Cambridge International
AS & A Level Mathematics
Pure Mathematics 1
STUDENT’S BOOK: Worked solutions
Helen Ball, Chris Pearce
Series Editor: Dr Adam Boddison
Pure Mathematics 1 International Students Book Title page.indd 1
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1 Mathematics 1.indd 1
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PM
1
WORKED SOLUTIONS
Worked solutions
1 Quadratics
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
2x2 – 5x – 3 = 0
(2x + 1)(x – 3) = 0
x=−
2
y
15
10
5
1
or x = 3
2
0
–25 –20 –15 –10 –5
–5
3x2 + x – 7 = 0
a = 3, b = 1, c = −7
x=
−1 ± 12 − (4)(3)(–7)
6
3
−1 ± 85
6
1
2
4
2
3
4
5
6
7
Exercise 1.1A
1
a x2 + 4x = (x + 2)2 − 4
b 2x2 – 8x = 2(x2 – 4x)
= 2[(x − 2)2 − 4]
= 2(x – 2)2 – 8
c x2 + 8x + 7 = (x + 4)2 – 16 + 7
= (x + 4)2 − 9
2
2
2
2
x < 4
1
2
2
3x – 5 < 7
3x < 12
0
( 32 ) − 94 − 7
3
37
= (x + ) −
2
4
3x 9
c 2x + 3x + 9 = 2 ( x +
+
2 2)

3
9 9
= 2 ( x + ) −
+ 
4
16
2


3
63 
= 2 ( x + ) +

4
16


5
25
4 x – 5x + 7 = ( x − ) −
+7
2
4
5
3
y = (x − ) +
2
4
2
5x – y = 13
2x + y = 1
Add 1 and 2 .
7x = 14
–1
a x2 – 8x – 5 = (x − 4)2 – 16 − 5
b x2 + 3x – 7 = x +
x = 2, y = −3
4
x
= (x − 4)2 − 21
x = 1.37 or − 1.70
3
10 15 20 25
–10
−1 ± 1 + 84
x=
6
x=
5
x2 – 10x + 11 = (x − 5)2 – 25 + 11
= (x − 5)2 − 14
2
2
2
At the turning point y has a minimum value
5
5
3
i.e. when x = . When x = , y = , the
2
2
4
coordinates of the turning point.
y
25
20
15
10
5
0
–25 –20 –15 –10 –5
–5
5
10 15 20 25
x
1
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1 Quadratics
5
6
(
2
13
13
2 ( x + 2) −  = 2 x 2 + 4x + 4 −

2 
2
)
9
5
3
4x 2 + 5x − 3 = 4  x 2 + x − 

4
4 
= 2x2 + 8x + 8 − 13
= 2x2 + 8x − 5
1
x 3
−
a 5x 2 + x − 3 = 5 x 2 +
2
10 5
(
)
( )

1
241 
= 5 ( x +
−
20 )
400 

1
241
= 5( x +
−
20 )
80
18
b 5x – 18 = 5 ( x − 5 ) but can go no further.
7x 5
c 5 – 7x – 3x = −3 ( x +
−
3 3)

7
49 5 
= −3 ( x + ) −
− 
6
36
3


7
109 
= −3  ( x + ) −
6
36 

7
109
= −3( x + ) +
6
12
3x 11
7 2x – 3x + 11 = 2 ( x −
+ )
2
2

3
9 11 
= 2 ( x − ) −
+ 
4
16
2


3
79 
= 2 ( x − ) +

4
16



3
79 
2 ( x − ) +
>0
4
16


2
2
2
2
2
2
2
2
(
)
(
=4 x+
2
2
) ()
2

5
73 
= 4 x +
−

8
64


2

1
1
3
= 5 x +
−
−
20
400 5 

2
(
2
2

5
5
3
= 4 x +
−
− 
8
8
4

2
2
2
Consequently b2 – 4ac < 0 because the curve of
this equation will not intersect the x-axis and
so there are no real roots.
y
25
20
5
8
) − 1673
2
When 4x2 + 5x – 3 = 0
(
) − 1673 = 0
5
73
4(x + ) =
8
16
( x + 85 ) = 6473
4 x+
5
8
2
2
2
x+
5
73
=±
8
64
x=
−5
±
8
x=
−5 ± 73
8
73
64
2

b 
b2
10 ax 2 + bx + c =  ax +
−
+c

4a

2 a
ax2 + bx + c = 0
2

b 
b2
 ax +
 − 4a + c = 0
2 a
2

b 
b2
b 2 − 4ac
 ax +
 = 4a − c =
4a
2 a
ax +
b
b 2 − 4ac
=±
4a
2 a
ax +
b
± b 2 − 4ac
=
2 a
2 a
ax = −
15
10
5
–20 –15 –10 –5
8
0
5
10 15 20 25
b 2 − 4ac
2 a
ax =
−b ± b 2 − 4ac
2 a
x=
−b ± b 2 − 4ac
a ×2 a
x=
−b ± b 2 − 4ac
2a
x
(1 – 2x) [(x + 4)2 − 1] = (1 – 2x)(x2 + 8x + 16 − 1)
= (1 – 2x) (x2 + 8x + 15)
= x2 + 8x + 15 – 2x3 – 16x2 – 30x
= 15 – 22x – 15x2 – 2x3
b
±
2 a
2
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WORKED SOLUTIONS
Exercise 1.2A
1
b2 – 4ac = 0 so two equal real roots.
c
a x2 + 2x + 13 = 0
3x2 + 12 – 5x = 0
b2 – 4ac = (−5)2 – (4)(3)(12)
b2 – 4ac = (2)2 – (4)(1)(13)
= −119
= −48
b2 – 4ac < 0 so no real roots.
b2 – 4ac < 0 so no real roots.
b 2x2 + 3x + 1 = 0
6
2x2 + 5x – c = 0
a No real roots:
b2 – 4ac = (3)2 – (4)(2)(1)
b2 – 4ac < 0
=1
(5)2 – (4)(2)(−c) < 0
b2 – 4ac > 0 so two distinct real roots.
25 + 8c < 0
c x2 + 4x + 4 = 0
b2 – 4ac = (4)2 – (4)(1)(4)
c<−
=0
b One repeated real root:
b2 – 4ac = 0 so two equal real roots.
2
b2 – 4ac = 0
2x2 – 5x + 3 = 0
b2 – 4ac = (−5)2 – (4)(2)(3)
= 25 − 24
=1
(5)2 –(4)(2)(−c) = 0
25 + 8c = 0
c=−
b2 – 4ac > 0 so has two distinct real roots.
3
b2 – 4ac > 0
b2 – 4ac = (−3)2 – (4)(1)(−5)
(5)2 – (4)(2)(−c) > 0
= 29
25 + 8c > 0
b2 – 4ac > 0 so two distinct real roots.
4
c>−
7
x=
13 − 3x
3
x2 =
13 − 3x
3
25
8
b2 – 4ac < 0 so no real roots.
c x2 – 12x + 36 = 0
b2 – 4ac = (−12)2 – (4)(1)(36)
=0
2
b – 4ac = 0 so two equal real roots.
3x2 = 13 – 3x
x2 + (2k + 2)x + (5k − 1) = 0
As b2 – 4ac > 0 there are 2 distinct solutions.
b2 – 4ac = 0 for one repeated real roots.
(2k + 2)2 – (4)(1)(5k – 1) = 0
Expanding and simplifying.
k2 – 3k + 2 = 0 or – k2 + 3k – 2 = 0
Both quadratic equations have the solutions k = 1
or k = 2.
When k = 1, x2 + 4x + 4 = 0 and x = −2;
when k = 2, x2 + 6x + 9 = 0 and x = −3
5
25
8
c Two distinct real roots:
a x2 – 3x – 5 = 0
b 3x2 – 2x + 7 = 0
b2 – 4ac = (−2)2 – (4)(3)(7)
= −80
25
8
a 4 + 2x – 3x2 = 0
b2 – 4ac = (2)2 – (4)(–3)(4)
= 52
b2 – 4ac > 0 so two distinct real roots.
b 49 + x2 – 14x = 0
b2 – 4ac = (−14)2 – (4)(1)(49)
=0
3x2 + 3x – 13 = 0
b2 – 4ac = 9 + 4 × 3 × 13 = 165 > 0
8
x2 – 3x + a2 = 0
No real roots when b2 – 4ac < 0
b2 – 4ac = 9 – 4a2
9 – 4a2 < 0
9 < 4a2
a2 >
a<
9
9
4
−3
3
or a >
2
2
y = 3 – x meets y = x2 – 5x + 7 when 3 – x = x2 – 5x + 7
ie 0 = x2 – 4x + 4
b2 – 4ac = 16 – 16 = 0
One repeated root, so the line touches, but
does not cross the curve.
3
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1 Quadratics
iii (x − 3)2 – 9 = 0
10 3x – xy + 1 = 0
3x + 1
x
substitute into 3y – xy + 1 = 0
y=
x = 0 or 6
2
( 3xx+ 1 ) − x ( 3xx+ 1 ) + 1 = 0
3x + 1
3(
) − 3x = 0
x
3
Quadratic formula:
›› can be used to solve any equation with real roots
including ones that don’t factorise
3x2 – 9x – 3 = 0
b2 – 4ac = 81 + 4 × 3 × 3 = 117 > 0
›› cumbersome and consequently easy to make a
As b2 – 4ac > 0 there are two distinct real roots
∴ the curves intersect in two places.
mistake
›› can solve: x2 − 5x – 7 = 0
›› cannot solve: x2 − 5x + 7 = 0.
Exercise 1.3A
1
x2 – 9 = 0
a i
Completing the square:
›› can be used to solve any equation with real roots
including ones that don’t factorise
(x + 3)(x − 3) = 0
x = 3 or −3
ii x =
›› can be cumbersome manipulations if b is odd and
0 ± 0 − ( 4 )(1)( −9)
2
a > 1, and consequently easy to make a mistake
›› can solve: x2 − 5x – 7 = 0
›› cannot solve: x2 − 5x + 7 = 0.
x = 3 or −3
iii x2 = 9
x = 3 or −3
b i
3
x2 + 3x + 2 = 0
(x + 1)(x + 2) = 0
x = −1 or −2
−3 ± 9 − ( 4 )(1)( 2)
ii x =
2
x = −1 or −2
( 32 ) − 14 = 0
iii x +
2
x = −1 or −2
c i x2 – 5x – 7 = 0
Cannot be factorised.
ii x =
5 ± 25 − ( 4 )(1)( −7 )
2
5 ± 53
x=
2
( 52 ) − 534 = 0
iii x −
2
x=
5 ± 53
2
d i x(x − 6) = 0
x = 0 or 6
ii x =
6 ± 36 − ( 4 )(1)( 0 )
2
x = 0 or 6
4
Factorisation:
›› can only be used on equations that factorise
›› sometimes spotting factors can be difficult
›› can solve: x2 + 3x + 2 = 0
›› cannot solve: x2 − 5x – 7 = 0.
a (x − 4)2 = 13
x = 4 ± 13
Completing the square. The equation was
already in form of a completed square.
b 2x2 – 5x – 11 = 0
x=
5 ± 25 − ( 4 )( 2)( −11)
4
x=
5 ± 113
4
Does not factorise so used the quadratic
formula. Alternatively, could have completed
the square but chose not to as a > 1.
c 3x2 = 5 – 14x
3x2 + 14x – 5 = 0
(3x − 1)(x + 5) = 0
1
x = or −5
3
Equation factorises so easy to do this.
Alternatively, could have completed the
square but chose not to as a > 1.
d 4x2 – 16x + 7 = 0
(2x − 1)(2x − 7) = 0
1
7
x = or
2
2
Equation factorises so easy to do this.
Alternatively, could have completed the
square but chose not to as a > 1.
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1
WORKED SOLUTIONS
4
x2 + bx + c = 0
8
Subtract any extra values that completing the
b2
square has produced, in this case− .
4
( )
x+
b
2
2
−
b
2
x+
x+
2
=
b 2 − 4c
4
b2 – 4ac = 16 – (4)(3)(−11) = 148
148 not a square number so does not
factorise.
9
6
7
−4 ± 148
6
x=
−2 ± 37
3
(x − 3)(x − 5) = 8
x2 – 8x + 15 – 8 = 0
x2 – 8x + 7 = 0
(x − 7)(x − 1) = 0
x = 1 or 7
ax2 + bx + c = 0
bx c
a x2 +
+
=0
a a
(
(
(
(
(
)
b
x+
2a
)
)
)
)
2
x+
b
2a
2
x+
b
2a
2
x+
b
2a
x+
2
b2
c
− 2+ =0
a
4a
=
b2
c
−
4a 2 a
=
b2
4ac
−
4a 2 4a 2
=
2
b − 4ac
4a 2
b
± b 2 − 4ac
=
2a
2a
−b ± b 2 − 4ac
x=
2a
a 9x2 + 55x – 56 = 0
b2 – 4ac = 552 + 4 × 9 × 56 = 5041
b 2 − 4ac = 71 so the quadratic can be
factorised.
b (9x – 8)(x + 7) = 0
8
x = or x = – 7
9
Exercise 1.4A
1
a x2 +3x – 4 > 0
x2 + 3x – 4 = 0
(x + 4)(x − 1) = 0
b 3x2 + 4x – 11 = 0
x=
74x + 33
7
The equation can be solved by factorising a
quadratic expression.
x=
5
x2 =
(7x + 3) (x – 11) = 0
b ± b 2 − 4c
=
2
2
−b ± b 2 − 4c
2
a
Cannot be factorised if b2 − 4ac is not a square
number.
74x + 33
7
7x2 – 74x – 33 = 0
b2
+c =0
4
Manipulate the equation to make x the subject.
( )
x=
x < −4 or x > 1
b x2 – 6x + 8  0
(x − 4)(x − 2) = 0
2x4
c x2 – 9  0
(x + 3)(x − 3)  0
x  −3 or x  3
d x2 – 6x < 0
x(x − 6) = 0
0<x<6
2
a 2x2 + 7x < −3
2x2 +7x + 3 < 0
(2x + 1)(x + 3) = 0
−3 < x < −
1
2
b −x2 – 3x + 4 > 0
(−x + 1)(x + 4) = 0
−4 < x < 1
c x2 > 4
x2 − 4 > 0
(x − 2)(x + 2) > 0
x < −2 or x > 2
d 3x2  5x
3x2 – 5x  0
x(3x − 5)  0
0x
5
3
5
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1 Quadratics
3
a 6 – 5x – x2 > 0
c 12 + 9x +x2 > 3(x + 1)
(6 + x)(1 − x) = 0
12 + 9x + x2 > 3x + 3
−6 < x < 1
9 + 6x + x2 > 0
and – x + 6 > 0
(3 + x)2 > 0
6>x
Inequality true for all values of x.
x<6
b
c
d x + 1 > 3 – 4x – 3x2
{−5, −4, −3, −2, −1, 0}
0 > 2 – 5x – 3x2
2x2 + 9x – 5  0
3x2 + 5x – 2 > 0
(2x − 1)(x + 5) = 0
1
−5  x 
2
or 2x < 5
5
x<
2
5
So x <
2
(3x – 1)(x + 2) > 0
x < –2 or x 1 [all integers excluding –2, –1
3
and 0]
6
(x + 6)(x – 1) = 0
–6 < x < 1 and x2 + 3x – 4 < 0
{all integers less than 2.5}
(x + 4)(x – 1) = 0
(2x + 1)2 – 9  0
–4 < x < 1
2x + 1 = ±3
So –4 < x < 1
−2  x  1 and 2x < 6
{−3, −2, −1, 0}
b x2 + 5x < 6 or x2 + 3x < 4
x<3
−2  x  1
So –6 < x < 1 or – 4 < x < 1
{−2, −1, 0, 1}
{−5, −4, −3, −2, −1, 0}
d 4x2 < 3x
4
c x2 – 6 < −5x and x2 – 4 > −3x
4x2 – 3x < 0
−6 < x < 1 and x < −4 or x > 1
x(4x −3) < 0
0 < x < 3 or 1 − 4x > 0
4
4
1 > 4x
1
x<
4
So x < 3
4
{all integers less than 0.75}
So −6 < x < −4
{−5}
d x2 + 5x – 6 > 0 or x2 + 3x – 4 < 0
x < −6 or x > 1 or −4 < x < 1
{all integers excluding −6, −5, −4 and 1}
7
x2 – 4x – 3  0, 1 – 2x2  0
r < 10
First consider x2 – 4x – 3  0
πr 2 > 250
x2 – 4x – 3 = 0 when x =
r > 8.9
So 8.9 < r < 10
5
a x2 + 5x – 6 < 0
a x2 + 4 < 7x − 2
x2 −7x + 6 < 0
(x − 1)(x − 6) = 0
1<x<6
{2, 3, 4, 5}
b 2x2 – 3x – 15 < 2x − 3
2x2 – 5x – 12 < 0
(2x + 3)(x − 4) = 0
3
− <x<4
2
{−1, 0, 1, 2, 3}
4 ± 16 + 12
=2± 7
2
when x = 0 x2 – 4x – 3 = –3
2− 7 x2+ 7
Now consider 1 – 2x2  0
1
1
−1
x
1  2x2, x 2 , so
2
2
2
The range of x values for which x2 – 4x – 3  0,
and 1 – 2x2  0
is 2 − 7 x 1
2
which can be written as 2 − 7 x 2
2
6
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1
WORKED SOLUTIONS
8
Exercise 1.5A
1 − 3x
5
x
1
1 − 3x
5
5x2 + 3x – 1  0
x2 x=
x=
−3 ± 3 − ( 4 )( 5)(−1)
10
x = 1 and y = −6
−3 + 29
10
3
x = 4 and y = 5
1
0
–1
–1
2
3
4
5
6
x
2
1
−x + 3y = −15
2
So for 5x2 + 3x – 1  0 either x −3 − 29
or
10
18x – 6y = 132
Add 1 and 3 .
23x = 138
x=6
Substitute x into 2 .
18 – y = 22
−3 + 29
x
10
4x – 7x2 – 8 > 0
Consider 4x – 7x2 – 8 = 0
7x2 – 4x + 8 = 0
b2 – 4ac = 16 – 224 = – 208 < 0
There are no solutions, ∴ the function does not
cross the x-axis, ie it is all above or below the
x-axis.
When x = 0, 4x – 7x2 – 8 = – 8
There are no x values for which 4x – 7x2 – 8 > 0
3x2 + x – 5 = 0 when x =
−1 ± 1 + 60 −1 ± 61
=
6
6
−1 + 61
6
49 < 61 < 64
−1 + 49 −1 + 7 6
=
= =1
6
6
6
−1 + 61
∴
>1
6
1
2
1
3
y = −4
3
If there are n unknowns then you need n distinct
equations involving the n unknowns.
4
3x – 4y = 3
1
6x + 4y = 3
2
Add 1 and 2 .
9x = 6
10 – 3x2  x – 5
3x2 + x – 5  0
y = −2 and x = 9
5x + 6y = 6
3x – y = 22
Multiply 2 by 6.
5x + 6y = 6
–3
The higher value is
c x – 2y = 13
Add 1 and 2 .
1
–2
Consider
2
13x = 52
2
9
5x + y = 25
Add 1 and 3 .
3
–2
1
Multiply 2 by 2.
4
–3
b 3x – 2y = 2
10x + 2y = 50
y
–4
2
3x = 3
−3 − 29
x=
10
–5
1
5x + y = −1
Subtract 1 from 2 .
2
Or
–6
a 2x + y = −4
x=
5
2
1
and y = −
4
3
x + y = 50
x − 1 = 15(y − 1)
x − 15y = −14
Subtract 2 from 1 .
16y = 64
1
2
y = 4 and x = 46. Helen was 42 when her
daughter was born.
No, the solution set for – 3x2  x – 5 is not a
subset of x  1.
7
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1 Quadratics
6
7
2x − y = −14
1
1h + 1m = 33.9
3y − 2z = 16
2
Multiply A by 2
z−x=3
Multiply 1 by 3.
3
4m + 2p = 67.2
2A
2h + 2p = 42.6
B
6x − 3y = −42
4
Work out 2A – B
3y − 2z = 16
Add 4 and 2 .
2
4m – 2h = 24.6
6x − 2z = −26
Multiply 3 by 2.
2z − 2x = 6
Add 5 and 6 .
4x = −20
x = −5
Substitute x into 3: z = −2
Substitute x into 1: y = 4
5
x
x-axis has equation y = 0
6m = 92.4
m=
92.4
= 15.4
6
1 maths textbook costs $15.40 (remember
to add the ‘0’ to make the money notation
correct).
Exercise 1.5B
a x + y = 3
2x2 − y = 25
(2x − 7)(x + 4) = 0
7
or −4
2
7 – 3x = 2x + 8, so x = –0.2, y = 7.6
y=−
( )
1
× base × height
2
1
7
1
= × 4+
× 7.6 = 24
square units
2
3
15
1
or 7
2
b 2x − y = 20
x2 + xy = −12
2x − 20 = y
Substitute into 2 .
x2 + x(2x − 20) = −12
3x2 − 20x + 12 = 0
y = 4 – 5x meets y = x + 2 when 4 – 5x = x + 2, so,
7
1
1 7
x = ,y =
,
3
3
3 3
x=
2
or 6
3
y=
−56
or −8
3
y = 7 – 3x meets y = x + 2 when 7 – 3x = x + 2, so
x = 1.25, y = 3.25
The coordinates of the vertices are (–1.5, 11.5),
1 7
,
, and (1.25, 3.25).
3 3
1
2
From 1 .
y = 4 – 5x meets y = 7 – 3x when 4 – 5x = 7 – 3x, so,
x = –1.5, y = 11.5 (–1.5, 11.5)
( )
2
2x2 + x − 28 = 0
x=
Area of the triangle =
1
Add 1 and 2 .
Find the coordinates of the point of
intersection of the lines y = 2x + 8 and y = 7 – 3x
The intersections of the lines with the line y = 0
7
are when 0 = 2x + 8, x = –4 and 0 = 7 – 3x, x =
3
2C
Work out D + 2C
1
Required area
9
2h + 2m = 67.8
6
y = 2x + 8
y = 7 – 3x y
8
D
Multiply C by 2
A sketch diagram helps see what is required.
0
C
(3x − 2)(x − 6) = 0
c y = 4x
( )
5 − x2 = y
Use the information to make 3 equations.
5 − x2 = 4x
Let m be the cost of a maths textbook, h the
cost of a history textbook, p the cost of a pen.
x2 + 4x − 5 = 0
2m + 1p = 33.6
A
x = −5 or 1
2h + 2p = 42.6
B
y = −20 or 4
1
2
1 = 2
(x + 5)(x − 1) = 0
8
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1
WORKED SOLUTIONS
2
2x − 3(x2 − 7) = 13
3x2 − 2x − 8 = 0
(3x + 4)(x − 2) = 0
−4
x=
or 2
3
aSubstitution or elimination: for elimination the
first equation would need to be multiplied by 2
so that y can subsequently be eliminated.
b Substitution only: neither addition or
subtraction of the equations will eliminate
a variable.
y=
c S
ubstitution only: neither addition or
subtraction of the equations will eliminate
a variable.
3
a 2x2 + y = 14
1
x − 2y = 11
2
So the coordinates of the points of intersection of
(
5
3
4x2 + x − 39 = 0
(4x + 13)(x − 3) = 0
−13
or 3
4
y=
−57
or −4
8
1
x2 + y2 = 64
2
−2 ± 4 − (4)(2)(−63)
4
x = 5.135 or −6.135
y = 6.135 or −5.135
x=
b xy − x = −4
1
x + y = 1
2
From 2 .
So the coordinates of the points of intersection
of this line and this circle are (5.135, 6.135) and
(−6.135, −5.135).
6 y − x = 10
y=1−x
x2 + y2 = 50
Substitute into 1 .
From 1 .
y = x + 10
Substitute into 2 .
x2 + (x + 10)2 = 50
2x2 + 20x + 50 = 0
x2 + 10x + 25 = 0
(x + 5)2 = 0
x = −5, y = 5
x(1 − x) − x = −4
x − x2 − x = −4
x = ±2, y = −1 or 3
c x − y = 10
1
xy = 140
2
From 1 .
7
Substitute into 2 .
2
Let the numbers be x and y.
Then x + y = 129, x2 + y2 = 8433
y(y + 10) = 140
The two numbers must satisfy both equations.
y2 + 10y − 140 = 0
y=
1
The line is a tangent to the circle at (−5, 5).
x = y + 10
x + y = 129 so y = 129 – x
−10 ± 100 − (4)(1)(−140)
2
Substitute into x2 + y2 = 8433
x2 + (129 – x)2 = 8433
y = −10 ± 660
2
x2 + 16 641 – 258x + x2 = 8433
2x2 – 258x + 8208 = 0
y = −5 ± 165 x = 5 ± 165
4
y − x = 1
From 1 .
y=x+1
Substitute into 2 .
x2 + (x + 1)2 = 64
2x2 + 2x − 63 = 0
Add 2 and 3 .
x=
)
4 47
this line and this curve are − 3 , − 9 and (2, −3)
Multiply 1 by 2.
4x2 + 2y = 28
−47 or −3
9
x2 – 129x + 4104 = 0
2x − 3y = 13
1
x2 − y = 7
From 2 .
x2 − 7 = y
Substitute into 1 .
2
(x – 57)(x – 72) = 0
x = 57, x = 72
The two numbers are 57 and 72.
9
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1 Quadratics
8
The line y = 9 – 2x meets the circle
(x – 3)2 + (y + 2)2 = 25
when (x – 3)2 + (9 – 2x + 2)2 = 25
(x – 3)2 + (11 – 2x)2 = 25
x = −5 square root sign missed
No real solutions.
3
x4 – 7x2 + 1 = 0
Let y = x2
x2 – 6x + 9 + 121 – 44x + 4x2 = 25
y2 – 7y + 1 = 0
5x2 – 50x + 105 = 0
x2 – 10x + 21 = 0
y=
7 ± 49 − 4
2
y=
7 ± 45 7 ± 3 5
=
2
2
(x – 3)(x – 7) = 0
x = 3, x = 7
y = 3, y = –5
y = x2
There are two intersections, and their
coordinates are (3, 3) and (7, –5).
9
3x2 − 7x + 2 > 5x – 6
3x2 – 12x + 8 > 0
∴x = −
4
102x + 10x − 2 = 0
Let y = 10x
y2 + y − 2 = 0
(y + 2)(y − 1) = 0
So y = −2 or 1
So 10x = −2 or 10x = 1
So 10x = 1
5
2x8 = 5x4 – 1
3x2 – 12x + 8 = 0 when
12 ± 144 − 96 12 ± 48
2 3
=
=2±
6
6
3
2
When x = 2, 3x – 12x + 8 = –4,
x=
∴ the solution set of 3x2 − 7x + 2 > 5x – 6 is
x <2−
2 3
2 3
,2+
<x
3
3
10 2x + 3y = 7, so y =
2x8 – 5x4 + 1 = 0
7 − 2x
3
Let y = x4
2y2 – 5y + 1 = 0
Substitute into x2 + y2 = 8
x2 +
( 7 −32x ) = 8
2
y=
9x2 + 49 – 28x + 4x2 = 72
y = x4 = 0.21922359…, x = ± 0.684261… = ± 0.684
(3 s.f.)
13x2 – 28x – 23 = 0
6
2q2 + q − 3 = 0
(2q + 3)(q − 1) = 0
3
So q = − or 1
2
Exercise 1.6A
x2 − 1 = 0
x2 – 1 = 0
x2 = 1
x=±1
2
x + 10 x = −25
x + 10 x + 25 = 0
( x + 5)( x + 5) = 0 square root sign missed
( x + 5) = 0 square root sign missed
2
p 3
=
2
2
Let q = p
The points of intersection are (2.79, 0.474) and
(–0.635, 2.76).
3
p+
2p + p − 3 = 0
y = 0.474, y = 2.756
1
5 ± 25 − 8 5 ± 17
=
4
4
y = x4 = 2.280776406, x = ± 1.22891… = ± 1.23
(3 s.f.)
9x2 + (7 – 2x)2 = 72
2
x = 28 ± 28 + 4 × 13 × 23 = 28 ± 1980
26
26
x = 2.788, or x = – 0.6345
7+3 5 7+3 5
7−3 5 7−3 5
,
,−
,
2
2
2
2
3
So p = − or 1
2
p = 1 so p = 1
(52x + 5x + 4)3 = 343
52x + 5x + 4 = 7
52x + 5x − 3 = 0
Let y = 5x
2
y + y − 3 = 0
a = 1, b = 1, c = −3
7
10
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1
WORKED SOLUTIONS
−1 ± 1 − 4.1. − 3
2
y=
=
−1 ± 13
2
5x =
−1 + 13
−1 − 13
or 5x =
2
2
So 5x =
8
3
x2 − 9x − 10  0
(x − 10)(x + 1) = 0
x = 10 or −1
x  −1 or x  10
4
3
13
f ( x ) = 2x 2 + 3x + 13 = 2  x 2 + x + 
2
2 

−1 + 13
2
(
x4 − 19x2 + 16 = 0
( 34 ) − 89 + 13
3
7
= 2 ( x + ) + 11
4
8
Let y = x2
2
y − 19y + 16 = 0
a = 1, b = −19, c = 16
y=
=
=2 x+
7
and always greater than 0,
8
i.e. f(x) has no real roots.
Minimum value is 11
19 ± 297
2
19 + 297
19 − 297
or x 2 =
2
2
So, if f(x) = p has no real roots, then
(
) + 11 78 = p
3
7
and 2 ( x + ) + 11 − p = 0 has no roots.
4
8
2 x+
x = ± 4.26 or ± 0.940
9
2x6 + x3 – 1 = 0
2y2 + y – 1 = 0
y=
5
1
, y = −1
2
(
=4 x−
a = 2, b = −3, c = −7
3 ± 9 − (4)(2)(−7)
x=
4
3 ± 9 − (4)(3)(−11)
6
x < −1.48 or x > 2.48
2
7
.
8
(
7
8
=−
x = 2.77 or x = −1.27
b 3x2 − 3x − 11 > 0
) − 1681
When x =
y=4 x−
3 ± 65
4
3 ± 141
6
x = −1.48 or x = 2.48
7
8
b At the stationary point y will have a
7
minimum value i.e. when x = .
8
b 2x2 − 3x − 7 = 0
=
)
( )
a y = 2x2 – 3x − 7
a x=
7x 1
−
4 2
2

7
49 1 
= 4 x −
−
− 
8
64
2

b2 − 4ac = 9 − (4)(2)(−7) = 65 > 0 two distinct
real roots.
2
(
= 4 x2 −
Exam-style questions
=
7
8
a y = 4x2 − 7x − 2
So p < 11
1
So, x = 3 or x = – 1
2
1
2
3
4
2
Let y = x3
−1 ± 12 + 8 −1 ± 3
y=
=
4
4
2
2
19 ± 361 − 4.1.16
2
x2 =
)
2

3
9 13 
= 2 x +
−
+ 
4
16
2

) − 1681
2
81
16
So coordinates of stationary point are
6
a 3x − 2y = 7
x2 − y2 = 8
Rearrange 1 .
7 + 2y
x=
3
( 78 ,− 1681 ).
1
2
Substitute into 2 .
11
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1 Quadratics
2
 7 + 2y  − y 2 = 8
 3 
9
4 x = 1 + 21− 2x = 1 + 21 × 2−2x = 1 +
4x = 22x
49 + 28y +4y2 − 9y2 = 72
5y2 − 28y + 23 = 0
(5y − 23)(y − 1) = 0
27 23
,
and ( 3,1)
5 5
(
22x = 1 +
)
3x − 7
2
Substitute into x2 – y2 = c
b 3x – 2y = 7, so y =
x2 −
( 3x2− 7 ) = c
22x × 22x = 22x + 2
∴ 42x – 4x – 2 = 0
Let y = 22x
y2 – y – 2 = 0
(y + 1) (y – 2) = 0
y = –1, y = 2
4x2 – (3x – 7)2 = 4c
4x2 – (9x2 – 42x + 49) = 4c
22x = – 1 not possible, so 22x = 2
– 5x2 + 42x – 49 – 4c = 0
2x = 1, so x =
b2 – 4ac
= 1764 – 4 × 5 × (49 + 4c)
10 3x2 + 5x − 2 > 7x + 15
3x2 − 2x − 17 > 0
2 ± 4 − (4)(3)(−17)
6
x=
= 784 – 80c
For no intersection b2 – 4ac < 0
2 ± 208
6
x = −2.070 or x = 2.737
8(98 – 10c) < 0
x < −2.07 or x > 2.74 (3 s.f.)
98 – 10c < 0
=
11 a px2 + 5x – 6 = 0
98 < 10c
b2 – 4ac < 0
c > 9.8
25 – 4 × p × (–6) < 0
6x2 > 3 − 7x
25 + 24p < 0
6x2 + 7x − 3 > 0
(2x + 3)(3x − 1) > 0
x<−
8
1
2
= 1764 – 980 – 80c
= 8(98 – 10c)
7
2
22x
24x – 22x – 2 = 0
2
5x2 – 42x + 49 + 4c = 0
3 or x > 1
3
2
p<−
p(x + p)2 + 5(x + p) – q = p3
px2 + (2p2 + 5)x + 5p – q = 0
(x + 3)(x + 7) > 0
One repeated root when b2 – 4ac = 0
x < −7 or x > −3 and 3x − 5 > 0
5
x>
3
b2 – 4ac = (2p2 + 5)2 – 4p(5p – q)
4p4 + 20p2 + 25 – 20p2 + 4pq = 0
4p4 + 4pq + 25 = 0 as required.
So x >
= 3x2 + 30x + 63 – a
= 3(x2 + 10x) + 63 – a
= 3 ( x + 5) − 75 + 63 − a
2
= 3 ( x + 5) − (12 + a )
2
25
24
b f(x + p) = p3
a x2 + 10x + 21 > 0
5
3
b y = gf (x) = 3(x2 + 10x + 21) – a
2
22x
12 a
a 2x 2 + 3bx + c = 0

3bx
c
a2  x 2 + 2 + 2  = 0

a
a 
2
 x + 3b  − 9b 2 + c = 0


2a 2 
4a 4 a 2
y = gf(x) has 2 real roots when 12 + a > 0,
i.e. a > –12
12
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1
WORKED SOLUTIONS
2
 x + 3b  = 9b 2 − c


2a 2 
4a 4 a 2
x+
=
9b 2 4a 2 c
−
4a 4
4a 4
=
9b 2 − 4a 2 c
4a 4
± 9b 2 − 4a 2 c
3b
2 =
2a
2a 2
−3b ± 9b 2 − 4a 2 c
x=
2a 2
b
y = a2x2 + 3bx + c has a repeated root when c = 1
16 hA = t + 4, t  0
hB = 8 + 3t – t2
hB > hA
8 + 3t – t2 > t + 4
4 + 2t – t2 > 0
0 > t2 – 2t – 4
t = 2 ± 4 + 16 = 2 ± 20 = 1 ± 5
2
2
When t = 0 hB > hA ∴ 0 t 1 + 5 seconds.
17 A sketch diagram often helps, it doesn’t need to be
drawn accurately.
A
x
B
i.e. b2 – 4ac = 0
9b2 – 4a2 = 0
i.e. a 2 = 9b
4
x = −3b ±
2
D
b 2 − 4ac = −3b = − 3b × 4 = − 2
2 9b 2
3b
2a 2
2a 2
13 a Pythagoras’ theorem AB2 = BD2 – AD2
C
Area = x(a + 4x) = 4x2 + ax
When a = 3, Area  45 m2
4x2 + 3x  45
AB2 = (5x2 + 14xy + 10y2) – (2x + 3y)2
4x2 + 3x – 45  0
AB2 = 5x2 + 14xy + 10y2 – (4x2 + 12xy + 9y2)
Consider 4x2 + 3x – 45 = 0
AB2 = x2 + 2xy + y2
AB2 = (x + y)2
AB = x + y
b So length = 19 + −12 = 7 or −1 + 3 = 2.
14 13πr  260
r  6.4
πr2 > 60
r > 4.4
4.4 < r  6.4
15 a
a + 4x
x=
−3 ± 32 − 4 × 4 × (−45)
2×4
−3 ± 27 −30 24
=
or
8
8
8
24
= 3 is the only solution.
As x > 0, x =
8
x=
The values of x for which the area is at least
45 m2 are x  3 m
18 6574t2 + 776t + 100 > 1 000 000
( p + p − 1) = 125
3
p + p −1 = 5
p +p−6=0
p+ p −6=0
b Let y = p
y2 + y − 6 = 0
(y + 3)(y − 2) = 0
So y = −3 or y = 2.
6574t2 + 776t – 9 99 900 > 0
Consider 6574t2 + 776t – 999 900 = 0
−776 ± 776 2 + 4 × 6574 × 9 99 900
2 × 6574
−776 ± 162154.163
t=
13148
t=
Reject the negative value, as t  0.
t = 12.27°C
The values of t for which the number of bacteria
is great than 1 000 000 is t > 12.3° C (3 s.f.).
p = −3 or p = 2
So p = 4. (No solution from p = −3.)
13
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1 Quadratics
Mathematics in life and work
1
2x2 − 3x −
3
=0
2
a = 2, b = −3, c = −
3
2
x=
3± 9+ 4×2×3
2
4
x=
3 ± 21
4
So x =
3 + 21
3 − 21
or
.
4
4
2
3 − 21
3 + 21
<x<
4
4
3
2x 2 − 3x −
3
3
3
= 2 x 2 − x − 
2
2
4 

( )
2


= 2 x − 3 − 9 − 3
4
16
4


( )
2

3
21 
= 2 x −
−

4
16


3
−21
So coordinates of turning point are x = , y =
.
4
8
3
Line of symmetry x = .
4
4
21
units
8
14
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2
WORKED SOLUTIONS
2 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 f(0) = 2 − 0 = 2
b f(1) = 2 − 1 = 1
c f(−3) = 2 + 3 = 5
d f
2
e This is an ellipse where an input value can
map to two output values. Consequently,
this is not a function.
f This is not a function because some input
values do not map to any output value because
there is a discontinuity at these values.
( 12 ) = 2 − 12 = 32
a 3x − 5 = 1
x=2
b 3x − 5 = −17
x = −4
c 3x − 5 = 0
x=
2
5
3
So the range of f(x) is {−17, −12, −7, −2, 3, 8}.
d 3x − 5 = 6
x=
3
11
3
2 − x = 3x − 5
4x − 7 = 0
x=7
4
4
a f(−3) = −17
f(−2) = −12
f(−1) = −7
f(0) = −2
f(1) = 3
f(2) = 8
3
b This is a linear relationship whereby every
input value is related to only one output value
and every output value is related to only one
input value. This is a one–one function.
a f(x) = 2 − x for x ∈ 
y
4
3
g(2) = 1
2
fg(2) = 2 − 1 = 1
f(2) = 0
gf(2) = 0 − 5 = −5
1
–6 –5 –4 –3 –2 –1 0
–1
a
f(x) = 5 – 3x is a linear relationship whereby
every input value is related to only one output
value and every output value is related to only
one input value. This is a one–one function.
b
g(x) = x2 + x is a quadratic relationship
whereby every input value is related to only
one output value. As an output value may be
related to more than one input value, this is
a many–one function.
c h(x) = (x + 1) is not a function because it is
not defined for x < –1.
d This is not a function because an input value
of x = 0 does not map to any output value
because there is a discontinuity at x = 0.
2
3
4
5
6 x
–2
Exercise 2.1A
1
1
–3
f(x) ∈ 
f(x) can have any real number value.
b g(x) = x3 for {x: x  1, x ∈ }
y
4
3
2
1
–11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0
–1
1
x
–2
–3
15
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2 FUNCTIONS
c 3 – 5b – 2b2 = 6
g(x) {g(x): g(x) ⩽ 1, g(x) ∈ }
2b2 + 5b + 3 = 0
All the values of g(x) are less than or equal to 1
(2b + 3)(b + 1) = 0
c h(x) = (x − 1)2 for x ∈ 
y
b=−
5
4
3
2
1
–4
–3
–2
0
–1
–1
1
2
3
4
5
6
7
8
x
5
d This is a quadratic relationship whereby
every input value is related to only one
output value. As an output value may be
related to more than one input value, this is
a many–one function.
1
a f(x) = for {x: x ⩾ 1, x ∈ }
x
y
5
h(x) {h(x): h(x) ⩾ 0, h(x) ∈ }
4
All the values of h(x) are greater than or
equal to zero (when you square a number it
is either zero or a positive number).
3
2
1
d j(x) = 1 for {x: x ≠ 0, x ∈ }
x
y
–1
4
–4
–3
–2
1
2
3
4
5
6
7
8
9
x
10 11 12 13
b g(x) = x2 for x ∈ 
2
–5
0
Range {f(x): 0 < f(x) ⩽ 1, f(x) ∈ }
3
–6
3
or b = −1
2
1
y
0
–1
–1
6
1
2
3
4
5
6
x
5
–2
4
–3
3
2
j(x){j(x): j(x) ≠ 0, j(x) ∈ }
1
j(x) can have any value except 0.
4
–6
a g(−2) = 3 + 10 − 8 = 5
y
b
0
–30 –20 –10
–10
10
–5
–4
–3
0
–1
–1
–2
1
2
3
4
5
6
3
4
5
6
7
x
Range {g(x): g(x) ⩾ 0, g(x) ∈ }
x
10 20 30 40
c h(x) = (x − 1)3 for x ∈ 
y
–20
20
4
–30
30
3
g(−5) = 3 + 25 – 50 = −22
2
Turning point:
1
( )
5
5
49
x = − , g (− ) =
4
4
8
2

5
49 
−2  x +
−
=0
4
16


49
Range of g(x) is −22 g(x) 8 .
–6
–5
–4
–3
–2
0
–1
–1
1
2
x
–2
–3
Range {h(x): h(x) ∈ }
16
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WORKED SOLUTIONS
d j(x) =
Exercise 2.2A
1 for {x: x > −2, x ∈ }
x+2
y
1
f(1) = (4 × 1) + 3 = 7
3
b f(1) = (4 × 1) + 3 = 7
2
g(7) = 72 = 49
c g(−2) = (−2)2 = 4
g(4) = 42 = 16
1
1
2
3
4
5
6
7
8
9
10
x
d f(−2) = (4 × −2) + 3 = −5
f(−5) = (4 × −5) + 3 = −17
–2
–3
2
Range {f(x): f(x) ⩾ 0, f(x) ∈ }
6
The relationship y =
1
isn’t currently a
(x + a)2
3
b { x: x < 0, x ∈ } Although the curve is
(
4x2 + 6x – 15 = 2x + 3
2
) − 94 − 15 = (2x + 32 ) – 694
2
fg(a) = 4
a2 − 10a + 25 = 4
a2 − 10a + 21 = 0
(a − 3)(a − 7) = 0
So a = 3 or a = 7.
4
5
2
3
This has a minimum value of – 69 when x = − .
4
4
3
f(x) is one-one for x − .
4
fg(x) = 5(x2 + 3) – 2 = 5x2 + 13
So fg ≠ gf.
a x + 2 = ff(x)
b x2 + 2x − 6 = gf(x)
6
f(x) = 2x − 3, g(x) = x3 and h(x) =
hg(x) =
1
x3 + 1
}
fhg(x) =
2
−3
x3 + 1
10 f(x) = px5 + qx3 + 2
ffhg(x) =
Hence f(x) is one-one for the set of values
x: x ∈ , x − 3 .
4
{
f(–1) = – p – q + 2 = 7
f 2hg =
–p–q–5=0 A
f(2) = 32p + 8q + 2 = 10 32p + 8q – 8 = 0
4p + q – 1 = 0 B
Work out A + B
3p – 6 = 0, so p = 2, q = – 7
11 2x2 – 4x + 13 = 2(x2 – 2x) + 13 = 2(x – 1)2 – 2 +
13 = 2(x – 1)2 + 11
This is a one-one function for values
{x: x ∈ , x 1} , so p = 1.
2
2
gf(x) = (5x − 2)2 + 3 = 25x2 − 20x + 7
continuous for x < 3, the question specifies
{ x ∈ − }
9
2
(a − 5)2 = 4
The relationship could become a function if the
domain was limited to either
{x < −a, x ∈ − } or {x > −a, x ∈ + }
a { x: x > 3, x ∈ } As the curve continuous for x > 3.
( 1x ) − 1 = 1 −x x
c gg(x) = (x2 − 1)2 − 1 = x4 − 2x2
1
d ff(x) = 1 = x
x
function because there is a discontinuity at x = −a.
8
a
Cannot be found as the range of g(x){x : x  – 1} is
not fully included in the domain of f(x){x : x ≠ 0}.
b gf(x) =
h(2) = d − 2
j(2) = 2 + 2 = 1
4
d–2=1
d=3
7
a g(1) = 12 = 1
4
0
–1
–1
2
7
1
x +1
4
−6−3
x3 + 1
4
−9
x3 + 1
fg(x) = 2x − 3
gf(x) = 2x − 3
fg = gf
2x – 3 = 2x − 3
2x
= 2x − 3
23
2x = (8)(2x) − 24
24 = (7)(2x)
17
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2 FUNCTIONS
y
( )
ln 24
7
x=
ln 2
= 1.78 (3 s.f.)
6
5
4
fg(x) can be found as the range of g is included in
the domain of f, but gf(x) cannot be found as f(x) > 0
for all x and the domain of g is x: x < 0.
8
f(x) =
2
x −1
3
2
1
–4
2
2
=
2 −1
2 − x −1
x −1
x −1 x −1
2
2(
x
−
1)
=
=
3− x
(3 − x)
x −1
The domain is { x : x ∈ , x > 3} .
–3
–2
ff(x) = f ( f(x)) =
9
1
2
3
f−1 (x) = 3 + x {x ⩾ 0, x ∈ }
y
= 4x6 – 14x4 + 16x2 + 1
g(x) = x + 5 , or g(x) = x + 5
4 4
4
Exercise 2.3A
a This is a linear function so it is one-one.
f(x) = 5 − 4x {x ∈ }
y = 5 − 4x
5− y
x=
4
x
y = (x − 3)2
fg(x) = f(g(x)) = 4(x2 – 1)3 – 2(x2 – 1)2 + 7
)
6
b This is one-one for the given domain.
f(x) = (x − 3)2 {x ∈ , x ⩾ 3}
x=3+ y
(
5
–3
f(x) = 4x3 – 2x2 + 7, g(x) = x2 – 1
10 f(x) = 4x – 5, for fg(x) = x ; g(x) must involve 1 x in
4
1
order for 4 × x = x .
4
5
5
It must also involve in order for 4 × = +5 .
4
4
x
5
Check: If g(x) = + :
4 4
x
5
–5=x+5–5=x
fg(x) = 4 ×
+
4 4
4
–2
= 4(x6 – 3x4 + 3x2 – 1) – 2(x4 – 2x2 + 1) + 7
1
0
–1
–1
6
5
4
3
2
1
–2
0
–1
–1
1
2
3
4
5
6
x
c The cube of any number is unique so the
function is one-one.
f(x) = x3 {x ∈ }
y = x3
x= 3 y
f−1 (x) = 3 x {x ∈ }
y
4
3
5−x
f−1(x) =
{x ∈ }
4
2
1
–5
–4
–3
–2
0
–1
––1
1
2
3
4
5
x
–2
–3
–4
18
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2
WORKED SOLUTIONS
2
f(x) = 7x − 2
5
Let y = 7x − 2.
y+2
=x
7
a f(x) =
x+2
7
2+2 4
−1
f (2) =
=
7
7
2
+ 5 , where {x ∈ , x > 1}.
x −1
Let y =
2
+ 5.
x −1
y−5=
2
x −1
f −1(x) =
3
Let y = 5 − 4x.
x=
f−1(x) =
5− y
4
f(x) =
y
5−x
{x ∈ }
4
6
5
y = f (x) 4
4
2
3
0
–2
–2
2
–4
4
6
8
1
0
–1
–1
–8
b
2
3
4
x = (y − 4)2
4x2 − 24x + 12 = 0
f(x) = (x − 4)2 {x ∈ , x ⩾ 4}
x2 − 6x + 3 = 0
y
20
10
y = f –1(x)
5
2
4
6
8 10 12 14 16 18 20 22 24 x
–10
6
8
9
10 11
x
f(x) = 2x2 + 5x − 13
Let y = 2x2 + 5x − 13.
( )
2

5
129 
y = 2 x +
−
4
16 

8y + 129 5
−
16
4
f(x) = x2 + 7x − 11
Let y = x2 + 7x − 11.
x=
2


y =  x + 7 − 49 − 11
2
4


f −1(x) =
( )


y = ( x + 7 ) − 93 
2
4


2
ff −1(x) =
y + 93 − 7
4
2
x + 93 − 7
4
2
{
8x + 129 5
−
16
4
 8x + 129 5 
 8x + 129 5 
ff −1(x) = 2 
−  + 5
−  − 13
16
4
16
4


2
f −1(x) =
7
(Only 3 + 6 is valid as 3 − 6 is not included
in the domain of either function.)
y = f(x)
15
x=
6
x=3± 6
25
–6 –4 –2 0
–5
5
2
2
+5=
+1
x −1
x−5
2(x − 5) + 5(x − 1)(x − 5) = 2(x − 1) + (x − 1)(x − 5)
Let y = x + 4.
4
1
c f(x) = f−1(x)
x + 4 {x ∈ , x ⩾ 0}
f−1(x) =
y = f –1(x)
2
x
y = f –1(x)
–6
y = f(x)
7
6
–4
2
+ 1, where {x ∈ , x > 5}.
x−5
b
y
–6
2
+1
y−5
x=
a f−1(x) = 5 − 4x {x ∈ }
8x + 129
8x + 129 25
8x + 129 25 52
−5
+
+5
−
−
8
16
8
16
4
4
ff −1(x) = x +
}
The domain is x : x ∈ , x > −93 .
4
129 25 50 104
+
−
−
8
8
8
8
ff−1(x) = x
19
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2 FUNCTIONS
7
3x − 1
f(x) = 3x − 1 Let y =
x−2
x−2
3
= 4 x + 1 − 2x + 5
The domain is { x : x ∈ , x > −1}.
Then (x – 2) y = 3x – 1
xy – 2y = 3x – 1
10 f(x) = x6 – 2x3 – 7
xy – 3x = 2y – 1
Let y = x6 – 2x3 – 7
x(y – 3) = 2y – 1
Then y = (x3 – 1)2 – 1 – 7
2y − 1
x=
y−3
y = (x3 – 1)2 – 8
y + 8 = (x3 – 1)2
f −1(x) = 2x − 1
x−3
x3 − 1 =
The domain is { x : x ∈ , x > 3}.
8
f(x) = 2x2 – 4x + 13
Let y = 2x2 – 4x + 13
Then y = 2(x2 – 2x) + 13
y + 8 +1
x=3
y + 8 +1
The domain is { x : x ∈ , x −8} .
y = 2(x – 1)2 + 11
Exercise 2.4A
y − 11
= (x − 1)2
2
1
y − 11
+1
2
f −1(x) =
x3 =
f −1(x) = 3 x + 8 + 1
y = 2[(x – 1)2 – 1] + 13
x=
y+8
a f(x + 3) one solution
y
25
x − 11 + 1
2
20
15
10
2




ff −1(x) = 2 ×  x − 11 + 1 − 4  x − 11 + 1 + 13
2
2




 x − 11

 x − 11

x − 11
= 2
+2
+ 1 − 4 
+ 1 + 13
2
2
 2



= x − 11 + 4
x − 11
x − 11
+2−4
+9
2
2
5
–9 –8 –7 –6 –5 –4 –3 –2 –1
1
2
3
x
4
5
6
x
8
9 10 11 x
b f(x) + 3 no solutions
y
25
20
=x
15
f −1f ( x ) =
10
2
2x − 4x + 13 − 11
+1
2
=
2x 2 − 4x + 2
+1
2
=
x 2 − 2x + 1 + 1
=
( x − 1) + 1
5
–6 –5 –4 –3 –2 –1 0
3
y
20
15
10
ff–1(x) = f–1f(x) = x as required.
5
f(x) = 4x3 – 2x2 + 7, g(x) = x2 – 1
–4 –3 –2 –1 0
Let y = x2 – 1; then x =
fg −1(x) = 4 x + 1 − 2
2
25
2
3
1
c f(x − 4) one solution
=x–1+1=x
9
0
y + 1 , i.e. g −1 ( x ) =
1
2
3
4
5
6
7
x + 1,
( x + 1) + 7
2
3
= 4 x + 1 − 2x − 2 + 7
20
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2
WORKED SOLUTIONS
d f(x + 4) one solution
d
y
y
3
25
20
2
15
1
10
5
–6
–11–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0
2
a
1
2
3
–4
–3
–2
–3
y
–3
3
–4
0
–1
–1
1
2
3
4
5
6
x
3
12
–3
10
–4
8
4
5
4
6
8
10 x
6
x
6
P (0, 3)
P (0, 2)
3
–10 –8
2
–6
–4
–2
0
–1
–1
4
2
–2 0
–2
2
–4
1
–3
3
y
–2
y
–4
2
14
b
–5
1
One solution
One solution
–6
0
–1
–1
–2
–2
1
–5
–4
4 x
2
–6
–5
1
2
3
4
5
6
–6
x
–2
f(x + 1) = f(x) + 1 has one solution
4
–3
–4
a, b f(x − 180), where −180° ⩽ x ⩽ 180°; asymptotes
at x = −90 and 90.
Axis intercepts at (−180, 0), (0, 0) and (180, 0).
y
One solution
c
y
3
2
–90°
1
–6
–5
–4
–3
–2
0
–1
–1
–2
1
2
3
4
5
6
0
x
90°
x
c, d f(x) + 2; asymptotes at x = −90 and 90
–3
3
y
–4
4
One solution
2
–90°
90
0
90°
90
x
21
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2 FUNCTIONS
5
a, b f(x + 2) = 2x + 3
1 
Translation, vector  
 2
f(x) + 2 = 2x + 1 + 2
9
f(x + 2)
y
25
20
f(x) = (x + 2)2 – 3 drawn in blue,
g(x) = (x – 1)2 – 2 in red
y
6
f(x)+ 2
15
4
10
2
5
–5
–4
–3
–2
–1
0
1
2
3
4
5
–8
x
–6
–4
0
–2
–2
2
4
6
x
–4
–6
c i If g(x) = 2x, then g(x + 3) = 2x + 3 = f(x + 2).
6
 3
a Translation vector  
1
ii If g(x) = 2x, then g(x + 1) + 2 = 2x+1 + 2
= f(x) + 2.
f(x) = x2 + 6x – 15 = (x + 3)2 – 24
 − 3
f(x) is a translation of y = x2 with vector 
 −24
 −3
b Translation vector  
 −1
 −2 
c Translation vector  
 −3 
So, the transformation which maps f(x) to x2 is a
 3
translation with vector   .
 24
7
1 
d Translation vector  
 −2 
f(x) = x2 – 2, g(x) = x – 1, ∴ g–1(x) = x + 1
fg–1(x) – 3 = (x + 1)2 – 2 – 3 = (x + 1)2 – 5
y
–8
–6
–4
10 f(x) = x3, ∴ f −1 ( x ) = 3 x
8
f(x + 2) = (x + 2)3, drawn in blue ∴ f −1 ( x ) = 3 ( x + 2)
6
drawn in red
4
y
2
6
0
–2
–2
2
4
6
4
x
2
–4
–8
–6
8
–6
–4
0
–2
–2
= (x – 2)2 + 2
y
10
1
a f(3x)
one solution
y
8
8
7
6
6
5
4
2
–6
–4
0
–2
–2
x
Exercise 2.4B
4
–8
6
–6
hfg(x) = hf(x – 1) = (x – 1)2 –2(x – 1) + 1 + 2
= x2 – 4x + 6 = (x – 2)2 – 4 + 6
4
–4
f(x) = x2 – 2x + 1, g(x) = x – 1, h(x) = x + 2
= x2 – 2x + 1 – 2x + 2 + 3
2
3
2
4
f(x) = x2 – 2x + 1 = (x – 1)2
6
2
x
1
–3
–2 –1 0
–1
1
2 x
22
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2
WORKED SOLUTIONS
b 3f(x)
b 2f(x)
one solution
y
y
10
8
8
7
6
6
5
4
4
2
3
0
–4 –3 –2 –1
–2
2
1
–3
–2 –1 0
–1
2
3
4
5 x
–4
2 x
1
1
–6
–8
c f(2x)
c f(−x)
one solution
y
y
3
8
2
7
1
6
5
4
–6
–5
–4
–3
–2
3
1
2 x
1
4
5
6
x
d −f(x − 2)
y
10
8
6
y
1
4
–4 –3 –2 –1 0
–1
1
2
3
4 x
2
–2
–4
–3
–4
–2
0
–1
–2
–6
–6
–7
–8
a f(2x)
3
y
1
2
3
4
5
x
4
8
3
6
2
4
1
2
0
–1
–2
–4
( )
1
f(−2x); P (1, 0), R − ,0 , Q (0, −2)
2
f(−2x) = 0 has two solutions.
y
10
–2
–3
–4
–5
–3
3
–8
one solution
–4
2
–3
–2 –1 0
–1
1
d − f(x)
2
2
1
–2
2
–3
0
–1
–1
1
2
3
4
5
x
0
–6 –5 –4 –3 –2 –1
–1
1
2
3
4
5
6 x
–2
–3
–6
−2f(x); P (−2, 0), R (1, 0), Q (0, 4)
–8
−2f(x) = 0 has two solutions.
23
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2 FUNCTIONS
6
y
5
4
a, b, d f(2x) = 22x + 1 no solutions and
2f(x) = (2)(2x + 1) no solutions
y
3
7
2
6
1
0
–6 –5 –4 –3 –2 –1
–1
5
1
2
3
4
5
6
x
4
–2
4
y = 2f(x)
y = f(−x) and y = −f(x) in both cases asymptote at
y = 0.
For y = f(−x) P is at (0, 1).
For y = −f(x) P is at (0, −1).
f(−x) = −f(x) has no solutions (they do not
intersect).
–7
–5
–4
–3
–2
0
–1
–1
1
2
3
x
7
f(x) = x2 – 2, g(x) = x – 5, ∴ g–1(x) = x + 5
2
g–1(–x) = – x + 5
1
2fg–1(– x) = 2[(5 – x)2 – 2] = 2(5 – x)2 – 4
This is the most useful form to use to produce
a sketch graph.
y
0
–6 –5 –4 –3 –2 –1
–1
–2
–3
1
2
3
4
5
6
x
y = –f(x)
10
–4
5
–6
y = f(2x)
c i If g(x) = 2x, g(2x + 1) = 22x + 1 and
ii 2g(x + 1) = (2)(2x+1)
4
3
2
1
y
y = f(–x)
3
a, b y = f(2x), where −180°  x  180°; asymptotes at
x = −135°, −45°, 45° and 135°.
5
Axes intercepts at (−180°, 0), (−90°, 0), (0, 0),
(90°, 0) and (180°, 0).
0
–5
y
5
x
–5
8
f(x) = x2 – 2x + 1, or f(x) = (x – 1)2
g(x) = 2x,
–180
0
–90
180 x
90
h(x) = – x
hfg(x) = – [(2x)2 – 2(2x) + 1] = – 4x2 + 4x – 1
Or, using f(x) = (x – 1)2
hfg(x) = – (2x – 1)2
c, d y = −2f(x), where −180°  x  180°;
asymptotes at x = −90° and 90°.
Axes intercepts at (−180°, 0), (0, 0) and (180°, 0).
Horizontal stretch with scale factor 1 and a
2
reflection in the x-axis
y
2
y
–2
0
2
x
–2
–180
–90
0
90
180
x
–4
–6
24
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2
WORKED SOLUTIONS
9
y = f(x) drawn in red, y = g(x) drawn in green.
y
–5
–4
–3
–2
3
a
{f(x) ∈ , f(x) < 17 }
4
y
3
0.15
2
0.1
1
0.05
0
–1
–1
1
2
3
x
5
5.5
–0.05
–2
–0.1
–3
–0.15
–4
b
Let y =
1
.
x+2
x+2=
1
y
–5
–6
a One-way stretch vertically with scale factor − 4
3
b One-way stretch vertically with scale factor − 3
4
c One-way stretch vertically with scale factor −3
x=
1
−2
y
=
1 − 2y
y
f(−1)(x) =
1 − 2x
x
d One-way stretch vertically with scale factor 4
10 f(x) = x3, so f −1(x) = 3 x
f(2x) = (2x)3 (red), f −1(2x) = 3 (2x) (green)
y
4
4
2
–5
–4
–3
0
–1
–1
–2
1
2
3
x
–3
–4
b fg(x) =
Exam-style questions
gf(x) =
) − 474
47
1
y+
= 3( x − )
4
2
2
1+
2
f −1(x) =
2
5
y 47
+
3 12
1
+
2
x 47
+
3 12
 −2
a Translation with vector  
0
b Reflection in the y-axis
 0
c Translation with vector  
1 
9.5
x
1
range of f(x) not all included
x 2 + 5x − 13
within domain of g(x) so not an appropriate
composition of functions.
2
x=
9
2
b y = 3(x2 – x) – 11
1
2
8.5
( 1x ) + x5 − 13 range of g(x) included
within domain of f(x)
(
8
1
=x
1
x
a {f(x) : f(x)  –11.75, f(x) ∈ }
y=3 x−
7.5
g2(x) = gg−1(x)
The inverse of g(x) is the same as g(x).
Consequently g2(x) = gg−1(x) become the same
composition of functions.
–2
1
7
1
x
gg−1(x) =
1
6.5
a g2(x) = 1 = x
1
x
g −1(x) =
3
6
Let y =
1
1
1
; then 3 − x = and x = 3 −
3−x
y
y
f −1 ( x ) = 3 −
6
1
x
a fg(x) = 2(x − 3) − (x − 3)2
= 2x − 6 − (x2 − 6x + 9)
= 2x − 6 − x2 + 6x − 9
= −x2 + 8x −15
25
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2 Functions
y
b
x + 7 = h2fg(x)
10 a
1
−1
b x 2 + 7 = ghf(x)
0.5
–6
0
–2
–0.5
–4
2
4
6
8
10
x
1
c x 2 + 7 = gh 2f(x)
–1
7
11 a Let y =
a f(x) = x2 − 4x + 7
x2 − 4x + 7 = (x − 2)2 + 3
When x > 2, (x − 2)2 + 3 > 0.
b
y
8
(2x + 3)(x + 2)
(x + 2)(x − 1)
=
2x + 3
x −1
y (x − 1) = 2x + 3
4
yx – 2x = y + 3
2
x (y – 2) = y + 3
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
x
f −1(x) =
c Let y = (x − 2)2 + 3.
x = 2+
y−3
x+3
=7
x−2
af(x) = 2x2 + 8x – 14 = 2(x2 + 4x) – 14
x + 3 = 7x − 14
= 2[(x + 2)2 – 4] – 14 = 2(x + 2)2 – 22
f(x) = 2(x + 2)2 – 22
6x − 17 = 0
x=
b f(x) = 2(x + 2)2 – 22
Let y = 2(x + 2)2 – 22. Then (x + 2)2 =
y + 22
2
(
= 3 x+5
2
 − 2
with vector 
.
 −22
2
22x
= 22x − x = gf(x)
2x
Let y = 2x.
2 × 2x × 2x – 2x = 2y2 – y = fg(x)
30
25
c 8x4 – 8x3 + x = 8x4 – 8x3 + 2x2 – 2x2 + x
20
= 8x4 – 8x3 + 2x2 – (2x2 – x)
15
= 2(4x4 – 4x3 + x2) – (2x2 – x)
10
= 2(2x2 – x)2 – (2x2 – x)
= ff(x)
5
–4
2
13 a
b 2x(2x + 1 – 1) = 2x(2 × 2x – 1) = 2 × 2x × 2x – 2x
y
–6
2
b f(5) = 75 + 75 – 10 = 140, so range is f(x)  140.
c One-way stretch scale factor 2, translation
–10 –8
) − 754 − 10
115
Minimum value of − 4 .
x + 22 − 2
2
a
17
6
12 af(x) = 3x2 + 15x – 10 = 3(x2 + 5x) – 10
y + 22
−2
2
f −1 ( x ) =
x+3
x−2
b f −1(x) = 7
f −1(x) = 2 + x − 3
x=
y+3
y−2
x=
{f(x) ∈ , f(x) > 3}
9
=
6
2
8
2x 2 + 7x + 6
x2 + x − 2
0
–2
–5
2
4
6
8
x
b f(−x) reflection in y-axis
26
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WORKED SOLUTIONS
14 a gf(x) =
1
1
1
=
=
1 − 3(2x − 5) 1 − 6x + 15 16 − 6x
2
c y = g(x) in red, y = h(x) in blue
y
10
b 2g(x) = f(x)
2
= 2x − 5
1 − 3x
5
2 = (2x – 5)(1 – 3x)
= 2x – 5 – 6x2 + 15x
–10
0 = – 6x2 + 17x – 7
0
–5
x = −17 ± 121 = 1 or 2 1
−12
2
3
15 a f(x) = x3 + x2 + x + 1
g(x) = (x – 1)3 + (x – 1)2 + (x – 1) + 1 – 3
g(x) = (x – 1)3 + (x – 1)2 + x – 3
b
h(x) = – 2x3 + 2x2 – 2x + 1
–5
 −1
d Translation vector  
0
(
One-way stretch vertically scale factor 2,
5y
1+ y
b
2a
q c −
b2
4a
) − 4ba + c
2
2
a – b + c = –9
A
a+b+c=1
B
4a + 2b + c = 15
C
2a + 2c = –8, ie a + c = –4
D
2A + C
b ff(x) = f(x),
x
x
5− x
5− x =
5− x
= x ×
5(5 − x) − x 5 − x 25 − 6x
5− x
5−x
5− x
6a + 3c = –3, ie 2a + c = –1
E
E–D
a=3
x
x
=
25 − 6x 5 − x
5x – x2 = 25x – 6x2
=a x+
A+B
f −1(x) = 5x
1+ x
x(5 – x) = x(25 – 6x)
)
b f(–1) = –9, f(1) = 1, f(2) = 15
5y = x + xy = x(1 + y)
x=
(
)
b
x +c
a
2

b
b2 
= a x +
− 2 + c
2
a
4a 

= 2[(–x)3 + (–x)2 + (–x) + 1] – 1
y(5 – x) = x
(
18 a f(x) = ax2 + bx + c = a x 2 +
= 2(–x)3 + 2(–x)2 + 2(–x) + 1
0
reflection in y-axis, translation with vector   .
 −1
x
16 a Let y =
5−x
x
5
c = –7
b=5
c
y
5
5x2 – 20x = 0
5x(x – 4) = 0
x = 0 or x = 4
17 a x2 + 6x + 4 = (x + 3)2 – 9 + 4 = (x + 3)2 – 5
 −3
b Translation vector  
 −5
–5
0
5
x
–5
–10
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2 Functions
19
296 – 20x
296 – 20x
10x
10x
10x
10x
214 – 20x
214 – 20x
10x
10x
a v(x) =
(296 – 20x)(214 – 20x)10x
= 40(148 – 10x)(107 – 10x)x
b Domain is {x: x ∈ ,0 < x < 10.7}. Range is
{v(x): v(x) ∈ , 0  v(x) < 1158}.
c x = 3 cm
v = 236 × 154 × 30 = 1 090 320 mm3, or 1090 cm3
(3 s.f.)
Mathematics in life and work
1
2
{t ⩾ 0, t ∈ }
f(t)
14 000
12 000
10 000
8000
6000
4000
2000
0
1
2
3
4
5
6
7
8
3
As the car ages its value depreciates.
4
 t 
fg(t ) = 15 000 − 1500  210 


5
9
10
t
Car Model A has the greater rate of depreciation, as
illustrated on the following graph.
f(t)
14 000
12 000
10 000
8000
6000
4000
2000
0
1
2
3
4
5
6
7
8
9
10 t
28
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3
WORKED SOLUTIONS
3 Coordinate geometry
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
3
x − 3y − 21 = 0
9−5
4
−2
.
=
=
( −2) − 8 −10 5
1
Gradient =
2
Gradient of first line = −3.
b Multiply by 5: 5y = − 2x + 30, and rearrange.
2x + 5y − 30 = 0
1
.
3
Equation of perpendicular line through (5, 8) is
given by y − 8 = 13 (x − 5).
c Multiply by 3: 3y = 4x +
Perpendicular gradient =
3y − 24 = x − 5
x − 3y + 19 = 0
3
)
x+
4
x − 4 = 17
3
8x – 6y + 21 = 0
4
5
The line in the form ax + by + c = 0, where a, b and
c are integers is:
5y
=x−4
3
5y = 3(x – 4)
= 3x − 12
3x − 5y − 12 = 0
6
When the line intersects the x-axis, y = 0.
Substituting y = 0
−3x + 2 = 0
Hence point of intersection = (9, 4).
Complete square on x2 − 14x + 23.
x2 − 14x + 23 = (x − 7)2 − 49 + 23
2 = 3x
2
x=
3
So the coordinates of the x-intercept are 2 , 0 .
3
When the line intersects the y-axis, x = 0.
(x − 7)2 − 26
( )
Exercise 3.1A
The lines not in the form ax + by + c = 0, where a, b
and c are integers are:
Substituting x = 0
c, because it does not equal zero
d, because it does not equal zero
f, because a, b and c are not integers.
a Rearrange y = 4 + 5x.
5x − y + 4 = 0, where a = 5, b = −1 and c = 4
−5y + 2 = 0
2 = 5y
2
y=
5
So the coordinates of the y-intercept are 0, 2 .
5
y 1
x + − = 0.
3 2
Multiply by 6: 6x + 2y – 3 = 0, a = 6, b = 2, c = –3
( )
e, because c is not an integer
2
( 32 )
has coordinates 0,
7x = 63
1
3
2
gradient = 4 and the y-intercept
7
x = 21
3
4
Rearrange: 8x + 3 = 2y
Rearrange and divide by 2: y = 4x +
2
x + 2 x − 2 = 17
3
x=9
2
When x = 9, y = 3 × 9 − 2 = 4.
21
.
2
Multiply by 2: 6y = 8x + 21, and rearrange.
Substitute y = 2 x − 2 into x + 2y = 17.
3
(
a Multiply by 3: 3y = x − 21, and rearrange.
7
b Rearrange y = 3 − 2x.
2x + y – 3 = 0, where a = 2, b = 1 and c = −3
29
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3 Coordinate geometry
8
9
4x 7 5y
+ =
5
3
4
Multiply by 60: 48x + 140 = 75y,
so 48x – 75y + 140 = 0
5
1
= − , 5 × 4y = – 1 × 7x, 20y = – 7x,
7x
4y
Numerator and denominator confused. Should be:
m=
4
m=
b m=
n = 0 is the only value.
(x2, y2) = (4, 9)
y − y1
a m= 2
x 2 − x1
9−3 6 =2
=
4 −1 3
Do not lie on a straight line with a gradient
of −3.
m=
(x1, y1) = (2, 3)
(x2, y2) = (7, 8)
8−3 5
= =1
7−2 5
y − y1
b m= 2
x 2 − x1
c m=
m=
(x2, y2) = (4, −16)
−16 − −7 = −9 = −3
3
4 −1
Lie on a straight line with a gradient of −3.
m=
(x2, y2) = (5, 9)
9−3 6
= =3
5−3 2
d m=
y − y1
c m= 2
x 2 − x1
(x2, y2) = (3, 1)
(x2, y2) = (3, −9)
1 − 7 −6 = −3
=
3−1
2
Lie on a straight line with a gradient of −3.
m=
m=
(x1, y1) = (3a, 3a)
5a − 3a 2a 2
=
=
8a − 3a 5a 5
y − y1
b m= 2
x 2 − x1
(x1, y1) = (a, a)
m=
(x2, y2) = (3a, −5a)
−5a − a −6a
=
= −3
3a − a
2a
y − y1
Should be m = 2
.
x 2 − x1
m=
3
y-coordinate incorrect in (x2, y2). It should be (5, −8).
y 2 − y1
x 2 − x1
(x1, y1) = (1, 7)
(x1, y1) = (1, −3)
(x2, y2) = (8a, 5a)
y 2 − y1
x 2 − x1
(x1, y1) = (1, −7)
(x1, y1) = (3, 3)
2
y 2 − y1
x 2 − x1
(x1, y1) = (1, 3)
Exercise 3.1B
−9 − −3 −6
= −3
=
3−1
2
y − y1
a m= 2
x 2 − x1
−8 − 1 = −9
= −3
3
4 −1
Lie on a straight line with a gradient of −3.
This gives a term 2x × 3y × ( – 3n) = – 18nxy, which
is non-zero if n ≠ 0.
m=
y 2 − y1
x 2 − x1
(x2, y2) = (4, −8)
1
5
− 3n =
in the required form
2x
3y
multiply by the denominators, 2x, 3y.
1
a m=
(x1, y1) = (1, 1)
20y + 7x = 0, a = 7, b = 20, c = 0
10 To write
−8 − 2 −10 −5
=
=
5−1
4
2
5
Ascent:
y − y1
m= 2
x 2 − x1
(x1, y1) = (−10, 0)
( x2, y 2 ) = 0, 12
( )
1
1 −0
1
2
= 2 =
0 + 10 10 20
Descent:
y − y1
m= 2
x 2 − x1
m=
( 12 )
( x , y ) = ( 2, 83 )
(x1, y1) = 0,
2
2
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3
WORKED SOLUTIONS
Exercise 3.1C
3
− 1 −1
1
m= 8 2 = 8 =−
16
2−0
2
The descent is the steepest because 1 > 1 .
16 20
6
m=
1
y 2 − y1
x 2 − x1
( )
( x , y ) = ( 34 , − 32 )
( x1, y1 ) = 12 , 13
2
m=
7
b m = 3 and (x1, y1) = (0, 3)
y − y1 = m(x – x1)
y − 3 = 3(x − 0)
= 3x
y = 3x + 3
2
−2 − 1
3 3 = −1 = −4
3−1
1
4 2
4
c m = 2 and (x1, y1) = (3, 4)
y − y1 = m(x − x1)
y − 4 = 2(x − 3)
= 2x − 6
y = 2x − 2
1
5
− 3n =
2x
3y
Multiply by 3y × 2x.
3y – 3n × 2x × 3y = 5 × 2x
d m = −5 and (x1, y1) = (2, 3)
y − y1 = m(x − x1)
y − 3 = −5(x − 2)
= −5x + 10
y = −5x + 13
For this to be a straight line n = 0.
1
3y = 10x. Gradient = 3 .
3
8
1
1
y1 − y 2 a + 5 a + 5
1
=
=
= −6 a +
= −6a − 1.2
5
x1 − x 2 1 − 1
−1
6
3 2
( )
2
Gradient = –2 ∴ – 6a – 1.2 = – 2, 6a = 0.8,
0.8
8
2
a=
=
=
6
60 15
9
4x 7 5y
+ =
. Multiply by 60 : 48x + 140 = 75y,
5
3
4
so y =
48
48
, and 0 <
< 1 as required.
75
75
10 For AB:
1+1
5
y1 − y 2
5 40
200
= 6 9 = 18 = −
×
=−
x1 − x 2 − 1 − 7 − 43
18 43
774
5 8
40
−1 < −
3
200
<0
774
For CD:
27
7 1
y1 − y 2 8 − 5
27 18
486
= 40 = −
×
=−
=
40 1
40
x1 − x 2 1 − 1 − 1
18
9 6
486
−
< −12
40
AB is shallowest, as it has a gradient closer to 0.
a m = −4 and (x1, y1) = (−2, −5)
y − y1 = m(x − x1)
y − −5 = −4(x + 2)
y + 5 = −4x − 8
y = −4x − 13
b m = −1 and (x1, y1) = (2, −2)
y – y1 = m(x – x1)
y − − 2 = −1(x – 2)
y + 2 = −x + 2
y = −x
48x 140
+
75
75
Gradient =
a m = 2 and (x1, y1) = (3, 0)
y – y1 = m(x – x1)
y − 0 = 2(x − 3)
y = 2x − 6
m = −2 and (x1, y1) = (0, 3)
y − y1 = m(x − x1)
y − 3 = −2(x − 0)
= −2x
2x + y − 3 = 0
4
First find the point of intersection.
2x + 4 = 7 − x
3x = 3
x=1
y=2+4=6
m = 3 and (x1, y1) = (1, 6)
y – y1 = m(x – x1)
y – 6 = 3(x – 1)
= 3x − 3
y = 3x + 3
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3 Coordinate geometry
5
Find the equation of the first line.
9
m = 3 and (x1, y1) = (1, 1)
Line A: y + 7 =
(
y − 1 = 3(x − 1)
y = −2x −
= 3x − 3
m = −1 and (x1, y1) = (4, 6)
−6x −
y − y1 = m(x − x1)
= −x + 4
y = −x + 10
y + 6.3 = 3x – 9.3
y = 3x – 15.6
Find the point of intersection.
1+1
5
10 y1 − y 2 = 6 9 = 18 = 5 × − 40 = − 100
x1 − x 2 − 1 − 7 − 43 18
43
387
5 8
40
3x − 2 = −x + 10
4x = 12
x=3
y = −3 + 10 = 7
Exercise 3.2A
Point of intersection is (3, 7).
1
5 and (x , y ) = (2, 3)
1 1
2
2
5
y − 3 = ( x − 2)
2
When you substitute in x = 0, y will equal 0 if the
line goes through the origin.
y−3=
=
Line A:
(x1, y1) = (1, 7)
(x2, y2) = (3, 11)
m=
y 2 − y1 11 − 7
=
=2
3−1
x 2 − x1
5
0 − 2)
2(
Line B:
5
−2
2( )
(x1, y1) = (2, −3)
(x2, y2) = (5, 3)
= −5
m=
y = −2
The line does not go through the origin.
(
3 1
Line with gradient −2 passing through − ,
2 3
1
3
has equation y − = −2 x + .
3
2
(
)
)
3
b Arrange the equations in the form y = mx + c:
y = 3x + 1, y = 3x − 3
m = 3 in both equations so the lines are parallel.
1
8
−3=−
3
3
equation y −
aArrange the equations in the form y = mx + c:
y = 4x + 2, y = −4x − 3
m is not equal so the lines are not parallel.
1
3
y − = −2 ×
3
2
Line with gradient
y 2 − y1 3 − −3
=2
=
x 2 − x1
5−2
m = 2 for both lines so the lines are parallel.
Intersects the y-axis when x = 0.
4
(
)
1
4 1
passing through
,
has
2
5 10
(
)
1
1
4
= x− .
10 2
5
Passes through point with x-coordinate 2 when
y−
a m = 2 in both equations so the lines are parallel.
b m = −3 in both equations so the lines are parallel.
y − y1 = m(x − x1)
8
3
3
= x − 22, 22 −
= 7x, x = 3.1, y = −6.3
10
10
Line with gradient 3 passing through (3.1, –6.3) has
equation y + 6.3 = 3 (x – 3.1).
y − 6 = −1(x − 4)
y=
)
1
A and B meet when 3 −2x − 10 = x − 22 ,
Find the equation of the second line.
7
1
10
(
y = 3x − 2
m=
)
2
1
10
Line B: y − 5 = −2 x + 4 , 5y − 2 = −10x − 4 ,
y − y1 = m(x − x1)
6
1
(x – 1), 3y + 21 = x – 1, 3y = x – 22
3
( )
1
1
4
= 2− .
10 2
5
From −4x + y + 7 = 0, y = 4x − 7, so m = 4.
The y-intercept of 2x – y + 3 = 0 is when x = 0, so is
at (0, 3).
(x1, y1) = (0, 3)
y − y1 = m(x − x1)
y − 3 = 4(x − 0)
y = 4x + 3
y = 0.7 ≠ 3, ∴ the line does not pass through (2, 3).
32
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3
WORKED SOLUTIONS
5
aArrange the equations in the form y = mx + c:
5
7
5
1
y = x − ,y = x +
3
3
3
2
m=
Exercise 3.2B
1
5
in both equations so the lines are parallel.
3
bArrange the equations in the form y = mx + c:
4
8
7
1
y= x− ,y= x− .
3
3
3
3
m is not equal so the lines are not parallel.
6
Arrange the equations in the form y = mx + c:
2x − y + 6 = 0, y = 2x + 6, m = 2
1
1
2x + 4y − 44 = 0, 4y = 44 − 2x, y = 11 − x, m = −
2
2
2x − y − 4 = 0, y = 2x − 4, m = 2
1
1
2x + 4y − 24 = 0, 4y = 24 − 2x, y = 6 − x, m = −
2
2
So 2x − y + 6 = 0 and 2x − y − 4 = 0 are parallel lines
as m = 2 for both lines.
And 2x + 4y − 44 =0 and 2x + 4y − 24 = 0 are parallel
1
lines as m = − for both lines.
2
7
2
y – y1 = m(x – x1)
(x1, y1) incorrectly substituted. Correcting this and
using correct value of m2 gives:
y – 0 = 2(x – 1)
So y = 2x − 2
3
5y x
3x 42
x 5y
−
+ 7 = 0,
= + 7, y =
+
4
6
6
4
10
5
3
Gradient =
10
2x + 4y − 44 = 0, 4y = 44 − 2x, y = 11 −
2x + 4y − 24 = 0, 4y = 24 − 2x, y = 6 −
1
1
x, m = −
2
2
So 2x − y + 6 = 0 and 2x + 4y – 44 = 0 are
perpendicular lines as m1 m2 = −1.
And 2x − y + 6 = 0 and 2x + 4y − 24 = 0 are
perpendicular lines as m1 m2 = −1.
y = 0.3x + 2.7
10y = 3x + 27
And 2x − y − 4 = 0 and 2x + 4y − 44 = 0 are
perpendicular lines as m1 m2 = −1.
16x 28
4x 7 5y
16
+ =
,y =
+
, gradient =
5
3
4
25 15
25
25y
25y
8x −
− 7 = 0,
= 8x − 7,
2
2
2
16x 14 , gradient = 16
2
y=
−
× 8x − 7 × , y =
25
25
25
25 25
As the two lines have the same gradient they are
parallel.
4
x
= − 1 , 16y = −11x , y = − 11
11x
4y
16
11
passing through (0, 1) has
Line with gradient −
16
equation
y −1= −
10
1
1
x, m = −
2
2
2x – y – 4 = 0, y = 2x − 4, m = 2
y – 3 = 0.3x – 0.3
11
( x − 0)
16
16y + 11x – 16 = 0
Arrange the equations in the form y = mx + c:
2x – y + 6 = 0, y = 2x + 6, m = 2
y – 3 = 0.3(x – 1)
9
b m1 = −3 and m2 = 1 so m1 m2 = −1, so the
3
lines are perpendicular.
1
m1 = − for the given line but the gradient of a line
2
perpendicular to this will be m2 = 2.
Let (x1, y1) = (1, 0).
Line with gradient 0.3 passing through (1, 3) has
equation
8
1
so m1m2 ≠ −1, so the lines are
2
not perpendicular.
a
m1 = 2 and m2 =
(
)
7y
2x 8
2
2x 8
4
+
=
y= ×
+
, gradient =
21
3 11
2,
7
3 11
15y
15y
2
2x −
− 9 = 0,
= 2x − 9, y =
× (2x − 9),
2
2
15
4
gradient =
15
4
4
≠
the lines are not parallel.
As
21 15
And 2x – y – 4 = 0 and 2x + 4y − 24 = 0 are
perpendicular lines as m1 m2 = −1.
4
From 8x − 2y − 7 = 0, 8x − 7 = 2y, 4x −
7
= y, m1 = 4.
2
m1 m2 = −1
1
So m2 = − .
4
The y-intercept of 4x − 2y + 5 = 0 is when x = 0 so is
5
at 0, .
2
( )
( )
( x1, y1 ) = 0, 52
y − y1 = m(x – x1)
y−
5
1
= − ( x − 0)
2
4
5 1
y= − x
2 4
x + 4y = 10
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3 Coordinate geometry
5
5
m1 = and m2 = − 3 so m1 m2 = −1 and the lines
a
3
5
are perpendicular.
9
10
and m2 =
so m1 m2 ≠ −1 and the
b m1 =
10
9
lines are not perpendicular.
6
As the two lines have the same gradient they are
parallel, so cannot be perpendicular.
9
Describe as line A
(x1, y1) = (0, 6)
y −1=
(x2, y2) = (4, 4)
y − y1 4 − 6
2
1
m= 2
=
=− =−
x 2 − x1 4 − 0
4
2
Describe as line B
10
y 2 − y1 10 − 6 4
=
= =2
x 2 − x1
2−0
2
(x1, y1) = (6, 8)
1
(x2, y2) = (7, 8)
(x2, y2) = (6, 8)
y − y1 8 − 4 4
=
= =2
m= 2
x 2 − x1 6 − 4 2
( 2 +2 7 , 3 +2 8 )
= (4.5, 5.5)
So A is parallel to C, B is parallel to D, A is
perpendicular to B and D and C is perpendicular
to B and D.
length =
=
So the quadrilateral can only be a square or a
rectangle.
(x2, y2) = (3, 3)
( 5 +2 3 , 9 +2 3 )
= (4, 6)
length =
( x2 − x1 )2 + ( y 2 − y1 )2
(3 − 5)2 + (3 − 9)2
=
= 4 + 36
3y – 9 = – 10x + 10
3y + 10x – 19 = 0
25y
25y
− 7 = 0,
= 8x − 7,
2
2
2
2
16x 14 , gradient = 16
−
y=
× 8x − 7 × , y =
25
25
25
25 25
( 7 − 2 )2 + ( 8 − 3 )2
(x1, y1) = (5, 9)
10
( x − 1)
3
8x −
( x2 − x1 )2 + ( y 2 − y1 )2
= 25 + 25 = 5 2
x + x 2 y1 + y 2 
,
b mid-point =  1
2 
 2
5y x
x 5y
3x 42
−
+ 7 = 0,
= + 7, y =
+
4
6
6
4
10
5
3
Gradient = .
10
10
Perpendicular line has gradient − .
3
10
passing through (1, 3) has
Line with gradient −
3
equation
4x 7 5y
16x 28
16
+ =
,y =
+
, gradient =
5
3
4
25 15
25
x + x 2 y1 + y 2 
,
a mid-point =  1
2 
 2
(x1, y1) = (2, 3)
(x1, y1) = (4, 4)
8
2y
2y
15 ( 2
− 9 = 0,
= −2x + 9, y =
× − x + 9),
15
15
2
Exercise 3.2C
Describe as line D
y−3=−
)
gradient = – 15
4
20
As
× −15 = −
≠ −1 the lines are not
21
7
perpendicular.
Describe as line C
7
(
7y
2
2x 8
2x 8
4
, y= ×
+
=
+
, gradient =
2
7
3 11
21
3 11
2x +
(x2, y2) = (2, 10)
(x2, y2) = (2, 10)
y − y1 10 − 8
2
1
=
m= 2
=
=−
2−6
x 2 − x1
2
−4
16
(x − 0)
11
11y – 11 = 16x
(x1, y1) = (0, 6)
m=
11y
4
1
=−
, 16y = −11x, y = −
16
11x
4y
16
Perpendicular line has gradient .
11
16
passing through (0, 1) has
Line with gradient
11
equation
= 2 10
2
( )
Gradient of line through (2, 2), 5, 1
2
2− 1
1
2 =−
m=
2−5
2
1
Equation of line through (2, 2), 5,
2
( )
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3
WORKED SOLUTIONS
y−2=−
1
x − 2)
2(
4
d=
1
y =− x+3
2
(11, −25 ) on line? Substitute into equation.
=
d=
So all points on a straight line.
( x2 − x1 )2 + ( y 2 − y1 )2
(x1, y1) = (2, 2)
(
(x2, y2) = 11, −
5
2
(
= 9 + (− 9 )
2
2
=
)
2
5
Line A:
y−5=
2
Mid-point of A:
( 3 +2 4 , 5 +2 9 )
7
= ( , 7)
2
Mid-point of A:
−1 + 3 3 + −1
,
2
2
9−5
(x − 3)
4−3
y = 4x − 7
= 101.25
≈ 10.1
(
[4 + 4]
Line A is the longest.
= 81 + 81
4
3
( 7 − 5)2 + (−6 − −4)2 


= 8
)
(11 − 2)2 + − 52 − 2
[9 + 81]
Line B length:
11 6
5
+ =−
2 2
2
length =
(5 − 2)2 + (16 − 7)2 


= 90
x = 11
y=−
Line A length:
)
Line B:
y − −3 =
= (1, 1)
y = −7x + 4
Gradient of A:
Point of intersection:
3 − −1
4
=
= −1
−1 − 3 −4
Mid-point of B:
m=
(
)
(− 52 , 52 )
4x − 7 = −7x + 4
x = 1, y = −3
−2 − 3 2 + 3
,
2
2
Gradient of B:
3−2
1
m=
=
= −1
−3 − −2 −1
Line through the mid-points:
 5 −1 
y − 1 =  25
(x − 1)
 − 2 − 1 
−31 − −3
(x − 1)
5−1
The point of intersection is not the mid-point of A.
6
(
 2 + p −1 + q
1 1
a  7 2 , 52  = − 3 , − 2


)
2+p
1
7
=−
2
3
20
p=−
21
1
− +q
1
5
=−
2
2
4
5
3
10
y=− x+
7
7
q=−
Gradient of line through mid-points is −3 . This is
7
not perpendicular to A or B (m = −1).
20 4
So B = − 21 , − 5 .
(
)
Note: The coords of B are the values of p and q,
20 4
So B = − 21 , − 5 .
(
)
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3 Coordinate geometry
Length of AB:
d=
(
 20 2
 − 21 − 7

9
) + (− 45 + 15 ) 
2
2
 11 1
1 2
 9 − 5 − 8 − 7  ie 23 , − 23
45 112
 2 ,
2 

y
(
( ) ( )
=
2
 26 2
3 
 − 21 + − 5 


=
20869
11025
(119 , − 18 ) and (− 15 , − 27 ) have mid-point.
11
1
64
+
=
9
5
45
63
, so the gradient of
130
the perpendicular is − 130 . The equation
63
of the perpendicular through M is
1
130
1 from which
y+ =−
x+
2
63
3
378y + 780x + 449 = 0.
bGradient of AB =
(
)
3 1
Gradient −3, through − 4 , − 3 , x-intercept has
coordinates (x, 0).
1
y − y2 0 + 3
=
= −3
Gradient = 1
x1 − x 2 x + 3
4
1
3
− = −3 x +
3
4
(
x=−
Length =
( 6445 ) + ( 569 ) = 1.43 3 s.f.
2
2
Exercise 3.3A
1
a centre = (−5, 8), radius = 6
b centre = (19, 33), radius = 20
c centre = (0, −4), radius = 3 5
d centre = (−3, −10), radius = 2 7
31
1 3
− =−
9 4
36
2
2
Length =  1  +  31 − 27  = 1 +  1 
 3   36 
9  9
2
a
(x + 5)2 + (y − 9)2 = 49 and
x2 + y2 + 10x − 18y + 57 = 0
b (x + 11)2 + (y + 1)2 = 169 and
x2 + y2 + 22x + 2y − 47 = 0
= 10 = 0.351 (3 s.f.)
81
8
11 –1
,
9 8
–1 –2
,
5 7
)
2
x
0
9
1
2
–
=
7
8 56
( )
7
)
c (x − 3)2 + y2 = 48 and x2 + y2 − 6x − 39 = 0
d
(x − 14)2 + (y − 6)2 = 44 and
x2 + y2 − 28x − 12y + 188 = 0
a Consider the y-coordinates of the points.
1
1
5
1
>
>
as
2 10
10 10
3
a (x − 5)2 + (y + 3)2 = 16
b (x + 4)2 + (y − 2)2 = 9
1 1
3 2
> as >
2 3
6 6
( )
3 1
is not the mid-point
,
Therefore, the point
5 2
of AB.
b Gradient of AB is − 7
9
1
7
1
Equation is y − = − x −
3
9
2
18y − 6 = −14x + 7
(
)
14x + 18y − 13 = 0.
c Gradient of perpendicular is 9
7
13 13
,
Midpoint of AB is
20 60
Equation of perpendicular bisector is
(
(
)
13 9
13
y−
= x−
60 7
20
420y − 91 = 540x − 351
540x − 420y = 260.
)
c (x + 1)2 + (y − 4)2 = 20
4
a (x − 9)2 − 81 + (y + 7)2 − 49 − 14 = 0
(x − 9)2 + (y + 7)2 = 144
centre = (9, −7), radius = 12
b (x + 4)2 − 16 + y2 = 9
(x + 4)2 + y2 = 25
centre = (−4, 0), radius = 5
c x2 + y2 + 10x + 18y + 79 = 0
(x + 5)2 − 25 + (y + 9)2 − 81 + 79 = 0
(x + 5)2 + (y + 9)2 = 27
centre = (−5, −9), radius = 3 3
d x2 + y2 + 30x − 6y +
855
=0
4
(x + 15)2 − 225 + (y − 3)2 − 9 +
855
=0
4
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3
WORKED SOLUTIONS
81
(x + 15)2 + (y − 3)2 = 4
centre = (−15, 3), radius =
5
( x + 76 ) + ( y − 65 ) = 366 + ( 76 ) + ( 65 )
2
9
2
2
2
=
d2 = (x1 − x2)2 + (y1 − y2)2
r2 = (−4 − 1)2 + (7 − 3)2
2
6
49 25 80
+
+
=
36 36 36 36
80
2
Area = π r = 36 π
= (−5)2 + (4)2 = 41
The circle has the equation (x + 4)2 + (y − 7)2 = 41.
10 Smaller circle radius 3 units centre (7, 3) has
equation (x – 7)2 + (y – 3)2 = 9
6
(x − 9)2 + (y − 2)2 = 81
Larger circle radius 7 units centre (7, 3) has
equation (x – 7)2 + (y – 3)2 = 49
7
d2 = (x1 − x2)2 + (y1 − y2)2
Area of path = area of larger circle – area smaller circle
= (4 − 1)2 + (7 − 3)2
= 49π – 9π = 40π
= (3)2 + (4)2 = 25
The circle has the equation (x − 1)2 + (y − 3)2 = 25.
The locations on the x-axis will lie on the
circumference of the circle.
Exercise 3.3B
1
= (−17 − 27)2 + (25 − (−8))2
= (−44)2 + (33)2 = 3025
When y = 0, (x − 1)2 + (−3)2 = 25
d = 3025 = 55
(x − 1)2 + 9 = 25
(x − 1)2 = 16
2
x − 1 = ±4
The required coordinates are (5, 0) and (−3, 0).
Let the centre (8, 1) = (x1, y1) and D(2, 9) = (x2, y2).
( x + 12 p) − 14 p + (y + 3) − 9 = 96
( x + 12 p) + (y + 3) = 14 p + 9 + 96
2
2
d2 = (x1 − x2)2 + (y1 − y2)2
2
2
r2 = (8 − 2)2 + (1 − 9)2
= (6)2 + (−8)2 = 100
2
r = 100 = 10
1
Since r2 = 121, p 2 + 9 + 96 = 121
c The circle has the equation
(x − 8)2 + (y − 1)2 = 100.
4
1 2
p + 9 + 96 = 121
4
3
1 2
p = 16
4
2
p = 64
Since p is a positive constant, p = 64 = 8 .
The centre of the circle has the coordinates
1
− p, − 3 .
2
(
radius is 37 .
2
b (x + 0.5)2 + (y − 17)2 =
Since p = 8, the centre has the coordinates (−4, −3).
9
1
2
7
5
1
x2 + y2 + x − y =
3
3
6
2
x + 9 y + 11 
,
= (5, −3)
a i 
2 
 2
x=1
( x + 76 ) − ( 76 ) + ( y − 65 ) − ( 65 ) = 366
2
4
( 372 )
x + 9 = 2 × 5 = 10
3x2 + 3y2 + 7x – 5y =
2
a
The student has correctly worked out that
AB2 = 1369, but the equation of the circle needs
the square of the radius, not the square of the
diameter. The diameter is 1369 = 37, so the
)
This is a distance of 5 units from the origin.
( 2 +214 , 9 −2 7 ) = (8, 1)
b The radius is the distance between the
centre and a point on the circumference.
Rewrite the equation of the circle by completing
the square.
2
a The centre is the midpoint between D and E.
Centre =
x = 1 ± 4 = 5 or −3
8
d2 = (x1 − x2)2 + (y1 − y2)2
2
2
y + 11 = 2 × −3 = −6
y = −17
G = (1, −17)
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3 Coordinate geometry
ii d2 = (x1 − x2)2 + (y1 − y2)2
7
( )
= (4)2 + (8)2 = 80
r = 80 = 4 5
(−3 − 5)2 + (1 + 3)2 = (−8)2 + (4)2 = 80
( 52 − 32 ) + ( 32 − 12 ) = 1 + 1 = 2
3
1
Equation of circle is ( x − ) + ( y + ) = 2
2
2
Since both sides of the equation are
satisfied, H lies on C.
2
Substitute x = 2, y = − into the LHS of the
5
equation
iii (x − 5)2 + (y + 3)2 = 80
2
)
2
1
1
+
= 0.26 ≠ 2
4 100
lie on the circle.
1 + p 14 + q 
= (−2, 7).
,
2
2 
2
The point with coordinates x = 2, y = −
8
x2 – 10x + (x + 2)2 – 14(x + 2) + 73 = 0
p = −5
x2 – 10x + x2 + 4x + 4 – 14x – 28 + 73 = 0
14 + q = 2 × 7 = 14
2x2 – 20x + 49 = 0
x=
r2 = (−9 − (−2))2 + (4 − 7)2
The circle has the equation (x + 2)2 + (y − 7)2 = 58.
The equation of the circle can be rewritten as
(x + 3)2 + (y − 8)2 = 125, so (−3, 8) is the centre of the
circle.
2
does not
5
x2 – 10x + y2 – 14y + 73 = 0 meets y = x + 2 when
1 + p = 2 × −2 = −4
= (−7)2 + (−3)2 = 58
6
2
=
q=0
b
2
( x − 32 ) + ( y + 12 ) = (2 − 32 ) + (− 52 + 12 )
The midpoint of UV will also be the centre of
the circle, (−2, 7).
Hence 

2
2
a The centre of the circle is the mid-point of
−9 + 5 4 + 10
,
ST =
= (−2, 7).
2
2
(
2
=
b Substitute the coordinates of H into the
equation of the circle.
5
Centre of circle is the mid-point of AB and has
 1 5 1 3
3 −1
coordinates  2 + 2 2 − 2  i.e. ,
2 2
 2 , 2 
Radius is distance from centre to one of the points
r2 = (9 − 5)2 + (5 − (−3))2
20 ± 400 − 392 20 ± 8
2
=
=5±
4
4
2
x =5−
2
2
, y =7−
2
2
x =5+
2
2
, y =7+
2
2
A sketch diagram helps:
5 + √ 2 , 7 + √2
2
2
Substitute y = 2x + 14 into x2 + y2 + 6x − 16y = 52.
x2 + (2x + 14)2 + 6x − 16(2x + 14) = 52
Expand and simplify.
y =x+2
x2 + 4x2 + 56x + 196 + 6x − 32x − 224 = 52
2×
5x2 + 30x − 80 = 0
x2 + 6x − 16 = 0
5–
Solve.
√2 ,
√2
7–
2
2
2×
(x + 8)(x − 2) = 0
x = −8 or 2
Use Pythagoras:
M and N are the two intersection points, it doesn’t
matter which is which.
Diameter =
When x = −8, y = 2 × −8 + 14 = −2, so M = (−8, −2).
When x = 2, y = 2 × 2 + 14 = 18, so N = (2, 18).
−8 + 2 −2 + 18
,
The midpoint of MN =
= (−3, 8).
2
2
Since the midpoint of M and N is the centre of the
circle, MN is a diameter of the circle.
(
)
9
√2 =
√2
2
√2 =
√2
2
( 2 ) + ( 2 ) = 2 units.
2
2
A sketch diagram helps:
–5 , 3
2
5
1
3
17
+
=
4
5
20
3
5
+ =4
2
2
3 , –1
2 4
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3
WORKED SOLUTIONS
( )
17
20
Diameter
2
d2 = (x1 − x2)2 + (y1 − y2)2
+ 4 2 = 4.0893 ,
r2 = (3 − 11)2 + (2 − 8)2
= (−8)2 + (−6)2 = 100
1
1
Area of triangle = bh = × 360 × 40 = 60.
2
2
Area of circle = πr2 = π × 102 = 100π.
radius = 2.04466.
Area = πr2 = 13.1 square units (3 s. f.)
1

10 Centre of the circle has coordinates  3 + 2 −1 
 2 , 2


7 1
i.e. , −
6 2
2
2
1
7
Distance of centre from the origin is
+
2
6
= 1.27 units (3 s.f.)
(
)
() ()
Green area = (100π − 60) m2
c 100π − 60 = 254.16 m2
254.16 ÷ 5 = 50.8, so 51 tins required.
51 × 4 = $204
3
Exercise 3.4A
1
Since AB is a diameter, ACB = 90° and the lengths
of the sides satisfy Pythagoras’ theorem.
AC 2 + BC 2 = AB2
(4x − 9)2 + (14 − x)2 = (2x + 5)2
a d2 = (x1 − x2)2 + (y1 − y2)2
DE 2 = (−1 − 1)2 + (11 − 5)2
16x2 − 72x + 81 + 196 − 28x + x2 = 4x2 + 20x + 25
= (−2)2 + (6)2 = 40
13x2 − 120x + 252 = 0
(13x − 42)(x − 6) = 0
42
x=
or 6
13
EF 2 = (1 − 13)2 + (5 − 9)2
= (−12)2 + (−4)2 = 160
DF 2 = (−1 − 13)2 + (11 − 9)2
When x =
= (−14)2 + (2)2 = 200
Since 40 + 160 = 200, DE 2 + EF 2 = DF 2.
1 149 149
×
=
2 13
26
When x = 6, d = 2 × 6 + 5 = 17
Radius = 1 × 17 = 17
2
2
Radius =
b Gradient of DE = 11 − 5 = −3.
−1 − 1
Gradient of EF =
9−5 1
= .
13 − 1 3
Since m1 × m2 = −1, DE and EF are
perpendicular.
c DF is a diameter of the circle.
Centre =
( −1 2+ 13 ,112+ 9 ) = (6, 10).
For the radius.
(1 + 2)2 + (0 − 1)2 = r2
(3)2 + (−1)2 = r2
r2 = 10
b Substitute x = −3 and y = q into (x + 2)2
+ (y − 1)2 = 10
(q − 1)2 = 9
q − 1 = ±3
The equation of the circle is
(x − 6)2 + ( y − 10)2 = 50.
a
Proof using Pythagoras’ theorem (could
compare gradients).
PQ 2 = (−5 − 11)2 + (−4 − 8)2 = (−16)2 + (−12)2 = 400
PR2 = (−5 − 13)2 + (−4 − 2)2 = (−18)2 + (−6)2 = 360
QR2 = (11 − 13)2 + (8 − 2)2 = (−2)2 + (6)2 = 40
Since PR2 + QR2 = PQ2, Pythagoras’ theorem is
satisfied and the triangle PQR is right-angled.
)
−5 + 11 −4 + 8
,
= (3, 2).
b Centre =
2
2
For the radius.
a Substitute x = 1 and y = 0 into (x + 2)2 + (y − 1)2 = r2
(−1)2 + (q − 1)2 = 10
= (5)2 + (5)2 = 50
(
4
(−3 + 2)2 + (q − 1)2 = 10
d2 = (x1 − x2)2 + (y1 − y2)2
r2 = (6 − 1)2 + (10 − 5)2
2
42
42
149
,d = 2×
+5=
13
13
13
q=1±3
q = −2 or 4
Since q > 0, q = 4
c Gradient of AB =
2−0
1
=− .
−5 − 1
3
Gradient of AC =
2−4
= 1.
−5 − (−3)
Gradient of BC =
0−4
= −1 .
1 − (−3)
d m1 × m2 = −1 so AC and BC are perpendicular.
Since the angle in a semicircle is a right
angle, AB is a diameter of the circle.
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3 Coordinate geometry
5
6 −1
1
a Gradient of L1 = 5 − (−10) = 3.
Gradient of L2 =
For the radius.
d2 = (x1 − x2)2 + (y1 − y2)2
4 − (−2)
= −3.
9 − 11
r2 = (8 − 1)2 + (4 − 5)2
= (7)2 + (−1)2 = 50
m1 × m2 = −1 so L1 and L2 are perpendicular.
The equation of C is given by
(x − 8)2 + (y − 4)2 = 50.
1
b Equation of L1 is given by y – 6 = (x − 5),
3
b JL2 = JK 2 + KL2
1
from which y = (x − 5)+ 6.
3
JL2 = (−11 − 7)2 + (6 − 22)2 = (−18)2 + (−16)2 = 580
JK 2 = (−11 − (−1))2 + (6 − (p − 8))2 = (−10)2 +
(14 − p)2 = p2 − 28p + 296
Equation of L2 is given by y − 4 = −3(x − 9),
from which y = −3(x − 9) + 4.
KL2 = (−1 − 7)2 + ((p − 8) − 22)2 = (−8)2 +
(p − 30)2 = p2 − 60p + 964
At W, 1 (x − 5) + 6 = −3(x − 9) + 4.
3
Hence 580 = p2 − 28p + 296 + p2 − 60p + 964.
x − 5 + 18 = −9(x − 9) + 12
0 = 2p2 − 88p + 680
x + 13 = −9x + 81 + 12
= p2 − 44p + 340
10x = 80
= (p − 10)(p − 34)
x=8
p = 10 or 34
When x = 8, y = −3(8 − 9) + 4 = 7.
Because L1 and L2 are perpendicular and
intersect at point W on the circumference, SV is
a diameter.
Centre =
7
y1 − y 2 2 − 1 1
=
=
x1 − x 2 6 + 1 7
23 − 13
y1 − y 2 2 − 5
=
= 5
26
x1 − x 2
6− 4
5
5
13 5
1
=− ×
=−
5 26
2
23 − 1 18
y − y2
18 5
Gradient BC = 1
= 5
= 5 =
× =2
x1 − x 2
4 +1
9
5 9
5
5
AC is perpendicular to BC (product of gradients
is −1).
Gradient AC =
( 5 +211 , 6 −2 2 ) = (8, 2).
For the radius.
d2 = (x1 − x2)2 + (y1 − y2)2
r2 = (5 − 8)2 + (6 − 2)2
= (−3)2 + (4)2 = 25
The equation of C is given by
(x − 8)2 + (y − 2)2 = 25.
∴ AB is the diameter.
Expanding and simplifying:
x2 + y2 − 16x − 4y + 43 = 0.
6
Gradient of AB =
Mid-point of AB has coordinates
i.e. (2.5, 1.5)
a PR2 = PQ2 + QR2
Length AB =
PR2 = (1 − (t + 17))2 + ((t + 7) − 3)2
= (−16 − t)2 + (t + 4)2 = 2t2 + 40t + 272
( 6 + 1 )2 + ( 2 − 1 )2 =
Radius of circle =
PQ 2 = ( 1 − 3)2 + ((t + 7) − (t + 11))2
= (−2)2 + (−4)2 = 20
( 6 2− 1 , 2 2+ 1 )
49 + 1 = 50
50
2
Equation of circle is
QR2 = (3 − (t + 17))2 + ((t + 11) − 3)2
= (−14 − t)2 + (t + 8)2 = 2t2 +44t + 260
(x – 2.5)2 + (y – 1.5)2 =
50
4
Hence 2t2 + 40t + 272 = 2t2 + 44t + 260 + 20.
x2 – 5x + 6.25 + y2 – 3y + 2.25 – 12.5 = 0
−8 = 4t
x2 + y2 – 5x – 3y – 4 = 0
t = −2
Hence the coordinates of P and R are (1, 5)
and (15, 3).
1 + 15 5 + 3
Centre of circle =
,
= (8, 4).
2
2
(
8
If PR is diameter and Q lies on the circle then PQ
and QR must be perpendicular to each other.
)
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3
WORKED SOLUTIONS
Gradient PQ =
Exercise 3.4B
17 1
y1 − y 2 6 − 3
=
= 3 =1
x1 − x 2 −1 + 4 1
3 3
1
9
c Gradient of FG = − 1 .
4
13
× 1 ≠ −1 PQ is not perpendicular to QR
2
A sketch diagram helps.
x=– 1
2
y
d Equation is given by y − 1 = −
y = 3x + 7
y = 1 − 1 + 2 = 2 as required.
f For the radius.
d2 = (x1 − x2)2 + (y1 − y2)2
C
r2 = (9 − 4)2 + (5 − 2)2
= (5)2 + (3)2 = 34
x
B
The equation is given by (x − 4)2 + (y − 2)2 = 34.
3y = –x – 3
2
The diagram shows the intersection of the line
1
x = − with the two lines, these are labelled as
2
A and B. AB is the diameter of the circle. C, the
centre of the circle, is the mid-point of AB.
Mid-point =
( )
1
1
1
5
Coordinates of B are x = − , y = − × ( − ) − 1 = −
2
3
2
6
When y = 8, 8 −
(
)
x = 9
)
y1 − y 2 −3 + 19 16
=
=
=1
x1 − x 2 −1 + 17 16
y1 − y 2 −3 + 23 20
=
=
=5
x1 − x 2
−1 + 5
4
As none of the gradients multiply to give −1 none
of the lines are perpendicular to each other.
Gradient AC =
(
9 7
13
= x−
.
2 5
2
63 7
= x
5
5
10 If one of the line segments is a diameter then the
other two line segments must be perpendicular to
each other.
y − y 2 −23 + 19
4
1
Gradient AB = 1
=
=−
=−
x1 − x 2
−5 + 17
12
3
Gradient BC =
)
7 7
91
= x−
2 5
10
1 7
1
Equation of the circle with centre − , , r = 3 is
2 3
6
2
2
2
1
7
19
x+
+ y−
=
2
3
6
) ( )
( 3 +210 , 7 +2 2 ) = (132 , 92 ) .
(
1
1
Radius = AB = 3
6
2
) (
7−2
5
=− .
3 − 10
7
Equation of bisector is given by
y − 9 = 7 x − 13 .
2 5
2
1
11 5 38
+ =
=6
6
3
2 6
y-coordinate of C is 11 − 3 1 = 7
6 3
2
Gradient of JK =
7
Gradient of perpendicular bisector = .
5
1
1
11
Coordinates of A are x = − , y = 3 × − + 7 =
2
2
2
(
1
(x − 8)
4
from which 4y + x = 12
1
e When x = 4, y – 1 = − (4 − 8).
4
A
Distance AB =
( 9 +2 7 , 5 −2 3 ) = (8, 1)
5 − (−3)
b Gradient of DE = 9 − 7 = 4 .
17 + 3
26
y1 − y 2
3
3 = − 26 × 3 = − 13
=
=
Gradient QR = x − x
4
3
4
2
1
2
− −0 −4
3
3
As −
a
Centre of circle = (9, 8).
For the radius.
2 = (x1 − x2)2 + (y1 − y2)2
d
r 2 = (9 − 3)2 + (8 − 7)2
= (6)2 + (1)2 = 37
The equation is given by (x − 9)2 + (y − 8)2 = 37.
3
10 + x 11 + y 
,
= (6, 3).
a Mid-point = 
2 
 2
10 + x = 2 × 6
x=2
11 + y = 2 × 3
y = −5
Coordinates of V are (2, −5).
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3 Coordinate geometry
11 − 3
b Gradient of UV = 10 − 6 = 2 .
d The quadrilateral MCNX has ∠XMC = ∠XNC
= 90° because MX and NX are perpendicular
bisectors. ∠MXN = 90° from c. Hence ∠ABC
= ∠MBN = 360° − (3 × 90°) = 90°.
Gradient of perpendicular bisector = − 1 .
2
1
Equation of bisector is given by y − 3 = − (x − 6).
2
1
When x = 8, y − 3 = − (8 − 6).
2
y=2
Centre of circle = (8, 2).
e AC is a diameter.
6
A(12, 8), B(11, 1) and C(20, 4)
12 + 11 8 + 1
23 9
Mid-point of AB =
,
=
, .
2
2
2 2
(
8 −1
= 7.
12 − 11
1
Gradient of perpendicular bisector = − .
7
9
1
23
.
Equation of bisector is given by y − = − x −
2
7
2
1
43
y=− x+
7
7
12 + 20 8 + 4
,
= (16, 6).
Mid-point of AC =
2
2
Gradient of AB =
For the radius.
d2 = (x1 − x2)2 + (y1 − y2)2
(
r2 = (8 − 10)2 + (2 − 11)2
= (−2)2 + (−9)2 = 85
The equation is given by (x − 8)2 + (y − 2)2 = 85.
4
(
a
Equation of circle can be rewritten as
(x + 3)2 + (y + 5)2 = 185.
Gradient of AC =
Hence centre of circle = (−3, −5).
b When y = 6, (x + 3)2 + (6 + 5)2 = 185.
(x + 3)2 = 64
x + 3 = ±8
x = 5 or −11
Length of chord = 5 − (−11) = 16.
c Height of triangle = 6 − (−5) = 11.
Base of triangle = PQ = 16.
Area of triangle = 1 bh = 1 × 16 × 11 = 88 .
2
2
(
)
3x + 2y = 7
8−4
1
=− .
12 − 20
2
= (15 − 12)2 + (4 − 8)2
r2 = (3)2 + (−4)2 = 25
Final answer is x2 – 30x + 225 + y2 – 8y + 16 = 25
7−3
2
= .
Gradient of AB =
2 − (−4) 3
3
Equation of bisector is given by y − 5 = − (x + 1).
2
2y − 10 = −3x − 3
)
d2 = (x1 − x2)2 + (y1 − y2)2
a Mid-point of AB = −4 + 2 , 3 + 7 = (−1, 5).
2
2
3
Gradient of perpendicular bisector = − .
2
x2 + y2 – 30x – 8y + 216 = 0
7
a (x + 6)2 + (y – 6)2 = 145 (x – 8)2 + (y + 1)2 = 40
Multiply out the brackets and collect terms.
x2 + 12x + 36 + y2 – 12y + 36 = 145
x2 + 12x + y2 – 12y = 73 A
( 2 +210 , 7 −2 5 ) = (6, 1).
x2 – 16x + 64 + y2 + 2y + 1 = 40
7 − (−5)
3
=− .
2 − 10
2
Find A – B
2
Gradient of perpendicular bisector = .
3
2
Equation of bisector is given by y − 1 = (x − 6).
3
3y − 3 = 2x − 12
2x – y = 7
2x = 3y + 9
x2 – 16x + 4x2 – 28x + 49 + 4x – 14 = – 25
b Mid-point of BC =
Gradient of BC =
)
Gradient of perpendicular bisector = 2.
Equation of bisector is given by y − 6 = 2(x − 16).
y = 2x − 26
1
43
For point of intersection, − x +
= 2x − 26 .
7
7
225 15
=
x
7
7
x = 15
When x = 15, y = 2 × 15 − 26 = 4.
Centre of circle = (15, 4).
For the radius:
(x + 3)2 + 121 = 185
5
) ( )
x2 – 16x + y2 + 2y = – 25 B
28x – 14y = 98
y = 2x – 7
Substitute this into equation B.
x2 – 16x + (2x – 7)2 + 2(2x – 7) = – 25
c 90° because gradients are perpendicular.
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3
WORKED SOLUTIONS
5x2 – 40x + 60 = 0
Mid-point of AC =
x2 – 8x + 12 = 0
(x – 6) (x – 2) = 0
Gradient of AC =
x = 2, y = – 3 i.e. (2, –3)
8
Equation of bisector is given by
3
23
.
y − = −3 x −
2
2
(
4 2 + 8 2 = 16 + 64 = 80 = 4 5
A(3, 8), B(1, 4) and C(−6, 5)
Mid-point of AB =
Gradient of AB =
( 3 2+ 1 , 8 +2 4 ) = (2, 6).
1
For point of intersection, − x + 11 = −3x + 36.
2
5
x = 25
2
1
Gradient of perpendicular bisector = − .
2
x = 10
When x = 10, y = −3 × 10 + 36 = 6.
1
Equation of bisector is given by y − 6 = − (x − 2).
2
1
y =− x+7
2
3−6 8+5
3 13
Mid-point of AC = 2 , 2 = − 2 , 2 .
) (
Centre of circle = (10, 6).
For the radius:
)
d2 = (x1 − x2)2 + (y1 − y2)2
r2 = (10 − 17)2 + (6 − 0)2
8−5
1
= .
Gradient of AC =
3 − (−6) 3
= (−7)2 + (6)2 = 85
Gradient of perpendicular bisector = −3.
Equation of bisector is given by
13
3
y−
= −3  x − −  .
2
2 

( )
y = −3x + 2
For point of intersection, − 1 x + 7 = −3x + 2.
2
5
x = −5
2
x = −2
When x = −2, y = −3 × −2 + 2 = 8.
Centre of circle = (−2, 8).
For the radius:
d2 = (x1 − x2)2 + (y1 − y2)2
r2 = (−2 − 1)2 + (8 − 4)2
= (−3)2 + (4)2 = 25
As all three points lie on a circle with centre (−2, 8)
with radius 5 none of the line segments could be a
radius, but all three line segments are chords.
9 A(19, 4), B(17, 0) and C(4, −1)
(
The equation is given by (x − 10)2 + (y − 6)2 = 85.
Exercise 3.4C
1
2
a 22 + (−2 − 1)2 = 4 + 9 = 13
b Centre of circle = (0, 1).
Gradient of radius =
1 − (−2)
3
=− .
0−2
2
2
.
3
The equation of the tangent is given by
2
y + 2 = (x − 2).
3
3y + 6 = 2x − 4
Gradient of the tangent =
)
4−0
= 2.
19 − 17
1
Gradient of perpendicular bisector = − .
2
Equation of bisector is given by
1
y − 2 = − (x − 18).
2
1
y = − x + 11
2
Gradient of radius = 6 − 8 = 2 .
4−7 3
3
Gradient of the tangent = − .
2
The equation of the tangent is given by
y − 8 = − 3 (x − 7).
2
2y − 16 = −3x + 21
3x + 2y = 37
Since both sides of the equation agree, T lies
on the circle.
19 + 17 4 + 0
Mid-point of AB =
,
= (18, 2).
2
2
Gradient of AB =
)
y = −3x + 36
8−4
= 2.
3−1
(
4 − (−1) 1
= .
19 − 4
3
Gradient of perpendicular bisector = −3.
x = 6, y = 5 i.e. (6, 5)
b Length =
(192+ 4 , 4 2− 1 ) = ( 232 , 32 ).
Hence 2x = 3y + 10.
3
a Centre = (3, 6).
b Radius = 8.
Let centre = X, point P = (15, 7) and point
where tangent meets circle = T.
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3 Coordinate geometry
b AXBQ is a kite. Split into congruent triangles
AXQ and BXQ.
c The length of PT is required, where
PT 2 + XT 2 = PX 2.
For PX:
Base of triangle AXQ = 11 −
Perpendicular height of triangle AXQ = 3.
PX 2 = (−12)2 + (−1)2 = 145
Area of triangle AXQ =
PT 2 + XT 2 = PX 2
Area of AXBQ = 2 ×
PT 2 + 82 = 145
PT 2 = 81
5
PT = 81 = 9 m
4
19 25
=
.
4
4
d2 = (x1 − x2)2 + (y1 − y2)2
PX 2 = (3 − 15)2 + (6 − 7)2
1 25
75
×
×3=
.
2 4
8
75 75
.
=
8
4
Rewrite the equation of the circle as
(x + 4)2 + (y − 1)2 = 81.
Centre = (−4, 1); radius = 9.
a For A:
Gradient of radius =
Let centre = X and point where tangent meets
circle = T.
11 − 7 4
= .
7−4 3
PT 2 + XT 2 = PX 2.
3
Gradient of the tangent = − .
4
For PX:
The equation of the tangent is given by
3
y − 7 = − (x − 4).
4
Hence y = − 3 (x − 4) + 7.
4
For B:
Gradient of radius =
PX2 = (7 − (−4))2 + (4 − 1)2
= (11)2 + (3)2 = 130
PT 2 + XT 2 = PX 2
PT 2 + 92 = 130
11 − 7
4
=− .
7 − 10
3
Gradient of the tangent =
PT 2 = 49
PT = 49 = 7
3
.
4
6
The equation of the tangent is given by
3
y − 7 = (x − 10).
4
Hence y =
d2 = (x1 − x2)2 + (y1 − y2)2
a
Rewrite the equation of the circle as
(x − 15)2 + (y − 23)2 = 400.
Centre = (15, 23), radius = 20.
Gradient of radius =
3
(x − 10) + 7.
4
4
Gradient of the tangent = − .
3
Put tangents equal.
3
3
− (x − 4) + 7 = (x − 10) + 7
4
4
The equation of the tangent is given by
4
y − 35 = − (x − 31).
3
3y − 105 = −4x + 124
3
3
− (x − 4) = (x − 10)
4
4
−(x − 4) = x − 10
4x + 3y = 229
−x + 4 = x − 10
When x = 10, y = q.
14 = 2x
40 + 3q = 229
x = 7 (this could also be deduced by
symmetry since AB is horizontal)
When x = 7, y =
3
(7 − 10) + 7.
4
3q = 189
q = 63
b Use d2 = (x1 − x2)2 + (y1 − y2)2
For PT:
3
y = (−3) + 7
4
= 19
4
35 − 23 3
= .
31 − 15 4
PT 2 = (10 − 31)2 + (63 − 35)2
= (−21)2 + (28)2 = 1225
( )
19
Hence the coordinates of Q are 7, 4 .
For PX:
PX 2 = (10 − 15)2 + (63 − 23)2
= (−5)2 + (40)2 = 1625
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3
WORKED SOLUTIONS
TX is the radius, so TX 2 = 400.
3y – 69 = – x – 8
Hence PT 2 + TX 2 = PX 2 because 1225 + 400 = 1625.
3y + x – 61 = 0
This line meets the tangent when
c Since triangle is right-angled, area = 1 × TX × PT.
2
TX = 20
3(3x + 5) + x – 61 = 0
9x + 15 + x – 61 = 0
PT = 1225 = 35
Area =
7
10x – 46 = 0
1
× 20 × 35 = 350.
2
x = 4.6, y = 18.8
12.6 2 + 4.22 = 176.4
Gradient of tangent = 1.
Radius, r =
Gradient of radius = −1.
(x + 8)2 + (y – 23)2 = 176.4
quation of line with gradient −1 through
E
(5, 11) is given by y − 11 = −(x − 5).
10 Draw a sketch
T1
y = 16 − x
P
r
ind point of intersection of y = 16 − x and
F
y = x + 2.
X
16 − x = x + 2
r
14 = 2x
T2
x=7
Let the tangents from P meet the circle at T1 and T2
and the centre of the circle be X.
y=7+2=9
ence the point (7, 9) lies on the
H
circumference.
PXT1 is a right angled triangle (tangent meets
radius at 90˚)
For the radius:
PXT2 is a right angled triangle (tangent meets
radius at 90˚)
2 = (x1 − x2)2 + (y1 − y2)2
d
r2 = (7 − 5)2 + (9 − 11)2
PX is common to both triangles and is the
hypotenuse in each triangle.
= (2)2 + (−2)2 = 8
he equation of the circle is given by
T
(x − 5)2 + (y − 11)2 = 8.
8
a Gradient of CT =
In triangle PXT1, by Pythagoras’ theorem
y1 − y 2
4 + 6 10
5
=
=
=−
x1 − x 2 −3 − 1 −4
2
PT1 = PX 2 − r 2
In triangle PXT2, by Pythagoras’ theorem
Gradient of the tangent at T = 2
5
2
Equation of tangent is y + 6 = (x − 1)
5
5y + 30 = 2x – 2
PT2 =
– 2x + 5y + 32 = 0
so PT = 145 − 16 = 129 for both T1 and T2.
b Equation of diameter
5
Gradient − through the point (1, – 6)
2
5
y + 6 = − (x − 1)
2
2y + 12 = – 5x + 5
2y + 5x + 7 = 0
9
T1X = T2X = radius of the circle.
Gradient of tangent is 3, so gradient of radius to
1
point of contact is − .
3
Equation of this radius is
1
y − 23 = − (x + 8)
3
PX 2 − r 2
Therefore PT1 = PT2
In this question, r = 4, PX = 145 ,
Exercise 3.4D
1
a Substitute y = x − 10 into x2 + y2 = 50.
x2 + (x − 10)2 = 50
x2 + x2 − 20x + 100 = 50
2x2 − 20x + 50 = 0
x2 − 10x + 25 = 0
(x − 5)2 = 0
The equation has repeated roots so y = x − 10
is a tangent of the circle.
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3 Coordinate geometry
b Substitute x = 7y − 50 into x2 + y2 = 50.
(1 + m2)x2 + (28m − 36)x + 468 = 0
b2 − 4ac = 0
(28m − 36)2 − 4 × (1 + m2) × 468 = 0
784m2 − 2016m + 1296 − 1872 − 1872m2 = 0
−1088m2 − 2016m − 576 = 0
34m2 + 63m + 18 = 0
(17m + 6)(2m + 3) = 0
6
3
m=−
or −
17
2
6
For the tangent y = − x + 24.
17
17y = −6x + 408
6x + 17y − 408 = 0
For the tangent y = − 3 x + 24.
2
2y = − 3x + 48
3x + 2y − 48 = 0
(7y − 50)2 + y2 = 50
49y2 − 700y + 2500 + y2 = 50
50y2 − 700y + 2450 = 0
y2 − 14y + 49 = 0
(y − 7)2 = 0
The equation has repeated roots so 7y = x + 50
is a tangent of the circle.
2
Substitute y = 4x − 5 into x2 + y2 + 10x − 18y + 38 = 0.
x2 + (4x − 5)2 + 10x − 18(4x − 5) + 38 = 0
x2 + 16x2 − 40x + 25 + 10x − 72x + 90 + 38 = 0
17x2 − 102x + 153 = 0
x2 − 6x + 9 = 0
(x − 3)2 = 0
The equation has repeated roots so y = 4x − 5 is
a tangent of the circle.
3
Substitute y = 3x + c into x2 + y2 − 10x − 6y − 6 = 0.
6
x2 + (3x – 2)2 + 10x – 18(3x – 2) + 38 = 0
x2 + (3x + c)2 − 10x − 6(3x + c) − 6 = 0
x2 + 9x2 + 6cx + c2 − 10x − 18x − 6c − 6 = 0
10x2 + (6c − 28)x + (c2 − 6c − 6) = 0
b2 − 4ac = 0
2
2
(6c − 28) − 4 × 10 × (c − 6c − 6) = 0
36c2 − 336c + 784 − 40c2 + 240c + 240 = 0
−4c2 − 96c + 1024 = 0
c2 + 24c − 256 = 0
(c − 8)(c + 32) = 0
c = 8 or −32
4
x2 + 9x2 – 12x + 4 + 10x – 54x + 36 + 38 = 0
10x2 – 56x + 78 = 0
5x2 – 28x + 39 = 0
b2 – 4ac = 784 – 780 = 4
7
1
Substitute y = − x + c into
2
As b2 – 4ac > 0 there are two intersections of the
line with the circle, ∴ the line is not a tangent to
the circle.
1
x2 + y2 – 8x – 4y – 6 = 0 has 2 tangents y = − x + c
2
Substitute the equation of the tangent into the circle.
(
x2 + y2 − 12x − 26y + 125 = 0.
1
x2 + − x + c
2
1
x2 + ( − 1 x + c)2 − 12x − 26(− x + c) + 125 = 0
2
2
x2 +
x2 +
1 2
x − cx + c2 − 12x + 13x − 26c + 125 = 0
4
2
1 2
x – cx + c2 – 8x + 2x – 4c – 6 = 0
4
For the line to be a tangent b2 – 4ac = 0
5
b2 – 4ac = (c + 6)2 – 4 × × (c2 – 4c –6) = c2 + 12c + 36
4
– 5c2 + 20c + 30
5x2 + (4 − 4c)x + (4c2 − 104c + 500) = 0
b2 − 4ac = 0
2
2
(4 − 4c) − 4 × 5 × (4c − 104c + 500) = 0
16 − 32c +16c2 − 80c2 + 2080c – 10 000 = 0
−64c2 + 2048c − 9984 = 0
c2 − 32c + 156 = 0
(c − 6)(c − 26) = 0
c = 6 or 26
1
1
The tangents are y = − x + 6 and y = − x + 26.
2
2
Substitute y = mx + 24 into (x − 18)2 + (y − 10)2 = 52.
(x − 18)2 + (mx + 24 − 10)2 = 52
(x − 18)2 + (mx + 14)2 = 52
x2 − 36x + 324 + m2x2 + 28mx + 196 = 52
) − 8x − 4(− 12 x + c ) − 6 = 0
5 2
x – (c + 6)x + c2 – 4c – 6 = 0
4
5 2
x + (1 − c)x + (c2 − 26c + 125) = 0
4
5
Substitute y = 3x – 2 into x2 + y2 + 10x – 18y + 38 = 0.
= –4c2 + 32c + 66
–4c2 + 32c + 66 = 0
−b ± b 2 − 4ac −32 ± 1024 + 1056
=
2a
−8
−32 ± 45.607
=
−8
c=
The values of c are –1.70 and 9.70.
8
x2 + y2 – 16x – 22y + 115 = 0 or (x – 8)2 + (y – 11)2 = 70
Circle has centre (8, 11).
1
so gradient of the
3
diameter between the points of contact is –3.
Gradient of tangent(s) is
46
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3
WORKED SOLUTIONS
b
Equation of diameter:
y
y – 11 = –3(x – 8), so y = – 3x + 35
10
Substitute this in the equation of the circle:
8
(x – 8)2 + (–3x + 35 –11)2 = 70
6
Multiply out the brackets and simplify:
4
10x2 – 160x + 570 = 0
2
x2 – 16x + 57 = 0
–2 –1 0
–2
x = 5.35, y = 18.9
–4
x = 10.6, y = 3.06
–6
For the tangent through x = 5.35, y = 18.9
1
y = x + 17.2
3
(x – 7)2 + (y + 4)2 = 57; tangents have gradient –2,
and circle has centre (7, –4).
5
6
7
8 x
2
Roots at x = 1 and x = 3 so factors are (x − 1) and
(x − 3).
(x − 1)(x − 3) = x2 − 4x + 3, so the equation is
y = x2 − 4x + 3
3
a
y
9
8
7
1
(x – 7)
2
6
5
y = 0.5x – 7.5
4
Diameter meets the circle when
3
(x – 7)2 + (0.5x – 7.5 + 4)2 = 57
2
x2 – 14x + 49 + 0.25x2 – 3.5x + 12.25 – 57 = 0
1
1.25x2 – 17.5x + 4.25 = 0
–1
0
1
2
3
x
1
2
3
x
2
−b ± b − 4ac 17.5 ± 306.25 − 21.25
=
2a
2.5
17.5 ± 16.882
=
2.5
b
y
3
x = 13.8, y = –0.624, the point is (13.8, –0.624)
2
x = 0.246, y = – 7.38, the point is (0.246, –7.38)
1
Exercise 3.5A
–1
1
4
(4, –9)
Equation of diameter
x=
3
–8
Gradient of the diameter joining the points of
1
intersection is .
2
y+4=
2
–10
For the tangent through x = 10.6, y = 3.06
1
y = x – 0.0486
3
9
1
a
–2
y
–3
10
–4
8
–5
6
4
4
2
0
–7 –6 –5 –4 –3 –2 –1
–2
(–3, –4)
–4
–6
0
–1
1
2
x
3 , –9
2 2
The sketch is not correct.
x2 − 4x − 21 = (x − 7)(x + 3)
So should cut x-axis at (−3, 0) and (7, 0).
a > 0 so graph should be U-shaped.
When x = 0, y = −21 so y-intercept at (0, −21).
x2 − 4x − 21 = (x − 2)2 − 25
So turning point at (2, −25).
47
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3 Coordinate geometry
5
a
8
y
Red graph is of y = – 2x2 – 9x + 7.
24
Blue graph is of y = 2x2 – 9x + 7.
y
20
20
16
12
15
8
4
–1
10
0
1
2
3
4
5
6
7
8
x
5
b
y
–5 , 51
4 8
–10
0
–5
5
3
9
2
a
y
4
1
–3
–2
0
–1
–1
x
–5
4
–4
5
2
1
2
(–6.416, 0)
x
–8
–6
–4
(0.416, 0)
0
–2
–2
2
4
6
(0, –1.333)
8
x
–4
6
a 4m
c Roots at t = −1 and t = 3 so factors are (t + 1)
and (t − 3). ∩ – t2
–(t + 1)(t − 3) = –t2 + 2t + 3, the equation is
h = –t2 + 2t + 3
7
–6
(–3, –5.833)
b 3s
(
y = 3x 2 − 7x + 11 = 3 x −
(
7
6
=3 x−
7
6
(–2.416, 0)
(–6.416, 0)
4
2
(0.416, 0)
–8
2
–6
–4
(4.416, 0)
8 x
0
–2
2
4
6
–2 (0, –1.333)
(–1, –3.883)
–4
) + 1283
2
(–3, –5.833)
–6
(0, –5.333)
(1, –5.833)
10 Let the equation of the curve be y = a(x – b)2 + c
12
Minimum point is at
10
(
8
y=a x−
6
4
5
2
( 52 , 72 ) so b = 52 , c = 72
) + 72
2
When x = 0, y = 16
2
0
y
) − 1249 + 11
y
–2
b
2
4
x
( 52 ) + 72
16 = a ×
Intersects the y-axis at y = 11.
Turning point is  7 , 83  .
 6 12 
2
16 = a 0 −
25 7
+
4 2
a = 12.5 ×
(
4
=2
25
∴y=2 x−
5
2
) + 72
2
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3
WORKED SOLUTIONS
Exercise 3.5B
1
13
21
c x+ 4 =−x+
4
a x2 + x − 2 = x + 2
2x = 2
x2 − 4 = 0
x = 1, y =
x = ± 2 so two points of intersection.
b x2 + x − 2 = x + k
5
2x2 + 5x – 3 = x + k
2x2 + 4x – (3 + k) = 0
b2 – 4ac = 16 – 4.2. – (3 + k) = 0
k=−5
Tangent: y = x − 5
Point where curve meets tangent:
2x2 + 5x – 3 = x − 5
x2 + 2x + 1 = 0
(x + 1)2 = 0
x = − 1, y = −6
Equation of perpendicular:
y + 6 = −1(x + 1)
y = −x − 7
Points where curve meets perpendicular to
tangent:
2x2 + 5x – 3 = − x − 7
2x2 + 6x + 4 = 0
x2 + 3x + 2 = 0
(x + 2)(x + 1) = 0
Intersect at x = − 1, y = − 6
and x = − 2, y = − 5
6
a For y = x + k: 3x2 – 2x + 5 = x + k
x2 – 2 – k = 0
x2 – (2 + k) = 0
b2 − 4ac = 0 – 4.1. – (2 + k) = 8 + 4k
i intersects: b2 – 4ac > 0
8 + 4k > 0
k > −2
ii is a tangent to: b2 − 4ac = 0
8 + 4k = 0
k = −2
iii does not touch: b2 – 4ac < 0
8 + 4k < 0
k < −2
2
x2 + x − 12 = x − 13
x2 = −1
No real solutions to this equation consequently the
line and the curve do not intersect.
3
a x2 − 2x + 5 = 5 − x
x2 − x = 0
3x2 – 3x + (5 – k) = 0
b2 − 4ac = 1 − 0 = 1 > 0 so the curve and the
line intersect
b x2 − 2x + 5 = x +
5
2
5
x2 − 3x + = 0
2
4.1.5
= −1 so the curve and the
2
line do not intersect
b2 − 4ac = 9 −
c x2 − 2x + 5 = x + 3
x2 – 3x + 2 = 0
b2 – 4ac = 9 − 4.1.2 = 1 so the curve and the
line intersect
4
a
−x2 + 2x + 3 = x + k
−x2 + x + 3 – k = 0
b2 – 4ac = 1 – 4. – 1(3 – k) = 0
13
k=
4
b −x2 + 2x + 3 = −x + k
−x2 + 3x + 3 – k = 0
b2 – 4ac = 9 – 4. – 1(3 – k) = 0
k=
21
4
( )
17
, the point is 1, 17
4
4
b2 – 4ac = 9 – 4.3.(5 – k) = −51 + 12k
i intersects: b2 – 4ac > 0
−51 + 12k > 0
k>
51
12
ii is a tangent to: b2 – 4ac = 0
−51 + 12k = 0
k=
51
12
iii does not touch: b2 – 4ac < 0
−51 + 12k < 0
k<
51
12
For y = –x + k: 3x2 – 2x + 5 = −x + k
3x2 – x + (5 – k) = 0
b2 – 4ac = 1 – 4.3.(5 – k) = −59 + 12k
i intersects: b2 – 4ac > 0
−59 + 12k > 0
k>
59
12
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3 Coordinate geometry
ii is a tangent to: b2 – 4ac = 0
8
−59 + 12k = 0
k=
59
12
x=
7
3x 2 −
0 = 4x 2 −
( 343 ) − 4 × 4 × ( 949 − k )
b 2 − 4ac = −
8
12
=
1
3
3
x − (7 + m) = 0
2
For the line to be a tangent b2 – 4ac = 0
k=
9
2y – 6x – 31 = 0, so y = 3x + 15.5
– 2x2 – 7x + 3 = 3x + 15.5
– 2x2 – 10x – 12.5 = 0
2x2 + 10x +12.5 = 0
A tangent when b2 – 4ac = 0
1
x − 7 meets the line y = – x + n when
2
b2 – 4ac = 100 – 4 × 2 × 12.5 = 100 – 100 = 0
1
x − 7 = −x + n
2
The line is a tangent to the curve.
10 y = 2x2 – 7x + 11 meets the line y = –2x + k
1
x − (7 + n) = 0
2
when 2x2 – 7x + 11 = – 2x + k
2x2 – 5x + 11 – k = 0
For the line to be a tangent b 2 − 4ac = 0
Tangent when b2 – 4ac = 0
2
1
1
b − 4ac =
+ 4 × 3 × (7 + n) = + 84 + 12n
4
2
b2 – 4ac = 25 – 4 × 2 × (11 – k) = – 63 + 8k = 0
= 84.25 + 12n = 0
337
n=−
48
k=
The tangents meet when x − 115 = −x − 337
16
48
337 115
1
−2x =
−
=−
48
16
6
63
8
To find the coordinates:
2x2 – 7x + 11 = –2x + 63
8
25
2x2 – 5x +
=0
8
337
y = −x −
48
x=
29
12
Meets the curve y = – 2x2 – 7x + 3 when
115
y=x−
16
()
29
12
2
= 86.25 + 12m = 0
115
m=−
16
2
348
9
Line is not a tangent when k ≠
( 32 ) + 4 × 3 × (7 + m) = 94 + 84 + 12m
b 2 − 4ac = −
3x 2 +
348
+ 16k = 0
9
16k =
1
x −7= x +m
2
2
1156 1504
−
+ 16k
9
9
=−
1
x − 7 meets the line y = x + m when
2
3x 2 −
3x 2 −
34
94
x+
−k
3
9
Tangent when b2 – 4ac = 0
51
59
= −x +
12
12
y = 3x 2 −
2
4x 2 − 28 x + 49 + 5 = 2x + k
3
9
59
12
y = 3x 2 −
) + 5 when
2
b For line of symmetry:
2x =
7
3
(2x − 73 ) + 5 = 2x + k
−59 + 12k < 0
x+
(
y = 2x −
iii does not touch: b2 – 4ac < 0
k<
The line y = 2x + k meets the curve
16x2 – 40x + 25 = 0
(4x – 5)2 = 0
(
1
341
,y =−
. Point of intersection is 1 , − 341
12
48
12
48
)
x=
(
5
5 63 43
5 43
, y = −2 × +
=
. The point is ,
4
4 8
8
4 8
)
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3
WORKED SOLUTIONS
Exam-style questions
1
=
338
4
=
169
2
=
13 2
2
a m = 3, (x1, y1) = (1, 6)
y – y1 = m(x – x1)
y – 6 = 3(x – 1)
y – 6 = 3x − 3
y = 3x + 3
4
so 3x – y + 3 = 0
AB is a diameter, i.e. from A, through the centre
to B
a = 3, b = –1
14 – 9 = 5, and 4 – (–8) = 12
1
b m = − , (x1, y1) = (1, 6)
3
B has coordinates (9 – 5, –8 – 12), i.e. (4, –20)
y – y1 = m(x – x1)
b Although the coordinates of B have been
found it is better (safer) to use those given in
the question.
y − y 2 4 − (−8) 12
Gradient AB is = 1
=
=
.
x1 − x 2
14 − 9
5
1
y – 6 = − (x – 1)
3
1
1
y–6=− x +
3
3
1
19
y=− x+
3
3
2
Perpendicular bisector has gradient −
1 y1 − y 2 k − 3
=
=
2 x1 − x 2 5 − 2
5
(x − 9)
12
12y + 96 = – 5x + 45
12y + 5x + 51 = 0
y+8=−
5
3
k − 3 = , k = 4.5
2
b Equation of BC:
Gradient
1
through (2, 3)
2
6
2y = x + 4
c Length AB = 4 2 + 22 = 20 = 2 5
3
5
2
Ratio AB : BC is 2 : 1.5 i.e. 4 : 3
3
a m = –1, (x1, y1) = (0, 5)
y – y1 = m(x – x1)
y – 5 = –1(x – 0)
b D(0, 2) and E(0, 12)
Mid-point (0, 7)
y = –x + 5, or f(x) = –x + 5
b –x + 5 = x – 8
x = 13 , y = − 3
2
2
Point of intersection at 13 , − 3 .
2
2
13 3
, − and (0, 5)
c
2
2
(
d=
(
)
(132 ) + ( −213 )
2
2
)
x2 + 6x + y2 – 14y + 22 = 0
(x + 3) 2 – 9 + (y – 7) 2 – 49 + 22 = 0
(x + 3)2 + (y – 7)2 = 36
The circle has centre (–3, 7) and radius 6.
a On the x-axis, y = 0.
(x + 5)2 + (−7)2 = 50
(x + 5)2 = 1
x + 5 = ±1
x = −5 ± 1 = −6 or −4
Coordinates for A and B are (−6, 0) and (−4, 0).
On the y-axis, x = 0
(5)2 + (y − 7)2 = 50
(y − 7)2 = 25
y − 7 = ±5
y = 7 ± 5 = 2 or 12
Coordinates for D and E are (0, 2) and (0, 12).
1
y − 3 = ( x − 2)
2
Length BC = 32 + 1.52 = 11.25 =
5
and
12
passes through (9, –8).
1
19
so m = − , c =
3
3
y − y2 7 − 3
a Gradient of AB = 1
=
= −2
x1 − x 2 0 − 2
Gradient BC =
a Centre of the circle is (9, –8), A(14, 4).
Perpendicular bisector of AB has equation y = 7.
7
x2 – 5x + 4 = x + k
x2 – 6x + (4 – k) = 0
b2 – 4ac = 36 – 4.1.(4 – k) = 20 + 4k
does not touch: b2 – 4ac < 0
20 + 4k < 0
k < −5
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3 Coordinate geometry
8
a Circle has centre (1, 3) and radius 5.
It meets the line x = 4 when
32 + (y – 3)2 = 25
(y – 3)2 = 16
y=3±4
The points of intersection are (4, 7) and (4, –1).
b Gradient AC =
y1 − y 2 7 − 3 4
=
= .
x1 − x 2 4 − 1 3
3
Gradient of tangent = − .
4
Equation of tangent:
3
y − 7 = − (x − 4)
4
4y – 28 = – 3x + 12
y1 − y 2 3 − (−1)
4
=
=
x1 − x 2
1−4
−3
Gradient of tangent = 3 .
4
Equation of tangent:
3
y + 1 = ( x − 4)
4
4y + 4 = 3x – 12
4y = 3x – 16
c The tangents meet at D.
– 3x + 40 = 3x – 16
6x = 56
3x = 28
28
x=
,y=3
3
28
25
CD = 3 − 1 = 3
1
d Area triangle = base × height
2
25
CD =
units
3
1
Area = × 4 × 25 = 50 square units.
3
2
3
9
y – 5 = 1(x – 8)
y=x–3
b Gradient of y = f(x) y1 − y 2 = a − 5 = 5 − a = 1,
x1 − x 2 −12 − 8
20
∴ a = – 15
y = g(x) has gradient = –1,
y + 15 = –1(x – 0)
y + x + 15 = 0, y = –x – 15
11 a d2 = (x1 − x2)2 + (y1 − y2)2
r2 = (180 − 60)2 + (40 − 130)2
= (120)2 + (−90)2 = 22 500
Radius = 22500 = 150 cm
4y = – 3x + 40
Gradient BC =
10 a y = f(x) has gradient 1 and passes through (8, 5)
a m = – 1 , (x1, y1) = (12, 4)
2
y – y1 = m(x – x1)
1
y – 4 = – (x – 12)
2
1
=– x+6
2
1
1
y = – x + 10, or f(x) = – x + 10
2
2
b When x = 0, y = 10 (0, 10).
When y = 0, x = 20 (20, 0).
1
1
Area of triangle = bh = × 20 × 10 = 100
2
2
units2.
b The circle has the equation
(x − 60)2 + (y − 130)2 = 22 500.
3
130 − 40
=− .
c Gradient of radius =
60 − 180
4
4
.
3
The equation of the tangent is given by
4
y − 40 = (x − 180).
3
3y − 120 = 4x − 720
Gradient of the tangent =
4x − 3y − 600 = 0
d d2 = (x1 − x2)2 + (y1 − y2)2
r2 = (5940 − 180)2 + (4360 − 40)2
= (5760)2 + (4320)2 = 51 840 000
Radius = 51840000 = 7200 cm = 72 m
12 Mid-point of AB =
( 9 +213 ,102− 2 ) = (11, 4).
For the radius, d2 = (x1 − x2)2 + (y1 − y2)2.
r2 = (11 − 9)2 + (4 − 10)2
= (2)2 + (−6)2 = 40
The circle has the equation
(x − 11)2 + (y − 4)2 = 40.
10 − 4
= −3.
Gradient of radius =
9 − 11
1
Gradient of the tangent = .
3
The equation of the tangent is given by
1
y − 10 = (x − 9).
3
3y − 30 = x − 9
x − 3y + 21 = 0
13 Centre of circle, C is (–9, 4)
Radius = 49 = 7
AC2 = (8 – ( – 9)) 2 + (21 – 4) 2 = 17 2 + 172 = 578
52
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3
WORKED SOLUTIONS
AB2 = AD2 = AC2 – r2 = 578 – 49 = 529
y–3=
AB = AD = 23
Area = r × AB = 7 × 23 = 161 square units
y=
14 a −x2 + 6x + 5 = x + k
−x2 + 5x + 5 – k = 0
(x1, y1) = (−2, 8)
(x2, y2) = (4, 5)
y − y1
m= 2
x 2 − x1
45
4
−x2 + 6x + 5 = −x + l
−x2 + 7x + 5 – l = 0
=
b2 – 4ac = 49 – 4 – 1(5 – l) = 0
Line BC
45
69
= −x +
4
4
(x1, y1) = (−2, 8)
2x = 6
x = 3, y =
(x2, y2) = (−10, −8)
( )
57
57
, the point is 3,
.
4
4
b
y
–2
y 2 − y1
x 2 − x1
=
−8 − 8
−10 + 2
m2 = 2
15
m1 m2 = −
10
perpendicular to each other and ABC is a
right-angled triangle.
0
–1
–5
1
2
3
4
–10
–15
–20
15 (x1, y1) = (2 2, 2)
(x2, y2) = ( 2, 2 3)
m=
m=
20
5
–3
5−8
4+2
m1 = − 1
2
l = 69
4
x+
6 +1
x+3
5
16 a Line AB
b2 – 4ac = 25 – 4 – 1(5 – k) = 0
k=
2
( x − 0)
2 3− 2
y 2 − y1
x 2 − x1
= 2 3− 2
2−2 2
= 2 3− 2
− 2
Gradient perpendicular to this:
2
m=
2 3− 2
(x1, y1) = (0, 3)
y – y1 = m(x – x1)
5
6
7
8
x
( 12 )(2) = −1 so AB and BC are
OR
Length of AB:
d2 = (4 − −2)2 + (5 − 8)2
= 45
Length of BC:
d2 = (−2 − −10)2 + (8 −−8)2
= 320
Length of AC:
d2 = (4 − −10)2 + (5 −−8)2
= 365
AB 2 + BC 2 = AC 2
bCentre of circle has coordinates
(
ie −3, −
3
2
Radius =
)
AC
=
2
365
2
( 4 −210 , 5 −2 8 )
( ) = 3654
Circle has equation ( x + 3) + y + 3
2
2
2
17 x + 4y – 8 = 0
x = 8 – 4y
3(8 – 4y) + 5y + 15 = 0
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3 Coordinate geometry
2x2 + 16x − 18 = 0
39
Solving gives y = 7 .
Substituting this value back into the original
100
equation gives x = −
.
7
(
)
100 39
,
.
Point of intersection (x1, y1) = −
7 7
3x + 5y + 15 = 0
Rearranging gives m1 = − 3 .
5
5
So m2 = .
3
y – y1 = m(x – x1)
39 5
100
y−
= x−−
7
3
7
(
)
Rearranging gives
y=
5
617
x+
3
21
18 x2 + y2 – 8x + 10y + 5 = 0,
(x – 4)2 + (y + 5)2 – 16 – 25 + 5 = 0
(x – 4)2 + (y + 5)2 = 36
Centre (4, −5), radius 6.
After translation centre has coordinates (1, 2) and
radius 6 units.
19 a r2 = (31 − 7)2 + (12 − 2)2
= (24)2 + (10)2 = 676
Equation is given by (x − 7)2 + (y − 2)2 = 676.
b Mid-point of XA =
Gradient of XA =
( 7 +231 , 2 +212 ) = (19, 7).
12 − 2
5
.
=
31 − 7 12
12
.
5
The equation of the perpendicular bisector is
12
(x − 19), i.e. 12x + 5y = 263.
given by y − 7 = −
5
c Let P equal the midpoint of XA.
Gradient of the perpendicular = −
XM = XA = AM because XM and XA are radii
and since MN is the bisector of XA it is a line of
symmetry for triangle MAX (hence XM = AM).
MP 2 = MX 2 − XP 2
= 262 − 132 = 507
MP = 507 = 13 3
MN = 2 ×13 3 = 26 3
20 a (x + 4)2 + (y − 11)2 = 50
b y=7−x
(x + 4)2 + (7 − x − 11)2 = 50
(x + 4)2 + (−4 − x)2 = 50
x2 + 8x + 16 + x2 + 8x + 16 = 50
x2 + 8x − 9 = 0
(x + 9)(x − 1) = 0
x = −9 or 1
When x = −9, y = 7 − (−9) = 16.
When x = 1, y = 7 − 1 = 6.
Since the x-coordinate of A is less than the
x-coordinate of B, A is (−9, 16) and B is (1, 6).
c Tangent at A:
Gradient of radius =
16 − 11
= −1.
−9 − ( −4 )
Gradient of the tangent = 1.
(Note that since the centre of the circle
lies on x + y = 7, AB is a diameter, so the
tangent is perpendicular to x + y = 7.)
The equation of the tangent is given by
y − 16 = 1(x + 9).
Hence y = x + 25.
Tangent at P :
Gradient of radius =
18 − 11
= 7.
−3 − ( −4 )
1
Gradient of the tangent = − .
7
The equation of the tangent is given by
1
y − 18 = − (x + 3).
7
1
Hence y = − (x + 3) + 18.
7
1
From which, x + 25 = − (x + 3) + 18.
7
1
3
x + 25 = − x − + 18
7
7
7x + 175 = −x − 3 + 126
8x = −52
x=−
13
2
13
37
When x = − 13 , y = −
+ 25 =
.
2
2
2
( 132 , 372 ).
Coordinates of T = −
21 a Equation is given by (x − 6)2 + (y − 3)2 = 9.
The centre is (6, 3). The radius is 3.
Method one:
Similar triangles:
XP TX
=
.
OP OA
54
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3
WORKED SOLUTIONS
0 = 92m2 − 120m + 28
Let length of XP = a.
a
62 + (3 + a )
2
=
0 = 23m2 − 30m + 7
3
6
b (23m − 7)(m − 1) = 0
7
23
2a = 36 + ( 3 + a )2
m = 1 or
4a2 = 36 + (3 + a)2
When m = 1, the tangent equation is given
by y = x.
= 36 + 9 + 6a + a2
Substitute y = x into the circle equation.
3a2 − 6a − 45 = 0
(x − 5)2 + (x − 3)2 = 2
a2 − 2a − 15 = 0
2x2 − 16x + 32 = 0
(a − 5)(a + 3) = 0
x2 − 8x + 16 = 0
a = 5 (can’t be −3)
(x − 4)2 = 0
Method two:
x=4
Since y = mx at T, (x − 6)2 + (mx − 3)2 = 9.
y=4
so coordinates of point where tangent meets the
circle are (4, 4)
x2 − 12x + 36 + m2x2− 6mx + 9 = 9
(1 + m2)x2 + (−12 − 6m)x + 36 = 0
Since y = mx is a tangent, b2 − 4ac = 0.
Mathematics in life and work
(−12 − 6m)2 − 4 × (1 + m2) × 36 = 0
1
144 + 144m + 36m2 − 144 − 144m2 = 0
4 + 4m + m2 − 4 − 4m2 = 0
3m2 − 4m = 0
m(3m − 4) = 0
m = 0 or
Since m =
4
3
4
AP 4
, then
= .
OA 3
3
3 + XP 4
=
6
3
3(3 + XP) = 24
y=
2
3 + XP = 8
XP = 5
(x − 5)2 + (mx − 3)2 = 2
x2 − 10x + 25 + m2x2 − 6mx + 9 = 2
(1 + m2)x2 + (−10 − 6m)x + 32 = 0
3
x
2
44
2
= 14
3
3
The coordinates of the entrance to tunnel B are
2
14 , 12 .
3
x=
(
Area of triangle OAX = 1 × 6 × 3 = 9.
2
22 a Substitute y = mx into the equation of the circle C.
3
x − 10
2
The entrance to tunnels A and B are at y = 12.
3
12 = x − 10
2
22 =
b Quadrilateral OAXT is made from two
congruent right-angled triangles each with a
base of 6 and a height of 3.
Area of OAXT = 2 × 9 = 18.
The equation for tunnel A is:
3
y − 12 = (x − 8)
2
2y − 24 = 3x − 24
3
y= x
2
Tunnel B is 10 vertically below, so the y-intercept
will be −10. Therefore, an equation for tunnel B is:
3
)
2
The gradient of tunnel C is − .
3
The mid-point of OP is
( 8 +2 0 , 122+ 0 ) = (4, 6).
Therefore, an equation for tunnel C is:
Given that y = mx is a tangent to the circle C,
b2 − 4ac = 0:
y − 6 = − 2 (x − 4)
3
3y − 18 = −2x + 8
(−10 − 6m)2− 4 ×(1 + m2) × 32 = 0
2x + 3y − 26 = 0
100 + 120m + 36m2 − 128 − 128m2 = 0
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3 Coordinate geometry
4
Rearranging the equations for tunnels B and C, we
can get:
Tunnel B:
9x − 6y − 60 = 0 (1)
Tunnel C:
4x + 6y − 52 = 0 (2)
(1) + (2)
13x − 112 = 0
13x = 112
112
x=
13
Substituting back to get y:
3 112
y= ×
− 10
2 13
y=
38
13
The point of intersection would be
38
.
,
(112
13 13 )
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4
WORKED SOLUTIONS
4 Circular measure and 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
30 sin 35° (or 30 cos 55°) = 17.2 cm
2
tan C =
b 150 × π = 5π
180 6
c 270 × π = 3π
180 2
8
6
d 315 × π = 7π
180 4
e 720 × π = 4π
180
8
= 53.1°
6
Alternatively, sin A = 0.8 or cos C = 0.6
C = tan−1
3
4
Factorise: (2x − 1)(x + 1) = 0
1
Either 2x − 1 = 0 → x =
2
or x + 1 = 0 → x = −1
1
x = or −1
2
a and b
y
3
d 3.5 × 180 = 630°
e 0.3 × 180 = 54°
y = x2 – 4
2
0
–5
–2
a 180 ÷ 5 = 36°
b 180 + 30 = 210°
13
× 180 = 292.5°
c
8
5
4
a 0.644
5
CB
= sin 0.85
42
CB = 42 × sin 0.85
= 31.6 cm
3.7
= tan 1.05
b
XZ
3.7
XZ =
tan1.05
x
y = (0.5x)2 – 4
a
= 2.12 cm
–4
6
a
iii 6.5 × 4 = 26 cm
b
Exercise 4.1A
2
a 30 × π = π
180 6
i 27 + 2 × 15 = 57 cm
ii 64.8 + 2 × 24 = 112.8 cm
a 0.5 × 180 = 28.6°
π
180
b 1.2 ×
= 68.8°
π
c 0.1 × 180 = 5.7°
π
d 4 × 180 = 229.2°
π
e 5.5 × 180 = 315.1°
π
i 15 × 1.8 = 27 cm
ii 24 × 2.7 = 64.8 cm
b This is a stretch of y = x2 – 4 from the y-axis
with a factor of 2.
1
b 0.454 c 0.0500 d 0.809 e 2.41
iii 26 + 2 × 6.5 = 39 cm
c
7
8
1
× 152 × 1.8 = 202.5 cm2
2
1
ii × 242 × 2.7 = 777.6 cm2
2
1
iii × 6.52 × 4 = 84.5 cm2
2
i
1
× 202 × 0.6 = 1140 cm2 to 3 s.f.
2
π
a 2 × 152 × 1.6 – 3π × 102 × 1.6 = 100 cm2
4
b 10 × 1.6 + 15 × 1.6 + 2 × 5 = 50 cm
π × 202 −
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4 Circular measure and trigonometry
9
1 2
r θ = 500 and rθ = 20
2
16 a C
D = CE; EBC is an isosceles
right-angled triangle so EB = r;
CE = CB 2 + EB 2 = 2r 2 = 2r = CD
π 3π
b Angle ECD = π − =
4
4
π
c AE = × 10 = 15.71; DE = 3π × 2 × 10 = 33.32;
2
4
AD = 20 + 2 × 10 = 34.14; the perimeter is
15.71 + 33.32 + 34.14 = 83.2 cm to 3 s.f.
1
d The area of triangle EBC = × 10 × 10 = 50
2
Area of shape = 1 × π × 10 2 + 50 + 1 × 3π × 200 =
2 4
2 4
39.27 + 50 + 235.62 = 325 cm² to 3 s.f.
1
r × 20 = 500
2
r = 50
50θ = 20
θ = 0.4
10 a Let r be the radius in metres.
1
Then 100rθ = r 2θ
2
200rθ = r2θ
r = 200 m
bSubstituting back into the first equation gives
θ = θ, so θ can be any angle from 0 to 2π
11 The area of sector OCD is 1 × 6 2 × 2 = 36 cm²
2
The area of sector OAB is 1 × (6 + x)2 × 2
2
= 36 + 12x + x 2 cm²
Exercise 4.2A
1
a 0.5
d −0.5
b −0.5
e 0.5
c −0.5
2
a −0.966
d −0.966
b 0.966
e 0.966
c 0.966
3
a 0.940
d 0.940
b 0.940
e −0.940
c 0.940
4
a
Hence area of ABDC = OAB – OCD = 12x + x2 = 64
Rearrange: x2 + 12x – 64 = 0; (x – 4)(x + 16) = 0
The positive solution is x = 4
12 Shaded area = area of sector – area of triangle
= 1 × 6.4 2 × 1.2 − 1 × 6.4 2 sin1.2 = 5.49 cm² to 3 s.f.
2
2
13 First find the angle θ.
iisin−1 (−0.35) = −20.5° which is out of the
range 0° ⩽ x ⩽ 360°
The solutions are 180° + 20.5° = 200.5° and
360° − 20.5° = 339.5°.
Area of sector = 1 × 20 2 × θ = 200θ ; 200θ = 164;
2
164
θ=
= 0.82
200
1
Area of triangle = × 20 2 × sin 0.82 = 146 cm²
2
to 3 s.f.
iii sin−1 (−0.812) = −54.3°
The solutions are 180° + 54.3° = 234.3°
and 360° − 54.3° = 305.7°.
iv No solution because −1 ⩽ sin x ⩽ 1.
14 a The angle of the sector is 2 radians.
The area of the segment is
1
1
× r 2 × 2 − × r 2 × sin 2 = 24
2
2
Hence r2 (2 – sin 2) = 48 and
48
r2 =
= 44.01 and r = 6.63 cm to 3 s.f.
2 − sin 2
b From triangle OAB, AB = 2r sin 1 = 11.16 cm
b
icos−1 (−0.25) = 1.82 and the other solution is
2π – 1.82 = 4.46
ii cos−1 0.1 = 1.47 and 2π – 1.47 = 4.81
iii π and 3π
2
2
iv π is the only solution.
5
a
The perimeter = 11.16 + 2 × 6.63 = 24.4 cm to 3 s.f.
15 a CB = 12 × 0.8 = 9.6 cm
1
b Area of sector = × 0.8 × 122 = 57.6;
2
DB = 12 tan 0.8 = 12.36
1
area of triangle = × 12 × 12.36 = 74.1
2
area of BCD = 74.1 – 57.6 = 16.5 cm2
12
c CD = cos 0.8 − 12 = 5.22;
perimeter = 5.22 + 9.6 + 12.36
= 27.2 cm to 3 s.f.
i sin−1 0.95 = 71.8° and 180° − 71.8° = 108.2°
i77.3°, 360° − 77.3° = 282.7°, 360° + 77.3° =
437.3°, 720° − 77.3° = 642.7°
ii180° + 12.7° = 192.7°, 360° − 12.7° = 347.3°,
540° + 12.7° = 552.7°, 720° − 12.7° = 707.3°
iii 0°, 180°, 360°, 540°, 720°
iv 90°, 450°
b
i −2π, 0, 2π
π 5π 7π 11π
ii − , − , ,
6 6 6 6
iii −5.49, −0.795, 0.795, −5.49
5π
iv 3π ,
2 4
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WORKED SOLUTIONS
6
x = −315°, −135°, 45°, 225°
7
 90
a For example  
0 
8
12 a If θ = sin–1 x = cos–1 0.5 then θ = 60°. This is the
principal value.
Hence x = sin 60° = 0.866 to 3 d.p.
b Rotation of 180° about (−45, 0)
π
aπ = 60° so one solution is 30° or . The other is
6
3
π 5π
π− =
6
6
b sin θ = − cos 7π ; cos 7π = cos 2π − 7π
4
4
4
(
)
= cos π = sin π
4
4
b If θ = sin–1 x = cos–1 x then sin θ = cos θ = x
If sin θ = cos θ, then θ = 45° and
x = sin 45° = 0.707 to 3 d.p.
Exercise 4.3A
1
5π
The equation is sin θ = − sin π so θ =
or 7π
4
4
4
9
11π
4θ = π or 5π or 7π or 3 or …
3
3
3
2
5π
π
or 12 or 7π or 11π . There are four
12
12
12
solutions in the interval.
Hence θ =
10 a and b
y
−0.5
= −0.577
0.866
sin θ
sin θ
b
tan θ =
→ sin θ =
→ −4.121 =
−0.236
cosθ
−4.121 × −0.236 = 0.973
a tan θ =
a cos2 x = 1 – sin2 x
= 1 – 0.36
= 0.64
cos x = ± 0.8
sin x
b tan x =
cos x
=
0.6
±0.8
= ± 0.75
2
0
0.5π
π
1.5π
2π x
3
305° − 180° = 125° and 305° − 360° = −55°
4
a tan−1 0.05 = 0.050 and π + 0.050 = 3.19
b tan−1 (−0.5) = –0.464 which is outside the interval.
–2
The solutions are π – 0.464 = 2.68 and
2π – 0.464 = 5.82.
–4
c tan−1 5 = 1.373 and π + 1.373 = 4.52
c4 (sin x)2 = 1; (sin x ) =
2
x = π , 5π , 7π or 11π
6 6 6
6
11
1
1
1
; sin x =
or − ;
4
2
2
π
5
π
5
0
sin π
5
1
π
5
–1
The diagram shows that one possible solution is
θ = π − π = 3π .
2 5 10
The solutions are π – 1.551 = 1.59 and
2π – 1.551 = 4.73.
5
1
–1
d tan−1 (−50) = –1.551 which is outside the interval.
Another solution is θ = 2π − 3π = 17π .
10
10
6
7
sin x
a
= tan x = 4 → x = 76.0° and
cos x
180° + 76.0° = 256.0°
sin x −3
→ tan x = −0.6
b 5 sin x = − 3 cos x →
=
cos x
5
tan−1 (−0.6) = −0.540 so the solutions are
π – 0.540 = 2.60 and 2π – 0.540 = 5.74
 180
a For example, 
.
 0 
b For example, rotation of 180° about the
origin or rotation of 180° about (180, 0)
sin θ
a tan θ =
cos θ
For 0° < θ < 90°, cos θ < 1
sin θ
= tan θ > sin θ
so
cos θ
59
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4 Circular measure and trigonometry
12 Write t for tan θ to get t (t – 2) = 3; rearrange as
t2 – 2t – 3 = 0.
Factorise: (t – 3) (t + 1) = 0; tan θ = 3 or –1; hence
θ = 1.25 or 2.36.
y
b
1
Exercise 4.4 A
0.5
–90
–45
0
45
90
q
1
a
1
2
3
2
b sin 120° = sin 60° =
–0.5
c sin 225° = −sin 45° = −
–1
8
1
; 3θ = 18.4° or 198.4° or 378.4° or…
3
Hence θ = 6.1° or 66.1° or 126.1° or 186.1° …
The only obtuse angle is 126.1°
a tan 3θ =
b 4θ + 30° = 78.7° or 258.7 ° or 438.7° or 618.7°
or …
Hence θ = 12.2° or 57.2° or 102.2° or 147.2° or …
The two obtuse solutions are 102.2° and
147.2°.
9
d
( )
sin x
cos x
2
= 4; tan x = ± 2
e cos 135° = −cos 45° = −
2
3
11 a x2 – 2x + 1 = 0; (x – 1)(x – 1) = 0; x = 1 is the only
solution.
b
3π = − tan π = −1
c tan
4
4
3
2
4
a
5
a θ=
6
a
Now ((tan θ)2 – 1)2 = 0 so (tan θ)2 = 1.
b −1
2
π 2π
,
3 3
b θ=
π 7π
,
6 6
2
2
2
2


=  1  +  − 3  +  − 1 
 2

 2 
2
b sin
=
7
c − 3
(sin 34π ) + (sin 43π ) + (cos 54π )
2
=
2
1 3 1
+ + = 3
2 4 2 14
π
2π
4π
5π
− sin 2π
− sin
+ sin π − sin
+ sin
3
3
3
3

3 
3
3
3
+ −
−
+ (0) −  −
−0 =0
2
2
 2   2 
a tan 210° + tan 240° = tan30° + tan60°
=
1
3
4 3
+ 3=
+ 3=
3
3
3
b cos30° + cos45° + cos60° =
This factorises to (tan θ – 1)2 = 0 so tan θ = 1
and so θ = 45° or 225°.
c
a sin 3π = sin π = 1
4
4
2
b cos 3π = − cos π = − 1
4
4
2
Possible values are y = 12.5° or 132.5° or
192.5° or 312.5° or 372.5°.
The second equation has no solution. Hence
z = 0° or 180° or 360° are the only solutions.
f cos 300° = cos 60° = 1
2
1
a tan 30° =
3
c tan 225° = tan 45° = 1
Either 2y + 35° = 60° or 300° or 420° or 660° or
780° or …
c In this case sin z (cos z – 2) = 0 and hence
either sin z = 0 or cos z = 2.
1
2
b tan120° = − tan 60° = − 3
If tan x = 2, then x = 1.11 or 1.11 − π = −2.03
If tan x = – 2, then x = −1.11 or −1.11 + π = 2.03.
sin x
10 a2 sin x = tan x ⇒ 2sin x =
cos x
⇒ 2 sin x cos x – sin x ⇒ sin x (2 cos x – 1) = 0
Either sin x = 0 or cos x = 1 . Hence x = 0°,
2
180° 360°, 60° or 300°
b Either sin (2y + 35°) = 0 or cos ( 2y + 35° ) = 1 .
2
Either 2y + 35° = 0° or 180° or 360° or 540°
or 720°… so y = 72.5° or 162.5° or 252.5° or
342.5° are the values in the interval given.
3
2
1
2
=
(
3
1
1
+
+
2
2 2
3
2 1 1
+
+ = 1+ 2 + 3
2
2
2 2
)
tan θ = 1 or −1 and so θ = 45°, 135°, 225° or 315°.
60
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WORKED SOLUTIONS
b
1+ 3
2 = 2 1 + 3 
c sin 30° + sin 60° = 2
 2
sin 45°
1
2 
2
1
1
2+ 6
=
2+ 2× 3 =
2
2
(
8
)
(
y
10
)
0
5π
9π
π
2θ + 3π = π or 4 or 4 or … ⇒ 2θ = − or 7π
8
4
8
8
0.5π
x
–10
π
7π
15π
or 15π or … ⇒ θ = −
16 or 16 or 16 or …
8
c period = 2π = 4π; amplitude = 2
0.5
The two solutions in the interval are 7π or 15π .
16
16
9
d
y
a
BD = a sin 60° = 3a; AB = a cos60° = ;
2
2
1
a
a
=
BC = AB tan 30° = ×
2
3 2 3
Hence CD = BD – BC = 3a − a
2
2 3
6
4




= a  3 − 1  = a  3 − 1 = a × 2 = a
2
3 2 3  2
3
3
2
10 a In triangle ABC, angle BAC is 30°.
0
The length of AC = 2 × AB cos30°
5
3
= 3a .
2
2 × AB cos30° = 2a ×
π
2π
3π
4π
x
It is a sine wave of amplitude 4 and period 360
−45
translated by 
.
 0 
b The area of triangle ABC is
y
1
1
1
3 2.
× AB × AC sin 30° = a × 3a × =
a
2
2
2
4
4
The area of triangle ACE is
1
× AC × AE sin 60°
2
2
= 1 × 3a × 3a × 3 = 3 3 a 2 .
2
2
4
–180
–135
–90
–45
0
45
90
135
180 x
–2
The area of the kite is 3 a 2 + 3 3 a 2 =
4
4
2
3a .
–4
Exercise 4.5A
1
2
a y = 15 sin x
a y = −2 cos x
3
a period =
b
b y = tan 2x
b y = 10 + 5 sin x
360
= 180, amplitude = 4
2
6
a y = 2 + tan (x − 1)
b
y
4
2
y
0
–2
4
π
2π
3π
4π
5π
6π
x
–4
0
90
180
270
360 x
–4
4
a period = 2π = π , amplitude = 10
4
2
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4 Circular measure and trigonometry
7
Exercise 4.6A
a and b
y
y = 2 – 2cosx
y = 2 – 2cos0.5x
4
3
1
Calculator operation
2
a cos θ = ± 1 − 0.68 2 = ± 0.733 two possible values
b tan θ =
2
3
1
–360
–270
–180
–90
0
90
180
270
360
4
8
a Start with the graph of y = cos x.
Reflect it in the y-axis. The equation is
y = cos (–x) but the graph is symmetrical and
unchanged.
90
Translate by   so that it is now identical
 0 
to the graph of y = sin x.
The equation is y = cos (–(x – 90°)) or
y = cos (90° – x).
b cos (90° – x) = 2 cos x ⇒ sin x = 2 cos x ⇒ tan x = 2.
Hence x = 63.4° or 243.4°.
a sin θ = ± 1 − 0.44 2 = ± 0.898
b tan θ =
x
sin θ
0.68
=±
= ± 0.927
cosθ
0.733
sin θ
0.898
=±
± 2.041
cosθ
0.44
5
5 4
so cos2 x = 1 – sin2 x = 1 − = ;
9
9 9
2
cos x =
3
asin2 x =
The value is positive because x is acute.
b tan x =
5
6
7
sin x
5 2
5
=
÷ =
cos x
3
3
2
1
.
2
2
2
(cos x + 1)(cos x − 1) = cos x – 1 = −(1 – cos x) =
−sin2 x
1 – sin θ cos θ tan θ = 1 – sin θ × cos θ × sin θ
cosθ
= 1 – sin θ × sin θ
One is a reflection of the other in the line y =
= 1 – sin2 θ = cos2 θ
The amplitude is 30 so start with y = 30 sin x
1
1
sin x cos x sin 2 x + cos2 x
=
+
=
=
8 a tan x +
Stretch by 4 parallel to the x-axis to get the correct
cos x sin x
tan x cos x sin x
cos x sin x
1
1
1 sin x cos x sin 2 x + cos2 x
period. The equation is y = 30sin x .
tan x +
=
+
=
=
4
cos
sin x
tan
x
cos
x
sin
x
cos
x
sin
x
x
π
 
Finally translate by   .
1
1
2
2
 0
b tan x + tan x = sin x cos x = 2sin x cos x = sin 2x .
1
The equation becomes y = 30sin ( x − π ) or
4
The range of values for sin 2x is – 1 ⩽ sin 2x ⩽ 1.
1
π .
y = 30sin x −
Hence 1 1 or ⩽ – 1 and hence
4
4
sin 2x
π
Hence a = 30, b = 1 and c = − .
4
tan x + 1 2 or ⩽ −2.
4
tan x
Another possible value for c is 7π corresponding
9 (cos x + sin x)2 + (cos x – sin x)2
4
 −7π
to a final translation of 
.
= cos2 x + 2 cos x sin x + sin2 x + cos2 x – 2 cos x sin x +
 0 
sin2 x = 2(cos2 x + sin2 x) = 2
π
10 a The period is
and so b = 2.
2
10 sin2 x + cos2 x = 1 so(sin2 x + cos2 x)2 = 1
The graph crosses the y-axis at 0.4 and so c = 0.4.
Therefore sin4 x + 2 sin2 x cos2 x + cos4 x = 1 and so
The equation is y = a tan 2x + 0.4.
sin4 x + cos4 x = 1 – 2 sin2 x cos2 x.
Substitute the coordinates π , 0.5 ;
8
1
sin 2 x
cos2 x
π
≡
11 tan 2 x + 2 +
0.5 = a tan + 0.4.
2
2 +2+
4
tan x cos x
sin 2 x
Hence 0.5 = a + 0.4 and so a = 0.1.
sin 4 x + 2sin 2 x cos2 x + cos4 x
≡
b The curve will look the same but cross the
sin 2 x cos2 x
y-axis at (0, −0.4).
2
sin 2 x + cos2 x
1
The equation is y = 0.1 tan 2x – 0.4.
≡
≡
sin 2 x cos2 x
sin 2 x cos2 x
9
(
)
(
)
(
)
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4
WORKED SOLUTIONS
(
)(
sin 2 θ + cos2 θ sin 2 θ − cos2 θ
sin 4 θ − cos4 θ
≡
12
sin θ − cosθ
sin θ − cosθ
≡
)
2x = −0.775, −π + 0.775, π + 0.775, 2π – 0.775 …
π
x = –0.388, − π + 0.388, 2 + 0.388, π – 0.388 …
2
1 × (sin θ + cosθ )(sin θ − cosθ )
≡ sin θ + cos θ
sin θ − cosθ
13 cos2 A = 1 − sin 2 A = 1 − 9 = 16 ; hence
25 25
4
cos A = ±
5
25 144
2
cos B = 1 − sin 2 B = 1 −
=
; hence
169 169
12
cos A = ±
13
4 12 3 5
+ ×
Hence cos A cos B + sin A sin B = ± ×
5 13 5 13
The four solutions in the interval –π ⩽ x ⩽ π are
x = –0.388, x = –1.18, 1.96 and 2.75.
3
Either cos x = 0 → x = 90° or 270°
or 3 cos x – 2 = 0 → cos x =
1
sin 2 θ − 1
cos2 θ
=
=−
sin θ
sin θ
sin θ
cosθ −
1
cos2 θ − 1
sin 2 θ
=
=−
cosθ
cosθ
cosθ
tan θ +
1
sin θ cosθ sin 2 θ + cos2 θ
=
+
=
tan θ cosθ sin θ
cosθ sin θ
1
cosθ sin θ
cos2 θ  sin 2 θ 
1
The product is −
× −
×
sin θ  cosθ  cosθ sin θ
= cosθ sin θ = 1
cosθ sin θ
=
4
3
. One solution is 2x = π . Other
2
3
π
π
π
π
solutions are π − ,2π + ,3π − ,4π + , etc.
3
3
3
3
π
2
π
7
π
8
π
13
π
That is 2x = , , , ,
etc.
3 3 3 3 3
asin 2x =
So x = π , π , 7π and 4π . There are four solutions.
6 3 6
3
The next value, 13π is outside the interval.
6
b4 tan 0.5x = −3 and so tan 0.5x = −0.75
tan−1 (−0.75) = −36.9°.
Therefore 0.5x = –36.9° , –216.9° , 143.1°, etc.
Therefore x = –73.7° or 286.3°. These are the only
values in the interval −360° ⩽ x ⩽ 360°.
2
a cos (x + 75)° = 0.5 and cos−1 0.5 = 60°
So x + 75 = 60, 300, 420, 660, … and x = –15, 225,
345, 585, ….
The solutions in the interval 0° ⩽ x ⩽ 360° are
x = 225° or 345°.
bRearrange as 50 sin 2x = −35 and sin 2x = −0.7;
sin−1 (−0.7) = −0.775.
cos2 x = 0.25 → cos x = ±0.5
π
5π
or
3
3
2π
If cos x= −0.5, x =
or 4π . These are the four
3
3
solutions.
If cos x = 0.5, x =
5
(3 sin x − 1)(2 sin x − 1) = 0
1
→ x = 19.5°, 160.5°
3
Either 3 sin x – 1 = 0 → sin x =
or 2 sin x – 1 = 0 → sin x =
6
Exercise 4.7A
1
2
→ x = 48.2° or 311.8°.
3
There are four solutions.
63
33
48 15
+
= 65 or − 65
= ±
65 65
14 sin θ −
3 cos2 x – 2 cos x = 0 → cos x (3 cos x − 2) = 0
1
→ x = 30°, 150°.
2
4 sin2 x – 7 sin x – 2 = 0 → (4 sin x + 1) (sin x − 2) = 0
1
Either 4 sin x + 1 = 0 → sin x = − → x = –0.253
4
which is not in the range so x = π + 0.253 and
2π − 0.253 = 3.394and 6.03.
or sin x – 2 = 0 → sin x = 2.
This has no solution.
7
sin x + 1 = 1 – sin2 x → sin x + sin2 x = 0 →
sin x (1 + sin x) = 0
Either sin x = 0 → x = 0°, 180°, 360°
or 1 + sin x = 0 → sin x = −1 → x = 270°.
There are four solutions.
8
aDiameter = 67.5 – 67.5 cos180
= 67.5 – 67.5(−1)
= 135 m
b First solve 67.5 – 67.5 cos θ = 100
67.5 cos θ = −32.5
cos θ = −0.481
θ = 180 – 61.2, 180 + 61.2
= 118.8°, 241.2°
The wheel is above the ground for
122.4° out of 360°.
Length of time =
122.4
× 30 mins
360
= 10.2 mins
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4 Circular measure and trigonometry
9
aThe graphs cross twice between 0° and 360° so
there are 2 solutions.
sin x
b cos x = tan x → cos x =
cos x
Multiply by cos x: cos2 x = sin x → 1 − sin2 x = sin x.
Rearrange: sin2 x + sin x − 1 = 0
Use the quadratic formula.
sin x =
Exam-style questions
1
Area =
2
−1 ± 1 + 4
= 0.618 or −1.618
2
The smallest solution of sin x = 0.618 is 38.2°.
10 20(1 − sin2 x) + 27 sin x = 29
This can be factorised as (5 sin x − 3)(4 sin x − 3) = 0
3
b sin x = ±
11 8(1 − cos² x) = 2cos x + 5
(4cos x + 3)(2cos x – 1) = 0
4
12 cos x = 1 − cos² x + 0.3
cos² x + cos x – 1.3 = 0
−1 ± 1 + 5.2
= 0.745 or −1.745
2
cos2θ =
x = 0.730, 5.55
This is a quadratic in tan x and the
discriminant is (–2)2 –4 × 1 × k = 4 – 4k.
There is no solution if 4 – 4k < 0 ⇒ k > 1.
b tan2 x – 2 tan x – 8 = 0; (tan x – 4)(tan x + 2) = 0
tan x = 4 or –2; x = 76.0°, –104.0°, –63.4 or 116.6°
π
1
π
so 2θ =
or 5π ; θ =
or 5π
3
6
2
3
6
are the two solutions in the domain 0 ⩽ θ ⩽ π.
5
Hence x = 30°, 150°, 199.5° or 340.5°.
Hence y = 45°, 135°, 225° or 315°
14 a tan2 x + k = 2 tan x; tan2 x – 2 tan x + k = 0
a 2 sin2 x = 3 cos x; 2(1 – cos2 x ) = 3 cos x ; 2 cos2 x
+ 3 cos x – 2 = 0
1
(2 cos x – 1)(cos x + 2) = 0; cos x = or −2
2
1
cos x = −2 has no solution. If cos x = then
2
π.
x=
3
b Use the solution to part a
cos x = −1.745 has no solution.
b If you write sin2 y = sin x then you find
1
1
or −
sin 2 y =
3
2
1
2
sin y = − has no solution.
3
1
1
1
2
then sin y =
If sin y =
or −
2
2
2
sin x
13
3
13
=±
÷−
=±
.
cos x
4
4
3
This could be written as ± 39 .
3
x = −138.6°, 138.6°, −60°, 60°
13 a 6(1 – sin2 x) + sin x = 5 ⇒ 6 sin2 x – sin x – 1 = 0
1
1
⇒ (2 sin x – 1)(3 sin x + 1) = 0 ⇒ sin x = or −
3
2
13
4
Then tan x =
8cos²x + 2cos x – 3 = 0
cos x =
asin2 x + cos2 x = 1 therefore
2
solutions to 3 s.f. are x = 0.644, 2.50, 0.848 or 2.29.
3
1
or cos x =
4
2
aThe angle of the sector is 2π – θ so the area is
1 a2 (2π − θ) or πa2 − 1 a2 θ.
2
2

3
3 13
sin 2 x = 1 −  −
=1−
=
.
16 16
 4 
Hence sin x = 3 or sin x = 3
4
5
cos x = −
1
× 2.25 × 162 = 288 cm2.
2
b The area of the sector containing the
1
triangle is a 2θ and the area of the triangle
2
1 2
1
1
is a sin θ and so a 2θ = 2 × a 2 sin θ and
2
2
2
hence θ = 2 sin θ.
sin x = −1.618 has no solution.
Rearrange as 20 sin2 x − 27 sin x + 9 = 0.
If θ is the angle of the sector, 16 θ = 36 so θ = 2.25
aThe amplitude is a = 5; b = 7π as only the
4
positive value is required.
b Again the amplitude is c = 5 ; d =
5π
as only
4
the positive value is required.
6
a 2 sin2 x − sin x = 0
Factorise: sin x (2 sin x − 1) = 0.
Either sin x = 0 or 2 sin x − 1 = 0.
If sin x = 0 then x = 0 or π.
If 2 sin x − 1 = 0 then sin x = 1 so x = π or 5π .
2
6
6
There are 4 values for x.
θ
b Using the solution to part a, either sin 2 = 0
θ 1
or sin =
2 2
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4
WORKED SOLUTIONS
θ
θ
= 0 , then = 0 and θ = 0 is the only
2
2
solution in the domain 0 ⩽ θ ⩽ π.
θ 1
If sin = , then θ = π or 5π or … and
2 2
2 6
6
the only solution is θ = π .
3
1 2
7 Area of triangle = a sin 60° .
2
1
Area of sector = a 2θ so θ = sin 60° = 0.866.
2
The perimeter of the triangle is 3a and of the sector
is 0.866a + 2a = 2.866a.
The sector perimeter is 2.866 × 100 = 95.53% of the
3
triangle’s, about 4.5% less.
If sin
8
a2y + 40 = 30° or 180 − 30 = 150° or 360 + 30
= 390°, etc
2y + 40 = 30 → y = − 5° which is outside the
interval.
2y + 40 = 150 → y = 55°; 2y + 40 = 390 → y = 175°;
2y + 40 = 510 → y = 235°; 2y + 40 = 750
→ y = 355°. There are four solutions.
1
b sin ( 2y + 40° ) = ; cos2 (2y + 40°)
2
1 3
= 1 – sin2 (2y + 40°) = 1 − =
4 4
So cos ( 2y + 40° ) = ±
11 a Divide by cos θ: tan θ = 4 → θ = 76.0° or 76.0
− 180 = −104.0°.
b This is the equation in part a with 2x + 30° = θ.
So tan (2x + 30°) = 4; 2x + 30° = 76° or 256° or
−104° or −284°.
The solutions are x = 23° or 113° or −67° or −157°.
12 cos A = 6 = 0.6 ; A = 0.9273 and the reflex angle at
10
A is 2π – 0.9273.
The length of the major arc on the left is
6(2π – 0.9273) = 32.135.
cos B =
8
= 0.8 ; A = 0.6435 and the reflex angle at
10
B is 2π – 0.6435.
The length of the major arc on the right is
8(2π – 0.6435) = 45.117
The total length is 77.3 cm.
13 a
sin θ
= 2 sin θ → sin θ = 2 sin θ cos θ
cosθ
Rearrange: 2 sin θ cos θ − sin θ = 0.
sin θ (2cos θ − 1) = 0
Either sin θ = 0 so θ = 0, π or 2π
3
2
sin ( 2y + 40° ) 1
3
1
c tan ( 2y + 40° ) =
= ÷±
=±
2
cos ( 2y + 40° ) 2
3
9
b sin θ tan θ = 4 sin θ In the domain, sin θ ≠ 0 so
tan θ = 4 and θ = 1.33 to 3 s.f.
a cos (x + 30°) = − 0.8
π
or cosθ = 1 so θ =
or 5π .
3
3
2
The solutions are 0, π , π, 5π and 2π.
3
3
b Using the answer to part a, sin (4x – π) = 0 or
1
cos(4x − π ) =
2
If sin(4x – π) = 0 then 4x – π = 0, ± π, ± 2π,
± 3π …
x + 30° = 143.1° or 360 − 143.1 = 216.9°
x = 113.1° or 186.9°
b
y
So 4x = π, 0, 2π, −π, 3π, −2π, 4π, …
9
3π
Hence x = 0, π , π ,
, π, 5π , 3π , 7π or 2π.
4 2 4
4 2 4
These are the values in the given domain.
5
If cos(4x − π ) =
0
–60 –1
60
240
x
–5
10 a
1
1 − cos2 θ sin 2 θ sin θ
1
− cosθ =
=
=
×
cosθ
cosθ
cosθ
cosθ cosθ
1 − cos2 θ sin 2 θ
sin θ
− cosθ =
=
= sin θ ×
= sin θ tan θ
cosθ
cosθ =
cosθ
1
π π
then 4x – π = − , ,
3 3
2
5π , 7π , 11π , …
3 3
3
So 4x = 2π , 4π , 8π , 10π , 14π , …
3
3
3 3 3
π
π
2
π
5
π
7
π
and x = , ,
,
,
, 4π , 5π
6 3 3 6 6 3 3
= sin θ tan θ
or 11π .
6
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4 Circular measure and trigonometry
17 a The perimeter is 2r + r θ cm and hence
2r + r θ = 50.
50 − 2r
Rearrange: r θ = 50 – 2r; θ =
r
14 a
y
1
b The area is
0.5
(
)
1 2
r θ = 150 ; 1 r 2 50 − 2r = 150 ;
2
2
r
25r – r2 = 150; r2 – 25r + 150 = 0.
0
90
180
270
360
x
c (r – 10)(r – 15) = 0; Hence r = 10 or 15 and
1
50 − 2r = 50 − 20
50 − 30
= 3 or
=1 .
3
r
10
15
θ
θ 2r
= 2 ; = 1.107
18 a If angle AOB = θ, then tan =
2
2
r
and θ = 2.21 to 3 s.f.
1
b Area of sector AOB = r 2 × θ = 1.107r 2 ; area
2
of ACBO = r × 2r = 2r 2.
θ=
The graphs of y = cos x and y = tan x cross
twice between 0° and 360°.
b tan x = cos x therefore
sin x
= cos x
cos x
Therefore sin x = cos2 x therefore
sin x = 1 − sin2 x.
Rearrange: sin2 x + sin x − 1 = 0
Use the quadratic formula:
−1 ± 1 + 4 −1 ± 5
=
2
2
= 0.618 or −1.618
sinx = −1.618 has no solutions
x = 38.2° or 141.8°
1
= 5.
15 a tan x +
tan x
sin x =
Multiply by tan x: tan2 x + 1 = 5 tan x.
Rearrange: tan2 x − 5 tan x + 1 = 0.
b Use the quadratic formula.
5 ± 25 − 4 = 4.791 or 0.2087. Therefore
tan x =
2
x = 1.37 or 0.206.
1
c tan x +
= k ; tan2 x − k tan x + 1 = 0. This
tan x
only has a solution if k2 − 4 ⩾ 0.
Hence k2 ⩾ 4 so that k ⩾ 2 or k ⩽ − 2.
16 a 6(1 − cos2 x ) + cos x = 5; 6 − 6cos2 x + cos x = 5;
6cos2 x − cos x − 1 = 0
1
b Factorise: (2 cos x − 1)(3 cos x + 1) = 0 so cos x =
2
1
or − 3 .
1
π
If cos x = then x = or 5π ; if cos x = − 1
2
3
3
3
then x = 1.91 or 4.37.
Area of ACBO outside the circle is
2r2 – 1.107r 2 = 0.893r2.
0.893r 2
× 100 = 44.6%.
c Percentage =
2r 2
19 a 6 cos2 x + sin x = 5; 6(1 – sin2 x) + sin x = 5;
6 – 6 sin2 x = – sin x + 5;
6 sin2 x – sin x – 1 = 0
b Factorise: (3 sin x + 1)(2 sin x – 1) = 0;
1
1
sin x = − or .
2
3
x = 30°, 150°, 199.5° or 340.5°
c 6 sin4 y + 5(1 – sin2 x) = 6
6 sin4 y + 5 – 5 sin2 y = 6
6 sin4 y – 5 sin2 y – 1 = 0
d This is a quadratic in sin2 y:
(6 sin2 y + 1)(sin2 y – 1) = 0
1
or 1
6
1
sin 2 y = − has no solution.
6
If sin2 y = 1 then sin y = 1 or −1; y = 90° or 270°.
sin 2 y = −
20 cos AXY = 0.75; Angle AXB = 2 × 0.7227 = 1.4455
1
The area of sector AXB = × 1.4455 × r 2 = 0.7227r2.
2
1
The area of triangle AXB = × sin AXB × r 2 =
2
0.4961r2.
The difference is 0.2267 r2. The area in common is
2 × 0.2267 r² = 0.4533 r2.
0.4533
The percentage of one circle is
× 100 = 14.4%.
π
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4
WORKED SOLUTIONS
Mathematics in life and work
1
The first oscillation is completed when 512πt = 2π
Hence t = 2 = 1 . There are 256 cycles in one
512 256
second so the frequency is 256 Hz.
2
The frequency is 256 ×
3
= 384.
2
1
If the equation is y = a sin kt , then kt = 2π when
2
1
t=
so k = 2π × 384 = 768π.
384
1
The equation is = a sin 768πt.
2
3
y
a
0.5a
0
1
256
t
–2
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5 Series
5 Series
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
 5
 5
c (x + 4)5 = x5 + 5 × x4 × 4 +   × x3 × 42 +   ×
 2
 3
1
× 5 × 6 = 15
2
b If
x2 × 43 + 5x × 44 + 45
1
n(n + 1) = 54, then n2 + n − 108 = 0.
2
 5   5 5 × 4
Now   =   =
= 10 so (x + 4)5 = x5 +
 2  3 2 × 1
This cannot be factorised with whole
numbers so 54 is not a triangular number.
2
20x4 + 160x3 + 640x2 + 1280x + 1024
4
(x + 1)3 − (x − 1)3 = x3 + 3x2 + 3x + 1 − (x3 − 3x2 +
3x − 1)
= x4 − 4x3y + 6x2y2 − 4xy3 + y4
b
=x
3 +3x2 + 3x + 1 − x3 + 3x2 −
3x + 1 = 6x2 + 2 or 2(3x2 + 1)
3
a (x − y)4 = x4 + 4x3(−y) + 6x2(−y)2 + 4x(−y)3 + (−y)4
(1 − 2y)4 = 1 + 4(−2y) + 6(−2y)2 + 4(−2y)3 + (−2y)4
= 1 − 8y + 24y2 − 32y3 + 16y4
c (2 − y)5 = 25 + 5 × 24(−y) + 10 × 23(−y)2 + 10
The multiplier for a 5% increase is 1.05.
× 22(−y)3 + 5 × 2(−y)4 + (−y)5
$24 000 × 1.055 = $30 630.76.
= 32 − 80y + 80y2 − 40y3 + 10y4 − y5
Exercise 5.1A
1
 6
6!
6 × 5!
a  =
=
=6
 1  5!1! 5!× 1
 10 10! 10 × 9
b  =
=
= 5 × 9 = 45
2×1
 2  2!8!
 5
aThe x3 term is   (3x)3 = 5 × 4 × 27x 3 = 10 × 27x 3 = 270x 3
2×1
 3
 5  5 
 5
5×4
3
3
3
3
 3 (3x) = 2 × 1 × 27x = 10 × 27x = 270x . The value of  3 =  2 could
5
also be found directly with a calculator.
 6
b The x3 term is   23(4x)3 = 20 × 8 × 64x3 =
 3
 6 6 × 5 × 4
.
10 240x3. Here   =
 3 3 × 2 × 1
 12
12!
12 × 11
c  =
=
= 6 × 11 = 66
2
 10 10!2!
 12 12! 12 × 11 × 10
d  =
=
= 2 × 11 × 10 = 220
3× 2×1
 3  3!9!
6
a
(2 + x2 ) = 2 + 3 × 2 × x2 + 3 × 2( x2 ) + ( x2 ) = 8 + 6x + 32 x + 18 x
3
3
2
2
3
2
 8
8!
8 × 7 × 6 × 53
2
3
e  =
=
3 2 1 3
x = 7 3× 2 × 5 =2 70x
x
x
 4 4!4! 4 × 3 2× +2 2× 1 = 2 + 3 × 2 × 2 + 3 × 2 2 + 2 = 8 + 6x + 2 x + 8 x
( )
2
() ()
 10 10 × 9 × 8 × 7
 10
= 210
The coefficient is   or   =
4 × 3× 2×1
 6
 4
b (5x − 4y)4 = (5x)4 + 4(5x)3(−4y) + 6(5x)2(−4y)2 +
4(5x)(−4y)3 + (−4y)4
0 10 × 9 × 8 × 7
= 210. This can also be found directly with a
=
4 × 3× 2×1
6 
calculator.
3
a (2 + x)4 = 24 + 4 × 23x + 6 × 22x2 + 4 × 2x3 + x4
= 16 + 32x + 24x2 + 8x3 + x4
b (3 + 5x)3 = 33 + 3 × 32 × 5x + 3 × 3 × (5x)2 + (5x)3
= 27 + 135x + 225x2 + 125x3
7
= 625x4 − 2000x3y + 2400x2y2 − 1280xy3
+ 256y4
 10
aThe term involving x3 is   × 27 x 3, where
3 
 10 10! 10 × 9 × 8
 3  = 7!3! = 3 × 2 × 1 = 120, so the coefficient
is 120 × 27 = 15 360.
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3
5
WORKED SOLUTIONS
14 (1 + x2)3 (1 – x2)3 = {(1 + x2) (1 – x2)}3 = (1 – x4)3
 9
b The term involving x3 is   × 26(3x)3, where
 3
 9
9!
9×8×7
 3 = 6!3! = 3 × 2 × 1 = 84, so the coefficient
= 1 – 3x4 + 3x8 – x12
15 a (1 + x)3 + (1 – x)3 = 1 + 3x + 3x2 + x3 + 1 – 3x
+ 3x2 – x3 = 2 + 6x2
is 84 × 64 × 27 = 145 152.
8
b (1 + 3y)3 + (1 – 3y)3 = 2 + 6(3y)2 = 2 + 54y2
 7
 7 7 × 6 × 5
= 35
The x3 term is   × 24(bx)3 and   =
 3
 3 3 × 2 × 1
c 2 + 54y2 = 11 ⇒ 54y2 = 9 ⇒ y 2 = 1 and so
6
1
y=±
6
 8
2
16 The terms in x2 and x3 are   a 6 ( bx ) and
 2
 8 5
3
a ( bx ) .
 3
so the term is 35 × 16 × b3x3 = 560b3x3. Hence 560b3
= 70 000 and b3 = 125 so b = 3 125 = 5 .
9
a(x2 + 2)4 = (x2)4 + 4 × (x2)3 × 2 + 6(x2)2 × 22 + 4x2 ×
23 + 24
Hence 28a6 b2 = 56a5 b3; hence 28a = 56b and
a 56 2 and so a : b = 2 : 1.
=
=
b 28 1
= x8 + 8x6 +24x4 + 32x2 + 16
b (1 − y3)5 = 1 + 5(−y3) + 10(−y3)2 + 10(−y3)3 +
Exercise 5.2A
5(−y3)4 + (−y3)5
= 1 −5y3 + 10y6 − 10y9 + 5y12 − y15
1
 8
 8
10 (1 + 2x)8 = 1 + 8(2x) +   (2x)2 +   (2x)3 +…, where
 2
 3
b The 15th term is 25 + 14 × 33 = 487.
 8 8 × 7
 8 8 × 7 × 6
 2 = 2 = 28 and  3 = 3 × 2 × 1 = 56 .
c The 30th term is 25 + 29 × 33 = 982.
2
So (1 + 2x)8 = 1 + 16x + 28 × 4x2 + 56 × 8x3 + …
= 1 + 16x + 112x2 + 448x3 + …
c d = 55.1 – 48.7 = 6.4; u20 = 48.7 + 19 × 6.4 = 170.3
d d = 192 – 215 = – 23; u15= 215 – 14 × 23 = –107
The x3 term is 1 × 448x3 + x × 112x2 = (448 + 112) x3
3
= 560x3. The coefficient is 560.
( 2 ) + 4( 2 ) + ( 2 )
2
3
4
4
Hence a = 17 and b = 12.
10
{2 × 20 + 9 × 7} = 515.
2
180
so n = 26.7.
c If 20 + 7(n − 1) = 200 then n − 1 =
7
4
2 So the term required is the 26th.
b S10 =
2
3
5
The series is 2 + 6 + 10 +... (20 terms). S20 =
6
+ 19 × 4) = 800.
a a = 15, and d = 6
()
 4
3
b   ( x ) ( a ) = 6 × 1 a hence a = 96 ;
2
4
 2 2 x
a2 = 64; a = ±8
20
(2 × 2
2
un = 6n + 9
u7 = 6 × 7 + 9 = 51
Alex read 51 pages on Saturday.
b Sn = n (30 + 6(n – 1))
2
S7 = 7 (30 + 6 × 6) = 7 × 66 = 231 pages
2
2
3
 6
3 1
 3 ( 2x ) x = 20 × 8 × 1 = 160
2
⇒
On Saturday, n = 7 ⇒
13 a The constant term is
2
a a = 20 and d = 7; 10th term = 20 + 9 × 7 = 83
( 2 ) + 4( 2 ) + ( )
= 1 + 4 2 + 12 + 8 2 + 4 = 17 + 12 2
2
10
(2 × 20 + 9 × 5) = 425
2
12
(2 × 120 + 11 × (−9)) = 846
2
15
c S15 = (2 × 9 + 14 × 2.5) = 397.5
2
The x10 term is 56x10 × (−1) = −56x10.
(1 + 2)4 = 1 + 4 2 + 6
a S10 =
b S12 =
11 (x + 1)(x − 1) = x2 − 1 so (x + 1)8(x − 1)8 = (x2 − 1)8
 8
The x10 term is   (x 2)5(−1)3 and
 5
8
  8×7×6
 5 = 3 × 2 × 1 = 56 .
b If x = 2 , then (1 + 2)4 = 1 + 4 2 + 6
a d = 57 – 45 = 12; u12 = 45 + 11 × 12 = 177
b d = 191 – 200 = –9; u14 = 200 – 9 × 13 = 83
So (1 + x)(1 + 2x)8 = (1 + x)(1 + 16x + 112x2
+ 448x3 + …).
12 a (1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4
a58 – 25 = 33; 91 – 58 = 33; 124 – 91 = 33; an
arithmetic progression with d = 33.
2
7
a = 50, d = 5 and n = 24; S24 =
24
(2 × 50 + 23 × 5) =
2
2580. So she will save $2580.
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5 Series
8
u5 = a + 4d = 27.6 and u10 = a + 9d = 24.1.
Subtract to get 5d = −3.5 so d = −0.7.
9
14 The 10th term is twice the 4th term so
a + 9d = 2(a + 3d); a + 9d = 2a + 6d; a = 3d.
Substitute into the first equation: a − 2.8 = 27.6
so a = 30.4.
The 18th term is 50 so a + 17d = 50; 3d + 17d = 50;
20d = 50; d = 2.5.
Therefore u20 = 30.4 − 19 × 0.7 = 17.1.
Then the first term a = 3d = 7.5.
aThe multiples form an arithmetic progression
with a = 6 and d = 6.
S10 =
15 If n = 1, then 1 + 4 = 5 = a.
If n = 2, then 4 + 8 = 12 = a + a + d = 2a + d = 10 + d
and so d = 2.
10
(2 × 6 + 9 × 6) = 330
2
n
n
(12 + (n − 1) × 6) = (12 + 6n − 6)
2
2
n
= (6n + 6) = 3n(n + 1)
2
10 aThe odd numbers form an arithmetic progression
with a = 1 and d = 2 so Sn = n (2 + (n − 1) × 2)
2
n
n
= (2 + 2n − 2) = × 2n = n2.
2
2
The nth term is 5 + 2(n – 1) = 2n + 3.
Hence the 100th term is 203
b Sn =
16 If n = 1, then the first term is 6 + 8 = 14.
If n = 2, then the first + the second term is 24 + 16 = 40
so the second term is 40 – 14 = 26.
If an arithmetic sequence starts 14, 26 … , then
a = 14 and d = 12 and the sum of n terms is
1
1
n 28 + 12(n − 1)} = n(28 + 12n − 12)
2 {
2
1
2
= n(12n + 16) = 6n + 8n, so the sequence is
2
arithmetic.
b Each even number is 1 more than the
corresponding odd number so Sn = n2 + n.
n
(20 + 4(n − 1)) = 792
2
n
Therefore (16 + 4n) = 792.
2
11 Sn =
8n + 2n2 = 792
Exercise 5.3A
1
n2 + 4n − 396 = 0
(n + 22)(n − 18) = 0
You want the positive root; there are 18 terms.
b Here are two possible methods:
2
3
b x = a + 7d and hence x = a + 7(y – x);
x = a + 7y – 7x; a = 8x – 7y.
0.01
= 2;
0.005
4
−6
= −2; u10 = 3 × (−2)9 = −1536
3
a a = 5 and r = 10 ÷ 5 = 2; u15 = 5 × 214 = 81 920
b S15 =
5(215 − 1)
= 163 835
2−1
a S10 =
2(310 − 1)
= 59 048
3−1
b S8 =
Each is an arithmetic progression with d = 4.
Add 50 terms of each one.
13 a The difference is y – x and the next term is
y + (y – x) = 2y – x.
24
= 1.2; u6 = 20 × 1.25 = 49.7664
20
d a = 3 and r =
Method 2: Split the sequence into two: 10,
14, 18, … and 11, 15, 19, …
Sum = 25 {20 + 49 × 4} + 25 {22 + 49 × 4}
= 5400 + 5450 = 10 850.
b a = 20 and r =
u15 = 0.005 × 214 = 81.92
Method 1: Add pairs of terms to get this
sequence: 21, 29, 37, …
This is an arithmetic progression with a = 21
and d = 8. The sum required is 50 terms of
this sequence. S50 = 25 {42 + 49 × 8} = 10 850.
0.2
= 5; u8 = 0.04 × 57 = 3125
0.04
c a = 0.005 and r =
n = –22 or 18
12 aThe differences between terms are 1, 3, 1, 3,
1, 3, … They are not constant so it is not an
arithmetic sequence.
a a = 0.04 and =
100(1 − 0.98)
= 569.533
570 (3 s.f.)
1 − 0.9
c a = 4 and r = –2 so S12 =
4((−2)12 − 1)
= −5460
−2 − 1
d a = 20 and r = 1.1 so S20 =
20(1.120 − 1)
= 1150
1145.5
(3 s.f.)
1.1 − 1
The total amount is a geometric series with
a = 12 000, r = 1.03 and n = 10.
S10 =
12 000(1.0310 − 1)
= $137 566.55 (3 s.f.)
0.03
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5
WORKED SOLUTIONS
5
6
1
1
; S∞ =
=2
2
1− 1
2
1
2
b a = 2 and r = ;
S∞ =
=3
3
1− 1
3
15
5
1
=
S∞ =
c a = 5 and r = − ;
4
3
1+ 1
3
3
1
d a = 1 and r = ;
S∞ =
=4
4
1− 3
4
18
2
−4
6
e a = 6 and r =
=− ;
S∞ =
=
5
6
3
2
1+
3
a a = 1 and r =
aThe amounts are a geometric progression with
a = 0.01 and r = 2.
The 21st term is 0.01 × 220 = $10 485.76.
b S21 =
0.01(221 − 1)
= $20 971.51
2−1
c On his 65th birthday he would receive $0.01 × 264
which is more than $1.8 × 1017. This is an absurdly
large amount.
7
aThe amounts each year are a geometric
progression with a = 2000 and r = 1.25.
c Distance north = 800 + 200 + 50 + …
= 800 = 800 = 3200
3
3
1− 1
4
4
Distance east = 400 + 100 + 25 + …
= 400 = 400 = 1600
3
3
1− 1
4
4
By Pythagoras, distance from start
1600 = 1193 m
+
( 3200
3 ) ( 3 )
2
=
2
10 This is a geometric series with a = 0.45 and r = 0.01.
S∞ =
0.45
0.45 45 5
=
=
=
1 − 0.01 0.99 99 11
11 a = 80 and
1−r =
80
a
= 200 so 1 − r = 200.
1−r
80
= 0.4 and therefore r = 0.6.
200
S11 =
80(1 − 0.611)
= 199.27 but
1 − 0.6
2000(1.258 − 1) = 39 684
$
b After 8 years she has S8 =
1.25 − 1
$39 684 which is less than $50 000.
S10 =
80(1 − 0.610)
= 198.79 so 11 terms are needed
1 − 0.6
2000(1.259 − 1)
= $51 605
1.25 − 1
9
2000(1.25 − 1)
S9 =
= $51 605 so it does take 9 years.
1.25 − 1
( ) = 165
a
1 1
1
1 1
+ of = +
4 4
4 4
4
b
5
1
+
4
16
2
( ) = 6421
3
1
c This is a geometric series with a = r = so
4
1
1
2
3
1
1
1
1
S∞ = +
+
+…= 4 = 4 = .
1
3
4
4
4
3
1−
4 4
() ()
One-third of the square is shaded.
9
This is a geometric series with a = 800 and r = 1 .
2
800
S∞ =
= 1600 so she walks 1600 m.
1 − 12
The amount in the third year is $2000 × 1.252
= $3125.
After 9 years she has S9 =
8
b The distance walked is 800 + 400 + 200 + ... m.
a The distance north is 800 – 400 + 200 – 100 + ... m.
This a geometric series with a = 800 and r = − 1 .
2
800
2
1600
1
= 533 m north
= 800 × =
So S∞ =
3
3
3
1 + 12
before the total is greater than 199.
12 Split the sum into two separate series: 1 + 2
1 1 1 1
+ 3 + 4 + … and + + +
+…
2 4 8 16
The first is an arithmetic series, a = d = 1 and
n
1
Sn = {2 + (n – 1)} = n(n + 1).
2
2
1
The second is a geometric series with a = r =
2
n
0.5(1 − 0.5 )
n
.
= 1 − 0.5
and S n =
1 − 0.5
So for the series given Sn = 1 n(n + 1) + 1 − 0.5n .
2
a
20
= 200 ; 1 − r =
= 0.1 ; r = 0.9
1−r
200
1% of 200 = 2 so the total must be at least 198.
13 a = 20 and
S43 =
(
)
(
)
20 1 − 0.943
= 197.84;
1 − 0.9
20 1 − 0.944 = 198.06
1 − 0.9
This shows that the n must be at least 44
S44 =
of the starting point.
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5 Series
14 The second term is ar and the sum to infinity is
a .
1−r
a
= 4ar ; 1 = 4r(1 – r); 4r2 – 4r + 1 = 0;
Hence
1−r
(2r – 1)(2r – 1) = 0.
1
The only solution is r = .
2
15 a From her aunt she receives amounts in an
arithmetic progression, a = 30 and d = 10.
On her 10th birthday, n = 6 and she receives
30 + 5 × 10 = $80.
From her uncle she receives amounts in a
geometric progression, a = 0.1 and r = 2.
On her 10th birthday, n = 6 and she receives
0.1 × 25= $3.20.
The total is $83.20.
b Anna has had 17 birthday presents so n = 17.
17
From her aunt 2 {2 × 30 + 16 × 10} = $1870.
(
17
a
a
=
=
1−r 1± 2
a
The term independent of x, where the x terms
4
 6
cancel out is   (x 2)2 × 1 = 6 × 5 x 4 × 1 4 = 15
2
x
2×1
16
16x
 2
4
a S20 = 10(2a + 19d ) and S30 = 15(2a + 29d )
Hence 15(2a + 29d ) = 20(2a + 19d ); 6a + 87d = 8a + 76d; 2a = 11d; a = 11d
2
b
15(2a + 29d) = 400; substitute to get
15(11d + 29d) = 400; 15 × 40d = 400;
10 2
d = 15 = 3
Then a = 11 × 2 = 11
2 3 3
5 a An arithmetic sequence with a = 1200 and d = 70.
12
S12 = (2 × 1200 + 11 × 70) = 19 020. So he
2
earns $19 020.
b In the second month he earns 1200 × 1.05 =
$1260 which is less than 1200 + 70 = $1270.
6
Hence r 2 =
and a =
a a
a ±2
 10
a (1 + 2x)10 = 110 + 10 × 2x +   × (2x)2 + …
2 
= 1 + 20x + 180x2 + …
7
b (3x − 4)(1 + 2x)10 = (3x − 4)(1 + 20x + 180x2 + …) =
The term x2 is 3x × 20x − 4 × 180x2 = 60x2 − 720x2
= −660x2.
The coefficient is −660.
a
32 8 32 3 4
÷ =
× =
27 3 27 8 9
8 4
÷ = 6.
3 9
4
2
, then r = ± .
9
3
2
6
If r = , then S∞ =
= 18.
3
1− 2
3
2
3
If r = − , then S∞ = 6 = 6 ÷ 5 = 18 or 3 .
3
5
2
3
5
1+
3
6
aThe term in x3 is   ( 2x )3 ( −k )3
 3
= – 20 × 8x3 k3 = – 160k3 x3.
b If r 2 =
= 1 + 20x + 45 × 4x2 + …
2
1200(1.0512 − 1)
= $19 100.55,
1.05 − 1
which is more than $19 020.
8
32
a ar 2 = and ar 4 =
3
27
In the year he earns
Exam-style questions
1
( )
3(2a + 29d ) = 4(2a + 19d );
)
From her uncle 0.1 2 − 1 = $13107.10.
2−1
The total is $14977.10.
16 a S∞ = a = x = 8 ; x = 8(1 – 3x); x = 8 – 24x;
1 − r 1 − 3x
8
25x = 8; x =
or 0.32
25
4
2
2
b ar = 4; r = ; r = 4 ; r = ± 2
a
a
a
S∞ =
3

(1 − 3x)4 = 1 + 4(−3x) +
4
2
3
4
 2 (−3x) + 4(−3x) + (−3x)
= 1 − 12x + 6 × 9x2 − 108x3 + 81x4
= 1 − 12x + 54x2 − 108x3 + 81x4
b (2 + ax2)(1 − 3x)4 = (2 + ax2)(1 −12x + 54x2 −
108x3 + 81x4)
 6
5
The term in x5 is   ( 2x ) (−k) = – 6 × 32x5 k
 5
= – 192k x5.
The coefficients are equal so – 160k3 = – 192k;
192
= 1.2 ; k = 1.2 .
hence k 2 =
160
 6
4
2
b The term in x2 is   ( 2x ) ( −k )
 2
= 15 × 4x2 × 1.22 = 86.4 x2.
The coefficient is 86.4.
The coefficient of x3 is 2 × − 108 + a × −12 =
−216 − 12a.
If −216 − 12a = −132 then −12a = 84 and a = −7.
72
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5
WORKED SOLUTIONS
8
a a = 24 and a + d = 18 so d = – 6
n
n
2a + ( n − 1) d } = {48 − 6 ( n − 1)}
2
2{
n
= ( 48 − 6n + 6 ) = 27n – 3n2
2
If Sn < – 30 then 27n – 3n2 < – 30;
3n2 – 27n – 30 > 0; n2 – 9n – 10 > 0
Sn =
(n – 10)(n + 1) > 0; hence n > 10 or n < – 1.
9
Since n is positive, n > 10.
b a = 24 and ar = 18; r = 18 = 3 ;
24 4
a
24 = 96
S∞ =
=
1−r 1− 3
4
a a + 2d = 51 and a + 10d = 187; hence
8d = 187 – 51 = 136; d = 136 ÷ 8 = 17.
Hence a + 34 = 51 and a = 17; the terms of the
sequence are 17, 34, 51, … the multiples of 17.
b 1000 = 58.8, so there are 58 multiples of 17
17
between 0 and 1000.
With a = 17 and d = 17 then
S58 = 26(34 + 57 × 17) = 26 078
 7
10 a The term in x3 is 3x ×   × 2x 2 × (−b)6
1
= 3 × 7 × 2b6 x3 = 42b6 x3.
Hence 42b6 = 2688; b6 = 64; b = ± 2.
 7
b The term in x2 is a ×   × 2x 2 × (−b)6
1
= 14ab6 x2.
2688
Hence 14ab6 = 2688; a =
= 3.
14 × 64
11 a $400 has 1 month’s interest, another $400 has
2 month’s interest and so on.
Total = 400 × 1.005 + 400 × 1.0052 + … + 400 ×
1.00512.
This is a geometric series with a = 400 ×
1.005 = 402 and r = 1.005.
12
b
S12 = 402(1.005 − 1) = 4958.90 , she has
1.005 − 1
$4960 (3 s.f.).
12 a(a + 5d) + (a + 6d) + (a + 7d) = 12; hence
3a + 18d = 12; a + 6d = 4; a = 4 – 6d
b a + 5d = 12; hence (4 – 6d) + 5d = 12;
4 – d = 12; d = –8
Hence a = 4 – 6d = 4 + 48 = 52.
13 a = 20, a + d = 24, d = 4
k
k
2a + ( k − 1) d } = {40 + 4 ( k − 1)}
2{
2
k
= {4k + 36} = 2k2 + 18k
2
Sk =
If Sk = 504, then 2k2 + 18k = 504; k2 + 9k – 252 = 0;
(k + 21)(k – 12) = 0.
Hence k = −21 or 12. The value of k is positive.
Hence k = 12.
14 aThe total length is 20 + 10 + 5 + ... which is a
geometric series with a = 20 and r = 0.5 so the
20
= 40 .
sum S∞ =
1 − 0.5
b The x-coordinate is 20 – 5 + 1.25 − ... which is
a geometric series with a = 20 and r = −0.25
20
so the x-coordinate is S∞ =
= 16 .
1 + 0.25
The y-coordinate is 10 – 2.5 + 0.625 – ... which is
a geometric series with a = 10 and r = −0.25 so
10
= 8.
the y-coordinate is S∞ =
1 + 0.25
The coordinates are (16, 8).
15 aThe ratios of successive terms are equal so
6p + 2 4p + 4
=
.
4p + 4 3p + 3
6p + 2 4
Hence 4p + 4 = 3 and 3(6p + 2) = 4(4p + 4);
18p + 6 = 16p + 16;
2p = 10 and so p = 5.
b The first term is 6 × 5 + 2 = 32, the second is
24 3
4 × 5 + 4 = 24 and r =
= .
32 4
S∞ = 323 = 32
1 = 128
1− 4
4
16 aIf the common difference is d then b = a + 2d
1
and d = (b − a).
2
5
The sixth term is a + 5d = a + (b − a)
2
5
5
3
1
5
= a + b − a = b − a or (5b − 3a).
2
2
2
2
2
b
b If the common ratio is r, then ar2 = b; r 2 = ;
a
b
r=±
a .
The fifth term is ar4 = a ×
()
b
a
2
=
b2
ab 2
2 = a .
a
63
17 a 64
n
b The nth term is 2 n− 1 .
2
n
2
−
1
1
c
=1− n .
2n
2
( )( )( )
Hence Sn = 1 − 1 + 1 − 1 + 1 − 1 + ...
2
4
8
1
+… + 1 − n  = n −  1 + 1 + 1 + ... 1n  .

2 4 8
2 
2 
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5 Series
The terms in the bracket form a geometric
sequence with a = r =
1 and n terms.
2
()
Mathematics in life and work
1
()
()
()
18 If the common ratio is r, then S∞ =
a
= ka .
1−r
Total value = 5000 + 50002 + … + 5000
.
1.02 1.02
1.0215
This is a geometric progression with a = 5000 ,r = 1
1.02
1.02
1 and n = 15.
5000
a=
,r =
1.02
1.02
5000
( k1 ) = a − ka .
2
12
a
12
S∞ =
= r =
= 50
1 − r 1 − r r(1 − r )
25r2 – 25r + 6 = 0
(
15
1 15
1.02
−0.02
R
, then in this case it is a
100
5000
1
geometric progression with a =
and r = .
c
c
Then
5000
5000
c
(1 − c −15) ×
S15 =
1 − c −15 1 − 1 =
c
c
c
c −1
If you write c = 1 +
)(
(
Hence 12 = 50r(1 – r); 6 = 25r – 25r2;
20 a If the shortest piece is a then the longest is
a + 9d = 4a.
1
Hence 3a = 9d and d = a
3
The sum is 100 so 10 ( 2a + 9d ) = 100 ;
2
1
5 2a + 9 × a = 100; 25a = 100 so a = 4.
3
The shortest piece is 4 cm.
1
= 64 246. The value is $64 200 (3 s.f.)
The second term is ar = a 1 −
3
2
b (5r – 2)(5r – 3) = 0; r = or
= 0.4 or 0.6
5
5
12
12
12
= 30 or
= 20
so a =
c a=
r
0.4
0.6
3
3
The 4th term is ar = 30 × 0.4 = 1.92 or
20 × 0.63 = 4.32.
(( ) − 1) = 5000 (( ) − 1)
n
1.02 1.02
S15 = a(r − 1) =
1
r −1
1.02 − 1
1
1
1
Therefore
= k; = 1 − r ; r = 1 − .
1−r
k
k
19 a The first term is 12 ;
r
5000
,
1.02
5000
the value of the second after 2 years is $
and
1.022
so on.
n
1 1 − 1 


n
2
2
1 1 1
1
 =1− 1
+ + + ... n = 
2
2 4 8
2
1− 1
2
n
n

1
So Sn = n − 1 − 1  = n − 1 +
2
2


The value of the first payment after 1 year is $
=
3
(
)
)
5000
500000
× 1 − c −15 =
× (1 − c −15) .
c −1
R
If the annuity is for n years, then
500 000
Sn =
× (1 − c −n).
R
Now limn→∞ c−n = 0 so the value of a perpetuity,
S∞ =
500 000 .
R
)
b If the shortest piece is a, then the longest is
ar9 = 4a; r9 = 4 ; r = 9 4 = 1.1665
(
)
a r 10 − 1
100
= 100 ; a =
=
r −1
22.015
4.54 cm to 3 s.f.
The sum is
21 The fifth term is r4. The sum of the subsequent
5
terms is ar5 + ar6 + ar7+ ... = ar .
1−r
5
Hence ar = 9ar 4 ; r = 9 ; r = 9(1 – r); 10r = 9;
1−r
1−r
r = 0.9.
a
a
=
= 10a .
Then S∞ =
1 − r 0.1
74
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6
WORKED SOLUTIONS
6 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
2
a 7
1
c
2
6
a f(x) = x2 − 6x + 5 so f ′(x) = 2x − 6
b f ′(1) = −4 and f ′(11) = 16
b −0.3
c
d −2
y
8
a 2x2 − 8x
6
x3 + 3x2
4
b
c x2 + 4x − 5x − 20 = x2 − x − 20
3
d
9x2 + 3x + 3x + 1 = 9x2 + 6x + 1
a
x−3
d
1
x3
b
1
x2
e
3
x2
2
c
–4
−1
x 2
0
–2
–2
3
4
b 6x
b 2x − 4
2
4
6
x
4
c 6x + 4
c 4 − 2x
2
dy
= 2x + 6
dx
dy
b y = x2 − 2x − 3,
= 2x −
c2
dx
dy
c y = 4x2 + 12x + 9,
= 8x + 12
dx
–6
–4
0
–2
–2
–4
–6
–8
8
–10
dy
= 2x + 1
dx
ii If x = 0,
dy
= 1.
dx
7
a f ′(x) = − 2x
8
b f ′(−1) = 2 and f ′(3) = −6
c
y
3
dy
= 30x − 30
dx
a If x = 1, then
c At (2, 0) and (−3, 0), the gradients are 5 and −5.
–2
x
–6
a y = x2 + 6x,
–3
8
y
b i at (0, −6)
5
6
6
a 3
a 2x + 4
a
4
–4
Exercise 6.1A
1
2
2
dy
=0
dx
2k × 4 − 4k = 2
1
8k − 4k = 2
–2
1
2
3
x
dy
= 120
dx
y = kx(x − 4) = kx2 − 4kx
dy
= 2kx − 4k
dx
dy
When x = 4,
= 2.
dx
2
0
–1
–1
b If x = 5, then
4k = 2
1
k = or 0.5
2
–3
–4
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6 Differentiation
9
dy
a = 2x − 6 ; where the curve crosses the y-axis
dx
13 a
x = 0 and then the gradient is −6.
Using the result given with a =
b The curve crosses the x-axis where
x2 – 6x – 16 = 0; (x – 8)(x + 2) = 0; x = 8 or −2.
dy
= 16 − 6 = 10;
If x = 8,
dx
( h8 ) = 2 + h4 .
(
The gradient of
PQ =
3
1.1 − 1
= 3.31.
1.1 − 1
b (1 + h)3 = 1 + 3h + 3h2 + h3 so B is
(1 + h, 1 + 3h + 3h2 + h3)
The gradient of PB is
1 + 3h + 3h 2 + h3 − 1 3h + 3h 2 + h3
=
= 3 + 3h + h 2.
1+h −1
h
11 a
A is (1, 2) and B is (3, 12). The gradient of
12 − 2
AB =
=5
3−1
b The y-coordinate of Q is (2 – h)2 + (2 – h)
= 4 – 4h + h2 + 2 – h = 6 – 5h + h2.
dy
14 a = 18 − 4x ; at A 18 – 4x = – 2; 4x = 20; x = 5 and
dx
y = 45; A is (5, 45)
b At A the equation of the tangent is
y – 45 = – 2(x – 5) or y + 2x = 55.
This crosses the y-axis at (0, 55) and the
x-axis at (27.5, 0).
The area of the triangle is 55 × 27.5 ÷ 2 = 756.25.
Exercise 6.2A
1
a 2 × 3x2 = 6x2
c 0.1 × 5x4 = 0.5x4
b 0.5 × 4x3 = 2x3
d 50 × 3x2 = 150x2
2
a 3x2 + 8x − 8
b 6x2 − 10x + 6
3
a 4x3 + 16x
b 5x4 − 30x2 + 2
4
a 3x2 − 4x
The y-coordinate of P
is (2 + h)2 + (2 + h) =
4 + 4h + h2 + 2 + h
= 6 + 5h + h2.
The gradient of PQ =
(6 + 5h + h ) − (6 − 5h + h ) = 10h = 5.
2
(2 + h) − (2 − h)
b i When x = 2, 3x2 − 4x = 12 − 8 = 4.
2h
dy
= 2x + 1 so the gradient at (2, 6) is 2 × 2 + 1
dx
= 5 = the gradient of PQ.
ii When x = −1, 3x2 − 4x = 3 + 4 = 7.
5
dx
At (−3, 0),
dy
= ( 2 × −3) − 1 = −7 ; at (4, 0),
dx
dy
= ( 2 × 4 ) − 1 = 7.
dx
b At (5, 8)
dy
= 2 × 5 − 1 = 9 so the equation of
dx
the tangent is y – 8 = 9(x – 5) or y = 9x – 37.
dy
3
c If
= 2, then 2x – 1 = 2 and x = = 1.5 ;
2
dx
y = (1.5 + 3)(1.5 – 4) = – 11.25.
a f ′(x) = 0.5 × 4x3 − 2 × 2x + 1 = 2x3 − 4x + 1
b i When x = −1, 2x3 − 4x + 1 = −2 + 4 + 1 = 3.
ii When x = 1, 2x3 − 4x + 1 = 2 − 4 + 1 = −1.
12 a If y = 0, then (x + 3)(x – 4) = 0 and so x = −3 or 4
dy
y = x2 – x – 12 and so
= 2x − 1.
h
4+h −2 2+ 4 −2 1
≈
= ; the gradient
4
4+h−4
h
1
of the curve at P is .
4
c As h approaches 0, 3 + 3h + h2 approaches 3
so the gradient = 3.
2
)
b P and Q are (4, 2) and 4 + h, 4 + h .
P is (1, 1) and A is (1.1, 1.13) so the gradient of PA
is
h
, if h is
4
small then 4 + h ≈ 2 1 +
if x = −2, dy = −4 − 6 = −10.
dx
10 a
( h4 ) = 4 1 + h4 = 2 1 + h4
4+h = 4 1+
iii When x = 2, 2x3 − 4x + 1 = 16 − 8 + 1 = 9.
6
a y = 2x3 + 5x2
dy
= 6x2 + 10x
dx
b y = x(x2 − 8x + 16) = x3 − 8x2 + 16x
dy
= 3x2 − 16x + 16
dx
c y = x 3 + x2 + x + 1
dy
= 3x2 + 2x + 1
dx
The point is (1.5, −11.25).
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6
WORKED SOLUTIONS
7
a f(x) = x3 + 3x2
13
f ′(x) = 3x2 + 6x
b i f ′(−3) = 27 − 18 = 9
ii f ′(−2) = 12 − 12 = 0
iii f ′(1) = 3 + 6 = 9
8
9
y = x(x2 − 6x + 8) = x3 − 6x2 + 8x
dy
= 3x2 − 12x + 8
dx
dy
If x = 0,
= 8.
dx
dy
If x = 2,
= 12 − 24 + 8 = −4.
dx
dy
If x = 4,
= 48 − 48 + 8 = 8.
dx
dy
= 3x 2 − 4x − 5 . The gradient of the straight line
dx
is 10.
dy
If
= 10, then 3x2 – 4x – 5 = 10;
dx
3x2 – 4x – 15 = 0; (3x + 5)(x – 3) = 0
5
So x = − or 3 and the points on the curve are
3
(−1.67, 4.15) and (3, 0)
The second of these is on the straight line and so
the line is the tangent at that point.
14 a(x + a)(x – a) = x2 – a2 so f(x) = (x2 – a2)2
(x2 – a2)2 = x4 – 2a2 x2 + a4 hence f' (x) = 4x3 – 4a2 x
dy
dy
= 1;
= 3x 2 − 4x + 1; at A x = 0, y = 3 and
dx
dx
the equation of the tangent is y – 3 = x or y = x + 3.
b If x = 2, then x3 – 2x2 + x + 3 = 5 and so (2, 5)
is on the curve.
b If f' (x) = 0, then 4x3 – 4a2 x = 0; x3 – a2 x = 0;
x(x2 – a2) = 0;
a
Also if x = 2, then x + 3 = 5 so the point (2, 5)
is also on the tangent.
10 a
y = 8x2 – x4 ⇒
x(x + a)(x – a) = 0; x = 0, a or –a; then y = a4, 0 or 0
The points are (0, a4), (a, 0) and (−a, 0)
Exercise 6.3A
1
dy
= 16x − 4x 3 ; if x = −3, then
dx
1
c y = x 3;
dy
= −48 + 108 = 60
dx
b If 16x – 4x3 = 0, then 4x – x3 = 0; x(4 – x2) = 0;
x(2 + x)(2 – x) = 0
So x = 0, −2 or 2; The three points are (0, 0),
(−2, 16) and (2, 16)
d
2
dy 1 − 32
= x
dx 3
3
dy
5 5 −1
= 4 × x 2 = 10x 2
2
dx
1
a s = 10t 2 so
1 −1
5
ds
= 10 × t 2 =
2
dt
t
b s = 50t–1 + 10 so
11 a f(x) = x4 – 2cx2 + c2 so f ′(x) = 4x3 – 4cx
b If f ′(x) = 0, then 4x3 – 4cx = 0; x(x2 – c) = 0;
x = 0 or x2 = c
If c  0 the only solution is x = 0 and there is
one point where the gradient is 0.
If c > 0, there are three solutions, 0,
dy
2
= −2x−3 = − 3
dx
x
dy
6
b y = 2x−3 and so
2 × (−3)x−4 = − 4
dx
x
a y = x−2 and so
c and
− c and hence there are three points.
12 y = x(x2 – 6x + 9) – 8 = x3 – 6x2 + 9x – 8
dy
Hence
= 3x 2 − 12x + 9
dx
If the gradient is −3, then 3x2 – 12x + 9 = – 3; 3x2 –
12x + 12 = 0
x2 – 4x + 4 = 0; (x – 2)2 = 0; x = 2
If x = 2, then y = 2(2 – 3)2 – 8 = – 6 so P is (2, −6)
The equation of the tangent is y + 6 = – 3(x − 2)
or y = – 3x and this passes through the origin.
50
ds
= 50 × −1 × t −2 = − 2
dt
t
c s = 10t2 – 10t–2 so
3
20
ds
= 20t + 20t −3 = 20t + 3
dt
t
a f(x) = 24x−1
f ′(x) = −24x−2 = −
24
x2
b i f ′(6) = − 242 = − 2
3
6
24
ii f ′(4) = − 2 = − 3
2
4
24
iii f ′(−2) = −
= −6
(−2)2
iv f ′(24) = − 242 = − 1
24
24
c If x ≠ 0, then x² is positive and the gradient
24
− 2 is negative.
x
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6 Differentiation
4
5
5
b i f ′(2) = − 2 = −
4
2
1
5
= −0.05
ii f ′(10) = − 2 = −
20
10
1
a y = x2
dy 1 − 12
1
= x =
dx 2
2 x
i If x = 4,
dy
1
1
=
= .
dx 2 4 4
c f ′(x) = −
ii If x = 9,
dy
1
1
=
= .
dx 2 9 6
x2 = 1
iii If x = 100,
b
1
2 x
=
x = 1 or −1
dy
1
1
.
=
=
dx 2 100 20
1
2
Points are (1, 7) and (−1, −3).
7
x = 1 and the coordinates are (1, 1).
c
a y=
1
2
c For the reflection
( )
1
x + 2x−1
2
dy 1
1 2
= − 2x −2 = − 2
2 x
dx 2
If x = 2,
dy 1 2 1 1
= −
= − = 0.
dx 2 22 2 2
dy 1
2
1 1
= −
= − = 0.
If x = −2,
dx 2 (−2)2 2 2
b If x = 0.5,
If x = 4,
c
dy 1
3
= × 3x 2 = x 2 ; Q is
8
dx 8
the point (10, 125).
The gradient is
of
8
2
.
75
3
75
× 100 =
= reciprocal
8
2
a f(x) = x4 + 6x2 + 9 so f'(x) = 4x3 + 12x
b h(x) = x3 (x4 + 6x2 + 9) = x7 + 6x5 + 9x3
So h'(x) = 7x6 + 30x4 + 27x2
9
y=
x 2 + 2ax + a 2
= x + 2a + a2 x–1; hence
x
dy
a2
= 1 − a 2x −2 = 1 − 2 .
dx
x
dy 1 2 1 1 3
= −
= − = .
dx 2 4 2 2 8 8
a2
= 0, then x2 = a2 and x = ±a
x2
(2a)2
= 4a. If x = −a, then y = 0. The
If x = a, then y =
a
points are (a, 4a) and (–a, 0).
dy 1 2
= −
dx 2 x 2
5
= 2 + 5x−1
x
5
f ′(x) = −5x−2 = − 2
x
a f(x) = 2 +
1 3
x
8
dy 1
1
2
1
= −
= − 8 = −7 .
2
dx 2 0.52 2
If x is large, 22 is a small positive number
x
1
and the gradient is close to .
2
1
The larger x is, the closer the gradient is to .
2
6
dy 2 − 32
= x ; if x = 125 then
dx 3
b The equation is x = 3 8y ; 8y = x3; y =
1
x = and the coordinates are 1 , 1 .
4
4 2
5
1
a y = 2x 3 ⇒
−2
dy 2
2 1
2
= 125 3 = × 2 =
3 5
75
dx 3
1
=1
2 x
x =
5
= −5
x2
If 1 −
( 44 ) = 5 = 25 so (4, 25) is on the
f (4) = 4 +
10 a
2
2
curve.
b f(x) = x2 + 8 + 16x–2 so
32
x3
32
1 15
c f ′ (4) = 8 −
=8− =
64
2
2
f'(x) = 2x – 32x–3 = 2x −
The equation of the tangent is
15
y − 25 = (x − 4) ; 2y – 50 = 15x – 60;
2
2y = 15x – 10.
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6
WORKED SOLUTIONS
Exercise 6.4A
1
2
3
a
dy
= 2(4x + 2) × 4 = 8(4x + 2) or 16(2x + 1)
dx
b
dy
= 3(4x + 2)2 × 4 = 12(4x + 2)2
dx
−1
dy 1
= × (2x + 5) 2 × 2 =
dx 2
c
dy
= 5(4x + 2)4 × 4 = 20(4x + 2)4
dx
b If x = 2, then y = 4 + 5 = 3.
a
dy
= 10(8 − x)9 × −1 = −10(8 − x)9
dx
b
dy
= 10(1 + 3x2)9 × 6x = 60x(1 + 3x2)9
dx
c
dy
= 10(6x − 3x2)9 × (6 − 6x) = 60(6x − 3x2)9(1 − x)
dx
a f(x) = (
1
1
x − 3 2 so f'(x) =
)
2
(
)
(
)
(
1
2
)
7
1
.
2 x−3
b y = 600(x2 + 50)–1
dy
= –600(x2 + 50)–2 × 2x
dx
−1
2
) × 2x = x x− 3
2
=−
dy
20
= 20 × −1 × (2 + x)−2 × 1 = −
dx
( 2 + x )2
dy
20
=−
= −1.25
16
dx
Therefore (2 + x)2 = 100.
8
20 = −0.2.
2
+
( x )2
−2
×1 = −
f′(x ) =
g′ ( x ) = 20 × ( −1)( x + 3)
−
g′ ( −1) = −
20
= −5
22
×1 = −
20
(x + 3)2
 −5
c A translation of y = f(x) by   is
0
20
20
=
= g(x).
(x + 5) − 2 x − 3
1
(
)
1
−
−x
1
100 − x 2 2 × (−2x) =
2
100 − x 2
The gradient of OP =
20
= −5
22
−2
)
So the gradient of the tangent = f ′ ( −6 ) =
20
.
(x − 2)2
b g(x) = 20(x + 3)–1 so
8
(x − 10), 8x + 15y = 140
15
a
f ( −6 ) = 100 − (−6)2 = 64 = 8 so (−6, 8) is on
the curve.
(
af(x) = 20(x – 2)–1 so
f ′ ( x ) = 20 × ( −1)( x − 2)
dy
12000
12000
24
8
=−
=−
=−
=−
22500
45
15
dx
150 2
b f ( x ) = 100 − x 2 2 ;
Therefore x = 8 and y = 2. The point is (8, 2).
f ′ (4) = −
1200x
(x 2 + 50)2
So y − 4 = −
Take the positive root: 2 + x = 10.
5
600
600
=
=4
10 2 + 50 150
so (10, 4) is on the curve.
a When x = 10, y =
c At x = 10,
b Where the gradient is –0.2, −
1
1
=
2x + 5 y
dy 1
= and the equation of the
dx 3
1
tangent is y − 3 = (x − 2).
3
1
On the x-axis y = 0 and −3 = (x − 2);
3
x – 2 = – 9; x = – 7. The point is (−7, 0).
a y = 20(2 + x)−1
If x = 2, the gradient
1
a y = (2x + 5)2 so
c At (2, 3)
−1
x−3 2 =
1 2
c f(x) = x 2 − 3 so f'(x) = 2 x − 3
1
−
1 2
x
.
x − 3 2 × 2x =
2
x2 − 3
(
6
1 2x − 3 − 12 × 2 =
1
.
)
2(
2x − 3
1
b f(x) = ( 2x − 3) 2 so f'(x) =
4
The gradient at x = 4 on the first is the
gradient at x = 4 – 5 = −1 on the second.
9
6 3
= .
8 4
8
4
=−
−6
3
4 3
× = −1 so the lines are perpendicular.
3 4
dy
dy
= 3(2x − 3)2 × 2 = 6(2x − 3)2 ; if
= 24, then
dx
dx
6(2x – 3)2 = 24; (2x – 3)2 = 4;
2x – 3 = 2 or −2; x = 2.5 or 0.5; then y = 8 or – 8;
points (2.5, 8) and (0.5, −8)
10 a 1 −
1
x +1
1
x +1−1
x
=
−
=
=
x +1 x +1 x +1
x +1
x +1
b f'(x) is the derivative of
1 − ( x + 1)
−1
= ( x + 1)
−2
=
1
.
(x + 1)2
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6 DIFFERENTIATION
x
1
=
+ 1 and the derivative
x −1 x −1
−2
1
is −1 × ( x − 1) = −
.
(x − 1)2
−2x
11 a f(x) = (x2 + 4)–1; f'(x) = –1 × (x2 + 4)–2 × 2x =
2
x2 + 4
c g(x ) =
(
b g(x ) =
2
dy
= 6 − 4x
dx
a
At a stationary point,
6 − 4x = 0
)
x =1
x2
x2 + 4 − 4
4
=
=1− 2
= 1 − 4f(x)
x +4
x2 + 4
x +4
2
So g′ ( x ) = −4f ′ ( x ) =
12 a If x = 2, then y =
8x
( x + 4)
2
2 .
(in fact it is always negative) so the point is a
maximum point.
c
y
9.5
dy
160
= − 3 = −2.5.
If x = 2, then
dx
4
5
The equation of the tangent is y – 5 = –2.5(x – 2)
or y + 2.5x = 10.
10
= 4; the area
2.5
0
3
a
1
× 10 × 4 = 20.
2
10
dy
dx
= 6 – 2x
dy
= 2x − 12
dx
8
y = x (6 – x)
6
4
At a stationary point,
2
dy
= 0.
dx
–4
2x − 12 = 0
b
x
1.5
a and b
Exercise 6.5A
1
2
d 2y
= −4 which is negative when x = 1.5
dx 2
b
b y = 80(x + 2)–2; then
dy
−3
160
= 80 × ( −2) × ( x + 2) = −
.
dx
(x + 2)3
of the triangle is
( ) = 9.5.
1
1
1
and y = 5 + 6 × 1 − 2 × 1
2
2
2
Coordinates are (1.5, 9.5).
80
= 5.
42
If x = 0, y = 10; if y = 0, x =
dy
= 0.
dx
–3
–2
0
–1
–2
x = 6 and y = 62 − 12 × 6 − 20 = −56
–4
Coordinates are (6, −56).
–6
1
2
3
4
5
6
7
8
x
–8
d 2y
= 2 which is positive when x = 6
dx 2
y = 6x − x2
dy
= 6 − 2x
dx
(in fact it is always positive) so the point is a
minimum point.
dy
crosses the x-axis
dx
gives the x-coordinate of a turning point.
c
c Where the graph of
y
d
0
6
x
d 2y
= −2, a constant value
dx 2
y
–20
0
–56
x
–2
80
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6
WORKED SOLUTIONS
4
a f ′(x) = 6x2 − 18x + 12
At a stationary point: f'(x) = 0
6
6x2 − 18x + 12 = 0
4x3 − 4x = 0
Divide by 6 and factorise: x2 – 3x + 2 = 0
4x(x − 1)(x + 1) = 0
(x − 1)(x − 2) = 0;
x = 0, 1 or −1
x = 1 or 2.
The stationary points are (0, 0), (1, −1) and (−1, −1).
f(1) = 2 – 9 + 12 + 8 = 13 and
f(2) = 16 – 36 + 24 + 8 = 12.
d 2y
= 12x2 − 4
dx 2
The stationary points are (1, 13) and (2, 12).
d 2y
= −4 and is negative so (0, 0) is a
dx 2
maximum point.
If x = 0,
b f ″(x) = 12x − 18
f ″(1) = 12 – 18 = −6
d 2y
= 8 and is positive so (1, −1) is a
dx 2
minimum point.
(1, 13) is a maximum point.
If x = 1,
f ″(2) = 24 – 18 = 6
(2, 12) is a minimum point.
c
13
12
8
5
a
2
If x = −1, d y2 = 8 and is positive so (−1, −1) is a
dx
minimum point.
y
0
7
1
2
dy
= 4x3 − 4x
dx
At a stationary point:
b f ′(x) = 3(x2 + 4)2 × 2x = 6x(x2 + 4)2. This is
negative if x is negative and positive if x is
positive. The function is neither increasing
nor decreasing.
300
c f(x) = 100x−3 so f'(x) = 100 × −3x−4 = − 4 . Now
x
300
x ≠ 0 so x4 is always positive and so − 4 is
x
always negative. The function is decreasing.
x
dy
= 3x2 − 12x − 180
dx
At a stationary point,
dy
= 0.
dx
3x2 − 12x − 180 = 0
Divide by 3 and factorise.
x2 − 4x − 60 = 0
a f ′(x) = 3x2 + 6x + 3 = 3(x2 + 2x + 1) = 3(x + 1)2
which, because of the square, is  0 for all
values of x. The function is increasing.
8
y = 10x−1 +
1
x
4
dy
1
= −10x−2 +
4
dx
(x − 10)(x + 6) = 0
x = 10 or −6
(10, −1400)
At a stationary point:
1
−10x−2 + = 0
4
x = −6, y = (−6)3 − 6 × (−6)2 − 180 × (−6) = 648
Rearrange.
The stationary points are (10, −1400) and
(− 6, 648).
10x−2 =
If x = 10, y = 103 − 6 × 102 − 180 × 10 = −1400.
d 2y
b
= 6x − 12
dx 2
d 2y
= 48 > 0, so (10, −1400) is a
If x = 10,
dx 2
minimum point.
d 2y
= −48 < 0, so (− 6, 648) is a
If x = − 6,
dx 2
maximum point.
1
4
40 = x2
x = 40 = 2 10
9
dv
=6−t
dt
At a maximum point:
dv
= 0.
dt
6−t=0
t=6
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6 DIFFERENTIATION
d 2v
= −1 which is negative for any value of t so
dt 2
this is a maximum point.
Exercise 6.6A
1
a This parabola crosses the x-axis at 1 and −3.
y
If t = 6, v = 10 + 6 × 6 − 0.5 × 62 = 10 + 36 − 18 = 28.
The maximum speed is 28 m s−1.
10 a
dy
= 23 − 10t
dt
0
–3
When the height is a maximum:
dy
=0
dt
–4
4
t = 2.3
b y = x2 + 2x − 3
d 2y
= −10 < 0 so this is a maximum point.
dt 2
dy
= 2x + 2
dx
When x = 2.3, y = 23 × 2.3 − 5 × 2.32
c If x = 3,
= 26.45.
The maximum height is 26.45 m.
dy
= 2 × 3 + 2 = 8.
dx
The equation of the tangent is y − 12 = 8(x − 3)
b Air resistance will reduce the maximum
height of the ball.
y − 12 = 8x − 24
y = 8x − 12
dy
11 a
= 3x 2 + 12x − 15 . At a stationary point
dx
d The gradient of the normal is −
dy
= 0 so 3x2 + 12x – 15 = 0;
dx
1
8
The equation of the normal is
y − 12 = − 1 (x − 3).
8
8y − 96 = −x + 3
x2 + 4x – 5 = 0; (x + 5)(x – 1) = 0; x = −5 or 1.
If x = −5, then y = – 125 + 150 + 75 – 90 = 10 so
(−5, 10) is a stationary point.
x + 8y = 99
dy
= 2 × −3 + 2 = −4, the tangent is
dx
y − 0 = −4(x + 3)
2
d y
d y
= 6x + 12, and if x = −5 then
= −18
dx 2
dx 2
and so (−5, 10) is a maximum point.
b If y = x3 + ax2 + bx + c, then
x
–3
23 − 10t = 0
2
1
e If x = −3,
y + 4x + 12 = 0
dy
= 3x 2 + 2ax + b .
dx
1
The gradient of the normal is and the
4
equation is y = 1 (x + 3) or 4y = x + 3.
4
There are two stationary points if the
quadratic equation has two distinct roots.
That is when the determinant ‘b2 – 4ac’ > 0;
(2a)2 – 4 × 3b > 0; 4a2 – 12b > 0; 4a2 > 12b; a2 > 3b.
12 a f′(x) = 3x2 + 2ax, and if f′(x) = 0 then 3x2 + 2ax = 0;
x(3x + 2a) = 0;
2a
x = 0 or − . There is a stationary point if x = 0.
3
b f ″ (x) = 6x + 2a; if x = 0, then f ″ (x) = 2a; if a > 0
then f ″ (x) > 0 and the point is a minimum point.
c There is another stationary point where
2a
24
x=−
=−
= −8.
3
3
Then f(–8) = (–8)3 + 12 × (–8)2 + 6 = 262; the
point is (−8, 262).
2
a If x = −1, y = (−1)3 − 4 × (−1)2 = −5 so (−1, −5) is
on the curve.
dy
= 3x2 − 8x
dx
dy
c If x = −1,
= 3 × (−1)2 − 8 × (−1) = 3 + 8 = 11.
dx
The equation of the tangent is y + 5 = 11(x + 1).
b
y + 5 = 11x + 11
y = 11x + 6
d The equation of the normal is y + 5 = − 1 (x + 1).
11
11y + 55 = −x − 1
11y +x = −56
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6
WORKED SOLUTIONS
3
y = 12x−2
dy
= −24x−3
dx
If x = 1, dy = −24.
dx
The tangent crosses the x-axis at 3.5 and the
1
area of the triangle is × 14 × 3.5 = 24.5.
2
6
a
The gradient of the normal is 1 .
24
y
The equation of the normal is y − 12 = 1 (x − 1).
24
24y − 288 = x − 1
2
24y = x + 287
4
1
a y = 12x 2
1
−1
dy 1
= (x + 4) 2 .
dx 2
dy 1 1 1
= × = .
At P, x = 0 and so y = 2 and
dx 2 2 4
b y = (x + 4)2 so
dy
−1
6
2
=
dx 6x = x
6
dy
If x = 4,
= 3.
=
dx
4
1
Gradient of normal = − .
3
1
Equation of normal is y − 24 = − (x − 4).
3
3y − 72 = −x + 4
The equation of the tangent at P is
1
y − 2 = (x − 0) or just y − 2 = 1 x.
4
4
At Q, y = 0 and so x = –8 and the coordinates
of Q are (–8, 0).
x + 3y = 76
dy
6
3
b If x = 100,
= .
=
dx
100 5
3
Gradient of tangent = .
5
3
Equation of tangent is y − 120 = (x − 100).
5
5y − 600 = 3x − 300
7
a
a
dy 2 − 13
= x
dx 3
−1
dy
2
2 1 1
If x = 8,
= ×8 3 = × = .
3
3 2 3
dx
The equation of the tangent is y − 4 =
3y − 12 = x − 8
1
(x − 8).
3
3y = x + 4
5y = 3x + 300
5
x
0
–4
dy
= −2x
dx
dy
If x = 2,
= −4 and the equation of the
dx
tangent is y − 6 = −4(x − 2).
b You need to find the coordinates of the point.
If
dy
−1
= 1, 2x 3 = 1.
dx
6 3
6
−1
4x 3 = 1
y − 6 = −4x + 8
1
4 = x3
y + 4x = 14
x = 43 = 64
b Draw a diagram. Only the tangent is required.
2
y = 64 3 = 42 = 16
y
Equation of tangent is y − 16 =
1
(x − 64).
6
6y − 96 = x − 64
6y = x + 32
10
8
a y = x(10 − x) so it crosses the x-axis at 0 and 10.
y
–10
10
x
0
10
x
10
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6 DIFFERENTIATION
b
Where it meets the x-axis y = 0
100
100
= − 2 × ( x − a ) so a = x – a and x = 2a
and −
a
a
1 200
× 2a = 200.
The area of the triangle is ×
2
a
dy
= 10 − 2x
dx
dy
If x = 3,
= 10 − 2 × 3 = 4.
dx
Equation of tangent is y − 21 = 4(x − 3).
y − 21 = 4x − 12
11 a
y = 4x + 9
1
Equation of normal is y − 21 = − (x − 3).
4
4y − 84 = −x + 3
The equation of the tangent is y – a2 = 2a(x – a);
at the point T, x = 0 and y – a2 = – 2a2; y = – a2;
so OT = a²
4y + x = 87
A is the point (0, a²) and so OA = a² = OT.
1
b The gradient of the normal is −
and the
2a
1
equation is y − a 2 = − (x − a)
2a
Sketch the tangent and the normal (the
curve is not necessary).
y
1
At N, x = 0 and y − a 2 = − 1 × (−a); y = a 2 + ;
2
2a
(3, 21)
21.75
dy
dy
= 2x ; at P
= 2a and the gradient of the
dx
dx
tangent is 2a.
hence ON = a 2 +
9
12 The normal is y =
0
The tangent meets the y-axis at 9. The
normal meets the y-axis at
87
3
= 21 .
4
4
The area of the triangle is
(
1
−1
dy 1
1
= (x + 9) 2 =
; if
dx 2
2 x+9
dy
1
= .
x = 16, then
dx 10
b y = (x + 9)2 so
y = a(1 + x2)−1
Using the chain rule,
If x = 1 then y =
dy
= a × −1(1 + x2)−2 × 2x
dx
−2ax
.
=
(1 + x 2)2
at A y = 0 and −5 =
a
a
= − (x − 1).
2
2
Where this meets the y-axis, x = 0 and then
a a
y − = so y = a and this is the point where the
2 2
curve crosses the y-axis.
100
10 f(x) = 100x–1 so f ′ ( x ) = −100x −2 = − 2 ;
x
100
f ′ (a ) = − 2
a
The equation of the tangent is
100
100
y−
= − 2 (x − a)
a
a
Where it meets the y-axis, x = 0 and
y−
The equation of the tangent is y − 5 =
dy −2a −a
a
and
= 2 =
.
2
2
dx
2
The equation of the tangent is y −
84
Then y = 9 – 12 = – 3; P is (−3, −3).
13 a If x = 16 then y = 16 + 9 = 25 = 5 so P is on
the curve.
)
1
3
1
21 − 9 × 3 = 19 or 19.125.
2
4
8
9
1
3
x − which has a gradient of
2
2
1
so the gradient of the tangent at P is −2
2
dy
= 2x + 4; if 2x + 4 = – 2,
If y = x2 + 4x, then
dx
then 2x = – 6 and x = –3.
x
3
1
1
and AN = ON – OA = .
2
2
200
100
100
100
= − 2 × ( −a ) =
so y =
.
a
a
a
a
1
(x − 16);
10
1
(x − 16); – 50 = x – 16; x = −34
10
and A is (−34, 0).
The gradient of the normal is −10; The equation
of the normal is y – 5 = –10(x – 16); at B y = 0 and
1
– 5 = –10(x – 16); = x − 16; x = 16.5 and A is
2
(16.5, 0).
The length of AB is 34 + 16.6 = 50.5.
Exercise 6.7A
1
When the depth is h cm, the volume V = π × 302 × h
= 900πh cm3.
dV
dV
= −400 cm3 s−1 and
= 900π
dt
dh
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6
WORKED SOLUTIONS
dV dV dh
dh
=
×
so −400 = 900π ×
dt
dh dt
dt
50
dV dV dx
dx
dx
=
×
=
so 50 = 3x2 ×
and
dt 3x 2
dt
dx dt
dt
dh
400
=−
= −0.141.
900π
dt
The depth is decreasing at a rate of 0.141 cm s−1.
The rate of increase of the surface area is
Therefore
2
The area is A = πr2 and
dA
= 2πr.
dr
dr
dA dA dr
= 0.3 and
=
×
= 2πr × 0.3 = 0.6πr
dt
dt
dr dt
4
7
b When r = 10,
dA
= 0.6π × 10 = 18.8 cm2 s−1.
dt
h=
c When r = 15,
dA
= 0.6π × 15 = 28.3 cm2 s−1.
dt
Therefore
If the radius is r then the surface area A = 4πr2.
dr
dA
= 0.5 and
= 8πr
dt
dr
dA dA dr
=
×
= 8πr × 0.5 = 4πr
dt
dr dt
dA
When r = 2.8,
= 4π × 2.8 = 35.2 cm2 s−1.
dt
dr
= 7.5 = 0.0746 cm s−1.
dt 2π × 16
b When r = 8,
dr
= 7.5 = 0.0187 cm s−1.
dt 2π × 64
When the depth is h the radius of the circle
8
h = 0.4h.
is
20
1
The volume of water V = π × (0.4h)2 × h = 0.1676h3
3
dV
= 0.5027h2.
and
dh
dV
dV dV dh
dh
= 25 and
=
×
so 25 = 0.5027h2 ×
dt
dt
dh dt
dt
dh 49.74
=
.
and
dt
h2
dh 49.74
When h = 12,
=
= 0.345 cm s−1.
dt
122
If the side of the cube is x cm then the volume
V = x3 and the surface area A = 6x2.
dV
dA
= 50, dV = 3x2 and
= 12x
dt
dx
dx
dh = − 2000 × 0.2 = − 400 .
dt
x3
x3
The height is decreasing at a rate of 0.4 cm s−1
8
2
a
u = 150w 3 so
2 − 1 100
du
= 150 × w 3 = 3 ; if w = 8,
3
dw
w
du 100
= 3 = 50
dw
8
4
dV
dV
= 4πr2.
The volume V = πr3,
= 30 and
3
dr
dt
a When r = 4,
1000
2000
dh
= 1000x−2 so
= −2000x−3 = − 3 .
dx
x2
x
When the shape is a cube, x = h and x3 = 1000 so
dh
400
=−
= −0.4.
1000
dt
dr = 30 = 7.5 .
dt 4πr 2 πr 2
6
dx
= 0.2 and the volume is x2h = 1000 and
dt
dh dh dx
=
×
.
dt dx dt
dV dV dr
dr
=
×
so 30 = 4πr2 ×
and
dt
dr dt
dt
5
dA 200
=
= 40 cm2 s−1.
5
dt
When x = 5,
dA
= 0.6π × 5 = 9.42 cm2 s−1.
dt
a When r = 5,
3
dA dA dx
50
200
=
×
= 12x × 2 =
.
x
dt
dx dt
3x
b
dw
= 0.05
dt
c
du du dw 100 dw
5
=
×
= 3 ×
= 3 ; when w = 64,
dt dw dt
dt
w
w
du
5
5
= 3
= = 1.25
dt
64 4
9
dV dA dV dV
dA
a ×
=
;
= 4πr 2 and
= 2πr so
dA dr
dr dr
dr
dV
× 2πr = 4πr 2 ; dV = 2r
dA
dA
1000
= 100; the volume is 100 cm³
5+5
dV
= −1000(t + 5)−2 ; when t = 5,
c V = 1000(t + 5)–1;
dt
dV
1000
=−
= 10 cm³s-1
dt
10 2
b V =
1
10 a
10 + 0.4t 2;
−1
dp
0.2
= 0.2t 2 =
; when t = 100,
dt
t
dp 0.2
=
= 0.02
10
dt
b
dv dv dp
1200
dv
1200
;v=
=
×
so
=− 2
p
dt dp dt
dp
p
When t = 100 p = 10 + 0.4 100 = 14;
dv dv dp
1200
=
×
=−
× 0.02 = −0.122 to 3d.p.
dt dp dt
14 2
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6 DIFFERENTIATION
Exam-style questions
1
5
1
a y = (a + bx 2)2 and
a y = 2x + 8x−1
−1
dy 1
= (a + bx 2) 2 × 2bx =
dx 2
dy
8
= 2 − 8x−2 = 2 − 2
dx
x
2x2 − 10x + 8 = 0.
6
Divide by 2: x2 − 5x + 4 = 0;
factorise: (x − 1)(x − 4) = 0.
a y = x4 – px2;
If x = 1, then dy = 2 − 8 = −6; if x = 4, then
dx
dy
= 2 − 0.5 = 1.5.
dx
x
1
2 2
(a + bx )
=
1 dy .
×
b dx
dy
= 4x 3 − 2px
dx
dy
= 4x 3 − 16x
dx
b If p = 8, then
So x = 1 or 4
1
(a + bx 2)2
1
b Since y = (a + bx 2)2 , x =
y
8
b If y = 10, then 2x + = 10; 2x2 + 8 = 10x;
x
2
bx
At a stationary point 4x3 − 16x = 0.
Divide by 4 and factorise.
x(x2 − 4) = 0
The gradient is 0 when x = −1 or 4.
x(x − 2)(x + 2) = 0
y
x = 0, 2 or −2
If x = 0, y = 0.
If x = 2, y = 24 − 8 × 22 = −16.
–1
3
0
4
If x = −2, y = (−24) − 8 × (−22) = −16.
x
Stationary points are (0, 0), (2, −16) and (−2, −16).
c
a y = 3x2 – 3x + ax – a
dy
= 6x − 3 + a
dx
b 6x – 3 + a = a
d 2y
= 48 − 16 = 32 > 0 so (2, −16) is
dx 2
a minimum point.
If x = 2, then
6x = 3
d 2y
= 32 > 0 so (−2, −16) is a
dx 2
minimum point.
If x = – 2, then
1
x=
2
y=
( 32 + a )(− 12 )
=−
3 a
−
4 2
(
4
a
1 3 a
,− −
2 4 2
d 2y
= 0 and this does not
dx 2
determine the type of stationary point at (0, 0).
If x = 0, then
However the other two stationary points are
below (0, 0) so it is a maximum point.
)
1
3
1
y = x 2(x + 4) = x 2 + 4x 2
−1
dy 3 12
= x + 2x 2
dx 2
b
−1
3 1
At a stationary point x 2 + 2x 2 = 0.
2
1
Multiply by x 2 :
x=−
d 2y
= 12x 2 − 16
dx 2
y
d
–2
0
2
x
–16
3
3
x + 2 = 0; x = −2;
2
2
4
1
or −1 .
3
3
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6
WORKED SOLUTIONS
7
dy
= 6x2 − 12x − 12
dx
If the gradient is 6, then 6x2 − 12x − 12 = 6.
x2 − 2x − 3 = 0
(x − 3)(x + 1) = 0
x = 3 or −1
If x = 3, y = 2 × 27 − 6 × 9 − 12 × 3 + 4 = −32.
If x = −1, y = − 2 − 6 + 12 + 4 = 8.
The points are (3, −32) and (−1, 8).
8
dh
When t = 10, dt = 4 − 3 = 1.
Speed is 1 m s−1 upwards.
dh
= 0.
dt
0.4t − 0.03t2 = 0
t(0.4 − 0.03t) = 0
Either t = 0, which is when the drone starts,
or 0.4 − 0.03t = 0.
t=
0.4
1
= 13
0.03
3
2
of 20 seconds, showing the ascent
3
is twice as long as the descent.
This is
9
b If p = 1 the stationary points are given by
3x2 − 8x + 5 = 0.
x = 1 or 5 .
3
If x = 1, y = –9 and
if x = 5 , y = – 247 or –9.148.
27
3
The coordinates of the stationary points are
(1, –9) and 5 , − 247
3
27
(
a dh = 0.4t − 0.03t2
dt
b At the highest point,
The roots of the equation p(16p – 15) = 0 are
15
p = 0 or
16
There are two distinct stationary points if
15
p < 0 or p >
16
a y = x2 + 250x−1
dy
= 2x − 250x−2
dx
b At a stationary point, 2x − 250x−2 = 0.
2x3 = 250
x3 = 125
x=5
d 2y
= 2 + 500x−3
dx 2
d 2y
If x = 5,
= 2 + 500 × 5−3 = 6 > 0 so this is a
dx 2
minimum point.
250
= 75.
If x = 5, y = 52 +
5
The minimum is at (5, 75).
dy
10 a y = px3 – 4px2 + 5x – 11 so
= 3px 2 − 8px + 5
dx
At a stationary point 3px2 – 8px + 5 = 0.
This quadratic in x has two distinct solutions
if b2 – 4ac > 0.
That is (– 8p)2 – 4 × 3p × 5 > 0; 64p2 – 60p > 0;
16p2 – 15p > 0; p(16p – 15) > 0
)
1
11 af(x) = (5x + 4)2 and so using the chain rule
f ′(x) =
1
5
1 (5x + 4)− 2
×
×5=
.
2
2 5x + 4
If x  0, then 5x + 4 is always positive. So is
f ′(x) and this means the function is increasing.
5
5
b When x = 1, f ′(x) =
= The gradient of
2×3 6
5
the tangent is .
6
Hence the gradient of the normal is − 6 and
5
the equation is y − 3 = − 6 (x − 1) which can be
5
rearranged as 5y − 15 = −6x + 6 or 5y + 6x = 21.
c f′(x ) =
5
5
so f ′ (1) = and the
6
2 5x + 4
tangent at (1, 3) is y − 3 =
5
(x − 1).
6
5
Where it meets the y-axis, x = 0 and y − 3 = − ,
6
y=
13 .
6
Where it meets the x-axis, y = 0 and
5
18
13
−3 = (x − 1), x − 1 = − ; x = − .
6
5
5
1 13 13 169
The area of the triangle is ×
×
=
2 6
5
60
or 2.82 to 3 s.f.
12 a The other side is = 240 − 2x.
The area A = x(240 − 2x).
b A = 240x − 2x2
dA
= 240 − 4x
dx
For maximum area, dA = 240 − 4x = 0.
dx
x = 60
d2A
= −4 < 0 so this will be a maximum.
dx 2
Maximum area = 60 × 120 = 7200 m².
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6 Differentiation
13 a
dy
= 2x + 4
dx
If x = 0, y =
dy
= 2 × 2 + 4 = 8.
dx
The equation of the tangent is y − 7 = 8(x − 2);
y = 8x − 9.
If x = 2,
b
dy
dy dy dx
= 2.4;
=
×
so at A,
dt
dt dx dt
dx dy dy
=
÷
= 2.4 ÷ 8 = 0.3 units s−1.
dt dt dx
dy
c The gradient is
so call this z.
dx
Then the rate of change of the gradient is
dz dz dx
=
×
.
dt dx dt
But
2
dz
dz d y
=
= 2 and so
= 2 × 0.3 = 0.6 units s−1.
dt
dx dx 2
14 a Volume of a cone, V =
h = 60 − r so V =
1 2
πr h.
3
1 2
πr (60 − r).
3
1
π (60r2 − r3)
3
dV = 1 π (120r − 3r2)
3
dr
b V=
When the volume is a maximum, dV = 0 .
dr
1
π(120r − 3r2) = 0
3
120r − 3r2 = 0
Divide by 3 and factorise.
r(40 − r) = 0
r = 0 or 40
d 2V = 1 π
(120 − 6r)
dr 2 3
d 2V = 1 π
(120 − 240) < 0 so
When r = 40,
dr 2 3
this is a maximum point.
c The maximum volume is
1
π 402(60 − 40)
3
= 32 000π cm 3 .
3
15 y = (x + 1)(x – p) = x2 + x – px – p and hence
dy
= 2x + 1 − p .
dx
dy
At (p, 0), x = p and
= 2p + 1 − p = p + 1.
dx
1
The gradient of the normal is −
and the
p +1
1
equation is y = −
(x − p).
p +1
y-axis.
p
and this is the intercept on the
p +1
If y = 0, x = p and this is the intercept on the x-axis.
p2
p
1
×p =
.
The area of the triangle is ×
2 p +1
2(p + 1)
4
16 When the radius is r, the volume V = πr3 and the
3
surface area A = 4πr2.
dV
dA
dV
= 4πr2,
= 8πr and
= 36
dr
dr
dt
dV dV dr
dr
=
×
so 36 = 4πr2 ×
and dr = 9 2
dt
dr dt
dt
dt πr
dA dA dr
9
72
=
×
= 8πr × 2 =
.
r
dt
dr dt
πr
1500
3
4
When V = 2000, πr 3 = 2000 and r = π and
3
r = 7.816.
Then
Then dA = 72 = 9.21.
dt 7.816
The area is increasing at a rate of 9.21 cm2 s−1.
17 af(x) = 10x – x2; f(3) = 21; f(3.1) = 21.39; f(3.5) = 22.75;
f(4) = 24
24 − 21
The gradient of AD is
= 3.
1
22.75 − 21
The gradient of AC is
= 3.5.
0.5
21.39 − 21
The gradient of AB is
= 3.9.
0.1
b The answers suggest that f'(3) is greater than
3.9 and could be 4.
18 aWhere the curve crosses the x-axis,
x(x + 2)(x – 5) = 0 so x = 0, −2 or 5; at P, x = −2.
dy
y = x3 – 3x2 – 10x and so
= 3x 2 − 6x − 10; if
dx
dy
x = −2, then
= 12 + 12 − 10 = 14.
dx
The gradient at P is 14.
b Where the gradient is 14, 3x2 – 6x – 10 = 14;
3x2 – 6x – 24 = 0; x2 – 2x – 8 = 0;
(x – 4)(x + 2) = 0; x = 4 or −2. Q is the point
where x = 4 and then y = 4 × 6 × −1 = −24
Q is (4, −24).
19 aThe height of the box is x cm, the length is
30 – 2x cm and the width is 20 – 2x cm.
Multiply these three to get v = x(30 – 2x)(20 – 2x).
b v = x(600 – 60x – 40x + 4x2) = 600x –100x2 + 4x3
dv
= 600 − 200x + 12x 2; where the volume
dx
dv
has a maximum value,
= 0;
dx
2
600 – 200x + 12x = 0; divide by 4: 3x2 – 50x + 150 = 0
88
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WORKED SOLUTIONS
50 ± 2500 − 1800 50 ± 700
=
;
6
6
x = 12.74 or 3.924. The first value is impossible;
it would give a negative volume. The volume is a
maximum when x = 3.92.
x=
d 2v
= −200 + 24x = −105.8 < 0 when x = 3.92
dx 2
which confirms that the volume is a maximum.
20 f ( x ) =
a
a ; At P, f ′ ( p ) = − a and the
;
x f′(x ) = − x 2
p2
a
a
= − 2 (x − p).
p
p
tangent is y −
At Q, y = 0 and −
If the side of the triangle is x cm the area is
1 x2 sin 60° = 0.433x2.
2
Then the volume is 0.433x2l = 1000 and so
2309
.
x2
The surface area is A = 2 × 0.433x2 + 3xl
= 0.866x2 + 6928 .
x
l=
, and if dA = 0, then
Then dA = 1.732x − 6928
dx
dx
x2
x3 = 6928 = 4000.
1.732
a
a
= − 2 (x − p); p = x – p; x = 2p.
p
p
Hence x = 3 4000 = 15.9.
a
a
a a
= − 2 × (−p); y − = ;
p
p p
p
The length is l =
At R, x = 0 and y −
y=
2
6
2309
= 9.2 cm.
15.9 × 15.9
2a
.
p
1
2a
× 2p ×
= 2a.
2
p
a
A is (p, 0) so OA = p; B is  0, a  so OB = ; the area
p
 p
a
of OAPB is p × = a. This proves the result.
p
The area of triangle OQR =
Mathematics in life and work
1
a Volume = x2h = 1000
Rearrange: h =
1000
x2
b The total surface area is 2x2 + 4xh
= 2x2 + 4x ×
c If a = 2x2 +
4000
1000
.
= 2x2 +
x
x2
4000
= 2x2 + 4000x−1, then
x
da
= 4x − 4000x−2 = 4x − 4000
.
dx
x2
When the surface area has a minimum value,
da = 0.
dx
=0
4x − 4000
x2
x3 = 1000
x = 10
2
d a
= 4 − 8000x−3
dx 2
2
8000
= 12 which is
When x = 10, d a2 = 4 +
103
dx
positive, so the surface area has a minimum
value. In fact, in this case the cuboid is a cube.
89
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7 Integration
7 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
3
−1
10 3
x +c
3
c 4x2 −
1
4x 2
2
The graph crosses the x-axis at 3 and −5.
4
2 5 5 2
b 5 x − 2 x + 10x + c
a 2x3 − 2x2 + c
a
y
1
3
2
∫ x 2 dx = 3 x 2 + c
1
b 2×
4 5
2 52
x + c = x2 + c
5
5
1
c 3 × 2x 2 + c = 6x 2 + c
16
−1
1
d
∫ 10x 2 dx = 20x 2 + c or 20 x + c
5
–5
3
–1
0
3
6
x
a 12x2 − 6
b y=
1
x + 2x−1
2
dy 1
1
2
= − 2x−2 or − 2
2 x
dx 2
4
∫ (x − 8x ) dx = 4 x − 3 x + c
c
∫ 10x
a
∫ 4x 2 dx = 4 × 3 x 2 + c = 3 x 2 + c
b
∫ 4x dx = ∫ 2 x dx = 2 × 3 x 2 + c = 3 x 2 + c
c
∫ 4x 2 dx = 4 × 2x 2 + c = 8x 2 + c or 8 x + c
3
2
−3
dx =
2
8
3
5
10 −2
x + c = −5x−2 + c or − 2 + c
−2
x
1
2
3
3
8
3
2
−1
1
4
3
1
8
a
You cannot integrate a product by integrating
each term separately and multiplying the results.
b First multiply out the brackets:
−2
dy
= 2x 3
dx
2
1
∫ (2x − x − 1) dx = 3 x − 2 x − x + c
2
9
Exercise 7.1A
6 2
a
x + c = 3x2 + c
2
b 2x2 + 2x + c
7 2
x − 5x + c
c
2
d 3x − 2x2 + c
1
3
2
a
∫ (4 − 2x )dx = 4x + x−2 + c
b
∫ (2x 2 + 8x 2 )dx = 2 × 3 x 2 + 16x 2 + c = 3 x 2 + 16x 2 + c
−1
−3
2
1
−1
3
1
2
4
3
3
1
4
3
1
1
2 × x 2 + 16x 2 + c = x 2 + 16x 2 + c
∫ (2x 2 + 8x 2 )dx = 3
3
c
2 5
x +c
5
b
2 −3
2
x = − x −3 + c
−3
3
c 5 x4 + c
4
d
5 −2
5
x + c = − x −2 + c
−2
2
a
1
b
2
f(x) = ∫ (2x 3 − 2x)dx = 2 x 4 − x 2 + c = 1 x 4 − x 2 + c
4
2
1
2
3
∫ (x − 4x) dx = 3 x − 2x + c
7
c y = 6x 3
1
1
a
∫ (4x
−2
−3
−1
+ 2x 2 ) dx= − 4x−1 − 4x 2 + c
10 The derivatives of f(x) and g(x) are the same so the
derivative of f(x) − g(x) is 0.
This means that f(x) − g(x) is ∫ 0 dx = a constant.
90
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7
WORKED SOLUTIONS
1
∫ ( x − 1) dx = ∫(x − 2x + 1) = 3 x − x + x + c
2
11 a
2
3
(
2
)
1
1 3
3
2
b 3 (x − 1) + c = 3 x − 3x + 3x − 1 + c
17 ∫(x + 2)2 dx = ∫(x 2 + 4x + 4) dx = 1 x 3 + 2x 2 + 4x + c
3
so Ari is correct.
However, (x + 2)3 = x3 + 3x2 × 2 + 3x × 22 + 23
= x3 + 6x2 + 12x + 8
Hence 1 (x + 2)3 = 1 x 3 + 2x 2 + 4x + 8 and this is the
3
3
3
= 1 x3 − x2 + x − 1 + c
3
3
(
)
12 ∫ ( x + 1)( x – 3)( x + 5) dx = ∫ x 2 – 2x – 3 ( x + 5) dx
=
∫ ( x + 3x –13x –15) dx
3
2
= 1 x 4 + x 3 − 13 x 2 − 15x + c
4
2
same as Ari’s answer except for the constant. This
means that 1 (x + 2)3 + c is the same solution written
3
in a different way. Both students are correct.
1
3
2 5
18 a ∫ x × x 2dx = ∫ x 2dx = x 2 + c
5
2
13 a x − 1 dx = (x + 1)(x − 1) dx
∫ x −1
∫ x −1
= ∫ ( x + 1) dx = 1 x 2 + x + c
2
b
4 − x2
∫ 2 + x dx = ∫
b
1
∫6x ( x + 1) dx = ( x + 1) + c and hence
2
2
(
1
2
dy
= 4 × (x + 7)3 × 1 = 4(x + 7)3.
dx
1
Hence ∫ (x + 7)3 dx = (x + 7)4 + c .
4
b If y = (x + 7)4, then
1
3
2
)
3
1 2
1
x + 1 + c , including the in the
6
6
dy
= 6 × (x + 7)5 × 1 = 6(x + 7)5.
dx
Hence ∫ (x + 7)5 dx = 1 (x + 7)6 + c .
6
c If y = (x + 7)6, then
arbitrary constant.
c
15 a
1 2
(x + 1)6 + c
12
2

1
−2
dy
= 4(6x + 1)3 × 6 = 24(6x + 1)3.
dx
1
(6x + 1)4 + c .
Hence ∫ (6x + 1)3 dx =
24
1 −1
x +c
−1
=x− 1 +c
x
=x+
b
 x 2 + 1
2

1
b If y = (6x + 1)4, then
dy
= 5(0.5x − 4)4 × 0.5
dx
= 2.5(0.5x − 4)4.
2
c If y = (0.5x − 4)5, then
∫  x 2  dx = ∫  1 + x 2  dx
(
)
= ∫ 1 + 2x −2 + x −4 dx = x +
= x − 2 − 13 + c
x 3x
(
)
2 −1 1 −3
x +
x +c
−1
−3
1
16 a y = x 2 + a 2 2 ; using the chain rule for
differentiation,
∫
x
dx =
x 2 + a2
Hence ∫ ( 0.5x − 4 )4 dx = 1 (0.5x − 4)5 + c
2.5
= 0.4(0.5x − 4)5 + c.
3
a
y = (10x + 1)−1 thus
−1
dy 1 2
= x + a 2 2 × 2x
dx 2
(
=
b
dy
= 3(2x − 3)2 × 2 = 6(2x − 3)2.
dx
1
Hence ∫ (2x − 3)2 dx = (2x − 3)3 + c .
6
aIf y = (2x − 3)3, then
∫ x 2 dx = ∫  1 + x 2  dx = ∫ (1 + x ) dx
x2 + 1
2
)
x
.
2
x + a2
2
8
2
 5
−1 
dx = ∫  x 3 − x 3  dx = 3 x 3 − 3 x 3 + c
8
2


dy
= 3 × (x + 7)2 × 1 = 3(x + 7)2.
dx
1
Hence ∫ (x + 7)2 dx = (x + 7)3 + c .
3
∫ x ( x + 1) dx = 6 ( x + 1) + 6 c or just
2
1
x3
a If y = (x + 7)3, then
3
2
x2 − 1
Exercise 7.2A
(2 − x)(2 + x) dx
2+ x
= ∫(2 − x)dx = 2x − 1 x2 + c
2
14 a f(x) = 3(x2 + 1)2 × 2x = 6x(x2 + 1)2
b
∫
x +a +c
4
3
dy
= −1 × (10x + 1)−2 × 10
dx
−10
= (10x + 1)2 .
3
0.3
b
∫ (10x + 1)2 dx = −10 (10x + 1) + c = − 10x + 1 + c
a
∫ x + 1 dx = ∫ (x + 1)2 dx
−1
1
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7 Integration
3
d
If y = (x + 1)2 , then
1
1
dy 3
3
= × (x + 1)2 × 1 = (x + 1)2
2
dx 2
Hence ∫ x + 1 dx =
b
6
6
3
2
(x + 1)2 + c .
3
5
5
7
6
1
1
= 7 ( x + 2) − 2 × 6 ( x + 2) + c
= 1 ( x + 2)7 − 1 ( x + 2)6 + c
7
3
1
3
If y = (2x + 1)2 , then
7
1
1
dy 3
= × (2x + 1)2 × 2 = 3(2x + 1)2 .
dx 2
c
5
= ∫ ( x + 2) dx − 2∫ ( x + 2) dx
∫ 2x + 1 dx = ∫ (2x + 1)2 dx
Hence ∫
∫ x ( x + 2) dx = ∫ {( x + 2) − 2 ( x + 2) } dx
1
x
x +1−1
x +1
=
=
−
(x + 1)3
(x + 1)3 (x + 1)3
a (x + 1)3
=
3
1
2x + 1 dx = (2x + 1)2 + c .
3
b
1
∫ 3 4x − 2 dx = ∫ 3(4x − 2)2 dx
3
1
1
−
(x + 1)2 (x + 1)3
x
1
1
∫ (x + 1)3 dx = ∫ (x + 1)2 dx − ∫ (x + 1)3 dx
−2
−3
= ∫(x + 1) dx − ∫(x + 1) dx
If y = (4x − 2)2 , then
1
1
dy 3
= × (4x − 2)2 × 4 = 6(4x − 2)2 .
dx 2
= 1 (x + 1)−1 − 1 ( x + 1)−2 + c
−1
−2
3
3
1
1
1
1
Hence ∫ 3 4x − 2 dx = 3 × (4x − 2)2 + c = (4x − 2)2 + c = −
+
+c
2
6
x + 1 2 ( x + 1)2
3
3
1
1
1
∫ 3 4x − 2 dx = 3 × 6 (4x − 2)2 + c = 2 (4x − 2)2 + c .
8 a∫ 10x + 5 dx = ∫(10x + 5)2 dx
5
a
1
∫ 3 0.6x + 5 dx = ∫ (0.6x + 5)3 dx
3
3
= 1 × 2 (10x + 5)2 + c = 1 (10x + 5)2 + c
15
10 3
4
If y = (0.6x + 5)3 , then
1
1
dy 4
= × (0.6x + 5)3 × 0.6 = 0.8(0.6x + 5)3 .
dx 3
Hence ∫ 3 0.6x + 5 dx =
1
4
b
4
1
(0.6x + 5)3 + c = 1.25(0.6x + 5)3 + c
0.8
4
4
∫ 3 0.6x + 5 dx = 0.8 (0.6x + 5)3 + c = 1.25(0.6x + 5)3 + c .
b
1
−1
∫ 3 0.6x + 5 dx = ∫ (0.6x + 5) 3 dx
−1
1
∫ 10x + 5 dx = ∫(10x + 5) 2 dx
1
1
× 2(10x + 5)2 + c
10
1
=
10x + 5 + c
5
=
1 10x + 5 − 1
(
) 2
x
10
=
1
2
+
+
10
x
5
10
x
5
− +1 =
3
3
1
1
1
=
10x + 5 − ×
2
2
10
10
x+5
3
If y = (0.6x + 5) , then
Hence
−1
−1
dy 2
= × (0.6x + 5) 3 × 0.6 = 0.4(0.6x + 5) 3 .
x
1
1
1
dx 3
∫ 10x + 5 dx = 10 ∫ 10x + 5 dx − 2 ∫ 10x + 5 dx
2
2
1
1
3
dx =
(0.6x + 5)3 + c = 2.5(0.6x + 5)3 + c 1
Hence ∫ 3
1
1 1
0.4
0.6x + 5
=
10x + 5 + c
× (10x + 5)2 − ×
2 5
10 15
2
2
1
1
3
∫ 3 0.6x + 5 dx = 0.4 (0.6x + 5)3 + c = 2.5(0.6x + 5)3 + c .
1
1
=
(10x + 5)2 −
10x + 5 + c
5
6
1
150
10
6 a ∫ ( x + 2) dx = ( x + 2) + c
6
9 f ′ ( x ) = ∫(ax + b)3 dx = 1 (ax + b)4 + c
b ∫ ( x + 2)6 dx = 1 ( x + 2)7 + c
4a
7
where c is a constant.
1
c (x + 2)6 = (x + 2) (x + 2)5 = x (x + 2)5 + 2 (x + 2)5
f ( x ) = ∫ f'(x)dx = ∫ (ax + b)4 + c
4a
Hence x(x + 2)5 ≡ (x + 2)6 – 2(x + 2)5.
1
1
×
(ax + b)5 + cx + d
=
4a 5a
c
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WORKED SOLUTIONS
=
1
(ax + b)5 + cx + d, where d is a constant.
20a 2
4
∫
There are two arbitrary constants.
2
2
1
3
2=5−
5
1
(2x + 5)3 dx
∫ 2x + 5 dx = ∫
b
(
a If x ≠ 0, then x2 is positive. This means that
always positive.
)
4
4
3
= 1 × 3 (2x + 5)3 + c = 8 3 2x + 5 + c
2 4
=
b y= ∫
c=5
The equation is y = − 2 + 5 = 5x − 2 .
x
x
a y = ∫ (4x − 2) dx
6
a y = ∫ (3x 2 − 3) dx
= x3 − 3x + c
b When x = 0, y = 3.
2=c
3 = 2 × 02 − 2 × 0 + c
The curve is y = x3 – 3x + 2.
=c
b The turning points are where 3x2 – 3 = 0
The equation is y = 2x2 − 2x + 3.
a y=∫
=
x2 = 1.
1
x 2 dx
x = 1 or −1
There are 2 stationary points.
2 32
x +c
3
If x = 1, y = 1 – 3 + 2 = 0 so one t.p. is (1, 0).
If x = −1, y = − 1 + 3 + 2 = 4 so the other t.p. is (−1, 4).
b If x = 9, y = 25
25 =
3
2
× 92 + c
3
=
2
× 27 + c
3
7
a f (x ) = ∫
4=−
= 18 + c
The equation is y =
2 32
x + 7.
3
= 0.2x2 + 3x + c
a 0=0+c
Hence the equation is y = 0.2x2 + 3x.
b 5=c
y = 0.2x2 + 3x + 5
c 0 = 0.2 × 52 + 3 × 5 + c
y = 0.2x2 + 3x − 20
10
+c
4
c = 4 + 2.5 = 6.5
y = ∫ (0.4x + 3)dx
c = −20
−3
20
dx = ∫ 20x dx = −10x −2 + c
x3
f(2) = 4
c=7
3
2
dx = ∫ 2x −2 dx
x2
4=−1+c
= 2x2 − 2x + c
2
2
is
x2
= −2x−1 + c
3 4
y +c
8
Exercise 7.3A
1
10
+c
5
c = −1
10
The equation is y = x −
− 1.
x
5
3
= (2x + 5)3 + c
10
3
x 2 + 10
dx = ∫ 1 + 10x −2 dx
x2
= x − 10x−1 + c
5
∫ ( 3 2x + 5 ) dx = ∫(2x + 5)3 dx = 2 × 5 (2x + 5)3 + c
10 a
7
f(x) = −10x−2 + 6.5 or f ( x ) = 6.5 −
10
x2
b
y
8
6
4
2
0
–2
2
4
6
8
10 12 x
–4
–6
–8
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7 Integration
∫ 0.03x ( x − 10 ) dx = ∫ 0.03x ( x − 20x + 100 ) dx
dy
12 dy = 4 − 100 ; at a stationary point
=0
2
d
x
dx
x
4
3
2
5
4
3
= ∫ (0.03x − 0.6x + 3x )dx = 0.006x − 0.15x + x + c
so 4 − 100 = 0; a2 = 25; a = 5.
a2
= ∫ (0.03x 4 − 0.6x 3 + 3x 2)dx = 0.006x 5 − 0.15x 4 + x 3 + c
 dx
The equation of the curve is y = ∫  4 − 100

When x = 0, y = 100, so c = 100.

x2 
5
4
3
y = 0.006x – 0.15x + x + 100
= 4x + 100 + c .
x
b After 10 seconds, x = 10 and
100
(5, 10) is on the curve so 10 = 20 +
+ c; c = −30.
y = 0.006 × 105 – 0.15 × 104 + 103 + 100
5
= 600 – 1500 + 1000 + 100 = 200
The curve is y = 4x + 100 − 30 .
x
The distance from A is 200 m.
13 f′(x) = mx + c; f′(–2) = – 2m + c = 7
1
2
3
2
f ( x ) = mx 2 + cx + d; f ( 0 ) = d = 18; f(– 2) = 2m – 2c
9 y = ∫ (3x − 12x + 8)dx = x − 6x + 8x + c
2
+ 18 = 8; 2m – 2c = – 10
When x = 0, y = 0 so c = 0.
8
a
2
2
2
2
Add the two equations: –c = – 3 so c = 3
y = x3 – 6x2 + 8x
Where the curve crosses the x-axis, x3 – 6x2 + 8x = 0.
Then – 2m + 3 = 7 so m = −2.
x(x2 – 6x + 8) = 0;
The equation is y = – x2 + 3x + 18.
dy
dy
= −0.5
14 a = k2 for some k. When x = 4,
dx
dx x
x(x − 2)(x − 4) = 0
x = 0, 2 or 4.
so k = −0.5 and k = −8.
16
The curve crosses the axes at (0, 0), (2, 0) and (4, 0).
−2
dy
8
= − 2 ; y = −∫ 8x −2 dx; y = 8 + c
x
dx
x
1
10 a y = ∫ 4x 3dx = 4 × 3x 3 + c
1
When x = 4, y = 6 so 6 = 8 + c and c = 4;
4
8
y = + 4.
x
dy
8
If the gradient is −2,
= − 2 = −2; x2 = 4;
dx
x
x = 2 or −2.
y = 12x 3 + c
When x = 8, y = 30.
1
30 = 12 × 8 3 + c
30 = 12 × 2 + c
c = 6
8
If x = 2, y = 8 + 4 = 8; if x = −2, y =
+ 4 = 0.
2
−2
1
y = 12x 3 + 6
The gradient is −2 at (2, 8) and (−2, 0).
1
b When x = 20, y = 12 × 20 3 + 6 = 38.57…
b
y
The radius is 38.6 cm to 3 s.f.
8
11 a f′(2) = a × 22 + b × 2 = 4a + 2b = – 0.8
---------(1)
f′(5) = a × 52 + b × 5 = 25a + 5b = 2.5
----------(2)
6
(1) × 5: 20a + 10b = – 4
4
(2) × 2: 50a + 10b = 5
Subtract: 30a = 9 so a = 0.3; then 1.2 + 2b = – 0.8
and b = −1.
b f′(x) = 0.3x2 – x so f(x) = 0.1x3 – 0.5x2 + c
f(2) = 4 so 0.1 × 8 – 0.5 × 4 + c = 4; c = 5.2
The equation of the curve is y = 0.1x3 – 0.5x2 + 5.2.
2
–6
–4
–2
0
2
4
6 x
–2
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WORKED SOLUTIONS
Exercise 7.4A
1
2
(
2
dx =  x 3  = 8 − 1 = 7
1
a
∫1 3x
b
4
 x 3  = 64 − 8 = 56
 2
2
1
1
dy 3 2
= x + 9 2 × 2x = 3x x 2 + 9 2 = 3x x 2 + 9 .
dx 2
3 4
4


Hence ∫ x x 2 + 9 dx =  1 x 2 + 9 2  =  125  − [ 9] = 32 2
 3 
3
3
0

0 
)
(
)
(
)
4
3
1 2

2
2
 125 
2
x
x
+
9
d
x
=
x
+
9

 =  3  − [ 9] = 32 3 .
∫0
3

0
(
4
3
c  x 3  = [ 27 ] −  −1 = 28
−1
6
Area =
2
)
2
5 2
∫−2(2 + 0.1x )dx =  2x + 0.02x −2 = [ 4.64] − [ −4.64] = 9.28
4
a  1 x 5  =  32  −  1  = 31 = 6.2
 5 1  5   5  5
2
2
(2 + 0.1x 4)dx =  2x + 0.02x 5  = [ 4.64 ] − [ −4.64 ] = 9.28
∫
4
−
−
2
2
b  2.5x 4  = [ 640 ] − [ 40 ] = 600
2
2
10
−2
−1 5
c ∫1 10x dx =  −10x 1 =  − 5  − [ −10 ] = −2 + 10 = 8 7


5
10
−2
−1  5



∫1 10x dx =  −10x 1 =  − 5  − [ −10 ] = −2 + 10 = 8
5
8
4
 2x 3 − 2x 2  = [128 − 32] − [ 54 − 18 ] = 96 − 36 = 60

3
2
5
2
3
2
∫ 3x + 6x dx =  x + 3x 
c
2
a
=
5
2
1
= 180
This shows that area of OAP is 2 of the area of OAPB.
3
4
1
4 3 
4
4
28
∫1 2x 2dx =  3 x 2  =  3 × 8  −  3  = 3 or 9 13
1
4
∫1 20x
b
−
1
1 4

2 dx = 40x 2
=



1
[80 ] − [ 40 ] = 40
4
1
4
4

 4
c
8x 3 dx = 6x 3  = 6 × 4 3  −
1

1 

4
4
4
∫



[6 ] = 32.10

1
x2 + 9 =
a Try y =
Then
(
4
0
4
3
2
1
3
3
2
3
3
1
6
∫0 (6x − x )dx = 3x − 3 x 0 = [108 − 72] − [0 ] = 36
6
2
1
2
3
6
∫0 (6x − x )dx = 3x − 3 x 0 = [108 − 72] − [0 ] = 36
6
2
2
3
c
y
10
1
x
2
x +9
8
y = 9 – x2
.
y = x(6 – x)
6
4
4
x
dx =  x 2 + 9  =  25  −  9  = 2
2
0

x +9
4
x
3
∫−3(9 − x )dx = 9x − 3 x −3 = [ 27 − 9] − [ −27 + 9] = 36
)
)
1
∫−3(9 − x )dx = 9x − 3 x −3 = [ 27 − 9] − [ −27 + 9] = 36
a
1
x2 + 9 2 .
−
dy 1 2
= x + 9 2 × 2x =
dx 2
Hence ∫
∫0
(
9
b
∫1 8x 3dx = 6x 3  = 6 × 4 3  − [6 ] = 32.10
5
3
The area of OAPB = OA × OB = a × a = a × a 2 = a 2.
4
a
2 32
2 3
a − 0 = a2 .
3
3
= [125 + 75] − [ 8 + 12]
= [ 200 ] − [ 20 ]
4
Suppose the coordinates of P are (a, a ).
a 1
 3
Then the area of OPA = ∫ x 2 dx =  2 x 2 
0
 3 0
2
b  x + 2x + 3x  = [ 8 + 8 + 6 ] − [ 0 ] = 22
0
3
1
= 32 + 42 2 − 48  = 26 2 m 2
3
3


4
a  2x 3 − 2x 2  = [128 − 32] − [ 54 − 18 ] = 96 − 36 = 60
3
3
4
4
1
3
4


Area = ∫ (8 + 8x 2 − 6x) dx = 8x + 16 x 2 − 3x 2 
3
0

0
2
–4
–3
–2
–1
0
1
2
3
4
5
6
7
8
–2
dx =  x 2 + 9  =  25  −  9  = 2.
0

x +9
3
1
1 One graph is a translation of the other and
dy 3 2
b Try y = x 2 + 9 2 . Then
= x + 9 2 × 2x = 3x x 2 + 9 2 =the
3xareas
x 2 + are
9 the same.
dx 2
x
2
(
)
(
)
(
)
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7 Integration
1
10 a If y = 2x − 2 , then y2 = 2x – 2 and x = y 2 + 1 .
2
(
)
Sketch of the curve:
v
4
4
4
The area is ∫ x d y = ∫ 1 y 2 + 1 d y =  1 y 3 + y 
0
2
0
 6

0
= 64 + 4 = 14 2 .
6
3
b The area is the area of OAPB − 14 2 = 36 − 14 2
3
3
1
= 21 .
3
11 a
b
1
∫0 x dx =  n + 1 x
1
n
1
1
n +1 
0
1
=
∫0 x (1 − x)dx = ∫0 x − x
n
n
1
1
−0=
n +1
n +1
n +1
–a
dx
The area is ∫
1
c
1
n
2
1
n
n+2
( a − x ) dx = a x − 13 x 
2
2
2
3
a
−a
=  a − 1 a 3  −  −a 3 + 1 a 3 
3  
3 

= 2 a3 + 2 a3 = 4 a3.
3
3
3
14 aWhere the curve meets the x-axis,
dx
3
(2x + 16)4 = 0 so 2x + 16 = 0 and x = −8;
1
=  1 x n +1 − 1 x n + 3 
 n + 1
0
n+3
the coordinates are (−8, 0).
2
= 1 − 1 = n + 3− n −1 =
n + 1 n + 3 (n + 1)(n + 3) (n + 1)(n + 3)
y = 16 4 = 23 = 8 so the coordinates are (0, 8).
Where it meets the y-axis, x = 0 and
3
12 a
v = 6t2 – t3; dv = 12t − 3t 2 ; when the speed is a
dt
3
0
b The area is ∫ (2x + 16)4 dx
−8
0
7
7
2



=  1 × 4 × (2x + 16)4  =  × 16 4  − [ 0 ]
7
2
7


 −8 
maximum, dv = 0 .
dt
12t – 3t2 = 0; 4t – t2 = 0; t(4 – t) = 0; t = 0 or 4
When t = 0, v = 0 and that will be a minimum.
When t = 4, v = 32.
The maximum speed is 32 ms–1.
b
t
3
1
1 = n + 2− n −1
1
−
n + 1 n + 2 (n + 1)(n + 2) = (n + 1)(n + 2)
∫0 x (1 − x )dx = ∫0 x − x
a
−a
=  1 x n +1 − 1 x n + 2 
n+2
 n + 1
0
=
a
v
= 2 × 128 = 256 or 36 4 .
7
7
7
Exercise 7.5A
1
32
2
2
1
8
4
a ∫ (x 2 − 2x)dx =  x 3 − x 2  =  − 4  − 0  = −

0  3
 3
3
0
4
The area is .
3
3
3 2
1 3
8
4
4
2
b Area = ∫2 (x − 2x) dx =  3 x − x  = [ 9 − 9] −  3 − 4  = 0 −  − 3  = 3
2

 


0
4
(
6
t3
1
3
8
4
4
∫2 (x − 2x) dx =  3 x − x 2 = [9 − 9] −  3 − 4  = 0 −  − 3  = 3 .
2
)
6
2
3
 3 1 4
c The distance is ∫0 6t − t dt =  2t − 4 t 
= [ 432 − 324 ] − 0 = 108.
3
2
c Since the curve is symmetrical, this area is
4
also , the same as part b.
3
6
0
2
3
81
1
∫2 x dx =  4 x 2 =  4  − [ 4 ] = 16 4
ii
1 4
81
1
3
∫−2 x dx =  4 x −2 =  4  − [ 4 ] = 16 4
The distance is 108 m.
13 Where the curve crosses the x-axis, a2 – x2 = 0
so x = ± a.
1
a i
3
3
3
4
3
2
2
1
iii ∫ x 3 dx =  x 4  = [ 4 ] − [ 4 ] = 0
−2
 4  −2
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7
WORKED SOLUTIONS
b
4
4
1
64 
−
Area under curve = ∫ (4x − x 2) dx =  2x 2 − x 3  = 32 −
3 1 
3 
1

y
40
1
4
4
20
2
2
3
–3
–2
0
–1
–10
1
2
3
4
x
Area of triangle =
–20
1
1
×3×3= 4 .
2
2
Area between the curve and the straight line
1
1
=9−4 = 4 .
2
2
–30
–40
5
The integral from −2 to 2 in part iii above is zero
because the graph is symmetrical and the areas
above and below the x-axis are the same.
3
1
2
2
= 10  − 1  = 9 .
 3   3 
10
–4
64
∫1 (4x − x ) dx =  2x − 3 x 1 = 32 − 3  −  2 − 3 
30
a The graphs look like this and cross at (0, 0)
and (1, 1).
y
6
a Where they intersect, x2 = 2 – x.
5
x2 + x – 2 = 0
4
(x + 2)(x − 1) = 0
3
x = −2 or 1
2
If x = –2, then y = 4
1
If x = 1, then y = 1
–3
The points are (–2, 4) and (1, 1).
–2
0
–1
–1
1
2
3
4
5
6
7
x
1
1
1 2
1 2 1 3 –2
b The area between them is ∫ (2 − x) dx − ∫ x dx =  2x − 2 x − 3 x 
−2
−2
 −–3

2
1
1
1 2
1
1
2
3

∫−2(2 − x) dx − ∫−2 x dx =  2x − 2 x − 3 x −2
The area between them is
4
a Where the lines cross, 4 – x = x(4 − x).
4 – x = 4x – x2.
x2 – 5x + 4 = 0
Factorise: (x − 1)(x − 4) = 0.
x = 1 or 4
Points are (1, 3) and (4, 0).
1
n +1
1 1
 n

1
1
n
x n −
x n +1  =
−
b Area = ∫ (x n − x n) dx = 
1
1
n
+
1
n
+
+
+1
n
n
0

0
1
n +1
1 1
 n
1
n
1
n −1
n
n +1 
∫0 (x n − x ) dx =  n + 1 x n − n + 1 x  = n + 1 − n + 1 = n + 1 .
0
c When n is large the area is close to, but less
than, one.
b
6
y
A
3
x2 – 4x + 3 = 0.
Factorise: (x − 3)(x – 1) = 0.
2
1
0
–1
–1
–2
2
Where they cross, 4x – x2 = x2 – 4x + 6.
Therefore 2x2 – 8x + 6 = 0.
4
–2
1
1
2 3 1 3
2 1 1
2
∫0 (x 2 − x ) dx =  3 x 2 − 3 x  = 3 − 3 = 3 .
0
1
=  2 − 1 − 1  −  −4 − 2 + 8  = 7 + 10 = 4 1 .
2 3  
3  6 3
2

C
1
The curves cross where x = 1 or 3.
B
2
3
4
5
6
7
x
Area between = ∫
3
2
3
1
((4x − x ) − ( x − 4x + 6)) dx = ∫ 8x − 2x − 6)dx
2
2
∫1 (( 4x − x ) − ( x − 4x + 6 )) dx = ∫1 8x − 2x − 6)dx
3
–3
3
The area required is the difference between
the area under the curve between A and B,
and the area of triangle ABC.
2
3
1
2
3
2
=  4x 2 − x 3 − 6x  = [ 36 − 18 − 18 ]
3
1

2
8
8
2
−  4 − − 6  = [ 0 ] −  −  = or 2 .
3
3


 3  3
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2
7 Integration
7
Where the curves cross, (x – 2)2 = (x – 10)(2 – x).
9
Hence x2 – 4x + 4 = − x2 + 12x – 20.
a
Where they cross 0.5x2 + 6 = 2x2; 1.5x2 = 6; x2 = 4;
x = 2 or −2.
If x = 2 or −2, y = 8; the curves cross at (2, 8) and
(−2, 8).
Hence 2x2 – 16x + 24 = 0.
Divide by 2: x2 – 8x + 12 = 0
b
Factorise: (x – 2)(x – 6) = 0.
y
Hence x = 2 or 6.
8
The required area is
6
6
10
∫2 ( x − 2) dx + ∫6 ( x − 10 )( 2 − x ) dx .
2
1
6
64
1
∫2 ( x − 2) dx =  3 ( x − 2) 2 =  3  − [0 ] = 21 3
6
2
3
4x
0
–4
2
2
2
2
c Area
10 = ∫ (0.5x + 6)d x − ∫ 2x d x
−2
−2
1 3
2


∫6 ( x − 10 )( 2 − x ) dx = ∫6 −x + 12x − 20 dx =  − 3 x + 6x − 20x 6
2
2
=  1 x 3 + 6x  −  2 x 3 
10
10
 6
 −2  3  −2
1
x − 10 )( 2 − x ) dx = ∫ −x 2 + 12x − 20 dx =  − x 3 + 6x 2 − 20x 
6
6
 3
40
40
  −  −  −  16  −  − 16  = 16.
 3   3 
 3   3 
1000
2
2
= 42
= −
+ 600 − 200  − [ −72 + 216 − 120 ] = 66  − [ 24 ]10
3
aIf
3
3 y = 0, then x – 5x2 + 6x = 0; x(x2 – 5x + 6) = 0;


 3 
x(x – 2)(x – 3) = 0;
2
2
− [ −72 + 216 − 120 ] = 66  − [ 24 ] = 42
 3 
3
x = 0, 2 or 3 and the coordinates are (0, 0), (2, 0)
and (3, 0).
1
2
The required area = 21 + 42 = 64.
3
3
y
b
10
10
2
{
8
}
aWhere they cross, x2 + 1 = 2x + 9; x2 – 2x – 8 = 0;
(x – 4)(x + 2) = 0.
x = 4 or −2. If x = 4, then y = 17; if x = −2, then
y = 5. The points are (4, 17) and (−2, 5).
b Here is a sketch:
y
0
2
x
3
c Area between 0 and 2
(
2
)
2
= ∫ x 3 − 5x 2 + 6x dx =  1 x 4 − 5 x 3 + 3x 2 
3
0
 4
0
0
–2
4
x
Area under straight line =
4
∫−2 ( 2x + 9) dx =  x + 9x −2 = [52] – [–14] = 66
(Alternatively, this is a trapezium and the area is
1
5 + 17 ) × 6 = 66 , the same answer.)
2(
4
(
∫2 ( x − 5x + 6x ) dx =  4 x − 3 x + 3x 2
3
4
2
=  4 − 40 + 12  − [ 0 ] = 8


3
3
)
2
Area under curve = ∫−2 x + 1 dx
3
2
1
4
5
3
2
3
=  81 − 45 + 27  −  8  = 9 − 8 = − 5
12
 4
  3  4 3
1
The total area is 8 + 5 = 37 or 3 .
12
3 12 12
4
=  1 x 3 + x  =  76  −  − 14  = 30.
 3
 −2  3   3 
The area between is 66 – 30 = 36.
98
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7
WORKED SOLUTIONS
11 a
y = x(x – 4) is a parabola that crosses the x-axis
at 0 and 4.
At P, x =
7
1 7 5 17
and y = × + =
; P is
2
2 2 2 4
( 72 , 174 ) or (3 12 , 4 14 ) .
The straight lines cross where 2x = 8 – 2x; 4x = 8;
x = 2; cross at (2, 4).
The lines look like this:
c The lines look like this:
y
y
4
4
3
0
2
4
2
x
2
1
0
–1
b
∫0 x(x − 4)dx = ∫0 ( x − 4x ) dx
4
4
2
Area under curve = ∫
2
3.5
1
3
4
x
( x − 4x + 6) dx
2
3.5
=  1 x 3 − 2x 2 + 6x 
1
 3
4
=  1 x 3 − 2x 2  =  64 − 32  − [ 0 ] = − 32
 3
0  3

3
= [10.792] – [4.333] = 6.4583.
Area of trapezium = 1 × ( 3 + 4.25) × 2.5 = 9.0625
2
The area below the x-axis is 32 .
3
The area of the triangle above the x-axis is
1
× 4 × 4 = 8.
2
(or by integrating, area = ∫
3.5
1
(0.5x + 2.5) dx
3.5
= 0.25x 2 + 2.5x  = [11.8125] – [2.75] = 9.0625

1
The total area is 8 + 32 = 18 2 .
3
3
12 a dy = 2x − 4 ; at a stationary point 2x – 4 = 0;
dx
x = 2 and then y = 4 – 8 + 6 = 2.
d 2y
= 2 > 0 so (2, 2) is a minimum point.
dx 2
dy
b At (1, 3)
= 2 − 4 = −2 ; the gradient of the
dx
tangent is −2; the gradient of the normal is 1 .
2
1
The equation of the normal is y − 3 = (x − 1) or
2
1
5
y= x+ .
2
2
1
which is the same answer).
Area enclosed = 9.0625 – 6.4583 = 2.60 to 3 s.f.
Exercise 7.6A
1
=  20 x 
2
25
1
25
−1


dx = ∫ 10x 2 dx = 10 × 2x 2  = [100 ] − [ 0 ] = 100
0
x

0
25 10
Area = ∫0
The area = ∫
25
0
∞ 20
= [100 ] − [ 0 ] = 100 .
∞
∞
dx = ∫ 20x −3 dx =  −10x −2 
2
2
x3
= [ 0 ] − [ −2.5] = 2.5.
2
Where the curve and the normal meet,
∞
∞
−2
−1 ∞
8
dx = ∫ 8 ( x + 2) dx =  −8 ( x + 2)  = [ 0 ] −
3 The area = ∫
5 ; 2x2 – 8x + 12 = x + 5;
1
2
0 (x + 2)2
0
0
x − 4x + 6 = x +
2
2
∞
∞
−2
−1 ∞
8
 8
=
7 2 dx = ∫0 8 ( x + 2) dx =  −8 ( x + 2)  = [ 0 ] −  − 2  = 4 .
2x2 – 9x + 7 = 0; (x – 1) (2x – 7) = 0; x∫0= (1xor
0
 
+ 2)
2
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7 INTEGRATION
4
a
b
y
y
10
0
5
x
2
The integral is represented by the unshaded
area.
2
c To evaluate  − 502  is impossible because
 x 0
0
you cannot divide by 0. The area between 0
and 2 is infinite.
x
2
4
4 20
20
20
20
a11
∫ 2 dx =  −  =  −  −  − 
10 dx = 1110 x − 2 − 12 dx = 10 × 2 x − 2 12  =2x20 x − 211x  2  4   2 
(
) 
∫2 ( )

x−2
= – 52 + 10 = 5
2 
8
11
b Area = ∫
2
11
1
11
11
−1


dx = ∫ 10 ( x − 2) 2 dx = 10 × 2 ( x − 2) 2  =  20 x − 2 
2
2

2
=  20 9  − ( 0 ) = 60.
b
−3
−1 

2 d x = −48x 2
∞
∞
The area is 1 and so
∞
20
2 20
20
2
dx =  −  =  − 20  −  − 20 
[0 ] −  − 48  ∫=−2 48
 x  −2  2   −2 
x2
 48 
 48  48
 =  − x  = [0 ] −  − a  = a .


0
a 


20
c The student calculated the answer like this:
∞
∞
∞
−3
−1 

 48 
24x 2 dx =  −48x 2  =  −
=
a
x 0

a 
The area is = ∫
5
∞ 20
∫4 x 2 dx =  − x 4 = [0 ] −  − 4  = 5

a
a
= – 10 – 10 = – 20
However, the graph looks like this:
y
48
= 1.
a
Hence a = 48 and a = 48² = 2304.
Area = ∫
6
∞
x
2 2
Then
(
∞
0
(
)
(1 + x )
2
∞
)
2
because  − 20  cannot be evaluated.
 x 0
2.
9
a If x = 0, then y = 242 and the point is  0, 242  .
 a 
a
y
∞
1 
1
1
 1
= [0 ] −  −  = .
2 dx =  − 2 ×
 2  2
1 + x 2 0

1 + x2
(
x
7
)
a
∞
100
x
2
2x
1 
1
1
 1
= [0 ] −  −  =
2 dx =  − 2 ×
 2  2
1 + x 2 0

1 + x2
x
0
The area between x = 0 and x = 2 is infinite
−2
dy
= −1 × 1 + x 2 × 2x = −
dx
Hence ∫
∫0
–2
2 −1
2
∞
dx .
(1 + x )
1
= (1 + x ) .
Try
1+ x
0
∞
 50 
∞
∫2 100x dx =  −2 x 2 =  − x 2 2
−3
−2
= 0 − [ −12.5] = 12.5
x = –a
–a
0
x
100
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7
WORKED SOLUTIONS
∞
∞
0
∞
0
b The area is ∫ 24(x + a)−2 dx =  −24(x + a)−1 
24
=  − 24  = [ 0 ] −  −  = 24 .
 a  a
 x + a 0
4
1
1
−1
−1
1
1
−1
−1
(
)
2
1
(
)
Volume = ∫ πy 2dx = ∫ π x 2 + 3 dx = ∫ π x 4 + 6x 2 + 9 dx
(
)
2
1
(
−1
)
= ∫ πy 2dx = ∫ π x 2 + 3 dx = ∫ π x 4 + 6x 2 + 9 dx
c If a is negative the graph looks like this:
y
−1
(
) (
) (
=  π ( 1 x + 2x + 9x )  =  π ( 1 + 2 + 9)  −  π ( − 1 − 2 − 9)  = 11.2 π + 11.2π = 22.4π.
5
 5
 
 5


5
)
1
=  π 1 x 5 + 2x 3 + 9x  =  π 1 + 2 + 9  −  π − 1 − 2 − 9  = 11.2 π +
5
 
 5
 −1  5

1
3
−1
5
x=a
If y = 0.4x2, then x2 = 2.5y.
4
4
4
2
2
2
Volume = ∫ πx 2 d y = ∫ π × 2.5y d y = 1.25πy 2 
4
∫2 πx
a
0
2
4
4
d y = x π × 2.5y d y = 1.25πy 2  =
2
2
∫
6
In this case the area is infinite because 24
x+a
cannot be evaluated when x = – a.
[ 20π ] − [5π ] = 15π.
The curve is a parabola that looks like this:
y
∞
8
64
10 a Area = ∫ x d y and x = y so x = 2 ;
4
y
∞
∞
∫4 x d y = ∫4 64y d y
−2
0
x
4
∞
=  − 64  = [0] – [–16] = 16
 y  4
∞
∞
4
4
−1
b If the area if finite, it is ∫ y dx = ∫ 8x 2 dx
∞
The curve meets the x-axis at 0 and 4.
1

∞
= 16x 2  = 16 x 0 .
0
4
4
0
0
0
0
0
Exercise 7.7A
4
2
4
5
5
5
2
2
2
Volume = ∫ πy 2 dx = ∫ π × 0.09x 2 dx = 0.03π x 3 
5
5
2
2
( )
2
2
Volume = ∫ πy dx = ∫ π
2
0
0
7 2
2
2
2

y 2dx = ∫ π 1 x 3 dx = ∫ π x 6dx =  π × x 
8
0
0 64
 64 7 
0
)
2
( )
(
)
(
)
4
(
)
=  π 1024 − 512 + 1024  − [ 0 ] = 34 2 π or 34.13π.
5
3 
15

7
a Volume =
∞
∞
−1 ∞
∫a πy dx = ∫a πx dx =  −πx a
2
−2
∞
dx = ∫ π × 0.09x 2 dx = 0.03π x 3  = [ 3.75π ] − [ 0.24π ] = 3.51π.
3
)
0
= ∫ π x 4 − 8x 3 + 16x 2 dx =  π 1 x 5 − 2x 4 + 16 x 3 
3
 5
0
0
2
π 2
1 Volume = ∫0 πy dx = ∫0 πx dx =  2 x  = [ 8π ] − [ 0 ] = 8π
0

4
4
4
2
 π x 2  = [ 8π ] − [ 0 ] = 8π .
π
y
d
x
=
π
x
d
x
=
∫0
∫0
 2 0
4
(
4
2
However, this cannot be evaluated because ∞
2
4
4
4
2
cannot be found. The area is infinite.∫ πy 2dx = ∫ π ( x ( x − 4 )) dx = ∫ π x 2 − 4x dx
4
(
Volume = ∫ πy 2dx = ∫ π ( x ( x − 4 )) dx = ∫ π x 2 − 4x dx
2
π
π
π
= −  = [0] − −  = .
 a  a
 x  a
2
1 x 3 dx = 2 π x 6dx =  π × x 7 
a
∫0 64
 64 7  b Volume = a πy 2 dx =  − π  .
8

0
∫0
 x 0
=  π × 128  − [ 0 ] = 2 π .
7 
7
 64
π
However, if x → 0, then → ∞ and the
x
volume is infinite.
101
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7 Integration
8
b The volume formed by the curved line is
y2
x2
+
= 1.
225 100
Hence
4.5
2 4.5
4.5
∫0 π x d y = ∫0 π × 2y d y =  πy 0 =20.25π.
y2
x2
4
=1−
and y 2 = 100 − x 2 .
100
225
9
2
The volume formed by the straight line is
∫
4.5
πx 2 d y
4
4

3 
2
Volume = ∫ πy dx = ∫ π 100 − x dx =  π 100x −
x 
y = 0.5x + 3 so 2y = x + 6; x = 2y – 6 and
9
27 where
−15
−15

 −15
2 = 4y2 – 24y + 36.
hence
x
15
15
4.5
4
4

2
3 
2
dx = ∫ π 100 − x dx =  π 100x −
x 
The volume is ∫ π x 2 d y
9
27
−15
3

 −15
15
(
)
(
15
2
(
)
(
)
)
= ∫
= [π(1500 − 500)] − [π(−1500 + 500)]
4.5
3
= 1000π + 1000π = 2000π.
(
)
π 4y 2 − 24y + 36 d y
4.5
= π  4 y 3 − 12y 2 + 36y 
 3
3
The volume of the ball is 6280 cm³ to 3 s.f.
9
150
= 40.5π – 36π = 4.5π
(or use the fact that it is a cone with radius = 3
and height = 1.5 so the volume is 1 πr 2h
The volume of the ring is the difference between
3
= 1 π × 32 × 1.5 = 4.5π).
the volumes of the shapes formed by rotating each
3
of the lines.
The
volume of the shape is 20.25π – 4.5π =
3
3
Volume formed by the arc of the circle = ∫ πy 2 dx = ∫ π 25 − 15.75π
x 2 dx
−3
−3
Where the circle and the ring cross, y = 4 and
x2 + 42 = 25 hence x2 = 9 and x = ±3.
(
3
3
−3
−3
(
)
= ∫ πy 2 dx = ∫ π 25 − x 2 dx
(
)
3
=  π 25x − 1 x 3  =  π ( 75 − 9)  −  π ( −75 + 9) 
3

 −3
12 a If x = 2 then 1 x 3 = 4 and (x – 4)2 = 4 so both
2
curves pass through (2, 4).
b The volume is in two parts.
For the curve y = 1 x 3 the volume
2
2
2
2
1
2
6
1
= ∫ πy dx = π ∫ 4 x dx = π  x 7 
0
0
 28 0
= 66π + 66π = 132π.
The volume formed by the straight line is
3
3
∫−3 πy dx = ∫−316π dx
2
= 128 π = 32 π.
28
7
= [16π x ]−3 = [48π] − [−48π] = 96π.
3
The curve y = (x – 4)2 meets the x-axis at 4 so
for this part the volume is
(Alternatively, as this is a cylinder, the volume is
πr2 h = π × 42 × 6 = 96π.)
4
10 aThe cylinder has a radius of 5 and height of 6.
Volume = πr2h = π × 52 × 6 = 150π
2
(
2
4
4
5
4
= π [ 0 ] − π  − 32  = 32 π .
 5  5
b When the curve is rotated the volume is
2
1
∫2 πy dx = ∫2 π ( x − 4 ) dx = π  5 (x − 4) 2
The volume of the ring is 132π – 96π = 36π.
6
)
The total volume is
∫2 πy dx .
32 π + 32 π = 10 34 π or
7
5
35
11.0π (3 s.f.).
y = 2 x−2
) = 4(x – 2) = 4x – 8
2
Exam-style questions
6
6
2
2
2
Volume = ∫ π ( 4x − 8 ) dx = π  2x − 8x 
= π[24] – π[–8] = 32π.
The volume of the shape is the difference
between these answers = 150π – 32π = 118π.
11 aWhere they cross 0.5x2 = 0.5x + 3;
0.5x2 – 0.5x – 3 = 0; x2 – x – 6 = 0
(x – 3) (x + 2) = 0; x = 3 or −2; the coordinates
are (3, 4.5) and (−2, 2)
1
y = ∫ (8x 3 − 12x 2)dx
= 2x4 – 4x3 + c
When x = 2, y = –3
32 – 32 + c = −3
So y = 2x4 – 4x3 – 3.
102
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7
WORKED SOLUTIONS
2
10
Distance = ∫
0
10
x 
(15 − 0.1x dx ) = 15x − 0.1
3

2
10
3
dV
b When t = 20, dt = 0.6 × 20 – 10 = 2.
0
This is positive and implies that the volume
is increasing. This cannot be the case. If
dV
will be negative.
there is a leak,
dt
∫0 (15 − 0.1x dx ) = 15x − 3 x 0 = 150 − 33 3  − [0 ] = 116 3 m.
10
0.1
2
3
3
1
2
⌠
−3
−1
a   5x −2 + x 2  dx = −5x −1 − 2x 2 + c

⌡
b
∞5+
∫1
x2
x
∞
∞
∞
Area = ∫ 24 ( x + 2)−2 dx =  −24 ( x + 2)−1  =  − 24  = [ 0 ] − [ −12]

0  x + 2 0
0
8
y = ∫ (2x + 6)dx
 5
2  = 0 − −5 − =
2] 7
[ ] [
dx =  − x −
∞
∞
x 1 ∞

−2
 = [ 0 ] − [ −12] = 12 .
24
x
+
2
) dx =  −24 ( x + 2)−1  =  − x24
∫0 (
+ 2 0
0

4 y = ∫ −16x −2 dx = 16x −1 + c
16
=
+c
x
= x 2 + 6x + c
At a turning point, dy = 0
dx
2x + 6 = 0
When x = 4, y = 6.
16
+c
4
6=
x = −3
c=6−4=2
The turning point is on the x-axis so the
coordinates are (−3, 0).
The equation is y = 16 + 2 .
x
5
Put these into the equation for y.
a Find some points on the curve:
0 = 9 – 18 + c
x
0
1
4
9
c=9
y
−2
−1
0
1
The equation of the curve is y = x2 + 6x + 9.
y
9
2
0
(
)
3
(
1
2
3
4
5
6
7
8
9
(
)
3
(
∫0 
1
 2 32


x − 2x
x 2 − 2 dx =

3

4
16
 =  3 − 8  − [ 0 ]

0 
= − 8 so the area below the x-axis is 8 .
3
3
9
6
9
(
)
1
2
2
2
4
1
dy 3
= 4x 2 + 5 2 × 8x = 12x 4x 2 + 5
dx 2
) × 8x = 12x 4x + 5 .
Hence ∫ x 4x + 5 dx = 1 ( 4x + 5) + c .
12
dy 3
= 4x 2 + 5
dx 2
–2
b
1
= 12x 4x 2 + 5
b If y = 4x 2 + 5 2 , then
x
)
a If f ( x ) = 4x 2 + 5 2 , then f ′ ( x ) = 3 4x 2 + 5 2 × 8x
2
1
–1
∞
7
2
3
2
10 Where the curve crosses the x-axis, x(4 − x) = 0.
x = 0 or 4
(
)
4
4
1
The area = ∫ 4kx − kx 2 dx =  2kx 2 − kx 3 
3
0

0
1
4

2 3

4
1 kx 3  = 32k − 21 1 k  10 2 k
=
.
∫4  x 2 − 2 dx =  3 x 2 − 2x  ∫0 4kx − kx 2 dx =  2kx 2 − 
3
3 
3
0 
4
2
= [18 − 18 ] −  − 8  = 8 so the area above the
10 k = 32
3
 3  3
x-axis is 8 .
k=3
3
11 Area beneath the curve between A and B is:
Hence the total area is 8 + 8 = 16 or 5 1 .
3 3
3
3
a
a 2
1 3
1 3
1 3
2 3
∫−a x dx =  3 x −a =  3 a  −  − 3 a  = 3 a
a V = ∫ (0.6t − 10) dt = 0.3t 2 − 10t + c
(
)
When t = 0, V = 100 so c = 100.
The area of rectangle ABCD = 2a × a2 = 2a3.
V = 0.3t2 – 10t + 100
The area between AB and the curve is
2
4
2
2a 3 − a 3 = a 3 and this is of the area of the
3
3
3
rectangle.
103
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7 INTEGRATION
12 a First, find the coordinates of the points where
the lines cross.
5
14 Volume = ∫ πy 2dx.
2
y = 4x + 1
Where x2 – x + 4 = 2x + 8
x2 – 3x – 4 = 0
Thus y 2 = 4x + 1.
(x − 4)(x + 1) = 0
(
5
)
2
Volume = ∫ π ( 4x + 1) dx =  π 2x + x 
x = −1 or 4
2
If x = −1, y = 1 + 1 + 4 = 6.
= [ 55π ] − [10π ] = 45π .
1 dy
−1
1
15 a y = (4x − 8)2;
= (4x − 8) 2 × 4 =
dx 2
If x = 4, y = 16 − 4 + 4 = 16.
The points are (−1, 6) and (4, 16).
y
if x = 6, then
B
16
2
;
4x − 8
dy
2
1
=
= ; the gradient of
dx
16 2
y
6
D
0
C −1
4
A
4
x
b You want the difference between the area of
the trapezium ABCD and the area under the
curve between A and B.
∫−1( x − x + 4 ) dx
4
Area under curve =
2
the normal is −2 and the equation is
y – 4 = – 2(x – 6) or y = –2x + 16.
b
A
5
P
x
R
6
2
4
1
1
=  x 3 − x 2 + 4x 
 −1
 3
2
At P 4x – 8 = 0 so x = 2; at R –2x + 16 = 0 so x = 8.
6
Volume formed by rotating curve = ∫ πy 2 dx
2
64
1 1
1
5
1
=  − 8 + 16  −  − − − 4  = 29 + 4 = 34 .
3
6
6
 3
  3 2

= π ∫ (4x − 8)dx = π  2x 2 − 8x 
6 + 16
× 5 = 55.
2
1
5
The required area is 55 − 34 = 20 .
6
6
= [24π] – [– 8π] = 32π.
Area of trapezium =
6
6
2
2
Volume formed by rotating the normal is a
cone with volume 1 π × 4 2 × 2 = 32 π
3
3
8
1
1 3
x = 2x 2 and either x = 0 or
13 Where they cross,
16
5
x 2 = 32 so x = 4.
Area between y =
4 1
1
1 3
x and the x-axis is
16
4
∫0 16 x dx =  64 x 0 = 4 − 0 = 4 .
3
4
Area between y = 2 x and the x-axis is
4
∫0
4
1
 4 32 
32
2
2x 2 dx =
x
=
− 0 = 10 .
3


0
3
3
Area between curves = 10 2 − 4 = 6 2 .
3
3
8
1
(or) ∫ π(16 − 2x)2 dx = π  − (16 − 2x)3 
6
 6
6
= [ 0 ] − π  − 64  = 32 π .
 6  3
The total volume is 32π +
32 π = 42 2 π.
3
3
16 a Where the curves cross, x2 – 10x + 25 = 5 + 4x – x2;
2x2 – 14x + 20 = 0;
(x – 2)(x – 5) = 0; x = 2 or 5.
When x = 2, y = 9 so A = (2, 9).
When x = 5, y = 0 so B = (5, 0).
b The volume is
∫2 π (5 + 4x − x ) dx − ∫2 π ( x − 10x + 25) dx
5
2 2
5
2
2
= 129.6π – 48.6π = 81π = 254.
104
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7
WORKED SOLUTIONS
17 a Where they cross, x4 + 2 = x2 + 14;
x4 – x2 – 12 = 0; (x2 – 4)(x2 + 3) = 0
x2 = 4 or −3
x2 = − 3 has no solution; if x2 = 4, then x = 2
or −2; in either case, y = 24 + 2 = 18 and the
coordinates are B (2, 18) and A (−2, 18).
b The minimum points are where the curves
cross the y-axis.
y = x4 + 2 is at (0, 2) and y = x2 + 14 is at (0, 14)
c If y = x4 + 2, then
y − 2 = x 2 and the volume
18
generated by y = x4 + 2 is ∫ πx 2 d y
b The areas are given by the formula in part a
with a = 1, 2, 4, 8, etc.
Total area = 50 + 50 + 50 + 50 + ....
1
2
4
8
1
This is a geometric series with a = 50 and r = .
2
a
50
=
= 100 .
Then S∞ =
1−r 1− 1
2
2
20 a Where it crosses the x-axis, y = 0; x 2 = 1; x2 = a2;
a
x = ± a; points (a, 0) and (– a, 0).
Where it crosses the y-axis, x = 0;
2
1
3 18
18


π( y − 2)2 d y = π  2 ( y − 2)2 
3
2

2
= ∫
=  128 π  − [ 0 ] = 128 π.
3
 3 
If y = x2 + 14, then y – 14 = x2 and the volume
18
generated by y = x2 + 14 is ∫ πx 2 d y
14
18
= ∫ π ( y − 14 ) d y = π  1 y 2 − 14y 
14
 2
14
18
= [–90π] – [–98π] = 8π.
128
104
The volume required = 3 π − 8π = 3 π
or 34 2 π.
3
18 a f ( x ) = ∫
1
dx = ∫ (
ax + 3
−1
ax + 3 2 dx
)
1
2
= (ax + 3)2 + c = 2 ax + 3 + c
a
a
b f (0) = 0 →
f (x ) =
2 × 3 + c = 0;
2 3 and
c=−
a
a
2
2 3
ax + 3 −
a
a
f (a ) = 2 2 − 2 →
2
2 3
a2 + 3 −
= 2 2 − 2;
a
a
From which
b Volume = ∫
a
−a
2
2
πy 2 dx , where y 2 = 1 − x 2 and
b
a

x2 
y 2 = b2 1 − 2  .

a 
2
a

So volume = ∫ πb 2  1 − x 2  dx
−a

a 
a
x3 
2
= πb  x − 2 
3a  −a

a
a
2
2
= πb  a − 3  − πb  −a + 3 




= πb 2 × 2 a + πb 2 × 2 a = 4 πab 2 .
3
3
3
Mathematics in life and work
Assume that the segment is from x = −a to x = a.
a
(
)
2
The volume is = ∫ π 8 − 0.009x 2 dx
−a
a
(
)
= ∫ π 64 − 0.144x 2 + 0.000 081x 4 dx
−a
= π 64x − 0.048x 3 + 0.000 016 2x 5 
a
−a
= 2π 64a − 0.048a 3 + 0.000 016 2a 5  .
2
If the length is 15.6 cm then a = 7.8 and
2
2a − 2a 2 + 2 6 − 2 3 = 0
a=
a volume of 2996 cm³ is just under 3 litres.
2 3−2 6
= 3
2−2 2
2a 100
a
y2 = b2; y = ± b; points (0, b) and (0, – b).
a2 + 3 = a
( 2 − 1) + 3
a + 3 = a ( 3 − 2 2 ) + 2a 3 ( 2 − 1) + 3
19 a
∫
y2
= 1;
b2
2a
dx =  − 100  =  − 100  −  − 100 
 x  a
 2a   a 
x
2
= − 50 + 100 = 50
a
a
a
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Summary REVIEW
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.
Warm-up Questions
1
3
a i 27, 38
 7
For ′x2′:   25(ax)2 = 672a 2x 2 .
 2
ii
Add consecutive odd numbers to get the
next term.
2
b 1, 5, 9
5x + 2y = 16
3x − 4y = 7
① × 2:
10x + 4y = 32
② + ③:
13x = 39
x=3
In ①:
15 + 2y = 16
2y = 1
①
②
448a = 672a2
0 = 672a2 − 448a
= 3a2 − 2a
= a(3a − 2)
③
a=
4
5 tan 2x = 2
2x = 63.4…, 243.4…
f′(x) = 3x2 − 7
Integrating: f(x) = x3 − 7x + c
Substitute f(3) = 5 ⇒ 27 − 21 + c = 5 ⇒ c = −1
x = 31.7°, 121.7°
6
2
2
k + cos θ = 1
2
2
cos θ = 1 − k
Therefore 64a = 8π
α=π
8
ii BA = 8a and OCA = 4π
P = 8a + 4π + 8
P = 8 × π + 4π + 8
8
2
cosθ = − 1 − k
Negative because  is obtuse.
sin θ
k
=−
cosθ
1 − k2
106
1
× 82a = 32a.
2
2
i sin  = k ⇒ sin2  = k2
sin2  + cos2  = 1
iii sin ( + π) = −k
i Area OAB =
Area OCA = π × 4 = 8π
2
∴ f(x) = x3 − 7x − 1
ii tan θ =
( 32π ,−1). After
the translation, this becomes (π, –1). After
the stretch, this becomes (π, –3). So the
minimum point of g(x) is (π, –3).
r = 7.5 cm
A Level Questions
2
− π
iA translation  2  and a stretch in the
 0 
ii The minimum point of f(x) is
Area of circle = π × 7.52 = 177 cm2.
1
2
3
y-direction, scale factor 3.
y = 0.5
3
 7
For ′x′:   26(ax)1 = 448ax.
 1
P = 5π + 8
dy
= −8x−3 − 1
dx
y = 4x−2 − x + c
Substitute (2, 4) 4 = 1 − 2 + c
c = 5
4
y = 2 − x + 5
x
7
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WORKED solutions
8
Substituting into original equation gives y = ±8.
So the stationary points are (2,8) and (−2, −8).
iLet C be the centre of the circle. So A( –7, –7)
and C( –1, 6). B is at the opposite side of the
circle, so you can conclude B(5, 19).
When x = 2,
ii AC is a radius of the circle. By Pythagoras,
AC = r = 6 2 + 132 = 205
When x = −2,
iii Gradient of AC is 13 ⇒ gradient of
6
perpendicular bisector is − 6 and it passes
13
⇒ maximum
x2 + 4x + 4 = 0
(x + 2)2 = 0
x = –2
13y – 78 = – 6x – 6
CB
4
⇒
Area of triangle =
y=2
ii The second tangent is at the opposite side
of the circle and passes through (2, –2). The
equation is
CB = 4 tan a
1
× 4 × 4 tan a = 8 tan a.
2
1
× 22a = 2a.
2
Shaded area = 8 tan a − 2a.
4
AC = 4
ii cosα =
cosα
AC
Sector area =
DC =
4 −2
cosα
4
− 2 + 2a + 2 + 4tan a
cos α
4
= cos α + 4 tan a + 2a
Perimeter =
10 i y − 2t = −2(x − 3t)
y = −2x + 8t
At A, y = 0 ⇒ x = 4t.
At B, x = 0 ⇒ y = 8t.
4t × 8t
Area of triangle OAB =
= 16t 2 .
2
1
ii
Equation of perpendicular line is y − 2t = (x − 3t ).
2
2y = x + t
At C, y = 0 ⇒ x = −t.
So C(−t, 0) and P(3t, 2t).
Mid-point of CP is
( 3t 2− t , 22t ) = (t, t), which
is on the line y = x.
11 i y = 8x −1 + 2x
dy
= −8x −2 + 2 = − 82 + 2
dx
x
y + 2 = 1 (x – 2)
13
Arc length = 2a
d 2y
−3
= 16
2 = 16x
x
d
x −3
⇒
The line and the circle intersect once only at
P( –2, 2) so the line is a tangent.
6x + 13y – 72 = 0
ii 0 = −82 + 2
x
= −8 + 2x2
x=±2
d 2y
= −2 < 0
d 2x
x2 + (x + 4)2 = 8
y − 6 = − 6 (x + 1)
13
i tan α =
⇒ minimum.
12 iThe centre of the circle is the origin O(0, 0). The
line and the circle intersect when
through C.
9
d 2y
=2>0
d 2x
⇒
y=x–4
4 sin2 x + 8 cos x − 7 = 0
4(1 − cos2 x) + 8 cos x − 7 = 0
4 cos2 x − 8 cos x + 3 = 0
(2 cos x − 1)(2 cos x − 3) = 0
3
⇒ no solutions
cos x =
2
1
⇒ x = 60°, 300°
cos x =
2
14 if(x) = x2 + 1
y = x2 + 1
y − 1 = x2
x = y − 1
f−1(x) = x − 1
(x > 1)
ii f(x2 + 1) = (x2 + 1)2 + 1 =
(x2 + 1)2 =
185
16
169
16
13
4
3
x=
2
x2 + 1 = ±
Only one solution, since x  0
 6
 6
 6
15 i
(2x − x2)6 =   (2x)6 +   (2x)5(−x2) +  
 0
1 
 2
(2x)4(−x2)2 + ... = 64x6 − 192x7 + 240x8 + ...
ii (2 + x)(2x − x2)6
For ‘x8’:
2 × 240x8 = 480x8 and x × −192x7 = −192x8
So the coefficient of x8 is 288.
107
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Summary REVIEW
16 Completing the square: x2 + y2 – 2x – 16y + 40 = 0
becomes (x – 1)2 + (y – 8)2 = 25
20 i f(x) = 4(x – 3)2 + 2
ii
y
50
This is a circle with centre (1, 8) and radius 5.
 −6
The translation   moves the centre to ( –5, 3)
 −5
and the radius remains as 5.
40
17 i−x2 + 6x − 5 = −1(x − 3)2 + 4 by completing the
square
(0, 38)
30
ii m = 3
iii y = −(x − 3)2 + 4
(x − 3)2 = 4 − y
x=3+ 4− y
20
f−1(x) = 3 + 4 − x
The domain is x ⩽ 0
18 i S =
10
a
a
and 3S =
1−r
1 − 2r
3a
a
=
1 − r 1 − 2r
3(1 − 2r) = 1 − r
2
r=
5
0
x
The minimum point is (3, 2)
iii f(2x) moves (3, 2) to (1.5, 2)
f(2x – 3) moves (3, 2) to (4.5, 2)
ii 84 = 7 + (n − 1)d ⇒
d=
77
n −1
①
245 = 7 + (3n − 1)d ⇒
d=
238
3n − 1
②
Equating ① and ②:
238(n − 1) = 77(3n − 1)
238n − 238 = 231n − 77
7n = 161
5
n = 23
19 iPerimeter = 400 ⇒ 400 = 2πr + 2x
⇒ 2x = 400 − 2πr
①
A = πr2 + 2xr
Substituting ①:
A = πr2 + r(400 − 2πr)
A = πr2 + 400r − 2πr2
A = 400r − πr2
dA
= 400 − 2πr
dr
0 = 400 − 2πr
2πr = 400
Substituting into ①:
2x = 400 − 400
x = 0
Therefore there are no straight sections.
ii
d2A
= −2π < 0 ⇒ maximum
dr 2
–f(2x – 3) moves (3, 2) to (4.5, –2)
So the maximum point of – f(2x – 3) is (4.5, –2)
21 i ar2 = −108 ①
ar5 = 32
②
② ÷ ①:
8
r3 = −
27
2
r=−
3
ii Substituting value of r into ① gives:
4
a = −108
9
a = −243
−243
iii S∞ =
1− −2
3
= −145.8
22 i x = x2 − 4x + 4
0 = x2 − 5x + 4
0 = (x − 1)(x −4)
So x = 1, y = 1
A(1, 1).
And x = 4, y = 4
B(4, 4).
Therefore, mid-point of AB is (2.5, 2.5).
dy
= 2x − 4
dx
2
x − 4x + 4 = (2x − 4)x
x2 = 4
ii
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WORKED SOLUTIONS
x = ±2
x = 2 ⇒ y = 0 ⇒ m = 0 Ignore this
solution since m = 0.
x = −2 ⇒ y = 16 ⇒ m = −8
So the tangent meets the curve at (−2, 16).
ii
y
8
2
23 i 4x2 − 24x + 11 = 4(x − 3)2 − 25
Vertex is at (3, −25).
ii g(1) = 4 − 24 + 11 = −9
Therefore g(x)  −9.
iii y = 4(x − 3)2 − 25
y + 25
(x − 3)2 =
4
x =3+
y + 25
4
=3+
1
y + 25
2
g−1(x) = 3 +
1
x + 25
2
The domain is x  −9.
24 By Pythagoras:
i
AB = 122 + 10 2 = 244
Similarly, AC = 12 + 112 = 122 and
BC = 112 + 12 = 122
Therefore, (AC)2 + (BC)2 = (AB)2 ⇔ ABC is a
right-angled triangle.
ii (AC)2 + (BC)2 = (AB)2 ⇒ AB is the hypotenuse
⇒ AB is the diameter of the circle
∴ The centre of the circle is the midpoint
of AB. The centre is (–4, 2) and the radius is
244 = 61
2
So the equation of the circle is
(x + 4)2 + (y – 2)2 = 61
4 − 1.75 = 4.5 and gradient of
25 i Gradient of CE =
4 − 3.5
DE = 4 − 3.51 = 4.9
4 − 3.9
ii The sequence of chords have gradients: 3,
4, 4.5, 4.9. This indicates that as the chords
approach E, the gradients are converging
towards 5. It indicates that f’(4) = 5.
x
26 i 5 + 3 cos 2 = 7
()
2
x
cos ( ) =
3
2
x
= 0.841…
2
x = 1.68
2π
x
iii f has an inverse because it is a one−one
function (or it has no turning point).
x
iv
y = 5 + 3 cos
2
()
()
y−5
x
= cos
3
2
2 cos−1 

y − 5
=x
3 
f−1(x) = 2 cos−1
( x 3− 5 )
27 i y = 8 −2x − x2
dy
= −2 − 2x
dx
2 = −2 − 2x
x = −2
y=8
Therefore
8 = −4 + c
c = 12
ii 8 − 2x − x2 = 2x + 11
0 = x2 + 4x + 3
= (x + 1)(x + 3)
x = −1 or x = −3
−1
−1
−3
−3
Area = ∫ (8 − 2x − x 2)dx − ∫ (2x + 11)dx .
−1
−1

x3 
4
Area = 8x − x 2 −
−  x 2 + 11x  = .
3 
3
−3

−3
28 i Completing the square: x2 + y2 – 8x – 6y + 21 = 0
⇒ (x – 4)2 + (y – 3)2 = 4.
Therefore, the centre is (4, 3).
ii The radius is 2.
iii When x = 0 ⇒ (y – 3)2 = –12 ⇒ no real
solutions ⇒ does not intersect y-axis.
When y = 0 ⇒ (x – 4)2 = – 5 ⇒ no real
solutions ⇒ does not intersect x-axis.
iv The diameter lies on the line AB.
The gradient of the tangent is 1 , so the
3
gradient of the diameter (and hence AB) is –3.
AB also passes through the centre. So the
equation is:
109
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Summary REVIEW
y – 3 = – 3(x – 4)
y – 3 = – 3x + 12
3x + y – 15 = 0
29 i
1
2
ii Area = ∫ (3 − 2x)3 dx − ∫ (−24x + 20)dx
0
0 = 3u + 3u−1 − 10
f ′′
32 The graphs intersect when 6x – x2 = 8
So x2 – 6x + 8 = 0
(x – 2)(x – 4) = 0
x = 2 or x = 4
⇒ minimum
3
1
−7 = 2(4)2 + 6(4)2 − 10(4) + c
(x + 2)2 = 16
ii Let C be the centre of the circle. The gradient
of CP is 7 − 4 = 3 .
2 − −2 4
V = π[[204.8] – [54.4]] – 128π
112π
V = 22.4π or
5
(
1
⇒
dy
= 3(3 − 2x)2(−2)
dx
= −6(3 − 2x)2
dy
When = 1 ,
= −6(3 − 1)2 = −24.
2 dx
So the equation of the tangent is:
1
y − 8 = −24 x −
2
110
Area of kite ABDC = 3 (BD)
Therefore, 12 = 3 (BD)
⇒
BD = 4 cm
By Pythagoras: (AD)2 = 32 + 42
⇒
AD = 5 cm
ii The locus of points is a circle, centre (5, 8) and
radius 5 cm. So the equation of the locus is:
By Pythagoras, PQ = 5.252 + 7 2 = 8.75
y = (3 − 2x)3
iRadius and tangent meet at B, so angle ABC is a
right-angle and ABC is a right-angled triangle.
Area of triangle ABD = 3(BD)
2
iii For the tangent line, y = 0 ⇒ 4x = 29 ⇒
x = 7.25 ⇒ Q(7.25, 0)
So, the area of triangle CPQ is
5 × 8.75 = 21.875 units2
2
)
Extension Questions
The tangent is perpendicular to CP, so it has
4
gradient − .
3
The equation of the tangent is
y − 7 = − 4 (x − 2) ⇒ 4x + 3y – 29 = 0
3
y − 8 = −24x + 12
y = −24x + 20
2
4
V = π 12x 3 − 3x 4 + 1 x 5  − π [ 64x ]2

5  2
⇒ x = 2 or x = – 6
( )
4
4
Therefore, P(2, 7).
31 i
2
2
1
So x + 2 = ± 4
2
V = π∫(36x 2 − 12x 3 + x 4) dx − π∫64 dx
f(x) = 2x 2 + 6x 2 − 10x + 5
(x + 2)2 + 9 = 25 ⇒
4
4
c = 5
⇒
4
V = π∫ (6x − x 2)2 dx − π∫ 8 2 dx
3
1
iii f(x) = 2x 2 + 6x 2 − 10x +c
30 i y = 7
=9
8
( 19 ) = −36 < 0 ⇒ maximum
3
= 81 − 9
8
2
4
f′′(9) = > 0
9
1
1
−  −12x 2 + 20x  2
=  − (3 − 2x)
0 
 8
0
3 −1 3 −3
f ′′(x) = x 2 − x 2
2
0
1
42
0 = 3u2 − 10u + 3
0 = (3u − 1)(u − 3)
u=1
x=1
9
3
u = 3 x = 9
ii
1
2
(x – 5)2 + (y – 8)2 = 5
2
i Sometimes. The result is not valid for a = 0.
ii Sometimes. The result is only true when
a = b. The circle is in the positive quadrant
and touches both the x-axis and the y-axis.
If a > b, the circle extends beneath the x-axis.
If a < b, the circle touches the y-axis only and
does not reach the x-axis.
iii Sometimes. Solving the equation:
tan θ = 1 ⇒ θ = π , 5π , …
4 4
In general, the result is true when
θ = (4n − 3)π , where n ∈ Z .
4
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WORKED solutions
3
iv Always. This is an identity and should be
written as sin2 θ + cos2 θ ≡ 1.
 5
4  4
2
2
6
3
 1  ( x )( 2) ×  2 ( 2x ) ( −3) × ( −4 ) = 70778880x
Solving the quadratic: (3x + 5)(x – 2) = 0
5
⇒ x = − or x = 2.
3
2
 5
4
4  6 x
4
3
 1  ( x )( 2) × ( −3) ×  2 2 ( −4 ) = 6 220800x
For the cubic:
()
()
dy
= 3x 2 − 8x + 6 . The discriminant
dx
 5
4  4
3  6 x
5
3
 1  ( x )( 2) ×  1  ( 2x )( −3) ×  1  2 ( −4 ) = 53084160x
⇒ no turning points.
At the intersection of the two curves: x3 – 4x2 + 6x
+ 5 = 3x2 – x – 10 ⇒ x3 – 7x2 + 7x + 15 = 0
x = – 1 is a root ⇔ (x + 1) is a factor
By inspection: (x + 1)(x2 – 8x + 15) = 0
⇔ (x + 1) (x – 3)(x – 5) = 0
This information allows a sketch to be drawn.
is 82 – 4(3)(6) = – 8
 4
 3
( 2)5 ×   ( 2x )3 ( −3) × ( −4 )6 = −12582912x 3
 4
 2
 6
1 
 4
1 
 6
 2
()
( 2)5 ×   ( 2x )2 ( −3)2 ×   x2 ( −4 )5 = −21233664x 3
()
2
( 2)5 ×   ( 2x )( −3)3 ×   x2 ( −4 )4 = −6635520x 3
y
(5, 60)
()
 6
 3
3
( 2)5 × ( −3)4 ×   x2 ( −4 )3 = −414720x 3
Summing: the coefficient of x3 is 11 802 624.
5
40
20
(3, 14)
(0,5)
(−1.667, 0)
−5
0
(2, 0)
x
5
(−1, −6)
The area between the curves from (3, 14) to (5, 60)
is the area we need to find.
5
((
))
) (
Shaded area = ∫ 3x 2 − x − 10 − x 3 − 4x 2 + 6x + 5 dx 6
3
5
(
)
Shaded area = ∫ −x 3 + 7x 2 − 7x − 15 dx
3
The errors in this solution include division by
sin2 x (which loses some solutions) and sinx = 1
2
has more solutions in the given range than shown.
The correct solutions is:
sin2x – cos2x = 4 sin3x – 1
sin2x – (1 – sin2x) = 4sin3x – 1
2sin2x – 1 = 4sin3x – 1
2sin2x = 4sin3x
sin2x = 2sin3x
2sin3x – sin2x = 0
sin2x (2sinx – 1) = 0
1
sin x = 0 or sin x =
2
i ax2 + bx + c = 0
x2 + b x + c = 0
a
a
x2 + b x = − c
a
a
5
=  − 1 x 4 + 7 x 3 − 7 x 2 − 15x  = 20 units2
 4
3
3
2
3
4
All of the different combinations that contribute to
the coefficient of x3 are:
 5
3
2
4
6
3
 3 ( x ) ( 2) × ( −3) × ( −4 ) = 13271040x
 5
2
3  4
3
6
3
 2 ( x ) ( 2) ×  1  ( 2x )( −3) × ( −4 ) = −70778880x
()
 5
2
3
4  6 x
5
3
 2 ( x ) ( 2) × ( −3) ×  1  2 ( −4 ) = −19906560x
x = π , 5π
6 6
x = 0, π, 2π
ii
( x + 2ba ) − 4ba = − ac
( x + 2ba ) = 4ba − ac
( x + 2ba ) = 4ba − 44aca
( x + 2ba ) = b 4−a4ac
2
2
2
2
2
2
2
2
2
2
2
2
2
111
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Summary REVIEW
The gradient either side of zero is negative
and the curve is continuous. This suggests
that x = 0 is a point of inflection.
2
iii x + b = ± b − 42ac
2a
4a
b 2 − 4ac
2a
x=− b ±
2a
b 2 − 4ac
2a
2
x = −b ± b − 4ac
2a
7
iii Not always. A point of inflection occurs when
the sign of the curvature changes. This can
occur without the need for a stationary point.
()
)
ii fgh ( x ) = f sin 2x = 1 − sin 2x = cos x
0 ⩽ x ⩽1 and 0 ⩽ fgh(x) ⩽ cos(1)
iArea of circle 1 is 100π and area of circle 2 is 25π.
So a = 100π and r = 1 . The area is reduced by
4
a scale factor of 4 for each consecutive circle.
The radius is halved for each consecutive circle.
Therefore, the 4th circle has area 25π cm2 from
16
which its radius is 5 cm.
4
The distance of the centre of the 4th circle
from O is the radii of circles 1 and 4 and the
diameters of circles 2 and 3 ⇒ 10 + (2 × 5) +
(2 × 2.5) + 1.25 = 26.25.
So the centre of circle 4 is (26.25, 0) with
radius 5 cm.
4
100
π 400π cm2
ii S∞ =
=
1
3
1−
4
iii The total area is found by doubling the sum
to positive infinity and subtracting the area
of the original circle as this will have been
double-counted.
400π
500π
Total area = 2
− 100π =
units
cm2 2
3
3
( )
9
i
dy
d 2y
= 4x 3 − 9x 2
= 12x 2 − 18x
dx
dx 2
At stationary points: 0 = 4x3 – 9x2 ⇒ x = 0 or
x = 9.
4
2
9 d y 81
> 0 ⇒ minimum
When x = , 2 =
4 dx
4
d 2y
ii When x = 0,
= 0 ⇒ may be a point of
dx 2
inflection.
dy
= −0.094 and when
When x = – 0.1,
dx
x = 0.1,
()
10 cm
8 cm
(
)
3 = π−x
2
2
π
x= − 3
2
2
0 ⩽ gh(x) ⩽1
8
)
cos π = π − x
6
2
igh(x) = g(sin x) = sin2 x, 0 x π and
2
(
(
10 cosx = sin π − x ⇒ sin  cos π  = sin π − x
2
6 
2

x = π− 3
2
11
5 cm
x+ b = ±
2a
a
10 cm
b
8 cm
Use the cosine rule to find α and β.
52 = 102 + 102 – 2(10)(10) cos α
25 = 200 – 200 cos α
cos α =
175
200
α = 0.505 radians
52 = 82 + 82 – 2(8)(8) cos β
25 = 128 – 128 cos β
cos β =
103
128
β = 0.636 radians
The shaded area can be found in two parts. In each
case, subtract the area of the isosceles triangle
from the area of the sector.
( 12 × 10 α − 12 × 10 sinα )
1
1
+ ( × 8 β − × 8 sinβ ) = 2.404…
2
2
2
Shaded area =
2
2
2
The required area is the area of each circle minus
twice the shaded area.
Required area = 100π + 64π − 2 × 2.404 … = 510.41 cm2
dy
= −0.086
dx
112
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WORKED solutions
12 i The terms can be written as: 2, 2 + 2d, 2 + 5d.
Consecutive terms in a geometric sequence
2 2
⇒ + d = 2 + 5d
2
2 + 2d
Solving: (2 + 2d)2 = 2(2 + 5d)
4 + 8d + 4d2 = 4 + 10d
4d2 – 2d = 0
2d2 – d = 0
d (2d – 1) = 0
The non-trivial value of d = 1
2
We know that
r = 2 + 2d = 2 + 1 = 3
2
2
2
ii For the arithmetic progressions, the 4th term
1 = 7.
is 2 + 3
2
2
()
For the geometric progression, the 4th term
( ) = 278 a
is a 3
2
3
Equating the terms:
27 a = 7
8
2
54a = 56
a = 28
27
113
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