IB DIPLOMA PROGRAMME
Raw
7%
^
' x\
,
\
JjS
WORKED
SOLUTIONS
Jennifer Cr
Suzanne D
4Li /jl
Jane Forre
David Harri
Nadia Stoy
Worked solutions
Measuring space: accuracy and 2D
geometry
1
Skills check
1
C 2 r  2 5.3 33.3cm
2
A   r 2    6.5 133cm2
2
3 a Using Pythagoras’ theorem, the length of the base is L  2 15.42  5.52  28.769 cm .
L 5.5
Therefore the triangle has area A 
 79.1 cm2
2
b A square with side 6.4cm has area A  41.0cm2  40.96 
c A trapezoid with bases 12cm , 20cm and height 6cm has area
1
A   12  20  6.5  104cm2
2
Exercise 1A
1 a
9.48m
b
5.32g
c
1.81cm
d
7.00in
2 a
9.48m
b
5.32g
c
1.81cm
d
7.00in
3 a 3
e
4580km
f
13200kg
b 4
c 3
d 4
e 1, 2, 3, 4, 5, 6 or 7
b 3.75
c 30000
d 0.0064
e
4 39.7
Exercise 1B
1 a 8890
2
Number
Round to the
nearest ten
Round to the
nearest hundred
Round to the
nearest thousand
a
2815
2820
2800
3000
b
75391
75390
75400
75000
c
316479
316480
316500
316000
d
932
930
900
1000
e
8253
8250
8300
8000
3 a 502 EUR
4
62  7.874   7.9(to 1 dp)
b 1000 USD
c 10 BGN
d 1400 JPY
C 2 r  2 33  207.345cm  207cm (to 3 s.f.)
5 A circle of area A  20cm2 has radius r 
A

 2.523  cm =2.5cm (to 2 s.f.)
6 The volume of a cube with side l  4.82cm is V  111.98 cm3  110cm3 (to 2 s.f.)
Exercise 1C
1 a Upper bound = 24 – 0.5 = 23.5mm. Lower bound = 24 + 0.5 = 24.5mm. Therefore, the
range of possible values of x is 23.5mm  x  24.5mm .
b Upper bound = 3.2 – 0.05 = 3.15m. Lower bound = 3.2 + 0.05 = 3.25m. Therefore, the
range of possible values of x is 3.15m  x  3.25m .
c Upper bound = 1.75 – 0.005 = 1.745kg. Lower bound = 1.75 + 0.005 = 1.755m. Therefore,
the range of possible values of x is 1.745kg  x  1.755kg
2 The height h is in the range 4.15cm  h  4.25cm and the base b is in the range
3.05cm  b  3.15cm . The lower bound for the area is therefore
© Oxford University Press 2019
1
Worked solutions
1
 4.153.05  6.32875 cm2 , and the upper bound for the area is
2
1
  4.253.15  6.69375cm2 .
2
ALB 
AUB
3 The upper bound for the number of bicycles is 72.5 million bicycles, and the upper bound for
the average distance per day is 2.5 km per day.
The upper bound for the total distance travelled by all the bicycles in Japan is 72.5 x 2.5 =
181.25 million km per day.
4 The upper bound for sales is $345 000 and the lower bound is $335 000. The upper bound for
costs is $225 000 and the lower bound is $235 000.
The maximum relative profit PUB is found using the upper bound on sales and lower bound on
345000  225000
 0.35 (to 2 s.f.). The minimum relative profit PLB is found
345000
335000  235000
using the lower bound on sales and upper bound on costs, i.e. PlB 
 0.30
335000
(to 2 s.f.).
costs, i.e. PUB 
Exercise 1D
1 Percentage error =
  3.14
3.14
 100%  0.0507% (to 3 s.f.)
2 Percentage error =
8840  8848
 100%  0.0904% (to 3 s.f.)
8848
3 Percentage error =
46620  40030.2
 100%  16.5% (to 3 s.f.)
40030.2
4 a Actual temperature =
b Percentage error =
5
1
 50  32 10 C , approximate temperature =  50  32  9 C .
9
2
10  9
 100%  10%
10
Exercise 1E
23 23  233  26
b
52 51 521 53
c
67
 67 5  62
65
d
2
42
 42 52   4  5  202
52
e
86 83  86 3  83
f
3 
g
34
1
1
1



 315
3 39 34 32 39 34 2 9 315
h
27 24
27
 4 3  27 7  20
3
2
2
i
54 34  53  154
j
203  20 
3

 5
43  4 
1 a
2 a
b
2
3
4
 32  4  38
3
4
  
2

x 2  x2  x2  x 6  x 2  x 6  x2  x 6  x 26 26  x12
 
x 0  x2
2
x x
3
3

1 x 6
1 x 6  x 3

 x 6  3 2  x 7
2
3
x x
x2
Exercise 1F
1 a
1.61103 g/cm3
b 1.971010 m
c
9.461012 km
d 1.6751024 g
2 a
1.2101
b
5.04 107
c
4.005 105
d 1103
© Oxford University Press 2019
2
Worked solutions
3 Real size of speck =
4 a
4.6109 years
Size in microscope 1.2102 1.2


 0.24mm (or 2.4101 mm)
Magnification
5102
5
b
5000
5 1 millilitre has 15 drops, each of which has 1.671021 molecules.
So each millilitre has 151.671021  2.5051022 molecules.
Therefore, 1 litre(103 millilitres) of water has 2.5051022 103  2.5051025 molecules.
6 a
b
Mass of 'Planet Nine'  50000.013031024  6.5151025 kg.
Planet Nine is
6.5151025
 11times larger than Earth (to the nearest digit).
5.971024
7 a The population of entire world in 2005 divided by the population of China in 2005 is
6.4109
5(to the nearest integer). Therefore, the ratio of population of entire world to
1.3109
that of China in 2005 was 5:1.
b
c
6.4  109  4.9109  1.5109 .
Country
Percentage Change (relative to 1985 value)
India
1.1109  7.6 108
100%  44.7%(increase)
7.6 108
United States
3.0108  2.4108
100%  25%(increase)
2.4108
China
1.3109  1.1109
100%  18.2% (increase)
1.1109
Bulgaria
8.9106  7.2106
100%  19.1%(decrease)
8.9106
d The data does not support the hypothesis that “it is always the case that the country with a
bigger percentage change has also a bigger population increase”. For example, the
percentage increase is larger in United States than China, but the net increase is larger in
China (2 x 108 vs. 6 x 107).
Exercise 1G
1 a
By Pythagoras' theorem, W2  132  12.22. Therefore, W = 132  12.22  17.8 m(to 1 d.p)
b
Re-arranging Pythagoras' theorem gives b  9.72  6.12  7.5mm(to 1 d.p.)
c
a  7.52  3.52  6.6mm (1 d.p.).
d
c  142  9.52  10.3mm (1 d.p.).
2 a
b
c
Since 92  402  1681  412, a triangle with sides 9 cm, 40 cm and 41 cm is right angled.
Since 102  242  676  262, a triangle with sides 10 cm, 24 cm and 26 cm is right angled.

200

2
 200  102  102 , so an isosceles triangle with two sides of 10 and one of
200 is right angled.
d
3 a
A triangle with sides 11.2, 7.5and 8.3 is not right angled as 11.22  7.52  8.32  0.3  0.
The hypotenuse of a right-angled triangle with both legs of 1 unit has length
l  12  12  2  1.41 units (to 3 s.f.) .
b
The second triangle has legs of length 1 unit and 2 units.
The hypotenuse of this triangle has length 12 
c
12 
 3
2
 2
2
 1  2  3 1.73 units (to 3 s.f.)
 4  2 units.
© Oxford University Press 2019
3
Worked solutions
d The hypotenuse of the nth triangle (counted anti-clockwise from the first) has length
Therefore, as there are 15 triangles in total, the length of x is
n  1.
16  4 units.
4
The length of L is
8.22  4.32  6.98 cm. Therefore, the area is
6.98 4.3
 15.0 cm2 (3 s.f.)
2
Exercise 1H
1 a
As sin 61.2 
y
z
, cos(61.2)  , then y  9 sin 61.2   7.89 cm (3 sf),
9
9
z  9cos(61.2)  4.34cm.
b
tan 34 
10.2
10.2
, so x 
 15.1. Then, using Pythagoras' theorem,
x
tan 34
z  10.22 15.12  18.2 (3 s.f.)
c
y 
21.3
21.3
 23.2, x 
 9.31.
sin 66.4
tan 66.4
2 a KL  8.5 cos 30  7.36m (3 s.f.)
b The ladder reaches a distance up the wall equal to the length LM  8.5 sin 30  4.25m.
c When angle LKM = 55°, the ladder reaches a height LM =8.5 × sin(55°) = 6.96m. The
maximum height up the wall the ladder can reach is 6.96m.
3 a
58°
b
Cliff height = 1.6 + 50 × tan(58°) = 81.6 m (3 s.f.)
Exercise 1I
1 a
b
tan 73  
450
450
, so x 
 138m (Nearest metre)
x
tan 73 
© Oxford University Press 2019
4
Worked solutions
2 The awning should be at least as long as L, where tan 20 
3 Let d be the depth of the crater. tan65 
L
, so L = 1.02m (2 d.p.).
2.8
d
 d  606  tan65  1300 m.
606
4 Let  denote the angle of elevation. tan  
72
72
   tan1
 52.6 (3 s.f.).
55
55
5 a
b Let c denote the height of the cliff and l the height of the lighthouse. Using the
c
trigonometric ratios, tan18 
 c  160  tan18  52.0 m. Further,
160
 c  l   l  160  tan22  c  12.7 m (3 s.f.)
tan22 
160
Exercise 1J
1 a Arc length = 2 r 
b Arc length = 2 r 
c Arc length = 2 r 

360

360

360
 2  5 
70
 6.11 cm.(2 d.p.).
360
 2  4 
45
 3.14 cm (2 d.p.).
360
 2  10.5 
130
 23.82 cm (2 d.p..).
360
5
of the
12
total circle. The total length of the circumference is c   d    25  25 cm. The length of the
5
arc between markings 12 and 5 is
 c  32.7 cm (3 s.f.)
12
2 The clock has twelve markings, so the region between markings 12 and 5 occupies
3 a The total circumference of the wheel of diameter d  120 m is c   d  120. Therefore, in a
200
rotation of 200, the capsule travels through a length d 
 209 m (3 s.f.).
360
b Time taken (in seconds) = total distance travelled/distance per second =
206
 805.5 s. In
0.26
minutes this is 13.4 minutes.
c The previous calculation was for a proportion
therefore take
200
of a revolution. A whole revolution would
360
360
360
 time for 200 
 13.4  24 minutes.
200
200
4 a The opposite of the angle has size   360  225  135. Using the trigonometric identities,
 4.2 
 2 
 , where r is the radius of the circle. Re-arranging gives
we have sin  
2
r

 4.2 
 2 
  2.27 m, or 227 cm (to the nearest cm).
r  
 
sin  
2
© Oxford University Press 2019
5
Worked solutions
b The arc has length 2 r 
135
 5.36 m, or 536 cm (to the nearest cm).
360
Chapter Review
1 Answers will vary.
2 a 9.23
3 a
b 0.511
c 2.12
40  m = 4  105 m.
b
40000  103
 1  1012
40  106
4 The lower bound on the target is 129.5 grams of CO2 per km. The upper bound on the target is
129.5 grams of CO2 per km. Therefore, the lower bound of total CO2 is
40000  129.5  5.18  106 g and the upper bound of total CO2 is 40000  130.5  5.22  106 g
5 The area of a semicircle with radius r is
 r2
2
. Therefore, the areas of the semicircles are
1.52  22  2.52 
,
,
; these satisfy the ‘Pythagoras relationship’, suggesting that this rule does
2
2
2
also hold for semi-circles on the sides of right-angled triangles.
6 a Using Pythagoras’ theorem, the diagonal has length d  7.52  152  16.77 m (nearest cm)
b Percentage error =
| actual – error|
16.77  16
 100% 
 100%  4.59% (3 s.f.).
| actual|
16.77
7 Let x denote the length of the side of the square, so, by Pythagoras’ theorem,
100 2 
2
 x2  x2  2x 2  x  100 ft. Therefore, the perimeter is p  4x  400 ft.
8 Let the distance of ABC above line BC be denoted by x. Using Pythagoras’ theorem,
2
 14.5 
x 2  8.52  
  19.68 m. Therefore, x  4.44 m (3 s.f.).
 2 
9 Using Pythagoras’ theorem, the distance is
3.652  0.62  3.60 m (3 s.f.).
10 a
b Using Pythagoras’ theorem, the distance x between the satellite and the horizon is
x
6370  800
2
 63702  3291 km (4 s.f.)
11 a Using trigonometric identities, the length l of the ladder reaching a height 20m up the wall
20
20
satisfies sin50 
l 
 26.11 m (nearest cm)
l
sin50
b A ladder of length 15 m can reach a height of h  15  sin50  11.49 m up the wall.
Therefore, if extended from a fire truck of height 4.15 m, the ladder can reach a height of
4.15  h  15.64 m up the wall.
12 The distance from the base of the tower to the blue car is db 
from the base of the tower to the red car is dr 
150
 86.6 m. The distance
tan60
150
 150 m. The total distance between
tan 45
the cars is db  dr  237 m (nearest m).
© Oxford University Press 2019
6
Worked solutions
13 a
b (Using the notation of the diagram in the solutions). Applying trigonometry to the triangle
y
BCD: tan 40 
 y  tan 40x, and for the triangle ACD:
x
y
tan33 
 y  tan33  x  0.5 . Solving these two equations simultaneously:
x  0.5
y  tan40x  tan33  x  0.5   tan40  tan33 x  0.5 tan33  x  1.711 km and
y 1.436 km.
2
 80 
2
14 The radius r of the arc is calculated using Pythagoras’ theorem: r 2  
  136  r  141.7
 2 
136
 73.6. By symmetry,
80
2
IMG  FMH , so HMI  180  2IMG  32.8, this is the angle at the centre of the arc.
cm. If M denotes the midpoint of FG, then the tan FMH 
Therefore, the arc has length 2 r 
15 a
b
16 a
b
HMI
32.8
 2  141.7 
 81.1 cm (3 s.f.)
360
360
AY 2  52  62  AY  7.81 cm
6
ˆ  37.5
ˆ 
 XAY
tan XAY
7.810
V    1.82  14.5  148 cm3

SA  2  1.8  14.5  2  1.8
2
M1A1
M1A1
  184
M1A1
cm3
M1A1
40
 2    7   4.88 cm
360
M1A1
Perimeter  4.88  7  7  18.9 cm
M1A1
17 Arc AB 
18 a
b
c
3
4
7
10x  3x
30x

2x 6
2x 6
M1A1
 15x13
x 2  4x 3 4x 1
 1  4
x 1
x
A1
M1A1A1
3x3  12x0  4x5  144x8
M1A1
 12x 4
x 
x 
2
d
3
A1
5
4

x 10
x 12
M1A1
 x2
A1
19 23.5  1.6  21.9
A1
© Oxford University Press 2019
7
Worked solutions
tan  
21.9
100
M1A1
  12.4
A1
20
355
113  100
M1A1
 8.49  106%
A1
 

90
x
21 tan19 
M1A1
Yacht / kittiwake is at an horizontal distance of
90
 261.38 m from Sharon
tan19
Vertical height of kittiwake from cliff is . 261.38 tan15  70.04 .m
Required distance is 90  70.04  160 m
A1
M1A1
A1
22 7.1sin70  6.672 m
M1A1
7.1sin80  6.992 m
M1A1
So the minimum height is 6.67 m and the maximum height is 6.99 m
23 a
b
c
time 
1.496  108  103
 498.7 seconds
3  108
498.7
 8.3 minutes
60
4.014  1013  103
time 
 1.338  108 seconds
3  108
1.338  108
 4.24 years
60  60  24  365
M1A1
3  108  2.5  106  365  24  60  60  2.3652  1022 m
M1A1


2.3652  1022 m  2.4  1019 km
24 a
b
25 a

8
1  3  10

2
16
 9  10
J
9  10
 1.5  1015 seconds
60
1.5  1015
 4.76  107 years
60  60  24  365
5.5
RMIN 
 22
0.25
6.5
 43.3
0.15
120
120
x 
e.g. tan31 
x
tan31
27 a
VMIN 
A1
M1A1
A1
M1A1
M1A1
26 Valid attempt to find a horizontal distance.
Distance required is
A1
A1
RMAX  30
 100  44.4%
30
Other horizontal distance is
A1
M1A1
16
RMAX 
b
M1A1
M1A1
A1
120
tan39
120
120

 51.5 m
tan31
tan39
4  3.453
 172 cm3
3
A1
M1A1
M1A1
© Oxford University Press 2019
8
Worked solutions
4  3.553
 187.4 cm3
3
VMAX 
b Re-arranging V 
28 a
4 r 3
gives r 
3
c
d
3V
4
M1A1
3
3  495
 4.91 cm
4
M1A1
rMAX 
3
3  505
 4.94 cm
4
M1A1
  1.3  106   4.5  106  2.39  1017 m3
2
M1A1
2.39  1017  2  1017
 100  16.3%
2.39  1017
M1A1
1.675  10   1.673  10   9.109  10   1.12  10
27
b
3
rMIN 
b Percentage error 
29 a
M1A1
27
31
27

 
3
 
2 1.675  1027  1.673  1027  9.109  1031
kg
  6.70  10
27
27
1.675  10
 1840 times heavier
9.109  1031
1  1030  9.109  1031
 100  9.78%
9.109  1031
30 Arc length AD 
Arc length . BC 
kg
M1A1
M1A1
M1A1
60
 2  5  5.236 cm
360
M1A1
60
 2  13  13.61 . cm
360
M1A1
Perimeter  5.236  13.61  8  8  34.8 cm
31 a Taking R1  7.25 and R2  3.65
b
M1A1
M1A1
M1
1
1
1


RTOT 7.25 3.65
A1
So upper bound  2.428
A1
Taking R1  7.15 and R2  3.55 .
M1
1
1
1


RTOT 7.15 3.55
A1
So lower bound  2.372
A1
2.428  2.40
 100  1.17%
2.40
2.372  2.40
 100  1.17%
2.40
M1A1
A1
So range of percentage errors is anything from 0 to 1.17%
© Oxford University Press 2019
A1
9
Worked solutions
Measuring space: non-right angled
triangles and volumes
2
Skills Check
1 a 15.1 cm
b 4.90 cm
2 a
b
34.7
45.6
2
3
37.5cm
4
20
Exercise 2A
1 a The sine rule applied here gives:

sin  A sin  B  sin  68
sin  68  


 A  sin1 12.5
  48.8. (3 s.f.)

a
b
15.4
15.4 

b The sine rule applied here gives:

sin  B  sin  C  sin  41 
sin  41  


 B  sin1  42.8
  66.6. (3 s.f.)

b
c
30.6
30.6 

c The sine rule applied here gives:

sin  C  sin  A sin 70
sin 70  


 A  sin1  5.1
  41.7. (3 s.f.)

c
a
7.2
7.2 

2 First use that the angles in a triangle total to 180  to find A  180  101  32  47.
Therefore, by the sine rule:
BC
AC
AB
BC
8
AB
8





 BC  sin  47 
 11.0 cm
sin  A sin  B  sin  C 
sin  47  sin 32  sin 101 
sin 32 
and AB  sin 101
8
14.8 cm (both 3 s.f.)
sin 32 
3 Given the information in the question was can produce the diagram (not to
scale), with A = 125 , D = 36 , b = 12km. Therefore,
B =180  125  55 and therefore C  180  55  36  89.
Then the sine rule gives:
c
12
12

 c  sin  89 
14.6 km (3 s.f.).
sin  89 sin 55 
sin 55 
Exercise 2B
1 Throughout these solutions, the notation in below diagram is used:
a Using the cosine rule,
 6.52  72  122 
b2  c2  a2 6.52  72  122
cos A 

 A  cos1 
  125.42...  125 and
2bc
2  6.5  7
 2  6.5  7 
cos B 
 122  72  6.52 
a2  c2  b2 122  72  6.52

 B  cos1 
  26.19...  26.2 .
2ac
2  12  7
 2  12  7 
Then C  180  125.42...  26.19... 28.38... 28.4 (all 3 s.f.).
© Oxford University Press 2018
Teacher notes
1
Worked solutions
b Using the cosine rule,
 142 182  112 
b2  c 2  a2 142 182  112
cos A 

 A  cos1 
  37.7 and
2bc
2  14  11
 2  14  11 
cos B 
 112 182  142 
a2  c2  b2 112 182  142

 B  cos1 
  51.0 .
2ac
2  11  18
 2  11  18 
Then C  180  37.7  51.0  91.3 (all 3 s.f.).
c Using triangle rules, C  180  25  83  72. Next use the sine rule:
c
22
a
b



sin  C  sin 72  sin 25  sin 83 
 a  sin 25  
22
22
 9.78m, b  sin 83  
 23.0 m. (3 s.f.).
sin 72
sin 72 
2 a Let x denote the angle opposite the 70 mm side. By the sine rule,

sin 52  sin  x 
sin 52  

 x  sin1  70
  41.0. Using the sum of angles in a triangle, the

84
70
84 

remaining angle is y  180  52  41.0  87.0. Using the cosine rule, the final side has
length L  702  842  2  70  84  cos 87   106 mm.
b The remaining angle has size 180  66  48  66. Using X and Y to denote the lengths
sin  48 sin 66 sin 66 


of the remaining sides,
14
X
Y
 X  14 
sin  66
sin  48
 17.2 cm,Y  14 
sin  66
sin  48
 17.2 cm (3 s.f.)
c Find x using sum of angles in a triangle: 2x  3x  4x  9x  180  x  20, so the angles
are 40,60,80. Using X and Y to denote the lengths of the remaining sides,
sin 80
25

 X  25 
sin 60
X
sin  60 
sin  80 

sin  40 
Y
 22.0m,Y  25 
sin  40 
sin  80 
 16.3 cm (3 s.f.).
3 First find ABC  180  45  55  80.
Then use the sine rule to find the length x from A to B:
x
450
450

 x  sin 55 
 374 m (to the nearest meter).
sin 55 sin 80 
sin  80 
4 The diagram represents the situation. We are given that  A = 58,
 B = 103 . The vertical lines are parallel so
D  180  A  122  C  360  103  122  135. We are also given
that x = 5km, y = 8km, so by the cosine rule:
z2  52  82  2  5  8  cos 135  z  12.1 km (3 s.f.).
5 Using the cosine rule:
 2232  1522  2852 
BAC  cos1 
  97.1,
2  223  152


2
2
2
 223  285  152 
ABC  cos1 
  32.0,
2  223  285


2
2
2


285

152

223
BAC  cos1 
  50.9 (3 s.f.)
2

285

152


6 One can use either Kristian or Velina’s suggestion.
Kristian’s suggestion is the easier of the two:
© Oxford University Press 2019
2
Worked solutions
Let x denote the length of the unknown side. The cosine rule gives
252  402  x2  2  40  x  cos 35 . This equation has solutions x  22.8 or 42.7, both of
which are possible. So AC  22.8m or 42.7m .
Velina’s suggestion:
 AB sin(A) 
C  sin1 
  66.59...  66.6 , then B  180  A  C  180  35  66.6  78.4
BC


thus giving AC 
BC sin(B)
 42.7cm (3 s.f.) alternatively,
sin(A)
 AB sin(A) 
C  180  sin1 
  113.40...  113 , then
BC


B  180  A  C  180  35  113  32 thus giving AC 
BC sin(B)
 22.8cm (3 s.f.)
sin(A)
Exercise 2C
1 a Area =
b
Area 
1
1
ab sin C   16  28 sin23  87.5 cm2 (3 s.f.)
2
2
1
1
ab sin C   11  7.25 sin56  33.1cm2 (3 s.f.)
2
2
c First find the missing angle: C  180  67  35 78. Then
1
area   10  7  3 5  sin C  67.0 cm2 (3 s.f.)
2


2 a We have two lengths and an angle between them, so the area formula immediately applies:
area 
1
 13.6  9.2  sin 49  47.2 cm2
2
b First find the missing angle C  180  33  42.5  104.5. Next use the sine rule to find
length a: 𝑎sin33°=𝑐sin∠𝐶=19sin104.5°⇒𝑎=10.7 m. Finally, the area of the triangle is then
1
1
ac sin B   19  10.7  sin 42.5  68.6 m2 (3 s.f.).
2
2
2
 2

 a2  b2  c 2 
1 25  (14 2)  59
c First use the cosine rule to find C  cos1 
  14.6.
  cos 
2bc


 2  25  14 2 
Then use the formula for the area: A 
1
1
ab sin C   25  14 2  sin14.6  62.4 cm2 (3
2
2
s.f.).
3 a First use the cosine rule to find
2
2
2
 AC 2  AD2  DC 2 
1  15  23.5  29 
A  cos1 
  cos 
  95.2. Then use the formula for
 2  AC  AD 
 2  15  23.5 
the area: A 
1
1
AD  AC sin A   15  23.5  sin95.2  176 m2 (3 s.f.).
2
2
b Mass of seeds =
Area (m2 )
176
 0.5 
 0.5  0.585kg (3 s.f.).
150
150
4 The missing angle in triangle ABC is B  180  91  44.5  44.5. Next use the sine rule to
7.4
 10.56m. Hence the area of ABC is
find the length AB: AB  sin91 
sin 44.5
1
 7.4  10.56  sin 44.5  27.39m2. The area of the parallelogram is twice the area of the
2
triangle ABC, i.e. area of parallelogram = 2  27.39  54.8m2 (3 s.f.).
5 The triangles (e.g. ALB) making up the hexagon have equal length sides and angles of
1
 9  9  sin60  35.07 cm2.
60  ALB  60. Therefore, the area of triangle ALB is
2
Therefore, the hexagon has area 6  35.07  210 cm2 (3 s.f.).
© Oxford University Press 2019
3
Worked solutions
6 Let  
360
. Splitting the heptagon into 7 triangles, we find MNS  180  2. Using the
7
cosine rule, the length MN  252  252  2  25  25  cos  . Using these two measurements (and
the fact that NS  MN ), we find MS  MN2  MN2  2  MN  MN  cos 180    . Using triangle
relationships, find MSQ  180  2. Then the area of triangle MSQ is found as
1
1
 MS2  sin MSQ. Finally, the whole heptagon has area A  7   252  sin  1710.3m2 .
2
2
Therefore the garden has area A  MSQ  1710.3  744.9  965m2.
Exercise 2D
1 a
Area 
b
Area 
c
Area 
2 a
Area 

360

360

360

360
 r2 
70
   62  22.0 cm2 (3 s.f.).
360
 r2 
49
   32  3.85 cm2 (3 s.f.).
360
 r2 
122
   10.52  117 cm2 (3 s.f.).
360
 r2 
140
   92  99.0 cm2 (3 s.f.).
360
b Using the area formula: 48 
c Using the area formula: 8 
X
48
   102  X  360 
 173 (3 s.f.).
360
100
50
360
   X2  X  8 
 7.59 cm (3 s.f.)
360
50
90
1
 r 2   r 2. The triangle
360
4
r2
 1 2
has area
. Therefore, the shaded section has area A  r      r .
2
 4 2
3 Let r denote the radius of the circle. The whole sector has area
a
A 8 18.3cm2 (3 s.f.)
4 The whole sector has area A1 
b
A 12  41.1cm2 (3 s.f.)
55
1
  42 , and the triangle ABC has area A2   4  4  sin55.
2
360
Therefore, the area of the shaded region is A  A1  A2  1.13cm2 (3 s.f.).
5 Consider a single ‘slice’, which has angle 60 at the centre. The area of the slice is
 360 
 6 
   1.52  1   1.52. The area of the grey triangle in the slice has area
As  
360
6
1
 360 
At   1.5  1.5  sin 
 . The grass area of the slice is As  At . Therefore,t he total area of
2
 6 
grass is Ag  6  As  At   1.22m2 (3 s.f.).
Exercise 2E
1
2
3 a
© Oxford University Press 2019
b
4
Worked solutions
Exercise 2F
1 a Using 3D version of Pythagoras’ theorem: DF  62  62  10.52  13.5 m.
b Let x denote the angle between DF and base ABCD. From trigonometry,
FB 10.5
 10.5 
sin x 

 x  sin1 
  51.1 (3 s.f.)
DF 13.5
 13.5 
c Let y denote the angle between DF and face DCGH. Since AD = GF = 6 m, then
sin y 
6
 6 
 y  sin1 
  26.4 ( 3.s.f.)
13.5
 13.5 
2 a Length DF = AC = 3 cm, then the length DI  32  1.52  2.60 cm . Then use trigonometry
12
to find the angle x between DH and base DEF: tan x 
 x  77.8 (3 s.f.)
2.60
b Base ABC is equilateral so, as G and H are midpoints of AB and AC, triangle BGH is
equilateral with side length 1.5 cm. Length HI is 12 cm, therefore length GI =
12.1 cm
c DG =
1.52  122 
1.52  122  12.1 cm.
3 a Angle between AF and the base of the cone is equal to the angle between CF and the base of
5
5
the cone. Let x denote this angle, then tan x   x  tan1    68.2.
2
2
b The slant height S is worked out by Pythagoras’ theorem: S  52  22  5.39 cm.
 120 
c First use trigonometry to find the length AC  2  2 sin 
  3.46 cm. Then use the cosine
 2 
 5.392  5.392  3.462 
rule to find AFC  cos1 
  37.5 (3 s.f.).
2  5.39  5.39


4 a Using trigonometry, AC = 2  3  cos20  5.64 cm (3 s.f.)
AC2  AB2  112  5.642  12.4 cm (3 s.f.)
b By Pythagoras’ theorem, BC =
c
BCA  tan1
11
 62.9 (3 s.f.)
5.64
Exercise 2G
1 The total volume of the pool is V  7  4  3  84m3 . The total cost of filling the pool =
cost per m3  V  0.15  84  $12.60.
2 The cross section has area Ac  2 
1
 3  6  18 cm2 . The volume is therefore
2
V  10.4  Ac  187.2 cm3. Mass = 0.71 × V = 132.91 = 133 g (3 s.f.).
3 The volume of the cylinder is Vc    3.22  8.5  273.4 cm3. Each cube has volume
Vcube  23  8 cm3. Therefore,
integer less than
273.4
 34 complete cubes can be made (taking the largest
8
237.4
)
8
4 Let h denote the height of the water. The volume of water in the vase is
310
V  310    42  h  h 
 6.17 cm (3 s.f.)
16
2800
 200 cm2. A regular hexagon is made up of six
14
A 200

cm2. Let L denote the
equilateral triangles; in this case each triangle has area At 
6
6
5 The hexagonal base has area A 
© Oxford University Press 2019
5
Worked solutions
length of the side of each of these triangles, then At 
2 At
1 2
 L  sin60  L2 
 L  8.77
2
sin60
cm (3 s.f.)
Exercise 2H
1 a
V   r 2h    32  23  650 cm3 (3 s.f.)
b V 
4 3 4
 r     23  33.5 cm3 (3 s.f.)
3
3
c
1 2
1
 r h     2.12  7.3  33.7 cm3 (3 s.f.)
3
3
V 
d A sphere with radius 3.1cm has volume V 
with radius 3.1 cm has volume
4
   3.13  124.7 cm3. Therefore a hemisphere
3
V
 62.4 cm3 (3 s.f.)
2
e The base area is Ab  8  5  40 cm2. Therefore, the pyramid has volume
V 
1
 Ab  17  227 cm3 (3 s.f.).
3
2 The volume of a cylinder is proportional to its radius squared, therefore doubling the width of
the boiler (effectively doubling the radius) would increase the volume by a factor of 2 2= 4. The
new volume would be 400L. If the width of the boiler is tripled, the volume would increase by
32=9, the new volume would be 900L.
3 The sphere has mass M  4.1  106 tonnes. Using the fact that 1 cubic meter of copper weighs
M
 459.1m3. Using the formula for the
8930 kg, we find that the sphere has volume V 
8930
volume of the sphere, we find the radius r of the sphere is r  3
3V
 4.79m (3 s.f.).
4
Exercise 2I
1 a The curved surface has area Ac  2  2.5  7.3  114.7 cm2 , the ends of the cylinder each
have surface area Ae    2.52  19.64 cm2 . The total surface area is A  Ac  2Ae  154 cm2
(3 s.f.).
b Compute the length of the slanted edge l using Pythagoras’ theorem:
l  3.52  122  12.5cm. The surface area of curved section is
Ac   rl    3.5  12.5  137 cm2 . The surface area of the base is Ab    3.52  38.5cm2.
The total surface area is Ac  Ab  176 cm2 (3 s.f.)
2 a The cuboid is made of 2 faces of area A1  8  2.5  20m2 , 2 faces of area
A2  4  2.5  10m2 and 2 faces of area A3  8  4  32m2. The total area is
A  2A1  2A2  2 A3  124m2.
b The cuboid lower section contributes a surface area
Ac  11  6  11  6  15  6  15  6  11  15  447m2 . The triangular sections each have a
hypotenuse of length l  52  5.52  7.433 m and area At  2 
1
 5  5.5  27.5m2. The
2
titled roof sections each have surface area Ar  l  15  111.5m2. The total area is
Ac  2 At  2 Ar  755m2 (3 s.f.).
© Oxford University Press 2019
6
Worked solutions
3
c The cone section has surface area Ac       4  6 cm2. The curved cylinder section has
2
2
9
3
surface area Acy    3  9  27 cm2. The base has area Ab      
cm2 . The total
4
2
surface area is Ac  Acy  Ab  35.25  111cm2 (3 s.f.).
3 A hemisphere with volume V  30 cm3 has radius r  3
3V
 2.4286 cm. The curved section of
2
this hemisphere has area Ac  2  r 2. The flat base has surface area Ab    r 2. The total
surface area is Ac  Ab  3 r 2  55.6 cm2 (3 s.f.)
4 a The volume of the cylindrical part is Vc    32  12  108 m3. The volume of the
hemispherical cap is Vh 
1 4
   33  18 m3. The total volume is Vc  Vh  126  396m3 .
2 3
b The surface is made of two parts: the curved part of the cylinder, that has area
Ac  2  3  12  72 m2 , and the hemispherical cap that has surface area
Ah  2  32  18 m2 . The total surface area is Ah  Ac  90  283m2 (3 s.f.) . Amount of
paint needed is
283
 33.3lt .
8.5
Chapter Review
1
A
1
1
ab sin C   90  65  sin105  2830m2 (3 s.f.)
2
2
2 Let x denote the distance between the two aircraft. Using the cosine rule,
x2  582  752  2  58  75  cos52 . Therefore, x  60.3 km. (1 d.p.)
3 a The angle between the paths AB and BC is 180  20 160. Also, length AB = 60km,
BC = 30km. Therefore, by the cosine rule,
AC 2  AB2  BC 2  2  AB  AC  cos160  AC  302  602  2  30  60  cos160  89 km.
(nearest integer)
b Using the sine rule:
sin CAB sin160
sin160 


 CAB  sin1  30 
  6.64 (3 s.f.)
BC
AC
89 

4 The sector has area A 
l 
45
   52  9.82 cm2. The length of the arc is
360
45
 2  5  3.93 cm . (3 s.f.)
360
5 Let  denote the angle at the centre of the sector. By the arc length formula,

360
 2 r  1.396    1.396 
 20. Therefore, the sector has area
360
2  4

20
A
   r2 
   42  2.79 cm2.
360
360
6 a Each sector is one sixth of the respective circle. AG  2  10 
60
 10.5 cm (3 s.f.).
360
b The radius of the circle containing BH is r  20 cm.
Therefore, BH  2  r 
60
 20.9 (3 s.f.).
360
c The radius of the circle containing CI is r  30 cm.
Therefore, BH  2  r 
60
 31.4 (3 s.f.).
360
© Oxford University Press 2019
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Worked solutions
d Arc length is proportional to radius, so BH is double AG , CI is three times AG .
DJ is four times AG , i.e. DJ  4  10.5  41.9 cm (3 s.f.) and EK is five times AG,
i.e. EK  52.4 cm (3 s.f.).
e Area of OFL 
7 a i
60
   602  1885 cm3
360
Side S
1 cm
2 cm
4 cm
6
24
96
Volume : S 3
1
8
64
Surface area to volume
6
ratio, SA:V :
S
6:1
3:1
1.5:1
Diameter D
1cm
2cm
4cm
3.14
12.6
50.3
0.523
4.19
33.5
Surface area to volume
6
ratio, SA:V :
:1
D
6:1
3:1
1.5:1
Diameter D x Length L
1cm × 1 cm
1cm × 2 cm
1cm × 4 cm
Surface area :
4.71
7.85
14.1
D
Volume :    L
2
0.78
1.57
3.14
Surface area to volume
2D  4L
ratio, SA:V :
:1
DL
6:1
5:1
4.5:1
Base side (b) x Length
(L)
1cm × 1 cm
1cm × 2 cm
1cm × 4 cm
Surface area : 2b2  4bL
6
10
18
Volume : b2L
1
2
4
Surface area to volume
2b  4L
:1
ratio, SA:V :
bL
6:1
5:1
4.5:1
Surface area : 6S
ii
2
2
D
Surface area : 4  
2
Volume :
iii
4

3
3
D
 
2
2
D
2     DL
2
2
iv
b Cells that are small have better SA:V ratio no matter the shape. However, if the cell has to
be larger then it’s better to have the shape of a cylinder or a prism rather than cube or
sphere, and to grow in length rather than in width.
8 a The cube has volume Vc  6  8  10  480 cm3 . The cutout has volume
Vcut    22  8  32 cm3. Therefore, the volume of the bolt is
Vb  Vc  Vcut  480  32  379 cm3 (3 s.f.)
© Oxford University Press 2019
8
Worked solutions
b The surface area of the whole cube is Ac  2  6  8  2  8  10  2  6  10  376 cm2. The cutout circles each have area Acut  2  22  8 cm2. The curved section of the cylinder has area
Acyl  2  2  10  32 cm2 . The total surface area is
A  Ac  Acyl  Acut  376  32  8  451cm2 (3 s.f.)
9 Building A has volume VA  8  10  7.5  600m3. Building B has volume
VB  8  22  11  1936m3. The overlap region has volume Vc  8  52  200m3. Therefore, the
composite building has volume V  VA  VB  Vc  600  1936  200  2336m3.
10 Let M denote the midpoint of AC . Then the line MB has length LMB  62  22  32 (using
Pythagoras’ theorem). Then calculate the length of MD as LMD 
ADC is therefore
L2MB  82  96 . The area of
1
 4  96 19.6 cm2 .
2
11 Each pentagon of the dodecahedron is made up of 5 equal triangles, which has one side of
360
length 2cm, and two sides of length x which are either side of a central angle
 72. Use
5
1
trigonometry to determine x 
 1.70 cm.
sin36
1
 1.702  sin72  1.376 cm2 . Each
2
pentagon is made of five of these triangles and there are 12 pentagons; the total surface area
is A  5  12  At  82.6 cm2.
The area of one of these triangles is therefore At 
12 The large cube has volume Vc  63  216 cm3 . The pyramid to be chopped off has volume
1
1
9
 IF  JF  FK   33  cm3. Therefore the remaining volume is
6
6
2
27
V  216 
 211.5 cm3.
2
Vp 
The surface area of the large cube (before removal) is Ac  6  62  216 cm2. The areas of faces
FJK, IFK and IJF are the same and equal to Af 
1
 32  4.5 cm2 . We have to add on the area
2
of IJK, which is an equilateral triangle of side l  32  32  18 . The area of IJK is therefore
1
Acorner   18  18  sin60  7.79 cm2. Therefore, the remaining surface area after pyramid
2
removed is A  Ac  3Af  Acorner  210 cm2 (3 s.f.).
13 a Attempt to use cosine rule:
cos A 
2
2
2
195  170  210
2  195  170
M1
 0.3443
A1
A1
A  69.9
1
b Area   170  195  sin69.9
2
 15600 m2
2
2
M1A1
A1
2
14 AC  560  1200  2  560  1200  cos110
M1A1
AC  1487.7 m
A1
Distance required is 560  1200  1487.7  272 m
15 Grazed area 
70
   852  4410 m2
360
Area of triangle 


1
 85  300  sin70  11981 m2
2

M1A1
M1A1

© Oxford University Press 2019
M1A1
9
Worked solutions
Ungrazed area 
16 Uses

360


360
   162 or
   162 
16

360

360
   82
M1A1
A1
   82  200
M1A1

2
360
1
70
 85  300  sin70 
   852  7570 m2.
2
360
 82  200
  375
 
17 a
375

M1A1
 119
BD2  92  162  2  9  16  cos33
(  BD  9.771 cm)
M1A1
AD  9.7712  72  6.816 cm
So perimeter  7  9  16  6.816  38.8 cm
1
 1

b Total area    6.816  7     9  16  sin33 
2
2

 

18 a
b
 63.1 cm2
sin A sin67

5.7
6.9
M1A1
A1
M1A1
A1
M1A1
 A  49.50
A1
B  180  67  49.50  63.5
1
Area   6.9  5.7  sin63.5  17.6 cm2
2
A1
ˆ  40
19 ABD
M1A1
A1
BD
5.6

sin115
sin 40
M1A1
 BD  7.90 cm
A1
AB
5.6

sin25
sin 40
M1
 AB  3.68 cm
A1
Total length required  2  3.68  5.6  7.90  26.5 cm
A1
20 a Volume 

2
 43
3

272
3
 
2 3 1 2
r  r h
3
3
1
128 144

  42  9 
3
3
3
M1A1
 
A1
A1
b Surface area hemisphere  2 r 2  2  42  32
Surface area cone   rl    4  9  4  4 97
2
Total surface area  32  4 97  224
2
cm2
© Oxford University Press 2019
M1A1
M1A1
A1
10
Worked solutions
21 a
A1A1
1

b Area of one hexagonal face  6   4  4  sin60   24 3
2


M1A1
Therefore volume  24 3  12  288 3  499 cm3
A1
c Total surface area  2  24 3  6  12  4  48 3  288  371
cm2
ˆ  138  Angle AXB
ˆ  15
22 a Angle ABX
AX
115

sin138
sin15
 AX  297.312
AX  297 m
b
M1A1
A1
M1A1
A1
297 sin27  135 m
M1A1
23 a If X is the foot of the vertical line from E , then letting AX  x , you have
x2  x2  162
M1
x2  128
x 8 2
cos  
A1
8 2
20
M1
  55.6
A1
2
2
b Let Y be the mid-point of BC . Then EY  20  8  336
8
cos  
336
  64.1
M1A1
M1
A1
ˆ  180  90  55.6  34.4
c From part a, AEX
ˆ  2  34.4  68.8
Therefore AEC
2
Ah 16  20 sin55.6

 1410 cm3
d V 
3
3
1

e A  4   16  202  82   162
2


 843 cm2
M1
A1
M1A1
M1A1
A1
© Oxford University Press 2019
11
Worked solutions
Representing and describing data:
descriptive statistics
3
Skills check
1 a  = 1 child
0

1

2

3

4

b
c
2 Mean = 7, Median = 8, Mode = 9, Range = 6.
3 Major gridlines correspond to 1cm.
Exercise 3A
1 a Discrete
2
3
b Continuous
c Discrete
d Discrete
e Discrete
f
Continuous
g Continuous
h Discrete
i
j
Continuous
Continuous
Number of sweets
Frequency
21
4
22
6
23
5
24
5
25
4
26
1
Height, in metres, h
Frequency
2≤h<3
6
3≤h<4
6
4≤h<5
5
5≤h<6
3
© Oxford University Press 2019
1
Worked solutions
4
Weight, in kilograms, w
Frequency
0 ≤ w < 10
11
10 ≤ w < 20
8
20 ≤ w < 30
3
30 ≤ w < 40
3
Exercise 3B
1 a
Mean = 10.1
Median = 9.0
The most appropriate measures for this case are the mean
and the median, because the data is continuous.
Mode = 8.6
b
Mean = 8.64
Median = 8.5
Mode does not
exist
c
Mean =32.62
Median = 30
The most appropriate measures for this case are both mean and
median. This is an example of a continuous data set where mode does
not exist.
The most appropriate measure for this case is both mean and median
as mode is clearly too small.
Mode = 15
2 Find the modal class by determining which of the modal classes has the highest frequency. To
calculate the mean and median, use mid values of the class intervals.
a
The modal class: 150 ≤ n < 180
an approximation for the mean: 112
an approximation for the median: 105
b
The modal class: 50 ≤ s < 55
an approximation for the mean: 54.4
an approximation for the median: 52.5
c
The modal class: 7 ≤ t < 8
an approximation for the mean: 5.86
an approximation for the median: 5.5
As it can be seen from the frequency table, the
data in this case is skewed, which is also
indicated by the modal class being at the
highest end of the range. In this situation, an
approximation for the median is the most
appropriate measure of the central tendency as
it is less affected by the skewed values.
The data set is well centred with all three
measures agreeing well. Hence, the best
measure of the central tendency in this case is
the approximate mean which minimises the
error for the guess of the next value.
The data does not have clear tendency. Best
measure to use is the approximation for the
median because the modal class is very high.
Exercise 3C
1 a Mean = 6.1, median = 5.2, mode = 7.5.
Possible outliers of the data set are 17.8 and 25. Excluding them from the calculation of
mean and median gives the following values: mean = 4.7, median = 4.8, which are closer
together than when the outliers are included.
b Mean = 3.5, median = 3.6, mode = 2.5.
Possible outlier is 6.1 as it is further away from the rest of the data points. Excluding 6.1
from the data set changes the mean to 3.4 but leaves median unchanged. The mean is now
further from the median. Hence, 6.1 is not an outlier.
c Mean = 65, median = 62, mode = 62.
Possible outlier is 22 as it is further away from the rest of the data points. Excluding 22 from
the data set changes the mean to 67 but leaves the median unchanged. Hence, 22 is not an
outlier.
© Oxford University Press 2019
2
Worked solutions
Exercise 3D
1 a Discrete
b It is useful to produce a frequency table of the data.
Number of daisies
Frequency
2
1
3
1
4
1
5
1
6
2
8
2
9
1
11
1
12
3
13
2
15
5
16
2
17
1
18
1
21
1
22
1
24
1
25
1
26
1
34
1
Mean = 13.9
Mode = 15
Median = 14
Since this is a discrete data set, mode is
more appropriate measure of the central
tendency, although all three measures
agree well in this case.
c Find standard deviation using the formula σ n 
2
1 n
  xi  x  :  n  7.3 .
n 
i 1
Since the standard deviation shows how the data values are related to the mean, the result
indicates that the data points are quite spread out.
d Range = 34 – 2 = 32.
Interquartile range:
Q1 is the
Q3 is the
30  1
 8th number (a quarter of the data points are below this number): Q1 = 8
4
3 30  1
4
 23th number (three quarters of the data points are below this number):
Q3 = 17
IQR = 17-8 = 9.
2 a Modal class: 30 ≤ c < 40.
b Take middle values of each interval to obtain the estimates:
an estimate for the mean: 34
an estimate for the median: 35.
c 10.12
Since the standard deviation shows how the data values are related to the mean, the result
indicates that the data points are very spread out.
3 60  1
60  1
 46th
 15th number, Q1 = 25, Q3 is the
4
4
number, Q3 = 35. IQR = 35 – 25 = 10. They are all estimates because they use the mid
value of the class intervals.
d Variance = 102.3, Q1 is the
© Oxford University Press 2019
3
Worked solutions
3 The new mean is US $3600, the new standard deviation is US $250. When every data point is
shifted by an equal amount, the mean shifts by the same amount while standard deviation does
not change (data is spread out by the same amount).
4 a Mean = 8.9, median = 10, mode = 12. Since mode is quite a bit higher than the other two
measures, the mean and the median are the most appropriate to use.
b Standard deviation = 4.10 which shows that the data points have medium spread in relation
to the mean.
3 36  1
36  1
 28th
 9th number, Q1 = 6, Q3 is the
4
4
number, Q3 = 12. IQR = 12 – 6 = 6. Range and IQR are relatively big implying that the data
is reasonably spread out.
c Range = 14 – 0 = 14. Q1 is the
5 The new mean is 60, the new standard deviation is 6. When each data point is modified by a
multiplier, both the mean and the standard deviation is modified by the same amount, i.e. both
the location and the spread of the data is changed.
6 a Modal class = 180 ≤ x < 190.
b Use mid interval values to estimate both measures: mean  180.2, standard deviation 
11.0, so the data are strongly spread out.
7 a Mrs Ginger’s new mean = 84, Mrs Ginger’s new standard deviation = 16;
Mr Ginger’s new mean = 80, Mr Ginger’s new standard deviation = 20;
Miss Ginger’s new mean = 76, Miss Ginger’s new standard deviation = 24.
See questions 3 & 5 for explanations.
b New grades:
Mrs Ginger
Mr Ginger
Miss Ginger
y
44
30
16
Zoe
70
62.5
55
Ans
92
90
88
8 a Basketball players: mean = 200.8, standard deviation = 10.5; males: mean = 172.3,
standard deviation = 10.3.
b Basketball players are much taller on average, however both samples have nearly the same
standard deviation, so the spread of the data is nearly the same.
9 a Males: mean = $2546.30, standard deviation = $729.78; Females: mean = $2114.58,
standard deviation = $635.25.
b Male salaries have higher mean and higher spread than female salaries.
Exercise 3E
1 a Mean height of the whole school = 155.4.
b There are several ways to sample this data. For example, use stratified sampling technique
taking an appropriate number of students from each grade: there are 160 students in total,
so if a sample of 50 students should be chosen, take 9,9,8,8,8,8 students from grades
7,8,9,10,11,12 respectively. The final answer will differ depending on the random numbers.
This method will be unbiased.
2 a Mean age of the 100 people = 25.92 which is well below 60 and the manager should not
lose much revenue.
b Use a random number generator to obtain 35 numbers between 1 and 100 and average the
ages represented by those numbers. The answers will differ.
c The result depends on the starting point. Starting with the first number the mean = 26.21.
d Usually the systematic sample mean will be closer to the population mean than the random
sample mean. However, that depends heavily on the set of the random numbers used.
3 a Mean number of goals scored in all 50 matches = 3.58.
b Generate a set of 24 random numbers between 1 and 50. Calculate the mean.
c The estimated mean should be reasonably close to the actual mean but the answers can
differ.
© Oxford University Press 2019
4
Worked solutions
Exercise 3F
1 Theo’s data:
Number
Frequency
1
8
2
8
3
6
4
7
5
5
6
6
Millie’s data:
Number
1
Frequency
7
2
7
3
8
4
5
5
7
6
6
Comment: the distribution of the numbers are similar for both Theo and Millie and the
frequencies for each number thrown are very similar.
2 Female data:
Number of goals
Frequency
0
4
1
5
2
5
3
5
4
3
5
2
6
1
Male data:
Number of goals
Frequency
0
3
1
4
2
5
3
3
4
3
5
2
6
3
7
1
8
1
The range of the number of goals is bigger for males. The female data is more uniform than
male data and more females scored no goals.
© Oxford University Press 2019
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Worked solutions
3 Frequency table and histogram:
a,b
Height, in centimetres,
h
Frequency
140<h≤145
2
145<h≤150
17
150<h≤155
9
155<h≤160
2
160<h≤165
2
c
d
Use the box-and-whisker plot to determine whether
the data is symmetric or not. From the box and
whisker diagram, the data is not symmetric. When
outliers are excluded, the data is symmetric.
4 a Histogram:
bTime in minutes:
To find approximate values of the mean, median, LQ
and UQ, use the midpoints of the given groups.
These are approximations only because the original
data have not been given.
Mean = 42
Median = 35
LQ = 35
To find median, identify the midpoint of the interval
which contains the data point above the 100th point.
UQ = 45
Range = 70
Outliers: data points in 60≤x≤90.
To find LQ, identify the midpoint of the interval
which contains the data point above the 50th point.
To find UQ, identify the midpoint of the interval
which contains the data point above the 150th point.
Data points are spread out between 20 and 90
minutes, hence the approximate range is 90-20=70
mins.
To determine the outliers, calculate IQR = UQ-LQ =
45-35 = 10. Then, outliers are the data points below
LQ – 1.5  . IQR =20 and above UQ + 1.5  . IQR
=60.
© Oxford University Press 2019
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Worked solutions
c Outliers are omitted from the box-and-whiskers graph. Note that median and LQ are equal.
d Marcus did worse than 75% of participants, so he may not be satisfied.
5 Boys and girls
a
Boys:
Mean = 6
Median = 6
LQ = 5
UQ = 7
To find the means, multiply the score by its frequency, add up and divide
by the number of girls or boys.
To find the median, find what score the 26th data point in each of the
sets represents.
Range = 7
To find the LQ, find what score the 13th data point in each of the sets
represents.
Girls:
To find the UQ, find what score the 38th data point in each of the sets
represents.
Mean = 6
Median = 6
LQ = 5
UQ = 7
Range = 8
Note, range is the only different parameter between the girls and boys
so far.
To find the outliers, calculate IQR = UQ − LQ = 7 – 5 = 2. Then, outliers
are the data points below LQ – 1.5  . IQR = 2 and above UQ + 1.5  .
IQR = 10. Hence, there are no outliers.
There are no
outliers.
b Plots are almost identical apart from the minimum value being different (boys have higher
minimum value).
c Whilst the boys data is not as symmetrical as that of the girls, both are clearly quite
symmetrical.
6 a
Time in minutes:
Mean = 17
Median = 16
LQ = 12
UQ = 21
Range = 37
Order the data points from the smallest to the largest to find the
quartiles. Median is the 18th data point, LQ is the 10th data point and
UQ is the 28th data point.
Range = 43-6 = 37
To determine the outliers, calculate IQR = UQ-LQ = 21-12=9. Then,
outliers are the data points below LQ – 1.5  . IQR =-1.5 and above
UQ + 1.5  . IQR =34.5. Hence, the outliers are 35 and 43.
Outliers: 35, 43.
b .
c Since 16 is the median, there are 17 students who took longer to complete the puzzle.
(Note, 16 students is also a valid answer, as there are two data points with the value of 16).
7 a Median = 3.
© Oxford University Press 2019
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Worked solutions
b IQR = 4 − 2 = 2.
c The data is almost symmetrical: the data between UQ and the maximum value are slightly
more spread out than between the LQ and the minimum value.
8
Boys
Girls
a Median = 55
a Median = 65
b IQR = 75-45 = 30
b IQR = 80-50 = 30
c 25% of boys scored between 45 and 55
d because
50% of the
50%
girls
of the
scored
boys
between
scored 65
below
and 95
because 100% of the girls scored below 95
e 15 boys scored below 45 (25% of 60).
f
45 girls scored above 50 (75% of 60).
g Neither of the data sets are perfectly symmetrical. Girls data are symmetrical in the
interquartile range, but more spread out between the UQ and the maximum value than
between the LQ and the minimum value. Boys data are more spread out between the
median and the maximum value than between the median and the minimum value.
9 a Median = 120.
b Range = 150 − 80 = 70.
c 10 pandas weigh less than 90kg (LQ = 90kg, there are 25% of pandas that weigh less than
90 kg and 25% of 40 is 10).
d 50% of pandas weigh between 120 and 160 kg (all the pandas above the median).
e 20 pandas weigh between 90 and 130 kg (LQ = 90, UQ = 130, there are 50% of pandas
between these values and 50% of 40 is 20).
f
Since the average weight corresponds with the median of the sample, we can deduce that
the distribution of the weight of the pandas is skew with respect to the average weight.
g Pandas in the sample must be mostly males because the box plot overlaps more with the
range of male pandas weights.
Exercise 3G
1 a
Upper boundary
Cumulative frequency
2
5
4
16
6
39
8
70
10
89
12
97
14
100
b
c
Median ≈ 6.75
IQR ≈ 3.5
To determine median, LQ and UQ, read off the time for cumulative
frequency 50, 25 and 75 respectively. IQR = UQ – LQ = 8.5 − 5 = 3.5
mins.
© Oxford University Press 2019
8
55 (median) and
(maximum val
Worked solutions
(In minutes)
d
10th percentile = 3
minutes
e
5
people
2 a
It is possible to use either the frequency table and the midpoints of
the intervals or the cumulative frequencies graph to estimate the
value of 10th percentile.
Use the graph to read off that 11 minutes corresponds to 95th percentile.
Upper boundary
Cumulative frequency
4
5
8
37
12
78
16
106
20
128
24
140
28
147
32
150
b
c
Median ≈ 12
IQR ≈ 10
To determine median, LQ and UQ, read off the number of words for
cumulative frequencies 80, 40 and 120 respectively.
IQR = UQ – LQ = 18 – 8 = 10 words.
d
e
There are no outliers.
22
Outliers would be found below LQ – 1.5  IQR = 8 – 15 = −7
and above UQ + 1.5  IQR = 18 + 15 = 33.
90th percentile corresponds to 160  0.9 = 135 cumulative
frequency data point. Read off the number of pages from the
graph.
f
g Adult books because the mean 15 falls inside the interquartile range while the mean 8 does
not.
3 a Cumulative frequency graph.
© Oxford University Press 2019
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Worked solutions
b
Median ≈ 450
IQR ≈ 205
c
To determine median, LQ and UQ, read off the number of visitors for
cumulative frequencies 175, 88 and 263 respectively. IQR = UQ – LQ =
555 – 350 = 205 visitors.
There are no outliers.
Outliers would be found below LQ – 1.5  IQR = 43 and above
UQ + 1.5  IQR = 863.
d
e
90 days.
Read off the cumulative frequency for 350 visitors.
4 a
Both, the cumulative frequency and
box-and-whiskers graphs are useful for
analysing this situation.
Use cumulative frequency graph to
determine the median, LQ and UQ for
both boys and girls.
Boys:
Median=32, LQ=29, UQ= 37.
Girls:
Median=33, LQ=31, UQ=37.
Assume 21 as the
lowest and 45 as
the highest
scores for both
girls and boys.
b Both graphs suggest fairly similar results. From the cumulative frequency graph it is seen
that more boys than girls acquired the lower range scores. From the box-and-whiskers
graph it is seen that both data sets are not perfectly symmetric. It is seen that the
interquartile range of the boys’ scores is wider than the girls’ scores. The median score of
the girls is just above the median score of the boys.
© Oxford University Press 2019
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Worked solutions
c Martin’s score is the lower quartile for the boys, so Martin did better than 25% of the boys.
however, Mary’s score is below the lower quartile for the girls, so more than 75% of girls did
better than Mary.
5 a
Median ≈ 95
IQR ≈ 60
To determine median, LQ and UQ, read off data points for
cumulative frequencies 170, 85 and 255 respectively.
90th percentile ≈ 150.
IQR = UQ – LQ = 130 – 70 = 60.
0.9*340 = 306.
b
6 a
There are no outliers.
Outliers would be found below LQ – 1.5  IQR = −20 and
above UQ + 1.5  IQR = 220.
Median ≈ 300
To determine median, LQ and UQ, read off data points for
cumulative frequencies 50, 25 and 75 respectively.
LQ ≈ 225
UQ ≈ 380
b
c
Length, l, in cm
Cumulative frequency
50 ≤ l < 100
5
100 ≤ l < 150
5
150 ≤ l < 200
10
200 ≤ l < 250
12
250 ≤ l < 300
18
300 ≤ l < 350
15
350 ≤ l < 400
13
400 ≤ l < 450
8
450 ≤ l < 500
4
500 ≤ l < 550
4
550 ≤ l < 600
3
600 ≤ l < 650
2
650 ≤ l ≤ 700
1
d Mean ≈ 312, Standard deviation ≈ 132. Estimates found by using the midpoint of each
interval.
Exercise 3H
1 a Negative, moderate.
b Positive, strong.
c Negative, strong.
d Positive, weak.
e No positive or negative correlation.
f
No positive or negative correlation (but could split into intervals of strong positive and
strong negative correlations).
2 a
© Oxford University Press 2019
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Worked solutions
b There is a strong positive correlation between the heights and the weights of the football
players. Hence, the taller the football player, the heavier he or she is expected to be.
c The correlation can indicate causation in this case because taller people have more body
mass.
3 a
b There is a weak positive correlation between the size of the laptop screen and the cost of
the laptop.
c Since the correlation is weak, the size of the screen has little influence on the cost of the
laptop. This can be expected as there are many other factors that contribute towards the
price, for example brand, type of the processor, whether or not the screen is touch
sensitive.
4 a
b There is no visible positive or negative correlation.
c The grade for the English test does not influence the grade for the Mathematics test and vice
versa.
5 a
b There is strong negative correlation between the team’s position in the league and the goals
scored.
c Position in the league can be influenced by the goals scored.
© Oxford University Press 2019
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Worked solutions
Chapter review
1 a Discrete
Number of apples
Frequency
7
4
8
4
9
4
10
3
11
1
12
1
b Continuous. Group the data into convenient intervals to produce a frequency table.
Lengths, in cm, l
Frequency
7≤l≤9
5
9 ≤ l ≤ 11
2
11 ≤ l ≤ 13
2
13 ≤ l ≤ 15
4
c Discrete
Size
Frequency
33
1
34
2
35
4
36
3
37
4
38
3
2 a Mean = 54.8, median = 56, mode = 32, median and mean are best because this is
continuous data.
b Mean = 122.25, median = 87.5, mode = 62. Data is very skewed hence it is best to use
median in this case.
c Mean = 6.42, median = 6, mode = 6. All measures agree well hence it is fine to use any of
them.
3 a 50 ≤ l < 60 since frequency is highest for this class.
b Use mid values of the intervals to estimate median  55, mean  54.42, standard deviation 
11.28.
c Histogram:
4 The new mean is 76, the new standard deviation is 14. When each data point is modified by a
multiplier, both the mean and the standard deviation is modified by the same amount, i.e. both
the location and the spread of the data is changed.
5 The new mean is 13, the new standard deviation is 1. When every data point is shifted by an
equal amount, the mean shifts by the same amount while standard deviation does not change
(data is spread out by the same amount).
6 a Mean = 32.8, standard deviation = 7.51 which indicates that the data points are reasonably
spread out.
© Oxford University Press 2019
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Worked solutions
b Range = 45 – 20 = 25, IQR = Q3 − Q1 = 39 – 27 = 12.
c Min = 20, LQ = 27, Median = 34, UQ = 39, max = 45. Then, outliers are the data points
below LQ – 1.5  IQR = 9 and above UQ + 1.5  IQR = 57. Min and max are within this
range, so there are no outliers.
d
7 a Cumulative frequency table:
Upper boundary
Cumulative
frequency
10
8
20
24
30
65
40
119
50
155
60
177
70
194
80
200
b
c Read off cumulative frequency 50, 100 and 150 for LQ = 27, Median = 36 and UQ = 48
respectively to give IQR = 21.
d
8 a
b Strong and positive correlation
c It is unlikely that weather is the cause of the increase/decrease of the number of eggs.
9 a 40  x  60
b Use of mid-point
correct method
48.3%
A1
M1
M1
A1
© Oxford University Press 2019
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Worked solutions
c 22.6
M1A1
d The second class had slightly lower marks on average.
R1
Their standard deviation was also lower, meaning their marks were more consistent, or
less spread out than the first class.
R1
398
10 a
M1A1
 15.92 g
25
b 13th mouse has weight of 16 g
M1
So median is 16 g
A1
c
d
Q1  14
A1
Q3  19
A1
Interquartile range is Q3  Q1  19  14 = 5
M1A1
Q3  1.5  IQ range  19  1.5  5  26.5 g
M1A1
11 a Using GDC, mean  23.58 C
M1A1
b Using GDC, SD  3.38 C
M1A1
c Using GDC, mean  22.83 C
M1A1
d Using GDC, SD  5.52 C
M1A1


e Tenerife has a higher mean temperature 23.58 C , so on average, temperatures could
be said to be higher in Tenerife.


R1R1
The standard deviation of Tenerife temperatures 3.38 C is lower than that in Malta,
therefore the temperatures can also be said to be more consistent.
12 a 5.25  3  8.25 years
b 1.2 years
13 Total population is 65.12 million.
54.8
Number required from England:
 5000  4208
65.12
3.10
Number required from Wales:
 5000  238
65.12
5.37
Number required from Scotland:
 5000  412
65.12
1.85
Number required from Northern Ireland:
 5000  142
65.12
14 a 45  15  30
b 37  22  15
c 75%
d Interquartile range is 15
Q3  1.5  IQ range  37  1.5  15  59.5 minutes
15 a
b
c
d
e
Yes, it would count as an outlier
69
50
83
170
Interquartile range is 83  50  33
Q3  1.5  IQ range  83  1.5  33  132.5
R1R1
M1A1
M1A1
A1
M1A1
A1
A1
A1
M1A1
M1A1
A1
M1A1
A1
M1A1
M1A1
M1A1
M1A1
A1
M1
Q1  1.5  IQ range  50  1.5  33 = 0.5
M1
Therefore 180 is an outlier
A1
© Oxford University Press 2019
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Worked solutions
f
16 a
b
c
M1A1A1
A1
M1A1
45  x  60
45  x  60
M1A1A1
d
e From graph, 37.5 on y-axis gives median of 55
Interquartile range is approximately 70  40  30
17 a Using GDC, mean = $ 43 600
b Median =
th
1
11  1 value  6th value, which is $ 25 000
2
1
11  1  3rd value, which is $ 25 000
4
th
3
UQ = 11  1  9th value, which is $ 80 000
4
IQR = 80 000  25 000  55 000
M1A1
M1A1
M1A1
M1A1
M1A1
c LQ =
M1
A1
d Analyst 8 would use the mean average
R1
in order to suggest that their salary of $ 25 000 was significantly below the ‘average’ of
$ 43 600 .
R1
e The managing director would use the median (or mode)
R1
and say that Analyst 8 was already earning the ‘average’ salary of $ 25 000.
R1
f Either: In this case, the median would be fairest
R1
as there is no one earning a salary close to the mean value of $ 43 600 , and the majority
of workers earn the median salary.
R1
18 a A random sample is a subset of a population where each member of the subset has an
equal probability of being chosen
R1
© Oxford University Press 2019
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Worked solutions
A stratified sample is where the population is divided into strata, based on shared
characteristics.
R1
Random samples of each strata are chosen in the same proportion as the strata found in
the population.
R1
A systematic sample is where members are chosen from a random starting point and a
fixed periodic interval.
R1
The interval is the population size divided by the sample size.
R1
b i Population size is unknowable, so specific periodic choice of interval is difficult to
determine.
R1R1
ii Random sampling.
R1
It is difficult to split rats into ‘strata’ as they all tend to look (and behave) the same.
R1
19 a
M1A1A1
Mean height  35.8 cm
M1A1
Variance  63.9 cm2
M1A1
SD  7.99 cm
M1A1
On average, the neighbour’s garden’s flowers had a slightly lower height compared to
Eve’s.
R1
However, their standard deviation was smaller, indicating they were grown to a more
consistent length.
R1R1
611
20 a
M1A1
 718824  366001
1200
b Geographical location of residents; leisure / work travellers; age (anything appropriate
or connected to travel)
R1R1
b
c
d
e
© Oxford University Press 2019
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Worked solutions
Dividing up sapce: coordinate
geometry, lines, Voronoi diagrams
4
Skills check
1 a
x 
7
2
b
x 
b

5
3
2
3 a 8
1
2
4 8cm
Exercise 4A
1 a
b
1


 2  2 3  3.5 
M
,
  M 1.71, 1.92  (3 s.f.)
2
 2



 1  7 5  8.5 4.2  11 
M
,
,
  M 3,6.75, 7.6 
2
2
 2

3
 2  2 1  4 

,
2 Midpoint of A  2,1 and B 2, 4 is M 
  M  0,  
2
2
2




 3  1 4  4 
,
3 Midpoint of A 1, 4 and B  3, 4 is M1 
  M1  1,0 . C is the midpoint of A and M1 ,
2 
 2
1  1 4  0 
,
i.e. C 
  C 0,2
2 
 2
4 a
c
B 6,7,0
b
A 6,7, 4
6  0 7  0 4  0
 7 
M
,
,
  M  3, ,2 
2
2 
 2
 2 
Exercise 4B
1 a
AB 
 4  1.5
b
AB 
1  4
2
2
  3  5  5.52  82  9.71 (3 s.f.)
2
  2  0  10  3  52  22  72  8.83 (3 s.f.)
2
2
2 By Pythagoras’ theorem, PQ  2 
 x  3
2
 7  5 
2
 x  3
2
 4. By squaring both sides of
this expression, we find  x  3  4  4 , so  x  3  0,i.e. x  3.
2
2
© Oxford University Press 2019
1
Worked solutions
3 We calculate the lengths (and leave as roots for now): AB 
BC 
 6  8
 2  4  232, AC 
2
2
 4  8
2
 4  6
2
 6  2  116.
2
 6  4  116. Therefore, as
2
AB2  AC 2  BC 2 , this triangle is a right angles triangle.
4 The diameter of the circle is AB 
1  2
2
 7  1  9  36  45. The circumference is
2
therefore C    AB  21.1(3 s.f.)
5 x = 4 + (4 – 1) = 7 and y =
63
 4.5
2
6 a Distance between the aircraft is
20  26
2
 25  31  11  12  62  62  12  73  8.54 km (3 s.f.)
2
2
b The radar will be able to detect both aircraft if they are both a distance less than 40km from
the station at O  0,0,0 . Using Pythagoras’ theorem, we calculate the aircraft to be
distances
d1  202  252  112  33.9 , and d2  262  312  122  42.2. Therefore, the radar will be
able to detect one, but not both, of the aircraft.
Exercise 4C
1 a A has the form A  x, 8 ,where x is any number.
b A has the form A  3, y  ,where y is any number.
2 a
M 
34
1

2   1 3
3 Stair 1 gradient =
b
M
27
5

11  2 13
c
M 
2.2   0.3 2.5

 1
1
2.5
1.3 — 1
5
12
10
15
 0.5, stair 3 gradient =
 0.48, stair 2 gradient =
 0.833
20
25
18
a Stair 3 has the greatest gradient
b Stair 1 has the least gradient.
4 See solutions
5 a
M
1  2 3

 0.75
1  3 4
b
M 
2.5  5
2.5

 7.5
1
 1
0   
3
 3
1
units)
3
1
over a single unit distance travelled horizontally, compared to the slope with gradient  .
3
6 The slope with gradient −3 is steeper, as the skier changes height more (3 units vs
7 Using the gradient formula, M  0.35 
h  10
h  10
. Re-arranging, we find that

490  90
400
h  10  4000.35 150 m.
Exercise 4D
1 The point A has co-ordinates A 30,2.5 , so the gradient of the ramp is M 
2.5  0
 0.0833 (3
30  0
s.f.). Therefore, the ramp does not conform to safety regulations.
2 a,b
See solutions
c Students should make a decision on the basis of the number of days they plan to work. If
they plan to work fewer days than 6 then they should take the city guide job. If they plan to
work longer than 6 days, the flower shop will pay more and they should take that job.
3 a The roof run is half the width, i.e 3.5 m.
© Oxford University Press 2019
2
Worked solutions
b The roof gradient is M 
rise 1.6

 0.457 (3 s.f.)
run 3.5
c The roof does not satisfy the requirements, as the gradient is steeper (larger) than the
maximum specified by the regulations (0.457>0.17).
4 a The road has gradient M 
5
 0.25, so “25%” would be written on the sign.
20
b The road with a sign indicating 15% is the steeper road.
c A 15% sign corresponds to a gradient of M  0.15. In the road has horizontal change (run)
of 5km, then the rise is M 5  0.75 km.
5 a i
ii
M
280
 0.622 (3 s.f.), so P1 is classified as black (62.2% incline).
450
M
50
 0.333 (3 s.f.), so P2 is classified as red (33.3% incline).
150
iii M 
48  0
 0.155 (3 s.f.), so P3 is classified as green (15.5% incline).
310  0
iv M 
56  0
 0.244 (3 s.f.), so P4 is classified as blue (24.4% incline).
230  0
b
c Use the trigonometric ratio: tan   
opposite 50
1
1

 . So   tan1    18.4 (3 s.f.).
adjacent 150 3
3
Exercise 4E
1 a
y  4  3  x  1
b
y  4  5  x  7
d This line has gradient M 
2 a
c
y 3  
1
 x  1
2
48
 2. Therefore, the line has equation y  4  2  x  1 .
1  1
b
c
© Oxford University Press 2019
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Worked solutions
3 a
4
4
y 5
 y  4x  13
x 2
b
2 y 2
2x

y 
4
7 x 7
7
c The gradient is M 
10
1
1
y
1
2
, so the line has equation


 y  x 
3  2 5
5 x 2
5
5
d The gradient is M 
2  0 2
2
y
2
4

y  x
 , so the line has equation
3 x 2
3
3
1  2 3
A  4, 5 has x 4, y 5. Therefore, as – x  9  4  9 5  y, A does sit on the line
y x  9. However, 5x  4y  54  45  9, so A does not sit on the line 5x  4y  9.
5 a
1
y  6  4  3  4  7
2
b From the equation of the line, y  9 
1
x  4. Hence, x  2 9  4 10
2
Exercise 4F
1 a
y  3x  2
b This line has equation M  4 
y 5
, which is re-arranged to gradient-intercept form
x 1
y 4x  1.
3  1 2
2 y  1
,

. Therefore, the equation of the line is

2 1
3
3 x 1
2
5
which is re-arranged to gradient intercept form: y 
x .
3
3
c The gradient of the line is M 
2 a
b
x-intercept  0,0 , y-intercept = (0,0).
y-intercept = 0,3 . Re-arranging the equation of the line gives x  
x-intercept is (
5
 y  3 , so the
2
15
,0) .
2
c When x  0, then y  1; the y-intercept is 0,1 . When
y  0, then -0.2 x  1,so the x-intercept is  5,0 .
3 a
b i
Setting x  0in the line equation gives y  1 1 4 4, so the y-intercept is 0, 3 .
Setting y  0 gives 1  1  x  4 , so the x-intercept is  3,0 .
ii We write (i) as y  1  x  4 1 and equate with (ii): 0.5x  3 1  x  4 1  x  3 . This
re-arranges to give 1.5 x 6,with solution x 4. The corresponding
y co-ordinate is y  0.5  4  3  1; the intersection point is  4,1 .
© Oxford University Press 2019
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Worked solutions
c The y-intercept of (iii) is 5 (by solving y  5  0). The line (iv) has equation y  3  2  x  1 ,
which has y-intercept at y  3  2  5 , i.e. the y-intercept is  0,5 therefore they both have
the same intercept.
4 a
212  32 9
 . Therefore, the line has
100  0
5
9 F  32
5
160
equation
.

, which may be re-arranged to give C  F 
5
C
9
9
b The gradient of the line between the points is M 
c The gradient of the equation in b is
5
9
d The gradient is the amount that the temperature changes, measured in C when the
temperature measured in F changes by 1.
e The y-intercept is at F 
f
160
9
The temperature in °C at 0°F.
g Using the expression in 4ii) with F  83, we find C =
5
160 85
83 

 28.3C (3 s.f.)
9
9
3
h Using the expression in 4ii) with C  10, we find -10 =
5
160
F
. This re-arranges to give
9
9
9 
160 
F    10 
  14.
5 
9 
i
160
160 5
to both sides of the expression in 4ii) gives C 
 F . Then multiplying both
9
9 9
9
9
sides by
gives the expression F  C  32 .
5
5
Adding
5 We see that L1 has a gradient M 
equation y 
rise 1
 . This line goes through O  0,0 and therefore has
run 2
1
x. L2: x 1 L3: y  5 .
2
L4 This line passes through
A 0, 4 and B  4,0 ,so has gradient M 
4
 1. The equation of the line is
4
y  0  1  x  4 ,i.e. y  x  4.
6 a
A  30  0.02x
b The gradient is 0.02, which represents the amount Maria ears (in USD) per dollar spent in
the restaurant.
c The y-intercept is 30. It represents the amount Maria would earn in a day if no food was
sold.
© Oxford University Press 2019
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Worked solutions
d
e Using the formula in 6i) with x  2400, we find A  30  0.022400 78 USD.
Exercise 4G
1 a −1
b Gradient is undefined (line is parallel to the y axis)
c This line can be re-arranged to y 
1
5
x  . The gradient of any line parallel to it is therefore
3
3
1
.
3
d
2
5
2 a The line has a gradient of -3, so any line perpendicular to it has gradient M 
b By re-arranging as y 
1 1
 .
3 3
x
1
, the gradient is seen to be . Therefore, any line perpendicular to
8
8
it has gradient
M 
1
 8.
1
8
 
c The line y = −3 has a gradient of 0, so any line perpendicular to it has an undefined
gradient.
d
3 a
2
1 3
The line has gradient M  , so any line perpendicular to it has gradient M 
 .
3
M
2
L1has gradient M1 
4  7
2  0 1
 3, L2 has gradient M2 
 .Since M1  M2  1, L1and L2 are
0 1
3  3 3
perpendicular.
b The equation of L1 can be re-arranged to y 
L2 has gradient 0.25 
c
1
3
1
x  : L1has gradient M1  . As
4
2
4
1
, these lines are parallel.
4
2
. L2 can be re-arranged to y  3x  12, which has gradient M2  3.
5
Therefore, these lines are neither parallel nor-perpendicular.
L1 has gradient M1 
y
 3 has gradient M  2. These
2
y
two lines are not parallel, so must have a point of intersection. Putting x  5 in x   3
2
gives y  4 : the intersection point is  5, 4 .
4 The line x  5 has an undefined gradient and the line x 
© Oxford University Press 2019
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Worked solutions
5 Using the gradient formula M 
1 5 — 3
2


. The solution of this equation is s  4.
3
s 2
s 2
6 The line of the new street has gradient M 
so the line has equation
1 7

. The new street has a point B  1, 0.2 ,
2 2
7
 
7 y  0.2
7
37

x
(or y 
in gradient-intercept form)
2
x 1
2
10
1  0 1
30
23 1
2 1
 , MBC 
 1, MCD 
 , MDA 
 1.
3  2 5
52
0 5 5
03
Therefore, the lines between AB an CD are parallel and those between BC and DA are parallel:
the quadrilateral is a parallelogram.
7 The gradient of the sides are MAB 
Exercise 4H
1 a Bernard is at a point with x  50, y  75. Since
on the line with equation y 
1
x  100  25  100  75  y, then Bernard is
2
1
x  100.
2
1
x  100. The solution to the
2
second equality is x  340, from which we use the first equality to find y 340  410  70.
b At the point of intersection  x, y  ,we have y   x  410 
2 a The line joining
A 2,2 and B  4,6  has gradient M 
62
2  4 2  6
 2 and midpoint m 
,
  m 3, 4 . The
42
2 
 2
perpendicular bisector therefore has equation y  4 
y 
1
 x  3 or (equivalently)
2
1
11
x .
2
2
b The line through
C and D has gradient -1, so the equation of the perpendicular bisector through 2,1 is
y 1  y 1 
1
 x  2  x  1
1
3
8
3
x  ; the gradient is
. Therefore, the line
5
5
5
5
through 2, 4 intersecting 3x  5y  8  0 at a right angle has equation y  4   x  2 . Lines
3
3 The point gradient form of the line is y 
intersect at (−1, −1) and distance to hotel is
2  1
4 The gradient of the line between A and B is m 
m  
2
  4  1  34  5.83 .
2
7
, so the perpendicular bisector has gradient
6
6
 3
. The midpoint of A and B is M   5,  . Therefore, the perpendicular bisector of A and
7
 2
3
6
   x  5 . Since every point on this line is equidistant to A and B, then
2
7
we find the location of the school finding the intersection of this line with the line on which the
school sits: 7y  x  4 . By solving these equations simultaneously, one finds the point
B has equation y 
6.36,0.337 .
© Oxford University Press 2019
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Worked solutions
5 Every point on the perpendicular bisector of AB is equidistant to points A and B, and every
point on the perpendicular bisector of AC is equidistant to points A and C. Therefore, the
intersection of these bisectors is equidistant to points A, B and C.
The equation of the perpendicular bisector of line through AB has gradient
1
4 1
3 3
MAB   
 1, (as MAB 
 1). The midpoint of AB is mAB  ,  . Therefore, the
MAB
4 1
2 2
perpendicular bisector of AB has equation y 
3
3

   x    y   x  3.
2
2

Similarly, for AC: Note that A and C have the same y co-ordinate, so the perpendicular
bisector has the form
xk 
1  7
 3.
2
These two perpendicular bisectors intersect at D 3,0 , which is equidistant to A, B and C.
Exercise 4I
1 a
b
2 a and c
b Bookstore A
d Bookstore D
Exercise 4J
1 a The intersection of the perpendicular bisectors will be a vertex in the Voronoi diagram. It will
be equally distant from A, B, and C and the centre of a circle passing through the three
points. Hence the position of the solution to ‘the toxic waste problem’.
© Oxford University Press 2019
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Worked solutions
b y = −5x + 21 and y = 2.5
c (3.7, 2.5)
d 2.75km
2 a i
y = x – 10
ii y = -2x + 140
c i
0.24
ii 0.33
d i
(50,40)
ii 31.6m
b
Chapter Review
1 a For example, 2x  2y  4,y  2  x, 3x  3y  6  0
b For example, y  2x  19  0,2y  6 4  x  8 , y 2x  19
c For example, 2x  4y 
1
1
1
 0,12x  24y  2, y  x 
3
2
12
2 a At the intersection point  x, y  , we have y  
1
x  7  5x  10. Re-arranging the second
2
equality gives
9
34
1 34
17
80
x  17  x 
. Then y   
7 
 7 
2
9
2 9
9
7
b Setting y  0 in y  1   x  4 re-arranges to give x  5 : the intersection is at 5,0 .
c The equation of the second line is equivalent to (multiplying whole expression by -4):
x  4y  8. Adding this expression to the equation of the first line gives
–x  3y  x  4y  2  8  10  7y  10  y  
10
. The corresponding value of
7
16
 10 
x is x  8  4y  8  4  
 
7
 7 
3 Calculate gradients of the edges:
3 1 4
0  3 3
4  0 4
4  1 3
mML 
 , mLT 

, mTH 
 , mHM 

. This quadrilateral has two
12 3
5 1
4
2 5
3
22
4
sets of parallel edges which are perpendicular to one another, so it’s a square or a rectangle.
The lengths of the sides are ML 
1  2
2
 3  1  25  5, LT 
2
 0  3
2
 5  1  5 : two
2
perpendicular edges have equal length so this is a square.
1
21
1
21
x
intersect where  x  3  x 
 x  5. The
8
8
8
8
corresponding y value is y 2 : these lines intersect at P1 5, 2 .
4 a Lines with equations y   x  3, y 
5
1
5
1
21
x  intersect where  x  3  x   x 
. The
6
2
6
2
11
12
 21 12 
: these lines intersect at P2 
,
corresponding y value is y 
.
11
 11 11 
Lines with equations y   x  3, y 
5
1
1
21
5
1 1
21
x  ,y  x 
intersect where x   x 
 x  3.
6
2
8
8
6
2 8
8
The corresponding y value is y 3 : these lines intersect at P3  3, 3 .
Lines with equations y 
© Oxford University Press 2019
9
Worked solutions
2
2
2
2
21 
12 


 34 
 34 
b Using Pythagoras’ theorem: the length P1P2   5 
   2 
  
 
  4.37
11
11
11






 11 
(3 s.f)
The length P1P3 
5  3
2
  2  3  82  12  8.06 (3 s.f)
2
2
2
2
2
21 
12 


 54 
 45 
The length P2P3   3 
   3 
  
 
  6.39 (3 s.f)
11 
11 


 11 
 11 
c The perimeter of the triangle is the sum of the lengths of the sides:
4.37  8.06  6.39  18.8 (3 s.f. )
5 a km per hour =
b Maria walks at
rise 6  3

 0.75 (using the points 8,6 and  4,3 )
run 8  4
5
 3.125 km per hour; Maria walks faster as she covers more distance per
1.6
unit hour.
3
and passes through
4
3
B  4,3 , so the equation of the line is y  x.
4
c Petya’s line has gradient
Maria’s line has gradient 3.125 and also passes through
B,so the equation of the line is y  3.125x  9.5.
6 a Incorrect. The line has y-intercept of -3, but gradient M 
equation is y  
rise 2
1

  . The correct
run 10
5
1
x  3.
5
b Incorrect. The correct equation is y  5.
c The gradient intercept form of the equation is y 
4
2
x  . The line in the figure has
11
5
83
 5 , which is consistent with the given equation. The sketched line also
4 1
2
has a y-intercept of , so the equation 20x  55y  22 does describe the line in the figure.
5
gradient m 
7 a line has gradient of
1
1
1
 , and passes through 0,0 , so has equation y  0    x  0  y   x.
4
4
4
b Line has gradient
0.5 1
1
x 1
 , and goes through  3, 1 , so has equation y  1   x  3  y   .
3
6
6
6 2
8 a The line x  3y  0 has gradient
1
1
M   . Therefore, any line perpendicular has gradient 
 3. The specific perpendicular
3
M
line through  3,2 has equation y  2  3  x  3  y  3x  11
1
1
 . A line with this gradient
3 3
1
1
7
passing through 1,1.5 has equation y  1.5   x  1  y  x  .
3
3
6
b Any line perpendicular to y 3x  0.75 has gradient M  
9 a i
120  20 100 5

 . A line
220  60 160 8
5
5
35
.
with this gradient passing through A has equation y  20   x  60  y  x 
8
8
2
The line through A 60,20 and B 220,120 has gradient MAB 
© Oxford University Press 2019
10
Worked solutions
ii
 60  220 20  120 
mAB 
,
  mAB 140,70 
2
2


iii The perpendicular bisector has gradient
1
8
and passes through

MAB
5
mAB , so has equation
y  70  
b i
ii
8
8
x  140  y   x  294

5
5
40  20
20 1

 . A line
240  60 180 9
1
1
40
.
with this gradient passing through A has equation y  20   x  60  y  x 
9
9
3
The line through A 60,20 and C 240, 40 has gradient MAC 
 60  240 20  40 
mAC 
,
  mAC 150,30 
2
2


iii The perpendicular bisector has gradient
1
 9 and passes through
MAC
mAC , so has equation
y  30 9  x  150  y 9x 1380
c i
The line through B 220,120 and C 240, 40 has gradient MAC 
40  120 80

 4. A
240  220 20
line with this gradient passing through A has equation
y  40  4  x  240  y  4x  1000.
c ii
 220  240 120  40 
mAC 
,
  mAC 230,80 
2
2


c iii The perpendicular bisector has gradient
1
1
and passes through

MAC
4
mAC , so has equation
y  80 
1
1
45
 x  230  y  4 x  2
4
d T sits at the intersection of any two of the perpendicular bisectors. Solving
1
45
5430
5430
2190
y  9x  1380  x 
 1380 
gives x 
, and y  9
4
2
37
37
37
e At T, a wind turbine is equidistant to A, B and C. The length
2
2
5430 
2190 


MAT   60 
   20 
  95.2 m (3.s.f). Therefore, a wind turbine at point T
37
37 



would meet the regulations.
f
Area of function =   3  25  17671  17700 m2.
2
10 a
© Oxford University Press 2019
11
Worked solutions
42 1
 . The perpendicular bisector of A and D has
82 3
1
8  2 4  2
 3 and passes through m 
gradient 
,
  m 5,3 . Therefore, the
M
2 
 2
b Line between A and D has gradient M 
perpendicular bisector has equation y  3  3  x  5 . Setting y  4 in this gives
4  3  3x  15  x 
c i
21
1
3
ii 19
14
.
3
5
6
iv 23
iii 15
5
6
Yes, B can support
11 Let B  (x, y, z)
M1
x4
y 6
 7  x  10 ,
 3  y  12 ,
2
2
z  10
 5  z  20
2
B  10,12, 20
A1A1A1
A1
12 L has gradient of 3
i Neither, gradient is
M1
A1
1
3
ii Parallel, gradient is 3
iii Neither, gradient is 2
iv Perpendicular, gradient is
1
3
A1
A1
A1
v Perpendicular, gradient is
1
3
A1
13 a
11
 3  5 8  3 
M
,
)
  (1,
2
2
2


b gradient 
M1A1
3  8 5

53
8
M1A1
8
5
c i
ii y 
A1
8
11
11 8
39
8
39
x  c through (1,
)
 c c 
, equation is y  x 
5
2
5
10
2
5
10
14 a Length of lift BP= 5002  4002  3002  707m(3 s.f.)
b Length of lift PQ=
900  500
2
So total distance is 1307  1.31  103 m(3 s.f.)
15 a V  500000  50000t
b 125000  500000  50000t  t  7.5
Time is 3:30 p.m.
c 500000  50000t  800000  10000t  t  6
Time is 2:00 p.m.
a
16 a L1 is 3y  ax  9 with gradient of
3
a 3
9

a
Require
3
2
2
b Intersection is (3.23,1.85)(3 s.f.)
17 a
2  6  5  2  6  3  2  3  5  126
2
2
2
3  5  6  70  8.37(3 s.f.)
b
c i
M1A1
 600  400  700  300  600m
2
M  1.5,2.5,3
M1A1
2
M1A1
A1
A1
M1A1
A1
M1A1
A1
A1
R1A1
A1A1
M1A1
M1A1
A1
ii Triangle AMB is isosceles. Let Q be the midpoint of AB
70
QB  2.5 , BM 
2
© Oxford University Press 2019
R1
A1A1
12
Worked solutions
sin QMB 
2.5
70
 QMB  36.69...
2
A1
AMB  73.4 (3 s.f.)
18 a
1200  a r
500  a r
 765
A1A1
 315
1200  a 765

 378000  315a  382500  765a  450a  4500
500  a
315
M1
a  10
A1
r 
765
9

 0.643
1190 14
3 s.f.
M1A1
9
 1086.43 (2 d.p.) Euros
14
19 a Minimum distance is the perpendicular distance
Gradient of road is 1 so line ST has gradient 1
y  x  c through 80,140  c  220
b
(1200  500  10) 
Intersection of y  x  220 and y  x  80 is S  150,70 .
b
M1A1
ST 
150  80
2
 70  140  99.0 (3 s.f.) km
2
M1A1
R1
R1
M1A1
M1A1
M1A1
20 a
A2
b
  102
4
2
 78.5 m (3 s.f.)
M1A1
c
d Square of side 5 has area of 25 m2
  102 25

 72.3m2 (3 s.f.)
e
4
4
f 4
© Oxford University Press 2019
A4
M1A1
M1A1
A1
13
Worked solutions
5
Modelling constant rates of change:
linear functions
Skills check
1 a
10  2x  x
multiply by 2
10  3x
add 2x 
x
b
x
2
5x 
10
3
 divide by 3
x  x  4
2x  4
add x 
x 2
 divide by 2
2
3
y  2x  4
Exercise 5A
1 a One-to-many
c Many-to-one
b One-to-one
d One-to-many
e Many-to-many
2 a R1 is a function as every independent variable is mapped to one and only one dependent
variable.
b R2 is not a function as the input value 1 is mapped to two distinct output value (6 and 4).
c R3 is a function as every input value is mapped to one and only one output value.
3 a R(−1)=(−1) + 1 = 2, R(0) = −0 + 1 = 1, R(0.5) = −0.5 + 1 = 0.5, R(1) = −1 + 1 = 0,
R(2) = −2 + 1 = −1.
B  1,0,0.5,1,2 .
b R(−1) = 2(1 + −1) = 2(0) = 0, R(0) = 2(1 + 0) = 2(1) = 2, R(0.5) = 2(1 + 0.5) = 2(1.5) = 3,
R(1) = 2(1 + 1) = 2(2) = 4, R(2) = 2(1 + 2) = 2(3) = 6.
B  0,2,3, 4,6 .
c R(−1) = (−1)2 + 1 = 1 + 1 = 2, R(0) = 02 + 1 = 1, R(0.5) = 0.52 + 1 = 0.25 + 1 = 1.25,
R(1) = 12 + 1 = 2, R(2) = 22 + 1 = 4 + 1 = 5
B  1,1.25,2,5.
4 a
x
1
.
x
b R is a function as every input value is mapped to one and only one output value
c
R 2  
1
.
2
© Oxford University Press 2019
1
Worked solutions
d
R  a 
1
3
a
multiply by a
1  3a
a
5 a
b
1
3
 divide by 3
y   1  1.
3
The output is  1 when the input is  1.
y  x3  64
x  3 64
take the cube root 
x  4
If the output is −64 then the input is −4.
c x = −1, y = (−1)3 = −1, x = 0, y = 03 = 0, x = 1, y = 13 = 1, x = 2, y = 23 = 8,
x = 3, y = 33 = 27, x = 4, y = 43 = 64.
B  1, 0, 1, 8, 27, 64 .
d The mapping y = x3 from A to B is a function since every input value is mapped to one and
only one output
Exercise 5B
1 a x = −1, y = −1 + 3 = 2, x = 0, y = 0 + 3 = 3, x = 1, y = 1 + 3 = 4, x = 2, y = 2 + 3 = 5.
Range is 2,3, 4,5.
b
Range is y : 2  y  5 .
c
Range is R .
2 a i
ii
b
x  1, y  2  1  3  2  3  5 .
x  3, y  2 3  3  6  3  3.
y  2  2x  3
1  2x
 subtract 3
2x  1
multiply by  1
x
1
2
divide by 2 
c
d
Range of the function is y : 3  y  5.
3 b and d are both functions as every input value is mapped to one and only once output. a, c
and e all have at least one input which is mapped to at least two distinct output elements so do
not correspond to functions.
4 a
The domain is 5, 4, 3, 2, 1, 0, 2, 4 and the range is 2, 0, 2, 4, 6, 8 .
b
The domain is x : 8  x  6 and the range is y : 4  y  3 .
c
The domain is x : 7  x  9 and the range is{y : 0  y  4} .
d
The domain is x : 7  x  7 and the range is{y : 4  y  3} .
e
The domain is x : 2  x  1 and the range is y : 2  y  2 .
© Oxford University Press 2019
2
Worked solutions
5 a i
True
ii False
iii True
b i
True
ii False
iii True
c i
True
ii True
iii False
d i
True
ii True
iii False
e i
False
ii True
iii True
6 a If −3 = 2x + 1 then 2x = −4 and x = −2. If −2 = 2x + 1 then 2x = −3 and x = −1.5.
If −1 = 2x + 1 then 2x = −2 and x = −1. If 0 = 2x + 1 then 2x = −1 and x = −0.5.
If 1 = 2x + 1 then 2x = 0 and x = 0. If 2 = 2x + 1 then 2x = 1 and x = 0.5.
The domain is 2,  1.5,  1,  0.5, 0, 0.5
b
x
x
x
x
then x  6. If  2 
then x  4. If  1  then x  2. If 0 
then x  0.
2
2
2
2
x
x
If 1 
then x  2. If 2  then x  4.
2
2
If  3 
The domain is 6,  4,  2, 0, 2, 4.
c
If  3  x  2 then x  5. If  2  x  2 then x  4. If  1  x  2 then x  3.
If 0  x  2 then x  2. If 1  x  2 then x  1. If 2  x  2 then x  0.
The domain is 0, 1, 2, 3, 4, 5.
7 a The relation is a function since every input value is mapped to one and only one output
value.
b
The domain is
t : 0  t  2 the
range is
V : 0  V
 4. The domain is the period of
time in which the persons temperature is not optimal. The range is the various differences in
temperature between the persons temperature and the optimal temperature.
c
0.2,3.6 , 0.8,2.9 , 1.5,2 , 2,0.
d
40o C .
e 2 hours.
f
The temperature increased in the first half an hour. The temperature of the person
decreased between half an hour and 2 hours.
Exercise 5C
1 a
The average temperature on the second of January was 25o C.
b
The domain is 1, 2, 3, , 31 .
c
An estimate for the range is y : 20  y  28 .
2 a
The independent variable is x.
b i
ii
f 2  10  4 2  10  8  2.
 1
 1
f     10  4     10  2  12.
2


 2
c
f 2.5  10  4 2.5  10  10  0.
d
f  x   6  10  4x
3 a
4x  16
subtract10
x4
divide by  4
g  10  100  2  10  100  20  80.
b i
g 0  100  2 0  100.
0,100 .
© Oxford University Press 2019
3
Worked solutions
ii
g  x   100  2x  0.
2x  100
subtract100
x  50
divide by2
 50,0
iii g  x   100  2x  50
2x  50
subtract100
x  25
divide by2
 25,50
c
4 a i
b i
c
5
ii −3
x 0
ii
x 3
{x  3}
5 a i
iv
b
ii
iii
v
Graphs i,ii,iiiand v are functions.
c Any straight line that has a gradient, so is not vertical, can represent a function. A vertical
line cannot represent a function.
6 a
Every value of t is mapped to one and only one value of N.
b
The independent variable is t. The dependent variable is N.
c 4
d 8
e 12
f
t :16  t  20
g
The range is N : 4  N  128
© Oxford University Press 2019
4
Worked solutions
7 a
C 30  150  5 30  150  150  300 ,
$300
b
C 72  150  5 72  150  360  510
$510
c
C  k   1275  150  5k
5k  1125
subtract150
k  225
 divide by5
225 kilometres.
8 a
v 0  50  8 0  50ms1. This is the initial speed.
b
v 2  50  8 2  50  16  34ms1.
c
v t   15.6  50  8t
d
34.4  8t
subtract50
t  4.3s
divide by  8
v t   0  50  8t. So 8t  50. t  6.25. The body comes to rest at 6.25 seconds.
Exercise 5D
1 a Linear
2 a i
b Linear
c Non-linear
d
Linear
The independent variable is time, which is measured in hours, the dependent variable is
price, which is measured in dollars.
ii A linear relation does exist between the time, t, and the price, P(t). The rate of change is
$12.50 per hour.
b i
The independent variable is time, measured in years, while the dependent variable is the
population of fish.
ii No linear relation, rate of change is not constant.
c i
The independent variable is price of purchase, in euros, while the dependent variable is
amount of VAT, in euros.
ii A linear relation does exist. The rate of change is 0.22 cents per euro.
d i
ii
3 a i
b i
iii
The independent variable is temperature, measured in °C, the dependent variable is the
number of passes sold.
A linear relationship does exist, the rate of change is  8 passes per degree.
Increasing
ii Increasing
iii Decreasing
iv
Neither
ii
iv
© Oxford University Press 2019
5
Worked solutions
4
Calculate the rate of change between successive values of x :
Between 1 and 3 :
Between 5 and 7 :
m
m
62 4
  2.
3 1 2
15  10 5
  2.5.
75
2
These are different so the function is not linear as it does not have a constant rate of change.
5 a 8
6.5  5 1.5

 0.75
3 1
2
b
m
c
f 1  0.75 1  c  5
 find c
c  5  0.75  4.25
by subtracting0.75
f 0  0.75 0  4.25  4.25
6 a
m
1150  1000 150

 50.
2019  2016
3
N 1  1000  50 1  c
So c  950.
A model for the number of students in the school is N t   50t  950.
b
7 a
N 0  50 0  950. So the number of students in the school in 2015 is 950.
m
1  5 4

 1.
5 1
4
f 1  1 1  c  5.
c  6.
So the formula for f  x   x  6.
b
8 a
f  3    3  6  3  6  9.
120ms1
b
8seconds
0  120 120

 15. The rate of change of velocity with respect to time is  15ms 2.
80
8
c
m
d
v 0  m 0  c  c  120.
A model for v is v t   15t  120.
Exercise 5E
1 a Any point, e.g. (9,12)
c
u  0 
4
0  c  c  0.
3
u x 
4
x
3
d i
b
m
12  0 4
4
 . Gradient is .
90
3
3
The exchange rate is that £1  $1.33
© Oxford University Press 2019
6
Worked solutions
ii
100 
4
x.
3
x  75
$100  £75
2 a
b
8
 1.6.
5
a
k 0  0  m  c  0.
k  m  1.6m.
Exercise 5F
1 a
x  1  f  x   g  x   2x
x  1
subtract x 
y  1  1  2.
substitute x
to find y 
Point of intersection is  1, 2 .
b
2x  3  f  x   g  x   1  4x
6x  3  1
add 4x 
6x  2
subtract3
x
y 
1
3
7
3
divide by6
substitute x
to find y 
 1 7
Point of intersection is   , 
 3 3
2 a
b
800  2p  3p.
800  5p  p  160.
Equilibrium point when p  $160. 160  S 160  480.
c D(100) = 800 – 200 = 600. S(100) = 3(100) = 300. When the price is $100 the demand is
twice the supply.
d D(250) = 800 −2(250) = 300. S(250) = 750. When the price is $250 the supply is 2 and a
half times the demand.
3 a i
5  100  500. $500
ii 150  2.30  100  150  230  380. $380.
b
For option A the total cost is A  x   5x. For option B the total cost is B  x   2.3x  150
c
5k  A  k   B  k   2.3k  150.
2.7k  150.
27  1500
© Oxford University Press 2019
7
Worked solutions
k 
4 a
500
.
9
k is
500
kilometres .
9
g  5  9  5m  3.
12  5m  m  2.4.
b
g  x   0  2.4x  3
24x  30  0
x  1.25
g  x intersects the x  axis at the point  1.25,0 .
Exercise 5G
1 a All three functions have an inverse as every distinct input value is mapped to a distinct
output value, so they are one-to-one.
b
The inverse to function b is k 
p
2.2
The inverse to function c is R 
100
P.
85
The inverse to function d is W  0.9V.
2 a Does not have an inverse as two numbers, −2 and 2, are mapped to 4, so it is not one-toone.
b Has an inverse as every distinct input is mapped to a distinct output, so it is one-to-one.
c Has an inverse as every distinct input is mapped to a distinct output, so it is one-to-one.
d Does not have an inverse as both −1 and 1 are mapped to 2, so it is not one-to-one.
3 a
f 1 exists as f is one to one.
b i
ii
1
5
iii 3
iv −3
c
4 a B, C and D are all one-to-one as every distinct value of x is mapped to a distinct value of y.
A is not one-to-one as every value of x is mapped to the same value of y.
b A linear function with a non-zero gradient will have an inverse, while any linear function with
a gradient of zero will not have an inverse.
5 a
A  x : 2  x  1
b
B  y : 1  y  2
c
f is a one to one function as it has a positive gradient.
d i
2
ii 0
iii 0
iv −2
6 a False: a many to one function sends two inputs to the same output. An inverse function
would only be able to send this output to one of the inputs, so an inverse does not exist.
Only one to one functions have an inverse.
b False: The linear function y = 3 on the domain {x: 0 ≤ x ≤ 1} is a many to one function
that does not have an inverse.
Exercise 5H
1
C 1 8 
4

. This is the radius of a circle with a circumference of 8.
© Oxford University Press 2019
8
Worked solutions
2 a
b
c
3 a
b
b  2.5.
c
Function
Domain
Range
f
0 x 3
2.5  y  5
f 1
2.5  x  5
0y 3
d
e
f 1  x   0.4x  2  f  x   2.5x  5
2.1x  3
x 
10
7
y 
4
10
2 
7
7
 10 10 
The point on both lines is 
,
.
 7 7 
4 a
The domain is x : 1  x  2 and the range is y : y  0
b 1
x  f 0  2.
c Decreasing
d
b 13, 21, 34
c 17, 23, 30
5 6 7
, ,
6 7 8
g 0, -25, -50
Exercise 5I
1 a 23, 28, 33
e −1, 1, -1
2 a 2, 3, 4
f
b 4, 7, 10
c 2, 4, 8
© Oxford University Press 2019
d 16, 32, 64
d 3.5, 3, 2.5
9
Worked solutions
3 a
un  n2
4 a
u6  3  4  6  1  3  4  5  23
b
b
un  n3
c
un  n
d
un 
2
n
un  3  4   n  1  207
4  n  1  204
n  1  51  n  52
c
un  111  3  4  n  1
108  4  n  1
27  n  1
n  28
d
400  3  4  n  1
397  4  n  1
397 is not a multiple of 4, so 400 is not a term in the sequence.
Exercise 5J
1 a
6d  1   10  9
d  1.5
b
2 a
b
u15  10  14  1.5  10  21  11
b1  1 2  1  3, b2  2  4  1  10, b3  3 6  1  21.
b2  b1  7, b3  b2  11. The difference between consecutive terms is not the same, so it is not
arithmetic.
3 a
un  5  4  n  1
b 116  5  4  n  1
111  4  n  1
Not a term as 111 is not a multiple of 4.
4 a
b
12  u1  2d.
7d  28.
40  u1  9d .
d  4.
 Take the difference of both equations in part a
u1  12  2  4  12  8  4. d  4
c
u100  4  99  4  400.
5 a During the second year there were 95 employees and during the third year there were 105
employees.
b 175
c
285  85  10  n  1
200  10  n  1
n  1  20
The company will have 285 employees after 20 years
6 a
un  55  3  n  1
b
u12  55  3 12  1  55  33  22.
c
0  55  3  n  1
$22
n  1  18.3333
© Oxford University Press 2019
10
Worked solutions
Pablo makes his last payment in the20th month
d
u19  55  3 19  1  55  54  1
The final payment is for $1.
7 a
an  2.6  1.22  n  1
b
a28  2.6  1.22 28  1  2.6  32.94  35.54.
c
84  2.6  1.22  n  1
35.54m.
81.4  1.22  n  1
n  1  66.7
The tree will reach this heigh in the year 2065
8 a −25
b
a10  1000  9  25  1000  225  775.
c
an  1000  25  n  1  225
1225  25  n  1
n  1  49
There are 50 terms in the sequence.
9 a The difference between successive terms is constant.
b 2
c −2
d
un  2  2  n  1 .
e
u20  2  2 20  1  2  2  19  2  38  36
20,36 does lie on this graph.
10 a Pattern 5 requires 16 sticks and pattern 6 requires 19 sticks.
b
un  4  3  n  1 .
u20  4  3 20  1  4  57  61. Pattern 20 needs 61 sticks.
c
un  127  4  3  n  1
123  3  n  1
n  1  41
Pattern 42 uses 127 sticks.
11 a
The common difference is 2. un  1  2  n  1 . 61  1  2  n  1 .
60  2  n  1
30   n  1
There are 31 terms in the series.
Exercise 5K
1
S17 
17
30  8  17 19  323
2  15  17  1 0.5  17
2
2
2
S20 
20
2  6  20  1  3  10 12  57  10  45  450
2
3
S30 
30
2  8  30  1 8  15 16  232  3720
2
© Oxford University Press 2019
11
Worked solutions
4 a
d  62  52  10.
n 1 
462  52
 41.
10
There are 42 terms.
b
5 a
b
42
52  462  10794
2
S42 
Sn 
n
2  7  n  1
2

5500 

n
2  7  n  1
2
11000  7n2  5n
7n  275  n  40  0
So S40  5500
Sn 
n
8  3  n  1
2
b
S10 
10
8  3 10  1  5 8  27  5  19  95
2
c
250 
6 a


n
11  3n
2
3n2  11n  500  0
n
11  2 112  4  3  500
 14.87 
23
n  15 is the smallest n with Sn  250
7 a
b
8 a
u10  3  0.5 10  1  3  4.5  7.5.
7.5km
15
2  3  0.5 15  1  97.5.
2
S15 
2d  14  5  9.
97.5km
d  4.5
b
a  5  4.5  9.5.
b  14  4.5  18.5
c
S10 
10
2  5  4.5 10  1  252.5
2
d
Sn 
n
10  4.5  n  1  500
2
9n2  11n  2000  0
n
11  2 112  4  9  2000
 14.308 
29
n  15 is the smallest such value of n.
9 a
2k  1  d  k  10.
k  10  d  k  1
3k  9  2k  11
5k  20
k 4
b 9, 6, 3
c −3
d
S20 
20
2  9  3 20  1  390
2
10 a 10, 13, 16
b
x 2
© Oxford University Press 2019
12
Worked solutions
c
x 2
10  2  3  x  3
2
520 
1040   x  2 3x  11
3x2  5x  1062  0
3x  59  x  18  0
x  18
Exercise 5L
9000  0.05  3  1350.
$1350
b 10000  0.085  1.5  1275.
$1275
1 a
c
2
6500  0.07 
41
 1554.58  $1554.58
12
9000  x  0.075  7
x  17142.86 
$17142.86
3
8000  r  5  1840
r  0.046
4.6% interest
4
8600  0.065  t  8600
t  15.38 
16 years
5 Interest paid after n years is 1000 × 0.003n = 3n. So, the final amount is, 1000 + 3n.
Therefore, the statement is false.
Exercise 5M
1 a
b
X
9
5
2
0.5
0.1
y
1
5
8
9.5
9.9
y  x   10  x
x : 0  x  10
c i
ii
y : 0  y  10
d
e The rate of change is −1. This means that for every centimetre the base increases, the
height decreases by the same amount.
2 a
Width(xm)
5
8
20
34
Length(ym)
60
54
30
2
b
© Oxford University Press 2019
13
Worked solutions
Linear modal
c
y  70  2x
d It is not reasonable as a width of 36m would require more than 72m fence.
e
The domain is x : 0  x  35 and the range is{y : 0  y  70}
3 a
b Linear model because the three points seem to lie on a straight line.
c From GDC, S  0.888x  6.14
d The model fits the data well.
e i
S 7  0.8875  7  6.225  12.4 
$12.4 million
ii
S 15  0.8875  15  6.225  19.5 
$19.5million
Chapter review
© Oxford University Press 2019
14
Worked solutions
1 a
R
 3, 1 ,  2, 2 ,  1,1 , 0,2 , 1,2 , 2,0
b Every value of x is mapped to one and only one value of y.
c
3, 2, 1,0,1,2
d
2, 1,0,1,2
2 a Not a function: fails vertical line test.
b A function: every value of x is mapped to one and only one value of y.
c A function: every value of x is mapped to one and only one value of y.
d Not a function: fails vertical line test.
One to one
ii
x : 5  x  5
iii
y : 5.5  y  9.5
b i
Many to one
ii
x : x  4
iii
y : 8  y  8
c i
Many to one
ii
x : 3  x  4
iii
y : 2  y  3
d i
Many to one
ii
x : 2  x  1
iii
y : 2  y  2
3 a i
4 a
12  2m  b.
10.5  5m  b
1.5  3m
m  0.5.
12  1  b.
b  13.
m  0.5.
b
f 3  0.5 3  13  1.5  13  11.5
c
3  0.5c  13.
c  20
5 a
x : 3  x  4
b
f  3  2  3  5  6  5  11
c
4  2x  5
2x  1
x  0.5
d
e
6
y : 3  y  11
5, 4, 0, 2.5, 0
7 a
R
b
c
R
d i
f 1  0.5 1  3  2.5.
A is on the graph.
© Oxford University Press 2019
15
Worked solutions
ii
8 a i
b i
f 100  0.5 100  3  50  3  47. B is not on the graph.
−1
ii 1
iii 1
4 and −4
ii 0
iii −7
c −8
d
9 a
{x : 5  x  5}
T  x   150  2.5x
b T(6) = 150 − 2.5(6) = 150 – 15 = 135. This is the amount on the card after the card has
been used 6 times.
c
105  150  2.5x
2.5x  45.
x  18. Therefore it has been used 18 times.
d i
0  150  2.5x.
x  60
The domain of the function is 0, 1, 2, 3, ,60
ii
10 a
The range of the function is 150, 147.5, 145, 0
m
80  200
 40.
10  7
d 10  40 10  c  80.
c  480.
d t   40t  480
b 480m
c
0  40t  480.
t  12.
12 minutes.
d
11 a
f 2  0  2m  c.
1  2c.
f  2  1  2m  c
c  0.5.
m  0.25.
f  x   0.25x  0.5
b
12 a
f 5  0.25 5  0.5  1.25  0.5  0.75
70  1.73  121.1.
£70  AUD121.10
b
c
a  p  1.73p
d
a 100  1.73 100  173. This is the amount of AUD equivalent to £100
e
a1 50 
50
 28.90 
1.73
This is the amount of UK£ equivalent to AUD50
13 x  5  f  x   g  x   0.5x  4
1  1.5x
© Oxford University Press 2019
16
Worked solutions
2
.
3
x 
2
13
2
f     5 
.
3
3
3
 2 13 
Intersection point is  ,

3 3 
14 a No inverse function as -1 and 1 are mapped to the same output, 1, so it is a many to one
function and many to one functions do not have inverses.
b This has an inverse function as it is a linear function with a non-zero gradient.
c No inverse function as all inputs are mapped to 3 and so this is a many to one function and
many to one functions do not have inverses.
d This has an inverse function as it is a one to one function.
15 a
c  x   70x
b
t  x   60x  200
c
70x  c  x   t  x   60x  200
130x  200
x
d
1400
 20 
 20  1400
. At this time they are both
km from sun city.

  70 

13
13
13
13




16 a i
b i
c
20
20
. They are the same distance from sun city after
hours
13
13
8
3
8
ii
−1
ii −1
The domain is
x : 7  x  8 and the range is y : 1  y  4.
d
17 a
u2  5  2.5  7.5.
b
un  5  2.5  n  1
c
un  5  2.5  n  1  377
u3  7.5  2.5  10
2.5  n  1  372
 n  1  148.8
n  150is the smallest such value of n
18 a
20.4  20  0.4  d.
n 1 
37.6  20
 44
0.4
45 terms in the series
b
S
45
37.6  20  1296
2
© Oxford University Press 2019
17
Worked solutions
19 a
Sn 
n
2  105   5  n  1  2n 215  5n
2
b 1140 
n
215  5n
2
5n2  215n  2280  0
5n  95  n  24  0
n  19or 24
20 a
un  3  2  n  1
u7  3  2 7  1  15
b
251  3  2  n  1
248  2  n  1
n  1  124.
c
S100 
n  125
100
(2  3  2 100  1  10200
2
21 a The difference between successive terms is
an1  an  3  10n  10  3  10n  10. This is constant, so the sequence is arithmetic.
b
ak  3  10k  1000
10k  1003
k  100.3
Smallest value of k is 101
22 a
1800  0.18  324.
b 1800 
r 
£324
r
 3  54r  324
100
324
 6.
54
23 a
b Linear model
c From GDC P  x   28.6x  144 . The model fits the data well.
d i
ii
P 27  28.6  27  144  $916.20
P  40  28.6  40  144  $1288
e The 27km is more appropriate as 27 is within the given data range, while 40km is less
appropriate as 40 is outside the data range.
24  f  x   26
A1A1
b
f  x   4, 2,0,2, 4,6
A1A1
c
0  f  x   100
A1A1
d 125  f  x   250
A1A1
24 a
© Oxford University Press 2019
18
Worked solutions
25 a
b The standing charge for a single journey
c The cost per km of travel
d 2.8  C  15.6
e Approximately £4.50
26 a u1  23
b
M1A1A1
A1
A1
A1A1
M1A1
A1
M1A1
u50  23  7  50  373
c Solving 23  7n  1007 gives n  140.6
n is not an integer, therefore 1007 is not a term in this sequence
27 a C  430  14.5P
b C  430  14.5  25  $792.50
c 1000  430  14.5P
1000  430
P 
 39.3
14.5
She can therefore invite a maximum of 39 people
d C  430  14.5  16  $662
662
 41.375
16
Denise will therefore need to charge a minimum of $41.38 per head
28 a Solving 3x  10  5 and 3x  10  50
Domain is 5  x  20
b Range is 5  f 1  x   20
M1A1
R1
M1A1
M1A1
M1
A1
A1
M1
A1
A1
M1A1A1
M1A1
A1A1
29 a P  1200T  850
b 5000  1200T  850  T  4.875
So a total of 5 months
30 a
M1A1
M1A1
A1
M1A1A1
b €60
M1A1
c
M1A1
$37
31 a NOT a function, since, eg. the value of x  5 is related to more than one co-ordinate on
the y-axis
A1R1
b This is a function. Each value of x is related to only one value for y
A1R1
c This is a function. Each value of x is related to only one value for y
A1R1
d This is a function. Each value of x is related to only one value for y
A1R1
32 Consider V  25600  1150t
M1A1
and V  18000  480t
M1A1
Plot respective graphs and read off intersection point, or solve the equation
25600  1150t  18000  480t
M1
t  11.3
A1
£12555
A1
© Oxford University Press 2019
19
Worked solutions
3
33 a f  x   128    15  177
2
b
M1A1
f  3  128  3  15  399
M1A1
f 15  128 15  15  1905
A1
Range is 399  f  x   1905
A1
c Solving 128a  15  1162.6
a  9.2
34 a £1200
b Using GDC
a  1200
b  75
c V  1200  75  50  £4950
d This is probably not a realistic model,
as it indicates the value of the painting will increase indefinitely
35 a Domain is 3  x  3
M1
A1
M1A1
M1
A1
A1
M1A1
A1
R1
A1A1
Range is 1  f  x   1
A1A1
b Domain is 1.5  x  5
A1A1
Range is 5  f  x   4
A1A1
c Domain is 0  x  24
A1A1
Range is 0  f  x   12
A1A1
d Domain is 3  x  3
A1A1
Range is 0  f  x   9
A1A1
a  4.5
M1A1
A1
b  25
b 4.5  23  25  128.5 mg.
M1A1
c The relationship may no longer be linear outside the range of children’s weights.
R1
Using the relationship involves extrapolation to adults’ weights, which is not
mathematically sound.
R1
36 a
37 a Substituting C  0 gives 32 F
b Solving C 
A1
9C
 32
5
M1A1
4C
 32
5
C  40

A1
c Attempting to make C the subject of F 
F  32 
C 
d
9C
 32
5
9C
5
M1
A1
5  F  32
A1
9
46.4 F  8 C
M1A1
84.2 F  29 C
A1
The range is therefore 8 C to 29 C
38 a Solving 30  12.5d  70  8.35d
d  9.64
So Abel’s holiday lasts a minimum of 10 days
b Using C  70  8.35d
70  8.35  14  £186.90
70  8.35  21  £245.35
So £186.90  C  £245.35
© Oxford University Press 2019
M1
A1
A1
M1
A1
A1
A1
20
Worked solutions
© Oxford University Press 2019
21
Worked solutions
Modelling relationships: linear
correlation of bivariate data
6
Skills check
1 a The gradient is 3. For every unit x increases, y increases by 3.
b The gradient is 
1
. For every unit x increases, y decreases by a half.
2
2 a Strong and negative
b No correlation
c Weak and positive
Exercise 6A
1 a i
positive
ii nonlinear
b i
Positive
ii Linear
c i
Positive
ii Linear
d i
Positive
ii Linear
e i
No correlation
f
Positive
i
ii Non-linear
2 a Zero correlation
b Positive perfect linear
c Strong negative linear
d Weak negative linear
e Weak positive linear
f
Perfect negative linear
g Moderate negative linear
Exercise 6B
1 a No
b Yes
c No
2 a
b
Sx  10.679, Sy  2070.803. Sxy  10164.84
r  0.945
c The correlation is a strong positive correlation.
3 a
b
Sx  973.519, Sy  11.657, Sxy  11206.25.
r  0.987
c A strong positive correlation.
.
© Oxford University Press 2019
1
Worked solutions
4 a
b
Sx  0.2738, Sy  35.2205,
Sxy  8.3196 , r  0.863
c Strong positive correlation.
Exercise 6C
1 a 0.9
b 1
c −0.6
2 a −0.9
b 0
c 0.7
3 a −1
b −0.3
c 0.5
4 a
Ss  29.086,
Sg  5.840,
Ssg  150 , r  0.883
b Strong positive correlation
c The more hours spent studying the better the grade achieved.
5 a
Sx  271.827,
Sy  15.681,
Sxy  4206. , r  0.987
b Strong positive correlation
Exercise 6D
1 a
b
Sx  8.0113,
Sy  7.7578, Sxy  24.182.
r  0.389
c The outlier is (4,12). It does not follow the trend.
d If the outlier is removed then
Sx  7.975,
Sy  4.111, Sxy  29.2 and r  0.891.
2 a
b
Sx  9.3901, Sy  7.7577, Sxy  61.8182,
r  0.849.
c The outlier is (10,12).
d If the outlier is removed then
Sx  7.9750, Sy  4.1110, Sxy  29.2, r  0.891.
© Oxford University Press 2019
2
Worked solutions
3 a
b
Sx  35.166, Sy  632.869, Sxy  26.667,
r  0.01.
c The two outliers are (15,20) and (30,800).
d With the two outliers removed: Sx  34.322, Sy  281.013, Sxy  4370,
r  0.865
Exercise 6E
1 a i, iv
ii Positive strong correlation
iii Mean of x is 14.6 and mean of y is 23.4
b i, iv
ii Strong negative correlation
iii Mean value of x is:
Mean value of y is:
3  2  1    6
 1.5
10
15  13  10    0
 6.5
10
2 a, e, f
b Moderate positive correlation.
c Mean area is
2.7  2.2  1.8  2.6  1.8  2.2  2.7  2  1.4  1
 2.04 million sqft
10
© Oxford University Press 2019
3
Worked solutions
d Mean number of visitors is
28  27  26  25  23  22  22  21  20  18
 23.2 million people.
10
3 a, e, f
b Strong positive linear correlation
c The mean GDP is
7904  10326  7616    50169
 26845
15
d Mean number of books is
2.2  2.9  3.0    17.0
 8.9
15
Exercise 6F
1 a
Sx  79.866, Sy  16.719, Sxy  1025.525,
b
Totals
y
r  0.768. Strong positive correlation
x
y
xy
X2
128
25.95
3321.6
16384
150
40
6000
22500
102
24.85
2534.7
10404
140
31.8
4452
19600
140
30.2
4228
19600
98
28.95
2837.1
9604
75
21.85
1638.75
5625
130
34.5
4485
16900
80
23.25
1860
6400
132
26
3432
17424
1175
287.35
34789.15
144441
287.35 1025.525 
1175 

x 
  y  0.161x  9.84
10
79.8662 
10 
c Mean travel times is
1175
 117.5 minutes.
10
d The mean for the price is
287.35
 28.74 euros.
10
e
© Oxford University Press 2019
4
Worked solutions
2 a
Sx  8.922, Sy  0.7642, Sxy  5.62,
b
Totals
r  0.824 . Strong positive correlation.
x
y
xy
x2
12
1.9
22.8
144
16
2.2
35.2
256
9
1.7
15.3
81
10
2
20
100
14
2
28
196
18
2.5
45
324
12
2.3
27.6
144
15
2.2
33
225
17
2.4
40.8
289
15
2.4
36
225
138
21.6
303.7
1984
21.6  5.62  
138 
y

x
  y  0.0706x  1.19
2 
10
10 
 8.922  
c Mean number of objects is
d Mean time is
138
 13.8 .
10
21.6
 2.16 minutes.
10
e
3 a
Sx  0.2045, Sy  20.601, Sxy  3.918,
b
Totals
y
r  0.930. Strong positive linear correlation.
x
y
xy
x2
1.9
275
522.5
3.61
1.83
267
488.61
3.3489
1.81
260
470.6
3.2761
1.79
257
460.03
3.2041
1.74
258
448.92
3.0276
1.91
272
519.52
3.6481
1.93
273
526.89
3.7249
1.86
268
498.48
3.4596
1.81
261
472.41
3.2761
1.95
273
532.35
3.8025
18.53
2664
4940.31
34.3779
2664
3.918 
18.53 

x 
  y  93.7x + 92.8
10
0.20452 
10 
© Oxford University Press 2019
5
Worked solutions
c Mean height is
18.53
 1.853 m
10
d Mean weight is
2664
 266.4 kg
10
e
4 a
Sx  5.292, Sy  0.9150, Sxy  4.7,
b
Total
y
r  0.971.
Strong negative correlation.
x
y
x2
xy
12
4.2
144
50.4
13
4
169
52
14
3.9
196
54.6
15
3.5
225
52.5
16
3.4
256
54.4
17
3.4
289
57.8
18
3.2
324
57.6
105
25.6
1603
379.3
25.6
4.7

2
7
5.292


105 

x 
  y  0.168x  6.18
7 

 105 25.6 
,
c Mean point is 
  15,3.66 
7 
 7
d
e An 11 year old might have 0.168 11  6.18  4.332 . 4 absences. Not a reliable estimate
as 11 is outside the range of given data.
Exercise 6G
1 a
b Strong positive linear correlation. Can use the regression line as the correlation is linear and
strong.
© Oxford University Press 2019
6
Worked solutions
c
x  190, y  11.6, Sxy  1540, Sx  231.95.
 1540 
y  11.6  
x  190  y  0.0286x  6.16
2 
 231.95 
d
y 280  0.0286 280  6.16  14.168.
Around 14 errors.
e It would not be reliable as 400 is outside the given data range.
2 a
Sx  7.0781, Sy  11.954, Sxy  65.3,
r  0.772 .
b There is a strong negative correlation
c
Total
x
y
xy
x2
3
10
30
9
4
15
60
16
5
14
70
25
3
12
36
9
7
7
49
49
7
12
84
49
8
6
48
64
9
5
45
81
9
6
54
81
8
4
32
64
63
91
508
447
91
65.3 
63 
y 

x 
  y  1.30x  17.3
10
7.0781
10



d
y 6  1.30 6  17.3  9.5 minutes
3 a Because the correlation is moderate and linear.
b Because x = 0 is quite far out of the given date range.
c Because x = 10 is inside the given data range.
d The regression line of y on x is used to predict the value of y for a given value of x and not
the other way round.
Exercise 6H
1 a i
x : 3  x  10.
ii
iii f 3  4, f  3  2.
b i
R
ii
iii f 3  3,
f  3  3.
© Oxford University Press 2019
7
Worked solutions
2 a
 x  5, 4  x  0
f x  
 5, 0  x  3
b
 3  x, 2  x  2
f x  
,2  x  4
x  1
3 a Can be modelled by a piecewise linear model as the points can be split into two parts each
of which each show a linear correlation.
b This correlation is not linear and cannot be modelled well by a piecewise linear function.
c This correlation can be modelled by a single linear model and so does not need a piecewise
linear model.
4 a
b Around 15.
5 a
b For the first section:
Sx  10.086, Sxy  27.463,
y
29.8
27.463

8
10.0862
52.5 

x 

8 

y  0.270x  5.50
For the second section
Sx  5.292,
y
Sxy  57
58.8
57

7
5.2922
105 

x 

7 

y  2.04x  22.1
0.270x  5.50, 0  x  12
f x  
 2.04x  22.1, 12  x  20
c
d i
f 8 0.27 8  5.5  3.34.
ii
f 15.5  9.52.
Exercise 6I
1 a False, if the gradient is positive then the correlation is positive.
b True.
c False.
© Oxford University Press 2019
8
Worked solutions
2 a i
Gradient is 1.04. This means that every mark gained in the last year in of high school
increases the marks in the first year of university by 1.04.
ii The y intercept is −2.50. This has no meaning, as it is not possible to get negative
marks.
b i
Gradient is 0.87. This means for every centimetre of height increase corresponds to an
increase in 0.87kg of weight.
ii The y intercept is −70. This has no meaning as it is impossible to have a negative weight
or no height.
c i
The gradient is −250. This means for every year after a car is bought, it is worth $250
less.
ii The y intercept is 9000. This means that when the car is bought it is worth $9000.
3 a
Sx  264.575,
Sy  24.142,
Sxy  6100,
r  0.955.
b Strong positive correlation.
c i
a
Sxy
2
x
S

6100
 0.0871 For every gram the weight goes up the length goes up
264.5752
0.0871.
by
ii
   279 
x Sxy
Sx2
7
6100
1750

 18.1 .
264.5752
7
This is not relevant as it is outside the
data range.
Chapter review
1 a I
b V
c III
d II
2 a 0
b 0.86
c −1
d 1
e −0.99
f
−0.9
3 a
b The mean point M is denoted by the cross.
c Approximately 62cm
d This is reliable as 6.5 months is inside the data range.
4 a
Sx  5.033, Sy  0.4157, Sxy  1.652,
r  0.790
b There is a negative strong correlation between the two variables.
c
y
34.7
1.652 
252.2 

x 
  y  0.0652x  5.11
10
5.0332 
10 
d It is not reliable as 40 is outside the data range.
5 a i
ii
f 30  100  30  130
f 80  2 80  250  160  250  90
b
c The range is y : 50  y  150
© Oxford University Press 2019
9
Worked solutions
6 a
Sx  9.0676,
Sy  68862.87, Sxy  561093.6, r  0.899
b It shows a strong negative correlation.
c i
m
561093.6
 6824.11
9.06762
ii For every position lower in the league the average attendance goes down by 6824.2.
327682 561093.6  50 


  74321.
9
9.06762  9 
d
c  y  mx 
e
6824.2 5  74321  40200.
f
Because 15 is outside of the given data range.
7 a
Sx  63.290, Sy  54.577, Sxy  3399.333,
r  0.984
b There is a strong positive linear correlation between the scores in the two tests.
c It is appropriate to find the regression line of y on x because the correlation between the
two variables is strong and linear.
d
y
644 3399.333 
670 

x 

12
63.2902 
12 
y  0.849x  6.28
Defined for 25  x  85
e
0.849 55  6.28  52.975. 53%
f
They should leave this data point out of their analysis as it is an outlier and so including it
will weaken the correlation that the data shows.
8 i
ii
iii
iv
v
9 a
perfect positive
strong negative
weak positive
weak negative
zero
A1
A1
A1
A1
A1
b Strong, negative
c i x  4.625
ii x  5.875
iii See above
d See above
e 3.2, see above for lines drawn on
A1 scales A3 points (A2 6 points, A1 3 points)
A1A1
© Oxford University Press 2019
A2A2A1
M1 thru average A1
A1A1
10
Worked solutions
10 a 100  70m  c
140  100m  c
40  30m
4
3
m
,c
20
3
(M1)A1A1
b Positive
c Line goes through  x, y 
y 
4
2
2
90  6  126
3
3
3
d Estimate is
11 a
b
c
d
A1
(R1)
(M1)A1
4
2
2
60  6  86
3
3
3
(M1)A1
40 C
A1
A1
A1
70 C
100 C
i
A1
ii T  80
40  2t  80  t  20
130  t  80  t  50
M1
A1A1
Interval is 20  t  50
12 a
A3 (A2 for 5 A1 for 3)
b r  0.0695 (3 s.f.)
A2
c Very weak (negative) correlation so line of best fit is meaningless
R1
25-year-old would be extrapolation
13 i
Gradient m 
0.6
 0.2
3
R1
M1A1
ii l  0.6
iii k  3
A1
A1
© Oxford University Press 2019
11
Worked solutions
iv a  5
v b  0.6
A1
A1
vi Gradient p 
0.9  0.6
 0.1
8 5
M1A1
vii 0.6  0.1  5  q  q  0.1
viii r  8
14 a r  0.358 (3 s.f.)
b Q1  12.3 Q3  12.7
IQR  0.4
M1A1
A1
A2
A1A1
12.3  1.5  0.4  11.7
So 10.8 is an outlier
c
d
e
f
15 a
b
r  0.860 (3 s.f.)
Changed from weak negative to strong negative
t  0.0627y  138.63
0.0627  2010  138.63  12.6s (3 s.f.)
0.51  120  7.5  68.7
The line of best fit goes through  x, y 
y  0.51  100  7.5  58.5
c
d
16 a
b
c
d
e
f
17 a
b
Strong, positive
x on y
r  0.979 (3 s.f.)
Strong, positive
i y  1.23x  21.3
ii x  0.776y  20.8
1.23  105  21.3  108
0.776  95  20.8  95
It is extrapolation
i 0.849 (3 s.f.)
ii Strong, positive
iii y  0.937x  0.242
i 0.267 (3 s.f.)
ii Weak, positive
iii The r value is too small for this to be particularly meaningful
© Oxford University Press 2019
R1
A1
A2
A1
A1A1
M1A1
M1A1
R1
A1
A1A1
A1
A2
A1A1
A1A1
A1A1
A1
A1
R1
A2
A1A1
A1A1
A2
A1A1
R1
12
Worked solutions
Quanitifying uncertainty: proability,
binomial and normal distributions
7
Skills check
1 There are 12 numbers in total.
a 2,5,11,17 are prime, so the probability that the number is prime 
4
1
 .
12 3
b 1,5,9,11,17,25,27 are odd, so the probability that the number is odd 
7
.
12
c 1,4,9,16,25 are square, so the probability that the number is square 
5
.
12
2 Total number of people = 116.
a Total number of females is 57, so the probability is
b
57
.
116
12
3

.
116 29
c Total number of non-smokers is 98, so the probability is
3 Mean
98
49

.
116 58
9  1  7  2  3  3  2  6  1  11 55

 2.5.
9 7 3 2 1
22
Exercise 7A
1
1
of being picked. 2 of these letters are
. (Each letter from RANDOM has probability of
6
3
1
also in MATHS. Hence, p   2 ).
6
1 p=
2 a
p
17
. All numbers except for 1,2,3 can be hit.
20
b
p
14
7

. There are 14 numbers above 6.
20 10
c p = 1. The range of numbers is 1−20.
d
p
14
7

. All numbers 1 − 14 can be hit.
20 10
e
p
8
2
 . 2,3,5,7,11,13,17,19 are prime.
20 5
f
p
4
1
 . 1,4,9,16 are square.
20 5
g p = 0. There are no solutions to this equation in this set of positive integers.
3 a
p
6
.
11
b
p
3
(1,4,9).
11
c
p
5
(2,3,5,7,11).
11
d
p
2
(1,9 are square and odd).
11
f
p
4
(2 is not odd).
11
e p = 0 (no square numbers are prime).
g
p
1
(2 is prime and even).
11
.
© Oxford University Press 2019
1
Worked solutions
4 There are 10000 possible PINs because each digit can take 10 values.
a
p
1
.
10000
b
p
99
.
10000
c
d
1
(last digit has to be 0 and other digits can take any value, assuming 0000 is
10
considered to be divisible by 10).
p
p
9987
(13 combinations do not fall in this range.).
10000
Exercise 7B
1 The total of 239 shoppers were surveyed.
a
2 a
p
73
.
239
p
b
37
.
136
c
1300 
100
 544 shoppers.
239
3
(there are 4  4 = 16 possible pairs 3,3; 4,4; 5,5; 6,3; and 4,2 give a natural
8
number).
p
b
320 
3
= 60 (only three possibilities of positive difference: 4 − 3, 5 − 3 and 5 − 4.
16
3 a
154 
6
= 77 (1,2,3,4,6,12 are factors of 12).
12
b 154 
c
154 
5
 64 (2,3,5,7,11 are prime).
12
2
 26 (2,3 are prime factors of 12).
12
4 207  (0.15 + 0.25 + 0.12)  108.
5 531 
6
79 
222
 340.
347
5
 36.
11
7 a 573  (0.005 + 0.012)  10.
b Assume the distribution of the cars is the same every month of the year.
8 67  0.0137 + 313  0.0041  2.
9
31 
6
 8.
24
10 Expected values of each dice: A
16
18
20
18
,B
,C
,D
, so dice C is most likely to win.
6
6
6
6
Exercise 7C
1 a Find x such that 127 = 81 + 70 - 29 + x, so x = 5
70  29
41

.
127
127
b
p
c
10000 
81  29
 4094.
127
2 Create a Venn diagram.
© Oxford University Press 2019
2
Worked solutions
a
b
c
p
9
. Find x: 20 = (12 − x) + (15 − x) + x + 2, x = 9.
20
9
3
 . Divide the number of students studying both subjects by the total number of
12 4
students studying biology.
p
60 
9
= 27.
20
3 a
b
p
21  20  3 44

91
91
c
p
14  1  17 32

91
91
4 a
b 94 = (27 + x) + (7 − x) + (16 + x) + (6 − x) + x + (11 − x) + (12 + x) + 11,
94 = 90 + x, x = 4.
c
p
7  x  6  x  11  x  x 16
8


.
94
94 47
Exercise 7D
1 Draw a sample space diagram.
1
4
9
16
2
1
2
7
14
3
2
1
6
13
5
4
1
4
11
7
6
3
2
9
11
10
7
2
5
13
12
9
4
3
a
p
10
5

(orange outcomes).
24 12
b
p
8
1

(green outcomes).
24 3
2 a Draw a sample space diagram, then p 
1
1
2
3
4
5
6
1
2
3
4
5
6
16 4
 (green + orange areas).
36 9
© Oxford University Press 2019
3
Worked solutions
b
2
2
2
3
4
5
6
3
3
3
3
4
5
6
4
4
4
4
4
5
6
5
5
5
5
5
5
6
6
6
6
6
6
6
6
945 
27
 709 (orange + brown areas).
36
3 Draw a sample space diagram.
1
2
3
4
5
6
7
8
1
1
2
3
4
5
6
7
8
2
2
4
6
8
10
12
14
16
3
3
6
9
12
15
18
21
24
4
4
8
12
16
20
24
28
32
a
p
8
1
 .
32 4
b
p
6
3

.
32 16
c
p
4
1
 .
32 8
Draw a sample space diagram for Bethany’s case.
1
2
3
4
5
6
1
1
2
3
4
5
6
2
2
4
6
8
10
12
3
3
6
9
12
15
18
4
4
8
12
16
20
24
5
5
10
15
20
25
30
6
6
12
18
24
30
36
d
p
9
1
 .
36 4
e
p
13
.
36
f
p
20 5

p = 20/36 = 5/9 (orange area).
36 9
g E.g. M and N are even.
4 a
Chromosome inherited
from mother
Chromosome
inherited from
father
X
X
X
XX
XX
Y
XY
YX
b There are 4 outcomes in total, 2 of which result in XX pair. Hence, p 
2
 0.5.
4
Exercise 7E
1 Draw a Venn diagram. There are 10 artists with no choice of formats of their albums. Hence,
10 5
p
 .
14 7
© Oxford University Press 2019
4
Worked solutions
2 Draw a sample space diagram. Then, add probability of each factor: 1, 343, 1679616 and
1
1 2 1 2 7
5764801. This gives p   1      .
3
3 3 3 3 9
0
3
8
6
1
216
1679616
7
1
343
5764801
3 Total number of outcomes is 24  16. Favourable outcomes can be written out as: MMFF,
MFMF, MFFM, FMMF, FMFM, FFMM, i.e. there are 6 favourable outcomes. Hence, the probability
6
3
p
 .
16 8
4 Probability that the number chosen is a multiple of 6 (30 from set X or 30, 60, 90 from set Y) is
1 1 1 3 2
2
p      . Hence, the expected number of points is 54 
= 15.4.
2 7 2 7 7
7
5 Draw the sample space diagrams. P(R = 5) =
4
1
3

and P(T = 5) =
so therefore T = 5 is
36 9
20
the more likely event.
R
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
T
1
2
4
8
1
2
3
5
9
2
3
4
6
10
3
4
5
7
11
4
5
6
8
12
5
6
7
9
13
6 A vs B: the winner table shows that the probability of dice B winning is P(B) =
probability of dice A winning is P(A) =
12 1

and the
36 3
24 2
 .
36 3
C vs D: the winner table shows that that the probability of dice D winning is P(D) =
and the probability of dice C winning is P(C) =
12 1

36 3
24 2
 .
36 3
B vs A
0
0
4
4
4
4
D vs C
2
2
2
2
6
6
3
B
B
A
A
A
A
1
C
C
C
C
C
C
3
B
B
A
A
A
A
1
C
C
C
C
C
C
3
B
B
A
A
A
A
1
C
C
C
C
C
C
© Oxford University Press 2019
5
Worked solutions
3
B
B
A
A
A
A
5
D
D
D
D
C
C
3
B
B
A
A
A
A
5
D
D
D
D
C
C
3
B
B
A
A
A
A
5
D
D
D
D
C
C
Exercise 7F
1 a P(A) =
3
4
2
1
6
3
, P(B) =
 , P(A  B) =
 .
and P(A  B) =
10
10 5
10
10 5
b P(A) + P(B) − P(A ∩ B) =
3
4
1
6
3
= P(A ∪ B).




10 10 10 10 5
c Events A and B are not mutually exclusive because P(A ∩ B) is not 0.
2 a P(C) =
7
5
12
, P(D) =
, P(C ∩ D) = 0 and P(C ∪ D) =
 1.
12
12
12
b P(C) + P(D) =
7
5
12


 1 = P(C ∪ D).
12 12 12
c Events C and D are mutually exclusive because P(C ∩ D) = 0.
3 a Draw a Venn diagram.
b The events are not mutually exclusive because the intersection of the sets S and F (see
Venn diagram) is not empty, i.e. both of the events can occur at the same time.
c
p
15 5
 .
24 8
4 a Draw the diagram
b A & D, B & D and C & D form mutually exclusive pairs of events.
Exercise 7G
1 a Independent.
b Neither.
c Neither.
d Independent.
e Mutually exclusive.
f
g Independent because P(T|S) 
Neither.
7
 P(T).
10
2 P(A ∪ V) = P(A) + P(V) − P(A ∩ V), P(A ∩ V) = P(A) P(V) because the events are independent.
Hence, P(A ∪ V) = 0.07 + 0.61 – 0.07 × 0.61 = 0.6373. P(A ∪ V) represents the probability of
either the event A, or B, or both A and B happening.
28  14 14  3
14

= 0.275 , P(S ∩ M) =
= 0.275. Since P(S∩M) = P(M) ×
51
51
51
P(S|M), P(S|M) must be equal to P(S), hence the events are independent.
3 a P(S) × P(M) =
15  45
15
= 0.61 and P(S|M) =
= 0.63. They are not equal hence the events are
99
24
not independent.
b P(S) =
© Oxford University Press 2019
6
Worked solutions
4 a Draw a Venn diagram. Since n(U) is given, n(A) = 7, n(R) = 16. The events are
1 2
1
.
independent, so P(A∩R)   
8 7 28
p
b
5 a i
35  5  14
 0.964.
56
P(A) =
2
11
ii P(M|A) = P(M) =
2
(independent events)
11
iii P(A ∩ M) = P(A) × P(M) =
b i
P(A) =
4
121
2
11
ii P(M|A) =
2
1

(now a card is drawn from a set of 10 cards),
10 5
iii P(A∩M) = P(A) × P(M|A) =
2
55
6 a P(B|A) =
83
= 0.52.
4836
b P(C|A) =
63
= 0.43.
4836
c P(C|B) =
x 3
9
, so x = 7. P(A|C) =
, so y = 11. Hence, z = 6.
x 358
x 36y
d P(A) =
4836
= 0.42, so both A and B, and A and C are pairs of dependent events.
50
Exercise 7H
1 a
b p = 0.2 × 0.05 + 0.8 × 0.5 = 0.41
2
p
13 10  12 10 13  12 12 10  13





 0.66
35
35
35
35
35
35
3 Best way to think about this problem is to consider the probability of no 6 occurring in the four
throws. That is, (5/6)4. Then, the probability to obtain at least one 6 is 1 − (5/6)4. Hence, the
best presentation is the last choice. Even if drawing a tree diagram with 1296 branches might
take a while, the worst representation is the second one, as it is an incorrect probability
calculation.
© Oxford University Press 2019
7
Worked solutions
4 a P(QCI) = 0.7 × 0.9 + 0.3 × 0.95 = 0.915.
b P(D|QCI) 
P  D  QCI 
P QCI 
 P QCI | D  
P D
P QCI 
 0.95 
0.3
 0.31
0.915
c 2000 × (1 − 0.915) = 170.
d Solve for x in (1 − x) × 0.9 + x × 0.95 = 0.93, x = 0.6 i.e. 60%.
5 Probability of throwing a double six in one throw is
1
. Probability of not throwing a double six
36
24
in one throw is
35
 35 
. Probability of not throwing a single double six in 24 throws is 
 .
36
 36 
24
 35 
Hence, Probability of throwing at least one double six in 24 throws is 1 − 

 36 
≈ 0.491.
6 Draw a Venn diagram labelling unknown quantities x, y, z. Then, construct the following
simultaneous equations: x + z = 0.5, z + y = 0.3, x + y + z = 0.6. Solve the equations to
obtain x = 0.3, y = 0.1, z = 0.2. Then, the probability that the car is not a blue car with five
doors is 1 – z – y – x = 0.4.
Draw a tree diagram.
Then, construct the following simultaneous equations: 0.5(x + z) = 0.3, 0.5(1 + z) = 0.6, z =
0.2, x = 0.4, y = 0.6, w = 0.8, so the probability that the car hasn’t got 5 doors and is not blue
is 0.5w = 0.4 as before.
15 14 13 12



= 0.128 because four choices are made
24 23 22 21
from a decreasing in size set of girls while the total number of people also decreases. Hence,
probability that at least one boy is selected is 1 – 0.13 = 0.872.
7 Probability that no boy is selected is:
Exercise 7I
1 Table a does not represent a discrete probability distribution because the probabilities of all
possibilities don’t add up to 1. Table b does not represent a discrete probability distribution
because P(B = 2) = −0.2 is negative. Table c, however, could represent a discrete probability
distribution because all values 0 ≤ P(C = c) ≤ 1, and the probabilities add up to 1.
123 456
t 4
 1 , f t  
defines a discrete probability distribution on a given
21
21
domain.
2 Since
t
5
6
7
8
9
10
f(t)
1
21
2
21
3
21
4
21
5
21
6
21
3 Find k such that
157 k
 1 , i.e. k = 6.
19
4 a Sample space U = {MMM, MMF, MFM, FMM, MFF, FMF, FFM, FFF}.
b To obtain P(F = f), divide the number of sequences which correspond to the particular
outcome by the total number of sequences.
f
0
1
2
3
P(F = f)
1
8
3
8
3
8
1
8
© Oxford University Press 2019
8
Worked solutions
5 a P(A = 12) = 1 − 0.5 − 0.05 − 0.04 − 0.1 − 0.2 = 0.11.
b P(8 < A ≤ 10) = P(A = 9) + P(A = 10) = 0.14.
c P(A is no more than 9) = P(A = 5) + P(A = 8) + P(A = 9) = 0.59.
d P(A is at least 10) = P(A = 10) + P(A = 11) + P(A = 12) = 0.41.
P  A  8  A 11
e P(A > 8|A ≤ 11) =
P  A 11
0.04  0.1  0.2
= 0.38.
0.89

6 a Probabilities don’t add up to 1.
b Find p such that 0.28 + 0.2 + p + 3p = 1, i.e. p = 0.13.
7 Let P(T = 5) = p. Then, 0.2 + 0.15 + 0.1 + 4p + p = 1, so p = 0.11.
t
1
2
3
4
5
P(T = t)
0.2
0.15
0.1
0.44
0.11
Exercise 7J
1 To find k, note that P(B = b) sums to 1: k((4 − 0) + (4 − 1) + (4 − 2) + (4 − 3)) = 1, k =
0.1.Then, E(B) = 0 × 0.4 + 1 × 0.3 + 2 × 0.2 + 3 × 0.1 = 1.
2 The same probability distribution table applies for M = m. Then, E(M) =
1
3
3
1 12
0   1  2   3 
 1.5. Expected number of male and female births in a set of
8
8
8
8
8
triplets is expected to be equal, and E(M) + E(F) = 3.
3 Construct a probability distribution table of a discrete variable K defined as the number of keys
taken out of the handbag.
k
0
1
2
P(K = k)
7
6
7
 
10 9 15
3 7 7 3
7
 
 
10 9 10 9 15
3 2
1
 
10 9 15
Hence, the expected number of keys is E(K)  0 
7
7
1
3
 1
 2
 .
15
15
15 5
4 Note, in this problem coins and keys can be treated as the same object making the calculations
easier. Construct a probability distribution table of a discrete variable M defined as the number
of mints taken out of the handbag.
m
0
1
2
P(M = m)
9
8
9


17 16 34
8
9
9
8
18




17 16 17 16 34
8
7
7


17 16 34
Hence, the expected number of mints is E(M)  0 
9
18
7
32 16
 1
 2


.
34
34
34 34 17
1
1
1
1
+ US$7 ×
+ US$5 ×
+ US$2 ×
= US$3.25. The expected
4
8
8
2
prize is not US$5, hence the game is not fair.
5 Expected prize E = US$3 ×
6 a P(US$5000) = 0.0001, P(US$1000) = 0.0005, P(US$200) = 0.001.
b Since the price of the ticket is US$10, E = − US$10 + (US$5000 × 0.0001 + US$1000 ×
0.0005 + US$200 × 0.001) = − US$10 + US$1.20 = − US$8.80.
c Expected value should be 0, so the price of a ticket should be US$1.20.
7 a There are 16 outcomes in total, and the probability distribution table is:
d
1
2
3
4
6
8
9
12
16
P(D = d)
1
16
2
16
2
16
3
16
2
16
2
16
1
16
2
16
1
16
© Oxford University Press 2019
9
Worked solutions
b P(D is a square number|D < 8) =
P  is a square number D  8
P 8  8
1
3

2
16
16

 .
10
5
16
c To make the game fair, the price of the ticket should be equal to the expected value of the
12 1
2 32 2 2 1
prize. E(D) = US$12 ×
+ US$6 ×
= US$7.5.
16
16
8 a P(B = 1) = 0.0001, P(B = 2) = 0.0001 × 0.9999, P(B = 3) = 0.0001 × 0.99992
b P(B = n) = 0.0001 × 0.9999n − 1, for n integer, because n − 1 bags of crisps don’t contain
the golden ticket and nth bag does, f(b) = P(B = b) = 0.0001(0.9999)b−1.
c f(b) is defined for b positive integers, b  1 .
d p = P(B = 1) + P(B = 2) + … + P(B = 10) = 0.0001(1 + 0.9999 + … + 0.99999) =
0.0009996.
Exercise 7K
1 a X ~ B(7,
1
)
2
b The ‘success’ probability is not constant because the die is not replaced.
c X ~ B(4,
1
)
2
d X ~ B(4,0.3)
e No ‘success’ probability.
2 Model this as X ~ B(6,
1
) and use technology to find the following probabilities.
2
a P(X = 3) = 0.313
b P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.891
c P(3 ≤ X < 6) = P(X = 3) + P(X = 4) + P(X = 5) = 0.641
3 Model this as X ~ B(30,0.005)
a P(X = 1) = 0.130
b P(X = 0) = 0.860
c P(X > 3) = 1 − P(X ≤ 3) = 0.0000154.
4 a Model this as X ~ B(5,0.17), P(X > 3) = 1 − P(X ≤ 3) = 0.00361.
b This is  P  X  3  = 0.0000130
2
5 a Model this as X ~ B(10,0.085) and find the probability of less than 5 panels failing P(X ≤ 4)
= 0.999.
b (P(X ≤ 4))6 = 0.995.
 3
6 Model this as X ~ B  8, 
 8
a P(X = 5) = 0.101.
b P(X
< 5) = 0.863.
c P(X ≤ 5) = 0.964.
7 Model this as X ~ B(5,0.964), P(X = 4) = 0.15545 while Y ~ B(5,0.5), P(Y = 4) = 0.15625.
Exercise 7L
1 a P(X = 4) = 0.0535.
b P(X ≤ 4) = 0.991.
c P(1 ≤ X < 4) = P(X ≤ 3) − P(X ≤ 1) = 0.809.
d P(X
≥ 2) = 1 − P(X ≤ 1) = 0.558.
© Oxford University Press 2019
10
Worked solutions
e P(X ≤ 4|X ≥ 2) =
f
P  X  4  X  2
P  X  2

P  X  4  P  X  1
1  P  X  1
= 0.983.
Events are dependent because P(X  4)  P(X ≤ 4|X ≥ 2).
g E(X) = np = 1.74.
h Variance of X = np(1 – p) = 1.24.
 1
2 Model this as Q ~ B  7,  because 2,3,5,7 are prime.
 2
a P(Q  3) = 1 − P(Q  2) = 0.773.
b E(Q) = np = 3.5.
c Variance of Q = np(1 – p) = 1.75.
3 a Model this as R ~ B(10,0.78)
i
This is equivalent to 7 reds being thrown, P(R = 7) = 0.224
ii P(3 < R < 7) = P(R  6) − P(R  3) = 0.157.
b P(A) = P(R > 7) = 1 − P(R  7) = 0.617,
P(B) = P(R < 3) = P(R  2) = 0.000160,
P(A|B) =
P  A  B
P B
= 0 because P  A  B  = 0.
c Events A and B are not independent because P(A|B)  P(A).
d Events A and B are mutually exclusive because they cannot occur together.
4 Model this as R ~ B(10,0.1), assuming equal probability of the ball falling through the holes.
a P(R  5) = 1 – P(R  4) = 0.00163
b First, find the probability David scores no points in one game: P(R = 0) = 0.349. Next,
model this as G ~ B(6,0.349), and find P(G  2) = 1 – P(G  1) = 0.679.
5 a R ~ B(5,0.964) and B ~ B(5,0.5), so E(R) = 5  0.964 = 4.82, E(B) = 2.50. On average, R
scores higher than B.
b Variance of R is np(1 – p) = 0.174, variance of B is np(1 – p) = 1.25. The results for R are
less well spread.
6 a To model the random variable A, use binomial distribution: A ~ B(25,0.2).
b P(A  5) = 0.617.
c P(A  7) = 1 − P(A  6) = 0.220.
d P(A  3) = 0.234.
e E(A) = np = 25 × 0.2 = 5, on average, Alex can expect to get 5 answers right by randomly
guessing.
f
P(A  5) = 1 − P(A  5) = 1 − 0.62 = 0.383.
g Alex is expected to score 5  4 − 20 = 0 points.
h For one student, the probability of answering at least 7 questions correctly is 0.220, so
model this as X ~ B(4,0.220). Then, P(X  2) = 1 − P(X  1) = 0.212.
7 a Binomial distribution T ~ B(538,0.91). Assume that whether an individual passenger turns
up on time is independent of any other passenger.
b P(T = 538) = 9.21 1023 – it is close to impossible for everyone to turn up on time.
c P(T  510) = 1 − P(T  509) = 0.000672
d Increase n and check P(T  510), for example n = 551 gives P(T  510) = 0.11 but n =
550 gives P(T  510) = 0.09, so n = 551.
e E(T) = np, so choose n =
f
538
= 591.
0.91
Using n = 591, P(T = 538) = 0.0573 and P(T > 538) = 0.468, so it is quite likely that more
people than there are seats would show up and not very likely that exactly as many people
as there are seats would show up.
© Oxford University Press 2019
11
Worked solutions
8


f  p  np 1  p  n p  p2 .The function has its maximum when f   p  n 1  2p  0 , i.e. when
p  0.5 (check it’s indeed a maximum and not a minimum by substituting e.g. f 1 = 0 <
f 0.5  0.25n ).
Exercise 7M
1 a
b
c
2 a
b
© Oxford University Press 2019
12
Worked solutions
c
3 a
b Since approximately 68% of normal distribution data points are within a standard deviation
from the mean and because    = 249ml − 3ml = 246ml, we expect 0.5  (100 − 68)%
= 16% of the shampoo bottles to contain less than 246ml.
c P(S < 246) = 0.159.
d In one bottle of shampoo, P(S  250) = 1 − P(S < 250) = 0.37. Model the sample of 200
shampoo bottles using binomial distribution. Let X be the number of bottles that will contain
at least 250ml, so X ~ B(200, 0.37). Expected number of X E(X) = 200  0.37 = 74.
4 a
b Since approximately 95.5% of normal distribution data points are within two standard
deviations from the mean and because   2 = 186s + 28s = 214s, we expect 0.5  (100
− 95.5)% = 2.25% of the commuter trains to take at least 214s to board all the
passengers.
c P(T  214) = 1 − P(T < 214) = 0.02275  2.28%.
d For one commuter train, P(T > 200) = 1 − P(T  200) = 0.16. Model the sample of 176
commuter trains using binomial distribution. Let X be the number of trains that will take
longer than 200 seconds to be fully boarded, so X ~ B(176,0.16). Expected number of X
E(X) = 176  0.16 = 28.
5 a P(T < 17.1) = 0.5 (half of the data below the mean).
b P(T < 14) = 0.16 (68% of the data within standard deviation from the mean).
c P(T > 20.2) = 0.16 (68% of the data within standard deviation from the mean).
d First, P(T  23.3) = 0.0225 (95.5% of the data within two standard deviations from the
mean), then P(14 ≤ T < 23.3) = P(T < 23.3) − P(T  14) = 1 − 0.0225 − 0.16 = 0.8175
e P(T < 7.8) = 0.0015 (99.7% of the data within three standard deviations from the mean).
f
P(T < 23.3|T > 20.2) =
P(20.2  T  23.3) 1  0.0225  (1  0.16)

 0.859 .
P(T  20.2)
0.16
6 a P(Q < 4) = 0.483
b P(Q < 3.4) = 0.184
c P(Q > 5) = 1 − P(Q  5) = 0.0829
d P(3.5 ≤ Q < 4.5) = P(Q < 4.5) − P(Q < 3.5) = 0.525
© Oxford University Press 2019
13
Worked solutions
e P(Q < 4.9|Q > 2.9) =
P(2.9  Q  4.9)
= 0.887
P(Q  2.9)
7 a A2 (mean is in the middle and data is reasonably spread out), B4 (mean is in the middle and
data is of similar density on both sides), C5 (mean is in the middle and most of the data is
located at the centre), D1 (mean is in the middle but most data located at the edges), E3
(mean is more towards the left; data to the left of the mean is more dense than to the right)
b Histogram C follows the normal distribution (can approximate with a bell-shaped curve).
c p is true because symmetric histogram has the mean in the middle and quartiles and range
is located symmetrically to both sides of the mean; q is true because the normal distribution
has got a symmetric histogram; r is not true – histograms A and B provide a perfect
counterexample.
Exercise 7N
1 Use inverse normal function to find r = 990g.
2 Use inverse normal function noting that 83% of the packs weigh less than t g: t = 384 g.
3 a s: false, t: true, u: true.
b Use inverse normal function and statement u to find Q3 = 22331.3… g. Then, IQR = (Q3 –
Q2)  2 = 405 g.
4 a Let S ~ N(115.7,102), then find P(110 < S < 120) = P(S < 120) − P(S < 110) = 0.382.
b Model this using the binomial distribution X ~ B(8,0.38), then the E(X) = 8  0.382 = 3.06.
c Find the probability P(X > 5) = 1 − P(X  5 ) = 0.0400, assuming that the speeds of the cars
are mutually independent.
5 a Let T ~ N(182,102), then P(T > 190) = 1 − P(T  190) = 0.212.
b Model this using the binomial distribution X ~ B(7,0.21), then P(X  3) = 0.959.
c Find P(T < 165) = 0.0446.
d Model this using the binomial distribution Y ~ B(10000,0.04), then E(Y) = 10000  0.0446
= 446.
6 a Let D ~ N(16,52), then P(13 < D < 15.3) = P(D < 15.3) − P(D < 13) = 0.170.
b Use inverse normal function noting that 87% of employees travel at most x km to find that x
= 21.6 km.
c First, find how many employees travel further than 14km to work: P(D > 14) = 1 − P(D 
14) = 0.66, so there are 23109  0.66 = 15252 employees living further away than 14km.
Hence, 0.91  15252 = 13783 ≈ 13800 employees will fail to get to work on a snow day.
7 a Route A takes a shorter time on average, although has a larger deviation while Route B
takes a longer time on average but has a very small standard deviation, so is more reliable.
b Let A ~ N(42,82) and B ~ N(50,32), find P(A  45) = 0.646 and P(B  45) = 0.0478, so
choose route A.
c Model this using the binomial distribution X ~ B(5,0.646)
i
P(X = 5) = 0.113
ii P(X  3) = 1 − P(X  2) = 0.759
iii P(X = 3) = 0.338, but we are only interested in 3 consecutive days. There are three ways
in total to choose three consecutive days out of five (starting day 1, starting day 2 and
54
starting day 3), but there are
ways to choose three days out of five in total. Hence,
2
3
 0.338 = 0.101.
probability to arrive by 9am on exactly three consecutive days is
10
8 a Use inverse normal function noting that the standard deviation is 5 to find Q3 = 73.4.
b Half the length of the box is the difference between Q3 and the mean, 73 − 70 = 3.4, so the
length of the box is less than 10 years.
Chapter review
© Oxford University Press 2019
14
Worked solutions
1 a There are 12 square numbers between 1 and 150 (122 = 144) so the probability is p =
12
2

150 25
b There are 51 numbers which are at least 100 and at most 150 so the probability is p =
51
150
c There are 30 numbers divisible by 5 (all the numbers with last digit 0 or 5) between 1 and
30
1

150 so the probability is p 
150 5
d There 11 numbers which are at least 1 and at most 11 so the probability is p =
11
150
2 To find  , note that the probabilities for all possible values of k have to add up to 1. Hence,
1
. Then, E(K) =
15
9
4
1
0
1 10 2
0
 1
 2
 3
 4 

15
15
15
15
5 15 3
 32  22  12  02  12   1 , 15  1 ,  
3 a Construct simultaneous equations using the fact that probabilities add up to 1 and the
expression for E(D): 0.3 + p + q + 0.15 + p − q + p + 2q = 1, i.e. 3p + 2q = 0.55 and 0 
0.3 + 1  (p + q) + 2  0.15 + 3  (p − q) + 4  (p + 2q) = 1.7, i.e. 4p + 3q = 0.7.
Solve the equations to obtain p  0.25, q  − 0.1.
b P(D = 3|D ≥ 1) =
P  D  3  D  1
P  D  1

P  D  3
1  P  D  0
=
0.35
= 0.5.
0.7
4 a Probability that the seed grows: p = 0.65  0.85 + 0.35  0.74 = 0.8115.
b Conditional probability: P(Green & Grows) = P(Green)  P(Grows|Green) = 0.65  0.85 =
0.5525.
c P(Red or Grows) = P(Red) + P(Green & Grows) = 0.35 + 0.5525 = 0.9025. Alternatively,
P(Red or Grows) = 1 − P(Green & Doesn’t grow) = 1 − 0.65  0.15 = 0.9025 leads to the
same answer.
5
chance of throwing no sixes at all. Hence, when the die is thrown
6
n times, P(Throw at least one six in n throws) = 1 − P(Throw no sixes in n throws) = 1 −
5 a In one throw, there is a
n
5
  .
6
n
n
log0.005
5
5
b 1 −    0.995 corresponds to    0.005 , n 
= 29.06... so take n = 30.
5
6
6
log
6
6 Draw a Venn diagram to visualise the situation. P(B’∪C) = P(B’) + P(B ∩ C) = 1 − P(B) +
P(B∩C) = 1 − 0.4 − 0.1 + 0.1 = 0.6.
7 a Construct the table noting there are 36 outcomes in total and counting ways to obtain each
of the value of t.
t
4
5
6
7
8
9
10
P(T = t)
1
36
4
36
8
36
10
36
8
36
4
36
1
36
b Prime numbers: 5,7, square numbers: 4,9. For game to be fair, E(T) = 0. This gives the
5  4 7  10
4 
 1

 x

following equation:
 = 0, i.e. 90 − 5x = 0, x = 18.
36
36
 36 36 
8 Let W ~ N(65,112). Then:
© Oxford University Press 2019
15
Worked solutions
a P(W > 70) = 0.325.
b Use inverse normal function to find UQ = 72.42kg and LQ = 57.58kg, so IQR = 14.8kg.
c Use inverse normal function for 92.7% to find 81.0kg.
d Use binomial distribution to model this: X ~ B(8,0.325), then P(X  3) = 0.758.
e P(W < 60) = 0.3247, so 1000  0.3247  325.
9 a There are at most 5 turns before a green balls is definitely picked. P(Judith wins) = P(Judith
wins on her first go) + P(Judith wins on her second go) + P(Judith wins on her third go).
3
4 3 3
6
P(Judith wins on her first go) = , P(Judith wins on her second go)    
,
7
7 6 5 35
4 3 2 1
4
109
P(Judith wins on her third go)     
. Hence, P(Judith wins) 
 0.629.
7 6 5 4 175
175
b Now that the ball chosen is replaced after each turn, it might take infinitely many turns until
the green ball is picked. To find the new probability of Judith winning p, note that after
Judith and Gilles both had an unsuccessful turn each, the probability of Judith winning from
that point resets to the original value p and the following equation can be constructed:
3 4 4p
21
p  
,p 
 0.636. Hence, Judith is more likely to win in this set up of the
7 7 7
33
game.
10 a
 3 13   13 3 





 16 15   16 15 

M1A1
78  13 
39
39




240  40 
240 240
A1
 13 12 

b 1

 16 15 
M1A1
84  7 
156



240 240  20 
0.7  0.4  0.8  0.224 ~
M1A1
b
0.7  0.6  0.2  0.3  0.4  0.2  0.3  0.6  0.8
M1A1
c
 0.252
0.4  0.8  0.32
A1
M1A1
d 1  0.3  0.6  0.2
M1A1
1
11 a
A1
 0.964
A1
12 a
M1A1A1
b 1  0.65  0.55  0.6425
c
13 a
b
c
P  Jake and Elisa solve
P Elisa solve

M1A1
0.35  0.6
0.35  0.6  0.65  0.45
 0.418
0.4
0.6
0.75
M1A1A1
A1
M1A1
M1A1
M1A1
© Oxford University Press 2019
16
Worked solutions
14 a
b
c
32  25  48 = 9
32  9
48
23

48
9
32 2
P E U  

and P  E  
32
48 3
P  E U   P  E  , so not independent.
15 a
b
c
d
e
43
50
7
25
5
11
0
5
34
M1A1
M1A1
A1
A1A1
R1
M1A1
M1A1
M1A1
M1A1
M1A1
P  A  B   P  A P  B   0.3  0.15  0.045
M1A1
b
P  A  B   P  A  P  B   P  A  B   0.3  0.5  0.045  0.755
M1A1
c
P  B  A  P  A  P  A  B   0.3  0.15  0.15
M1A1
d
P  B A  
16 a

P  B  A 
P  A 
M1
P  B   P  A  B  0.5  0.045

P  A 
0.7
 0.65
17 a
M1
A1
1  P no scoring  1  0.72
4
M1A1
 0.731
A1
b
4  0.28  0.763
 0.492
M1
A1
c
1  P no goals  P exactly one goal
M1
4
A1
A1
 1  0.72  0.492
 0.24
18 a Let X be the discrete random variable ‘number of boys’.
So X ~ B 10,0.512
10 
4
P  X  6     0.5126 1  0.512 
6
 
 0.215
M1A1
A1
b
10 
10
P  X  0    0.5120 1  0.512   0.000766
0
M1A1
c
P  X  4  0.348
M1A1
19 a
k
k
 k  k 2  2k 2   1
2
2
M1
3k 2  2k  1  0
A1
3k  1  k  1  0
M1
k 
1
3
A1
© Oxford University Press 2019
17
Worked solutions
b
E X  0 
E X  
k
k
 0.5  k  1  k 2  1.5  2k 2  2 
2
2
M1
k
3k
 k 2  3k 2  k  4k 2 
2
2
A1
2
3  1  4 1 17
1
 4       
3
2
 
 3  9 2 18
c
P  X  1.25  2k 2 
M1A1
k
2
M1
2
2 1
7
1
1
 

 2  
9 6
18
6
3
M1A1
20 a Let X be the discrete random variable ‘time taken for Blossom to walk to her cafe’.

So X ~ N 35,3.42

P  X  37  0.278
b
c
M1A1
P  X  36.5  P  X  34
A1
 0.670  0.384
A1
 0.286
A1
P  X  30  0.071
M1
0.071  25  1.77
M1
So approximately two occasions.
A1
21 a Let X be the discrete random variable ‘mass of a can of baked beans’.

Then X ~ N 415,122

Using GDC
P  X  m  0.65
M1A1
 m  410.4
A1
b You require P  X  422.5 | X  420
P  X  422.5 | X  420 
P  X  422.5
P  X  420
M1
M1
0.266
0.338
A1
 0.787
A1

c Using GDC
P  X  413.5  0.450
M1A1
Now using Y ~ B 144,0.450
M1
P Y  75  0.0524
A1
© Oxford University Press 2019
18
Worked solutions
Testing for validity: Spearman’s, hypothesis
testing and  test for independence
8
2
Skills check
1 P(S) 
2 1
1
 while P(S|E) = . P(S) = P(S|E) so the events are independent.
6 3
3
2 D ~ N(35,32), P(D < 36) = 0.631
3
x
0
1
2
3
4
P(X = x)
0.0016
0.0256
0.1536
0.4096
0.4096
Exercise 8A
1 a 1 (data monotonically increasing)
b 1 (data monotonically increasing)
c − 1(data monotonically decreasing)
d 0(data is not consistently increasing or decreasing)
2 Put the ranked data into a GDC and obtain PMCC rs = 0.2 so there is only weak positive
correlation between the taste and value for money.
3 a The ranks are (note that when more than one piece of data have the same value, the
average of the rank given is used):
x
7
6
5
4
3
2
1
y
1
2
3
4
6
6
6
Use GDC to find the PMCC for the ranked data: rs = − 0.964.
b The ranks are:
x
3
2
4
6
7
1
5
y
2
3
5
7
6
1
4
Use GDC to find the PMCC for the ranked data: rs = 0.893.
4 a PMCC is used for linear relationships and the scatter plot shows the relationships is not
linear.
b The ranks of A − L are:
v
11
10
9
8
7
6
5
4
3
1
12
2
F
1
2
4
5
6.5
6.5
8
11
9
10
3
12
Use GDC to find the PMCC for the ranked data: rs = − 0.942.
c There is a strong negative correlation between velocity and force, as can be expected from
the scatter plot. Since the force does not change significantly at high values of the velocity,
the value of rs could be affected significantly by small changes in data.
5 a Use GDC to find the PMCC for the data: rs = 0.670.
b Scatter plot:
PMCC indicates that there is a positive correlation of medium strength between the English
and Maths scores. The scatter plot shows strong but non-linear positive correlation between
the scores..
© Oxford University Press 2019
1
Worked solutions
c The ranks are:
Maths
11
10
9
8
7
6
5
3.5
3.5
1.5
1.5
English
9.5
8
11
7
6
9.5
5
4
3
2
1
Use GDC to find the PMCC for the ranked data: rs = 0.883. This indicates a strong positive
correlation between the scores which is a more realistic result given the scatter plot.
d Spearman’s rank correlation because the data points are not linear.
6 a Because instead of quantitative data the ranks of the taste are given.
b The ranks are:
Taste rank
1
2
3
4
5
6
Cost rank
1
3
2
5
4
6
Use GDC to find the PMCC for the ranked data: rs = 0.886, so there is a strong positive
correlation between the price and taste of coffee.
7 a Data table:
x
0.82
1.28
1.78
1.46
2.46
2.48
2.02
3.02
2.98
7.46
y
0.86
1.56
1.22
0.62
0.84
1.76
1.82
1.42
0.62
4.98
Use GDC to find the PMCC
i
b i
with the outlier J: r = 0.874
ii without the outlier J: r = 0.0776
The ranked data table with the outlier J:
x
10
9
7
8
5
4
6
2
3
1
y
7
4
6
9.5
8
3
2
5
9.5
1
Use GDC to find the PMCC for the ranked data with the outlier J: rs = 0.304.
ii The ranked data table without the outlier J:
x
9
8
6
7
4
3
5
1
2
y
6
3
5
8.5
7
2
1
4
8.5
Use GDC to find the PMCC for the ranked data without the outlier J: rs = 0.0418.
c
Even though both of the measures are affected by the outlier, Spearman’s rank correlation
coefficient is affected less.
Exercise 8B
1 a Probability that a person chosen at random likes black cars best is
22
. Probability that a
80
38
. If the two events are independent, the expected
80
22 38
number of males who prefer black cars is 80 
= 10.45.

80 80
person chosen at random is male is
b Probability that a person chosen at random likes white cars best is
18
. Probability that a
80
42
. If the two events are independent, the expected
80
18
42
number of males who prefer black cars is 80 
= 9.45.

80 80
person chosen at random is male is
c Use GDC to find the value  2  4.69.
2 a Probability that a person chosen at random buys small coffee is
34
. Probability that a
110
54
. If the two events are independent, the expected
110
34
54
number of males who prefer black cars is 110 
= 16.7.

110 110
person chosen at random is male is
© Oxford University Press 2019
2
Worked solutions
b Probability that a person chosen at random buys large coffee is
46
. Probability that a
110
56
. If the two events are independent, the expected
110
46
56
number of males who prefer black cars is 110 
= 23.4.

110 110
person chosen at random is female is
c Use GDC to find the value  2  5.21.
3 Find totals:
Pet
Rabbits
Guinea
pigs
Hamsters
Totals
Lettuce
16
16
28
60
Carrots
34
18
18
70
Totals
50
34
46
130
a Probability that a pet chosen at random eats carrots is
70
. Probability that a pet chosen at
130
50
. If the two events are independent, the expected number of
130
70
50
rabbits who eat carrots is 130 
= 26.9.

130 130
random is a rabbit is
b Probability that a pet chosen at random eats lettuce is
60
. Probability that a pet chosen at
130
46
. If the two events are independent, the expected number of
130
60
46
rabbits who eat carrots is 130 
= 21.2.

130 130
random is a hamster is
c Use GDC to find the value χ 2  8.05.
4 Find totals:
Transport
Car
Bus
Bicycle
Walk
Totals
Male
12
12
28
8
60
Female
21
13
15
11
60
Totals
33
25
43
19
120
a Probability that a person chosen at random comes by bicycle is
43
. Probability that a
120
60
. If the two events are independent, the expected
120
43
60
number of males who come by bicycle is 120 
= 21.5.

120 120
person chosen at random is a male is
b Probability that a person chosen at random comes by car is
33
. Probability that a person
120
60
. If the two events are independent, the expected
120
33
60

number of females who come by car is 120 
= 16.5.
120 120
chosen at random is a female is
c Use GDC to find the value χ 2  6.90.
© Oxford University Press 2019
3
Worked solutions
Exercise 8C
1 a Contingency table:
Sport
Cycling
Basketball
Football
Totals
Males
7
10
6
23
Females
9
8
10
27
Totals
16
18
16
50
b H0: favourite sport is independent of gender. H1: favourite sport is not independent of
gender.
c v = (rows − 1)(columns − 1) = (2 − 1)  (3 − 1) = 2.
d Use GDC to find the values χ 2  1.16 and p = 0.560.
e Expected values calculated by E(F/M liking C/B/F) = P(F/M) × P(likes C/B/F) × total:
Sport
Cycling
Basketball
Football
Males
7.36
8.28
7.36
Females
8.64
9.72
8.64
All expected values are greater than 5.
f
χ 2  1.16 < 4.605 so H0is accepted.
g Find p value for χ 2 c = 4.605: pc = 0.10, p = 0.560 > 0.10 so H0 is accepted and p value
supports the conclusion.
2 a H0: favourite bread is independent of gender. H1: favourite bread is not independent of
gender.
b E(Female liking white bread) = P(Female) × P(likes white bread) × total =
41 31

 80
80 80
= 15.8875  15.9
c v = (rows − 1)(columns − 1) = (2 − 1)  (4 − 1) = 3.
d Use GDC to find the values χ 2  2.12 and p = 0.548.
e
χ 2 < χ 2 c so H0 is accepted.
3 a H0: favourite genre of film is independent of age. H1: favourite genre of film is not
independent of age.
b E(20− 50 year-olds prefer horror films)
= P(20− 50 year-olds) × P(prefers horror films) × total =
130 77

 300 = 33.367  33.4.
300 300
c v = (rows − 1)(columns − 1) = (3 − 1)  (3 − 1) = 4.
d Use GDC to find the values χ 2  45.2 and p = 3.6 109 .
e p < 0.10 (for 10% test) so H0 is rejected.
4 a H0: favourite flavour of dog food is independent of breed. H1: favourite flavour of dog food is
not independent of breed.
b Expected values calculated by E(F/M liking C/B/F) = P(F/M) × P(likes C/B/F) × total:
Flavour
Beef
Chicken
Lamb
Boxer
11.7
7.00
9.33
Labrador
15.8
9.50
12.7
Poodle
14.6
8.75
11.7
Collie
7.92
4.75
6.33
Not all expected values are greater than 5.
© Oxford University Press 2019
4
Worked solutions
c New contingency table:
Flavour
Beef
Chicken
Lamb
Totals
Boxer
14
6
8
28
Labrador
17
11
10
38
Poodle/Collie
19
13
22
54
Totals
50
30
40
120
2
d Use GDC to find the values χ  3.14 and p = 0.535.
e p = 0.535 > 0.05 so H0 is accepted.
5 a H0: favourite flavour of chocolate is independent of gender. H1: favourite flavour of
chocolate is not independent of gender.
b v = (rows − 1)(columns − 1) = (2 − 1)  (3 − 1) = 2.
c Use GDC to find the values χ 2  9.52 and p = 0.00856.
d
χ 2  9.52 > 9.210so H0 is rejected.
6 a H0: GPA is independent of number of hours spent on social media. H1: GPA is not
independent of number of hours spent on social media.
b E(0− 9 hours and high GPA) = P(0− 9 hours) × P(high GPA) × total =
85
99

 270
270 270
= 31.167  31.2.
c v = (rows − 1)(columns − 1) = (3 − 1)  (3 − 1) = 4.
d Use GDC to find the values χ 2  78.5 and p = 3.6 1016 .
e
χ 2  78.5 > 7.779 so H0 is rejected.
7 State the null hypothesis and the alternative hypothesis: H0: the number of people walking
their dog is independent of the time of the day. H1: the number of people walking their dog is
not independent of the time of the day.
Noting that the number of degrees of freedom is v = 4, use GDC to find the values χ 2 
5.30 and p = 0.257.
Since 0.257 > 0.05 and 5.30< 9.488, H0 is accepted.
8 a H0: the number of bottles of water sold is independent of temperature. H1: the number of
bottles of water sold is not independent of temperature.
b v = (rows − 1)(columns − 1) = (3 − 1)  (3 − 1) = 4.
c Use GDC to find the values χ 2  3.30 and p = 0.509.
d Since 0.509 > 0.01 and 3.30 < 13.277, H0 is accepted.
9 a H0: annual salary is independent of the type of degree. H1: annual salary is not independent
of the type of degree.
b v = (rows − 1)(columns − 1) = (3 − 1)  (3 − 1) = 4.
c Use GDC to find the values χ 2  24.4 and p = 6.53 105 .
d Since 6.53 105 < 0.05 and 24.4 > 9.488, H0 is rejected.
Exercise 8D
1 a Expected frequencies:
Colour
Frequency
Yellow
120
Orange
120
Red
120
Purple
120
Green
120
b v = (n – 1) = 5 − 1 = 4
© Oxford University Press 2019
5
Worked solutions
c H0: The data satisfies a uniform distribution.
H1: The data does not satisfy a uniform distribution.
Use GDC to find the values χ 2  10.45 and p = 0.0335.
d Since 0.0335 < 0.05 and 10.45 > 9.488, H0 is rejected.
2 a Expected frequencies:
Month
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
Freq
5
5
5
5
5
5
5
5
5
5
5
5
b v = (n – 1) = 12 − 1 = 11.
c H0: The data satisfies a uniform distribution.
H1: The data does not satisfy a uniform distribution.
Use GDC to find the values χ 2  6 and p = 0.873.
d Since 0.873 > 0.10and 6 < 17.275, H0 is accepted.
3 a E(number of calls) =
840
= 120.
7
b v = (n – 1) = 7 − 1 = 6.
c H0: The data satisfies a uniform distribution.
H1: The data does not satisfy a uniform distribution.
Use GDC to find the values χ 2  86.1 and p = 1.97 1016
d Since 1.97 1016 < 0.05 and 86.1 > 12.592, H0 is rejected.
4 a Expected frequencies
Last digit
0
1
2
3
4
5
6
7
8
9
Frequency
49
49
49
49
49
49
49
49
49
49
b v = (n – 1) = 10− 1 = 9.
c H0: The data satisfies a uniform distribution.
H1: The data does not satisfy a uniform distribution.
Use GDC to find the values χ 2  9.06 and p = 0.432.
d Since 0.432 > 0.10 and 9.06 < 14.684, H0 is accepted.
Exercise 8E
1 a H0: The lengths are normally distributed with mean of 19 cm and standard deviation of 3
cm.
H1: The lengths are not normally distributed with mean of 19 cm and standard deviation of 3
cm.
b Let L ~ N(19,32), then use GDC to find P(9 < L < 12) = 0.00939
c Expect 250  0.00939 = 2.35 fish.
d Repeat the same procedure to obtain the expected frequency table:
Length of
fish, x cm
Probability
Expected
frequency
9 ≤ x < 12
0.009386
2.35
12 ≤ x < 15
0.081396
20.35
15 ≤ x < 18
0.278230
69.56
18 ≤ x < 21
0.378066
94.52
21 ≤ x < 24
0.204702
51.18
24 ≤ x < 27
0.043960
10.99
27 ≤ x < 30
0.003708
0.927
e Combine the rows with expected frequencies less than five with the rows next to them, i.e.
the top row with the second row and the last row with the second to last row.
© Oxford University Press 2019
6
Worked solutions
f
Updated table:
Length of
fish, x cm
Frequency
Expected
Frequency
9 ≤ x < 15
27
22.70
15 ≤ x < 18
71
69.56
18 ≤ x < 21
88
94.52
21 ≤ x < 24
52
51.18
24 ≤ x < 30
12
11.92
g v = (n – 1) = 5 − 1 = 4.
h Use GDC to find the values χ 2  1.31 and p = 0.860.
i
Since 0.860 > 0.05 and 1.31 < 9.488, H0 is accepted.
2 Let W ~ N(52,32), and calculate the probabilities using GDC. Multiply them by 200 (total number
of girls).
a Table with the expected frequencies:
Weight, w kg
w < 45
45 ≤ w < 50
50≤ w < 55
55 ≤ w < 60
w ≥ 60
Expected
frequency
1.96
48.54
117.77
30.96
0.77
b Merge the first and the last columns with their neighbouring columns to obtain an updates
table:
Weight, w kg
w < 50
50≤ w < 55
55 ≤ w
Observed
frequency
56
82
62
Expected
frequency
50.50
117.77
31.73
c v = (n – 1) = 3 − 1 = 2.
d H0: The weights are normally distributed with mean of 52 kg and standard deviation of 3 kg.
H1: The weights are not normally distributed with mean of 52 kg and standard deviation of 3
kg.
Use GDC to find the values χ 2  40.3 and p = 1.74 109 .
e Since 1.74 109 < 0.05 and 40.3 > 5.991, H0 is rejected.
3 Let X ~ N(65,7.52), and calculate the probabilities using GDC. Multiply them by 300 (total
number of students).
a Table with expected frequencies:
Grade, x%
x < 50
50 ≤ x < 60
60 ≤ x < 70
70 ≤ x < 80
80 ≤ x
Expected
frequency
6.83
68.92
148.50
68.92
6.83
b v = (n – 1) = 5 − 1 = 4.
c H0: The grades are normally distributed with mean of 65% and standard deviation of 7.5%.
H1: The grades are not normally distributed with mean of 65% and standard deviation of
7.5%.
Use GDC to find the values χ 2  0.705 and p = 0.951.
d Since 0.951 > 0.1 and 0.705 < 7.779, H0 is accepted.
4 Let H ~ N(250,112), and calculate the probabilities using GDC. Multiply them by 250 (total
number of elephants).
a Table with expected frequencies.
Height, h
cm
h < 235
235 ≤ h < 245
245 ≤ h < 255
255 ≤ h < 265
265 ≤ h
Expected
frequency
21.59
59.59
87.64
59.59
21.59
b v = (n – 1) = 5 − 1 = 4.
© Oxford University Press 2019
7
Worked solutions
c H0: The heights are normally distributed with mean of 250cm and standard deviation of 11
cm.
H1: The heights are not normally distributed with mean of 250cm and standard deviation of
11 cm.
Use GDC to find the values χ 2  8.02 and p = 0.0910.
d Since 0.0910 > 0.05 and 8.02 < 9.488, so H0 is accepted.
5 Let H ~ N(1200,1002), and calculate the probabilities using GDC. Multiply them by 400 (total
number of light bubs).
a Table of expected frequencies.
Lifespan,
h hours
Freq
h<
1000
1000 ≤ h <
1100
1100 ≤ h <
1200
1200 ≤ h <
1300
1300 ≤ h <
1400
1400 ≤
h
9.1
54.36
136.54
136.54
54.36
9.1
b v = (n – 1) = 6 − 1 = 5.
c H0: The lifespan is normally distributed with mean of 1200 hours and standard deviation of
100 hours.
H1: The heights are not normally distributed with mean of 1200 hours and standard
deviation of 100 hours.
Use GDC to find the values χ 2  78.7 and p = 1.5 1015 .
d Since 1.5 1015 < 0.05 and 78.7 > 11.070, so H0 is rejected.
Exercise 8F
3
1 a Expected probabilities for S ~ B(3, 0.75): P(S = 0) =   0.7500.253  0.015625, P(S = 1) =
0
3
3
1
2
2
1
  0.75 0.25  0.140625, P(S = 2) =   0.75 0.25  0.421875, P(S = 3) =
1
2
 
 
3
3
0
  0.75 0.25  0.421875.
3
b Table of expected frequencies: multiply the probabilities by total number of seeds 50.
Number of seeds germinating
0
1
2
3
Expected frequency
0.78
7.03
21.09
21.09
c Expected value of no seeds germinating is less than 5. Hence, combine this with the data of
1 seed germinating:
Number of seeds germinating
0,1
2
3
Observed frequency
15
15
20
Expected frequency
7.81
21.09
21.09
d v = (n – 1) = 3 − 1 = 2.
e H0: The number of germinating seeds follows a binomial distribution.
H1: The number of germinating seeds does not follow a binomial distribution.
Use GDC to find the values χ 2  8.43 and p = 0.0147.
f
Since 0.0147 < 0.05 and 8.43 > 5.991, H0 is rejected.
3
2 a Expected probabilities for S ~ B(3, 0.5): P(S = 0) =   0.500.53  0.125, P(S = 1) =
0
3
3
3
1
2
2
1
3
0
  0.5 0.5  0.375, P(S = 2) =   0.5 0.5  0.375, P(S = 3) =   0.5 0.5  0.125.
1
2
3
 
 
 
Table of expected frequencies: multiply the probabilities by total number of families: 100.
Number of boys
0
1
2
3
Expected frequency
12.5
37.5
37.5
12.5
b No expected frequency values below 5.
© Oxford University Press 2019
8
Worked solutions
c v = (n – 1) = 4 − 1 = 3.
d H0: The number of boys follows a binomial distribution.
H1: The number of boys does not follow a binomial distribution.
Use GDC to find the values χ 2  10.77 and p = 0.0130.
e Since 0.0130> 0.01 and 10.77 < 11.345, H0 is accepted.
0
3 a Expected probabilities for S ~ B(2,
1
2
2  1   5 
1
): P(S = 0) =        0.694, P(S = 1) =
6
0  6   6 
2
1
0
2  1   5 
2  1   5 
       0.278, P(S = 2) =        0.0278.
2  6   6 
1  6   6 
Table of expected frequencies: multiply the probabilities by total number of tosses: 250.
Number of 6s
0
1
2
Expected frequency
173.61
69.44
6.94
b No expected frequency values below 5.
c v = (n – 1) = 3 − 1 = 2.
d H0: The number of 6s follows a binomial distribution.
H1: The number of 6s does not follow a binomial distribution.
Use GDC to find the values χ 2  28.1 and p = 7.7 107 .
e Since 7.7 < 0.05 and 28.1 > 5.991, H0 is rejected.
4 a Since there are four answers, P(getting a question right) = 0.25.
b,c Model this using the binomial distribution: X ~ B(5,0.25).
Number correct
0
1
2
3
4
5
Probability
0.2373
0.3955
0.2637
0.0879
0.0146
0.0010
Expected frequency
118.65
197.75
131.84
43.95
7.30
0.50
d Expected frequency of 5 correct answers is less than 5, hence need to combine the data for
4 and 5 correct answers. The new table is:
Number correct
0
1
2
3
4,5
Observed frequency
38
66
177
132
87
Expected frequency
118.65
197.75
131.84
43.95
7.8
e v = (n – 1) = 5 − 1 = 4.
f
H0: The number of correct answers follows a binomial distribution.
H1: The number of correct answers does not follow a binomial distribution.
Use GDC to find that χ 2  1139 and p = 0.
g H0 is rejected as 0 < 0.05 and 1138 > 9.488.
Exercise 8G
1 a H0: x
1
= x
2
(there is no difference in the weights)
H1: x 1 < x 2 (there is a difference in the weights of the apples: apples from the shade
weigh less).
b This is a one-tailed test as Petra is trying to find out if the trees in the shade weigh less.
c Use GDC to find t = − 0.687 and p = 0.251.
d Since 0.251 > 0.10, H0 is accepted.
2 a H0: x
1
= x
2
(there is no difference between the weights of town and country babies),
H1: x
1
> x
2
(babies born in the country weigh more than babies born in the town).
b This is a one-tailed test as Fergus is trying to find out if babies born in the country weigh
more than babies born in the town.
c Use GDC to find t = − 0.1913 and p = 0.575.
d Since 0.575 > 0.10, H0 is accepted. Comment: author used t-test at the 5% significance
level, not sure why the answers are different.
© Oxford University Press 2019
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Worked solutions
3 a H0: x
1
= x
2
(there is no difference between the lengths of the beans),
H1: x
1
 x
2
(there is a difference between the lengths of the beans).
b This is a two-tailed test as Jocasta is interested in finding out if the lengths are different.
c Use GDC to find t = − 3.126 and p = 0.00584.
d Since 0.00584 < 0.05, H0 is rejected.
4 a H0: x
1
= x
2
(there is no difference between the lifetimes of the bulbs),
H1: x
1
 x
2
(there is a difference between the lifetimes of the bulbs).
b This is a two-tailed test.
c Use GDC to find t = 0.3 and p = 0.769
d Since 0.769 > 0.05, H0 is accepted.
5 a H0: x
1
= x
2
(there is no difference between the weights of the girls and the boys),
H1: x
1
 x
2
(the boys weigh less than the girls).
b This is a one-tailed test (testing if the boys weighed less than the girls).
c Use GDC to find t = − 2.45 and p = 0.015.
d Since 0.015 < 0.05, H0 is rejected.
6 a H0: x
1
= x
2
(there is no difference between the weight loss with and without the remedy),
H1: x
1
 x
2
(people lose more weight with the remedy than without it).
b This is a one-tailed test (testing if the weight loss is higher with the remedy).
c Use GDC to find t = 2.84 and p = 0.00539.
d Since 0.00539 < 0.01, H0 is rejected.
7 a H0: x
1
= x
2
(there is no difference in the lengths of the sweetcorn cobs),
H1: x
1
 x
2
(there is a difference in the lengths of the sweetcorn cobs).
b This is a two-tailed test.
c Use GDC to find t = 0.535 and p = 0.600.
d Since 0.600 > 0.10, H0 is accepted.
Chapter review
1 a The ranks are:
Height,
12
10.5
10.5
9
8
7
6
4.5
4.5
3
2
1
2
1
3
4
5
8.5
6
7
8.5
10
11.5
11.5
cm
Time, s
Use GDC to find the PMCC for the ranked data: rs = − 0.953.
b It indicates strong and negative correlation between the height and the time it took to run,
i.e. the taller the person, the faster they are.
2 a Instead of the quantitative data, the ranks of tennis players are given.
b The ranks are:
Tennis ranks
1
2
3
4
5
6
7
8
Aces
1
2
6
3
4.5
4.5
7
8
Use GDC to find the PMCC for the ranked data: rs = 0.850.
c There is a strong and positive correlation between the tennis rank of the player and the
number of aces they hit, i.e. the higher the rank, the more aces they are likely to hit.
3 a H0: colours of the eggs laid are independent of the type of the hen. H1: colours of the eggs
laid are not independent of the type of hen.
© Oxford University Press 2019
10
Worked solutions
30
. Probability that an egg chosen at
90
b Probability that a hen chosen at random is Leghorn is
42
. If the two events are independent, the expected number of white
90
30 42
eggs laid by Leghorn hens is 90 
= 14.

90 90
random is white is
c v = (rows − 1)(columns − 1) = (2 − 1)  (3 − 1) = 2.
d Use GDC to find the values χ 2  21.7 and p = 1.94 105 .
e Since 21.7 > 5.991, H0 is rejected.
4 a H0: favourite colour of car is independent of gender; H1: favourite colour of car is not
independent of gender.
b Probability that a person chosen at random likes white cars is
20
. Probability that a person
100
48
. If the two events are independent, the expected
100
20
48

number of males who like white cars is 100 
= 9.6.
100 100
chosen at random is a male is
c v = (rows − 1)(columns − 1) = (2 − 1)  (4 − 1) = 3.
d Use GDC to find the values χ 2  9.43 and p = 0.0241.
e Since 9.43 > 6.251, H0 is rejected.
5 a The table of expected values:
Day
Sun
Mon
Tue
Wed
Thu
Fri
Sat
Expected frequencies
20
20
20
20
20
20
20
b v = (n – 1) = 7 − 1 = 6.
c H0: The data satisfies a uniform distribution.
H1: The data does not satisfy a uniform distribution.
Use GDC to find the values χ 2  10and p = 0.1247, since 10 < 12.592, H0 is accepted.
6 a The table of expected values:
Number on die
1
2
3
4
5
6
Expected frequencies
15
15
15
15
15
15
b v = (n – 1) = 6 − 1 = 5.
c H0: The data satisfies a uniform distribution.
H1: The data does not satisfy a uniform distribution.
Use GDC to find the values χ 2  0.67 and p = 0.98, since 0.67 < 15.086, H0 is accepted.
7 a Let S ~ B(2,
1
), then P(S = 0) = 0.25 and expected value is 0.25  60 = 15.
2
b Use binomial distribution defined above to find the expected frequencies:
Number of tails
0
1
2
Expected frequency
15
30
15
c All expected frequencies are higher than 5.
d v = (n – 1) = 3 − 1 = 2.
e H0: The number of tails follows a binomial distribution.
H1: The number of tails does not follow a binomial distribution.
Use GDC to find the values χ 2  1.2 and p = 0.549.
f
Since 1.2 < 5.991, H0 is accepted.
8 Let X ~ N(158,42), and calculate the probabilities using GDC. Multiply them by 500(total number
of girls).
© Oxford University Press 2019
11
Worked solutions
a Table with expected frequencies:
Height, x
cm
x < 152
152 ≤ x <
156
156 ≤ x <
160
160≤ x <
164
164 ≤ x
Expected
frequency
33.40
120.87
191.46
120.87
33.4
b There are no expected values less than 5.
c v = (n – 1) = 5 − 1 = 4.
d H0: The heights are normally distributed with mean of 158 cm and standard deviation of 4
cm.
H1: The heights are not normally distributed with mean of 158 cm and standard deviation of
4 cm.
Use GDC to find the values χ 2  20.60and p = 0.00038.
e Since 0.00038 < 0.10and 20.6 > 7.770, H0 is rejected.
9 a H0: x
H1: x
1
= x
2
(there is no difference in the test scores)
1
 x
2
(there is a difference in the test scores).
b This is a two-tailed test.
c Use GDC to find t = −0.421 and p = 0.678.
d Since 0.678 < 0.05, H0 is accpted.
10 a
H0 : 1  2
A1
H1 : 1  2
A1
b One-tailed test
c t  value  0.706
p  value  0.244
A1
M1A1
A1
d
M1R1
11 a
0.244  0.05 so accept H0
i.e. there is no significant difference in the results of the two schools.
H0 : 1  2
A1
H1 : 1  2
A1
b
t  value  1.735
p  value  0.0499
M1A1
A1
c
0.0499  0.05 so reject H0 .
M1R1
i.e. there is significant evidence to suggest that older students read fewer books
12 a H0 : 1  2
H1 : 1  2
A1
A1
b
t  value  1.942
p  value  0.0725
M1A1
A1
c
0.0725  0.1 so reject H0 .
M1R1
i.e. there is significant evidence to suggest that there is a difference in average battery
length
13 a
M1A1A1
b
rS  0.707
c
rS is positive and reasonably close to 1 ,
M1A1
indicating a fairly strong positive correlation between a person’s age
and their reaction time.
d Include a greater number of participants
Ensure the participants were spread equally throughout the age range
© Oxford University Press 2019
R1
R1
R1
R1
12
Worked solutions
14 a
M1A1A1
b
M1A1
rS  0.596
c Neeve is incorrect.
A1
A value of rS  0.596 indicates a small but significant negative correlation between a
person’s age and the hours per week they watch TV.
However, you cannot say this is causal.
R1
R1
i.e. You cannot conclude that your age affects the amount of TV you watch
15 A2
A1
B1
A1
C3
A1
D6
A1
E4
A1
F5
A1
16 a
H0 : Type of burger favoured is independent of age
A1
H1 : Type of burger favoured is not independent of age
A1
b
3  1   4  1  6
A1
c
 2calc  12.314
M1A1
p  value  0.0553
d

2
calc
A1
 12.314  12.59 , therefore we accept H0.
A1R1
i.e. the type of burger favoured is independent of age.
17 a
M1A1A1
b
H0 : Smoking is independent of age
A1
H1 : Smoking is not independent of age
A1

2
calc
 9.408
M1A1
p  value  0.00216

2
calc
A1
 9.408  6.64 , therefore we reject H0 and accept H1
i.e. smoking is not independent of age
18 a H0 : Movie preference is independent of gender
A1R1
A1
H1 : Movie preference is not independent of gender
A1
b
5  1  2  1  4
A1
c

2
calc
 11.111
M1A1
p  value  0.0253
d
A1
 2calc  11.111  9.49 , therefore we reject H0 and accept H1
A1R1
i.e. Movie preference is not independent of age.
19 a
© Oxford University Press 2019
13
Worked solutions
M1A1
b
H0 : The data satisfies a binomial distribution
A1
H1 : The data does not satisfy a binomial distribution
A1

2
calc
 7.583
M1A1
p  value  0.108
A1
 2calc  7.583  9.49 , therefore we accept H0
A1R1
i.e. The observed data fits a binomial distribution
20 a
M1A1A1
b Re-writing:
M1A1
Degrees of freedom  3
A1
H0 : The data satisfies a normal distribution
A1
H1 : The data does not satisfy a normal distribution
A1
The critical value is  5% 3  7.82
A1
2

2
calc
 10.47
M1A1
p  value  0.015

2
calc
A1
 10.47  7.82 , therefore we reject H0 and accept H1
A1R1
i.e. The observed data does not fit a normal distribution with mean 62 and standard
deviation 16.
© Oxford University Press 2019
14
Worked solutions
9
Modelling relationships with functions:
power functions
Skills check
1 a
2x  3  x  4
 2x2  3x  8x  12  2x2  5x  12
b
7x  5 2x  3
 14x2  21x  10x  15
 14x2  11x  15
2 a
2x2  5x  2
 2x2  4x  x  2   x  2 2x  1
b
5x2  13x  6
 5x2  15x  2x  6   x  3 5x  2
3 a
 
1  2 22  3 2  c  8  6  c  2  c
c  3
b
 
5  5 12  1  c  5  1  c  6  c
c  11
Exercise 9A
1 a
Vertex is (−1.25, −1.125), x intercepts are (−0.5,0), (−2,0), y intercept is 2. Domain is
and range is y : y  1.125 .
b
Vertex is (3,16), x intercepts are (7,0), (−1,0), y intercept is (0,7), Domain is
range is y : y  16
and the
c
© Oxford University Press 2019
1
Worked solutions
49 
1
4 
Vertex is  , 
 , x intercepts are (−1,0) and  , 0  , y intercept is (0,−4) domain is
6
12


3 
49 

and the range is y : y  
.
12


d
Vertex is (−0.4,12.8), x intercepts are (1.2,0) and (−2,0), the y intercept is (0,12), the
domain is
and the range is y : y  12.8 .
2 a
0  0.4x2  2x  8
x 
2  22  4 0.4  8
2 0.4
 2.5 
5
16.8
4
Graph cuts the x axis at 7.62,0 and  2.62,0 .
b Y intercept is (0,−8)
5
5


 2.5  4 16.8  2.5  4 16.8 


c Equation of symmetry is x 
2
 2.5
d Point of intersection is (1.26,−9.88).
Exercise 9B
1 a
a  1, b  4,c  2
b
a  2, b  2,c  3
c
a  2, b  1,c  1
d
a  1, b  2,c  3
e
a  5, b  1,c  10
2
9  a 0  b 0  c  c.
2
9  a  1  b  1  9.
2
18  a  b.
As the graph is symmetric in the line x=4, the reflection of the point ( −1,9), (9,9), is also on
the graph.
9  a  9  b  9  9
2
2  9a  b . Solving these simultaneous equations:
18  2  20  a  b  9a  b  10a
a  2, b  2  18  16, c  9
Exercise 9C
1 a
b
h
70  2x
 35  x.
2
A  x 35  x   35x  x2
© Oxford University Press 2019
2
Worked solutions
c
d The x intercepts are (0,0) and (35,0)
e These two values are the values that x must lie between.
f
The line of symmetry is
x 
35  0
 17.5 . This tells us that the largest area occurs when
2
the width is 17.5cm
2 a
a  6,d  10  6  4.
n
2  6  4  n  1
2
 2n  n  1  6n  2n2  4n
Sn 
880  2n2  4n.
2n2  4n  880  0.
b
The vertex is (−1,−882), the x intercepts are (−22,0) and (20,0). The y intercept is −880.
c The positive x intercept is (20,0).
d The positive x intercept tell us the that the sum of the first 20 terms is 880.
3 a

 
P  x   0.12x2  30x  0.1x2  400

 0.22x2  30x  400
b
The vertex is (68.2, 623), The x intercept is (15.0,0) and (121,0) and the y-intercept is
−400.
c The x intercepts are the upper and lower bounds for the number of books that can be
produced per week such that the company makes a product.
d The axis of symmetry is x=68.2. This tells us that in order to maximize profit 68 books
should be produced per week.
© Oxford University Press 2019
3
Worked solutions
4 a
The vertex is (26,5), the x intercepts are (6,0) and (46,0) and the y intercept is (0,−3.45)
b The x intercepts are (6,0) and (46,0). This means that the ball is in the air between the
horizontal distances 6m and 46m.
6  46
 26 . This means the ball is at the maximum height
2
when it’s at the horizontal distance of 26m
c The line of symmetry is
x 
5 a
The vertex is (1.5,9), the t intercepts are (0,0) and (3,0) and the h intercept is (0,0).
b The t intercepts are (0,0) and (3,0). The ball is in the air between 0 seconds and 3 seconds.
c The axis of symmetry is x 
30
 1.5 The ball is at the maximum height after 1.5 seconds.
2
Exercise 9D
1 a
f  x   x 2  4x  2
i
f 0  2. 2  f  x2   x22  4x2  2.
f  x2   x22  4x2  x2  x2  4
x2  4.
 4 0  4  0 
Vertex is 
,f 
   2, 2 .
 2 
 2
ii The line of symmetry is x=2.
b
f  x   2x 2  2x  3
i
f 0  3. 3  f  x2   2x22  2x2  3.
f  x2   2x22  2x2  2x2  x2  1
x2  1.


Vertex is 0.5, f 0.5  0.5,2.5 .
ii Line of symmetry is x 
c
10
 0.5.
2
f  x   2x2  x  1   x  1  2x  1
i
f 0  1. 1  f  x2   2x22  x2  1.
0  2x22  x2  x2  2x2  1 . x2  2.
© Oxford University Press 2019
4
Worked solutions


Vertex is 0.5, f 0.5  0.5,2.5 .
0.25, f 0.25   0.25,1.125 .
ii Line of symmetry is x 
d
1  0.5
 0.25.
2
f  x    x2  2x  3
i
f 0  3
f  x2   3  x22  2x2  3.
0  x22  2x2  x2  x2  2 . x2  2.
Vertex is (1, f 1  1, 2 .
ii Line of symmetry is x 
e
20
1
2
f  x   5x2  x  10
i
f 0  10
f  x2   10  5x22  x2  10
5x22  x2  0  x2 5x2  1 . x2  0.2


The vertex is 0.1, f  0.1   0.1, 10.05 .
ii Line of symmetry is x 
2 a
0  0.2
 0.1.
2
f 0  1.
f  x2   0.018x22  0.54x2  1  1
0  0.018x2  x  30 . x2  30.
 30  0 
Maximum height is f 
  f 15  5.05 metres.
 2 
b
x
0.54  0.542  4 1  0.018
2  0.018
 31.7 metres.
This is the distance from Zander that the ball hits the ground.
c
3 a
f 0  1. The y intercept is (0,1), this is the height at which the ball is hit.
f 0  1.5
f  z   1.5  1.5  0.75z  0.05z 2
z2  15z  z  z  15  0 , z  15.
 15  0 
Maximum height is f 
  f 7.5  4.3125.
 2 
b The axis of symmetry is x = 7.5.
c
0  1.5  0.75x  0.05x2
x 
0.75  0.752  4  0.05 1.5
2  0.05
 16.8
This is the horizontal distance travelled by the shotput while it traveling through the air.
d The y intercept is 1.5. This is the height from which the shot put is thrown.
4 a
f  x   0.06x2  1.2x  0.06x 20  x  .
© Oxford University Press 2019
5
Worked solutions
 20  0 
f
  f 10  6 metres.
 2 
Maximum height is
b Axis of symmetry is x=10
c The x intercepts are x=0 and x=20. The ball is kicked at 0m and lands on the ground at
20m.
5 a
f 0  10.67.
f  z   10.67  1.67z  0.0417z 2
0  0.0417z 2  1.67z
 0.0417z  z  40.05
 40.05  0 
Maximum depth is f 
  6.05 metres.
2


b
x 
1.67  1.672  4 0.0417 10.67 
2 0.0417
x  32.1 and x  7.98 are the x intercepts. These mean that the parabola part of the ramp
starts at 7.98m and ends at 32.1m.
6 a
f  0  1
f  z   1  1  7.25z  1.875z 2
0  1.875z 2  7.25z
 1.875z  z  3.973
Maximum height is
 3.973  0 
f
  8.01 m
2


b
0  1  7.25t  1.875t 2
t 
7.25  7.252  4  1.875 1
2  1.875
= 4 seconds.
7 a
b
h
100  2x
 50  x
2
A  x 50  x 
c
d The x intercepts are x = 0 and x = 50.
e The y intercept is y=0.
f
The axis of symmetry is x 

50  0
 25 .
2

g The vertex is 25, f 25  25,625 .
h The maximum area of the picture frame is 625m2.
© Oxford University Press 2019
6
Worked solutions
8 a 96m
b
f  z   96  0.0147z2  2z  96
0.0147z 2  2z
 0.0147z  z  136.05  0
Maximum height is
 136.05  0 
f
  164 m
2


c
0  0.0147x2  2x  96
x 
2  22  4  0.0147  96 
2  0.0147
 173.66
174m
Exercise 9E
i
a
5  9  45 , (0,45)
b
x  5 and x  9 , (5,0), (9,0)
 9  5  9  5 
c The vertex is at 
,f 
   7, 4
 2 
 2
d The axis of symmetry is x = 7
e
ii a
0,2  3  1  0, 6
b (−3,0), (1,0)
 3  1  3  1  
c The vertex is at 
,f 
    1, 8
 2 
 2
d The axis of symmetry is x = −1
e
iii a
0, 3  1  3  0, 9
b (1,0), (3,0)
 3  1  3  1
c The vertex is at 
,f 
   2,3
 2 
 2
d The axis of symmetry is x = 2
e
© Oxford University Press 2019
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Worked solutions
iv a
0, 2  1  2  0, 4
b (−1,0), (2,0)
2 1
c The vertex is at 
,f
 2
 2  1

   0.5, 4.5
 2 
d The axis of symmetry is x=0.5
e
v a
0,1  1  0, 1
b (1,0), (−0.5,0)
 1  0.5  1  0.5  
c The vertex is at 
,f 
   0.25, 1.125
 2 
 2
d The axis of symmetry is x = 0.25
e
© Oxford University Press 2019
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Worked solutions
Exercise 9F
1
4  a 0  b 0  c  c
2
0  a  4  b  4  4  0  4a  b  1
2
0  a 1  b 1  4  a  b  4
2
5a  5  0.
a 1
b  4  1  3.
y  x2  3x  4
2 6  a  0  b  0  c  c
2
0  a  1  b  1  6  a  b  6
2
0  a 2  b 2  6  4a  2b  6
2
6a  18  0
a  3
b  a6  3
y  3x2  3x  6
3
4  a 0  b 0  c  c
2
0  a  4  b  4  4  4a  b  1  0
2
2
1
1
0  a    b    4  a  2b  16  0
2
2
9a  18  0
a2
b  4 2   1  7
y  2x2  7x  4
4
6  a 2  b 2  c  4a  2b  c
2
6  a  4  b  4  c  16a  4b  c
2
4  a 3  b 3  c  9a  3b  c
2
8  18a  6b  2c
12  8  2a
a2
2  2b  c
26  4b  c
2b  24
b  12
c  22
y  2x2  12x  22
5
− 13  a 0  b 0  c  c
2
13  a  10  b  10  13
2
 13  100a  10b
10a  b  0
© Oxford University Press 2019
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Worked solutions
12  a  5  b  5  13  25a  5b  13
2
5  5a  b
5a  5
a  1.
b  10
y  x2  10x  13
6
4  a  0  b  0  c  c
2
4  a 2  b 2  4  4a  2b  4
2
2a  b  0
6  a 1  b 1  4  a  b  4
2
2  ab
a  2
b4
y  2x2  4x  4
Exercise 9G
1
x intercept is (−3,0), y intercept is (0,27)
2
x intercepts are (−1.15,0) and the y
intercept is (0,3). The maximum is
(−0.215,3.11). The minimum is
(1.55,0.369)
© Oxford University Press 2019
10
Worked solutions
3
y intercept is (0,0), x intercepts are (−2,0), (0,0) and (3,0), Maximum point is (−1.12,8.12)
and the minimum point is (1.79, −16.4)
4
y intercept is (0,20), x intercept is (−0.899,0), No minimum or maximum points.
5
y intercept is (0,0), x intercepts are (−1,0), (0,0) and (4,0), maximum point is (−0.528,3.39),
minimum point is (2.53,−39.4)
6 a
The equation of the inverse y  x3  3
x3  y  3
Inverse is f 1  x  
3
x 3
© Oxford University Press 2019
11
Worked solutions
b
y  4x 3
x3 
y
4
Inverse function is f 1  x  
3
x
4
3
x 1
.
2
c
y  2x 3  1
y  1  2x 3
x3 
 y  1
2
Inverse function is f 1  x  
7
a The x intercepts are (5,0), (2,0),
(−1,0). The y intercept is (0,10)
b The vertices are (3.73,−10.4) and (0.270,10.4)
 3.73  2.70 10.4  10.4 
,
c The point of rotational symmetry is 
  2,0 .
2
2


d x is sent to  x by this reflection, so the new equation is
g  x   f  x   x3  6x2  3x  10
Exercise 9H
1 a
V  6  2x  8  2x  x
 4x 3  x   4  x 
© Oxford University Press 2019
12
Worked solutions
b
c The x intercepts are (0,0), (3,0) and (4,0)
d The value of x must be between 0 and 3
e The local minimum is (3.54,−3.52) and the local maximum is (1.13,24.3).
f
V cannot take the value of the local minimum as volume cannot be negative.
2 a
b The local maximum is (1.92,10.0) and the local minimum is (6.08,3.99).
c The difference in height is 6.01m
3 a
b The number of fish after 12 years is N 12  51 .
c The minimum number of fish is 43
4 a
b The graph does not have a local maximum or minimum
c
f 20  20  15  4000  4125,
3
f  40   40  15  4000  19625
3
d It is not a suitable model: this predicted number of cases from the model keeps increasing
as x increases, while in reality at some point the number of cases will start decreasing.
© Oxford University Press 2019
13
Worked solutions
5 a
b The local minimum is −2.76, the local
maximum is 11.6
c
f 10  6.89
6 a 5.17
b
The local maximum is (15.2,13.2) and the local minimum is (33.0,10.9)
c Not a suitable model as after 33 minutes the number of bacteria continues to increase and
never stops increasing.
7 This is a suitable model for the first 5 seconds, however after this the graph continues
increasing indefinitely and so is no longer a suitable model after 5 seconds.
8 a
The lowest temperature is 19.2. The highest temperature is 24.7.
b
f 29  19  f 10  20.7
c
The intersections of the two lines occurs when x=0.984, 12.03, 22.71. So the air conditioner
must be switched on for 0.984  22.71  12.03  11.7 hours.
d This model would not be useful as after 24 hours the graph just continues to decrease while
it is very unlikely that the temperature will do this.
© Oxford University Press 2019
14
Worked solutions
Exercise 9I
1 a
Point of intersection is (1.18,1.82)
b
The points of intersection are (1,6) and (−1,6)
c
Points of intersection are (3,10), (0,1) and (−1,−2).
2
The ball is at a height of 8 metres when t = 0.528 and t = 3.35 seconds
3
The arrow is at a height of 10 metres when t = 0.337 and t = 5.60 seconds
© Oxford University Press 2019
15
Worked solutions
4
Number of units sold to make a profit of $300 is 18.8 or 19 units and 38.9 or 39 units
Exercise 9J
1
pn
p  kn
21  12k
k 
7
4
p
7
n
4
Cost for traveling 40km is p 
2 a
7
 40  70 .
4
$70 AUD
d  t2
d  kt 2
9  4k  k  2.25
d  2.25t 2
b
d  2.25 5  56.25 metres
c
26.01  2.25t 2
2
t2 
26.01
2.25
t  3.4 seconds
3 v  r3
v  kr 3
113.1  27k
k  4.189
v  4.189 5  524cm3
3
© Oxford University Press 2019
16
Worked solutions
Exercise 9K
1 a
m  w 1
k
w
m
150 
k
 k  150  20  3000
20
m
3000
.
w
b
m
3000
 200
15
2 a
h
1
,
p
h
k
p
2
k
 k  12
6
h
12
p
b
h
12
 1.2 hours.
10
c
4
12
p
p
12
 3 people
4
3 a
y  x 2 ,
k
x2
y 
3
k
 k  48
16
y 
48
x2
b
y 
48 48 4


62
36 3
c
12 
48
x2
x2 
48
4
12
x  2.
4 V 
1
p
V 
k
p
180 
k
20
k  3600
V 
3600
p
© Oxford University Press 2019
17
Worked solutions
5
V 
3600
 40m3.
90
c
1
n
c 
k
n
10 
k
16
k  160
c 
160
n
c 
160
 8 pieces.
20
6 a
I  d 2
I 
k
d2
I1 
k
252
I2 
k
502
k
2
I1
502
 25 
4.
k
I2
252
502
The intensity decreased by a factor of 4.
b
I3 
k
402
k
2
I2
502
50


 1.5625.
k
I3
402
2
40
The intensity increases by a factor of 1.5625
7 a
yx

1
2
k
y 
1
x2
3
k
4
k  12
y 
12
1
x2
y 
12
4
b
1
2

12
6
2
3 12
 1
2
x2
1
x2 
12 24

8
3
3
2
© Oxford University Press 2019
18
Worked solutions
x  64
Exercise 9L
1
xy  x  x  6
 6  0  6  0 
Vertex of f  x   x(x−6) is at 
,f 
   3, 9 .
 2 
 2
Minimum value is −9
2 a
V  x  9  2 x   6  2x 


 x 54  30x  4x2  4x3  30x2  54x
b
Maximum value of V is 28.5cm3 .
3
Maximum daily profit is $120 and 50 cakes are needed to be baked to make it.
4
The maximum profit is $391.11
5
The local maximum is 13.2 million, the local minimum is 10.9 million
6
S  2 r 2  2 rh  5000
V   r 2h
© Oxford University Press 2019
19
Worked solutions
5000  2 r 2 
2V
r
V  2500r   r 3
The radius that gives the maximum volume is 16.3cm, the height is 32.6cm and this gives a
maximum volume of 27145cm3.
Chapter review
1 a
The vertex is  0,2 , the y intercept is  0,2 . The x intercepts are



2,0 ,  2,0 .
b
The y intercept is (0,4), the x intercepts are (1,0) and (4,0) and the vertex is (2.5,−2.25)
2 a
b
h
400  2x
 200  x
2
A  x 200  x 
c
d The x intercepts are (0,0) and (200,0). These represent the lower and upper bounds for the
width of the picture.
3 a
a  1, b  6,c  7
b (0,−7)
c (−7,0) and (1,0)
d The axis of symmetry is x 

1  7
 3
2

e The vertex is 3, f  3   3, 16 
© Oxford University Press 2019
20
Worked solutions
4 a
b
f 20  0.00816 20  0.372 20  1.8  5.976 metres.
c
f 0  1.8
2
f  z   1.8  1.8  0.372z  0.00816z 2
0.00816 z 2  0.372z
 0.00816 z  z  45.59
z  45.59
 45.59  0 
Maximum height is f 
  6.04 metres
2


d
0.00816x2  0.372x  1.8  0
x 
0.372  0.3722  4  0.00816  1.8
 50 metres.
2  0.00816 
This is the horizontal distance at which the javelin hits the ground.
5 a The y intercept is 1.9. This is the height at which Anmol lets go of the stone.
b
h 0  1.9
h  z   2.2625z 2  8.575z  1.9  1.9
2.2625z 2  8.575z
 2.2625z  z  3.79  0
z  3.79
 3.79  0 
The maximum height is f 
  10.0 metres
2


c
d
f 1.9  f 1.8
1.9  1.8
 0.204ms1 . It is increasing.
0  2.2625x2  8.575x  1.9
x 
6 a i
8.575  8.5752  4 1.9  2.2625
2  2.2625
= 4 seconds.
(2,0), (4,0)
ii The axis of symmetry is x 

42
 3.
2

iii The vertex is 3, f 3  3, 3
b i
(−1,0), (5,0)
ii The axis of symmetry is x 

5 1
 2.
2

iii The vertex is 2, f 2  2, 36 
7 a
6  a 0  b 0  c  c
2
0  a  2  b  2  6  4a  2b  6
2
© Oxford University Press 2019
21
Worked solutions
0  a 3  b 3  6  9a  3b  6
2
36a  18b  54  36a  12b  24
 30b  30  0
b  1
a 1
y  x2  x  6
b
0  a  1  b  1  c  a  b  c
2
0  a  4  b  4  c  16a  4b  c
2
50  a  6  b  6  c  36a  6b  c
2
0  15a  5b
0  3a  b
50  35a  5b
7a  b  10
10a  10
a  1
b3
c4
y  x2  3x  4
c
5  a 0  b 0  c  c
2
10  a 2  b 2  5  4a  2b  5
2
10  a  8  b  8  5  64a  8b  5
2
40  16a  8b  20
50  80a  25
a
15
16
15 
15
 2b
4
b
45
8
y 
15 2 45
x 
x 5
16
8
8 a The volume of a cylinder is  r 2h .
400
 r2
b
h
c
A   r 2  2 rh
d
800
 400 
A   r 2  2 r  2    r 2 
r
 r 
© Oxford University Press 2019
22
Worked solutions
e
f
The minimum area is 239cm2 when r  5.03cm.
9 a
The y intercept is (0,−1) and the x intercept is
3
1
.
2
b
The y intercept is (0,3) and the x intercepts are (1,0), (−1,0) and (3,0)
10 a
0  a 1  b 1  c  a  b  c
2
10  a 0  b 0  c  c
2
0  a  5  b  5  c  25a  5b  10
2
30a  10  50  0
a2
5b  40
b8
y  2x2  8x  10.
b
0  a  2  b  2  c  4a  2b  c
2
0  a  4  b  4  c  16a  4b  c
2
0  24a  3c
24  a 0  b 0  c  c
2
a  3
© Oxford University Press 2019
23
Worked solutions
2b  12  24  12
b6
y  3x2  6x  24
11 a
b
f 5  0.04 5  0.9 5  7 5  70
3
2
 52.5. = 53 lilies.
c
f 12  0.04 12  0.9 12  7 12  70  46.48 = 46 lilies.
2
2
d The maximum number of lilies is 70 in 2000.
e The minimum number of lilies is 0 in 2018.
f
There are 60 lilies in 2001
12 a, b
c The lights meet the handrail at (4.39,1) and (25.65,1).
d The length of lights required is 25.65−4.39 = 21.3 metres.
13 a
ds
d  ks
b 100  1.25k
k  80
d  80s.
d  80 2  160 km
c
14 a
300  80s
s
300
 3.75 hours
80
p
1
n
p
k
n
6
k
 k  12
2
p
12
n
© Oxford University Press 2019
24
Worked solutions
b
p
12
 4 pieces
3
c
3
12
n
n
12
 4 children.
3
15 a From GDC,
maximum height is 2.38 m
b
M1
A1
x  1.07  H 1.07  1.073
A1
 1.07 . So ball just goes over the net.
c Choose X as being the value when y  0
R1
M1
From GDC X  14.86
A1
This is sensible, since if X  14.86 , then H  0 , making no sense with regard to the model.
R1
1
16 F  2
r
k
F  2
r
So k  Fr 2
 980  63702
980  63702
i.e. F 
r2
11 km above the earth’s surface, r  6370  11  6381 km
980  63702
So F 
 977 N
63812
17 a Substituting t  0 into P  x  gives P 0  £10850
b Maximum salary occurs when t  40 and is £53990
c From GDC,
P  35  t  7, 24, 32 years
d Salary decreases for 28.4  13.6  14.8
years.
14.8
 0.37
40
18 a y  0.34x2  5.04x  4.82
b 12.1  2.72  9.38 months
M1
A1
A1
A1
M1A1
A1A1
A3
M1A1
A1
A3
M1A1
c The maximum height of the quadratic curve is only 23.5 C , compared to 26 C in real life.
R1
When x  1 , the quadratic curve gives a reading of 9.5 C , some way off the actual 13 C .
R1
19 a
A2
© Oxford University Press 2019
25
Worked solutions
b 50 ferrets
c 0  t  9.2 42.8  t  72.2
A1
A2A2
1
k
l 
f
f
k  lf  400  13  5200
When l  10 ,
k 5200
f 

 520 Hz
l
10
M1A1
M1
20 l 
A1
© Oxford University Press 2019
26
Worked solutions
10
Modelling rates of change:
exponential and logarithmic functions
Skills check
1 a
42  49  411.
b
58
 56
52
2 a
0.03  24  0.72
b
0.28  150  42
4
2i  2  4  8  16  30
3 a
b
i 1
6
  k  3  4  5  6  7  8  9  39
.
k 1
Exercise 10A
1 a 2
b
u15  3  214  3  16384  49152
2 a 1.2
b
l12  0.5  1.211  3.72m
c Unlikely to be accurate, as the plant cannot grow indefinitely or live forever.
3  27r 2
3 a
r2 
r 
1
9
1
3
x  27 
1
3
b
4 a
1
9
3
c
u8  27 
1
1

7
3
81
50
10
 5.
 5. The ratio between successive terms is the same, so it is a geometric
10
2
sequence.
b
f7  2  56  31250
c
f365  2  5364. This answer is unlikely to be correct as it is likely to be far larger than the
population of the town.
5
u2  u1r  32. u4  u1r 3  2.
r2 
1
16
1
4
r 
u1 
u2
 128.
r
Exercise 10B
1 a
p1  200000. p3  200000r 2  264500
r2 
264500
 1.3225.
200000
r  1.15
p2  200000  1.15  230000.
b
p5  200000  1.154  349801
c Probably as the population in the world is also increasing.
© Oxford University Press 2018
Teacher notes
1
Worked solutions
2
p1  2.20, r  1.0265.
p5  2.20  1.02654  2.44.
US$2.44
3 v1  45000, r  0.95
v7  45000  0.956  33079.14.
33079.14 Euros.
4 a
v1  15000, r  0.88
v4  15000  0.883  10222.08.
b
10222.08 euros.
5000  15000  0.88n
0.88n 
1
3
n  8.59  years.
Exercise 10C
1 a
2 a
b
6
 3.
2
Common ration is 3.
5
p6  2500  1.02  2760.20.
S12  2500
1.02  1
 30421.79
1.02  1
p8  12000  1.0127  13045
4 a
a3  a1r 2  8.
S10 

  59048
2 310  1
3 1
EGP 2760.20
12
3 a
b
EGP 33530.22
1.0128  1
 100130
1.012  1
b
S8  12000
c
1 8
4 1
S8  2
 10922.5
4 1
a5  a1r 4  128.
r 2  16
r  4.
b
5 a
b
6 a
7
a1 
8
1
 .
16 2
1
a8   47  8192
2
f2  270000  1.04  280800.
S6  270000


f3  280800  1.04  292032
6
1.04  1
 1790903.38
1.04  1
f2  75  1.02  76.50 euros
b
S5  75
1.025  1
 390.30 euros
1.02  1
36
 1.5  r
24
1.5
9
S9  24
8 a
  1797.28
1
1.5  1
6000 3
  r.
8000 4
m6  8000  3 / 4  1898.44 CAD
5
b
9
S6  8000
3
 3  r.
1
0.756  1
 26304.69 CAD
0.75  1
19683  3n 1
© Oxford University Press 2019
2
Worked solutions
n  10
S10 
10 a
310  1
 29524
3 1
c16  500  1.0115  580
b
S15  500
1.0115  1
 8048
1.01  1
Exercise 10D
1 Merel: 5000  5000  0.045  15  8375
Misty: 5000  1.04415  9538.44 .
Misty has more money at the end of 15 years
2 a
b
a9  500000000  1.0328  643291158.90
2  1.032n
n  22.01 years
3 a
5066.55  4500 1  r 
5
(1  r )5  1.1259
r  2.400
2.4%
b
8000  4500  1.024n
1.024n 
16
9
n  24.26
25 years
4 1500  1000  1.075n
1.075n  1.5
n  5.61
It will take 6 years.
5 a
b
h6  5000  1.0125  5307.29
5307.29 euros
5675.33  5307.19  1  r 
5
1  r 
5
 1.06935
r  1.35
1.35%
Exercise 10E
127
1 a
0.046 

FV  8000 1 

12 

 11032.28
MYR11032.28
12 n
b
0.046 

2  1 

12 

12n  181.167
n  15.1 years
2 a
b
0.034 

FV  3500 1 

4 

0.034 

4000  x 1 

4 

46
 4288.34 euros
46
© Oxford University Press 2019
3
Worked solutions
x  3265 euros.
c
0.034 

2  1 

4 

4 n
4n  81.89
n  20.47
21 years
24
3 a
0.05 

FV  40000 1 

2 

 48736.12
SGD48736.12
12 4
b
0.049 

FV  48736.12 1 

12 

 59265.14
0.038 

4 Mr Chen: FV  20000 1 

4 

SGD59265.14
45
 24163.31
Mrs Chang: FV  20000 1  0.039  24216.30
5
Mrs Chang earns more.
1210
5 a
0.04 

FV  400 1 

12 

 596.33
£596.33
12 n
b
0.04 

2  1 

12 

12n  208.29
n  17.35
18 years
120
6
r 

2  1 

1200 

1
r
 1.00579 
1200
r  6.95
0.026 

7 a Colin: FV  1500 1 

4 

46
 1752.35
126
0.0255 

Ryan: FV  1500 1 

12 

 1747.70
Kyle: FV  1500 1  0.0275  1765.15
6
Kyle has the most.
12 n
b
0.0255 

2500  1500 1 

12 

12n  240.64
n  20.05
21 years
c
2  1  0.0275
n
n  25.55
26 years
Exercise 10F
© Oxford University Press 2019
4
Worked solutions
1
FV 

  77434.02 .
1500 1.035
30
1
0.035
guaranteed interest rate for 30 years.
£77434.02 It is a good investment as she has a
2
0.028 60


2500 1 
)  1
12

  160806.57 .
FV 
0.028
12
3
P 
4
0.048
6000
12
P 
 £112.68 Any sensible reason for either choice.
60
0.048 

1  1 
12 

0.06  25000
1  1  0.06 
5 500 
MXN160806.67.
 3396.70 euros
10
0.008  PV 
60
1  1  0.008
0.008PV  190.017
PV  23752.11 . £23752.11
6
0.009  20000
P 
1  1  0.009
120
 273.24
US$273.24
Saving for pension or any other sensible reason.
Exercise 10G
2012
1 a
0.05 
0.05 
1

12 
12 
A  150000
 989.93
2012
0.05 

1
1  12 


989.93 euros
1012
b
0.05 
0.05 
1
12 
12 
A  150000
 1590.98
1012
0.05 

1
1  12 


1590.98 euros
c The difference in monthly payments is 601.05 euros. In the 20 years case, the total
amount payed is 240  989.93  237583.20 euros so the total amount of interest payed is
87583.20 euros. In the 10 year case the total amount payed is 120  1590.98  190917
and so the interest paid is 40917.60 euros.
3012
0.023 
0.023 
1
12 
12 
2 A  350000
3012
0.023 

1

1

12 

amount for a mortgage.
3 A
4

2000 0.04 1  0.04 
5
1  0.04
A  28000
5
1
  449.25 , 449.25 euros
125
 0.1 

0.1 


1


 12 

12



125
0.1 

1  12 


 1346.80 , £1346.80 per month. This is a reasonable
 594.91 , US$594.91. This seems a reasonable amount per
1
month.
© Oxford University Press 2019
5
Worked solutions
Exercise 10H
1 a (0,2)
b y=1
c Increasing
2 a (0,−2)
b y = −3
c Decreasing
3 a (0,1)
b y=0
c Increasing
4 a (0,3)
b y=2
c Increasing
5 a (0,−2)
b y = −5
c Increasing
6 a (0,7)
b y=3
c Decreasing
7 a (0,4)
b y = −1
c Increasing
8 a (0,1)
b y = −1
c Decreasing
Exercise 10I
1 a
f 0  2 1.1  2.
0
b
f 10  2 1.1
c
2 1.1  10.
10
2m
 5.187 
5.187m
t
t  16.89 weeks
f 0  130 0.85  130 grams
b
f  4  130 0.85  67.86 g
y 0
d
1
 0.85x  x  4.265 years
2
b
T 0  70  21  91
c
T 5  70 1.45
d
50  70 1.45
f
21oC
2 a
c
0
4
3 a
1.45
x

x
5
 21  31.92
 21
70
29
x  2.37 minutes
e y = 21
4 a
36000  34200
 100  5%
36000
b V  36000  0.956  26463.31.
c
£26463.31
n
20000  36000  0.95
0.95n 
5
9
n  11.46 years
d y=0
5 a
55000000  50000000
 100  10 %
50000000
b
K  50000000  1.13  66550000 Kangaroos
c
2  1.1n
n  7.273
2025
© Oxford University Press 2019
6
Worked solutions
6 a
b Exponential
c
f  x   mbx
f 1  mb  5.
b16 
f 17  mb17  92.
92
92
 b  16
 1.19964...  1.20
5
5
5  m  1.19964..  m  4.16791...  4.17
y  4.17  1.20 .
x
d
4.17  1.20
10
 25.8
This is a reasonable estimate.
7 a
b Linear decay
c From GCD: y  4.61x  49.6
d
4.61 10  49.6  3.5 minutes
Exercise 10J
1 a
b
f 0  e0  1.
f  x   10
x  2.30
2 a
b
f 2  102  100
f  x   500
2.70 weeks
c
f  x   2000
x  3.30 weeks
3 a
2  log3  2.48
b 101  x  10.
c
4 a
103  x  1000
0.5  ln3  1.60
b
x  e4.2  66.7
c
x  log3 5  1.46
5 a
b
f 0  4  e0  4  1  5
f  x   150  4  ex
© Oxford University Press 2019
7
Worked solutions
ex  146
x  ln 146  4.98. Around 5 minutes.
Chapter review
1 a
2
5
 2.5.
2
Common ratio is 2.5
b
a8  2  2.57 
c
S8 

78125
.
64
  130123
2 2.58  1
2.5  1
64
5
350  0.88  184.71.
3 a
$184.71
4
a3  a1r 2  363000
a5  a1r  439230.
r 2  1.21
r  1.1 so 10% per year
363000
 300000.
1.21
b
a1 
c
a9  300000  1.18  643076.64 euros.
4 a
She paid 300000 euros.
6
 3  r.
2
118098
 r n 1  59049
2
n  11
b
5 a
b
S11  2
311  1
 177146
3 1
9500  1.0116  9610.20 .
9610.20  1.0114  9719.76.
£9610.20
£9719.76
126
6 a
0.0235 

FV  PV 1 

12 

 3453.80
SGD 3453.80
12 n
b
0.0235 

5000  3000 1 

12 

12 n
3 
0.0235 
 1 

5 
12 
12n  261.10
n  21.76
22 years
24
r 

7 5179.27  4500 1 

400 

24
r 

1.15094   1 

400 

r 

1 
  1.00587 
400 

r  2.35 %
© Oxford University Press 2019
8
Worked solutions
0.04
 4000
12

 73.67 euros
5 x12
0.04 

1  1 
12 

rPV
8
P 
9
60
 0.025 
0.025  

6500 
1

 12 
P r 1  r 
12  

A

 115.36 euros
n
60
0.025 

1  r   1
1


1

12 

1  1  r 

10 a
b
n
n

f 5  24000  1.125  42296
2  1.12n
n  6.12
7 years
11 a
T 0  21  74 1.2  95
0
b
T 10  21  74 1.2
c
40  21  74 1.2
1.2
x

10
= 32.95
x
74
19
x  7.46 minutes.
d 21oC
12 a y = 16
b
f 0  4  2.30  16  4  16  20.
0,20
13 a
b
f 7  4  e7  1100 people
25000  4  en
n  ln24996  10.13
11 days
14 a
b
h  4  0.25  log 2  4  1.15 m
2  0.25  log 2t 
101.75  2t  56.23
t  28.1
29 weeks
15 6  k   k 2  12  k   6  k 
M1
k  k  3  0
M1
k  0 or k  3
For k  3 the sequence is constant
Hence, k  0
an  6n  6
16 a
100000  0.05  10  US$50000
b 100000  1.0510  100000  US$62889.46
A1A1
R1
A1
M1A1
M1A1
1012
c
5


100000  1 

100  12 

17 a i
 100000  US$64700.95
Use Financial APP with
M1A1
M1
n = 20
© Oxford University Press 2019
9
Worked solutions
i% = 12.8
PV = 20000
PMT =
FV = 0
PpY = 4
CpY = 4
Answer PMT =SOL 1369.29 (2 d.p.)
ii
PMT  20  SOL 27385.89 (2 d.p.)
iii SOL 7385.89 (2 d.p.)
A1
M1A1
A1
b i 400  60  0.1  20000  SOL 26000
ii 10% deposit = SOL 2000; borrowed SOL 18 000
Use Financial APP with
M1A1
R1
M1
n = 60
i% =
PV = 18000
PMT = -400
FV = 0
PpY = 12
CpY = 12
R =11.96 (2 d.p.)
A1
c If Carlos has 10% of the money the option B is better as he can save SOL 1385.89.
A1R1
(If he does not have the deposit amount then he must choose option A.)
18 a
A1
b 2 A1
c 12 A1
d y  16
A1
e x  2
A1
f y  16
A1
19 a
b
A1 shape A1 domain
T 6  25  e0.46
36.1°C (1 d.p.)
A1
c Solve 25  e0.4t  100
M1
t  10.793...
A1
10 hours 47 minutes 37 seconds
A1
20 a
M1A1
b
© Oxford University Press 2019
10
Worked solutions
Scales A1A1 Points A1 Labels A1
R1
M1A1
c Linear model
d log N  a  bt  N t   10a bt

N  t   10a  10b
A

t
M1A1
B
e Determine coefficients of regression line
log N  a  bt  a  1.9181..., b  0.06716...
1.9181...
A  10
 82.8
A1
 1.17
A1
0.06716...
B  10
21 a
b
M1A1A1
N 0  35
A1
N  4  410
M1A1
c
N t   1000
M1
d
t  5.449...
0  t  5.45
A1
A1
22 Option 1: 10000  10  $100000
A1
Option 2: 2500  2  2000  9  5  US$115000
Option 3: 100 
10
2 1
 US$102300
2 1
Option 2 has the greatest total value
23 a i
ii
b
T5  73.205 thousand taxis
M1A1
M1A1
A1
M1A1
Tn  100  n  9
M1
2019
A1
M1A1
P10  2.1873705...
2.187 million people
A1
c Adjust units
R1
i
P5  106
 28.4 people per taxi;
T5  103
M1A1
ii
P10  106
 18.6 people per taxi.
T10  103
A1
d The model predicts a reduction of the number of people per taxi which may mean that the
taxis are in use for less hours or less taxis are used everyday.
R1
24 a i
ii
b
3
f  0   
2
2 2
2
 2.67
3
2.67,0
3
x  
2
3
 
2
1
M1
A1
y 1
2
M1
 x 2
M1
y 1
© Oxford University Press 2019
11
Worked solutions
y  log3  x  2  1
M1A1
2
g  x   log3  x  2  1
2
c i
ii
d
y 2
A1
x 1
A1
x 1
A1
e Use GDC solver or intersection of graphs
M1
A1
x  2.16
25 a i
2 min 16 sec
A1
ii 2 minutes 6 seconds = 126 seconds
126  1.042  136.28...sec  2min16 sec
b Katharina´s time: 126  2  9  5  5  1485sec  24min45sec
Carolina´s time: 126 
10
1.04  1
 1512.77 sec  25min13 sec
1.04  1
A1
M1A1A1
M1A1
M1A1
c 28 seconds
A1
26 a Each year, the new annual salary is obtained by multiplying the previous salary by the
same constant.
R1
R  1.05
A1
20  1.052  COP 22.05 million
1.0510  1
 COP 251.5578... million
c 20 
1.05  1
COP 251 558 000 (nearest thousand pesos)
b
d
M1A1
M1A1
A1
2  20  9x
 10  251.5578...
2
M1A1
x  1.14573001... million pesos
M1A1
x  1145730 pesos (nearest peso)
A1
n 1
e Original offer annual salary: un  20  1.05
A1
Alternative scheme annual salary: vn  20  1.14573  n  1
A1
Use GDC to tabulate values or solve vn  un
M1
vn  un  n  7
A1
© Oxford University Press 2019
12
Worked solutions
11
Modelling periodic phenomena:
trigonometric functions
Skills check
1
a2  b2  c2
52  b2  132
b2  169  25  144
PR  12 m
ˆ  sin1 5  22.6o.
P
13
ˆ  sin1 12  67.4o .
Q
13
2
h  73 tan  43  68.1
3 a
b
Solution for
Solution for
x  0 is 11.5
t  0 is 2.08
Exercise 11A
1 a
30
 25.4. So the values of  are
70
25.4, 180  25.4, 180  25.4, 360  25.4.
  sin1
All possible values are 25.4o , 155o , 205o , 335o.
b   sin1
20
 16.6. The values are: 90  16.6  73.4o and 270  16.6  287o.
70
2 a i
ii
Solutions are
o
o
Solutions are
1  45 , 225 .
b i
ii
1  45,
iii
Solutions are
2  53.1 ,126.9 .
o
o
3  95.7o ,264.3o.
 x, y    70sin  45 ,70 cos 45   49.5, 49.5
1  225
 x, y   (70sin 225 ,70 cos 225   49.5, 49.5
2  53.1
 x, y    70 sin 53.1 ,70 cos 53.1   56.0, 42.0
2  126.9
iii 3  95.7
3  264.3
 x, y    70 sin 126.9 ,70 cos 126.9   56.0, 42.0
 x, y    70 sin 95.7 ,70 cos 95.7   69.7, 6.95
 x, y    70 sin 264.3 ,70 cos 264.3  69.7, 6.95
© Oxford University Press 2019
1
Worked solutions
3 a
T  4  11cos 30  4  7.5  13
b
The temperature will be zero in the middle of February and November.
4 a
D 5.5  1.8 sin 30  5.5  12.3  12.8 m
b
Depth of 10.9 after 7.7 hours, 10.2 hours, 19.7 hours and 22.3 hours.
Exercise 11B
1 a Amplitude is 3, period is 90o, y = 1 is principal axis and the range is y : 2  y  4
b Amplitude is 0.5, period is 720o, principal axis is y = −3 and range is y : 3.5  y  2.5
c Amplitude is 7.1, period is 120o, principal axis is y = 1 and the range is y : 6.1  y  8.1
d Amplitude is 5, period is 720o, principal axis is 7 = 8.1 and the range is y : 3.1  y  13.1
2 a a=2
b a = 1, b = 2
c a = 3, b = 0.5, d = −1
3 a b=2
b a = −3, d = 1
c a = 2.5, b = 0.5, d = 0
4 a
b i
5 a
b
5.2  1
 3.1.
2
t 
t
d 
s
5.2  1
 2.1.
2
ii s
r 
iii
4.6  6.2
4.6  6.2
 0.8. a 
 5.4.
2
2
90
 45.
2
360
8
r
b
180
 90.
2
a is the highest amount above sea level. b is the number of cycles in 6 hours. d is sea
level.
6 a
b r = 10, s 
360
 72
5
© Oxford University Press 2019
2
Worked solutions
7 You can draw a sample space diagram for the sum of a and d, the only parameters to affect the
maximum and minimum of the function, as follows:
a
d
1
2
3
4
1
2
3
4
5
2
3
4
5
6
3
4
5
6
7
There are three possibilities in a sample space of 12 with a maximum greater than five,
3
hence the probability required is
= 0.25
12
Exercise 11C
1 a
b
a
13.4  4.1
 4.65.
2
d 
13.4  4.1
 8.75.
2
b
360
 45
20  12
y  4.65 sin  45x   8.75
2 a
b
a
10.1  4
 3.05,
2
d 
10.1  4
 7.05,
2
b
360
 30.
12
The data is modelled by y  3.05 cos 30x   7.05
c
1
 21 
cos1  
  3.67 , between 0000 and 0340 hours and between 0820 and 1200 hours.
30
 61 
3 a
© Oxford University Press 2019
3
Worked solutions
b
a  0.9961, b 
360
 150000,
6  104  4
d  0.
D t   0.9961sin 150000t 
c
D 0.00011  0.9961sin 150000  0.00011  0.283 decibels
d
D 0.002  0.9961sin 150000  0.002  0.863 decibels
e Part c is more reliable as 0.00011 falls within the given data range while 0.002 is outside of
it.
4 a
a
10.1  5.31
 2.395,
2
d
10.1  5.31  7.705,
b 1
2
y  2.395 cos    7.705.
b
AD2  AB2  BD2  2  AB  BD cos 180     
AD  65.05  36.96 cos 180     
c Model is a reasonable fit, but cannot be accurate as the cosine rule depends on two angles
and not just one.
5 a
Month(t)
Rise
Set
0
11.33
15.7
1
10.15
17.22
2
8.62
18.72
3
6.78
20.27
4
5.02
21.82
5
3.4
23.45
6
3.08
23.95
7
4.57
22.57
8
6.15
20.78
9
7.58
19.00
10
9.15
17.22
11
10.73
15.82
This is necessary so that they can be graphed on the cartesian plane, which uses decimal
notation.
b
r:
a
11.33  3.08  4.125,
b
2
360
 30,
12
d
11.33  3.08
 7.205
2
r t   4.125cos 30t   7.205
s:
a
23.95  15.7  4.125,
2
b
360
 30,
12
d 
23.95  15.7
 19.825
2
s t   4.125cos 30t   19.825
© Oxford University Press 2019
4
Worked solutions
c
i
Around 3 months or 90 days
ii Around 6 months have at least 12 hours while around 4 have no more than 18. So
1
around
of the year has at least 12 hours but no more than 18
4
6 a
a
6.87  4.88
 0.995,
2
b
360
 30,
12
d 
6.87  4.88
 5.875
2
r t   0.995cos 30t   5.875
b Karim can arrive at least 1 hour before sunset without arriving earlier than 0500 between
the dates of 7th October through to the following year on 24th March.
7 a
a
30  0
 15,
2
b
360
 6000,
0.06
d 
30
 15
2
y  15 cos 6000t   15
b The dot travels 30 cm in 0.06s. So the speed of the fan is
30
 500 cms 1 .
0.06
Chapter review
1
Solutions are t = 499 and 581
2 a
3
y   cos 3x   1.
b
y  3 sin 0.5x   1
c
y   sin  x   3


a  x  is not periodic. b  x  is not periodic. c  x   3 2 cos  x   6 cos  x  .
o
Period is 360 , amplitude is 6 and the principal axis is y = 0.
d  x   2cos 3x  . Period is 120o, amplitude is 2, the principal axis is y = 0.
4 a
b
a
3  1
2
 2, b 
360
270  180  2
 2, d 
3 1
 1. y  2 cos 2x   1.
2
  1
5 a Maximum value is 18.77m and the minimum value is 17.83m
b
First time after 19s to reach 18m is 19.63s.
© Oxford University Press 2019
5
Worked solutions
6 a
 30 
6  16  6
360
30
6  16  6
x   14 or
 8, b 

,d 
 14. y  8 sin 
2
2  102 17
2
 17 
a
 30 
y  8 sin 
x   14
 17 
b F = (204,14), A = (255,22)
7 a Minimum is 1.8
A1
Maximum is 6.4
A1
b Using GDC
M1
x  268
A1
x  452
A1
5.5  1.5
  3.5 . Hence p  3.5
2
8 The principal axis is
The amplitude is
5.5  1.5
 2 . Hence q  2
2
The period is 120 : 120 
360
r
M1A1
M1A1
M1
Hence r  3 , So y  3.5  2 cos3x
A1
9 a
M1A1A1
b
y 0
A1
c
f  x   5 when x  48.6 , 86.4 , 137 , 176
M1A1
Solution is therefore 48.6  x  86.4 and 137  x  176
A1A1
10 The amplitude is 3
M1
Hence p  3
A1

The period is 81  9
90 
  90
M1
360
q
A1
Hence q  4
A1
So y  3 sin  4x  r 
y  0 when x  36 (equidistant from 9 and 81 )

So 0  3 sin 144  r


M1

So sin 144  r  0 and the first positive root is when r  36

Therefore y  3 sin 4x  36
11 a
b
M1
A1

0.3
A1
yMIN  5.4
A1
First occurs when 12.5x  180
M1
© Oxford University Press 2019
6
Worked solutions
x 
180
 14.4
12.5
c Period 
A1
360
= 28.8
12.5
M1A1
12 a
M1A1
b The principal axis is
The amplitude is
16  4
 10 . Hence p  10
2
M1A1
16  4
  6  .Hence q  6
2
M1A1
The period is 2  60  120
120 
M1
360
r
A1
Hence r  3 . So y  10  6 sin3x
A1
13 Amplitude  2 , so b  2

A1

At 60 ,5 , 5  a  2
M1A1
So a  3
A1
Therefore y  3  2 sin cx


By symmetry, the curve goes through the point 180 ,1
M1
So 1  3  2 sin 180c 
A1
1  sin 180c 
180c  270
Therefore c 
A1
3
2
A1
 3x 
Therefore y  3  2 sin 

 2 
© Oxford University Press 2019
7
Worked solutions
14 a
D
22  12
= 17
2
M1A1
A
22  12
=5
2
M1A1
The period is
360
 24
B
M1
Therefore B  15
A1
So T  5 sin 15 t  C    17
At 3,12 , 12  5sin 15 t  C    17

1  sin 15 3  C 
M1

15 3  C   90
A1
C 9
A1
Therefore T  5 sin 15 t  9   17
b Solving T  5 sin 15 t  9   17 and T  20 by GDC
Solutions are T  18.54 and T  11.46
M1
A1A1
18.54  11.46
 7.08 hours 7 hours 5 minutes
© Oxford University Press 2019
A1
8
Worked solutions
12
Analysing rates of change:
differential calculus
Skills check
1 a
2
 x  5 3x  2 3x2  13x  10
y 
b
6x3  7x2  5x
1
11  x 
2
3 Volume: 32 cm3, surface area: 16  8 13 cm2
4 a
x 5
b
x 1
5 20
Exercise 12A
a i
ii
y  0
y   0 at x  1 (the tangent to y is stationary at x  1 ).
iii The function is stationary for any value of x.
b i
ii
y  4
y   4 at x  1 (the tangent to y is increasing at x  1 ).
iii Since y  4  0, the function is increasing for any value of x.
c i
ii
df
 3  2x 2 1  6 x
dx
df
 6 at x  1 (the tangent to f  x  is increasing at x  1 ).
dx
iii The function is increasing when 6x  0. This is equivalent to x  0.
d i
ii
df
 5  2x 2 1  3  10x  3
dx
df
 10  1  3  7 at x  1 (the tangent to f  x  is increasing at x  1 ).
dx
iii The function is increasing when 10x  3  0  x 
e i
ii
3
.
10
df
 3  4x 4 1  7  12x 3  7
dx
df
 12  13  7  19 at x  1 (the tangent to f  x  is increasing at x  1 ).
dx
iii The function is increasing when 12x3  7  0. This inequality can re-arranged to x 3  
or x  3 
f
7
,
12
7
 0.836 (3 s.f.)
12
i
df
 5  4x 4 1  3  x2 1  2  20x3  6 x  2
dx
ii
df
 20  13  6  1  2  16 at x  1 (the tangent to f  x  is increasing at x  1 ).
dx
iii The function is increasing when 20x3  6x  2  0. The function g  x   20x3  6x  2 has a
single root at x 0.670 (which is found by solving g  x   0 ). Therefore, the derivative
f   x  is increasing when x  0.670 .
© Oxford University Press 2019
1
Worked solutions
g i
ii
First note that y  2x2  3x 1 , so y   x   2  2x21  3   1 x 11  4x  3x 2  4x 
y  4  1 
3
 4  3  7 at x  1 (the tangent to y is increasing at x  1 ).
12
iii The function is increasing when 4x 
3
3
 0. Solving 4x  2  0 (for x  0 ) shows that the
2
x
x
equation y   0 has a single root at x  xr   3
x  xr , 4x 
3
. When
4
3
3
 0, when x  xr , 4x  2  0 (this can be verified using, for example, a
x2
x
graphical calculator). Therefore, the function y  2x 2 
h i
ii
3
3
is increasing when x   3 .
4
x
First write y  6x 3  4x  3 . Therefore, y   6   3 x 31  4  18x 4  4  
y  
3
x2
18
4
x4
18
 4  14 at x  1 (the tangent to y is decreasing at x  1 ).
14
iii The function g  x   
18
 4 has roots ( g  x   0 ) at x 1.46 . When x 1.46, g  x  0 ,
x4
when 1.46  x  1.46, g  x   0 , and when x  1.46, g  x   0 . Therefore, the function
y 
i
6
 4x  3 is increasing when x  1.46 and when x  1.46.
x3
i
First expand: y  2x  1 3x  4  6x2  5x  4 , so y 12x  5 .
ii
y 12  1  5  17 at x  1 (the tangent to y is increasing at x  1 ).
iii y is increasing when y 12x  5  0 , which may be re-arranged to x  
j
5
.
12
i
Expand: f  x   2x 4  8x2  10x , so f   x   8x3  16x  10 .
ii
f  1  8  13  16  1  10  18 (the tangent to f  x  is decrease\ng at x  1 ).
iii The only solution of f   x   0 is x 1.66 . Therefore, f   x   0 ( f is increasing) when
x  1.66.
k i
First write y  7x 3  8x 4  6x2  2 . Then
y 3  7  x 31  8  4x 41  6  2x21 21x 4  32x3  12x .
ii
y  
21
 32  13  12  1  1 at x  1 (the tangent to y is decreasing at x  1 ).
14
iii y   0 has a single solution x  1.01 , with y   0 when x  1.01 and y   0 when x  1.01.
Therefore y is increasing when x  1.01.
Exercise 12B
1 a
dA
 2 r 2 1  2 r
dr
2 a
dP
 0.056  2c21  5.6  0.112c  5.6
dc
b When c  20,
b
dA
 2  2  4 when r  2.
dr
dP
dP
 0.112  20  5.6  3.36 , when c  60,
 0.112  60  5.6  1.12 .
dc
dc
c At the larger number of sales, selling more cupcakes will actually decrease profit, whilst it
will increase profit at the lower value.
3 a
f  t   80  2t 21  160 160t  160  160 t  1
© Oxford University Press 2019
2
Worked solutions
b The function f   t  represents the velocity of the bungee jumper.
c
f  0.5  160 0.5  1 80, f  1.5 160 1.5  1  80. At these times, the bungee jumper is
travelling at the same speed, but in opposite directions (moving away from start point at
t  0.5 and towards the start point at t  1.5 ).
d
f 2  160 2  1 160  f 0 . The bungee jumper passes through the start point at the
same speed that he left at – this is unrealistic; some energy will be lost overcoming, for
example, air resistance.
4
f   x   3x2  2x  2 . The gradient at A and B is 3, so the x -co-ordinates of these points
satisfy 3  f   x   3x2  2x  2  3x2  2x  1  0. This equation has solutions x1  1, x2 
The
1
.
3
corresponding y co-ordinates are y1  f  1   1   1  2  1 2 , and
3
3
2
2
2 22
1 1
1
y2  f          
.
3 27
3 3
3
 1 22 
The co-ordinates of points A and B are  1, 2 and  ,
.
 3 27 
5 The pink line has a gradient of m  tan1 45  1. Also note that h  x   2  0.2x . Therefore, the
pink and purple lines meet where the gradient of the purple line equals the gradient of the pink
line: m  2  0.2x  x  5. The point of intersection has y co-ordinate of
h 5  2  5  0.1  52  7.5 - the point is 7.5 m above the ground.
Exercise 12C
1 At x  3, y  f 3  2  32  4  14 . Also, f   x   4x so the gradient of the tangent at x  3 is
m  f  3 12 . Therefore, the tangent at x  3 has equation y  14  12  x  3  y  12x  22.
2 The y -co-ordinate of the point of contact is y  f 1  12  2  1 . Also, f   x   2x  2 , so the
gradient of the tangent at x  1 is f  1  2  2  0 . Therefore, the tangent at x  1 (i.e. the
equation of the plank) is simply the constant function y  1.
3 a
f   x  4x  f  1 4; the gradient of the wheel at x  1 is −4.
b The gradient of the spoke is therefore 
1
1
 .
4 4
4 First find the gradient m of the tangent to f  x  at x  1 : f   x   6x  4, so m  f  1  2 . The
normal at this point has gradient of 
y 4 
1
1
  . Therefore, the equation of the normal at 1, 4 is
m
2
1
x 9
 x  1  y   2  2 .
2
5 At x  2, y  24  6  2  3  7. The derivative of y is y  x   4x3  6 , so the gradient of the
 
tangent at x  2 is m  4 23  6  26 . The tangent, therefore, has equation
y  7  26  x  2  y  26x  45 and the normal (which has gradient 1 / m ) has equation
y  7 
6
1
92
x
 x  2  y  13  26 .
26
f   x   2x , so the gradient of the tangents at x  2, x  2 are m1  4, m2 4 (respectively).
The equations of the normals are, therefore, y  22  
1
1
9
 x  2  y   4 x  2 and
4
1
1
9
 x  2  y  4 x  2 . The normals therefore meet at x  0 (by setting the two
4
9
 9
normals equal to each other); at this point y 
: the fountain will be placed at  0,  .
4
 4
y — 22 
© Oxford University Press 2019
3
Worked solutions
7 a
b
f 15 58. 8 = f 35 . Also, f   x  0.224x  5.6 , so f  15  2.24, f  35 2.24. Therefore,
the normal at x  15 has equation y  f 15  
normal at x  35 has equation y  f 35  
1
 x  15  y  65.5  0.446x and the
f  15
1
 x  35  y  43.1  0.446x .
f  35
c The normal meet where 43.1  0.446x  65.5  0.446x 22.4  0.892x  x  25.1 (3 s.f.).
At this point, y  43.1  0.446  25  54.3.
d Yes, position is within the park.
8
f   x   2ax  3 . Since f  2  7, then 4a  3  7  a  1. Then b  f 2  a  22  3  2  1  9.
9 First find k using the fact that f  1  2  1  k  3  k  1 (since f   x   2x  k ). Then
b  f 1 12  k  3  5 .
10 Since y  2 when x  1, then we must have 2  a  b  1. Also, the gradient of the tangent at
x  1 is y ' 1  2a  b (since y  x   2ax  b ). Therefore, the normal at x  1 has gradient

1
1
1
and hence 1  
 2a  b  1. We need to simultaneously solve

2a  b
y  1
2a  b
2  a  b  1 and 2a  b  1; the solution is a  2, b  5 .
Exercise 12D
1 a
d
3
4
47
y   3 
196
y   3 
55
36
b
y 3  1  ln3  2.10 (3 s.f.).
c
f   3 
e
y 3  7e6  2824
f
g 3  672
Exercise 12E
1 a
f  t   7.25  2  1.875t  7.25  3.75t
b At the stationary point, f  t   7.25  3.75t  0  t  1.93 (3 s.f.). At this time,
f t   1  7.25  1.93  1.875  1.93  8.01 (3 s.f.): the stationary point is at 1.93,8.01 .
2
c When t  1.93, then 3.75t  7.25 so f  t   0 and when t  1.93, then 3.75t  7.25 , so
f  t   0 , hence t  1.93 s is a maximum.
2 a
© Oxford University Press 2019
4
Worked solutions
b
P 2  0.08  23  1.9  22  15.2  2  18.04, P 3  0.08  33  1.9  32  15.2  3  22.56. The
22.56  18.04
 4.52 ;
32
the average rate of change between x  2 and x  3 is 4.52 thousands of dollars per
million units sold.
gradient of the chord between 2, P 2  and 3, P 3  is therefore m 
c
P  x   0.24x2  3.8x  12.5 , so P 3  3.26, P 8 2.54, P 13  3.66. These represent the
instantaneous rate of change of profit with respect to units sold.
d The instantaneous rate of change is negative when P   x   0 and positive when P   x   0.
The equation P   x   0 has solutions x  4.66,11.2 . Therefore, the instantaneous rate of
change is negative when 4.66  x  11.2 and instantaneous rate of change is positive when
x  4.66 and x  11.2.
This means that profit increases with more sales when x  4.66 and x  11.2 but profit will
decrease with more sales when 4.66  x  11.2.
e The instantaneous rate of change is zero when x  4.66,11.2 .
f
At the points where P   x   0 (i.e. x  4.66,11.2 ), then P  x   25.1,14.1 (respectively). We
can see from the sketch that the gradient function P '  x  changes sign at these points, i.e.
they are indeed (local) maxima and minima.
3 The maximum height is f  4 at t  6.
4 Find the stationary points of the profit function by solving P  n 0.112n  5.6  0  n  50.
(Note that when n 50, P   n 0 and when n  50, P  n  0 so this is indeed a maximum). At this
point, the profit is P 50 US$120 .
5 a i
4
: the stationary points of P  n occur (for 0  n  5 ) at n  1.82.
n 1
This is a local minimum and there are no other stationary points for 0  n  5 , so the
maximum profit occurs at either n  0 or n  5 (in this range). Since
P  n  0.5n  1.5 
P 0 5.5, P 5  4.67 then, under this model, they should buy no parts to maximise
profit! (The profit will be 55000 EUR).
ii
n3 5n2

 6n  4 : the stationary points of P  n occur (for 0  n  5 ) at n  2
3
2
2
1
(local maximum) and n  3 (local minimum), with P 2  , P 3  . Also,
3
2
31
P 0  4, P 5 
. Therefore, under this model, the profit is maximised by buying 5000
6
parts, which gives a profit of 51667 EUR.
P  n 
n3 5n2

 3n : the stationary point of P  n occurs (for 0  n  5 ) at n  4 (local
24
8
14
maximum), with P  4 
. There are no other turning points in 0  n  5 so this local
3
maximum is at the maximum value of P  n on 0  n  5 . Therefore, under this model,
iii P  n 
the factory should by 4000 parts, which gives a profit of 46666 EUR.
b They should adopt the first strategy.
6
y  x  0.324x3  2.67x2  5.74x  3 . By solving y   x   0, we find stationary points at
x  0.776 (local max), x  5.15 (local max) and x  2.31 (local min). Since
y 0.776  0.986,y 5.15  3.92 then we determine that the maximum height on the route is
392.
© Oxford University Press 2019
5
Worked solutions
Exercise 12F
1 a The volume of a cylinder is the product of its cross sectional area (in this case  r 2 ) and its
height h , therefore, as the volume is 400 cm3 , we have 400   r 2h .
b The surface area of the curved surface is Ac  2 rh and the area of the base is Ab   r 2 .
Hence the total surface area is A  Ac  Ah  2 rh   r 2 .
c Using part a, we can expressed Ac as Ac 

2  r 2h
r
  2  400  800 . Using this form, the
r
r
800
total surface area is A   r 2 
.
r
d
e The minimum area is min A  239 at r  5.03.
f
This can be verified graphically.
2 Using the same method as q1: the surface area of a closed cylinder of radius r and height h is
A  2 rh  2 r 2 , and the volume is V   r 2h. If we’re given that the total surface area is 5000
5000  2 r 2
cm2, we can express h in terms of r : A  5000  2 rh  2 r 2  h 
. Hence, the
2 r
volume can be expressed only in terms of the radius as V  r  

r 5000  2 r 2
2
.
V  r  has a maximum of V  27145 at r  16.3 , at 16 the gradient is positive and at 17 it is
negative, so r  16.3 is a maximum.
3 a The perimeter is p  100  2x  2l , where l is the length of the garden. Therefore,
l  50  x.
b The area is the product of the length and the width, i.e. A  xl  x 50  x  m2.
c
dA
 50  2x
dx
d
dA
dA
dA
 0 when x  25 . This is indeed a maximum as
 0 when x  25 and
 0 when
dx
dx
dx
x  25 . Therefore, the maximum area of the grass is A  252  625m2 , which occurs when
the garden is a 25 m 25 m square.
4 The volume of a cone with radius r and height h  18  r is V 
1 2
1
 r h   r 2 18  r  .
3
3
1
 36r  3r 2 , so V   r   0 has solutions r  0, 12 . We restrict ourselves to r  0, so
3
the only turning point is r  12 . This is a maximum because V   r   0 when r  12 and
V  r  


V   r   0 when 0  r  12.
Therefore, the maximum volume of the cone is V  V 12  288  905cm3 , when r  12 (3
s.f.).
5 a We can imagine that, after removing the squares, the sides of the rectangle are split into
three pieces, which have length x, x and 20  2x on one side (the first two correspond to
the removed sections, and the latter to the remaining section), and x, x and 24  2x on the
other side. The resulting box therefore has a base of size 20  2x 24  2x and height x ;
the volume of the box is V  x 20  2x  24  2x  .
© Oxford University Press 2019
6
Worked solutions
b Expanding, we have V  4x3  88x2  480x , and hence V   x  12x2  176x  480 . The
24
at
2
x  3.62,11.0 . The second of these corresponds to a negative volume, so is ignored. The
stationary points of V are at V   x   0 , which has two solutions in the range 0  x 
former is a local maximum with V 3.62  774.16 . Since V 0  0  V 24 , this is a
maximum on the interval of interest: 0  x  12.
The value of x which maximises the volume is x  3.62cm which provides a volume of
V  774 cm3.
6 a The volume is V   r 2h
b The surface area of the curved part is Ac  2 rh and the surface area of the ends are each
Ae   r 2 . Therefore, the total surface area is A  Ac  2Ae  2 r 2  2 rh .
We can use the volume constraint to write the r in terms of h : 300   r 2h  h 
A  2 r 2 
c
300
, so
 r2
600 r
600
 2 r 2 
 r2
r
150
600
, so the only stationary point is at r  r1  3
, and this is a local

r2
minimum (this can be seen by, for example, plotting the graph of A  r  ). Therefore, the
A  r   4 r 
minimum surface area is A  r1   248 cm2 which occurs at r  r1  3.63 cm and h  7.25 .
7
33.3
 180.9 .
0.184
This is a maximum point of the function P (a quadratic with a negative leading co-efficient has
a single global maximum at its turning point), but the quantity n can only take integer values.
The maximum profit is therefore attained at the next largest or next smallest integer to
n  180.9 . We calculate P 180  2700.2 2700.29  P 181 : the maximum profit is $2700.29,
P  n 0.184n  33.3 , so the only stationary point of P in n  0 occurs at n 
when n  181 goods are sold per day.
8
f   x   1.8x  52 , so the only turning point of f occurs when x  x1 
52
 28.9. This is a
1.8
maximum because f   x   0 for x  x1 and f   x   0 for x  x1. However, x is an integer, so
the maximum is attained at x  28 or x  29. Since f 28  390.4  391.1 f 29 , we
the maximum profit is f 29  391.10 USD, when 29 units are sold.
conclude
Chapter Review
1 a
f   x   0 , the tangent to f at x  1 has gradient m  0.
b
y   x   3 , the tangent to y at x  1 has gradient m  3.
c
g  x   4x  4 , the tangent to f at x  1 has gradient m  0.
d
y   x  18x2  6x  1 , the tangent to f at x  1 has gradient m  13.
e
f  x 
2
 3 , the tangent to f at x  1 has gradient m  2  3  1.
x2
f
f  x 
18
 4x , the tangent to f at x  1 has gradient m  18  4  14.
x4
2 First note f  4  0.5  42  3  4  2 2 , so a point on both the normal and the tangent is
P  4, 2 . Since f  t   t  3 , then f   4 1 ; the gradient of the tangent at P is 1 and the
gradient of the normal at P is −1. Hence, the normal at P has equation
y  2 1  x  4  y  2  x and the tangent at P has equation y  2  x  4  y  x  6.
© Oxford University Press 2019
7
Worked solutions
3 Since f   x   2x  5 , to find the x-coordinate of the point A, when the gradient of the tangent to
f is 1, we need to solve f   x   2x  5  1  x  3 . The corresponding y co-ordinate is
f 3  10 , so A has co-ordinates of 3, 10 .
4 Let  x1, y1  be the co-ordinates of B. Note that f   x   6x  4 so the normal to the curve at B
has gradient of 
1
1
, and we’re given that this has to equal , so 6x1  4  2  x1 1.
2
6 x1  4
Then y1  f  x1   3  4  3  4.
5 a
b
f  t  1.667  0.0834t
f  12  0.666 travelling downhill, f  32  1.00 travelling uphill
c The stationary points are where f  t   0  0.0834t  1.667  t  20.0 s (this is the time at
which Jacek is at the minimum point on the track). This is a minimum point because
f  t   0 when t  20 and f  t   0 when t  20.
6 a The volume of a cylinder of radius r and height h is V   r 2h . We’re given that V  300
cm3, so 300   r 2h.
b
h
300
, The surface area of the curved part is Ac  2 rh and the surface area of each of
 r2
the ends is Ae   r 2 . Hence, the total surface area is S  Ac  2Ae  2 rh  2 r 2.
c Using the expression from part a, S  2 r 
d
300
600
 2 r 2  2 r 2 
.
 r2
r
dS
600
 4 r  2
dr
r
e The only stationary point of S is where 4 r 3  600  r  3
corresponding surface area is S  248 cm3 and h 
150

 3.62 cm . The
300
= 7.26 cm. (all 3 s.f.)
 r2
dy
 0.1x  1.5
dx
dy
b Setting
0
dx
7 a
M1A1
M1
0.1x  1.5  0
x  15
A1
y  0.05  152  1.5  15  82  93.25 m
A1
dy
at x  14.5 and x  15.5
dx
M1
c Evaluating
8 a
dy
dy
 0.05 and
 0.05
dx
dx
A1
Sign goes from positive to negative, therefore a maximum point
R1
V  x  40  2x  30  2x 

 x 1200  60x  80x  4x2
b
M1A1


 x 1200  140x  4x2

 1200x  140x2  4x3
dV
 1200  280x  12x 2
dx
A1
M1A1
© Oxford University Press 2019
8
Worked solutions
dV
0
dx
M1
1200  280x  12x2  0
A1
c Setting
12x2  280x  1200  0
3x2  70x  300  0
x2 
70
3
x  100  0
d Using GDC to solve x2 
70
3
x  100  0
M1
x  5.66 cm
A1
Vmax  1200  5.657  140  5.6572  4  5.6573
M1
 3032 cm3 3030 to 3 s.f.
A1
9 a Use of GDC (demonstrated by one correct value)
a  0.0534
b  1.09
c  7.48
T  0.0534h2  1.09h  7.48
dT
b
 0.107h  1.09
dh
dT
c Setting
0
dh
M1
A1
A1
A1
M1A1
M1
A1
h  10.2
d The maximum temperature usually occurs after midday, whereas this is only 10 hours
after midnight.
R1
10  1.59, 13.2
A1A1
0.336,3.81
A1A1
1.17,1.97
A1A1
11 a
dy
 2x 2  7x  2
dx
M1A1
dy
 3
dx
M1
2x2  7x  2  3
2x2  7x  5  0
2x  5  x  1  0
x 
M1
5
35
y 
24
2
A1
25
6
0.314  x  3.19
A1
x 1 y 
b
12 At x  1 , y 
A1A1
3
2
A1
dy
 2x 3
dx
At x  1
M1A1
dy
 2
dx
A1
gradient of the normal is therefore
1
2
© Oxford University Press 2019
M1
9
Worked solutions
Equation of the normal is therefore y 
y
3 1
1
 x
2 2
2
y 
1
x 1
2
3 1
  x  1
2 2
M1
A1
13 Substituting 2, 1 gives
M1
A1
1  4a  2b  3
4a  2b  4
2a  b  2
dy
 2ax  b
dx
M1
8  4a  b
A1
Solving simultaneously gives
M1
a  5 A1
A1
b  12
14 a
b
y  20  x
xy  x 20  x   20x  x2
M1
Differentiate and set to zero:
M1
20  2x  0
A1
x  10
A1
So xyMAX  100
A1
x2  y 2  x2  20  x 
2
M1
 x2  x2  40x  400
 2x2  40x  400
Differentiate and set to zero:
M1
4x  40  0
A1
x  10

2
So x  y
c
A1
2

MAX
2
2
 10  10  200
4  9.5  40  2
A1
A1
4  10.5  40  2
A1
The derivative goes from negative to positive, therefore this is a maximum
R1
© Oxford University Press 2019
10
Worked solutions
Approximating irregular spaces:
integration
13
Skills check
1 a 18 cm2
2 a
3x 2 
b 14.1 cm2
5
2
b
8x  1
Exercise 13A
1 The distance travelled between t = 0 to t = 20 is d1 
1
2050  500 m (area of triangle). The
2
distance travelled between t = 20 50 t = 50 is d2  50  20 50  1500 m (area of square). The
distance travelled between t = 50 and t = 60 is d3 
1
1050  250 m. The total distance is
2
d  d1  d2  d3  2250 m.
2 a Distance travelled is the area A under the line, between t= 1 and t = 4: A =
1
  4  1  5  11  24 units.
2
b The line v  t  1 intercepts
v  0 at t = 1. The area under the curve is therefore A 
1
1
  4  1  v  4   3  3  4.5 km.
2
2
c At t  0,v  9. At v  0,t  3. Therefore, the triangle has area A 
1
27
 3  0   9 
. The
2
2
distance travelled is 13.5m.




3 a The ends of the line are at A 0.5, f 0.5  A 0.5,7 and B 3.5, f 3.5  B 3.5,1 . Therefore,
using the formula for area of a trapezium, the area is
1
 1  7  3.5  0.5 12.
2
b Two points on the line are A 0,2 and B 1,0 . Therefore, the gradient of the line is -2, and
at x  2,y  2  2  2  6. Therefore, the base of the triangle is 3, and its height is 6; the
area is
1
36  9 .
2
c Let C  x1, y1  be the intersection point between the lines.
Then y1  1.5x1  3 0.5x1  7  x1  2, y1  6. The area is
A  Ag  Af , where Ag is the area under g  x between x  0 and x  x1  2 and
Af is the area under f  x  between x  x1  2 and x  8. Using the area of a trapezium
formula, Ag 
1
1
  g 0  6   2  0  9 and Af   6  f 8   8  2  27. The total area is
2
2
36.
d The lines intersect where 6 x  7  x  1.
The area is A  Ag  Af , where Ag is the area under g  x between x  3and x  1 and
Af is the area under f  x  between x  x1  1and x  5. Using the area of a


square/trapezium formula, Ag  g  3  g 1  1  3  24 and
Af 
1
 6  f 5   5  1 16. The total area is 40.
2
© Oxford University Press 2019
1
Worked solutions
4 a&b
c First calculate f 1  3.5and f 6  6, then use the trapezium area formula to find:
Area 
1
  f 1  f 6    6  1  23.75.
2
5 a&b
c The line intersects the x-axis at A 3,0 , so the area enclosed is
1
 f 0  0  3  0  9.
2


6 a&b
c The area can be split into a triangle for 0  x  5 and a rectangle of height 5 for 5  x  9.
1
 5  5  12.5, and the rectangle has area 4  5  20. The total area of
2
the shaded region is 32.5.
The triangle has area
Exercise 13B
4
1 a i
 5 dx
2
b i
3.5
ii
3
  2x  8 dx
iii
0.5
  3x  10 dx
1
4
iv
2x dx
0
Area =  4  2  5  30
ii Using the area of a trapezium: A =
1
1
  f 0.5  f 3.5   3.5  0.5   7  1  3  12
2
2
iii Using the area of a trapezium: Area =
iv Using the area of a triangle: Area =
1
1
  f 1  f 3   3  1   7  1  2  8
2
2
1
1
 f  4  4   4  8  16
2
2
2 a
Using the formula for the area of a trapezium: Area =
1
 6  2  7  3  20
2
© Oxford University Press 2019
2
Worked solutions
b
Using the formula for the area of a triangle: Area =
1
  4  0  8  0 16
2
c
Using the formula for the area of a trapezium: Area =
1
 2  0  5  4  9
2
Exercise 13C
2
1 a i
  x  1
2
 0.5 dx
ii
34
3
ii
65
64
2
1.5
b i

x 3dx
1
3
c i

x  1 dx
ii 5.33
1
0
d i
x
3
 4x dx
ii 4
2
e i
Find the root unknown intersection by solving 0  x2  2x  x  2. The area is therefore
2
x
2
 2x dx
0
f
ii
4
3
i
The roots (found by solving 0    x  1  x  3 ) are x 1,3. Therefore the area is
3
   x  1  x  3 dx
1
ii
32
3
4
2 a i
x dx
2
ii
2
56
3
1
b i
 2x dx
ii 0
1
1
c i
1
 1 x
2
dx
ii 1.57
1
3
d i
1
dx
0.5 x

ii 1.79
© Oxford University Press 2019
3
Worked solutions
e i
1
   x  3  x  2 
37
6
ii
dx
0
0
f
i
  x  3  x  2 dx or

2
3
   x  3  x  2 
0
3
g i
   x  3  x  2 
h i

 x 2  2x  15 dx
22
or 13.5
3
125
6
ii
dx
2
4.5
ii
dx
ii 80.71
2
i
i
The solutions of the equation –x2  2x  15  0 are x 3,5. Therefore, the area bounded
by f  x   x2  2x  15 and y  0 is
5
 x
2
 2x  15 dx .
3
j
ii
256
3
i
The equation f  x   3  e x has solution x  ln 3 . Therefore, the area bounded by the xaxis, the curve f  x  and the line x  1 is
ln3
 3e
x
dx
1
ii
k i
ln27 
1
e
The equation  x  2  5  0has a single root at x 2  3 5  0. The area between
3
y   x  2  5 and the co-ordinate axes is therefore
3
0
  x  2
3
 5 dx.
1
2 53
ii
14 
153 5
4
3 a Re-arrange to x2  4  x 2,2.
b The curve cuts the y-axis at x  0. At this point, y  4, i.e. the co-ordinated are  0, 4 .
c Using the answer from part b, this area is 54  20.
2
d
x
2
 4 dx 
0
16
3
e Total area is the sum of the two parts found in dii and b
16
76
 20 
3
3
Exercise 13D
a
1 Use the solver to find a  0 such that
   x  1  x  5 dx  24  a  3.
0
a
2 Use the solver to find a  3 such that
2
x
dx  9. The solution is
3
a  log(8  9log2) / log 2 0.8169 (3 s.f.)
3 a
© Oxford University Press 2019
4
Worked solutions
3
b Use the solver to find 0  a  3 such that x 
a
1
dx  6. The solution is a  0.5693 (4 s.f.)
x
4 a a = −1
b The integral describes the area between the curve y  x 2 and the x-axis between
x  2 and x  1.
5 a b=1


b The integral describes the area between the curve y  1  x 3 and the x-axis, between
x 1 and x  1.
6 a t=3
b The integral describes the area between the curve y 
x 1 and x  3.
x  1 and the x-axis, between
Exercise 13E
1 Using the notation of the p18, b  9, a  1, n  4 
area as:
ba
 2, so the trapezium rule estimates the
n
1
 2  5  2 7  6  10  4 55.
2
2 Using the notation of the p18, b  6, a  0, n  4 
the area as:
ba
 1.5, so the trapezium rule estimates
n
1
 1.5  1  2  4  2  5.5  0 18.
2
3 Using the notation of the p18, b  4, a  1, n  4 
estimates the area as:
ba
 1.25, so the trapezium rule
n
1
 1.25  5  2 7  3.5  6   8  28.75.
2
4 First set out the table of points:
x
0
0.5
1
1.5
2
2.5
3
y
2
2.7
3
3
3.3
4
5.4
Then, using the notation of the p18, b  3, a  0, n  6 
estimates the area as:
ba
 0.5, so the trapezium rule
n
1
 0.5  2  2 2.7  3  3  3.3  4  5.4  9.85
2
5 First set out the table:
x
1
1.6
2.2
2.8
3.4
4
y
5
4.32
3.87
3.57
3.37
3.25
Then, using the notation of the p18, b  4, a  1, n  5 
estimates the area as:
ba
 0.6, so the trapezium rule
n
1
 0.6  5  2  4.32  3.87  3.57  3.37  3.25 11.6 (3 s.f.)
2
6 a The x-co-ordinates of the edges of the trapeziums are
x  0,0.8,1.6,2.4,3.2, 4  a  0, b  4, n  5 . We have a formula ( y  x ) for the height of the
edges of these trapeziums. The formula therefore gives an approximation of the area as:
1 4
  0  2 0.8  1.6  2.4  3.2  4 5.198 (4 s.f.)
2 5




b The x-co-ordinates of the edges of the trapeziums are
x  1,0.25,1.5,2.75, 4  a  1, b  4, n  4 . We have a formula ( y  2x ) for the height of the
© Oxford University Press 2019
5
Worked solutions
edges of these trapeziums. The formula therefore gives an approximation of the area as:
1 5
  21  2 20.25  21.5  22.75  24  23.74 (4 s.f.)
2 4




c The x-co-ordinates of the edges of the trapeziums are
10
 1) for the height of
x
the edges of these trapeziums. The formula therefore gives an approximation of the area as:

1 1  10
10
10
10
10
 10
 10
 
 1 2 
 1
1
1
1
 1 
 1 12.21 (4 s.f.)
2 2  2
2.5
3
3.5
4
4.5
5



x  2,2.5,3,3.5, 4, 4.5,5 a  2, b  5, n  6 . We have a formula ( y 
d The x-co-ordinates of the edges of the trapeziums are x  0,1,2,3, 4,5 a  0, b  5, n  5 . We
have a formula ( y 0.5x  x  5  x  1) for the height of the edges of these trapeziums. Put
these in a table:
x
0
1
2
3
4
5
f(x)
0
4
9
10
12
0
The formula therefore gives an approximation of the area as:
1
 1  0  2  4  9  10  12  0  35
2
7 a
6
b i
  2  x  3  x  6 dx
ii 9
3
c Apply the trapezium rule method with 6 trapezoids: the edges of these are at
x  3,3.5, 4, 4.5,5,5.5,6. Construct a table (below).
x
3
3.5
4
4.5
5
5.5
6
y
0
2.5
4
4.5
4
2.5
0
The trapezium rule formula  a  3, b  6, n  6 gives an approximation of the area as
1 1
  0  2 2.5  4  4.5  4  2.5  0  8.75.
2 2
d Percentage Error =
Actual - Estimated
9  8.75
 100% 
 100%  2.78% (3 s.f.)
Actual
9
8 a
2
b i
1  e
x
dx
ii 1  e2  8.389 (3 s.f.)
0
© Oxford University Press 2019
6
Worked solutions
c Apply the trapezium rule method with 5 trapezoids: the edges of these are at
x  0,0.4,0.8,1.2,1.6,2.
Using the trapezium rule formula with a  0, b  2, n  5, gives an approximation of the area
as


1
 0.4  1  e0  2 1  e0.4  1  e0.8  1  e1.2  1  e1.6  1  e2  8.474. (3 s.f. )
2
d Percentage Error =


Actual - Estimated
8.474  8.389
 100% 
 100%  1.01% (3 s.f.)
Actual
8.389
Exercise 13F
1 a i
The derivative of F  x   x is F   x  1  f  x  , so F  x   x is an antiderivative of
f  x   1.
ii The derivative of F  x   x  3 is F   x  1  f  x  , so F  x   x  3 is an antiderivative of
f  x   1.
iii The derivative of F  x   x  6 is F   x  1  f  x  , so F  x   x  6 is an antiderivative of
f  x   1.
b For example, F1  x   x  10, F2  x   x  20.
c
F  x   x  c , where c is any number.
2 a i
The derivative of F  x   x 2 is F   x   2x  f  x  , so F  x   x 2 is an antiderivative of
f  x   2x.
ii The derivative of F  x   x2  1 is F   x   2x  f  x  , so F  x   x2  1 is an antiderivative
of f  x   2x.
iii The derivative of F  x   x2  4 is F   x   2x  f  x  , so F  x   x2  4 is an antiderivative
of f  x   2x.
b For example, F1  x   x2  1, F2  x   x2  2.
c
F  x   x2  c , where c is any number.
3 a i
The derivative of F  x  
x2
x2
is F   x   x  f  x  , so F  x  
is an antiderivative of
2
2
f  x   x.
ii The derivative of F  x  
x2
x2
 2.5 is F   x   x  f  x  , so F  x  
 2.5 is an
2
2
antiderivative of f  x   x.
iii The derivative of F  x  
x2
x2
 12 is F   x   x  f  x  , so F  x  
 12 is an
2
2
antiderivative of f  x   x.
b
F x 
4 a i
x2
 c , where c is any number.
2
The derivative of F  x   x2  x is F   x   2x  1  f  x  , so F  x   x 2  x is an
antiderivative of f  x   2x  1.
ii The derivative of F  x   x2  x  3.2 is F   x   2x  1  f  x  , so F  x   x2  x  3.2 is an
antiderivative of f  x   2x  1.
© Oxford University Press 2019
7
Worked solutions
iii The derivative of F  x   x2  x  4 is F   x   2x  1  f  x  , so F  x   x2  x  4 is an
antiderivative of f  x   2x  1.
b
F  x   x2  x  c , where c is any number.
5 For example, g  x   3x
6 For example, g  x   2x
7 For example, g  x  
x2
4
8 For example, F  x  
x3
is an anti-derivative of f  x   x 2
3
x
9 a
A  x   2.5t dt . Since the function f t  1.25t 2 has f  t   2.5t , then it is an anti-derivative
0
of F t   2.5t and A  x   f  x   f 0 1.25x2 .
x
b
A  x     0.5t  4 dt . Since the function f t  0.25t 2  4t has f  t   0.5t  4 , then it is an
0
anti-derivative of F t   0.5t  4 and A  x   f  x   f 0 0.25x2  4x .
x
c
A  x   t 2 dt . Since the function f  t  
0
F t   t 2 and A  x   f  x   f 0 
t3
has f  t   t 2 , then it is an anti-derivative of
3
x3
.
3
x
d
A  x   t 2  1dt . Since the function f  t  
0
t3
 t has f  t   t 2  1 , then it is an anti3
derivative of F t   t 2  1 and A  x   f  x   f 0 
x3
x.
3
x
10  4 dt is equal to the area between the line y  4 and the x-axis, over the interval 0  t  x;
0
x
this rectangle has area of 4x. Therefore
4 dt  4x.
0
11 Using the formula for the area of trapezoid (the function f t   2t  1 is a straight line),
x
1
 2t  1 dt  2  1  2x  1   x  0  x  x  1 .
0
12 Using the formula for the area of trapezoid (the function f t   4t is a straight line),
x
1
4t dt  2  8  4x    x  2   4  2x   x  2 2  x  2  x  2  2x
2
 8.
2
Exercise 13G
1 The family of anti-derivatives of 2 are functions of the form f  x   2x  c , where c is any real
number.
2 The family of anti-derivatives of x  1 are functions of the form f  x  
x2
 x  c , where c is
2
any real number.
3 Any function whose derivative is equal to -1 is of the form f  x    x  c , where c is any real
number.
4 Any function whose derivative is equal to 2x is of the form f  x   x2  c , where c is any real
number.
© Oxford University Press 2019
8
Worked solutions
5 a
x  c, c  R
b
6x  c,c  R
c
x
 c, c  R
2
6 a
x2
 c, c  R
2
b
x2  c,c  R
c
5x 2
 c, c  R
2
d
x2
 c, c  R
4
x3
 c, c  R
3
b
x3  c,c  R
c
4x 3
 c, c  R
3
d
x3
 c, c  R
6
e
x3
 c, c  R
9
f
x3
 c, c  R
2
g
ax 3
 c, c  R
3
8 a
x4
 c, c  R
4
b
x 4  c,c  R
c
x4
 c, c  R
2
d
x4
 c, c  R
5
e
x4
 c, c  R
12
f
–
x4
 c, c  R
4
g
ax 4
 c, c  R
4
b
0.2x3  c,c  R
c
x6
 c, c  R
6
ax 2
 c, c  R
2
e
7 a
Exercise 13H
1 Use the power law:
a
10x c,c  R
d
 7  2x  dx   7 dx   2x dx  7x  c1  x 2  c2  7x  x 2  c c  c1  c2  R 
e
 1  2x  dx   1dx   2x dx  x  c1  x2  c2  x  x2  c c  c1  c2  R 
f

x2 
x2
x2
x3
 5  x 
dx  5x  c1 
 c2 
 c3 
 dx   5dx   x dx   
3 
3
2
6

5x 
x2 x3

 c  c  c1  c2  R 
2
6
g


3x 2
3x 2
x2 x3
  –x 
 0.5  dx    x dx  
dx   0.5 dx  

 0.5x  c
4
4
2
4


h

x3 
x3
x2
x4
 1  x 
dx  x  c1 
 c2 
 c3 
 dx   1dx   x dx  
2 
2
2
8

x
x2 x 4

 c (c  c1  c2  c3  R)
2
8
x
x
x3
x2


  x 2   4  dx   x 2 dx   dx   4 dx 
 c1 
 c2  4x  c3 
2
2
3
4


i
x3 x2

 4x  c  c  c1  c2  c3 
3
4
j

1 1
x c
2
k

2
2
dx   2x 4 dx   x 3  c
x4
3
l
5 
5

  4x  3  dx   4x dx   5x 3 dx  2x 2  x 2  c
x
2


2 a If f  x   x 2 
b
3

x
1
 4, then f   x   2x  .
3
3
 f  x  dx 
x3 x2

 4x  c (using the power rule)
3
6

t2
 t3  c
2
 t  3t 2 dt   t dt   3t 2 dt 
© Oxford University Press 2019
9
Worked solutions
4


 4t 3  3t  1 dt   4t 3 dt   3t dt   dt  t 4 
5 Any function F which has gradient 3  x 
3t 2
t c
2
x2
is of the form
4

x2 
x2 x3
F  x    3  x 

 c c  R 
 dx  3x 
4 
2 12

6 Any function F which has gradient  x  0.5x2 is of the form
x2 x3
F  x     x  0.5x 2 dx  

 c c  R 
2
6

7 a

 x  1  x  2  x2  2x  x  2  x2  x  2
dy
  x  1  x  2 has the form
dx
x3 x2
y  f  x    x 2  x  2 dx 

 2x  c , where c is any real number.
3
2
b Any function y  f  x  with

8

2
x4
x4 2


 g  x  dx    x 3  2  1  dx 
  2x 2 dx  x  c1 
 xc
x
4
4
x


Exercise 13I
1 As the derivative of f  x  is f   x   3x  4x2 , we can use the power-rule to find that
f x 
3x 2 4 x 3

 c for some value of c. As  1,0 lies on the graph, then
2
3
f  1  0  0  3 
 1
2
2
 1
3
 4
3
c c 
4 3
1
3x 2 4 x 3 1
   . Therefore, f  x  


3 2
6
2
3
6
x
x2 x3
 2 has the form

 2x  c where c is a real number;
5
2 15
therefore y has this form, and it remains to find c : since y  3 when x  4 , then
2 Any anti-derivate of x 
3
2
42 43
64
64
259
x2 x3
259

24  c  8 
 8  c  c  3  16 

. Therefore y 

2x 
2 15
15
15
15
2 15
15
x2
 c , where c is a real number. Therefore,
2
x2
22
f  x   3x 
 c , where c is to be determined; as f 2 1, then 1  3  2 
 c  c  3.
2
2
x2
Hence f  x   3x 
 3.
2
3 Any anti-derivative of 3  x is of the form 3x 
4 Any anti-derivative of 3x 2 
x4
x5
is of the form x 3 
 c , where c is a real number. Therefore,
15
3
 
05
x5
 c  c  2.
 c , where c is to be determined; as f 0  2, then 2  03 
15
15
x5
Hence f  x   x 3 
 2.
15
f  x   x3 
5 First find y  x  : since y  x   0.2x  3x2  1, then y  x   0.1x2  x3  x  c for some c which is
to be determined. As y  1 when x  1, then 1  0.1  12  13  1  c  c 3  0.1 3.1.
Therefore, y  x   0.1x2  x3  x  3.1.
Next, use this expression to find y when x  0.5 : y 0.5  0.1  0.52  0.53  0.5  3.1 2.45.
© Oxford University Press 2019
10
Worked solutions
6 a Since f   x    x  3, then f  x   
x2
 3x  c for some c which is to be determined. The
2
point  2,0 lies on the graph of f so f  2  0  0  
Therefore, f  x   
b
f 2   
 2
2
2
 3   2  c  c  2  6  8.
x2
 3x  8.
2
22
 3  2  8  12.
2
7 The cost function C  x  is an antiderivative of the marginal cost function:
x2 2
  x  c . To find c, use the fixed cost: C 0 145  145  c. Hence,
2
x
x2 2
the cost function is C  x  
  x 145.
2
x
C  x    M  x  dx 
8 The cost function C  x  is an antiderivative of the marginal cost function:
x3
 c . To find c, use the fixed cost: C 0 1000  c  1000.
3
x3
Hence, the cost function is C  x   3x  2x 2 
 1000.
3
C  x    M  x  dx  3x  2x 2 
9 a The revenue function is an antiderivative of the marginal revenue function:
t3
R t    t 2  80 dt 
 80t  c . The factory earns 567000$ from selling 120 refrigerators,
3
1203
so R 120 567000  567000 
 80  120  c  c  600. Therefore,
3
t3
R t  
 80t  600.
3


b The revenue from selling 150 refrigerators is R 150 
1503
 80  150  600  1113600.
3
Chapter Review
4
1 a i
 0.5x  4 dx
1
ii This area is trapezoidal, so we do not need to use the power rule to integrate. Use the
formula for the area of a trapezoid with edges of length
0.5  1  4  3.5 and 0.5  4  4  6 .
4
So,
1
 0.5x  4 dx  2  6  3.5   4  1 23.75.
1
b i
First we need to find the x-intercept by solving 3x  9  0  x  3. Therefore, the limits
of the integral are x  0, x  3. The area is
3
  3x  9 dx.
0
ii Using the area for a triangle (with base of length 3 and height 9): the area is
3
1
0  3x  9 dx  2  3  9  13.5.
2 a i
64
 1, and therefore has
2  0
equation y  4  x. The limits of integration are x 2, x  3. The area is written as a
First find the equation of the line: it’s gradient is m 
3
definite integral as
  4  x  dx .
2
© Oxford University Press 2019
11
Worked solutions
ii We can use the formula for area of a trapezoid with base of length 5 and edges of height
6 and 4  3  1 (the second is found using the equation of the line). Therefore, the area is
3
1
  4  x  dx  2  1  6  5 
2
4
b i
The area is

35
 17.5 .
2
16  x 2 dx.
0
ii The line represents a quarter of a circle, with radius 4 and centre  0,0 . Therefore, the
area of the shaded region is
0
3 a i
The area is
x
2
1
   42  4  12.6 (3 s.f.)
4
dx.
4
t
ii The area function A  t    x 2 dx 
4
4
b i
The area is
 ( x  1
2
64
t 3  4
.

. Therefore, the shaded area is A 0 
3
3
3
3
 3) dx.
1
ii The antiderivative of  x  1  3 is f  x  
2
f  4  f  1 
 x  1
3
3
 3x  c . Therefore, the area is
33
23
80
.
 3 4 
3 
3
3
3
2
c i
The area is
    x  4  x  1 dx.
0
 x 3 3x 2


 4x   c .
ii The antiderivative of –  x  4 x  1  x2  3x  4 is f  x    
2
 3



 23
 34
22
3
 4  2 
.
Therefore, the area is f 2  f 0  f 2   
3
2
3


1.5
d i
The area is
 8  x  dx.
3
1
ii The antiderivative of 8  x3 is f  x   8x 
f 1.5  f  1  8  1.5 
1.54 
(1)4 
  8  (1) 
  19.0 (3.s.f.).
4
4 

2
e i
The area is
x4
 c . Therefore, the area is
4
  x  3  x  2 
2
dx.
2


ii First expand the integrand:  x  3  x  2   x  3 x2  4x  4  x 3  x 2  8x  12.
2
The antiderivative of x3  x2  8x  12 is f  x  
area is f 2  f  2 
3
f
i
The area is
4
3
x
x

 4x 2  12x  c . Therefore, the
4
3
128
 42.7 (3 s.f.).
3
  x  3  x  2 
2
dx.
0


ii First expand the integrand:  x  3  x  2   x  3 x2  4x  4  x 3  x 2  8x  12.
2
The antiderivative of x3  x2  8x  12 is f  x  
area is f 3  f 0 
x 4 x3

 4x 2  12x  c . Therefore, the
4
3
45
(3 s.f.).
4
© Oxford University Press 2019
12
Worked solutions
g i
The area is
2
3
 1 x
2
ii Using GDC, find this area is 5.68 (3 s.f.)
dx.
1
3
h i
The area is
 1 

2
i
i
The area is
e
x
ii Using GDC, find this area is 19.3 (3 s.f.)
3  x dx.
4
ii Using GDC, find this area is 7.25 (3 s.f.)
dx.
2
4 a i
3
ii Using the power rule, an anti-derivative of
4

x dx  f  4  f 0 
0
3
x2
2x 2

. Therefore,
x is f  x  
3
3
2
3
2
16
 42 
.
3
3
b i
ii Using the power rule, an anti-derivative of x 2 is f  x  
2
x
2
2
dx  f 2  f  2  
x3
. Therefore,
3
23 23 16
.


3
3
3
c i


ii Note that –  x  2 x  4  x2  6x  8 . Therefore, the antiderivative of –  x  2  x  4
x

 3x 2  8x  . Hence
is (by power rule) f  x   – 
3


3
3
   x  2  x  4 dx  f 3  f 2.5  0.458
(3 s.f.)
2.5
d i
4
ii Using the anti-derivative found in 4cii,
4
   x  2  x  4 dx  f 4  f 2   3  1.33
(3
2
s.f.)
© Oxford University Press 2019
13
Worked solutions
e i
5
ii Use the GDC to find
10
 x  1 dx  6.93 (3 s.f.). Alternatively, note that an anti-derivative of
2
10
is f  x  10ln  x  1 . Therefore,
x 1
5
10
 x  1 dx  f 5  f 2  10ln 6  10ln 3  10ln 2  ln 1024  6.93
(3 s.f.)
2
f
i
1
ii Use the GDC to find
3
x
 2 dx  6.43 (3 s.f.). Alternatively, since 3x  2  ex ln3  2, we
1
can find an anti-derivate of 3x  2 as f  x  
1
3
x
1
 2 dx 
e x ln 3
 2x. Therefore,
ln3
eln 3
e ln 3
8
2
2  4
 6.43 3 s.f. .
ln3
ln3
ln27
g i
ii Using the power rule, an anti-derivative of x2  2x  3 is f  x  
3
x
2
2
 2x  3 dx  f 3  f  2  
x3
 2x  3. Therefore,
3
65
.
3
© Oxford University Press 2019
14
Worked solutions
5 a The x-intercepts of the graph of y   x  1  x  4 are x  1, 4; these are the limits of
4
integration. The area enclosed is therefore
   x  1  x  4 dx.
1


b First expand the integrand:   x  1  x  4  x2  3x  4 . The anti-derivative of
x

3x
 x2  3x  4 is f  x    

 4x  , therefore
2
 3


3

2
4
   x  1  x  4 dx  f  4  f  1 
1
125
 20.8 (3 s.f.).
6
6 a The x-intercepts are solutions of the equation 0  x  x  4 . These are x  0, 4.
2
4
b
x  x  4
2
dx
0


c First expand the integrand: x  x  4  x x 2  8x  16  x 3  8x 2  16x. The anti-derivative
2
of x3  8x2  16x is f  x  
4
x  x  4
2
4
3
x
8x
16x


, therefore
4
3
2
dx  f  4  f 0  f  4 
0
64
 21.3 (3 s.f.).
3
7 a The x-intercept is at 0.
b
2
c
x
3
dx
0
d The anti-derivative of x 3 is f  x  
2
24
x4
 4.
. Therefore x 3 dx  f 2  f 0 
4
4
0
8 a
b
b The area is written in integral form as
  x  2
2
 1dx. The integrand is expanded as
0
 x  2
2
 1  x2  4x  5, which has anti-derivative of f  x  
b
integral is
  x  2
2
x3
 2x 2  5x. Therefore, the
3
 1dx  f  b   f 0  f  b  . Given this area is equal to 42, the solution b
0
solves f  b  
b3
 2b2  5b  42. Using an equation solver, the solution is b  3.
3
9 Applying the trapezium rule with end-points a  1, b  10 and n  4 trapezoids. The
approximation is
1 10  1

 3  2  9  7  12  5  72.
2
4
© Oxford University Press 2019
15
Worked solutions
10 5 equal width trapezoids over the interval 2  x  4 have edges at x  2,2.4,2.8,3.2,3.6, 4.
Using the formula for the trapezium rule (with a  2, b  4, n  5 ): area

1 42


2
5

2  2 2



2.4  2  2.8  2  3.2  2  3.6  2  4  2  1.838 (3 s.f.)
11 a
b
f t   t 2  2t has x-intercepts at t  0,2. Therefore, when 0  t  x  2, the area under the
x
graph of f over the interval 0  t  x is
  t
2
 2t
0

dt  
x3
 x 2. However, when x  2,
3
f  0 over some part of 0  t  x , for which the integral represents the negative of the area
between the graph of f and the x -axis. In this case, the area under the graph of f over
2
the interval 0  t  x is
  t
2
 2t
0
x
4t dt
12 The integral

x

dt   t 2  2t
2

dt.
with x  3 is equal to the area of a trapezium with base length  x  3 and
3
x
heights 4x and 4 3  12 . Therefore,
1
4t dt  2  4x  12  x  3  2  x
3
13 Using the power rule,  2  x  dx  2x 
2

9 .
x2
 c , where c  R is a constant of integration.
2

x3 
x2 x 4

 c , where c  R is a constant of
14 Using the power rule,  1  x 
 dx  x 
4
2 16

integration.
15 a Using the co-ordinates of the points  0.25,0 ,1,10 , the area under the graph of L for
0.25  x  1 is equal to the area of a triangle with base of length 1   0.25 1.25 and
height 10  0 ; i.e. this area is
   x  4
6
b
2
1
1
 1.25  10  6.25.
2

 1 dx
c First expand the integrand as  x  4  1  x2  8x  17, which has anti-derivative
2
f x 
6
x3
 4x 2  17x . Therefore, the area under the curve between 1  x  6 is
3
  x  4
1
2
6
 1dx  x 2  8x  17 dx  f 6   f 1 
1
50
 16.7 (3 s.f.). Hence, the total area under
3
the graph of L is 6.25  16.7  22.9 (3 s.f.)
16 a The three lines have the gradient m. Since the points A 0,0 and B 2,6  lie on the purple
line, we can compute m 
60
 3. A point on the graph of the green line is C 1, 4 and
20
therefore the green line has equation y  4  3  x  1  y  f1  x   3x  1. A point on the
graph of the purple line is A 0,0 and therefore the purple line has equation y  f2  x   3x.
A point on the graph of the pink line is D 0, 2 and therefore the green line has equation
y  2  3x  y  f3  x   3x  2.
b
fi  x  are anti-derivatives of y  t  x  and hence t  x   fi'  x   3.
© Oxford University Press 2019
16
Worked solutions
c
 t  x  dx  3x  c , where c is an arbitrary constant of integration.
17 Begin by putting the data in a table:
x
1.5
2
2.5
3
3.5
4
4.5
5
y
0.43
1.5
2.82
4
4.69
4.5
3.04
0
Using the trapezium rule formula ( a  1.5, b  5, n  7 ), we estimate the area as
1 5  1.5

 0.43  2 1.5  2.82  4  4.69  4.5  3.04  0 10.4 (3 s.f.)
2
7
18 a Splitting the interval 0  x  1 into 5 equal strips means divisions at
x  0,0.2,0.4,0.6,0.8,1.0. Therefore, the trapezium rule with a  0, b  1, n  5 approximates
the area under the graph of f  x   e x as
2
A
1
b i




2
2
2
2
1 10

 e0  2 e0.2  e0.4  e0.6  e0.8  e1  0.7444 (4 s.f.)
2
5
e
 x2
dx
0
1
ii Using the GDC, find e x dx  0.7468 (4 s.f.).
2
0
c % error =
19 5x 
actual – estimated
actual
 100% 
0.7468  0.7444
 100%  0.321% (3 s.f.)
0.7468
12x 3 4x 4

 c 
3
4
M1A1A1
 5x  4x3  x 4  c
A1
x3 x2
3


 3x  c
20   x 2  x  3  dx 
2
2
2

Substituting x  1 and equating to
M1A1A1
13
2
1 1
13
 3c 
2 2
2
19
c 
2
x3 x2
19
f x 

 3x 
2
2
2

21 a
M1
A1
A1
6x  x2  10  x
M1
x2  7x  10  0
 x  2  x  5  0
M1
Coordinates are 2,8
A1
and 5,5
A1
b
A1A1
© Oxford University Press 2019
17
Worked solutions
5
5

x3 
c Area under curve   6 x  x 2 dx  3x 2 

3 2

2

100 28 72


 24
3
3
3
M1A1
1
39
 5  8  3 
2
2
39 9

units2
Required area  24 
2
2
Area under line 
3
M1A1
M1
Valid attempt to integrate
M1
2
 4t  1 dt
2
3
 0.2t  2t  t 
2
A1
0.2  33  2  32  3  0.2  23  2  22  2
 14.8 m
M1
3
b
A1

22 a Require 
 0.6t
M1A1
2
A1
23 a
M1A1A1
10
b

0
c
x2

 x  10 dx
10
A1
x2
x3
x  10  x 2 

10
10
10
 2 x3 
0  x  10  dx

M1
10
 x3 x 4 



 3 40 0
A1A1
1000 10000

3
40
1000 1000 1000  250 
2




 units
3
4
12 
3 

24 a Use of GDC
 1, 1 , 0,0 , 1,1
M1
A1
M1
A1A1A1
b
A1A1
© Oxford University Press 2019
18
Worked solutions
1
1
c
 x4 
1
3
0 x dx   4   4
0
M1A1
1
 3x 3 
3
0 x dx   4   4

0
1
4
1
3
A1
3 1 1
 
4 4 2
1
Therefore total area is 2   1 unit2
2
Positive area is therefore
25 a
M1
M1A1
A 1,0
A1
B  6,0
A1
C 3, 4
A1
6
b Area 
1
36
 2  4   2 dx
2
3 x
M1A1A1
6
6
 1
 4  36 x 2 dx  4  36   
 x 3
3
A1
 1  1 
 4  36       
 6  3 
1
 4  36   10 units2
6
M1
A1
26 a
M1A1A1
b
M1A1
5

x3 
1 5  0
Area    5 
 dx  
 5  0  2  4.96  4.68  3.92  2.44
25 
2 5 
0
M1A1
 18.5 units2
A1
c The curve is concave down in the interval 0  x  5 , so each trapezium will be an
underestimate.
Therefore the sum of the trapezia will also be an underestimate.
5
d
R1
R1
5


x3 
x4 
0  5  25  dx  5x  100 
0
M1A1
625
 18.75 units2
100
18.5  18.75
 100  1.33%
e Percentage error 
18.75
 25 
© Oxford University Press 2019
M1A1
M1A1
19