < x < –2.
f(5) = 25 – 15 – 13 = –3
f(6) = 36 – 18 – 13 = 5
Sign change so there is a root in the interval
5 < x < 6.
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6
WORKED SOLUTIONS
b i
xn +1 =
Exercise 6.4A
xn2 − 13
3
1
x0 = 5
x1 = 4
x2 = 1
x3 = –4
x4 = 1
When finding x2 need to square root a
negative so iterations stop.
b x=
x0 = 6
x1 = 7.666 67
x2 = 15.259 259
x3 = 73.281 66
Not convergent to a root.
(
ii xn +1 = 3xn + 13
)
(
xn +1 = xn2 + 4
)
1
4
x0 = 1
x1 = 1.495 348 781
x2 = 1.580 256 954
x3 = 1.596 547 191
x4 = 1.599 716 901
x5 = 1.600 335 232
Convergent to root at x = 1.600 (3 d.p.).
(
)
1
5
c x=
1
10 xn +1 = 2 (xn2 + 8)
x0 = 1
x1 = 1.350 960 039
x2 = 1.374 869 048
x3 = 1.376 688 186
x4 = 1.376 827 506
Convergent to root at x = 1.377 (3 d.p.).
11 − x 2
5
Converges to a different root x = 1.65 to 2 d.p.
2


xn + 1 =  6 
 xn − 1 
x0 = 5
x1 = 1.224 745
x2 = 5.166 908
x3 = 1.199 965
x4 = 5.477 702
x5 = 1.157 572
x6 = 6.170 722
x7 = 1.077 209
x8 = 8.815 382
x9 = 0.876 195
1
2
x0 = 5
x1 = 5.291 503
x2 = 5.373 501
x3 = 5.396 341
x4 = 5.402 687
x5 = 5.404 448
x6 = 5.404 937
x7 = 5.405 073
Convergent to root at x = 5.405 (3 d.p.).
x0 = 6
x1 = 5.567 764
x2 = 5.450 073
x3 = 5.417 584
x4 = 5.408 581
x5 = 5.406 084
x6 = 5.405 391
x7 = 5.405 199
Convergent to root at x = 5.405 (3 d.p.).
11
−5
x
Converges to root x = –6.65 to 2 d.p.
Not convergent to a root.
9
a x = 11 − 5x
This combination of iteration formula and starting
value does not result in a convergent sequence of
results and eventually the iterations stop because
one iteration results in finding the square root of a
negative number.
3
xn + 1 =
(7 − x )
2
n
4
x0 = −5
x1 = −4.5
x2 = −3.3125
x3 = −0.993 164
x4 = 1.503 41
x5 = 1.184 94
x6 = 1.398 98
x7 = 1.260 72
x8 = 1.352 65
x9 = 1.292 58
x10 = 1.332 31
x11 = 1.306 24
x12 = 1.323 43
x13 = 1.312 13
x14 = 1.319 58
x15 = 1.314 68
x16 = 1.317 91
x17 = 1.315 78
Converges to root x = 1.32 to 2 d.p.
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Numerical solution of equations
xn + 1 = 7 − 4
xn
x0 = −5
x1 = −5.4
x2 = −5.2963
x3 = −5.321 68
x4 = −5.315 37
Converges to root x = –5.32 to 2 d.p.
4
5
2xn3 + 1
5
Many solutions are possible for example: x0 = −3
x1 = −10.6
x2 = −476.2064
x8 = –2.405 974 505
x9 = –2.404 746 82
Yes, this iterative formula can be used. Convergent
to root at x = –2.405 (3 d.p.).
1
9
x1 = 0.707107
x2 = undefined
No this iterative formula cannot be used as cannot
square root values less than 0.
10 xn +1 =
7
(
x4 = 1.376 813 68
)
Yes, this iterative formula can be used. Convergent
to root at x = 1.377 (3 d.p.).
Exam-style questions
1
f(2.3) = 0.046
Sign change between f(2.1) and f(2.3) so f(x)
has a root in the interval 2.1 < x < 2.3.
b xn + 1 = 3 − sin ( xn )
1
3
x0 = 2.1
x1 = 2.1368
x2 = 2.15 59
x3 = 2.16 64
x4 = –1.732 053 695
Yes, this iterative formula can be used. Convergent
to root at x = –1.732 (3 d.p.).
a f(x) = x + sin (x) – 3
f(2.1) = –0.037
x3 = –1.731 994 802
13
xn − 3
= 2.17 to 2 d.p.
2
a f(x) = x2 + 6x – 13
f(x) = 0
x2 + 6x – 13 = 0
x0 = –3
x2 + 6x = 13
x1 = –2.166 666 667
6x = 13 – x2
x2 = –2.516 129 032
x3 = –2.356 725 146
x4 = –2.426 855 895
x5 = –2.395 493 864
1
5
x3 = 1.376 507 73
x2 = –1.733 134 316
xn +1 =
)
+ 8)
x2 = 1.372 510 297
x1 = –1.709 975 947
8
2
n
x1 = 1.319 507 911
xn + 1 = 4xn2 − xn4
x0 = 2
x1 = 0
x2 = 0 etc.
x0 = 3
x1 = −45
x2 = −409 252 5
So first integer value is x0 = 3.
xn +1 = xn2 + 3xn − 3
x0 = –2
( 12 (x
x0 = 0
x0 = 1.5
x1 = –0.131 944
x2 = –17.684 31
x3 = –32 601.166
x4 = –3.7654 × 1017
6
n
x0 = –3
xn + 1 =
7 − xn5
Yes, for example: xn + 1 = 3x
n
xn +1
 4 2
= 2
 x − 1 
x=
13 − x 2
6
b xn + 1 =
x6 = –2.409 417 993
x0 = 1
x7 = –2.403 216 024
x1 = 2
(13 − x )
n
2
6
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6
WORKED SOLUTIONS
x2 = 1.5
y
x3 = 1.792
2
x4 = 1.632
3
a
–2
y
–2
–1
–1
5
6
7
8
1
2
3
4
–14
Negative root –1 < x < 0.
–3
Positive root 7 < x < 8.
There is one point of intersection so the
equation 3 = 2x has one root.
x
b x0 = 1
3
b xn + 1 = 7 + x
n
x0 = 7
x1 = 1.5
x1 = 7.428 571
x2 = 1.060 66
x2 = 7.403 846
x3 = 7.405 195
x4 = 7.405 121
The positive root is x = 7.41 to 2 d.p.
x3 = 1.438 238
= 1.438 to 3 d.p.
a
x
x
–12
x
–2
4
4
–10
1
–3
3
–8
2
–4
2
–6
3
–5
1
–4
4
0
0
–1
–2
y
5
xn + 1 = 5 + 4 − 7 2
xn xn
x0 = 5
x1 = 5.52
x2 = 5.494 907
x3 = 5.496 113
x4 = 5.4960 55
The solution is x = 5.50 to 2 d.p.
6
a f(x) = x5 + 2x – 7
−1
5
0
−3
1
−9
2
−13
3
−15
4
−15
f(1) = –4
5
−13
f(2) = 29
6
−9
7
−3
8
5
9
15
x5 = 7 – 2x
10
27
x = (7 − 2x)5
A sign change between f(1) and f(2) so a
root of the equation x5 + 2x – 7 = 0 lies in the
interval 1 < x < 2.
b x5 + 2x – 7 = 0
1
1
xn + 1 = ( 7 − 2xn ) 5
a = 7, b = 2
c x0 = 2
x1 = 1.245 731
x2 = 1.351 472
x3 = 1.338 549
x4 = 1.340 155
x = 1.34 to 2 d.p.
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NUMERICAL SOLUTION OF EQUATIONS
d
b f(x) = x3 – 3x + 4
y
f(–2.5) = –15.625 + 7.5 + 4 = –4.125
1.8
1.6
f(–2) = –8 + 6 + 4 = 2
y=x
Sign change so there is a root in the interval
–2.5 < x < –2.
1.4
1.2
1
c xn + 1 = ( 3xn − 4 ) 3
1
0.8
0.6 0.8
7
8
1
1.2 1.4 1.6 1.8
2
x
(
)
x3 = −2.194 03
x4 = –2.195 45
x5 = –2.195 75
Solution is x = –2.20 to 2 d.p.
1
10 A = ab sin C
2
24 = 1 (x)(2x + 1)sin 150 2
2x2 + x – 96 = 0
x 2 = 96 − x
2
xn+1 = 48 − x
2
x0 = 7
x1 = 6.670 832 030
1
3
x 2 = 6.683156 738
x3 = 6.682 695 686
x0 = 2.2
x1 = 2.213 144
x2 = 2.217 085
x3 = 2.218 268
x4 = 2.218 624
x5 = 2.218 731
The solution is x = 2.219 to 3 d.p.
9 a
–5
x 4 = 6.682 712 934
x = 6.683 to 3 d.p.
2x + 1 = 14.3
37 to 2 d.p.
So the side lengths are 6.68 an
nd 14.37 units to 2 d.p.
(
11 a xn +1 = xn 3 + 6
y
–4
–3
–2
x1 = −2.154 43
x2 = −2.187 21
xn + 1 = cos xn
x0 = 0
x1 = 1
x2 = 0.540 30
x3 = 0.857 55
x4 = 0.654 29
x5 = 0.793 48
x6 = 0.701 37
x7 = 0.763 96
x8 = 0.722 10
x9 = 0.750 42
x10 = 0.731 40
x11 = 0.744 24
x12 = 0.735 60
x13 = 0.741 43
Solution is x = 0.74 to 2 d.p.
Coordinates (0.74, 0.74)
xn + 1 = xn 2 + 6
x0 = −2
1
4
x0 = 2
10
x1 = 1.934 336
8
x2 = 1.907 447
6
x3 = 1.896 633
4
x4 = 1.892 318
2
x5 = 1.890 601
0
–1
–2
)
1
2
3
4
x
–4
The equation x3 – 3x + 4 = 0 has only one real
root.
x6 = 1.889 919
x7 = 1.889 648
x8 = 1.889 541
x9 = 1.889 498
x10 = 1.889 481
Solution is x = 1.89 to 2 d.p.
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6
WORKED SOLUTIONS
b
14 xn +1 = e −xn
y=x
2
2
With x0 = –1
the formula converges to the root x = 0.653 (3 d.p.)
1.8
1.6
15 f(x) = x5 – x4 + 2x3 – 3x2 + 1
2
1.4
1.2
f(–1) = – 1 – 1 – 2 – 3 +
1
1
= −6.5
2
f( 0 ) = 1
2
Sign change so there is a root in the interval
–1 < x < 0
0.8
0.6
0.4
0.2
0.2 0.4 0.6 0.8
1
1.2 1.4 1.6 1.8
2
2.2
12 a xn+1 = − 4 − 1 (for negative root)
xn
x0 = −2
x1 = −2.1
121 320 343
x 2 = −2.114 569 583
x3 = −2.114 925 407
x 4 = −2.114 906 597
x = −2.11 to 2 d.p.
b xn + 1 = 4 − 12
xn xn
1 = −1
2
2
Sign change so there is a root in the interval
0<x<1
1
1
f(1) = 1 – 1 + 2 – 3 + = −
2
2
1
f(2) = 32 – 16 + 16 – 12 + = 41
2
2
Sign change so there is a root in the interval
1 < x < 2.
f(1) = 1 – 1 + 2 – 3 +
(
16 xn +1 = xn 2 + 1
x0 = 2
)
1
4
x1 = 1.495 348 781
x3 = 1.293 442 522
x1 = −5
x4 = 1.278 643 508
x2 = −0.84
x3 = −6.179 14
The iteration formula in a converges
quickly to the negative root, whereas the
one in b does not converge with x0 = −1, so
a is preferred for the negative root. With
x0 = 2, the formula in b converges on the
positive root 1.86 (other convergent iterative
formulae may also be possible).
(
f( 0 ) = 1
2
x2 = 1.341 233 552
x0 = −1
5x − 1
13 xn + 1 =
2
x
)
1
3
x0 = −1
x1 = −1.442 249 570
x2 = −1.601 252 135
x3 = −1.651 346 356
x4 = −1.666 515 004
x5 = −1.671 054 053
x6 = –1.672 407 530
x7 = –1.672 810 693
Negative root is x = –1.67 to 2 d.p.
x5 = 1.274 066 916
x6 = 1.272 652 327
x7 = 1.272 215 161
x8 = 1.272 080 066
Convergent to root at x = 1.272 (3 d.p.).
x0 = –2
x1 = 1.495 348 781
x2 = 1.341 233 552
x3 = 1.293 442 522
x4 = 1.278 643 508
x5 = 1.274 066 916
x6 = 1.272 652 327
x7 = 1.272 215 161
x8 = 1.272 080 066
Convergent to root at x = 1.272 (3 d.p.).
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NUMERICAL SOLUTION OF EQUATIONS
(
17 xn +1 = 5x − 1
2
)
1
3
x0 = 4
x1 = 2.117 911 792
x2 = 1.686 253 559
x3 = 1.548 855 749
x4 = 1.499 576 084
x5 = 1.481 087 01
x6 = 1.474 029 625
x7 = 1.471 317 877
x8 = 1.470 273 245
x9 = 1.469 870 43
x10 = 1.469 715 044
Convergent to root at x = 1.470 (3 d.p.).
Mathematics in life and work
1
y
14
12
10
8
6
4
2
–4
2
–3
–2
xn + 1 =
–1
0
1
2
3
x
e xn − 5
3
Negative root:
x0 = −1
x1 = –1.544 040 187
x2 = −1.595 494 431
x3 = −1.599 063 924
x4 = −1.599 304 801
Negative root is x = –1.60 to 2 d.p.
xn + 1 = ln (3x + 5)
x0 = 2
x1 = 2.397 895 273
x2 = 2.500 918 262
x3 = 2.525 949 003
x4 = 2.531 937 093
Positive root is x = 2.53 to 2 d.p.
3
The first iteration formula always converges to
the negative root or diverges. The second iteration
formula always converges to the positive root.
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7
WORKED SOLUTIONS
7 Vectors
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering
the question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge
1
2
2
2
 −2 
iii 
 −11
2
a
(6 − 1) + (−10 − 2) = 5 + (−12) = 13
b
(−5 − 17)2 + (−4 − 7)2 = (−22)2 + (−11)2
(
)
= 112 22 + 12 = 11 5
()
5
8 = 32.0°
2
tan–1
3
c2 = 102 + 172 – 2 × 10 × 17 cos 142°
= 656.92
4
ii
 11 
 4.5
 −20
iii 
 12 
iv
 72 
 −24


 −40
ii
 1
 18
 29 
iii 
 4.25
 4
 −9
 −5
a   +   =  
 −3
 8
 5
2
 4
 −9
 12 
 −9
 21 
b 3  –   =   –   = 
 −3
 8
 −9
 8
 −17
4 2 + (−3)2 = 5
c
c i 10i + 20j
d i 3i – 11j
= 656.92 = 25.6 cm
iv 5i – 10j + 4k
1
2
5
5 + 3t = 9 – u
–1 + 4t = 11 + 2u
Multiply 1 by 2.
10 + 6t = 18 – 2u
Add to 2 .
9 + 10t = 29
10t = 20
t = 2
Substitute t = 2 into 2
–1 + 4(2) = 11 + 2u
7 = 11 + 2u
2u = –4
u = –2
 3
 4
a i
 4
ii
It is the vector which joins the start of  
 1
 −1
to the end of   .
 3
 5
b i  
 −2
ii
Exercise 7.1A
1
a i 2i + 12j
 13 
iii  
 −4
b i
–4i – 2j
ii 7i – 4j
iv
 5
 −3
 
 6 
ii
 0
 10
iii –b is the opposite vector, so is parallel
to (but in the opposite direction to) b.
 5
 −2 is the vector which joins the start of
 4
 −1
 1  to the end of – 3  .
 
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Vectors
 8
c i  
 2
b 2
ii
 5
 7
 19
–u
=2
 −2
 4
 6 
 10 − 7u 
 38
=
 −4 − 4u 
 12 
10 – 7u = 38
iii It is the vector twice as long as and parallel to a.
3
7u = –28
u = –4
 5
 7
 −20 + 21
ua + 3b = –4
+3
=
 −2
 4
 8 + 12 
1
=
 20
1
2
a 5p + 8q = 26 10 – 10q = 15 From 2 , 10q = –5
1
2
1
Substitute q = – into 1 .
2
5p – 4 = 26
q=–
5
5p = 30
p = 6
 6
 2 
 3
k  w + 47 – 4  v  =  31
 


 
 w 
 −4 
 2
2k – 12 = 6
b 2p + 4q = 10 1
7p + 5q = 53 2
k=9
9(–4) – 8 = w
1 × 7:
14p + 28q = 70
3
9(–44 + 47) – 4v = 31
2 × 2:
14p + 10q = 106
4
w = –44
6
v = –1
a p(4i – j) + 2(i + 2j) = k(5i + j)
Equating i coefficients.
3 – 4 :
4p + 2 = 5k
18q = –36
Equating j coefficients.
q = –2
Substitute q = –2 into 1 .
–p + 4 = k
2p – 8 = 10
Substitute k = –p + 4 into first equation.
4p + 2 = 5(–p + 4)
2p = 18
4p + 2 = –5p + 20
p = 9
c 4p + 3q = 18
9p = 18
1
p = 2
2
–p + 5q = 7
b 3(4i – j) + q(i + 2j) = k(i – j).
2 × 4:
–4p + 20q = 28
Equating i coefficients.
3
12 + q = k
1 + 3 :
Equating j coefficients.
23q = 46
–3 + 2q = –k
q = 2
Substitute q = 2 into 1 .
Substitute k = 12 + q into second equation.
4p + 6 = 18
–3 + 2q = –(12 + q)
–3 + 2q = –12 – q
4p = 12
3q = –9
p = 3
4
a
4a – 2b = 4
=
 5
 7
 20 − 14
–2
=
 −2
 4
 −8 − 8 
 6 
 −16
7
q = –3
a Equating i coefficients.
2a + 8b = 6
1
Equating j coefficients.
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7
WORKED SOLUTIONS
–a – 3b = –4 (a – 3)2 = 0
2
2 × 2:
–2a – 6b = –8 a=3
a(a + 3) – 15 = 3b
3
3(3 + 3) – 15 = 3b
1 + 3 :
3b = 3
2b = –2
b = –1
Substitute b = –1 into 1 .
9
2a – 8 = 6
p
b=1
 3
 1
 5
+q
=k
 −1
 3
 3
3p + q = 5k
1
a = 7
–p + 3q = 3k
2
b If a vector is parallel to the x-axis then it has
 1
the form k   .
 0
 1
If a vector of the form k   has a magnitude
 0
of 12, then k = 12.
k = 3p + q = −p + 3q
5
3
2a = 14
3(3p + q) = 5(–p + 3q)
9p + 3q = –5p + 15q
14p = 12q
7p = 6q
Equating j coefficients.
–a – 3b = 0
a = –3b
10 a 8
Equating i coefficients.
 2
 4
k
+3
=
 n 
 13
 79
16 + 12 = k
2a + 8b = 12
Substitute a = –3b into 2a + 8b = 12.
b 8n + 39 = 79
–6b + 8b = 12
2b = 12
b=6
a = –3(6) = –18
c 4
k = 28
n=5
 2
 4
 1
+h
=m
 5
 13
 2
8 + 4h = m
c If a vector is parallel to y = –x then it has the
 −1
form k   .
1
20 + 13h = 2m
20 + 13h = 2(8 + 4h)
20 + 13h = 16 + 8h
Equating i coefficients.
5h = –4
2a + 8b = –k
h = –4
5
Equating j coefficients.
–a – 3b = k
Substitute k = –a – 3b into 2a + 8b = –k.
2a + 8b = –(–a – 3b)
Exercise 7.2A
1
ST = SR + RT = –4a + 6b = 2(–2a + 3b)
R
2a + 8b = a + 3b
a = –5b
Any pair of values for which a = –5b will
satisfy the requirement that the vector is
parallel to y = –x.
8
a
 a + 3
 3
b
–5
=3
 a − 1
 a 
 −3
6b
4a
U
S
V
T
a(a – 1) – 5a = –9
a2 – 6a + 9 = 0
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Vectors
2
RU = 2a
RV = 3b
UV = UR + RV = –2a + 3b
Hence ST = 2UV.
Since the two vectors have a common factor, they
are parallel.
5
From question 2, TU = 3a – 2c
AX = 6a
TX = TA + AX = 3a – 6c + 6a = 9a – 6c = 3(3a – 2c).
TU and TX have a common factor, so are parallel.
Since they also share a common point, T, U and X
are collinear.
a i OB = 6a + 6c
iii OU = 6a + 4c
6
BA + AD + DC = –15a – 14b + 5a = –10a – 14b
BC =
= 2(–5a – 7b)
BE = BA + AE = BA + 1.5AD = –15a + 1.5(–14b)
= –15a – 21b = 3(–5a – 7b)
The two vectors have a common factor so they are
parallel.
ii
AC = –6a + 6c
iv TA = 3a – 6c
v OS = 3a + 3c
vi US = –3a – c
vii UT = –3a + 2c
viii ST = 3c
b Parallel.
ST = 3c and AB = 6c, so AB = 2ST.
3
The two vectors are parallel and include a
common point, so all three points are collinear.
B
a EF = ED + DF = –12a + 8b
MN = MD + DN = –4a + 8 b so EF = 3MN
3
Since EF is parallel to MN, FEMN is a
trapezium.
C
b The ratio 2 : 1 means that the scale factor
DM : DE = 1 : 3 and the area of DEF is 9 times
the area of DMN.
If the area of DEF = 72 units2, then the area of
DMN = 8 units2 and the area of FEMN = 64 units2.
C
B
4
c
E×
a
O
D
×
A
F
a OB = OC + CB = c + a
DE = DA + AE = 1 OA + 1 AB = 1 c + 1 a = 1 (c + a)
2
2
2
2
2
b CF = CO + OF = CO + 2OA = –c + 2a
CE = CB + BE = CB + 1 BA = a + 1 (–c) = – 1 c + a
2
2
2
=
1
(–c + 2a)
2
1 CF
2
The two vectors have a common factor so
they are parallel.
So CE =
Since the two vectors are parallel and include
a common point, all three points are collinear.
E
D
7
Since OABC is a parallelogram, OA = CB = a and
OC = AB = c.
1
So DE = OB.
2
The two vectors have a common factor so
they are parallel.
15a
5a
a OM =
14b
A
2
2
OA = a
3
3
2
MB = MO + OB = – a + b
3
MX = λ  − 2 a + b 

 3
OX = OM + MX =
(
)
2
a + λ  − 2 a + b 
3

 3
= 2 − 2 λ a + λb
3 3
2
2
ON = OB = b
3
3
2
NA = NO + OA = – b + a
3
NX = µ  − 2 b + a 
 3

OX = ON + NX =
2
 2

b + µ − b + a
3
 3

2 2 
=  − µ  b + µa .
3 3 
Coefficients of a:
2 2
– λ = µ, so 3µ + 2λ = 2
3 3
1
Coefficients of b:
2 2
– µ = λ, so 2µ + 3λ = 2
3 3
2
Putting 1 = 2
3µ + 2λ = 2µ + 3λ
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7
WORKED SOLUTIONS
µ=λ
λ +
2λ + 3λ = 2
λ(p + q) + pλ
p
=
p+q
p+q
5λ = 2
λ=
2
2
,µ=
5
5
λ(p + q) + pλ = p
2
2
2
Hence OX =  2 − 2 × 2  b + a = a + b
5
5
5
 3 3 5
pλ + qλ + pλ = p
λ(2p + q) = p
2
(a + b).
5
p
p
OA =
b OM =
a
p+q
p+q
=
MB = MO + OB = –
OX = OM + MX =
p
a+b
p+q
8
p
p


a + λ −
a + b
p+q
 p+q

NM = 9b.
P could be dividing NM in the ratio 1 : 2, so
MP = –6b.
p
p
ON = p + q OB = p + q b
p
NA = NO + OA = – p + q b + a
LP = LM + MP = 6a – 6b.
Alternatively, N could be dividing MP in the
ratio 1 : 1,
so MP = –18b.
p


b + a
NX = µ  −
 p+q

LP = LM + MP = 6a – 18b.
9
p
p


b + a
OX = ON +NX = p + q b + µ  −
 p+q

p
 p

−
µ b + µa.
= 
 p + q p + q 
p
p
–
λ = µ,
p+q p+q
p
p
λ=
p+q
p+q
Coefficients of b:
Putting 1 = 2
p
p
µ+λ
µ +
λ=
p+q
p+q
µ(p + q) + pλ
pµ + λ(p + q)
=
p+q
p+q
µ(p + q) + pλ = pµ + λ(p + q)
pµ + qµ + pλ = pµ + pλ + qλ
qµ = qλ
µ=λ
1
p
p
–
µ = λ, so
p+q p+q
p
p
µ+λ=
p+q
p+q
a
JK and LM are parallel, so the angle at J is equal
to the angle at M (alternate angles), the angle at
K is equal to the angle at L (alternate angles), and
angle JNK is equal to angle MNL (opposite angles).
Therefore JKN and MLN are similar triangles.
b JN = 6b
p
 p

−
λ a + λb
= 
 p + q p + q 
so µ +
p
λ=
2p + q
and the required result follows.
p


a + b
MX = λ  −
 p+q

Coefficients of a:
p
p
λ= p+q
p+q
2
JG = 4b
GF = 12a
GK = 4a
JK = 4b + 4a = 4(b + a)
Since JQ is parallel to JK, JQ can be given by
JQ = n(b + a).
EF = 8b
Since FQ is parallel to EF, FQ can be given by FQ = kb.
JQ = JF + FQ = 4b + 12a + kb = (k + 4)b + 12a
Hence (k + 4)b + 12a = n(b + a).
n = 12
k + 4 = 12
k = 8
DQ = DE + EF + FQ = 12a + 8b + 8b = 12a + 16b
10 PQ = OR = 4r.
Since PQU is a straight line, QU is parallel to PQ ,
so QU = kr.
OS = 1 (4r + 6p) = 2r + 3p
2
OT = 4r + 5p
ST = SO + OT = –(2r + 3p) + (4r + 5p) = 2r + 2p
= 2(r + p)
Since STU is a straight line, TU is parallel to ST,
so TU = n(r + p).
QU = QT + TU = –p + n(r + p)
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Vectors
Hence kr = –p + n(r + p).
Equating coefficients for p.
–1 + n = 0
n=1
Equating coefficients for r.
k=n
k=1
OU = OP + PQ + QU = 6p + 4r + r = 6p + 5r.
3
 −a 
b   where a > 0
 a
4
11, 3 is possible because 112 + 32 = 130. 92 + 72 also
equals 130. Apart from the negatives of these, there
are no other integer pairs with this property.
All these are possible:
11, 3; 11, –3; –11, 3; –11, –3; 3, 11; 3, –11; –3, 11; –3, –11;
9, 7; 9, –7; –9, 7; –9, –7; 7, 9; 7, –9; –7, 9; –7, –9
There are 16 possible pairs of values.
5
a
= 13
1
Unit vector = (–5i + 12j).
13
1
52 × (–5i + 12j) = –20i + 48j
13
b
632 + 16 2 = 65
1
Unit vector =
(63i + 16j).
65
1
(63i + 16j) = 189i + 48j
195 ×
65
Exercise 7.3A
1
()
7
b tan ( 10 ) = 35.0°
4
a tan–1 7 = 29.7°
–1
aBoth are correct. If a is positive, then the
bearing is 045° but if a is negative then the
bearing is 225°.
180° – 35.0° = 145.0°
()
c tan–1 5 = 39.8°
6
180° – 39.8° = 140.2°
Angle = –140.2°.
9
d tan–1 4 = 66.0°
()
Angle = –66.0°.
2
a i Magnitude = 1822 + ( −120 ) = 218.
2
ii
c
122 + ( −8 ) + ( −9) = 17
2
2
12 ×
2
52 + ( −4 ) + 32 = 5 2
v
( −7 )2 + ( −7 )2 + ( −7 )2
iii
1
(143i – 24j).
145
9 26 ×
1
(–72i + 33j + 56k).
97
iv 112 + 132 + 8 2 = 354
Unit vector =
1
(11i + 13j + 8k).
354
5 + ( −4 ) + ( −2) = 3 5
2
2
Unit vector =
32 + (−1)2 + 4 2 = 26
Unit vector =
= 97
2
1
(5i – 4j – 2k).
3 5
6
1
(3i – 6j + 2k).
7
1
(3i – 6j + 2k) = 6i – 12j + 4k
7
2
( −72)2 + 332 + 562
Unit vector =
v
e
1432 + ( −24 ) = 145
Unit vector =
32 + (−6)2 + 22 = 7
14 ×
1
(–4i + 3j).
5
1
(i – 2j + 2k).
3
1
(i – 2j + 2k) = 4i – 8j + 8k
3
Unit vector =
=7 3
=5
Unit vector =
ii
d
2
( −4 )2 + 32
b i
12 + (−2)2 + 22 = 3
Unit vector =
iii 46 2 + ( −46 ) + 232 = 69
iv
( −5)2 + 122
1
(3i – j + 4k).
26
1
(3i – j + 4k) = 27i – 9j + 36k
26
aAssume that the position vector for Room 168
can be written as (2i + 9j + 7k) and for the two
toilet blocks as (7i + j + 5k) and (8i + 3j + 10k).
The displacement of the first toilet block
from Room 168 is given by (7i + j + 5k) –
(2i + 9j + 7k) = 5i – 8j – 2k.
Distance = 52 + ( −8 ) + ( −2) =
2
2
93 > 9 .
The displacement of the second toilet block
from Room 168 is given by (8i + 3j + 10k) –
(2i + 9j + 7k) = 6i – 6j + 3k.
Distance = 6 2 + ( −6 ) + 32 = 81 = 9.
2
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7
WORKED SOLUTIONS
Since the second toilet block is not further
than 9 units from Room 168, the manager’s
rule is satisfied for Room 168.
 4
 −7
 −3


 


For MX,  4  –  −2 =  6  .


 1 
 8 
 7
b It has been assumed that the units are
equally spaced for all three axes.
Length of MX =
( −7 )2 + 62 + 12
= 86 .
Hence both L and M lie on the surface of S2.
 14 + 9 
2
2

2  

 9 
9
OS =  14 +
km
7 a OS = 
 +  −
 = 21.34
10 (u – 3)2 + (u + 1)2 + (u – 2)2 = 172

2
2
9 
 −


2 
u2 – 6u + 9 + u2 + 2u + 1 + u2 – 4u + 4 = 289
2
2


 9 
9
OS =  14 +
 +  −
 = 21.34 km
3u2 – 8u – 275 = 0

2
2
(3u + 25)(u – 11) = 0

9 
9
b OS =  14 +
j
i −

2
2
u = – 25 or 11
3




1
Exercise 7.4A
= −17.4o bearing is
c tan −1  − 9 ×
 14 + 9  
2






2
 3  5 
 −2




o
1 a a – b = 8 – −1 =  9 
90 + 17.4 = 107.4
   
 
 2  3 
 −1
8 Vectors a and b are given by (–3i + 4j + 12k) and
(7i + 39j – 2k) respectively.
a
( −3)
2
 −4  3
 −7 




b c – a = 3 – 8 =  −5 
   


 −9  2
 −11
+ 4 2 + 122 = 13
b b – a = (7i + 39j – 2k) – (–3i + 4j + 12k)
= 10i + 35j – 14k
 2
c b – a = –(a – b) =  −9
 
 1 
10 2 + 352 + ( −14 ) = 39
2
c 3(–3i + 4j + 12k) + c(7i + 39j – 2k)
= –23i – 66j + 40k
 5
 −4
 9
d b – c =  −1 –  3  =  −4
 
 
 
 3 
 −9
 12 
3(–3) + 7c = –23
c = –2
9
2
2
aDistance of A from centre = 7 + ( −3) + 3
2
2
rH – rT = (–i + 7j) – (3i – 9j) = (–4i + 16j) km
= 67 .
Distance of B from centre
rH – rM = (–i + 7j) – (–2i – 4j) = (i + 11j) km
rS – rT = (5i + 2j) – (3i – 9j) = (2i + 11j) km
= 22 + ( −8 ) + ( −1) = 69 .
2
rH – rS = (–i + 7j) – (5i + 2j) = (–6i + 5j) km
2
rS – rM = (5i + 2j) – (–2i – 4j) = (7i + 6j) km
Since B is further than A from the centre, B
lies outside the surface of S1.
  −4  −2 
 −3
 
1  6  +  2 
b Centre of S2, X =       =  4  .
2    
 8 
  −1  17  
rT – rM = (3i – 9j) – (–2i – 4j) = (5i – 5j) km
a i supermarket and town hall
ii supermarket and museum
iii town hall and hospital
b (i + 11j) km
For radius of S2, d2 = (–3 – (–4))2 + (4 – 6)2
+(8 – (–1))2 = 12 + (–2)2 + 92 = 86.
Radius of S2 = 86 .
c magnitude of (5i – 5j) = 52 + ( −5) , so town
hall and museum
2
3
 −3
 2
 −5
 4
 10
 
For LX,   –   =  −6 .
 8 
 3 
 5 
ED = (11i + 5j – 4k) – (7i – 3j + 4k) = (4i + 8j – 8k)
EF = (15i + j + 12k) – (7i – 3j + 4k) = (8i + 4j + 8k)
DF = (15i + j + 12k) – (11i + 5j – 4k) = (4i – 4j + 16k)
ED 2 = 4 2 + 8 2 + ( −8 ) = 144
2
Length of LX =
( −5)2 + ( −6 )2 + 52
= 86 .
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Vectors
8
EF 2 = 8 2 + 4 2 + 8 2 = 144
DF 2 = 4 2 + ( −4 ) + 16 2 = 288
2
Magnitude of TU = 32 + 4 2 + ( −11) = 146 .
2
Hence ED 2 + EF 2 = DF 2
TV = v – t = (3i + 27j + 2k) – (10i + 8j – 3k) =
–7i + 19j + 5k
Therefore DEF is a right angle
4
a 1
2
 6 
 20   −8  
 123 



 
 23  +  100  =  2 

 −15  29   

  7 
2
2
Magnitude of TV = ( −7 ) + 19 + 5 = 435 .
UV = v – u = (3i + 27j + 2k) – (13i + 12j – 14k)
= –10i + 15j + 16k
2
 −8   20 
 −28
b b – a =  100 –  23  =  77 

 



 29   −15
 44 
( −28 )2 + 772 + 442
5
Magnitude of UV =
= 93
c Since AD and BC are parallel but different
lengths, ABCD is a trapezium
 −11  2   −13
DE = e – d =  3  –  −8 =  11 

   

 −7   5   −12
( −13)2 + 112 + ( −12)2
= 434 .
( −8 )2 + 72 + ( −3)2
(c + 1)2 + (–5 – c)2 + (c + 10)2 = 192
c2 + 2c + 1 + 25 + 10c + c2 + c2 + 20c + 100 = 361
3c2 + 32c + 126 = 361
3c2 + 32c – 235 = 0
b (3c + 47)(c – 5) = 0
− 47
c=
or 5
3
Since c > 0, c = 5
 5 − 1  4 

  
=
OH  −4  =  −4
 5 + 2  7 
10 aPQ = q – p = (–65i – 11j + 32k) – (43i + 145j +
383k) = –108i – 156j – 351k
= 122 .
Magnitude of EF = 52 + ( −4 ) + 92 = 122 .
Since DF = EF ≠ DE, the triangle is isosceles.
2
a ST = (–2i + 3j) – (4i – 6j) = –6i + 9j
( −6 )2 + 92
 c − 1   −2 
 c +1 
a GH = h – g =  −4  –  c + 1 =  −5 − c 

 



 c + 2  −8 
 c + 10 
2
 −6  −11  5 
EF = f – e =  −1 –  3  =  −4
  
  
 2   −7   9 
7
9
4 2 + ( −4 ) + 7 2 = 9
 −6  2   −8
DF = f – d =  −1 –  −8 =  7 
     
 2   5   −3
Magnitude of DF =
= 581 .
b Since TU 2 + TV 2 = UV 2, the triangle is right-angled.
 18  6   12 
 4
AD = d – a =  1  –  7  =  −6 = 3  −2
a
     
 
 1   −2  3 
 1 
 
Magnitude of DE =
( −10 )2 + 152 + 162
Since TU ≠ TV ≠ UV, the triangle is scalene.
 9
 −11
 20 
 4
b
BC = c – b =  −2 –  8  =  −10 = 5  −2
 




 
 12 
 7 
 5 
 1 
6
a
TU = u – t = (13i + 12j – 14k) – (10i + 8j – 3k)
= 3i + 4j – 11k
PQ =
( −108 )2 + ( −156 )2 + ( −351)2
= 399
b PR = r – p = (169i + 61j – 72k) – (43i + 145j
+ 383k) = 126i – 84j – 455k
PR 2 = 1262 + (–84)2 + (–455)2 = 229 957
QR = r – q = (169i + 61j – 72k) – (–65i – 11j
+ 32k) = 234i + 72j – 104k
QR2 = 2342 + 722 + (–104)2 = 70 756
Since 3992 + 70 756 = 229 957,
PQ 2 + QR 2 = PR 2 and PQR is a right angle
= 3 13
b ST = (–2i + 3j) – (qi – 6j) = (–2 – q)i + 9j
(–2 – q)2 + 92 = 152
(–2 – q)2 = 225 – 81 = 144
c QR = 70756 = 266
1
Area = × 266 × 399 = 53 067
2
–2 – q = ±12
q = 10 or –14
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7
WORKED SOLUTIONS
Exercise 7.5A
Note: there is more than one correct solution
for Questions 1, 2 and 3.
1 a r = 5i + 3j + t(–7i + j)
b r = –2i + 8k + t(–3i + 2j – k)
5
b (10i + 5j – 7k) + t(3i – i – 2k) = 19i + 2j – 13k
t = 3 for all three, so (19i + 2j – 13k) lies on
the line.
 9
 4
c r=   +t 
 −2
 3
 10 
 0
 
 
d r =  3  + t  −1
 −6
 3 
2
c i a=8
ii b = 11, c = 5
iii d = 4, e = 7
 2  3
 −1 
 −4  7 


a GH =   –   =  −11
 3   −9
 12 
iv f = 13, g = –9
d Point on the x–z plane has y coordinate = 0
 3   10 + 3t 
 10 
So  5  + t  −1 =  0 

  
 
 2   −7 + 2t 
 −7
 3
 −1 
 7


r =   + t  −11
 −9
 12 
 −26
 −30
 4
 −12
 −6 
 −6
b PQ = 
 – 
 =  
 33 
 30 
 3 
Therefore t = 5 and the point has coordinates
(10 + 3(5), 0, – 7 + 2(5)) = (25, 0, 3)
6
 −30
 4
 −6 
 −6
r= 
 +u 
 30 
 3 
3
4
 7
 3
 31 
 −2
 −4 


When t = 8, r =   + 8   =  −34 .
 19 
 −2
 3 
7
 11
 7
 v 
c   +t  = 
5
−
2
 
 
 −3v 
11 + 7t = v
5 – 2t = –3v
5 – 2t = –3(11 + 7t)
a
UV = (2i + 5j – 3k) – (pi – 3j – k)
= ((2 – p)i + 8j – 2k)
r = (2i + 5j – 3k) + t((2 – p)i + 8j – 2k)
 2 + ( 2 − p )t 
 7
b  5 + 8t  =  1 


 
 −3 − 2t 
 q 
 11
 39 
 7
a   +t  =  
 5
 −3
 −2
11 + 7t = 39
t=4
5 – 2t = –3
t=4
 39 
Since t = 4 for both x and y,
lies on the line
 −3
 11
 7
 −10
b r=   –3  = 
 5
 −2
 11 
A is the point on l for which t = 2. B is the point on
l for which t = 5. So for point C, t − 2 = 2 t − 5 .
Point C has either t = 4 or t = 8
 7
 3
 19 
 −2
 −4 


When t = 4, r =   + 4   =  −18 .
 19 
 −2
 11 
 1   11 
 −10
 4   −6


AB =   –   =  10 
 29  9 
 20 
 −1
 
Simplified direction vector =  1 
 2 
 11 
 −1
 
 
r =  −6 + t  1 
 9 
 2 
5 – 2t = –33 – 21t
19t = –38
t = –2
v = 11 + 7(–2) = –3
a r = (10i + 5j – 7k) + t(3i – i – 2k)
5 + 8t = 1
t = –1
2
2 – 1 (2 – p) = 7
2
p = 12
1
c –3 – 2 − = q
2
q = –2
( )
8
a
c – b = (14i + j + 5k) – (–4i + 13j – k) = 18i – 12j +
6k = 6(3i – 2j + k)
r = (14i + j + 5k) + t(3i – 2j + k)
 2
 14 + 3t 
 p  =  1 − 2t 
 


 q 
 5 + t 
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Vectors
2 = 14 + 3t
BC 2 = 32 + (–36)2 + 192 = 1666
t = –4
Since 352 + 441 = 1666, AB2 + AC 2 = BC 2 and
angle BAC is a right angle
p = 1 – 2(–4) = 9
q=5–4=1
d Area =
 14 + 3t 
 0
b  1 − 2t  =  y 


 
 5 + t 
 z 
Exercise 7.6A
14 + 3t = 0
t = –14
3
( )
( )
y = 1 + − 14 (–2) = 31
3
3
14
1
=
z = 5 + −
3
3
31 1
0, ,
3 3
(
)
a=
2
 15 
 3  8
 2
 10 
 2   12 
 3
a

 = 5   ,   = 4   : not parallel,
 −20
 −4  24
 6
since they do not have a common factor
 27 
 9 
 −42
 −14
b 
 =3
 : not parallel,
 −36
 −12
AC = 122 + ( −8 ) + 4 2 = 224 = 4 14
2
t2 + 6t + 9 + 4t2 – 8t + 4 + 9t2 – 36t + 36 = 841
 −3
 −6
 


c −4 = 2  −2 : not parallel. For same
 
 5 
 10 
14t2 – 38t + 49 = 841
14t2 – 38t – 792 = 0
7t2 – 19t – 396 = 0
coefficient of i, the other coefficients are
(7t + 44)(t – 9) = 0
44
t=−
or 9
7
44
When t = − , point is  − 23 ,− 102 ,− 174  .
7
 7
7
7 
When t = 9, point is (12, 16, 21).
 −3
 −3
 −2
 
different,   and  −1
 5 
 13 
 12 
 3   21 
 3
 −8 
 −2  −14
 −2
d   =4  , 
 = 7   : parallel,
 32 
 8   56 
 8 
10 a
r = (–10i + 4j + k) + t(3i + 6j – 2k) = 5i + 34j – 9k
–10 + 3t = 5
they have a common factor
t=5
3
4 + 6t = 34
 9 
 −14

 is not
 −12
 9 


the same as  −16
 −12
(3 + t)2 + (–2 + 2t)2 +(–6 + 3t)2 = 292
4
8
× –2 = −
3
3
1
c AC = (14i + j + 5k) – (2i + 9j + k) = 12i – 8j + 4k
9
1
735
× 35 × 21 =
2
2
Equating i coefficients: 3 – t = –30 + 4u.
Equating j coefficients: 7 – 11t = –6 – 6u.
t=5
Equating k coefficients: –9 + 12t = 30 + 3u.
1 – 2t = –9
Adding equations: 1 = –6 + u.
t = 5
t = 5 in all three cases, so B lies on the line.
b AB = (5i + 34j – 9k) – (–10i + 4j + k)
= 15i + 30j – 10k
AB = 152 + 30 2 + ( −10 ) = 35
2
c AC = (8i – 2j + 10k) – (–10i + 4j + k)
= 18i – 6j + 9k
AC 2 = 182 + (–6)2 + 92 = 441
BC = (8i – 2j + 10k) – (5i + 34j – 9k) = 3i – 36j + 19k
u=7
Point of intersection = (–2, –48, 51).
4
Equating i coefficients: –2 + t = 4 – u.
1
Equating j coefficients: 2 + 2t = 4 + 3u.
2
Equating k coefficients: –1 – t = –7 + u.
3
Double equation 1 : –4 + 2t = 8 – 2u.
Subtract from 2 : 6 = –4 + 5u.
u=2
Point of intersection = (2, 10, –5).
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7
WORKED SOLUTIONS
5
c RT = (12i – 23j + ak) – (–8j – k)
= 12i – 15j + (a + 1)k
r = –8j – k + v(12i – 15j + (a + 1)k)
 −3
a Lines intersect at  11 
 
4 
b Lines are skew
c Lines are parallel
6
1
Equating j coefficients: 2 – 5t = –8 – 15v
2
Equating k coefficients: 4 – t = –1 + v(a + 1) 3
a
AB = (–2i + 10j – 3k) – (–8i + 6j + k) = 6i + 4j – 4k
= 2(3i + 2j – 2k)
Multiply 1 by 5:
r = (–8i + 6j + k) + t(3i + 2j – 2k)
–25 + 15t = 60v
b CD = (–4i + 3j + k) – (–3i – 2j + 3k) = –i + 5j – 2k
6 – 15t = –24 – 45v
c Equating i coefficients: –8 + 3t = –3 – u
1
Add 4 and 5 :
Equating j coefficients: 6 + 2t = –2 + 5u
2
Equating k coefficients: 1 – 2t = 3 – 2u
3
–19 = –24 + 15v
v= 1
3
t=3
Substitute for v and t in equation 3 :
4 – 3 = –1 + 1 (a + 1)
3
a=5
Add 2 and 3 :
7 = 1 + 3u
u=2
t=1
E(–5, 8, –1)
d AE = (–5i + 8j – k) – (–8i + 6j + k) = 3i + 2j – 2k
EB = (–2i + 10j – 3k) – (–5i + 8j – k) 3i + 2j – 2k
Ratio = 1 : 1
a PQ = (–2i – 3j + 3k) – (–5i + 2j + 4k) = 3i – 5j – k
r = –5i + 2j + 4k + t(3i – 5j – k)
RS = (12i – 23j + 8k) – (–8j – k) = 12i – 15j + 9k
8
5
 −5 + 3( 3)
 4 


=  −13
d
 2 − 5( 3) 


 4 − 3 
 1 
a AB = (–2i + 11k) – (–3i – j + 12k) = i + j – k
Equation of l2: r = 11i + 9j + 12k + t(i + j – k)
b CD = (15i + 11j + 15k) – (11i + 9j + 12k)
= 4i + 2j + 3k
Equation of l 3: r = 15i + 11j + 15k + u(4i + 2j + 3k)
= 3(4i – 5j + 3k)
r = –8j – k + u(4i – 5j + 3k)
b Equating i coefficients: –5 + 3t = 4u
4
Multiply 2 by 3:
r = (–3i – 2j + 3k) + u(–i + 5j – 2k)
7
Equating i coefficients: –5 + 3t = 12v
Equation of l 1: r = –2i + 11k + v(i + j – k)
1
Equating i coefficients: –2 + v = 15 + 4u
Equating j coefficients: 2 – 5t = –8 – 5u
2
Equating j coefficients: v = 11 + 2u 2
Equating k coefficients: 4 – t = –1 + 3u
3
Equating k coefficients: 11 – v = 15 + 3u
Multiply 3 by 3
12 – 3t = –3 + 9u
1
3
Add 1 and 3 :
4
9 = 30 + 7u
Add 1 and 4 :
u = –3
7 = –3 + 13u
v=5
u = 10
13
t = 35
13
Check these values for u and t in equation 2
2 – 5 35 = –149
13
13
(3, 5, 6)
( )
–8 – 5 10 = –154
(13 ) 13
Since these values are not the same, the lines
are skew.
 −4  −6 
 2
 1
9 QS =  6  –  14  =  −8 = 2  −4
  

 
 
 −3  −19
 16 
 8 
 −4
 1
Equation of l 1: r =  6  + t  −4
 
 
 8 
 −3
 1
 4
Equation of l 2: r =  9 + t  7 
 
 
 9
 4
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Vectors
Equating i coefficients: –4 + t = 1 + 4u
1
Equating j coefficients: 6 – 4t = 9 + 7u
2
Equating k coefficients: –3 + 8t = 9 + 4u
3
 −11
 8
 −3




d CG = 5 – 9 =  −4


 
 
 15 
 −8
 7 
CG 2 = 82 + (–4)2 + (–8)2 = 144
 −5
 −11
 6 
CF =  9  –  9  =  0 
 




 3 
 15 
 −12
Subtract 1 from 3 :
1 + 7t = 8
t=1
u = –1
CF 2 = 62 + (–12)2 = 180
 −3
 −5
 2
 
 
 
FG =  5  –  9  =  −4
 7 
 3 
 4 
U(–3, 2, 5)
 1
 −4
 5
=   –   =  
6
9
3
ST  
 
 
 9
 −3
 12
FG 2 = 22 + (–4)2 + 42 = 36
ST 2 = 52 + 32 + 122 = 178
 −3
 −4
 1
 
 
 
SU =  2  –  6  =  −4
 5 
 −3
 8 
Since 36 + 144 = 180, FG 2 + CG 2 = CF 2 and
angle CGF is a right angle
 −5
 −1
 −4
e EF =  9  –  4  =  5 
 
 
 
 3 
 5 
 −2
SU 2 = 12 + (–4)2 + 82 = 81
 −3
 1
 −4




TU =  2  –  9 =  −7
 5 
 9
 −4
EF =
=3 5
 −3
 2   −5 
 
DG =  5  –  −5 =  10 
 7 
 17   −10
TU 2 = (–4)2 + (–7)2 + (–4)2 = 81
Since SU = TU ≠ ST, the triangle is isosceles
 −1
 −11
 10 
 2
10 a CE =  4  –  9  =  −5  = 5  −1
 




 
 5 
 15 
 −10
 −2
DG =
225 = 15
 −3
 −1
 −2
EG =  5  –  4  =  1 
 
 
 
 7 
 5
 2
 2
 −1
Equation of l 1: r =  4  + t  −1
 
 
 −2
 5 
EG = 9 = 3
Area of quadrilateral CDEF = Area of triangle
EFC + Area of triangle ECD
 −5
 2
 −7 
1
b DF =  9  –  −5 =  14  = –7  −2
 
 


 
 3 
 17 
 −14
 2 
=
1
1
× 15 × 6 + × 15 × 15 = 315 = 157.5
2
2
2
Exercise 7.7A
 −5
1


Equation of l 2: r = 9 + u  −2
 
 
 3 
 2 
1
a • b = (4i – 2j + 5k) • (3i + 4j – 2k)
a • b = 4 × 3 + –2 × 4 + 5 × –2
c Equating i coefficients: –1 + 2t = –5 + u
1
Equating j coefficients: 4 – t = 9 – 2u
2
Equating k coefficients: 5 – 2t = 3 + 2u
3
Add 1 and 3 :
4 = –2 + 3u
u=2
t = –1
 −3
G 5 
 
7
( −4 )2 + 52 + ( −2)2
a • b = 12 – 8 – 10 = –6
Magnitude of (4i – 2j + 5k) = 4 2 + (−2)2 + 52 = 45 .
Magnitude of (3i + 4j – 2k) = 32 + 4 2 + (−2)2 = 29 .
a•b
cos θ = a b =
−6
45 × 29


−6
θ = cos–1 45 × 29  = 99.6°
Acute angle = 180° – 99.6° = 80.4°.
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7
WORKED SOLUTIONS
2
3
7
The scaler product of the direction vectors
= 35 – 57 + 22 = 0.
Magnitude of BA = (−8)2 + 4 2 = 4 5 .
Therefore cos θ = 0, θ = 90˚, so the lines are
perpendicular.
BC = (14i – 15j) – (10i – 7j) = 4i – 8j
(2i – j + 4k) – (–3i + 5j + 7k) = 5i – 6j – 3k
Magnitude of BC = 4 2 + (−8)2 = 4 5
(6i – 5j + 3k) – (i + 2j – k) = 5i – 7j + 4k
BA • BC = (–8i + 4j) • (4i – 8j) = –32 + –32 = –64
(5i – 6j – 3k) • (5i – 7j + 4k) = 25 + 42 – 12 = 55
cos θ =
Magnitude of (5i – 6j – 3k) = 52 + (−6)2 + (−3)2
2
2
ii sin θ = 1 − − 4
5
2
Magnitude of (5i – 7j + 4k) = 5 + (−7) + 4 = 90 .
a•b
cos θ = a b =
55
70 × 90
Area of ABC =
=
3
5
1
3
× 4 5 × 4 5 × = 24 unit2.
2
5
Magnitude of DE = (−3)2 + 4 2 = 5.
AO = –3i – 5j + 4k
DG = (6i + 7j) – (4i – 2j) = 2i + 9j
AB = (7i + 4j – 3k) – (3i + 5j – 4k) = 4i – j + k
AO • AB = (–3i – 5j + 4k) • (4i – j + k)
Magnitude of DG = 22 + 92 = 85 .
AO • AB = –3 × 4 + –5 × –1 + 4 × 1 = –12 + 5 + 4 = –3
DE • DG = (–3i + 4j) • (2i + 9j) = –6 + 36 = 30
Magnitude of AO = (−3)2 + (−5)2 + 4 2 = 5 2 .
2
2
cos D =
2
Magnitude of AB = 4 + (−1) + 1 = 3 2 .
cos θ =
 6 
7
sin D = 1 − 
=
 85 
85
AB • AC = 4 × 9 + 9 × 1 + 1 × 4 = 36 + 9 + 4 = 49
2
Area of DEG =
2
Magnitude of AB = 4 + 9 + 1 = 7 2 .
AB • AC
1
49
=
=
AB AC
7 2×7 2 2
sin θ = 1 −
( 12 )
Area of ABC =
2
=
3
2
1
3 49 3
×7 2 ×7 2 ×
=
.
2
2
2
a (i + 3uj – 2k) • (–ui + uj + 2k) = –u + 3u2 – 4
–u + 3u2 – 4 = 10
3u2 – u – 14 = 0
(3u – 7)(u + 2) = 0
7
u = or –2
3
b Since u < 0, u = –2.
a = (i – 6j – 2k) and b = (2i – 2j + 2k)
Magnitude of a = 12 + (−6)2 + (−2)2 = 41 .
Magnitude of b = 22 + (−2)2 + 22 = 12 = 2 3 .
cos θ = a • b =
a b
1
7
35
× 5 × 85 ×
=
.
2
2
85
Area of parallelogram DEFG =
Magnitude of AC = 92 + 12 + 4 2 = 7 2 .
cos θ =
DE • DG
6
30
=
=
DE DG 5 × 85
85
2
AO • AB
−3
1
=
=−
10
AO AB
5 2×3 2
2
6
2
b DE = (i + 2j) – (4i – 2j) = –3i + 4j


55
θ = cos–1 70 × 90  = 46.1°
5
BA • BC
−64
=
=−4
5
BA BC 4 5 × 4 5
( )
= 70 .
4
a i BA = (2i – 3j) – (10i – 7j) = –8i + 4j
8
35
× 2 = 35 unit2.
2
 4   −4
 8




a AB = 17 – 13 =  4 
   
 
 14  22 
 −8
 2   −4
 6 




 −42
=
–
=
−29
AD

  13 


 7   22 
 −15
 8  6 
 4   −42
  • 
 = 48 – 168 + 120 = 0, hence
 −8  −15
AB and AD are perpendicular.
b OC = OB + BC = OB + AD = OB + OD – OA
4 
 2   −4  10 


= 17 +  −29 –  13  =  −25
 

   

 14
 7   22   −1 
5
10
=
123
41 × 2 3
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Vectors
d (–2i + k) • (2i + 5j + 4k) = –4 + 0 + 4 = 0
c Magnitude of AB = 8 2 + 4 2 + (−8)2 = 12.
therefore vectors are perpendicular and θ = 0
Magnitude of AD = 6 2 + (−42)2 + (−15)2 = 45.
e AP = (3i – 2j + 7k) – (15i – 2j + k) = –12i + 6k
Area of rectangle = 12 × 45 = 540.
d The centre of the rectangle is the mid-point of BD.
 4   2    3 
1  17 +  −29  

  = −6
2   
 14  7    10.5


 
9
 a  2 
 b   −4
  •   = 0.
 c   −7
1
× 6 5 × 9 5 = 135.
2
Equating j coefficients: 16 – 3t = 31 – 6u.
2
Equating k coefficients: 1 – 6t = –26 + 7u.
3
Add to equation 1 and 2 : 11t = 11.
P(–1, 13, –5)
t=1
 −8
 7
For line L1, when t = 3, r =  16  + 3 −3
 1 
 −6
Add the equations:
3a – 9c = 0
a = 3c
Q(13, 7, –17)
Substitute a = 3c into the first equation:
 13 
 −1
 14 
 7 
 13 
 −6 
PQ = 
 –   = 

 −17
 −5
 −12
3c + 4b – 2c = 0
c + 4b = 0
c = –4b
Magnitude of PQ = 14 2 + (−6)2 + (−12)2 = 376 .
Let b = 1. Then
c = –4
 −10
 −9 + 3u 
 3   −1
 31  + u  −6  13 
 18 − 6u 
PR = 


  –   = 
 −26
 −21 + 7u 
 7   −5
a = –12
 −12


Therefore  1  is perpendicular to both
 −4 
 2
 
and  −4 .
 −7
1
Double equation 1 : –16 + 14t = –20 + 6u.
2a – 4b – 7c = 0
1
 4
 
 −2
Since PQ = PR:
(–9 + 3u)2 + (18 – 6u)2 +(–21 + 7u)2 = 376
9u2 – 54u + 81 + 36u2 – 216u +
324 + 49u2 – 294u + 441 = 376
94u2 – 564u + 470 = 0
10 a
AB = (11i – 2j + 3k) – (15i – 2j + k) = –4i + 2k
= 2(–2i + k)
u2 – 6u + 5 = 0
(u – 1)(u – 5) = 0
Equation of L1 is (15i – 2j + k) + t(–2i + k).
b CD = (–i – 12j – k) – (9i + 13j + 19k)
u = 1 or 5
Since u > 3, u = 5.
= –10i – 25j – 20k = –5(2i + 5j + 4k)
Hence R(5, 1, 9).
Equation of L2 is (9i + 13j + 19k) + u(2i + 5j + 4k).
c Equating i coefficients: 15 – 2t = 9 + 2u.
1
Equating j coefficients: –2 = 13 + 5u.
2
Equating k coefficients: 1 + t = 19 + 4u.
3
P(3, –2, 7)
Magnitude of CP = (−6)2 + (−15)2 + (−12)2 = 9 5
11 a Equating i coefficients: –8 + 7t = –10 + 3u.
a + 4b – 2c = 0
From 2 , u = –3.
CP = (3i – 2j + 7k) – (9i + 13j + 19k) = –6i – 15j – 12k
f Area =
 a
 
Let the vector be  b  .
 c 
 a  1 
   
Hence  b  •  4  = 0 and
 c   −2
Magnitude of AP = (−12)2 + 6 2 = 6 5
 6 
 −9 + 3(5) 




PR = 18 − 6(5) =  −12



 −21 + 7(5)
 14 
PQ • PR = 14 × 6 + –6 × –12 + –12 × 14
= 84 + 72 – 168 = –12
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7
WORKED SOLUTIONS
b CD = d – c = (–6i – 2j) – (3i + 6j) = –9i – 8j
FE = e – f = (–5i – 14j) – (4i – 6j) = –9i – 8j
Magnitude of PR = 6 2 + (−12)2 + 14 2 = 376 .
−12
3
=−
94
376 × 376
cos θ = PQ • PR =
PQ PR
( 943 )
b sin P = 1 − −
Area of ABC =
2
=
DE = e – d = (–5i – 14j) – (–6i – 2j) = i – 12j
CF = f – c =(4i – 6j) – (3i + 6j) = i – 12j
This confirms that the shape has two pairs
of parallel sides
8827
8836
1
× 376 × 376 ×
2
Also, since |CD| = |DE | (because 92 + 82 =
12 + 122), the shape is a rhombus
8827
8836
= 2 8827 .
c DF = f – d = (4i – 6j) – (–6i – 2j) = 10i – 4j
CB = b – c = (8i + 4j) – (3i + 6j) = 5i – 2j
Exercise 7.7B
FB = b – f = (8i + 4j) – (4i – 6j) = 4i + 10j
 1  2
 −1
1 TU =
–
=
 1  5
 −4
DC = c – d = (3i + 6j) – (–6i – 2j) = 9i + 8j
There are no identical sides, but DF = 2CB.
 −3
 1  =  −4
UV =   – 
 
 2
 1   1 
 −2
 −3
 1
VW =   –   =  
 6
 2
 4
This confirms that there is one pair
of parallel sides. Hence the shape is a
trapezium.
5
 2  −2
 4
WT =   –   =  
 5  6 
 −1
OH = OI + IH = OI + JK = OI + JO + OK
 −12  − 6
 3
9
= OI + OK – JO =  −7 +  9  –  − 1  =  3
 
 
 13   28 
 3
 18 
All sides are 17 in length
 −1  −4
TU • UV =   •   = 4 – 4 = 0
 −4  1 
Since angle TUV is a right angle and all the
sides are the same length, TUVW is a square.
2
Coordinates of the fourth exhibit = (3, 3, 3)
6
12 – 2a + 108 = 0
3
KN = n – k = (41i + 58j – 40k) – (17i + 50j – 28k)
= 24i + 8j – 12k
a = 60
l1 can be rewritten as r = 5i + 3j – 2k + t(2i – j + 5k).
LM = m – l = (56i + 28j – 30k) – (32i + 20j – 18k)
= 24i + 8j – 12k
This confirms that the shape has two pairs of equal
length parallel sides.
KL • KN = (15i – 30j + 10k) • (24i + 8j – 12k)
= 15 × 24 – 30 × 8 + 10 × –12
= 360 – 240 – 120 = 0.
Since KL • KN = 0, LKN is a right angle, and hence
KLMN is a rectangle.
l4 can be rewritten as r = 6i – k + u(2i – j + 5k).
Hence l1 and l4 are parallel.
For l3 and l5, (3i – j + 2k) • (–4i – 2j + 5k)
= –12 + 2 + 10 = 0.
Hence l3 and l5 are perpendicular.
l2 is neither parallel nor perpendicular to any of
the other lines.
4
a AB = b – a = (8i + 4j) – (–i – 4j) = 9i + 8j
DC = c – d = (3i + 6j) – (–6i – 2j) = 9i + 8j
BC = c – b = (3i + 6j) – (8i + 4j) = –5i + 2j
AD = d – a =(–6i – 2j) – (–i – 4j) = –5i + 2j
This confirms that the shape has two pairs
of parallel sides
Also, since AB ≠ BC, the shape is a
parallelogram (but not a rhombus)
KL = l – k = (32i + 20j – 18k) – (17i + 50j – 28k)
= 15i – 30j + 10k
NM = m – n = (56i + 28j – 30k) – (41i + 58j – 40k)
= 15i – 30j + 10k
(4i + aj – 12k) • (3i – 2j – 9k) = 0
Let vertices be HIJK, where H is the vertex to be
found.
7
a
AB = b – a = (3i – j – k) – (6i – 11j + 4k)
= –3i + 10j – 5k
Magnitude of AB =
( −3)2 + 102 + ( −5)2
= 134 .
BC = c – b = (10i + j + 8k) – (3i – j – k) = 7i + 2j + 9k
Magnitude of BC = 7 2 + 22 + 92 = 134 .
CD = d – c = (23i – 17j + 27k) – (10i + j + 8k)
= 13i – 18j + 19k
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Vectors
2
2
Magnitude of CD = 13 + ( −18 ) + 19 = 854 .
2
AD = d – a = (23i – 17j + 27k) – (6i – 11j + 4k)
= 17i – 6j + 23k
Magnitude of DA = 17 2 + ( −6 ) + 232 = 854 .
Since adjacent sides are the same length, the
quadrilateral is a kite.
2
b Since AB = BC and CD = DA, BD is the line of
symmetry.
Hence the diagonals intersect at the midpoint of AC.
1
((6i – 11j + 4k) + (10i + j + 8k)) = 8i – 5j + 6k
2
8
 12   6  6 
a QP =   −   =  
 −1  7  −8
B and D lie on l1. Since the direction vectors
of the two lines have the same magnitude,
the positions of B and D are
 −2
 2
 4
 −8
 7
 −1
 4
 10 
  ± 3   =   and  
 10 
 −2
 4
 16 
= (8i + 3j – 2k) + (–2i – 4j – 4k) + (–10i + 10j – 5k)
+ (–6i – 3j + 6k)
= –10i + 6j – 5k
QP ⋅QR = 6 × – 8 + – 8 × – 6 = 0
Therefore QP and QR are perpendicular
b AC = AB + BC = AB + AD
 12   −8  4 
b PS = QR and s = p + QR =   +   =  
 −1  −6  −7
s = 4i – 7j
c Centre of square X is mid-point of PR
= (2i + 4j + 4k) + (–10i + 10j – 5k)
= –8i + 14j – k
AG = AB + BC + CG = AB + AD + AE
= (2i + 4j + 4k) + (–10i + 10j – 5k) + (–6i – 3j + 6k)
1
1  12 − 2  5
=
X = (p + r ) = 
2
2  −1 + 1  0
X = 5i
= –14i + 11j + 5k
AC • AG = (–8i + 14j – k) • (–14i + 11j + 5k)
= 112 + 154 – 5 = 261
d QP = 6 2 + ( −8 ) = 10
2
Magnitude of AC = (−8)2 + 14 2 + (−1)2 = 3 29 .
Area of PQRS = 100 unit2
 2  1
 −1  −2
a
  •   = 2 + 2 – 4 = 0, so the lines are
 −2  2 
Magnitude of AG = (−14)2 + 112 + 52 = 3 38 .
261
= 0.8736
cos A = AC • AG =
AC AG 3 29 × 3 38
A = cos–1(0.8736) = 29.1°
perpendicular.
Equating i coefficients: 8 + 2t = –9 + u.
1
Equating j coefficients: 2 – t = 21 – 2u.
2
Equating k coefficients: –2t = –4 + 2u.
3
Add equations 2 and 3 : 2 – 3t = 17.
t = –5, u = 7
Point of intersection = (–2, 7, 10).
b The only way that all four vertices can lie
on l1 or l2 is if they are equidistant from the
point of intersection.
Let the point of intersection be X.
 −2
 −5
 3
1
 
 
 
AX =  7  –  13  =  −6 = 3 −2
 
 10 
 4 
 6 
 2 
 −2  3 
OC = OX + XC = OX + AX =  7  +  −6
   
 10   6 
 1
 1
=  
 16
10 a OH = OB + BA + AD + DH = OB + BA + AD +AE
 −2  6  −8
QR =   −   =  
 1   7  −6
9
Hence A lies on l2, as does C.
c EF = AB = 2i + 4j + 4k
EB = EA + AB = (6i + 3j – 6k) + (2i + 4j + 4k)
= 8i + 7j – 2k
EF • EB = (2i + 4j + 4k) • (8i + 7j – 2k)
= 16 + 28 – 8 = 36
Magnitude of EF = 22 + 4 2 + 4 2 = 6.
Magnitude of EB = 8 2 + 7 2 + (−2)2 = 3. 13
36
cos E = EF • EB =
= 0.5547
EF EB 6 × 3 13
E = cos–1(0.5547) = 56°
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7
WORKED SOLUTIONS
1
d XD = XE + EH + HD = BA + AD + EA
2
1
= (–2i – 4j – 4k) + (–10i + 10j – 5k)
2
+ (6i + 3j – 6k)
12 a AD = BC so d − a = c – b
 27 − 5   16 + 6 
 p − 3  =  47 − 25 ⇒ p − 3 = 22 ⇒ p = 25

 

 −10 − 1  12 − 23 
HG = AB so g – h = b − a
 −6 − 5   −6 − 5
 58 − 36  =  25 − 3  ⇒ −10 − q = 22 ⇒ q = −32

 

 −10 − q   23 − 1 
= –5i + 11j – 13k
XC = XF + FG + GC =
1
AB + AD + EA
2
1
= (2i + 4j + 4k) + (–10i + 10j – 5k) + (6i + 3j – 6k)
2
= –3i + 15j – 9k
XD • XC = (–5i + 11j – 13k) • (–3i + 15j – 9k)
= 15 + 165 + 117 = 297
Magnitude of XD = (−5)2 + 112 + (−13)2 = 3 35 .
Magnitude of XC = (−3)2 + 152 + (−9)2 = 3 35
b Centre of cube X is mid-point of AG
 5 − 6
 −1
1
1
1
a + g ) =  3 + 58 =  61 
(
2
2
 2 
 1 − 10
 −9
1
X = (–i + 61j −9k)
2
c Side of cube = |AB| = | b – a|
X =
(note that triangle DXC is isosceles).
33
297
=
cos X = XD • XC =
3 35 × 3 35 35
XD XC
1
× 4 × 5 = 10.
2
1
Volume of tetrahedron = × area of base ×
3
height.
1
= × 10 × 6
3
= 20 unit3
11 a Area of OAB =
=
Volume =
1
= 12i – 6j + 4k
2
= 24 – 36 + 12 = 0
= 9i + 6j – 18k
= 108 – 36 – 72 = 0
iii Magnitude of PR = 122 + (−6)2 + 4 2 = 14.
1
Area of PQR = × 14 × 7 = 49.
2
1
Volume of tetrahedron = × area of base ×
3
height.
1
= × 49 × 21
3
OP = (16 + 5t)i + (11 + 7t)j + (–3 – 3t)k
[(16 + 5t)i + (11 + 7t)j + (–3 – 3t)k] • (5i + 7j – 3k) = 0
5(16 + 5t) + 7(11 + 7t) – 3(–3 – 3t) = 0
80 + 25t + 77 + 49t + 9 + 9t = 0
83t + 166 = 0
t = –2
Hence OP = (16 + 5(–2))i + (11 + 7(–2))j
+ (–3 – 3(–2))k.
 3 + 4t   4 
 −25 + 5t    = 0

  5
 13 − t   
 −1
 3 + 4( 3 ) 
 15 
OV =  −25 + 5( 3) =  −10


 10 
 13 − 3 
PR • PS = (12i – 6j + 4k) • (9i + 6j – 18k)
Magnitude of PS = 92 + 6 2 + (−18)2 = 21.
35 937 unit3
12 + 16t – 125 + 25t – 13 + t = 0
t=3
ii PS = (8i + 10j – 16k) – (–i + 4j + 2k)
Magnitude of PQ = 22 + 6 2 + 32 = 7.
= 1089 = 33
OP = 6i – 3j + 3k
PQ = (i + 10j + 5k) – (–i + 4j + 2k)
PR • PQ = (12i – 6j + 4k) • (2i + 6j + 3k)
333 =
Exercise 7.7C
b i
PR = (11i – 2j + 6k) – (–i + 4j + 2k)
= 2i + 6j + 3k
( −11)2 + 222 + 222
3
 −13  20
 −33
 3

  


 
a DE =  8  –  19  =  −11 = –11  1
 −15  40
 −55
 5
 20
 3
 19 
 
r =   + t  1
 40
 5
 20 + 3t   −3
 23 + 3t 
b GF =  19 + t  –  8  =  11 + t 

  


 40 + 5t   7 
 33 + 5t 
= 343 unit3
101
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Vectors
GF L = 0
–20 + 25t + 8 + t – 90 + 25t = 0
 23 + 3t   3
 11 + t   1 = 0

   
 33 + 5t   5
t=2
 −4 + 5( 2) 
 6


OF =
=  10 
 8+2 
 
 −8
 −18 + 5( 2)
69 + 9t + 11 + t + 165 + 25t = 0
t=7
 23 + 3( 7 )
 44


OF =  11 + 7  =  18 
 
 68 
 33 + 5( 7 )
4
5
 26
 6
 20
c FC =  14  –  10  =  4 
 
 
 
 12 
 −8
 20
OS = (4 + 6t)i + (26 + 5t)j + (21 + 4t)k
TS = ( 4 + 6t)i + (26 + 5t)j + (21 + 4t)k – (i + j – k)
= (3 + 6t)i + (25 + 5t)j + (22 + 4t)k
[(3 + 6t)i + (25 + 5t)j + (22 + 4t)k] • (6i + 5j + 4k) = 0
6(3 + 6t) + 5(25 + 5t) + 4(22 + 4t) = 0
18 + 36t + 125 + 25t + 88 + 16t = 0
231 + 77t = 0
t = –3
OS = (4 + 6(–3))i + (26 + 5(–3))j + (21 + 4(–3))k
OS = –14i + 11j + 9k
| FC |= 20 2 + 4 2 + 20 2 = 816 = 4 51
7
= –20i + 35j + 15k = 5(–4i + 7j + 3k)
r = (–3i + 16j + 8k) + t(–4i + 7j + 3k)
b OP  l = 0
 −3 − 4t   −4
 16 + 7t   7  = 0

   
 8 + 3t   3 
a
RS = (11i – 9j + 11k) – (27i – 17j – k)
= –16i + 8j + 12k = 4(–4i + 2j + 3k)
12 + 16t + 112 + 49t + 24 + 9t = 0
b OQ = (27 – 4t)i + (–17 + 2t)j + (–1 + 3t)k
t = –2
[(27 – 4t)i + (–17 + 2t)j + (–1 + 3t)k]
• (–4i + 2j + 3k) = 0
P(5, 2, 2)
 5  17 
 −12
 −4
c AP =  2 –  −19 =  21  = 3  7 
  



 
 2  −7 
 9 
 3 
–4(27 – 4t) + 2(–17 + 2t) + 3(–1 + 3t) = 0
–108 + 16t – 34 + 4t – 3 + 9t = 0
–145 + 29t = 0
 −3
 5
 −8
 −4






 7
–
=
2
=
=
PB
16
14
2
 
 
 
 
 8 
 2
 6 
 3 
t=5
Hence OQ = (27 – 4(5))i + (–17 + 2(5))j +
(–1 + 3(5))k
AP : PB = 3 : 2
OQ = 7i – 7j + 14k
d OP = 52 + 22 + 22 = 33
 26
 −4 
 5
a  14  =  8  + t  1
 


 
 12 
 −18
 5
26 = –4 + 5t
t=6
14 = 8 + t
t=6
12 = –18 + 5t
t=6
b OF l = 0
 −4 + 5t   5
 8 + t   1 = 0

   
 −18 + 5t   5
1
× 10 2 × 4 51 = 20 102
2
AB = (–3i + 16j + 8k) – (17i – 19j – 7k)
a
Area =
r = (27i – 17j – k) + t(–4i + 2j + 3k)
6
2
6 2 + 10 2 + ( −8 ) = 10 2
| OF | =
PB = 2
( −4 )2 + 72 + 32
= 2 74
1
× 33 × 2 74 = 49.4
2
XY = (3i + 2j – 6k) – (9i + 5j – 12k) = –6i – 3j + 6k
= 3(–2i – j + 2k)
r = (9i + 5j – 12k) + t(–2i – j + 2k)
Let the foot of the perpendicular be F.
OF = (9 – 2t)i + (5 – t)j + (–12 + 2t)k
ZF = (9 – 2t)i + (5 – t)j + (–12 + 2t)k – (–2i + 7j + 14k)
= (11 – 2t)i + (–2 – t)j + (–26 + 2t)k
[(11 – 2t)i + (–2 – t)j + (–26 + 2t)k] • (–2i – j + 2k) = 0
–2(11 – 2t) – (–2 – t) + 2(–26 + 2t) = 0
–22 + 4t + 2 + t – 52 + 4t = 0
–72 + 9t = 0
t=8
Area =
8
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WORKED SOLUTIONS
9
ZF = (11 – 2(8))i + (–2 – (8))j + (–26 + 2(8))k
ZF = –5i – 10j – 10k
CF = (–42 + 10(3))i + (–25 + 15(3))j
+ (–48 + 6(3))k
Magnitude of ZF = (−5)2 + (−10)2 + (−10)2 = 15.
CF = –12i + 20j – 30k
a Equating i coefficients: –5 + 3t = 2 + 2u
Equating j coefficients: 7 + t = 8
2
Equating k coefficients: 1 + 4t = –1 – 3u
3
From 2 :
t=1
u = –2
Magnitude of CF = (−12)2 + 20 2 + (−30)2 = 38.
1
b Magnitude of AB = 40 2 + 60 2 + 24 2 = 76.
Area of triangle =
Exam-style questions
Point of intersection (–2, 8, 5)
1
 3−v 
OF =  19 + 3v 


 10 + v 
 43   59   −16

 
 

a PQ = q – p =  36  –  −19 =  55 
 −26  13   −39
Magnitude of PQ =
 3−v 
 −2
 5−v 
TF =  19 + 3v  –  8  =  11 + 3v 


 


 10 + v 
 5 
 5 + v 
 19   59   −40
 

 

PR = r – p =  −4  –  −19 =  15 
 −11  13   −24
 5 − v   −1
 11 + 3v   3  = 0

   
 5 + v   1 
Magnitude of PR =
 19   43   −24
 

 

QR = r – q =  −4  –  36  =  −40
 −11  −26  15 
 3 − (−3) 
OF =  19 + 3(−3)


 10 − 3 
Magnitude of QR =
F(6, 10, 7)
Since QR = PR ≠ PQ, the triangle is isosceles.
b PR 2 + QR 2 = 2401 + 2401 = 4802 = PQ 2
Since Pythagoras’ theorem is satisfied, the
triangle is right-angled.
2
8 +2 +2 =6 2
AB = (21i + 24j + 10k) – (–19i – 36j – 14k)
10 a
= (40i + 60j + 24k) = 4(10i + 15j + 6k)
r = (–19i – 36j – 14k) + t(10i + 15j + 6k)
Let the foot of the perpendicular be F.
OF = (–19 + 10t)i + (–36 + 15t)j + (–14 + 6t)k
CF = (–19 + 10t)i + (–36 + 15t)j + (–14 + 6t)k
– (23i – 11j + 34k)
= (–42 + 10t)i + (–25 + 15t)j + (–48 + 6t)k
2
a (1, 4, –5)
b a•b=0
–6 × 5 + (q + 5)(q – 6) + 3 × –4 = 0
q2 – q – 72 = 0
(q – 9)(q + 8) = 0
q = 9 or –8
3
a (p + 3)2 + (–4)2 = 82 + (p – 5)2
p2 + 6p + 9 + 16 = 64 + p2 – 10p + 25
[(–42 + 10t)i + (–25 + 15t)j + (–48 + 6t)k]
• (10i + 15j + 6k) = 0
6p + 25 = –10p + 89
10(–42 + 10t) + 15(–25 + 15t) + 6(–48 + 6t) = 0
p=4
–420 + 100t – 375 + 225t – 288 + 36t = 0
–1083 + 361t = 0
t=3
( −24 )2 + ( −40 )2 + 152
= 2401 = 49.
 5 − (−3)   8
b TF =  11 + 3(−3) =  2
  

 5 − 3   2
2
( −40 )2 + 152 + ( −24 )2
= 2401 = 49.
v – 5 + 33 + 9v + v + 5 = 0
v = –3
2
( −16 )2 + 552 + ( −39)2
= 4802 .
TF  l3 = 0
|TF |=
1
(76)(38) = 1444.
2
16p = 64
()
()
4
1
b tan–1 7 – tan–1 8 = 22.6°
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Vectors
4
5
2
2
a
OP = OA + AP = 8a + 5 AC = 8a + 5 (7c – 8a)
14
16
1
2
= 8a + c – a = (24a + 14c) = (12a + 7c)
5
5
5
5
 −4
 2
 


b Equation of l1 is r =  −5 + s  8 
 1 
 1 
b OB = OC + CB = 7c + 12a.
2
so OP = OB.
5
OP and OB have a common factor, so are parallel.
Equating i coefficients: 2 – 4s = 19 + 3t.
Since they also share a common point O, P
and B are collinear.
2
c Since OP = OB, OP : PB = 2 : 3.
5
Multiplying equation 3 by 2: 2 + 2s = 4 + 2t.
Equating k coefficients: 1 + s = 2 + t.
b
s = –2
Point of intersection: (10, –21, –1).
c (8i + aj – 2k) = –2(–4i + 8j + k)
Hence a = –2 × 8 = –16.
(5 − 2k)2 + (−8)2
d (–4i + 8j + k) • (5i + 3j – 6k) = –20 + 24 – 6 = –2
Since a • b is not 0, the lines are not
perpendicular.
(5 − 2k)2 + (−8)2 > 2 17
2
2
Solve (5 − 2k) + (−8) = 2 17 .
(5 – 2k)2 + 64 = 68
8
Find the magnitude of each line segment.
4k2 – 20k + 21 = 0
BA = (–8i – 4j) – (–i + 10j) = (–7i – 14j),
(2k – 3)(2k – 7) = 0
magnitude AB = (−7)2 + (−14)2 = 245
CB = (–i + 10j) – (3i + 8j) = (–4i + 2j),
3
7
For magnitude > 2 17 , k < 2 or k > 2 .
a BA = (5i – 2j + 4k) – (8i + 4j + 10k) = –3i – 6j – 6k
magnitude BC = (−4)2 + 22 = 20
Area = 245 × 20 = 70.
BC = (14i + 7j + 4k) – (8i + 4j + 10k) = 6i + 3j – 6k
BA • BC = (–3i – 6j – 6k) • (6i + 3j – 6k)
= –18 – 18 + 36 = 0
AB and BC are perpendicular
b i OD = OA + AD = OA + BC
OD = (5i – 2j + 4k) + (6i + 3j – 6k)
= 11i + j – 2k
1
ii ((5i – 2j + 4k) + ((14i + 7j + 4k))
2
19
5
= i + j + 4k)
2
2
iii Magnitude of BA = (−3)2 + (−6)2 + (−6)2 = 9.
Area = 9 × 9 = 81 units2.
7
aGiven three vertices of a rectangle, one must be
the right angle.
4k2 – 20k + 25 + 64 = 68
3
7
k = or
2
2
6
 −2  2   −4
     
a AB =  3  –  −5 =  8 
 2   1   1 
Magnitude of AB = (−4)2 + 8 2 + 12 = 9.
3
Subtracting from 2 : –7 + 6s = –19.
 5
 k   5 − 2k 
a i   – 2  = 
.
 −6
 1   −8 
ii
1
Equating j coefficients: –5 + 8s = –15 + 2t. 2
9
b Fourth vertex = (–8i – 4j) + [(3i + 8j) – (–i + 10j)]
= (–4i – 6j)
Since ADP is a straight line, DP is parallel to AD, so
DP = kd.
XP = XC + CD + DP = 6d – 10b + kd.
XY = 6d – 4b
Also, XP = nXY = n(6d – 4b).
Hence 6d – 10b + kd = n(6d – 4b).
Equating coefficients for b and d.
b: –10 = –4n
5
n=
2
d: 6 + k = 6n
5
6+k=6×
2
6 + k = 15
k=9
BP = BC + CD + DP = 10d – 10b + 9d = 19d – 10b
 −2 − 5λ   −3 + 2µ 
10 a At X  −5  =  −17 + 4µ 

 

 9 + 7λ   5 − µ 
– 5 = – 17 + 4µ µ = 3
– 2 – 5λ = – 3 + 2(3) λ = – 1
2
1
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WORKED SOLUTIONS
9 + 7(– 1) = 5 – 3
✓
Coordinates of X (3, –5, 2)
 −5
b 0
 
 7 
2
• 4  =
 
 −1
3
XD 2 = (–2 – (–5))2 + (–5 – (–6))2 = 10
CD2 = (–4 – (–5))2 + (1 – (–6))2 = 50
( −5)2 + 72
22 + 4 2 + ( −1) cosθ
(– 5 × 2) + (0 × 4) + (7 × –1) = 74
– 17 = 74
2
21 cosθ
21 cosθ
−17 
θ = cos 
= 115.5
 74 21 
acute angle 180 – 115.5 = 64.5°
magnitude of XC = 40 = 2 10 .
Since CG and DX have a common factor,
they are parallel and DXGC is a trapezium.
13 a AB = (3i + 7j + k) – (4i + 5j – k) = –i + 2j + 2k
Equation of l1: r = (4i + 5j – k) + t(–i + 2j + 2k).
2
 1
l1 =  4  + λ  −4
 
 
7
 −5
UV = (–3i + 7j – 11k) – (13i – j + 5k)
= –16i + 8j – 16k = 8(–2i + j – 2k)
Equation of l2: r = (13i – j + 5k) + u(–2i + j – 2k)
 −4  1   −5
     
AC = 1 − 4 = −3
     
 3   −5  8 
(–i + 2j + 2k) • (–2i + j – 2k) = 2 + 2 – 4 = 0.
Since a • b = 0, the lines are perpendicular.
b Let the point of intersection of l1 and l2 be X.
1 
 −5
l2 =  4  + µ  −3
 
 
 −5
8
Equating i coefficients: 4 – t = 13 – 2u.
1
Equating j coefficients: 5 + 2t = –1 + u.
2
Equating k coefficients: –1 + 2t = 5 – 2u.
3
θ
( −5)2 + ( −3)2 + 82 cos
Subtracting
 2   −5
2
2
2
AB ⋅ AC =  −4 ⋅  −3 = 22 + ( −4 ) + 7 2 ( −5) + ( −3) + 8 2 cosθ




7 8
– 10 + 12 + 56 = 69 98 cosθ
58
69 98
58
=
, θ = 45.1°
7 138
1
AB AC sin θ
2
1
= 2 69 98 sin 45.1
= 29.1 unit2
c
e Magnitude of DE = magnitude of CF = 3 10 ,
f CG = g – c = (2i + 3j) – (–4i + j) = (6i + 2j)
= 2(3i + j).
 3  1   2 
11 a AB =  0 −  4  =  −4
     
 2  −5  7 
 2   −5
   
2
b AB ⋅ AC = −4 ⋅ −3 = 22 + ( −4 ) + 7 2
   
7 8
Since XD2 + XC 2 = CD 2, XC is perpendicular
to XD.
Area = 3 10 × 2 10 = 60.
−1 
cosθ =
d XC 2 = (–2 – (–4))2 + (–5 – 1)2 = 40
Area =
12 a CF = f – c = (5i + 4j) – (–4i + j) = (9i + 3j)
92 + 32 = 90 = 3 10 .
3 from 2 : 6 = –6 + 3u.
u=4
Point of intersection: X(5, 3, –3).
Note that the point of intersection is also the
mid-point of U and V.
UX = (5i + 3j – 3k) – (13i – j + 5k)
= –8i + 4j – 8k = 4(–2i + j – 2k)
Since UVW is isosceles with a right angle
at W, XW and UX are of equal length, so
OW = OX +XW = (5i + 3j – 3k) ±4(–i + 2j + 2k)
OW = i + 11j + 5k or 9i – 5j – 11k
Coordinates of W are (1, 11, 5) or (9, –5, –11).
DE = (10i + 3j – 3k) – (18i + j – 9k) = –8i + 2j + 6k
14 a
= 2(–4i + j + 3k)
Equation of l1 is (18i + j – 9k) + t(–4i + j + 3k).
b CD = d – c = (–5i – 6j) – (–4i + j) = (–i – 7j)
FE = e – f = (4i – 3j) – (5i + 4j) = (–i – 7j)
FG = (8i – 29j + 31k) – (–6i + 20j – 4k)
c DE = e – d = (4i – 3j) – (–5i – 6j) = (9i + 3j)
Equation of l2 is (–6i + 20j – 4k) + u(2i – 7j + 5k)
DX = 1 DE = 1 (9i + 3j) = (3i + j)
3
3
OX = OD + DX = (–5i – 6j) + (3i + j) = (–2i – 5j)
= 14i – 49j + 35k = 7(2i – 7j + 5k)
(–4i + j + 3k) • (2i – 7j + 5k)
= –4 × 2 + 1 × –7 + 3 × 5 = 0.
105
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Vectors
b Since FJ is a chord of circle C, F and J are
equidistant from l1
AB • AC = (4i – 4j + 6k) • (–4i – 12j + 2k)
= –16 + 48 + 12 = 44
Let the point of intersection of the lines be X.
Magnitude of AB = 4 2 + (−4)2 + 6 2 = 68 .
Equating i coefficients: 18 – 4t = –6 + 2u. 1
Equating j coefficients: 1 + t = 20 – 7u.
Magnitude of AC = (−4)2 + (−12)2 + 22 = 164 .
2
Equating k coefficients: –9 + 3t = –4 + 5u. 3
cos A =
Multiply 2 by 3: 3 + 3t = 60 – 21u.
AB • AC
=
AB AC
44
= 0.4167
68 × 164
A = cos–1(0.4167) = 65.38°
1
Area = × 68 × 164 × sin 65.38° = 48 unit2.
2
Subtract from 3 : –12 = –64 + 26u.
u=2
16 a OR = (28 + 3t)i + (20 – t)j + (4 + 3t)k
[(28 + 3t)i + (20 – t)j + (4 + 3t)k] • (3i – j + 3k) = 0
Point of intersection = X(–2, 6, 6).
FX = XJ
OX – OF = OJ – OX
3(28 + 3t) – (20 – t) + 3(4 + 3t) = 0
OJ = 2OX – OF
OJ = 2(–2i + 6j + 6k) – (–6i + 20j – 4k)
= 2i – 8j + 16k
84 + 9t – 20 + t + 12 + 9t = 0
19t + 76 = 0
t = –4
J(2, –8, 16)
15 Let the first line be L1, the second L2 and the third L3.
Let the intersection of L1 and L2 be A, of L1 and L3
be B and of L2 and L3 be C.
To find A:
Equating i coefficients: 3 + 2t = –1 + 2u.
1
Equating j coefficients: 21 – 2t = –7 + 6u.
2
Equating k coefficients: –7 + 3t = 3 – u.
3
Adding 1 and 2 : 24 = –8 + 8u.
Equating i coefficients: 3 + 2t = 17 + 2v.
1
Equating j coefficients: 21 – 2t = 19 + 2v.
2
Equating k coefficients: –7 + 3t = 8 + v.
3
Adding 1 and 2 : 24 = 36 + 4v.
r = (–2i + 2j) + u(i – 2j – 3k)
When t = 9, –1 + 2(9) = 17.
c Equating i coefficients: –2 + u = –40 + 7t. 1
Equating j coefficients: 2 – 2u = –2 + 2t.
2
Equating k coefficients: –3u = –1 + 2t.
3
u = –3
Point of intersection: B(11, 13, 5).
Point of intersection: P(–5, 8, 9)
To find C:
Equating i coefficients: –1 + 2u = 17 + 2v.
1
Equating j coefficients: –7 + 6u = 19 + 2v.
2
Equating k coefficients: 3 – u = 8 + v.
3
AC = (3i + 5j + k) – (7i + 17j – k) = –4i – 12j + 2k
17 a QR = (–2j – 6k) – (–2i + 2j) = (2i – 4j – 6k)
= 2(i – 2j – 3k)
Subtracting 3 from 2 : 2 + u = –1.
v = –3
AB = (11i + 13j + 5k) – (7i + 17j – k) = 4i – 4j + 6k
= 162 + 24 2 + (−8)2 = 896 = 8 14 = 29.9
When t = 9, –2 + 2(9) = 16.
To find B:
Point of intersection: C(3, 5, 1).
b Shortest distance between O and L = OR
t=9
Point of intersection: A(7, 17, –1).
u=2
OR = 16i + 24j – 8k
b Equating i coefficients: –40 + 7t = 23.
u=4
Subtracting 1 from 2 : –6 + 4u = 2
Hence OR = (28 + 3(–4))i + (20 – (–4))j
+ (4 + 3(–4))k.
d PR = (–2j – 6k) – (–5i + 8j + 9k) = 5i – 10j – 15k
PS = (23i + 16j + 17k) – (–5i + 8j + 9k)
= 28i + 8j + 8k
PR • PS = (5i – 10j – 15k) • (28i + 8j + 8k)
= 140 – 80 – 120 = –60
Magnitude of PR = 52 + (−10)2 + (−15)2
= 350 .
Magnitude of PS = 28 2 + 8 2 + 8 2 = 912 .
106
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7
WORKED SOLUTIONS
cos P =
PR • PS
=
PR PS
−60
= –0.1062
350 × 912
P = cos–1(–0.1062) = 96.10°
1
e Area = × 350 × 912 × sin 96.10°
2
= 281 unit2.
18 a AB = (–4i + 5j + 2k) – (4i + 3j – k) = –8i + 2j + 3k
AC = (6i – j) – (4i + 3j – k) = 2i – 4j + k
AB • AC = (–8i + 2j + 3k) • (2i – 4j + k)
= –16 – 8 + 3 = –21
Magnitude of AB = (−8)2 + 22 + 32 = 77 .
2
2
2
Magnitude of AC = 2 + (−4) + 1 = 21 .
cos A = AB • AC =
AB AC
−21
= – 33
11
77 × 21
2


sin A = – 1 −  33  = 2 22
11
 11 
Area = 1 × 77 × 21 × 2 22 = 7 6 .
2
11
b EA = (4i + 3j – k) – (i + 2j + k) = 3i + j – 2k
ED = (10i + 11j + 19k) – (i + 2j + k) = 9i + 9j + 18k
EA • ED = (3i + j – 2k) • (9i + 9j + 18k)
= 27 + 9 – 36 = 0
Since EA • ED = 0, ∠AED = 90°.
c ED is the height of the tetrahedron.
Magnitude of ED = 92 + 92 + 18 2 = 9 6 .
1
Volume of tetrahedron = × area of base
3
× height.
=
1
×7 6 ×9 6
3
= 126 unit3
19 a GH = (15i + 11j + 4k) – (7i – 5j + 6k) = 8i + 16j – 2k
Magnitude of GH = 8 2 + 16 2 + (−2)2 = 18.
b 8i + 16j – 2k = 2(4i + 8j – k)
Equation of l2 is r = (7i – 23j + 24k) + t(4i + 8j – k).
Magnitude of EG = 18 2 + (−18)2 = 18 2 .
cos q =
EF • EG
648
=
36 × 18 2
EF EG

q = cos–1
 1 
648 
 = cos–1 2  = 45°
36 × 18 2
d FH = (15i + 11j + 4k) – (23i + 9j + 20k)
= –8i + 2j – 16k
(4i + 8j – k) • (–8i + 2j – 16k) = 4 × –8 + 8 × 2
+ –1 × –16 = 0
e Magnitude of FH = (−8)2 + 22 + (−16)2 = 18.
Area of trapezium =
1
(18 + 2 × 18) × 18 = 486.
2
 8  −7
 15 
 5
20 a AB =  5 –  −4 =  9  = 3  3 
   
 
 
 3  9 
 −6
 −2
 −7
 5
r =  −4 + u  3 
 
 
 9 
 −2
 6−t 
 8
b  11 + 3t  =  5


 
 7 + 2t 
 3
6–t=8
t = –2
11 + 3t = 5
t = –2
7 + 2t = 3
t = –2
 5   −1
c  3    3  = –5 + 9 – 4 = 0
   
 −2  
 2
 6   −7
 13 
d AC =  11 –  −4 =  15 
   
 
 7   9 
 −2
When t = 4, –23 + 8t = –23 + 8(4) = 9.
 13   15 
AC  AB =  15    9  = 13 × 15 + 15 × 9 +
   
 −2  −6
When t = 4, 24 – t = 24 – 4 = 20.
(–2) × (–6) = 342
Equating i coefficients: 7 + 4t = 23.
t=4
c EF = (23i + 9j + 20k) – (7i – 23j + 24k)
= 16i + 32j – 4k
EG = (7i – 5j + 6k) – (7i – 23j + 24k) = 18j – 18k
EF • EG = 16 × 0 + 32 × 18 + –4 × –18 = 648
Magnitude of EF = 16 2 + 322 + (−4)2 = 36.
| AC | = 132 + 152 + (−2)2 = 398
| AB | = 152 + 92 + (−6)2 = 3 38
342
cosθ =
398 × 3 38
θ = 22.0°
107
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Vectors
 6   8
 −2




e BC = 11 – 5 =  6 
   
 
 7   3
 4 
| BC | = (−2)2 + 6 2 + 4 2 = 2 14
Area = 1 × 3 38 × 2 14 = 6 133 = 69.2
2
f CB = BD
Mathematics in life and work
1
MP = (16.4i + 23.1j + 0.5k) – (–2.2i + 21.5j + 0.7k)
= (18.6i + 1.6j – 0.2k) km
2
Let a = (18.6i + 1.6j – 0.2k) and b = (13i – 14j – k).
a • b = (18.6i + 1.6j – 0.2k) • (13i – 14j – k)
= 18.6 × 13 + 1.6 × –14 – 0.2 × –1
= 219.6
OD = 2OB – OC
|a| = 18.6 2 + 1.6 2 + 0.22 = 348.56
 8  6 
OD = 2  5 –  11
   
 3  7 
|b| = 132 + 14 2 + 12 = 366
cos q =
D(10, –1, –1)
 −19 + t 
 −2
 −17 + t 




21 a TF = 14 − 3t – 5 =  9 − 3t 


 


 −5 + at 
 8 
 −13 + at 
TF  l = 0
 −17 + t   1 
 9 − 3t    −3 = 0

  
 −13 + at   a 
219.6
= 0.6148
348.56 × 366
q = cos–1(0.6148) = 52.1°
3
Let the foot of the perpendicular be F.
Shortest distance = PF = 348.56 sin 52.1 = 15 km.
4
PF = [(–2.2i + 21.5j + 0.7k) + t(13i – 14j – k)]
– (16.4i + 23.1j + 0.5k)
= [(–18.6 + 13t)i + (–1.6 – 14t)j + (0.2 – t)k]
Hence [(–18.6 + 13t)i + (–1.6 – 14t)j + (0.2 – t)k]
• (13i – 14j – k) = 0.
–17 + t – 27 + 9t – 13a + a2t = 0
13(–18.6 + 13t) – 14(–1.6 – 14t) – (0.2 – t) = 0
(a2 + 10)t = 13a + 44
–241.8 + 169t + 22.4 + 196t – 0.2 + t = 0
13a + 44
t= 2
a + 10
13a + 44
b 5= 2
a + 10
5a2 + 50 = 13a + 44
366t = 219.6
t = 0.6
OF = [(–2.2i + 21.5j + 0.7k) + 0.6(13i – 14j – k)]
= (5.6i + 13.1j + 0.1k) km
5a2 – 13a + 6 = 0
(5a – 3)(a – 2) = 0
a=
3
or 2
5
3
When a = ,
5


 −19 + 5 
 −14
 14 − 3( 5) 


OF = 
 =  −1 
 −2 
 −5 + 3 5 
()

5 
When a = 2,
 −19 + 5 
 −14
 14 − 3 5 
( )  =  −1 
OF = 
 5 
 −5 + 2( 5)
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8
WORKED SOLUTIONS
8 Differential equations
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge
1
dy a − 12
a .
= x =
dx 2
2 x
a
This is a solution with k = .
2
1
c If y = a x = ax 2 then
a Using the chain rule
2x cos(x2 + 1).
b Using product rule u = e–x, v = cos 2x and
du
dv
= −e −x ,
= −2sin 2x
dx
dx
5
2
3
3x + 2
a
∫ 4x
−0.5
c Substitute t = 0, 50 = a × 1 so a = 50 and
y = 50e0.005t.
dx = 8 x 0.5 + c = 8 x + c
1 e 4x + c
4
c tan x + c
When t = 20, y = 50e0.1 = 55.3 so an estimate
of the population is 55.3 million.
b
6
d ln |3x + 2| + c
e
ln|2x3
f −
3
dy
a dx = xy
1 x2
b If y = e 2
+ 4| + c
1
cos(4x) + c
4
dy
= 0.005y
dt
b If y = ae0.005t then
dy
= a × 0.005e0.005t = 0.005y
dt
so giving –e–x cos 2x – 2e–x sin 2x.
c
a
7
a
a 2 sin 2x
When x = 5,
Therefore
c e2x
b dx = 2
dt x
Exercise 8.1A
2
3
4
Therefore
dy
= 4x 3 + 4x + 2, hence
If y = x4 + 2x2 + 2x + 4 then
dx
d = 2.
dy
= 1.5 × x 0.5 = 1.5 × x , hence
If y = x1.5 + c then
dx
k = 1.5.
dy
= kx 2 where k is a constant.
dx
dy
= 3x 2 and this is a
b If y = x3 + 4 then
dx
solution with k = 3.
dy k
=
dx y
dy 1 − 12
1
= x =
then
and
b If y = x =
dx 2
2 x
this is a solution with k = 1 .
2
dx 2
dx
= or x
= 2.
dt x
dt
d 2x
2
= − 2.
dt 2
x
c The particle will reduce in speed but in
theory will never become stationary.
8
a
dr
c
=
dt r 2
b 0.5 =
r2
1
x2
dx
c
= 0.4 so 0.4 = so c = 2.
5
dt
When x = 5 the acceleration is
2
−
= −0.08 m s−2
25
a
a
1 x2
1 x2
dy
= e 2 × x = xe 2 = xy.
dx
dx c
=
dt x
b –cos 2x
1
then
dr 50
c
so c = 50 and dt = 2 or
100
r
dr
= 50.
dt
1
2
dr 1 − 3
= at and
dt 3
2
dr
1 −2 1
r2
= a 2t 3 × at 3 = a 3.
3
3
dt
c r = at 3 so
109
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Differential equations
This is a solution if
a = 3 150
9
1 3
a = 50 or a 3 = 150 or
3
1
b y = sin ( 3x ) + c
3
Substitute y = 4 and x =
1
a dx = kx 3 .
dt
( )
1
1
π
4 = 3 sin 3 × 3 + c = 3 sin(π) + c
1
3
4 = k ( 8 ) = 2k
Which gives 4 = 0 + c, so c = 4.
k=2
1
so dx = 2x 3
dt
(
4t + 4
b x=
3
)
1
Thus y = sin ( 3x ) + 4.
3
3
2
c
(
dx = 4 × 3 × 4t + 4
3
dt 3 2
(

4t
dx
= 2×
+4
dt
 3
)
1
2
1
3 3

2
) 
dy 2 + x 2 − 4x 5 2 x 2 4x 5
=
= +
−
x
x
x
x
dx
=
=
1
2x 3
2
+ x − 4x 4
x
So y = 2 ln|x|+
as required
So 2 = 2 ln|1| +
Thus y = 2ln|x| +
2
1 2 4 5 23
.
x − x +
10
2
5
dy
= 8ax 3
dx
y = 2ax4 + c
3
dx
= y −0.5
dy
x = 2y0.5 + c
y 0.5 =
( )
x−c
x−c
or y =
2
2
2
Alternatively separate the variables to get
y –0.5 dy = dx.
So ∫ y −0.5 d y = ∫ 1dx and 2y0.5 = x + c.
y=
10
( x 2+ c ) . Here the constant has changed sign
2
but the solutions are equivalent.
8
4
6
4
Rewrite as
dx
1
=
= 10y −2 so
dy 0.1y 2
x = ∫ 10y −2 d y = −10y −1 + c or x = −
2
2
4
6
8
10 12 14
a y = 2x − 0.3 x 3 + c = 2x − 0.1 x 3 + c
3
Substitute y = 2 and x = 2, so
2 = 2 × 2 – 0.1 × (2)3 + c.
Which gives 2 = 4 – 0.8 + c, so c = −1.2.
Thus y = 2x – 0.1x3 – 1.2.
This can be rewritten as
t
Exercise 8.2A
1
1
4
× (1)2 − × (1)5 + c.
2
5
23
4
Which gives 2 = 0 + 1 − +c, so c = 10 .
2 5
−1
c) 2
dP 1
= (2k)(2kt +
2
dt
k
=
2kt + c
k
= as required.
P
dP
k
= 2 so 2 =
b When P = 2,
2
dt
k=4
So P = 8t + c
When t = 0, P = 2, Then 2 = c
c=4
So P = 8t + 4
At t = 2, P = 8(2) + 4 = 2 5 (thousand)
So P = 4500 (2s.f.)
P
c
1 2 4 5
x − x + c.
2
5
Substitute y = 2 and x = 1.
c If x = 0 at the start then the snowball would
not start to move.
dP k
=
10 a Rate of growth
P
dt
If P = 2kt + c , then
0
π so
,
3
5
10
+ c.
y
10
10
.
= c − x or y =
c−x
y
dy
6
= 8x + 3 = 8x + 6x −3
dx
x
y = 4x2 – 3x–2 + c
Substitute x = 1 and y = 9, so
9 = 4 × (1)2 – 3 × (1)–2 + c.
Which gives 9 = 4 − 3 + c, so c = 8.
Thus the equation of the curve is y = 4x2 – 3x–2 + 8.
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8
WORKED SOLUTIONS
6
(t +10100 ) .
7 + 100
107
=
b After 1 week, t = 7, so y = (
10 ) ( 10 )
a ẋ = t 3 – 6t 2+ at or dx = t 3 − 6t 2 + at
dt
1
a
4
3
2
x = t − 2t + t + c
4
2
Which rearranges to y =
5
The particle starts at the origin, so c = 0.
1
a
Thus x = t 4 − 2t 3 + t 2
4
2
c The colony will continue to grow without
restrictions, this is clearly not an accurate
model, and so the model must only be valid
within a certain time period.
8
c At the origin, the displacement is zero, so x = 0.
1
a
Hence we solve 0 = 4 t 4 − 2t 3 + 2 t 2
0 = t 4 – 8t 3 + 2at 2
0 = t 2 (t 2 – 8t + 2a)
So t = 0 or t 2 – 8t + 2a = 0
So t =
( −8 )2 − 4 × 1 × 2a
8±
no roots if (–
this will have
2
– 4 × 1 × 2a < 0
8)2
64 – 8a < 0
64 < 8a
8<a
7
a
(
Hence t =
)
2 0.2
y + c = 10y 0.2 + c.
0.2
This needs to be written as a function of t, so
t − c = 10y0.2
y
0.2
5
Substitute in y = 100 000 at t = 0.
100 000 =
(–c)5
5
Substitute u = 10 – 0.5v.
du
dv
= −0.5 and
So
= −2.
dv
du
1
dv du = 1 × −2du = −2 ln u + c
t=∫
∫u
10 − 0.5v du
v = 20 – 2e–0.5t + 0.5c
You can write this as v = 20 – 2e–0.5t × e0.5c or
v = 20 – ke–0.5t.
( −c )5
100000
Now v = 0 when t = 0.
= 100 000 × 100 000
–c = 5 10 000 000 000 = 100, so c = −100
(
)
5
t + 100
.
giving a final result of y =
10
Alternatively, we could find c earlier in the
method.
t = 10y0.2 + c
at t = 0, y = 100 000
0 = 10 × (100 000)0.2 + c, so c = −10 × 5 100000
= −10 × 10 = −100
So t = 10y0.2 – 100.
1
∫ 10 − 0.5v dv.
0.5v = 10 – e–0.5(t – c)
( )
0−c
100 000 =
10
 1 2 dx = ∫ sec2x dx = tanx + c
t =⌠
⌡ cos x
At t = 0, x = π
So 0 = tan π + c
0=0+c
So c = 0
t = tan x
Giving x = arctan t or x = tan−1 t
b As t increases x will get closer to π
2
dt
1
.
9 a You can write
=
dv 10 − 0.5v
t = –2 ln |10 – 0.5v| + c
1
ln 10 − 0.5v = − (t − c )
2
10 – 0.5v = e–0.5(t – c)
( )
t −c
t −c
=
giving y =
10
10
a dx = cos2x
dt
Then t =
dy 1 0.8
dt 1 1 0.8
= y so
= ÷ y
= 2y −0.8
2
dt 2
dy
5
= (10.7)5 = 140 255.173 07 = 140 000 (3 s.f.).
b dx = t 3 − 6t 2 + at
dt
d 2x = x = 3t 2 − 12t + a
dt 2
5
Therefore 0 = 20 – k so k = 20.
Therefore v = 20 – 20e−0.5t.
b As t increases, e–0.5t → 0.
Therefore the speed approaches a limiting
value of 20 − 0 = 20 m s−1.
c The speed of the vehicle is limited by the
power of the car in overcoming air and other
resistances.
The solution is only valid so long as the
road is long enough, and the car powerful
enough, to reach the limiting value.
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Differential equations
10 Separate the variable to give
−4 dx = −4arcsinx + c
t =⌠

⌡ 1 − x2
at t = 0, x = 1
0 = – 4arcsin (1) + c
0 = −4 × π + c
2
So c = 2π
t = – 4arcsinx + 2π
t − 2π
= arcsinx
Giving
−4
(
)
(
x = sin t − 2π or x = sin 2π − t
−4
4
∫e
e–y = –ex – c
−y = ln| – ex – c|
⇒ y = −ln|ex + c|
e dy = − 3x2
dx
y
Separate the variables: y 2 dy = −3xdx
∫y
)
d y = ∫ −3x dx
y 3 = − 9 x 2 + 3c
2
(
2
y = − 9 x 2 + 3c
2
1.5
8π
1
0.5
2π
2
4
5
8
–1
26
t
2
1
ln y = x3 + c
3
−1
8+8+c
a
dy
y
=−
x
dx
1
1
Separate the variables: y dy = − x dx.
1
1
∫ y d y = − ∫ x dx
ln |y| = –ln |x| + c
ln |y| + ln |x| = c
ln |xy| = c
|xy| = ec.
d y = ∫ 3x dx
This could be written as |xy| = k.
b xy = k
3 2
x +c
2
0.5
y = 0.75x2 + k where k = 0.5c
2y 0.5=
c You cannot use the separation of variables
method.
dy
d
= e x + y = ex × ey
dx
Separate the variables: e–y dy = ex dx.
x3
Thus c = −6.
−1
y= 3
x + 4x − 6
This can be written as y = ke x where k
replaces ec.
dy
b
= 3x y
dx
1
Separate the variables:
dy = 3xdx
y
1
∫ y dy = ∫ 3x dx
y = (0.75x2 + k)2
)
)
+ 4 dx
When x = 2, y = − 0.1 so − 0.1 =
3
.
=
2
(
1
dy = 3x 2 + 4 dx .
y2
+ 4x + c
−1
y= 3
x + 4x + c
dy
= 3x 2 y
dx
1
Separate the variables: dy = 3x 2dx
y
1
2
∫ y d y = ∫ 3x dx
∫y
)
∫ y 2 d y = ∫ (3x
Exercise 8.3A
−0.5
(
dy
= 3x 2 + 4 y 2
dx
–y–1
3 +c
= 3 3c − 9 x 2
2
Separate the variables:
4π
So y = e x
)
1
3
f You cannot use the separation of variables
method.
6π
10 12 14 16 18 20 22 24
–0.5
a
2
1 y3 = − 3 x2 + c
3
2
x
1
d y = ∫ e x dx
–e–y = ex + c
Sketch:
0
−y
24
x
a Separate the variables: 3y dy = –2x dx.
6 × 4 = 24 so xy = 24, or y =
4
∫ 3y d y = − ∫ 2x dx
3 2
y = −x 2 + c
2
3
3
x 2 + y 2 = c or x 2 + y 2 = a , replacing c with a.
2
2
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8
WORKED SOLUTIONS
b Substitute x = 5 and y = 4.
1
b When half the element remains, x = a and
2
1
so a = ae −λt .
2
1
e −λt =
2
1
−λt = ln
2
25 + 24 = a so a = 49.
3 2
2
The equation is x + 2 y = 49 .
Where it crosses the x-axis, y = 0 so x2 = 49
and x = 7 or −7.
The points are (7, 0) and (−7, 0).
5
y
⌠
 y − 3 d y = ∫ 1dx
⌡
y −3+3
⌠
 y − 3 d y = ∫ 1dx
⌡
⌠  y − 3 + 3  d y = 1dx
  y − 3 y − 3
∫
⌡
8
1
= 1 dt
50 − 3I 4 dI
1
1
∫ 50 − 3I dI = ∫ 4 dt
− 1 ∫ −3 dI = ∫ 1 dt
4
3 50 − 3I
− 1 ln 50 − 3I = t + c ’
3
4
ln 50 − 3I = − 3t + c where c = −3c’
4
⌠
⌠ 3 d y = 1 dx
1 dy + 
∫
⌡y−3
⌡
1
∫ 1d y + 3⌠⌡ y − 3 d y =∫ 1dx
y + 3ln y − 3 = x + c
6
⌠ 1 dy =
 2
⌡y −4
1
∫ x dx
1 =
1
= A + B
y 2 − 4 ( y + 2)( y − 2) y + 2 y − 2
So 1 = A (y – 2) + B (y + 2)
1
If y = 2 then 1 = 4B, so B = 4
If y = −2 then 1 = −4A, so A = − 1
4
−
1
1
1
So 2
=
+
y − 4 4( y + 2) 4( y − 2)
⌠  −1

1
1
  4 ( y + 2) + 4 ( y − 2)  d y = ∫ x dx
⌡
− 1 ln y + 2 + 1 ln y − 2 = ln x + c '
4
4
−ln y + 2 + ln y − 2 = 4ln x + c where c = 4c’
–λt = –ln 2
ln 2
.
So t =
λ
50 = 4dI + 3I
dt
dI
50 – 3I = 4
dt
− 3t + c
9
50 − 3I = e 4 = Ae
− 3t
I = 50 − A e 4
3
3
dy
= ycosx
dx
⌠ 1 d y = cosx dx

∫
⌡y
y = esin x
2.5
1
ln x + c
λ
ln x = λc – λt
x = eλc – λt which you can write as x = ae–λt
where a = eλc.
When t = 0, x = a.
So a is the initial mass of the element.
t=−
1.5
1
y−2
A( y + 2)
dt
1
=−
dx
λx
t = − ∫ 1 dx
λx
y = e sinx
2
So y − 2 = Ax 4
y+2
a
y
3
( )
7
where A = ec
ln y = sinx + c
At x = 0, y = 1 so 0 = 0 +c so c = 0
ln y = sinx
y−2
= ln x 4 + ln ec = ln Ax 4 where A = ec
ln
y+2
Or x = ± 4
− 3t
4
0.5
–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0
1 2 3 4 5 6 7 8 9 10 x
When sin x < 0 then esin x < 1
When sin x > 0 then esin x > 1
As sin x ⩽ 1 for all x the two lines will never
intersect.
10 a
∫ y dy = ∫ 2sin
2
x dx =
∫ (1 − cos(2x)) dx
y2
= x − 1 sin ( 2x ) + c '
2
2
y2 = 2x – sin (2x) + c where c = 2c'
A x = 0, y = 0 so 0 = 0 – sin (0) + c so c = 0
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=
Differential equations
y2 = 2x – sin (2x)
b y=
y = 2x − sin ( 2x )
b
y
∫
(
x
x2 + 1
1
= 4∫
6.5
6
5.5
y = √2x–sin(2x)
5
=
4.5
4
6
3.5
3
2.5
1
0.5
2
4
6
8
10
12
14
16
18
20
22
7
()
2
3
4
()
−2
a cos 2 x = cos2 x – sin2 x
= cos2 x – (1 – cos2x)
= 2cos2 x – 1
So 1 + cos 2x = 2cos2x
b cos2(x) = 0.5(cos(2x) +1)
gives tan ( y ) = 1 sin(2x) + x + c When x = 0, y = π ,
2
4
−1 1
so c = 1 and y = tan 2 sin(2x) + x + 1 .
(
2x 3 + 2x − 2x 3 + 2x
(
x2 + 1
)
2
=
(
(
4x
x2 + 1
)
)
2
)
(
dx =
1 x2 − 1
×
4 x2 + 1 + c
x −1 +c
4 x2 + 1
(
∫ cos y d y = ∫ 4
)
x dx = ∫ 4x 0.5 dx
)
( 83 x )
1.5
a dx = 0.6cos 2t + 0.8sin 2t and
dt
2
4x = 1.2sin 2t − 1.6cos2t = − d x2
dt
2
so d x2 + 4x = 0.
dt
b The particle will oscillate, with a time period of π.
a v ∝ s, so v = ks.
ds
So dt = ks.
b ∫ 1 ds = ∫ k dt
s
ln(s) = kt + c
s = Aekt
When s = 30, v = 15.
So v = ks, giving k = 0.5.
30 = Ae0.5 × 3.6 gives A ≈ 5.0 to 1 d.p.
So s = 5.0e0.5t.
8
9
)
a Using the quotient rule, the derivative is
2x x 2 + 1 − 2x x 2 − 1
4x
2x 3 + 2x − 2x 3 + 2x
=
=
2
2
2
x2 + 1
x2 + 1
x +1
(
)
2
2
= −4 x = − x .
4
4
dx
b When t = 3,
= −4
dt
9
ds
a
= −120cos(4t)sin(4t)
dt
b 3.75 cos2(4x) + C
x3
0.5
−0.5
2
∫ y d y = ∫ x dx gives 2y = 3 + c When x = 0,
2
 3

y = 2, so c = 2 2 and y =  x + 0.5c 

 6
2
 x3

or 
+ 2
 6

−2
∫ cos ( y ) d y = ∫ 2 × 0.5(cos 2x + 1)dx
5
+1
2
d 2x = −1.2sin 2t + 1.6cos2t
dt 2
a x = 4t−1, so dx = −4t−2, but t = 4 giving
x
dt
dx = −4 4
x
dt
2
So y = sin −1
x
Exam-style questions
1
4x
2
(
1.5
0
dx
2
8
sin y = 3 x 1.5 + c
y = sin−1 8 x 1.5 + c
3
0 = sin−1(0 + c), giving c = 0
y = √2x
2
(x
)
)
(
)
(
)
2
2
a dv = − v
100
dt
dt
100
100
+ c.
So dv = − 2 and therefore t =
v
v
When t = 0, v = 20
100
+ c.
So 0 =
20
100
− 5.
Therefore c = −5 and t =
v
100
Rearrange: v = t + 5 .
b In this formula the speed is never actually 0
however large t is.
In practice the car will stop at some point
and the solution will no longer be valid.
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8
WORKED SOLUTIONS
5
A
B
=
+
(2x + 1)(x − 2) (2x + 1) (x − 2)
5 = A(x − 2) + B(2x + 1).
10 a
⌠ 1 dy =

⌡ sec2 y
∫ cos
Letting x = −0.5 and x = 2 in turn gives
A = −2, B = 1.
−2
1
+
(2x + 1) (x − 2)
−ln|2x + 1| + ln|x − 2| = −e–y + c
ln
x−2
= −e–y + c
2x + 1
ln
−3 − 2
= −e–0 + c
−6 + 1
1 sinycosy + y = tanx + c
′
)
2(
sinycosy + y = 2tan x + c where c = 2c'
(
y = e x + 3ln x − 3 + c
So c = 1.
x−2
ln
= −e–y + 1
2x + 1
ln ( x − 3)3
y = ex × e
× ec
y = Aex (x – 3)3 where A = ec
2x + 1
= e−y − 1
x−2
15
e − y = ln 2x + 1 + 1
x−2
11 u = sin x
u du
So ∫ cos x e cos x
= ∫ eu du = eu + c = esin x + c
∫e
y
d y = ∫ cos x esin x dx
ey = esin x + c
e0
=
e0
dy
= 8 – 4y
dx
1
∫ 2 − y d y = ∫ 4 dx
−ln|2 − y| = 4x + c
ln|2 − y| = −4x − c
|2 − y| = e–4x – c = Ae–4x
y = 2 − Ae–4x
3 = 2 − Ae0
A = −1
So y = 2 + e–4x
+c
16 a a ∝(20 − v)
dv ∝ (20 − v)
dt
dv = k(20 − v)
dt
So c = 0.
ey = esin x
Giving: y = sin x.
12
)
x−3
3
⌠ 1 dy = ⌠

+
dx

⌡y
⌡ x−3 x−3
⌠
 1 d y = ∫ 1dx + ∫ 3 dx
x−3
⌡y
ln y = x + 3ln x − 3 +c
ln|1| = −1 + c
ln
y dy = ∫ sec2 x dx
14 ⌠
 1 d y = ∫ x dx
x−3
⌡y
1
⌠ d y = x − 3 + 3 dx

∫ x−3
⌡y
5
−y
∫ (2x + 1)(x − 2) dx = ∫ e d y
b
2
2
∫ (1 + tan x )dx
(
dy
= cos2 y cos2x + sin 2x
dx
dy
= cos2 y
dx
)
b
1
∫ 20 − v dv = ∫ k dt
− ln 20 − v = kt + c
ln 20 − v = −kt − c
∫ cos12 y dy = ∫ dx
20 − v = Ae −kt
v = 20 − Ae −kt
∫ sec dy = ∫ dx
2
c As time increases, the speed increases and
approaches 20.
tan y = x + c or y = arctan (x + c)
13 dy = 1 + tan 2 x + tan 2 y + tan 2 x tan 2 y
dx
dy
= 1(1 + tan 2 x) + tan 2 y 1 + tan 2 x
dx
dy
= (1 + tan 2 y) 1 + tan 2 x
dx
1
⌠
d y = ∫ 1 + tan 2 x dx

⌡ (1 + tan 2 y)
(
(
)
)
(
)
17
dy
= xe xe y
dx
∫e
−y
dy = ∫ xe xdx
−e − y = xe x − ∫e xdx (integration by parts)
–e – y = xex – ex + c
So −c = xex – ex + e – y = constant, as required.
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Differential equations
18 a f(x, y) = x2 (1 – e–y) – (1 – e–y) = (x2 – 1)(1 – e–y)
∫ (x
1
b
∫ 1 − e−y dy =
∫ e y − 1 dy = ∫ ( x
ey
2
2
)
− 1 dx
Mathematics in life and work
1
)
− 1 dx
ln e y − 1 = 1 x 3 − x + c
3
1 x3− x
e y − 1 = Ae 3
ey =
2
1 x3− x
Ae 3
y = ln
+1
1 x3− x
Ae 3
+1
 1

= ( x − 1)  2 − y  = ( x − 1) y −2 − y
y

dy
−2
b
= ( x − 1) y − y
dx
Separate the variables:
(
(
)
⌠ y2
dy =

3
⌡ 1− y
(
–10ce–ct = –c(10e–ct + r – r)
)
–10ce–ct = –c(10e–ct)
Which is true.
)
⌠
1
dy =
 −2
⌡ y −y
)
This shows that y = 10e–ct + r is a solution.
3
∫ ( x − 1)dx
20 = 10 + r
∫ ( x − 1)dx
Therefore r = 10 and y = 10e–ct + 10.
)
x ( 3y − 2 )
2x ( 2y − 1)
x
2x
20 (2y − 1) − ( 3y − 2) = ( 2y − 1)( 3y − 2) − ( 3y − 2)( 2y − 1)
3xy − 2x
4xy − 2x
−
=
−xy
( 2y − 1)(3y − 2) (3y − 2)( 2y − 1) (3y − 2)( 2y − 1)
=
−y
(3y − 2)( 2y − 1)
So
When t = 0, y = 20.
Substitute these values into the solution.
1 ⌠ −3y 2
− 
d y = ∫ ( x − 1) dx
3 1 − y3
⌡
− 1 ln 1 − y 3 = 1 x 2 − x + c′
3
2
3
3
ln 1 − y = − x 2 + 3x + c where c = −3c′
2
(
If y = 10e–ct + r, you need to show that this satisfies
the differential equation.
dy
= −10ce−ct
dt
Substitute these into the differential equation,
dy
= −c( y − r ):
dt
 1
  1

19 a x  2 − y  −  2 − y 
y
 y

(
dy
∝ ( y − r ) where r is the outside temperature.
dt
Introduce a constant of proportionality c:
dy
= −c( y − r )
dt
The − sign shows that the building is cooling.
When t = 5, y = 15. Substitute these values.
15 = 10e–5c + 10
Rearrange:
5 = 10e–5c
e −5c = 5 = 0.5
10
–5c = ln0.5
ln0.5 = 0.1386
c=
−5
The solution is y = 10e–0.1386t + 10
Substituting t = 20 gives y = 10.6° C
×x
dy
x
=
− 2x
dx (2y − 1) ( 3y − 2)
dy
−y
=
×x
dx ( 3y − 2)( 2y − 1)
⌠ ( 3y − 2)( 2y − 1) d y = xdx

∫
−y
⌡
2
⌠ 6y − 7y + 2 d y = xdx

∫
−y
⌡
⌠  −6y + 7 − 2  d y = xdx

∫
y
⌡
1
−3y 2 + 7y − 2ln y = x 2 + c
2
116
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9
WORKED SOLUTIONS
9 Complex numbers
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge
1
2
2
Im
a 12x2 – 27x + 8x – 18 = 12x2 – 19x – 18
16
b 16x2 – 36x – 36x + 81 = 16x2 – 72x + 81
a (7x – 8)(7x + 8) = 0
x = 8 or – 8
7
7
b x2 – 4x – 60 = 0
14
12
10
h
(x – 10)(x + 6)
8
6
a
2
c 5x2 + 11x – 12 = 0
–16–14–12–10 –8 –6 –4 –2 0
–2
(5x – 4)(x + 3) = 0
x = 4 or –3
5
a (2i – 7j) + (–4i + 11j) = –2i + 4j
g×
5
( )
–10
c
1
a Re z = –6, Im z = 4
b Re z = 13, Im z = – 5
c Re z = –8, Im z = –9
d Re z = 2, Im z = 15
e Re z = 3, Im z = –10
f Re z = 11, Im z = 4
g Re z = –16, Im z = –2
h Re z = –1, Im z = 7
8 10 12 14 16
Re
b
–12
e
–14
–16
3
a 4 – 2i
b 9+i
c –15 – 3i
Gradient of line segment = 2 − (−8) = 10 = – 5 .
2
7 − 11
−4
Perpendicular gradient = 2 .
5
Equation of line: y – (–3) = 2 (x – 9).
5
y + 3 = 2 (x – 9)
5
5y + 15 = 2x – 18
Exercise 9.1A
6
–8
Mid-point = (9, –3).
–2x + 5y + 33 = 0
4
–6
c 4(2i – 7j) = 8i – 28j
Magnitude = 52 + 10 2 = 5 5 .
Angle = tan–1 10 = 63.4°.
5
2
–4
b (2i – 7j) – (–4i + 11j) = 6i – 18j
4
f
4
x = 10 or –6
3
d
d 8 + 7i
e 12 – 17i
f –3 + 25i
g 20 + 20i
h –8 – 3 i
4
a | z | = 6 2 + 52 = 61
5
arg z = tan–1 6 = 0.695
()
b | z | = 10 2 + (−7)2 = 149
7
arg z = –tan–1 10 = –0.611
( )
c | z | = (−4)2 + 92 = 97
()
9
arg z = π – tan–1 4 = 1.99
d | z | = (−2)2 + (−3)2 = 13
()
3
arg z = –(π – tan–1 2 ) = –2.16
117
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Complex numbers
5
a | z | = 52 + 52 = 5 2
5
3π
arg z = π – tan–1 5 = 4
()
9
b | z | = (7 3)2 + 7 2 = 14
 7 
arg z = –tan–1 7 3  = – π
6
c z = 11 + 11i
| z | = 112 + 112 = 11 2
arg z = tan–1
( )
11
π
11 = 4
 4 
 1 
π
arg(z1) = –tan–1 4 3  = –tan–1 2  = –
6
π
arg(z2) =
6
π
For z2, x = 40cos = 20 3
6
π
y = 40sin 6 = 20
z2 = 20 3 + 20i
10 a
Im
5√2
d z = 6 – 6 3i
| z | = 6 2 + (6 3)2 = 12
Re
−2
6 3
π
arg z = –tan–1 6  = –
3
6
7
π
a x = r cos  = 9 2 cos = 9
4
π
y = r sin  = 9 2 sin = 9
4
9 + 9i
π
b x = r cos  = 10 cos = 5
3
π
y = r sin  = 10 sin = 5 3
3
5 + 5 3i
5π
c x = r cos  = 6 cos
= –3 3
6
5π
y = r sin  = 6 sin
=3
6
–3 3 + 3i
3π
d x = r cos  = 4 2 cos(– ) = –4
4
3π
y = r sin  = 4 2 sin(– ) = –4
4
–4 – 4i
b |z | = (−2)2 + (5 2)2 = 54 = 9 × 6 = 3 6
5 2
c arg z = π – tan–1 2  = π – 1.295 = 1.85
11 aSince arg z1 = tan–1(2), z1 can be written in the
form x + 2xi
x2 + (2x)2 = (4 5 )2
5x2 = 80
x2 = 16
x = 4 (x must be positive)
z1 = 4 + 8i
Z3
a z* = –8 – 8i
b12 × 11 – (
7
Re
Z2
1
1
1
× 9 × 7 + × 12 × 4 + × 3 × 11) = 60
2
2
2
Exercise 9.2A
1
z = –10 + 10i
4
–3
a
x = 10 2 cos
b
1
–5
()
( 34π ) = 10 2 × – 22 = –10
3π
2
= 10
y = 10 2 sin( ) = 10 2 ×
4
2
Z1
5
|z*| = (−8)2 + (−8)2 = 128 = 8 2
8
3π
b arg z* = –(π – tan–1 8 ) = –
4
8
Im
a 9 + 3i
b 3 – 7i
Im
c 4 + 13i
10
d 16 – 14i
–10
Re
e –2 + 47i
f 9 – 16i
g –24 + 14i
c
3π
2
h –4 – 33i
118
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9
WORKED SOLUTIONS
2
a 15 + 5i + 6i + 2i2 = 13 + 11i
b 28 – 14i – 32i +
c 4 + 12i – 3i –
16i2
9i2
d 50 + 10i – 5i –
c z13 = (4 – 3i)3
= 12 – 46i
= (7 – 24i)(4 – 3i) = 28 – 21i – 96i – 72
= –44 – 117i
= 13 + 9i
i2
| –44 – 117i | = 44 2 + 117 2 = 125
117
arg(–44 – 117i) = –(π – tan–1 44 ) = –1.93
= 51 + 5i
( )
e 16 + 20i + 20i + 25i2 = –9 + 40i
f 81 – 18i – 18i + 4i2 = 77 – 36i
d z14 = (4 – 3i)4
= (–44 – 117i)(4 – 3i) = –176 + 132i – 468i
– 351
= –527 – 336i
g (14 – 4i + 7i – 2i2)(3 + 5i) = (16 + 3i)(3 + 5i)
= 48 + 80i + 9i + 15i2 = 33 + 89i
h (4 – 8i – 6i + 12i2)(2 – 3i) = (–8 – 14i)(2 – 3i)
= –16 + 24i – 28i + 42i2 = –58 – 4i
3
| –527 – 336i | = 527 2 + 336 2 = 625
336
arg(–527 – 336i) = –(π – tan–1 527 ) = –2.57
a 7 + 3i
( )
b 3 + 11i
c 10 – 20i + 14i – 28i2 = 38 – 6i
4
Im
a a = –9, b = 1
b a = 2, b = 12
0
c a = 15, b = 4
Re
4
×a
d a = –8, b = 12
e a = –1, b = –2
5
(7 + 2i)(7 + 2i)(7 + 2i) = (49 + 14i + 14i + 4i2)(7 + 2i)
= (45 + 28i)(7 + 2i) = 315 + 90i + 196i + 56i2
= 259 + 286i
6
a (4 – i) + 3(2 + 3i) = 10 + 8i
b (2 + 3i) – (4 – i) = –2 + 4i
c (4 – i)(2 + 3i) = 8 + 12i – 2i –
×b
–24
3i2
= 11 +10i
d (2 + 3i)(2 + 3i)(2 + 3i) = (4 + 6i + 6i + 9i2) (2 + 3i)
Im
(–5 + 12i)(2 + 3i) = –10 – 15i + 24i + 36i2
=
= –46 + 9i
e (4 – i)(4 – i) – (2 + 3i)(2 + 3i) = (16 – 4i – 4i +
– (–5 + 12i) = (15 – 8i) – (–5 + 12i) = 20 – 20i
7
8
z(3 + 2i) = 28 – 3i
(a + bi)(3 + 2i) = 28 – 3i
3a + 2ai + 3bi + 2bi2 = 28 – 3i
⇒ 3a + 2bi2 = 3a – 2b = 28
and 2a + 3b = – 3
⇒ a = 6, b = – 5
z = 6 – 5i
a z1 = 4 – 3i
| 4 – 3i| = 4 2 + 32 = 5
3
arg(4 – 3i) = –tan–1 4 = –0.644
()
b z12 = (4 – 3i)2
= (4 – 3i)(4 – 3i) = 16 – 12i – 12i – 9
= 7 – 24i
| 7 – 24i | = 7 2 + 24 2 = 25
24
arg(7 – 24i) = –tan–1 7 = –1.29
( )
–527
–44
i2)
×
c
d
9
×
0
Re
–117
–336
a − 6i = b +
4i
1 − 2i
a − 6i = (b + 4i)(1 − 2i)
a − 6i = b − 2bi + 4i + 8
= (b + 8) + (4 − 2b)i
Re: a = b + 8
Im : −6 = 4 − 2b
So b = 5 and a = 13
10 a For z1, x = 6cos
π
=3 2
4
π
=3 2
4
z1 = 3 2 + 3 2 i
y = 6sin
119
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Complex numbers
π
For z2, x = 4 3 cos 6 = 6
Exercise 9.2B
π
y = 4 3 sin 6 = 2 3
1
z2 = 6 + 2 3 i
a
b
z1z2 = (3 2 + 3 2 i)(6 + 2 3 i)
= (18 2 – 6 6 ) + (18 2 + 6 6 )i
 18 2 + 6 6 
b arg z1z2 = tan–1 18 2 − 6 6 
c
3 2 + 6
= tan–1 3 2 − 6 
(
(
)(
)(
 3 2+ 6 3 2+ 6
= tan–1 3 2 − 6 3 2 + 6

) 
) 
 18 + 3 12 + 3 12 + 6 

= tan–1
18 − 6
(
)
π π
tan +
=
4 6
=
=
1+
3
3
2
= 164 + 287i = 4 + 7i
41
 
 
tan  π + tan  π 
4
6
π
 
1 − tan  tan  π 
4
6
c
3+ 3
3− 3
3
d
12 + 6 3
6
= 2 + 3 Therefore tan −1 2 + 3 = 5π
12
=
(
)
11 a(p – 2i)4 = p4 + 4p3(– 2i)1 + 6p2(– 2i)2 + 4p1(– 2i)3
+ (– 2i)4
p4
4
12 + 5i 2 − 3i 24 − 36i + 10i + 15
×
=
2 + 3i
2 − 3i
4+9
39 − 26i
= 3 – 2i
=
13
a 2a – ai + 2bi + b = 5 + 5i
2a + b = 5
2b – a = 5
–
24p2
+ 16 = –119
Multiply second equation by 2 and add to
first equation:
–
24p2
+ 135 = 0
4b – 2a = 10
b
19 − 77i 7 + 5i 133 + 95i − 539i + 385
×
=
7 − 5i
7 + 5i
49 + 25
518 − 444i
= 7 – 6i
=
74
a z1 + z2 = (12 + 5i) + (2 + 3i) = 14 + 8i
c z1z2 = (12 + 5i)(2 + 3i) = 24 + 36i + 10i + 15i2
= 9 + 46i
9 + 3 + 3 3 + 3 3)
9−3
p4
3 + i 30 + 10i
10
×
=
=3+i
3−i 3+i
9+1
b z1 – z2 = (12 + 5i) – (2 + 3i) = 10 + 2i
(3 + 3)(3 + 3)
=
(3 − 3)(3 + 3)
=
−51 + 17i 3 − 5i −153 + 255i + 51i + 85
×
=
3 + 5i
3 − 5i
9 + 25
−68 + 306i
= –2 + 9i
=
34
d
3
3
1−1×
−2 + 6i −108 + 324i − 4i − 12
54 + 2i
× −2 + 6i =
−2 − 6i
4 + 36
−120 + 320i
= –3 + 8i
=
40
53 − 31i 7 + 3i 371 + 159i − 217i + 93
×
=
7 − 3i
7 + 3i
49 + 9
464 − 58i
=8–i
=
58
48 + 19i 5 + 4i 240 + 192i + 95i − 76
×
=
a
5 − 4i
5 + 4i
25 + 16
b
5π π π
= +
12 4 6
−5 + 14i 3 − 2i −15 + 10i + 42i + 28
× 3 − 2i =
3 + 2i
9+4
13 + 52i
= 1 + 4i
=
13
d
 24 + 12 3 
–1
= tan–1
12
 = tan (2 + 3 )
c
3 + 7i −1 − i −3 − 3i − 7i + 7 4 − 10i
×
=
=
= 2 – 5i
−1 + i −1 − i
1+1
2
=
(p2
–
p4
9)(p2
–
8p3i
–
24p2
– 15) = 0
p2 = 9 or p2 = 15
p = ±3 or ± 15
–8p3 + 32p = –120
+ 32π + 16
5b = 15
b=3
a=1
b (a + bi)(1 – i) = 5 – 5i
p3 – 4p – 15 = 0
a – ai + bi + b = 5 – 5i
(p – 3)(p2 + 3p + 5) = 0
p=3
a+b=5
b – a = –5
120
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9
WORKED SOLUTIONS
2b = 0
| z6 | = 4 2 + 12 = 17
b=0
arg z6 = tan–1 ( 4 ) − π = −1.816
a=5
c (3 – 2i)(a + bi) = 6 – 17i
7
3a + 3bi – 2ai + 2b = 6 – 17i
()
Since arg z1 = tan–1 1 , z1 can be written as 2y + yi.
2
(2y)2
3a + 2b = 6
5y2
3b – 2a = –17
+
y2
= (5
5 )2
= 125
y2 = 25
Multiply first equation by 2 and second
equation by 3, then add:
y = 5 (y must be positive)
6a + 4b = 12
z1 = 10 + 5i
9b – 6a = –51
10 + 5i (10 + 5i)(3 − 4i)
=
3 + 4i
(3 + 4i)(3 − 4i)
13b = –39
=
b = –3
a=4
30 − 40 i+ 15i+ 20
9 + 16
a + 3ai + bi – 3b = –11 + 7i
50 − 25i
25
=2–i
a – 3b = –11
p = 2, q = –1
3a + b = 7
Since 2 – 1 = 1, p + q = 1.
=
d (a + bi)(1 + 3i) = –11 + 7i
Multiply second equation by 3 and add to
first equation:
8
1168 − 9i 8 + 9i 9344 + 10512i − 72i + 81
×
=
8 − 9i
8 + 9i
64 + 81
9a + 3b = 21
=
10a = 10
65 + 72i = 652 + 722 = 97
a=1
b=4
9
5
3 4i
2 − i 2 − i 4 − 2i − 2i − 1
×
=
= 5− 5
4 +1
2+i 2−i
6
z
21 + i
a z4 = 2 =
z1 2 − 3i
(u − 9i)(3 − i) 3u − 9 − (u + 27)i
=
(3 + i)(3 − i)
9+1
=
3u − 9 −(u + 27)
+
i
10
10
( 3u10− 9 ) + ( −(u10+ 27)) = 5
2
21 + i 2 + 3i 42 + 63i + 2i − 3 39 + 65i
×
=
=
2 − 3i 2 + 3i
4+9
13
= 3 + 5i
| z4 | = 32 + 52 = 34
5
arg z4 = tan–1 3 = 1.03
2
2
(3u – 9)2 + (u + 27)2 = 2500
9u2 – 54u + 81 + u2 + 54u + 729 = 2500
10u2 + 810 = 2500
()
b z5 =
9425 + 10440i
= 65 + 72i
145
10u2 = 1690
u2 = 169
1
1
=
z 3 17 − 17i
17 + 17i
17 + 17i
1+ i
1
×
=
=
34
17 − 17i 17 + 17i 289 + 289
1
1
12 + 12 =
2
34
34
1
π
arg z5 = tan–1 1 =
4
z 3 17 − 17i
c z6 = z * =
3 + 5i
4
| z5 | =
u = –13
10 a
()
17 − 17i 3 − 5i 51 − 85i − 51i − 85
×
=
3 + 5i
3 − 5i
9 + 25
−34 − 136i –1 –
=
4i
=
34
b
z1 a + 2i (a + 2i)(5 − bi)
=
=
z 2 5 + bi (5 + bi)(5 − bi)
=
5a − ab i + 10i + 2b
25 + b 2
=
5a + 2b + (10 − ab)i
25 + b 2
10 − ab
= –2
25 + b 2
10 – ab = –2(25 + b2)
10 – ab = –50 – 2b2
121
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Complex numbers
2b2 = ab – 60
b2 =
c (x + iy)2 = 5 – 12i
1
(ab – 60)
2
x2 + 2xyi – y2 = 5 – 12i
x2 – y2 = 5
1
= (23b – 60)
c
2
2
2b – 23b + 60 = 0
b2
2xy = –12
y = −6
x
(2b – 15)(b – 4) = 0
( ) =5
d q=
x4 – 36 = 5x2
5a + 2b
25 + b 2
When b = 4,
When b =
x4 – 5x2 – 36 = 0
(x2 + 4)(x2 – 9) = 0
5(23) + 2(4)
=3
25 + (4)2
15
,
2
x = ±3
( )
( )
5(23) + 2 15
2
8
2 = 5
15
25 +
2
y = 2
Square roots are 3 – 2i and –3 + 2i
d (x + iy)2 = –7 – 24i
x2 + 2xyi – y2 = –7 – 24i
Exercise 9.2C
1
x2 – y2 = –7
a (x + iy)2 = 3 – 4i
2xy = –24
x2 + 2xyi – y2 = 3 – 4i
y = −12
x
x2 – y2 = 3
( ) = –7
2xy = –4
x4 – 144 = –7x2
( ) =3
2
x2 – −2
x
x4 + 7x2 – 144 = 0
(x2 + 16)(x2 – 9) = 0
x4 – 4 = 3x2
x = ±3
x4 – 3x2 – 4 = 0
y = 4
(x2 + 1)(x2 – 4) = 0
x = ±2
Square roots are 2 – i and –2 + i.
b (x + iy)2 = –15 + 8i
x2 + 2xyi – y2 = –15 + 8i
x2 – y2 = –15
2xy = 8
y= 4
x
x4
Square roots are 3 – 4i and –3 + 4i.
2
y = 1
()
2
x2 – −12
x
y = −2
x
4
x2 – x
2
x2 – −6
x
15
b=
or 4
2
a (x + iy)2 = 85 + 132i
x2 + 2xyi – y2 = 85 + 132i
x2 – y2 = 85
2xy = 132
66
y=
x
x2 –
( 66x ) = 85
2
x4 – 4356 = 85x2
2
– 16 =
= –15
–15x2
x4 + 15x2 – 16 = 0
(x2 + 16)(x2 – 1) = 0
x = ±1
y = ±4
Square roots are 1 + 4i and –1 – 4i.
x4 – 85x2 – 4356 = 0
(
(x
) ( ) – 4356 = 0
85
24649
− ) =
2
4
x2 −
2
85
2
2
–
85
2
2
2
x2 –
85 157
=
2
2
x2 =
242
= 121
2
122
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9
WORKED SOLUTIONS
x = ±11
(p – q)(p + q) = 45
y = ±6
Factors of 45 are 1, 45 and 3, 15 and 5, 9
Square roots are 11 + 6i and –11 – 6i
p – q = 1, p + q = 45,
b (x +
iy)2
from which p = 23, q = 22, m = 2 × 23 × 22 = 1012
= 336 + 320i
x2 + 2xyi – y2 = 336 + 320i
p – q = 3, p + q = 15,
x2 – y2 = 336
from which p = 9, q = 6, m = 2 × 9 × 6 = 108
2xy = 320
p – q = 5, p + q = 9,
160
y=
x
from which p = 7, q = 2, m = 2 × 7 × 2 = 28
x2 –
(160x ) = 336
c Since (9 + 6i)2 = 45 + 108i, (9 – 6i)2 = 45 – 108i
2
x4 – 25600 = 336x2
x4 – 336x2 – 25600 = 0
(x2 – 168)2 – 1682 – 25600 = 0
(x2 – 168)2 = 53824
x2 – 168 = 232
x2 = 400
x = ±20
Square roots are 9 – 6i and –9 + 6i
6 a (x + iy)2 = 7 + 24i
x2 + 2xyi – y2 = 7 + 24i
x2 – y2 = 7 and 2xy = 24
2
12
⇒ x2 −
=7
x
( )
⇒ x4 – 7x2 – 144 = 0
⇒ (x2 + 9)(x2 – 16) = 0
⇒ x = ± 4, y = ± 3)
Roots are (4 + 3i) and (–4 – 3i)
y = ±8
3
arg(4 + 3i) = tan −1
(x + iy)2 = 55 – 48i
x2 + 2xyi – y2 = 55 – 48i
c 7 + 42i = 7 2 + 24 2 = 25
x2 – y2 = 55
3 + 4i = 32 + 4 2 = 5
2xy = –48
y=
−24
x
x2 –
( )
7
2
−24
= 55
x
x2 – y2 = 105
x4 – 55x2 – 576 = 0
y=
–
64)(x2
+ 9) = 0
– 64 = 0
x = ±8
y = 3
Square roots are 8 – 3i and –8 + 3i
a = 8, b = –8, c = 3
Angelene should have used the coefficients of i in
her equations, but instead she included i which
initiallly made the algebra more complex.
Angelene solutions to the quadratic equation are
incorrect. The correct answer is b2 = 16 or –81
So b = ±4 and a = ±9 and the square roots are
9 – 4i and –9 + 4i
5
x2 + 2xyi – y2 = 105 + 88i
2xy = 88
x2
4
a (x + iy)2 = 105 + 88i
x4 – 576 = 55x2
(x2
( 247 ) = 1.287
( 34 ) = 0.6435
b arg(7 + 24i) = tan −1
Square roots are 20 + 8i and –20 – 8i
a (p + qi)2 = p2 + 2pqi – q2
b p2 – q2 = 45
x2 –
44
x
( 44x ) = 105
2
x4 – 1936 = 105x2
x4 – 105x2 – 1936 = 0
(x
(x
2
2
) – (1052 ) – 1936 = 0
105
18769
−
=
2 )
4
−
x2 –
105
2
2
2
2
105 137
=
2
2
242
= 121
2
x = ±11
x2 =
y = ±4
Square roots are 11 + 4i and –11 – 4i
123
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Complex numbers
b 945 + 792i = 32(105 + 88i)
c (x + iy)2 = 40 – 42i
x2 + 2xyi – y2 = 40 – 42i
Square roots are 33 + 12i and –33 – 12i
8
x2 – y2 = 40
−262 + 130i (−262 + 130i)(7 − 5i)
=
(7 + 5i)(7 − 5i)
7 + 5i
2xy = –42
−1834 + 1310i + 910i + 650
=
49 + 25
y=
−1184 + 2220i
74
= –16 + 30i
x2 –
=
2
x4 – 40x2 – 441 = 0
x2 + 2xyi – y2 = –16 + 30i
(x2 – 49)(x2 + 9) = 0
x2 – y2 = –16
x2 = 49
2xy = 30
x = ±7
15
x
x2 –
( −x21 ) = 40
x4 – 441 = 40x2
(x + iy)2 = –16 + 30i
y=
−21
x
y = 3
Since Re z2 > 0, Re z2 = 7 and Im z2 = –3
(15x ) = –16
2
z2 = 7 – 3i
x4 – 225 = –16x2
d z3 = –7 + 3i
x4 + 16x2 – 225 = 0
| z3| = (−7)2 + 32 = 58
(x2 – 9)(x2 + 25) = 0
3
arg z3 = π – tan–1 7 = 2.74
x2
()
=9
x = ±3
Exercise 9.3A
y = ±5
Square roots are 3 + 5i and –3 – 5i
1
a 4 – 11i
b z = 4 ± 11i
362 − 153i 2 + 3i 724 + 1086i − 306i+459 1183 + 780i
=
= 91z+–60i
×
=
4 = ±11i
2 − 3i
2 + 3i
4+9
13
53i 2 + 3i 724 + 1086i − 306i+459 1183 + 780i
(z – 4)2 = –121
×
=
=
= 91 + 60i
i
2 + 3i
4+9
13
z2 – 8z + 16 = –121
(x + iy)2 = 91 + 60i
z2 – 8z + 137 = 0
x2 + 2xyi – y2 = 91 + 60i
2
2
k = 137
x – y = 91
2xy = 60
2 a (–14)2 – 4(1)(58) = 196 – 232 = –36 < 0
y = 30
b (z – 7)2 – 49 + 58 = 0
x
9
x2 –
( 30x ) = 91
2
(z – 7)2 + 9 = 0
x4 – 900 = 91x2
x4 – 91x2 – 900 = 0
(x2 – 100)(x2 + 9) = 0
x = ±10
y = ±3
Square roots are 10 + 3i and –10 – 3i.
362 − 153i
= 10 2 + 32 = 109
2 − 3i
10 a |z1| = 40 2 + (−42)2 = 58
( )
42
b arg z1 = –tan–1 40 = –0.810
(z – 7)2 = –9
z – 7 = ±3i
z = 7 ± 3i
3
a z = ±5i
b z2 = –144
z = ±12i
c z2 + 10z + 26 = 0
(z + 5)2 – 25 + 26 = 0
(z + 5)2 = –1
z + 5 = ±i
z = –5 ± i
124
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9
WORKED SOLUTIONS
( 52 ) = − 92
( z − 52 ) = − 94
d z2 – 14z + 53 = 0
(z –
7)2
– 49 + 53 = 0
(z –
7)2
= –4
2
z – 7 = ±2i
z = 7 ± 2i
e
z2
z–
+ 8z + 80 = 0
z=
(z + 4)2 = –64
z + 4 = ±8i
9z2 – 48z + 68 = 0
(
z = –4 ± 8i
f
9 z2 −
+ 12z + 37 = 0
(z +

8
9 z −
3

= –1
z + 6 = ±i
g z2 + 104 = 20z
2
– 20z + 104 = 0
(z – 10)2 – 100 + 104 = 0
(z –
10)2
2
= –4
z – 10 = ±2i
z = 10 ± 2i
h z2 + 18z + 202 = 0
(z + 9)2 – 81 + 202 = 0
(z + 9)2 = –121
z + 9 = ±11i
z = –9 ± 11i
i z2 + 41 = 10z
z2 – 10z + 41 = 0
(z – 5)2 – 25 + 41 = 0
(z – 5)2 = –16
z – 5 = ±4i
j z2 – 12z + 4936 = 0
(z – 6)2 – 36 + 4936 = 0
(z – 6)2 = –4900
z – 6 = ±70i
z = 6 ± 70i
k 2z2 + 17 = 10z
2z2 – 10z + 17 = 0
2(z2 – 5z) + 17 = 0
( )
2
4
z–
8
2
=± i
3
3
z=
8 2
± i
3 3
a z = 2 ± 5i
z – 2 = ±5i
(z – 2)2 = –25
z2 – 4z + 4 = –25
z2 – 4z + 29 = 0
b z = 7 ± 4i
z – 7 = ±4i
(z – 7)2 = –16
z2 – 14z + 49 = –16
z = 5 ± 4i

5
2 z −
2

2
2
z = –6 ± i
z2
)
16
z + 68 = 0
3
( ) − 649  + 68 = 0
8
9 ( z − ) − 64 + 68 = 0
3
8
9 ( z − ) = −4
3
( z − 83 ) = −94
(z + 6)2 – 36 + 37 = 0
6)2
5
3
=± i
2
2
5 3
± i
2 2
2
l 9z + 68 = 48z
(z + 4)2 – 16 + 80 = 0
z2
2
2 z−
25 
−  + 17 = 0
4
z2 – 14z + 65 = 0
c z = –8 ± 20i
z + 8 = ±20i
(z + 8)2 = –400
z2 + 16z + 64 = –400
z2 + 16z + 464 = 0
d z = –3 ± 2i
z + 3 = ±2i
(z + 3)2 = –4
z2 + 6z + 9 = –4
z2 + 6z + 13 = 0
125
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Complex numbers
5
a x3 + x2 + 15x – 225 = 0
x3
x2 – 2x + 5 = 0
x2 + 6x + 45 = 0
x4 + 3x3 + x2 + 13x + 30
= (x2 – 2x + 5) (x2 + 5x + 6)
(x + 3)2 – 9 + 45 = 0
x2 + 5x + 6 = 0
(x + 3)2 = –36
(x + 2)(x + 3) = 0
x + 3 = ±6i
Roots are –2, –3, 1 ± 2i.
+
x2
+ 15x – 225 = (x –
5)(x2
+ 6x + 45)
6
x = –3 ± 6i
Roots are 5, –3 ± 6i.
b
x3
+
7x2
– 13x + 45 = (x +
f(8) = 0, so (z – 8) is a factor.
9)(x2
– 2x + 5)
z3 – 8z2 + 9z – 72 = (z – 8)(z2 + 9)
x2 – 2x + 5 = 0
z2 + 9 = 0
(x – 1)2 – 1 + 5 = 0
z2 = –9
(x – 1)2 = –4
z = ±3i
x – 1 = ±2i
Roots are 8, ±3i.
b z3 + 4z + 10 = 5z2
x = 1 ± 2i
c
a z3 – 8z2 + 9z – 72 = 0
Roots are –9, 1 ± 2i.
z3 – 5z2 + 4z + 10 = 0
x3
f(–1) = 0, so (z + 1) is a factor.
+
10x2
+ 29x + 30 = 0
x = –2 ± i
z3 – 5z2 + 4z + 10 = (z + 1)(z2 – 6z + 10)
x + 2 = ±i
z2 – 6z + 10 = 0
(x +
2)2
= –1
(z – 3)2 – 9 + 10 = 0
x2 + 4x + 4 = –1
(z – 3)2 = –1
x2 + 4x + 5 = 0
z – 3 = ±i
x3 + 10x2 + 29x + 30 = (x2 + 4x + 5)(x + 6)
z=3±i
Roots are –6, –2 ± i.
Roots are –1, 3 ± i.
c 2z3 – 8z2 – 13z + 87 = 0
d x = 3 ± 5i
x – 3 = ±5i
(x –
3)2
f(–3) = 0, so (z + 3) is a factor.
= –25
2z3 – 8z2 – 13z + 87 = (z + 3)(2z2 – 14z + 29)
x2 – 6x + 9 = –25
2z2 – 14z + 29 = 0
x2 – 6x + 34 = 0
2(z2 – 7z) + 29 = 0
x3 – 10x2 + 58x – 136 = (x2 – 6x + 34)(x – 4)

7
2 z −
2

Roots are 4, 3 ± 5i.
e 3x3 – 38x2 + 135x – 74 = 0
x=6±i
x – 6 = ±i
(x –
x2
6)2
2
2
2
= –1
– 12x + 36 = –1
x2 – 12x + 37 = 0
3x3 – 38x2 + 135x – 74 = (x2 – 12x + 37)(3x – 2)
Roots are
( ) − 494  + 29 = 0
2 ( z − 7 ) − 49 + 29 = 0
2
2
7
9
2( z − ) + = 0
2
2
7
9
2( z − ) = −
2
2
( z − 72 ) = − 94
2
, 6 ± i.
3
f x4 + 3x3 + x2 + 13x + 30 = 0
x = 1 ± 2i
x – 1 = ±2i
(x – 1)2 = –4
2
2
3
z– 7 =± i
2
2
7 3
z= 2 ± i
2
Roots are –3,
7 3
± i.
2 2
x2 – 2x + 1 = –4
126
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9
WORKED SOLUTIONS
7
a i z3 – 5z2 – 4z + 60 = 0
b z = 1 ± 7i
f(–3) = 0, hence (z + 3) is a factor.
z – 1 = ±7i
z3
(z – 1)2 = –49
–
5z2
– 4z + 60 = (z +
3)(z2
– 8z + 20)
z2 – 8z + 20 = 0
z2 – 2z + 1 = –49
(z – 4)2 – 16 + 20 = 0
z2 – 2z + 50 = 0
(z –
4)2
2z3 + kz2 + 102z – 50 = (z2 – 2z + 50)(2z – 1)
= –4
z – 4 = ±2i
k = –5
z = 4 ± 2i
Roots are
Roots are –3, 4 ± 2i.
9
ii
Im
×
2
×
0
–3
Re
×
–2
b i z4 – 2z3 + z2 + 2z – 2 = 0
f(1) = 0 and f(–1) = 0, hence (z – 1) and
(z + 1) are factors.
z4 – 2z3 + z2 + 2z – 2 = (z2 – 1)(z2 – 2z + 2)
z2 – 2z + 2 = 0
(49 − 59i)(2 + 13i)
z1 = 49 − 59i =
(2 − 13i)(2 + 13i)
2 − 13i
98 + 767 + 637i − 118i
=
4 + 169
865 + 519i
=
173
z1 = 5 + 3i
Two roots of the equation are 5 ± 3i
z = 5 ± 3i
z – 5 = ±3i
(z – 5)2 = –9
z2 – 10z + 25 = –9
z2 – 10z + 34 = 0
z3 + 94z = 16z2 + 204
z3 – 16z2 + 94z – 204 = 0
(z – 1)2 – 1 + 2 = 0
(z2 – 10z + 34)(z – 6) = 0
(z – 1)2 = –1
z=6
10 a (x + iy)2 = –27 – 36i
z – 1 = ±i
z=1±i
x2 + 2xyi – y2 = –27 – 36i
Roots are ±1 and 1 ± i.
x2 – y2 = –27
ii
2xy = –36
Im
1
×
0
–1
–1
8
1
, 1 ± 7i.
2
()
2
a c=3 3
3
y=
×
×
1
×
()
2
+ 10 3
Re
x2 –
( −x18 ) = –27
2
x4 – 324 = –27x2
2
()
2
+ 16 3 = 16
(3x3 + 10x2 + 16x – 16) ÷ (3x – 2) = x2 + 4x + 8
x2 + 4x + 8 = 0
(x + 2)2 – 4 + 8 = 0
(x + 2)2 = –4
x + 2 = ±2i
x = –2 ± 2i
Roots are
−18
x
2
, –2 ± 2i.
3
x4 + 27x2 – 324 = 0
(x2 – 9)(x2 + 36) = 0
x2 = 9
x = ±3
y = 6
Since Im z1 > 0, z1 = –3 + 6i
b If –3 + 6i is a root, then so is the complex
conjugate, –3 – 6i
z = –3 ± 6i
z + 3 = ±6i
127
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Complex numbers
Exercise 9.4A
(z + 3)2 = –36
z2 + 6z + 9 = –36
1
z2 + 6z + 45 = 0
a z = 52 + 52 = 5 2
( 55 ) = 34π
5 2 ( cos 3π + isin 3π ) and 5 2e
4
4
arg z = π – tan −1
z4 + 59z2 + 3330 = 2z(2z2 + 3)
z4 + 59z2 + 3330 = 4z3 + 6z
z4 – 4z3 + 59z2 – 6z + 3330 = 0
(z2 + 6z + 45)(z2 – 10z + 74) = 0
b z =
z2 – 10z + 74 = 0
(
(z – 5)2 = –49
z = 5 ±7i
arg z = tan −1
11 a z1 = (10 – i) – (2 – 7i) = 10 – i – 2 + 7i = 8 + 6i
b z – 8 = 6i
= –36
( )
arg z = –  π − tan −1 10  = –2.11
6 

(z2 – 16z + 100)(z – 4) = 0
2 34 (cos 2.11 – i sin 2.11) and 2 34 e–2.11i
z = 4 or 8 ± 6i
2
c Let z = x2
=4
x2
= 8 ± 6i
Square roots of 8 + 6i
(a + ib)2 = 8 + 6i
a2 + 2abi – b2 = 8 + 6i
a2 –
( a3 ) = 8
2
a4 – 9 = 8a2
a4 – 8a2 – 9 = 0
(a2 – 9)(a2 + 1) = 0
a2 = 9
a = ±3
a3(cos 2 + i sin 2) × 7(cos 3 + i sin 3)
= 21(cos 5 + i sin 5)
(
) (
) (
3
3
3
3
3
3
3
3
12 ( cos + i sin ) × 4 ( cos − i sin ) = 12 ( cos + i sin ) × 4  cos ( − ) + i sin ( − )
4
4
2
2
4
4
2
2 

3
3
= 48  cos ( − ) + i sin ( − )
4
4 

b 12 cos
a2 – b2 = 8
2ab = 6
3
b=
a
( 73 ) = 0.405
d z = 6 2 + 10 2 = 2 34
– 16z + 100 = 0
x = ±2
π
58 (cos 0.405 + i sin 0.405) and 58 e0.405i
z3 – 20z2 + 164z – 400 = 0
x2
)
c z = 7 2 + 32 = 58
Other three roots are –3 – 6i and 5 ±7i
z2
+ 22 = 4
− i
4 cos π − i sin π and 4e 6
6
6
z – 5 = ±7i
(z –
2
π
 2 
arg z = –tan −1 
=–6
 2 3 
(z – 5)2 – 25 + 74 = 0
8)2
(2 3 )
3π i
4
πi
)
πi
8π i
c 5 e 3 × 6 e 5 = 30 e 15
(cos 103π + i sin 103π ) × (cos 25π + i sin 25π ) = (cos 107π + i sin 107π )
(cos 103π + i sin 103π ) × (cos 25π + i sin 25π ) = (cos 107π + i sin 107π )
π
π
π
π
π
π
e 11 ( cos − i sin ) × 2 ( cos + i sin ) = 11  cos ( − ) + i sin ( − ) × 2 ( cos
6 
6
6
2
2
6

11 ( cos π − i sin π ) × 2 ( cos π + i sin π ) = 11  cos ( − π ) + i sin ( − π ) × 2 ( cos π + i sin π )
6
6
2
2
6
6 
2
2

π
π
= 22 ( cos + i sin )
3
3
d
f 4(cos 0.573 + i sin 0.573) × 5(cos 0.228 +
i sin 0.228) = 20(cos 0.801 + i sin 0.801)
b = ±1
Roots are 3 + i and –3 – i
Similarly, the square roots of 8 – 6i are 3 – i and
–3 + i
3
a3(cos 2 + i sin 2) ÷ 7(cos 3 + i sin 3)
= 3 ( cos ( −1) + isin ( −1))
7
x = ±2, 3 ± i, –3 ± i
128
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57743_P117_142.indd 128
( )
3
3
3
3
3
3
3
+ i sin × 4 cos − i sin = 12 cos + i sin × 4  cos − +
4
4
2
2
4
4
2

6/28/18 8:09 PM
9
WORKED SOLUTIONS
(
) (
)
)
) (
( ) ( )
( ()
(
)
( ) )
)
(
( )
)
2π + isin
3
3
3
3
3
3
3  2π
18 cos 2π + isin 2π
183 cos
+ i sin ÷ 4 cos − i sin = 12 cos + i sin ÷ 4 zcos −
+ i sin
3
3
3 − 2  3
4
4
2
2
4
4 b  1 = 2
=
z2
π
π
5
5
π
5

3
3
3
3
3
3
3
− isin
3 cos
+ isin − 5π 
3 cos −
+ i sin ÷ 4 cos − i sin = 12 cos + i sin ÷ 4  cos − + i sin − 
12
12
12
12 

4
2
2
4
4
2
2 

18 cos 2π + isin 2π
18 cos 2π + isin 2π
9
9
z1
3
3
3
3
= 3 cos + i sin
=
=
4
4
z2
π
π
5
5
π
5

− isin
3 cos
+ isin − 5π 
3 cos −
12
12
12
12 

2π i
πi
πi
5
c 5e 3 ÷ 6e 5 = 6 e 15
= 6 cos 13π + isin 13π
12
12
π
π
3
3
2
2
π
π
π
π
π


d cos
+ i sin
÷ cos
+ i sin
=  cos −
+ i sin −
i
sin
= cos π11−

10  = 6 cos10 π − i sin
1011π
10
10
5
5
10

12
12
i sin 3π ÷ cos 2π + i sin 2π =  cos − π + i sin − π  = cos π − i sin π
i
θ
10
5
5
10
10 
10
10

z1 r1e 1
=
c
z 2 r2e iθ2
π
π
π
π
π
π
π
π


e 11 cos − isin ÷ 2 cos + isin
== 11 cos − + isin −
÷ 2 cos + isin
6
6
2
2
6
6 = r1 e iθ1 − iθ22
2

r2
π
π
π
π
π 
π
π

isin ÷ 2 cos + isin
== 11 cos − + isin −
÷ 2 cos + isin
2
6
2
2
6
6 
2

r
= 1 e i(θ1 −θ2)
r2
= 11  cos − 2π + i sin − 2π  = 11 cos 2π − i sin 2π
3
2
3
3 
2
3
π
π
π
π
6 a 8 cos 4 − isin 4 × 8 cos 4 + isin 4
f 4(cos 0.573 + i sin 0.573) ÷ 5(cos 0.228 + i sin 0.228) =
4
π
π
π
π
(cos 0.345 + i sin 0.345)
= 8  cos − + isin −  × 8 cos + isin
5
4
4 
4
4

4 a x = r cos θ = 9 cos − π = 9 2
= 64
4
2
b
r(cos θ – i sin θ) × r(cos θ + i sin θ)
y = r sin θ = 9 sin − π = − 9 2
4
2
= r(cos(–θ) + i sin (–θ)) × r(cos θ + i sin θ)
b 12 cos
) (
) (
(
(
)(
)
) ( ) ( ) (
(
) (
)
)
( ) ( ) (
( ) ( ) (
)(
) (
( )
( )
(
(
)
)
)
( )
)
) )
(
( ) ( ) ((
)
( ) ( ) (
)
)
)
(
) (
)
( ) ( ) (
9 2 9 2
–
i
2
2
5π
b x = r cos θ = 2 3 cos 6 = –3
5π
y = r sin θ = 2 3 sin 6 = 3
( )
π
y = r sin θ = 14 sin ( − 3 ) = –7 3
π
d x = r cos θ = 14 cos − 3 = 7
= r2
7
( ) ( )
2π
5π
π
+ −
=
3 ( 12 ) 4
π
π
z z = (18 × 3)( cos + isin )
4
4
π
π
= 54 ( cos + isin )
4
4
2π
5π
13π
− −
=
3 ( 12 ) 12
5π
5π 
+ isin −
a z 2 = 3  cos −
12
12 

1 2
b
13π
11π
− 2π = −
12
12
z1 18 
11π
11π 
cos −
=
+ isin −
z2
3
12
12 
7 − 7 3i
) (
)
= 18 ( cos 2π + i sin 2π ) × 3  cos ( − 5π ) + i sin ( − 5π )
3
3
12
12 

= 54  cos ( 2π − 5π ) + i sin ( 2π − 5π )
3 12
3 12 

= 54 ( cos π + i sin π )
4
4
)
= r2(cos 0 + i sin 0)
–3 + 3 i
π
c x = r cos θ = 3 cos 2 = 0
π
y = r sin θ = 3 sin 2 = 3
3i
5
(
( )
( )
(
(
= 6 cos
a z1z2 = 18 cos 2π + i sin 2π × 3 cos 5π − i sin 5π
3
3
12
12
c
( )
11π
11π
− isin
12
12
)
( )
2π 2π 4π
+
=
3
3
3
4π
2π
− 2π = −
3
3
2π
2π 
z12 = 18 2  cos −
+ isin −
3
3 

( )
( )
2π
2π
= 324 ( cos
− isin )
3
3
129
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Complex numbers
(
) (
3 +1 + 3
8
2π
2π 
2π
2π
= 18 2  cos −
+ isin −
× 18 cos
+ isin
3
3 
3
3

d z13 = z12 × z1
( )
( )
3
(
)
2π 2π
−
+
= 0.
3
3
z13 = 183(cos 0 + i sin 0).
(
)
2
z2 = 42 + 42 = 4 2
9


b arg z1 = tan −1  3 3  = π
 3  3
arg z2 = tan −1 4 = π
4
4
π
π
c z1 = 6 cos 3 + i sin 3
(
)
(
z2 = 4 2 cos π + i sin π
4
4
)
(
(
)
(
)
(
)
(
(
2
)
12 1 − 3  + 12 1 + 3 




–4√3
)
 12
arg 12 1 − 3 + 12 1 + 3 i = π − tan −1 
 12
 12 3 + 1 
7π
12 1 − 3 + 12 1 + 3 i = π − tan −1 
=
 12 3 − 1  12
(
)
(
)
))
(
(
(
))
(
)
)
)
(
(
)
2
= 1152 = 24 2
( (
2
2π
3
( )
2π
y = 8sin( − ) = –4 3
3
2
2
12 1 − 3  + 12 1 + 3b x = 8cos − 2π = –4
3




)
12 1 − 3 + 12 1 + 3 i =
12 1 − 3 + 12 1 + 3 i =
)
3 −1 

Re
–4
= 12(1 – 3 ) + 12(1 + 3 )i
)
(
Im
d i(3 + 3 3 i)(4 + 4i) = 12 + 12i + 12 3 i + 12 3 i2
(
3
2
4
2
)
ii 6 ( cos π + i sin π ) ÷ 4 2 ( cos π + i sin π )
3
3
4
4
6
π
π
3
π
=
cos + i sin ) =
2 ( cos + i sin π )
12
12
4
12
12
4 2(
2π
2π
a
z = 8(cos
– isin ) = 8  cos ( − 2π ) + isin ( − 2π )
3
3
3
3 

=6
()
=
)
3
3 +1  + 

 8
(
e Since 5 and 12 are coprime, n = 12.
z1 = 32 + 3 3
3
 8
3 2 cos π + i sin π
4
12
12
= 5832, which is real.
a
(
=




arg  3 + 3 3i  = tan −1  3 − 1  = π
 3 + 1  12
 4 + 4i 
= 5832(1 + 0).
8
)
3 −1 i
(
(
(a + ib)2 = –4 – 4 3 i
3 + 1  7πa2 + 2abi – b2 = –4 – 4 3 i
=
3 − 1  12a2 – b2 = –4
)
)
2ab = –4 3 i
b=
−2 3
a
2
24 2 cos 7π + i sin 7π
 −2 3 
12
12
a2 –  a  = –4


ii 6 cos π + i sin π × 4 2 cos π + i sin π = 24 2 cos 7π + i sin 7π
12
3
3
4
4
12
4
a – 12 = –4a2
π
π
π
7
π
7
π
i sin × 4 2 cos + i sin
= 24 2 cos
+ i sin
a4 + 4a2 – 12 = 0
3
4
4
12
12
(a2 + 2)2 – 16 = 0
3 + 3 3i 4 − 4i
e i
×
(a2 + 2)2 = 16
4 + 4i
4 − 4i
)
(
(
)
(
=
=
=
3
(
)
)
12 − 12i+12 3i+12 3
16 + 16
12
3
(
(
)
3 + 1 + 12
32
) (
3 +1 + 3
8
) (
3 +1 + 3
8
)
3 −1 i
(
)
(
)
a2 + 2 = ±4
a2 = 2 or –6
)
3 −1 i
a=± 2
)
b = 
3 −1 i
2 3
= 6
2
Roots are 2 – 6 i and – 2 + 6 i
=
3
 8
(
)
2
3
3 +1  + 
 8

(
)
3 −1 

2
130
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9
WORKED SOLUTIONS
10 a | z1 z2| = 12 × 4 = 48
( )
c z1 × i = –2 – 4i; rotation by π anti-clockwise
2
about the origin
( )
b z1 = 12  cos − 5π + isin − 5π 
6
6 

d z1 – (2 – 3i) = –6 + 5i; translation by (–2 + 3i)
4cos β = 2 2
e z1 ÷ i = 2 + 4i; rotation by π clockwise about
2
the origin
2
2
 2 π
β = cos–1 2  =
4
cos β =
( )
( )
z2 = 4(cos β – isin β) = 4  cos − π + isin − π 
4
4 

( )
3
z
5π
π
7π
– − =–
arg  1  = –
4
6
12
 z2 
c
Im
−7π
12
a i rotation by π anti-clockwise about the origin
2
ii rotation by π about the origin
iii rotation by π clockwise about the origin
2
iv maps onto itself
) (
(
)
(
4
d i
5
b i
ii
6
()
3π

−1 2 
a i −  π − tan 2  = − 4
ii
b i rotation by π clockwise about the origin
2
ii maps onto itself
22 + 22 = 2 2
5 × − 3π = − 15π
4
4
π
5
arg z = 4
(
z5 = 2 2
)
5
= 128 2
( )
a arg z = π − tan −1 10 = 1.86
3
b arg z2 = 1.86 × 2 = 3.72
Since 3.72 > π, arg z2 = 3.72 – 2π = –2.56
c arg z = 1.86 ÷ 2 = 0.931
iii modulus divided by 3
iv rotation by π clockwise about the origin
2
b z1* = –4 – 2i; reflection in x-axis (real axis)
2π
2π
a cos 3 − i sin 3
c 5i
ix modulus doubled, rotation by π anti-clockwise
2
about the origin
a z1 + (2 – 3i) = –2 – i; translation by (2 – 3i)
)
b –1
viii rotation by π clockwise about the origin
2
x
modulus multiplied by 0.6, rotation by π about
the origin
)
e z1* = 5 cos π − i sin π ; reflection in x-axis
4
4
(real axis)
v rotation by π clockwise about the origin
2
vi rotation by π clockwise about the origin
2
vii maps onto itself
)
π
π
π
π
− isin
= 35 cos + isin ;
12
12
3
3
modulus multiplied by 7, rotation by π
12
anti-clockwise about the origin
3π
3π
π
π
d z1 ÷ 5 cos 4 + i sin 4 = cos 2 − i sin 2 ;
modulus divided by 5, rotation by 3π
4
clockwise about the origin
Exercise 9.5A
2
(
c z1 × 7 cos
Re
1
) (
b z1 ÷ 4(cos 2.4 + i sin 2.4) = 1.25(cos 1.61 − i sin 1.61);
modulus divided by 4, rotation by 2.4 clockwise
about the origin
z1 12
=
=3
z2
4
3
(
az1 × 3 cos π + i sin π = 15 cos 9π + i sin 9π ;
5
5
20
20
π
modulus multiplied by 3, rotation by 5 anticlockwise about the origin
d arg 3 z = 1.86 ÷ 3 = 0.621
7
2π
= –3
3
2π
y = 6sin
=3 3
3
z1 = –3 + 3 3 i
a x = 6cos
131
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COMPLEX NUMBERS
b Rotation of 90° anticlockwise about the origin
2π π 7π
c
+ =
2
3
6
9
a
Im
B
7
7π
5π
– 2π = –
6
6
d z1 = –3 + 3 3 i, z2 = –3 3 – 3i
z1 – z2 = (–3 + 3 3 i) – (–3 3 – 3i)
= (3 3 – 3) + (3 + 3 3 )i
A
| z1 – z2| = (3 3 − 3)2 + (3 + 3 3)2
0
5
10
Re
= 27 − 18 3 + 9 + 9 + 18 3 + 27
= 72
8
–4
=6 2
−20 + 21i × −2 − 5i
a z2 =
−2 + 5i
−2 − 5i
40
+
100
−
+ 105
i
42i
=
4 + 25
= 145 + 58i
29
= 5 + 2i
b
b AB = (10 + 7i) – (–1 + 2i) = 11 + 5i
AB = 112 + 52 = 146
AC = (5 – 4i) – (–1 + 2i) = 6 – 6i
AC = 6 2 + 6 2 = 72
BC = (5 – 4i) – (10 + 7i) = –5 – 11i
BC = 52 + 112 = 146
Im
×
–2
Since AB = BC , the triangle is isosceles.
5
2
0
C
×
5
Re
 8 
 1 
π
10 a arg z = –tan–1 8 3  = –tan–1 3  = –
6
π
π
arg z2 = 2 × – = –
6
3
π
π
b Since 2π = 12 × , arg z33 = arg z9 = 9 × –
6
6
3π π
=–
=
2
2
c |z| = (8 3)2 + (−8)2 = 256 = 16
|z4.5| = 164.5 = 262 144
c Gradient OA = – 5 .
2
2
Gradient OB = 5 .
Gradient OA × Gradient OB = –1.
Exercise 9.6A
1
x + iy − 2 + 8i = 13
( x − 2) + i ( y + 8 )
Or i(5 + 2i) = 5i – 2, hence i z2 = z1 and
multiplication by i is a rotation of 90°
anticlockwise.
d Since OAB is a right angle, the centre of the
circle is the mid-point of AB.
( −22+ 5 , 2 +2 5 ) = ( 32 , 72 ).
z − 2 + 8i = 13
(x –
2)2
+ (y +
8)2
= 13
= 132
Centre = (2, –8), radius = 13.
2
z −3 = z −2−i
x + iy − 3 = x + iy − 2 − i
( x − 3) + iy
(x –
3)2
+
y2
= ( x − 2 ) + i ( y − 1)
= (x – 2)2 + (y – 1)2
x2 – 6x + 9 + y2 = x2 – 4x + 4 + y2 – 2y + 1
–6x + 9 = –4x + 4 – 2y + 1
–2x + 2y + 4 = 0
–x + y + 2 = 0
or y = x – 2
132
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9
WORKED SOLUTIONS
3
a x + iy = 10
d x + iy − ( 5 − 6i ) = 2
x2 + y2 = 102
Circle, centre (0, 0), radius 10.
( x − 5) + i ( y + 6 )
=2
(x – 5)2 + (y + 6)2 = 22
Im
Circle, centre (5, –6), radius 2.
Im
10
0
0
–10
10
Re
Re
×
–6
–10
b x + iy − 9 = 4
( x − 9) + iy
5
e x + iy + 4 − 3i = 7
=4
( x + 4 ) + i ( y − 3)
=7
(x – 9)2 + y2 = 42
Circle, centre (9, 0), radius 4.
(x +
Im
Circle, centre (–4, 3), radius 7.
5
4)2
+ (y –
3)2
= 72
Im
4
3
2
1
0
–1
1
2
3
4
5
6
7
8
9 10 11 12 13
Re
–2
–3
×
–4
3
–5
c
x + iy + 2i = 8
–4
0
Re
x + i ( y + 2) = 8
x2 + (y + 2)2 = 82
Circle, centre (0, –2), radius 8.
4
Im
a x + iy − 5 = x + iy − 3i
( x − 5) + iy
(x –
5)2
+
y2
= x + i ( y − 3)
= x2 + (y – 3)2
x2 – 10x + 25 + y2 = x2 + y2 – 6y + 9
0
–2
Re
–10x + 25 = –6y + 9
10x – 6y = 16
5x – 3y = 8
133
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COMPLEX NUMBERS
d x + iy − 2 − 4i = x + iy + 5 − 2i
Im
( x − 2) + i ( y − 4 ) = ( x + 5) + i ( y − 2)
(x – 2)2 + (y – 4)2 = (x + 5)2 + (y – 2)2
x2 – 4x + 4 + y2 – 8y + 16 = x2 + 10x + 25
+ y2 – 4y + 4
3
2
–4x + 4 – 8y + 16 = 10x + 25 – 4y + 4
1
14x + 4y + 9 = 0
0
–1
1
2
3
4
Im
Re
5
4
×
–2
2
×
–3
b x + iy + i = x + iy − 4
0
–5
Re
2
x + i ( y + 1) = ( x − 4 ) + iy
x2 + (y + 1)2 = (x – 4)2 + y2
x2 + y2 + 2y + 1 = x2 – 8x + 16 + y2
2y + 1 = –8x + 16
e x + iy + 5 + 2i = x + iy − 7 − 6i
8x + 2y = 15
( x + 5) + i ( y + 2)
Im
= ( x − 7) + i ( y − 6)
(x + 5)2 + (y + 2)2 = (x – 7)2 + (y – 6)2
x2 + 10x + 25 + y2 + 4y + 4 = x2 – 14x
+ 49 + y2 – 12y + 36
×
10x + 25 + 4y + 4 = –14x + 49 – 12y + 36
24x + 16y – 56 = 0
3x + 2y – 7 = 0
Re
×
Im
6
c
x + iy + 3i = x + iy − 7i
×
x + i ( y + 3) = x + i ( y − 7 )
x2 + (y + 3)2 = x2 + (y – 7)2
x2 + y2 + 6y + 9 = x2 + y2 – 14y + 49
6y + 9 = –14y + 49
–5
20y = 40
×
0
7
Re
–2
y=2
Im
4
2
–5
0
2
Re
134
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9
WORKED SOLUTIONS
5
a
6
Im
a
Im
1 radian
Re
b
Re
Im
b
Im
π
4
–3
c
0
8
Re
–8
8
Im
Re
–8
0
c
1
7π
10
–2
d
Im
Re
Im
2
Re
0
Re
–1
3 radian
7
a
x + iy − 3 − 2i = 5
( x − 3) + i ( y − 2)
(x –
3)2
+ (y –
2)2
=5
= 52
Circle, centre (3, 2), radius 5.
x + iy − 6i = x + iy − 7 + i
x + i ( y − 6 ) = ( x − 7 ) + i ( y + 1)
x2 + (y – 6)2 = (x – 7)2 + (y + 1)2
x2 + y2 – 12y + 36 = x2 – 14x + 49 + y2 + 2y + 1
135
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COMPLEX NUMBERS
14x – 14y = 14
(x – 12)(x + 5) = 0
y=x–1
x = 12 because half-line only exists for x  4 and y = 8.
Im
9
Complex number is 12 + 8i
a |z – 3| = |z + 2i|
|(x – 3) + iy| = |x + i(y + 2)|
(x – 3)2 + y2 = x2 + (y + 2)2
x2 – 6x + 9 + y2 = x2 + y2 + 4y + 4
–6x + 9 = 4y + 4
×
2
6x + 4y – 5 = 0 1
|(x + 3) + i(y – 1)| = |(x – 1) + i(y + 5)|
0
–1
1
Re
3
(x + 3)2 + (y – 1)2 = (x – 1)2 + (y + 5)2
x2 + 6x + 9 + y2 – 2y + 1 = x2 – 2x + 1 + y2 + 10y + 25
8x – 12y – 16 = 0
2x – 3y – 4 = 0
b Substitute y = x – 1 into the equation for the
circle:
(x – 3)2 + (x – 1 – 2)2 = 52
7
13
2x – 3(–
2(x – 3)2 = 52
7
)–4=0
13
31
26
31 7
– i
26 13
b |z – 3| > |z + 2i| is true when z = 0
x=
25
2
x–3=±
1 – 2 13y + 7 = 0
y=–
(x – 3)2 + (x – 3)2 = 52
(x – 3)2 =
6x – 9y – 12 = 0 2
5 2
2
|z + 3 – i| < |z – 1 + 5i| is true when z = 0
5 2
2
5
y=2± 2
2
Complex numbers are

5 2 
5 2 
 3 + 2  +  2 + 2  i  and


x=3±
Im
5
4

5 2 
5 2 
 3 − 2  +  2 − 2  i 


8
5
6
2
Re
Cartesian equation for | z – 3i| = 13
|x + iy – 3i| = 13
–4
|x + i(y – 3)| = 13
3
x2 + (y – 3)2 = 132
Cartesian equation for arg(z – 4) =
Gradient = 1, passes through (4, 0)
y=x–4
x2 + (x – 4 – 3)2 = 132
x2 + (x – 7)2 = 132
x2 + x2 – 14x + 49 = 169
π
4
10 a i arg(z + 2 – 5i) = π is a half-line starting at (–2, 5)
4
ii arg(z + 2 – 5i) = – π is a half-line starting at
2
(–2, 5)
iii |z + 2 – 5i| = 29 is a circle, centre (–2, 5),
radius 29 . It passes through the origin.
2x2 – 14x – 120 = 0
x2 – 7x – 60 = 0
136
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9
WORKED SOLUTIONS
b
2
a
2
The complex numbers use the appropriate
combinations of these solutions in the form x + yi
12 a z* = 5 – 2i
x = –a 
Im
z
5 + 2i (5 + 2i)(5 + 2i)
=
=
z * 5 − 2i (5 − 2i)(5 + 2i)
5
–2
=
Re
25 + 10i + 10i − 4
25 + 4
21 + 20i
1
=
(21 + 20i)
29
29
2
b arg z = tan–1 5
=
()
2
arg z* = –tan ( 5 )
11 a |z + a| = 2a is a circle, centre (–a, 0), radius 2a,
equation (x + a)2 + y2 = a2
–1
Im
()
Also arg
–3a
–a
()
()
2
2
z
= arg z – arg z* = tan–1 5 + tan–1 5
z*
2
= 2tan–1 5
arg
a
( )
20
z
= tan–1 21
z*
c z + z* = 5 + 2i + 5 – 2i = 10
Re
Re
b |z – ai| = |z + a(2 + i)|.
A
2
|x + i(y – a)| = |(x + 2a) + i(y + a)|
B
x2 + (y – a)2 = (x + 2a)2 + (y + a)2
0
x2 + y2 – 2ay + a2 = x2 + 4ax + 4a2 + y2 + 2ay + a2
5
10
Im
C
–2
0 = 4ax + 4ay + 4a2
0=x+y+a
x + y = –a
Rhombus
13 a w3 = (6 + i)3 = 63 + 3(6)2(i) + 3(6)(i)2 + (i)3
= 216 + 108i – 18 – i
Im
= 198 + 107i
b tan w =
1
6
()
1
1
1
arg w = tan ( 6 ) + tan ( 6 ) + tan ( 6 )
1
= 3tan ( 6 )
107
From a, arg w = tan ( 198 )
1
arg w = tan–1 6
–a
3
Re
–1
–1
–1
–a
3
c (x + a)2 + y2 = a2
Let x = –a – y
(–a – y + a)2 + y2 = a2
2y2 = a2
y=± 2a
2
–1
–1
Exam-style questions
1
a –1 – 7i
b
12 + 7 2 = 5 2
52 + 52 = 5 2
137
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COMPLEX NUMBERS
()
c arg z = − tan −1 5 = − π
5
4
(
z = 5 2 cos π − i sin π
4
4
4
)
b |w| = 22 + 52 = 29 = 5.39
c zw = (17 – i)(2 + 5i) = 34 + 85i – 2i + 5 = 39 + 83i
d (–1 + 7i)(5 – 5i) = –5 + 5i + 35i + 35
e z + w = (–1 + 7i) + (5 – 5i) = 4 + 2i
2
a
( 42 )= 0.464
14 − 31i 3 + 2i
×
3 − 2i
3 + 2i
5
42 + 28i − 93i + 62
=
9+4
=
z 17 − i (17 − i)(2 − 5i)
=
=
w 2 + 5i (2 + 5i)(2 − 5i)
34 − 85i − 2i− 5
=
4 + 25
29 − 87 i
=
29
= 1 – 3i
a (3 + 2i)(3 + 2i) = 9 + 6i + 6i – 4 = 5 + 12i
d
= 30 + 40i
arg(z + w) = tan −1
( )
1
a arg z = –tan–1 17 = –0.0589
(5 + 12i)(3 + 2i) = 15 + 10i + 36i – 24 = –9 + 46i
b z1 + z2 = (19 – 9i) + (3 + 2i) = 22 – 7i
104 − 65i
= 8 – 5i
13
z1 + z 2 = 222 + 7 2 = 533
b w = 8 ± 5i
c
w – 8 = ±5i
19 − 9i 3 − 2i
×
3 + 2i 3 − 2i
(w – 8)2 = –25
w2 – 16w + 64 = –25
57 − 38i − 27i − 18
13
39 − 65i
=
= 3 – 5i
13
=
w2 – 16w + 89 = 0
c = –16, d = 89
c
z1 19 − 9i
=
3 + 2i
z2
d z − (19 − 9i) = z − (3 + 2i)
Im
x + iy − 19 + 9i = x + iy − 3 − 2i
(x − 19) + i( y + 9) = (x − 3) + i( y − 2)
(x – 19)2 + (y + 9)2 = (x – 3)2 + (y – 2)2
x2 – 38x + 361 + y2 + 18y + 81 = x2 – 6x + 9
+ y2 – 4y + 4
×
5
–38x + 361 + 18y + 81 = –6x + 9 – 4y + 4
–32x + 22y + 429 = 0
0
8
Re
6
a
1
3
+
i =
2
2
b tan −1
3
a (3 + 5i)(8 – 7i) = 24 – 21i + 40i + 35
= 59 + 19i
w = 59 ± 19i
w – 59 = ±19i
2
2
 3
+
=1
 2 
( 3 ) = π3



c  1 + 3 i  1 + 3 i = 1 + 3 i + 3 i − 3 = − 1 + 3 i
2 2
2  4
4
4
4
2
2
2
1
3 1
3  1
3
3
3
1
3
 2 + 2 i   2 + 2 i  = 4 + 4 i + 4 i − 4 = − 2 + 2 i
(w – 59)2 = –361
w2 – 118w + 3481 = –361
w2 – 118w + 3842 = 0
b tan −1
()
1
2
( 1959 ) = 0.31
1
3  1
3 
1 3
 2 + 2 i   − 2 + 2 i  = − 4 − 4 = −1
d Roots are –1, 1 ± 3 i
2
2
c w + 2z2 – z1 = (59 + 19i) + 2(8 – 7i) – (3 + 5i)
= 59 + 19i + 16 – 14i – 3 – 5i
= 72
138
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9
WORKED SOLUTIONS
e
3
1
× 65 × 65 ×
5
2
39
=
2
=
Im
√3
2
9
0
–1
a r = 18 ÷ 3 = 6
θ = 2π − − 7π = 11π
5
10
10
( )
11π − 2π = − 9π
10
10

w = 6  cos − 9π + i sin − 9π 
10
10 

Re
1
2
( )
( )
w = 6 ( cos 9π − i sin 9π )
10
10
−√3
2
b
Im
7
a Centre of circle = (0, 7) and radius = 7.
z − 7i = 7
b r = 72 + 72 = 7 2
()
θ = π − tan −1 7 = 3π
7
4
(
w1 = 7 2 cos 3π + i sin 3π
4
4
(
)
)
8
c w18 = 7 2 (cos 6π + i sin 6π)
c wz = –12
= 92 236 816(1 + 0)
= 92 236 816
8
9π
10
6
( )
( )
( )
( )
6  cos − 9π + i sin − 9π  × z = 12(cos(−π) + i sin(−π))
10
10 

6  cos − 9π + i sin − 9π  × z = 12(cos(−π) + i sin(−π))
10
10 

a OM = 12 + 8 2 = 65
z=
2
ON = 4 + 7 = 65
2
( )
( )
12(cos(−π) + isin(−π)) = 2  cos − π + isin − π 
10
10 


6 cos − 9π + isin − 9π 
10
10 

( ) ( )
MN = z – z = (4 + 7i) – (1 – 8i) = 3 + 15i
z = 12(cos(−π) + isin(−π)) = 2  cos ( − π ) + isin ( − π )
MN = 3 + 15 = 234
10
10 

6  cos ( − 9π ) + isin ( − 9π )
2
Re
1
2
2
10

Two sides are the same, so OMN is isosceles.
b Cosine rule: cos O =
(
65
) +(
2
65
) −(
2
2 × 65 × 65
65 + 65 − 234
=
2 × 65
=
−104
4
=−
130
5
c sin2 O + cos2 O = 1
( )
sin2 O + −
4
5
2
( )
)
2
10 |z – 6| = |z – 4 – 6i|
|(x – 6) + iy| = | (x – 4) + i(y – 6)|
(x – 6)2 + y2 = (x – 4)2 + (y – 6)2
x2 – 12x + 36 + y2 = x2 – 8x + 16 + y2 – 12y + 36
0 = 4x – 12y + 16
0 = x – 3y + 4
=1
sin2 O = 9
25
234
10 
−πi
z = 2e 10
4
3
When y = 0, x = –4
When x = 0, y =
2
3
sin O =
5
Area = 1 ab sin c
2
Line also passes through (5, 3)
|z – 5 – 3i| = 3 is a circle of radius 3 centre (5, 3)
For the complex number z = 0, |0 – 6| > |0 – 4 – 6i| is
false, so the region is not on the same side of the
line as z = 0 and |z – 5 – 3i| < 3 is also false, so the
region is inside the circle
139
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COMPLEX NUMBERS
z2 – 4z + 125 = 0
Im
z3
– 2z2 + az + 250 = (z + 2)(z2 – 4z + 125)
(z + 2)(z2 – 4z + 125) = z3 – 2z2 + 117z + 250
a = 117
3
c Real root is –2.
4
3
11 a z =
=
2 − 11i 4 − 3i
×
4 + 3i 4 − 3i
d w=
–4
5
Re
26 + 29i 6 − i
×
6+i
6−i
=
8 − 6i − 44i − 33
16 + 9
=
−25 − 50i
25
= –1 – 2i
14 a x + iy − 8 = x + iy + 2i
156 − 26i + 174i + 29
36 + 1
( x − 8 ) + iy
= x + i ( y + 2)
185 + 148i
=
37
(x –
= 5 + 4i
x2 – 16x + 64 + y2 = x2 + y2 + 4y + 4
b iz = i(5 + 4i) = –4 + 5i
arg(–4 + 5i) = π – tan −1
8)2
+
y2
= x2 + (y + 2)2
–16x + 64 = 4y + 4
()
4x + y = 15
5
= 2.25
4
b arg (z + 3 – 6i) = − π passes through (–3, 6)
4
with a gradient of –1.
c z = 5 ± 4i
z – 5 = ±4i
y – 6 = –1(x + 3)
(z – 5)2 = –16
y=3–x
z2 – 10z + 25 = –16
4x + (3 – x) = 15
z2 – 10z + 41 = 0
3x + 3 = 15
5z3 – 49z2 + 195z + 41 = (5z + 1)(z2 – 10z + 41)
x=4
1
z = − , 5 ± 4i
5
When x = 4, y = –1.
49
1
.
Sum of roots = − + (5 + 4i) + (5 – 4i) =
5
5
12 a 53e– 0.557i
4–i
c
Im
16
b 148877
15
14
c i 7 – 2i and –7 + 2i
ii
13
Im
12
4
11
10
2
–8
–6
–2 0
–2
–4
9
2
4
–4
6
8
Re
8
7
6
5
4
13 a z = (2 –
i)3
= (2 – i)(2 – i)(2 – i)
= (3 – 4i)(2 – i)
3
2
1
–6 –5 –4 –3 –2 –1 0
–1
= 2 – 11i
1 2 3 4 5
Re
–2
b z = 2 ± 11i
z – 2 = ±11i
(z – 2)2 = –121
z2 – 4z + 4 = –121
–3
–4
d arg(z + 3 – 6i) = π passes through (–3, 6) with
4
a gradient of 1.
140
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9
WORKED SOLUTIONS
( )
y – 6 = 1(x + 3)
y=x+9
4x + (x + 9) = 15
5x = 6
6
x=
5
51
6
When x = , y =
.
5
5
( )
6 51
Let the vertices be A(4, –1), B(–3, 6) and C 5 , 5 .
Triangle is right-angled with the right angle at B.
AB =
( −3 − 4 )2 + (6 − ( −1))
BC =
(
) (
2
6
51
− ( −3) +
−6
5
5
)(
(
)
2
)
=7 2
2
= 12(cos 0.9 + i sin 0.9)
21
=
2
5
b
and 2(cos (π – 0.2) – isin (π – 0.2))
c 3m × 4n = 432
Im
P×
w = 4 ( cos0.4 + isin 0.4 )
= 2(cos 0.2 + isin 0.2)
Area = 1 7 2 21 2 = 147 .
2
5
5
15 a
2
sin2 Q = 9
25
3
sin Q =
5
1
Area of PQR = ab sin c
2
1
= × 85 × 85 × 3
5
2
= 51
2
Area of OPQR = 51 × 2 = 51.
2
16 a z2 = 3(cos 0.5 + i sin 0.5) × 4(cos 0.4 + i sin 0.4)
432 = 24 × 33
7
m = 3, n = 2
×Q
5
arg(z3w2) = 3 × 0.5 + 2 × 0.4 = 2.3
17 a z = –2 ± 4i
z + 2 = ±4i
0
–6
3
Re
9
×R
–2
(z + 2)2 = –16
z2
+ 4z + 4 = –16
z2 + 4z + 20 = 0
z4 – 6z3 + 14z2 – 64z + 680
= (z2 + 4z + 20) (z2 – 10z + 34)
b | OP | = 6 2 + 7 2 = 85
| OR | = 92 + 22 = 85
z2 – 10z + 34 = 0
RQ = (3 + 5i) – (9 – 2i) = –6 + 7i
(z – 5)2 – 25 + 34 = 0
PQ = (3 + 5i) – (–6 + 7i) = 9 – 2i
(z – 5)2 = –9
Since OP (z1) = RQ and OR (z3) = PQ, sides
are parallel and the same length.
z – 5 = ±3i
z = 5 ± 3i
Since adjacent sides are also the same
length, OPQR is a rhombus.
Other three roots are –2 – 4i, 5 ± 3i.
b, e
Im
c PR = (9 – 2i) – (–6 + 7i) = 15 – 9i
| PR | = 152 + 92 = 306
Cosine rule: cos Q =
(
85
( )
4
Q = cos–1 − 5 = 143.1°
d sin2 Q + cos2 Q = 1
( )
4
5
2
85
=1
) −(
2
2 × 85 × 85
85 + 85 − 306
2 × 85
4
−136
=
=−
170
5
=
sin2 Q + −
) +(
2
306
)
×
4
2
×
2
–4
–2
0
2
4
6
Re
–2
×
×
–4
c ABCD is a trapezium.
Area = 1 (8 + 6) × 7 = 49.
2
141
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COMPLEX NUMBERS
d By symmetry, the circle that passes through
the roots –2 ± 4i and –5 ± 3i has its centre on
the Re axis, say at the point (a, 0) representing
the real number a. Then the equation of the
circle is (x − a)2 + y2 of the form (x − a)2 + y 2 =
b2,where b > 0 is the radius.
Since (−2, 4) and (5, 3) lie on the circle,
5 3
– i
2 2
2z = 5 – 3i
d z=
2z – 5 = –3i
(2z – 5)2 = –9
4z2 – 20z + 25 = –9
(−2 − a)2 + 4 2 = b 2
4z2 – 20z + 34 = 0
(5 − a)2 + 32 = b 2
2z2 – 10z + 17 = 0
so (−2 − a)2 + 16 = (5 − a)2 + 9
6z3 + 11z + 68 = 22z2
6z3 – 22z2 + 11z + 68 = 0
4 + 4a + a 2 + 16 = 25 − 10a + a 2 + 9
14a = 14
a =1
b=5
Therefore the equation of the circle is |z – 1| = 5
(3z + 4)( 2z2 – 10z + 17) = 0
4 5 3
z=– , + i
3 2 2
20 a (2 + pi)4 = 24 + 4(2)3(pi) + 6(2)2(pi)2
+ 4(2)(pi)3 + (pi)4
18 a z* = –8 – 8i
= 16 + 32pi – 24p2 – 8p3i + p4
2
2
|z*| = (−8) + (−8) = 128 = 8 2
b 16 – 24p2 + p4 = 41
()
8 
3π
b arg z* = –π +  tan –1
=–
8 
4

p4 – 24p2 – 25 = 0
(p2 – 25)(p2 + 1) = 0
c |(–8 + 8i) + (a + 2i)| = 26
p2 = 25
|(a – 8) + 10i| = 26
p = ±5
(a – 8)2 + 102 = 676
c p = –5
(a – 8)2 = 576
Im z = 32(–5) – 8(–5)3
a – 8 = ±24
= –160 + 1000 = 840
a = 32 or –16
Mathematics in life and work
Since a < 0, a = –16
19 a
1
7 + 6i (7 + 6i)(1 − 3i)
=
1 + 3i (1 + 3i)(1 − 3i)
29 – 5 = 24
Z = 24 2 + 10 2 = 26 Ω
7 − 21i + 6i + 18
1+9
25 − 15i
=
10
5 3
= – i
2 2
3
b arg z = –tan–1 5 = –0.540
=
2
( )
24
tan–1 10 = 1.18
26(cos 1.18 + i sin 1.18)
()
c z + 2i =
5 1
+ i
2 2
Im
π
3
1
2
5
2
Re
142
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WORKED Solutions
Summary Review
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering
the question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees,
unless a different level of accuracy is specified in the question.
ii Gradient of AB = 0 − 2 = − 1
3 − −3
3
6−0
=3
Gradient of BC =
5−3
Warm-up Questions
1
2
A = πr2 ⇒ dA = 2πr
dr
dr
=3
dt
dA dA dr
=
×
= 2πr × 3 = 6πr
dt
dr dt
dA
= 6π ( 50 ) = 300π m2h −1
When t = 0, r = 50 ⇒
dt
1 
i OA ⋅ OC =  2p 
 
q
(
(
 − 4p 2 + q 2

⋅
2p

q

2
= − 4p + q
(
2
− 4p + q
ii AC =
2
2
) + 4p
)−1
2p − 2p
2
( −4p
CA > 0
iii BA =
2
⇒
− q2 − 1
)
2
 −3  2  −1
OD = OA + AD =   +   =  
 2   6  8 



⇒
D(–1, 8)
AD = 22 + 6 2 = 40
2
+q =0
A Level Questions:
Pure Mathematics 2
2
−4p − q − 1
0
0
q−q
=
 5 − 3   2
iii AD = BC = 
=
 6 − 0  6
)
2
=
(Gradient of AB) × (Gradient of BC ) = –1
⇒ The gradients are perpendicular.
1
(
= ± −4p 2 − q 2 − 1
)
i h = 0.5 (width of strips)
x0 = 0 ⇒ y0 = 0.125
x1 = 0.5 ⇒ y1 = 0.107 556…
x2 = 1 ⇒ y2 = 0.087 438…
1
1
∫ 6 + 2e x dx ≈
CA = AC = 4p 2 + q 2 + 1
0
0.5
0.125 + 2 × 0.107556 + 0.087438 ]
2 [
≈ 0.11
−x
ii x + 4e − 2e−2x + c
1
2p − q
q − 2p
p = 3 and q = 2 ⇒
1
BA = 4 = 12 + 4 2 + 8 2
8
2
a i
ii
⌠ e 2x + 6 dx =

⌡ e 2x
∫3cos
2
3
cos2x + 1) dx
2∫(
3 1
=
sin 2x + x + c
2 2
x dx =
= 81 = 9
3
1 
So the unit vector is 1  4
9 
 8
=
b h=
1
6−2
=
iGradient of AC =
5 − −3 2
⇒ Gradient of MB = –2
−3 + 5 2 + 6 ⇒ M(1, 4)
M
,
2
2
(
x1 = 1.5
Equation of line through MB is y – 4 = – 2(x – 1)
⇒ y = – 2x + 6
When y = 0, 2x = 6 ⇒
57743_P143_153.indd 143
x=3 ⇒
B(3, 0)
x2 = 2
2
(
)
3
3x
sin 2x +
+c
4
2
2−1 1
=
2
2
x0 = 1
)
−2x
−2x
∫ (1 + 6e ) dx = x − 3e + c
⇒
⇒
⇒
y0 = 6
ln3
6
ln3.5
y2 = 6
ln4
y1 =
6
1 6
6
6 
⌠
 ln(x + 2) dx ≈ 4  ln3 + 2 × ln3.5 + ln4  = 4.84 (2 dp)
⌡
1
143
6/28/18 2:38 PM
SUMMARY REVIEW
16
3
∫
i
6
16
16
6
2
dx = 3 ∫
dx = 3[ ln 2x − 7 ]6
2x − 7
2x − 7
7
0 = − 8 + 4a + b
4a + b = 8
①
Similarly, g(−1) = −18
Remainder theorem
−18 = −1 + b − a
a − b = 17
②
① + ②:
5a = 25
a=5
In ②: 5 − b = 17
b = −12
ii g(x) − f(x) = (x3 − 12x2 − 5) − (x3 + 5x2 − 12)
= − 17x2 + 7
Maximum occurs when x = 0.
Maximum = 7.
6
= 3 ( ln 25 − ln 5) = 3 ln 5 = ln 125
17 − 1
=4
4
x0 = 1 ⇒ y0 = 0
x1 = 5 ⇒ y1 = log 5
x2 = 9 ⇒ y2 = log 9
x3 = 13 ⇒ y3 = log 13
x4 = 17 ⇒ y4 = log 17
ii h =
17
∫ log x dx ≈ 2 (0 + 2( log 5 + log 9 + log13) + log17 )
4
1
= 13.5 (3 sf )
A Level Questions:
Pure Mathematics 2 and Pure Mathematics 3
4
5
5x + 3 = 7x − 1
ln 5x + 3 = ln 7x − 1
(x + 3)ln 5 = (x − 1)ln 7
3 ln 5 + ln 7 = x ln 7 − x ln 5
3 ln 5 + ln 7 = x(ln 7 − ln 5)
3 ln 5 + ln 7
x=
ln 7 − ln 5
x = 20.1 (this works equally well using
log10)
y = 6 sin x − 2 cos 2x
dy
= 6 cos x + 4 sin 2x
dx
When x = π , dy = 6 cos π + 4 sin π
6 dx
6
3
dy
3
3
=6
+4
=5 3
2
2
dx
( y − 2) = 5 3 x − π6
i (x + 2) is a factor ⇔ f(−2) = 0 Factor theorem
8
i
y
15
10
y = 15 – x3
5
y = 3lnx
( )
0
y = 8.66x − 2.53
6
i
∫ 4 cos
2
double
(θ2 )dθ = ∫ 4( cosθ2 + 1 ) dθ (using
angle formulae)
= 2∫ (cosθ + 1)dθ
6
ii
1
ii f(x) = x3 + 3 ln x – 15
= 2sin  + 2 + c
f(2.0) =−4.9205…
1
2
∫ 2x + 3 dx = 2 ∫ 2x + 3 dx
−1
−1
=
6
1
ln 2x + 3 ]−1
2[
1
= [In 15 − In 1]
2
=
x
The solution to the equation is the
intersection of the graphs. They intersect
once, so there is only one root.
= 2(sin  + ) + c
6
5
f(2.5) = 3.373…
Change of sign ⇔ Root in the interval
[2.0, 2.5].
iii x0 = 2
x1 = 2.346 53…
1
ln15
2
144
57743_P143_153.indd 144
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WORKED Solutions
x2 = 2.317 15
Converting back into x:
3x = 0 ⇒ no solutions, since 3x is always positive
x3 = 2.319 49
1− 5
⇒ no solutions, since 3x is always
2
positive
1+ 5
3x =
2
1 + 5
ln 3x = ln 
 2 
3x =
x4 = 2.319 31
x5 = 2.319 32
x6 = 2.319 32
x7 = 2.319 32
9
( )
Therefore, x = 2.319
a 2 cosec 2 tan  ≡ 2tan θ
sin 2θ
2 sin θ
cosθ
≡
2sin θ cosθ
2sin θ
≡
2sin θ cos2 θ
1
≡
cos2 θ
b i
≡
1 + 5
x In 3 = In 
 2 


ln  1 + 5 
 2 
x=
ln 3
x = 0.438
12 i sin 2 sec  ≡ 2sin α cos α
cos α
≡ 2 sin 
sec2 
2 cosec 2 tan  = 5
ii 3 cos 2 + 7 cos  = 0
sec2  = 5
1
cos2 θ =
5
cosθ = ±
3(2 cos2  − 1) + 7 cos  = 0
6 cos2  + 7 cos  − 3 = 0
(3 cos  – 1)(2 cos  + 3) = 0
1
5
cos β =
 = 1.11 rad or  = 2.03 rad
ii
π
6
π
6
0
0
∫ 2cosec4x tan 2x dx = ∫ sec
10
=
3
2
4x2 – 20x + 25 > 36x2 + 36x + 9
11 3x + 32x = 33x
If u = 3x then: u + u2 = u3
0 = u3 − u2 − u
= u(u2 − u − 1)
So u = 0 or u2 − u − 1 = 0
1 ± 1 − 4 (1)( −1)
u=
2
=
1± 5
2
2x dx
π
=  1 tan π  −  1 tan 0 
3   2

 2
4x2 – 20x + 25 > 9(4x2 + 4x + 1)
0 > 4x2 + 7x – 2
–2<x < 1
4
Therefore cos β =
6
=  1 tan 2x 
 2
0
(2x – 5)2 > 9(2x + 1)2
0 < 32x2 + 56x – 16
2
1
3
or cos β = − ⇒ no solutions
3
2
13 i
1
3
1
= (cos t )−3
cos3 t
dx
3sin t
= −3(cos t)−4 × −sin t =
dt
cos4 t
x=
y = tan3 t
dy
= 3(tan t)2 sec2 t = 3 tan2 t sec2 t
dt
3sin 2 t × 1
dy
2
2
dy
cos2 t = 3sin t = sin t
= dt = cos t
3sin t
3sin t
dx dx
dt
cos4 t
1 
ii y − tan3 t = sin t x −


cos3 t 
y = x sin t −
sin t
sin 3 t
3 +
cos t cos3 t
 sin 3 t − sin t 
= x sin t + 


cos3 t
 sin t sin 2 t − 1 
= x sin t + 
×
 cos t
cos2 t 
 sin t − cos2 t 
= x sin t + 
×
 cost
cos2 t 
y = x sin t − tan t
145
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Summary REVIEW
14 i 2 tan 2x + 5 tan2 x = 0
2−
2tan x 
+ 5 x = 0
2 
 1 − tan 2 x 
6 – x = – 6x – 18
5x = – 24
24
x=−
5
4t
2
2 + 5t = 0
1 − t
4t + 5t2 (1 − t2) = 0
Therefore f(x)  g(x) ⇒
4t + 5t2 − 5t4 = 0
or x −
t(4 + 5t − 5t3) = 0
t = 0 or 4 + 5t − 5t3 = 0
t = 3 t + 0.8
ii Let f(t ) = 3 t + 0.8 − t
f(1.2) = 0.059 92…
f(1.3) = −0.0194…
Change of sign ⇒ 1.2 < t < 1.3
iii t0 = 1.3 (or can use 1.2)
( )
( )
x 
3
ln ( x − ) ln  
2
 8
t1 = 1.280 58
17 ln x − 3 2 ln x − 3 ln 2
2
3
ln x −
ln x 2 − ln 23
2
t2 = 1.276 62
t3 = 1.275 81
t4 = 1.275 64
2
t5 = 1.275 61
t6 = 1.275 60
3 x2
2
8
8x – 12  x2
0  x2 – 8x + 12
0  (x – 6)(x – 2)
x  2 or x  6
x−
Therefore t = 1.276
d We know that tan x = 0 or tan x = 1.276
So x = 0, −π, π or x = 0.906, −2.24
y
15 i
18 i Let f (x) = x4 + 2x – 9
20
y = g(x)
f (1.5) = – 0.9375 and f (1.6) = 0.7536
15
Change of sign ⇒ solution is in the interval
[1.5, 1.6]
10
5
–10
–5
ii When f(x) = g(x):
x
2 − = 2x + 6
3
0
(–3, 0)
ii At P, x4 + 2x – 9 = 0 ⇒ x4 = 9 – 2x
9
⇒ x3 = 9 − 2 ⇒ x = 3 x − 2
x
iii x0 = 1.6
(0, 6)
y = f (x)
(0, 2)
–15
12
7
5
(6, 0) 10
x
x1 = 1.5362
x2 = 1.5685
x3 = 1.5520
x4 = 1.5604
6 – x = 6x + 18
7x = – 12
x5 = 1.5561
12
7
x7 = 1.5572
x=−
or
24
5
12
−1
5x
(5x)2  12 – 5x
(5x)2 + 5x – 12  0
(5x + 4)(5x – 3)  0
Therefore: 0  5x  3
5x > 0 for all values of x, so the only condition is
5x  3
log 3
⇒ log 5x  log 3 ⇒ x log 5  log 3 ⇒ x log 5
⇒ x  0.683 (3 sf )
t3 = t + 0.8
x−
16 5x 0.8 + t − t3 = 0
x
= −2x − 6
3
x6 = 1.5583
x8 = 1.5577
Therefore, at P, x = 1.56 (to 2dp)
146
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WORKED Solutions
19 i 5sin 2θ + 2cos 2θ ≡ Rsin (2θ + α)
5 − 4 = 1 + 5μ
5sin 2θ + 2cos 2θ ≡ R(sin 2θ cos α + cos 2θ sin α)
From ①:
2μ = 4 − 14
5sin 2θ + 2cos 2θ ≡ θ (Rcos α)sin 2θ
+ (Rsin α) cos 2θ
In ②:
Therefore:
1 − 3 = 1 + (4 − 14)
Rcos α = 5 (1)
7 = 14
Rsin α = 2 (2)
=2
(2) ÷ (1) ⇒
tan θ = 0.4 ⇒
Square and add ⇒
α = 21.80°
R2 = 52 + 22
⇒
In ①:
R = 29
4=7+μ
μ = −3
29 sin ( 2θ + 21.80 ) = 4
ii
Check in ③:
4
29
2θ + 21.80 = 47.97°, 132.03°, 407.96°, 492.03°
sin ( 2θ + 21.80 ) =
5 − 8 ≠ 1 + 5(−3)
θ = 13.1°, 55.1°, 193.1°, 235.1°
iii
③
1
(10sin 2θ + 4cos 2θ )2
≡
≡
(2
1
)
29 sin (2θ + 21.80)
2
Therefore the lines do not intersect and are
skew.
1 
ii A vector in the direction of the x-axis is  0
 
 0
Using the scalar product:
1
116sin (2θ + 21.80)
1 1
 0  2 = 12 + 0 2 + 0 2 12 + 22 + 52 cosθ
  
 0  5
2
To minimise the expression, we need to
1
maximise the denominator ⇒
116
20 i 3cosθ + sinθ ≡ Rcos(θ – α)
cosθ =
3cosθ + sinθ ≡ R(cosθ cosα + sinθ sinα)
3cosθ + sinθ ≡ (Rcosα)cosθ + (Rsinα)sinθ
Therefore:
Rsinα = 1 (2)
1 ⇒ α = 18.43°
3
Square and add ⇒ R2 = 32 + 12 ⇒ R = 10
4 + 12x + x2 ≡ A(1 + 2x)2 + B(3 − x)(1 + 2x)
+ C(3 − x)
tan α =
10cos ( 2x − 18.43) = 2
ii
4 + 12x + x 2
A
B
C
≡
+
+
(3 − x)(1 + 2x)2 3 − x 1 + 2x (1 + 2x)2
4 + 12x + x 2
A(1 + 2x)2 + B(3 − x)(1 + 2x) + C(3 − x)
≡
(3 − x)(1 + 2x)2
(3 − x)(1 + 2x)2
Rcosα = 3 (1)
(2) ÷ (1) ⇒
22 i
1
= 79.5°
30
cos ( 2x − 18.43) =
2
10
2x – 18.43 = 50.76°, 309.23°, 410.76°, 669.23°
θ = 34.6°, 163.8°, 214.6°, 343.8°
A Level Questions:
Pure Mathematics 3
When x = 3
4 + 36 + 9 = 49A
49 = 49A
A=1
When x = −
1
2
1 7C
=
4
2
16 − 24 + 1 = 14C
4−6+
14C = −7
 0
 2
 7
1
21 i L1: r =  1  + λ  −3 and L2: r =  1  + µ  2
 
 
 
 
 5
 −4 
1
 5
At ‘intersection’:
2 = 7 + μ
①
1 − 3 = 1 + 2μ
②
1
C=−
2
When x = 0
4 = A + 3B + 3C
3
2
8 = 2 + 6B − 3
= 1 + 3B −
6B = 9
147
57743_P143_153.indd 147
6/28/18 7:24 PM
SUMMARY REVIEW
When θ = π
4
π 1
ln x + 2 = − sin π + ln 2
8 8
ln x + 2 = π + ln 2
8
3
2
Therefore
B=
3
−1
4 + 12x + x 2
1
2 +
2
≡
+
(3 − x)(1 + 2x)2 3 − x 1 + 2x (1 + 2x)2
( )
ii (3 − x)−1 = 3−1 1 − x
3
π + ln 2
( )

(−1)(−2) − x
3
1
x
= 1 + (−1) − +
3
3
2!


( )
=
π + ln 2
x + 2 = e8
−1
2


+ 


x = e8
−2
= 0.962
22 + 4i = 22 + 4i
24 i w =
(2 − i)2 4 − 4i + i 2
= 22 + 4i
3 − 4i

1
x x2
1+ +
+ 

3
3 9

= 22 + 4i × 3 + 4i
3 − 4i 3 + 4i
2
= 66 + 88i + 12i2 + 16i
9 − 16i
2
= 1 + x + x +
3 9 27

3
3
(−1)(−2)(2x)2
(1 + 2x)−1 = 1 + (−1)(2x) +
+ 
2
2
2!

3
= 1 − 2x + 4x 2 + 
2
3
= − 3x + 6x 2 + 2

1
1
(−2)(−3)(2x)2
− (1 + 2x)−2 = − 1 + (−2)(2x) +
+ 
2
2
2!

= 50 + 100i
25
= 2 + 4i
Alternative method:
Let w = x + iy
(x + iy)(2 – i)2 = 22 + 4i
(x + iy)(4 – 4i + i2) = 22 + 4i
(x + iy)(3 – 4i) = 22 + 4i
1
= − 1 − 4x + 12x 2 + 
(3x – 4xi + 3yi – 4yi2) = 22 + 4i
2
(3x + 4y) + ( – 4x + 3y)i = 22 + 4i
1
= − + 2x − 6x 2 + Equating real and imaginary parts:
2
3x + 4y = 22 ⇒ 12x + 16y = 88
(1)
4 + 12x + x 2
1 3 1
1
1
=
+ −
+ −3+2 x +
+ 6 − 6 x 2 +–
4x + 3y = 4 ⇒ – 12 x + 9y = 12 (2)
3 2 2
9
27
(3 − x)(1 + 2x)2
(1) + (2)
1
12x + x 2
1 3 1
1
2
25y = 100 ⇒ y = 4
2 = 3 + 2 − 2 + 9 − 3 + 2 x + 27 + 6 − 6 x + )(1 + 2x)
Substituting into 3x + 4y = 22
4 8
1 2
3x + 16 = 22 ⇒ x = 2
= − x+
x +
3 9
27
Therefore w = 2 + 4i
π  arg((2 + p) + 4i)  3
23 dx = (x + 2)sin 2 2θ
ii
dθ
4
4
1
2
Im
∫ x + 2 dx = ∫ sin 2θ dθ
1
2
= ∫ (1 − cos 4θ )dθ
4i
2
1
1
ln x + 2 = θ − sin 4θ + C
2
4
(
(
) (
)(
(
)(
)
) (
)
)
θ − 1 sin 4θ + C
2 8
When x = 0,  = 0
Therefore C = ln 2
So:
θ 1
ln x + 2 = − sin 4θ + ln 2
2 8
ln x + 2 =
π
4
π
4
–4
4
Re
From the Argand diagram:
−4  2 + p  4
−6  p  2
148
57743_P143_153.indd 148
6/28/18 2:38 PM
WORKED SOLUTIONS
iii S: w = 2 + 4i
I = ∫ e ysin y d y
T: w* = 2 − 4i
Integration by parts:
Im
u = ey ⇒
dv
= sin y
dy
S
4i
2
Re
a
I =
4a = 20
a=5
Therefore we have a circle with centre (5, 0)
and radius 5, so z − 5 = 5
4
1
2I = – ey cos y + ey sin y
e y(sin y − cos y)
2
Therefore:
(a − 2)2 + 42 = a2
∫
Integration by parts:
du
u = ey ⇒
= ey
dy
dv
= cos y ⇒ v = sin y
dy
I = – ey cos y + ey sin y – I
Pythagoras’ theorem on the triangle gives:
25 I =
4
−1
ln x
dx = ∫ x 2 ln x dx
x
1
e y(sin y − cos y)
+ c = tan −1 x
2
 e y(sin y − cos y)

x = tan 
+ c
2


28 i
Integrate by parts:
−1
dv
=x 2
u = ln x and
dx
4
4
4
4
1
−1
 1

=  2x 2 ln x  − ∫ 2x 2 dx

1 1
ii
4
1
 1

=  2x 2 ln x − 4x 2 

1
= [4 ln 4 − 8] − [0 − 4]
I = 4 ln 4 − 4
) ddxy = 1
1
∫ e sin y d y = ⌠⌡ x 2 + 1 dx
y
⌠ 1 dx = tan −1 x

⌡ x2 + 1
3
2
2
4
2
2
2i
(( x − 1) − i 2 )((x − 1) + i 2 )
= (x – 1)2 – 2i2
= (x – 1)2 + 2
= x2 – 2x + 3
)
()
2
2
3
2
x
ii ⌠
dx = ⌠
dx = tan −1
+c
 2
3
3
3
⌡x +9
⌡ x 2 + 32
)(
4
( x − (1 + i 2 ))( x − (1 − i 2 ))
=
2x
26 i ⌠
dx = ln x 2 + 9 + c
 2
⌡x +9
(
4
2
 1

2x 2
I =  2x 2 ln x  − ∫
dx

1 1 x
(
(1 + i 2 ) = 1 + 4 × 1 ( i 2 ) + 6 × 1 ( i 2 ) +
4 × 1( i 2 ) + ( i 2 ) = −7 − 4 2i
(1 + i 2 ) = 1 + 2 × 1( i 2 ) + ( i 2 ) = −1 + 2
∴ p ( x ) = ( −7 − 4 2i ) + ( −1 + 2 2i )
+ 2 (1 + 2i ) + 6 = 0
A second root of p(x) is (1 − 2i )
3
1
du 1
= and v = 2x 2
dx x
27 e ysin y x 2 + 1
v = – cos y
I = −e ycos y + e ysin y − ∫e ysin y dy
T
–4i
⇒
I = −e ycos y + ∫e ycos y dy
a
0
du
= ey
dy
By polynomial long division:
x4 + x2 + 2x + 6 = (x2 – 2x + 3)(x2 + 2x + 2)
x2 + 2x + 2 = 0
−2 ± 22 − 4(1)(2)
2
−2 ± − 4
x=
2
−2 ± 2i
x=
2
x = −1 ± i
x=
149
57743_P143_153.indd 149
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Summary REVIEW
 7 − − 8   15 
 −1 − − 1 =  0 
29 i

 

 −4 − 8   −12
∴ – 2a = – 4a3
5
0
 
 −4
⇒
4a3 – 2a = 0
2a(2a2 – 1) = 0
 −8
5
⇒ l1:  −1 + λ  0 
 
 
8
 −4
 5 − −1   6 
 −6 − 4 =  −10

 

 −7 − 7   −14
⇒
5
3


⇒ l2: −6 + µ  −5
 
 
 −7
 −7
(1)
– 1 = – 6 – 5μ
(2)
8 – 4λ = – 7 – 7μ
From (2) ⇒
In (1) ⇒
(3)
5μ = – 5 ⇒
μ=–1
– 8 + 5λ = 2 ⇒
5
iii  0 
 
 −4
a>0
⇒
a=
5λ = 10 ⇒ λ = 2
ii  1 + 1 x 

2 

1 
x
 1 +
2 
−2
15 + 0 + 28 = 41 83 cos θ
30 i
1
(1 − 4x )− 2
( )
1
( −4x ) +
2
3π
4
2
2
2!
−1
1 + 2x
1 + 2x
1 + 2x
ii
=
× (1 − 4x ) 2 =
2
2
4 − 16x
−1
1 + 2x
1
+
2
x
=
× (1 − 4x ) 2 =
1 + 2x + 6x 2 + …
2
2
The coefficient of x2 is obtained from:
1
2x
6x 2 +
( 2x ) = 5x 2
2
2
The coefficient of x2 is 5.
)
( )
31 i (1 + ax)−2 = 1 + ( −2)( ax ) +
3
(−2)(−3)(ax)2
2!
+ (−2)(−3)(−4)(ax) + …
3!
The coefficient of x is – 2a and the
coefficient of x3 is – 4a3
3 2
x +…
2
(1)
(2)
(− 12 )(− 32 )(−4x) + …
–i
= 1 + 2x + 6x 2 + …
(
= 1 − 2x +
2i
θ = 42.5° or 0.742 radians (3sf)
=1+ −
 1 
x
= 1 + ( −2) 
 2 
(−2)(−3) 1 x 
 2 
+…
+
2!
Im
ii
43
= 0.737…
41 83
1
1
2
= −10 − 10i = −2 − 2i
1+ 4
Substituting in (1)
– 2 – 2i + 2v = 2i
2v = 2 + 4i
v = 1 + 2i
when λ = 2 A(2, –1, 0)
(1 − 4x )− 2
a=±
2
u = −6 + 2i × 1 + 2i = −6 −2 10i + 24i
1 − 2i 1 + 2i
1 − (2i)
3
⋅  −5 = 52 + 0 2 + 4 2 32 + 52 + 7 2 cos θ .
 
 −7
cos θ =
⇒
1
2
32 i u + 2v = 2i
2ui + 2v = 6
(1) – (2)
u – 2ui = 2i – 6
u(1 – 2i) = 2i – 6
u = −6 + 2i
1 − 2i
8 – 4(2) = – 7 – 7( – 1)
⇒
2a 2 = 1
−2
Check in (3)
0=0
or
2
3
 −5
 
 −7
ii – 8 + 5λ = 5 + 3μ
a=0
Re
–2i
z − w MIN represents the shortest distance
between the circle and the line. This can be
found by considering the line perpendicular
to the half-line (locus of points for w) and
passing through the centre of the circle. The
shortest distance from the line to the centre
of the circle, x, occurs when:
3
x ⇒
x=
3
2
The radius of the circle is 1.
3
−1
⇒ z − w MIN =
2
cos 45° =
150
57743_P143_153.indd 150
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WORKED Solutions
4
33 i [x 2(2ln x – 1)] 2 = 56 ln 2 – 12
Therefore:
π
 24
3
1
ii  sin 4x + sin12x
4
0
 16
x3
7x + 6
≡ ( x − 3) +
(x + 1)(x + 2)
(x + 1)(x + 2)
11
96
7x + 6
A
B
ii (x + 1)(x + 2) ≡ x + 1 + x + 2
Extension Questions
7x + 6 ≡ A(x + 2) + B(x + 1)
=
1
i
y
When x = – 2
⇒
–8 = –B
When x = – 1
⇒
–1 = A
⇒
⇒
B=8
A = –1
x3
1
8
Therefore: (x + 1)(x + 2) ≡ ( x − 3) − x + 1 + x + 2
3
Let y = sin x + cos x
dy
= cos x − sin x
dx
⇒
2
⇒ d y2 = − sin x − cos x
dx
We want minimum and maximum values of y so
solve
(0, 5)
π 5π
x = , ,…
4 4
0
( 5, 0)
When x =
x
⇒
ii The graphs intersect when 3x − 1 = x 2 − 5
(3x –
9x2 –
1)2
=
(x2 –
6x + 1 =
5)2
x4 –
10x2 +
25
By inspection: 0 = (x +
x = 4 is a root ⇔
1)(x3 –
x2 –
18x + 24)
So the roots of the quartic equation (and
x-coordinates of the points of intersection)
are:
−3 − 33 x = −3 + 33
x = – 1, x = 4, : x =
,:
2
2
f(x) < g(x) occurs when
2
e−
4
x 3 + 3x 2 + 2x
− 3x 2 − 2x
− 3x 2 − 9x − 6
7x + 6
tan x = 1
y MAX = 2
2
5π d y = 2 + 2 = 2
2
2
4 , dx 2
minimum ⇒ y MIN = − 2
2
<a<e
2
i ln (eπ) + ln (e2π) + ln (e3π) + … + ln (e10π)
= π ln e + 2π ln e + 3π ln e + … + 10π ln e
= π (1 + 2 + 3 + … + 10)
=π
−3 − 33
< x < −1
2
 −3 + 33 
or 
 < x < 4
2

i (x + 1)(x + 2) = x2 + 3x + 2
x−3
x 2 + 3x + 2 x 3 + 0x 2 + 0x + 0
⇒
The cyclic nature of y = sin x + cos x is such
that other values of x at minimum and
maximum points will yield the same values
for yMIN and yMAX.
(x – 4) is a factor
By inspection: 0 = (x + 1)(x – 4)(x2 + 3x – 6)
Solving the quadratic: x = −3 ± 33
2
⇒
maximum
⇒
(x + 1) is a factor
sin x = cos x
2
π d y =− 2− 2 =− 2
2
2
4 , dx 2
When x =
0 = x4 – 19x2 + 6x + 24
x = – 1 is a root ⇔
⇒
0 = cos x – sin x
(0, 1)
(– 5, 0)
dy
=0
dx
(10 ×2 11 )
= 55π
ii Using the same logic as part (a):
Sn = π(1 + 2 + 3 + … + n)
Sn =
5
n(n + 1)
π
2
y = tan–1 x
y=
⇒
dy
1
=
dx 1 + x 2
1 3
x − 4x + ln5
3
⇒
dy
= x2 − 4
dx
1
> x2 − 4
1 + x2
151
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Summary REVIEW
1 > (x2 – 4)(1 + x2) This is valid since 1 + x2 > 0
1 > x2 + x4 – 4 – 4x2
x4 – 3x2 – 5 < 0
Quadratic formula:
x2 =
π 5π
, ,…
4 4
⇒ θ = π , 5π , … ⇒ θ = (4n + 1)π
16 16
16
Sometimes true.
tan 4θ = 1
3 ± 9 − 4(1)(−5) 3 ± 29
=
2
2
7
x=±
3 − 29
2
⇒
No real solutions
x=±
3 + 29
2
⇒
Two real solutions
⇒
4θ =
n ∈Z
icos ((A + B) + C)
≡ cos (A + B)cos C – sin (A + B) sin C
cos ((A + B) + C) ≡ (cos A cos B – sin A sin B) cos C
– (sin A cos B + cos A sin B) sin C
cos ((A + B) + C) ≡ cos A cos B cos C – sin A sin B
cos C – sin A cos B sin C – cos A sin B sin C
y
ii Let A = π , B = π , C = π
2
3
4
(1312π ) = cos( π2 )cos( π3 )cos( π4 )
π
π
π
− sin ( )sin ( ) cos ( )
2
3
4
π
π
π
− sin ( ) cos ( ) sin ( )
2
3
4
π
π
π
− cos ( ) sin ( ) sin ( )
2
3
4

13π
3
2 
1
− 1× ×
cos (
= 0 − 1 ×
×
12 )
2
2  
2

( 6 + 2)
13π
6
2
cos (
=−
−
=−
4
12 )
4
4
cos
x
3 + 29 , 0
–
2
3 + 29 , 0
2
(0, –5)
−
6
3 + 29
<x<
2
3 + 29
2
i tan2 θ ≡ sec2 θ – 1 and cosec2 θ = 1 + cot2 θ
8
∴ cosec2 θ + tan2 θ ≡ 1 + cot2 θ + sec2 θ – 1
y=
Always true.
)
⇒
1
1
y=⌠
×
dx
 2
⌡ x + 1 tan −1 x
1
∫ u du = ln u + c = ln tan
−1
x +c
The conditions for x are such that the modulus
function is not required.
cosθ
2sin θ cosθ ≡ 2cos2 θ ≡ 1 + cos2θ
sin θ
When x = 1, y = ln π
2
π
ln = ln tan −1 1 + c
2
So the statement is only true when
1 + cos 2θ = 1 – cos 2θ
(
⇒ cos 2θ = – cos 2θ
2cos 2θ = 0
cos 2θ = 0
π 3π
2θ = , , …
2 2
π 3π
θ = , , … ⇒ θ = (2n − 1)π
4 4
4
(
Integrating by substitution with u = tan–1 x
cosec2 θ + tan2 θ ≡ sec2 θ + cot2 θ
ii cot θ sin 2θ ≡
dy
1
= 2
dx
x + 1 tan −1 x
2
−0
2 
)
π
π
= ln + c
2
4
π
π
c = ln − ln = ln 2
2
4
ln
n ∈Z
y = ln (tan–1 x) + ln 2
y = ln (2tan–1 x)
Sometimes true.
iii sin2 2θ + cos 4θ ≡ sin2 2θ + (cos2 2θ – sin2 2θ)
≡ cos2 2θ
Always true.
iv Comparing with part (c), this statement is
only true when sin 4θ = cos 4θ
9
z = 3 + 2i is a root ⇔ z = 3 – 2i is a root.
(z – (3 + 2i))(z – (3 – 2i)) is a factor
((z – 3) – 2i)((z – 3) + 2i))
(z – 3)2 – 4i2
z2 – 6z + 13 is a factor.
By polynomial long division you get:
152
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WORKED Solutions
z5 – 4z4 – 6z3 + 72z2 – 115z + 52 = (z2 – 6z + 13)
(z3 + 2z2 – 7z + 4)
z = 1 is a root ⇔ (z – 1) is a factor.
By inspection:
z5 – 4z4 – 6z3 + 72z2 – 115z + 52 = (z2 – 6z + 13)(z – 1)
(z2 + 3z – 4)
z5 – 4z4 – 6z3 + 72z2 – 115z + 52 = (z2 – 6z + 13)(z – 1)
(z + 4)(z – 1)
So the roots are: z = – 4, 1, 3 + 2i, 3 – 2i (note that
z = 1 is a repeated root).
10 Im
1
11 AB =  −5
 
 2
 4
AC =  0 
 
 −1
AB = 30
AC = 17
1  4
 −5 .  0  = 30 17cosθ
   
 2   −1
2 = 30 17cos θ
cos θ =
(3, 7)
B
(3, 5)
cos2 θ =
(5, 5)
4
2
=
510 255
1 − cos2 θ =
sin θ =
A
2
30 17
253
255
253
255
Area of ABC =
1
253 1
30 17
=
506
2
255 2
12 i f(1) = –1.2817…..
Re
The yellow region on the sketch is the locus
of possible points for z. The minimum and
maximum values of |z| are on the line passing
the origin and the centre of the circle, at A and B
respectively.
5
The equation of the line is y = x and the
3
equation of the circle is (x – 3)2 + (y – 5)2 = 4
2
At A and B: (x − 3)2 + 5 x − 5 = 4
3
(
)
25 2 50
x −
x + 25 = 4
9
3
9x2 – 54x + 81 + 25x2 – 150x + 225 = 36
x 2 − 6x + 9 +
34x2 – 204x + 270 = 0
x=
and
f(2) = 18.0855….
Change of sign ⇒ there is a solution in the
interval 1 < x < 2
ii e2x – 1 = 6 – 2x
2x – 1 = ln (6 – 2x)
ln ( 6 − 2x ) + 1
2
ln ( 6 − 2xn ) + 1
xn +1 =
2
Use x0 = 1 since this is near to the solution.
x=
x1 = 1.1931…, x2 = 1.1423…, x3 = 1.1562…,
x4 = 1.1524…, x5 = 1.1534…, x6 = 1.1532…,
x7 = 1.1532…,
Therefore: x = 1.15 correct to 3 significant
figures.
204 ± 204 2 − 4(34)(270)
68
204 ± 4896
68
At A: x = 1.971
y = 3.285
At B:
y = 6.715
x=
z =
x = 4.029
x2 + y2
z MIN = 3.831 z MAX = 7.831
153
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