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.

Answers
1 Problem solving From the table we see that the greatest number of
questions answered correctly is 12.
1 The basket was half full at 10 : 59. 9 Let the even number be 2k, where k is an integer.
2 $3.50 ( 2k ) 2 = 4 k 2
Work backwards, use guess and check or algebra. Since any multiple of 4 is even, the square of any
3 Let x be the number of consecutive shots. even number is even.
3 + x > 16 + x 10 Let the odd number be 2k + 1, where k is an integer.
7 + x 25 + x
(3 + x )(25 + x ) > (16 + x )(7 + x ) ( 2k + 1) 2 = 4k 2 + 4k + 1
75 + 28 x + x 2 > 112 + 23x + x 2 = 4(k 2 + k ) + 1
5 x > 37 This is an even number + 1, which is odd.
x > 7.4 Further practice (Page 4)

So he needs 8 consecutive successful shots. 1 One friend has x marbles.

4 1 = 4⇒ x +5= 1 The second has x − 2 marbles.

x +5 4
x + 1 = 2(x − 3)
x + 8 = 31
4
x=7
x +8 = 13
4 x–2=5
1 = 4
x + 8 13 The friends have 12 marbles in total.
pd 2 Looking at all of the possible sets of three ages to
5
p+r give a product of 72:
6 The difference in weight represents half the contents Ages Sum
so 1 c = 24 − 13 ⇒ c = 22 kg . 1 2 36 39
2
1 3 24 28
So the suitcase weighs 2 kg.
1 4 18 23
7 The area of the cross-section of the water will
1 6 12 19
be the same.
1 8 9 18
Area ADC = 1 × 50 × 40 = 1000 cm 2 2 2 18 22
2
Let the height be h. 2 3 12 17
2 4 9 15
100h = 1000
2 6 6 14
h = 10 cm
3 3 8 14
8 Clearly the number of correct answers must be even,
3 4 6 13
and greater than 10 or the total would be less than 48.
Since there is some doubt about the answer after the
Correct Wrong Unanswered Total clue about Alma’s house number, the sum of the ages
10 1 9 48 must be 14.
12 6 2 48 The fact that Peta says she has an oldest child means
14 11 X X the answer must be 3, 3, 8.

1 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 2 Algebra

3 There is no reason to add the $270 and $20 here. You 8 Let the distance be d.
can do the calculations this way:
33
Actual cost of suite: $250 d =d+ 4
4 6 60
Money returned: $30 to men
15d = 10d + 3 3
$20 to lift attendant 4
Total $300 5d = 3 3
4
3
4 Speed = distance d = km
4
time
For the ‘against the wind’ journey:
9 After the first refilling there is 9 litres of water and
Distance = 300 × 2.5 = 750 km 36 litres of juice.
For the ‘with the wind’ journey: The fraction that is juice is 36 = 4 .
45 5
Time = 750 = 1.875 hours = 112.5 minutes When 9 litres are removed, the amount of juice
400
5 Sanil’s watch is moving at 5 of the speed of a normal
removed is 4 × 9 = 36 = 7.2 litres
5 5
6
watch. Volume of juice in the final mix is
Hence a normal watch is going 6 times faster than 45 − 9 − 7.2 = 28.8 litres
5
Sunil’s watch.
Volume of water in the final mix is
Sunil’s watch moves through 8 hours between 9 a.m. 45 − 28.8 = 16.2 litres
and 5 p.m. Ratio is 16.2 : 28.8 = 9 :16
Hence a normal watch will move through 10 Let a = 2k + 1 and b = 2m + 1 where k and m are
6 × 8 = 9 3 hours = 9 hours 36 minutes integers.
5 5
Hence the actual time is 6.36 p.m. a 2 − b 2 = (a + b)(a − b)
= (2k + 1 + 2m + 1)(2k + 1 − (2m + 1))
6 If Kyle, Leon or Ryan committed the crime then
more than one suspect would be telling the truth. = (2k + 2m + 2)(2k − 2m)
= 2(k + m + 1)2( k − m)
If Dave committed the crime then Dave is lying, Kyle
is lying, Ryan is telling the truth and Leon is lying. = 4(k + m + 1)(k − m)
If k and m are either both even or both odd, k − m
Answer: Dave
is even and if one of these numbers is odd and the
7 Let the total volume of the bath be b. other is even, then k + m + 1 is even. In both cases,
Hot tap fills b of the bath per minute. the number (k + m + 1)(k − m) is even.
12
So (k + m + 1)(k − m) = 2p, where p is an integer.
Cold tap fills b of the bath per minute.
6 So 4(k + m + 1)(k − m) = 4 × 2p = 8p,
Together they fill b + b = b per minute. which is a multiple of 8.
12 6 4
Together they fill 1 of the bath in one minute so it

4 2 Algebra
takes 4 minutes to fill the bath.
2.1 Background algebra (Page 5)
OR
1 (i) 2(a − 3b) − 3(b − 3a)
The hot tap will fill 1 bath in 12 minutes.
= 2a − 6b − 3b + 9a
The cold tap will fill 2 baths in 12 minutes. = 11a − 9b
Together they will fill 3 baths in 12 minutes. (ii) 7cd(d2 − 2) − 3cd2(8d + 5c3)
= 7cd3 − 14cd − 24cd3 − 15c4d2
So 1 bath in 4 minutes.
= −17cd3 − 14cd − 15c4d2

2 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 2 Algebra

8 f 4 9g 3 7 (i) 2 = 2 ×
5=2 5
(iii) × 5 5 5 5
3 g 12 fg
3 3 × 7 +1
3
= 2 f g (ii) =
7 −1 7 −1 7 +1
(iv) x − 1 + 5 − x
4 5
5( x − 1) 4(5 − x ) = 3 7 +3
= + 7 −1
20 20
= 3 7 +3
= 5 x − 5 + 20 − 4 x 6
20
= 7 +1
= x + 15 2
20
(iii) 1 + 3 = 1 + 3 × 1 + 3
2 (i) 12mn2 + 9mn3 = 3mn2(4 + 3n) 1− 3 1− 3 1+ 3
(ii) p2 − p − 12 = (p − 4)(p + 3) = 1+ 2 3 + 3
1− 3
3 (i) 2(x + 5) = x − 7
= 4+2 3
2x + 10 = x − 7 −2
  x = −17 = −2 − 3
(ii) 1 ( 6 x + 8 ) − 3 = 9 − 3 ( 4 − 10 x ) 8 (i) ( 2 + 1)( 2 − 1) = 2 − 1 = 1
2 2
3x + 4 − 3 = 9 − 6 + 15 x (ii) (1 − b )( 2 + b ) = 2 − b − b
3x + 1 = 3 + 15 x
− 12 x = 2 2.2 Quadratic equations (Page 7)

x = 2 = −1 2
1 (i) x + 5 x = 0
−12 6
x ( x + 5) = 0
4 Let the distance from Auckland to Hamilton be x. x = 0 or − 5
Hakim’s speed is x km/h
2 (ii) x 2 − 2 x − 8 = 0

2 ( )
Ravi’s speed is x − 4 km/h

( x − 4)( x + 2) = 0
x = 4 or − 2
Distance = speed × time so

( )
x ×2= x −4 ×2 1 (iii) 2x 2 − x − 3 = 0
2 2 10 (2 x − 3)( x + 1) = 0
x = 1.05 x − 8.4
x = 3 or − 1
8.4 = 0.05 x 2
x = 168 km (iv) 3x 2 − 6 x = 0

v −c = d 3x ( x − 2) = 0
5 (i)
b e x = 0 or 2
v − bc = d
b e (v) x 2 − 3x − 40 = 0
b =e ( x + 5)( x − 8) = 0
v − bc d
x = −5 or 8
e = bd
v − bc
(vi) 6x 2 + 7 x − 3 = 0
(ii) km 2 + n = p − wk
(3x − 1)(2 x + 3) = 0
km 2 + wk = p − n
x = 1 or − 3
k(m 2 + w ) = p − n 3 2
p−n
2
2 (i) Let t = xLet. t = x2
k= 2
m +w t 2 + 3t − 4 = 0
6 (i) 18 = 9 × 2 = 9 × 2 = 3 2 (t + 4)(t − 1) = 0
75 = 25 × 3 = 25 × 3 = 5 3 t = −4 or 1
(ii)
x 2 = −4 or x 2 = 1
(iii) 1 7 = 16 = 4
9 9 3 x = ±1

3 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 2 Algebra

(ii) Multiply by x. (ii) 3x 2 + 12 x − 4

2
5x − 2 = 2x = 3( x 2 + 4 x ) − 4
2
0 = 2x − 5x + 2 = 3 ( x + 2) 2 − 4  − 4
0 = (2 x − 1)( x − 2)
= 3( x + 2) 2 − 12 − 4
x = 1 or 2 2
2 = 3( x + 2) − 16
(iii) Let t = Let
x. t = x. (iii) 4 x 2 + 24 x − 16
= 4( x 2 + 6 x − 4)
t 2 + 2t =t 82 + 2t = 8
t 2 + 2t −t 82 =
+ 02t − 8 = 0 = 4 ( x + 3) 2 − 13 
2
(t + 4)(t (−t + = 0t − 2) = 0
2)4)( = 4( x + 3) − 52
t = −4 ort 2= −4 or 2 (iv) 9 x 2 − 6 x
x = −4 or
x=4 x=4
x = x−4=or
2 x =2
(
= 9 x2 − 6x
9 )
3 t = x3
(iv) Let t = xLet
2
. (
= 9 x2 − 2x
3 )
( ) − 19 
t + 8 = 9t  2
= 9 x − 1
2
t − 9t + 8 = 0  3

( 3 )
(t − 1)(t − 8) = 0 2
= 9 x − 1 −1
t = 1 or 8
x 3 = 1 or x 3 = 8 4 3 − 8 x − x 2
x = 1 or x = 2 = −( x 2 + 8 x − 3)
= −[( x + 4) 2 − 19]
2.3 Completing the square (Page 8)
= −( x + 4) 2 + 19
2
1 (i) ( x − 3) − 8 = 19 − ( x + 4) 2
2
(ii) ( x + 2) − 4
a = 19, b = 4
2
2 (i) x − 6 x + 1 = 0
2.4 The graphs of quadratic functions (Page 9)
( x − 3) 2 − 8 = 0 y
1 (i)
( x − 3) 2 = 8 5
4
x −3= ± 8
3
x = 3± 8
2
2
(ii) x + 4x = 0 1
2
( x + 2) − 4 = 0
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x
( x + 2) 2 = 4 –1
x +2=± 2 –2
x = 0 or − 4 –3
–4
3 (i) 2 x 2 − 4 x + 7
–5
= 2( x 2 − 2 x ) + 7
Vertex (−1, 0)
= 2 ( x − 1) 2 − 1 + 7
= 2( x − 1) 2 − 2 + 7
= 2( x − 1) 2 + 5

4 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 2 Algebra

y
(ii) (ii) 2x 2 − 5x = 6
5
4
2x 2 − 5x − 6 = 0
2
3 x = 5 ± 5 − 4 × 2 × −6
2
2× 2
1 = 5 ± 73
4
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x
= −0.886 or 3.39 (3 s.f.)
–1
–2 2 (i) Discriminant = (−2) 2 − 4 × 1 × 4 = −12
–3
No solutions
–4
(ii) Discriminant = 4 2 − 4 × 4 × 1 = 0
–5
Vertex (−4, −2) One solution

(iii) y 3 (i) (a) 4 2 − 4 × k × −1 = 0

5 16 + 4 k = 0
4
k = −4
3 2
(b) k − 4 × 1 × (k − 1) = 0
2
1 k 2 − 4k + 4 = 0
(k − 2)(k − 2) = 0
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x
–1 k=2
–2 2
(ii) (a) (−2) − 4 × 1 × k > 0
–3
4 − 4k > 0
–4
4 > 4k
–5
Vertex (2, 1) 1> k

k <1
(iv) y
5 (b) (− k ) 2 − 4 × k × 1 > 0
4
k 2 − 4k > 0
3
k(k − 4) > 0
2
1
k < 0 or k > 4
(iii) (a) (−2k ) 2 − 4 × 2 × 1 < 0
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x
–1
4k 2 − 8 < 0
–2
4k 2 < 8
–3
–4 k2 < 2
–5 − 2<k< 2
Vertex 1 , −4
2 ( )
2
(b) (−4) − 4 × k × 2k < 0
16 − 8k 2 < 0
2 (i) y = ( x − 4) 2 − 2
16 < 8k 2
(ii) y = −( x − 1) 2 + 5
2 < k2
2.5 The quadratic formula (Page 10)
k > 2 or k < − 2
1 (i) x 2 + x − 5 = 0
4 (i) Roots of 6 and −2 ⇒ ( x − 6 )( x + 2 ) = 0
−1 ± 12 − 4 × 1 × −5
x= x 2 − 4 x − 12 = 0 ⇒ m = −4, n = −12
2 ×1
= −1 ± 21
2
= 1.79 or − 2.79 (3 s.f.)

5 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 2 Algebra

(ii) x 2 − 4 x − 12 = p ⇒ x 2 − 4 x − 12 − p = 0 (ii) 1 − 3x > 7

2 4
b − 4ac = 0
4 − 3x > 28
(−4) 2 − 4 × 1 × (−12 − p) = 0
−3x > 24
16 + 48 + 4 p = 0
x < −8
64 + 4 p = 0
p = −16 (iii) −4 x − 1 < − x + 5   −4 x − 1 < − x + 5
  −3x < 6 or 6 < 3x
2.6 Simultaneous equations (Page 12) x > −2 −2 < x
1 (i) x = 3, y = −1
(ii) x = −2, y = 5 2 (i) y
5
2 (i) x 2 + 4x = x − 2 4
x 2 + 3x + 2 = 0 3
( x + 2)( x + 1) = 0 2
x = −1 or −2 1
y = −3 or − 4
(−1, − 3) or (−2, − 4) –5 –4 –3 –2 –1 0 1 2 3 4 5 x
–1
–2
(ii) x 2 + 4 x − 2 = −2 x − 2 –3
x 2 + 6x = 0 –4
x ( x + 6) = 0 –5
x = 0 or − 6 x < 0 or x > 1
y = −2 or 10
(ii) y
(0, − 2) or (−6, 10) 5
4
2
(iii) x 2 + (−x − 1) = 25 3
2 2
x + x + 2 x + 1 = 25 2
2
2 x + 2 x − 24 = 0 1
2
x + x − 12 = 0 –5 –4 –3 –2 –1 0 1 2 3 4 5 x
( x + 4)( x − 3) = 0 –1

x = −4 or 3 –2

y = 3 or −4 –3

(−4, 3) or (3, − 4) –4
–5

(iv) x (7 − 2 x ) = 3 0  x  1
(iii) y
7 x − 2x 2 = 3 5
0 = 2x 2 − 7 x + 3 4
0 = (2 x − 1)( x − 3) 3
x = 1 or 3 2
2
y = 6 or 1 1

( )
1 , 6 or (3, 1)
2
–5 –4 –3 –2 –1 0
–1
1 2 3 4 5 x

–2
2.7 Inequalities (Page 13) –3
–4
1 (i) 3( x + 4)  −15
–5
x + 4  −5
x  −9 x < −1 or x > 1

6 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 2 Algebra

(iv) y
(iii) 3x 2 − 5 x − 2 = 0
5
4
(3x + 1)( x − 2) = 0
3 x = − 1 or 2
3
2
1
(iv) 5 x 2 + 13x − 6 = 0
(5 x − 2)( x + 3) = 0
–5 –4 –3 –2 –1 0 1 2 3 4 5 x
–1 x = 2 or − 3
5
–2
4
–3
(v) Multiply by x :
–4 3 − 11x 2 = 4 x 4
–5 0 = 4 x 4 + 11x 2 − 3
–6
0 = (4 x 2 − 1)( x 2 + 3)
–7
–8 x 2 = 1 or x 2 = −3
4
–9
x = ±1
–10 2
x2 < 9 (vi) Multiply by x:
2
x −9<0 2 − x = 4x − 9 x
−3 < x < 3
0 = 5x − 9 x − 2
Further practice (Page 14) Let t = x .
1 (i) (3q − 1)(q + 2) 0 = 5t 2 − 9t − 2
(ii) t (s + p ) − 2u(s + p ) 0 = (5t + 1)(t − 2)
= (t − 2u)(s + p) t = − 1 or t = 2
5
2 (i) 1 =1+1
f u v x = − 1 or x = 2
5
1 = v +u
f uv x = 4
uv = f ( v + u ) 4 (i) 2 x 2 + 12 x + 11
uv = fv + fu = 2( x 2 + 6 x ) + 11
uv − fv = fu = 2 ( x + 3) 2 − 9  + 11
v (u − f ) = fu
= 2( x + 3) 2 − 18 + 11
fu
v= = 2( x + 3) 2 − 7
u− f
p (ii) 5 x 2 − 40 x + 72
(ii) d − 3e = 1
2π w = 5( x 2 − 8 x ) + 72
 p
d − 3e = 1 2   = 5 ( x − 4) 2 − 16  + 72
4π  w 
= 5( x − 4) 2 − 80 + 72
 p
d − 1 2   = 3e 2
4π  w  = 5( x − 4) − 8
p
e=d− 5 7 + 8x − 2x 2
3 12 π 2 w
= −2 x 2 + 8 x + 7
3 (i) x 2 − 25 = 0
= −2( x 2 − 4 x ) + 7
x 2 = 25
= −2[( x − 2) 2 − 4] + 7
x = ± 25
x = 5 or − 5 = −2( x − 2) 2 + 8 + 7
2
(ii) x + 5 x − 14 = 0 = −2( x − 2) 2 + 15
( x + 7)( x − 2) = 0 = 15 − 2( x − 2) 2
x = −7 or 2 a = 15, b = 2, c = 2

7 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 2 Algebra

6 (i) y = x ( x + 2) = ( x + 1) 2 − 1 (b) 3 − 2kx 22 = 6 x

2
(ii) y = −( x − 3) 2 + 4 2kx 2 + 6 x − 3 = 0
2
7 (i) 1 − 3x = 5 x 622 − 4 × 2k × −3 > 0
36 + 24 k > 0
0 = 3x 2 + 5 x − 1
24 k > −36
−5 ± 5 2 − 4 × 3 × −1
x=
2×3 k > −36
24
−5 ± 37
= k > −3
6
2
= 0.180 or − 1.85 (3 s.f.)
2
(iii) (a) (−1) 2 − 4 × 9 × k < 0
(ii) 12 x = 6 x − 5
1 − 36k < 0
0 = 6 x 2 − 12 x − 5 1 < 36k
12 ± 12 2 − 4 × 6 × −5 1 <k
x=
2×6 36
=
12 ± 264 k> 1
12 36
= −0.354 or 2.35 (b) kx 2 + 2 x − 1 = 0
8 90 = 9 × 10 = 9 × 10 = 3 10 2 2 − 4 × k × −1 < 0
so a = 3
4 + 4k < 0
2
9 (i) Discriminant = (−3) − 4 × −1 × −2 = 1 4 k < −4
Two solutions k < −1
2
(ii) Discriminant = 0 − 4 × 1 × 4 = −16 11 (i) x = 3, y = 12
No solutions (ii) x = 6, y = −3
2
(iii) Discriminant = (−1) − 4 × 5 × 2 = −39 12 (i) x 2 + 9 = 4x + 5
No solutions x 2 − 4x + 4 = 0
2
Discriminant = 3 − 4 × −4 × 0 = 9
(iv) ( x − 2)( x − 2) = 0
Two solutions x = 2, y = 13
(2, 13)
10 (i) (a) k 2 − 4 × 1 × 4 = 0
k 2 − 16 = 0 (ii) ( x − 2) 2 + (9 − 2 x ) 2 = 5

k 2 = 16 x 2 − 4 x + 4 + (81 − 36 x + 4 x 2 ) = 5
k = 4 or − 4 5 x 2 − 40 x + 80 = 0
(b) kx 2 − kx − 1 = 0 x 2 − 8 x + 16 = 0
(− k ) 2 − 4 × k × −1 = 0 ( x − 4) 2 = 0
k 2 + 4k = 0 x = 4, y = 1
k(k + 4) = 0 (4, 1)
k = 0 or − 4 (iii) x 2 + 2(1 − x ) 2 = 9
2
(ii) (a) (− k ) − 4 × 3 × 3 > 0 x 2 + 2(1 − 2 x + x 2 ) = 9
k 2 − 36 > 0 x 2 + 2 − 4 x + 2x 2 = 9
k 2 > 36 3x 2 − 4 x − 7 = 0
k < −6 or k > 6 (3x − 7)( x + 1) = 0
x = 7 or − 1
3
y = − 4 or 2
3

(73 , − 43 ) or ( −1, 2)
8 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 2 Algebra

y
(iv) 2x + 3 y = 7 ⇒ y = 7 − 2x (iii)
3 10

3x 2 = 4 + 4 x 7 − 2 x
3 ( ) 9
8
2
3x = 4 + 28 x − 8 x
2 7
3 3
6
9 x = 12 + 28 x − 8 x 2
2
5
2
17 x − 28 x − 12 = 0 4

(17 x + 6)( x − 2) = 0 3
2
x = − 6 or 2
17 1

y = 131 or 1 –4 –3 –2 –1 0 1 2 3 4 5 6 x
51
–1
(−176 , 131
51 )
or (2, 1) –2
–3

13 (i) y –4
16 –5
15 –6
14 –7
13 –8
12 –9
11 –10
10 −2  x  4
9 (iv) y
8 14
7 12
6 10
5 8
4 6
3 4
2 2
1
–10 –8 –6 –4 –2 0 2 4 6 8 x
–2
–5 –4 –3 –2 –1 0 1 2 3 4 5 x
–1 –4

–2 –6

–3 –8

–4 –10

–5 –12
−4  x  4 –14
(ii) y –16
3 –18
2 –20
1 –22
–24
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 x
–1
x < −7 or x > 2
–2
–3
–4
–5
–6

  x < −5 or x > 0

9 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 2 Algebra

y
(v) (viii) 3x 2 > 6 x ⇒ 3x 2 − 6 x > 0
y
20
4
15
3
10
2
5
1
–6 –5 –4 –3 –2 –1 0 1 2 3x
–5 –2 –1 0 1 2 3 x
–1
–10
–2
–15
–3

x < −5 or x > 3 –4
2
x < 0 or x > 2
(vi) y
30 14 (i) 4 x 2 + 32 x + 70
25 = 4( x 2 + 8 x ) + 70
20
= 4 ( x + 4 ) 2 − 16  + 70
15
10 = 4 ( x + 4 ) 2 − 64 + 70
5
= 4 ( x + 4)2 + 6
–4 –3 –2 –1 0 1 2 3 4 5 6 x Vertex is (−4, 6)
–5
–10 (ii) 4 ( x + 4 ) 2 + 6 < 22
4 ( x + 4 ) 2 < 16
−1 < x <3
2 ( x + 4)2 < 4
(vii) y x 2 + 8 x + 12 < 0
10 ( x + 6)( x + 2) < 0
8 −6 < x < −2
6
Past exam questions (Page 15)
4
2 1 ( x − 2)( x + 1) > 0
x < −1 or x > 2
–6 –5 –4 –3 –2 –1 0 1 2 3 x
2 (i) x + 6 x + 2 = ( x + 3) − 7
2 2
–2
–4
(ii) ( x + 3) − 7 > 9
2

–6
–8
( x + 3) 2 > 16
–10 x + 3 > 4 or x + 3 < −4
x > 1 or x < −7
−3  x  1 3 (i) 2 x 2 − 10 x + 8 = 2( x 2 − 5 x ) + 8
3

( ) − 254  + 8
2
= 2 x − 5
 2

= 2 ( x − 5 ) − 25 + 8
2

2 2

= 2( x − 5 ) − 9
2

2 2

10 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 3 Coordinate geometry

(ii) 2 x 2 − 10 x + 8 = kx Rearranging, you get

3
α 3 + β 3 = (α + β ) − 3αβ (α + β )
2 x 2 − 10 x − kx + 8 = 0
= 2 3 − 3(3)(2) = −10
2 x 2 + (−10 − k )x + 8 = 0
(−10 − k ) 2 − 4 × 2 × 8 < 0 (αβ ) 3 = 33 = 27
100 + 20k + k 2 − 64 < 0 The equation is x 2 + 10 x + 27 = 0

k 2 + 20k + 36 < 0 2    9 x − 3 x +1 − 54 = 0
(k + 18)(k + 2) < 0 (3 x ) 2 − 3 × 3 x − 54 = 0
−18 < k < −2 Let t = 3 x .
t 2 − 3t − 54 = 0
Stretch and challenge (Page 16)
(t + 6)(t − 9) = 0
1 (i) ax 2 + bx + c = 0 ⇒ x 2 + b x + c = 0
a a t = −6 or 9
Given the roots, you can write the equation as x
3 = −6 or 9
( x − α )( x − β ) = 0
2
3 x = −6 has no solutions so 3 x = 9 ⇒ x = 2
x − α x − β x + αβ = 0
3 k 2 x + 2 − x = 8
x 2 − (α + β ) x + αβ = 0
Let t = 2 x ; the equation can then be written as
Equating, you get
kt + 1 = 8
− (α + β ) = b ⇒ α + β = − b t
a a 2
kt + 1 = 8t
αβ = c
a kt 2 − 8t + 1 = 0
(ii) 4 x + ( k + 2 ) x + 72 = 0
2
For a single solution, b 2 − 4ac = 0
The roots are α and 2α . (−8) 2 − 4 × k × 1 = 0 ⇒ k = 16
α + 2α = − k + 2 and α × 2α = 72 16t 2 − 8t + 1 = 0
4 4
Solving the second equation, ( 4t − 1) 2 = 0 ⇒ t = 14
2α 2 = 18
2 x = 1 ⇒ x = −2
α2 =9 4
α = ±3
Substituting into the first equation you get 3 Coordinate geometry
3 + 2 × 3 = − k + 2 ⇒ k = −38 3.1 The length, gradient and midpoint of a line
4
or (−3) + 2 × (−3) = − k + 2 ⇒ k = 34 (Page 17)
4
1 (i) m = 3
(iii) α + β = 4 and αβ = 7
3 3 (ii) m = 0
4 1
1 + 1 = α + β = 3 = 4 and 1 = 3 (iii) m = −
4
α β αβ 7 7 αβ 7
3 2 (i) Length = AB = 4 2 + 2 2 = 20 ≈ 4.47
So the equation is Midpoint = (1, 0)
x 2 − 4 x + 3 = 0 or 7 x 2 − 4 x + 3 = 0 Gradient = 1
7 7 2
(iv) α + β = 2 and αβ = 3 (ii) Length = CD = 4 2 + 10 2 = 116 ≈ 10.8
3
(α + β ) = α + 3α 3 2 2
β + 3αβ + β 3
Midpoint = (10, 2)

= α + β + 3αβ (α + β )
3 3
Gradient = − 10 = − 5 = −2.5
4 2

3 −1 + m = 2 ⇒ m = 5
2
5+n = 5 ⇒ n = 5
2

11 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 3 Coordinate geometry

4 (i) x = 0 3 (i) 3 y = − x − 6
(ii) x = −4 x + 3y + 6 = 0
(iii) x = 1 (ii) 15 y = 12 x + 5

5 (i) m(Line 1) = 1 ⇒ m1 = tan −1 1 = 26.6°

2 (2) 12 x − 15 y + 5 = 0
4 A: y = − x + 1 or x + y = 1
m(Line 2) = 2 ⇒ m2 = tan −1 (2) = 63.4°

B: y = 2 x − 3
m(Line 3) = −1 ⇒ m3 = tan −1 (−1) = −45° = 135°
1
C: y = x
(2)
(ii) θ = tan −1 (2) − tan −1 1 = 36.9° (1 d.p.)

3
D: y = −3

3.2 The equation of a straight line (Page 19) E: x = 6

2
F: y = − x − 2
1 3
Gradient y-intercept
(i) 3 −1 5 Gradient of line Gradient of perpendicular
(ii) −2 3 1 −1
3 1
(iii) 1 –4
2 4
(iv) −1 −2
2 −2 1 = − 7 3
3 3 7
y
5 −3 2
2 3
4
3 0.3 = 3 − 10
10 3
2
(iv)
1 6 (i) 2 x − y = 1 ⇒ y = 2 x − 1
The gradient is 2 so the equation is
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x
–1 y = 2x + c
(ii)
–2 1 = 2 × 4 + c ⇒ c = −7
–3 y = 2x − 7
(iii) (i)
–4
(ii) The gradient of the perpendicular is −
1
–5
2
The equation is y = − 1 x + c
2 2
Equation Gradient y-intercept
1
1 = − × −3 + c ⇒ c = − 1
(i) y = −x + 2 –1 2 2 2
(ii) y = 3x − 2 3 –2 1
y =− x − 1
2 2
(iii) y = −1 x + 9 −1 9 y
2 4 2 4
5
(iv) y = 3x −4 3 −4 4
2x – y = 1
2 2
3
y = −0.5x – 0.5
y
2
5 2x – y = 7
(ii) (iv) 1
4
(iii)
3 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x
–1
2
–2
1
–3
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x –4
–1
–5
–2
(i)
–3
–4
–5

12 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 3 Coordinate geometry

3.3 The intersection of two lines (Page 22)

4 ax − 3 y + 1 = 0 ⇒ 3 y = ax + 1 ⇒ y = a x + 1
y 3 3
1
5 2 x + by − 6 = 0 ⇒ by = −2 x + 6 ⇒ y = − 2 x + 6
4
b b
If the lines are perpendicular,
3
2
m1 × m2 = −1
1 a × − 2 = −1
3 b
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x a=3
–1 b 2
–2 a:b =3: 2
–3 5 y
–4
Tangent
–5
Normal
The x-intercept is when y = 0, so
C
3x = 8 ⇒ x = 8
3
The y-intercept is when x = 0, so
2y = 8 ⇒ y = 4
B A
Area = 1 × 8 × 4 = 16 = 5 1 0 x
2 3 3 3
2 y
5
4
When x = 3, y = 3 × (4 − 3) = 3
so the point C is (3, 3)
3
The x-intercept of the line y = 9 − 2 x
2
9
1 is when 0 = 9 − 2x, so x = 2
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x The normal at (3, 3) has gradient 1
–1 2
The equation of the normal is
–2
y = 1 x + c ⇒ 3 = 1 × 3+ c ⇒ c = 3
–3 2 2 2
–4 y=1x+3
2 2
–5
The x-intercept of this line is when
(
The midpoint of AB is 2 + −6 , 4 + 0 = (−2, 2)
2 2 ) 0 = 1 x + 3 ⇒ x = −3
2 2
1
The gradient of AB is so the gradient of the
2 Length of AB = 7 1
2
perpendicular line is –2. Area of triangle ABC = 1 × 7 1 × 3 = 11 1
The equation is 2 2 4
y = −2 x + c ⇒ 2 = −2 × −2 + c ⇒ c = −2 3.4 The circle (Page 24)

y = −2 x − 2
1 (i) Centre (1, −3), radius 4
3 Midpoint of BC is (2 2 )
2 + 4 , 3 + −5 = (3, − 1)
(ii) Centre (−3, 1), radius 3
Call the midpoint M. 2 (i) ( x − 4) 2 + ( y + 1) 2 = 36 or
The gradient of AM is − 3 . x 2 + y 2 − 8 x + 2 y − 19 = 0

8
The equation of AM is (ii) ( x + 2) 2 + ( y − 7) 2 = 8 or
y = −3 x +c
8 x 2 + y 2 + 4 x − 14 y + 45 = 0

3 1 (iii) Radius = (2 − (−1)) 2 + (4 − 0) 2 = 25 = 5
2 = − 8 × −5 + c ⇒ c = 8
y = −3 x + 1 ( x − 2) 2 + ( y − 4) 2 = 25 or
8 8
x 2 + y 2 − 4x − 8 y − 5 = 0

13 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 3 Coordinate geometry

2
(iv) ( x − 3) 2 + ( y + 3) 2 = 9 or (ii) x + 4 x − 2 = −2 x − 2
x 2 + y 2 − 6x + 6 y + 9 = 0
x 2 + 6x = 0
x ( x + 6) = 0
3 A: ( x + 1) 2 + y 2 = 9 or x 2 + y 2 + 2 x − 8 = 0
x = 0 or − 6
( − 2) 2 + ( y − 4) 2 = 16 or
B: x y = −2 or 10
x 2 + y 2 − 4x − 8 y + 4 = 0
(0, −2) or (−6, 10)
4 (i) x 2 + y 2 − 2 x + 4 y − 20 = 0
(iii) x (7 − 2 x ) = 3
( x − 1) 2 + ( y + 2) 2 = 25 7 x − 2x 2 = 3
Centre C is (1, −2) 0 = 2x 2 − 7 x + 3

4.5 = 3
(ii) Gradient of DC is 0 = (2 x − 1)( x − 3)
1.5
Gradient of AB is − 1 x = 1 or 3
2
3
Equation of AB is y = 6 or 1

2.5 = − 1 × 2.5 + c ⇒ c = 10
3 3
(
2
) ( )
1 , 6 or 3, 1

So y = − 1 x + 10 (iv) ( x − 2) 2 + (9 − 2 x ) 2 = 5
3 3
x + 3 y − 10 = 0 x 2 − 4 x + 4 + (81 − 36 x + 4 x 2 ) = 5

5 x 2 − 40 x + 80 = 0
2 2
(iii) (10 − 3 y ) + y − 2(10 − 3 y ) + 4 y − 20 = 0 x 2 − 8 x + 16 = 0
100 − 60 y + 9 y 2 + y 2 − 20 + 6 y + 4 y − 20 = 0 ( x − 4) 2 = 0
10 y 2 − 50 y + 60 = 0 x = 4, y = 1
(4, 1)
y 2 −5y +6 = 0
( y − 2)( y − 3) = 0 2 x 2 + 3x + 9 = x + k
2
y = 2 or 3 x + 2x + 9 − k = 0
x = 4 or 1 For one solution, discriminant = 0
2 2 − 4 × 1 × (9 − k ) = 0
A(1, 3), B(4, 2)
y 4 − 36 + 4 k = 0
8 −32 + 4 k = 0
6 4 k = 32
4 A k=8
2 B
x + 3y – 10 = 0
3 2 − x 2 = 3 − kx
–8 –6 –4 –2 0 2 4 6 8 x 0 = x 2 − kx + 1
–2
–4 For two points of intersection, discriminant > 0
–6
x2 + y2 – 2x + 4y – 20 = 0
(− k ) 2 − 4 × 1 × 1 > 0
–8
k2 − 4 > 0
3.5 The intersection of a line and a curve (Page 26) k2 > 4
1 (i) x 2 + 4x = x − 2 k > 2 or k < −2
x 2 + 3x + 2 = 0 4 (i) Centre of circle is (−1, 0)
( x + 2)( x + 1) = 0 Gradient of radius = − 4
3
x = −1 or −2
Gradient of tangent = 3
y = −3 or −4 4
(−1, −3) or (−2, −4)

14 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 3 Coordinate geometry

Equation of tangent is When k = 4,

4 = 3 × −4 + c ⇒ c = 7 2 + 4x 2 = 4x + 1
4
4x 2 − 4x + 1 = 0
So y = 3 x + 7
4 ( 2x − 1)( 2x − 1) = 0
3
m= , c=7 x=1
4 2
y y = 4 × 1 +1 = 3
2
( )
8
B 1
Point is , 3
6
y= 3 x+7 2
4 4
2 Further practice (Page 28)
A
–10 –8 –6 –4 –2 0 2 4 6 8 10 x 1 (i) Length = EF = 8 2 + 4 2 = 80 ≈ 8.94
–2
Midpoint = (−1, 1)
–4
–6 (x + 1)2 + y2 = 25 Gradient = −4 = − 1
8 2
–8
(ii) Length = GH = 6 2 + 12 2 = 180 ≈ 13.4
(ii) A is −
3 (
28 , 0 , B is 0, 7
( ) ) Midpoint = ( −7, − 3)

Area AOB = 1 × 28 × 7 = 98 Gradient = 12 = 2

2 3 3 6
5   y + kx = 8 ⇒ y = 8 − kx 2 m(line A) = tan −1 2 = 63.4°
2 x 2 +23xx2 + 10
2 x 2 +23xx2 + kx
3x =+810− =kx8 − kx
3x + 2kx= +0 2 = 0

( )
m(line B) = tan −1 − 1 = −18.4°
3
2
x 2++k(3 Angle between lines = 63.43° + 18.43° = 81.0° (1 d.p.)
2 x +2(3 )x++k2)x= +0 2 = 0
3 (i) Gradient = 1 , y-intercept = −3
2 2
For aFor
tangent, b − 4bac
a tangent, −=40ac = 0
4
) 2+−k4) 2× −2 4× ×2 2= ×0 2 = 0
(3 + k(3
(ii) Gradient = − 2 , y-intercept = 5
9 + 6k9++ k62k −+16
k 2 =−016 = 0 3
k 2 + 6kk2 −+ 76k= −0 7 = 0 y

( k − 1()(k k−+1)(7 )k=+07 ) = 0

10
k = −k7 =or−71 or 1
y = 5 – 2x
6 (i) 2 + kx 2 = kx + 1 3 5
2
kx − kx + 1 = 0
For no common points, b 2 − 4ac < 0 –10 –5 0 5 10 x
2
(− k ) − 4 × k × 1 < 0 y = 1 x –3
2 –5 4
k − 4k < 0
k(k − 4) < 0
–10
0<k<4
(ii) For a tangent, b 2 − 4ac = 0 4 (i) y = 1 x + 1
2 2 2
k − 4k = 0
k(k − 4) = 0 (ii) y = − x − 2
1
2 5
k = 0 or 4 5 (i) 3x − 2 y + 8 = 0
Since k ≠ 0, k = 4
(ii) 8 x + 5 y + 5 = 0

15 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 3 Coordinate geometry

6 (i) Line is of the form 4 x + 3 y = k y

3
Substituting (-2, 0) gives
2
4 × −2 + 3 × 0 = k 1
k = −8
4 x + 3 y = −8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 x
–1
4x + 3y + 8 = 0
–2
(ii) Rearranging the equation gives –3
–4
4x + 3y = 1
3 y = −4 x + 1 –5
–6
y = −4x+1
3 3 –7
The gradient of the line is − 4 so the gradient of –8
3
–9
the perpendicular line is .3
4 –10
3
The equation is y = x + c
–11
4 –12
Substituting the point (3, −1) gives
8 (i) x − 3 y = 6 ⇒ 3 y = x − 6 ⇒ y = 1 x − 2
3
−1 = 3 × 3 + c 1
4 The gradient is so the gradient of the
3
c = −13 perpendicular line is −3.
4
Equation of the perpendicular line is
3
y = x − 13
4 4 y = −3 x + c ⇒ 2 = −3 × −3 + c ⇒ c = −7
4 y = 3 x − 13
y = −3 x − 7
3 x − 4 y − 13 = 0
y (ii) Solving simultaneously,
5
y = −3 x − 7
4
3 x − 3 y = 6
2 Substitution gives
4x + 3y = – 8 3x – 4y = 13
1 x − 3(−3 x − 7) = 6
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x x + 9 x + 21 = 6
–1
10 x = −15
–2
–3 x =−3
4x + 3y = 1 2
–4
y = −3 × − 3 − 7 = − 5
–5 2 2

(
7 (i) Midpoint is 4 + −1 , 1 + −9 = 3 , −4
2 2 ) (2 ) (
The distance between − 3 , − 5 and (−3, 2) is
2 2 )
(ii) m(AC) = 2 so m(⊥ ) = − 1
(−3 − − 32 ) + (2 − − 52 )
2 2
2 d = = 4.74 (3 s.f.)
y = mx + c ⇒ −2 = − 1 × 6 + c ⇒ c = 1
2

Equation is y = − 1 x + 1
2

16 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 3 Coordinate geometry

y
(−3) 2 − 4 × 2 × k < 0
5
9 − 8k < 0
4
9 < 8k
3
P (–3, 2) 9<k
2 8
1
k>9
8
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x
–1 14    x 2 − x (kx − 8) = k
–2
x 2 − kx 2 + 8x − k = 0
–3
(1 − k )x 2 + 8x − k = 0
–4
–5
( 8 ) 2 − 4 × (1 − k ) × − k = 0
8 + 4k − 4k 2 = 0
9 (i) Centre (0, −5), radius 8
2 + k − k2 = 0
(ii) Centre (−2, 1), radius 3 (2 − k )(1 + k ) = 0
10 (i) ( x − 5) 2 + ( y + 1) 2 = 49
k = 2 or − 1

(ii) ( x − 3) 2 + ( y + 3) 2 = 9 Past exam questions (Page 29)

1 (i) Gradient of perpendicular bisector =

3
11 (i) x 2 + 9 = 4x + 5 2
2
x − 4x + 4 = 0 Gradient of AB = − 2
( x − 2)( x − 2) = 0 3
x = 2, y = 13 Equation of AB:
(2, 13) 6 = − 2 × −2 + c ⇒ c = 14
3 3
(ii) x 2 + 2(1 − x ) 2 = 9 2 14
x 2 + 2(1 − 2 x + x 2 ) = 9 So y = − 3 x + 3 or 2 x + 3 y − 14 = 0
x 2 + 2 − 4 x + 2x 2 = 9 (ii) Solving simultaneously,
3x 2 − 4 x − 7 = 0 2 y += 53xor+ 53xor
2y = 3x 3x=−-5
- 2y 2 y = −5 ①
(3x − 7)( x + 1) = 0
2 x + 3 y = 14 2x + 3y = 14 ②

x = 7 or − 1 MultiplyMultiply
① by 3, ② 1 byby3,2:2 by 2
3
9 x − 6 y = −15
y = − 4 or 2
3 4 x + 6 y = 28
7
3 3
4
(
, − or (−1, 2) ) 13x = 13
x = 1, y = 4
12 2 − x 2 = 3 − kx
Midpoint is (1, 4)
0 = x 2 − kx + 1 By symmetry, B is (4, 2)
For two points of intersection, discriminant > 0 y
(− k ) 2 − 4 × 1 × 1 > 0
10
k2 − 4 > 0
A(–2, 6)
k2 > 4
5
k > 2 or k < −2
B(4, 2)
13 2x + k = 3 –10 –5 0 5 10 x
x
2
2 x + k = 3x 2y = 3x + 5
–5
2
2 x − 3x + k = 0
–10
No points of intersection ⇒ discriminant < 0

17 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 3 Coordinate geometry

(
2 (i) Midpoint is 5 + 9 , 7 + −1 = ( 7, 3)
22 ) (ii) Gradient of 2x + 3y = 9 is − 2
3
Gradient of AB is (−1) − 7 = −2 Gradient of AB is 3
9−5 2
3p 3
=
Gradient of perpendicular line is 1 9− p 2
2
Equation is 6 p = 27 − 3 p
9 p = 27
3= 1 ×7+c ⇒c = −1
2 2 p=3
1
So y = x − 1
2 2 4 x 2 − 4 x + c = 2x − 7
(ii) Equation of CX is x 2 − 6 x + (c + 7) = 0
2 = −2 × 1 + c ⇒ c = 4 b 2 − 4ac = 0
So y = −2 x + 4 (−6) 2 − 4 × 1 × (c + 7) = 0

36 − 4c − 28 = 0
Solving simultaneously,
4c = 8
−2 x + 4 = 1 x − 1 c=2
2 2
−4 x + 8 = x − 1 5 Solving simultaneously,
9 = 5x y = 3x y = 3x

x = 9, y = 2 4 y = x + 11 ⇒4 y = x1 x+ 11 ⇒ y = 1 x + 11
+ 11
5 5 4 4 4 4
1 11 1 11
5 5 ( )
So X is 9 , 2 or (1.8, 0.4 ) 3x = x + 3x = x +
4 4 4
12 x = x + 11 12 x = x + 11
4
2 2
BX = (9 − 1.8) + (−1 − 0.4 )
2
11x = 11 11x = 11
BX = 7.33
y x = 1, y = 3 x = 1, y = 3
10 A is (1, 3)
9
By symmetry, C is (12, 14).
8
A(5, 7) Equation of BC is
7
6 44 yy =
5
= xx +
+ cc
44 × 14 =
× 14 12 +
= 12 + cc ⇒
⇒ cc =
= 44
44
4
y = 0.5x – 0.5 44 yy = x + 44
3 = x + 44
C(1, 2)
2
SolvingSolving simultaneously
simultaneously
Solving to findto
simultaneously B,find
to find B
B
1 ( )
X 9,2
2 5 4(3xx )) =
4(3 = xx +
+ 44
44
–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 x
–1 11xx =
11 = 44
44
B(9, −1) xx =
–2
= 4, yy =
4, = 12
12
–3 B is (4, 12)
–4 B is (4, 12) 12)
B is (4,
–5
By symmetry, D is (9, 5).
–6
–7 6 3x 2 − 4 x + 7 = mx + 4
–8 3x 2 + (−4 − m)x + 3 = 0
–9
–10 For two distinct points of intersection, b 2 − 4ac > 0
(−4 − m) 2 − 4 × 3 × 3 > 0
3 (i)    (9 − p ) 2 + (3 p ) 2 = 13 2
2 2
16 + 8m + m 2 − 36 > 0
(81 − 18 p + p ) + 9 p = 169
2
m 2 + 8m − 20 > 0
10 p − 18 p − 88 = 0 (m + 10)(m − 2) > 0
5 p 2 − 9 p − 44 = 0 m < −10 or m > 2
(5 p + 11)( p − 4) = 0
p = − 11 or 4
5

18 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 3 Coordinate geometry

Stretch and challenge (Page 30) y

2 x − = 4
1 The diagram shows the relationship between the a b
tangent and normal at the point A. Multiplying every term by ab:

y bx − ay = 4ab
5 bx − 4ab = ay
4
y = bx − 4ab
a
3 B
b
y = x − 4b
A 2 a
1 From this form the gradient of the line is b ,
a
b 1
so = ⇒ 2b = a
–4 –3 –2 –1 0 1 2 3 4 x a 2

The x-intercept is when y = 0 so
You need to find the coordinates of both A and
B. To find where the line meets the curve, solve x − 0 = 4 ⇒ x = 4 ⇒ x = 4a , so M is ( 4a, 0 )
a b a
simultaneously: The y-intercept is when x = 0 so
4 x + y + 24= y + 2y =
x +0 ⇒ x −y2= −4 x − 2
= 0−4⇒
0 − y = 4 ⇒ − y = 4 ⇒ y = −4b, so N is 0, − 4b
2 x 2 = −42xx−2 2= −4 x − 2 a b b
( )
2 x 2 + 4 x 2+x22 =+04 x + 2 = 0 By Pythagoras’ theorem,
x 2 + 2 x +x12=+02 x + 1 = 0
MN = (4a) 2 + (4b) 2 = 16a 2 + 16b 2
( x + 1)( x +
( x1)+=1)(0x + 1) = 0
x = −1 x = −1 16a 2 + 16b 2 = 720
y = 2(−1)y2 == 2(
2 −1) 2 = 2 16a 2 + 16b 2 = 720

So A is (−1, 2) 16(2b) 2 + 16b 2 = 720

The gradient of the tangent line is −4 so the gradient 16 × 4b 2 + 16b 2 = 720

of the normal is 1 64b 2 + 16b 2 = 720
4
The equation of the normal: 80b 2 = 720
y = mx + c ⇒ 2 = 1 × −1 + c ⇒ c = 9 b2 = 9
4 4 b=3
So the equation is y = 1 x + 9 a = 2b = 6
4 4
y
To find B, solve simultaneously: 16
2x 2 = 1 x + 9 14
4 4 12
8 x 2 = x + 9   ⇒  8 x 2 − x − 9 = 0 10
(8 x − 9)( x + 1) = 0 8
x = 9 or − 1 6
8 4
Substitute x = 9 into the equation of the curve to find 2
8 M
the y-coordinate: –6 –4 –2 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 x
–2

()
2
y=2 9 = 2 × 81 = 81
–4

8 64 32 –6

( )
So B is 9 , 81
8 32
–8
–10
Finally, use Pythagoras’ theorem to find the length of –12
N
AB: –14
–16

( ) ( )
2 2
AB = −1 − 9 + 2 − 81 = 2.19 (3 s.f.) –18
8 32

19 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 4 Sequences and series

3 Midpoint of the line joining A ( −6, 1) and B ( k , − 3) is (iii) The x-intercepts are x = a and b.
So the line must pass through ( a, 0 ) or (b, 0 ).
(−62+ k , 1+2−3) = ( k −2 6 , −1) y = mx + c ⇒ 0 = k × a + c ⇒ k = − c
a
Gradient of AB is 1 − (−3) = 4 y = mx + c ⇒ 0 = k × b + c ⇒ k = − c
−6 − k −6 − k b
Gradient of perpendicular is − −6 − k = 6 + k 5 2 y = kx + 1 ⇒ y = k x + 1
4 4 2 2
Equation of perpendicular bisector is 2x 2 + x +1 = k x + 1
2 2
y = mx + c ⇒ −1 = 6 + k × k − 6 + c
4 2 2
4 x + 2 x + 2 = kx + 1
2
−1 = k − 36 + c 4 x 2 + (2 − k )x + 1 = 0
8
2
c = − k − 36 − 1 For two points of intersection, b 2 − 4ac > 0
8
(2 − k ) 2 − 4 × 4 × 1 > 0
Since the y-intercept is −9,
4 − 4 k + k 2 − 16 > 0
2
− k − 36 − 1 = −9 k 2 − 4 k − 12 > 0
8
2 ( k + 2)( k − 6) > 0
− k − 36 = −8
8 k < −2 or k > 6
2
k − 36 = 8
8
2
k − 36 = 64
4 Sequences and series
k 2 = 100 4.1 Arithmetic progressions (Page 31)
k = ±10 1 (i) 5, 11, 17, 23, …
y
6 a = 5, d = 6, n = 12
4 u12 = 5 + (12 − 1) × 6 = 71
2 (ii) −3, −8, −13, …

–14 –12 –10 –8 –6 –4 –2 0 2 4 6 8 10 12 14 x

a = −3, d = −5, n = 15
–2 (2, –1) u15 = −3 + (15 − 1) × −5 = −73
(–8, –1)
–4
(iii) 2 , 6 , 26 ,
–6 3 5 15
2 8
–8
(0, –9) a = 3 , d = 15 , n = 7
–10
2 8 58
u7 = 3 + (7 − 1) × 15 = 15
4 (i) x − 4 x + 3 = ( x − 1)( x − 3) so the x-intercepts
2
(iv) 0.85, 0.55, 0.25, …
are x = 1 and 3.
a = 0.85, d = -0.3, n = 11
So the line must pass through (1, 0 ) or ( 3, 0 ). u11 = 0.85 + (11 − 1) × −0.3 = −2.15
y = kx + 1 ⇒ 0 = k × 1 + 1 ⇒ k = −1 2 (i) 21, 15, 9, …
or y = kx + 1 ⇒ 0 = k × 3 + 1 ⇒ k = − 1 a = 21, d = −6, n = 9
3
(ii) The x-intercepts are x = a and b. S9 = 9 [ 2 × 21 + (9 − 1) × −6 ] = −27
2
So the line must pass through (a, 0) or (b, 0).
(ii) −8, 0, 8, …
y = kx + 1 ⇒ 0 = k × a + 1 ⇒ k = − 1 a = −8, d = 8, n = 18
a
y = kx + 1 ⇒ 0 = k × b + 1 ⇒ k = − 1 S18 = 18 [ 2 × −8 + (18 − 1) × 8] = 1080
b 2

20 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 4 Sequences and series

(iii) 7, 9, 11, … , 79 7 a = 4u4 = 4 ( a + 3d ) ⇒ a = 4a + 12d

a = 7, d = 2, n = ?, un = l = 79 3a + 12d = 0
un = 7 + (n − 1) × 2 = 79 a + 4d = 0
2(n − 1) = 72 u6 = u4 − 4 ⇒ a + 5d = a + 3d − 4
n − 1 = 36 2d = −4
n = 37 d = −2

S37 = 37 [7 + 79] = 1591 a + 4 × −2 = 0 ⇒ a = 8
2
(iv) 0.05, 0.15, 0.25, … , 4.05
u8 = 8 + 7 × −2 = −6

a = 0.05, d = 0.1, n = ?, un = l = 4.05 4.2 Geometric progressions (Page 33)

un = 0.05 + (n − 1) × 0.1 = 4.05
1 (i) 1 , 1, 2, 4, …
0.1(n − 1) = 4 2
n − 1 = 40 a = 1 , r = 2, n = 10
2
n = 41 1
S 41 = 41 [0.05 + 4.05] = 84.05
u10 = × 210−1 = 256
2 2

(ii) -16, 8, -4, 2, …
3 34 + 5d = 43
a = -16, r = − 1 , n = 8
5d = 9 2

( )
8−1
9 9 u8 = −16 × − 1 = 0.125 = 1
d = d ==1.8= 1.8 2 8
5 5
a = 34 a =−344 ×−1.8
4 ×=1.8
26.8
= 26.8 (iii) 0.1, 0.01, 0.001, …
20 20 a = 0.1, r = 0.1, n = 9
S20 =S20 =[ 2 × [26.8 + (20+−(20
2 × 26.8 1) ×] 1.8
1) ×−1.8 ] = 878
= 878 9−1
2 2 u9 = 0.1 × (0.1) = 0.000 000 001 = 1 × 10−9
4 d = (2m + n) − m = m + n (iv) 6, 9, 13.5, …
u10 = m + (10 − 1) × (m + n) a = 6, r = 1.5, n = 12
= m + 9(m + n) u12 = 6 × (1.5 )
12−1
= 519 (3 s.f.)
= 10m + 9n
2 (i) 3, 6, 12, 24, …
5 a = 40, S30 = 3375, n = 30, d = ?
a = 3, r = 2, n = 8
S30 = 3375 ⇒ 30 [ 2 × 40 + (30 − 1)d ] = 3375
2
15(80 + 29d ) = 3375 S8 =
(
3 1 − 28 ) = 765
1− 2
1200 + 435d = 3375
(ii) 81, 27, 9, 3, …
435d = 2175
d =5 a = 81, r = 1 , n = 20
3

()
u5 = 40 + (5 − 1) × 5 = $60  20 
81 1 − 1 
6 (i) u7 = 32 ⇒ a + 6d = 32  3 
S 20 = = 121.5
5 1− 1
S5 = 130 ⇒ [ 2a + 4d ] = 130 3
2
5a + 10d = 130 (iii) 2, −5, 12.5, −31.25, …
a + 2d = 26 a = 2, r = −2.5, n = 14
Solving simultaneously, d = 1.5, a = 23
(ii) un = 56 ⇒ 23 + (n − 1) × 1.5 = 56 S 14 =
(
2 1 − (−2.5)
14
) = −212 874.3028
1 − (−2.5)
23 + 1.5n − 1.5 = 56
= −213 000 (3 s.f.)
1.5n + 21.5 = 56
(iv) −240, 48, −9.6, 1.92, …
1.5n = 34.5
a = −240, r = −0.2, n = 10
n = 23

S10 =
(
−240 1 − ( −0.2 )
10
) = −200
1 − (−0.2)

21 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 4 Sequences and series

3 The sequences in parts (ii) and (iv) have a sum to (ii) Plan B: a = 1, d = t, n = 20
infinity: 20
81
S20 = [2 × 1 + (20 − 1) × t ]
(ii) S∞ = = 121.5 2
1− 1 = 10(2 + 19t )
3
−240 = −200 = 20 + 190t
(iv) S ∞ = So to be the same,
1 − (−0.2)
20 + 190t = 1260
4 x =2
32 x t = 6.53 minutes (3 s.f.)
x 2 = 64
9 (i) For the A.P., a = 18
x = ±8
u 4 = a + (4 − 1)d = 18 + 3d
5 a = 8, S ∞ = 12 ⇒ 8 = 12 u6 = a + (6 − 1)d = 18 + 5d
1− r
8 = 12(1 − r ) 8 = 12(1 − r ) First three terms of G.P. are 18, 18 + 3d, 18 + 5d
8 8 18 + 3d 18 + 5d
=1− r =1− r so r = =
12 12 18 18 + 3d
8 1 r =1− =
8 1 (18 + 3d ) = 18 (18 + 5d )
2

r = 1 − 12 = 3 12 3
5 −1
324 + 108d + 9d 2 = 324 + 90d
5 −1
 1 8  1 8
u5 = 8 ×   = u5 = 8 ×  3  =
81 9d 2 + 18d = 0
 3 81
9d(d + 2) = 0
2
6 3, 6x, 12x , … d = 0 or − 2

6 x 12 x 2 Since d ≠ 0, d = −2
r= = = 2x
3 6x
When d = −2, r = 18 + 3d = 18 + 3 × −2 = 12 = 2
For the sequence to have a sum to infinity, 18 18 18 3
−1 < r < 1 so −1 < 2x < 1 (ii) S∞ = 18 = 54
1 1 1− 2
− <x< 3
2 2 (iii) You want Sn = 0 with a = 18, d = –2
5x + 1 2 x + 2 2x + 4 n 2 × 18 + (n − 1) × −2 = 0
×r = x + 2 ⇒ r2 = = 2[ ]
7 (i)
2 5x + 1 5x + 1
2 n 36 − 2n + 2 = 0
x 2
[ ]
x x
( x + 2) × r 2 = ⇒ r 2 = 2 = n 38 − 2n = 0
[ ]
2 x + 2 2x + 4 2
2x + 4 x n = 0 or 19
so =  So n = 19
5x + 1 2x + 4
( 2x + 4 )2 = x (5x + 1) 4.3 Binomial expansions (Page 36)
2 2
4 x + 16 x + 16 = 5 x + x 1 (i) ( x + 2 )
4

0 = x 2 − 15 x − 16  4  4  4
= x 4 +   x 3 × 2 +   x 2 × 22 +   x 1 × 23 + 24
0 = ( x − 16)( x + 1) 1   2  3
x = −1 or 16 4 3 2
= x + 8 x + 24 x + 32 x + 16
(ii) r 2 = x = 16 = 4 ⇒ r = 4 = 2 (ii) (1 − 3x )
3
2 x + 4 36 9 9 3
 3  3
5 x + 1 5 × 16 + 1 = 13 +   12 × (−3x ) +   1 × (−3x ) 2 + (−3x )3
a= = = 40.5 1  2
2 2
2 3
40.5 = 1 − 9 x + 27 x − 27 x
S∞ = = 121.5
1− 2 2 (i) (2 x − 3)8
3
 8  8
8 (i) Plan A: G.P. with a = 2, r = 1.3, n = 20 = (2 x )8 +   (2 x )7 × (−3)1 +   (2 x )6 × (−3) 2 + 
1  2
S 20 =
(
2 1 − (1.3)
20
) = 1260 minutes (3 s.f.)
= 256 x 8 − 3072 x 7 + 16 128 x 6 − 
1 − (1.3)

22 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 4 Sequences and series

6 The x3 term in the expansion of

(ii) ( 3a − b )7
(1 + bx )( x + 2)5 will be
 7 7 6 1  
7 5 2
= (3a) +   (3a) × (−b) +   (3a) × (−b) +  1 × x 3 term + bx × x 2 term
1   2
 5  5
= 2187a 7 − 5103a 6b + 5103a 5b 2 −  ( x + 2 )5 =  +   x 3 × 2 2 +   x 2 × 2 3 + 
 2  3
7
(iii)  1 + x 3   =  + 40 x
3
+ 80 x 2 + 
x2 
7 6 5
The x3 term is then
   7    7  
=  12  +    12  × ( x 3 )1 +    12  × ( x 3 )2 +  1 × 40 x 3 + bx × 80 x 2 = ( 40 + 80b ) x 3
 x  1   x   2  x 
In order that there is no term in x3,
= 114 + 79 + 214 +  40 + 80b = 0 ⇒ b = − 1
x x x 2
 10
()
r
7 (i) ( 3 + u )
10 − r 1 6
3 General term is   ( 2 x )
r x
 6  6
( x )10−r = 3 6 +   35 × u1 +   3 4 × u 2 + 
x terms are =x 10 − r − r
=x 10 − 2r  
1  2
xr
You want 10 − 2r = 2 = 729 + 1458u + 1215u 2 + 
2
r=4 (ii) u = x − x

()
 10 4 = 729 + 1458( x − x 2 ) + 1215( x − x 2 ) 2 + 
So the x2 term is   ( 2 x ) 1
6 2
= 13 440 x
 4 x = 729 + 1458 x − 1458 x 2 + 1215( x 2 − 2 x 3 + x 4 ) + 

So the coefficient is 13 440 The x2 term is −1458 x 2 + 1215 x 2 = −243x 2
r
 15 2
4 General term is 
 r 
( )
x 15 − r  5
 3 
x
The coefficient is −243
 7
8 The general term is   4
7−r
( −ax )r
x terms are
(x )
2 15 − r
= x 30− 2r − 3r = x 30−5r
r

(x ) 3 r
For the x3 term you need r = 3 so
You want 30 − 5r = 0  7 4
 3  4 ( −ax ) = -8960a x
3 3 3
r=6
So the term independent of x is −8960a 3 = −1120 ⇒ a 3 = 1 ⇒ a = 1
8 2
 15 2
6 The x2 term is when r = 2

 6  x( ) 9 5
 3  = 78 203125 7
( )
  4 5 − 1 x = 5376 x 2
2
x
 2 2
 6
5 (i) General term is   2 6− r (−3x )r The coefficient of the x2 term is 5376
r
You need r = 2, 9 (i) First three terms of 3 − 2 x 2 ( ) are
7

 6 4  7 7

2
 2 2 (−3x ) = 2160 x
2
(
37 +   36 −2 x 2
1 
) +  2 3 ( −2x )
1 5 2 2

So the coefficient is 2160. = 2187 − 10 206 x 2 + 20 412 x 4

(
(ii) ( 3 + 2 x )  ___ x + 2160 x +  2
) (ii) (1 + x 2 )(3 − 2 x 2 )7
You need the x term in the expansion of ( 2 − 3x ) .
6
= (1 + x 2 )(2187 − 10 206 x 2 + 20 412 x 4 )
Here r = 1 so
The x 4 term is
 6 5 1
 1  2 (−3x ) = −576 x 1 × 20 412 x 4 + x 2 × −10 206 x 2

= 20 412 x 4 − 10 206 x 4
The x2 term is
4
2 x × −576 x + 3 × 2160 x 2 = 5328 x 2 = 10 206 x
Hence the coefficient is 5328 The coefficient of the x 4 term is 10 206

23 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 4 Sequences and series

( )
 6 r Now using the term formula,
10 The general term is   k 6− r − 1 x
r 2 u 21 = 15 + (21 − 1) × −1.5 = −15
3
The x term is when r = 3 Alternatively, you can form two equations from the
6  3 1 3

 3 ( ) 3
  k − 2 x = 20 × k × − 8 x
1 3

information and solve them simultaneously:
3rd term = 12 ⇒ 12 = a + 2d
= − 5 k 3x 3 7th term = 6 ⇒ 6 = a + 6d
2
5 3 Subtracting the equations gives 4d = -6
− k = 160
2 ⇒ d = −1.5
k 3 = −64 The rest of the answer follows from there.
k = −4 8
5 r = 2 = = 2
2 2
Further practice (Page 39)
u6 = 2 × ( 2 )
6 −1
=8
u3 = 12 ⇒ a + 2d = 12   €
1  
6 , , , , -40, , , , , -20, …
S8 = 168 ⇒ 8 [ 2a + (8 − 1)d ] = 168 +d +d +d +d +d
2
⇒ 8a + 28d = 168   −40 + 5d = −20

Solving simultaneously, 5d = 20
d=4
8a + 16d = 96  €×8 = ‚
Subtracting 4 from −40 four times gives
12d = 72   − ‚
a = −56
d=6
You want Sn = 0.
a=0
n 2 × −56 + (n − 1) × 4 = 0
 
u14 = 0 + (14 − 1) × 6 = 78 2
[ ]
n −112 + 4n − 4 = 0
2 a = 25, d = ?, u9 = 5, un = l = − 45
2
[ ]
25 + 8d = 5 n 4n − 116 = 0
2
[ ]
8d = −20
d = −2.5 2n 2 − 58n = 0

un = l = −45 ⇒ −45 = 25 + (n − 1) × −2.5 2n(n − 29) = 0

−45 = 25 − 2.5n + 2.5 n = 0 or 29
2.5n = 72.5 Since n = 0 is not a possible answer, the answer is n = 29
n = 29 7 , 9, , , 11 , …
8
S 29 = 29 [ 25 + −45 ] = −290 ×r ×r ×r ×r ×r
2
4
3 128r = 40.5 From this, write 9r 3 = 1 1 ⇒ r 3 = 1 ⇒ r = 1
8 8 2
So the first term, a = 18
r 4 = 40.5
128 Alternatively, write ar = 9 and ar 4 = 1 1
8
r = 4 40.5 = 0.75 Dividing the equations gives
128
11
u5 = 128 × 0.75 2 = 72 ar 4 = 8 so r 3 = 1
ar 9 8
a = 1282 = 227.5
0.75 and the answer is the same.
 S ∞ = a = 18 = 36
S ∞ = 227.5 = 910.2 1 − r 1 − 0.5
1 − 0.75
4 , , 12, , , , 6, … 8 r = 0.95
+d +d +d +d
u6 = 45 000 × ( 0.95 )8 = $29 853.92
From the above, you can write 12 + 4d = 6 ⇒ d = −1.5
= $29 900 (3 s.f.)
Subtracting d from 12 twice gives the first term, a = 15

24 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 4 Sequences and series

( x1 + 2x )
6
 4 2 2  5 3
 2  3 ( − x ) +  2 2 ( ax )
2
9

= ( 1 ) +   ( 1 ) × (2 x ) +   ( 1 ) × (2 x )
 6
6  6
5 4
1 2
= 54 x 2 + 80a 2 x 2
x 1 x  2 x
= 16 + 124 + 602 + 
(
= 54 + 80a 2 x 2 )
x x x 54 + 80a = 554 2

 8 x
() ( ) 80a 2 = 500
8− r r
10 General term is   −3
 r 2 x
a 2 = 6.25
(x ) 8− r
a = 2.5
x terms are = x 8− r − r = x 8− 2r
xr
8 − 2r = 4 Past exam questions (Page 39)
r=2
( x)
5
1 (i) 2 x − 1
 8
( )( )
6 2
So the x4 term is   x − 3 = 63 x 4
 2 2 x 16  5
( )
r
General term is   ( 2 x )5− r − 1
11 General term is   ( )
 12 2 12 − r  r x
r x
(x ) 2 r

( x )5−r
To find the value of r that gives us the term x terms are = x 5− r − r = x 5− 2r
xr
independent of x, it is possible to guess and check 5 − 2r = 1
until one that works is found OR use an algebraic    r = 2
 5
( ) = 80x
2
approach.
So the x term is   ( 2 x ) − 1
3

 2 x
(x ) 2 r
2r
= x12− r = x 2r −(12− r ) = x 3r −12 Coefficient of x is 80
12− r
x x
)( x)

(
5
2
(ii) 1 + 3x 2 x −
1
To get the term independent of x, the power of x
must be 0, so 3r − 12 = 0 ⇒ r = 4
( )
5
x −1 term in 2 x − 1 :
Substituting into the general term, x

()
 12 2 12− 4 2 4
( ) 5 − 2r = −1
8
 4  x x = 495 × 2 8 × x 8 = 126 720
x r=3

( )
5  3
 5
() So the x-1 term is  ( 2 x ) − 1
2
12 General term is   ( kx )
5− r 2
r
= −40 x −1
x 3
  x
r
( )( )
5
5− r
x terms in 1 + 3x 2 2 x − 1 are
x terms are x r = x 5− 2r x
x
( ) + 3x ( )
5 5
To get the x-1 term: 1 × x term in 2 x − 1 2
× x −1 term in 2 x − 1
x x
5 − 2r = −1
= 1 × 80 x + 3x 2 × −40 x −1
r = 3
= 80 x − 120 x
 5
( ) = 80k x
3
2 2
 3 ( kx ) x
2 −1
= −40 x
Coefficient of x is −40
80k 2 = 720
k2 = 9 2 a + ar = 50 ⇒ a(1 + r ) = 50

k = ±3 ar + ar 2 = 30 ⇒ ar (1 + r ) = 30
 4 Dividing the equations,
13 The general term for ( 3 − x ) is   3 ( − x )
4−r 4 r
r  ar (1 + r ) 30
=
a(1 + r ) 50
 5
The general term for ( 2 + ax )5 is   2 5− r ( ax )r r = 3 = 0.6
r 5
In both cases, r = 2 is the value needed to get the x2 term. a = 31.25
S ∞ = 31.25 = 78.125
1 − 0.6

25 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 4 Sequences and series

3 (i) (a) A.P. with a = 10, d = 2 (b) u 25 = −15 + (25 − 1) × 3 = 57

u30 = 10 + (30 − 1) × 2 = 68 litres (c) The first six terms are
(b) Sn = n [ 2 × 10 + (n − 1) × 2 ] = 2000 −15, − 12, − 9, − 6, − 3, 0
2
n [ 20 + 2n − 2] = 2000 So the sum of the 7th to 25th terms is needed.
2 S7 = −15 + −12 + −9 + −6 + −3 + 0 = −45
18n + 2n 2 = 4000
S = S 25 − S7 = 525 − (−45) = 570
+
2n 2 + 18n − 4000 = 0
n 2 + 9n − 2000 = 0 (ii) (a)  2022 is 10 years later, or the 11th term:

Solving gives u11 = 4000 × 1.0510 = $6515.58 or $6516
11
n = 40.4 or − 49.4 (b)  S = 4000(1.05 − 1)
= $56 827 or $56 800
11 1.05 − 1
n = 41
(ii) G.P. with a = 10, r = 1.1 Stretch and challenge (Page 40)

S30 =
(
10 1 − (1.1)30 ) = 1644.94 L 1 u7 = 400 ⇒ a + 6d = 400
1 − (1.1) S30 = 1800 ⇒ 30 [ 2a + (30 − 1) × d ] = 1800
2
Percentage left = 2000 − 1644.94 × 100% = 17.8% 15[ 2a + 29d ] = 1800 ⇒ 30a + 435d = 1800
2000
2a + 29d = 120
4 (i) (a − x ) 5 Solving simultaneously,
 5  5  5
= a 5 +   a 4 (−x )1 +   a 3 (−x ) 2 +   a 2 (−x )3 + d = -40 and a = 640
1   2  3
2 General term for the first expansion is
= a 5 − 5a 4 x + 10a 3 x 2 − 10a 2 x 3 + r
 6
( ) 3 6− r  7
(ii) (1 − ax )( a − x )
5
 r  kx  − 3 
x
(
= (1 − ax ) a 5 − 5a 4 x + 10a 3 x 2 − 10a 2 x 3 +  ) Term independent of x is when r = 3
3
3
 6
( )
Terms in x : 3 6− 3  7 3 9 −343
 3 kx  − 3  = 20k x × 9

−10a 2 x 3 + ( −ax ) 10a 3 x 2 ( ) x x
= −6860k 3
= −10a 2 x 3 − 10a 4 x=3−10a 2 x 3 − 10a 4 x 3
General term for the second expansion is
( ) (
= −10a 2 − 10a 4 x=3 −10a 2 − 10a 4 x 3 )  8 r
( )
4 8− r m
−10a − 10a 4 = −200
2
−10a 2 − 10a 4 = −200  r  kx  4 
x
10a 4 + 10a 2 − 200 10
= 0a 4 + 10a 2 − 200 = 0 Term independent of x is when r = 4
4 2
a + a − 20 = 0 a 4 + a 2 − 20 = 0 8  4 8− 4  4
( )
4
 m4  = 70k x × m16 = 70k m
4 16 4 4
  kx
(a 2
)( ) (
− 4 a 2 + 5 = 0a 2 − 4 a 2 + 5 = 0 )( ) 4 x  x
2
a = −5 (ignore) a = −5 (ignore) 2 So 70k 4m 4 = −6860k 3
2 2 k = − 984
a = 4 ⇒ a = ±2 a = 4 ⇒ a = ±2 m
5 (i) (a) a = −15, n = 25 3 a + ar = −3 ⇒ a(1 + r ) = −3
S 25 = 525 ar 5 + ar 6 = 729 ⇒ ar 5 (1 + r ) = 729
⇒ 25 [ 2 × −15 + (25 − 1) × d ] = 525 ar 5 (1 + r ) 729
2 =
25 −30 + 24d = 525 a(1 + r ) −3
2
[ ]
r 5 = −243
−375 + 300d = 525
r = 5 −243 = −3
d=3
a (1 + (−3)) = −3 ⇒ −2a = −3 ⇒ a = 1.5

26 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 5 Functions and transformations

4 u12 = 3 × u6 ⇒ a + 11d = 3( a + 5d ) 5 Functions and

a + 11d = 3a + 15d
2a + 4d = 0
transformations
a + 2d = 0 5.1 The language of functions (Page 41)
30
S30 = 450 ⇒ [ 2a + 29d ] = 450
2 1 (i) f(x) = x + 3
30a + 435d = 450
(ii) f(x) = x2 + 1
6a + 87d = 90
(iii) f(x) = −2x + 1 = 1 − 2x
Solving simultaneously,
a = −2.4, d = 1.2 (iv) f(x) = 10x + 2

5 Stamp 1: 2 All are functions. (ii) is not one—one; the rest are
one—one.
55 000, 52 600, …
A.P. with a = 55 000, d = −2400 3

Stamp 2: Function One—one

G.P. with r = 0.96 (i) ✔ ✗

For Stamp 1, u10 = 55 000 + (10 − 1) × −2400 = 33 400 (ii) ✗ ✗
a × 0.9610 = 33 400 (iii) ✔ ✔
33 400 (iv) ✔ ✗
a= = $50 238
0.9610
4 (i) f (−2) = (−2) 2 − 2(−2) + 1 = 9
 10 10− r r
6 (i) The general term is   ( x ) (a) (ii) f ( x + 1) = ( x + 1) 2 − 2( x + 1) + 1
r
The expansion is = x 2 + 2x + 1 − 2x − 2 + 1
x 10 + 10ax 9 + 45a 2 x 8 + 120a 3 x 7 + 210a 4 x 6 + 252a 5 x 5 = x2

+ 210a 6 x 4 + 120a 7 x 3 + 45a 8 x 2 + 10a 9 x + a10 5 (i) Domain: x∈

Comparing the coefficients of the terms either Range: f ( x )  −2
side of the x3 term: (ii) Domain: x ∈ 
120a 7 > 210a 6 ⇒ a > 7 Range: f (x )  3
4
7 8 8 (iii) Domain: x  1
120a > 45a ⇒ a <
3 Range: f (x )  2
7
So < a < 8
4 3 (iv) Domain: x ∈ 
Range: f ( x ) ∈ 
(ii) The terms that are of interest are
 100  100 6 Range is f(x)  −2
  ( x ) 44 ( a )56 +   ( x ) 43 ( a )57
 56   57 
y
 100
+ ( x ) 42 ( a )58  4
 58  3

 100 57  100 56 2
57
 57  a >  56  a ⇒ a > 44 1

 100 57  100 58 –4 –2 0 1 2 3 4 5 6 7 x
and  a > a ⇒ a < 58 –2
 57   58  43
–4

So 57 < a < 58
44 43
7 Domain: 0x4
Range: 0  f (x )  2

27 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 5 Functions and transformations

8 y 2 ff ( x ) = f (ax + b )

2 = a(ax + b) + b
= a 2 x + ab + b
–4 –2 0 1 2 3 4 5 6 x
–2 so a 2 x + ab + b = 9 x − 4
–4
∴ a 2 = 9 ⇒ a = ± 3
–6
When a = 3 3b + b = −4
–8
4b = −4
–10
b = −1

Domain: x ∈  When a = -3 −3b + b = −4

−2b = −4
Range: f ( x )  1
b=2
9 Given that x  −1, the range (from the diagram) is a = 3, b = -1 or a = -3, b = 2
f ( x )  −3.
k(x) ( x)
3 (i) fg( x ) = f 4 + 2

= 3( 4 + 2 ) + 2
7
6 x
5
= 12 + 6 + 2
4 x
3 = 14 + 6
x
2
(ii) gf ( x ) = g(3x + 2)
1
= 4+ 2
–6 –5 –4 –3 –2 –1 0 1 2 x 3x + 2
–1 4(3x + 2) + 2
=
–2 3x + 2
–3 = 12 x + 8 + 2
–4
3x + 2
–5 = 12 x + 10
3x + 2
–6
(iii) ff ( x ) = f (3x + 2)
5.2 Composite functions (Page 43)
= 3(3x + 2) + 2
1 (i) f (−2) = 1 − (−2) = 3 = 9x + 6 + 2
(ii) g(−2) = 1 − (−2) 2 = −3 = 9x + 8
(iii) fg(−2) = f (−3) = 1 − (−3) = 4 9 x + 8 = −3
(iv) gf (−2) = g(3) = 1 − (3) 2 = −8 9 x = −11
(v) fg( x ) = f (1 − x 2 ) = 1 − (1 − x 2 ) = x 2 x = − 11
9
(vi) gf ( x ) = g(1 − x )
(iv) 4 + 2 = kx
= 1 − (1 − x ) 2 x
= 1 − (1 − 2 x + x 2 ) 4 x + 2 = kx 2
= 2x − x 2 0 = kx 2 − 4 x − 2
(vii) ff ( x ) = f (1 − x ) = 1 − (1 − x ) = x For two solutions, b 2 − 4ac > 0
2
(viii) gg( x ) = g(1 − x )
(−4) 2 − 4 × k × −2 > 0
= 1 − (1 − x 2 ) 2
16 + 8k > 0
= 1 − (1 − 2 x 2 + x 4 )
8k > −16
= 2x 2 − x 4
k > −2

28 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 5 Functions and transformations

4 g(1) = −1 ⇒ b − a = −1 (ii) f −1 ( x ) = 5 − x
gg(1) = 5 ⇒ g(−1) = 5 ⇒ b + a = 5 y
6
Solving simultaneously gives
a = 3, b = 2 5
4
5 (i) fg( x ) = f ( x 2 + 4 x ) 3

( 2
= 3 (x + 4x ) + 2 ) 2
1
= 3x 2 + 12 x + 6
3x 2 + 12 x + 6  x 2 + 4 x –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x
–1
2x 2 + 8x + 6  0 –2
2
x + 4x + 3  0 –3
–4
( x + 3)( x + 1)  0
–5
x  −3 or x  −1
–6
(ii) gf ( x ) = g ( 3( x + 2))
(iii) f ( x ) = 1 − 5 x = −5 x + 1
= g ( 3x + 6 )
f −1 ( x ) = x − 1 = 1 − x
= ( 3x + 6 ) + 4 ( 3x + 6 )
2 −5 5
y
= 9 x 2 + 36 x + 36 + 12 x + 24
6
= 9 x 2 + 48 x + 60 5
4
gf ( x )  45
3
9 x 2 + 48 x + 60  45 2
2
9 x + 48 x + 15  0 1
2
3x + 16 x + 5  0 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x
–1
(3x + 1)( x + 5)  0
–2
−5  x  − 1 –3
3
–4
5.3 Inverse functions (Page 45) –5

  –6
1 (i) f −1 ( x ) = x − 1
2 −1
(iv) f ( x ) = 2( x + 1) = 2 x + 2
y y
6 6

5 5
4
4
3
3
2
2
1
1
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x –1
–1 –2
–2 –3
–3 –4
–4 –5

–5 –6

–6

29 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 5 Functions and transformations

2 (i) k = 0 Domain f −1 ( x ): x  −4
y
6
Range f −1 ( x ): f −1 ( x )  2
5 (iii) k = −1
4
3   y = x 2 + 2 x − 4 = ( x + 1) 2 − 5
2 y
1 6
5
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x
–1 4
–2 3
–3 2
–4 1
–5
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x
–6 –1
Inverse is x = y 2 − 4 –2
–3
    x + 4 = y2
–4
   x+4 = y –5
−1 –6
    f (x ) = x + 4
Domain f ( x ): x0 Inverse is x = ( y + 1) 2 − 5
Range f ( x ): f ( x )  −4 x + 5 = ( y + 1) 2
−1
Domain f ( x ): x  −4 x + 5 = y +1
−1 −1
Range f ( x ) : f (x )  0 x + 5 −1 = y
(ii) k = 2 f −1 ( x ) = x + 5 − 1
  y = x 2 − 4 x = ( x − 2) 2 − 4 Domain f ( x ): x  −1
y Range f ( x ): f ( x )  −5
6
−1
5 Domain f ( x ): x  −5
4 Range f −1 ( x ): f −1 ( x )  −1
3
(iv) k = −2
2
1 y = 2 x 2 + 8 x + 7 = 2( x + 2) 2 − 1
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x y
–1 8
–2 7
–3 6
–4 5
–5 4

–6 3
2
Inverse is x = ( y − 2) 2 − 4 1
2
x + 4 = ( y − 2) –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 x
–1
x+4 = y−2
–2
x+4+2= y –3
f −1 ( x ) = x + 4 + 2 –4

Domain f ( x ): x2
Range f ( x ): f ( x )  −4

30 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 5 Functions and transformations

Inverse is x = 2( y + 2) 2 − 1 5 k = π

x + 1 = 2( y + 2) 2 y

x + 1 = ( y + 2) 2 1
2
x +1 = y + 2
2
x +1 − 2 = y
2
f −1 ( x ) = x + 1 − 2 0 p p 3p 2p x
2 2 2
Domain f ( x ): x  −2
Range f ( x ): f ( x )  −1
−1
Domain f ( x ): x  −1
–1
Range f −1 ( x ): f −1 ( x )  −2

( 2 ) − 54
2
3 g( x ) = x 2 + 3x + 1 = x + 3 5.4 Transformations (Page 48)

so x  − 3 means g will be one–to–one 1
Transformation Detail
2
The largest possible value of k is − 3 For y = f(x) + b, the effect • If b > 0 it moves
2 of b is to translate the
y upwards.
6 graph of y = f(x) vertically • If b < 0 it moves

5 through b units. downwards.

4
3 For y = f(x – a), the • If a > 0 it moves to the
2 effect of a is to translate right.
1 the graph of y = f(x) • If a < 0 it moves to the
horizontally through a left.
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x units.
–1
For y = f(x – a) + b, the You say it has been
–2
graph of y = f(x) has been translated by the vector
–3
translated horizontally  a
–4  b 
by a units and translated
–5 vertically by b units.
–6
For y = pf(x), p > 0, the • If p > 1 it moves points
effect of p is to vertically on y = f(x) further away
4 Inverse is x = 2( y + 1)3 − 1 stretch the graph of from the x-axis.
  y = f(x) by a factor of p. • If 0 < p < 1 it moves
x + 1 = 2( y + 1)3
points on y = f(x) closer
x + 1 = ( y + 1)3 to the x-axis.
2
For y = f(kx), k > 0, • If k > 1 it moves points
3 +1 = y +1
x
2 the effect of k is to on y = f(x) closer to the
horizontally stretch y-axis.
3 x +1 −1 = y
2 the graph of y = f(x) by a • If 0 < k < 1 it moves

factor of 1 . points on y = f(x)

f −1 ( x ) = 3 x + 1 − 1 k further away from the
2
y-axis.
For y = −f(x), the effect on
the graph of y = f(x) is to
reflect it in the x-axis.
For y = f(−x), the effect on
the graph of y = f(x) is to
reflect it in the y-axis.

31 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 5 Functions and transformations

2 f(x) (iii) f(x)

5
5
4
4
3
3
2
y = f (x + 1) – 2 2
1
1

–5 –4 –3 –2 –1 0 1 2 3 4 x x
–1 –5 –4 –3 –2 –1 0 1 2 3 4 5
–1
–2
y
5 –3
3 (i)
4 –4

3 –5
y = f(–x) y = f(x)
2 (iv) f(x)
1 5
4
–4 –3 –2 –1 0 1 2 3 4 x
–1 3

–2 2
y = –f(–x) y = –f(x)
–3 1

–4 x
–5 –4 –3 –2 –1 0 1 2 3 4 5
–5 –1
–2
 0
(ii) Translating by   is the same as y = f(x) + 2 –3
 2
–4
Equation is y = ( x − 1) 2 + 2 –5
2
y = x − 2x + 1 + 2
(v) f(x)
y = x 2 − 2x + 3 5
4
4 (i) f(x)
3
5
2
4
1
3
x
2 –5 –4 –3 –2 –1 0 1 2 3 4 5
–1
1
–2
x
–5 –4 –3 –2 –1 0 1 2 3 4 5 –3
–1
–4
–2
–5
–3
–4 (vi)
f(x)
–5 5
(ii) f(x) 4
5
3
4
2
3
1
2
x
1 –5 –4 –3 –2 –1 0 1 2 3 4 5
–1
x
–5 –4 –3 –2 –1 0 1 2 3 4 5 –2
–1
–3
–2
–4
–3
–5
–4
–5

32 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 5 Functions and transformations

 3 (ii) The definition of an inverse is ff −1 ( x ) = x

5 Translation by   is the same as f(x − 3) + 1
1 or f −1f ( x ) = x . Since ff ( x ) = x here, the
function f must be its own inverse so
f ( x − 3) + 1 = ( x − 3) 2 − 3( x − 3) + 1
f −1 ( x ) = x + 2 , x ∈ , x ≠ 1 .
= x 2 − 6 x + 9 − 3x + 9 + 1 4x − 1 4
= x 2 − 9 x + 19 OR
f(x) The original function is y = x + 2 so the
4x − 1
inverse is
5
y+2
4 x=
4y −1
3
2
x (4 y − 1) = y + 2
1 4 xy − x = y + 2
4 xy − y = x + 2
x
–1 0 1 2 3 4 5 6
–1 y (4 x − 1) = x + 2
y = x+2
–2
4x − 1
6 • Reflection in the y-axis So f ( x ) = x + 2 , x ∈ , x ≠ 1
−1
4x − 1 4
• Vertical stretch, scale factor 2
3 (i) f (−1) = 1 ⇒ a(−1) + b = 1 so − a + b = 1
Further practice (Page 50) f (2) = 7 ⇒ a(2) + b = 7 so 2a + b = 7
1 Completing the square: 3a = 6 ⇒ a = 2
f(x) = x2 – 4x + 2 = (x – 2)2 − 2 −2 + b = 1 ⇒ b = 3
So the vertex is at (2, −2), as shown on the graph. (ii) Given that f ( x ) = 2 x + 3
f(x)
ff ( x ) = f (2 x + 3) = 2(2 x + 3) + 3 = 4 x + 9
4 4 x + 9 = 1 ⇒ x = −2
3
4 Inverse is x= 2
2 3− y
1 x (3 − y ) = 2
–2 –1 0 1 2 3 4 5 x 3− y = 2
–1 x
2
–2 3− = y
x
–3
j −1 ( x ) = 3 − 2 = 3x − 2
The domain of f ( x ) is x ∈ . x x
The range of f ( x ) is y  −2. 5 (i) When you are asked this kind of question, what is

The domain of f −1 ( x ) is x  -2. meant is what is the value of k so that x  k means

g( x ) has an inverse function, or said another way,
The range of f −1 ( x ) is y ∈ .
that the inverse of g( x ) is a function.

) (( ))
x+2 +2
(
f(x)
x + 2 4x − 1
2 (i) ff ( x ) = f = 2
4x − 1
4 x + 2 −1
4x − 1 1
x + 2 + 2(4 xx−+1)2 + 2(4 x − 1) 0 x
–7 –6 –5 –4 –3 –2 –1 1
= 4x − 1 4x −1 –1
4( x + 2) −=(44(
x x− +
1)2) − (4 x − 1) –2
4x − 1 4x −1 –3
9x x + 2 + 8x − 2
= –4
= 4x − 1 4x + 8 − 4x +1
9 –5
4 x − 1 = 9x –6
9
= 9x × 4 x − 1
4 x − 1 =9x –7

=x

33 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 5 Functions and transformations

Completing the square: Since b 2 − 4ac = (−1) 2 − 4 × 1 × 1 = −3, the

2
y = ( x + 3) − 7 equation has no real solutions.
This form shows us the vertex is at (−3, −7) as
shown.
(iv) f −1 g( x ) = f −1
x+2
2x + 7 ( )
The value of x that divides the parabola down
=
2− x +2
2x + 7 ( )
the middle is the value of x needed. This is −3. 3
So k = −3. 2( 2x + 7 ) − ( x + 2)
= 2x + 7
(ii) y = ( x + 3) 2 − 7 3
x = ( y + 3) 2 − 7 4 x + 14 − x − 2
= 2x + 7
x + 7 = ( y + 3) 2 3
x +7 = y +3 3x + 12
x +7 −3= y = 2x + 7
3
g −1 ( x ) = x + 7 − 3 = 3 x + 12
6 x + 21
(2 − 3x ) + 2 4 − 3x 3( x + 4 )
6 (i) gf ( x ) = g(2 − 3x ) = = =
2(2 − 3x ) + 7 11 − 6 x 3( 2 x + 7 )
gf ( x ) = x ⇒ 4 − 3x = x
11 − 6 x = x+4
2x + 7
4 − 3x = x (11 − 6 x )
(v) −1
4 − 3x = 11x − 6 x 2 Point on y = f ( x ) Point on y = f ( x )
6 x 2 − 14 x + 4 = 0 (0, 2) (2, 0)
2
3x − 7 x + 2 = 0
(3x − 1)( x − 2) = 0 (1, − 1) (−1, 1)
x = 1 or x = 2 (−1, 5) (5, − 1)
3

(ii) f −1 ( x ) is x = 2 − 3 y ⇒ 3 y = 2 − x ⇒ y = 2 − x
f(x)
3 8
6
so f −1 ( x ) = 2 − x
3 4
y+2 2
g −1 ( x ) is x =
2y + 7
x
–8 –6 –4 –2 0 2 4 6 8
x (2 y + 7) = y + 2 –2
2 xy + 7 x = y + 2 –4

2 xy − y = 2 − 7 x –6

–8
y (2 x − 1) = 2 − 7 x
 −1
y = 2 − 7x 7 Translation by  is the equivalent of
2x − 1  2 
g (x ) = − 7x
−1 2
2x − 1 (
f ( x + 1) + 2 = ( x + 1) + 1 + 2
3
)
−1 2 − 7x
(iii) g ( x ) = x − 4 ⇒ 2 x − 1 = x − 4
( 3
= x + 3x + 3x + 1 + 1 + 2 2
)
3 2
2 − 7 x = ( x − 4)(2 x − 1) = x + 3x + 3x + 4

2 − 7 x = 2x 2 − 9x + 4 8 • Reflection in the x-axis

2 • Horizontal stretch, scale factor 1
0 = 2x − 2x + 2 2
2
0 = x − x +1

34 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 5 Functions and transformations

Past exam questions (Page 51) 3 (i) Since g is defined for x  −1, the domain of f

1 (i) fg( x ) = f (5x4+ 2 ) must be x  −1.

3x + 1  −1
= 4 −2 3x  −2
4
5x + 2 x −2
4(5 x + 2) 3
=
4
−2 Largest value of a is − 2
3
= 5x
(ii) fg( x ) + 14 = 0
Range of fg is y  0
(ii) g −1 ( x ) is given by x = 4 f (−1 − x 2 ) + 14 = 0
5y + 2
x (5 y + 2) = 4 3(−1 − x 2 ) + 1 + 14 = 0
−3 − 3x 2 + 15 = 0
5y + 2 = 4
x
3x 2 = 12
5y = 4 − 2
x x2 = 4
y= 4 −2 x = ±2
5x 5
But since x  −1, x = −2
y = 4 − 2x
5x
(iii) gf ( x )  −50
Range of g is 0 < y  2 so
g(3x + 1)  −50
domain of g-1 is 0 < x  2
−1 − (3x + 1) 2  −50

1− 5y 49  (3x + 1) 2
2 (i) f(x) is given by x =
2y
2 xy = 1 − 5 y (3x + 1) 2  49
2 xy + 5 y = 1 3x + 1  7 or 3x + 1  −7
y (2 x + 5) = 1 x  2 or x  − 8
3
y= 1
2x + 5 But since x  −1, x  − 8
3
f (x ) = 1 2
4 (i) gf ( x ) = g(2 x + 3)
2x + 5
= 3(2 x 2 + 3) + 2
Range of f−1 is y  − 9
4 = 6 x 2 + 11
Domain of f is x  − 9
4 fg( x ) = f (3x + 2)
−1
(ii) f g( x ) = f (x)
−1 1

= 2(3x + 2) 2 + 3
= 2(9 x 2 + 12 x + 4) + 3
1 − 5( 1 ) = 18 x 2 + 24 x + 11
x
=
2( )
1 (ii) (fg)−1(x) is
x
x = 2(3 y + 2) 2 + 3
1− 5
x x − x3 = 2
=
2 2(3 yy +
= 2(3 + 2)
2) 2 + 3
x ± xx −−33 = 32(3 y +y2+ 2) 2
( )( )
= 1− 5 x
x 2
(fg
Since
2
± ) −x1x(−x
3 = 31 y +x2− 3 2
2 ) =0,3only2use−positive root:
3
= x−5 −1 1 x −3 − 2
( fg ) ( x ) = 3
2 2
2 3
a = 1,b = −5
2 2

35 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 5 Functions and transformations

Domain of (fg)−1(x) is range of fg(x) 6 (i) f (2) = 6 + a

Since x  0, range is y 11 ff (2) = f (6 + a) = 3(6 + a) + a = 18 + 4a
y
14
18 + 4a = 10 ⇒ a = −2
12 g −1 (2) = 3 ⇒ g(3) = 2
10
b − 2 × 3 = 2 ⇒ b = 8
8
(ii) f ( x ) = 3x − 2 and g( x ) = 8 − 2 x
6
4 fg( x ) = f (8 − 2 x ) = 3(8 − 2 x ) − 2
2 = 24 − 6 x − 2
= 22 − 6 x
–1.5 –1 –0.5 0 0.5 x

Stretch and challenge (Page 52)
(iii) gf(2x) = fg(x)
g(1) = f (0) = 1
1  
6(2 x ) 2 + 11 = 18 x 2 + 24 x + 11
g(2) = fg(1) = f (1) = 3
24 x 2 + 11 = 18 x 2 + 24 x + 11 g(3) = fg(2) = f (3) = 7
2
6 x − 24 x = 0 g(4) = fg(3) = f (7) = 15
6 x ( x − 4) = 0 A pattern can be seen here: every number is 1 less
x = 0 or x = 4 than a power of 2 so g(n) = 2 n − 1.
5 (i) Range of f is 0  f  4 h(0) = g(2) = 3
(ii) f(x) h(1) = gh(0) = g(3) = 7
6
h(2) = gh(1) = g(7) = 2 7 − 1 = 127
4
2 f ( x + 1) + f ( x − 1) = af ( x )

2 3 x +1 + 3 x −1 = a3 x
x +1 x −1
 a = 3 + 3 x −1 = 3 (3 + 3 ) = 3 + 3 −1 = 10
3 x
3 x 3
0 2 4 6 x

3 f (2n − 1) = a(2n − 1) + b = 2an − a + b

(iii) For the first function, the inverse is f (2n) − 1 = a(2n) + b − 1 = 2an + b − 1

x = 1 y2 2f (n) − 1 = 2(an + b) − 1 = 2an + 2b − 1
2
2x = y 2 2an + b is common to all three expressions
so the three expressions can be simplified to
y = 2 x for 0  x  2
−a, −1 and b − 1
For the second function, the inverse is
Since −1 is one of the terms, the possible consecutive
x = 1 y +1 terms are (i) − 3, − 2, − 1 (ii) − 2, − 1, 0 or (iii) − 1, 0, 1
2
2 x = y + 2 (i) a = 3, b = −1 ⇒ f (n) = 3n − 1
y = 2 x − 2 for 2 < x  4 a = 2, b = −2 ⇒ f (n) = 2n − 2
 2 x for 0  x  2 (ii) a = 2, b = 1 ⇒ f (n) = 2n + 1
f −1 ( x ) =  (iii) a = −1, b = 1 ⇒ f (n) = −n + 1
2 x − 2 for 2 < x  4

36 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 6 Differentiation

6 Differentiation 5 f ( x ) = x 3 + x 2 + 2 x − 1 ⇒ f ′( x ) = 3x 2 + 2 x + 2

3x 2 + 2 x + 2 = 3
6.1 Basic differentiation (Page 53)

dy 3x 2 + 2 x − 1 = 0
1 (i) y = 10 x 2 − 3x + 1 ⇒ = 20 x − 3
dx
(3x − 1)( x + 1) = 0
dy
(ii) y = 4 + x − 5 x 4 ⇒ = 1 − 20 x 3
dx x = 1 or − 1
3

() () ()
dy
(iii) y = 52 = 5 x −2 ⇒ = −10 x −3 = − 103
3 2

x dx x y = 1 + 1 + 2 × 1 −1 = − 5
3 3 3 27
3 dy 6 x 2
(iv) y = 2 x ⇒ = = 2x 2 y = (−1)3 + (−1) 2 + 2 × (−1) − 1 = −3
3 dx 3
4 x
(v) y = − = 4 x − x
x 4
−1 1
4

1
( 5
)
Points are 3 , − 27 and (−1, − 3)

dy f(x)
⇒ = −4 x −2 − 1 = − 42 − 1 7
dx 4 x 4 6
−3
x dy −3x −4 5
(vi) y = 1 3 = ⇒ = = − 34 4
4x 4 dx 4 4x 3
1
2
(vii) y = 2 x − 3x = 2 x 2 − 3x 1
dy −1
⇒ = x 2 −3= 1 −3 –2 –1 0
–1 1 2 3 x
dx x –2
−1 1 –3
(viii) y = 3 x + 2 = x 3 + 2 x –4
3x 3 –5
dy 1 − 23 −2 x −2 –6
⇒ = x + = 31 2 − 2 2 –7
dx 3 3 3 x 3x –8

2 (i) f ( x ) = 2 x + 1 2x 1
2 = 2 + 2 = 2x
−1
+ x −2 6 (i) Chord AE BE CE DE
x x x
Gradient of chord 6 7 7.4 7.9
f ′( x ) = −2 x −2 − 2 x −3 = − 22 − 23
x x (ii) f ′(2) = 8
5
3 5
(ii) f ( x ) = 6 x = 6x 3 6.2 Tangents and normals (Page 55)
2
f ′( x ) = 6 × 5 x 3 = 10 3 x 2 1
dy
= 8x 3 + 2x
3 dx
1 dy
3 f ( x ) = 6 + 5 x = 6 x 2 + 5 x
−
At x = −1, = 8 × (−1)3 + 2 × (−1) = −10
x dx
−3 Equation of the tangent:
f ′( x ) = −3x 2 +5 = − 3 +5
x y = mx + c ⇒ 2 = −10 × −1 + c ⇒ c = −8

f ′(9)= − 3 + 5 = 4 y = −10 x − 8
9
1
2 g( x ) = 4 − 3 = 4 x 2 − 3
−
dy
4 y = 3 − 8 x − x 2 ⇒ = −8 − 2 x x
dx
−8 − 2 x = 2 g(4) = 4 − 3 = −1
4
−2 x = 10
−3
x = −5 g ′( x ) = −2 x=− 23
2
x
y = 3 − 8 × −5 − (−5) 2 = 18
Point is ( −5, 18 ) g ′(4) = − 2 3 = − 1
4 4
So the gradient of the normal is 4.
Equation is
y = mx + c ⇒ −1 = 4 × 4 + c ⇒ c = −17

y = 4 x − 17

37 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 6 Differentiation

g(x)
9 4 y = x + k = x + kx −1
x
8
7 dy
6 = 1 − kx −2 = 1 − k2
dx x
5
4 Gradient of the tangent at x = −1 is
3
2 1− k 2 = 1− k
1 (−1)
–4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 11 12 x Gradient of the normal is − 1
–1 1− k
–2
–3 − 1 1
= ⇒ −4 = 1 − k ⇒ k = 5
–4 1− k 4
–5
–6 6.3 Maximum and minimum points and
increasing and decreasing functions (Page 56)

3 (i) h( x ) = 6 x − x 2
1 (i) y = 2 x 3 − 3x 2
h ′( x ) = 6 − 2 x
dy
h ′(2) = 6 − 2 × 2 = 2 = 6x 2 − 6x
dx
y = mx + c ⇒ 8 = 2 × 2 + c ⇒ c = 4 Stationary points when

y = 2x + 4 ⇒ 2x − y + 4 = 0
6x 2 − 6x = 0
(ii) Gradient of normal is − 1 6 x ( x − 1) = 0
2
Equation of normal is x = 0 or 1
y = 0 or − 1
y = mx + c ⇒ 8 = − 1 × 2 + c ⇒ c = 9
2 (0, 0) or (1, − 1)

1
y = − x +9 y
2
− 1 x + 9 = 6x − x 2
2
1
− x + 18 = 12 x − 2 x 2
2 x 2 − 13x + 18 = 0

(2 x − 9)( x − 2) = 0 –2 –1 0 1 2 3 x

x = 9 or 2
2
–1
So the y-coordinate of Q is
−1 × 9 +9= 63
2 2 4

( )
–2
1
Q is 4 , 6 3
2 4
d2y
= 12 x − 6
y dx 2
12 d2y
11 When x = 0, < 0 so ( 0, 0 ) is a local maximum
10 dx 2
9 P d2y
8 When x = 1, > 0 so (1, − 1) is a local minimum
7 Q dx 2
6
5 (ii) f ( x ) = 4 x + 1 = 4 x + x −1
4
x
3
2 f ′( x ) = 4 − x −2 = 4 − 12
1
x
For stationary points,
–5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 x
–1
–2 4 − 12 = 0 ⇒ 4 = 12 ⇒ x 2 = 1
x x 4
–3
x = 1 or − 1
2 2

38 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 6 Differentiation

2 ()
f 1 = 4×1 + 1 = 4
2 1
d2 y
dx 2
= −
4
3

4
5
1 x −2 + 3 x −2 = − 1 + 3
4 x3 4 x5
2

( )
f − = 4 × − + 1 = −4
1
2
1
2 −1 When x = 1,
d2 y 1
= > 0 so x = 1 is a local min.
dx 2 2
2
1
( ) (
Points are , 4 and − 1 , − 4
2 2 ) 2 g( x ) = 3x 2 − 2 x 3

y g ′( x ) = 6 x − 6 x 2 = 6 x (1 − x )

30 Increasing when 6 x (1 − x ) > 0

25
Solving the quadratic inequality,
20
15 0<x<1
10
y
5
2
–2 –1 0 1 2 3 x
–5
–10
–15 1
–20
–25
–30
–1 0 1 2 x
f ′′( x ) = 2 x = 23
−3

x
f ′′ 1
2() ( ) 1
> 0 ⇒ , 4 is a local minimum
2
–1

( ) ( )
f ′′ − 1 < 0 ⇒ − 1 , − 4 is a local maximum
2 2
dy
1 1
1 = x 2 + x − 2 (note that x ≠ 0) 3 y = − x + x 2 − x 3 ⇒ = −1 + 2 x − 3x 2
(iii) y = x + dx
x
dy 1 − 12 1 − 32 Decreasing when −1 + 2 x − 3x 2 < 0
= x − x = 1 − 1 3
dx 2 2 2 x 2 x For this quadratic,
1 − 1 3 = 0 ⇒ 1 = 1 3 b 2 − 4ac = 2 2 − 4 × −3 × −1 = −8
2 x 2 x 2 x 2 x
Since the discriminant is < 0, there are no solutions
⇒ 2 x3 =⇒ 2 2x x 3 = 2 x when −1 + 2 x − 3x 2 = 0 so −1 + 2 x − 3x 2 is always < 0,

⇒ x3 − ⇒ x = 0x 3 − x = 0 as shown in the diagram.
x = 1 (x ≠ x0)= 1 (x ≠ 0) y
2
y = 1 + 1 y == 2 1 + 1 = 2
1 1
Point is (1, Point
2 ) is (1, 2 )
1
y
7

6 x
–1 0 1 2
5

4 –1

3

2

–3 –2 –1 0 1 2 3 4 5 6 7 8 x

39 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 6 Differentiation

4 (i) y = 2 x 3 − 3x 2 1
(iii) y = 3 + 4 x = (3 + 4 x ) 2
y
dy 1 1
⇒ = ( 3 + 4 x )− 2 × 4
5 dx 2
4 2
=
3 3 + 4x
2 2 = 2( x + 3)−1
(iv) y =
1 x +3
dy
–1 0 1 2 x ⇒ = −2( x + 3)−2 = − 2 2

–1 dx ( x + 3)
–2 4
2 = 4 (1 − 2 x )
−2
(v) y =
–3 (1 − 2x )
–4 dy
–5 ⇒ = −8(1 − 2 x )−3 × −2 = 16 3

dx (1 − 2x )
(ii) g( x ) = 3x 2 − 2 x 3
( 6)
4
2 f ( x ) = 3 1 − x
y

f (12) = 3(1 − 12 ) = 3
4

4
6
f ′( x ) = 12(1 − x ) × − 1 = −2(1 − x )
3 3

6 6 6
2
f ′(12) = −2(1 − 12 ) = 2
3

6
–2 –1 0 1 2 x
Equation of the tangent is
–2 y = mx + c ⇒ 3 = 2 × 12 + c ⇒ c = −21

y = 2 x − 21
–4
3 y =
8 = 8( x 2 − 4 x ) −1
5 (i) Stationary point at x = 1 means that f ′(1) = 0 x 2 − 4x
3 × 12 + k × 1 − 8 = 0 dy
= −8( x 2 − 4 x ) −2 × ( 2 x − 4 )
dx
3+ k − 8 = 0
−8 ( 2 x − 4 )
k −5 = 0 = 2
( x − 4 x )2
k=5 −8 ( 2 x − 4 )
(ii) Stationary points when 3x 2 + 5 x − 8 = 0 Stationary point when 2 =0
( x − 4 x )2
(3x + 8)( x − 1) = 0    (3x + 8)( x − 1) = 0
−8( 2 x − 4 ) = 0
x = − 8 or 1 x = − 8 or 1
3 3 x=2
The other stationary point is when x = − 8 = −2 2 y= 8 = −2
3 3 (2) 2 − 4 × 2
(iii) f ′( x ) = 3x 2 + 5 x − 8 ⇒ f ′′( x ) = 6 x + 5
Point is ( 2, − 2 )
f ′′(1) = 6 × 1 + 5 = 11 > 0
y
⇒ x = 1 is a minimum

( ) ( )
f ′′ − 8 = 6 × − 8 + 5 = −11 < 0
3 3
10

⇒ x = − 8 is a maximum 5
3

6.4 The chain rule (Page 58)

–4 –2 0 2 4 6 8 10 x
7 dy
1 (i) y = ( x − 4) ⇒ = 7( x − 4)6
dx
–5
dy
(ii) y = (5 − 2 x )9 ⇒ = 9(5 − 2 x )8 × −2
dx
= −18(5 − 2 x )8 –10

40 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 6 Differentiation

4 g ′( x ) = 9( x + 2 )
2
So maximum point A is ( 0, − 2 ), minimum point
B is ( 2, 6 ).
Since g ′( x ) > 0 for all values of x, the curve is always
increasing so there are no stationary points. Hence y
the function is one–to–one so it has an inverse. 9
8
7
The graphs of y = g( x ) for x > −2 and its inverse are
6
shown below. 5 B
4
y 3
3 2
1
2
–2 –1 0 1 2 3 x
1 –1
–2 A
–3
–6 –5 –4 –3 –2 –1 0 1 2 3 x
–4
–1 –5
–6
–2

–3
6.5 Applications (Page 59)
–4
1 Let the two numbers be x and y.
–5
x + y = 18 ⇒ y = 18 – x
P = xy = x (18 − x ) = 18 x − x 2
dy
5 (i) = −2( x − 1)−2 + 2 = − 2 2 + 2 dP = 18 − 2 x
dx ( x − 1) dx
d2y Maximum when dP = 0
= 4( x − 1)−3 = 4 3 dx
dx 2 ( x − 1) 18 − 2 x = 0
dy x =9
(ii) Stationary points when
=0
dx y = 18 − 9 = 9
− 2 2 +2=0⇒ 2= 2 2 Maximum value of P is 9 × 9 = 81
( x − 1) ( x − 1) 36 − 6 x = 9 − 1.5 x
2 6 x + 4 y = 36 ⇒ y =
2
2 ( x − 1) = 2 4
A = xy = x (9 − 1.5 x ) = 9 x − 1.5 x 2
( x − 1) 2 = 1
dA = 9 − 3x
x − 1 = ±1 dx
x = 0 or 2
Maximum when dA = 0
2 + 2 × 0 = −2 dx
When x = 0, y =
0 −1 9 − 3x = 0 ⇒ x = 3 m

When x = 2, y = 2 + 2 × 2 = 6 y = 9 − 1.5 × 3 = 4.5 m
2 −1
Dimensions are 3 m by 4.5 m
When x = 0,
2
3 (i) V = πr 2h = 5000 ⇒ h = 5000
d y = 4 < 0 ⇒ x = 0 is a maximum 2
πr
dx 2 ( 0 − 1) 3
S = 2πr 2 + 2πrh
When x = 2,
 
= 2πr 2 + 2πr  5000
d2y
= 4 > 0 ⇒ x = 2 is a minimum  πr 2 
dx 2 ( 2 − 1)3 10 000
= 2πr 2 +
r

41 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 6 Differentiation

10 000 The cost of the open box is

(ii) S = 2πr 2 + = 2πr 2 + 10 000r −1
r C = 12 × 4 x 2 + 5( xy + xy + 4 xy + 4 xy )
dS = 4πr − 10 000r −2 = 4πr − 10 000 = 48 x 2 + 5(10 xy )
dr r2
= 48 x 2 + 50 xy
Stationary value when dS = 0
dr  
= 48 x 2 + 50 x  502 
 4x 
4πr − 10 000 = 0
r2 = 48 x 2 + 625
10 000 x
4πr =
r2 = 48 x + 625 x −1
2

3
4πr = 10 000 dC = 96 x − 625 x −2 = 96 x − 625

10 000 dx x2
r=3 = 9.27 cm
4π Minimum when dC = 0
(iii)
d 2 S = 4π + 20 000r −3 = 4π + 20 000 dx
dr 2 r3
96 x − 625 =0
2 20 000 x2
When r = 9.27, d S2 = 4π + >0
dr 9.27 3 96 x 3 − 625 = 0
So r = 9.27 is a minimum value 96 x 3 = 625
4 Let the length of the square end be x. x 3 = 625
96
Let the length be y.  
4x + y = 120 so y = 120 – 4x x = 3 625 = 1.87 m (3 s.f.)
96
V = x2y 2
When x = 1.87, d C2 = 96 + 1250 >0
= x 2 (120 − 4 x ) dx x3

= 120 x 2 − 4 x 3 So x = 1.87 is a minimum.
Dimensions are
dV = 240 x − 12 x 2
dx length = 1.87 m, width = 7.47 m, height = 3.56 m
Maximum volume when dV = 0
dx
6.6 Rates of change (Page 61)
240 x − 12 x 2 = 0
1 Let the length of the side of the square be x.
12 x (20 − x ) = 0
Let the area of the square be A.
x = 0 cm or 20 cm
Since x ≠ 0, x = 20 cm A = x 2 ⇒ dA = 2 x ⇒ dx = 1
dx dA 2 x
y = 120 − 4 × 20 = 40 cm
dx = dA × dx
Volume is 16 000 cm3 dt dt dA

5 = 8× 1
2x
y
= 8× 1
2 × 20
= 0.2 cm/s
2 V = πr 2h = π × 1.2 2 × h = 1.44πh
4x dV = 1.44π ⇒ dh = 1

dh dV 1.44π
x dh = dV × dh
dt dt dV
V = 4x 2 y
= 0.5 × 1
1.44π
4 x 2 y = 50 ⇒ y = 502 = 0.111 m/min (3 s.f.)
4x
= 11.1 cm/min (3 s.f.)

42 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 1 6 Differentiation
= 4000 ×
45h + 3600
dy = 4000 × 1
3 = 10(0.2 x − 1)9 × 0.2 = 2(0.2 x − 1)9 45 × 10.4... + 3600
dx
= 0.983 cm/min
dy dx dy
= ×
dt dt dx
Further practice (Page 62)
= 0.08 × 2(0.2 x − 1)9
dy
= 0.08 × 2(0.2 × 10 − 1)9 1 You want to find the points where = −1.
dx
= 0.16 units/s dy
= x2 + x −3
dx
4 x = 200θ ⇒ dx = 200 ⇒ dθ = 1  so x 2 + x − 3 = −1
dθ dx 200
dθ = dx × dθ x2 + x −2= 0

dt dt dx ( x − 1)( x + 2) = 0
= 10 × 1 x = 1 or x = −2
200
= 0.05 rad/s Substitute into the original equation to find the
y values:
5 (i) Using similar triangles:
y = 1 × 13 + 1 × 12 − 3 × 1 = − 13
3 2 6
y = 1 × (−2)3 + 1 × (−2) 2 − 3 × (−2) = 16
3 2 3
60 cm

120 cm
(
The points are 1, − 13 and −2,
6 ) ( 16
3 )
80 cm
xx 2 (i) y = 12 + x = x −2 + x
h
x
30 cm dy
y = −2 x −3 + 1 = − 23 + 1
dx x
dy
When x = 1, = − 23 + 1 = −1
y + 80 y dx 1
= ⇒ y = 80 cm
60 30 Equation of tangent is
h + 80 = 80 ⇒ 3h + 240 = 8 x ⇒ x = 3 h + 30
y = mx + c ⇒ 2 = −1 × 1 + c ⇒ c = 3
x 30 8
y = −x + 3
Area of trapezium cross section
x-intercept is 0 = -x + 3 ⇒ x = 3

=
( 8 )
3 h + 30 + 30
×h=
3 h 2 + 60h
8 = 3 h 2 + 30h
So Q is (3, 0)
2 2 16 y-intercept is y = 0 + 3 ⇒ y = 3
Volume = 120 × 3 h 2 + 30h
16 ( )

So R is (0, 3)
RQ = 3 2 + 3 2 = 18 = 4.24 (3 s.f.)
45
= h + 3600h cm3
2
2 (ii) m(tangent) = −1 so m(normal) = 1
45 h 2 + 3600h = 40 000 Equation of normal:
(ii)
2 2 = 1 × 1 + c ⇒ c = 1 so equation is y = x + 1
45h 2 + 7200h − 80 000 = 0 1 + x = x +1
Using the quadratic formula, x2
h = 10.4 cm or − 170.4 cm 1 =1
Since h cannot be negative, h = 10.4 cm. x2
(iii) 4 litres/min = 4000 cm3/min
x2 =1
x = ±1
V = 45 h 2 + 3600h ⇒ dV = 45h + 3600
2 dh x = 1 ⇒ y = 2, x = −1 ⇒ y = 0

⇒ dh = 1 So S is (−1, 0)
dV 45h + 3600
dh = dV × dh
dt dt dV
1
= 4000 ×
45h + 3600
= 4000 × 1
43 Cambridge International AS & A Level 45 × 10.4...
Mathematics + 3600
– Pure Mathematics 1 Question & Workbook © Greg Port 2018
= 0.983 cm/min
 6 Differentiation

y y
4 4
2
3
0
2 –3 –2 –1 –2 1 2 3 x
–4
1 –6
–8
–10
x –12
–4 –3 –2 –1 0 1 2 3 4 –14
–1
–2
5 f ′( x ) = 12 x − 3x 2 so the function is increasing when
–3
–4 12 x − 3x 2 > 0
3x (4 − x ) > 0
3 f ( x ) = x + 4 x −1 so f ′( x ) = 1 − 4 x −2 = 1 − 42
x From the graph, the solutions to 3x (4 − x ) > 0 are
4
f ′(1) = 1 − 2 = −3 so the gradient of the tangent is −3 the values of x where the graph is above the x-axis.
1 y

4
f (1) = 1 + = 5 so the point is (1, 5) 15
1
y = mx + c ⇒ 5 = −3 × 1 + c ⇒ c = 8 10
The equation of the tangent is
5
y = −3x + 8 or 3x + y − 8 = 0
The gradient of the normal is 1 –2 0 2 4 x
3
y = mx + c ⇒ 5 = × 1 + c ⇒ c = 14
1
–5
3 3
The equation of the normal is –10
y = 1 x + 14 or x − 3 y + 14 = 0
3 3 –15

dy
4 The stationary points are when =0 So the function is increasing for 0 < x < 4
dx
The graph above is the graph of the derivative.
dy
= 4 x 3 − 16 x so solve 4 x 3 − 16 x = 0 The graph of f ( x ) is
dx
y
4 x ( x 2 − 4) = 0 ⇒ 4 x = 0 or x 2 − 4 = 0
10
x = 0 or 2 or − 2
9
Finding the y-values, 8

4
y = 0 −8×0 +2= 2 2 7
6
y = 2 4 − 8 × 2 2 + 2 = −14 5

4 2
y = (−2) − 8 × (−2) + 2 = −14 4
3
The stationary points are (0, 2), (2, −14) and (−2, −14)
2
d2y 1
= 12 x 2 − 16
dx 2
d2y –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 x
At x = 0, 2 = −16 so (0, 2) is a maximum
dx
d2y dy
At x = 2, = 12 × (2) 2 − 16 = 32 6 (i) y = (3x + 2)8 ⇒ = 8(3x + 2)7 × 3
dx 2 dx
  = 24(3x + 2)7
so (2, −14) is a minimum
1
d2y (ii) y = 3 1 − 9 x = (1 − 9 x ) 3
At x = −2, 2 = 12 × (−2) 2 − 16 = 32
dx dy 1 2
⇒ = (1 − 9 x ) − 3 × −9
so (−2, −14) is a minimum dx 3
  =− 3
3 (1 − 9 x ) 2

44 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 6 Differentiation

Maximum area is when dA = 0

1
(iii) y = 5 = 5 (5 x + 3) − 2
5x + 3 dr
dy 3 400 − 4πr = 0
⇒ = − 5 (5 x + 3) − 2 × 5 = − 25
400 = 4πr
dx 2 2 (5 x + 3) 3
r = 400 = 31.8 m (3 s.f.)
7 (i) y = k + 1 + x = ( k + 1)( 2 x + 3) + x
−1
4π
2x + 3
dy b = 200 − π × 31.8... = 100 m
= − ( k + 1)( 2 x + 3) −2 × 2 + 1
dx 9 The y-intercepts of both lines are 1, so the equation
−2 ( k + 1) of the line that P lies on is y = −x + 1
= +1
( 2 x + 3) 2 The coordinates of P are ( x , 1 − x )
dy The area of the rectangle, A, is
When x = −1, =2
dx
A = 2 x (1 − x ) = 2 x − 2 x 2
−2( k + 1)
+1 = 2 dA = 2 − 4 x
( 2 × −1 + 3) 2 dx
  −2(k + 1) + 1 = 2 2 − 4 x = 0 ⇒ x = 0.5
  −2k − 2 = 1 y = 1 − 0.5 = 0.5
  −2k = 3
A = 2 × 0.5 × (1 − 0.5) = 0.5 units 2
k=−3 The largest area is 0.5 units 2
2
(ii) When k = 1, 10 Let S be the surface area of the cube, and L the length
dy −2(1 + 1) −4 + 1 of its side.
= +1 =
dx ( 2 x + 3) 2 ( 2 x + 3) 2 The equation that relates S and L is S = 6L2 so
−4 + 1 > 0 dS = 12 L
Increasing when dL
( 2 x + 3) 2
−4 > −1 dS = dL × dS
   dt dt dL
( 2 x + 3) 2 = 5 × 12 L
  −4 > − ( 2 x + 3) 2
= 5 × 12 × 20
   4 < ( 2 x + 3) 2 = 1200 mm 2/min
x 2 > 4 ⇒ x < −2 or x > 2 1
11 y = 1 + 6 x = (1 + 6 x ) 2
so 2 x + 3 < −2 or 2x + 3 > 2
dy 1 −1
x < − 5 or x > − 1 = (1 + 6 x ) 2 × 6 = 3
2 2 dx 2 1 + 6x
y
4 The rate you are given is dx = 0.02units/s
dt
2 dy
The rate you want is .
dt
dy dx dy
–4 –2 0 2 x Using the chain rule, = ×
dt dt dx
–2 dy
= 0.02 × 3
dt 1 + 6x
–4
= 0.02 × 3
–6
1+ 6 × 4

= 0.02 × 3
8 (i) 2b + 2πr = 400 25
   2b = 400 − 2πr = 0.012 units/s
b = 200 − πr
(ii) A = 2rb = 2r (200 − πr ) = 400r − 2πr 2

dA = 400 − 4πr
dr

45 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 6 Differentiation

2 3 ( )
12 (i) Volume = πr 2h + 1 4 πr 3 = πr 2h + 2 πr 3
3
dy
dx
= −12(2 − x )−2 × −1

πr 2h + 2 πr 3 = 100 = 12 2
3 (2 − x )
πr 2h = 100 − 2 πr 3 dy
3
When x = 6, = 12 = 3
dx (2 − 6) 2 4
100 − 2 πr 3 2 πr 3
h= 3 = 100 − 3 Equation of the tangent is
πr 2 πr 2 πr 2
y = mx + c
= 1002 − 2r 0 = 3 ×6+c ⇒ c = −9
πr 3 4 2
3 9
(
(ii) S = 2πrh + πr 2 + 1 4πr 2
2 ) y = x − or 3x − 4 y − 18 = 0
4 2

S = 2πr 1002 − 2r  + πr 2 + 2πr 2

 (ii) dx = 0.04
 πr 3 dt
2 dy dx dy
= 200 − 4πr + 3πr 2 = ×
dt dt dx
r 3

( )
12 = 3 units/s (0.12 units/s)
= 0.04 ×
= 200 + 3π − 4π r 2 (2 − 4) 2 25
r 3
= 200 5π
+ r 2 2 (i) y = 1 − 9
r 3 x −1 x − 5

(iii) S = 200r −1 + 5π r 2 y = ( x − 1)−1 − 9( x − 5)−1

3
dy
dS = −200r −2 + 10π r = −( x − 1)−2 + 9( x − 5)−2
dx
dr 3
=− 1 2 + 9 2
= − 200 + 10π r ( x − 1) ( x − 5)
r2 3 dy
When x = 3, = − 1 2 + 9 2 = 2
Stationary point is when dx (3 − 1) (3 − 5)
− 200 10π
2 + 3 r =0
Gradient of normal = − 1
r 2
Equation of normal:
−200 + 10π r 3 = 0
3 5 = − 1 × 3 + c ⇒ c = 13
2 2
r 3 = 200 = 19.098... 1 13
10π y =− x+
3 2 2
x -intercept is when y = 0
 r = 2.67cm (3 s.f.)
0 = − 1 x + 13 ⇒ x = 13
2 2 2
(iv) d S2 = 400r −3 + 10π = 400
3 +
10π
dr 3 r 3 y
2 10
When r = 2.67cm, d S2 = 4003 + 10π > 0
dr 2.67 3
8
So r = 2.67 cm is a minimum value.

6
Past exam questions (Page 64)
4
1 (i) y = 3 + 12
2− x
2
Curve crosses x-axis when
3 + 12 = 0 0
2− x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 x

3(2 − x ) + 12 = 0 –2
6 − 3 x + 12 = 0
x=6 –4

y = 3 + 12(2 − x ) −1

46 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 6 Differentiation

(ii)
dy
=− 1 2 + 9 2 =0 (ii) dS = 12 x − 768 = 0
dx ( x − 1) ( x − 5) dx x2
9 12 x 3 − 768 = 0
= 1
( x − 5) 2 ( x − 1) 2
x 3 = 768 = 64
9( x − 1) 2 = ( x − 5) 2 12
x = 4 cm
9 x 2 − 18 x + 9 = x 2 − 10x + 25
d 2 S = 12 + 1536
2
8 x − 8 x − 16 = 0 dx 2 x3
2
When x = 4, d S2 = 12 + 1536
2
x −x−2=0 >0
dx 43
( x + 1)( x − 2) = 0
So x = 4 is a minimum
x = −1 or 2
−1
2
d y 5 (i) y = 12(3 − 2 x )
= −2( x − 1) −3 − 18( x − 5) −3
dx 2 dy
= −12(3 − 2 x ) −2 × −2
2 − 18 dx
=
( x − 1)3 ( x − 5)3 = 24
d2 y (3 − 2 x ) 2
When x = −1, = − 1 < 0 ⇒ maximum
dx 2 6 dx dy
(ii) At A, = 0.15, = 0.4
d2 y 8 dt dt
When x = 2, = > 0 ⇒ minimum
dx 2 3 dy dy dt
= ×
dx dt dx
3 f ( x ) = 2 x + ( x + 1)−2 24 = 0.4 × 1
2 (3 − 2 x ) 2 0.15
f ′ ( x ) = 2 − 2( x + 1) −3 = 2 −
( x + 1)3 24
=8
f ″ ( x ) = 6( x + 1) −4 = 6 (3 − 2 x ) 2 3
( x + 1) 4 72 = 8(3 − 2 x ) 2
f ′ (0) = 2 − 2(0 + 1) −3 = 2 − 2 =0
9 = (3 − 2 x ) 2
(0 + 1)3
⇒ x = 0 is a stationary point ± 3 = 3 − 2x
x = 0 or 3
f ″ (0) = 6(0 + 1) −4 = 6 >0
(0 + 1) 4 6 y + 3 x = 9 ⇒ y = 9 − 3 x
⇒ x = 0 is a minimum u = x 2 y = x 2 ( 9 − 3x ) = 9 x 2 − 3x 3
du = 18 x − 9 x 2
4 (i) Let the height of the cuboid be h cm.
dx
V = 3x 2h = 288 18 x − 9 x 2 = 0
⇒ h = 2882 9 x (2 − x ) = 0
3x x = 0 or 2
S = 2(3x 2 ) + 2 xh + 2(3xh) Since x ≠ 0, x = 2
2
= 6 x + 8 xh
u = 9 × 2 2 − 3 × 2 3 = 12
 
= 6 x 2 + 8 x  2882  d 2u = 18 − 18 x
 3x 
dx 2
= 6 x 2 + 768 2
x When x = 2, d u2 = 18 − 18 × 2 < 0
dx
So u = 12 is a maximum

47 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 6 Differentiation

2 −1
7 yy =
= kk 2 (( xx + 2)−1 +
+ 2) + xx 2 SP = x 2 + 10 2 so cos α = x
dyy = − kk 2 + 1
d 2 x 2 + 10 2
= − 2 +1
dxx
d (( xx +
+ 2)
2) 2 PW = (30 − x ) 2 + 5 2
22
− kk 2 +
− + 11 =
= 00 = (900 − 60 x + x 2 ) + 25
(( xx +
+ 2)
2) 2
2 2
= 925 − 60 x + x 2
−
−kk 2 +
+ (( xx + 2) 2 =
+ 2) = 00
22 cos β = 30 − x
+ 2)
(( xx + 2) = = kk 22
925 − 60 x + x 2
xx +
+ 22 ==± ±kk 2 2 2
xx ==− −22 ±± kk Time = x + 10 + 925 − 60 x + x
10 5
2
d 2 yy = 22kk 2
d 2
x 2 + 10 2 + 2 925 − 60 x + x 2
2 = 3 =
+ 2)
dxx 2 (( xx +
d 2)3 10
2
d 2 yy = 22kk 2 = 22 > 0 ⇒ minimum
d 2
1 2 −1 
When x = −2 + k
When x = −2 + k , , 2 = 3 = k > 0 ⇒ minimum
dxx 2 kk 3 k
d 1  2 ( )
x + 100 2 × 2 x + 
T′ =  
d 22 y 2k 22 2 10 1
=−
When xx =
When − kk ,, d y2 =
−22 −
d x
= 2k 3 =
− k
= −2k <
< 00 ⇒ maximum
⇒ maximum  2 ( )
 2 × 1 925 − 60 x + x 2 − 2 × ( 2 x − 60 ) 

dx 2
−k −k
3

 2( x − 30) 
Stretch and challenge (page 65) = 1  2x + 2
10  x + 100 925 − 60 x + x 
1 (i) g ′( x ) = k − 2 x so if the gradient of the normal is
 30 − x 
1 , the gradient of the tangent is −2 = 1  2x −2 
2 10  x + 100 2
925 − 60 x + x 
k − 2 x = −2 = 1 [ cosα − 2cos β ]
10
k − 2 × 2 = −2

k − 4 = −2 T′ = 0 when 1 [ cosα − 2cos β ] = 0
10
k=2 cosα = 2cos β
(ii) The y-value when x = 2 is g(2) = 2 × 2 − 2 2 = 0
3 (i) V = πr 2h
Equation of the normal is
( ) ⇒r
2 2
2 2 h 2
= R2 − h
y = mx + c ⇒ 0 = 1 × 2 + c ⇒ c = −1 so y = 1 x − 1  R = r + 2 4
2 2
 2

To find the other point, solve simultaneously. V = π R2 − h  h
1 x − 1 = 2x − x 2  4 
2 3
= πR 2 h − πh
x − 2 = 4 x − 2x 2 4
2
2 x − 3x − 2 = 0 dV = πR 2 − 3πh 2
dh 4
(2 x + 1)( x − 2) = 0
dV = 0 when πR 2 − 3πh 2 = 0
x = − 1 or 2 dh 4
2 2
So the other point is x = − 1 πR 2 = 3πh
2 4

( ) ( )
2
2 4πR = h 2
g − 1 = 2 × − 1 − − 1 = −1 1 3π
2 2 2 4
The coordinates of the point P are − 1 , − 1 1
y
2 4 ( )
h = 2R
3

( ) =R −R
3 2
2R
2 2 2 2
2 h 3
1
2
r =R − =R −2 2
= 2R
4 4 3 3
–2 –1 0 1 2 3 4 5 6x r = 2R
–1 3
–2
–3
–4
–5

48 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 7 Integration

( )( ) L = ( 40 − x ) + 2 x 2 + 900
2
3 2 R 2R
(ii) Vcylinder 2 2 3 3
= πr h = 3r h3 = = ( 40 − 17.3...) + 2 (17.3...) 2 + 900
Vsphere 4 πR 3 4 R 4R 3
3   = 91.9615... km

=
 2
3 2R  2R
 3  3 ( ) = 92 km (3 s.f.)
4R 3
4R 3 7 Integration
  = 33 7.1 Reversing differentiation (Page 66)
4R
= 1 2x 4 + x 2 + c = x 4 + x 2 + c
∫
3
3 1 (i) (2 x + x ) d x =
4 2 2 2
The ratio is 1: 3 or 3:3 2
(ii) ∫ (1 + 3x ) dx = x + 3x + c
2 2
4 AE + 30 = 50 ⇒ AE = 40 km 2 2
 −2 
B (iii) ∫ x43 d x = 4 ∫ x −3 d x = 4  x−2  + c = − x22 + c
4
1 9 3 x4
∫6 ∫
3
50 km (iv) 3 x dx = 6 x 3 dx =6x +c = +c
30 km 4 2
3
1

∫( )
1
2 dx = 2 x − 2 dx = 2 x 2 + c = 4 x + c
A
40 − x D x E
2 (i)
x ∫ 1
2
3

30 km
∫3
(ii) 10 x dx = 10 ∫ x 2 dx
50 km 5
= 10 x + c
2
5
C
2
= 4 x5 + c
2 2 2 2
In  BDE, BD = x + 30 ⇒ BD = x + 900
2  2 
Total length of roads, L, is 3 ∫ 5xx 4−1 dx = ∫  5xx4 − x14  dx
L = AD + BD + CD
= ∫ (5 x −2 − x −4 ) dx
= ( 40 − x ) + x 2 + 900 + x 2 + 900
−1 −3
= ( 40 − x ) + 2 x 2 + 900 = 5x − x + c
−1 −3
= − 5 + 13 +c
1
(
= ( 40 − x ) + 2 x 2 + 900 ) 2
x 3x

∫ ( x 4 + 1)d x
−1

dx 2 (
dL = −1 + 2 × 1 x 2 + 900
) 2
× 2x 4
dy 3
=
dx x 4
+1⇒ y = 3

2x
∫ ( 3x + 1) d x
= −1 + = −4
x 2 + 900
−3
Minimum length is when = 3x + x + c
−3
−1 + 2x =0
x 2 + 900 = − 13 + x + c
x
2x =1 Substitute (−1, 2) into the equation:
x 2 + 900
2 x = x 2 + 900 2 = − 1 3 + (−1) + c ⇒ c = 2
(−1)
4 x 2 = x 2 + 900
So y = − 13 + x + 2
  3x 2 = 900 x

x 2 = 300
x = 300 ≈ 17.3 km

49 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 7 Integration

1 4
5 f′ ( x ) = 3 x − 2 x = 3x 2 − 2 x  3 4
2 x3 
= 2x  = 2
2

 3  
 3 0
∫ ( 3x 2 − 2 x ) d x
1
f (x ) =  2 0
3
 3  2 03  
= 2  2 4  − 
2
= 3x − 2 x + c
2

3 2 
3   3  
2
= 2 x3 − x 2 +c = 2 × 16 = 10 2
3 3
f (4) = 30 ⇒ 30 = 2 4 3 − 4 2 + c ⇒ c = 30 Area between the curve and the y-axis
So f ( x ) = 2 x 3 − x 2 + 30 = 4 × 4 − 10 2 = 5 1
3 3

∫( ) ∫−1 (3x )
2
6 dy = 1 − k ⇒ y = 3 A = 2
1 − k2 dx + 1 dx
dx 2
x x 2
=  x 3 + x 
∫ (1 − kx )dx
−2
= −1

−1 ( ) (
= 2 + 2 − (−1)3 + (−1)
3
)
= x − kx + c = 10 − (−2) = 12
−1
k
= x + +c 4 Points of intersection:
x
Substitute the two points into the equation: x 2 − 2x −1 = x −1

2 = 1+ k + c ⇒ k + c = 1 x 2 − 3x = 0
1
x ( x − 3) = 0
−6 = −3 + k + c ⇒ − k + c = −3 x = 0 or 3
−3 3

∫0 (( x − 1) − ( x − 2x − 1))dx
3
Subtract the equations: A= 2
4k = 4 ⇒ k = 3 3
= ∫ ( − x 2 + 3x ) dx
3
3 + c = 1 ⇒ c = −2 0

The equation is y = x + 3 − 2
3
 3 2

x =  − x + 3x 
 3 2 0
7.2 Finding areas (Page 67)  3 3 × 32   03 3 × 02 
= −3 + − − 3 + 2 
1 Finding the x-intercepts first,  3 2   
x 2 − 4 x + 3 = 0 ⇒ ( x − 1)( x − 3) = 0 = 4.5

⇒ x = 1 or 3 5 Finding points of intersection,

3 x 2 − 2x − 3 = 9 − x 2
A= ∫1 ( x 2
− 4 x + 3 dx) 2 x 2 − 2 x − 12 = 0
3
 3  x2 − x −6 = 0
=  x − 2 x 2 + 3x 
3 1 ( x + 2 )( x − 3) = 0
 3   3  x = −2 or 3
=  3 − 2 × 3 2 + 3 × 3 −  1 − 2 × 12 + 3 × 1
y
 3  3 
10
= 0 − 11 9
3 8
= −1 1 7
3 6
(−2, 5) 5
So the area is 1 1 4
3
3
2 Area between the curve and the x-axis: 2
1 (3, 0)
4 4 1
A= ∫0 2 x dx = 2 ∫0 x 2 dx
−5 −4 −3 −2 −1 0
−1
1 2 3 4 5 6 x

−2
 3 4 −3
4
 3
= 2 x  = 2 2 x 
2 −4
3 −5
 3 0
 2 0
 2 4 3  International
50 = 2Cambridge  3  AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
  −  2 0 
 3   3 
16 2
 7 Integration

7.3 The reverse chain rule (Page 69)

∫−2 ((9 − x ) − ( x − 2x − 3))dx
3
2 2
A= 4
( x − 2)
∫ (x − 2) d x =
3
3 1 (i) + c
= ∫ ( −2 x 2 + 2 x + 12 ) dx
4
−2

(1 − 6 x )10 1
 3

=  − 2 x + x 2 + 12 x 
3 (ii) ∫ 2(1 − 6 x )9 d x = 2
10
× +c
−6
3
  −2 (1 − 6 x )10
=− +c
30
 3

=  − 2 × 3 + 3 2 + 12 × 3
 3 
x + 3 dx = ( 4 )
4
x +3
 2 × (−2)3 
( )
3
− − + (−2) 2 + 12 × −2 (iii) ∫ × 1 +c
 3  4 4 1
4
3 ( )
= 27 − −14 2 = 41 2
3 ( )
= x +3 +c
4
4

6 Area between curve and the x-axis: 3

1
(2 x − 1) 2
(iv)
∫ (2x − 1) 2 dx = × 1 +c
∫ 1 ( 2x )
2
2 3 2
A= + 1 dx
2
2 3
 3 
=  2x + x  (2 x − 1) 2
 3 1 = +c
3
 3
  3

=  2 × 2 + 2 −  2 × 1 + 1
1

 3   3  2 (i) ∫ ∫
1 − x dx = (1 − x ) 2 dx

= 71 − 12 = 5 2
3 3 3 ( ) =
(1 − x ) 2
3
3

× 1 +c
−1
Area between curve and the y-axis 2
3
2 (1 − x ) 2
= 9 × 2 − 5 2 − 3 = 91 =− +c
3 3 3
k 2 (1 − x ) 3
7 ∫1 3 x dx = 14 =−
3
+c

k 1
∫1 3x 2 dx = 14 (ii) 2 2
∫ 3(x − 4)5 d x = 3 ∫ (x − 4)
−5
dx
k
 3
( x − 4) −4 1
 3x 2  = 14 =2 × +c
 3  3 −4 1
 2 1 1
=− +c
2 x 3 k = 14 6( x − 4) 4
 1
3 1
3
2 k − 2 1 = 14 3 (iii) ∫ 5 − 2x ∫
dx = 3 (5 − 2 x ) − 2 dx
3 1
2 k − 2 = 14 (5 − 2 x ) 2
=3 × 1 +c
2 k 3 = 16 1 −2
2
k3 = 8 = −3 5 − 2 x + c
k 3 = 64
−1
k=4 (iv) ∫ 12 dx = 12
3 1+ x
∫( 1+ x
2 ) 3
dx
2 2

= 12
( )
1+ x
2
3

× 1 +c
2 1
3 2

( )
2
= 36 3 1 + x +c
2

51 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 7 Integration

( )
7.4 Improper integrals (Page 70) 1 2 1
3
∫−2 ∫−2( 2 − x )
−2
2 V = π dx = 9π dx
1
1 1 2− x
∫ ∫0
−3
1 dx = (2 x + 1) dx 1
0 (2 x + 1)3  ( 2 − x ) −1 1 
1 1 = 9π  × 
 (2 x + 1) −2 1    −1 −1
= ×  = − 1   −2
 −2 2   4(2 x + 1) 
2
1
= 9π  1 
0 0

 1   1   2 − x  −2
= − − −
( )
2  2
 4(2 × 1 + 1)   4(2 × 0 + 1)   
= 9π  1 −  1 
= − 1 − −1 = 2
36 4 9 ( )  2 − 1  2 − (−2) 
= 9π  3 
4 
2 2
2 dx = 2 −1
2 ∫1 x −1 ∫1 ( x − 1) 2 dx 27
= π
4
2
 1 6
( x − 1) 2 
∫0 πx
2
3 V (solid) = V (cylinder) −
 = 2[ 2 x − 1 ]1 dy
2
=2
 1 
 2 1 V(cylinder) = π × 2 2 × 6 = 24π
= 2 ( 2 2 − 1 ) − ( 2 1 − 1 ) 
y y
y = 2( x 2 − 1) ⇒ = x 2 − 1 ⇒ x 2 = + 1
2 2
=4
6 6 y 
∫0 πx dy = π ∫0  2 + 1 dy
2
∞ ∞
3 dx = 3
∫0 ∫0 (1 + x ) dx
−2
3
(1 + x ) 2 6
 y2 
−1 ∞ ∞
= π + y
 (1 + x )   1  4
= 3  = 3 −  0
 −1  0  1 + x  0
 2   2 
 1 + ∞ (
= 3 − 1 − − 1 
1 + 0  )( )
= π   6 + 6 −  0 + 0 
  4   4 
=3 = 15π
V(solid) = 24π − 15π = 9π = 28.3 (3 s.f.)
7.5 Finding volumes by integration (Page 71)
4 (i) 4 x + 4 = 10
()
∞ 2 ∞ x
1 (i) V = ∫2 π 4 dx = 16π ∫2 x −2 dx
x 4 x 2 + 4 = 10 x
−1 ∞ 4 x 2 − 10 x + 4 = 0
 
= 16π  x 
 −1  2 2x 2 − 5x + 2 = 0
∞
= 16π  − 1  (2 x − 1)( x − 2) = 0
 x 2
x = 1 or x = 2
 ∞ ( )( )
= 16π  − 1 − − 1 
2 
2

= 16π  1 

2 ( )
A is 1 , 10 , B is ( 2, 10 )
2
= 8π = 25.1 (3 s.f.) To find M, find where dy = 0
dx

() y = 4 x + 4 = 4 x + 4 x −1
4 2
(ii) V = ∫3 π 4 dx x
x
dy
4
= 4 − 4 x −2 = 4 − 42
= 16π  − 1  dx x
 x 3
4 − 42 = 0
 4 ( )( )
= 16π  − 1 − − 1 
3 
2
x
4x − 4 = 0
= 16π  1 
12  4x 2 = 4
4
= π = 4.19 (3 s.f.)
3 x2 =1
x = 1 (since x > 0)

52 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 7 Integration

4
(ii) Volume of cylinder = π × 10 2 × 1.5 = 150π
∫ (2x + 1)3 d x = 4 ∫ (2x + 1)
(vi) −3
dx
 Volume under curve
(2 x + 1) −2 1
( ) dx
2
2 =4 × +c
= ∫ π 4x + 4 −2 2
0.5 x
=− 1 +c
2  
+ 32 + 162  d x (2 x + 1) 2
∫0.5 π  16x
2
=
x 
dy
= ∫0.5 (
2
π 16 x 2 + 32 + 16 x −2 d x ) 2 If
dx
= 2 x + 1 then y = ( 2 x + 1)dx ∫
y = x2 + x +c
2
 3

= π 16 x + 32 x − 16 x −1  Substitute (1, −2) into the equation to find c:
 3  0.5
2 −2 = 12 + 1 + c
 3

= π 16 x + 32 x − 16  c = –4
 3 x  0.5
The curve is y = x 2 + x − 4
 16 × 2 3
16  
 3 + 32 × 2 + 2  −  y
= π 
12
 16 × 0.5 3 16 
 + 32 × 0.5 +   10
 3 0.5  
8

 3 ( )( )
= π  344 − 146 
3 
6
4
2
= 66π
−6 −4 −2 0 2 4 x
Final volume = 150π − 66π = 84π ≈ 264 units 3 −2
−4 A(1, −2)
Further practice (Page 73) −6

2 5 3 Points of intersection:
1 (i) (2 x − 5 x 4 )dx = 2 x − 5 x + c = x 2 − x 5 + c
∫ 2 5
2 = 6 − x 2 ⇒ x 2 = 4 ⇒ x = ±2
∫ ( 2 + 2x )dx = 12 × x2 + 2x6
2 6
x 5

∫−2( (6 − x ) − ( 2)) dx = 2∫0 ( 4 − x )dx

(ii) +c 2
2 2
2
A=
2 6
= x + x +c
2
 3

4 3 = 2 4x − x 
 3 0
∫  2x 2 + 6x  dx = ∫ ( 2 x )
 1  1 −2
(iii) + 6 x dx  3  3
= 2  4 × 2 − 2  −  4 × 0 − 0  
−1 2  3   3 
= 1 x + 6x + c
2 −1 2 = 2 × 5 1 = 10 2
1 3 3
= − + 3x 2 + c
2x
4 Finding the x-intercept:
∫ (2x )
3
(iv) ∫ x (2 x − 3 x )d x = 2
− 3x 2 d x
2 − 2 = 0 ⇒ 2 = 2 ⇒ 3 x =1⇒ x =1
5 3 x 3 x
3
= 2x − +c 3x 2
Area between x = 8 and x = 1:
3 5
2 8
 2 − 2 dx = 8 −1
3
= 2x − 6 x + c
5 A= ∫ 1 x
3  ∫1 (2 x 3 − 2)dx
3 5 8
(3x + 1)6 1  2 
∫
5 8
(v) (3x + 1) d x = × +c  2 x 3
6 3 = − 2 x  = 3 3 x 2 − 2 x 
 2  1
(3x + 1)6  3 1
= +c
18
( ) (
= 3 3 8 2 − 2 × 8 − 3 3 12 − 2 × 1 )
= −4 − 1 = −5

53 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 7 Integration

Area between x = 1 and x = 0 16

1
Shaded area = ∫0 ( 4 x − x ) dx
= 3 x − 2 x 
3 2
16  1

∫0
0
=  4 x − x  d x
2

( ) (
= 3 3 12 − 2 × 1 − 3 3 0 2 − 2 × 0 ) 16
=1  3 2

=  4 x 2 x
− 
So total area = 5 + 1 = 6  3 2 
 2 0
5 To find this area, find the area ABGF between the
16
curve and the x-axis between x = 1 and x = 4, find 8 x 3 x 2 
= − 
the area of the rectangle OAFD and subtract these  3 2 
0
areas from the area of the rectangle OBGE.
 8 16 3 16 2   8 0 3 0 2 
y = − − − 
y = 2 x +1  3 2   3 2
E(0, 5)
= 42 2 − 0
G(4, 5)
3
= 42 2
D(0, 3) F(1, 3) 3
1
1 −1
 1
dx =  x  = [ 2 x ]0 = 2 1 − 2 0 = 2
2
∫
1
7 (i) x 2
C(0, 1) 0  1 
 2 0
r
O A(1, 0) B(4, 0) x r  −1  r
(ii) ∫ x −2 dx =  x  =  − 1 
1  −1 1  x 1
∫1
4
Area of ABGF = (2 x + 1) dx
4 1 

= − 1 − −=1−
 r ( ) ( )
1 1
1 r − −1 
= ∫  2x 2
1 
+ 1 d x
 = −1 +1 = −1 +1
r r
4
 32  As r → ∞As 1
, − r→ 0 1
=  2x + x  r → ∞, − r → 0
 3 
∫1 x −2Sod x∫→
∞ ∞
 2 1 So x1−2 d x → 1
1
4
 3 
= 4 x + x  2 2
∫0 π (3x ) dx = π ∫0 9x dx
2 2
3 8 V =
 1
  2
 3   3  = 9π ∫ x 2 dx
=  4 4 + 4  −  4 1 + 1
 3   3  0
2
 3
= 12 1 = 9π  x 
3  3 0

Area of DFGE = area of OBGE − area of ABGF −  3 3

= 9π  2 − 0 
area of OAFD 3 3
= ( 4 × 5 ) − 12 1 − ( 3 × 1) = 9π  8 
3 3
= 42 72
3 = π = 75.4 (3 s.f.)
3
6 First, find the points of intersection of the curves.
V (solid) = 1 π × 6 2 × 2 = 24π
2 2 3
x = 16 x ⇒ x − 16 x = 0 ⇒ x ( x − 16) = 0
⇒ x = 0 or x = 16

54 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 7 Integration

9 Volume = ∫ πy dx
2
Volume of the cone = 1 π × 12 × 1 = 1 π
3 3
= ∫ π( 3 ) dx
2
Final volume = 1 π − 1 π = 1 π
2

0 2x + 1 2 3 6
1
∫0 π ( 1 − x ) − (1 − x ) 2  d x
2
2
9 OR V =
=π ∫
0 (2 x + 1)
2 dx
1


∫0 (1 − x ) − (1 − 2x + x ) d x
2
2 =π
∫0
−2
= 9π (2 x + 1) dx
1
∫0 (x − x
2
 (2 x + 1) −1 1  =π 2
) d x
= 9π  × 
 −1 2 1
0  2 3

2 = πx − x 
= 9π  − 1   2 3 0
 2(2 x + 1)  0
 2 3  
= π  1 − 1  − 0 
= 9π  − 1 − −1   2 3  
 2(2 × 2 + 1) 2(2 × 0 + 1) 
= 1π
6
= 9π  − 1 + 1 
 10 2  (b)  y = 1 − x ⇒ y 2 = 1 − x

= 9π  2  = 5 π units3

18 x = 1 − y 2

5
( )
2
x 2 = 1− y 2
∫ 0 1 − x dx
1
10 (i) Area under curve =
x 2 = 1 − 2y 2 + y 4
= ∫ (1 − x ) 2 dx
1 1

∫ 0 πx 2 dy
1
0
V =
1
 3

= −
(1 − x ) 2

∫ 0 π (1 − 2y 2 + y 4 ) dy
1
 3  =
 2 0
1
 2 1 − x 3 1  2y3 y5 
( )  = πy − + 
= −  3 5 
3 0
 
0
 3 5
 
= π  1 − 2 × 1 + 1  − 0 
2 (1 − 1)  2 (1 − 0) 
3 3
 3 5  
=− − −
3  3 
  = 8π
15



( )
= 0 − − 2  = 2
3 3 Volume of the cone = 1 π × 12 × 1 = 1 π
3 3
Area under line = 1 × 1 × 1 = 1
2 2 Final volume = 8 π − 1 π = 1 π
Shaded area = area under curve − area of triangle 15 3 5

  = 2 − 1 = 1 OR y = 1− x ⇒ x = 1− y ⇒ x 2 = 1− 2y + y 2
3 2 6
∫ 0 (1 − 2 y 2 + y 4 ) − (1 − 2y + y 2 )dy
1
V =π
(ii) (a) Volume of curve rotated around the x-axis:

∫0 ( 2 y − 3 y )
1 1
∫0 πy d x
2 2
V= =π + y 4 dy
1
= π ∫ ( 1 − x ) dx
1 2
 y5 
0 = πy2 − y3 + 
 5 
= π ∫ (1 − x ) dx
1 0
0 = 1π
2 1
5

= π x − x 
 2 0
 2 
= π 1 − 1  − 0 = 1 π
 2   2

55 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 7 Integration

OR Volume of the cone = 1 π × 12 × 1 = 1 π

1
∫0 πy
2
11 (i) V = dx 3 3
3
Volume of y = x rotated around the y-axis:
( )
1
π ( x ) 2 − x 3
2
 dx
= ∫0   1  2

1
V= ∫0 π  y 3  dy
∫0  x − x  dx
2 6
=π
1
 53 
7 1 y
  = π 
3
= πx − x   5 
3 7 0
 3 0
 3 7  
= π  1 − 1  − 0 
1
3 3 y5 
  3 7   = π 
4  5  0
= π
21
 3 3 15  
OR Volume of the cone = 1 π × 12 × 1 = 1 π = π   − 0
3 3  5  
Volume of y = x3 rotated around the x-axis:
= 3π
5
∫ 0 π (x ) dx
1 3 2
V=
Final volume = 3 π − 1 π = 4 π units3
= π ∫ (x 6 )dx
1
5 3 15
0

 x 7 1 Past exam questions (Page 73)

= π 
 7 0
()
4 4 2
 7   1 Volume = π ∫1 (5 − x ) 2dx − π ∫ x4 dx
= π 1  − 0 1
 7   4
 (5 − x )3  4
=1π = π −  − π  − 16 
7  3   x 1
1

Final volume = 1 π − 1 π = 4 π  (5 − 4)3   (5 − 1)3  

3 7 21 = π  − − −
 3   3  
2

( )( )
(ii) If y = x 3 then x = 3 y ⇒ x2 = 3 y2 = y3
− π  − 16 − − 16 
If y = x then x = y ⇒ x 2 = y 2  4 1 
= 9π (or 28.3)
 2
1 
Volume =
∫π  y 3 − y 2  dy
0   8
1
2 Equation: y= ∫ (5 − 2 x ) 2 d x
 53 
y y3 
=π 
5
−
3 ∫
= 8 (5 − 2 x ) −2 d x
 3 0  (5 − 2 x ) −1 1 
= 8 × +c
3 3 y5 y 3 
1 −1 −2
 
= π − 
 5 3  = 4 +c
0 5 − 2x
 3 3 15 13   3 3 0 5 0 3   7= 4 +c⇒c =3
= π  −  − −  5−2×2
 5 3  5 3 
 So y = 4 + 3
5 − 2x

 5 3 ( )
= π  3 − 1 − 0
 3 (i) A is where 1 − 2 x = 0 or ( 2 x − 1) = 0
2

= 4π units 3 (0.838 units3 to 3 s.f.) ⇒x =1

15 2

56 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 7 Integration

1 1
dy
∫0 ∫0 ( 2x − 1) d x = −12 x −2 − 4 x + 11
2 2 2
(ii) Area = 1 − 2x d x − (iii)
dx
1
∫ ( −12x −2 − 4 x + 11)dx
1
 3 2  y= −2
(1 − 2 x ) 2 1 ( 2 x − 1) 3
1  2
= ×  − × 
 3 −2  3 2
 2 0  0 y = 12 − 2 x 2 + 11x + c
x
1 1 13 = 12 − 2 + 11 + c
 (1 − 2 x ) 32  2  ( 2 x − 1)3  2
=  −  −  c = −8
 3   6
 0 0 When x = 2,

 ( )
= 0 − − 1  − 0 − − 1 
3   6  ( )
y = 12 − 2(2) 2 + 11(2) − 8 = 12
2
P is (2, 12)
=1
6
6 (i) Intersection is where
8 = 9− x3
4 (i) Gradient of normal = − 1 so gradient of tangent at x3
3
P is 3. 8 = 9x 3 − x 6
dy 12 x 6 − 9x 3 + 8 = 0
  
So = =3
dx 4×2+a
12 = 3 Let t = x 3. Then the equation becomes
8+a
12 = 3 8 + a t 2 − 9t +t82 =− 09t + 8 = 0

4 = 8+a (t − 1)(t (−t 8−)1=)(0t − 8 ) = 0

16 = 8 + a t = 1 or t8= 1 or 8
33
a=8 x = 1 or x 3x= 1= or8 x3 = 8
12 dx x = 1 orx x= = 1 2or x = 2
(ii) y = ∫ 4x + 8 So a = 1, b =a 2= 1, b = 2
1
∫
= 12 ( 4 x + 8 )− 2 dx
(ii) A =
2
∫1  (9 − x
3 
) −  83   d x
 x 

( 4 x + 8) 2 1 
1

= 12  × +c  4

2
 1 4 = 9 x − x + 42 
 2   4 x 1
= 6 4x + 8 + c  4
  4

=  9 × 2 − 2 + 42  −  9 × 1 − 1 + 42 
14 = 6 4 × 2 + 8 + c ⇒ c = −10  4 2   4 1 

So y = 6 4 x + 8 − 10 = 15 − 12.75 = 2.25

d 2 y 24 dy
(iii) = −3x 2
2 = 3 − 4 = −1 < 0 ⇒ maximum
5 (i) dx
dx 2 dy
y = 83 = 8 x −3 ⇒ = −24 x −4 = − 244
∫( )
dy dx
= 24 x −3 − 4 dx x x
dx
= −12 x −2 − 4 x + c If the tangents are parallel, the gradients are the
same so
= − 122 − 4 x + c
x −3x 2 = − 244
dy x
When x = 2, =0 x6 = 8
dx
0 = − 122 − 4(2) + c x = 6 8 = 2 ≈ 1.41
2
c = 11 So c = 6 8 = 2 ≈ 1.41
dy
So = − 122 − 4 x + 11
dx x

57 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 7 Integration

Stretch and challenge (Page 75) k +1

3 Area = ∫k 2 x 2 d x = 109
6
∫−2 (( 2x + k ) − ( x + 2) )dx = 3
0
1
2 10
k +1
 2x 3 
= 109
∫−2(( 2x + k ) − ( x + 4 x + 4 )) dx = 3
0
2 10  3  6
k

 2(k + 1)3 2k 3  109

∫−2( − x − 2x + k − 4 )dx = 3
0
2 10 − =
 3 3  6

0
 x3 2  10  2(k 3 + 3k 2 + 3k + 1) − 2k 3  109
 − 3 − x + (k − 4)x  = 3  3 = 6
−2  
 (−2)3 
0−− − (−2) 2 + (k − 4) × −2 = 10  2k 3 + 6k 2 + 6k + 2 − 2k 3  109
 3  3  3  = 6

− 8 + 4 + 2k − 8 = 10  6k 2 + 6k + 2  109
3 3  3  = 6
20
− + 2k = 10
3 3 2(6k 2 + 6k + 2) = 109
2k = 10 12k 2 + 12k + 4 = 109
k=5
12k 2 + 12k − 105 = 0
h
2 Gradient of the sloping line is m =
R−r k = 5 or − 7
2 2
Since x > 0, k = 5
Equation is y = mx + c ⇒ 0 = h × r + c ⇒ c = − hr 2
R−r R−r
y= h x− hr 4 x 2 + y 2 = R 2 ⇒ x 2 = R 2 − y 2
R−r R−r

( ) +r
2 2
hr h
y+ = x Also w 2
= R2 ⇒ r 2 = R2 − w
R−r R−r 2 4
x = R−r y+r w /2
∫−w /2 πx dy − V (cylinder)
2
h V=

( )
2
2r ( R − r )
x2 = R − r y2 + y +r2
= 2 ∫ πx 2 dy − πr 2w
w /2
h h
0

= 2 ∫ π R 2 − y 2 d y − πr 2w
h w /2
∫0 πx
2
V= dy
0
w /2
 y3
( )
h 
2
2r ( R − r )
= π  R−r
∫
2
y + y + r 2  dy = 2π R 2 y −  − πr 2w
3
0 h h   0
h
= π ( R − r )
w 
( )
 2 3
y 2r ( R − r ) y  2  3
2
+ +r y  
 h 3 h 2  0

2 ( )
= 2π R 2 w − 2  − πr 2w
3

 
( ) h3 + 2r(Rh− r ) h2 + r h − 0
 2 3 2 
= π R − r 2
 h  2 3
= 2π  R w − w  − πr 2w
 2 24 
 h( R − r ) 2  3
= π + hr ( R − r ) + r 2h = πR 2w − πw − πr 2w
 3  12
3  2
 h( R 2 − 2 Rr + r 2 )  = πR 2w − πw − π  R 2 − w w
= π + hr ( R − r ) + r 2h  12  4 
 3  3 3
= πR 2w − πw − πR 2w + πw
12 4
= πh  R 2 − 2 Rr + r 2 + 3rR − 3r 2 + 3r 2 
3 = 1 πw 3
6
= πh  R 2 + Rr + r 2  Since the volume does not depend on R, the claim is
3
true.

58 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

8 Trigonometry 2 Using Pythagoras’ theorem gives a length of 15 for

the other side.
8.1 Trigonometrical functions (Page 76)

d = 8.5
1 (i)
sin 95° sin 50°
8.5 × sin 95° = 11.1cm
d = 4
sin 50°
√15
sin β sin 53°
(ii) =
35 32
sin β = sin 53° × 35 x
32
1
β = 60.9°
15
θ = 180 − 60.9 − 53 = 66.1° (i) sin x =
4
2 2 2
(iii) x = 15 + 12 − 2 × 15 × 12 × cos 82° (ii) tan x = 15
2
x = 318.897... ⇒ tan 2 x = 15
x = 17.9 cm
3 Using symmetry from the unit circle,
(iv) cosθ =
4.12 + 2.8 2 − 5.6 2 = −0.2922...
y
2 × 4.1 × 2.8
1
θ = cos −1 (−0.2922...) = 107 o
2 (Using rounded answers from Q1)

(i) Area = 1 × 8.5 × 11.1 × sin 35° = 26.9 cm 2

2
1 θ
(ii) Area = × 32 × 35 × sin 66.1° = 512 m 2 –1 k 1 x
2
1
(iii) Area = × 15 × 12 × sin 82° = 89.1cm 2
2
1
(iv) Area = × 4.1 × 2.8 × sin107° = 5.49 m 2
2 –1

3 (i) x = 8 ⇒ x = 8 × 7 = 11.2 cm (i) cos(−θ ) = k

7 5 5
y y+9 (ii) cos(π − θ ) = −k
(ii) =
4 10 (iii) sin θ = 1 − k 2
10 y = 4( y + 9)
(iv) cos(π + θ ) = −k
10 y = 4 y + 36
1 − k2
6 y = 36 (v) tan θ =
k
y = 6 cm
8.2 Trigonometrical functions for angles of any size 8.3 Trigonometrical graphs (Page 78)
(Page 77))
1 (i) y
1 First ratio Other ratio that Exact
has the same value value 3

1 2
sin 30° sin 150° 2
1

sin 210° sin 330° −1 0

2 x
90° 180° 270° 360°
1 −1
cos 60° cos 300° 2 −2
1
tan 30° tan 210° −3
3

cos 150° cos 210° − 3

2
tan 120° tan 300° − 3

59 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

(ii) y 7 (i) Since the amplitude is half the range,

2 b = 1 ( 8 − −2 ) = 5 . Also, 3 is halfway between
2
1 these two values so a = 3.
0 (ii)
p p 3p 2p x
−1 y
2 2
8
−2
7
−3
6
−4 5
−5 4
3
(iii) y 2
1
3 0
−1 p p 3p 2p x
2 2 2
−2
1

0 8.4 Identities (Page 80)

90° 180° 270° 360° x
−1 1 (i) sin x tan x ≡
1 − cos x
cos x
−2
−3 LHS = sin x × sin x
cos x
2
(iv) y
= sin x
8 cos x
7 2
6 = 1 − cos x
cos x
5
2
4
= 1 − cos x
3 cos x cos x
2
1 = 1 − cos x
cos x
0
p p 3p 2p x = RHS
−1 2 2

(ii) 1 + cos x ≡
sin x
2 A = 4, B = 1, C = −2 1 − cos x
sin x
3 A = 5, B = 2, C = −1 RHS = sin x × 1 + cos x
1 − cos x 1 + cos x
4 A = 2, B = 3, C = −2 sin x (1 + cos x )
=
1 − cos 2 x
5 A = 2, B = −1
sin x (1 + cos x )
=
6 (i) Period = 2π = 2 sin 2 x
3π 3
(ii) T = 1 + cos x
5
sin x
4 = LHS
3
2 (iii) tan 2 x − sin 2x ≡ tan 2x sin 2x
1 2
0
t LHS = sin 2x − sin 2 x
−1 1 2 3 4 cos x
−2 2 2 2
−3 = sin x − sin2 x cos x
cos x
−4
−5 sin x (1 − cos 2 x)
2
=
cos 2 x
2 2
(iii) From the graph, the highest points occur at = sin x ×2 sin x
cos x
t = 0.7, 1.3, 2, 2.7, 3.3 and 4 (1 d.p.) 2
= sin 2 x × sin 2 x
t = 12.40 a.m., 1.20 a.m., 2 a.m., 2.40 a.m.,
 cos x
3.20 a.m., 4 a.m. = tan 2 x sin 2 x
= RHS

60 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

8.5 Trigonometrical equations (Page 81)

1 ≡ cos x
(iv) tan x +
cos x 1 − sin x 1 (i) 2 sin x = −1
LHS : = sin x + 1
cos x cos x sin x = − 1
2
= sin x + 1
cos x ( )
x = sin −1 − 1 = −30°
2
= sin x + 1 × cos x x = 210° or 330°
cos x cos x
( sin x + 1) cos x (ii) cos 2 x = 1
=
cos 2 x 2

=
( sin x + 1) cos x
1 − sin 2 x

2 ()
2 x = cos −1 1 = 60° or 300° or 420° or 660°
x = 30° or 150° or 210° or 330°
( sin x + 1) cos x
=
(1 − sin x )(1 + sin x ) 2 2cos3x − sin3x =0
= cos x 2cos3x − sin3x = 0
1 − sin x cos3x cos3x cos3x
= RHS 2 − tan3x =0
1 1 tan3x =2
2 (i) + ≡ 2
1 − cos x 1 + cos x sin 2 x
3x = tan −1(2) = 63.4° or 243.4 ° or 423.4°
(1 + cos x ) + (1 − cos x ) x = 21.1° or 81.1° or 141.1 ° (1 d.p.)
LHS =
(1 − cos x )(1 + cos x )
= 2 3 (i) y
1 − cos 2 x 1

= 22
sin x
= RHS 0.25

(ii) cos x − cos x ≡ 2 –1 1 x

1 − cos x 1 + cos x tan 2 x

LHS = cos x − cos x

1 − cos x 1 + cos x
cos x (1 + cos x ) − cos x (1 − cos x ) –1
=
(1 − cos x )(1 + cos x )
x = 14.5° or 165.5° (1 d.p.)
2
x + cos 2 x
= cos x + cos x − cos
2
1 − cos x (ii) y
2 1
= 2 cos2 x
sin x
= 22
sin x
cos 2 x
–1 −0.7 1 x
= 22
tan x
= RHS

–1

x = 134.4° or 225.6°

61 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

( cos1 θ + tanθ ) ≡ 11 +− sin

(iii) y 2
1 6 (i) sinθ
θ

LHS = ( 1 + sinθ )
0.7
2

cos θ cosθ

= (1 + sinθ )
2

–1 − 0.8 0.8 1 x cosθ

(1 + sinθ ) 2
=
−0.7 cos 2 θ
(1 + sinθ ) 2
–1 =
1 − sin 2 θ
x = 41.2° or 221.2°
(1 + sinθ ) 2
(iv) y =
1
(1 − sinθ )(1 + sinθ )
= 1 + sinθ
1 − sinθ
= RHS

( cos1 θ + tanθ )
2
(ii) = 3 ⇒ 1 + sinθ = 3
7 1 − sinθ 7
–1 1 x
1 + sinθ =13+ sinθ = 3
1 − sinθ 17− sinθ 7
−2
5 7 (1 + sinθ 7) (=1 +
3(sin ) =θ 3) (1 − sinθ )
1 −θsin
7 + 7sinθ 7=+3 7sin θ θ= 3 − 3sinθ
− 3sin
–1
10sinθ = −10sin4 θ = −4
x = 203.6° or 336.4°
sinθ = − 4sin=θ−=2 − 4 = − 2
4 (i) 2sin x = − 3 10 5 10 5

sin x = − 3
2
( ) ( )
θ = sin −1 θ− 2= sin
5
− 2° = −23.6°
= −−123.6
5
θ = 203.6°θ or = 336.4
203.6° or 336.4°
x = sin −1  − 3  = − π
 2  3 2
7 (i) 2cos x = cos x
x = 4π or 5π 2cos 2x − cos x = 0
3 3
cos x ( 2cos x − 1) = 0
(ii) cos3x = −1
cos x = 0 or cos x = 1
3x = cos −1 ( −1) = π or 3π or 5π  2
x = 90°, 270°, 60°, 300°
x = π or π or 5π
3 3 2
2π (ii) 2tan x = tan x
5 (i) Period = = 40 seconds 0 = tan 2x − 2tan x
π
20 0 = tan x ( tan x − 2 )
(ii) Maximum height is when sin π t = 1 ( 20 ) tan x = 0 or tan x = 2
x = 0, π, 2π,1.11, 4.25
i.e. 54 + 53 = 107 m

( )
(iii) h = 54 + 53sin π t = 30
20
(iii) cos x = −3tan x
cos x = −3 sin x
cos x

20(
53sin π t = −24 ) cos 2x = −3sin x
1 − sin 2x = −3sin x
20(
sin π t = −24)
53 0 = sin 2x − 3sin x − 1

( ) ( )
2
π t = sin −1 −24
sin x = 3 ± 3 − 4 × 1 × −1
20 53 2 ×1

20( )
π t = −0.47 or 3.61 =
3 ± 13
2
= 3.30 or − 0.303
t = 3.61 × 20 = 23 seconds x = −0.308 or − 2.83
π

62 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

(iv) 4 + 5cos x = 2sin 2x 3 (i) P = 7 + 7 + 7 × 1.2 = 22.4 cm

(
4 + 5cos x = 2 1 − cos x 2
) A = 1 × 7 2 × 1.2 = 29.4 cm 2
2
4 + 5cos x = 2 − 2cos 2x
(ii) Angle of sector is θ = 360° − 130° = 230°
2cos 2 x + 5cos x + 2 = 0
Converting to radians,
( 2cos x + 1)(cos x + 2) = 0
cos x = − 1 or cos x = −2 230 × π = 23π = 4.014...
2 180 18
x = 120° or − 120° P = 12.5 + 12.5 + (12.5 × 4.014 )

 = 75.2 cm (3 s.f.)
8 (i) 4cos 2 x + 7 sin x − 2 = 0
A = 1 × 12.5 2 × 4.014... = 314 cm 2 (3 s.f.)
( )
4 1 − sin 2 x + 7 sin x − 2 = 0 2
2 4 12 = 7θ
4 − 4 sin x + 7 sin x − 2 = 0
− 4 sin 2x + 7 sin x + 2 = 0 θ = 12 = 1.71 = 98.2°
7
4 sin 2x − 7 sin x − 2 = 0 5 Perimeter:
( 4 sin x + 1)(sin x − 2) = 0
Length of arc AB = 5 × 2π = 10π ≈ 10.47...
sin x = − 1 or sin x = 2 3 3
4
By joining OC create two right-angled triangles AOC
Since sin x is never greater than 1, sin x = − 1 and BOC.
4
(ii) sin (θ − 20° ) = − 1 tan π = AC ⇒ AC = 5 × tan π
4 3 5 3
θ − 20° = sin −1 − 1
4 ( )    = 5 3 ≈ 8.66...

θ − 20° = −14.5°, 194.5°,  P = 10π + 5 3 + 5 3

3
θ = 5.5°, 214.5° = 10π + 10 3 = 27.8 cm (3 s.f.)
3
8.6 Circular measure (Page 84) Area:

1 Degrees Radians Area of triangle AOC = 1 × 5 × 5 3 = 25 3

2 2
120° 2π 1
Area of sector AOB = × 5 × 2 2π = 25π
3 2 3 3
36° π 25 3
5 Shaded area = 2 × − 25π
11π 2 3
330° 6 25π
= 25 3 −
5π 3
150° 6 2
= 17.1cm
4π
240° 3 6 (i) Length of arc, s = 10 × 0.8 = 8 cm
3π
54° 10 sin 0.8 = RP
10
5π RP = 10 × sin0.8 = 7.1735 cm
225° 4
7π cos0.8 = OP
315° 10
4
OP = 10 × cos 0.8 = 6.967  cm
2 Degrees Radians so QP = 10 − 6.967… = 3.0329… cm
12° 0.209 Perimeter = 8 + 7.1735 + 3.0329
10.9° 0.190 = 18.2 cm (3 s.f.)
145° 2.53
(ii) Area of shaded region
169.0° 2.95
= area of sector – area of triangle OPR
235° 4.10
= 1 r 2θ − 1 × OP × PR
288.8° 5.04 2 2
342.5° 5.98 = × 10 × 0.8 − 1 × 6.967... × 7.173...
1 2
2 2
78.5° 1.37
= 40 − 24.989...
= 15.0 cm2 (3 s.f.)

63 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

(iii) Shaded area

7 (i) 1 × 6 2 × θ − 1 × 6 × 6 × sin θ = 36
2 2 = area OAB − area OCD − area ACD

18θ − 18 sin θ = 36
= 1 × 16 2 × 1 π − 1 × 8 2 × sin 1 π − 1 × 8 2 × 2 π
  18(θ − sin θ ) = 36 2 3 2 3 2 3
      θ − sin θ = 2 = 128 π − 32 × 3 − 64 π
3 2 3
    θ = sin θ + 2 64
= π − 16 3
(ii) 6 + 6 + 6θ = 27.3 3

12 + 6θ = 27.3 Further practice (Page 88)

6θ = 15.3
1 (i) The amplitude of the graph is 2 and it repeats
θ = 2.55 3 times in 360° (the period is 120°) so the
8 equation is y = 2sin3x.
(ii) The graph repeats twice in 2π so c = 2.
8 cm 8 cm The range is -2  y  6 and the amplitude is
half of the total range so b = 4.
The graph has been shifted up 2 units so a = 2.
The negative sign in front of the cos term
means the graph is upside-down.
8 cm The equation is y = 2 − 4cos2 x.
2
Since each part of the band around a circle is 2 (i) tan x (1 − sin x ) ≡ sin x cos x
one-third the length of the circumference, the total LHS = tan x × cos 2 x
length is
= sin x × cos 2 x
(2 × π × 4) + 8 + 8 + 8 = 49.1 cm (3 s.f.) cos x
= sin x cos x
9 (i) tanθ = AC ⇒ AC = 8 tanθ
8 = RHS
Shaded area (ii) sin x − cos 4 x ≡ sin 2 x − cos 2 x
4

= area of triangle OAC – area of sector OAB (

 LHS = sin 2x − cos 2 x sin 2x + cos 2 x)( )
= 1 × 8tanθ × 8 − 1 × 8 2 × θ = ( sin 2 x − cos 2 x ) (1)
2 2
= 32 tanθ − 32θ = sin 2 x − cos 2 x
= 32(tanθ − θ ) = RHS
(ii) tan π = AC ⇒ AC = 8 tan π cos θ
(iii) ≡ 1 −1
3 8 3 tan θ (1 + sin θ ) sin θ
= 8 3 ≈ 13.86 cm cos θ
LHS =
sin θ (1 + sin θ )
cos π = 8 ⇒ OC = 8 = 16 cm cos θ
3 OC cos π
3 cos 2 θ
=
So BC = 16 − 8 = 8 cm sin θ (1 + sin θ )
Length of arc AB = 8 × π = 8π
3 3
=
(1 − sin θ )
2

sin θ (1 + sin θ )
Perimeter = 8π + 8 + 8 3 ≈ 30.2 cm (3 s.f.)
3 (1 − sin θ )(1 + sin θ )
=
10 (i) Since OCD is equilateral, ∠OCD = π .
sin θ (1 + sin θ )
3
π 2 = 1 − sin θ
So ∠ACD = π − = π sin θ
3 3
(ii) Perimeter = AD + BD + AB = 1 − sin θ
sin θ sin θ
= 8 × 2 π + 8 + 16 × 1 π = 1 −1
3 3 sin θ

32 = RHS
= π + 8 cm
3

64 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

3 (i) tan x + 1 = 3 5 (i) The diagram shows the relationship between

tan x = 2 sin x and sin(π − x ) :
y
x = tan −1 (2) = 63.4° or 243.4° (1 d.p.) 1

(ii) sin( x − 30°) = 1 k

x − 30° = sin −1 (−1)
p−x
x − 30° = −90°
x
x = 120° –1 1 x
4 (i) 2tan x − 1 = 1
2tan x = 2
tan x = 1
–1
x = tan −1 (1) = π or 5π sin x = sin(π − x ) = k
4 4
1 sin(2 x + 1) = 0.1 (ii) Looking at the right-angled triangle formed in
(ii)
2 the unit circle, we can calculate the length of
sin(2 x + 1) = 0.2 the unknown side with Pythagoras’ theorem:
2 x + 1 = sin −1 ( 0.2 )
x 2 + k 2 = 12
2 x + 1 = 0.2014 or 2.940
x = 1 − k2
or 6.485 or 9.223 or 12.77
x = 0.970 or 2.74 or 4.11 or 5.88 (3 s.f.) Hence cos x = 1 − k 2
(iii) Looking at the diagram, we can calculate the
(iii) 4sin 2 x + 4 sin x − 3 = 0 value of tan x from the answers to the first two
( 2sin x + 3)( 2sin x − 1) = 0 parts of the question.
sin x = − 3 or sin x = 1 y
1
2 2
π 5π √(1 – k2)
x= ,
6 6 k
p x
 −
(iv) 2sin 2 x = 3 − 3cos x 2
x

( )
2 1 − cos 2 x = 3 − 3cos x –1
– k 1 x

2 − 2cos 2 x = 3 − 3cos x
0 = 2cos 2 x − 3cos x + 1
0 = ( 2cos x − 1)(cos x − 1) –1
–

cos x = 1 or cos x = 1 tan x = k

2 1 − k2
x = π , 5π , 0, 2π
3 3 ( )
Looking at tan π − x we can see that the values
2
3sin22 x + 5cos22 x
= 9sin
(v) 3sin x + 5cos x = 9sin xx of sin x and cos x have swapped so
(
3sin22xx ++ 55 11 −− sin )
sin22xx == 9sin
( )
3sin 9sin xx 2
tan π − x = 1 − k

3sin22xx ++ 55 −− 5sin
5sin22xx == 9sin
9sin xx 2 k
3sin
2
OR the values can be found directly by looking at
00 == 2sin
2sin2 xx ++ 9sin
9sin xx −− 55
the unit circle:
00 == ((2sin
sin xx == 11 or
)(sin
2sin xx −− 11)( sin xx ++ 55))
sin xx == −−55
π
(
tan − x =
sin π − x
)
2
= 1− k
( 2 )
( )
sin or sin 2 k
22 cos π − x
π 5π 2
xx == π6 ,, 5π
6 66

65 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

6 Pythagoras’ theorem gives the length of the other Clearly there are four solutions.
side as 53.
cos2 x = 3
2
√53
2   2 x = cos −1  3 
2 
x
7 2 x = 30° or 330° or 390° or 690°

(i) From the triangle, cos x = 7 , but if x = 15° or 165° or 195° or 345°
53
90°  x  180°, then the value of cos x is (ii) Whenever there is an equation with a mixture
negative, so cos x = − 7 of sin and cos terms, you need to use either
53 cos 2 x = 1 − sin 2 x or sin 2 x = 1 − cos 2 x so that the
equation can be written as a quadratic with
( 253 ) = 534
2
(ii) sin 2 x = only sin x or cos x terms.
cos 2 x − 1 = sin x
7 LHS = 1 + tan 2 θ
(1 − sin 2 x ) − 1 = sin x
2
= 1 + sin 2 θ   0 = sin 2 x + sin x
cos θ
2 2  0 = sin x (sin x + 1)
= cos θ +2 sin θ
cos θ sin x = 0 or sin x = −1
1
= x = 0 or − π
cos 2 θ 2
= RHS If the domain for the solutions was − π  x  π,
sin x + 1 then you would have included x = π or − π .
8 LHS =
cos x sin x 2sin x tan x = 3
(iii)
cos x
= sin x + cos x 2sin x sin x = 3
cos x sin x cos x
sin 2 x + cos 2 x 2sin 2 x = 3
= cos x
sin x cos x
1   2sin 2 x = 3cos x
=
sin x cos x 2(1 − cos 2 x ) = 3cos x
= RHS
2 − 2cos 2 x = 3cos x
9 1 − cos x + sin x ≡ 2 0 = 2cos 2 x + 3cos x − 2
sin x 1 − cos x sin x
0 = (2cos x − 1)(cos x + 2)
(1 − cos x )(1 − cos x ) + sin 2 x
LHS =
sin x (1 − cos x ) cos x = 1 or − 2
2
(1 − 2cos x + cos 2 x ) + sin 2 x
=
sin x (1 − cos x ) x = π,− π
3 3
= 2 − 2cos x y
sin x (1 − cos x ) 1
= 2(1 − cos x )
sin x (1 − cos x )
= 2 0.5
sin x x
−1 −1
= RHS
10 (i) From the graphs of y = 2cos2 x and y = 3:
y −1
2

0 90° 180° 270° 360° x

−1

−2

66 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

11 Area of segment = area of sector − area of triangle

sinθ = BM ⇒ BM = r sinθ
r
= 1 r 2θ − 1 r 2 sinθ
2 2 So AB = 2r sinθ
= 1 × 8 2 × π − 1 × 8 2 × sin π OR
2 3 2 3
= 5.80 cm2 (3 s.f.) AB 2 = r 2 + r 2 − 2r 2 cos2θ

12 Shaded area = area of triangle OAB − area of sector AB = 2r 2 − 2r 2 cos2θ

OAC = 2r 2 (1 − cos2θ )
= 1 × OA × AB − 1 r 2θ Arc AB = 2rθ
2 2
Perimeter = 2r + 2r sinθ + 2rθ
= 1 × 10 × AB – 1 × 10 2 × π = 2r (1 + sinθ + θ )
2 2 3
(ii) Shaded area = a rea rectangle − area segment
tan π = AB so AB = 10 × tan π = 10 3
3 10 3 Area rectangle = r × 2r sinθ
So shaded area = × 10 × 10 3 – 1 × 10 2 × π
1
2 2 3 = 5 × 2 × 5 × sin 1 π
6
(
= 50 3 − 50 π cm 2
3 ) = 25
Area segment = 1 r 2θ − 1 r 2 sinθ
To find the perimeter of the shaded region, we need 2 2
to find the length of OB. = 1 × 5 2 × 1 π − 1 × 5 2 × sin 1 π
2 3 2 3
cos π = 10 ⇒ OB = 10 = 20 cm = 2.26
3 OB cos π Shaded area = 25 − 2.26 = 22.7 (3 s.f.)
3
BC = OB – OC = 20 cm – 10 cm = 10 cm 4 2 2
2 (i) cos x = (1 − sin x )
Perimeter of shaded region = arc AC + AB + BC
= 1 − 2sin 2 x + sin 4 x
= 10 × π + 10 3 + 10
3
(ii) 8sin 4 x + (1 − 2sin 2 x + sin 4 x ) = 2(1 − sin 2 x )
= 10π + 10 3 + 10
3 9sin 4 x = 1
9sin 4 x = 1
13 (i) s = rθ ⇒ 10 = 6θ ⇒ θ = 10 =5
6 3 sin 4 x = 1 sin 4 x = 1
9 9
(ii) Angle ODE is 90° so triangle ODE is a
sin x = ± 4 1 sin x = ± 4 1
right-angled triangle. 9 9
x =°35.3°, 144.7°, 215.3°, 324.7°
Angle DOE = 1 × angle DOF = 1 × 5 = 5 x = 35.3°, 144.7°, 215.3°, 324.7
2 2 3 6
tan 5 = DE ⇒ DE = 6 × tan 5 = 6.60 (3 s.f.) 3 (i) tan π = AC
6 6 6 3 2x
(iii) Shaded area = 2 × area triangle ODE − area AC = 2 x × tan π = 2 x × 3 = 2 3x
of sector DOF 3

(
2 2 )
= 2 × 1 × 6.60 × 6 – 1 × 6 2 × 5
3
AM 2 = AC 2 + MC 2
= (2 3x ) 2 + x 2
2
= 9.63 cm (3 s.f.)
= 12 x 2 + x 2
Past exam questions (Page 90)
= 13x 2
1 (i) Length DC = length AB
AM = 13x

θ θ
r cm

A M B

67 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

x = 1 (ii) At C, y = 0 so
(ii) tan MAC =
2 3x 2 3 a cos x − b = 0

∠MAC = tan −1  1  cos x = b
 2 3 a

∠BAC = 1 π
6
x = cos −1 b
a ()
θ = 1 π − tan −1  1  Since x = cos −1 c , c = cos x = b
6  2 3 a
At D, x = 0 so
4 (i) OC = r cos α , AC = r sin α d = a cos0 − b = a − b
Area OAC = 1 × r cosα × r sin α
2 6 (i) tan(BAO) = 5 ⇒ BAO = 0.3948
( )
1 1 r 2α = 1 × r cosα × r sin α 12
2 2 2 π
(ii) Angle ABO = − 0.3948 ≈ 1.176
1 r 2α = r 2 cosα sinα 2

2 Shaded area = a rea sector APQ + area sector
1 α = cosα sinα BOQ − area triangle ABO
2
= 1 × 12 2 × 0.3948 + 1 × 5 2 × 1.176 − 1 × 12 × 5
sinα cosα = 1 α 2 2 2
2
= 13.1cm 2
(ii) Perimeter OAC = r + r cosα + r sin α

( sin1 θ − tan1 θ ) = 11 +− cos

2
= r + r cos0.9477 + r sin0.9477 7 (i) cosθ
= 2.396r θ

LHS = ( 1 − 1 )
Perimeter ACB 2

= rα + r sin α + (r − r cosα ) sinθ tanθ

= r × 0.9477 + r sin0.9477 + (r − r cos0.9477)
( sin1 θ − cos
sinθ )
2
= θ
= 2.176r

= (1 − cosθ )
2
Ratio is 2.396r : 2.176r
2.396 : 2.176 sinθ
2.396 :1 (1 − cosθ ) 2
=
2.176 sin 2 θ
1.1:1 (1 − cosθ ) 2
=
(iii) ∠AOB = 0.9477 × 180 = 54.3 ° (1 − cos 2 θ )
π
(1 − cosθ )(1 − cosθ )
=
5 (i)
1 + 3sinθ tanθ + 4 = 0 (1 − cosθ )(1 + cosθ )
cosθ
1 + 3sinθ sinθ + 4 = 0 = 1 − cosθ
1 + cosθ
cosθ cosθ
= RHS
1 + 3sin 2 θ + 4 = 0
cosθ cosθ 1 − cosθ = 2
(ii)
1 + 3sin 2 θ + 4cosθ = 0 1 + cosθ 5

1 + 3(1 − cos 2 θ ) + 4cosθ = 0 5 (1 − cosθ ) = 2 (1 + cosθ )

1 + 3 − 3cos 2 θ + 4cosθ = 0 5 − 5cosθ = 2 + 2cosθ
3 = 7cosθ
0 = 3cos 2 θ − 4cosθ − 4
3 = cosθ
3cos 2 θ − 4cosθ − 4 = 0 7
(3cosθ + 2)(cosθ − 2) = 0 θ = 64.6° or 295.4°
cosθ = − 2 or cosθ = 2
3

( )
θ = cos −1 − 2 = 131.8  or 228.2 
3

68 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

π = 6× 3 = 3 3 (ii) Period = 2 π =
2π = 4 hours
8 (i) Distance from D to AB is 6sin
3 2 B π
2
Distance from E to AB is 10sinθ (iii) 1 a.m.

(8 )
Equating these distances gives
3 10.5 = 9 − 3cos π t
10sinθθ ==33 33
10sin
sinθ = 33 33
sinθ =
10
10

8 ( )
3cos π t = −1.5

θθ ==sin
3 3 
sin−−11 3 3
 10
10 
π
cos t = − 1
8 2( )
(ii) θ = 31.3° (1 d.p.) or 0.546 radians
AB = 6 + 10 = 16 cm
π t = cos −1 − 1
8 2 ( )
π t = 2π , 4 π
π
DE = 16 − 6cos − 10cosθ 8 3 3
3
1
= 16 − 6 × − 10cos(31.3°,)
t = 16 , 32
2 3 3

= 4.46 cm t = 5 1 ,10 2
3 3
P = arc DX + arc EX + DE
Since t is hours after 9.30 a.m., refloating can occur
= 6 × 1 π + 10 × 0.546 + 4.46 between 5 1 and 10 2 after 9.30 a.m., so between
3 3 3
= 16.20 cm (4 s.f.) 2.50 p.m. and 8.10 p.m.
4 Sun Moon

Stretch and challenge (Page 92) r

2
1 3sin3θ + 3 = 2cos 3θ θ
(
3sin3θ + 3 = 2 1 − sin 3θ 2
) r
2
3sin3θ + 3 = 2 − 2sin 3θ
2sin 2 3θ + 3sin3θ + 1 = 0
( 2sin3θ + 1)( sin3θ + 1) = 0
sin3θ = − 1 or sin3θ = −1
(i)
2 (
A = 2 1 r 2 2θ − 1 r 2 sin2θ
2 )
2 1 πr 2 = r 2 2θ − r 2 sin2θ
3θ = −30°, − 150°, − 390°, − 510°, 210°, 330°, − 90°, 2

270°, − 450° π = 4θ − 2sin2θ
θ = −170°, − 150°, − 130°, − 50°, − 30°, − 10°, 70°, 2sin2θ = 4θ − π

90°,110° (ii)

2 (i) H (m)
8
( (
px + 3
y = 2 sin 
2
7 Amplitude = 2
6 Period = 2p/(p/2) = 4
r
5
4
3
2
1 θ
0
t (hours) r
−1 1 2 3 5 6 7 8
−2 2
−3
−4 r
cosθ = 2 ⇒ cosθ = 1 ⇒ θ = 60° = π
r 2 3

69 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018
 8 Trigonometry

(2
(iii) A = 2 1 r 2 2θ − 1 r 2 sin2θ
2 )
= 2r 2θ − r 2 sin2θ
= 2r 2 π − r 2 sin 2 π
3 3
2 π 2
= r − 3r2
3 2
 2π 
=  − 3 r 2
 3 2 

% covered =
(
2π − 3 r 2
3 2
2
× 100%
)
πr

=
2π − 3
3 (
2
× 100%
)
π
= 39.1%

70 Cambridge International AS & A Level Mathematics – Pure Mathematics 1 Question & Workbook © Greg Port 2018