WORKED
SOLUTIONS
Worked solutions
1
From patterns to generalizations: sequences
and series
Exercise 1C
Skills check
1 a
x = −3
b
a=2
c
x =4
2 a
5
12
b
53
48
c
−
3 a 128
b 9
1 a
6
5
b
c −81
c
d
e
2 a
b
d
125 125 125
or 62.5, 31.25, 15.625
,
,
2
4
8
5 6 7
5
6
7
f
, ,
,
,
6 7 8
243 729 2187
u=
10 × 5
n
un =
−6n + 47 , arithmetic
c
un =
( −1)
n +1
3
∑n ( n + 1) = 2 + 6 + 12 = 20
5
∑
n =3
2 a
( −1)
n +1
=1 +
n−2
∞
∑ 4n
b
n =1
d
8
∑−2
e
n=1
, geometric
b
6
∑4n − 3 = 5 + 9 + 13 + 17 + 21 = 65
n =1
49,64,81
30,36, 42
n −1
n
n =2
c
−20, −23, −26
∑ ( −1) ( n + 1) =−2 + 3 − 4 + 5 =2
n =1
Exercise 1A
1 a
4
f
11
n −1
or
n =7
1
, geometric
3n
n
∑
n =3 n + 1
5
∞
∑n
1
un =−4 × 3n , geometric
u9= 5 + ( 8 ) ( 8 )
n −1
u9 = 69
, geometric
n −1
2
,
u11 =
40 + (10 ) ( −8 )
Exercise 1B
u=
40 − 80
11
1 a 1, -4, 16, -64, 256
u11 = −40
2
2
,3, − ,3
3
3
3
c -1, 2, 8, 128, 32768
d m, 3m + 5, 9m + 20, 27m + 65,
81m + 200
b un 4=
un −1,u1 1
=
un = u1 + ( n − 1) d
u11 = 40 + (11 − 1) ( −8 )
geometric
un =
un −1 − 2,u1 =
−2
mn
u9= 5 + 64
3 a 100,200,300, … , un =
100n, arithmetic
2 a
2
un = u1 + ( n − 1) d
f
3, −
1
n
n =6
u9 =5 + ( 9 − 1) ( 8 )
b
n
+1
10
e
c 70, 77, 84.7,…, un = 70 (1.1)
∑ ( −1)
n =1
∑ ( n + 1)
2n − 1
, neither
un =
2n
3
1
6,3, , … , un =
6 
2
2
100
Exercise 1D
d =
un un −1 + un −2 , neither
b
2
c
n =2
∑n m
2
−1 1 5
+ =
2 3 6
un = u1 + ( n − 1) d
u7= 5.05 + (7 − 1) ( 0.32 )
=
u7 5.05 + ( 6 ) ( 0.32 )
=
u7 5.05 + 1.92
u7 = 6.97
un −1
c un
,u1 52
=
=
10
d
un = un −1 + 5,u1 = 14
e
un =
un −2 × un −1,u1 =
2,u2 =
3
f un n=
=
(un−1 ) ,u1 1
© Oxford University Press 2019
Worked solutions
1
Worked solutions
4
un = u1 + ( n − 1) d
u6 =
3
2 = 11 + ( n − 1) ( −3)
1
1
+ ( 6 − 1)  
2
3
−9 =−3n + 3
1
1
u6=
+ (5 )  
2
3
1 5
u6=
+
2 3
u6 =
5
6
−12 =
−3n
n=4
4
u6 → u4
un = u1 + ( n − 1) d
u14 → u12
u9 = x + 2 + ( 9 − 1) (3)
u4 = u1 + ( 4 − 1) d
u9 = x + 2 + ( 8 ) (3)
184= 4 + (3) d
u9 = x + 2 + 24
180 = 3d
u9= x + 26
d = 60
4 + (12 − 1) ( 60 )
u12 =
un = u1 + ( n − 1) d
u12= 4 + (11) ( 60 )
u12= 4 + 660
u=
3a + (11) (3a)
12
u12 = 664
u=
3a + 33a
12
u12 = 36a
So 14th term of the given series is 664
5
−49 =
−7n
un = u1 + ( n − 1) d
n=7
65 = u1 + (21 − 1) ( −2 )
65 =
u1 + (20 ) ( −2 )
−36 =6 + ( n − 1) ( −7 )
−42 =
−7n + 7
Exercise 1E
2
un = u1 + ( n − 1) d
u3 → u1
13
6
u12 =3a + (12 − 1) (3a)
1
un = u1 + ( n − 1) d
It is the 7th term.
6
u12 = 30 + (12 − 1) (2 )
65
= u1 − 40
u=
30 + (11) (2 )
12
u1 = 105
u=
30 + 22
12
un = u1 + ( n − 1) d
u12 = 52 seats
u5 → u1
7 Let u1 2010
=
=
and d 4 .
u15 → u11
2050
= 2010 + ( n − 1) 4
u19 → u15
40
= 4n − 4
44 = 4n
u11 =u1 + (11 − 1) d
n = 11
−52.3 =
−3.7 + (10 ) d
Since they are held in 2050 (since n is a
natural number), the next time they will
be held is 2054.
−48.6 =
10d
d = −4.86
u15 = u1 + (15 − 1) ( −4.86 )
u15 =
−3.7 + (14 ) ( −4.86 )
u15 =
−3.7 + (14 ) ( −4.86 )
u15 = −71.74
8
82 = 40 + ( n − 1) ( 6 )
42
= 6n − 6
48 = 6n
n=8
In 7 weeks
© Oxford University Press 2019
2
Worked solutions
Exercise 1G
Exercise 1F
1
u6 = 9 (3)
6 −1
1 a
303.75 = 40r 6 −1
u6 = 9 (3)
5
r 5 = 7.59375
u6 = 2187
r = 1.5
b Not geometric
u=
40 × 1.511−1
11
u7 = 6 ( 0.75)
7 −1
c
u=
40 × 1.510
11
u7 = 6 ( 0.75)
6
=
u11 2306.60156 …
=
u7 1.0678 …
u11 ≈ 2307
u7 ≈ 1.068
2
u8 =−4 ( −1.5)
8 −1
d
15 −1
7
14
 4
u15 =
−1280  − 
 5
u8 = 68.34375
u8 ≈ 68.3
 268 435 456 
u15 = −1280 

 6103515 625 
13 −1
u13
1
= 500  
5
u13
1
= 500  
5
u15 = −56.295.. ≈ −56.3
12
3
Not geometric
g
u12 = 3 ( m )
un = u1r n −1
1 = 16r 3 −1
500
4
=
=
u13
244140625 1953125
f
u6 → u1 u20 → u15
 4
u15 =
−1280  − 
 5
u8 =−4 ( −1.5)
e
u5 → u1 u10 → u6 u15 → u11
r2 =
12 −1
r = ±
u12 = 3 ( m )
11
r =
u12 = 3m11
u30 = 1 (2 )
30 −1
x =2
u30 = 2
or
u30 = 536 870 912 cents or $5 368 709.12
3 Use an r value that is a factor of 64. For
example, r = 2 :
u
=
4
u5 32
= = 16
r
2
−1 x + 2
=
4
16
−4 = x + 2
x = −6
4 1536 = 6 (2 )
n−1
256 = 2n−1
28 = 2n−1
u4 16
u
=
= = 8
3
2
r
n −1 =
8
n=9
u3 8
u=
= = 4
2
2
r
u=
1
u2
u1
4= x + 2
29
u6
64
= = 32
2
r
1
4
1 x +2
=
4
16
2 1,2, 4… u30 = ?
u
=
5
1
16
5
32 = 2 ( r )
5 −1
16 = r 4
u2
4
= = 2
r
2
∴ One possible sequence is 2, 4,8,16,32,…
24 = r 4
r =2
© Oxford University Press 2019
3
Worked solutions
6
4 Geometric because a rate implies you are
multiplying.
u10 = 232 (1.03)
10 −1
u10 = 232 (1.03)
9
324 = 6 ( r )
21−1
=
u10 302.70737 …
54 = r 20
u10 = 303 students
7 a
r =
u30 = 1 (2 )
30 −1
54
=
r 1.220730 …
r ≈ 122%
u30 = 229
u30 = 536 870 912 grains
b 512 = 1 (2 )
20
Exercise 1I
n−1
1 a
=
d un − un −1 =
29 = 2n−1
n −1 =
9
=
S7
7  1
 1 
2   + (7 − 1)   
2  5
 3 
=
S7
7 2
 1 
 + (6 )  
2 5
 3 
=
S7
7 2

+ 2
2  5

=
S7
7  2 10 
+
2  5 5 
n = 10 th square
8 128 = 8 ( r )
5 −1
16 = r 4
r =2
8, 16, 32, 64, 128
Exercise 1H
1 Geometric because you are multiplying
1
each previous height by .
2
10 −1
1
u10 = 1  
2
b
10 −1
1
u10 = 1  
2
1
u10 =  
2
u10 =
7 12 
2  5 
S7 =
42
5
1
1
3
u1 =( −3) =
−
2
2
2
1
1
9
( −3) = 2 (9) = 2
2
9
un
9  2
2
=
= × −
=−3
r =
3 2  3 
un −1
−
2
1
meters
512
2 Arithmetic because you are adding more
money to your account every month.
6500= 2000 + (36 − 1) ( x + 5)
=
4500
S7 =
u2 =
9
8 1 1
− =
15 5 3
(35) ( x + 5)
128.57 = (𝑥𝑥 + 5)
x = $123.57
3 Arithmetic because you are adding from
year to year.
2017
= 1962 + ( n − 1) (12 )
=
55 12n − 12
S8 = −
3  1 − ( −3)
2  1 − ( −3)

S8 = −
3  1 − 6561 


2 1+3 
S8 = −
3  −6560 


2 4 
8




S8 = 2460
c=
r
un
0.05
=
= 0.5
un −1
0.1
 1 − ( 0.5)8 

S8 = 0.1 
 1 − 0.5 


67 = 12n
n = 5.583
Finland did not gain independence in the
year of the tiger.
S8 = 0.19921875
S8 ≈ 0.199
© Oxford University Press 2019
4
Worked solutions
d
3 The series is 1 + 2 + 4 + …
d = un − un −1 = 12 − 6 = 6
 1 − 26 
S6 = 1 

 1−2 
288 =6 + ( n − 1) ( 6 )
282
= 6 ( n − 1)
 1 − 64 
S6 = 

 1−2 
47= n − 1
n = 48
 −63 
S6 = 

 −1 
48
2 ( 6 ) + ( 48 − 1) ( 6 ) 
=
S48
2 
S6 = 63 family members
=
S48 24 12 + ( 47 ) ( 6 ) 
4 The series is 1 + 2 + 3 + …
S48 = 24 294
d = un − un −1 = 2 − 1 = 1
S48 = 7056
e
u1 = 4 , u2 = 8, d = 8 − 4 = 4
As 1000 is a multiple of 4, the largest
multiple of 4 less than 999 would be
996.
S12 6 2 + (11) (1) 
=
S12 = 6 13
996 = 4 + ( n − 1) ( 4 )
S12 = 78
= 4 ( n − 1)
992
But since there are two 12-hour cycles in
a 24-hour day:
248= n − 1
S24 = 78 × 2 = 156 chimes
n = 249
249
2 ( 4 ) + (249 − 1) ( 4 ) 
S
=
249
2 
=
S249
f
249
8 + 992
2 
5 The series is 5 + 9 + 13 + …
d = un − un −1 = 9 − 5 = 4
48
2 (5) + ( 48 − 1) ( 4 ) 
2 
=
S48
S249 = 124500
=
S48 24 10 + ( 47 ) ( 4 ) 
u1 =
2
( −1) (2) =
=
S48 24 10 + 188
0
1
u2 =
−4
( −1) (2) =
1
r =
2
un
−4
=
= −2
2
un −1
 1 − ( −2 )6 

S6 = 2 
 1 − ( −2 ) 


 1 − 64 
S6 = 2 

 3 
S48 = 24 198
S48 = 4752 line segments
Exercise 1J
1 a Not converging as r = 1.5.
b
S=
2 ( −21)
6
d = un − un −1 = 26 − 22 = 4
=
S30
30
2 (22 ) + (30 − 1) ( 4 ) 
2 
=
S30 15 44 + (29 ) ( 4 ) 
=
S30 15 44 + 116 
S30 = 15 160
S30 = 2400 seats
9
un
9
8
3
32
=
−
r = = =×
−3 32 −3
4
un −1
8
u1
S∞ =
=
1−r
S6 = −42
2
12
2 (1) + (12 − 1) (1) 
2 
=
S12
c
3
3
−
8 =
8 = −3 × 4 = −3
7
−3
8 7 14
1—
4
4
−
−5
un
−5 2
1
= 4 =
×
=
r =
−5
4 −5 2
un −1
2
5
5
−
−
u1
−5 2
2
S∞ =
=
=2 = × =
−5
1
1
1−r
2 1
1—
2
2
d Not converging as r = −2 .
© Oxford University Press 2019
5
Worked solutions
e
S28=
9 ( x − 1)
un
9x − 9
9
1
=
=
=
= =
r
un −1 27 x − 27 27 ( x − 1) 27 3
S28 = 14 (119 )
27 x − 27 27 x − 27
S∞
=
=
1
2
1−
3
3
81
 3  81
=
(27x − 27)  2  =2 x − 2
 
f
S28 = 1666
c
n
2 ( −8 ) + ( n − 1) (5)  > 2000
2
n  −16 + (5n − 5)  > 4000
1
un
1
2 1
r =
= 2 2 =
×
=
1
1
2
un −1
2 2
2
u1
S∞ =
=
1−r
1
2 =
1
1—
2
1
2 =
1
2
1
2
×
n (5n − 21) > 4000
5n2 − 21n − 4000 > 0
2
=
1
By GDC, n ≈ −26.3 or 30.5
2
Since n > 0, n =
31
2
2 Any infinite geometric series where
−1 < r < 1 , r ≠ 0
u1
=
1−r
S
=
∞
40
2u1 + ( 40 − 1) d 
2 
2 1900
=
=
1900 20 2u1 + (39 ) d 
3 Any infinite geometric series where
r < −1, or r > 1
4
n
2u1 + ( n − 1) d 
2
S
=
n
Not converging as r = 2 .
g
28
 −8 + 127
2 
=
95 2u1 + 39d
12
12
= = 30
3
2
1−
5
5
106 =u1 + ( 40 − 1) d
106
= u1 + 39d
Total distance = 2S∞ − u1 = 2 (30 ) − 12
=
d 106 − u1
39
= 48 ft.
95 = 2u1 + 106 − u1
u1
426
426
=
=
= 426 000
1 − r 1 − 0.999 0.001
5=
S∞
u1 = −11
39
=
d 106 − −11
426000 × 42 =
17 892 000 gallons
39d = 117
d =3
Exercise 1K
1 a
S∞ =
u1
1−r
20 =
u1
1 − 0.2
790 = 10  −16 + (19 ) d 
20 =
u1
0.8
=
79 19d − 16
u1 = 16
n
2u1 + ( n − 1) d 
2
S
=
n
790
=
20
2 ( −8 ) + (20 − 1) d 
2 
95 = 19d
d =5
b i
un = u1 + ( n − 1) d
u28 =−8 + (28 − 1) (5)
u28 =−8 + (27 ) (5)
u28 =−8 + 135
u28 = 127
b ii =
Sn
n
u1 + un 
2
3
4 3u1 =
u1
1−r
u
1 − r =1
3u1
1
1−r =
3
r= 1 −
r =
1
3
2
3
© Oxford University Press 2019
6
Worked solutions
=
u1 27 (1 − r )
5 Choose any r value −1 < r < 1, r ≠ 0 .
Example:
1
2
Let r =
8=
8=
u1
1−
1 − r2 
=
15 27 (1 − r ) 

 1−r 
15
= 1 − r2
27
1
2
r 2= 1 −
u1
1
2
r2 =
u1 = 4
Since the geometric series has only
positive terms,
u4 = u1r 4 −1
8u1 = u1r 3
r =
3
r =8
1 − rn 
Sn = u1 

 1−r 
1
u1 = 27  
3
 1 − 210 
2557.5 = u1 

 1−2 
u1 = 9
 −1023 
2557.5 = u1 

 −1 
9 a
2557.5 = 1023u1
u1 = 2.5
r =
6
m −1
r =
m+8
6
6
m+8
=
m −1
6
b i
u10 = 2.5 (2 )
10 −1
36 = m2 + 7m − 8
u10 = 2.5 (2 )
9
0 = m2 + 7m − 44
u10 = 2.5 (512 )
0=
( m + 11) ( m − 4)
u10 = 1280
m = −11 m = 4
1 − 5 
7 2375 = 5 

 1−5 
n
−11 + 8
3
1
r =
=
− =
−
6
6
2
b ii
 1 − 5n 
475 = 

 −4 
=
r
−1900 =
1 − 5n
By GDC,
=
n 4.69116 …
A minimum of 5 rounds are required.
4 + 8 12
= = 2
6
6
c i
Since the sum of an infinite series
can only be found when
1
−1 < r < 1, r ≠ 0 , r = − .
2
c ii
u1 =m − 1 =−11 − 1 =−12
5n = 1901
1 − r2 
S2 = u1 

 1−r 
S∞ =
1 − r2 
15 = u1 

 1−r 
27 =
2
3
2

b=
u1 27 1 − 
3

r =2
8 a
12
27
12
2 3
2
r =
±
=
±
=
±
27
3
3 3
∴ The series is 4 + 2 + 1 + …
6
15
27
S∞ =
u1
1−r
−12
1
1+
2
−12
3
2
S∞ = −8
© Oxford University Press 2019
7
Worked solutions
Exercise 1L
=
104r 0.067585 …
1 a =
A P (1 + nr )
=
r 0.000649855 …
(
=
A 1500 1 + 10 ( 0.06 )
Fernando will pay an annual simple
interest rate of 0.065%.
)
A = 1500 (1.6 )
ii
A = 2400
≈ 23846 Columbia Pesos.
I = 2400 − 1500 = $900
3=
A P (1 + r )
b=
A P (1 + nr )
12×5
A = 90000 (1.001875)
60
A = 32000 (1.1)
=
A 100705.8944966 …
A = 35200
A ≈ $100705.89
I = 35200 − 32000 = 3200 GBP
4=
A P (1 + r )
n
4×5
32546 = P (1.0105)
20
36
0.02 

=
A 14168000 1 +

12 

=
32546 1.232328 … P
=
A 14168000 (1.0617835 …)
=
P 26410.17651144 …
P ≈ $26410.18
=
A 15043348.839948 …
5=
A P (1 + r )
=
I 15043348.84 − 14168000
I ≈ 875348.84 Yen
n
12× n
0.0325 

=
10000 5000 1 +

12 

n
2×365
0.04 

=
A 300000 1 +

365 

=
2
(1.0027083 …)
12 n
Using the GDC,
=
n 21.3567 …
730
0.04 

=
A 300000 1 +

365 

n ≈ 21.4 years
6=
A P (1 + r )
=
A 324984.69581 …
n
12× (18 −5)
0.055 

50000
= P 1 +

12 

=
I 324984.70 − 300000
I ≈ 24 984.70 Mexican Pesos
e =
A P (1 + r )
n
0.042 

= P 1 +
32546

4 

3×12
0.02 

A 14168000 1 +
=

12 

d=
A P (1 + r )
n
0.0225 

=
A 90000 1 +

12 


 0.0125  
=
A 32000 1 + 32 

4



c =
A P (1 + r )
2480000
= 23846.15
104
50000 P (1.00458333 …)
=
156
n
=
50000 P (2.04085012 …)
12×25
0.0225 

=
A 250000 1 +

12 

=
P 24499.594316 …
P ≈ 24500 Brazilian Reals
300
0.0225 

=
A 250000 1 +

12 

7 Oliver:
=
A P (1 + r )
=
A 438532.634627 …
=
I 438532.63 − 250000
n
12×5
0.0125 

=
A 400 1 +

12 

I ≈ 188532.63 Swiss Francs
2 i =
A P (1 + nr )
=
A 425.783932 …
(
2480000
= 2323000 1 + (52 × 2 ) r
1.067585 …= 1 + 104r
)
A ≈ 425.78 GBP
Harry:
=
A P (1 + r )
© Oxford University Press 2019
n
8
Worked solutions
or are switching jobs, or looking for
jobs.
A 400 (1 + 0.0175)
=
5
A = 436.2466257…
3 a This means that it takes 1.23 years for
the substance to decrease to half of
the original mass.
A ≈ 436.25 GBP
Harry earned more than Oliver.
t −1
8 Savings account:
b
A P (1 + nr )
=
(
A 20000 1 + n ( 0.012 )
=
)
original mass, and h is the half-life.
=
A 20000 (1 + 0.012n )
7.2 −1
d
GIC:
=
A P (1 + r )
n
 1  1.23
A = 52  
2
5.0406504…
1
A = 52  
2
12×2
0.035 

A 20000 1 +
=

12 

=
A 1.57985 …
A ≈ 1.58 g
=
A 20000 (1.00291666 …)
24
=
A 21447.978280670 …
21447.978280670
=
… 20000 (1 + 0.012n )
1.0723989140 …= 1 + 0.012n
0. 0723989140 … = 0.012n
=
n 6.03324 …
Exercise 1N
1
x 4 + 20 x 3 + 150 x 2 + 500 x + 625
2
−b5 + 10b4 − 40b3 + 80b2 − 80b + 32
3
64 x 6 − 192 x 5 + 240 x 4 − 160 x 3
+60 x 2 − 12 x + 1
n ≈ 6.03 years
Exercise 1M
1=
a r
1 h
, where A is the amount
A = A0  
2
remaining after t years, A0 is the
11000
= 0.88
12500
4
256 x 4 + 256 x 3y + 96 x 2y 2 + 16 xy 3 + y 4
5
x 3 − 9 x 2y + 27 xy 2 − 27y 3
6
243x 5 + 1620 x 4y + 4320 x 3y 2 + 5760 x 2y 3
+3840 xy 4 + 1024y 5
=
C 12500 × 0.88t , where C represents
the white blood cell count and t is the
time every 12 hours.
Exercise 1O
b 3 days = 72 hours = 6 12-hour periods
1 a
6
C = 12500 × 0.88 = 5805.0510 …
2 a This is an arithmetic sequence since
the rate decreases by -0.2% each
month.
b
U =
7.9 − 0.2 ( t − 1) , where U
represents the unemployment rate and
t is the month starting with January.
c
U =
7.9 − 0.2 (12 − 1)
(3x ) ( −5)
7
4
(
= (330 ) 2187 x 7
C ≈ 5805 cells/mcL
c The limitation of the general formula is
that white blood cell count does not
continue to decrease infinitely. Once
the antibiotics killed the infection, the
patient’s white blood cell count would
return to normal.
11C4
) (625)
= 451068 750x 7
b
10C8
( x ) ( 6y )
2
8
( )(
= ( 45) x 2 1679616y 8
)
= 75582 720x 2y 8
c Since n = 6, there will be 7 terms in
the expansion. Hence the middle term
is the 4th term.
6C 3
(2) ( −3y )
3
3
(
=20 ( 8 ) −27y 3
)
= −4320y 3
=
U 7.9 − 0.2 (11)
d The constant term will contain x 0 ,
hence
U = 5.7%
d It is not realistic. There will always be
people you are not capable of working,
9C 9
© Oxford University Press 2019
( x ) ( −3)
4
0
9
9
Worked solutions
=
r =5
( −3)
9
7 −5
= −19 683
e
(
7C 7
−2x 2
7C 5
7
 −3 


 x 
)
0
2
7
 x 
x
21k 5 x15
= 168
4 x15
2 a 1, 4, 6, 4, 1
=4
(3x ) ( −2)
1
3
21k 5
= 168
4
(3x ) ( −8)
k 5 = 32
= −96
8C 5 ( x )
3
3 a
k =2
( −3)
5
6
=(56) ( x 3 ) ( −243)
( −2x )
8C 4
4
(x
33 − 3r
x 4 ( −3) =
−387 072 x 5
) ( −k )
3
=
−387 072 x 5
1792 ( −k ) =
−387 072
3
5
( −k )
3
)  x1  = x
3
3
r
(x )
3
5
(56) (32 ) ( −k ) =
−387 072
1
9
  =x
x
11− r
( 2 x ) ( −k )
(
4
=– 11340 x
8C 3
(56) 32 x 5
= −13 608x 3
b
5
 x8   k5 
21x 7  2   15  = 168
 2  x 
= − 2187
7
4C 3
5
 k 
 3  = 168
x 
 x4   k 
21x 7 
  3  = 168
 2  x 
=  −3 
b
 x4 
x7 

 2 
−k =
=
−216
−216
3
−k =−6
9
r
k =6
x 33 −3r
= x9
xr
7
6C 3
( a) ( −b2 )
3
3
(
)
x 33 −3r − r = x 9
20a3 −b6
x 33− 4r = x 9
−20a3b6
33 − 4r =
9
The coefficient is −20.
4r = 24
8
r =6
11C6
(x )
( )
 729 
 6 
 x 
462 x
5
15
3
3
3
3
+   21(0.52)2 +   20 ( 0.52 )
2
3
 
 
(2.52)
3
336 798x 9
( )
7−r
r
 1 
0
 3 = x
x 
 1 
x 7 x 28 − 4r  3r  = x 0
x 
(
)
=
8 + (3) ( 4 ) ( 0.52 ) + (3) (2 ) ( 0.52 )
2
+ ( 0.52 )
3
5 The constant term will contain x 0 , hence
x7 x 4
3
3
3
3
0
=
(2.52)   23 (0.52) +   22(0.52)1
0
 
1
6
 −3 


 x 
3
(2.52)= (2 + 0.52)
(2.52)
=
8 + 12 ( 0.52 ) + 6(0.52)2 + ( 0.52 )
(2.52)
=
8 + 12 ( 0.52 ) + 6 ( 0.2704 )
3
3
3
+0.140608
x 28 − 4r +7 −3r = x 0
(2.52)
=
8 + 6.24 + 1.6224 + 0.140608
x 35−7r = x 0
(2.52)
= 16.003008
35 − 7r =
0
7r = 35
3
3
(2.52)
3
© Oxford University Press 2019
≈ 16.003
10
Worked solutions
9 a
x 5 − 25x 4 + 250 x 3 − 1250 x 2 + 3125x
−3125
Exercise 1P
1
b Hence, the term containing x 4 will be
(2x ) (250x ) = 525x
3
(3 − 2x )
10 a
4
LHS
RHS
−2 ( a − 4 ) + 3 (2a + 6 ) − 6 ( a − 5)
−2 ( a − 28 )
−2a + 8 + 6a + 18 − 6a + 30
−2a + 56
4
−2 ( a − 28 )
= 16 x 4 − 96 x 3 + 216 x 2 − 216 x +81
( −2x + 3)
LHS ≡ RHS
4
4
2
3
LHS
2
= 16 x − 96 x + 216 x − 216 x +81
b No, when the exponent is odd, the
expansions will not be the same.
RHS
( x − 3)
2
x2 − 6x + 14
+5
x2 − 6x + 9 + 5
11 Since there are four terms, n must be 3.
x2 − 6x + 14
Let k be the coefficient of x and m be
the coefficient of y .
LHS ≡ RHS
Using the first term,
3
3
3
0
3
  ( kx ) ( my ) = 27 x
1
LHS
RHS
1
m
1
1
+
m + 1 m2 + m
k 3 x 3 = 27 x 3
1
1
+
m + 1 m (m + 1)
k 3 = 27
k = 3 27
m
1
+
m (m + 1) m (m + 1)
k =3
Using the last term,
m+1
m (m + 1)
3
0
3
3
  (3x ) ( my ) = −64y
3
( my )
3
1
m
= −64y 3
m3y 3 = −64y 3
RHS ≡ LHS
3
m = −64
m=
3
4 a
−64
m = −4
∴ ( a + b) =
(3x − 4y )
n
n
12 2=
(1 + 1)
3
LHS
RHS
x − 2 3x − 6
÷ 2
x
x +x
x +1
3
x − 2 x2 + x
×
x
3x − 6
n
 n n
 n  n −1 1  n  n −2 2
0
2n =  (1) (1) +   (1) (1) +   (1) (1)
0
1
2
 n  1 n −1  n  0
n
+ …
 (1) (1) +   (1) (1)
 n − 1
 n
Since 1x = 1 for any x ∈ R ,
x − 2 x ( x + 1)
×
x
3 ( x − 2)
x +1
3
LHS ≡ RHS
b
x ≠ −1,0,2
 n  n  n
 n   n
2n =   +   +   + … 
+ 
0 1 2
 n − 1  n 
© Oxford University Press 2019
11
Worked solutions
Chapter review
1 i
a This sequence is not arithmetic
since 18 − 6 ≠ 6 − 3. This sequence
18 6
is not geometric since
≠ .
6
3
b un = u1rn−1
un= 3 × (2 )
n −1
u5= 3 × (2 )
5 −1
c
u5= 3 × (2 )
ii a This sequence is arithmetic since
−12 − −14 = −14 − −16 = 2 .
u5= 3 × 16
b un = u1 + (n − 1) d
u5 = 48
un =
−16 + 2 (n − 1)
 1 − 210 
S10 = 3 

 1−2 
d
un =
−16 + 2n − 2
u=
2n − 18
n
 1 − 210 
S10 = 3 

 −1 
c
u10
= 2 (10 ) − 18
= 2
d
S
=
n
n
2u1 + (n − 1) d
2
S8=
8
2 ( −16 ) + ( 8 − 1) (2 )
2
(
(
S10 = 3069
)
(
S8 = 4 −32 + (7 ) (2 )
v a The first few terms of the sequence
are 105, 110, 115…
)
This sequence is arithmetic since
115 − 110 = 110 − 105 = 5 .
)
b un = 105 + (n − 1) 5
S8 = 4 ( −32 + 14 )
un = 105 + 5n − 5
S=
4 ( −18 )
8
u=
5n + 100
n
S8 = −72
iii a This sequence is geometric since
500
1000 1
= =
.
1000 2000 2
b un = u1r
c=
u7 5 (7 ) + 100
u=
35 + 100
7
u7 = 135
n −1
d
n −1
1
=
un 2000 ×  
2
1
=
u9 2000 ×  
2
S
=
n
=
S9
1
c=
u9 2000 ×  
2
9 −1
n
2u1 + (n − 1) d
2
(
)
9
2 (105) + ( 9 − 1) (5)
2
(
(
=
S9 4.5 210 + ( 8 ) (5)
)
)
=
S9 4.5 (210 + 40 )
8
S9 = 4.5 (250 )
u9 = 7.81
d
4
S9 = 1125
1 − r 
Sn = un 

 1−r 
n
2 Renaming the terms:

1
1 −  

2
S7 = 2000 
1
 1−

2

7
u6 = u1 = −5







u9 = u4 = −20
This means we now want to find S20 = S15.
We first need to find d:
S7 = 3968.75
iv a The first few terms of this sequence
is 3, 6, 12,…
This
12
=
6
is a geometric sequence since
6
= 2.
3
u4 = u1 + (n − 1) d
−20 =−5 + ( 4 − 1) d
−15 =
3d
d = −5
© Oxford University Press 2019
12
Worked solutions
6 a An infinite sum can only be found for a
converging geometric sequence.
S=
15
n
2u1 + (n − 1) d
2
S=
16
15
2 ( −5) + (15 − 1) ( −5)
2
(
)
(
(
S16 = 7.5 −10 + (14 ) ( −5)
S∞ =
)
)
S∞ =
S16 = 7.5 ( −10 − 70 )
u1
1−r
1
4
1− −
=
S16 7.5 ( −80 )
S16 = −600
S∞ =
3 u2 =−2 ( −4 ) + 3 =
8+3 =
11
u3 =−2 (11) + 3 =−22 + 3 =−19
u5 =−2 ( 41) + 3 =−82 + 3 =−79
b
The first five terms are
−4, 11, − 19, 41, − 79 .
1
2
1
1
4
=
3 6
2
0.12
= 2 > 1 so the series is not
0.06
converging
b=
7 un = u1 + (n − 1) d
4 We first need to find r:
−0.1
= −0.2
0.5
61 = 4 + (n − 1) (3)
61 =4 + 3n − 3
 1 − ( −0.2 )n−1 

0.416 = 0.5 
 1 − −0.2 


60 = 3n
n = 20
−0.0016 = − ( −0.2 )
n −1
0.0016 =
1+
S=
∞
u4 =−2 ( −19 ) + 3 =
38 + 3 =41
r=
1
4
1
2
8 u4 = 8u1
un = u1rn−1
( −0.2)
n −1
u4 = u1r 4 −1
By GDC, n = 5
5 Renaming the terms:
8u1 = u1r3
u=
u=
4.5
3
1
8 = r3
u=
u=
22.78125
7
5
r=2
This means we now want to find u1 = u−1
 1 − 29 −1 
765 = u1 

 1−2 
Finding r:
un = u1rn−1
u5 = u1r5 −1
22.78125
= 4.5 × r 4
5.0625 = r 4
4
r 4 = 4 5.0625
r = 1.5
 1 − 28 
765 = u1 

 −1 
 1 − 28 
765 = u1 

 −1 
 1 − 256 
765 = u1 

 −1 
u−1 = u1r −1−1
 −255 
765 = u1 

 −1 
u
=
4.5 × 1.5−2
−1
765 = 255u1
u−1 = 2
u1 = 3
u9 = u1r 9 −1
u9= 3 × 28
u9 = 768
© Oxford University Press 2019
13
Worked solutions
9
d u=
4n + 41
n
6
x+2
=
x −3
6
101
= 4n + 41
2
36 = x − x − 6
60 = 4n
2
0 = x − x − 42
n = 15
0=
( x − 7) ( x + 6 )
15x100g = 1500g or 1.5kg
24
x = 7 or x = −6
10 a
55, 51.15, 47.5695, 44.239635…
b It is a geometric sequence because
47.5695 51.15
= = 0.93 .
51.15
55
c
u11 = u1r11−1
u=
55 × 0.9310
11
u11 ≈ 26.6 litres left in the tank
d u16 = u1r16 −1
55 − 18.5 =
36.5 litres drained from
the tank
u
S∞ = 1
1−r
Each row in Pascal’s triangle is
symmetric.
b You can add each set of consecutive in
the 14th row to find the 15th row.
1, (1 + 13) , (13 + 78 ) , (78 + 286 ) ,
=
S∞ 785.714285 …
S∞ ≈ 785 minutes or 13 hours and
5 minutes
11 a
r
= 0.003750 …
12
1716, 1287, 715, 286, 78, 13, 1
u11 ≈ 18.5 litres
55
0.07
1.09401 …
13 a 1, 13, 78, 286, 715, 1287, 1716,
=
u11 18.5185474 …
S∞ =
24
r ≈ 4.5%
u16
= 55 × 0.93
55
1 − 0.93
r
1=
+
12
r = 0.0450009
15
S∞ =
24
r 

1.09401 …= 1 +

12 

r
1=
+
1.003750 …
12
=
u11 26.61902 …
e
r 

12
22960 20 987 1 +
=

12 

un = u1 + (n − 1) d
(286 + 715) , (715 + 1287) ,
(1287 + 1716 ) , (1716 + 1716 ) ,
(1716 + 1287) , (1287 + 715) ,
(715 + 286 ) , (286 + 78) , (78 + 13) ,
(13 + 1) , 1
1, 14, 91, 364, 1001, 2002, 3003, 3432,
3003, 2002, 1001, 364, 91, 14, 1
6
6
6
0
5
1
14   (3x ) ( −y ) +   (3x ) ( −y )
0
1
6
6
4
2
3
3
+   (3x ) ( −y ) +   (3x ) ( −y )
2
3
 
 
un = 45 + (n − 1) ( 4 )
un = 45 + 4n − 4
u=
4n + 41
n
6
6
2
4
1
5
+   (3x ) ( −y ) +   (3x ) ( −y )
 4
5
b=
= 10 weights
1kg 1000g
=
u10 4 (10 ) + 41
6
0
6
+   (3x ) ( −y )
6
 
u=
40 + 41
10
= 729x6 + 6 (243x5 ) ( −y ) + 15 ( 81x 4 )( y2 )
u10 = 81 cm
c Eventually the spring will hit the
ground or the surface it is sitting on,
so the length will become constant.
Also, the spring could break from too
much weight.
(
)( −y ) + 15 (9x )( y )
+6 (3x ) ( −y ) + y
+20 27x3
3
5
2
4
6
=
729x6 − 1458x5y + 1215x 4y2 − 540x3y3
+135x2y 4 − 18xy5 + y6
© Oxford University Press 2019
14
Worked solutions
5
8  3 
15    2  −4x 4
3  x 
(
9
)
3
k =2
 243 
56  10  −64x12
 x 
)
 243 
56  10  −64x12
 x 
)
(
(
19 (2x − 1) ( x − 3) − 3 ( x − 4 )
2
(
= 2x2 − 7x + 3 − 3 x2 − 8x + 16
2
=
− x 2 + 17 x − 45
The coefficient is -870 912.
16 The sixth term will have be in the form of
n −5
( −5x )
5
5
.
= ( x − 3) ×
( )
=
 1 
 2n−10  = 1
x

=x
b
0
x − 7x + 12
x
x ≠ −4, − 2, 0
9
5
4
  ( x ) ( −3)
4
 
2
( ) (81)
4
(
)
1−
=
−20412x6
18 To determine which term is the constant,
ignore the coefficients for now:
a
1
0
  =x
x
+
x36 −3a− a = x0
=
36 − 4a =
0
=
a=9
3
9
 x k 
112640
220 
 9  =
27
x
27

 
9
9
220k 9 112640
=
27
27
5x 4
x5
−
256 1024
A1
M1
5 × 0.1 5 × 0.12
0.9755 ≈ 1 −
+
4
8
=1 −
4a = 36
M1A1
5x 5x 2 5x 3
+
−
4
8
32
b Substituting x = 0.1
x36 −3a
= x0
xa
12   x3   k 
    
 3  3  x
5
5  x 
5  x 
+   11  −  +    − 
 4  4 
5  4 
10206x5
−2x 10206x5
3
5  x 
5  x 
+   13  −  +   12  − 
4
2


 
3  4 
126 x5
12 − a
QED
5
5  x 
=  15 +   14  − 
0
 
1  4 
n=5
 x3 
 
1
x
2
5
−2n =
−10
b
( x − 4)
x

21 a 1 − 
4

−2n + 10 =
0
17 a
( x − 3) ( x + 2 ) × ( x − 4 ) ( x + 4 )
x+4
x ( x + 2)
=
 1  25
= x25
 2n−10  x
x


x
x2 − x − 6 x2 − 16
× 2
x+4
x +2
20 a
Ignoring any coefficients:
−2n +10
)
2
= 2x − 7x + 3 − 3x + 24x − 48
−870 912x2
n  2 
  2 
5  x 
k 9 = 9 512
1
5
+
8 800
A1
800 100
5
−
+
800 800 800
705  141 
=


800  160 
22 a Using un =a + ( n − 1) d
143= a + 14d
183= a + 30d
Solving simultaneously
a = 108
d =
5
2
A1
M1
A1
A1
A1
A1
220k 9 = 112640
k 9 = 512
© Oxford University Press 2019
15
Worked solutions
b 100th term is a + 99d
M1
5
= 108 + 99 ×
2
A1
= 355.5
23 a Money in the account would be
3000 × 1.01510 ( =
$3482 )
4
 4  1 
 1

28 a 
x
−=
 

 2x

 0   2x 
3
 4  1 
+ 

 1   2x 
M1A1
A1
5500 × 1.02754
= $6130.42
b Consider 5500 × 1.0275n−1 =
12000
M1A1
12000
1.0275n−1 =
5500
Using GDC:
M1
n −1 =
28.76
A1
n = 29.76
So Brad must wait 30 years A1
25 Require ( 3 × coefficient of term in x 5 )
(
+ 1 × coefficient of term in x 4
)
8
8
5
4
3 ×   43 ( −2 x ) + 1 ×   44 ( −2 x ) M1A1A1
5
4
 
 
= 3 × ( −114688 ) + 1 × 286720
A1
= −57344
 n
2
26   1n −2 (3x ) = 495x 2
2
 
(
)
9n ( n − 1)
2
n2 − n − 110 =
0
A1
0
( n − 11) ( n + 10) =
M1A1
So n = 11 since n > 0
8
27 Require   x 3
2
( )
= 28 × ( −2 )
6
= 28 × 64
= 1792
2
 2
− 
 x
A1
6
M1A1
A1
A1
1
M1A1
=27 − 27 x + 9 x 2 − x 3
 1

− x

2
x


4
1
1
3

= 27 − 27 x + 9 x 2 − x 3  x 4 +
− 2x 2 −
+ 
16 x 4
2x 2 2 

(
)
3 9

Therefore required term is  27 ×  −
2

 2
A1
A1
= 36
29 120 =
120 =
a
1 − 0.2
M1A1
5a
4
a = 96
The 6th term is therefore
A1
96 × 0.25 =
0.03072
M1A1
20
9
Solving simultaneously
20
1
ar 5
=
= = r4
9 × 180 81
ar
30 ar = 180 and ar 5 =
Therefore r =
So
a
=
=
S∞
n ( n − 1) =
110
( −x )
(3 − x )
3
M1A1
= 495
2
A1
(3 − x )
= 3000 × 1.01511
24 a
0
( −x )
1
1
3
=x 4 +
− 2x 2 −
+
16 x 4
2x 2 2
3
b
M1
 1.01510 − 1 
+ (1200 × 1.015) 

 1.015 − 1 
M1A1
A1
= $16570
2
 4  1 
+  

 2   2x 
M1A1
3000 × 1.01510 − 3000 =
$482 A1
b Total amount is
+1200 × 1.01510 )
0
1
Therefore interest gained is
+(1200 × 1.015 +1200 × 1.0152 + K
1
( −x )
 4  1 
 4  1 
3
+ 
 ( −x ) +   

2
x
3

 
 4   2x 
M1A1
3000 × 1.01511
( −x )
4
1
3
180 180
540
= =
1
r
3
a
540
3 × 540
= =
= 810
1 − r 1 − 13
2
M1A1
M1
A1
A1
M1A1
31 First part is geometric sum, a = 1 ,
M1
r = 1.6 , n = 16
Second part is arithmetic sum, a = 0 ,
M1
d = −12 , n = 16
Third part is 16 × 1 =
A1
16
Geometric sum:
1.616 − 1
A1
=
S16
= 3072.791
1.6 − 1
Arithmetic sum:
16
S16 =
(2 × 0 + 15 × ( −12) ) = −1440
2
A1
© Oxford University Press 2019
16
Worked solutions
So
n =15
∑ (1.6
n
n =0
)
− 12n + 1
= 3072.791 − 1440 + 16 = 1648.8 A1
32 Required distance
5
5 5

 

= 20 +  2 × × 20  +  2 × × × 20  + L
6
6
6

 

M1M1A1
= 20 +
= 20 +
100
3
1−
5
6
A1
100
3
1
6
= 20 + 200
= 220 m
 n − 1  n − 1 
33 
+

 k   k − 1
( n − 1) ! +
( n − 1) !
=
k ! ( n − k − 1) ! ( k − 1) ! ( n − k ) !
( n − k ) ( n − 1) !+ k ( n − 1) !
k !( n − k ) !
n ( n − 1) !− k ( n − 1) !+ k ( n − 1) !
=
k !( n − k ) !
n ( n − 1) !
=
k !( n − k ) !
=
A1
M1A1A1
A1
A1
A1

  n
n!
=
 =
 k ! ( n − k ) !   k 


34 Consider multiples of 7:
504 is the first multiple and 1400 is the
final multiple
1400 = 504 + 7 ( n − 1)
M1
A1
⇒n=
129
So the sum of the multiples of 7 is
129
S
=
(2 × 504 + 7 × (129 − 1) )
129
2
M1A1
= 122808
Sum of the integers from 500 to 1400
(inclusive) is
901
S
=
(2 × 500 + 1 × (901 − 1) )
901
2
= 855950
Therefore require
855950 − 122808 =
733142
M1A1
A1
© Oxford University Press 2019
17
Worked solutions
2
Representing relationships:
introducing functions
b
Skills check
1
2 A (3, 0), B (-2, 4), E (-1, 1), R (2, -1)
3 a 4(2) −3(−3) = 17
b (2)2 – (−3)2 = −5
1
3
= −
2
2
c
2 + −3 + −
d
3
 1
−6  −  =
−
2
2


c
2
4 a 8x = 16 → x = 2
b 4x = 2 → x =
1
2
c x – 10 = 3 → x = 13
d 12x = −12 → x = −1
5 a
d
© Oxford University Press 2019
Worked solutions
1
Worked solutions
Exercise 2A
1 a If the marbles are identical of mass a
then the function takes x to ax . This
function satisfies the vertical test line.
c
 x , x ≥ 0
y =
 x, x ≥ 0
d
b This is a function because for s sides of
a polygon the sum of the interior
angles of the polygon is ( s − 2 ) ⋅ 1800
which satisfies the vertical test line.
c This is not a function. If a ticket for an
adult is 10 pounds and a ticket for a
student is 5
pounds then if we
have one adult and 2 students the
total cost is 20 pounds, for 3 students
the total cost is 25 pounds, therefore
we will have more than one y-value for
the same x = 1 coordinate (adult movie
tickets purchased).
d This is a function. Each x -coordinate
has a unique y -coordinate, y= x + 1 .
e This is not a function. It does not
satisfy the vertical line test for
example at x = 0 .
f
This is a function. We see that every
x -coordinate has a unique
y -coordinate.
g This is a function. Every x -coordinate
has a unique y -coordinate.
h This is a function. This is seen when
drawing the graph of the function
y =
−2 x + 6 and applying the vertical
line test.
i This is not a function. We see that for
x = 3 there are infinitely many y coordinate values.
j This is a function. This is seen when
drawing the graph of the function
y = x 2 and applying the vertical line
test.
k This is a function. Apply the vertical
line test.
l
This is a function. Apply the vertical
line test.
m This is not a function. This is showed at
x = −2 which has 2 distinct
y -coordinate values.
2 a
X
2
2
3
Y
0
1
4
3 In a function for every x -coordinate there
exists a unique y -coordinate which
satisfies the definition of a relation. On
the other hand, the examples from the
previous questions are all relations but
none of them are functions.
“All functions are relations, but not all
relations are functions.”
Exercise 2B
1 a
g ( −4 ) =− ( −4 ) + 2 =−16 + 2 =−14
2
b
f ( −9 ) =5 ⋅ ( −9 ) − 1 =−45 − 1 =−46
c
C (100 ) =20 ⋅ 100 + 250
= 2000 + 250 = 2250
d
h (5) = −4
e
f (2 ) = 3
f
5
f ( −3) =
g
f ( −1) =
1
2 a
f ( −3) =−3 ( −3) − 1
2
=−3 ( 9 ) − 1 =−27 − 1 =−28
b
g (15) =−4 (15) + 7 =−60 + 7 =−53
c
f (1) + g ( −1) = −3 (1) − 1 + −4 ( −1) + 7
2
=−3 − 1 + 4 + 7 =7
d 6
e
f ( x − 2) =
−3 ( x − 2 ) − 1
2
(
)
=
−3 x 2 − 4 x + 4 − 1
b
=
−3x 2 + 12 x − 12 − 1
=
−3x 2 + 12 x − 13
f
g ( n) =
−4n + 7
g
f (1) × h (1) =
−3 (1) − 1 ( 6 )
© Oxford University Press 2019
(
2
)
2
Worked solutions
=
−24
( −4) (6 ) =
h
5 a
f ( x + 1) =
−3 ( x + 1) − 1
2
(
)
=
−3 x 2 + 2 x + 1 − 1
=−3x 2 − 6 x − 3 − 1
b
g<0
c
C (14
=
= 165
) 10 (14) + 25
d
C ( g ) = 100,
10g + 25 =
100
=
−3x 2 − 6 x − 4
10g = 75
g ( x − 2) =
−4 ( x − 2 ) + 7
g = 7.5 gigs
=−4 x + 8 + 7
=
−4 x + 15
Exercise 2C
f ( x + 1) × g ( x − 2 )
(
C=
( g ) 10g + 25
1 a
)
=−3x 2 − 6 x − 4 ( −4 x + 15)
= 12 x 3 − 45x 2 + 24 x 2 − 90 x + 16 x − 60
= 12 x 3 − 21x 2 − 74 x − 60
3 a Yes, it is a function. Every value of t
will yield only one value of d.
b
d (t ) =
−75t + 275
c
d (0) =
−75 ( 0 ) + 275 =
275 km
d
0 < t < n, where n is the amount of
time it takes to drive to Perth.
4 a Yes, it is a function. Every
temperature in Celsius will only yield
one temperature in Fahrenheit.
b
F
b
F (17 ) is asking what temperature in
is equivalent to 17oC.
o
9
= 62.6 oF.
(17) + 32
5
F (17=
)
c
c
F ( C ) = 100 is asking what
temperature in oC is equivalent to
100oF.
9
C + 32 =
100
5
9
C = 68
5
=
C 37.7 ≈ 37.8 0C
9
(0) + 32= 32 oF
5
d
F ( 0 )=
e
F (100
=
)
f
F (38.75
=
)
g
9
C + 32 =
350
5
2 a
9
32 212 oF
(100) + =
5
9
+ 32 101.75 oF
(38.75)=
5
9
C = 318
5
C 176.6 ≈ 177 oC
=
© Oxford University Press 2019
3
Worked solutions
b
x-intercept: (-9.17,0),
c
y-intercept: (0, 23)
Exercise 2D
1 a Not a function
b Domain: {−5, −2,3}
Range: {4,6,14}
c Domain: {−12, −8, −5}
Range: {−8,7}
d Not a function
e Domain: x ∈ R ,  −∞, ∞  or ( −∞ , ∞ )
d
Range: y ∈ R,  −∞ , ∞  or ( −∞ , ∞ )
f
Domain: x ∈ R ,  −∞, ∞  or ( −∞ , ∞ )
Range: {4}
g Not a function
h Domain: x ∈ R ,  −∞, ∞  or ( −∞ , ∞ )
Range: 3 ≤ y ≤ 5, 3,5
i
Domain: x ≥ 0, 0, ∞ or  0, ∞ )
Range: y ≤ 0,  −∞, 0 or ( −∞, 0
j
3 a
Domain: 𝑥𝑥 ≥ 2.5, [2.5, ∞[ or [2.5, ∞)
Range: 𝑦𝑦 ≥ 0, [0, ∞[ or [0, ∞)
k Domain: 𝑥𝑥 ≤ −1 or 𝑥𝑥 ≥ 1, ]−∞, −1] ∪ [1, ∞[
or (−∞, −1] ∪ [1, ∞)
L
Range: 𝑦𝑦 ≥ 3, [3, ∞[ or [3, ∞)
It is not a function. For 𝑥𝑥 = 0, 𝑓𝑓(0) = 0
and 𝑓𝑓(0) = 4.
x-intercepts: (-9,0), (2.5,0)
y-intercepts: (0, -4.5)
minimum: (-3.25, -6.61)
b
© Oxford University Press 2019
4
Worked solutions
Domain: x ∈ R ,  −∞, ∞  or ( −∞ , ∞ )
2 a
Range: −1 ≤ y ≤ 5,  −1,5
3 Answer will vary.
a
Domain: x ∈ R ,  −∞, ∞  or ( −∞ , ∞ )
Range: y ∈ R ,  −∞, ∞  or ( −∞ , ∞ )
b
b
c
Domain: x ∈ R , x ≠ 1,  −∞,1 ∪  1, ∞ 
or ( −∞,1) ∪ (1, ∞ )
Range: y ∈ R , y ≠ 1,  −∞,1 ∪  1, ∞ 
or ( −∞,1) ∪ (1, ∞ )
c
d
Domain: x > 1, 1, ∞  or (1, ∞ )
Range: y ∈ ¡ ,  −∞, ∞  or ( −∞, ∞ )
d
4 a i
ii
© Oxford University Press 2019
f (6 ) =
6
+2 = 4
3
f (8) =−8 + 10 =2
5
Worked solutions
b
b The equation satisfies the vertical line
test.
c Domain: t ∈ ¡ , t ≥ 0
Range: [10000, ∞).
12t
0.025 

d=
20000 10000 1 +

12 

12t
c 0 ≤ x ≤ 10; 0 ≤ y ≤ 4
5
−3 ≤ x ≤ −1
 − x,

+
−1 < x ≤ 2
2
3,
f ( x=
x
) 
 7,
2<x ≤6

It is also possible to include −1 in the
second interval rather than the first and 2
in the third interval rather than the
second.
C (n
=
) 40 + 21n , where C is the cost
and n is the number of hours.
t = 27.755...
Exercise 2F
1 a i
b Domain: n ≥ 0, 0, ∞  or  0, ∞ )
c
12t = 333.0571
Javier needs 27 years and 10 months
to double his money.
Exercise 2E
1 a
0.025 

=
2 1 +

12 

0.025 

=
ln2 12t ln 1 +

12 

ln2
12t =
0.025 

ln 1 +
12 

(
)
(
)
− ( 4x − 2) + 5 ( 4x − 2)
f g (x) =
2
(
f g (x) =
− 16 x 2 − 16 x + 4
Range: 𝐶𝐶(𝑛𝑛) ≥ 40, [40, ∞[ or (40, ∞)
)
+ 5 ( 4x − 2)
C (4
=
) 40 + 21 ( 4)
(
)
f g (x) =
−16 x 2 + 16 x − 4
C (4
=
) 40 + 84
+ 20 x − 10
C ( 4 ) = $124
(
)
f g (x) =
−16 x 2 + 36 x − 14
2 a
ii
(
f (f ( x )) =
− (x
(
)
f f (x) =
− − x 2 + 5x
4
)
2
− 10 x 3 + 25x 2
(
2
+ 5 − x + 5x
(
(
)
+ 5 − x 2 + 5x
)
)
)
f f (x) =
− x 4 + 10 x 3 − 25x 2
− 5x 2 + 25x
(
)
f f (x) =
− x 4 + 10 x 3 − 30 x 2 + 25x
b Domain e.g: 10 ≤ f ≤ 80
Range e.g: 75.8 ≤ h ≤ 228
(
) + 5 ( x + 1)
f (h ( x )) =
− ( x + 2 x + 1) + 5 ( x + 1)
(
)
(
)
(
)
− x +1
iii f h ( x ) =
c=
h (51) 2.47 (51) + 54.10
h (51) = 180.07
2
f h ( x ) =− x − 2 x − 1 + 5 x + 5
h (51) ≈ 180 cm
f h ( x ) =− x + 3 x + 4
d=
161 2.47f + 54.10
iv g o h ( x=
) 4
2.47f = 106.9
=
f 43.27935 …
(
)
x +1 −2
g o h ( x=
) 4 x +4−2
f ≈ 43.3 cm
g o h=
(x) 4 x + 2
12t
0.025 

3 =
a I ( t ) 10000 1 +

12 

v
© Oxford University Press 2019
f ( −1) =
− ( −1) + 5 ( −1) =
−1 − 5 =
−6
2
6
Worked solutions
f o f ( −1) = f ( −6 )
4 a
= − ( −6 ) + 5 ( −6 )
2
2
( ) (
f ( g ( x ) ) = 3x
f ( g ( x ) ) = 3x
f o f o f ( −1) = f ( −66 )
= − ( −66 ) + 5 ( −66 )
2
=
−4356 − 330
= 4686
(
)
)
f g ( x ) = 3 x 2 − 8 x + 16 − 6
=
−36 − 30 =
−66
vi g ( h ( 9=
)) 4
(
f g ( x ) = 3 ( − x + 4) − 6
2
− 24 x + 48 − 6
2
− 24 x + 42
b
)
9 +1 −2
( )
g ( h=
(9) ) 4 ( 4) − 2
g ( h ( 9=
) ) 16 − 2
g ( h ( 9 ) ) = 14
g h ( 9 ) = 4 (3 + 1) − 2
(
)
vii. g o f (2 ) =4 − (2 ) + 5 (2 ) − 2
2
c Domain: x ∈ R or  −∞, ∞  or ( −∞, ∞ )
Range: y ≥ −6 or  −6, ∞  or  −6, ∞ )
g o f (2 ) = 4 ( −4 + 10 ) − 2
5 a
f ( x=
) x + 25
g o f (2
=
) 24 − 2
b
g ( x ) = 1.06 x
g o f (2 ) = 22
c
f g
=
( x ) 1.06 x + 25 ; this represents
g o f=
(2 ) 4 ( 6 ) − 2
(
− 4 (2 ) − 2
f o g (2 ) =
)
2
(
+ 5 4 (2 ) − 2
(
paying tax on both the price of the
fridge and the delivery fee.
− (6 ) + 5 (6 )
f o g (2 ) =
2
d
g o f (2 ) + f o g (2 ) = 22 − 6 = 16
b i
ii
(
f g (x)
1 a i
2 Answers will vary
3 a
b
(
)
f ( g ( x ) ) =−8 x + 2 + 5
f ( g ( x )) =
−8 x + 7
f ( g ( x ) ) =−8 x + 7 =
12
f g (x) =
−2 ( 4 x − 1) + 5
g ( −2 ) =
3
(
)
f g ( −2 ) = f (3)= 2
ii
iii x ≥ 0 or 0, ∞  or 0, ∞ )
iv x ≥ 0 or 0, ∞  or 0, ∞ )
)
Exercise 2G
x ∈ R or  −∞, ∞  or ( −∞, ∞ )
x ∈ R or  −∞, ∞  or ( −∞, ∞ )
)
g=
f (x)
1.06 ( x + 25) ; this represents
2
f o g (2 ) = −6
)
only paying tax on the price of the
fridge.
)
f o g (2 ) =
− (8 − 2) + 5 (8 − 2)
f o g (2 ) =
−36 + 30
(
f (5 ) = 6
(
)
g f (5) = g ( 6 ) = −3
iii g (3) = 11
(
)
g g
=
(3) g=
(11) 0
b For f ( x ) ,
domain: {2,3,5,10}
range: {-4, 1, 2, 6}
For g ( x ) :
−8 x =
5
Domain: {–2, 3, 6, 11}
5
x = −
8
Range: {–3, 0, 3, 11}
© Oxford University Press 2019
7
Worked solutions
2 a
g (3) = 7
b i
ii After tax, the TV cost
1.10 × 674.99 =
742.489
≈ $742.49
f o g (3) = f (7 ) = −2
b
f ( −1) =
9
3 a
f o f ( 9 ) = f ( −1) = 9
= 1.10 (525.99 − 25) + 49.99
f (0) = 0
= 601.079 ≈ $601.08
(
3 Answer will vary.
)
Exercise 2I
f (1) = 1
(
)
g f (1) = g (1) = −3
c
1 a
g ( −2 ) =
0
(
g ( −0 ) =
−4
)
f g ( 0 ) = f ( −4 ) = 4
with an x-coordinate of -1. ; f ( x ) must
3
 x −3
f ( g ( x )) 2  3
b=
 +3

2 

 x − 3
f g (x)
=
2
+3
 2 
(
have a point with a y-coordinate of 2.
(
)
f ( g ( x )) = x
f ( n=
) n − 100
g ( n ) = 2.20n
(
commission on every new person who
signs up after the first 100 people.
(
f (224 ) = 224 − 100 = 124
ii=
g (124 ) 2.20
=
(124) 272.80 GBP
3
f ( x=
) x − 25 ; this could represent
$25 off the price of the TV
g ( x ) = 1.10 x ; this could represent a
)
+3 −3
2x
2
3
3
(
)
g=
f (x)
x , these are
(
)
 x2 2 
+ −2
3
 3 3
(
)
x2 + 2 − 2
f g ( x=
)
(
) x
f ( g ( x )) = x
f g (x) =
x = 152 people
3
2
3
f g ( x )=
52= x − 100
(2 x
inverses.
c
ii 114.40
= 2.20 ( x − 100 )
2 a
3
( ) x
g (f ( x )) = x
Since f=
( g ( x ))
g f (x) =
2.20 GBP for each person (after the
first 100) who sign up.
= 2.20
=
(176 ) 387.20 GBP
)
g f (x) =
g ( n ) represents that you receive
d i =
S (276 ) 2.20 (276 − 100 )
)
g f (x) =
f ( n ) represents that you receive
c i
)
f g (x) = x − 3 + 3
Exercise 2H
b
)
inverses.
4 Answers will vary; g ( x ) contain a point
1 a
(
(
)
f ( g ( x ) ) =− x − 2
Since f ( g ( x ) ) ≠ x , these are not
)
(
1

−2  x + 2  + 2
f g (x) =
2

f g ( x ) =− x − 4 + 2
f g ( −2 ) = f ( 0 )= 0
d
P (525.99 )
ii
f ( 9 ) = −1
g f ( 0 ) = g ( 0 ) = −4
b
P (x
=
) 1.10 ( x − 25) + 49.99
c i
g o f ( −1) =
g (9) =
−4
c
You paid 699.99 − 25 =
$674.99 .
(
=
g (f ( x ))
2
3x − 2
3
)
2
+
2
3
tax of 10%.
© Oxford University Press 2019
8
Worked solutions
(
)
3x − 2 2
+
3
3
(
)
3x − 2 + 2
3
(
)
3x
3
(
)
f=
( g ( x ))
g=
f (x)
g f (x) =
g f (x) =
3 a i & ii
g f (x) = x
Since
(
)
g=
f (x)
x , these are
inverses.
d
3  4 x − 20 
− −
g h (x) =
+5
4
3

(
)
(
)
12 x − 60
+5
12
=
g h (x)
iii f ( x ) =
−4 x 2 + 4
( )
g (h ( x )) = x
g h (x) = x − 5 + 5
(
)
(
)
(
)
h g (x)
x =
−4y 2 + 4
4y 2 =− x + 4
 3

4  − x + 5  − 20
4


= −
3
h g (x) = −
h g (x) =
(
−4 x 2 + 4
y =
y2 =
−
−3x + 20 − 20
3
y 2 =± −
3x
3
)
f=
( g ( x ))
y =± −
h g (x) = x
Since
x
+1
4
(
)
g=
f (x)
x , these are
inverses.
x
+1
4
x
+1
4
f −1 ( x ) =± −
x
+1
4
b i & ii
2 x-intercept: 0 =
−4 x + 2
x =
1
2
1 
∴  ,0
2 
y-intercept: y =
−4 ( 0 ) + 2
y =2
iii g ( x ) =
−2 x + 5
∴ ( 0,2 )
Since you only need two points to graph a
line, you can switch the coordinates to
find two point that the inverse passes
 1
through:  0,  and (2,0 ) .
 2
y =
−2 x + 5
x =
−2 y + 5
2 y =− x + 5
y =
( )
y
2
−x + 5
2
2
 −x + 5 
=

 2 
2
 −x + 5 
g −1 ( x ) 
=
 ,x ≥ 0
 2 
© Oxford University Press 2019
9
Worked solutions
b The domain of the function becomes
the range of its inverse, and the range
of the function becomes the domain of
its inverse.
Note 1: The domain restriction is
needed since the original function
g (x) =
−2 x + 5 would have the
same restriction.
Note 2: The inverse, g −1 ( x ) , can
be simplified further if desired:
6 a
( − x + 5)
2
y =
5 Answers will vary. In order for the
function to be a one-to-one function, the
inverse must be a function.
2x − 5 =
11
4
2 x = 16
2
y =
x − 10 x + 25
4
y =
1 2 5
25
x − x+
,x ≥ 0
4
2
4
f ( x ) = 2 x − 5 = 11
x =8
b
f (x
=
) 2x − 5
=
y 2x − 5
c i & ii
=
x 2y − 5
2y= x + 5
y =
c
1
iii g : x → x + 6
2
d
x +5
2
f −1 (11) =
11 + 5
2
f −1 (11) =
16
2
f ( x ) = 11 gives the same answer as
f −1 (11) .
1
y +6
2
e
1
y= x − 6
2
7
f ( x ) = y ⇔ x = f −1 ( y )
f (x) =
−2 x − 1
y 2 x − 12
=
y =
−2 x − 1
( x=)
x =
−2y − 1
g
4 a i
f −1 ( x ) =
f −1 (11) = 8
1
=
y
x +6
2
=
x
x +5
2
−1
2 x − 12
2y =− x − 1
Domain: 𝑥𝑥 ≥ 2.875 or
[2.875, ∞[ or [2.875, ∞).
y =
Range:
y ≥ 1.25 or 1.25, ∞  or 1.25, ∞ )
−x − 1
2
f −1 ( x ) =
ii Domain: x ∈ R or  −∞, ∞ 
−x − 1
2
2
 −x − 1 
g o f −1 ( x ) = −3 

 2 
or ( −∞, ∞ )
Range: y ∈ R or  −∞, ∞  or ( −∞, ∞ )
iii Domain: x > 3 or 3, ∞  or (3, ∞ )
Range: y ∈ R or  −∞, ∞  or ( −∞, ∞ )
iv Domain: x ≤ 1 or  −∞,1 or ( −∞,1
g of
−1
(x) =
( − x − 1)
−3
2
4
g o f −1 ( x ) = −3
g o f −1 ( x ) =
(x
2
)
+ 2x + 1
4
2
−3x − 6 x − 3
4
Range: y ≤ 2 or  −∞,2 or ( −∞,2
© Oxford University Press 2019
10
Worked solutions
( x + 2) − 2 (5x + 1)
(5x + 1) (5x + 1)
( −5x − 10) + (5x + 1)
(5x + 1) (5x + 1)
−3 ( −1) − 6 ( −1) − 3
( −1) =
4
2
g of
−1
−3 (1) − 6 ( −1) − 3
g o f −1 ( −1) =
4
=x
x + 2 − 10 x − 2
5x + 1
=x
−9
5x + 1
−3 + 6 − 3
g o f −1 ( −1) =
4
0
g o f −1 ( −1) =
4
−9 x
=x
−9
0
g o f −1 ( −1) =
x =x
Exercise 2J
4
1
(
)
f f (x) = x
 2x − 4 
2
−4
 x+m
=x
 2x − 4 
 x+m+m


 4x − 8 
 x + m
 x + m  − 4 x + m



 =x
 2x − 4 
 x + m
 x + m  + m x + m




(
4 x − 8 − 4 x − 4m
x+m
=x
2 x − 4 + xm + m2
x+m
)
f f (x) = x
2 a
3 − (3 − x ) =
x
−4m − 8
=x
2 x − 4 + xm + m2
3−3+ x =
x
8 x 2 x − 4 + xm + m2
−4m −=
x =x
(
2
−2 − ( −2 − x ) =x
−2 + 2 + x =x
m = -2
x =x
)
(
Since there is no 𝑥𝑥 2 on the left-hand side:
2+m = 0
No solution because for m = -2 we get the
constant function y = 2, which has no
inverse.
)
f f (x) = x
c
1 1

−  − x =
x
2 2

Chapter review
1 1
− −x =
x
2 2
1 a Yes
x =x
f (x) =
n − x, n ∈ R is a self-inverse.
d
3
(
)
2
−4𝑚𝑚 − 8 = 2𝑥𝑥 − 4𝑥𝑥 + 𝑥𝑥 𝑚𝑚 + 𝑚𝑚 𝑥𝑥
−4𝑚𝑚 − 8 = (2 + 𝑚𝑚)𝑥𝑥 2 + (𝑚𝑚 2 − 4)𝑥𝑥
(
f f (x) = x
b
2
)
f f (x) = x
 −x − 2
−
 5x + 1
 −x − 2
5
 5x + 1

−2

=x

+
1


b No
c Yes
d No
e Yes
f
Yes
g Yes
h No
i
j
Yes
k Yes
l
Yes
No
m No
2 a Domain: {-5, -1, 0, 1, 4, 9}
Range: {-8, -1, 0, 1, 6, 9}
b Domain: {0, 2, 3, 4}
Range: {1, 2}
c Domain: {-8, -5, 0, 1}
Range: {-2, 2, 3}
d Domain: x ∈ R or ( −∞, ∞ ) or  −∞, ∞ 
© Oxford University Press 2019
11
Worked solutions
Range: y ∈ R or ( −∞, ∞ ) or  −∞, ∞ 
g −1 ( x ) =
e Domain: −3 ≤ x ≤ 3 or  −3,3
−3
g −1 ( −3) =
−2
Range: −3 ≤ x ≤ −1 or  −3, −1
f
3
g −1 ( −3) =
2
Domain: x ∈ R or ( −∞, ∞ ) or  −∞, ∞ 
h
Range: x ≥ −12.25 or 12.25, ∞ )
or 12.25, ∞ 
g Domain: x ≥ 0 or 0, ∞ ) or 0, ∞ 
h Domain: x ∈ R or ( −∞, ∞ ) or  −∞, ∞ 
Range: x ≥ 5 or 5, ∞ ) or 5, ∞ 
f (3=
) 32 − 6
f (3)= 9 − 6
2
4x 2 − 6
x
−1
f o g=
( x )  −2  − 6


=
f o g −1 ( x
)
x2
−6
4
f o g −1 ( x
=
)
x 2 24
−
4
4
f o g −1 ( x ) =
x 2 − 24
4
4 a
f (3) = 3
b
(
)
f ( g (=
x ))
f g (x) =
( −2x ) − 6
2
i
Range: y ≤ 1 or ( −∞,1 or  −∞,1
3 a
x
−2
f ( −2 ) =( −2 ) − 6
2
f ( −2 ) = 4 − 6
−2
f ( −2 ) =
c
g ( −6 ) =−2 ( −6 )
g ( −6 ) =
12
d
f (1) + h (2 )=
(1)
2
−6−4
Domain: x ∈ R or ( −∞, ∞ ) or  −∞, ∞ 
f (1) + h (2 ) = 1 − 6 − 4
Range: y ≥ −2 or  −2, ∞ ) or  −2, ∞ 
f (1) + h (2 ) =
−9
e
(
)
(
2f ( 0 ) − 2g ( −1=
) 2 02 − 6 − 2 −2 ( −1)
)
b
2f ( 0 ) − 2g ( −1) =
−12 − 2 (2 )
2f ( 0 ) − 2g ( −1) =
−12 − 4
2f ( 0 ) − 2g ( −1) =
−16
f
(
h ( 0 ) × f ( −1) =
−4 ( −1) − 6
2
)
h ( 0 ) × f ( −1) =
−4 (1 − 6 )
h ( 0 ) × f ( −1) =−4 ( −5)
h ( 0 ) × f ( −1) =
20
g
Domain: x ∈ R or ( −∞, ∞ ) or  −∞, ∞ 
g ( x ) = −2 x
Range: y ≥ −8.38 or  −8.38, ∞ )
y = −2 x
or  −8.38, ∞ 
x = −2y
y =
x
−2
5 a
 x − 2
−4  −
f g (x) =
+2
4 

(
© Oxford University Press 2019
)
12
Worked solutions
)
(
f ( g ( x )) = x
(
f g (x) = x − 2 + 2
)
(
)
4x − 2 + 2
4
(
)
4x
4
(
)
f=
( g ( x ))
g f (x) =
g f (x) =
)
 2x + 3 
3+

 3x − 1 
=
 2x + 3 
3
−2
 3x − 1 
)
 2x + 3 
3+

 3x − 1 
=
 6x + 9 
 3x − 1  − 2


)
9x − 3 2x + 3
+
= 3x − 1 3x − 1
6x + 9 6x − 2
−
3x − 1 3x − 1
(
)
11x
3
= x −1
11
3x − 1
(
)
11x
11
(
)
f=
( g ( x ))
(
g f (x)
( −4x + 2) − 2
(
g f (x) = −
)
f g (x) = x
4
(
g f (x)
g f (x) = x
Since
(
(
g f (x)
)
g=
f (x)
x , these are
inverses.
b
1  x − 2
f g ( x ) =−

−4
2
4 
(
)
(
)
−x + 2
−4
8
)
− x + 2 32
−
8
8
f =
g (x)
(
f =
g (x)
(
f g (x)
g f (x)
g f (x) =
g f (x) = x
− x − 30
=
8
)
Since
Since f ( g ( x ) ) ≠ x , these are not
inverses.
c
(
)
1
1
 2x +  + 4
2
4
(
)
1 2
1 
 4x + x +
+4
2
16 
(
)
f g (=
x)
f g ( x ) = 2x 2 +
)
inverses.
−4 ii 4
6 a i
b
2
f g (x) =
(
g=
f (x)
x , these are
−4 ≤ x ≤ 4 or  −4, 4
c
1
1
x+
+4
2
32
Since f ( g ( x ) ) ≠ x , these are not
inverses.
d
(
f g (x)
(
f g (x)
(
f g (x)
(
f g (x)
(
)
 3+ x 
2
+3
3x − 2 
= 
 3+ x 
3
 −1
 3x − 2 
)
6 + 2x
+3
= 3x − 2
9 + 3x
−1
3x − 2
)
6 + 2x 9x − 6
+
= 3x − 2 3x − 2
9 + 3x 3x − 2
−
3x − 2 3x − 2
)
11x
3
= x −2
11
3x − 2
)
f g (x) =
11x
11
7
(f
o g=
) (x)
6
−8 x=
3
3
−8 x 6
3
( g ( x ) + 2)
=
( g ( x ) + 2)
3
( g ( x ) + 2)
3
−2 x 2= g ( x ) + 2
g (x) =
−2 x 2 − 2
8
2 ) f ( h ( −2 ) )
( f o h ) ( −=
−1
−1
Since h (16 ) =
−2, h−1 ( −2 ) =
16
2 16 + 16
(
)
f ( h( ( −2 ) ) =
2 ( 4 ) + 256
f ( h( ( −2 ) ) =
8 + 256
f h(−1 ( −2=
) f (16=)
2
−1
−1
© Oxford University Press 2019
13
Worked solutions
(
)
f h(−1 ( −2 ) =
264
9 For a function to be a self-inverse, we
must show that f ( f ( x ) ) = x :
(
)
f f (x) = −
(
)
(
)
−x
3
c
−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
2
y =
x +2
4
f −1 ( x ) =
d
x +2
4
35
x 2 − 8 x + 15 =
M1A1
b
A1
= 50
A1
x + 10
f −1 ( x ) =
3
M1A1
16 a=
y
=
x
k
+1
y −1
M1
A1
M1
x ( y − 1) = k + y − 1
M1
xy − x = k + y − 1
A1
xy − y = k + x − 1
0
( x − 10) ( x + 2) =
M1
y ( x − 1) = k + x − 1
x = 10 or x = −2
A1A1
y =
3
f=
15 177
( x ) 128  2  −=
 
M1A1
f ( −3) =
128 ( −3) − 15 =
−399
M1A1
f (=
15) 128 (15) −=
15 1905
A1
=
y
13 a Domain is −3 ≤ x ≤ 3
Range is −1 ≤ f ( x ) ≤ 1
M1
A1
A1A1
A1A1
Range is −5 ≤ f ( x ) ≤ 4
A1A1
Range is 0 ≤ f ( x ) ≤ 12
k
+1
x −1
A1
k
+1
x −1
So f is self-inverse
b Range is f ( x ) > 1 , f ( x ) ∈ ¡
A1A1
c
A1A1
b Domain is −1.5 ≤ x ≤ 5
c Domain is 0 ≤ x ≤ 24
k + x −1
x −1
1
f −=
(x)
Range is −399 < f ( x ) < 1905 A1
c Solving 128a − 15 =
1162.6
a = 9.2
A1A1
A1R1
k
+1
x −1
x 2 − 8 x − 20 =
0
12 a
= f (20 )
A1R1
d This is a function. Each value of x is
related to only one value for y
g ( −2 ) =( −2 ) − 8 ( −2 ) + 15
=
x 4y − 2
A1
M1
A1R1
c This is a function. Each value of x is
related to only one value for y
f ( −2 ) =( 4 × −2 ) − 2 =−8 − 2 =−10
=4 + 16 + 15 =35
c =
y 4x − 2
)
15 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
M1A1
b
(
ff (10 ) = f f (10 )
d Range is 5 < f −1 ( x ) < 20
f f (x) = x
11 a
Domain is 5 < x < 20
b
3
 3
− x 


f f ( x ) =−3 ×
10 a
14 a Solving 3x − 10 =
5 and 3x − 10 =
50
M1A1A1
A1A1
A1A1
d Domain is −3 ≤ x ≤ 3
A1A1
Range is 0 ≤ f ( x ) ≤ 9
A1A1
A1A1
© Oxford University Press 2019
14
Worked solutions
17 a Range is f ( x ) ≠ −
2
, f (x) ∈ ¡
3
(
)
A1
b
She can therefore invite a maximum of
A1
39 people
e C = 430 + 14.5 × 16 = $662
M1
1 − 2x
y =
3x + 6
x =
662
A1
= 41.375
16
Katie will therefore need to charge a
minimum of $41.38 per head A1
1 − 2y
3y + 6
x (3y + 6 ) =1 − 2y
M1A1
3xy + 6 x =
1 − 2y
20 a
b
h ( x ) ≥ 2 , (h ( x ) ∈ ¡
y=
x
+2
3
x=
y
+2
3
2y + 3xy =
1 − 6x
y (2 + 3x ) =−
1 6x
y =
f
−1
h−1 ( x
=
) 3x − 6
A1
)
Range is f ( x ) ≠ −2 , ( f ( x ) ∈ ¡
A1
c
hh
=
(x)
)
=
A1
2
x= 2 x − 1
M1
x 2 − 2x + 1 =
0
2
=
0
A1
x =1
b
(2x − 1)
A1
gf (=
x ) 2x 2 − 1
A1
2
(2x − 1)
2
2
= 2x − 1
2
x − 2x + 1 =
0
0
( x − 1) =
M1
x =1
A1
M1A1
f ( p ) is a function since it passes the
vertical line test
c=
C 430 + 14.5p
C − 430
=p
14.5
x 2
+ +2
9 3
A1
x
+ 2 = 3x − 6
3
M1
x =3
e Because h ( x ) and h
−1
(x)
intersect on the line y = x
2
2x − 4x + 2 =
0
19 a =
C 430 + 14.5p
M1A1
21 x 2 + 4 x − 11 = ( x + 2 ) − 15
2
2
x
+2
3
+2
3
8x
=8
3
M1
4 x 2 − 4 x + 1= 2 x 2 − 1
b
d
A1
x 8
+
9 3
=
A1
fg (=
x)
M1A1
=
y 3x − 6
1 − 6x
( x ) = 2 + 3x
( x − 1)
A1
3x= y + 6
1 − 6x
2 + 3x
2
c Domain is, x ≠ − , ( x ∈ ¡
3
18 a
)
R1
M1
p − 430
A1
14.5
1000 − 430
d=
f −1 (1000 ) = 39.3
14.5
M1
A1
both
R1
M1A1
Therefore h ( x=
) x +2
A1
g ( x ) = x2
A1
f ( x=
) x − 15
A1
c
( f ( x ) ∈ ¡ ) , f ( x ) ≥ −4
(g ( x ) ∈ ¡ ) , g ( x ) ≠ 0
(h ( x ) ∈ ¡ ) , h ( x ) > 0
d
gf ( x ) =
22 a
b
(x
1
x2 − 3
1
=9
x2 − 3
=
f −1 ( p ) =
e
1
2
)
− 4 +1
A1
A1
A1
M1
A1
M1
1
x2 − 3 =
9
© Oxford University Press 2019
15
Worked solutions
x2 =
f
28
9
b y = ( x − 3) + 4
2
A1
1
gh ( x ) = x
2 +1
M1A1
1
1
>
2x + 1 17
M1
b
2x < 16
A1
x<4
A1
−8 ≤ p ( x ) ≤ 8
A1A1
p−1 ( x ) =
3
( y − 3)
=x − 4
y − 3=
x−4
y =3 +
x−4
2
f −1 ( x ) =3 +
2x + 1 < 17
23 a
x = ( y − 3) + 4
2
2 7
x = ±
3
x , −8 ≤ x ≤ 8 , ( x ∈ ¡
)
A1A1
c Using GDC, or ex., solving x = x
M1
3
x = −1, x = 0 , x = 1
A1
d
x−4
M1A1
A1
c The domain of f −1 ( x ) is x ≥ 4 ,
(x ∈ ¡ )
A1
The range of f −1 ( x ) is f ( x ) ≥ 3 ,
(f ( x ) ∈ ¡ )
A1
26 ( x − 3) = x 2 − 6 x + 9
M1
2
2 ( x − 3) = 2 x 2 − 12 x + 18
2
M1A1
2 ( x − 3) + 12 x = 2 x 2 + 18
2
Therefore g (=
x ) 2 x 2 + 12 x
A1
A1A1
24 a
y =
3x + 5
4x − 3
x =
3y + 5
4y − 3
x ( 4y − 3) = 3y + 5
M1A1
4 xy − 3x = 3y + 5
4 xy − 3y = 3x + 5
y ( 4 x − 3) = 3x + 5
y =
3x + 5
4x − 3
A1
So r ( x ) is self-inverse
b
25 a
rrrrrr
=
(5) rrrr=
(5) rr=
(5 ) 5
M1A1
x 2 − 6 x + 13 = ( x − 3) + 4
M1A1
Therefore k = 3
A1
2
© Oxford University Press 2019
16
Worked solutions
3
Modelling relationships: linear and
quadratic functions
Skills Check
b
1 a
x = −3
2 a
3m ( m − 5) b
( n + 1) ( n + 7)
d
9x ( x + 2)
f
e
(3x + 2) ( 4x − 1)
Exercise 3B
1 a They are not parallel, as their gradients
are not the same, and not
perpendicular, as both gradients are
positive.
9
t = ± 7 c a =−
2
( x + 6) ( x − 6)
( x + 1) ( 4x − 3)
c
b They are parallel, as m1 =−4 =m2 .
( a + 1) (2a − 5) g
h ( 4a + 7b ) ( 4a − 7b )
Exercise 3A
1 a Using the points (-1,0) and (1,-1) on
y2 − y1 ( −1) − 0 −1
the graph,
=
=
m =
x2 − x1 1 − ( −1)
2
b Using the points (–5,0) and (0,2) on
y2 − y1
2−0
2
=
m =
=
the graph,
x2 − x1 0 − ( −5) 5
b
y2 − y1 ( −4 ) − 2 −6
=
=
= −1
x2 − x1 4 − ( −2 )
6
m=
Exercise 3C
y2 − y1
8 −1
7
1
=
= =
x2 − x1 7 − ( −7 ) 14 2
m
c =
3 As the line joining the scatter plot (drawn
up with t on the x-axis and h on the yaxis) is linear, the gradient can be found
by using any two points in the scatterplot:
h − h1 ( 4.15) − 4.3 −0.15
m= 2
=
=
= −0.015
t2 − t1
30 − 20
10
. This is the rate of change of the height of
the candle, i.e. how fast it is burning down
in cm/s.
4 a You can use the Pythagorean theorem
to find the coordinate of B: as the
elevation of B above A is 70m and the
direct distance is 350m,
xB=
3502 − 702=
122500 − 4900
117600 ≈ 342.93
=
.
Coordinates of B are (342.93,100 ) .
=
b m
=
y2 − y1 100 − 30
≈
x2 − x1
342.93
1 a The gradient is 3, y-intercept is −7.
2
b The gradient is − , y-intercept is 4.
3
c This could be written as =
y 0x − 2 ;
thus, the gradient is 0 and the yintercept is −2.
1
1
and the
x + 1 as the gradient is
5
5
y-intercept is 1
2 =
y
3 a The gradient is equal to the gradient of
=
y 4 x − 3 , which is 4 , and the yintercept is -1. Thus =
y 4x − 1 .
12
10 .
= 3 and thus 3 (1) + a =
4
Therefore a = 10 − 3 = 7 .
Thus =
y 3x + 7
b =
m
4 a The x-coordinate remains constant so
the equation is x = 8 .
b The y-coordinate remains constant so
the equation is y = −10
70
≈ 0.20
342.93
rise
itself,
run
grade= gradient × 100% ≈ 20% .
c As the gradient is given by
3 a For the first segment, the gradient is
320 − 0 320
given as =
m1
= = 8 . The
40 − 0
40
gradient of the second segment
560 − 320 240
=
= = 12 .
m2
60 − 40
20
b This shows that Liam earns 8 dollars
per hour regular wage (for the first 40
hours) and 12 dollars per hour worked
overtime.
y2 − y1 11 − 8 3
= =
8−4
4
x2 − x1
2 =
a m
4
−3
=
−1, m = . Therefore
3
4
−3 5 − 2
3
, which is rearranged to
= =
4
x −3 x −3
3*4
x −3 =
=−4 , yielding x = −1 .
−3
2 As m *
c As horizontal lines are perpendicular to
vertical lines, the line is vertical and the
equation is x = 9 .
d The lines intersect at the point where
x = −2 and y = 7 .
© Oxford University Press 2019
Worked solutions
1
Worked solutions
↔ y =−3 ( x + 5) + 2 =−3x − 15 + 2
Exercise 3D
1 a The line goes through ( 0, 0 ) and
through (1, −3) .
=
−3x − 13
Exercise 3E
1 a =
y
1
x −3
6
−y +
1
x −3 =
0
6
−6y + x − 18 =
0
b
2
y =
− x+4
3
−y −
b The point ( −4,2 ) is on the line and so is
( −4 + 3,2 + 1) = ( −1,3) .
2
x+4=
0
3
3y + 2 x − 12 =
0
c
y − 2 =− ( x + 3)
y − 2 =− x − 3
y + x − 2 =−3
y + x +1 =
0
2 a
3x + y − 5 =
0
y − 5 =−3x
c The line is horizontal at y =
y =
−3x + 5
1
2
b
2 x − 4y + 8 =
0
1
0
x −y +2 =
2
=
y
d The line goes through ( 0,5) and
c
1
x +2
2
5x + 2y + 7 =
0
5
7
x+y+ =
0
2
2
through ( 4,2 ) .
5
7
− x−
y =
2
2
3 a
x -intercept:
x +2×0 +6 =
0
x +6 =
0
x = −6
2
The x -intercept is ( −6, 0 ) .
y − 6 =−3 ( x − 2 )
3 a
b
y -intercept:
6
m=
= −3
−2
0 + 2y + 6 =
0
y+4=
−3 ( x + 3) and y − 2 =−3 ( x + 5)
corresponding to the two points given.
c
y+4=
−3 ( x + 3)
↔ y =−3 ( x + 3) − 4 =−3x − 9 − 4
2y + 6 =
0
2y = −6
y = −3
The y -intercept is ( 0, −3) .
=
−3x − 13
y − 2 =−3 ( x + 5)
© Oxford University Press 2019
2
Worked solutions
e
(f
f
( h o f ) ( −7)=
g
o g ) ( 4 ) =− g ( 4 ) + 5 =−11 + 5 =−6
1
f ( −7 ) − 4
3
=
1
* 12 − 4 = 4 − 4 = 0
3
(f
og) ( x ) =
−g ( x ) + 5
=− (2 x + 3) + 5 =−2 x + 2
b
h
x -intercept:
2x − 6 * 0 + 8 =
0
1
1
f (x) − 4 =
( − x + 5) − 4
3
3
1
5
1
7
=
− x+ −4=
− x−
3
3
3
3
2x + 8 =
0
2 x = −8
2 a As any real number can be inserted for
x and any real number can be
obtained as 3x + 8 for an x , both
domain and range are all real numbers.
x = −4
The x -intercept is ( −4, 0 ) .
y -intercept:
b Just as above, domain and range are all
real numbers.
2 × 0 − 6y + 8 =
0
3 a The line y = 6 has range {6} as only 6
−6y + 8 =
0
can be obtained for y .
6y = 8
y =
(h o f ) ( x ) =
4
3
 4
The y -intercept is  0,  .
 3
b No vertical line is a function as the y
corresponding to the x -coordinate of
the x -intercept is not unique (in fact,
any y corresponds to it).
Exercise 3F
1 a (−2, −5)
c (−3.58, −8.19)
b (0.75, 2.5)
d (1.18, 1.12)
2 a 0.9
b −5.05
3 $1666.67
4 a =
x
Exercise 3G
1 a
2 x= y + 8
f (3) =−3 + 5 =2
f −1 ( x
=
) 2x − 8
b
g ( 0=
3 3
) 2 * 0 +=
c
h ( 6 ) − g (1
=
)  13 * 6 − 4  − (2 * 1 + 3)


b
(
f (2 ) + g ( −1) = ( −2 + 5) + 2 * ( −1) + 3
= 3+1 = 4
x =
−3y + 9
x − 9 =−3y
1
f −1 ( x ) =
− x +3
3
=−
( 2) − 5 =−7
d
1
y+4
2
)
Exercise 3H
1 a =
x 4y − 5
4y= x + 5
© Oxford University Press 2019
3
Worked solutions
=
y
f −1 (=
x)
b
gradient-intercept form:
y= 0.16 ( x − 1500 ) + 600= 0.16 x − 240 + 600= 0.16 x + 3
1
5
x+
4
4
.
1
5
x+
4
4
b The y -intercept represents Frank’s
basic weekly salary of £360. The
gradient shows that Frank’s commission
is 16% of his sales.
1
x =
− y +3
6
−
1
y =x − 3
6
c
=
y 0.16 * 900 +=
360 504 pounds.
y =
−6 x + 18
3 a Let y be the total cost in dollars and
the number of months of
x
membership.
f −1 ( x ) =
−6 x + 18
c x 0.25y + 1.75
=
For Plan =
A: y 9.99 x + 79.99
4 x= y + 7
For Plan B: y = 20 x
=
y 4x − 7
f −1 ( =
x ) 4x − 7
2 The graph of the inverse function is
obtained by mirroring the graph of f at
the line y = x .
3 a
b We would like to know after how many
months the amount paid under each
plan is the same (From then onwards,
Plan A will be more cost-effective). We
therefore solve:
9.99 x + 79.99 =
20 x
f (55
=
65 615
) 10 * 55 + =
79.99 = 10.01x
b=
x 10y + 65
=
x
10y= x − 65
Therefore, Plan A is more cost-effective
from 8 months onwards.
=
y 0.1x − 6.5
f −1=
( x ) 0.1x − 6.5
x here represents the money available
in CAD and f −1 ( x ) is the number of tshirts one can buy with x dollars.
c
y= 0.1 * 5065 − 6.5= 506.5 − 6.5= 500
Exercise 3I
1 a The gradient can be computed from any
two points on the line; in this case, a
force F of 160 Newtons leads to an
extension d of 5 centimetres, while no
force (i. e. a force of 0 Newtons) leads
to no extension ( 0 centimetres).
Therefore the y -intercept is ( 0, 0 ) and
5−0
1
the gradient is
. This gives
=
160 − 0 32
1
the model d =
F .
32
b d
=
79.99
≈ 7.99 .
10.01
1
* 370 11.5625 cm.
=
32
4 a In the first 40 hours, his pay in pounds
320
is given =
by p =
h 8h . From then
40
on, his pay is given by
560 − 320
p − 320 =
( h − 40) = 12 ( h − 40)
60 − 40
. In gradient-intercept form, this is
p 12h − 160 .
=
 8h,
p(h) = 
12h − 360,
b i
ii
5 a
0 ≤ h ≤ 40
40 < h ≤ 60
p =8 × 22 =176 pounds
p = 12 × 47 − 160 = 404 pounds.
q=
−6.5 × 200 + 3000 =
1700
b They will drop by 6.5 × 20 =
130
printers a month.
c
2000
= 48 p − 1600
48 p = 3600
=
p
3600
= 75 Euro.
48
2 a The gradient is given by
680 − 600
80
= = 0.16 . As
2000 − 1500 500
(1500, 600)
is on the graph, a point-
gradient form of the equation of the line
is y − 600
= 0.16 ( x − 1500 ) . We find the
© Oxford University Press 2019
4
Worked solutions
d
The graph is reflected about the x -axis
and shifted upwards by 4 units.
c
Use the “solve” function of the GDC.
e Solving −6.5p + 3000 = 48 p − 1600
84.40..
⇒p=
Axis: x = 0, vertex: (0, 0)
The graph is compressed vertically with
1
.
scale factor
4
So p = €84.40
Then q =
48 × 84.40.. − 1600 =
2451
printers
d
Exercise 3J
1 a
Axis: x = -3, vertex: (-3, 0)
The graph is translated to the left by 3
units.
b
Axis: x = 4, vertex: (4, -3)
The graph is first translated to the right
by 4 units, then stretched vertically
with scale factor 2 and finally translated
downwards by 3 units.
2 a It is compressed vertically by a scale
1
. Thus, the function is given
factor of
4
1
1 2
by=
g (x)
=
f (x)
x .
4
4
b It is stretched vertically by a scale
factor of 2 and reflected along the x axis. Thus, the function is given by
g (x) =
−2f ( x ) =
−2 x 2 .
c It is translated to the right by 3 and
upwards by 2 units. Thus, the function
is given by
g ( x ) = f ( x − 3) + 2 =
Axis: x = 0, vertex: (0, 4)
( x − 3)2 + 2 .
d It is stretched vertically by a scale
factor of 1.5, translated to the left by 3
and downwards by 5 units. Thus, the
function is given by
2
g (=
x ) 1.5f ( x + 3) −
=
5 1.5 ( x + 3) − 5 .
© Oxford University Press 2019
5
Worked solutions
Exercise 3K
1 a The graph is reflected about the
y -axis.
e The graph is translated to the left by
6 units.
f
The graph is translated downwards by
3 units.
b The graph is reflected about the
x -axis.
c The graph is compressed horizontally
1
.
with scale factor
2
2 a The graph of r is stretched by a scale
factor of 2. Thus r ( x ) = 2f ( x ) .
The graph of s is translated to the
right by 3 units and reflected about the
x -axis. Thus s ( x ) =
−f ( x − 3 ) .
b The graph of r is reflected about the
y -axis. Thus r ( x=
) f ( −x ) .
The graph of s is stretched horizontally
by a scale factor of 2 and translated
downwards by 4 units. Thus
1 
=
s (x) f  x  − 4
2 
d The graph is stretched vertically with
scale factor 3.
3 a
0≤y ≤6
b It is reflected about the y -axis.
c
2 ≤ − x ≤ 8 , which is equivalent to
−8 ≤ x ≤ −2 .
d The range of g is the same as the
range of f . 0 ≤ y − c ≤ 6 is equivalent
to c ≤ y ≤ 6 + c , so c = −4 . Thus
h=
(x) g (x) − 4 .
© Oxford University Press 2019
6
Worked solutions
e
h ( x ) = g ( x ) − 4 = f ( −x ) − 4
Exercise 3L
1 x-intercepts: (−2.81, 0), (0.475, 0);
y-intercept: (0, −4);
vertex: (−1.17, −8.08)
2 x-intercepts: none;
y-intercept: (0, −3);
vertex: (0.726, −0.785)
3 Domain: x ∈ ¡
Range: f(x) ≤ 9.125
Exercise 3M
x = 3 is the axis of symmetry and (3, 4 )
1 a
the coordinates of the vertex.
4 Domain: x ∈ ¡
Range: f(x) ≥ −30.752
(1, −5)
b
x = 1 is the axis of symmetry and
the coordinates of the vertex.
c
x = −3 is the axis of symmetry and
( −3,2)
d
the coordinates of the vertex.
x = −6 is the axis of symmetry and
( −6, −5) the coordinates of the vertex.
2 a The y -intercept is given by ( 0,5) , the
−8
axis of symmetry is at x =
4 and
−
=
2
the vertex is at
4, f ( 4 ) = ( 4,16 − 32 + 5) = ( 4, −11) .
(
5 Range: 0 ≤ f(x) ≤ 8.1
)
b The y -intercept is given by ( 0,2 ) , the
−6
axis of symmetry is at x =
−
=
1 and
6
the vertex is at
1, f (1) = (1,3 − 6 + 2 ) = (1, −1) .
(
)
c The y -intercept is given by ( 0, −11) ,
the axis of symmetry is at
−8
x =
−
=
−2 and the vertex is at
−4
6 Range: −6 ≤ f ( x ) ≤ 3
( −2, f ( −2) ) =( −2, −8 + 16 − 11) =( −2, −3) .
d The y -intercept is given by ( 0,3) , the
axis of symmetry is at x =
−
6
3
=
−
4
2
and the vertex is at
 3  3   3 9
  3 3
 − , f  −   = − , − 9 + 3 = − , − 
2
2
2
2

 
  2 2

.
© Oxford University Press 2019
7
Worked solutions
3 a The x -intercepts are at (2, 0 ) and
b
( 4, 0) . The axis of symmetry lies at
x
=
2+4 6
= = 3 . The vertex is at
2
2
(3, f (3) ) = (3,1 * ( −1) ) = (3, −1) .
b The x -intercepts are at ( −3, 0 ) and
(1, 0) . The axis of symmetry lies at
x =
−3 + 1 −2
=
= −1. The vertex is at
2
2
( −1, f ( −1) ) =( −1, 4 * 2 * ( −2) ) =( −1, −16 ) .
c The x -intercepts are at ( −5, 0 ) and
(3, 0) . The axis of symmetry lies at
x =
−5 + 3 −2
=
= −1. The vertex is at
2
2
c
( −1, f ( −1) ) =(−1, − ( 4 * ( −4) ) =( −1,16 ) .
d The x -intercepts are at ( −3, 0 ) and
( −2, 0) . The axis of symmetry lies at
x
=
−3 − 2
=
2
−5
. The vertex is at
2
 −5  −5    −5
1  1 
,f 
,2 * *  −  
=
 
2
2
2
2  2 

 

1
 −5
= 
,− 
2
2


4 a
d
© Oxford University Press 2019
8
Worked solutions
Exercise 3N
1 a
f (x) =
( x − 2) ( x + 9) . The x -intercepts
are (2, 0 ) and ( −9, 0 ) (from the
intercept form), and the y -intercept is
(0, −18)
b
5

3 ( x − 2)  x −  .
3

5 
The x -intercepts are (2, 0 ) and  ,0 
3 
(from the intercept form), and the y intercept is ( 0,10 ) (from the standard
b i
(
)
The x -intercepts are − (2,0 ) and
the y -intercept is ( 0, 4 ) (from the
The x -intercepts are at ( 4, 0 ) and
c The axis of symmetry is at
4−2
=
x
= 1 Thus the vertex is at
2
(1, f (1) ) = (1, ( −3) × 3) = (1, −9) .
5 a i
The vertex is at (3, −2 ) .
b
f ( x ) = x2 − 6 x + 7
c
B is the y -intercept of the graph, and
(
2
)
its coordinates are 0, ( −3) − 2 =
(0,7) .
d By symmetry, p = 6 as 6 − 3 = 3 − 0 .
6 a
f (x) =
− ( x − 4) ( 4x − 2)
h (x) =
( x − 2)2 − 2 ( x − 2) − 3
= x2 − 6 x + 5
1

=
−4 ( x − 4 )  x − 
2

b The axis of symmetry lies at x =
−
1 
The x -intercepts are ( 4, 0 ) and  ,0 
2 
(from the intercept form), and the y -
2 a
iii q = −2
p=4
ii The y -intercept is at ( 0, −8 )
(from the intercept form), and
standard form).
d
ii
ii The axis of symmetry is at x = 3 .
1 2
1
f ( x )=
x + 6x + 8 =
( x + 2) ( x + 4) .
2
2
( −4, 0)
a=1
( −2, 0)
form).
c
f (x) =
( x − 4) ( x + 2) . Thus
i
(from the standard form).
(3x − 5) ( x − 2 ) =
f (x) =
4 a
6
3
=
2
c The vertex is at
3, h (3) = (3, 9 − 18 + 5) = (3, −4 ) .
(
)
h (x) =
( x − 5) ( x − 1)
intercept is ( 0, −8 ) (from the standard
d
form).
e The graph is the same as that of h ( x ) ,
f ( x ) = 4 x + 16 x − 20 . The x 2
but reflected about the x -axis.
intercepts are (1, 0 ) and ( −5, 0 ) (from
the intercept form), and the y intercept is ( 0, −20 ) (from the standard
form).
b
f (x) =
−2 x 2 − 16 x − 14 . The x intercepts are ( −7, 0 ) and ( −1, 0 ) (from
the intercept form), and the y intercept is ( 0, −14 ) (from the standard
form).
3 a
f (x) =
−3x 2 − 6 x − 9 . The vertex is at
( −1, −6 )
(from the vertex form) and the
y -intercept is ( 0, −9 ) .
b
1 2
f ( x )=
x − 4 x + 11 . The vertex is at
2
( 4,3)
(from the vertex form) and the
y -intercept is ( 0,11) .
Exercise 3O
1 a The vertex is at (2, −16 ) and the y intercept is at ( 0, −12 ) . Thus
2
f ( x ) = a ( x − 2 ) − 16 , and
2
−12 = a ( −2 ) − 16 = 4a − 16 . Thus a = 1
. In standard form,
f (x) =
( x − 2)2 − 16 =
x 2 − 4 x + 4 − 16
= x 2 − 4 x − 12
© Oxford University Press 2019
9
Worked solutions
b
f ( x ) = a ( x − 1) ( x + 3) from the x -
3 a The vertex is at ( 4, 80 ) . The model
intercepts. 3 =−
a * ( 1) * 3 =
−3a . Thus
a = −1 . In standard form,
f (x) =
− ( x − 1) ( x + 3) =
− x2 − 2x + 3 .
c
rocket is predicted to reach a maximum
of 80 m, 4 s after it is launched.
b In intercept form, h=
(t ) at (t − 8) .
Inserting the coordinates of the vertex,
we obtain 80 = a × 4 × ( −4 ) =−16a .
f ( x ) = a ( x − 5) ( x − 1) from the x intercepts. −12 =
a * ( −1) * 3 =
−3a .
Thus a = −5 . Overall, h ( t ) =
−5t ( t − 8 )
Thus a = 4 . In standard form,
f ( x )= 4 ( x − 5) ( x − 1)= 4 x 2 − 24 x + 20 .
0≤t≤8
c
d The vertex is at (2, −6 ) . Thus
Therefore, the rocket is predicted to be
67.20 metres high.
2
f ( x ) = a ( x − 2 ) − 6 , and
2
6 = a (2 ) − 6 = 4a − 6 . Thus a = 3 . In
Exercise 3P
standard form,
1 a
2
f ( x )= 3 ( x − 2 )
e
2
− 6= 3x − 12 x + 6 .
f ( x ) = a ( x − 2 ) ( x + 5) from the x intercepts. 3 =
a * ( −1) * 6 =
−6a . Thus
1
. In standard form,
2
1
1
3
f (x) =
− ( x − 2 ) ( x + 5) =
− x2 − x + 5
2
2
2
a= −
f
The vertex is at ( −10, 60 ) . Thus
2
f ( x ) = a ( x + 10 )
+ 60 , and
2
3
. In standard form,
5
3
3
=
=
f ( x ) − ( x + 10
)2 + 60 − 5 x2 − 12x .
5
a= −
2 a In intercept form,
f ( x ) = a ( x − 3) ( x + 1)
axis of symmetry is at
Therefore, the
=
x
3 −1
= 1.
2
b The vertex is at (1, 4 ) as x = 1 is the
c Since the vertex is at (1, 4 ) , h = 1 and
b
c
also know that f (3) = 0 , 4a + 4 =
0 and
thus a = −1 . So f ( x ) =
− ( x − 1) + 4
2
(
)
=
− x 2 − 10 x + 25 − 1
x 2 − x − 20 =
( x − 5) ( x + 4) . Thus
x =5
x 2 − 8 x + 12 = ( x − 6 ) ( x − 2 ) . Thus
x = 2 or x = 6 .
d
x 2 − 121 =( x − 11) ( x + 11) . Thus x = 11
or x = −11 .
e
f
2 a
x 2 + x − 42 =
( x − 6 ) ( x + 7) . Thus
x =6
2
x 2 − 8 x + 16 = ( x − 4 ) . Thus x = 4 .
2 x 2 + x − 3=
(2x + 3) ( x − 1) . Thus
3
.
2
x = 1 or x = −
b
c
d
e
4
or x = −3 .
3
4 x 2 + 11x + 6 =
( x + 2) ( 4x + 3) . Thus
x = −2 or x = −
3
.
4
7 
7

7
9 x 2 − 49 =  x −   x +  . Thus x =
3 
3
3

7
or x = − .
3
4 x 2 −16 x + 7=
x =
f
(3x − 4) ( x + 3) . Thus
3x 2 + 5x − 12 =
x =
2
=
− ( x − 5) − 1
x =1
or x = −4 .
k = 4 . So f ( x ) = a ( x − 1) + 4 . As we
2
( x − 3) ( x − 1) . Thus
or x = 3 .
axis of symmetry and 4 the maximum
value.
g ( x )= f ( x − 4 ) − 5
x2 − 4x + 3 =
or x = −7 .
45 = a ( −5) + 60 =25a + 60 . Thus
d
h (2.4 ) =−5 × 2.4 × ( −5.6 ) = 67.2 .
(2x − 7) (2x − 1) . Thus
7
1
or x = .
2
2
12 x 2 + 11x − 5=
x =
(3x − 1) ( 4x + 5) . Thus
1
5
or x = − .
3
4
2
=
− x + 10 x − 26
© Oxford University Press 2019
10
Worked solutions
Exercise 3Q
1 a
(x
)
− x − 20 − (2 x + 8 ) = x − 3x − 28
2
2
=
( x − 7) ( x + 4)
(2 x
) (
2
− 3x − 8 − − x 2 + 2 x
= 3x − 5x − 8=
Thus x =
c
( 4x
2
(
)
8
or x = −1 .
3
) (
+ 20 − 3x 2 + 10 x − 4
2
1

4x +  =
4
2

2
1

1
x +  =
2

1
x =− ± 1
2
 3 3
f −  =
+ 5 = 6.5
 2 2
)
Thus x = 4 or x = 6 .
d
(3x
1
1
f   =− + 5 =4.5
2
2
)
+ 15x + ( x + 5)
= 3x 2 + 16 x + 5=
(3x + 1) ( x + 5)
1
or x = −5 .
Thus x = −
3
e
7 (1, 5)
8 (2.72, 7.64), (0.613, −0.0872)
10 x = −2.91, 0.915
(3x + 4 ) ( x − 3)
4
or x = 3 .
Thus x = −
3
f
6 (1.18, 7.35), (−1.96, 1.07)
9 x = −0.802, 1.80
3 ( x + 2 ) ( x − 2 ) − (5 x )
= 3x 2 − 5x − 12 =
Exercise 3S
1
−15
For x ≠ 0, x + 8 =
if and only if
x
x 2 + 8 x + 15 = ( x + 3) ( x + 5) and thus
x = −3 or x = −5 .
(f
og) ( x ) =
2
to ( x + 6 ) =
38 , giving x =−6 ± 38
2
(2x + 1)2 − 2
= 4x2 + 4x + 1 − 2 = 4x2 + 4x − 1
b
( 4x
2
) (
+ 4 x − 1 − x 2 + 5x + 3
= 3x 2 − x − 4=
Thus x =
)
(3x − 4) ( x + 1)
x
=
4
or x = −1 .
3
3
3 3 ± 17
+=
2
2
4
x2 − 6 x + 4 =
0 is equivalent to
2
consider x 2 − 6 x + 9 =−4 + 9 =
5 . This
2
x 2 − 8 x + 16 = ( x − 4 ) = 10 . Thus
2
x 2 + 20 x + 100 =( x + 10 ) =
15 . Thus
2
factorises to ( x − 3) =
5 , giving x= 3 ± 5
4
x 2 − 12 x + 4 =
0 is equivalent to
2
2
b
(−6)2 =
36 .
 =
2
x 2 − 12 x =
−4 . 
x 2 + 12 x + 36 = ( x + 6 ) = 12 . Thus
x =
± 12 − 6 .
± 17
b
(−3)2 =
9 . Therefore
x2 − 6 x =
−4 .   =
2
x =
± 15 − 10 .
3
2
2
x =
± 10 + 4 .
2
2
9
b
 3
  =
 −  =. Therefore consider
4
2
 2
9
9 17
. This factorises to
x 2 − 3x + = 2 + =
4
4
4
3
17

, giving
x −  =
2
4

Exercise 3R
1
2
b
2
6=
36 . Therefore consider
 =

2
x 2 + 12 x + 36 =2 + 36 =38 . This factorises
−15.
x2 + 8x =
2 a
)
4 x2 + x =
3
= x 2 − 10 x + 24 = ( x − 6 ) ( x − 4 )
2
4 x 2 + 3x + 2 =− x + 5
4x 2 + 4x =
3
(3x − 8) ( x + 1)
2
2
x 2 − 10 x + 25 = ( x − 5) = 27 . Thus
x =
± 27 + 5 .
5
Thus x = 7 or x = −4 .
b
4
Therefore consider
x 2 − 12 x + 36 =
−4 + 36 =
32 . This
© Oxford University Press 2019
11
Worked solutions
2
4
factorises to ( x − 6 ) =
32 , giving
2
x =±
6
32 =±
6 4 2
5
4
x 2 + 5x − 4 =
0 is equivalent to x 2 + 5x =
2
2
25
b
5
. Therefore consider
=

=

2
2
4
 
 
25
25 41
x 2 + 5x +
=4 +
= . This factorises
4
4
4
2
5
41

to  x +  =, giving
2
4

6
11
x 2 + x − 11 =
0 is equivalent to x 2 + x =
2
1
b
1
. Therefore consider
=

=

2
2
4
 
 
1
1 45
. This factorises to
x 2 + x + = 11 + =
4
4
4
2
1
45

 x +  =, giving
2
4


± 45
=
x
4
1 −1 ± 45 −1 ± 3 5
−=
=
2
2
2
1 2 x 2 + 16 x =
10 is equivalent to
2
b
2
4=
16 . Therefore
x2 + 8x =
5.  =

2
 
consider x 2 + 8 x + 16 =5 + 16 =21 . This
2
factorises to ( x + 4 ) =
21 , giving
b
. =

2
1
x2 + x
2
1
x =
3
2
2
1
1
. Therefore consider
=

16
4
1
1
49
=
3+
= . This factorises
16
16 16
+
2
1
49

to  x +  =, giving
4
16

=
x
± 49
1 −1 ± 49 −1 ± 7
−=
=
. This
4
4
4
16
3
or −2 .
2
6 2 x ( x + 8 ) + 12 =
0 is equivalent to
2
b
2
4=
16 .
x ( x + 8) =
x2 + 8x =
−6 .  =

2
Therefore consider
x 2 + 8 x + 16 =
−6 + 16 =
10 . This factorises
2
to ( x + 4 ) =
10 , giving x =−4 ± 10 .
7 a Revenue is equal to cost when
R ( x ) = C ( x ) , i. e. when
b This is equivalent to
2 5x 2 − 30 x =
10 is equivalent to
−0.25x 2 + 20 x =
300 , which is in turn
2
b
9 . Therefore
x2 − 6 x =
2.   =
( −3)2 =
2
2
consider x − 6 x + 9 = 2 + 9 = 11 . This
2
factorises to ( x − 3) =
11 , giving
equivalent to x 2 − 80 x =
−1200 .
2
b
1600 . Therefore
( −40)2 =
  =
2
 
consider
−1200 + 1600 =
x 2 − 80 x + 1600 =
400 .
x= 3 ± 11 .
2
This factorises to ( x − 40 ) =
400 ,
2
6 x − 12 x − 3 =
0 is equivalent to
2
1 b
1 . Therefore
.   =
x2 − 2x =
( −1)2 =
2 2
1
3
. This
+1 =
2
2
3
, giving
=
2
consider x 2 − 2 x + 1 =
2
factorises to ( x − 1)
3
.
2
x =−4 ± 18 =−4 ± 3 2 .
35x − 0.25x 2 =
300 + 15x .
x =−4 ± 21 .
x= 1 ±
2
to ( x + 4 ) =
18 , giving
means that x is either
Exercise 3T
3
x 2 + 8 x + 16 =2 + 16 =18 . This factorises
2
5 −5 ± 41
−=
2
2
4
2
b
2
4=
16 .
x ( x + 8) = x2 + 8x = 2 .  =

2
Therefore consider
5 2x2 + x − 6 =
0 is equivalent to x 2 +
± 41
=
x
6 x ( x + 8) =
12 is equivalent to
giving x = 40 ± 400 = 40 ± 20 = 20,60 .
c The break-even points lie at x = 20 and
x = 60 .
d We will want to find where the
maximum of the equation
P=
( x ) R ( x ) − C ( x ) lies. This will just
be the coordinates of the vertex, since
the leading coefficient is negative.
P=
(x) R (x) − C (x)
=
−0.25x 2 + 20 x − 300
© Oxford University Press 2019
12
Worked solutions
In vertex form, this is
2
P (x) =
−0.25 ( x − 40 ) + 100 . Thus, the
3 a x
=
maximal profit is reached at 40
subscribers.
e As seen from the vertex form above,
the vertex has coordinates (40,100)
and therefore the maximal profit is
equal to 100 thousand Euros.
Exercise 3U
1 a
−4 ± 16 + 8
24
x =
=−2 ±
=−2 ± 6
2
2
8 ± 64 − 60 8 ± 2
; that is, x = 1
=
6
6
5
or x = .
3
=
2 a
5 ± 25 + 16 5 ± 41
=
4
4
x 2 + 3x − 9 =
0 . Thus
x =
−3 ± 9 + 36
−3 ± 45
= −
2
2
4 ± 16 − 8
4± 8 2± 2
x =
=
−
=
4
4
2
c
x=
4 a
b
3x 2 − 4 x − 2 =
0 . Thus
− x2 + 2x + 2 =
0 . Thus
at (1, −4 ) .
c Using the quadratic formula:
4 ± 16 + 16
x=
= 1 ± 2 . Therefore
4
r = 1 and s = 2 .
b
d
2
∆ = ( −4 ) − 4 × 1 × 15 = 16 − 60 = −44 .
−10 ± 100 − 72 5 ± 7
=
−4
2
e
2 x ( x − 1) . This is
equivalent to x 2 + 4 x + 3= 2 x 2 − 2 x ,
which simplifies to x 2 − 6 x − 3 =
0 . Thus
6 ± 36 + 12
x =
=±
3
12 =±
3 2 3.
2
( −4)2 − 4 × 1 × 4 =
16 − 16 = 0 .
2
∆ = ( −1) − 4 × 5 × 10 = 1 − 200 = −199 .
Therefore the equation has no real
roots.
2x2 − 9x + 9 =
0 . Thus
9 ± 81 − 72
9±3
; that is,
= −
4
4
3
x = 3 or x = .
2
∆ =
Therefore the equation has two equal
real roots.
f
x =
∆= 42 − 4 × 3 × ( −8=
) 16 + 96= 112 .
Therefore the equation has two distinct
real roots.
−2 x 2 + 10 x − 9 =
0 . Thus
( x + 3) ( x + 1=)
∆= 72 − 4 × 6 × ( −3=
) 49 + 72= 121 .
Therefore the equation has no real
roots.
−4 ± 16 − 120 −4 ± −104
. As =
6
6
104 has no real square root, the
equation has no real solution.
g
2
∆ = ( −5) − 4 × 1 × 9 = 25 − 36 = −11 .
Therefore the equation has two distinct
real roots.
=
x
f
)
2
c
d 3x 2 + 4 x + 10 =
0 . Thus
=
x
(
2 x 2 − 4 x −=
2 2 x2 − 2x − 1
Therefore the equation has no real
roots.
−2 ± 4 + 8
= 1± 3
x=
−2
e
c = −2
= 2 ( x − 1) − 4 . Therefore the vertex is
1 a
4 ± 16 + 24
4 ± 40 2 ± 10
x =
=
−
=
6
6
3
c
−2 ± 4 + 16
= 1± 5
−2
Exercise 3V
−3 ± 3 5
=
2
b
−5 ± 13
2
3
; that is, x =
or x = − .
12
3
2
b
b x
=
c x
=
−5 ± 25 + 144 −5 ± 169
=
12
12
2 a
∆ = 32 − 4k = 9 − 4k . This is positive
9
whenever k < .
4
b =
∆ 202 − 20=
k 400 − 20k . This is
positive whenever k < 20 .
3 a
∆= 52 − 4 p= 25 − 4 p . This is 0 if and
only if p =
© Oxford University Press 2019
25
.
4
13
Worked solutions
b
2
∆ =
144 − 12 p . This is 0
( −12) − 12p =
if and only if p = 12 .
c
∆ =
( −2p )2 − 32 =
4 p2 − 32 . This is 0 if
and only if p2 = 8 , which holds for
p=
± 8=
±2 2 .
d
2
∆ =
9p + 8p =
p (9p + 8) .
( −3p ) + 8p =
2
This is 0 if and only if p = 0 or
8
p= − .
9
4 a
∆ =
( −2)2 − 4m =
As the parabola described by ∆ is concave
1
up, this is negative if and only if 0 < m <
9
5=
∆ 36k 2 − 4k ( k +
=
2 ) 32k 2 − 8k
= 8k ( 4k − 1) . The zeroes of this equation
1
. As the parabola
4
described by ∆ is concave up, this is
1
positive if and only if k < 0 or k > .
4
are at k = 0 and k =
∆= p2 − 48
6 a
4 − 4m . This is
b As the graph has no x -intercepts,
p2 − 48 < 0 . This means that
negative if and only if m > 1 .
b
∆ =
( −6 )2 − 12m =
− 48 < p < 48 , which can be
36 − 12m . This is
negative if and only if m > 3 .
c
simplified as −4 3 < p < 4 3 .
∆= 5 − 4 ( m − 2=
) 33 − 4m . This is
2
negative if and only if m >
33
.
4
c As 62 = 36 < 48 < 49 = 72 , m = 6 .
d
4

3x 2 + 6 x +=
4 3  x2 + 2x + 
3


1

2
2
= 3  ( x + 1) +  = 3 ( x + 1) + 1
3


Exercise 3W
1 a We need to find the x -intercepts. By
the quadratic formula,
−5 ± 25 + 24 −5 ± 7
. Since the
=
x
=
6
6
coefficient of x 2 is positive, the
parabola will be concave up. Thus the
inequality is satisfied whenever x ≤ −2
1
or x ≥
3
b
x 2 ≤ 5 if and only if − 5 ≤ x ≤ 5 .
c This is equivalent to x 2 + 4 x − 6 < 0 . By
the quadratic formula,
−4 ± 16 + 24
x =
=−2 ± 10.
2
As the parabola is concave up, the
inequality is satisfied whenever
−2 − 10 < x < −2 + 10 .
Exercise 3X
1 24
=
2h2 + 4h − 48 =
0
0
h2 + 2h − 24 =
(h + 6)(h − 4) =
0
h
= 4, −6
h must be positive
So h = 4 m
b = 2h + 4 = 12 m
2 a
h(3) =
2 + 20(3) − 4.9(32 ) =
17.9 m
2 + 20t − 4.9t 2 =
6
4.9t 2 − 20t + 4 =
0
2
− ≤ x ≤3
3
t =
c −0.890 ≤ x ≤ 1.26
3 a
1
h (2h + 4 )
2
48 2h2 + 4h
=
b
2 a x ≤ −0.245 or x ≥ 12.2
b
Thus a = 3 , h = −1 and k = 1
∆
= k 2 − 16 . This is positive whenever
k > 4 or k < −4 .
b =
∆ 4k 2 − 12 . This is positive whenever
k > 3 or k < − 3 .
4=
∆ 36m2 −=
4m m (36m − 4 ) . The zeroes
1
of this equation are at m = 0 and m = .
9
20 ± 400 − 78.4
9.8
t = 0.211 seconds, 3.87 seconds
c
20
b
Maximum height when: t =
−
=
2a 9.8
2
 20 
 20 
2 + 20 
h=
 − 4.9 

 9.8 
 9.8 
h = 22.4 metres
3 a Fare
= 5.50 − 0.05x
b Number of riders
= 800 + 10x
© Oxford University Press 2019
14
Worked solutions
c Revenue =
(5.50 − 0.05x ) (800 + 10x )
= 4400 − 40 x + 55x − 0.5x 2
Chapter review
1 a
= 4400 + 15x − 0.5x 2
d
4400 + 15x − 0.5x 2 =
4500
0= 0.5x 2 − 15x + 100
x = 10,20
10 or 20 decreases
e
4400 + 15x − 0.5x 2 > 0
Using GDC: x < 110
4 a
y =
− ( x − 2) + 4 =
− x(x − 4)
2
or y =
− x 2 + 4x
b If the center of the object is aligned
with the center of the archway, it spans
form x = 0.5 to x = 3.5 . Evaluating
the function at x = 0.5 and x = 3.5
gives 1.75. Since 1.6 < 1.75, the object
will fit through the archway.
5 a A ( x=
) x (155 − x=) 155x − x 2
b
b Maximum area occurs at:
−b 155
=
x
=
= 77.5
2a
2
310 − 2(77.5)
=
w
= 77.5
2
Dimension: 77.5 metres by 77.5 metres
c No; The touchline would not be longer
than the goal line and 77.5 metres is
less that the minimum of 90 metres for
the touchline.
d
90 ≤ x ≤ 120 (If the goal line
restrictions are also taken into
consideration the answer is
90 ≤ x ≤ 110 .
e Maximum occurs when x = 90
=
w
310 − 2(90) 310 − 180
= = 65
2
2
Area = 90 × 65 =
5850 m2
c
© Oxford University Press 2019
15
Worked solutions
2a m=
2 − −1
3
1
=
= −
−4 − 8 −12
4
c
1
(x + 4)
4
1
1
y −2 =
− x −1 ⇒ y =
− x +1
4
4
c=
m
y-intercept found from the function:
(0, 6 )
6 a
1
x −5
2
3x 2 + 18 x + 20 = 3(x 2 + 6 x ) + 20
= 3(( x + 3) − 9) + 20
2
−1
3
=
 2 2
− 3


= 3 ( x + 3) − 27 + 20
2
= 3 ( x + 3) − 7
2
3
( x − 2)
2
3
3
y − 4=
x − 3 ⇒ y=
x +1
2
2
i
y − 4=
d
−b
4
=
= −2
2a −2
x = −2
y − 2 =−
b =
y
Axis of symmetry:
a=3
ii h = −3
iii
k = −7
b ( −3, −7 )
c
y = −4
7 a
3 a f (1) = 3 , f (2) = 3
( −3 + 5, −7 − 3) =(2, −10)
( x − 3)
2
=
64
x − 3 =±8
x = −5,11
b
b
( x + 2)
2
7
=
x + 2 =± 7
x =−2 − 7, −2 + 7
c
x 2 + 14 x + 49 =
0
(x + 7)2 = 0 ⇒ x = −7
d
x 2 + x − 12 =
0
(x + 4)(x − 3) =
0⇒x =
−4,3
e 3x 2 + 4 x − 7 =
0
4 a Vertical stretch with scale factor 2,
horizontal translation right 3
b Vertical dilation with scale factor
1
, vertical translation up 5
2
8 Equal real root: b2 − 4ac =
0
2
9k=
− 16 0=
⇒ k2
c Reflection in the x-axis, horizontal
translation left 2, vertical translation
down 1
d Horizontal dilation with scale factor
1
3
e Reflection in the y-axis, vertical
translation up 6
5 a
x − intercepts: 2(x − 3)(x + 7) =
0
⇒ x = 3, −7 ∴ (3, 0 ) ,(−7, 0)
16
4 4
⇒k =
− ,
9
3 3
9 From the x-intercepts:
f (x ) = a(x + 4)(x − 2) = ax 2 + 2ax − 8a
From the y -intercept:
2
−8a =
−16 ⇒ a =
f (x ) = 2 x 2 + 4 x − 16
10 Using GDC solver
a −0.679 , 3.68
b −4.92 , 1.42
11 a t = 0, h = 18 m
b Maximum height occurs when:
Axis of symmetry occurs
at midpoint of x-intercepts
=
x
3 + −7
x =
⇒x =
−2
2
13
−b
=
2a 9.8
2
 13 
 13 
h=
18 + 13 
 − 4.9 

9.8


 9.8 
b Found from the function
Axis of symmetry: x = 4,
7
3
(3x + 7)(
=
x − 1) 0 ⇒ x = 1, −
Vertex: ( 4,2 )
h = 26.6 m
© Oxford University Press 2019
16
Worked solutions
c
18 + 13t − 4.9t 2 =
0
t = −1.00,3.66 as t > 0
d
x= 3 ± 52
A1
0 ≤ t ≤ 3.66
x= 3 ± 2 13
A1
b Using GDC
e 18 + 13t − 4.9t =
23
2
−4.9t + 13t − 5 =
0
t = 0.4667..,2.1863..
16 a
b
(1, −18)
c
d
x =1
f ( x ) ∈ ¡ , f ( x ) ≥ −18
e
g ( x=
) 3 ( x − 2) − 1
(
)
2
A1
A1A1
− 18 − 1
M1A1
2
(
Maximum area occurs when
)
= 3 x 2 − 6 x + 9 − 19
−b 14
= = 2
2a
7
A1
= 3x 2 − 18 x + 8
So dimensions are 4 cm by 3.5 cm
17 a
−7 x − 12y + 168 =
0
−7 x + 168
12y =
M1
7x
y =
−
+ 14
12
A1
A1A1
2
8x + 6 x − 5 =
0
0
( 4x + 5) (2x − 1) =
Area = 4 × 3.5 =
14 cm2
A (24, 0 ) and B ( 0,14 )
A1
= 3 ( x − 3) − 19
d Area =
2 p ( −1.75p + 7 ) =
−3.5p2 + 14 p
b
A1
= 3x − 6 x − 15
7−0
7
m=
= −
0−4
4
c 2p cm by −1.75p + 7 cm
13 a
)
2
7
7
y − 0 =− (x − 4) ⇒ y =− x + 7
4
4
f
(
3 ( x − 1) − 18
= 3 x 2 − 2 x + 1 − 18
2
M1
12 a A ( −4, 0 ) ; B ( 0,7 ) ; C ( 4, 0 )
=
p
M1A1
3 − 2 13 ≤ x ≤ 3 + 2 13
2.1863.. – 0.4667.. = 1.72 seconds
e
A1
Time taken = 3.66 seconds
2
b
6 ± 208
2
x =
4 x + 5 =0 ⇒ x =−
2x − 1 = 0 ⇒ x =
b
5
4
1
2
M1A1
A1
A1
8x 2 + 6 x − 5 − k =
0
No real solutions
1
c Area = × 24 × 14 =168 units2 M1A1
2
14 a
⇒ b2 − 4ac < 0
M1
36 − 4 × 8 × ( −5 − k ) < 0
A1
36 + 32 (5 + k ) < 0
5+k < −
k <−
18 a
36
36
⇒ k <−
−5
32
32
9 40
49
⇒ k <−
−
8
8
8
x 2 − 10 x + 27
= ( x − 5) − 25 + 27
M1A1
= ( x − 5) + 2
A1
2
2
M1A1
and ( −0.885, 0 )
b
(0,5.9)
c
−1.1 ≤ f ( x ) ≤ 7.35
15 a
x =
−b ± b2 − 4ac
2a
A1A1
A1
b Coordinates of the vertex is (5,2 )
A1
c Equation of symmetry is x = 5 A1
A1A1
M1
© Oxford University Press 2019
17
Worked solutions
19 a At (10, 0 ) , 0 = 102 + 10b + c , so
M1A1
10b + c =−100
Line of symmetry is x = −
b
, so b = −5
2
A1
Solving simultaneously gives
−50 + c =−100
So c = −50
A1
Therefore the equation is
y = x 2 − 5x − 50
b Setting x = 0 gives the y-intercept of
(0, −50)
A1
Setting y = 0 and solving gives the xA1
intercept of ( −5, 0 )
20 a
f ( x )= 2  x 2 − 2 x − 4
2
= 2 ( x − 1) − 1 − 4


M1
A1
2
= 2 ( x − 1) − 5


= 2 ( x − 1) − 10
2
A1
b A horizontal translation right 1 unit
A1
A vertical stretch with scale factor 2
A1
A vertical translation down 10 units
A1
21 a Two real roots ⇒ b2 − 4ac =
0 M1
36 − 4 (2k ) ( k ) =
0
A1
36 − 8k 2 =
0
2
k=
3
36 9
=
⇒ k = ±
2
8
2
A1A1
b Equation of line of symmetry is
b
6
3
M1A1
−
=
−
=
−
x =
2a
4k
2k
Therefore
c
3
3
=1 ⇒ k =
2k
2
A1
k = 2 ⇒ 4x 2 + 6 x + 2 = 0
2 x 2 + 3x + 1 =
0
0
(2x + 1) ( x + 1) =
1
x =
− or x =
−1
2
22 a
A1A1
A′ ( −6,10 ) , B′ ( 0, −16 ) , C ′ (1, 9 )
and D′ (7, −10 )
b
M1
A4
A (12,13) , B ( 0, −13) , C ( −2,12 )
and D ( −14, −7 )
A4
© Oxford University Press 2019
18
Worked solutions
4 Equivalent representations:
rational functions
Exercise 4B
Skills check
1 a
c
b
x = −5
1 a
x =6
5
2x = 5 ⇒ x =
2
2
b
Exercise 4A
1 a
1
3
d
−
1
=−1
1
g
−
9
8
b
1
5
c
−
e
5
3
f
7
22
1
h
=
3
2
4
3
1
2
2 a 1.5 = ⇒
=
2
1.5 3
c
1
2x
f
t
d
d
1
4y
g
4d
3
1
2
1
4
=
2 ⋅ 4 + 3 11
4
b
1
x
e
4
3x
c
h
x −3
x +2
3 a
b
4⋅
1 4
=
=1
4 4
7 11 7 ⋅ 11 77
⋅
=
=
= 1
11 7
7 ⋅ 11 77
c
2 x 2x
⋅ =
= 1
x 2 2x
d
x − 1 x − 2 (x − 1)(x − 2)
=
⋅
= 1
x − 2 x − 1 (x − 1)(x − 2)
© Oxford University Press 2019
Mixed review
1
Worked solutions
2
c
3 a
d The curves are in the opposite
quadrants. The negative reflects the
function in the x-axis.
4 x = 0, y = 0
Domain: x ∈ ¡ , x ≠ 0
Range: y ∈ ¡ , y ≠ 0
b
Exercise 4C
1 a
b
x =2 ⇒ y =
2 2
= =1
x 2
y =4
y =
2
x
2
=4
x
2
x =
4
x = 0.5
Chamse spends 30 seconds brushing
her teeth.
© Oxford University Press 2019
2
Worked solutions
2 a and c
b
y =
1
x −5
The vertical asymptote is at
x = −(−5) = 5 and the horizontal
asymptote at y = 0.
The domain is x ∈ ¡ , x − 5 ≠ 0 ⇔ x ≠ 5.
The range is y ∈ ¡ , ¡ − {0}.
c
y =
−1
x−4
The vertical asymptote is at
x − 4 = 0 ⇔ x = 4 and the horizontal
asymptote at y = 0.
The domain is x ∈ ¡ , x − 4 ≠ 0 ⇔ x ≠ 4.
b
l = 10 ⇒ v =
The range is y ∈ ¡ , ¡ − {0}.
1000
= 100 Hz
10
c A string 5 cm long has vibrations of
frequency 200 Hz.
3 a =
y
64
= 4 videos of length 16
16
minutes
b
the
64
is the equation that models
y =
x
number of videos of x minutes.
d
y =
The vertical asymptote is at
x +5 =
0 ⇔ x =−5 and the horizontal
asymptote at y = 0.
The domain is
x ∈ ¡ , x + 5 ≠ 0 ⇔ x ≠ −5.
The range is y ∈ ¡ , ¡ − {0}.
e=
y
c and d
5
x +5
12
+2
x +1
The vertical asymptote is at
x + 1 =0 ⇔ x =−1 and the horizontal
asymptote at y = 2.
The domain is
x ∈ ¡ , x + 1 ≠ 0 ⇔ x ≠ −1.
The range is y ∈ ¡ , ¡ − {2}.
f=
y
12
−2
x +1
The vertical asymptote is at
x + 1 =0 ⇔ x =−1 and the horizontal
asymptote at y = −2.
The domain is
x ∈ ¡ , x + 1 ≠ 0 ⇔ x ≠ −1.
The range is y ∈ ¡ , ¡ − {−2}.
1.33 minutes
g=
y
Exercise 4D
1
x +1
1 a
y =
and
The vertical asymptote is at x = −1
the horizontal asymptote at y = 0.
The domain is
x ∈ ¡ , x + 1 ≠ 0 ⇔ x ≠ −1.
4
+2
x −3
The vertical asymptote is at
x − 3 = 0 ⇔ x = 3 and the horizontal
asymptote at y = 2.
The domain is x ∈ ¡ , x − 3 ≠ 0 ⇔ x ≠ 3.
The range is y ∈ ¡ , ¡ − {2}.
The range is y ∈ ¡ , ¡ − {0}.
© Oxford University Press 2019
3
Worked solutions
h
=
y
−4
−4
x−4
d
x ∈ ¡ , x ≠ −5
e
x ∈ ¡ , x ≠ −0.5 y ∈ ¡ , y ≠ 0
f
x ∈ ¡ , x ≠ −2
y ∈ ¡ ,y ≠ 1
The vertical asymptote is at
x − 4 = 0 ⇔ x = 4 and the horizontal
asymptote at y = −4.
The domain is x ∈ ¡ , x − 4 ≠ 0 ⇔ x ≠ 4.
The range is y ∈ ¡ , ¡ − {−4}.
2 a
b
c
x ∈ ¡ , x ≠ −4
x ∈ ¡ , x ≠ −4
x ∈ ¡ , x ≠ −4
y ∈ ¡ ,y ≠ 0
y ∈ ¡ ,y ≠ 0
y ∈ ¡ ,y ≠ 0
y ∈ ¡ ,y ≠ 1
© Oxford University Press 2019
4
Worked solutions
g
x ∈ ¡ ,x ≠ 2
y ∈ ¡ ,y ≠ 2
1000
0.6c + 330
6(0.6c + 330) =
1000
3.6c + 1980 =
1000
6=
1000 − 1980
3.6
c = −272.22o
c =
5 a
c =
200
s −5
The vertical asymptote is at
s − 5 = 0 ⇔ s = 5 and the horizontal
asymptote at c = 0.
h
x ∈ ¡ ,x ≠ 2
y ∈ ¡ ,y ≠ 1
b 15 sessions.
6
3 a 2: Translation of 2 units right
b 5: Reflection in y = 0 and a translation
of 2 units right
c 1: Translation of 2 units right and 2
units up
d 4: Translation of 2 units right and 2
units down
e 3: Translation of 2 units right and
vertical stretch by a factor of 3
4 a
b 5.56
c
t =6
The linear function is a line of symmetry
for the rational function. The linear
function crosses the x-axis at the same
place as the vertical asymptote.
© Oxford University Press 2019
5
Worked solutions
Exercise 4E
1 a
2 i
x +1
y =
⇒a=
1, b =
1, c =
1, d =
−1
x −1
a=
1, b =
−3, c =
1, d =
2
d
=
−2
c
a
Horizontal asymptote: y= = 1
c
Vertical asymptote: x =
−
The vertical asymptote is at
d
(−1)
x =
− =
−
=
1 and the horizontal
c
1
asymptote at y=
ii A
a 1
= = 1
c 1
a 0,
b 4,
c 1,
d 0
=
=
=
=
d
=
0
c
a
Horizontal asymptote: y= = 0
c
Domain x ∈ ¡ , x ≠ 1.
Vertical asymptote: x =
−
Range y ∈ ¡ , y ≠ 1.
b
y=
2x + 3
⇒ a= 2, b= 3, c= 1, d= 1
x +1
iii D
The vertical asymptote is at
d
1
x =− =− =−1 and the horizontal
c
1
asymptote at y=
a=
−2, b =
3, c =
1, d =
2
d
=
−2
c
a
Horizontal asymptote: y =
= −2
c
Vertical asymptote: x =
−
a 2
= = 2.
c 1
iv C
Domain x ∈ ¡ , x ≠ −1.
a=
2, b =
−3, c =
1, d =
2
d
Vertical asymptote: x =
− =
−2
c
a
Horizontal asymptote: y= = 2
c
Range y ∈ ¡ , y ≠ 2.
c
6x − 1
⇒a=
−1, c =
y =
6, b =
2, d =
4
2x + 4
The vertical asymptote is at
d
4
x =
− =
− =
−2 and the horizontal
c
2
asymptote at y=
a 6
= = 3.
c 2
3
x−p
y =
⇒a=
1, b =
− p, c =
1, d =
−q
x −q
The vertical asymptote is at
d
(−q)
x =
− =
−
=
q and the horizontal
c
1
Domain x ∈ ¡ , x ≠ −2.
Range y ∈ ¡ , y ≠ 3.
d
B
2 − 3x
⇒a=
−3, b =
−4, d =
y =
2, c =
5
5 − 4x
The vertical asymptote is at
d
5
x =
− =
−
=
1.25 and the
c
(−4)
asymptote at y=
a 1
= = 1.
c 1
Domain x ∈ ¡ , x ≠ q.
Range y ∈ ¡ , y ≠ 1.
4 a
horizontal asymptote at
a −3
= 0.75.
y= =
c
−4
Domain x ∈ ¡ , x ≠ 1.25.
Range y ∈ ¡ , y ≠ 0.75.
e
9x − 2
9, b =
6
y =
⇒a=
−2, c =
−3, d =
6 − 3x
The vertical asymptote is at
d
6
x =
− =
−
=
2 and the horizontal
c
(−3)
asymptote at y =
9
a
=
= −3.
c (−3)
Domain x ∈ ¡ , x ≠ 2.
Range y ∈ ¡ , y ≠ −3.
© Oxford University Press 2019
6
Worked solutions
b
b
2x − 3 x + 6
=
x +1
x −2
(2 x − 3)(x − 2) = (x + 1)(x + 6)
2 x 2 − 3x − 4 x + 6 = x 2 + 6 x + x + 6
x 2 − 14 x =
0
x(x − 14) =
0
So x = 0 and x = 14.
c
7−
5
10
=
x −2 x +2
7(x − 2) − 5
10
=
x −2
x +2
7 x − 19
10
=
x −2
x +2
(7 x − 19)(x + 2)= 10(x − 2)
c
7 x 2 + 14 x − 19 x − 38 = 10 x − 20
7 x 2 − 15x − 18 =
0
(x − 3)(7 x + 6) =
0
So x = 3 and x = −
d
6
.
7
x +5
6
= 1+
x +8
x +1
x +5
x +1+ 6
==
x +8
x +1
x +5 x +7
=
x + 8 x +1
(x + 5)(x + 1) = (x + 8)(x + 7)
d
x 2 + 6 x + 5 = x 2 + 15x + 56
9 x + 51 =
0
51
17
−
=
−
x =
9
3
6 x = 3 is the extraneous solution.
Therefore
the solution to Will’s
equation is x = 2.
7 a
f (x) =
x +3
x −2
y +3
y −2
x(y − 2) = y + 3
xy − 2 x =y + 3
xy − y = 2 x + 3
x =
5 a
5
x +7
2
+
=
2x x + 4
5(x + 4) + 2 x(x + 7)
=2
2 x(x + 4)
5x + 20 + 2 x 2 + 14 x= 4 x(x + 4)
2 x 2 + 19 x + 20 = 4 x 2 + 16 x
2 x 2 − 3x − 20 =
0
2 x 2 − 8 x + 5x − 20 =
0
2 x(x − 4) + 5(x − 4) =
0
(x − 4)(2 x + 5) =
0
So x = 4 and x =
y(x − 1) = 2 x + 3
2x + 3
y =
x −1
2x + 3
−1
f (x) =
x −1
−5
.
2
© Oxford University Press 2019
7
Worked solutions
b
f (x) =
7 − 2x
x
7 − 2y
y
xy= 7 − 2y
7
y(x + 2) =
7
y =
x +2
7
f −1(x ) =
x +2
9 a
x =
c
f (x) =
20 + 10m
as 20 is the initial
m
cost and them for every month there is
another 10 AUD cost.
C (m) =
b
1 + 7x
9−x
1 + 7y
9−y
x(9 − y ) =1 + 7y
x =
9 x − xy =
1 + 7y
y(7 + x ) = 9 x − 1
9x − 1
y =
7+x
9x − 1
f −1(x ) =
x +7
d
f (x) =
5 − 11x
x +6
c 4 months
d The price will get closer to the
horizontal asymptote y = 10.
10 a
f (x=
) m+
6
x−n
m(x − n) + 6
x−n
mx − mn + 6
=
x−n
a=m
b= 6 − mn
c =1
d = −n
=
5 − 11y
y +6
x(y + 6) =5 − 11y
xy + 6 x =5 − 11y
x =
5 − 6x
y(x + 11) =
5 − 6x
y =
x + 11
5 − 6x
f −1(x ) =
x + 11
The vertical asymptote is at
(−n)
d
5.
x =
n=
− =
−
=
1
c
8 a and c
Hence n = 5.
b
f (7) = 7
6
6
=
m+
7−5
2
f (7) = m + 3 = 7
m=4
f (7) =
m+
c The vertical asymptote is at
a 4
x= = = 4.
c 1
11 a =
y
b 20
c M(s)
=
10s + 500
= 20
s
10s + 500 =
20s
500 = 10s
s = 50
4
4 + 3(x − 2) 3x − 2
+=
=
3
x −2
x −2
x −2
a=3
b = −2
c =1
d = −2
© Oxford University Press 2019
8
Worked solutions
The horizontal asymptote is at
a 3
y= = = 3.
c 1
c
2x + 1
=0
x −1
2x + 1 =
0
1
x = −
2
b The vertical asymptote is at
d
(−2)
x =
2.
− =
−
=
1
c
c The x-intercept is when y = 0.
3x − 2
=0
x −2
3x = 2
The point is
2
x =
3
The x-intercept of f is at point
1
(− , 0) =
(−0.5, 0).
2
13 a
g o f (x ) = g(f (x ))
x +2
(
)
= g=
x +3
2
( , 0) = (0.667, 0).
3
=
The y-intercept is when x = 0.
3⋅0 −2
=
0−2
f (x) = 0
x +3
x +2
1
x +2
x +3
b
−2
= 1= y
−2
The point is (0,1).
d
x = −2.5
Chapter review
12 a
1 a
y =
2
⇒ a = 0, b = 2, c = 1, d = 0
x
The horizontal asymptote is at
a 0
y= = = 0.
c 1
The vertical asymptote is at
d
0
x =
− =
− =
0.
c
1
Domain: x ∈ ¡ , x ≠ 0
Range: y ∈ ¡ , y ≠ 0
b
2x + 1
f (x) =
⇒a=
2, b =
1, c =
1, d =
−1
x −1
The horizontal asymptote is at
a 2
y= = = 2.
c 1
The vertical asymptote is at
d
(−1)
x =
− =
−
=
1.
c
1
b
y=
1
⇒ a= 0, b= 1, c = 1, d = 8
x +8
The horizontal asymptote is at
a 0
y= = = 0.
c 1
The vertical asymptote is at
d
8
x =
− =
− =
−8.
c
1
Domain: x ∈ ¡ , x ≠ −8
Range: y ∈ ¡ , y ≠ 0
© Oxford University Press 2019
9
Worked solutions
c
x
y =
⇒a=
1, b =
0, c =
2, d =
−10
2 x − 10
g
The horizontal asymptote is at
a −1
y =
=
= −1.
c
1
The horizontal asymptote is at
a 1
.
y= =
c 2
The vertical asymptote is at
d
4
x =
− =
− =
−4.
c
1
The vertical asymptote is at
d
−10
x =
− =
−
=
5.
2
c
Domain: x ∈ ¡ , x ≠ −4
Range: y ∈ ¡ , y ≠ −1
Domain: x ∈ ¡ , x ≠ 5
Range: y ∈ ¡ , y ≠
d =
y
h=
y
1
2
3
3 + 3(x − 2) 3x − 3
+=
3
=
x −2
x −2
x −2
2 x − 1 − 8 x − 24 −6 x − 25
=
2x + 6
2x + 6
⇒a=
−6, b =
−25, c =
2, d =
6
The horizontal asymptote is at
a −6
y =
=
= −3.
c
2
The horizontal asymptote is at
a 3
y= = = 3.
c 1
The vertical asymptote is at
d
6
− =
− =
−3.
x =
c
2
The vertical asymptote is at
−2
d
− =
−
=
x =
2.
c
1
Domain: x ∈ ¡ , x ≠ −3
Domain: x ∈ ¡ , x ≠ 2
Range: y ∈ ¡ , y ≠ −3
Range: y ∈ ¡ , y ≠ 3
y=
2x − 1
2 x − 1 − 4(2 x + 6)
=
−4
2x + 6
2x + 6
=
⇒a=
3, b =
−3, c =
1, d =
−2
e
1− x
⇒a=
−1, b =
y =
1, c =
1, d =
4
x+4
2 a
2x
⇒ a= 2, b= 0, c= 1, d= −9
x −9
The horizontal asymptote is at
a 2
y= = = 2.
c 1
The vertical asymptote is at
d
−9
x =
− =
−
=
9.
c
1
Domain: x ∈ ¡ , x ≠ 9
Range: y ∈ ¡ , y ≠ 2
f
8x − 5
y =
⇒a=
8, b =
−5, c =
2, d =
4
2x + 4
The horizontal asymptote is at
a 8
y= = = 4.
c 2
The vertical asymptote is at
d
4
x =
− =
− =
−2.
c
2
Domain: x ∈ ¡ , x ≠ −2
Range: y ∈ ¡ , y ≠ 4
© Oxford University Press 2019
10
Worked solutions
b
The vertical asymptote is at
d
−1
1.
x =
− =
−
=
1
c
1
b The x-intercept is ( , 0) = (0.5, 0) as:
2
f (x) = 0
2x − 1
=0
x −1
2x − 1 =
0
1
x =
2
The y-intercept is (0,1) as:
x =0
=
f (0)
c
c
d
3 a
2⋅0 −1
= 1
0 −1
4
f (x=
)
1
1 + 2(x − 1) 2 x − 1
+=
2
=
x −1
x −1
x −1
⇒a=
2, b =
−1, c =
1, d =
−1
The horizontal asymptote is at
a 2
y= = = 2.
c 1
x = −1.5, 1
5 a 1.29, 2.71 b 2.71
6 a
c 1.27
f (x) = 0
2x − 8
=0
1− x
2x − 8 =
0
8
x= = 4
2
© Oxford University Press 2019
11
Worked solutions
The x-intercept is therefore (4, 0).
b
f (x) =
2x − 8
1− x
 1 
9 a The x-intercept is  − , 0 
 2 
⇒a=
2, b =
−8, c =
−1, d =
1
The vertical asymptote is at
d
1
x =
− =
−
=
1.
c
−1
c The horizontal asymptote is at
2
a
y =
=
= −2.
c
−1
7 a
f (x) =
ax + b
x −d
b x = 2.5, y = 4
c 2.375
The vertical asymptote is at
−d
x =
−
=
d.
1
10 a
y(x − 2) = x + 1
x +1
y =
x −2
x +1
f −1(x ) =
x −2
Hence 3 = d and 2 = a .
f (1) =
a+b 2+b
=
= −4
1−d
1−3
2+b
= −4
−2
2+b =
8
b = 6.
b
f (1) =
8 a
f (=
x)
=
2y + 1
y −1
x(y − 1) = 2y + 1
xy − x = 2y + 1
The horizontal asymptote is at
a
y= = a.
1
b
x =
d 3.8
5
5 + n(x − m)
+=
n
x−m
x−m
nx − mn + 5
x−m
a=
n, b =
−mn + 5, c =
1, d =
−m
4=
−
b
c
d
−m
=
−
=
m
c
1
The vertical asymptote is at
d
−1
1.
x =
− =
−
=
1
c
f (0) = 7
n ⋅ 0 − 4n + 5 −4n + 5
= = 7
0−4
−4
4n − 5 =
28
4n = 33
33
n=
4
=
f (0)
c =
y
33
33
4
=
1
4
a = 2, b = 1, c = 1, d = −1
The horizontal asymptote is at
a 2
y= = = 2.
c 1
d
f (x) = 0
2x + 1
=0
x −1
2x + 1 =
0
1
x = −
2
The x-intercept is (−
© Oxford University Press 2019
1
, 0).
2
12
Worked solutions
e
c
f (x ) = f −1(x )
g(x ) = 0
1
0
+3 =
x −3
1
= −3
x −3
1
x − 3 =−
3
1 8
x =3 − =
3 3
2x + 1 x + 1
=
x −1
x −2
(2 x + 1)(x − 2) = (x − 1)(x + 1)
2 x 2 − 3x − 2 = x 2 − 1
0
x 2 − 3x − 1 =
x1,2
=
=
11 a
−b ± b2 − 4ac 3 ± 9 + 4
=
2a
2
3 ± 13
= -0.303,3.30
2
The x-intercept is (2.67, 0).
x =0
1
f (x) =
x −2
g(0) =−
1
y −2
1
xy − 2 x =
1 + 2x
y =
x
1 + 2x 1
)
f −1(x=
=
+2
x
x
1
8
+3 =
3
3
The y-intercept is (0,2.67).
x =
d
g(x
=
)
=
1
1 + 3(x − 3)
+=
3
x −3
x −3
1 + 3x − 9 3x − 8
=
x −3
x −3
⇒a=
3, b =
−8, c =
1, d =
−3
b
The vertical asymptote is at
d
−3
3.
x =
− =
−
=
1
c
The horizontal asymptote is at
a 3
y= = = 3.
c 1
e
c
1
1 + 2x
=
x −2
x
(1 + 2 x )(x − 2)
x =
x = x + 2x 2 − 2 − 4x
2x 2 − 4x − 2 =
0
x 2 − 2x − 1 =
0
=
x1,2
x >2
−b ± b2 − 4ac
2± 8
=
2a
2
13 a
f (x
=
) 2x + 3
=
x 2y + 3
2y= x − 3
x −3
y =
2
x −3
−1
f (x) =
2
Hence the solution is x = 2.41.
12 a
b
x −3
(
)
g=
o f −1(x ) g=
2
5
5
=
2(x − 3) 2 x − 6
=
b =
g(x )
1
+3
x −3
c
5
x −3
4
2
5
5
x =
0 ⇒ h(0) =
=
−
2⋅0 −6
6
© Oxford University Press 2019
13
Worked solutions
The y-intercept of h is
5
(0, − ) = (0, −0.833).
6
The range is y ∈ ¡ , ¡ − {2}.
c The x-intercept:
f (x) = 0
d
2x + 2
=0
x−4
2x + 2 =
0
x = −1
The point (−1, 0).
The y-intercept: f (0) =
2
= −0.5
−4
The point (0, −0.5).
d
e
h(x ) =
5
2x − 6
5
2y − 6
5
x(2y − 6) =
2 xy − 6 x =
5
5 + 6x
y =
2x
5 + 6x
−1
h (x) =
2x
x =
The x-intercept of h−1 is given by
h−1(x ) = 0
5 + 6x
=0
2x
5 + 6x =
0
5
x = −
6
e Horizontal shift of 4 units right and a
vertical shift of 2 units up.
15 a
The point is therefore
5
(− , 0) =
(−0.833, 0).
6
b
A1
3
f (x) ∈ ¡ , f (x) ≠
2
A1
20
c When x = 0 , f ( x ) =
−
=
−5 .
4
f =
=
=
=
a 6,
b 5,
c 2,
d 0
So one coordinate is ( 0. − 5)
The vertical asymptote is at
d
0
x =
− =
− =
0.
c
2
14 f (x ) =
2+
x ∈ ¡ , x ≠ −2
When y = 0 , x =
10
2(x − 4) + 10 2 x + 2
=
=
x−4
x−4
x−4
a = 2, b = 2, c = 1, d = −4
a The vertical asymptote is at
d
−4
x =
− =
−
=
4.
c
1
A1
20
3
 20 
So the other coordinate is 
,0
 3

A1
16 a Domain is x ∈ ¡ , x ≠ −2
Range is f ( x ) ∈ ¡ , f ( x ) ≠ 0
A1A1
b Domain is x ∈ ¡ , x ≠ −2
The horizontal asymptote is at
a 2
y= = = 2.
c 1
Range is f ( x ) ∈ ¡ , f ( x ) ≠ 4
A1A1
c Domain is x ∈ ¡ , x ≠ 0
b The domain is x ∈ ¡ , x − 4 ≠ 0 ⇔ x ≠ 4.
Range is f ( x ) ∈ ¡ , f ( x ) ≠ 4
© Oxford University Press 2019
A1A1
14
Worked solutions
d Domain is x ∈ ¡ , x ≠ 0
17 a
b
20 a
Range is f ( x ) ∈ ¡ , f ( x ) ≠ 0
A1A1
x =1
y =3
A1
A1
c
A1
6
b P
=
18 (1 + 0.82 × 12 )
3 + ( 0.034 × 12 )
c Solving 100 =
≈ 57
18 (1 + 0.82t )
3 + 0.034t
M1A1
M1
300 + 3.4t = 18 (1 + 0.82t )
300 + 3.4t =
18 + 14.76t
282 = 11.36t
282
= 24.8 months
11.36
=
t
A1
d A horizontal asymptote exists at
18 × 0.82
M1A1
P
=
= 434.12
0.034
Therefore for t ≥ 0 , P < 435
17 − 10 x 12 + 5 − 10 x
=
2x − 1
2x − 1
21=
a f (x)
A3
18 a
b
c
y = 10
A1
=
x =2
A1
10 (2 − x ) + 3
3
f (x) =
10 +
=
2−x
2−x
M1A1
−10 x + 23
A1
=
−x + 2
19 a Vertical asymptote occurs when
M1
c + 8x =
0
c + 8 ( − 34 ) =
0
A1
c =6
a + bx
b y =
6 + 8x
Substituting the first coordinate:
=
12 + 5 (1 − 2 x )
2x − 1
A1
2x − 1
5 (2 x − 1)
12
−
2x − 1
(2x − 1)
=
12
−5
2x − 1
c
M1A1
12 − 5 (2 x − 1)
=
b
R1
1
2
y = −5
x =
A1
A1
A1
d
M1
2 a + 12 b
=
5
10
4= a + 12 b
=
8 2a + b (1)
A1
Substituting the second coordinate:
3
a + 4b
−
=
38
38
−3 = a + 4b (2)
A3
A1
Solving (1) and (2) simultaneously:
a=5
A1
b = −2
A1
© Oxford University Press 2019
15
Worked solutions
22
Asymptotes are x = −
3
and y = 2
2
A1A1
 1
Intersections with axes are at  0,  and
 3
 1 
 − ,0
 4 
A1A1
A2
© Oxford University Press 2019
16
Worked solutions
5
Measuring change: differentiation
Skills check
1 a
b
since lim − = −∞ and lim + = ∞
x →− 3
and
12
−1 − 2
= −
4 − ( − 34 )
19
Horizontal asymptote at y = −1
since lim g(x ) = −1
lim− = ∞ and
3
1
= x −2
x2
8
c
5 x
3
=
x→ 3
x →±∞
7 x = 7x 2
b
lim+ = −∞
x→ 3
1
2 a
x →− 3
−3 − 0 3
=
−4 − 0 4
Vertical asymptote at x = 1
since lim− f (x ) = ∞ and lim+ f ( x ) = −∞
x →1
x →1
Horizontal asymptote at y = −1
since lim h(x ) = −1
8 − 23
x
5
x →±∞
3
4
Vertical asymptotes at x = ± 2
since
and
lim − = ∞ and
lim + = −∞
x →− 2
lim− = ∞ and
x →− 2
lim+ = −∞
x→ 2
x→ 2
Horizontal asymptote at y = 0
since
5x 

lim  − 2
0
=
 x − 2
x →±∞
Exercise 5C
4
Since | 12 |< 1,
n
n
∑ 5 =

2
f=
'(x ) 7=
x 7 −1 7 x 6
2=
f '(x ) 18
=
x18 −1 18 x17
∞
5(1)
1
= 10
5∑ =

1 − 12
n 02
0=
∞
1
1
3
1 − 1 −1
1 −3
− x 2 =
− x 2
f '(x ) =
2
2
4
1
1 1 −1 1 − 4
5
f (x) =
x =
x 5 ⇒ f '(x ) =x 5 =x 5
5
5
lim− (5 − 2 x )= lim+ (5 − 2 x )= 3
5
1
1 − 1 −1
1 −3
−1
f (x ) ==
x 2 ⇒ f '(x ) =
− x 2 =
− x 2
2
2
x
3
 2x 2 − x 
 2x 2 − x 
lim− 
 = lim+ 
 = −1
x →0
x
x

 x →0 

6
3
3 3 −1 3 − 1
4
f (x) =
x3 =
x 4 ⇒ f '(x ) = x 4 = x 4
4
4
4
 x2 − x 
 x2 − x 
lim
=
=

 lim

 1
−
+
x →1
 x − 1  x →1  x − 1 
Exercise 5D
=
n
Exercise 5A
1
2
(
)
(
)
=
=
lim x 2 +
1
lim+ x 2 +
1 10
x →3−
x →1
x →3
x →1
1 a
Exercise 5B
1
Vertical asymptote at x =
b
1
6
x → 16
Horizontal asymptote at y =
since lim f ( x ) =
x →±∞
2
1
2
f (x )= 5x(x 2 − 1)= 5x 3 − 5x
∴ f '(x ) = 15x 2 − 5
since lim− f (x ) = −∞ and lim+ f ( x ) = ∞
x → 16
dy
= 4x 3 − x
dx
1
2
c
f=
'(x ) 24 x 3 − 6 x
d
ds
= 4t + 3
dt
e
dv
= −9.8
dt
f
dc
= 24
dx
Vertical asymptotes at x = ± 3
© Oxford University Press 2019
Worked solutions
1
Worked solutions
2 a
1
−1
3x 2
f=
(x ) 6=
x 6 x 2 ∴ f '(x ) =
3
c
f (x ) =4 x +
−2
5
b =
f (x ) 5=
x 3 5x 5 ∴ f '(x ) =
3x 5
c
x
1
=x 4 + 8 x
− 21
1 − 34
−3
x − 4x 2
4
1
15
−3
−3
so f '(1) =
(1) 4 − 4(1) 2 =
−
4
4
∴ f '(x ) =
1
2
f (x) = − 3 x =
2 x −1 − 3x 2
x
∴ f '(x) =
−2 x −2 −
8
3 − 12
x
2
2
3 a
(x)
f=
3
3 −2
−3x −3
x ∴ f '(x ) =
=
2x 2 2
b
f=
(x)
3
3
3 −2
=
=
x
2
2
(2 x )
4x
4
f '(x ) = 2 x 2 − 9 x − 3 ∴ 2= 2 x 2 − 9 x − 3
⇒ 2 x 2 − 9 x − 5= (2 x + 1)(x − 5)= 0
3
∴ f '(x ) =
− x −3
2
1
so x =
or x =
5
−
2
1
 1  199
when x =
− , y =
f −  =
2
24
 2
c
f '(x ) = 12π x 2
when x = 5,
d
f (x ) = (x + 1)2 = x 2 + 2 x + 1
217
6
217 
 1 199 

so  − ,
 and  5, −

6 
 2 24 

∴ f '(x ) =
2x + 2
e
f (x) =
3
x3 + x − 3
= x2 + 1 −
x
x
2
= x + 1 − 3x
f
4 a
b
−1
Exercise 5F
∴ f '(x ) =
2 x + 3x
−2
1
f (x ) =(2 x − 1)(x + 3)
dy
=
−2 x −3 + 2 x −2
dx
∴
∴ f '(x ) = 6 x 2 − 2 x + 6
3

Therefore, the gradient at  2, − 
4

1
is − 2(2)−3 + 2(2)−2 =
4
So the gradient of the normal at
this point is − 4
3
y =
1+ x x =
1 + x2 ∴
y =
dy 3 12
=
x
dx 2
7
1
−1
−
=7 x −2 − x 2
2
x
x
y =3 x +
4
1
29
 3
∴ y −  −  =−4(x − 2) ⇒ y =−4 x +
4
4


1
x =x 3 + x 4
2
Exercise 5E
dy
= 2 x − 4 so the gradient at
dx
x = −1
b
1 − 2x
1
2
=
−
= x −2 − 2 x −1
x2
x2 x
= 2x 3 − x 2 + 6 x − 3
dy 1 − 23 1 − 34
∴
=
x + x
dx 3
4
1 a
y =
2
dy
1 −3
=
−14 x −3 + x 2
dx
2
c
y = f (5) = −
y=
∴
is 2(−1) − 4 =−6
5
2x − 5
5
= 2 x 4 − = 2 x 4 − 5x −1
x
x
dy
= 8 x 3 + 5x −2
dx
dy
=
−3x 2 + 4 x
dx
So the gradient at x =
−1 is − 7
dy
=−7 ⇒ 3x 2 − 4 x − 7
dx
= (3x − 7 ) ( x + 1) = 0
7
∴ x = or x =
−1 (i.e. the tangent itself)
3
22
7
y  = −
27
3
∴y +
22
7
419

=−7  x −  ⇒ y =−7 x +
27
3
27

so the gradient at (1, −3) is
8(1)3 + 5(1)−2 =
13
© Oxford University Press 2019
2
Worked solutions
3
Exercise 5G
dy
1
= 1− 2
dx
x
1 a =
y u5 where =
u 2x + 3
dy
1
1
=−3 ⇒ 1 − 2 =−3 ⇒ x =±
x
dx
2
5
 1
±
y ±  =
2
 2
Gradient of
dy dy du
= =
dx du dx
= 10 (2 x + 3)
b
1
normal is
3
1
 1 
 x −  ± 
3
 2 
c
− 12
3
−1
−
=
−3u 2
y =
2
2x − 1
)
= 6 x 2x 2 − 1
− 23
3
d
2

y = 2  x 2 −  = 2u3
x

2
x
where =
u x2 −
dy dy du
= =
dx du dx
1
−nx − n −1
= x − n ∴ g '(x ) =
xn
(6u )  2x + x2 
2
2
2
⇒ xg '(x ) + ng(x ) =
x(−nx − n −1 ) + nx − n
=
−nx − n + nx − n =
0
2 
1 

= 12  x 2 −   x + 2 
x
x

 

2
At
=
x 0,
=
y 6 so tangent passes
f '(x )= 15ax 2 − 4bx + 4c
through (0,6)
f '(x ) ≥ 0 ⇒ 15ax 2 − 4bx + 4c ≥ 0
y =
6 (1 − 2 x ) 3 =
6u 3 where u =
1 − 2x
⇒ x2 −
2
−20 x −1 + 1
f (x) =
1
1
(
dy dy du
−2
=
=2u 3
dx du dx
4b
4c
x+
≥0
15a
15a
2b 
4b2
4c

⇒ x −
+
≥0
 −
2
a
a
a
15
225
15


The LHS is valid for all real x and
2b
attains its minimum at x =
so
15a
4b2
4c
−
+
≥ 0 ⇒ b2 ≤ 15ac
225a2 15a
8
=
− (1 − 2 x )
(
2
2 
1
7
x − = x −  =
3
3 
3
9
Coordinates are
 1 + 7 7 − 14 7   1 − 7 7 + 14 7 
,
,

 , 

27
27
 3
  3

b
− 12
dy dy du  3 − 3 
= =  u 2  ( 4 x )
dx du dx  2

1± 7
∴x =
3
7 a
1
where=
u 2x 2 − 1
f '(x )= 3x 2 − 2 x − 2= 0
g(=
x)
y = 1 − 2 x =(1 − 2 x ) 2
=
−u
2
6
4
dy dy du  1 − 1 
= =  u 2  ( −2 )
dx du dx  2

f '(x
=
) 6 x − 2k
⇒ x2 −
10u 4
1
f '(1) =
6 − 2k =
10 ⇒ k =
−2
5
4
y= u 2 where u= 1 − 2 x
 5
∴ y − ± =

 2
1
5 1
x± m
⇒ y=
3
2 6
1
7
1
7
and y=
x+
x−
∴ y=
3
3
3
3
4
) (2 )
(5u=
=
−4 (1 − 2 x )
) ( −2) =−4u
− 23
− 23
=
x 0 is − 4
so the gradient at
∴ y − 6 =−4 ( x − 0 ) ⇒ y =−4 x + 6
for x > 0
∴ f '(x=
) 20 x −2 , g '(x=
) 5 for x ∈ ¡
=
f '(x ) g=
'(x ) when 20 x −2 5
⇒ x2 =
4⇒x =
±2
But x > 0 so x =
2 only
© Oxford University Press 2019
3
Worked solutions
3
When =
x 1, =
y 1 so =
a
y =
a (1 + bx )
− 12
c
1+ b
−1
and =
v
au 2 where u =
1 + bx
=
=
d
4
(3 − x )
= 4u −3 where u= 3 − x
(
= 12 (3 − x )
) ( −1) =12u
e
−4
−4
2
)
− x +1
2
y = (2 − 3x ) x + 2 = uv
dy du
dv
=
v +u
dx dx
dx


1
=( −3) x + 2 + (2 − 3x ) 

+
x
2
2


−10 − 9 x
=
2 x +2
2 a
Curve horizontal when
y = x + 1 (3 − x ) = uv where
2
u=
x + 1 and v =
(3 − x )
2
dy
=0
dx
dy
=
dx

= 
2
2
2
So x =
±
=
±
9
3
du
dv
v +u
dx
dx
2

1
 (3 − x ) +
x +1
(
x + 1 −2 (3 − x )
)
dy ( x − 3) + 4 ( x − 3) ( x + 1)
=
dx
2 x +1
( x − 3) ( x − 3 + 4(x + 1))
2
Exercise 5H
1 a =
y x 2 (2 x − =
1) uv where =
u x2
and =
v 2x − 1
dy du
dv
=
v +u
dx dx
dx
= (2 x ) (2 x − 1) + x 2 (2 )
∴
=
2 x +1
( x − 3) (5x + 1)
=
2 x +1
= 6 x 2 − 2 x = 2 x(3x − 1)
b Using the result from part a,
y = (2 x − 3) ( x + 3) = uv where
x −3 = 0 ⇒ x = 3
u=
2 x − 3 and v =
( x + 3)
or 5x + 1 =0 ⇒ x =−
3
3
dy du
dv
=
v +u
dx dx
dx
(2 ) ( x + 3)
3
=
(x
where u =
2 − 3x and v =
x +2
dy
=
−9 x 2 + 2
dx
b
2
2
=
dy 3
=
so=
at x 1,
and therefore
dx 4
4
the normal has gradient −
3
1
4
4
11
∴ y − =− ( x − 1) ⇒ y =− x +
2
3
3
6
5
)
(2) ( x 2 − x + 1)
+ 2 (2 x + 1) (2 x − 1) ( x 2 − x + 1)
2 x (5x − 1) ( x 2 − x + 1)
=
3
dy dy du
=
= −12u −4
dx du dx
(
y= (2 x + 1) x 2 − x + 1 = uv where
dy du
dv
=
v +u
dx dx
dx
3a2
3a2
⇒ a2 = 1 +
4
4
⇒ a = 2 ( a > 0 ) ∴ b = a2 − 1 = 3
y=
2 2 − 3x
u = 2 x + 1 and v =
⇒ a = 1+
4


3
2 − 3x + x  −

 2 2 − 3x 
4 − 9x
(1)
=
dy
3
= −
dx
8
−3
3
ab
b
⇒−
(1 + b ) 2 = − 2a2 = − 8
2
3a2
so b =
4
2 − 3x
dy du
dv
=
v +u
dx dx
dx
dy dy du  a − 23 
ab − 23
u
=
=
−
 − u  ( b) =
dx du dx  2
2

−3
ab
=
−
(1 + bx ) 2
2
At (1,1) ,
y = x 2 − 3x = uv where u = x
(
+ (2 x − 3) 3 ( x + 3)
2
1
5
)
=
( x + 3) ( 8 x − 3)
2
© Oxford University Press 2019
4
Worked solutions
3
y =x (1 − 2 x )
−1
=uv where u =x
(1 − 2x )
and v
=
dy du
dv
v +u
=
dx dx
dx
= (1) (1 − 2 x )
−1
d
−1
(
+ x 2 (1 − 2 x )
−2
=
2
1 + 3x u
=
5−x
v
x
u
=
2−x v
=
u
where
16
(5 − x )
2
c
1 + 2x
1− x
2
u ' = 2, v ' = −
u
v
where u= 1 + 2 x
=
2
3
2
2
−3x 2 + 4 x + 3
=
, f '(0) 3
2
( x + 1)
x3 + x2 + x + 1 u
=
x
v
vu '− uv '
v2
2
x 3x + 2 x + 1 − x 3 + x 2 + x + 1 (1)
(
) (
x
)
2
1
2
1 a
y =( x − 1) ( x + 3) =uv where
2
2
x
dy du
dv
=
v +u
dx dx
dx
= (x + 3)2 + 2 ( x − 1) ( x + 3)
1 − x2


x
1 − x 2 (2 ) − (1 + 2 x )  −

2
1− x 

1 − x2
x +2
(1 − x )
)
+1
u=
x − 1 and v =
( x + 3)
dy vu '− uv '
=
dx
v2
=
2
Exercise 5J
1 − x2
and =
v
(x
1
∴x =
3
2
x
=
2
f '(x ) = 1 ⇒ 2 x 3 + x 2 − 1 = x 2 ⇒ x 3 =
2+x
y=
+1
2x 3 + x 2 − 1
=
x2
( x ) ( −1)
2
)
2
3 − 2 x − 3x 2
3 f (x)
=
=
, v ' = −1
2 (2 − x )
(x
f '(x ) =
2 x
dy vu '− uv '
=
v2
dx
 1 
(2 − x ) 
−
2 x 
=
2
(2 − x )
=
)
+ 1 (3) − (1 + 3x ) (2 x )
where u = x 3 + x 2 + x + 1 and v = x
x
and v= 2 − x
1
2
so normal at this point has
1
gradient −
and passes through (0, − 2)
3
1
∴ y =− x − 2
3
where u= 1 + 3x
dy vu '− uv '
=
dx
v2
5 − x ) (3) − (1 + 3x ) ( −1)
(=
2
(5 − x )
u' =
=
(x
=
2 f '(x )
and v= 5 − x , u ' = 3, v ' = −1
b
=
y
where u= 1 + 3x
dy vu '− uv '
=
dx
v2
) = (1 −12x )
Exercise 5I
y=
1 + 3x u
=
x2 + 1 v
and =
v x2 + 1
so the gradient at (0,0) is 1 and
the normal therefore has gradient − 1
∴ y =− x
1 a
y=
( x + 3) ( x + 3 + 2(x − 1))
=+
( x 3) (3x + 1)
=
b Most easily done using the product
(and chain) rule:
y = ( x + 1) 1 − 2 x = uv
where u =
1 − 2x
x + 1 and v =
dy du
dv
v +u
=
dx dx
dx
3x
x +1
1 − 2x −
=
=
−
1 − 2x
1 − 2x
© Oxford University Press 2019
5
Worked solutions
c Most easily done using the quotient
rule:
x +1 u
where u= x + 1
=
x −1 v
and v= x − 1
dy vu '− uv '
=
dx
v2
( x − 1) − ( x + 1) = − 2
=
2
2
( x − 1)
( x − 1)
2 a
Increasing: nowhere
Decreasing: ∀x ∈ ¡
y=
b
)
c
−1
Decreasing: x ∈ (1, ∞ )
dy dy du  2 
3
==
−
 4x − 2
dx du dx  u 2 
(
2
(
)
( x − 2x + 1)
f (x )=
1+ x
u
=
x −1
v
)
d
where u= 1 + x
(
x −1
− 1+ x
2 x
2
( x − 1)
(
2
)
Exercise 5L
1 a
)
2
f '(x ) = 2 x , f '(x ) = 0 ⇒ x = 0
f (x ) decreasing for x < 0 and
increasing for x > 0 so this is a
local minimum
1
=
−
24
∴ ( 0, −2 ) is a local minimum point
Graphically, this is a positive parabola
so the turning point must be a local
1
1
=
−
( x − 9)
2
24
7
x
−
8 24
1
Normal:y − = 24 ( x − 9 )
2
431
⇒ y= 24 x −
2
−2
 1

Decreasing: x ∈ 
,∞
16


x +1
x +2 x +1
=
−
=
−
2
2
2 x ( x − 1)
2 x ( x − 1)
(3 + 1)
f '(9) =
−
2
2 (3) ( 9 − 1)
1
2 x
1 

Increasing: x ∈  0,

 16 
(note the function is only valid here
for x > 0)
2
and v= x − 1
Tangent:y −
f=
'(x )
3
vu '− uv '
f '(x ) =
=
v2
2 x −1
for x > 1)
where u = x − 2 x + 1
4
1
Increasing: nowhere
4
4 1 − 2x
f '(x ) = −
(note the function is only valid here
y= 2 x 4 − 2 x + 1 = 2u −1
=
f '(x ) = 4 x
Increasing: x > 0
Decreasing: x < 0
d Most easily done by chain rule
(quotient rule also valid)
(
f '(x ) = −3x 2
minimum (students should draw this)
⇒y =
b
f '(x )= 1 −
1
x
, f '(x ) = 0 ⇒ x = 1
f (x ) decreasing for 0 < x < 1
and increasing for x > 1 so
this is a local minimum
Exercise 5K
∴ (1, −1) is a local minimum point
1 a i
Graphically, the graph is continuous,
x >0
begins at ( 0, 0 ) and lim f ( x ) = ∞ so
ii Nowhere
b i
ii
c i
ii
d i
ii
x →∞
x ∈ (−∞, −1) U ( −1, 0 )
the turning point at (1, −1) must be a
local minimum point (and in fact this
case a global minimum).
(Students should draw this.)
x ∈ (0,1) U (1, ∞)
x ∈ (−∞, −0.215) U (1.55, ∞ )
x ∈ ( −0.215,1.55)
x ∈ ( −∞, −1) U (1, ∞ )
c
f '(x ) = 3x 2 − 12 x = 3x ( x − 4 )
x ∈ ( −1,1)
© Oxford University Press 2019
6
Worked solutions
f '(x ) = 0 ⇒ x = 0 or x = 4
2
Consider the point ( 0,2 ) ,
f (x ) increasing for x < 0 and
decreasing for 0 < x < 4 so this
is a local maximum
Consider the point (4, −30)
f (x ) decreasing for 0 < x < 4 and
increasing for x > 4 so this is a
local minimum
∴ ( 0,2 ) is a local maximum
and ( 4, −30 ) is local minimum
Graphically, this is a positive cubic,
so the first turning point is a
maximum and the second point
a minimum
(students should draw this)
2
f '(x )= 3ax 2 + 4 x= x (3ax + 4 )
f '(x ) =
∴ f ''(x ) =
x2 − 4
f ''(x ) =x 2 − 4 =
0⇒x =
±2
3
f '(x) =
−2 (5 − 4 x )
4
3
(terms of order x and constants
disapper upon differentiating twice)
∴ 2a2 =
8⇒a=
±2
Exercise 5N
1 a
Coordinates of point of inflexion are
(0,0)
b
c
so a − b + c =
4
p '(0) = 3 ⇒ c = 3 ⇒ a − b = 1
2 a
p '(−1)= 0 ⇒ − ( −3a + 2b ) + 3= 0
2b
− 1 − b =1 ⇒ b =−6 ⇒ a =−5
3
−5, b =
−6, c =
so a =
3, d =
1
There are no points of inflexion
d2y
b =
12 x 2 > 0
dx 2
Functions is concave up throughout its
domain
dy
= 3x 2 + 2ax = x (3x + 2a) = 0
dx
dy
2a
=
0 when x =
0 or x =
−
dx
3
2a
∴−
= 4 ⇒ a = −6
3
∴
c Function is never concave down
3 a
dy
= 3x 2 − 12 x − 12
dx
d2y
= 6 x − 12 = 0 at x = 2
dx 2
y ( 4) =
64 − 6 (16 ) + b =
b − 32 =
−11
Coordinates of point of inflexion are
(2, −38)
21
⇒b=
so the local maximum is at ( 0,21)
3
15 12
x
f '(x ) = 5x 2 ∴ f ''(x ) =
2
dy
= 4x 3 − 3
dx
d2y
=
12 x 2 > 0
dx 2
2b
−1
3
∴
1
d2y
= 6x < 0 ⇒ x < 0
dx 2
Function concave down on ] − ∞, 0[
p '(x=
c x (3ax + 2b ) + c
) 3ax 2 + 2bx + =
Exercise 5M
d2y
= 6x > 0 ⇒ x > 0
dx 2
Function concave up on ]0, ∞[
p ( −1) =−a + b − c + 1 =−3 4
dy
= 3x 2 − 1
dx
d2y
=
6 x ⇒ 0 at x =
0
dx 2
p(0)
= d= 1 so d= 1
⇒ a=
− 23
d2y
= 2a2
dx 2
4
3a
It is given that the turning point,
away from x 0,
occurs at x 1
=
=
4
4
⇒ a =−
3a
3
− 12
∴ f ''(x ) =
−4 (5 − 4 x )
f '(x ) = 0 when x = 0 or x = −
∴ 1 =−
x3
− 4x + 5
3
b
d2y
= 6 x − 12 > 0 ⇒ x > 2
dx 2
Function is concave up on ]2, ∞[
c
d2y
= 6 x − 12 < 0 ⇒ x < 2
dx 2
© Oxford University Press 2019
7
Worked solutions
Function is concave down on ] − ∞,2[
4 a
dy
= 3x 2 + 2 x
dx
b
Function is concave up for x > 0;
x < −0.25
2
dy
1
=
6x + 2 =
0 at x =−
2
dx
3
c
Coordinates of point of inflexion are
 1 25 
− ,−

 3 27 
b
8 a
d2y
1
= 6x + 2 > 0 ⇒ x > −
dx 2
3
dy
5 a = 12 x 2 − 4 x 3
dx
2
dy
=24 x − 12 x 2 =0 at x =0,2
dx 2
Coordinates of point of inflexion are
(0, 0) , (2,16 )
b
Coordinates of point of inflexion are
(0, −16 ) , (2, 0)
b
c
9 a
d2y
= 24 x − 12 x 2 > 0 ⇒ 0 < x < 2
dx 2
f '' ( 0 ) = 4
 4
f ''  −  =
−4
 3
0⇒x =
f '' ( x ) =
−
dy
= 3x 2 − 6 x + 3
dx
2
3
 2 43 
Non-horizontal inflexion at  − ,

 3 27 
d2y
= 6x − 6 > 0 ⇒ x > 1
dx 2
b
f '=
( x ) 3 ( x − 1)
2
3 ( x − 1) = 0 ⇒ x = 1
2
f '' ( x )= 6 ( x − 1)= 0 at x= 1
Function is concave up on ]1, ∞[
d2y
= 6x − 6 < 0 ⇒ x < 1
dx 2
Function is concave down on ] − ∞,1[
7 a
4
3
f '' ( x
=
) 6x + 4
d2y
= 24 x − 12 x 2 < 0 ⇒ 0 > x, x > 2
dx 2
Coordinates of point of inflexion are
(1, 0)
c
f ' (=
x ) 3x 2 + 4 x
3x 2 + 4 x = 0 ⇒ x = 0, −
d2y
= 6 x − 6 = 0 at x = 1
dx 2
b
d2y
= 12 x 2 − 24 < 0 ⇒ 0 < x < 2
dx 2
Function is concave down when
0<x<2
Function is concave down for
x < 0; x > 2
6 a
d2y
= 12 x 2 − 24 > 0 ⇒ x < 0, x > 2
dx 2
Function is concave up when x < 0,
x>2
Function is concave up for 0 < x < 2
c
dy
=4 x 3 − 12 x 2 + 16
dx
d2y
= 12 x 2 − 24 x = 0 at x = 0,2
dx 2
d2y
1
6x + 2 < 0 ⇒ x < −
=
2
dx
3
1
Function is concave up on ] − ∞, − [
3
d2y
= 24 x 2 + 6 x < 0 ⇒ −0.25 < x < 0
dx 2
Function is concave down for
−0.25 < x < 0
1
Function is concave up on ] − , ∞[
3
c
d2y
= 24 x 2 + 6 x > 0 ⇒ x > 0, x < −0.25
dx 2
dy
= 8 x 3 + 3x 2
dx
f '' (1.1
=
) 0.6 > 0
f '' ( 0.9 ) =
−0.6 < 0
Second derivative = 0 at x = 1, there
is a change in concavity at x = 1.
Therefore there is a horizontal inflexion
at (1, 0)
d2y
=24 x 2 + 6 x =0 at x =0, −0.25
dx 2
Coordinates of point of inflexion are
( −0.25, 0.992) , (0,1)
© Oxford University Press 2019
8
Worked solutions
c
f '(x) =
−12 x 3 − 24 x 2
decreasing for x < 1
iv Concave upward for x ∈ ¡
−12 x 3 − 24 x 2 = 0 ⇒ x = 0, −2
4
f '' ( x ) =
−36 x 2 − 48 x =
0 at x =
0, −
3
129
f '' ( 0.1) = −
25
111
f '' ( −0.1) =
25
c i
=
f ''
( x ) 36 x 2 + 24x
f '' ( −1) =
60
f '' ( 0 ) = 0
Therefore local min at (−1, −3)
First and second derivatives = 0 at
x = 0, and there is a change in
concavity at x = 0.
Therefore there is a horizontal inflexion
at (0, 2)
ii
iii Increasing x > −1
4
, and
3
there is a change in concavity at
4
x = −
3
Second derivative = 0 at x = −
d
1 −3
f '(x) = x 2
2
2
− ,0
f '' ( x ) =
36 x 2 + 24 x =
0⇒x =
3
Horizontal inflexion point at (0, −2)
Non-horizontal inflexion point at
 2 70 
− ,−

 3 27 
f ''(−2) =
−48
Therefore there is a non-horizontal
 4 310 
point of inflexion at  − ,

 3 27 
f '(x) =
12 x 3 + 12 x 2 =
0⇒x =
−1, 0
Decreasing x < −1
iv Concave upward x < −
Concave downward −
2
or x > 0
3
2
<x <0
3
Exercise 5O
1 a
First derivative has no roots, therefore
there are no points of inflexion.
10 a i
f ' ( x ) = 3x 2 − 6 x − 6
3x 2 − 6 x − 6 =
0 ⇒ x =−0.732,2.73
f ''(x
) 6x − 6
=
f '' ( −0.732 ) =
−10.392
f '' (2.73) = 10.38
Therefore local max at
(−0.732, 3.39) and local min at
(2.73, −17.4)
ii
f '' ( x ) = 6 x − 6 = 0 ⇒ x = 1
Non-horizontal inflexion at (1, −7)
iii Increasing: x < −0.732 or x > 2.73
decreasing for −0.732 < x < 2.73
iv Concave downward x < 1 and
concave upward for x > 1
b i
b
f ' ( x )= 2 ( x − 1) ⇒ x= 1
f '' ( x ) = 2
f (1) = 2
Therefore local min at (1, 0)
ii
f '' ( x ) = 2 therefore there are no
inflexion points
iii Increasing for x > 1
© Oxford University Press 2019
9
Worked solutions
c
2 a
b
d
Exercise 5Q
1 a
L=
100
x
100
assuming clarification to
x
the question is made, as written in the
comments
b =
P 2x +
Exercise 5P
1 a
c
P '(x )= 2 −
100
x2
P '(x ) = 0 ⇒ x = 50 = 5 2 (x > 0)
This must be a minimum because
∞ and lim P(x ) =
∞
lim+ P(x ) =
x →∞
x →0
and this is the only turning point
(and the function is continuous)
100
So x =
5 2∴P 5 2 =
2 5 2 +
5 2
(
b
)
(
)
= 10 2 + 10 2
= 20 2 (measured in metres)
d
2
dy
= 600 + 30 x − 3x 2
dx
dy
=0 ⇒ x 2 − 10 x − 200
dx
=( x − 20 ) ( x + 10 ) =0
© Oxford University Press 2019
10
Worked solutions
Exercise 5R
=
∴ x 20 since x > 0
= 600 (20 ) + 15 (20 ) − (20 )= 10000
y(20)
2
3
1 a
Therefore, maximum profit is $10 000
3 a
h=
216
s2
864
s
b
=
A s2 +
c
dA
864
=
2s − 2 =
0
ds
s
c
3
accelerating away from the origin
at 12m/s2
Therefore the total length of wire used
for the rectangle is 150 − 4s
Since the length is twice the length
of the width,
4s
the length of the rectangle is 50 −
3
2s
and the width of the rectangle is 25 −
3
so the total area enclosed by the
square and rectangle is
4s  
2s 

A =s2 +  50 −
  25 −

3
3 


d
t >1
f
s(0) = 1, s (1) = −1
so 20m travelled in this period
∴ Altogether distance travelled is 22m
2 a 1m
b
a maximum so 17m
c
v (=
t ) 12 − 3t 2
v ( 0 ) = 12, v (1) = 9, v (3) = −15
d 16 + 17 =
33 so 33m
the base be l , so the longer side
measures 2l
3
s '(t ) = 15 − 10t ⇒ t =
2l 2 (10 ) + 2 (2l ) ( h ) ( 6 ) + 2 ( l ) ( h ) ( 6 )
C =
the function is a negative parabola
2
45
3
3
3
smax =s   =15   − 5   =
4
2
2
2
4 a
b
s ( 0 ) = 10
c
so 10m
s ( t ) = 0 ⇒ t 2 − 5t − 10 = 0
=
∴t
1
3
3
2
clearly attains maximum here as
Therefore the area of the base is2l 2.
Let the height be h
5
V = 2l 2h = 10 ⇒ h = 2
l
So the total cost is
180
l
s '(t ) = 12 − 3t 2 = 0 ⇒ t = 2
s (2 ) = 17 and this is clearly
5 Let the length of the shorter side of
1
e
> 0)
19
s (1) =
−1, s (3) =
17s2 200s
=
−
+ 1250
9
3
dA 34
200
300
s−
=
=0⇒s=
ds
9
3
17
 9 3
 2 3
∴ Cmin
= 20   + 180  
2
9
= 164 (to nearest dollar)
(t
so travels 2m in this period
2s 

s2 + 2  25 −
=

3 


100s 4s2 
s2 + 2  625 −
=
+

3
9 

dC
180
9
= 40l − 2 = 0 ⇒ l =  
dl
l
2
The change of sign of v ( t )
occurs =
at t 1
2
= 20l 2 + 36lh = 20l 2 +
The particle is moving away from
the origin at 9m/s and is
432
The length of each side of the square is s
2
s(0) =
1, v(0) =
−3, a ( 0 ) =
0
At this instant, the particle is 1 metre
from the origin in the positive
direction, travelling towards the
origin at 3m/s, and is not accelerating
=
⇒ s3 432=
so s
4
b
v(t ) =
3t 2 − 3, a(t ) =
6t
5 + 65
2
≈ 6.53 ( t > 0 )
v ( t )= s ' ( t )= 5 − 2t
 5 + 65 
−1
v
 = −8.06 ms

2


a ( t ) = v ' ( t ) = −2 ms−2
© Oxford University Press 2019
11
Worked solutions
As both the velocity and acceleration
are negative, the diver is speeding up
as he/she hit the water.
5=
h0 0,
=
v0 50 ∴ h ( t ) =50t − 4.9t
h '(t ) = 50 − 9.8t = 0 ⇒ t =
(clearly maximum here)
4
6
dy
=
−10 (1 − 2 x ) (3x − 2 )
dx
+ 18 (1 − 2 x ) (3x − 2 )
5
2
=
(1 − 2x ) (3x − 2)
4
50
9.8
=
2 (1 − 2 x ) (3x − 2 ) (19 − 33x )
4
b
5 a
so
dy
1
= − 2
dx
x
y = 0, x = ±1
(
)
x 2 − 1 − x (2 x )
dy
=
2
dx
x2 − 1
ii Westward is negative → 3 < t < 6
(
)
2
x +1
=
−
< 0 for all x ∈ ¡
2
x2 − 1
ii t = 4.5
(
d t 1.5
=
=
and t 4.5
e
x −3
1
=
x ( x − 3)
x
b Using the quotient rule,
Eastward is positive → 0 < t < 3;
6 < t < 11
t = 1.5
y =
5
dy 1 − 12 4 − 23
c =
x − x
dx 2
3
6 a=
t 0,=
t 3,
=
t 6,
=
t 11
c i
5
( −10 (3x − 2) + 18 (1 − 2x ) )
 50 
=
hmax h=

 127.551
 9.8 
so maximum height is 127.6m to 1d.p.
50
= = 10.2041
t ground
4.9
so hits ground after 10.2s to 1d.p.
b i
5
)
c
Speeding up: t ∈ (0,1.5) ;
t ∈ (3, 4.5) ; t ∈ ( 6, 9 )
Slowing down: t ∈ (1.5,3) ;
t ∈ ( 4.5, 6 ) ; t ∈ ( 9,11)
Chapter Review
1 a
y =2
b
a=2
2 a
6
dy
= 3x 2 − 6 x − 9= 3 ( x − 3) ( x + 1)
dx
dy
0⇒ x =
3
=
−1 or x =
dx
y ( −1) =−1 − 3 + 9 + 2 =7
∴
27 − 27 − 27 + 2 =
y (3) =
−25
So y =
−25 and y =
7
7 Using the quotient rule,
b
3 a
dy 2 ( x − 1) − 2 x
2
=
= −
2
2
dx
x
−
x
−
1
( 1)
(
)
5x + x 2
5x + x 2
=
=
lim
lim
5
−
+
x →0
x →0
x
x
y = 6, x = ±3
b
y = 0, x = −3
∴ at (2, 4 ) the gradient is − 2
4 a Using the product rule,
© Oxford University Press 2019
12
Worked solutions
11 a
So the tangent at this point is
y −4=
−2 ( x − 2 ) ⇒ y =
−2 x + 8
c
1
The gradient at (3,3) is −
so
2
the gradient of the normal at this
point is 2
8
1 
1
(x) 9  + 2 
f=
x
x


2 
 1
⇒ f '(x ) = 9  − 2 − 3 
x 
 x
Therefore the normal at this point is
1
x 9
y − 3 =− ( x − 3) ⇒ y =− +
2
2 2
7
x 9
∴ − + = −2 x + 8 ⇒ x =
2 2
3
∴ f '(x ) =
−2
0⇒x =
9
−
y =
f (−2) =
4
6 
 2
f=
''(x ) 9  3 + 4 
x
x


10
7
 7 10 
⇒y =
−2   + 8 = ∴ P  ,

3
3
3 3 
 1 3 9
>0
f ''(−2) = 9  − +  =
 4 8 8
9

so  −2, −  is a local minimum
4

f '(=
x) 6 x 2 − 3
f '(1) = 3
So the normal to the curve at
d For f(x) to be concave up f’’(x) > 0
6 
 2
9  3 + 4  > 0 ⇒ 2x + 6 > 0
x
x


x > −3
1
3
1
x 1
∴ y − 0 =− (x − 1) =− +
3
3 3
this point has gradient −
9
b=
y 0,
=
x 0
f (x) =
0⇒x =
−1
e
dy
=
−3x 2 + 4 x
dx
dy
=−4 ⇒ 3x 2 − 4 x − 4 =0
dx
⇒ (3x + 2 ) ( x − 2 ) =
0
2
or x =
2
3
8
59
 2
4
+ 2   + 1=
y  − =
3
27
9
27


 
∴x =
−
y (2 ) =−8 + 8 + 1 =1
 2 59 
∴ − ,
 and
 3 27 
(2,1)
10 Using the quotient rule,
2
dy 2 x ( x + 1) − x
=
=
2
dx
( x + 1)
x 2 + 2x
=
2
( x + 1)
x ( x + 2)
( x + 1)
2
dy
=
−2
0⇒x =
0 or x =
dx
Students may either use first derivative
or second derivative test here
e.g. second derivative test:
(
2 ( x + 1) − 2 x 2 + 2 x
d2y
=
4
dx 2
( x + 1)
3
=
12 a
b
f '(x )= 1 −
( x + 1)
∴ ( −2, −4 ) is a local maximum
and ( 0, 0 ) is a local minimum
13 a
)
x − b =0
b
2 x
f '(x ) > 0 when x >
b2
4
f '(x ) < 0 when 0 < x <
f '' ( x ) =
3
The second derivative is negative at
x =
−2 and positive at x =
0
(
=
⇒ x 0 or
=
x b
ii
c
x
2
i
) ( x + 1)
2
f (x ) =0 ⇒
b2
4
b
3
4x 2
i
f '' ( x ) > 0 when b > 0
ii
f '' ( x ) < 0 when b < 0
v(=
t ) 49 − 4.9t
b v(t ) = 0 ⇒ t = 10
© Oxford University Press 2019
13
Worked solutions
=
h(10) 49 (10 ) − 2.45 (10 )
2
= 490 − 245 = 245
So 245m
14 a
v ( 0 ) = −2
b v ( t ) = 0 ⇒ (1 + t ) = 4t + 9
2
t 2 − 2t − 8 = ( t − 4 ) ( t + 2 ) = 0
So t = 4
c
2
a ( t )= 1 −
2
a ( 4) =
1−
d
15 a
b
d
3
=
5
25
Always speeding up since
R(x ) =
(320 + 10x ) (200 − 5x )
A1
acceleration is always positive
R ' ( x ) =400 − 100 x =0
M1
f ′ ( x ) = 4x 3 − 6 x 2 − 2x + 3
g '(x) =
g '(x) =
c
18 a Letting x represent the number of $10
increases above $320. Then rental
income is
4t + 9
(
)
−4 x 2 + 1 − ( −4 x ) ⋅ 2 x
(x
2
)
+1
2
4x 2 − 4
(x
2
M1A1
A1
)
+1
A1
2
h ' ( x ) =1 ⋅ ( x − 7 ) + ( x + 2 ) ⋅ 1
M1
h'(x
=
) 2x − 5
A1
i ' ( x ) = 3 ⋅ 2 ⋅ (2 x + 3)
M1
i=
' ( x ) 6 (2 x + 3)
A1
2
2
16 a Graph 1
as y increases as x increases
c Graph 3
19 a h ( 4 ) = 370
and h (5) = 438 (3 s.f.)
A1A1
b v=
(t ) h′=
(t ) 112 − 9.8t
M1A1
c
v (t ) =
0 ⇒ 112 − 9.8t =
0
M1
t = 11.4 (3 s.f .)
A1
A1
as the gradient of the tangent at any
point is non-positive and therefore
different from 1.
R1
b Graph 2
x =4
A1
Which corresponds to $360 rent
R1
A1
b i 200 − 5 × 4 =
180
ii 360 × 180 =
M1A1
$64800
A1
R1
d Double x-coordinate of maximum or
determine zero
M1
22.8 (3 s.f .)
A1
Shape
A1
Domain 0 ≤ x ≤ 22.9 (3 sf)
A1
Maximum 640 (3 sf)
A1
v (22.9 ) = −112 ms
M1A1
e
A1
as the other two functions are not
defined at infinity
R1
d Graph 1
A1
as the function is decreasing.
R1
17 a i
0 ≤ t ≤ 2 , 4.6 ≤ t ≤ 5
and 8.5 ≤ t ≤ 10
ii 2 ≤ t ≤ 4 and 5 ≤ t ≤ 7
iii 4.6 ≤ t ≤ 8.5
b
A1A1
A1
A1A1
A1
f (t ) = 2t , g(t ) = 2
h(t ) =
−3t + 14 , i(t ) = −1
=
f (t )
1
(2t − 17)
3
A4
c Up to two correct branches correct
A1; all branches correct A2; all branches
correct and labels and scale also correct A3
f
© Oxford University Press 2019
-1
14
Worked solutions
g
a ( t ) = v ′ ( t ) = −9.8
M1A1
which is constant
A1AG
f ′ ( 2 ) g ( 2 ) − f ( 2 ) g′ ( 2 )
 f ′
  (2 ) =
2
g
 
g (2 )
20 a i
(
)
( h′ o g ) ( x ) = h′ ( g ( x ) )
−1
= 2⋅
−1
x +2
x −1
A1
( h o g )′ ( x ) ≠ ( h′ o g ) ( x )
−1
M1
−1
AG
M1
 4
10 × 4 − 9 ×  − 
 3
=
42
52  13

= =
 = 3.25 
16  4

ii
A1
A1
( g o f )′ (1) = g′ ( f (1) ) f ′ (1)
M1
4
16
× 4 =−
3
3
False
A1
=−
b i
A1
as derivative changes sign. R1
b False
A1
as the derivatives at these points
are not negative reciprocals.
R1
N (3) − N (1)
21 a
3 −1
N (5 ) − N ( 4 )
5−4
= 1410
M1A1
= 2220
A1
the first period the number of cases
is increasing in average 1410 per
day; in the second period it
increases in average 2220 per day.
dN
b = 900t − 90t 2
dt
M1A1
c After 10 days (reaches 15 000 cases)
M1A1
d2N
= 900 − 180t
dt 2
d
M1A1
which gives the variation of the rate
at which the spread of the disease
spreads.
R1
22 a
x =
y +2
y −1
M1
x ( y − 1) = y + 2
M1
xy − y = x + 2
A1
1
g −=
(x)
(
)
x +2
= g (x)
x −1
(
A1AG
)( )
b h o g −1 ′ ( x ) = h′ g −1 ( x ) g −1 ′ ( x )
M1A1
= 2⋅
  6 ( x + 2) 
x + 2 
3
 = −

⋅ −
2
x − 1  ( x − 1)   ( x − 1)3 

 

A1
© Oxford University Press 2019
15
Worked solutions
6
Representing data: statistics for
univariate data
Skills check
1 a Mean
=
b Mean
=
2+3+4+5+6
= 4
5
13 + 9 + 7 + 12 + 15 + 19 + 2
= 11
7
2 a The number that occurs most often is 5
b The numbers that occur most often are
1 and 7 . The data is bimodal
3 a The median is the middle number, 6
b Arrange the data in order of size.
2,3,5,7,8,9 .
d The data is right or positively skewed
2 a Continuous
b The frequency table is given here (note
difference between this one and the one
in the solutions provided)
The median is in between 5 and 7 .
1
6
(5 + 7 ) =
2
Exercise 6A
1 a Discrete
Continuous
b
c The histogram is given here (note
difference between this one and the one
in the solutions provided)
c Continuous
d Discrete
2 a Stratified sampling
b Systematic sampling
c Simple random sampling
d Quota sampling
3 a Stratified sampling
b Stratified sampling
c Systematic sampling
d Simple random sampling
e Quota sampling
d The data is right or positively skewed
Exercise 6B
3 a Continuous
1 a Continuous
b The frequency table is given here (note
difference between this one and the one
in the solutions provided)
b Histogram from concise solutions
c The histogram is given here (note
difference between this one and the one
in the solutions provided)
© Oxford University Press 2019
Worked solutions
1
Worked solutions
c The data is neither right nor left
skewed, it has normal distribution
2 a
x
Mid value
f
4 a Frequency table from concise solutions
b The data is left or negatively skewed
5 a The frequency table is given here (note
difference between this one and the one
in the solutions provided)
fm
(m)
0 ≤ x ≤ 10
18
5
90
10 < x ≤ 20
14
15
210
20 < x ≤ 30
12
25
300
30 < x ≤ 40
9
35
315
40 < x ≤ 50
7
45
315
Σ f = 60
=
x
Σ fm = 1230
Σ fm 1230
= = 20.5
60
Σf
b
f
x
b The data is left or negatively skewed
Exercise 6C
1 a The number that occurs most often is
8
b The number that occurs most often is
4
c The number that occurs most often
is 13
d Each number occurs only once, so there
is no mode
e The numbers that occur most often are
2 and 4 . The data is bimodal
2 a The shoe size with the highest
frequency is 10
b The modal mark range is 60 < y ≤ 80
4
6
24
12 < x ≤ 24
0
18
0
24 < x ≤ 36
8
30
240
36 < x ≤ 48
15
42
630
48 < x ≤ 60
13
54
702
7
66
462
60 < x ≤ 72
Exercise 6D
1 a The mean is 3.65
b The mean is 12.8056
c The mean is 3.35
Σ fm = 2058
Σ f = 47
x
=
Σ fm 2058
=
= 43.7872 ≈ 43.8
47
Σf
c
x
f
Mid value
(m)
fm
1 ≤ x ≤ 1.5
4
1.25
5
1.5 < x ≤ 2
6
1.75
10.5
2 < x ≤ 2.5
7
2.25
15.75
2.5 < x ≤ 3
7
2.75
19.25
3 < x ≤ 3.5
5
3.25
16.25
Σf = 29
=
x
ii The modal range is 30 < x ≤ 35
ii Continuous data, since there is a
continuous scale of values on the xaxis.
fm
0 ≤ x ≤ 12
3 a i The mode is 3
b i Discrete data, since the scale on the
x-axis is given as discrete values.
Mid value
(m)
Σfm = 66.75
Σfm 66.75
=
= 2.30172 ≈ 2.30
29
Σf
3 Mean
0 × 6 + 1 × 5 + 2 × 4 + 3 × 7 + 4 × 10 + 5 × 4
=
6 + 5 + 4 + 7 + 10 + 4
=
94
= 2.6111 ≈ 2.6 cups of coffee
36
4 a Phil had
2 + 4 + 4 + 6 + 10 + 15 + 4 + 5
= 50 tomato plants
b The modal number of tomatoes per
plant was 8
c Mean =
3 × 2 + 4 × 4 + 5 × 4 + 6 × 6 + 7 × 10 + 8 × 15 + 9 × 4 + 10 × 5
50
=
177
= 7.08
25
© Oxford University Press 2019
2
Worked solutions
5 The mean number of fish caught per day
was
0 × 1 + 1 × 5 + 2 × 4 + 3 × 2 + 4 × 3 + 5 × 5 + 6 × 3 + 7 × 2 + 8 × 3 + 9 × 1 + 10 × 2
1+5 + 4 +2 +3+5+3+2 +3 +1+2
=
6 × 4 = 13 + x + y
141
≈ 4.55
31
6 The mean amount received per day is
5 × 6 + 15 × 14 + 25 × 15 + 35 × 8 + 45 × 2
6 + 14 + 15 + 8 + 2
=
197
= $21.89
9
7 a There are
7 + 12 + 10 + 9 + 7 + 6 + 6 + 3
= 60 families represented
b The data is right or positively skewed
c The mode of the data is 2 children per
family
d The mean number of children is
7 × 1 + 12 × 2 + 10 × 3 + 9 × 4 + 7 × 5 + 6 × 6 + 6 × 7 + 3 × 8
60
39
= 3.9
10
=
8 a There are 40 + 60 + 80 + 30 + 10
= 220 people in the village
b The modal class is 40 < a ≤ 60
c The mean age of the villagers is
40 × 10 + 60 × 30 + 80 × 50 + 30 × 70 + 10 × 90
220
9 In the set of numbers, each appears only
once, so therefore for 2 to be the mode,
a = 2 . Given that the mean is 5, we have
1 + 2 + 2 + 4 + 5 + 6 + b + 8 + 10 ,
5=
9
= 38 + b , =
5 ×=
9 38 + b , 45
b 45 − 38 ,
b=7
10 Given that the mean of the numbers is 23,
we have to find x
23 =
8 + x + 17 + (2 x + 3) + 45
23 × 5 = 73 + 3x
11= x + y
As x and y are positive integers less than
8 , the only possible solution is if x = 5
and y = 6 (or y = 5 and x = 6 ), so the
numbers are 2,3, 4, 4,5, 6
12 The mean mass of the students is
52 × 8 + 44 × 12 236
=
= 47.2 kg
20
5
Exercise 6E
1 a The median is the middle number, 18
b The middle number lies between 18
18 + 19 37
and 19 ,
= = 18.5
2
2
c Arranging the numbers in size order
1,2, 4,5,5 , the middle number is 4
d The numbers are already in size order
(reversed), so the middle number is the
median, the middle number lies
3+4 7
between 3 and 4 ,
= = 3.5
2
2
2, 4,5,5,6,7,7,8,8,10 . The middle
number is between 6 and 7 ,
6 + 7 13
= = 6.5
2
2
f
2,3,5,5,6,8 . The median is 5
2 Total number of days
= 2 + 4 + 3 + 7 + 11 + 18 + 6 + 2 = 53
th
5
th
 n + 1
 53 + 1 
th
Median
27
9
= 
=
=
=



 2 

2

3 a Mean
=
=
115
= 73 + 3x
2000000 × 1 + 1000 × 10 + 600 × 14 + 200 × 25
1 + 10 + 14 + 25
2023400
= $40468
50
b The mode is the most common number,
200
115 − 73 =
3x
42 = 3x
x =
24 − 13 =x + y
e
460
≈ 41.8
11
=
numbers, the mean is 4 , so
2+3+4+4+ x +y
4=
6
c The median number is the number in
42
,
3
th
x = 14
11 Starting with 2 and 3 , we know that as 4
is the mode, it must occur at least twice,
start by assuming that there are two 4 s
then x and y are the remaining two
 50 + 1 
th
the 
 = 25.5 position, that is
 2 
the number between 200 and 600 ,
200 + 600
= 400
2
© Oxford University Press 2019
3
Worked solutions
Exercise 6F
IQR
= Q3 − Q1 .
1 a The median is the middle number, 8
th
b
 n + 1
 11 + 1 
rd
=
=
=
Q1 
3=
7



 4 
 4 
th
 3 ( n + 1) 
=
c Q3 =


4


 3 (11 + 1) 


4


th
Q1 is the median of the lower half of the
numbers, 2
Q3 is the median of the upper half of the
IQR = Q3 − Q1 = 12 − 7 = 5
e Range = largest – smallest
numbers, it lies between 4 and 5 , 4.5
IQR = Q3 − Q1 = 4.5 − 2 = 2.5
= 15 – 3 = 12
2 In ascending order, the numbers are
2, 4, 5, 5, 5, 6, 6, 7, 8, 9,10, 15
a Median
th
6 a i
9+r
=
9.5
2
9+r
9.5 =
2
th
⇒
Q1 is the median of the lower half of
the numbers, 5
c
9.5 × 2= 9 + r
Q3 is the median of the upper half of
19 − 9 =
r
the numbers, it lies between 8 and 9 ,
8.5
d
IQR = Q3 − Q1 = 8.5 − 5 = 3.5
e
Range = largest − smallest = 15 − 2 = 13
r = 10
ii
Q3 is the median of the upper half of
the numbers, it is between s and 13
s + 13
13 =
2
3 Sorting the number of sit-ups into
ascending order,
2, 10, 10, 12, 14, 16, 16, 20, 25, 25,
21 × 3 = s + 13
26 − 13 =
s
28, 30, 37, 40, 45, 50
s = 13
a Median
th
 n + 1
= =

 2 
20 + 25
= =
2
b
b The value of t can be found as follows
th
 16 + 1 
8.5th
=

 2 
10 =
,
120 − 102 =
t
Q1 is the median of the lower half of
12 + 14
= 13
2
t = 18
Exercise 6G
Q3 is the median of the upper half of
the numbers,
5 + 6 + 7 + 7 + 9 + 9 + 10 + 10 + 13 + 13 + 13 + t
12
10 ×=
12 102 + t
22.5
the numbers,
th
 n + 1
 12 + 1 
Median =
6.5th
=

=

2


 2 
 n + 1
 12 + 1 
th
= 
=
=
=
6.5
6



 2 
 2 
b
th
 3 (11 + 1) 
th
=
9=
45


4


IQR = Q3 − Q1 = 45 − 25 = 20
5
th
th
 3 ( n + 1) 
Q
=
=


3
4


th
th
12
= 9=
d
th
 n + 1
 11 + 1 
rd
3=
25
Q1 
=
=
=



 4 
 4 
th
1
30 + 37
= 33.5
2
IQR = Q3 − Q1 = 33.5 − 13 = 20.5
c On 8 of the 16 days, Lincy did more
than 22.5 sit-ups.
d The ‘middle half’ of the number of situps Lincy did was between 13 and 33.5.
e On 4 of the 16 days, Lincy did more
than 33.5 sit-ups.
2 a The minimum time was 30.1
b The maximum time was 35
c The median time was 32.5
d The IQR was 33.1 − 31.9 =
1.2
4 Sorting the number of cars into ascending
order,
20, 20, 25, 30, 35, 35, 35, 35, 45, 45, 50.
© Oxford University Press 2019
4
Worked solutions
3 a
c
IQR = 13.6 − 8.2 = 5.4 minutes
d
k = 15.6 minutes
2 a The median is 40 minutes
b
b
IQR = 50 − 30 = 20
c 53
Q3 + 1.5(IQR)
3 a
= 2067.5 + 1.5 × 1272.5
= 3976.25 < 6000
so it is an outlier
b
c
d The outlier was removed because it
distorted the analysis
4 a
c The median is 73 km
d
b The morning exam
c This means that there is a bigger
difference between the 25% and the
75% of the scores
IQR = 82 − 60 = 22 km
e 3 cars
4 a
5 a
b
b
c The data is right or positively skewed
6 1A, 2C, 3B
Exercise 6H
1 a The longest time taken was 18 minutes
b The median is 11 minutes
c The median is 420
d
IQR = 510 − 300 = 210
© Oxford University Press 2019
5
Worked solutions
e 80 students
2 a
5 1C, 2B, 3A
=
Exercise 6I
1 a =
x
= 4.375
2
b
x =
σ2 =
=
σ
b
Σx
n
1 × 5 + 3 × 12 + 5 × 16 + 7 × 22 + 9 × 27 + 11 × 30 + 13 × 18
5 + 12 + 16 + 22 + 27 + 30 + 18
2
σ2 =
2 + 5 + 8 + 7 + 1 + 3 + 9 + 11 + 4 + 2
10
= 5.2
= 80.385 − 69.273 = 11.111
2
Σx 2  Σx 
σ =
−

n
 n 
σ
=
22 + 52 + 82 + 72 + 12 + 32 + 92 + 112 + 42 + 22
− 5.22
10
c
=37.4 − 37.04 ≈ 10.4
x =
=
σ 2 ≈ 3.22
σ 2 3.33
=
Σx
n
5 × 18 + 15 × 14 + 25 × 13 + 35 × 11 + 45 × 6
18 + 14 + 13 + 11 + 6
≈ 20.6
Σx −4 + −2 + 0 + 3 + −5
=
= −1.6
n
5
2
σ2 =
2
Σx 2  Σx 
−

n
 n 
(−4)2 + (−2)2 + 02 + 32 + (−5)2
− 5.22
5
= 10.8 − 2.56 = 8.24
σ2 =
=
σ
d =
x
= 602.419 − 426.223 = 176.197
=
σ
Σx 1 + 2 + 3 + 4 + 5
=
= 3
n
5
x =
2
2
σ2 =
=
12 × 2 + 22 × 2 + 32 × 4 + 42 × 10 + 52 × 12 + 62 × 2 + 72 × 2 + 182 × 1
− 4.6292
2 + 2 + 4 + 10 + 12 + 2 + 2 + 1
=
σ
b
x =
2
Σx 2  Σx 
−

n
 n 
12 + 22 + 32 + 42 + 52 + 5002
=
− 85.8332
6
= 41675.833 − 7367.36 ≈ 34308
σ 2 ≈ 185.2
Σx 2  Σx 
−

n
 n 
= 28.571 − 21.424 = 7.148
σ 2 ≈ 1.41
Σx 1 + 2 + 3 + 4 + 5 + 500
e =
x
=
≈ 85.8
6
n
=
σ
Σx
n
= 4.63
Σx 2  Σx 
−

n
 n 
12 + 22 + 32 + 42 + 52
=
− 32
5
= 11 − 9 = 2
σ2 =
σ 2 ≈ 13.3
1 × 2 + 2 × 2 + 3 × 4 + 4 × 10 + 5 × 12 + 6 × 2 + 7 × 2 + 18 × 1
2 + 2 + 4 + 10 + 12 + 2 + 2 + 1
=
σ2 =
=
σ
Σx 2  Σx 
−

n
 n 
52 × 18 + 152 × 14 + 252 × 13 + 352 × 11 + 452 × 6
− 20.6452
18 + 14 + 13 + 11 + 6
=
3 a
σ 2 ≈ 2.87
Σx 2  Σx 
−

n
 n 
12 × 5 + 32 × 12 + 52 × 16 + 72 × 22 + 92 × 27 + 112 × 30 + 132 × 18
− 8.3232
5 + 12 + 16 + 22 + 27 + 30 + 18
=
2
x =
σ 2 ≈ 1.33
≈ 8.32
Σx
n
=
c
x =
=
σ 2 ≈ 2.12
σ
=
1×3 + 2× 8 + 3× 6 + 4 × 6 + 5×7
= 3.2
3+8+6+6+7
Σx 2  Σx 
−

n
 n 
12 × 3 + 22 × 8 + 32 × 6 + 42 × 6 + 52 × 7
=
− 3.22
3+8+6+6+7
= 12 − 10.24 = 1.76
Σx 2  Σx 
−

n
 n 
2
2
2
4 + 6 + 7 + 72 + 52 + 12 + 22 + 32
− 4.3752
8
=23.625 − 19.1406 ≈ 4.48
σ2 =
Σx
n
2
Σx 4 + 6 + 7 + 7 + 5 + 1 + 2 + 3
=
n
8
σ
=
x =
=
=
σ 2 2.67
Σx
n
1 × 2 + 2 × 2 + 3 × 4 + 4 × 10 + 5 × 12 + 6 × 2 + 7 × 2
2 + 2 + 4 + 10 + 12 + 2 + 2
= 4.24
2
σ2 =
=
Σx 2  Σx 
−

n
 n 
12 × 2 + 22 × 2 + 32 × 4 + 42 × 10 + 52 × 12 + 62 × 2 + 72 × 2
− 4.2352
2 + 2 + 4 + 10 + 12 + 2 + 2
© Oxford University Press 2019
6
Worked solutions
= 19.882 − 17.938 = 1.945
=
σ
4
x =
2 a
=
σ
1.40
2
Σx
n
7 + 9 + 3 + 0 + 1 + 8 + 6 + 4 + 10 + 5 + 5
= 5.2727
11
=
2
σ2 =
93 + 86.2 + 80 + 64 + 60.6 + 50 + 50 + 47.3 + 46.6 + 46
10
=
= 62.37
Σx 2  Σx 
−

n
 n 
= 36.909 − 27.802 = 9.11
=
σ
932 + 86.22 + 802 + 642 + 60.62 + 502 + 502 + 47.32 + 46.62 + 462
− 62.372
10
b
= 4177.27 − 3890.02 = 287.248
=
σ
5
=
=
Σx
n
6 + 8 + 6 +3+1 =
24 months
Σx
n
15 × 6 + 25 × 8 + 35 × 6 + 45 × 3 + 55 × 1
6 + 8 + 6 +3+1
= 28.75
=
2
d σ2 =
=
Σx 2  Σx 
−

n
 n 
212 + 272 + 92 + 02 + 92 + 242 + 182 + 122 + 302 + 152 + 152
− 15.81822
11
σ
=
σ 2 9.054
=
c The mean is multiplied by 3, and since
the variance is multiplied by 9,
standard deviation (which is square
root of variance) is multiplied by 3.
3 mean
= 17.2 + 4
= 21.2
median = 17 + 4 = 21
standard deviation = 0.5
4 The mean, median and standard deviation
will double
5 The new variance is 92 x = 81x
Chapter Review
1 a
mode = 1
th
152 × 6 + 252 × 8 + 352 × 6 + 452 × 3 + 552 × 1
− 28.752
6 + 8 + 6 +3 +1
=
950 − 826.563 =
123.438
=
σ
Σx 2  Σx 
−

n
 n 
= 332.182 − 250.215 ; 82.0
=
σ 2 91.7
x =
21 + 27 + 9 + 0 + 9 + 24 + 18 + 12 + 30 + 15 + 15
11
σ2 =
2
2
b The modal range is 20 to 30 hours
c
Σx
n
2
Σx
 Σx 
−

n
 n 
1502 × 3 + 2502 × 6 + 3502 × 11 + 4502 × 5
=
− 3222
3 + 6 + 11 + 5
= 112100 − 103684 = 8416
=
σ
=
σ 2 3.02
= 15.8182
150 × 3 + 250 × 6 + 350 × 11 + 450 × 5
= 322
3 + 6 + 11 + =
5
σ2 =
6 a
x =
=
σ 2 16.9
x =
Σx 2  Σx 
−

n
 n 
72 + 92 + 32 + 02 + 12 + 82 + 62 + 42 + 102 + 52 + 52
− 5.27272
11
=
2
σ2 =
Σx
n
x =
=
σ 2 11.1
b
4+2
 10 + 1 
median = 
=
5.
=
5th
=3

2
 2 
c
mean =
=
Exercise 6J
2 + 8 +1+5+ 0 + 4 + 4 +1+1+ 6
10
16
= 3.2
5
d range = 8 − 0 = 8
1+3+5+5+8
1 a mean = 4.4
=
5
median = 5
mode = 5
2
9 × 420 + 3 × 740
= 500
12
3 a
5 + 7 + 9 + 9 + 12
= 8.4
5
median = 9
mode = 9
b
b mean
=
c
c Adds 4 to mean, median and mode
mode = 3
 50 + 1 
median = 

 2 
mean =
=
th
= 25.5th = 3
0 × 4 + 1 × 8 + 2 × 10 + 3 × 20 + 4 × 4 + 5 × 3 + 6 × 1
50
5
= 2.5
2
© Oxford University Press 2019
7
Worked solutions
4 The mean will increase by 4 and the
standard deviation will stay the same;
mean = 21.9 , standard deviation = 1.1
5 a
b
mean =
736
= 32
23
11 mode = 4
mean
2 × 3 + 3 × 4 + 4 × 10 + 5 × 3 + 6 × 2 + 7 × 2
=
3 + 4 + 10 + 3 + 2 + 2
= 4.125
 24 + 1 
median = 

 2 
736 + 24 + 15
mean =
= 31
23 + 2
6 a The mean will increase by 10 and the
standard deviation will stay the same;
mean = 58 , standard deviation = 5
b The mean will increase by a factor of 10
and the variance will increase by a
factor of 102; mean = 480 ,
σ2 =
12 a
 53 + 1 
median = 

 2 
b 60
50 = c − 40 ⇒ c = 50 + 40 = 90
d
IQR = 24 = 74 − d ⇒ d = 74 − 24 = 50
8 a 800 students
b 65 marks
IQR = 75 − 55 = 20
d 100 students
e No, because there are 100 students
who scored more than 80 marks, this is
not 10%
= 166.627 − 118.728 = 47.8996
b Because we are using the midpoint of
each range, as opposed to the actual
original data, which assumes that the
number of items is equally spread
throughout the class interval.
13 Given that the mean number of watches is
2.5, we have to find k
2.5 =
k = 40
105 − 101 =
(3 − 2.5)k
4 = 0.5k
15 + 12 + 22 + 30 + 25 + 7 + 19 + 33 + 19 + 41 + 53 + 12 + 3 + 8 + 6 + 17
16
= 20.125
k =
2
=
Σx 2  Σx 
−

n
 n 
14 a 80 bats
= 579.375 − 405.016 = 174.359
=
σ
13.2
2
b Write the numbers in size order;
3, 6, 7, 8, 12, 12, 15, 17, 19, 19,
22, 25, 30, 33, 41, 53
then find Q1 as the median of the first
half of the list,
 8 + 1
=
Q1 

 2 
th
==
4.5th
8 + 12
= 10 and
2
find Q2 as the median of the second
b 50 grams
5
c
=
0.0625
= 6.25%
80
d
a = 10 , c = 80 − 75 = 5
e
b = 75 − 55 = 20
f
x =
8 + 1
25 + 30

th
Q1 =
=
27.5
8 +
 = 12.5 =
2 
2

, so IQR= 27.5 − 10= 17.5
Σx
n
15 × 10 + 45 × 45 + 75 × 20 + 105 × 5
10 + 45 + 20 + 5
= 52.5
=
2
σ2 =
half of the list,
th
4
0.5
k =8
152 + 122 + 222 + 302 + 252 + 72 + 192 + 332 + 192 + 412 + 532 + 122 + 32 + 82 + 62 + 172
16
− 20.1252
=
σ
0 × 11 + 1 × 7 + 2 × 6 + 3 × k + 4 × 8 + 5 × 10
,
11 + 7 + 6 + k + 8 + 10
2.5 × (42 + k ) = 101 + 3k
105 + 2.5k =101 + 3k
Σx
n
σ2 =
= 27th = 12.5
σ = 6.92
9 1B, 2C, 3A
=
th
2.52 × 15 + 7.52 × 11 + 12.52 × 9 + 17.52 × 12 + 22.52 × 6
− 10.8962
15 + 1 1 + 9 + 1 2 + 6
σ2 =
c
x =
2.5 × 15 + 7.5 × 11 + 12.5 × 9 + 17.5 × 12 + 22.5 × 6
15 + 11 + 9 + 12 + 6
mean =
≈ 10.9
7 a 40
10 a
22 × 3 + 32 × 4 + 42 × 10 + 52 × 3 + 62 × 2 + 72 × 2
− 4.1252
3 + 4 + 10 + 3 + 2 + 2
=18.875 − 17.0156 =1.8593
variance = 5 × 100 =
2500
f
= 12.5th = 4
σ = 1.36
2
c
th
=
Σx 2  Σx 
−

n
 n 
152 × 10 + 452 × 45 + 752 × 20 + 1052 × 5
− 52.52
10 + 45 + 20 + 5
=3262.5 − 2756.25 =506.25
=
σ
© Oxford University Press 2019
=
σ 2 22.5
8
Worked solutions
15 a
50 =3 + 11 + 16 + m + 8
⇒ m = 50 − 38 = 12
n = 14 + 16 = 30
b
x =
=
Σx
n
20 a
10 × 3 + 15 × 11 + 20 × 16 + 25 × 12 + 30 × 8
3 + 11 + 16 + 12 + 8
= 21.1
2
c
σ2 =
Σx 2  Σx 
−

n
 n 
b
102 × 3 + 152 × 11 + 202 × 16 + 252 × 12 + 302 × 8
=
− 21.12
3 + 11 + 16 + 12 + 8
=477.5 − 445.21 =32.3
21 a
b
c
16 a Discrete
b
x =
=
Σx
n
1 × 41 + 2 × 60 + 3 × 52 + 4 × 32 + 5 × 15 + 6 × 8
41 + 60 + 52 + 32 + 15 + 8
≈ 2.73
d
2
c
Σx 2  Σx 
σ2 =
−

n
 n 
=
12 × 41 + 22 × 60 + 32 × 52 + 42 × 32 + 52 × 15 + 62 × 8
− 2.7312
41 + 60 + 52 + 32 + 15 + 8
=
9.25 − 7.4571 =
1.7929
=
σ
=
σ 2 1.34
d 1 standard deviation above the mean is
2.731 + 1.339 =
4.07 , so 15 + 8 =
23
families have more than one standard
deviation above the mean mobile
devices
17 a
b
c
d
18 a
b
19 a
b
c
Discrete
A1
Continuous
A1
Continuous
A1
Discrete
A1
As the mode is 5 there must be at least
another 5.
R1
So we have 1, 3, 5, 5, 6 with another
number to be placed in order R1
The median will be the average of the
R1
3rd and 4th pieces of data.
For this to be 4.5 the missing piece of
data must be a 4.
Thus a=5, b=4
A1 A1
1 + 3 + 4 + 5 + 5 + 6 24
=
x
= = 4
6
6
M1 A1
An outlier is further than 1.5 times the
IQR below the lower quartile or above
the upper quartile.
A1
i mode = 8
A1
ii median = 7
A1
iii lower quartile = 3
A1
iv upper quartile = 9
A1
IQR = 6
22 a
b
c
d
e
1.5 × IQR =9
19 – 9 = 10
19 is the (only) outlier
M1
A1
∑x
A1
10
=70 ⇒
∑ x =700
Let the new student’s mass be s.
∑ x + s = 72
M1
11
A1
700 + s =
792
So s = 92kg
A1
IQR = 10
A1
M1
76 + 1.5 × IQR = 76 + 15 = 91
So new student’s mass of 92 is an
outlier
R1
200
A1
35
A1
Using mid-points 5, 15, 25… as
estimates for each interval
M1
i Estimate for mean is 22.25 A2
ii Estimate for standard deviation is
11.6 (3sf)
A2
Median is approximately the 100th
piece of data which lies in the interval
A1
20 < h ≤ 30 .
Will be 15 pieces of data into this
interval
15
Estimate is 20 +
× 10 =
23 M1A1
50
Discrete
A1
5
A1
i 4.79 (3sf)
A2
ii 1.62 (3sf)
A2
i 5
A1
ii 4
A1
iii 5.5
A1
A1 general shape
A1 median
A1 quartiles
f
IQR=1.5
(A1)
1.5 × 1.5 =
2.25
5.5 + 2.25 = 7.75
4 - 2.25 = 1.75
M1
So the 2 (unhappy) candidates with
grade 1 are outliers
A1
© Oxford University Press 2019
9
Worked solutions
c
23 a
x
Frequency
Cumulative
frequency
0
10
10
1
7
17
2
11
28
3
13
41
4
15
56
5
15
71
6
12
83
7
10
93
8
4
97
9
2
99
10
1
100
A4 for
A3 for 4 or
A2 for 2 or
A1 for
b i
6
5
3
1
correct
correct
correct
correct
4
ii 2
iii 6
c i
ii
A1A1A1
4.05 (3sf)
(2.4140...)
2
= 5.83 (3sf) A1(M1)A1
d No. It is bimodal at x= 4 and 5. 24
A1R1
24 a
b
A1
80 < w ≤ 90
mass
cumulative
frequency
40 < w ≤ 50
5
50 < w ≤ 60
20
60 < w ≤ 70
45
70 < w ≤ 80
75
80 < w ≤ 90
125
90 < w ≤ 100
160
100 < w ≤ 110
185
110 < w ≤ 120
200
A1A1scales A3 points and curve
d i 85
ii 73
iii 97
25 a i 7.5
ii 6.125
b i 6
ii 6.9
c Sally’s had the greater median
d Rob’s had the greater mean
26 a
A1A1A1
M1 lines
A1A2
A1A2
R1
R1
A2 numbers A1 labelling
chart A1; scaleA2
b i 4
ii 4
iii 4
A1A1A1
c The values of the median and the mean
are the same due to the symmetry of
the bar chart.
A1R1
© Oxford University Press 2019
10
Worked solutions
7
Modelling relationships between two data
sets: statistics for bivariate data
4 a
Skills check
1 a 1296
b
64
c
343
2 a
b
4
c
3
b
y= 1 − 2 x
5
3 a =
y 4x − 2
d
3
Exercise 7A
1 a There is a strong, positive, linear
correlation
b There is a weak, negative, linear
correlation
c There is a strong, negative, linear
correlation
d There is a weak, positive, linear
correlation
e There is no correlation
2 i
a Positive
b There is a strong, positive, linear
correlation
5 a
b Linear
c Strong
ii a Positive
b Linear
c Moderate
iii a Positive
b Linear
c Weak
iv a No correlation b Non linear
c Zero
v a Negative
b Linear
c Strong
vi a Negative
b Non linear
c Strong
b There is a strong, positive, linear
correlation
3 a
c As the kitten gets older, it gets heavier
Exercise 7B
1 a
b There is a strong, negative, linear
correlation
c As the maximum depth increases the
time at that depth decreases
x
y
x2
y2
xy
20
250
400
62500
5000
24
300
576
90000
7200
30
350
900
122500
10500
40
360
1600
129600
14400
50
480
2500
230400
24000
75
580
5625
336400
43500
80
750
6400
562500
60000
90
840
8100
705600
75600
100
900
10000
810000
90000
120
1000
14400
1000000
Σx =
629
Σy =
5810
© Oxford University Press 2019
2
Σx
= 50501
2
Σy
= 4049500
Worked solutions
120000
Σxy
= 450200
1
Worked solutions
( Σx )
2
Sxx =
Σx 2 −
Sxx
Sxy
r =
n
6292 109369
= 50501 −
=
10
10
(S
r =
c The price of the motorbike can never
fall below 0.
n
629 × 5810
=
450200 −
84751
Sxy =
10
Sxy
r =
Sxx Syy
(
)
84751
= 0.987
109369
× 67 3 8 90
10
=
r
b There is a strong positive correlation
c As the floor area increases, house price
increases
2 a
x
y
x2
y2
xy
1
40000
1
1600000000
40000
2
36500
4
1332250000
73000
3
31000
9
961000000
93000
4
26658
16
710648954
106632
5
24250
25
588062500
121250
6
19540
36
381811600
117240
7
19100
49
364810000
133700
8
18750
64
351562500
150000
9
15430
81
238084900
138870
12600
100
158760000
10
Σx =
55
Σy =
243828
126000
Σxy =
1099692
2
2
Σx =
385
Σy =
6686990464
( Σx )
2
Sxx =
Σx 2 −
Sxx
n
=
Syy 6686990464 −
x
y
x2
y2
xy
148
34
21904
1156
5032
153
38
23409
1444
5814
165
42
27225
1764
6930
142
36
20164
1296
5112
155
42
24025
1764
6510
141
32
19881
1024
4512
171
40
29241
1600
6840
154
34
23716
1156
5236
170
40
28900
1600
6800
168
38
28224
1444
6384
Σx
= 1567
Σy
= 376
2438282
10
2
Σx
= 246689
2
Σy
= 14248
Σxy
= 59170
( Σx )
2
Σx 2 −
Sxx =
n
Sxx = 246689 −
15672 11401
=
10
10
( Σy )
2
Σy 2 −
Syy =
n
3762 552
=
Syy = 14248 −
10
5
Σx ) ( Σy )
(
Σxy −
Sxy =
n
1567 × 376 1254
=
Sxy =
59170 −
10
5
Sxy
r =
Sxx Syy
=
r
( Σy )
=
Σy 2 −
2
Syy
3 a
(
n
552 165
= 385 −
=
10
2
= −0.976
b There is a strong negative correlation
58102
= 673890
10
( Σx ) ( Σy )
Syy= 4049500 −
Σxy −
Sxy =
)
165 3708905528
×
2
5
( Σy )
n
Syy
−241362
2
Σy 2 −
Syy =
xx
)
1254
5
=
0.707
11401 552
×
10
5
b There is a moderate positive
correlation
3708905528
5
( Σx ) ( Σy )
Sxy =
Σxy −
n
55 × 243828
=
Sxy 1099692 −
10
= −241362
=
© Oxford University Press 2019
2
Worked solutions
4 a
( Σx )
−
2
x
y
x
y
6
78
36
6084
468
4
80
16
6400
320
2
7
86
49
7396
602
5
88
25
7744
440
1
66
1
4356
66
2
70
4
4900
140
4
78
16
6084
312
6
95
36
9025
570
8
97
64
9409
776
4
76
16
5776
Σx
= 47
Σy
= 814
2
Σx
= 263
Sxx =
Σx
xy
2
2
Σy
= 67174
Sxx
n
1062
= 335.5
Syy = 1740 −
8
( Σx ) ( Σy )
Σxy −
Sxy =
n
24.8 × 106
=
−18.7
309.9 −
Sxy =
8
Sxy
r =
Sxx Syy
304
(
( Σx )
r =
5 a
x
y
x
y
3.9
10
15.21
100
39
2.7
14
7.29
196
37.8
3.8
5
14.44
25
19
2.4
8
5.76
64
19.2
1.7
2
xy
24
2.89
576
40.8
2.6
17
6.76
289
44.2
4.0
21
16
441
84
7
13.69
49
2
2
3.7
Σy
= 24.8
= 106
Σx
= 82.04
Σy
= 1740
y
x2
y2
xy
52
60
2704
3600
3120
60
68
3600
4624
4080
62
66
3844
4356
4092
69
4225
4761
4485
75
4624
5625
5100
76
82
5776
6724
6232
77
83
5929
6889
6391
78
84
6084
7056
6552
80
88
6400
7744
7040
84
90
7056
8100
7560
85
93
7225
8649
7905
95
92
9025
8464
Σx
= 882
Σy
= 950
b There is a strong positive correlation
c Yes
x
65
861
5
=
0.878
421 4572
×
10
5
Σx
= −0.449
68
)
2
5.16 × 335.5
6 a
n
8142
4572
=
Syy = 67174 −
10
5
Σx ) ( Σy )
(
Σxy −
Sxy =
n
47 × 814 861
=
3998 −
Sxy =
10
5
Sxy
r =
Sxx Syy
=
r
−18.7
c Yes, because the correlation is only
weak
( Σy )
(
)
b There is a weak negative correlation
2
Σy 2 −
Syy =
( Σy )
2
Σxy
= 3998
n
472
421
= 263 −
=
10
10
Sxx
n
24.82
= 82.04 −
= 5.16
8
Syy =
Σy 2 −
2
Σx 2 −
Sxx =
2
2
Σx
= 66492
2
Σy
= 76592
8740
Σxy
= 71297
25.9
Σxy
= 309.9
© Oxford University Press 2019
3
Worked solutions
b
( Σx )
2
Sxx =
Σx 2 −
Sxx
n
8822
= 66492 −
= 1665
12
( Σy )
2
Syy =
Σy 2 −
n
9502
= 1383.67
Syy = 76592 −
12
( Σx ) ( Σy )
Σxy −
Sxy =
n
882 × 950
71297 −
1472
=
Sxy =
12
Sxy
r =
Sxx Syy
(
=
r
)
1472
= 0.970
1665 × 1383.67
b There is a strong positive correlation
c More practice questions will likely
increase the overall grade
c
2.7825 − 2
× (x − 5.75)
5.75 − 2
=
y 0.209 x + 1.583
y − 2.7825
=
d
y = 0.209 × 9 + 1.583 =
3.46 m
e
y = 0.209 × 120 + 1.583 =
26.6 m
f
Not reliable as it is known that giraffes
only grow to 6 m
3 a
mean =
36 + 55 + 42 + 35 + 58 + 65
6
b
mean =
3 + 6 + 10 + 12 + 15 + 20
6
mean =
60 + 45 + 32 + 28 + 18 + 15
6
= 33, 000 Rupees
= 48.5
b
mean =
= 11 km
Exercise 7C
1 a
y2 − y1
× (x − x1 )
x2 − x1
y −=
y1
17 + 30 + 23 + 11 + 44 + 51
6
c
= 29.333
c
2 a
mean age
1 + 2 + 3 + 4 + 6 + 8 + 10 + 12
=
= 5.75
8
33 − 50
× (x − 11)
11 − 5
y =
−2.833x + 64.1667
y −=
33
mean height
=
y2 − y1
× (x − x1 )
x2 − x1
d y − y1
=
1.78 + 1.98 + 2.17 + 2.40 + 2.82 + 3.26 + 3.71 + 4.14
8
e
y = −2.833 × 8 + 64.1667 =
41.5
⇒ 41,500 Rupees
= 2.7825
f
50 =
−2.833x + 64.1667
⇒x =
© Oxford University Press 2019
50 − 64.1667
= 5 km
−2.833
4
Worked solutions
b
g y −2.833 × 30 + 64.1667
=
= −20.8233
y= a + bx , where
Sxy
b=
⇒ −20, 823 Rupees
−241.75
= −7.084 and
34.125
=
Sxx
1367
92.5
+ 7.084 ×
= 202.227
10
10
so=
y 202 − 7.084 x
Not suitable to extrapolate, negative
rent is not correct
a = y − bx=
Exercise 7D
1 A student who plays no sport will spend
35 hours on homework and each day
spent playing sport reduces the hours of
homework by 30 minutes
c y 202.227 − 7.084 × 7.5 =
=
149 kmh−1
d As the time taken to accelerate from 0
to 90 increases by 1 second, the top
speed decreases by 7.08
2 A person who has no friends who are
criminals has 1 conviction and adding one
extra criminal friends leads to 6 extra
convictions
5 a
x
3 A brand new speaker is worth $300 and
as it gets older, its value decreases by
$40 per year
4 a
y
x2
y2
xy
80
74
6400
5476
5920
73
62
5329
3844
4526
95
93
9025
8649
8835
84
75
7056
5625
6300
67
73
4489
5329
4891
81
7744
6561
7128
x
y
x2
y2
xy
88
6
157
36
24649
942
69
58
4761
3364
4002
92
90
8464
8100
8280
84
8100
7056
7
155
49
24025
1085
8
147
64
21609
1176
8.5
142
72.25
20164
1207
9
138
81
19044
1242
9.5
132
90.25
17424
1254
10
134
100
17956
1340
11
127
121
16129
1397
11.5
120
132.25
14400
1380
12
115
144
13225
1380
Σx
= 92.5
Σy
= 1367
Σx 2
= 889.75
Σy 2
= 188625
90
Σx
= 738
Σx
Sxx =
Sxx
(
n
92.52
= 34.125
10
( Σy )
r =
−241.75
34.125 × 1756.1
= −0.988
( Σx )
−
n
7382
= 61368 −
= 852
9
(
n
)
Σy
= 54004
7560
Σxy
= 57442
( Σy )
n
6902
= 1104
Syy = 54004 −
9
( Σx ) ( Σy )
Σxy −
Sxy =
n
738 × 690
=
Sxy =
57442 −
862
9
Sxy
r =
Sxx Syy
Σxy
= 12403
13672
= 1756.1
Syy = 188625 −
10
( Σx ) ( Σy )
Σxy −
Sxy =
n
92.5 × 1367
=
−241.75
12403 −
Sxy =
10
Sxy
r =
Sxx Syy
Σx
= 61368
2
2
2
Σy 2 −
Syy =
2
Σy 2 −
Syy =
( Σx )
Sxx = 889.75 −
2
2
2
Σx 2 −
Sxx =
Σy
= 690
=
r
b
)
86 2
= 0.889
852 × 1104
y= a + bx , where
Sxy
862
= 1.012 and
b ==
Sxx 852
690
738
− 1.012 ×
= −6.30
9
9
so y 1.01x − 6.30
=
a = y − bx =
c y 1.012 × 75 − 6.30 = 69.6
=
© Oxford University Press 2019
5
Worked solutions
c
6 a
x
y
x2
y2
xy
x
y
x2
y2
xy
35
13
1225
169
455
0.5
30
0.25
900
15
38
18
1444
324
684
1
28
1
784
28
42
27
1764
729
1134
1.5
14
2.25
196
21
45
28
2025
784
1260
2
18
4
324
36
47
36
2209
1296
1692
2.5
10
6.25
100
25
48
34
2304
1156
1632
3
7
9
49
21
50
40
2500
1600
2000
4
1
16
1
Σx
= 305
Σy
= 196
Σxy
= 8857
Σx
= 14.5
Σy
= 108
2
2
2
Σx
= 13471
2
Σy
= 6058
( Σx )
2
Sxx =
Σx 2 −
Sxx
Sxx
( Σy )
2
=
r
Syy
Sxx
)
r =
317
=
= 1.744 and
181.714
a = −8.46 and b = 33.0
( Σx ) ( Σy )
)
−73.714
8.714 × 687.714
= −0.952
d Strong, negative
8 a
x
y
x2
y2
xy
28
66
784
4356
1848
33
70
1089
4900
2310
35
85
1225
7225
2975
42
94
1764
8836
3948
40
96
1600
9216
3840
38
80
1444
6400
3040
Σx
= 216
Σy
= 491
y = 1.74 × 40 − 48.0 =
21.6 cm
d For every cm that the cat grows in
length, it grows 1.74 cm in height
7 a
n
1082
= 2354 −
= 687.714
7
(
196
305
a = y − bx = − 1.744 ×
=
−48.0321
7
7
so y 1.74 x − 48.0
=
c
( Σy )
−
n
14.5 × 108
=
−73.714
Sxy =
150 −
7
Sxy
r =
Sxx Syy
y= a + bx , where
b=
2
Σxy −
Sxy =
317
= 0.985
181.714 × 570
Sxy
n
14.52
= 38.75 −
= 8.714
7
Syy =
Σy
There is a strong, positive correlation
b
( Σx )
2
n
1962
= 570
Syy = 6058 −
7
( Σx ) ( Σy )
Σxy −
Sxy =
n
305 × 196
=
8857 −
317
Sxy =
7
Sxy
r =
Sxx Syy
(
Σy
= 2354
2
Sxx =
Σx 2 −
n
3052
= 13471 −
= 181.714
7
Syy =
Σy 2 −
Σx
= 38.75
4
Σxy
= 150
2
Σx
= 7906
2
Σy
= 40933
Σxy
= 17961
b y 32.95 − 8.46 × 3.5 =
=
3.34
⇒ 3 mudbugs
© Oxford University Press 2019
6
Worked solutions
y= a + bx , where
( Σx )
2
Σx 2 −
Sxx =
S
15076.7
=
b = xy
= 6.416 and
Sxx
2350
n
2162
= 7906 −
= 130
6
Sxx
3040
453
− 6.416 ×
= 14.858
9
9
so y 6.416 x + 14.858
=
a = y − bx=
( Σy )
=
Σy 2 −
2
Syy
n
4912
= 752.833
Syy = 40933 −
6
( Σx ) ( Σy )
Σxy −
Sxy =
n
216 × 491
=
17961 −
285
Sxy =
6
Sxy
r =
Sxx Syy
(
=
r
b i
ii When no pizzas are made, there is
a cost of $14.86
c y 6.416 × 60 + 14.858 = $399.82
=
d i
Not reliable as 5000 is not close to
the domain used
ii=
100 6.416 x + 14.858
)
100 − 14.858
6.416
x = 13.27
x =
285
= 0.911
130 × 752.833
13 pizzas
Sxy 285
b ==
= 2.19 and
Sxx 130
b
Each additional pizza costs $6.42
10 a
491
216
a = y − bx =
− 2.192 ×
= 2.92
6
6
c If a student scores 1 mark better in
the IB diploma then they will do 2.19%
better in their first year at university
d y 2.19 x + 2.92
=
= 2.19 × 30 + 2.92 =
68.7%
9 a
x
y
x2
y2
xy
1
115
1
13225
115
2
110
4
12100
220
3
92
9
8464
276
4
89
16
7921
356
5
80
25
6400
400
8
63
64
3969
504
9
59
81
3481
531
x
y
x2
y2
xy
10
54
100
2916
540
25
200
625
40000
5000
40
260
1600
67600
10400
Σx
= 42
Σy
= 662
Σx 2
= 300
Σy 2
= 58476
Σxy
= 2942
65
350
4225
122500
22750
53
360
2809
129600
19080
46
260
2116
67600
11960
30
250
900
62500
7500
50
310
2500
96100
15500
74
600
5476
360000
44400
450
4900
202500
70
Σx
= 453
Σy
= 3040
Σx
2
Σy
= 25151
2
= 1148400
Sxx
Sxx
n
6622
= 3695.5
Syy = 58476 −
8
( Σx ) ( Σy )
Σxy −
Sxy =
n
42 × 662
=
−533.5
2942 −
Sxy =
8
Sxy
r =
Sxx Syy
31500
Σxy
= 168090
( Σx )
(
( Σy )
−
2
Syy =
Σy
Sxy =
Σxy −
Sxy
r =
n
2
3040
= 121256
9
( Σx ) ( Σy )
Syy= 1148100 −
n
453 × 3040
168090 −
15076.7
=
=
9
( Σy )
2
n
4532
= 25151 −
= 2350
9
2
n
422
= 300 −
= 79.5
8
Σy 2 −
Syy =
2
Sxx =
Σx 2 −
( Σx )
2
Σx 2 −
Sxx =
b
b=
Sxy
Sxx
)
−533.5
79.5 × 3695.5
=
−533.5
= −6.71 and
79.5
a = y − bx =
© Oxford University Press 2019
= −0.984
662
42
+ 6.711 ×
= 117.98
8
8
7
Worked solutions
c
y = 117.98 − 6.711 × 6 =
77.717
= ¥78000
Σy
= 35
= 169
1 a
x
y
x2
y2
xy
12
45
144
2025
540
15
44
225
1936
660
18
45
324
2025
810
18
42
324
1764
756
22
40
484
1600
880
25
34
625
1156
850
30
26
900
676
Σx
= 140
Σy
= 276
2
Σx
= 3026
Syy
Sxy
Sxy
Sxx =
Σx
780
Sxy
Σy
= 11182
3
( Σy )
n
2762
= 11182 −
= 299.714
7
( Σx ) ( Σy )
=
Σxy −
n
140 × 276
=
=
−244
5276 −
7
Sxx
=
=
y 61.0 − 1.08 × 20= 39.4 ⇒ 39 tickets
x= a + by , where
b=
S xy
Syy
=
y
x2
y2
xy
90
87
8100
7569
7830
88
57
7744
3249
5061
65
52
4225
2704
3380
92
76
8464
5776
6992
50
30
2500
900
1500
67
4489
4489
4489
100
96
10000
9216
9600
100
74
10000
5476
7400
73
65
5329
4225
4745
90
87
8100
7569
7830
−244
= −1.0797 and
226
276
140
+ 1.0797 ×
= 61.0226
7
7
so
=
y 61.0 − 1.08 x ,
x
67
a = y − bx =
b
n
1692
= 6609 −
= 1848.83
6
( Σx ) ( Σy )
=
Σxy −
n
35 × 169
1279 −
293.167
=
=
6
b=
x 1.35 + 0.159 × 50 = 9.3 mins
n
1402
= 3026 −
= 226
7
Sxy
( Σy )
35
169
− 0.159 ×
= 1.35 so
6
6
=
x 1.35 + 0.159y
y= a + bx , where
b=
= 1279
a = x − by =
Σxy
= 5276
2
2
Sxy
Σxy
S
293.167
=
b = xy
= 0.159 and
Syy
1848.83
( Σx )
−
Syy =
Σy 2 −
Syy
Σy 2
= 6609
x= a + by , where
2
2
Σx 2
= 251
2
Syy =
Σy 2 −
Exercise 7E
Sxx
Σx
83
78
6889
6084
6474
94
89
8836
7921
8366
78
6889
6084
83
Σx
= 1075
Σy
= 936
2
Σx
= 91565
2
Σy
= 71262
6474
Σxy
= 80096
( Σy )
2
Syy =
Σy 2 −
−244
= −0.814 and
299.714
140
276
a = x − by =
+ 0.814 ×
= 52.1
7
7
so
=
x 52.1 − 0.814y ,
=
y 52.1 − 0.814=
× 35 23.61 ⇒ $24
2 a
x
y
x2
y2
xy
2
6
4
36
12
3
10
9
100
30
5
22
25
484
110
7
33
49
1089
231
8
42
64
1764
336
10
56
100
3136
560
Syy
Sxy
Sxy
n
9362
= 71262 −
= 3870
13
( Σx ) ( Σy )
=
Σxy −
n
1075 × 936
80096 −
2696
=
=
13
x= a + by , where
Sxy
2696
b ==
= 0.697 and
Syy
3870
1075
936
− 0.697 ×
= 32.508
13
13
so
=
x 32.5 + 0.697y ,
a = x − by=
© Oxford University Press 2019
8
Worked solutions
2 a,d & f
=
x 32.508 + 0.697 × 52 =
68.752
⇒ 69 marks in mathematics
4 a
x
y
x2
y2
xy
1
180
1
32400
180
5
164
25
26896
820
9
148
81
21904
1332
12
120
144
14400
1440
14
118
196
13924
1652
19
90
361
8100
1710
21
85
441
7225
1785
24
82
576
6724
1968
30
65
900
4225
1950
60
1156
3600
34
Σx
= 169
Σy
= 1112
2
Σx
= 3881
b
=
2040
2
Σy
= 139398
Σxy
= 14877
Sxx =
Σx
c
Sxx
e=
y − y1
( Σy )
Σxy −
Sxy =
Sxy
=
y − 12
2
1112
= 15743.6
10
( Σx ) ( Σy )
n
169 × 1112
=
=
−3915.8
14877 −
10
3 a
y= a + bx , where
b=
S xy
S xx
=
−3915.8
= −3.821 and
1024.9
1112
169
+ 3.821 ×
= 175.775
10
10
so y 175.775 − 3.821x ,
=
a = y − bx=
=
y 175.775 − 3.821
=
× 7 149.028
= 149
b
x= a + by , where
b=
Sxy
Syy
=
−3915.8
= −0.249 and
15743.6
169
1112
+ 0.249 ×
= 44.589
10
10
so x 44.589 − 0.249y ,
=
=
x 44.589 − 0.249 × 100 =
19.7 km
a = x − by =
Chapter Review
1 a The PMCC lies between -1 and 1
b A -0.6, B 0.9, C 0.5, D 0, E -0.96
y1 − y2
(x − x1 )
x1 − x2
12 − 16
(x − 40)
40 − 45
y= 0.8 x − 32 + 12
=
y 0.8 x − 20
n
Syy = 139398 −
11 + 8 + 16 + 13 + 14 + 10 72
=
= 12
6
6
so 1200 Dirhams
2
Σy 2 −
Syy =
mean cost
=
( Σx )
−
n
1692
= 3881 −
= 1024.9
10
39 + 36 + 45 + 41 + 42 + 37 240
=
6
6
= 40 °C
2
2
mean temp
t
e
t2
e2
te
0
29
0
841
0
2
38
4
1444
76
4
27
16
729
108
6
19
36
361
114
12
64
144
8
Σt
= 20
Σe
2
Σt
= 120
= 125
2
Σe
= 3519
96
Σte
= 394
( Σt )
2
Stt =
Σt 2 −
n
202
Stt = 120 −
= 40
5
( Σe )
2
Σ e2 −
See =
n
1252
= 394
See = 3519 −
5
( Σ t ) ( Σe )
Σte −
Ste =
n
20 × 125
394 −
=
−106
Ste =
5
c Strong negative, linear
© Oxford University Press 2019
9
Worked solutions
b
=
e at + b , where
S
−106
= −2.65 and
a = te =
Stt
40
125
20
+ 2.65 ×
= 35.6
5
5
b = e − at =
x
y
x2
y2
xy
24
260
576
67600
6240
23.5
199
552.25
39601
4676.5
23
174
529
30276
4002
b=
e 35.6 − 2.65 × 5 = 22.35 ⇒ 22 eggs
22
162
484
26244
3564
c Because t = 40 is too far outside the
domain
21
149
441
22201
3129
20.3
135
412.09
18225
2740.5
20
118
400
13924
2360
18.2
115
331.24
13225
2093
17
102
289
10404
1734
246
676
60516
4 a
x
y
x2
y2
xy
28
3600
784
12960000
100800
46
5200
2116
27040000
239200
38
4400
1444
19360000
167200
34
3800
1156
14440000
129200
52
6000
2704
36000000
312000
50
5900
2500
34810000
295000
Σx
= 248
Σy
= 28900
Σx 2
= 10704
Σy 2
= 144610000
Σxy
= 1243400
26
Σx
= 215
Σy
= 1660
( Σx )
( Σy )
Σxy −
Sxy =
n
Sxx
=
( Σy )
n
16602
= 26656
10
( Σx ) ( Σy )
n
215 × 1660
=
36935 −
1245
Sxy =
10
Sxy
r =
Sxx Syy
(
=
r
y= a + bx , where
Sxy
2152
= 68.08
10
Syy = 302216 −
289002
144610000 −
= 5408333
S=
yy
6
( Σx ) ( Σy )
Σxy −
Sxy =
n
248 × 28900
1243400 −
48866.7
=
Sxy =
6
b=
n
2
Syy =
Σy 2 −
2
Syy =
Σy 2 −
Σy
= 302216
6396
Σxy
= 36935
( Σx )
= 4690.58 −
Sxx
n
2482
= 10704 −
= 453.333
6
Sxx
Σx
= 4690.58
2
2
Sxx =
Σx 2 −
2
Sxx =
Σx 2 −
2
48866.7
= 107.794 and
453.333
28900
248
− 107.794 × = 361.181
6
6
so
=
y 108 x + 361
a = y − bx
=
1245
= 0.924
68.08 × 26656
c There is a strong positive correlation.
The hotter the day, the more bottles
sold.
d
y= a + bx , where
b=
b Need to find the smallest x such that
120 x > 107.794 x + 361.181 ,
120 x > 107.794 x + 361.181
(120 − 107.794)x > 361.181
361.181
x >
12.206
x > 29.59
)
Sxy
S xx
=
1245
= 18.29 and
68.08
1660
215
a = y − bx =
− 18.29 ×
=
−227.235
10
10
so y 18.3x − 227
=
e y 18.29 × 19.6 − 227.235 = 131.249
=
⇒ 131 bottles
f
36 is far outside the domain that we
have
So the smallest number of chairs is 30
5 a
24 + 23.5 + 23 + 22 + 21 + 20.3 + 20 + 18.2 + 17 + 26
10
=
215
= 21.5
10
© Oxford University Press 2019
10
Worked solutions
6 a
( Σx )
2
x
y
3500
110000
12250000
x
12100000000
y
385000000
2000
65000
4000000
4225000000
130000000
5000
100000
25000000
10000000000
500000000
6000
135000
36000000
18225000000
810000000
5000
120000
25000000
14400000000
600000000
3000
90000
9000000
8100000000
270000000
2
2
xy
4000
100000
16000000
10000000000
400000000
8000
140000
64000000
19600000000
1120000000
Σx =
36500
Σy =
860000
2
2
Σy =
96650000000
Σx =
191250000
Sxx =
Σx 2 −
Sxx
Σxy −
Sxy =
Sxy
Σxy =
4215000000
36500
8
= 0.876
b The gradient indicates that a car
travelling one additional mile uses
0.0874 litres of fuel
n
= 4200000000
8600002
8
c y 0.876 + 0.0874 × 160 =
=
14.9 litres
d Not reliable as 5 is outside the domain
of the original data
( Σx ) ( Σy )
n
=
Sxy 4215000000 −
= 291250000
8 a
36500 × 860000
8
y= a + bx , where
b=
Sxy
Sxx
=
50.7
530
− 0.0874 ×
5
5
b = y − ax=
( Σy )
=
Syy 96650000000 −
Sxy =
Σxy −
n
530 × 50.7
6674 −
1299.8
=
=
5
S
1299.8
a=
= xy
= 0.0874 and
S xx
14870
2
2
Syy =
Σy 2 −
50.72
= 123.212
5
( Σx ) ( Σy )
=
y ax + b , where
n
= 24718750
n
Syy = 637.31 −
( Σx )
=
Sxx 191250000 −
( Σy )
2
Syy =
Σy 2 −
2
Sxx =
Σx 2 −
n
5302
= 71050 −
= 14870
5
291250000
= 11.78 and
24718750
860000
36500
− 11.78 ×
= 53753.8
8
8
so y 11.8 x + 53754
=
a = y −=
bx
x
y
x2
y2
xy
1
6
1
36
6
1.5
7
2.25
49
10.5
2
10
4
100
20
2.5
15
6.25
225
37.5
3
9
9
81
27
3.5
17
12.25
289
59.5
4
20
16
400
80
18
20.25
324
b=
y 11.8 × 7000 + 5375
=
4 $136354
4.5
c i and ii would change, iii would remain
the same
Σx
= 22
Σy
= 102
2
Σx
= 71
Σy
2
= 1504
81
Σxy
= 321.5
( Σx )
2
7 a
x
y
x2
y2
xy
30
3.2
900
10.24
96
65
7.5
4225
56.25
487.5
110
8.4
12100
70.56
924
140
15.1
19600
228.01
2114
185
16.5
34225
272.25
3052.5
Σx
= 530
Σy
= 50.7
2
Σx
= 71050
2
Σy
= 637.31
Σxy
= 6674
Sxx =
Σx 2 −
Sxx
n
222
=71 −
=10.5
8
( Σy )
2
Syy =
Σy 2 −
Syy
Sxy
Sxy
© Oxford University Press 2019
n
1022
= 1504 −
= 203.5
8
( Σx ) ( Σy )
=
Σxy −
n
22 × 102
321.5 −
41
=
=
8
11
Worked solutions
f
=
y ax + b , where
=
x ay + b , where
S
73200
=
a = xy
= 3.641 and
Syy
20103.4
S xy
41
a ==
= 3.90 and
Sxx 10.5
b = y − ax =
102
22
− 3.90 ×
= 2.01
8
8
2800
2123
b = x − ay =
− 3.641 ×
=
−704.263
7
7
so x 3.641y − 704.263 ,
=
b An increase in one gram of hormone
leads to just under 4 extra flowers
c A plant with no growth hormone will
produce 2 flowers
x = 3.641 × 300 − 704.263 =
388 g
10 a
d=
y 2.01 + 3.905 × 1.75 =
8.84
x
y
x2
y2
xy
15
26
225
676
390
e=
12 2.01 + 3.905x
25
30
625
900
750
35
25
1225
625
875
45
26
2025
676
1170
55
20
3025
400
1100
65
14
4225
196
910
Σx
= 240
Σy
= 141
9.99
3.905x = 12 − 2.01 ⇒ x =
= 2.56 g
3.905
f Not appropriate as 1000 is far outside
the domain of the data provided
9 a
x
y
x
y
100
204
10000
41616
20400
200
257
40000
66049
51400
300
292
90000
85264
87600
400
315
160000
99225
126000
500
330
250000
108900
165000
600
355
360000
126025
213000
370
490000
136900
700
Σx
= 2800
2
Σy
= 2123
2
xy
2
Σx
= 1400000
Sxx
Sxx =
Σx
Syy
( Σx )
−
( Σy )
2
2800
= 280000
7
Sxy
n
Sxy
a=
2
2123
= 20103.4
7
( Σx ) ( Σy )
Syy = 663979 −
Σxy −
Sxy =
n
2800 × 2123
922400 −
73200
=
=
7
n
1412
= 3473 −
= 159.5
6
S
73200
=
a = xy
= 0.261 and
Sxx 280000
2123
2800
− 0.261 ×
= 199
7
7
b Each additional gram increases the
length of the spring by 0.261 mm
c The spring was 199 mm long before
any weight was added
( Σx ) ( Σy )
n
240 × 141
=
=
−445
5195 −
6
Sxy
−445
= −0.254 and
1750
=
Sxx
141
240
+ 0.254 ×
= 33.7 ,
6
6
=
y 33.7 − 0.254 x
b = y − ax =
b=
y 33.7 − 0.254 × 50
= 21 decimal places
=
y ax + b , where
b = y − ax=
( Σy )
=
y ax + b , where
2
Syy =
Σy 2 −
( Σx )
Sxy =
Σxy −
n
Sxx= 1400000 −
c
r =
=
r
Sxy
(S
xx
Syy
)
−445
= −0.842
1750 × 159.5
d There is a strong, negative correlation
11 a
0.51 × 120 + 7.5 =
68.7
M1A1
y = 0.51 × 100 + 7.5 = 58.5
R1
A1
b The line of best fit goes through ( x, y )
d=
y 199 + 0.261 × 550 = 343 mm
e 2 kg is outside the domain of the data,
so extrapolation is unreliable
Σxy
= 5195
2
2
2
Σy
= 3473
n
2402
= 11350 −
= 1750
6
Syy =
Σy 2 −
259000
Σy
= 663979
Σx
= 11350
2
2
Sxx =
Σx 2 −
Σxy
= 922400
2
2
c Strong, positive
d x on y
© Oxford University Press 2019
A1A1
A1
12
Worked solutions
12 i
ii
iii
iv
v
13 a
b
c
perfect positive
strong negative
weak positive
weak negative
zero
r = 0.979 (3sf)
Strong, positive
i y 1.23x − 21.3
=
A1
A1
A1
A1
A1
A2
A1A1
A1A1
ii x 0.776y + 20.8
=
A1A1
130 − t = 80 ⇒ t = 50
M1
A1
Interval is 20 ≤ t ≤ 50 .
A1A1
110
d 1.23 × 105 − 21.3 =
95
e 0.776 × 95 + 20.8 =
f It is extrapolation
14 a
A1
R1
A1
ii
T ≥ 80
40 + 2t = 80 ⇒ t = 20
17 a
x
13
14
15
16
16
17
18
18
19
19
y
2
0
3
1
4
1
1
2
1
2
A3 (A2 for 5 A1 for 3)
b
r = −0.0695(3sf )
c
Very weak (negative) correlation so
line of best fit is meaningless R1
A2
25-year-old would be extrapolation
R1
(scales: A1; 3 points plotted correctly:
A2; all points plotted correctly: award a
further A1)
b strong, negative
c i
ii
A1A1
x = 4.625
y = 5.875
=
140 100m + c
40 = 30m
4
3
20
3
b Positive
(M1)A1A1
c =
c Line goes through ( x, y )
y=
4
2
2
90 + 6 = 126
3
3
3
d Estimate is
4
2
2
60 + 6 =
86
3
3
3
A1
(R1)
(M1)A1
(M1)A1
16 a
40 oC
A1
b
70 oC
A1
c 100 oC
d i
ii
Gradient=
m
0.6
= 0.2
3
l = 0.6
iii k = 3
A1
M1A1
A1
A1
iv a = 5
A1
v
A1
b = 0.6
0.9 − 0.6
= 0.1
8−5
vii 0.6= 0.1 × 5 + q ⇒ q= 0.1
vi Gradient
=
p
iii see above
A2A2A1
d see above
M1
line passes through the mean A1
e 3.2 see above for lines drawn on
A1A1
15 a 100
= 70m + c
m=
18 i
viii
r =8
19 a i 0.849 (3sf)
ii strong, positive
iii y 0.937 x + 0.242
=
M1A1
M1A1
A1
A2
A1A1
A1A1
b i 0.267 (3sf)
A2
ii weak, positive
A1A1
iii the r value is too small for this to
be particularly meaningful R1
20 a i no change
A1
r = 0.87
ii no change
15
A1
iii The scatter diagram has just been
moved down by 4 and to the right
by 5.
R1
iv Strong, positive
A1A1
b i
ii
© Oxford University Press 2019
no change
r = 0.87
A1
30
2 × 15 =
A1
13
Worked solutions
iii the scatter diagram has been
stretched vertically.
R1
c i r = −0.87
A1
15
= −5
A1
−3
iii The scatter diagram has been
stretched horizontally and reflected
in the y-axis.
R1R1
iv Strong, negative
A1A1
ii
© Oxford University Press 2019
14
Worked solutions
8
Quantifying randomness: probability
Skills Check
1
a
3
=
7
2 5
+ =
5 7
1−
b
f
P(9) = 0
7 3 4
− =
7 7 7
14 25 39
+
=
35 35 35
c
2 2 2×2
4
×=
=
5 3 5 × 3 15
d
3
53
1 3
1 −  ×  =1 −
=
56 56
7 8
6
1 − 0.375 =
0.625
0.65 + 0.05 = 0.7
c
7×6
42
0.7 × 0.6 =
=
= 0.42
100
102
d
0.25 × 0.64 = 0.64 ÷ 4 = 0.16
e
0.5 × 30 = 30 ÷ 2 = 15
f
0.22 × 0.22 =
7
222
484
=
= 0.0484
1002 10000
8
n({1,3,5,7,9})
n({1,2,3, 4,5,6,7,8,9,10})
3
=
P(chorus)
4
a
P(even)
=
b
P(multiple of 3) =
n({2, 4,6,8})
4 1
= =
n({1,2,3, 4,5,6,7,8}) 8 2
n({3,6})
n({1,2,3, 4,5,6,7,8})
2 1
=
8 4
P(multiple of 4) =
a
Every other number is even, so
1
P(even) =
2
b
P(contains digit 1)
7
n({1,10 − 19,21,31, 41}) 14
= =
50 25
50
Let x be the number of seats
minibus, then
3x
P(coach)
=
=
x + x + x + x + 3x
on a
3x 3
=
7x 7
Using P(green) = 2P(yellow) and
Number of people who buy raffle tickets
360
= = 180 , of these half bought 2 tickets and
2
the other half bought one, so there were
180
2×
+ 180 =
360 raffle tickets sold. Therefore
2
P(win) =
1
360
Exercise 8B
1
P(not a multiple of 4)
P(less than 4) =
9
n({4,8})
n({1,2,3, 4,5,6,7,8})
2 1
=
8 4
3
=
8
3
10
0.3
= P(yellow)
3
P(yellow) = 0.1
P(green) 2=
P(yellow) 0.2
=
20
20 4
= =
20 + 10 + 5 35 7
= 1 − P(multiple of 4) =1 −
e
n({ A, I , I})
n({S, T , A, T , I , S, T , I , C , S})
1 = 0.4 + P(yellow) + 0.3 + P(green)
1=
0.7 + P(yellow) + P(green)
1 −=
0.7 P(yellow) + 2P(yellow)
0.3 = 3P(yellow)
30
1
=
150 5
P(defective)
=
d
P(vowel) =
1 =P(red) + P(yellow) + P(blue) + P(green) ,
2
=
c
we see that
1 =P(red) + P(yellow) + P(blue) + P(green)
5
1
= =
10 2
c
As there is no P in the word
“STATISTICS”, P(P ) = 0
=
Exercise 8A
=
b
=
a
b
1 P(odd) =
n({C })
1
=
n({S, T , A, T , I , S, T , I , C , S}) 10
5 a P(C )
=
3
3 ÷ 20 3
20
e= =
7
7 ÷ 20 7
20
2
As there is no 9 on an 8 sided dice,
a
1 3
=
4 4
i
P(age 15) = 0.18
ii
P(age 16 or higher)
= P(age 16) + P(age 17) + P(age 18)
n({1,2,3})
n({1,2,3, 4,5,6,7,8})
= 0.22 + 0.27 + 0.13 = 0.62
b
Number of 15 year old students
= 1200 × P(age 15) = 1200 × 0.18 = 216
© Oxford University Press 2019
Worked solutions
1
Worked solutions
2
3
a
The relative frequency of getting a 1 is
frequency of 1
27
= = 0.27
total spins
100
b
The spinner is probably not fair because the
relative frequencies are not close to each
other, a 1 occurred nearly 4 times more than
a6
c
Estimated number of 4s = 3000 ×
a
10 of each
=
1 − P(card or present) =
1−
4
23
2
=
25 25
A = {P , R, O, B, A, I , L, T , Y } and
B = {C , O, M, P , L, E , N, T , A, R, Y }
a
15
= 450
100
b
Relative
0.1
frequency
0.1
0.15 0.138
0.138
0.15
0.138
0.0875
c
b
A∩B =
{P , R, O, A, L, T , Y }
c
A∪B =
{P , R, O, B, A, I , L, T , Y , C , M, E , N}
a
A∩B =
{6}
b
A∪B =
{2,3, 4,6,8,9,10}
Exercise 8C
c
A′ = {1,3,5,7,9}
1
d=
A′ ∩ B {1,3,5,7,9} ∩=
{3,6,9} {3,9}
Relative
0.0925 0.1225 0.1375 0.125 0.14 0.145 0.1075 0.13
frequency
d
There is a big difference between relative
frequency of getting a 1 and getting a 6.
This suggests that the dice is not fair.
5
a
e
A ∪ B′
= {2, 4,6,8,10} ∪ {1,2, 4,5,7,8,10}
= { 1, 2, 4, 5, 6, 7, 8, 10 }
f
b
2
A′ ∪ B′
= {1,3,5,7,9} ∪ {1,2, 4,5,7,8,10}
= { 1, 2, 3, 4, 5, 7, 8, 9, 10 }
From the Venn diagram,
8
4
P(neither)
= =
38 19
6
U={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
a
a
i
M = {3,6,9,12,15}
ii
F = { 1, 2, 3, 5, 6, 10, 15 }
i
P(M ∩ F ) =
ii
6
2
P((M ∪ F )′ ) =
=
15 5
i
P(only 9 pm) = 33%
b
b
From the Venn diagram,
19 + 21 + 16 + 7 =
63
c
i
P(badminton) =
ii
P(both)
=
iii
P(neither)=
iv
P(at least one)= 1 − P(neither)
=1 −
3
40
63
c
21 1
=
63 3
7
1
=
63 9
7
3
1
=
15 5
a
1 8
=
9 9
Let A = gave a card and B = gave a present
31 + 40 − 25 23
=
50
25
a
P(card =
or present)
b
P(card but no present)
=
c
P(neither card nor present)
31 − 25
3
=
50
25
b
© Oxford University Press 2019
2
Worked solutions
ii
P(only 6 pm) = 24%
= P( A) + P(B) − P( A ∩ B)
iii
P(no news) = 30%
=
9
so 4P( A ∩ B) =
16
Exercise 8D
1
n({2,3,5,7})
4
2
= =
10
10 5
a
P(prime)
=
b
P(prime or multiple of 3)
⇒ P( A ∩ B) =
n({2,3,5,7}) + n({3,6,9}) − n({3})
=
10
6
3
4 + 3 −1
=
= =
10
10 5
c
=
n({3,6,9}) + n({4,8})
n({10})
1
3+2
5
1
= =
10
10 2
2
P(camera owner or female)
3
9
3
−
=
16 64 64
a
No
b
Yes
c
No
e
No
f
No
g
No
so N and M are mutually exclusive
3
P( A ∩ B) = P( A ∩ C ) = P(B ∩ C ) = 0 because
a
a
P(=
A)
n( A)
8
4
= =
n(U ) 26 13
b
P
=
(B)
n(B)
9
=
n(U ) 26
c
4
2
n({E , I , M, T })
P( A ∩ B) =
=
=
26 13
n(U )
d
n( A) + n(B) − n( A ∩ B)
P( A ∪ B) =
n(U )
P( A ∪ B)= P( A) + P(B) − P( A ∩ B)
=
b
P( A ∪ B ∪ C=
) P( A) + P(B) + P(C )
=
c
1 1
7
+ −0 =
3 4
12
1 1 1 47
+ + =
3 4 5 60
Yes, because the probability of A, B or
C winning is not equal to 1.
Exercise 8F
8 + 9 − 4 13 1
= =
26
26 2
1
U = {HHH, HHT , HTH, THH, HTT ,
P(fiction or non-fiction)
THT , TTH, TTT }
P(fiction) + P(non-fiction) − P(both)
=
= 0.4 + 0.3 − 0.2 = 0.5
a
b
c
1 1 1 1
+ − =
4 8 8 4
b
1 3
P( X ∪ Y )′ =1 − P( X ∪ Y ) =1 − =
4 4
a
P( A ∩ B)= P( A) + P(B) − P( A ∪ B)
b
P( A′ ∪ B=
) P( A′) + P(B) − P( A′ ∩ B)
n({HHH, HHT , THH}) 3
=
n(U )
8
P(heads and tails alternately)
=
2
n({HHH, HHT , HTH, THH}) 4 1
= =
n(U )
8 2
P(at least two heads consecutively)
=
P( X ∪ Y )= P( X ) + P(Y ) − P( X ∩ Y )
=
P(more heads than tails)
=
P(no book)
= 1 − P(fiction or non-fiction)
1 0.5 =
0.5
=−
a
Yes
1 1
3
= +
−
=0
5 10 10
B = {T , R, I , G, O, N, M, E , Y }
a
d
P(N ∩ M )= P(N ) + P(M ) − P(N ∪ M )
only one school can win
b
6
P( A ∩ B′)= P( A) − P( A ∩ B)
Let A = {M, A, T , H, E , I , C , S} and
=
5
c
27 37
=
64 64
Exercise 8E
n(cam) + n(fem) − n(fem and cam)
n(U )
30 + 25 − 18 37
= =
55
55
4
P( A ∪ B)′ =1 − P( A ∪ B) =1 −
=
=
3
9
27
⇒ P( A ∪ B) =
64
64
b
P(multiple of 3 or 4)
=
2
3 3
+ − P( A ∩ B)
16 8
n({HTH, THT }) 2 1
= =
n(U )
8 4
a
= 0.2 + 0.4 − 0.5 = 0.1
=−
1 P( A) + P(B) − (P(B) − P( A ∩ B))
=−
1 0.2 + 0.4 − (0.4 − 0.1) =
0.9
7
a
3P( A ∩ B) = P( A ∪ B)
b
i
© Oxford University Press 2019
P(red is higher than blue)
3
Worked solutions
=
b
iii
c
n({(1,2),(2,3),(3, 4),(2,1),(3,2),(4,3)})
n(U )
6
3
=
16 8
=
P(red is odd and blue is even)
iv
3
Exercise 8G
P(sum is prime)
n({(1,1),(1,2),(1, 4),(2,1),(2,3),(3,2),(3, 4),(4,1),(4,3)})
n(U )
=
n({(1,2),(2,1),(2,3),(3,2),(3, 4),(4,3),(4,1),(1, 4)})
n(U )
8
2
=
36 9
=
4
1
=
16 4
=
n({(1,1),(2,2),(3,3),(4, 4)})
4
1
= =
n(U )
36 9
To be more than 1 but less than 2 meters
away, he must go to a corner
P(between 1 and 2 meters after 2 rolls)
=
n({(1,2),(1, 4),(3,2),(3, 4)})
=
n(U )
=
P(2 meters from start after 2 rolls)
=
P(difference between numbers is 1)
=
8
2
=
36 9
=
6
3
=
16 8
=
ii
n({(1,2),(1,3),(1, 4),(2,3),(2, 4),(3, 4)})
n(U )
9
16
1 1
1
× =
5 5 25
1
P(both purple) =
2
64
4
P(all 3 like pasta)
= =

125
5
3
P(loses both) = (1 − 0.75) × (1 − 0.85)
3
= 0.0375
a
4
a
P(B) = P( A ∩ B) + P( A ∪ B) − P( A)
=
0 + 0.4 − 0.2 =
0.2
P(B ∩ C )= P(B) + P(C ) − P(B ∪ C )
= 0.2 + 0.3 − 0.34
= 0.16
b
i
b
P(cards have same number)
=
ii
n({(2,2),(3,3)})
2
1
= =
n(U )
12 6
P(largest number is 3)
P(head and not 6) =
6
P(not hitting with 4 missiles)
4
7
a
P(E ) =−
1 P(E′) =−
1 0.6 =
0.4
b
i
P(E ) × P(F ) =0.24 =P(E ∩ F )
n({(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1)})
n(U )
iv
c
P(at least one even number)
n({(2,1),(2,2),(2,3),(3,2),(4,1),(4,2),(4,3),(5,2)})
n(U )
8
The only possible way to have a sum of 6
3
is if all dice show 2.
9
1
2
P(sum to 6) =   =
27
6
a
P( A ∩ B) = P( A) × P(B) = 0.9 × 0.3 = 0.27
b
P( A ∩ B′)= P( A) − P( A ∩ B)
=0.9 − 0.27 =0.63
c
P( A ∪ B′)= P( A) + P(B′) − P( A ∩ B′)
= 0.9 + 0.3 − 0.63 = 0.97
P(at start after 2 rolls)
=
P(E ∪ F ′)= P(E ) + P(F ′) − P(E ∩ F ′)
= 0.4 + 1 − 0.6 − 0.4 + 0.24= 0.64
8
2
= =
12 3
a
Because P(E ∩ F ) =
/ 0
= P(E ) + 1 − P(F ) − (P(E ) − P(E ∩ F ))
P(product is at least 8)
=
4
ii
9
3
=
12 4
n({(3,3),(4,2),(4,3),(5,2),(5,3)})
=
n(U )
5
=
12
v
Because
P(sum is less than 7)
=
1 5
5
× =
2 6 12
5
1
1
= =

6561
9
n({(2,3),(3,1),(3,2),(3,3)})
=
n(U )
4
1
= =
16 4
iii
Not independent as P(B ∩ C ) =
/ 0
n({(1,3),(2, 4),(3,1),(4,2),(5,6),(6,5),(5,5),(6,6)})
n(U )
© Oxford University Press 2019
4
Worked solutions
Exercise 8H
1
a
4
n(both subjects)
=
n(film) + n(theatre) − n(either)
= 15 + 20 − (27 − 4) = 12
b
i
P(theatre and not film)
a
P(male and left-handed)
=
b
P(right-handed) =
c
P(right-handed | female)
= P(theatre) − P(theatre and film)
20 12
8
=
−
=
27 27 27
ii
P(theatre or film)
= P(theatre) + P(film)
− P(theatre and film)
=
iii
=
2
a
P( J |=
K)
6
P(two boys | one is a boy)
=
Exercise 8I
1
P(even | not multiple of 4)
P(even and not multiple of 4)
P(not multiple of 4)
n({2,6,14})
3
8
= =
n({1,2,6,7,11,14,29}) 7
8
=
b
P(less than 15 and greater than 5)
P(greater than 5)
b
P(two picture cards) =3 ×
3
P(two broken pens) =
b
P(at least one broken pen)
4
=
a
b
a
P(male) =
b
P(one male and one female)
a
c
3
10
3 7
7
× =
10 9 15
P(at least one answers correctly)
= P(one answers correctly)
+ P(both answer correctly)
=
n({6,7,11,14})
4
8
=
n({6,7,11,14,24}) 5
8
b
P(V ∩ W ) =
0 because they are mutually
exclusive
P(V | W ) = 0 because they are mutually
exclusive
1
4
P(girl picks broken pen) =
P(1 ↔ 15 and 5 ↔ 25)
=
P(5 ↔ 25)
3
9
5 10 55
×
+
=
14 13 91 91
c
=2 ×
P(1 ↔ 10 | 5 ↔ 25)
5
4
20
10
×
=
=
14 13 182 91
a
P(less than 5 | less than 15)
P(less than 5 and less than 15)
P(less than 15)
n({1,2})
2 1
8
=
= =
n({1,2,6,7,11,14}) 6 3
8
12 11 40
×
×
52 51 50
132
1105
=2 ×
=
d
P(three picture cards) =
= P(one broken pen) + P(two broken pens)
n({6,7,11,14})
4 2
8
= =
n({6,7,11,14,24,29}) 6 3
8
=
c
2
12 11 10
11
×
×
=
52 51 50 1105
a
=
P(< 15 |> 5)
=
P(two boys and one is a boy)
P(one is a boy)
n({BB})
n({BB, BG, GB, GG}) 1
=
n({BB, BG, GB})
3
n({BB, BG, GB, GG})
=
12
12 4
27
= =
15 15 5
27
P( J ∩ K ) P( J) × P(K )
( J) 0.3
=
= P=
P(K )
P(K )
5
P(theatre | film)
P(theatre and film)
=
P(film)
43
50
11
P(right-handed and female) 50 11
= =
13 13
P(female)
50
=
20 15 12 23
+
−
=
27 27 27 27
5
1
=
50 10
5 4 2 5 5 5 55
× + × + × =
7 9 7 9 7 9 63
P(Luca correct | at least one correct)
P(Luca correct)
=
=
P(at least one correct)
c
5
9
7
=
55 11
63
P(Ian correct | at least one correct)
P(V ∪ W )= P(V ) + P(W ) − P(V ∩ W )
= 0.26 + 0.37 − =
0 0.63
© Oxford University Press 2019
5
Worked solutions
P(Ian correct)
=
P(at least one correct)
=
5
7
9
=
55 11
63
= 0.6 + 0.2 − 0.02
= 0.78
c
P( A ∪ B) − P( A ∩ B)
= 0.78 − 0.02 = 0.76
d
P(two correct | at least one correct)
P(two correct)
=
=
P(at least one correct)
5 5
×
5
7
9
=
55
11
63
d
5
a
n({10,15,20,...85,90,95})
n({10,11,...,98,99})
18 1
=
90 5
=
P(divisible by 3)
=
c
n({12,15,18,...93,96,99}) 30 1
= =
90 3
n({10,11,...,98,99})
c
P(greater than 50)
15 + 3 + 7 + 20 + 6 x + 2 x + x =
90
45 + 9 x =
90
9 x = 45
x =5
49
n({51,52,53,...,98,99})
=
=
90
n({10,11,...,98,99})
d
6x
b
P(divisible by 5)
=
b
a
P( A ∩ B)
P( A)
0.02
2
1
= =
0.6
60 30
=
Chapter review
1
P(B | A) =
P(a square number)
6
6
1
n({16,25,36, 49,64,81})
= =
=
90 15
n({10,11,...,98,99})
a
P(C ∩ D)= P(D) × P(C | D)
= 0.5 × 0.6 = 0.3
2
b
Not mutually exclusive as P(C ∩ D) ≠ 0
c
P(C ) × P(D) = 0.4 × 0.5 = 0.2
P(C ∩ D) ≠ P(C ) × P(D)
Therefore C and D are not
independent events.
d
From Venn diagram P(Cat and dog) =
3
P(C ∪ D)= P(C ) + P(D) − P(C ∩ D)
= 0.4 + 0.5 − 0.3
= 0.6
11
30
e
a
P(D | C ) =
=
7
a
P(C ∩ D)
P(C )
0.3 3
= = 0.75
0.4 4
P(properly) =
3
2
× 0.35 + × 0.55
5
5
= 0.21 + 0.22 = 0.43
b
P(Jill | Properly ') =
From Venn diagram P(C ∩ D ') =
0.55
b
P(C ∩ D) =
0.15
P(C ) × P(D) = 0.7 × 0.2 = 0.14
P(C ∩ D) ≠ P(C ) × P(D)
=
P(Jill ∩ Properly')
P(Properly')
2
× 0.45
5 = 0.18
= 0.316
0.57
0.57
Therefore C and D are not
independent events.
4
a
P( A ∩ B)= P(B) × P( A | B)
= 0.2 × 0.1 = 0.02
b
P( A ∪ B)= P( A) + P(B) − P( A ∩ B)
© Oxford University Press 2019
6
Worked solutions
8
a
c
42 36
P(F ) × P(C ) = ×
70 70
54
=
175
19
≠
= P(F ∩ C )
70
Therefore not independent.
11 a
0.3 × 0.3 =
0.09
ii
0.3 × 0.6 =
0.18
iii 0.3 × 0.3 + 0.6 × 0.6 + 0.1 × 0.1
c
Primes are 2,3,5:
d
4 or 5:
0.7 × 0.7 × 0.7 =
0.343
d
3 × 0.12 × 0.6 + 3 × 0.32 × 0.1
= 0.018 + 0.027 = 0.045
9
a
b
c
6
3
=
16 8
3 1
=
6 2
(M1)A1
2 1
=
6 3
Impossible: 0
(M1)A1
A1
A1
b
6
1
=
36 6
A1
c
1
36
A1
d
2×
e
(1,6 ) , (2,5) , (3, 4) , ( 4,3) , (5,2) , (6,1)
1
1
=
36 18
or using a lattice diagram
A1
6
1
=
36 6
(M1)A1
f
1
since independent
6
g
P ( R5 ∪ B5=
) P ( R5) + P(B5) − P ( R5 ∩ B5)
10 2
=
15 3
=
5
4
1 2
2
×
= × =
15 14 3 7 21
h
10
A1
1
36
= 0.09 + 0.36 + 0.01 = 0.46
c
3 1
=
6 2
2,4 or 6:
12 a
i
A1
b
e
b
1
6
13 a
A1
1 1
1
11
or using a lattice diagram
+ −
=
6 6 36 36
(M1)A1
2 1
Considering the list in (e)
=
6 3
or using conditional probability formula
(M1)A1
Independent ⇔ P(F ∩ R)= P(F ) × P(R)
R1
Both female and eating carrots = 19.
a
b
11
70
(
1 1 1
so not independent
≠ ×
6 3 4
A1
P(F ∪ R)= P(F ) + P(R) − P(F ∩ R)
M1
=
)
P F C
b=
P ( F ∩ C ) 19
=
P (C )
36
1 1 1
5
+ − =
3 4 6 12
A1
5 1 1
− = (M1)A1
12 6 4
Could also use a Venn diagram in (b) and (c)
c
P(exactly one team)=
d
P ( F ∩ R ) 16 1
P ([F ∩ R] F ) =
=
=
1
P (F )
2
3
M1A1
14 a
© Oxford University Press 2019
7
Worked solutions
1
60
so not independent
≠
4 100
A1
17 a
A2
b
3 3
9
× =
4 5 20
c
1−
d
P ( ship not hear tower has no reply )
=
=
e
M1A1
9
11
=
20 20
M1A1
P ( ship not hear ∩ tower has no reply )
P ( tower has no reply )
5
=
11
1
4
11
20
M1A1
P ( A ∩ B) =
0 so events are mutually
exclusive.
30 ×
3
=
18
5
(M1)A1
b
50 ×
2
=
20
5
(M1)A1
16 a
A1
c
30
80
6
×
=
100 100 25
A1
d
30
20
70
90
69
×
+
×
=
100 100 100 100 100
M1A1
e
P ( I′ =
C)
P ( I′ ∩ C )
=
P (C )
63
100
69
100
=
21
23
M1A1
f
′)
P ( I C=
P ( I ∩ C ′)
=
P (C ′)
24
100
31
100
24
31
M1A1
R1A1
15 a
c
b
A4 (A2 layout A2 numbers)
70
90
63
×
=
100 100 100
=
18 a
3
(M1)A1
= 30 ⇒ T = 50
5
Let x be the number speaking both English
T×
and French. ( 60 − x ) + x + (40 − x ) + 10 =
100
⇒ 110 − x= 100 ⇒ x= 10
(M1)A1
b
A4 (A1 shape, A3, 7 numbers, A2, 4 numbers,
A1 2 numbers)
b
200 − 140 =
60
c
i
30
3
=
200 20
A1
ii
122
61
=
200 100
A1
iii
92
23
=
200 50
A1
A3 (A1 shape A2 numbers)
c
i
50
1
=
100 2
A1
ii
90
9
=
100 10
A1
d
iii
40
2
=
100 5
A1
19 a
P ( E ∩ F ) 10 1
= =
P (F )
40 4
d
P (E =
F)
e
If independent then P E F = P ( E )
(
)
M1A1
R1
iv 1 −
82
118
59
=
=
200 200 100
(M1)A1
(M1)A1
20
5
or by using the formula A2
=
48 12
i
ii
5 4
5
× =
8 7 14
RG or GR
5 3 3 5 15
× + × =
8 7 8 7 28
© Oxford University Press 2019
(M1)A1
(M1)A1
8
Worked solutions
b
i
ii
20 a
i
5 5 25
× =
8 8 64
RG or GR
5 3 3 5 15
× + × =
8 8 8 8 32
P ( A B) =
0.4
⇒
=
(M1)A1
(M1)A1
P ( A ∩ B)
P (B)
P ( A ∩ B)
0.5
⇒ P ( A=
∩ B ) 0.2
M1A1
ii
P ( A )= P ( A ∩ B ) + P ( A ∩ B′ )
iii
P ( A ∪ B=
) P ( A) + P ( B ) − P ( A ∩ B )
= 0.2 + 0.4 = 0.6
= 0.6 + 0.5 − 0.2 = 0.9
iv
M1A1
M1A1
P ( A ∩ B′ ) 0.4
P ( A=
B′ )
= = 0.8
0.5
P ( B′ )
M1A1
b
P ( A B ) ≠ P ( A B′ ) so not independent
R1A1
© Oxford University Press 2019
9
Worked solutions
9
Representing equivalent quantities: exponentials
and logarithms
Skills Check
6
1 a 2 = 2 × 2 × 2 × 2 × 2 = 32
5
d
53 125
5
=
 =

63 216
6
( )
3
= 33 ⋅ r 3 ⋅ s3
3
= 27r 3s9
5
15
1
1
=
 =

35 243
3
3
= 27r 3 ⋅ s3⋅3
b 103 = 10 × 10 × 10 = 1000
c
(3 ⋅ rs )
7
3
( −2x yz )
4
5
3
( )
=−
( 2) ⋅ x 4
3
3
( )
y 3 z5
3
=−8 ⋅ x 4⋅3 ⋅ y 3 ⋅ z 5⋅3
= −8 x12y 3 z 15
2 a 2 = 8, x = 3
3
8 (
b 104 = 10000, x = 4
c 44 = 456, x = 4
x12y 8 2
x12 y 8
=
)
( 5 ⋅ 6 )2
5 6
x y
x
y
= (x12 −5y 8 −6 )2
3
= (x 7y 2 )2
= x 7⋅2y 2⋅2
= x14y 4
9
(5x )2 (5y 3 )
52 x 2 5y 3
= 3 3 3 4 3
3 4 3
(5x y )
5 (x ) (y )
=
125x 2y 3
125x 3⋅3y 4⋅3
=
x 2y 3
x 9y 12
x2 y 3
⋅
x 9 y 12
=
= x 2 −9 ⋅ y 3−12
= x −7 ⋅ y −9
1
= 7 9
x y
Exercise 9A
1
a5 ⋅ a3 ⋅ a7 = a5 +3 ⋅ a7
= a8 ⋅ a7
10
8 +7
=a
= a15
2 2x y ⋅ 7x y = 2 ⋅ 7 ⋅ x ⋅ x ⋅ y ⋅ y
3
2
4
6
3
4
2
=14 ⋅ x 3 + 4 ⋅ y 2 + 6
= 14 x 7y 8
3
4ab3 ⋅ 0.5a6c = 4 ⋅ 0.5 ⋅ a ⋅ a6 ⋅ b3 ⋅ c
=2 ⋅ a1+ 6 ⋅ b3 ⋅ c
= 2a7b3c
4
8m5
8
2 ⋅ m2
=⋅ m5 −3 =
4m3
4
5
6u5v 2
6 5 −3 2 −3
=
⋅u v
9u3v 3
9
6
9 x 3(y 3 )3
9 x 3y 3⋅3
=
−2 4 11
−81(x ) y
−81x −2⋅4y 11
9 x 3y 9
−81x −8y 11
1
= − x 3 −( −8)y 9 −11
9
1 11 −2
= − x y
9
x11
= −
9y 2
=
11 The area of a square of length l is l 2 .
Therefore the area of a square with side
length 3x 2y is (3x 2y )2 .
(3x 2y )2 = 32 (x 2 )2 y 2
= 9 x 2⋅2y 2
2 2 −1
⋅u ⋅v
3
2u2
=
3v
=
= 9 x 4y 2
© Oxford University Press 2019
Worked solutions
1
Worked solutions
12 The area of a rectangle with width w and
length l is w ⋅ l . Then the area of this
rectangle is
4a3b2 ⋅
5a
4 ⋅ 5 3 +1 2 1
=
⋅a ⋅b ⋅ 3
2b3
2
b
= 10 ⋅ a4 ⋅ b2 −3
= 10 ⋅ a4 ⋅ b−1
= 10
e
⇒ 2 x =1 ⇒ x =
1
2
7
=
7
=
3 ⋅ 2x = 48 = 3 ⋅ 16 = 3 ⋅ 24 ⇒ x = 4
g
+2
4x=
a4
b
3
2x +4
x + 2)
=
(22=
)x +2 22(
2=
2−6
−6
2x + 4 =
x = −5
h
1
5 3
b=
25 (2
=
)
(5=
2)3
3
1
5
4
1
4 5
2 3
c=
6 2 (6
=
)
(2=
6)3
4
d=
2
(2
=
)
(=
2)5
−1
5
23
2
63
4
25
1
1
f=
=
x )2 ]−3
(3x ) 2 [(3
=
=
(3x )
−3
2
3x
1
2 −3
1
2
= 3=
x3
x
(5−1 )x = 5− x
−3
2
2
2 a
3
c =
m7 (3=
m )7 (=
m )
m
2 x − 6 = 2(3x − 4) = 6 x − 8
1
2 −1
2 = 4x
−1
2
−1
5 4
−1
=
=
(2d )5
((2
d) )
(2d )
1
−5
4
x
=
3⋅ x
−1
1
2 5 x +2
=
=
95 x +2 (3
)
32(5 x +2)
2(5x + 2) = x − 11
10 x + 4 = x − 11
9 x = −15
−5
x =
3
−1
=
3 ⋅ (x 2 )−1 =
3x 2
Exercise 9C
1 a
2x = 16 = 24 ⇒ x = 4
b 10x = 1000000= 106 ⇒ x = 6
c
2x +1 = 64 = 26 ⇒ x + 1 = 6 ⇒ x = 5
d
32 x −1 = 27 = 33 ⇒ 2 x − 1 = 3
⇒ 2x = 4 ⇒ x = 2
1
2
1 11− x
−1 11− x
d =
95 x +2 (=
)
(3=
)
3x −11
3
3 x = 3x 2
3
3
2
x +3
x −2
2=
4=
(22 =
)x −2 22( x −2)
x =
4
3
= 22 ⇒ x =
3x −4
2 3x −4
c =
62 x −6 36
=
(6=
)
62(3 x − 4)
7
3
=
5x
[(5x=
) ]
(5x )
1
e =
4
(2d )5
1
2
x − 3 = 2(x − 4) = 2 x − 8
x =5
6
−1
1+
x −4
2 x −4
b =
5x −3 25
=
(5=
)
52( x − 4)
6
6
 51 
5
=
a
a 
a5
=
 
( )
1
2 x = 2 2 = 2 ⋅ 22 = 2
x + 3 = 2(x − 2) = 2 x − 4
x =7
3
2
1
3 7
g
j
−3
3
3
1 x
( =
)
25
= 52
5
5− x = 52
x = −2
3


10
10 
10
2 a =
=


f
i
3
1
2
1
d =
5x
1
5
−3
g =
x )
x
3x
3(=
3=
3 x
5 6
a
b =
2
3 4 = 32 x
1
= 2x
4
1
x =
8
1
(3x )3
−3
2
=
5−1
2
x
x
=
3 9=
(32 )=
32 x
4
1
2
2 −1
e =
5 2 (5=
)
(2 =
5)−1
−3
1
1
= = 2−6
64 26
4 = 22
7
2
1
2
f
Exercise 9B
1 a
31−2 x = 1 = 30 ⇒ 1 − 2 x = 0
Exercise 9D
1 a
y = 10x has exponential growth as
10 > 0 and 10 ≠ 1
© Oxford University Press 2019
2
Worked solutions
b
c
2
y = 6 − x has exponential decay as
6 > 0 , 6 ≠ 1 and the coefficient of
x is negative.
c i
3
y = ( )x has exponential decay as
5
3
<1
5
d
y = (0.45)x has exponential decay as
0.45 < 1
e
y = (1.5)x has exponential growth as
1.5 > 1
f=
(0) h=
(0) 1 So f and h are either
A or D.
h has exponential decay so it must be
line A.
Thus f is line D.
g(0) = 2
ii y = 2
4 H (65)2
=
−
t
2
iii
x∈¡ , y < 2
+ 25
a
t =0 ⇒ H =65 ⋅ 20 + 25 =65 + 25 =90oC
b
t =3 ⇒ H = 65 ⋅ 2
c
H ≤ 40
1
= 2=
j(0)
4
65 ⋅ 2
−2
65 ⋅ 2
Thus g is C, j is E and finally i is B.
2
3 a i
2
2
−
t
2
−
t
2
−
t
2
−
t
2
−
t
2
−
3
2
+ 25 = 48oC
+ 25 ≤ 40
≤ 15
15
3
=
65 13
1
≤
4
≤
≤ 2−2
t
≤ −2
2
t ≥4
−
d
ii y= 4
iii x ∈ ¡ , y > 4
5 a
b i
25oC is the temperature of the room
as that is the only constant in the
equation.
y = 30(0.9)x
x =0
0
=
y 30(0.9)
=
30
The value of a new car is $30 000.
b
x =3
y = 30(0.9)3
y = 21.87
The value of a 3 year old car is
$21 870.
ii y = − 3
iii x ∈ ¡ , y > −3
x
c =
y 30(0.9)
=
x
=
(0.9)
30
= 15
2
15 1
=
30 2
Using GDC we find x = 6.58 .
6 a
P = 40(1.5)t
t =0
0
=
P 40(1.5)
=
40
There were initially 40 squirrels.
© Oxford University Press 2019
3
Worked solutions
b
t =2
=
y 7(1 + 0.011)t billion.
2
=
P 40(1.5)
=
90
b
x
c P 40(1.5)
=
=
200
x
=
(1.5)
y =
7(1 + 0.011)14 =
8.16 billion.
200
= 5
40
c
−
A = A0 (2)
t
5730
From the graph t = 32.6 so in the
year 2043.
7 The value of a car decreases by 15%
every year. Tatiana buys a new car for
$25 000.
A0 = 100
t = 1000
−
A = 100(2)
A = 88.6
1000
5730
Use the formula for exponential decay
with a = $25000 and r = 0.15 .
b Use GDC to sketch the graph.
a y 25000(1 − 0.15)t in thousands
=
of dollars
t
−
5730
i A 100(2)
=
= 75
b
t
5730
75
3
2= =
100 4
−
t
3
$15400
= 25000(0.85)
=
c
100
= 50
2
5730
ii =
A 100(2) =
t =3
=
y 25000(1 − 0.15)3
That gives t = 2378 years.
−
y =+
7(1 0.011)t =
10
10
(1 + 0.011)t =
7
Using GDC we find x = 3.97 .
7 a
t = 2025 − 2011 = 14
y = 10 000
(0.85)𝑡𝑡 =
From the graph t = 5730 years.
10000
25000
=
2
5
Exercise 9E
e = 2.718
b
e2 = 7.389
c
e −2 = 0.135
d
3e = 8.155
e
e
= 1.359
2
f
5 e = 8.244
g
4e − 5 =
5.873
1 a
x
2 a =
y f (x=
) 1=
1 is a line.
b The graph of h(x ) is between the
graphs of g(x ) and i(x ) . They are all
exponential graphs.
3 The transformation that maps f (x ) onto
g(x ) is a reflection in the y − axis.
4 a
t =0
=
G(0) 4500
=
e0.3⋅0 4500
=
e0 4500cm2
b
t = 10
=
G(10) 4500
=
e0.3⋅10 4500e3
From the graph t = 5.64 years.
Exercise 9F
1 a The graph of g(x ) is a vertical
translation of 4 units up
b The graph of h(x ) is a horizontal
translation of 3 units right.
c The graph of i(x ) is a vertical stretch
of scale factor 2.
2 a
= 90 400 cm2 (3 s.f.)
5
PV = 5000
r = 0.05
t =6
FV = 5000 ⋅ e0.05⋅6 = 5000e0.3 = $6749
6 a The population a= 7 billion was
growing at a rate of r= 1.1% so the
exponential growth formula gives
© Oxford University Press 2019
4
Worked solutions
b
c
10x =
25 ⇒ log25 =
x
d
1
1
3−2 =
⇒ log3 =
−2
9
9
e
2
2
273 =
9 ⇒ log27 9 =
3
3 a
log7 1 ⇒ 7x =
1
7x = 70
x =0
log7 1 = 0
c
b log8 1 ⇒ 8x =
1
8x = 80
x =0
log8 1 = 0
c
log9 1 ⇒ 9x =
1
9x = 90
d
x =0
log9 1 = 0
d logx 1 ⇒
xy = 1
xy = x0
y =0
logx 1 = 0
4 a
log3 3 ⇒ 3x =
3
3x = 31
x =1
log3 3 = 1
3
b log4 4 ⇒ 4x =
4
4x = 41
x =1
log4 4 = 1
c
5x = 51
x =1
log5 5 = 1
x = −1.98, 2.72
Exercise 9G
1 a
logp q =r ⇒ pr =q
d logx x ⇒
xy = x
b log3 5 =r ⇒ 3 =5
r
c
x y = x1
y =1
logx x = 1
log7 q =6 ⇒ 76 =q
d logp 5 =3 ⇒ p3 =5
e
2 a
b
log5 5 ⇒ 5x =
5
log11 =
x ⇒ 10x =
11
rs =
t ⇒ logr t =
s
82 =⇒
64
log8 64 =
2
5 a
log3 9 ⇒ 3x =
9
3x = 32
x =2
log3 9 = 2
© Oxford University Press 2019
5
Worked solutions
b log2 32 ⇒ 2x =
32
f
2x = 25
x =5
log2 32 = 5
c
x
log5 125 ⇒ 5 =
125
d log4 256 ⇒ 4x =
256
4 x = 44
x =4
log4 256 = 4
k
1
3
l
log6 + 3log3 − log2 = log6 + log33 − log2
1
⇒
27
=log6 + log27 − log2
= log6 ⋅ 27 − log2
= log162 − log2
162
= log
2
= log81
1
27
3x = 3−3
x = −3
1
= −3
log3
27
3x =
Exercise 9I
1 a
Exercise 9H
log3 + log5
= log3 ⋅ =
5 log15
16
= log8
2
c
3
3log5
= log5
=
log125
d
3log 4 − 4log3 =log 43 − log34
= log64 − log81
64
= log
81
e
5log2 + 4log3 =log25 + log34
= log32 + log81
= log32 ⋅ 81
= log2592
8x = 8
1
x =
3
1
log8 2 =
3
b log16 − log2 = log
log2 + log3 + log=
4 log2 ⋅ 3 + log 4
= log12 − log 4 + log3
12
= log
+ log3
4
= log3 + log3
= log3 ⋅ 3
= log9
8x = 2
1 a
= log128 − log64
128
= log
64
= log2
j
log12 − 2log2 + log3 = log12 − log22 + log3
1
2
log8 2 ⇒
g log3
h log128 − 6 log2 = log128 − log26
= log6 + log 4
= log6 ⋅ 4
= log24
25x = 5
f
= log32 + log25 = log32 × 25
= log800
i
log25 5 ⇒
25x = 25
1
x =
2
1
log25 5 =
2
236
= log236
1
g 5log2 + 2log5 =log25 + log52
5x = 53
x =3
log5 125 = 3
e
log236 − log1
= log
log x + log1
= log x ⋅=
1 log x
log18
= log3 ⋅ 6
= log3 + log6
= x+y
b log2 = log
6
3
= log6 − log3
= y−x
c
log9 = log32
= 2log3
= 2x
d log27 = log33
© Oxford University Press 2019
6
Worked solutions
= 3log3
= 3x
e
= log5 52 ⋅ 4
= log5 52 + log5 4
2
log36 = log6
= 2log5 5 + n
= 2+n
= 2log6
= 2y
f
3
1
log
= log1 − log2
2
y = log2 B
z = log2 C
6
= 0 − log
3
=
−(log6 − log3)
=
−(y − x )
= x−y
2 a
A 4
A
) = 4log2 (
)
BC 3
BC 3
= 4(log2 A − log2 (BC 3 ))
log2 (
4(x − (log2 B + log2 C 3 ))
=
= 4(x − (y + 3log2 C ))
log=
log5 7 ⋅ 4
5 28
= 4(x − y − 3z )
= log5 7 + log5 4
4=
log3 P x=
,log3 Q y
= m+n
a
7
b log
log5 7 − log5 4
=
5
4
= 3x + y
P
b log
=
log3 P − log3 Q
3
Q
log5 49 = log5 72
= 2log5 7
1
= 2m
= log3 P 2 − log3 Q
1
log3 P − log3 Q
2
1
=
x−y
2
d log5 64 = log5 43
=
= 3log5 4
= 3n
e
3
log=
log3 P 3 + log3 Q
3 P Q
= 3log3 P + y
= m−n
c
x = log2 A
49
=
log
log5 49 − log5 4
5
4
5 a
log x − log(x − 5) =
log M
log
= log5 72 − log5 4
= 2log5 7 − log5 4
= 2m − n
f
x
= log M
x −5
x
=M
x −5
b log x − log(x − 5) =
1
7
log=
log5 7 − log5 16
5
16
= log10
2
= m − log5 4
log
= m − 2log5 4
= m − 2n
g log5=
112 log5 7 ⋅ 42
= log5 7 + log5 42
= log5 7 + 2log5 4
x
= log10
x −5
x
= 10
x −5
x 10(x − 5)
=
= 10 x − 50
50 = 9 x
= m + 2n
x =
h
log5 7 m
=
log5 4
n
i
log5 49 log5 72 2log5 7 2m
= = =
log5 64 log5 43 3log5 4
3n
j
log
=
log5 25 ⋅ 4
5 100
50
9
Exercise 9J
1 a =
=
e 3
ln e3 3ln
b =
ln e4 4ln
=
e 4
© Oxford University Press 2019
7
Worked solutions
c
1
1
1
=
ln e
2
2
1
1
1
=
ln e
3
3
ln=
e ln
=
e2
3
d ln=
e ln
=
e3
e
f
1
=
ln1 − ln e =
0 − ln e =
0 −1 =
−1
e
1
ln 2 =−
ln1 ln e2 =
0 − 2ln e =
−2
e
ln
2 a
eln 2 = 2
b
eln 3 = 3
c
eln x = x
d
4
4
2
e2 ln=
eln=
4=
16
ln x 3
e= e= x
f
ln 3
3
e −=
e=
3 ln x
ln
1
1
3
f
log
=
6 18
g
ln x = 2.7
x +1 =
e1.86
1.86
= e
x
−=
1 5.42
x ln a
ln a
e=
e=
ax
1 a
log3 8 = 1.89
b
c
log5 8 = 1.29
d` log3 30 = 3.10
e
log7
1
= −0.712
4
f
log6 24 = 1.77
log2
3
= −0.737
5
p = log3 A
b
x = 2.32
logx 3 r
=
=
log
63
logx 6 s
b
3x = 17
log3x = log17
x log3 = log17
c
logx 6
s
2
=
=
=
log
log
2log
2=
2
3 36
36
36
r
logx 3
logx 54
logx 3
2x = 5
log2x = log5
x log2 = log5
log5
x =
log2
logx 6 s
=
logx 3 r
d log3 54 =
log10
1
=
log e
log e
Exercise 9L
log3 B q
=
=
log
A B
log3 A p
log
=
36
logy y
1
=
logy x logy x
6 “Show that” is to use numbers to
demonstrate a certain property and that it
works for the numbers that you are using.
To prove it to use variables to prove that
the system works for all numbers.
1 a
q = log3 B
3 a
logx 6 − logx 3 s − r
=
logx 3
r
5 =
ln10
Exercise 9K
logx 18 logx (6 ⋅ 3)
=
logx 6
logx 6
logx 6 + logx 3 s + r
=
logx 6
r
4 log
=
x y
x
logx 6
s
s
s
= =
=
logx 9 logx 32 2logx 3 2r
6
logx 2 logx 3
log
=
=
32
logx 3 logx 3
=
b ln(x + 1) =
1.86
2
=
log
9 6
=
3
2.7
=
x e=
14.9
4
e
2
e
3 a
logx (6 ⋅ 9)
r
logx 6 + logx 32
=
r
s + 2logx 3
=
r
s + 2r
=
r
=
x =
log17
log3
x = 2.58
c
9x = 49
log9x = log 49
x log9 = log 49
log 49
x =
log9
x = 1.77
© Oxford University Press 2019
8
Worked solutions
d
3x = 69
2 a
log3x = log69
x log3 = log69
log69
x =
log3
x = 3.85
log24 x = log9
(4 x )log2 = log9
log9
x =
4log2
x = 0.792
b
e 16 = 67
x
x = 1.52
f
12 = 5
c
log7x = log 4
x log7 = log 4
x
1
d
2x +1 = 15
log2x +1 = log15
(x + 1)log2 =
log15
log 4
log7
x = 0.712
x =
log15
x +1 =
log2
=
x
h 19x = 2
log19x = log2
x log19 = log2
log2
log19
x = 0.235
e
j
x
e = 10
x
log e = log10
x log e = 1
1
log e
x = 2.30
x =
6
x −2
=4
log6 x −2 = log 4
(x − 2)log6 =
log 4
log 4
x −2 =
log6
log 4
x
=
+2
log6
ex = 5
log e x = log5
x log e = log5
log5
x =
log e
x = 1.61
log15
−1
log2
x = 2.91
x =
i
x
1
x log5 = log79
2
2log79
x =
log5
x = 5.43
log5
x =
log12
x = 0.65
7x = 4
1
52 = 79
log52 = log79
log12x = log5
x log12 = log5
g
63 x = 4
log63 x = log 4
(3x )log6 = log 4
log 4
x =
3log6
x = 0.258
log16 x = log67
x log16 = log67
log67
x =
log16
x
24 x = 9
x = 2.77
f
e
x −1
−4=
6
e x −1 = 10
log e x −1 = log10
(x − 1)log e =
1
1
x −1 =
log e
1
log e
x = 3.30
x= 1 +
© Oxford University Press 2019
9
Worked solutions
g
b
23 x −2 = 53
4 × 63 x =
16
16
= 4
4
log63 x = log 4
3x log6 = log 4
log 4
x =
3log6
x = 0.258
log23 x −2 = log53
(3x − 2)log2 =
log53
log53
3x − 2 =
log2
3x= 2 +
log53
log2
1
log53
(2 +
)
3
log2
x = 2.58
x
=
h
2 x +1
4
= 10
log 42 x +1 = log10
(2 x + 1)log 4 =
1
1
2x + 1 =
log 4
2x
=
1
−1
log 4
1 1
(
− 1)
2 log 4
x = 0.330
x
=
i
11x −8 − 11 =
48
11x −8 = 59
log11x −8 = log59
(x − 8)log11 =
log59
log59
x −8 =
log11
log59
x= 8 +
log11
x = 9.70
j
c
3 × 4e2 −2 x + 1 =
4
3 × 4e2 −2 x =
3
3
e2 −2 x =
12
1
e2 −2 x =
4
1
log e2 −2 x = log
4
1
(2 − 2 x )log e =
log
4
1
log
2 − 2x = 4
log e
1
log
4
2 x= 2 −
log e
1
log
1
4)
(2 −
x
=
2
log e
x = 1.69
d 10 − 2e7 x +5 =
3
2e7 x +5 = 7
7
e7 x + 5 =
2
9x +10 + 22 =
100
9x +10 = 78
x +10
= log78
log9
(x + 10)log9 =
log78
log78
x + 10 =
log9
log78
=
− 10
x
log9
x = -8.02
3 a
3x
6=
7
2
7
(7 x + 5)log e =
log
2
7
log
7x + 5 = 2
log e
log e7 x +5 = log
7
2 −5
=
7x
log e
log
6 × 2x =
14
2x =
7
log
1
2 − 5)
=
x
(
7 log e
x = - 0.535
14
6
14
6
14
x log2 = log
6
14
log
6
x =
log2
x = 1.22
log2x = log
© Oxford University Press 2019
10
Worked solutions
e
5 a
2x −1 = 3x +1
e2 x − 4e x − e x + 4 =
0
log2x −1 = log3x +1
(x − 1)log2 =
(x + 1)log3
x log2 − log2 = x log3 + log3
e x (e x − 4) − (e x − 4) =
0
(e x − 4)(e x − 1) =
0
x(log2 − log3) =log3 + log2
log3 + log2
x =
log2 − log3
x = - 4. 42
f
Hence e x = 4 or e x = 1 . Thus
ex = 4
log e x = log 4
x log e = log 4
32 x −1 = 5x
log32 x −1
(2 x − 1)log3
2 x log3 − log3
x(2log3 − log5)
x
=
= log5x
=
x log5
=
x log5
=
log3
ex = 1
log3
2log3 − log5
x = 1.87
g
4
log e x = log1
x log e = 0
x = 0.
1− 2 x
=6
log 43 x +1
(3x + 1)log 4
3x log 4 + log 4
x(3log 4 + 2log6)
= log61−2 x
=
(1 − 2 x )log6
= log6 − 2 x log6
=
log6 − log 4
log6 − log 4
x =
3log 4 + 2log6
x = 0.0524
h
log e
= log5
3ln2 + ln3 = ln23 + ln3
= ln8 + ln3 = ln(8 ⋅ 3) = ln24
b
b
6 ln2 − ln 4 =
− ln x
− ln x
ln26 − ln 4 =
26
= − ln x
4
64
= − ln x
ln
4
ln16 = − ln x
ln x = − ln16
ln
x = e − ln16
1
x =
16
e2 x − 2e x − 3 =
0
0
e2 x − 3e x + e x − 3 =
0
e x (e x − 3) + (e x − 3) =
(e x − 3)(e x + 1) =
0
Therefore, e x − 3 =
0 or e x + 1 =
0.
Hence
ex − 3 =
0
x −2
(x + 1)log e =
(x − 2)log5
x log e + log e = x log5 − 2log5
− log e − 2log5
x(log e − log5) =
− log e − 2log5
x =
log e − log5
x = 6.92
4 a
Hence x = 0,ln 4.
e x +1 = 5 x − 2
x +1
log 4
= ln 4
log e
or
x =
3 x +1
e2 x − 5e x + 4 =
0
ex = 3
ln e x = ln3
x ln3 = ln3
x = ln3
Or e x + 1 =
0 which has no real
solutions. Therefore x = ln3 .
c
0
e4 x + 4e2 x − 12 =
0
e4 x + 6e2 x − 2e2 x − 12 =
e2 x (e2 x + 6) − 2(e2 x + 6) =
0
(e2 x + 6)(e2 x − 2) =
0
Hence either e2 x + 6 =
0 or e2 x − 2 =
0.
e2 x + 6 =
0 has no real solutions thus it
remains e2 x − 2 =
0 which gives
0
e2 x − 2 =
e2 x = 2
ln e2 x = ln2
2 x ln e = ln2
2 x = ln2
ln2
x =
.
2
© Oxford University Press 2019
11
Worked solutions
6 a
b
10(1.075)k ⋅0
t = 0 ⇒ h(0) =
⇒ W(10) =
84 − 10ln(10 + 1)
0
= 10(1.075)
=
10 cm
b
=
84 − 10ln11 =
60
h(4) = 12
k ⋅4
10(1.075)4k
=
h(4) 10(1.075)
=
c
12 = 10(1.075)4k
12
(1.075)4k =
10
12
10
12
4k log(1.075) = log
10
12
log
1
10
k =
4 log(1.075)
k = 0.630
9 a
100
1
250
= 500(0.5)
=
499
t
25000
b A(t ) 500(0.5)
=
=
100
t
t
log0.525000 = log0.2
t
log0.5 = log0.2
25000
25000log0.2
t =
log0.5
t = 58048
log2
= 15.2
0.630log(1.075)
t =0
= 20000(0.9)0 + 1000
= 20000 + 1000 = 21000
10 2x = 3e4 x
P(3) = 16000
ln2x = ln3e4 x
3k
=
P(3) 20000(0.9)
+=
1000 16000
20000(0.9)3k = 15000
15000
3k
=
= 0.75
(0.9)
20000
log(0.9)3k = log0.75
3k log(0.9) = log0.75
=
k
c
log0.75
= 0.910
3log0.9
1 a
20000(0.9)t ⋅0.910 + 1000 =
5000
t ⋅0.910
20000(0.9)
= 4000
4000
t ⋅0.910
=
= 0.2
(0.9)
20000
log(0.9)t ⋅0.910 = log0.2
0.910t log0.9 = log0.2
8 a
log0.2
= 16.8
0.910log0.9
t =0
⇒ W(0) =
84 − 10ln(0 + 1)
= 84 − 10ln1
= 84 − 10 ⋅ 0 = 84
x ln2
= ln3 + ln e4 x
x ln2
= ln3 + 4 x ln e
x ln2
= ln3 + 4 x
ln3
x(ln2 − 4) =
ln3
x =
ln2 − 4
Exercise 9M
P(t ) = 5000
=
t
100 1
=
500 5
=
0.525000
=
⇒ P(0) 20000(0.9)k ⋅0 + 1000
b
t = 100
⇒ A(100) =
A0 (0.5)25000
t ⋅k
20 10(1.075)
10(1.075)t ⋅0.630
=
=
20
(1.075)t ⋅0.630
= = 2
10
log(1.075)t ⋅0.630 = log2
log2
t ⋅ 0.630log(1.075) =
7 a
100
= 50
2
=
t e3.4 − 1
t = 29
2 ⋅ h(0) =
20
h(t ) =
t
=
W=
(t )
84 − 10ln(t + 1) =
50
10ln(t + 1) = 84 − 50 = 34
34
ln(t + 1) =
10
t +1 =
e3.4
log(1.075)4k = log
c
t = 10
(7e x ) ' = 7e x
b (−
c
1 x
1
− ex
e )' =
4
4
1
(9ln =
x )' 9=
x
d (π ln x )' =
e
f
9
x
π
x
1
5
1
(5x=
)' =
5x
5x
x
(6 x )'
6
1
(ln 6 x=
)'
= =
6x
6x
x
(ln5x=
)'
© Oxford University Press 2019
12
Worked solutions
g (ln7 x=
)'
(7 x )'
7
= =
7x
7x
1
x
3 a
3
c
c
+ ln x )' = 2 x − e
1
− x
2
1
1
(− ) +
2
x
1 − 12 x 1
e
+
2
x
g ((x 2 + 1)e3 x )'
= (x 2 + 1)' e3 x + (x 2 + 1)(e3 x )'
= 2 xe3 x + 3(x 2 + 1)e3 x
h (xe ax
(e
5
− 5ln x + 6e )' =
− + 6e4 x (4 x )'
x
=
−
5
+ 24e4 x
x
4x
3
x
3
2
(e x )' e x + 3x e
=
ex x3
ex x3
+1
)'
= e ax
i
2
+1
+1
(ax 2 + 1)'
+ 2ax 2e ax
2
+1
( x=
ln x )' x 'ln x + x(ln x )'
= ln x + x
j
2
1
= ln x + 1
x
(=
x 3 ln x )' (x 3 )'ln x + x 3(ln x )'
x3
x
= 3x 2 ln x + x 2
= 3x 2 ln x +
x2 + 1
x3 − x x2 + 1
=
))'
(
)'
x3 − x
x2 + 1 x3 − x
= x 2 (3ln x + 1)
1 2 x(x 3 − x ) − (x 2 + 1)(3x 2 − 1)
x2 + 1
x3 − x
4
2
1 2 x − 2 x − 3x 4 + x 2 − 3x 2 + 1
= 2
x +1
x3 − x
4
2
1 − x − 4x + 1
= 2
x +1
x3 − x
− x 4 − 4x 2 + 1
= 2
(x + 1)(x 3 − x )
2 x(x 3 − x ) − (x 2 + 1)(3x 2 − 1)
(x 2 + 1)(x 3 − x )
=
e ax +1 + xe ax
x
x +3
3
= 1+
x
x
2x
3x 2 − 1
− 3
2
x +1 x − x
2
2
=
=
2x
(xe
=
)' x ' e2 x + x(e2 x )'
= 6 x 2 (1 − x )e −3 x
x 3 − x (x 2 + 1)'(x 3 − x ) − (x 2 + 1)(x 3 − x )'
= 2
(x 3 − x )2
x +1
=
3
(ln x )2
x
= 6 x 2e −3 x − 6 x 3e −3 x
1
= + 7e7 x − 7
x
g (ln(
5
x
(2
=
x 3e −3 x )' 2(x 3 )' e −3 x + 2 x 3(e −3 x )'
(7 x )'
=
+ e7 x (7 x )'− 7
7x
=
(3x 5 )' 15x 4 5x 4
=
= =
3x 5
3x 5
x5
= 6 x 2e −3 x + 2 x 3(−3x )' e −3 x
d (ln7 x + ln7 + e7 x − 7 x )'
x 3
x ))'
(ln(e=
+ 5)2
d ((ln x )3 )' 3(ln
=
=
x )2 (ln x )'
f
1
=
− − 5e −5 x + 3x 2
x
f
3
=
e2 x + 2 xe2 x =
(1 + 2 x )e2 x
(4 − ln9 x + e −5 x + x 3 )'
x
((4 x 3 + 5)2 )'
(2(4 x 3 + 5)(4 ⋅ 3x 2 ))
5
(ln(3x=
))'
e
(9 x )'
=
−
+ e −5 x (−5x )'+ 3x 2
9x
e
+ 5)2
+ 5)2
= 24 x (4 x 3 + 5)e(4 x
5
(5ln x − 2e )' =− 2e x
x
10
3
3
2
x
2x +
=
3
2
= e(4 x
5x
j=
(e5 x )' e=
(5x )' 5e5 x
1
− x
2
3
b=
(e(4 x +5) )' e(4 x
4x
i=
(e4 x )' e=
(4 x )' 4e4 x
b (x 2 − e
3
= 6 x 2 e2 x
2x
h=
(e2 x )' e=
(2 x )' 2e2 x
2 a
3
2x
(e=
)' e2 x (2
=
x 3 )' e2 x (2 ⋅ 3x 2 )
k
(x 2 ln(2 x + 3))'
= (x 2 )'ln(2 x + 3) + x 2 (ln(2 x + 3))'
(2 x + 3)'
2x + 3
2x 2
= 2 x ln(2 x + 3) +
2x + 3
= 2 x ln(2 x + 3) + x 2
l
(
e3 x
(e3 x )' x 2 − e3 x (x 2 )'
)' =
2
x
x4
3e3 x x 2 − 2 xe3 x
3xe3 x − 2e3 x
=
4
x
x3
e3 x (3x − 2)
=
x3
=
 2e4 x 
(2e4 x )'(1 − e x ) − 2e4 x (1 − e x )'
m 
'=
x 
(1 − e x )2
1 − e 
© Oxford University Press 2019
13
Worked solutions
ex + 1
)'
ex − 1
(e x + 1)'(e x − 1) − (e x + 1)(e x − 1)'
=
(e x − 1)2
n (
o
y − 2= 4(x − 0)
y −2 =
4x
=
y 4x + 2
8e4 x (1 − e x ) + 2e4 x e x 2e4 x (4 − 3e x )
=
(1 − e x )2
(1 − e x )2
=
=
e x (e x − 1) − e x (e x + 1)
(e x − 1)2
=
e2 x − e x − e2 x − e x
(e x − 1)2
=
−2e x
(e x − 1)2
(
p (
6
y(1) = 0
1
y' =
x
y '(1) = 1
The gradient at the point (1, 0) is 1.
Hence the equation of the tangent at the
point (1, 0) is:
y − 0 = 1(x − 1) = x − 1
y= x − 1
x
x 'ln x − x(ln x )' ln x − 1
)' =
=
ln x
(ln x )2
(ln x )2
The product of the tangent and the
normal is −1 . Thus the gradient of the
normal at (1, 0) is −1 Therefore the
equation of the normal at the point (1, 0)
is:
2 − ln x
(2 − ln x )' x − (2 − ln x )x '
)' =
x
x2
1
− x − 2 + ln x
−3 + ln x
x
= =
x2
x2
1 + ln x
(1 + ln x )' x 2 − (1 + ln x )( x 2 )'
q (
)' =
2
x
x4
1 2
x − 2 x(1 + ln x )
= x
x4
x − 2 x(1 + ln x )
=
x4
1 − 2(1 + ln x )
=
x3
y − 0 =−1(x − 1)
y =− x + 1
7 The point where x = 6 is f (6)
= e −6 + 4 .
f '(x ) =
(e − x + 4)' =
−e − x
f '(6) = −e −6
Hence we get that the value of the
gradient of the tangent of f (x ) at
(6, e −6 + 4) is −e −6.
4 A turning point has the first derivative
equal to 0 .
y ln x − x
=
1
y=
'
−1
x
8
f (x=
) x 2 + ln x
) 2x +
f '(x
=
1
x
f '(x ) = 3 ⇔ 2 x +
1
=3 ⇔
x
2x 2 + 1 =
3x
1
y '(x ) = 0 ⇔ − 1 = 0 ⇔ x = 1
x
y(1) = −1
5
y = ln x
2 x 2 − 3x + 1 =
0
Therefore (1, −1) is a turning point. To
check if it is a maximum or minimum we
need the second derivative.
2(x 2 − x ) − x + 1 =
0
2 x(x − 1) − (x − 1) =
0
(x − 1)(2 x − 1) =
0
(x − 1)(2 x − 1) =
0
1
1
y '' =
( − 1)' =
− 2 < 0 . Hence (1, −1) is a
x
x
maximum.
Hence the values for which the derivative
1
is 3 are x = 1 and x = .
2
f (x ) = 2e2 x
The y − coordinates are:
f (1) =
1 + ln1 =
1 and
x
(2 x )' 4e2 x
=
f '(x ) (2
=
e2 x )' 2e2=
=
f (0) 2=
e 2 ⋅0 2
2 ⋅0
'(0) 4=
4
f=
e
Equation of the tangent at the point (0,2)
is:
1
1
1 1
f ( ) = + ln = − ln2. Thus the points
2
4
2 4
1 1
are (1,1) and ( , − ln2).
2 4
© Oxford University Press 2019
14
Worked solutions
22 x −2 = 24
2x − 2 =
4
9 =
f (x ) ln(e x + e − x )
(e x + e − x )' (e x − e − x )
=
(e x + e − x ) (e x + e − x )
f '(x ) = 0.6
f '(x )
=
2x = 6
x =3
(e x − e − x )
= 0.6
(e x + e − x )
b
x −1
93 =
32(3 x −1) = 3−3
2(3x − 1) =
−3
x
0.6(e x + e − x )
e x − e −=
0.4e x − 1.6e − x =
0
6 x − 2 =−3
6 x = −1
1
x = −
6
0.4e2 x − 1.6 =
0
1.6
e2 x
=
= 4
0.4
ln e2 x = ln 4
c
2 x ln e = ln22
2 x = 2ln2
x = ln2
f (ln2) = ln(e
+e
− ln 2
1
5
) = ln(2 + ) = ln
2
2
5
Thus the point is (ln2,ln ).
2
10 f (=
x ) xe − e
2−x =
5
x = −3
d
The gradient of the tangent at x = 1 is
f '(1) = e while the gradient of the normal
is:
−15
= −1
15
1
log6 x = log
10
1
6x =
10
1
x =
60
log6 x =
y − 0= e(x − 1)
y = e(x − 1) = ex − e.
The equation of the normal is:
1
1
1
(x − 1) =− x + .
e
e
e
b log(−7 x ) =
3
log(−7 x ) =
log103
1000
−7 x =
−1000
x =
7
Chapter review
1 a
log2 16 =4 ⇒ 24 =16
b log5 125 =3 ⇒ 53 =125
c
log9 81 =2 ⇒ 92 =81
1
64
4 a 15log6 x = −15
1
. Hence the equation of the tangent
e
y − 0 =−
41−2 x =
41−2 x = 4−3
1 − 2x =
−3
2x = 4
x =2
x
f '(x ) = e x + xe x − e x = xe x
is −
32 − x = 243
32 − x = 35
ln 2
x
1
= 3−3
27
c
3log10 x = −6
d log12 144 =
2 ⇒ 12 =
144
log10 x = −2
e
log10 x = log10−2
1
10 x =
100
1
x =
1000
2
2 a
log10000 =
4 ⇒ 104 =
10000
34 =⇒
81
log3 81 =
4
b 152 =
225 ⇒ log15 225 =
2
c
1
1
812 =
9 ⇒ log81 9 =
2
d
a
e
4
e4 =
x ⇒ ln x =
3 a
14
=
c ⇒ loga c =
14
22 x −2 = 16
d
− log 4 x =
−2
log 4 x = 2
log 4 x = log102
4 x = 100
x = 25
© Oxford University Press 2019
15
Worked solutions
5 a
e6 x + 8 = 5
2
log
=
log
=
2
5 25
55
log e6 x + 8 = log5
(6 x + 8)log e =
log5
log5
6x + 8 =
log e
7
b log=
log
=
7
2 128
22
c
log21 21 = 1
d loga a = 1
log5
−8
log e
x =
6
x = -1.07
0
e =
log6 1 log
=
0
6 6
f
log3 (79)
3
= 79
g ln e = 19
eln 7 = 7
19
h
6 a
g
2x = 17
4e6 x + 9 − 3 =
30
4e6 x + 9 = 33
33
e6 x + 9 =
4
4.09
x log
=
=
2 17
b
66 x +3 = 19
33
4
33
(6 x + 9)log e =
log
4
33
log
4 −9
log e
x =
6
x = -1.15
log e6 x + 9 = log
6x + 3 =
log6 19
log6 19 − 3
6
x = -0.226
x =
c
2 × 123 x =
11
123 x =
11
2
11
2
11
3x log12 = log
2
11
log
2
x =
3log12
x = 0.229
log123 x = log
d
h 5 − 4e −5 x − 4 =
−16
4e −5 x − 4 = 21
21
e −5 x − 4 =
4
21
4
21
(−5x − 4)log e =
log
4
21
log
−5x − 4 = 4
log e
log e −5 x − 4 = log
6 × 8−5 x =
18
8−5 x = 3
log8−5 x = log3
−5x log8 =
log3
− log3
x =
5log8
x = -0.106
e
e x +5 = 13
log e x +5 = log13
(x + 5)log e =
log13
log13
x +5 =
log e
log13
=
x
−5
log e
x = -2.44
f
2e6 x + 8 = 10
21
4 +4
log e
x = −
5
x = -1.13
log
7 a
log3 + log 4
= log3 ⋅ 4
= log12
b log15 − log5= log
15
= log3
5
c
2log x − 5log y = log x 2 − log y 5 = log
d
8log5 x + 2log5 y =log5 x 8 + log5 y 2
x2
y5
= log5 x 8y 2
e
ln x +
1
1
1
1
ln y + ln z =ln x + ln y 2 + ln z
2
2
2
= ln x y
© Oxford University Press 2019
1
z
2
16
Worked solutions
f
4ln x − 3ln y − 2ln z = ln x 4 − ln y 3 − ln z 2
= ln
8 a
x4
y 3z2
d x=1
ln ab =
ln a + ln b =
p+q
13 f (x ) = 2x
b =
ln a
3ln
=
a 3p
3
c
2
3
2
g(x ) = -(2− x ) - 2
3
ln a b = ln a + ln b = 2ln a + 3ln b
= 2 p + 3q
d
b5
ln 4 = ln b5 − ln a4 = 5ln b − 4ln a
a
= 5q − 4 p
9 a
log
=
3 17
b log5 0.5 =
c
c A reflection in the y-axis and a
horizontal translation of 2 units to the
right.
14 f (x ) = ln x
=
g(x ) 3ln(x + 5)
15 a
(8e x + 7ln x )' =
8e x +
b=
(e3 x )' (3
x )' e3 x 3e3 x
=
log17
= 2.58
log3
c
log0.5
= −0.431
log5
(x ln x − x )'= ln x +
x
−1
x
= ln x + 1 − 1= ln x
2
d=
(e6 x +5 x )' e6 x
log200
log
=
= 2.55
8 200
log8
= (12 x + 5)e6 x
10 1.02, 5.65
11 a
e
(ln(x 2 + 8))'
=
f
(
2
2
+5 x
(6 x 2 + 5x )'
+5 x
(x 2 + 8)'
2x
=
(x 2 + 8) (x 2 + 8)
9e x + 1
)'
2e x + 1
=
(9e x + 1)'(2e x + 1) − (9e x + 1)(2e x + 1)'
(2e x + 1)2
=
9e x 2e x + 9e x − 9e x 2e x − 2e x
(2e x + 1)2
=
7e x
(2e x + 1)2
2)'
g (ln 3x −=
=
3x − 2 '
(3x − 2)'
=
3x − 2
2( 3x − 2)2
3
2(3x − 2)
h (e x =
ln x )' e x ln x +
b x=0
c Vertical stretch of scale factor 5 and a
vertical translation of 2 units down.
12 a
7
x
16 f (x ) = 4 xe x
2
−1
2
f '(x ) =
4e x −1 + 4 xe x
= 4e x
2
ex
x
−1
2
+ 8 x 2e x
−1
2
(x 2 − 1)'
−1
At point (1, 4) the tangent has gradient
f '(1) = 4 + 8 = 12 so the normal has
gradient −
1
.
12
Therefore the equation of the normal is
−1
(x − 1)
12
12y − 48 =− x + 1
0
x + 12y − 49 =
y −=
4
b x=2
© Oxford University Press 2019
17
Worked solutions
23 a
17 f (x=
) x 2 + ln x
) 2x +
f '(x
=
f '(2) = 4 +
18 a
1
x
1 9
= = 4.5
2 2
t =⇒
0
P(0) =
30e0.032⋅0 =
30e0 =
30
Hence the population in 2020 is 30000.
b
t =
5 ⇒ P(5) =
30e0.032⋅5 =
35.205
Hence the population in 2025 is
35205.
c P(t ) 30
=
=
e0.032⋅t 40
A1 shape A1 domain
40 4
=
30 3
4
0.032 ⋅ t =
ln
3
4
ln
3
t
9
=
=
0.032
⋅t
e0.032
=
b T (6=
) 25 + e
0.4×6
36.0 (1 d.p.)
c Solve 25 + e0.4t =
100
19 V (t ) 150
=
=
000e0.05875t 200 000
24 a
b
200 000 4
=
e
=
150 000 3
A1
A1
1
3
3
x −2
3
=1 ⇒ x = 5
x −2
x 3ln3 − 4 )
(or y =+
25 a
f (24) = 500(0.75)24
log2 3x − log2 ( x − 3)
log2
f (24) = 0.502
¡
A1
−2
A1
12
y < 16
A1
A1
x > −2
y = 16
A1
A1
N ( 0 ) = 35
A1
3x
x −3
2
26 a i
M1
M1
A1
A1
A1
b ln x 3 − ln ( x − 1) + ln e
ln
M1A1
A1
y − (3ln3 + 1) = x − 5
20 f (t ) = 500(0.75)t
ex 3
A1A1A1
A1
( x − 1)
2
T5 = 73.205 thousand taxis
M1A1
ii
Tn= 100 ⇒ n= 9
2019
M1A1
M1
A1
N ( t ) > 1000
M1
b
t > 5.449...
A1
A1
c Adjusting units in (i) or (ii)
0 ≤ t < 5.45
−
=
f (5) 3ln3 + 1
Hence the predicted year is 2025.
d
x >2
x =2
f '(x ) =1 ⇒
4
ln
3
=
t
= 4.90
0.05875
c
A1
d f ′(x) =
4
3
b N ( 4 ) = 410
10 hours 48 minutes
c 3ln ( x − 2 ) + 1 = 0 ⇒ x = 2 + e
0.05875t
21 a
b
c
d
e
f
22 a
A1
t = 10.793...
The population is expected to reach
40000 in 2029.
0.05875t = ln
A1
M1
P10 = 2.1873705...
M1A1
2.187 million people
A1
i
A1
6
P5 × 10
= 28.4 people per taxi
T5 × 103
M1A1
© Oxford University Press 2019
18
Worked solutions
P10 × 106
= 18.6 people per taxi
T10 × 103
A1
d The model predicts a reduction in the
number of people per taxi, which may
mean that the taxis are in use for
fewer hours or fewer taxis are used
every day.
R1
ii
27 a For example, for x = e
( )
( )
M1
ln x 2 =
ln e2 =
2 ≠ (ln e ) =
1
A1R1
2
M1
0
(ln x − 5) (ln x + 3) =
M1
ii
A1
x >2
A1
A1
x = 2.16
f ( 4) =
log2 4 + log2 (15) − log2 (5)
30 a
M1
4 × 15
log2 12
= log
=
2
5
A1
−3
A1A1
T ( 0 ) = 94 ⇒ 25 + a = 94
M1
a = 69
A1
A1AG
f ( x ) = log2 x + log2 ( x − 1) ( x + 1)
− log2 ( x + 1)
f ( x ) = log2
ln x −=
5 0,ln x +=
3 0
ln x = 5,ln x = −3
28 a i
A1
x =2
e Use GDC solver or intersection of
graphs
M1
b
⇒ (ln x ) − 2ln ( x ) − 15 =
0
2
x e=
,x e
=
ii
d
y =2
2
b (ln x ) − ln ( x 2 ) − 15 =
0
5
c i
x ( x − 1) ( x + 1)
x +1
=
f ( x ) log2 x ( x − 1)
(
=
f ( x ) log2 x 2 − x
)
M1
M1A1
M1
AG
T (20 ) = 29
29
⇒ 25 + 69e20b =
M1
b = −0.142 (3 s.f.)
A1
b T (30 ) = 26.0 (3 s.f.)
M1A1
A1
c y = 25
d The temperature of the room. R1
dT
M1A1
e
= −9.82...e−0.142...t
dt
d2T
f
M1A1
= 1.399...e−0.142...t
dt 2
g The rate of change is always negative
which means the temperature is
decreasing; as the second derivative is
always positive, the temperature will
not have a minimum but will approach
the value 25 given by the horizontal
asymptote.
A2
29 a i
3
f (0
=
)  2 
 
−1
2 2
+=
= 2.67 (3 s.f)
ii
(2.67, 0)
3
b x  
=
2
2
3
M1A1
A1
y −1
+2
M1
3
  = x −2
2
M1
y −1
=
y log3 ( x − 2 ) + 1
2
M1A1
g (=
x ) log3 ( x − 2 ) + 1
2
© Oxford University Press 2019
19
Worked solutions
10
From approximation to generalization:
integration
Skills Check
11 f =
(x)
1 a 30 × 60 = 1800 cm
2
c 0.5 × 42 × π = 8π mm2
2 a
c
3 a
c
3x 2 + 15x
b
9x 2 + 6 x + 1
d
2x 2 − 9x − 5
b
x7
d
( x − 3) 3
1
x3
1
(2 x + 5) 2
x 2 − 25
1
2
F ( x=
)
1
1 6
x 5 +1 + C=
x +C
5+1
6
3
F (=
x)
1
1 26
x 25 +1 +=
C
x +C
25 + 1
26
4
1
1
− x −5 + C
F (x) =
x −6 +1 + C =
−6 + 1
5
5
f (=
x)
1
=
x −8
x8
1
x −8 +1 + C
−8 + 1
1
1
=
− x −7 + C =
−
+C
7
7x 7
F (x)
=
1
+1
11
1
10 10
10
x
+=
C
x +C
1
+1
11
10
F (=
x)
9
1
1
4 3
F ( x=
) − 1 + 1 x − 4 +1 + C= 3 x 4 + C
4
F ( x=
)
1
∫x
2
∫ (6 x
4
1
1 5
x 4 +1 + C=
x +C
4 +1
5
dx=
2
)
+ 4 x + 5 dx
= 6 ∫ x 2 dx + 4 ∫ x dx + 5 ∫ dx
= 2 x 3 + 2 x 2 + 5x + C
∫ (15t
3
4
)
+ 12t 3 + 2t + 5 dt
= 15∫ t 4 dt + 12∫ t 3 dt + 2∫ t dt + 5∫ dt
= 3t 5 + 3t 4 + t 2 + 5t + C
4
x
∫ 8 d=
5
∫u
6
∫x
1
7
2
∫w
3
∫ (4
8
8x + C
1 −6
1
−7
du =
∫ u du =− 6 u + C =− 6u6 + C
−
dx =
2∫ x −5 dx =
3
)
∫
+ 3 w dw
1
dw + ∫ w 3 dw =
)
1
x + 3 dx
= 4 ∫ x 2 dx + 3 ∫ dx
9
=
x 5 dx
11 a
5
9
x dx
∫=
f ( x ) = x5 +
1
1 +1
1
2 23
x 2 + C=
x +C
+1
3
1 4 3 34
w + w +C
4
4
9 149
x +C
14
10 ∫ du= u + C
=
x x2
1
2
1 −4
1
−
+C
x +C =
2
2x 4
8 23
x + 3x + C
3
=
9
5
∫ (w
7
2 +1
1
3 53
x 3 + C=
x +C
2
+1
5
3
8
10 f (=
x)
1
7 67
− 1 +1
x 7 + C=
x +C
− 17 + 1
6
Exercise 10B
1
= x −2
x2
1
1
F (x) =
x −2 +1 + C =
− +C
x
−2 + 1
F ( x=
)
1
−1
= x 7
x
2
1
1 11
F (=
x)
x10 +1 +=
C
x +C
10 + 1
11
7
3
1
4 74
+1
x 4 + C=
x +C
+1
7
7
F ( x=
)
Exercise 10A
f (=
x)
3
4
x)
12 f (=
4
Note: throughout this chapter, C denotes an
arbitrary constant.
6
=
x3 x 4
F ( x=
)
b 0.5 × 4 × 9 = 18 m2
3
4
3
= x 5 + 3x −2
x2
f '(x ) =5x 4 − 6 x −3 =5x 4 −

∫  x
b
=
5
+
3 
 dx =
x2 
∫x
1 6
x − 3x −1 + C =
6
© Oxford University Press 2019
5
6
x3
dx + 3∫ x −2 dx
x6 3
− +C
6
x
Worked solutions
1
Worked solutions
∫ (3x
12
)
2
+ px + q dx = x 3 +
p 2
x + qx + C
2
= x 3 + 8x 2 + 7x + C
So comparing the coefficients,
p
= 8 ⇒ p = 16
2
q=7
1
6
1
dx 6 ∫ =
dx 6 ln x + c
∫=
x
x
2
∫ 5e
3
1
1 1
1
dx
=
dx
ln x + C
∫ 2=
x
2∫x
2
4
∫
5
∫ (3x + 2 )
=
du 5 ∫ e =
du 5e + C
u
u
ex
1 x
1 x
=
dx
e=
dx
e +C
3
3∫
3
2
dx=
dx
∫ 10x + 13=
4
∫e
5
4∫ (5x + 1) =
dx
∫ (9x
+ 12 x + 4 dx
2
7
∫ 3e
8
dx
∫ 7 (2 x − 9 ) =
7
∫ t ( t + 3 ) dt = ∫ ( t
2
∫t
=
3
3
)
+ 3t 2 dt
dt + 3∫ t 2 dt=
∫
9
=
dx 3∫ x=
dx
4
2
x + 3x + 2 x
dx =
x
3
)
= ∫ x dx + 3 ∫ x d x + 2 ∫ d x
1 4 3 2
x + x + 2x + C
4
2
∫
10
=
5
7
(2 x − 9 ) + C
10
1
dx ∫ ( 4 x + 3 )
∫ ( 4x + 3)=
9
dx
∫ (2x + 1) =
1
3
10
=
2
∫
12
6
1
3
=
dx
4x + 7
− 13
dx
2
2
1
3
4 x + 7 ) 3 + C=
4x + 7) 3 + C
(
(
2
4⋅ 3
8
4
∫ (3x + 10)
5
b
=
5
1
(7 x − 5) + C
7 ⋅5
14 a
x
d=
4
6
1
(3x + 10) + C
3⋅6
6
1
(3x + 10) + C
18
f=
(x)
(12x + 7)
−1
Using the chain rule,
7
1
dx =
−
( −3x + 7) + C
3⋅7
7
1
=
−
( −3x + 7) + C
21
∫ ( 4x + 7)
f '(x) =
5 ⋅ 3 (3x + 10 ) =
15 (3x + 10 )
5
1
(7 x − 5) + C
35
∫ ( −3x + 7)
4
1
2 x + 1) 3 + C
(
⋅2
13 a Using the chain rule,
1 u
e − 2u + C
2
4
4
3
1 5x 8
e + ln 5x − 3 + C
5
5
=
=
eu − 4
1 u
=
du
e du − 2 ∫ du
2
2∫
dx
∫ (7 x − 5) =
dx
4
3
2 x + 1) 3 + C
(
8
Exercise 10D
1
−2
2
+ 3 x + 2 dx
3
=
5
7
2x − 9) + C
(
2 ⋅5
8 
8

11 ∫  e5 x +
e5 x dx + ∫
dx
 dx =
∫
5
3
5
x
x
−
−3


3 2
x +C
2
∫ (x
3
dx =
− e4 − 2 x + C
2
4
=
1 4
t + t3 + C
4
ln(3 x )
∫e
8
4 −2 x
−1
1
=
− ( 4 x + 3) + C
4
1 2
x + x +C
2
x +1
2
ln 3x + 8 + C
3
dx
∫ 3x + 8 =
= 3x 3 + 6 x 2 + 4 x + C
∫ ln ( e ) dx= ∫ ( x + 1) dx=
4
4
(5x + 1) + C
5⋅4
6
= 9∫ x 2 dx + 12∫ x dx + 4∫ dx
6
1
− e−4 x +3 + C
dx =
4
4
1
(5x + 1) + C
5
=
)
2
−4 x + 3
1
ln 10 x + 13 + C
10
3
=
Exercise 10C
u
1
3
f '(x) =
−12 (12 x + 7 )
b
1
dx
∫ 12x =
+7
© Oxford University Press 2019
−2
1
ln 12 x + 7 + C
12
2
Worked solutions
Exercise 10E
1
h ( t )=
∫ (6t
2
3 a
)
+ 1 dt= 2t 3 + t + C
18 + C =
8⇒C =
h (2 ) =
−10
b=
A
∴ h ( t )= 2t + t − 10
3
2
y (x) =
∫ 8 ( 2 x − 3 ) dx =
3
(2 x − 3)
4
+C
y (2 ) = 6 = C + 1 ⇒ C = 5
∴ y (x) =
3 a
(2 x − 3)
4
4 a
+5
5 a
= 2t 2 + t + C
∴ v ( t )= 2t 2 + t + 2
s (=
t)
dt ∫ (2t
∫ v (t ) =
2
a=
(t )
s ( t )=
)
b
1
dv
= 8e2t + 1
dt
2t
)
=
f ( x ) dx
∫=
∫ 3dx +
2
t
+2
2
2
16
bh
=bh =4 ( 4 ) =
16
2
3

2

∫ (9 − 3x ) dx =6 + 9x −
Geometric: −
4
1
dx
8x − 7
3
3x 2 

2 2
1
1
bh =
− (2 ) (3) =
−3
2
2
5
∫ (9 − 3x ) dx + ∫ (3x − 15) dx
3
4
4
5

 3x 2

3x 
=9 x −
− 15x 
 +
2 3  2

4
2
27   75


= 12 −
− ( −36 )  =−3
 + −
2   2


3
15
1
21
− 3 + (2 ) ( 6 ) =
2
2
2
Geometric:
Integration:
Exercise 10F
7
6
2x
dx 12
1=
a A ∫=
0 3
1
1
=
bh
(6=
) ( 4) 12
2
2
5 dx
∫=
4
| x | dx
∫=
Integration:
∫
7e
7e
= ln1 + C = C ⇒ C =
8
8
7e
∴ f ( x=
) ln 8x − 7 + 8
2 =
a A
2
1 2 1
25π
=
πr
π=
5)
(
4
4
4
 27
 15
6+
=
− 12  =
2
2


t2
+
+C
2
f (1) =
2
25π
4
25 − x 2 dx =
15
a + b
3 + 2
Geometric: =

 h =
 (3)
2
2
2




2
1
ln ( 8 x − 7 ) + C
8
b=
A
5
0
Integration:
s (0) = 4 = 2 + C ⇒ C = 2
5 f (x)
=
∫
A=
2⋅
0
+ t dt = 2e2t
∴ s ( t )= 2e2t +
9π
2
Exercise 10G
2 3 1 2
t + t + 2t + 8
3
2
∫ ( 4e
=
2
−4
∴ a (3) =8e6 + 1
b
A=
+ t + 2 dt
s ( 0 ) =8 =C1 ⇒ C1 =8
4 a
=
2
6
=
6
a A
2 3 1 2
t + t + 2t + C1
=
3
2
∴ s (t ) =
2
x

A =∫  + 3  dx =24
3


0
b
=
A
v (0) = 2 = C ⇒ C = 2
b
π (3)
π r2
a + b
3 + 5
b A =
=
h  =
 ( 6 ) 24
 2 
 2 
) dt ∫ ( 4t + 1) dt
∫ a (t=
v=
(t )
3
9π
A =∫ 9 − x 2 dx =
2
−3
15
3
5
7
0
3
5
∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx + ∫ f ( x ) dx
0
15
=
−3+
2
=
7
7

9  3x 2
3
15
d
x
−
x
=
+
− 15x 
(
)
∫5
2  2
5
21
2
−1
b =
A bh
= 3 (=
5) 15
© Oxford University Press 2019
3
Worked solutions
6
1

∫1  2 f ( x ) − g ( x )  dx
4
6
3
6
4
10
10
5
1
10
6
10
1
1
6
6
g ( x ) dx ∫ g ( x ) dx + ∫ g ( x ) dx
∫=
6
7
∫ g ( x ) dx = 0
10
6
f ( x ) dx ∫ f ( x ) dx − ∫ f ( x ) dx
∫=
6
1
8
1
10
7
10
−2
1
12
1
1

g ( x ) dx
 g ( x − 2 )  dx =
2
2 ∫1


x ∫ f (x) d
x
=
∫ f ( x + 3) d=
∫
3
11
12
e5
10
1
3
10
1
e5
2
4
4
0
0
4
3
1
1
−3
−3
b
b−4
a
a− 4
dx
dx
∫ f ( x − 4)=
∫ f ( x )=
2
dx
∫=
x
11 a
k
2
dx
∫ x=
20
2
∴
4
4
−2
k
2
k
k
=
=
x 2 2ln
ln
ln9
2ln =
2
4
(k
> 0)
4
4
1
1
1
1
dx  ln ( 4 x − 2=
) 4 ln7
∫1 4x − 2 =
4
1
2
∫ (2t − 1)
Exercise 10H
 = 27 − 12 = 15
−2
0
2ln x + C
k2
=9 ⇒ k =6
4
1
2
0
∫ 6 x dx = 3x
0
Exercise 10I
20 + 6k = 32 ⇒ k = 2
3
4
 x2 
=
20 +   =
28
 2 0
−3
2
4
4
b
for a and b is
b−4=3⇒ b =7
a − 4 =−3 ⇒ a =1
2=
(10) 20
dx ∫ 2f ( x ) dx + ∫ x dx
∫ (2f ( x ) + x )=
0
3
∫ ( f ( x ) + k ) dx =
2
1
1
1
 1
dx =  −  =− − ( −1) =
x2
x
2
2

1
x ) dx 2 ∫ f ( =
x ) dx
∫ 2f ( =
b
f ( x ) dx ∫ f ( x ) dx − ∫ f ( x ) dx
∫=
3
2
)
8
4
+4=
3
3
A =∫
1
1
f (=
x ) dx =
(20) 10
2 −∫3
2
x ) dx
∫ 2 f (=
16
)
dx = ln x  e2 = 5 − 2 = 3
(
1
So a possible pair of values
2
(
16
1 
1
4 3 2 3 
−1
− t 2  dt = ∫ t 4 − t 2 dt =  t 4 − t 2 
3 0
3

0
2
 x3

A =∫ − x 2 + 2 x dx = −
+ x2 
 3
0
0
10 a
= 20 − 6 = 14
1
1
3
−3
3
1
4
∫x
b
= 12 + ( 40 − 4 ) = 48
d
1
∫  t
1
1
c
16
=−
x ∫ f ( x ) dx + 4 x 
∫ ( f ( x ) + 4) d=
b
∫
3
1
1  1 4
 1 
=−
− −  =
dx = −
2
x3
18  2  9
 2 x 1
10
10
12 a
3
9 a
1
=
(20) 10
2
=
2
e2
= 12 − ( −4 )= 16
9
2
32 128
=−
=
−32
3
3
6
10
)
+ 4 x − 2 dx =  x 3 + 2 x 2 − 2 x 
−1
 4

 4

∫1  x 2 + 1 dx = − x + x 1 = 0 − ( −3) = 3
0
6
8
2
= 5ln 4 − 5ln3 = 5ln 34
2
1
=6 + 14 =20
7
∫ (3x
4
3
= 12 − 3 = 9
− ∫ f ( x ) dx =
−12
∫ f ( x ) dx =
5
2
−1
1
= ( −4 ) − 6 =−8
2
1
5
∫ u du = 5ln u 
3
1
f ( x ) dx − ∫ g ( x ) dx
2 ∫1
1
=
4
3
2
∫e
−1
1
3
1
1
1
t  (2t − 1) =
d=
1 − ( −1=
) 3
6
6

0
2
(
1 10
1
+4 
=
dx  e3 x=
e −e

3

 −1 3
3x + 4
(
)
)
 ex 
ex
e4 − e
dx =

∫1 =
2
2
 2 1
© Oxford University Press 2019
4
Worked solutions
2
2
 2k + 1 
2ln3 =
ln9
⇒ ln 
=
 3 
2k + 1
⇒
= 9 ⇒ k = 13
3
2
∫ ( x + 2) ( x − 1) dx= ∫ ( x + x − 2) dx
4
0
0
2
 x3 x2

8
2
= 
+
− 2x  =
+2−4 =
3
2
3
3

0
2
∫ (3x − 3)
5
3
1
2
2
Exercise 10J
dx = 27∫ ( x − 1) dx
3
1
1
(
)
= 27∫ x 3 − 3x 2 + 3x − 1 dx
1
2
 x4

3x 2
= 27 
− x3 +
− x
4
2

1

 1   27
= 27  0 −  − =

4
 4 

4
∫
6
4
∫ ( 4x + 9)
4 x + 9dx =
0
1
2
dx
0
4
3 
 1
4x + 9) 2 
= 3
(
 2 ⋅ 4
 0
)
(
3
3
1
1
49
252 − 9 2=
(125 − 27=) 3
6
6
=
2
∫ (e
7
t
−2
)
2
f ( x )= g ( x ) ⇒ x 2 + 2 x − 3
et + e−t 
− e − t dt =
−2
(
) (
0⇒x =
1
=
−3, x =
( x + 3) ( x − 1) =
)
= e2 + e−2 − e−2 + e2 = 0
9
∫
8
3 t +2
t
4
9
(
dt =
∫ 3 + 2t
4
− 12
(27 + 12) − (12 + 8) =
=
9 a
(
3t 4t 
) dt =+


1
2
∴A=
9
1
∫ ( −2x + 3 − x ) dx
2
−3
1

x3 
5
32
=  − x 2 + 3x −
 = − ( −9 ) =
3
3
3

 −3
4
19
2
)
f (x) =
−2 x x 2 − 4 =
0
⇒x =
0 or x =
±2
So
b
(0, 0) , ( −2, 0) , (2, 0)
2
∫ 2 x ( 4 − x ) dx
2
0
c
2
2
∫ 2x ( 4 − x ) dx =∫ (8x − 2x ) dx
2
0
3
0
2

x 
= 4 x 2 −
 =8
2 0

4
10 a
k
1
∫ 2 x + 1 dx
x 2 − 5x + 4 = 4 ⇒ x ( x − 5) = 0
1
b
k
1
=
1
dx
∫ 2x=
+1
k
1
ln (2 x + 1) 
1
2
1  2k + 1 
=
ln 
 ln3
2  3 
or x 5
x 0=
=
5
∫ ( 4 − 4 + 5 x − x ) dx
A=
2
0
5
5
 5x 2 x 3 
=∫ 5x − x 2 dx =
−


3 0
 2
0
125
=
6
(
© Oxford University Press 2019
)
5
Worked solutions
3
−0.5x 2 + 8 =0.5x 2 − 8 ⇒ x =±4
∫ ( ( −0.5x
4
2
−4
4
))
) (
+ 8 − 0.5x 2 − 8 dx =
85.3
x 2 − 2 x= 4 x ⇒ x= 0, 4
∫
4
0
(4
))
(
8
x − x 2 − 2 x dx =
16
x+4
=− x − 7 ⇒ x =−7.36, 1.36
x −2
1.36  x + 4

∫−7.36  x − 2 − ( − x − 7) dx =27.5
1
x − 2 ⇒ x= 0.15, 8.21
2
8.21 
1

8.78
∫0.15  ln x −  2 x − 2  dx =
5 ln x=
9
The area enclosed is given by the integral
4
2
4

2

x ∫  + 2  dx
∫  x − ( −2)  d=
x

1
1
4
6
x 2 − 5x + 1 =6 − x 2 ⇒ x =−0.77, 3.27
∫
3.27
−0.77
( (6 − x ) − ( x
2
2
))
− 5 x + 1 dx =
21.8
= 2ln x + 2 x =
1
= 4ln2 + 6
⇒p
= 4, q
= 6
10 a
(2ln 4 + 8) − (2)
f (x)= g (x) ⇒
x=
x
2

x
∴ x 1 −
 = 0 ⇒ x = 0 or x = 4

2


∴ ( 0, 0 ) and ( 4,2 )
b i
4

∫ 
x −
0
7 2e =
3x + 4 ⇒ x =−1.11, 2
x
∫
2
−1.11
ii
3.68
( (3x + 4) − 2e )dx =
x
c i
4
4
2 3 x2 
x
4

∫0  x − 2  dx =  3 x 2 − 4  = 3
0
k

∫ 
0
© Oxford University Press 2019
x
 dx
2
x −
x
 dx
2
6
Worked solutions
Chapter Review
2 23 k 2
k −
3
4
ii
2
2 23 k
2
k −
= ⇒ k = 1.510 (3d.p.)
3
4
3
iii
Exercise 10K
1
∫ (( x
0
− 2 x − 10 x + x 2 − 3x 3
∫ ( (10x + x
2
+
) ) dx
) (
2
−2
2
0
))
) (
∫x
1 a
∫ (5 x
c
∫
d
∫x
∫ (( x
1
+
)
)
∫ ( ( x + 1) − ( x
2
3
∫
(( x
0
−1.51677
+
∫
3
1.51677
0
− 3x 2 + 3x + 1 dx
)
− 2 x − 3xe − x
(3xe
− x2
2
)
∫
−1.41421
−4
(( −x
+
∫
1.41421
+
∫
4
−1.41421
1.41421
(( x
4
4
(( −x
4
− x 3 − 2 x dx
4e x + C
g
∫ (6
)
) (
− 20 x 2 + 64 − − x 4 + 16 x 2
) ) dx
))
) (
=
x2
− 1 =−3x − 5.5
2
2
= 0
∴ P ( −3,3.5)
The gradient of h is − 3
(
Using GDC ⇒ Q ( −1.10, −2.21)
)
+ 2 dx
)
(
2
)
x5
+ 2x 3 + 9x + C
5
7
1
( 4 x + 5) + C
4⋅7
7
1
4 x + 5) + C
(
28
6 3 x +2
+=
C 2e3 x +2 + C
e
3
k
dx
∫ 6 x − 7=
l
dx ∫ 3 x =
dx
∫ ln ( e )=
3 x +2
x
d=
1
1
ln 6 x − 7 + C
6
3x 2
+C
2
3x
3
∫ (6 x − 1) dx = 3x
2
3
− x 
−2
c
3
3
 x3 
 1  28
∫−1 x dx =  3  = 9 −  − 3  = 3
−1
2
25
3

 dx

d
e4
5
dx
∫=
x
1
e
25
dx = 6  x  = 6 (5 − 3) = 12
 9
x
∫
9
 x2

− 1 − ( −3x − 5.5)  dx

 2

 x2
+ ∫3 3 3 
− 1 − −x2 − 1
−
2 2  2
)
∫ 6e
b
(
2
 dx
x
= (24 − 14 ) = 10
g ( x ) = h ( x ) ⇒ − x 2 − 1 = −3x − 5.5
0
1
2
j
)
h (x) =
−3x − 5.5
∫
+
+ 3 dx = ∫ x 4 + 6 x 2 + 9 dx
−2
y − 3.5 =
−3x − 9
d i
2
6
2 a
f '(x) =
x⇒m=
f ' ( −3) =
−3
y − 3.5 =−3 x − ( −3)
∫ (6 x
x + 2 dx =
dx
∫ ( 4 x + 5) =
i
=
h ( −3) =−3 ( −3) − 5.5 =
3.5
3 3 3
−
2 2
−3
∫ (x
+ 16 x 2 − x 4 − 20 x 2 + 64 dx
⇒x =
−3
c
x
h
))
( x + 3)
3
3
) (
f ( x ) =h ( x ) ⇒

∫  4x
dx
∫ 4e =
+ 16 x 2 − x 4 − 20 x 2 + 64 dx
⇒ x2 + 6x + 9 =
b
8x5 + 4x
=
dx
2x 2
f
≈ 440
5 a
1
−9
dx =
− x −8 + C
∫ 4 x dx =
2
9
= 4x 2 + 2x + C
))
(
13
10 10
x +C
13
3
10
dx
∫ x=
=
x 4 + 2ln x + C
dx
≈ 4.65
4
∫
e
= 0.5
)
− 6 x 2 + 7 dx = x 5 − 2 x 3 + 7 x + C
1
=
−
+C
2x 8
))
3
1
4
=
x 3 dx
4
− 3x 3 − x 2 − 2 x dx
− 3x 2 + 3x + 1 − ( x + 1) dx
3
0
10
x9
+C
9
x
d=
b
= 24
2
8
0
e4
x 1
5ln
=
e4 20
5ln=
∫ 8 (2 x + 3)
−2
3
0
4
dx =(2 x + 3) 

 −2
= 81 − 1 = 80
ii 1.81 (2 d.p.)
f
5
5
1 20
 1 4x 
4x
dx  =
e 
e − e12
∫3 e=
4
4

3
© Oxford University Press 2019
(
)
7
Worked solutions
4
4
) dx 2∫ f ( x=
) dx 2=
(10) 20
∫ 2f (x=
3 a
1
b
3
x=
3x − 2
4
3
= ( x − 1)
3
1
1
∴x =
−2
= 10 − 6 = 4
4
4
1
1
1
= 10 + 4 x 1 = 10 + (16 − 4 ) = 22
4
4
f (x) =
∫ ( 4x
3
)
+ 2 dx = x 4 + 2 x + C
f (2 ) = 20 + C = 24 ⇒ C = 4
∴ f ( x ) = x 4 + 2x + 4
0
∴ ( −2, −8 )
c
∫ (x
1
f ′(x) =
=
8
21π
b | −8π | + ∫ f ( x ) dx =
13π
⇒ ∫ f ( x ) dx =
0
8
∴
∫
f ( x ) dx =
−8
0
8
∫
−8
f ( x ) dx + ∫ f ( x ) dx
0
6
3
(
)
2 x 2 + 1 − 2x ⋅ 2x
(x
2
(x
∫x
10 a
a=
2
2
A1
)
+1
2
2x
dx
= ln x 2 + 1 + C
2
+1
(
)
∫ (
−4
)
( −4)
=0 +
3
s (t ) =
M1A1
∫ ( 40 − 3t ) dt =
⇒C =
−
+ 2 ( −4 )
2
3
64 32
= 32 −
=
3
3
40t −
∫
(0.5 e + 1 − ( x − 5x )) dx
+∫
( x − 5x − (0.5 e + 1)) dx
2.4765
x
4
4
11 a i
2.4765
= 7530.19
≈ 7530
b Lines intersect at x = −2, 0 and 3
∫
0
−2
(x
) ) dx
+ ∫ ( x − 3 x − ( x − 9 x ) ) dx
(
3
− 9 x − x 2 − 3x
3
2
3
0
253
= = 21.1
12
ii
b
∴ y − 1 = 3 ( x − 1) ⇒ y = 3x − 2
3t 2 57
−
2
2
2×2
= 2
2
M1A1
Aregion
=
2
2×3
= 3
2
M1A1
A1
Aregion3 = 3 − 2 = 1
12 a Use GDC to obtain value of definite
integral
M1
1
f ( x ) dx
∫=
=
A
1.1202
A1
0
b i
2.24 (3 s.f.)
13 a
A1
2
1
1
0
dx 2 ∫ f ( x ) =
dx
∫ 2f ( x − 1) =
2.24
2.24 (3 s.f.)
M1A1
ii
8 a=
y f=
(1) 1
f ' ( x ) =3x 2 ⇒ f ' (1) =3
M1
Aregion
=
1
2
x
2
3
+ C = 10
2
A1
−2.2808
9.7467
3t 2
+C
2
M1A1
57
2
s ( t ) = 40t −
7 a Lines intersect at x = −2.2808, 2.4765
and 9.7467
M1A1A1
dv
= −3 ms −2
dt
s (1) = 10 ⇒ 40 −
0
 x3

− x − 4 x dx =
−
+ 2x 2 
3

 −4
2
M1A1
)
+1
f (x) =
− x ( x + 4) =
0⇒x =
−4 or x =
0
0
)
− 3 x + 2 dx
2 − 2x 2
b
b
5π
=
−8π + 13π =
1
−2
1
( )
0
∫ (x
 x 4 3x 2

3
= 
−
+ 2x  =
− ( −6 )
2
4
 −2 4
27
= = 6.75
4
9 a
8
)
− (3x − 2 ) d=
x
3
−2
1
− π 42 =
−8π
∫−8 f ( x ) dx =
2
5 a
0
y =
3 ( −2 ) − 2 =−8
4
x ∫ f ( x ) dx + ∫ 4 dx
∫ ( f ( x ) + 4) d=
c
( x + 2) =
2
f ( x ) dx ∫ f ( x ) dx − ∫ f ( x ) d x
∫=
b
( x − 1) ( x 2 + x − 2)
⇒ x 3 − 3x + 2 =
1
4
x = 0, x = ±1
A1A1
b Either:
© Oxford University Press 2019
8
Worked solutions
(
)
f ′ ( x ) =1 ⋅ x 2 − 1 + 2 x 2 =3x 2 − 1 M1A1
Or:
f ( x=
) x3 − x
M1
f ′ (=
x ) 3x − 1
A1
2
c
f ′(x) =
0⇒x =
±
d
∫ f ( x ) dx = 0
1
=
±0.577
3
M1A1
1
0
2
 1 2
1 2 
=
− 2 x  +  2 x 

 −1 
0
A1
1
5

=  0 +  + (2 − 0 ) =
2
2


AG
17 a
M1A1
−1
e The function changes sign in the
interval  −1,1 , so the areas above and
below the x-axis cancel out.
1
∫ f (x)
f
R1
dx = 0.5
M1A1
−1
f (1) = 12 = 3 − 2 × 1 = g (1)
14 a
A1 for shape; A1 for domain; A1 for
end-points coordinates; A1 for
maximum point and its coordinates.
M1A1
f ( −3) =( −3) =3 − 2 × ( −3) =g ( −3)
2
A1
1
1

x3 
2
3x − x −

3  −3

∫ (3 − 2x − x )dx =
b
2
−3
M1A1A1
1 
27  32

A1
=  3 − 1 −  −  −9 − 9 +
=
3 
3 
3

( x − 2)
15 a
b
c m
=
d
+ 4 × ( −2 ) x + ( −2 )
3
AB contains the origin
R1
y = 0.0202 x
A1
(
)
f ′ ( x ) =3e− x + 3x −e− x =(3 − 3x ) e− x
A1
4
M1A1
A1
=x 4 − 8 x 3 + 24 x 2 − 32 x + 16
f
∫ ( x − 2 ) dx
4
∫ x − 8x + 24x − 32x + 16 dx
4
3
f ( 0.98201...) = 1.10345...
A1
y − 1.10
= 0.0202 ( x − 0.982 )
A1
Use of GDC to calculate
M1
5
∫ 3 xe
x
− 0.0202 x dx =
2.63
A2
0
2
M1
=
M1
M1A1AG
2
=
0.1011
= 0.0202...
5
e Solve =
f ′ ( x ) 0.0202...
=
⇒ x 0.98201...
4
= x 4 + 4 × ( −2 ) x 3 + 6 × ( −2 ) x 2
b
A1
0 ≤ y ≤ 1.104
18 a
x5
− 2 x 4 + 8 x 3 − 16 x 2 + 16 x + C
5
A1A1
16 Use GDC to obtain graph of y = x
M1A1
Attempt to calculate area of both triangles
M1
2
∫
x dx =
−1
1×1 2 × 2
+
= 2.5
2
2
1 correct: A1; all correct: A2
b
A1AG
OR:
∫
2
−1
x dx =∫
0
−1
2
( − x ) dx + ∫0 x dx
M1A1A1
© Oxford University Press 2019
9
Worked solutions
A1 for shape; A1 for domain; A1 for
intercepts
c Either:
1   15 1  
1   21 1 

A = 0 ×  + 
×  + 3 ×  + 
× 
2  8 2 
2  8 2

M1A1A1
15
=( 3.75)
4
=
A1
Or:
1   21 1  
1
 15 1  
A= 
×  + 3 ×  + 
×  + 0 × 
2  8 2 
2
 8 2 
M1A1A1
15
=( 3.75)
4
=
d
∫ (
2
0
A1
2
 2 x4 
4 x − x 3 dx =
2 x −

4 0

)
=8−4 =4
M1A1
A1
© Oxford University Press 2019
10
Worked solutions
11
Relationships in space: geometry and
trigonometry in 2D and 3D
Skills check
1 a
5 10 ≈ 15.8
2 a i 12600 cm
2
b
b
2 6 ≈ 4.9
ii
1.26 m2
b
4 π ≈ 12.6 m3 ≈ 12566 l
c
Exercise 11A
1 a
c
(3, 0, 0)
(3, 0,2)
(3, 4, 0)
(3, 4,2)
b
d
d
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
0 + 3 0 + 4 0 + 2
,
,
= =
 (1.5,2,1)
2
2 
 2
2 a
4 a
( x2 − x1 )
2
+ ( y2 − y1 ) + ( z2 − z1 )
=
(3 − 0 )
=
9 + 16 + 4=
2
2
+ ( 4 − 0 ) + (2 − 0 )
2
2
29 ≈ 5.4
d
c
( −3,3,7)
d
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
(0.5, −0.5, −6 )
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
 −5.1 + 1.4 −2 + 1.7 9 + 11 
=
,
,

2
2
2 

d=
1 a
=
( 4 − 2)
=
4 + 0 + 16=
2
25 + 9 + =
9
d=
( x2 − x1 )
=
(1 + 1)
2
20 ≈ 4.47
2
2
43 ≈ 6.56
+ ( y2 − y1 ) + ( z2 − z1 )
2
4 + 36 + 64 =
2
( −2 − 2)
2
2
( x2 − x1 )
+ ( y2 − y1 ) + ( z2 − z1 )
2
2
+ ( −6 − 2 ) + ( −7 − 3)
( −5 − 1)
=
36 + 64 + 100=
2
2
( x2 − x1 )
2
2
200 ≈ 14.1
+ ( y2 − y1 ) + ( z2 − z1 )
2
2
+ ( 0 + 4 ) + (5 − 2 )
2
16 + 16 + =
9
( x2 − x1 )
2
41 ≈ 6.4
2
(1 + 1)
=
4 + 9 + 16=
2
2
+ ( y2 − y1 ) + ( z2 − z1 )
=
d=
2
20 ≈ 4.47
=
2
2
+ (1 + 1) + (3 − 3)
d=
( 4 − 0)
104 ≈ 10.2
2
16 + 4 + =
0
2
2
+ ( y2 − y1 ) + ( z2 − z1 )
=
d=
2
2
( x2 − x1 )
d=
2
+ ( −3 − 3) + ( 4 + 4 )
2
2
+ (2 + 1) + (3 + 1)
2
( x2 − x1 )
2
(1 − 4)
=
9 + 0 + 16=
2
2
29 ≈ 5.39
+ ( y2 − y1 ) + ( z2 − z1 )
=
2
2
+ (1 − 1) + (1 + 3)
2
2
25 = 5
SA = x 2 + 2 xl = 202 + 2 × 20 × 26
2
b
SA = x 2 + 2 xl = 42 + 2 × 4 × 6.3 = 66.4 cm2
c
SA = x 2 + 2 xl = 52 + 2 × 5 × 13 = 155 cm2
d
SA = π r 2 + π rl = π × 52 + π × 5 × 13
+ (3 − 3) + (1 − 5)
2
+ ( 4 − 7 ) + ( −1 − 2 )
2
d=
2
= 1440 cm2
( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 )
2
=
2
Exercise 11B
=
( −1.85, −0.15,10)
3 a
(2 + 3)
=
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
 5 − 4 2 − 3 −4 − 8 
= 
,
,
=
2
2 
 2
=
=
(0.5,1.5,3)
 −4 − 2 4 + 2 5 + 9 
= 
,
,
=
2
2 
 2
c
b
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
 −4 + 5 4 − 1 3 + 3 
= =
,
,

2
2 
 2
b
2
+ ( y2 − y1 ) + ( z2 − z1 )
2
=
Distance of OF
d=
( x2 − x1 )
=
e Midpoint of OF
f
d=
2
= 283 cm2
e
SA = π r 2 + π rl = π × 62 + π × 6 × 14
= 377 cm2
f
SA = π r 2 + π rl = π × 42 + π × 4 × 12
≈ 201cm2
© Oxford University Press 2019
Worked solutions
1
Worked solutions
2 a
7 Volume of the water tank, VT
= 4π =
SA
r 2 4 π × 52 ≈ 314 cm2
4 3 4
V =
πr =
π × 53 = 524 cm3
3
3
1 2
=
VT π r 2hcyl + π r hcone
3
2
b
3
SA =4 π r 2 =4 π ×   =
28.3 cm2
2
= π × 12 × (13 − 2 ) +
= 36.7 m3
3
4
4
3
V = π r 3 = π ×   =14.1cm3
3
3
2
Conversion to litres
1000000 cm3
1
L
×
1m3
1000 cm3
= 36700L
1
3 a V
=
(base area × height )
3
36.7 m3 ×
1
( 4 × 4 × 12=) 64 cm3
3
=
8 Volume of each ball,
3
4 3 4
Vball =
πr =
π × (3.35) =
157.5 cm3 .
3
3
1
(base area × height )
3
1  10 × 13.1

=
=
× 11  240 cm3

3
2

b V
=
hcyl =×
3 6.7 =
20.1cm
Vcyl =
π r 2h =
π × 3.352 × 20.1 =
708.7 cm3
total free space is
VT = Vcyl − 3VB ≈ 708.7 − 3 × 157.5 =
236 cm3
1
(base area × height )
3
1
=
(9 × 7 × 5=) 105 cm3
3
c V
=
4
Exercise 11C
SA c =curved surface area,
1 a
SA T = total surface area
a
b
l = 52 + 122 = 25 + 144 = 169 = 13
SA c = π rl = π × 5 × 13 ≈ 204cm2
b
c
SA T =π r + π rl =π × 5 + 204.2
2
2
d
≈ 283 cm2
c
1 2
1
πr h =
π × 52 × 12 ≈ 314 cm2
3
3
V=
5 a =
SA c
b
SA=
T
2 a
4π r 2
4 π × 32
=
= 56.5 cm2
2
2
b
4π r 2
2
+ π r=
56.5 + π × 32
2
c
= 84.8 cm2
c
6 a
d
4 3 2
π r = π × 33 = 56.5 cm3
V = 6
3
3
1 2
14

VT =VC + VHS = π r h +  π r3 
3
23

=
b Sum of the curved surface area of the
cone, SA cc , and curved surface area of
the hemisphere, SA CHS
l=
19
19
, θ sin−1= 44.7°
=
27
27
33
33
, θ tan−1= 30.5°
tan θ = =
56
56
sin θ =
12
12
, θ tan−1= 67.4°
=
5
5
11
−1 11
, θ cos = 56.6°
cos θ = =
20
20
tan θ =
x
, x 27 =
cos 22° =
=
cos 22° 25.0
27
44
44
,x
= 42.5
tan 46° ==
tan 46°
x
7
7
tan 46° =
= ,x
= 6.76
x
tan 46°
x
, x 22 =
sin 43° ==
sin 43° 15.0
22
h
, h 12 =
tan25° =
=
tan25° 5.60
12
tan a =
1
14
3
3
π × 42 × 10 +  π × 4  = 302 cm
3
23

2
1
π × 12 × 2
3
5.6
5.6
, a tan−1 = 29.2°
=
10
10
q
,
55
55
=
q 55 tan30
=
°
= 31.75... ≈ 31.8
3
4
tan30° =
tan50
=
°
2
10 + 4 ≈ 10.8 cm
31.75
55
( − p)
(55 − p ) tan50° =31.75
SA T =
SA CC + SA CHS =
π rl + 2π r 2
= π × 4 × 10.8 + 2 π × 4 = 236 cm
2
q
=
55
( − p)
2
31.75
(55 − p ) =
tan50°
© Oxford University Press 2019
2
Worked solutions
p=
55 −
=
h 50 =
sin55° 41m
31.75
≈ 28.4
tan50°
5
7
7
cos 20° =
= ,x
= 7.45
x
cos 20°
6
6
6
, h = 10.4
tan30° =
=
h
tan30°
7 a
3 a
0.8
=
θ sin−1 = 15.5°
3
b
1
× 20 × 15 × 30 =
3000 cm3
3
4
b AC = 2NC = 202 + 152 = 25 . Then
NC = 12.5 cm
TC = 302 + NC 2 = 302 + 12.52
Then
1056.25 32.5 cm
= =
c
sin θ =
30
30
, θ sin−1 = 67.4°
=
32.5
32.5
d
TM =
TC2 − MC2 =
e
8 a
32.52 − 7.52 = 31.6
30
30
,
sin TMN
= =
TM 31.6
30
= sin−1 = 71.7°
TMN
31.6
c
d
EO 146.5
,
tan EMO
= =
OM 115.2
146.5
EMO
= tan−1
= 51.8° ≈ 52°
115.2
1
AT =4 × AF + AB , AF = × AB × EM
2
= 139000 cm2
EM + BM
EO 146.5
sin EBO
= =
EB 219.1
sin−1
146.5
= 42.0°
219.1
Exercise 11D
h
tan25° =
12
=
h 12 =
tan25° 5.60 m
2
=
H 3 tan70
=
° 8.24 m
6 a
sin55° =
h
50
cos 36° =
N
25
sin36° =
W
25
sin68° =
W
51
8 =
H HE + HD
tan23° =
HE
300
tan23° 127.3m
=
HE 300
=
tan30° =
HD
300
=
HD 300
=
tan30° 173.2 m
H = HE + HD = 301m
9 Let C be the bottom of the Eiffel tower.
Then
2
186.42 + 115.22 = 219.1
=
H
3
=
W 51 =
sin68° 47.3km
AT =
2 × 230.4 × 186.4 + 230.4 × 230.4
1
tan70° =
7
146.52 + 115.22 ≈ 186 cm
e =
EB
81.5
= 97.1m
tan 40°
=
W 25 =
sin36° 14.7 km
146.52 + OM 2
2
81.5
d
5 R and J are the same height, so it cancels
out. Then the calculation is
b
AB = 2 OM ,
=
tan 40° =
=
d
1
V=
× 230.4 × 230.4 × 146.5
3
=
EM
h = 32 − 0.82 =2.89 m
=
N 25 =
cos 36° 20.2km
= 2592276.5 cm3
b
0.8
3
sin θ =
=
AB ABC − BC
40° + 32° + θ= 90°
θ= 18°
tan (32° + 18° ) =
ABC
300
=
ABC 300
=
tan50° 357.53
tan18° =
BC
300
=
BC 300
=
tan18° 97.48
AB = ABC − BC = 260 m
10 tan75° =
© Oxford University Press 2019
D + 1.5
498
3
Worked solutions
=
D 498 tan75
=
° − 1.5 1857 m
Area = 173cm2
5
Exercise 11 E
1
Area =
a
1
ab sin C
2
Area =
1
× 8 × 6 × sin80°
2
6
=
Area 24 =
sin 80° 23.6 cm2
b
1
Area = × 2.5 × 3.9 × sin34°
2
Area =
Area = 5 ×
1
× 4 × 4 × sin72° = 38.0 m2
2
1
( x + 2) (2x + 1) sin60° =5 3
2
2 x 2 + 5x − 18 =
0
Either factorise, or use the quadratic
formula:
1
× 4 × 7 × sin96°
2
x1,2 =
Area=
1
× 12 × 20 × sin (180° − 80° − 40° )
2
=
Area 120
=
sin 60° 104cm2
f
Area=
1
× 14 × 18 × sin(180° − 78° − 60°)
2
Area 126
sin 42° 84.3 cm2
=
=
g
Area=
1
× 12 × 8 × sin(180° − 30° − 67°)
2
2
=
Area 48=
sin83° 47.6 cm
2
Area =
a
Then x = 2
Exercise 11F
sin θ
sin35°
=
23
45
1 a
1
× 8 × 5 × sin39°
2
Area= 2 ×
b
1
2
1
ab sin C
2
6 sin75°
18
=
θ sin−1
4 4 faces, so area is multiplied by 4
3 angles in an equilateral triangle are 60°
d
4 sin66°
= 27.2°
8
sin θ
sin75°
=
6
18
sin θ =
=
Area 240
=
sin 60° 208 cm2
1
ab sin C
2
4 sin 66°
8
=
θ sin−1
c
23 sin35°
= 17.0°
45
sin θ
sin66°
=
4
8
sin θ =
Area = 20 × 12 × sin60°
Area= 4 ×
23 sin35°
45
θ sin−1
=
1
C
= sin−1 = 30°
2
3
We take the positive value, as distances
cannot be negative.
1
× 8 × 8 × sin C
2
sin C =
2×2
−5 ± 13
4
sin θ =
Area 20 =
sin39° 12.6 cm2
=
b 16 =
x1,2 =
1
ab sin C
2
Area =
−5 ± 52 − 4 × 2 × ( −18 )
−5 ± 25 + 144
=
4
=
Area 14 =
sin 96° 13.9 cm2
e
10 3
= 20
3
2
2x 2 + x + 4x + 2 =
20
=
=
Area 4.875
sin34° 2.73 cm2
d
1
ab sin C
2
1)
( x + 2 ) (2 x + =
1
Area = × 10 × 15 × sin125°
2
Area 75 =
sin125° 61.4 cm2
=
c
Area= 5 ×
6 sin75°
= 18.8°
18
sin θ
sin 48°
=
22
63
sin θ =
Area =2 × 10 × 10 × sin60°
© Oxford University Press 2019
22 sin 48°
63
4
Worked solutions
=
θ sin−1
We substitute back into the first equation
22 sin 48°
= 15.0°
63
h=
sin θ
sin82°
=
20
29
e
sin θ =

sin70°  sin50°  
sin70°
−15 ×
1 −

 h =
sin20°  sin 40°  
sin20°

20 sin 82°
= 43.1°
29
sin70°
−15 ×
sin20°
18.1
h =
=
sin70°  sin50° 
1−
sin20°  sin 40° 
sin θ
sin78°
=
18
34
f
sin θ =
2 a
4
18 sin78°
34
θ sin−1
=
sin70°  sin50°

h − 15 

sin20°  sin 40°

so h
=
20 sin 82°
29
=
θ sin−1
sin70°
x
sin20°
sin74° sin 64°
=
10
x
=
x
18 sin78°
= 31.2°
34
sin99° sin (180° − 99° − 18° )
=
37
x
10 sin 64°
= 9.35km
sin74°
5 We have
sin 49° sin 41°
=
d
20 − b
sin 63°
=
= 33.4 cm
x 37
sin 99°
20 − b =
d
sin 41°
sin 49°
sin53° sin 44°
=
x
7
=
b 20 − d
sin 41°
sin 49°
b
and
sin53°
=
x 7= 8.05
sin 44°
c
b=d
sin23° sin77°
=
x
15
20 − d
sin33° sin108°
=
x
24
sin (180° − 100° − 22° )
x
=
d
=
sin22°
10
sin58°
=
x 10
= 22.6
sin22°
f
sin (180° − 52° − 56° )
x
sin52°
=
6
sin72°
=
x 6= 7.24
sin52°
3 base = 15 + x
We have that
and
sin70° sin20°
=
h
x
sin 40° sin50°
=
h
15 + x
sin50°
Then 15 + x =
h
sin 40°
=
x
sin50°
h − 15
sin 40°
sin 41°
sin52°
=
d
sin 49°
sin38°
 sin 41° sin52° 
20 d 
=
+

 sin 49° sin38° 
sin33°
=
x 24
= 13.7
sin108°
e
sin52°
sin38°
We equate both expressions for b
sin23°
x 15
=
= 6.02
sin77°
d
sin38° sin52°
=
d
b
20
= 9.31km
sin 41° sin52°
+
sin 49° sin38°
6 We have
sin74°
sin16°
=
16 + AD
DC
sin74°
DC
16 + AD =
sin16°
=
AD DC
and
sin74°
− 16
sin16°
sin 62° sin28°
=
AD
DC
AD = DC
sin 62°
sin28°
We equate both expressions for AD
DC
sin74°
sin 62°
− 16 =
DC
sin16°
sin28°
 sin74° sin62° 
DC 
−
16
=
 sin16° sin28° 
© Oxford University Press 2019
5
Worked solutions
16
= 9.96km
sin74° sin62°
−
sin16° sin28°
=
DC
7 We have
 10

=
B sin−1 
sin 45°=
 62.1°
8


and 180° − 62.1
=
° 117.9°
sin15° sin75°
=
h
10 + d
4
sin75°
=
+ d h = 3.73h
10
sin15°
=
sin C
=
d 3.73h − 10
and
and 180° − 53.5
=
° 126.5°
sin72°
=
d h= 3.08h
sin18°
5
Then 3.08
=
h 3.73h − 10
1
Area = × AB × BC × sin B =20
2
1
× 8 × 10 × sin B =
20
2
3.73h − 3.08h =
10
0.65h = 10
h = 15.3
sin B =
sin55° sin 90°
=
h
4
The obtuse angle is 180° − 30
=
° 150°
sin78° sin 47°
and
=
4
b
Exercise 11H
sin 47°
=
b 4= 2.99077..mm
sin78°
1 a
1
1
Area =
bh =
× 3.27660 × 2.99077 =
4.90 mm2
2
2
a = 11.1cm
b
8
sin64°
10
 8

=
A sin−1 
sin 64=
°  46.0°
10


and 180° − 46.0
=
° 134°
2
sin20° sin A
=
3
5
sin A
=
5
sin20°
3
b = 36.4 cm
c
3
sin 45° sin B
=
8
10
=
sin B
10
sin 45°
8
a2 = b2 + c 2 − 2bc cos A
a2= 142 + 222 − 2 × 14 × 22 × cos 80°
a = 23.9 m
d
c 2 = a2 + b2 − 2ab cos C
c 2= 102 + 92 − 2 × 10 × 9 × cos 66°
c = 10.4 m
e
a2 = b2 + c 2 − 2bc cos A
a2= 402 + 252 − 2 × 40 × 25 × cos 20°
a = 18.6 cm
f
5

=
C sin−1  sin20°=
 34.8°
3


and 180° − 34.8
=
° 145.2°
b2 = a2 + c 2 − 2ac cos B
b2= 152 + 282 − 2 × 15 × 28 × cos112°
Exercise 11G
=
sin A
a2 = b2 + c 2 − 2bc cos A
a2= 122 + 92 − 2 × 12 × 9 × cos 62°
Then
sin 64° sin A
=
10
8
20
40
20
= sin−1 = 30°
B
40
=
h 4 sin55
=
° 3.27660..mm
1
30
sin 40°
24
 30

=
C sin−1 
sin 40°=
 53.5°
24


sin18° sin72°
=
h
d
8 We have
sin 40° sin C
=
24
30
a2 = b2 + c 2 − 2bc cos A
a2= 212 + 302 − 2 × 21 × 30 × cos123°
a = 45.0 cm
2 a
cos θ =
10.42 + 182 − 21.92
2 × 10.4 × 18
cos−1 ( −0.1267 ) =
97.3°
θ =
b
cos θ =
© Oxford University Press 2019
8.62 + 3.12 − 9.72
2 × 8.6 × 3.1
6
Worked solutions
c
θ =−
cos−1 ( 0.197299 ) =
101°
Exercise 11I
652 + 552 − 1182
cos θ =
2 × 65 × 55
1
=
θ cos
d
cos θ =
−1
Area =
52 + 52 − 32
2×5×5
=
θ cos−1 0.82
=
= 34.9°
e
cos θ =
2
2
2
24 + 22 − 20
2 × 24 × 22
θ cos 0.625
=
= 51.3°
cos θ =
3.82 + 72 − 42
2 × 3.8 × 7
150 × 106
sin7° =
d
3 a
=
θ cos 0.891729
= 26.9°
sin55°
sin PRS
=
11.84..
14
1
ab sin C
2
sin PRS
= 14 ×
1
A=
× 12 × 9 × sin29.0°= 26.1cm2
2
4
sin PSR sin PRS
=
PR
PS
b
=
θ cos−1 0.875
= 29.0°
A=
PR2= PS 2 + RS 2 − 2 × PS × RS × cos S
PR = 11.8 m
92 + 122 − 62
cos θ =
2 × 9 × 12
b
150 × 106
= 1230.8 million km
sin7°
PR2= 142 + 112 − 2 × 14 × 11 × cos 55°
−1
3 a
1
bc sin A
2
1
× 20 × 12 × sin 43.5°= 82.6 cm2
2
=
d
−1
f
202 + 122 − 142
2 × 20 × 12
=
A cos−1 0.725
= 43.53°
−3337
= 159°
3575
2
cos A =
sin55°
11.84..
c 2 = a2 + b2 − 2ab cos C
ˆ = sin−1 14 × sin55°  = 75.4809..°
PRS
11.84.. 

c 2= 602 + 302 − 2 × 60 × 30 × cos160°
ˆ= 180° − 75.4809..
=
° 104.519...°
PRQ
c = 89km
5 a
tan33° =
ˆ= 180° − 50° − 104.519..
QPR
=
° 25.4809...°
h1
46
ˆ
ˆ
sin PQR
sin QPR
=
PR
QR
=
h1 46=
tan33° 29.9 m
and tan17° =
h2
28
sin50°
sin25.4809..°
=
11.84...
QR
=
h2 28 =
tan17° 8.56 m
QR =
11.84.. ×
b =
A 180° − 33° − 17
=
° 130°
b = 462 + 29.872 =
54.9
1
A = × QS × PS × sin S
2
c
c = 282 + 8.562 =29.3
=
a2 = b2 + c 2 − 2bc cos A
=
a2 54.92 + 29.32
− 2 × 54.9 × 29.3 × cos130°
4 a
a = 77.0 m
cos ADB =
DB2 + DA2 − BA2
2 × DB × DA
122 + 202 − 282
2 × 12 × 20
ADB = cos−1 − 0.5 = 120°
c 2 = a2 + b2 − 2ab cos C
c 2 = 92 + 152 − 2 × 9 × 15 × cos140°
1
× (11 + 6.68 ) × 14 × sin50°= 94.81m2
2
cos ADB =
6 =
C 210° − 70
=
° 140°
sin25.4809..°
=
6.65m
sin50°
b
c = 22.6 km
1
Area = × BD × DA × sin ADB
2
Area=
c
1
× 12 × 20 × sin120°= 104 m2 .
2
sin DCB sin BDC
=
BD
BC
© Oxford University Press 2019
7
Worked solutions
sin DCB sin60o
=
12
13
sin DCB =
12
sin60o
13
 5

=
°  48.3°
A sin−1 
sin 95=
6.67


−1 12
=
=
DCB sin
sin60o 53.1o
13
d
c
ACD
= 180° − 32° − 48.3
=
° 99.7°
d Let C=AD. Then
sin D
sin C
=
d
c
CBD = 180o − BCD − BDC
= 180o − 53.1o − 60o = 66.9o
sin 48.3.. ° sin 99.7..°
=
6.67..
c
sin BAD sin ADB
=
BD
AB
sin99.7..°
c 6.67..
=
= 8.8039..
sin 48.3..°
sin BAD sin120o
=
12
28
sin BAD =
5
sin 95°
6.67
=
sin A
12
sin120o
28
e
−1 12
sin120o 21.79o
BAD sin
=
=
28
1
× 4 × 5 × sin 95°= 9.96 cm2
2
AABC =
1
× 6.67.. × 8.80.. × sin32° =
AACD =
15.55.. cm2
2
Then
ABD = 180o − ADB − BAD
= 180o − 120o − 21.79o = 38.2o
and so
ABC = CBD + ABD = 66.9o + 38.2o
(3 s.f.)
= 105.1o ≠ 90o
5 a
7
sin ABC
sin ACB
=
AC
AB
22.5
sin 46o
48
h
10 − x =
tan60°
= 180 − 46 − 19.71 = 114.3
o
o
o
2
BC
= AC 2 + AB2
− 2 × AC × AB × cos BAC
=
BC 2 22.52 + 482
− 2 × 22.5 × 48 × cos114.3°
BC = 60.8 m
6 a Let AC=b
=
x 10 −
b2 = 52 + 42 − 2 × 5 × 4 × cos 95°
h
h
= 10 −
tan50°
tan 60°
1
1


h
+
10
=
tan50
°
tan
60
°


=
h
10
= 7.06 m
1
1


 tan50° + tan60° 


8 a 180° − 67
=
° 113°
b = 6.67 cm
113° + 123° + ABC
= 360°
b Let BAC = A. Then
sin 95° sin A
=
6.67
5
h
tan60°
We equate both expressions for x and get
b2 = a2 + c 2 − 2ac cos B
sin B sin A =
b
a
h
tan50°
(10 − x ) tan60° =h
BAC = 180o − ABC − ACB
o
h
x
h
and tan60° =
10 − x
−1 22.5
=
=
ACB sin
sin 46o 19.7o
48
b
tan50° =
x =
sin 46o sin ACB
=
48
22.5
sin ACB =
AABCD = AABC + AACD = 9.96.. + 15.55.. = 25.5cm2
ABC
= 124°
b
b2 = a2 + c 2 − 2ac cos B
b2= 802 + 1202 − 2 × 80 × 120 × cos124°
b = 178km
© Oxford University Press 2019
8
Worked solutions
c
cos C =
a2 + b2 − c 2
2ab
=
VD
=
1202 + 1782 − 802
= = 0.9289
2 × 120 × 178
b
= 21.7°
cos−1 0.9289
OD2 + VO2
2
7.072 + 20=
21.2 cm
tan α =
VO
OM
20
=
α tan−1= 76.0°
5
complement to 123° is 57° , then
360° − 57° − 21.73
=
° 281°
c Let K be the point that connects A with
OA perpendicularly. Let M denote the
midpoint of BA
9 We use lowercase letters for sides opposite
capital letter angles. Find p
Complementary angle to 84° :
OA2 + MA2 − OM 2
2 × OA × MA
21.22 + 52 − 20.62
=
2 × 21.2 × 5
cos OAM =
180° − 84=
° 96° .
ˆ = 360° − 210° − 96° = 54°
Then HPQ
We have two sides and one angle
a
OAM
= cos−1 0.236226
= 76.3 °
ˆ
p = q + h − 2hq cos HPQ
2
2
2
Then sin BAK =
2
3402 + 1602 − 2 × 340 × 160 × cos 54°
p
=
BK= AB sin BAK= 10 × sin76.3=
° 9.72
p = 278km
The angle between two sloping edges,
β is formed by two sides of length BK
and the diagonal of the base
b We find H as
cos H =
2
BK
AB
2
2
p +q −h
2 pq
2782 + 3402 − 1602
=
2 × 278 × 340
= 0.884913
cos β =
9.722 + 9.722 − 14.42
2 × 9.72 × 9.72
β cos−1
=
=
= 27.8°
H cos−1 0.884913
Then 84° + 27.8
=
° 111.8° is Bc , the
complement of the angle
complementary to the bearing B .
12
a cos B
=
9.722 + 9.722 − 14.42
= 95.6 °
2 × 9.72 × 9.72
−11
a2 + c 2 − b2 62 + 82 − 122
=
=
2ac
2×6×8
24
b The cosine of the angle is negative, so
ABC > 90° , i.e. we have an obtuse
angle.
c
B
=
=
° 68.2°
180° − 111.8
Then =
B 360° − 68.2
=
° 292°
10 Triangle ABC with sides a, b, c
=
B 360° − (180° − 30° ) − 100
=
° 110°
b2 = a2 + c 2 − 2ac cos B
Chapter Review
2
b
=
3202 + 5002 − 2 × 320 × 500 × cos110°
=
b = 680km
Then
l =
=
= 26.2°
A cos 0.897
Bearing: 360° − (180° − 30° − =
A ) 236°
11 a Let M be the midpoint of AD. Then
triangle OMD is right-angled with
OM=5, MD=5, then
Then
32 + 42 = 5
b = x 2 + 2 xl = 82 + 2 × 8 × 5 = 144 m2
−1
OM 2 + MD2 = 7.07 cm
1
(8 × 8 × 3=) 64 m3
3
Slant height l is the hypotenuse of the
triangle formed by the pyramid height and
the distance from the origin O to the
midpoint of a side of the base.
b2 + c 2 − a2
cos A =
2bc
6802 + 5002 − 3202
=
2 × 680 × 500
= 0.897
OD =
1
( base area × height )
3
1 a
=
2
a=
1 2
1
3
π r h=
π × 62 × 8= 96 cm
π
3
3
slant height l is the hypotenuse of the
triangle formed by the cone height and the
cone radius
l =
82 + 62 = 10
2
b = π r 2 + π rl = π × 62 + π × 6 × 10 = 96 cm
π
© Oxford University Press 2019
9
Worked solutions
4 3 32
=
πr
π
3
3
3
=
V
4 3 32
r =
3
3
=
r3 = 8
c
r =2
Then SA = 4π r 2 = 4π × 22 = 16 m
π 2
Vcone =
OC =
1 2
1
πr h =
π × 42 × 10
3
3
Then
VO2 + OC 2
=
VC
=32 + 4.242 =
5.20 cm
1 2
1
π rcut hcut =
π × 22 × (10 − 6 )
3
3
16 π
= = 16.8 cm3 3
as required.
Vcut =
d We split the triangle BVC into two rightangles triangles, BVM and MVC, M is the
midpoint of BC.
Vtr =168 − 16.8 =151cm3
Then
d = 2r
sin BMV
sin BVM
=
VB
BM
sin90° sin BVM
=
5.20
3
65 = 2r
r =
65
2
sin BVM = 3
=
r 32.5
=
mm 3.25 cm
b Each ball has a diameter of
2 × 3.25 =
6.5 cm
2BMV = BVC
e Slant height
=
VM
6 tennis balls fit in the cylinder
=
V=
Vcyl − 6Vball
air
8 a
1m3
1000 000 cm3
4
=
VT π r h + π r 3
3
2
=
π × 1.52 × 8.5 +
b
d=
( x2 − x1 )
d=
(1 − 1)
2
2
+ ( y2 − y1 ) + ( z2 − z1 )
2
2
+ (5 − 0 ) + ( 3 − 3 )
2
2
d =5
b Midpoint
 x1 + x2 y1 + y2 z1 + z2 
,
,


2
2 
 2
6 a =
VT Vcyl + Vsph
d 3
= = 1.5m
2 2
5.202 − =
32 4.25 cm
= 62 + 2 × 6 × 4.25 = 87.0 cm2
= 0.431 × 10−3 m3
=
r
VB2 − BM 2
SA
= x 2 + 2 xl
4

Vair= 1294.14 − 6  π × 3.253 
3

= 431.38..
= 431cm3 (3 s.f.)
431.38.. cm3 ×
sin90°
= 35.2°
5.20
so BVC =
2 × 35.2 =
70.5°
h
39
= = 6
6.5 6.5
d
sin90°
5.20
BVM
= sin−1 3
V =π r 2h =π × 3.252 × 39 =1294.14 cm3
c
62 + 62 = 8.49 cm
1
1
Then OC =AC =× 8.49 =
4.24 cm
2
2
b =
Vtr Vcone − Vcut
5 a
1
AC
2
AB2 + BC 2 =
AC =
160 π
= 168 cm3
3
=
1
× 6 × 6 × 3 = 36 cm3
3
b W = 12 × V = 12 × 36 = 432 grams
4r 3 = 32
4 a
1
( base area × height )
3
7 a V
=
1 + 1 5 + 0 3 + 3 
,
,
= =

2
2 
 2
4
π × 1.53 =
74.2 m3
3
SA =
h × 2π r + 4π r 2
= 8.5 × 2π × 1.5 + 4π × 1.52 = 108 cm2
c
(1,2.5,3)
d=
( x2 − x1 )
+ ( y2 − y1 ) + ( z2 − z1 )
d =
(7 − 1)
(
d=
62 + 15 + 72 = 10
© Oxford University Press 2019
2
2
+
2
15 − 0
)
2
2
+ (10 − 3)
2
10
Worked solutions
9
Area =
1
ab sin C
2
V = 3x 2
3
x 300 − 3x 2
4
9
x 100 − x 2
=
4
1
× 6 × 4 sin30°
2
12
= 12 sin30=
°
= 6 cm2
2
(
tan 45° =
4 3 4
πr =
π × 83 = 2144.66 cm3
3
3
Then
= 171573 cm3
Then
1 2
π r h 171573 cm3
=
3
tan30° =
=
OA
70
= 397 m
tan10°
65.4
m
1 km
60 min
×
×
=
3.92 km / h
min 1000 m
1h
15 We can form a triangle with the bottom of
the building, O
CD
10
° 58°
ACO= 90° − 32=
tan15° =
=
CD 10 =
tan60° 10 3 m
b
CD
OA
CO =
10 3
= 30 m
tan30°
and tan58° =
12 4 faces, equilateral triangles
=
°
tan58
62 − 32 = 3 3
A=
=
x
= 6 x 2 + 8 xh
82 + 72 − 62
= 0.6875
2×8×7
=
θ cos−1 0.6875
= 46.6°
6 x 2 + 8 xh =
600
b
8=
xh 600 − 6 x 2
600 − 6 x 2 300 − 3x 2
=
8x
4x
V= base area × height=
x
(80 + x ) tan15°
−80 tan58° tan15°
= 60.1m
tan58° tan15° − 1
16
a cos θ
=
as required.
c
x
°
tan15
=
80 + x
x ( tan58° tan15° − 1) =
−80 tan58° tan15°
13 a =
A 2 ( x ) (3x ) + 2 xh + 2 (3x ) h
h
=
CO
80 + x
tan58° tan15° ( 80 + x ) =
x
1
1
bh =
×6×3 3 = 9 3
2
2
AT =×
4 9 3 =
36 3 cm2
b
x
CO
x
tan15°
AB = OA − OB = 30 − 10 = 20 m
h=
70
OB
Δd 397 − 70
= = 65.4 m / min
Δt
5
1
π × 402 h =
171573
3
tan60° =
70
OA
distance at angle 10°
=
OB
171573
=
h 3= 102 cm
1600 π
11 a
)
70
= 70 m
tan 45°
=
OA
tan10° =
V=
80V=
80 × 2144.66
T
B
b=
Vcone
)
14 Distance at angle 45°
d 16
=
= 8 cm
2
2
VB =
(
=
Area =
10 a =
r
300 − 3x 2
4x
A=
1
1
ab sin C =
× 8 × 7 × sin 46.6°
2
2
= 20.3 cm2
(3x ) ( x ) h
17 a
2
CB=
AC 2 + AB2 − 2 × AC × AB × cos BAC
CB2= 152 + 342 − 2 × 15 × 34 × cos 25°
CB = 21.4 m
b
ACB= 180° − 25° − 85=
° 70°
© Oxford University Press 2019
11
Worked solutions
Then
sin ACB sin ABC
=
AB
AC
sin130° sin BCA
=
100
70
sin70° sin85°
=
34
AC
sin BCA = 70
sin85°
=
AC 34
= 36.0 m
sin70°
BCA
= sin−1 70
360° − 32.4° − (180° − 150=
° ) 298°
1
=
× 34 × 36.0 × sin25°= 259 m2
2
d
c
sin ACB sin ABC
=
AB
AC
=
BC 2 1002 + 702
− 2 × 100 × 70 × cos17.6°
BC = 39.4km
20 a
sin110° sin 45°
=
34
AC
b
sin QRS sin QSR
=
QS
QR
sin26.5°
=
PQ 119
= 57.1m
sin68.5°
sin 42°
=
= 20.2 m
QS 30
sin85°
c
° 53°
SQR= 180° − 85° − 42=
1
× 20.2 × 30 × sin53°= 242 m2
2
sin85°
=
PA 119
= 127.4 m
sin68.5°
sin PAG =
1
Area = × PQ × QS sin θ =141
2
sin θ =
=
PG 127.4
=
sin26.5° 56.8 m
242
= 86.7°
12 × 20.2
21 a
and the obtuse angle,
180° − 86.7
=
° 93.3°
b
S =4 ×
2
PS=
PQ2 + QS 2 − 2 × PQ × QS × cos θ
2
PS
=
122 + 20.22
− 2 × 12 × 20.2 × cos 93.3°
PS = 24.1m
ABC
= 360° − (180° − 100° ) − 150°
= 130°
b
sin ABC sin BCA
=
AC
AB
5×7
+ 52 =95 cm2
2
h = 72 − 2.52 =6.54 cm
V =
d We choose the obtuse angle,
=
θ 93.3°
19 a
PG
PA
PG
sin26.5° =
127.4
242
12 × 20.2
=
θ sin−1
sin QPA sin PQA
=
QA
PA
sin68.5° sin85°
=
119
PA
1
Area = × QS × QR × sin SQR 2
c
sin PAQ sin QPA
=
PQ
QA
sin26.5° sin68.5°
=
PQ
119
sin 42° sin85°
=
30
QS
Area=
° 85°
PQA= 180° − 95=
QPA
= 180° − 85° − 26.5
=
° 68.5°
sin 45°
=
AC 34
= 25.6 m
sin110°
b
CAB
= 180° − 130° − 32.4
=
° 17.6°
2
BC=
AC 2 + AB2 − 2 × AC × AB × cos CAB
ACB
= 180° − 70
=
° 110°
and so ABC= 180° − 110° − 25=
° 45°
18 a
sin130°
= 32.4°
100
Then the bearing is given by
1
A = × AB × AC × sin BAC
2
c
sin130°
100
M1A1
M1A1
1
× 52 × 6.538... = 54.5 cm3
3
M1A1
22 a
b
l=
2
M1A1
2
10 − 3 = 9.54 m
6 × 9.538...
S= 4 ×
2
= 114.47...
= 114 m2
c
d
2
M1A1
2
h=
42 + 9.538... − 3
= 51.1 m
 3 
arccos 
 = 72.5º
 10 
© Oxford University Press 2019
M1A1
M1A1
12
Worked solutions
e CP 6=
=
tan 60º 10.4 m
23 a
b
M1A1
M1A1
52 + 32 = 5.83 cm
l =
S = 2 × (π × 3 × 5.83...) = 110 cm2
M1A1
1
2 × × π × 32 × 5
3
31.9%
× 100% =
π × 3.052 × 10.1
M1A1
22 − 12
= 5
2
24
a x
=
A1
132 − 52= 12
22 + 12
b=
12 204 cm2
A
×=
2
 5 
c C =
90º + sin−1 
112.6º
=
 13 
M1A1
d
M1A1
h=
25 a
M1A1
M1A1
172 + 122 = 20.8 cm
AC =
^
A1
A B C = 135º
2
AC
=
^
d The angle C D B can either be acute or
obtuse
A1
and the two possible values add up to
180º.
A1
28 a
b
sin C
sin135º
=
20
41.61...
M1
26 a
b
c
d
𝐹𝐹𝐹𝐹 = √82 + 102 + 62 = 10√2
42 + 02 + 62 = 2 13
FM =
2
CM =
2
2
4 + 10 + 0 = 2 29
p = 2 13 + 2 29 + 10 2 cm
e
27 a
b
6
3
𝑡𝑡𝑡𝑡𝑡𝑡 𝑀𝑀 = 4 = 2
𝑐𝑐𝑐𝑐𝑐𝑐 𝑀𝑀 =
A=
1
9
� +1
4
=
M1A1
A1
M1A1
29 a
tan32º =
⇒
=
x
b
c
30 a
b
c
A1
30
x
30
= 48.0 metres
tan32º
𝐴𝐴𝐴𝐴 = 𝑦𝑦 = �(3 + 48.0)2 + 302
= 59.2 metres
M1A1AG
0 + 8 0 + 0 6 + 6
M
,
,
 = ( 4, 0, 6 ) M1A1
2
2 
 2
M1A1A1
1
× 2π × 32 + 2π × 3 × 7 + π × 32
2
= 60
=
π 188 cm (3 s.f.)
^
A1
C = 19.9º
Therefore the bearing of A from point C
M1A1
is 360 − 105 − 19.9 =
235.1o
S =
2
^
b
1 4π
×
× 33 + π × 32 × 7
2 3
= 81
=
π 254 cm3 (3 s.f.)
2
20 + 25 − 2 × 20 × 25 × cos135º
M1A1
A1
AC = 41.6 km
V =
30
M1A1A1
M1A1
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 �51.0� = 30.5º
M1A1
= 89.7 metres
M1A1A1
482 + 572 − 2 × 48 × 57 cos117º
BC
=
1
A = × 48 × 57 sin (117º )
2
= 1219 sq metres (3 s.f.)
𝑠𝑠𝑠𝑠𝑠𝑠 𝐵𝐵
48
=
𝑠𝑠𝑠𝑠𝑠𝑠 117º
89.7
⇒ 𝐵𝐵 = 28.5º
M1A1
M1A1A1
A1
M1A1
M1
2√13
13
1
25
× 5 × 10 sin30º =
2
2
M1A1AG
M1A1
BD2 = 52 + 102 − 2 × 5 × 10 cos 30º
M1A1
=
BD
125 − 50 3
A1
=
BD
25 5 − 2 3
M1
(
)
=
BD 5 5 − 2 3
AG
^
c
sin C D B
sin 45º
=
13
5 5−2 3
^
sin C D B =
13 2
10 5 − 2 3
M1A1
A1
© Oxford University Press 2019
13
Worked solutions
12
Periodic relationships: trigonometric functions
Skills check
1 a
2 a
b
3 a
b
2
2
b
3
c
3
2
( −0.618, 0) ,(1, 0) ,(1.62, 0)
(0.633, 0)
( −1.61, 0.199)
(2.21, 0.792)
i
j
b
c
d
b
c
g
h
i
j
2 a
π
18π
=
180 10
225π
5π
225
=
°
=
180
4
b
c
d
e
f
g
h
80π
4π
=
180
9
200π
10π
=
°
=
200
180
9
80
=
°
120π
2π
=
180
3
135π
3π
135
=
°
=
180
4
=
°
120
π
=×
6 6
π
π
180°
π
= ×
10 10
f
85π
≈ 1.48
180
12.8π
h 12.8
=
°
≈ 0.233
180
g
=
°
85
i
=
°
37.5
j =
1°
37.5π
≈ 0.654
180
π
180
≈ 0.0175
180°
4 a 1c =
1×
57.3°
=
π
180°
b 2c =
=
2×
115°
π
c
180°
0.63c =0.63 ×
d 1.41c =1.41 ×
=
18°
5π
5π 180°
150°
= ×
=
6
6
π
180°
3π =
3π ×
=
540°
180°
π
7π
7π 180°
= ×
=
63°
20 20
π
4π
4π 180°
= ×
=
144°
5
5
π
7π
7π 180°
315°
= ×
=
4
4
π
14π
14π 180°
=
×
=280°
9
9
π
e 1.55c =1.55 ×
180°
f
3c =
3×
g
0.36c =0.36 ×
π
π
=80.8°
180°
π
=88.8°
172°
=
h 1.28c =1.28 ×
i
0.01c = 0.01 ×
j
2.15c =2.15 ×
© Oxford University Press 2019
=36.1°
π
=°
30
180°
π
110π
≈ 1.92
180
75π
75
=
°
≈ 1.31
180
e 110
=
°
270π
3π
=
°
=
270
180
2
360π
360
=
°
= 2π
180
π
25π
≈ 0.436
180
300π
300
=
°
≈ 5.24
180
25
=
°
d
π
45π
=
°
=
45
180
4
π
60π
=
°
=
60
180 3
e 18
=
°
f
10π
≈ 0.175
180
40π
=
°
≈ 0.698
40
180
3 a 10
=
°
Exercise 12A
1 a
5π
5π 180°
= ×
=
300°
3
3
π
13π
13π 180°
=
×
=585°
π
4
4
180°
π
180°
π
180°
π
180°
π
=20.6°
=73.3°
= 0.573°
=
123°
Worked solutions
1
Worked solutions
Exercise 12C
Exercise 12B
1 a i
l = rθ = 14 ×
ii
π
1 a 130° is obtuse, hence we have a
negative cosine
= 7π cm
2
3π
= 9π cm
l = rθ =12 ×
4
b
5π
5π
= cm
6
2
14π
70π
iv l =
=
rθ =
15 ×
cm
9
3
c
iii l =rθ =
3×
b i
1 2
1
π
r θ = × 142 × = 49π cm2
2
2
2
1
1
3π
A = r 2θ = × 122 ×
=54π cm2
2
2
4
A=
ii
1 2
1
5π
15π
cm2
r θ = × 32 ×
=
2
2
6
4
1
1
14π
iv A = r 2θ = × 152 ×
=
175π cm2
2
2
9
iii A =
2
A=
r 2 = 3π × 2 ×
=
r
12
π
= 72
=
72 6 2 cm
3 a=
A
1 2
=
r θ 36π cm2
2
36π × 2 π
θ
=
=
144
2
b
l = rθ = 12 ×
π
2
sin36
=
° sin (180° −=
36° ) sin144°
b
sin50
=
° sin (180° −=
50° ) sin130°
c
sin85
=
° sin (180° − 85
=
° ) sin95°
d
sin=
460° sin ( 460° − 360
=
° ) sin100°
e
= 6π
Then P = 2r + l = 2 × 12 + 6π = 42.8 m
4 Area of sector:
1
1
As = r 2θ = × 102 × 1.5 =75
2
2
Area of triangle:
1 2
1 2
=
At
=
r sin θ
10=
sin1.5 49.9
2
2
note that the angle is in radians
5
l per second: l =rθ =4 ×
12
8π
2π
 8π

− 2π=
sin = sin 
 sin
3
3
3


cos110
=
° cos (360° −=
110° ) cos 250°
c
cos 300
=
° cos (360° − 300
=
° ) cos 60°
d
cos=
500° cos (500° − =
360° ) cos140°
e
15π
π
π

cos= cos  2π − =
 cos
8
8
8


f
π
π 
19π

cos = cos  2π − =
 cos
10
10 
10

3π
3π 
π

cos = cos  2π − =
 cos
2
2 
2

9π
π
 9π

− 2π=
h cos = cos 
 cos
4
4
 4

g
=
3
10800
1 π
=
2 6
angles
1
6 We have
° per minute, which in radians
60
1
π
π
is
=
°
=
60
60 × 180 10800
π
sin−1
π
π
5π

sin = sin  π − =
 sin
6
6
6

60 times in a minute:
Then l =
rθ =
6371 ×
cos
=
40° cos (360° −
=
40° ) cos 320°
b
π
60l = 20π m
1.85km
=
5π

 = sin
7

h
A = As − At = 75 − 49.9 = 25.1 units2
π
2π
2π

= sin  π −
7
7

sin
4 a
Area of shaded region:
π
π
2π

sin = sin  π − =
 sin
3
3
3


π
π
4π

sin = sin  π − =
 sin
5
5
5

g
3 a
1
× 122 × θ =
36π
2
sin225°
is negative sine
cos 225°
divided by negative cosine hence we
have a positive tangent
tan225° =
2 a
f
1 2
1
π
r θ = × r2 ×
= 3π
2
2
12
320° is obtuse, hence we have a
negative sine
b
cos−1
π 5π
6
,
6
2 π
=
2
4
π
π
7π

cos= cos  2π − =
 cos
4
4
4

angles
© Oxford University Press 2019
π 7π
4
,
4
2
Worked solutions
c
tan−1 3 =
2 sin θ − 1 =
0
π
3
sin θ =
π
4π
π

tan= tan  + π=
 tan
3
3
3

angles
5 a
π
5π
−1 1
=
=
θ sin
and
2 6
6
π 4π
3
,
4 a
3
2
2
 8 
sin2 θ= 1 − 

 17 
tan x =
2
 8 
sin θ =
1−

 17 
15
=
±
17
and 0.93 + π =
4.07
b
15
sin θ
15
17
tan
=
θ
= =
8
cos θ
8
17
sin x
1
= −
cos x
2
tan x = −
1 a=
θ cos−1 =
0.6 53.1°
b=
θ sin−1 0.15
= 8.63°
c
and 180° − 8.63
=
° 171.4 °
tan2 x − tan x − 2 =
0
0
( tan x − 2) ( tan x + 1) =
c=
θ tan−1 =
0.2 11.3°
tan x = 2 and tan x = −1
and 180° + 11.3
=
° 191.3°
−1
Then
=
x tan
=
2 1.107..
= 1.11 (3 s.f.)
d θ = tan−1 − 0.76 = 322.8°
+ π 4.249..
= 4.25 (3 s.f.)
and 1.107=
and 322.8° − 180
=
° 142.8°
θ = cos−1 − 0.43 = 115.5°
and=
x tan−1 ( −=
1) 5.50
and 360° − 115.5
=
° 244.5°
and 5.50 − π =
2.36
−1
2 a θ sin
=
=
0.82 0.96
d
and π − 0.96 =
2.18
2 cos2 x + sin x =
1
2(1 − sin2 x ) + sin x =
1
b θ = tan−1 − 0.94 = 5.53
−2 sin2 x + sin x + 1 =
0
and 5.53 − π =
2.39
−(sin x − 1)(2 sin x + 1) =
0
θ = cos−1 − 0.94 = 2.79
so sin x 1,
=
=
then x 1.57
and 2π − 2.79 =
3.49
1
and sin x =
− , then x =
5.76 and 3.67
2
−1
d θ cos
=
=
0.77 0.69
e
1
2
1
x= tan−1 − = 5.82 and 2.68
2
and 360° − 53.1
=
° 306.9°
c
2 sin x + cos x =
0
2 sin x = − cos x
Exercise 12D
e
4
3
−1 4
=
x tan
=
0.93
3
We take the positive value for θ acute
b
4 cos x = 3 sin x
sin x
4
=
cos x
3
sin θ + cos θ =
1
2
1
2
and 2π − 0.69 =
5.59
5 a
θ =sin − 0.23 =6.05
−1
=
θ cos
=
0.3 1.27
−1
and π − 6.05 =
−2.91 =
2π − 2.91 =
3.37
3 2 sin θ + 5 sin θ =
3
cos θ = 0.3
and 2π − 1.27 =
5.02
For 2π ≤ θ ≤ 3π
2
0
( sin θ + 3) (2 sin θ − 1) =
2π + 1.27 =
7.55
For −π ≤ θ ≤ 0
Then sin θ + 3 =
0
5.02 − 2π =
−1.27
sin θ = −3
=
θ sin−1 − 3
This is outside of the domain for the sine
function. Second equation gives us
© Oxford University Press 2019
3
Worked solutions
b
c
tan θ = 1.61
θ tan
1.61 1.01
=
=
−1
and 1.01 + π =
4.16
Because we have 3x , and
−180° ≤ x ≤ 180° , then we use
−540° ≤ 3x ≤ 540° .
4.16 + 2π =
10.4
sin θ = −2 cos θ
sin θ
= −2
cos θ
3x= 60°, −60°,300°, −300°, 420°, −420°
Then x= 20°, −20°,100°, −100°,
tan θ = −2
140°, −140°
θ=
tan−1 − 2 =−1.11
d
2.03
π − 1.11 =
2 tan2 θ + 5 tan θ =
3
x
+3 =
0
2
x
3
=
− =
−1
2
3
tan
1
−3 and tan θ =
tan θ =
2
Because we have
x
, and
2
−180° ≤ x ≤ 180° , then we use
x
−90° ≤
≤ 90° .
2
1
2
θ = −1.25
and θ =
0.46 =
0.46 − 2π =
−5.82
as well as −1.25 − π =
−4.39
and − 5.82 + π =
−2.68
2 a
6 3 cos x = 5 sin x
tan x =
3
5
sin−1
3
= 3θ
2
π 2π 7π 8π 13π 14π
3
,
3
then θ =
b
as well as 31° + 180
=
° 211°
,
3
,
3
,
,
3
3
π 2π 7π 8π 13π 14π
9
,
9
,
9
,
9
,
9
,
9
cos 3θ − 1 =
0
3θ = cos−1 1
Because we have 3θ , and 0 ≤ θ ≤ 2π ,
then we use 0 ≤ 3θ ≤ 6π .
Exercise 12E
3
= 2x
2
3θ = 0,2π , 4π ,6π then θ = 0,
Because we have 2x , and
−180° ≤ x ≤ 180° , then we use
−360° ≤ 2 x ≤ 360° .
2 x= 30°, −30°,330°, −330°
c
sin
θ
2
=
2π 4π
,
,2π
3 3
2
2
θ
2
= sin−1
2
2
Then x= 15°, −15°,165°, −165°
cos−1
x
−90°
=
−45° and so x =
2
3θ =
3
=
x tan−1 = 31°
5
cos−1
Then
Because we have 3θ , and 0 ≤ θ ≤ 2π ,
then we use 0 ≤ 3θ ≤ 6π .
sin x
3
=
cos x 5
b
3 tan
0
( tan θ + 3) (2 tan θ − 1) =
θ =
tan−1 − 3 andθ =
tan−1
1 a
1
2
3x = cos−1
2π + 1.01 =
7.30
d
1
2
cos 3x =
For 2π ≤ θ ≤ 4π
c
2 cos 3x − 1 =
0
3
= 3x
2
Because we have
Because we have 3x , and
−180° ≤ x ≤ 180° , then we use
−540° ≤ 3x ≤ 540° .
then we use 0 ≤
θ
=
4
,
3x= 30°, −30°,330°, −330°,390°, −390°
Then x= 10°, −10°,110°, −110°,
Then θ =
© Oxford University Press 2019
2
2
, and 0 ≤ θ ≤ 2π ,
≤π
π 3π
2
130°, −130°
θ
θ
4
π 3π
2
,
2
4
Worked solutions
d
sin2
sin
2θ
−1 =
0
3
sin2 θ =1 −
2θ
= ±1
3
sin θ =
2θ
, and 0 ≤ θ ≤ 2π ,
3
2θ
4π
then we use 0 ≤
.
≤
3
3
Because we have
π
2
×
3 3π
=
2
4
1 a
c
3
1
 1
−
cos 2θ =
cos2 θ − sin2 θ =−

 − =
4
2
 2
d
sin2θ
tan2
θ
=
=
cos 2θ
2
2 sin5 cos =
5 sin2 × =
5 sin10 by the
double angle formula
π
π
2
 1
2
1
 −  + cos θ =
 8
π
c
2 sin 4π cos 4π = sin2 × 4π = sin8π by the
double angle formula
cos2 θ =
1−
d
cos2 0.4 − sin2 0.4 = cos 2 × 0.4 = cos 0.8
cos θ = −
f
by the double angle formula
double angle formula
π
= 2×−
π
π
b
2
2
acute cos θ =
b
2 2
3
sin2θ = 2 sin θ cos θ = 2 ×
1 2 2 4 2
×
=
3
3
9
2
c
d
3 a
8 1
7
cos 2θ =
cos2 θ − sin2 θ =−   =
9 3
9
tan2
=
θ
sin2θ
=
cos 2θ
4 2
4 2
9
=
7
7
9
sin2 θ + cos2 θ =
1
63
32
cos
=
2θ cos2 θ − sin2 θ
=
1
2
1
  + cos θ =
3
We take only the positive value as θ is
1
63
×−
=
8
8
2
sin2 θ + cos2 θ =
1
1 8
1
cos θ =1 −   =1 − =
9 9
3
63
8
2

63 
 1
= −
− − 


8 
 8

2 a We use the Pythagorean identity
2
1
63
=
64 64
sin2θ = 2 sin θ cos θ
2 cos2 6 −=
1 cos 2 ×=
6 cos12 by the
1 − 2 sin2 = cos 2 × = cos by the
4
4
2
double angle formula
3
sin2 θ + cos2 θ =
1
cos = sin2 × = sin π by the
2
2
2
double angle formula
e
3
−
2
=
1
−
2
θ is obtuse, so we take the negative
value of the cosine
b
2 sin
3
1
3
×− = −
2
2
2
sin2θ = 2 sin θ cos θ = 2 ×
4 a
Exercise 12F
3
2
b
2θ
π
= sin−1 ±=
1
3
2
Then θ =
1 3
=
4 4
63
1
62 31
−
=
=
64 64 64 32
63
32
=
31
32
c
sin2θ
=
=
θ
tan2
cos 2θ
d
sin 4θ = 2 sin2θ cos 2θ
=
2×
5 a
63
31
63 31 31 63
×
=
32 32
512
sin2θ = sin θ
this is true for θ = 0,2π ,π
divide by sin θ
2 cos θ = 1
cos θ =
1
2
π 5π
−1 1
=
θ cos
=
,
2 3 3
2
 1
sin2 θ +  −  =
1
 2
© Oxford University Press 2019
5
Worked solutions
b
6 a
cos 2θ + sin θ =
0
32 sin x cos x
1 − 2 sin θ + sin θ =
0
=2 × 16 × sin x cos x =16 sin2 x
0
− ( sin θ − 1) (2 sin θ + 1) =
Then a=16, b=2
2
b 16 sin2 x = 8
sin θ = 1
2
and sin θ = −
θ = sin−1 −
π
and so x =
11π
6
=
6
7
sin2θ = 3 cos θ
1
Area =
× 15 x sin2θ =
10
2
15 x sin θ cos θ = 10
π 3π
2
,
2
15 x sin θ
divide by cos θ
2 sin θ = 3
sin θ =
d
π 2π
3
10
( 1 − sin θ ) =
with sin θ =
3
2
then θ =
,
3
cos θ = sin θ sin2θ
cos θ = sin θ 2 sin θ cos θ
true for θ =
,
2
1
we have that
4
15 x ×
2 
1 
1
× 1−   =
10
4 
4 


15 x ×
1
15
10
×
=
4
4
15
x = 10
16
π 3π
2
π 5π
,
12 12
1
15x × 2 × sin θ cos θ =
10
2
2 sin θ cos θ = 3 cos θ
this is true for θ =
8
1
=
16 2
Note that 0 ≤ 2 x ≤ 2π , so
1
π 5π
sin−1= 2=
,
x
2
6 6
1
2
1
π 7π
=π +
=
2
6
6
and θ = 2π −
c
so sin2=
x
π
−1
=
=
1
θ sin
2
=
x
divide by cos θ
160 32
=
15
3
2 sin2 θ = 1
sin2 θ =
Exercise 12G
1
2
sin θ = ±
1 a
1
2
π 3π 5π 7π
Then θ =sin−1 ± 1 / 2 = ,
,
,
4 4 4 4
e
cos 2θ = cos θ
2 cos2 θ − 1 =
cos θ
2 cos2 θ − cos θ − 1 =
0
(cos θ − 1)(2 cos θ + 1) =
0
cos θ = 1
cos−1 1= θ= 0,2π
2 cos θ + 1 =
0
cos θ = −
1
2
1
2
θ= cos−1 − =
2π 4π
,
3 3
© Oxford University Press 2019
6
Worked solutions
b
c period of 2π , amplitude of 2 , then
y = −2 cos x
d period of 2π , amplitude of 2 , then
=
y 2 sin ( − x )
4 a The amplitude is 6 = 6
b The period is
2π
=4
π /2
Exercise 12H
c
1 a amplitude 3, period
2π
= π . Option iv
2
b amplitude 2, period 2π and vertical
shift +1 . Option ii
1
π
π
, horizontal shift −
or
2
2
2
units to the right. Option i
c amplitude
d amplitude 1, period
2π
= 4π , vertical
1/2
shift 2. Option iii
d
2 a period 2π , amplitude
max − min 5 − 1
= = 2 , vertical shift +3
2
2
, =
y 2 sin x + 3
b period 2π , amplitude
max − min 2 − 0
= = 1 , vertical shift +1
2
2
horizontal shift π , =
y cos ( x − π ) + 1
2 a The amplitude is 1 = 1 and the period
is
2π
.
3
2π
=π .
2
c The amplitude is −4 =
4 and the period
is
2π
.
3
d The amplitude is −
period is
2π
, amplitude
3
max − min −0.5 + 1.5
= = 0.5 ,
2
2
vertical shift
−1 , y 0.5 sin3x − 1
=
d period
b The amplitude is 0.5 = 0.5 and the
period is
c period 2π , amplitude
max − min 0 + 4
= = 2 , vertical shift −2
2
2
=
y 2 cos x − 2
3 a vertical shift 2, amplitude 3, period π ,
plot given
1 1
= and the
2 2
2π
= 6π .
1
3
3 a period of π , amplitude of 1 , then
y = sin2 x
b period of π , amplitude of 3 , then
y = 3 cos 2 x
© Oxford University Press 2019
7
Worked solutions
b horizontal shift −
π
, amplitude 0.5 ,
3
period 2π , plot given
1
=
y , find zero at
2
π 5π 13π 17π
x = ,
,
,
6 6
6
6
6 Plot sin x −
7 Plot e x − cos x =
y between −2 and −1 .
Find zero at ( −1.29, 0 )
8 a c is the horizontal shift:
π
2
b The graph of y = cos x may be
c vertical shift −1 , amplitude 1 ,
period 2π , horizontal shift −π , plot
given
translated
π
horizontally to the right to
2
become the graph of y = sin x
9 a
d vertical shift 2, amplitude 2, period π ,
plot given
b
10 a
b
c
0.6075,1.571,2.534, 4.712
(2.36, −1)
2π
= 4π
1
2
x = 0,1.26,3.77, 4.19
Exercise 12I
1 a
e vertical shift 1 , amplitude 2 ,
2π
period
, horizontal shift −π ,
3
plot given
4
( −0.824, 0) ,(0.824, 0)
5 Plot 2 sin x − x − 1 =
y . Find zero at
( −2.38, 0)
© Oxford University Press 2019
8
Worked solutions
b
c
f
2 a 1.25, 4.39
b 1.13, 4.53
c
–π ,0,π
d
−0.903, 0.677,1.98,2.61
Exercise 12J
π

1 a Minimum when sin  ( t − 5)  =
−1 , i.e.
6


h ( t ) =−5 + 7 =2 m
π

Maximum when sin  ( t − 5)  =
1 , i.e.
6


h ( t ) = 5 + 7 = 12 m
d
b
π

sin  ( t − 5)  =
1
6


π
 π
⇒t = 8
 ( t − 5)  =
6

 2
High tide at 8 am
c
π

sin  ( t − 5)  =
−1
6


π
π

 ( t − 5)  =− ⇒ t =2
6
2


Low tide at 2 am
e
π

d =
h ( 9 ) 5 sin  ( 9 − 5)  + 7
6


2π
= 5 sin
+7
3
=
© Oxford University Press 2019
5 3
+7
= 11.3 m
2
9
Worked solutions
π
so −12 cos
4
4
+ sin−1   , − sin−1   ,
5
5
4
4
π + sin−1   , 2π − sin−1  
5
5
(t − 5) =−π
6
c
6 a
0:46 am, 3:14 am, 12:46 am, 3:14 pm
(multiply decimals by 60 to convert the
decimal number of hours into minutes)
2 a 13000 on February 1
7000 on August 9th
c
3000 cos 0.5 ( 4 − 1) + 10000 =
10212
)
b 10
cos
20
π
20
23 − 21
1
= −
10
−20
x =
cos−1 −
1
=x =10.64 s
10
60 − 40
= 10
2
period 2 × (1.8 − 0.3) =
3 , then b =
35 − 5
= 15m
2
vertical shift
d=
a 15, 20
=
c vertical shift
=
e horizontal distance between
maximums: 4 − 0 =
4 (period)
b is calculated as=
b
π
2
t + 20 . Find y = 30 at
60 + 40
= 50
2
 2π
50 + 10 cos 
y =
(t − 0.3) 
3


b
 2π
y =
50 + 10 cos 
43.3m
(17.2 − 0.3)  =
 3

maximum is at 2m + 24 m = 26 m
above the ground
minimum is at 2 m above the ground
26 − 2
= 12 , increases then
2
decreases, so a = −12
amplitude
max + min 26 + 2
= = 14 m
2
2
π
h (t ) =
−12 cos
x + 14
20
π
20
c Plotting the function and the line y = 59
gives t = 0.0847 s
Chapter review
1 a
b
30 ×
π
π
=
6
π
5π
150 ×
=
180
6
180
7π
=
4
π
2π
d 120 ×
=
180
3
2π
π
=
b =
40 20
b At t = p , −12 cos
2π
3
maximum at 0.3 instead of 0 , so
horizontal shift of 0.3
π
2π
=
4
2
5 a radius=12 m, so diameter=24 m
=
c
π
7 a amplitude
b 5m
t = 0.535
2π
π
=
40 20
π
b for h =
−20 cos
x + 21 ,
23 m =
20
d 16 fish
4 a 35 m
g=
plot y 15 cos
max = 20 × 2 + 1 = 41m , min = 1m .
Then amplitude is 20, a = −20
h (t ) =
−20 cos
x + 21
20
20 − 0
c amplitude
= 10 , vertical shift 10
2
. Then y 10 sin0.5x + 10
period ≈ 4π=
f
40
= 13.3 s
3
π
3 a 20
c
20
6 , i.e. cos
p=
2π
then=
p
3
p=
period
st
b
π
20
vertical shift =
max + min 41 + 1
= = 21m
2
2
t = 0.771, 3.229, 12.771, 15.229
(
π
1
π
p= − .
20
2
2π
Then the required angle is θ =
3
π
e=
h ( t ) 5 sin  ( t − 5
=
)  + 7 3
6


4
π
−1 
(t − 5=) sin  − 5 
6


p + 14 =
20
c
315 ×
e
−20 ×
f
π
180
π
=
−
π
9
4π
−240 ×
=
−
180
3
© Oxford University Press 2019
180
π
10
Worked solutions
3π
=
−
2
4π
π
h 144 ×
=
180
5
g
2 a
b
c
d
e
f
g
h
−270 ×
π
c
180
6 a
b
7π 180
×
=
−105°
12
π
π 180
×
=
20°
π
9
2x =
6
,
6
6
rad
5
A=
1 2
1
6
r θ = × 25 × = 15
2
2
5
6
,
6
π 5π 13π 17π
,
,
,
12 12 12 12
2
 12 
2
1
−
 + sin θ =
 13 
sin θ =
b
cos2 θ + 0.62 =
1
5
13
cos
=
2θ cos2 θ − sin2 θ
2
8 a
cos θ = ±0.8
2
5
sin (θ + π ) =
− sin θ =
−
13
2 sin2 θ + sin θ − 1 =
0
(sin θ + 1)(2 sin θ − 1)
We take the positive value for acute θ
Then sin θ = −1
cos θ = 0.8
and sin θ =
sin θ
0.6
= = 0.75
cos θ
0.8
b
2 sin x = tan x
144
25
=
169 169
119
 12 
 5 
=
−
 −
 =
169
 13 
 13 
c
cos2 θ =
1 − 0.36 =
0.64
holds for x = 0 . We divide by tan x to
get
1
2
3π
sin−1 − 1 =
2
and sin−1
2 cos x = 1
1 π 5π
= ,
2 6 6
2
1
2
holds for x =
,
7 a the angle is obtuse, so we need a
positive sine
cos2 θ + sin2 θ =
1
b
π 5π 13π 17π
sin2 θ =
1−
θ =
9 a
π
3
and x = −
π
7π 11π
,
6
6
3
2
1
  + cos θ =
4
cos2 θ =
1−
3
9
7
=
16 16
Obtuse angle, so we take the negative
1
π
π
=π +
and2π −
2
6
6
So θ =
1
= 2x
2
so x =
11π 180
×
=330°
6
π
34π 180
×
= 408°
15
π
sin−1 −
1
2
Note that 0 ≤ 2 x ≤ 4π , so
4 a We use the Pythagorean identity
5 a
4 sin2 x = 2
sin−1
7π 180
×
= 420°
3
π
11π 180
−
×
=
−66°
30
π
cos x =
5π
4
8 sin x cos x =
4 × 2 sin x cos x =
4 sin2 x
sin2 x =
6= 5 × θ
b
π
and π=
+
4
2 sin2 x = 1
−
and tan
=
θ
4
so a = 4 and b = 2
3π 180
×
=270°
π
2
7π 180
×
=210°
6
π
3 a We have that rθ = l so
b
π
−1
tan
1
=
cosine cos θ =
b
− 7
4
cos
2θ cos2 θ − sin2 θ
=
2 π
π 7π
−1
cos=
and 2=
π−
2
4
4
4
© Oxford University Press 2019
11
Worked solutions
2
15 a Plot
=
y cos x − x 2 and find zeros for
2

7
1
3
=
−
 −
 −   =
4
4
8
 


π
10 period:
, amplitude =
2
30 + 30
= 30 .
2
2π
= 4 and a = 30
b
Then=
π
2
11 a period is π − 0 =
π
b 3
12 a
b
2π
= 2
π
2
, then x = 0.824
b Plot =
y 4 sin π x − 4e − x + 3 for
0.5 ≤ x ≤ 1.5 and find zero at x = 1.14
1
1
× 10 × 8 × sin θ = 10
ab sin θ =
2
2
16 A =
sin θ =
1
4
Obtuse angle and positive sine,
π
π − 0.25 =
2.89
5
2π
π
−1 1
=
θ sin
=
0.25
4
c 3
d =
b
0≤ x ≤
17 Area of the sector is
1
1
Asector = r 2 θ = × 102 × 0.8 =40
2
2
=4
2
ˆ =
Then OAN
π − 0.8 =
0.77
c
Using the sine rule
sin
ˆ
ˆ
sin ANO
sin OAN
=
AO
ON
π
2 = sin 0.77
10
ON
=
=
ON 10
sin 0.77 6.96
1
Then Atriangle = × 10 × 6.96 × sin 0.8 =25
2
Then
Ashaded
= Asect − Atriangle
= 40 − 25 = 15 cm2
18 a Using the sine rule
Asector − Atriangle
13 A=
shaded
Ashaded
=
1 2
1
r θ − r 2 sin θ
2
2
Ashaded =
π 1
π
1
× 82 × − × 82 × sin
2
6 2
6
= 0.755
14 a =
A
1
2 × 4 × sin=
θ 2
2
2 2 sin θ = 2
sin
=
θ
1
3π
=
4
2
for obtuse θ
b Then CBD =π −
and so ABDC =
3π
π
=
4
4
1
π
× 42 × = 2π
2
4
sin ADO sin AOD
=
AO
AD
sin AOD
sin 0.8
=
AD AO = 8= 14.7 cm
sin ADO
sin 0.4
b
DAO =
π − ADO − AOD
=
π − 0.8 − 0.4 =
1.94
sin DAO sin ADO
=
OD
AO
=
OD
sin DAO
sin1.94..
AO
= 8= 19.1cm
sin ADO
sin 0.4
1 2
1
r θ = × 82 × 0.8 = 25.6 cm2
2
2
c
A=
d
1
Atriangle = × AD × OD × sin ADO
2
1
54.9 cm2
=× 14.7 × 19.1 × sin 0.4 =
2
=
AABCD Atriangle − Asector
= 54.9 − 25.6 = 29.3 cm2
© Oxford University Press 2019
12
Worked solutions
19 a
f ( x ) =2 × 10 × sin3x cos 3x =10 sin6 x
22 a
q
p = 3 ,=
b 10 sin 6 x = 0
sin 6 x = 0
−d =−
for 0 ≤ 6 x ≤ 2π . Then 6 x = 0,π ,2π
and x = 0,
20 a =
f (θ )
π π
b
,
6 3
(2 cos
2
c
d=
f ( x ) 3 sin
l = rθ = 15 ×
A=
π
1
2
Wheel then turns a further
π
6
2π
complete a total turn of
.
3
hA = 15 + 15 sin
π
6
= 15 ×
π
2
, A is 15
radians to
3
= 22.5 m
2
e
π 
h (0) =
15 − 15 cos 2   =
4.39 m
8
π

When cos 2  t +  is −1 , we have a
8

maximum. Then
t =
π
2
−
π
8
= 1.18 s
x +5
π
−(3 sin ( x − 3) + 5) =
fˆ ( x ) =
g (x)
4
A ( t ) = (2 cos t − 1) ( cos t − 1)
24 a
b
A1A1
0
(2 cos t − 1) ( cos t − 1) =
2 cos t − 1 = 0 ⇒ cos t =
1
2
π 5π
⇒ t =,
3 3
π 
π π 
h   = 15 − 15 cos 2  +  = 25.6 m
4
4 8
π

2 t +  =
π
8

4
reflected in the x axis
d
π

−1
cos 2  t +  =
8

π
π
fˆ ( x ) 3 sin ( x − 3) + 5
=
4
1 2
1
π
r θ = × 152 × = 58.9 m2
2
2
6
m above the ground.
8+2
=5
2
translated (3,0)
= 7.85m
6
2π
π
=
8
4
b 10 − 2 =
8 =period,=
b
cos θ = 0
⇒θ =
−90°, 90°
c When wheel turns through
f
23 a 3
c
⇒θ =
−120°,120°
b
4
3 cos
=
x sin2 x − 1
3
cos θ (2 cos θ + 1) =
0
2 cos θ + 1 =0 ⇒ cos θ =−
21 a
2
=−1 , so d = 1
2
The intersection points can be found as
x = 1.30,3.41,6.19
θ − 1) + cos θ + 1
= 2 cos2 θ + cos θ
b This is a quadratic equation in cos θ
with positive discriminant, so there are 2
distinct values of cos θ which satisfy it.
2π
4
2π
,=
=
b = 2
3π
3
π
2
M1A1
cos t − 1 = 0 ⇒ cos t = 1
M1A1
t = 0,2π
25 2 cos x = sin2 x
2
M1
⇒ 2 cos2 x − 2 sin x cos x =
0
2 cos x ( cos x − sin x ) =
0
=
cos x 0,
=
cos x sin x
x =
x =
π
2
π
4
M1
M1
, x =
3π
2
A1
, x =
5π
4
A1
26 a Correct attempt to at least one
parameter
=
a
=
b
=
c
© Oxford University Press 2019
M1
14 − 8
= 3
2
A1
2π
= 2
A1
14 + 8
= 11
2
A1
π
13
Worked solutions
b
ii θ =
29 a i
A1
2
A1
a = −2
A1
=
b
A1 for trigonometric scale and correct
domain,
A1 for correct max/min,
27 a
2
2
S ( x ) = sin
2x
2x + 2
x cos
144
4 2+ cos
4 4 43
1 sin2
4 44 2
4 42
43x
AG
b A1 for correct shape, A1 for 2 cycles,
A1 for correct max/min
M1A1
A1
−2 cos π x + 1 = 0 ⇒ cos π x =
 π π 5π 7π 
, ,
,
,
 3 3 3 3 
M1A1A1
= 1 + sin 4x
2π
= π
2
c =1
c
A1 for two complete cycles
A1
−1 ≤ y ≤ 3
ii
b
π
18
π x ∈ −
M1
 1 1 5 7 
x ∈  − , , , ,
 3 3 3 3 
M1
30 a i =
x 0,
=
x
ii
b
1
M1
2
π
=
,x π
2
π
A1
A1
2
iii ¡
A1
b=2
A1
d =1
A1
c The first point of inflexion occurs at
π
R1
x =
4
c i
ii
π
A1
2
0≤y ≤2
d <Please insert the graph of
=
y cos (2 x ) − 1 for 0 ≤ x ≤ 2π >
e i
ii
1
2
ii
b i
A3
A=
1
2π
4π
× 22 ×
=
2
3
3
2π
4π
l =
2×
=
3
3
rθ =
r
2
θ
2
π
3
=π
 π
−3
a tan  −  =
4
1 4 2 43 
A1
a=3
AG
−1
p = π (or any π + 2nπ , n ∈ ¢ )
A1
A1
π 
f   = −2
8
  π π 
⇒ f ( x ) =a tan  2  −   + 1 =−2
  8 4 
M1
A1
A1
q = −2
28 a i
d
e
π
A1
8
5π
9π
and
8
8
A1
31 a
A1
Solve simultaneously
M1
r = 6 cm
A1
−2 cos2 x + sin x + 3
(
)
=
−2 1 − sin2 x + sin x + 3
A1
A1
A1
= 2 sin2 x + sin x + 1
b
M1
A1
−2 cos2 x + sin x + 3 =
2
⇒ 2 sin2 x + sin x + 1 =
2
M1
2
2 sin x + sin x − 1 =
0
⇒ (2 sin x − 1) ( sin x + 1) =
0
© Oxford University Press 2019
M1
14
Worked solutions
sin x =
1
, sin x = −1
2
A1
π
7π π 5π 3π 
 11π
x ∈ −
,− ,−
, ,
,

6
2
6 6 6 2 

A2
Award A1 for two correct
solutions
© Oxford University Press 2019
15
Worked solutions
13
Modelling change: more calculus
( )
Skills Check
1 a
2
2
−
3
2
d
2 a
1
3
6 y sin
=
=
x sin x 3
2
b -1
c
−
e 0
f
−
2
2
3
2
dy  1 − 23 
1 − 23
= =
x  cos 3 x
x cos 3 x
dx  3
3

1
2
π
2
cos 2 x − sin 2 x =cos u − sin u
dy
=
−π x −2
dx
(
2u cos 4 x
= cos
=
b
6 sin x cos =
3 × 2 sin x cos x =
3 sin2 x
c
e x sin2 x + e x cos2 x =
e x (1) =
ex
d
dx
3 a
=
(
3
)
4 x 3 + 7=
x
d 
3
 4x + 7x
dx 
1
12 x 2 + 7 4 x 3 + 7 x
3
(
)(
(
)
−
2
3



(
3 4x 3 + 7x
d
3x 2=
e2 x
6 x e2 x + 6 x 2 e2 x
dx
c
1 2
x − 2 x ln x
d  ln x  x
 2 =
dx  x 
x4
=
1
3
g '(x) =
)
)
2
3
9
11 a
b
c
d
d
( g o f ) ( x )  =
cos 4 x 3
dx 
dx
( (
)(
(
(
r ( x ) = x 2 cos 4 x 3
(
(
= 2 x cos 4 x
12 a i
ii
1
x
=
cos + 3 sin (3x )
3
3
))
−12 x sin ( 4 x )
)) =
2
3
)
3
(
)
) − 12x
4
(
(
sin 4 x
3
)
))
f '(x ) = cos x
f ''(x ) = − sin x
iii f '''(x ) = − cos x
iv f (4)(x ) = sin x
−1
3
= 3 ( cos x )
cos x
f '(x) =
3 ( − sin x ) ( −1) ( cos x )
−2
3 sin x
=2
cos x
b n = 4x where x is an integer, therefore
n = 4, 8, 12
(
)
c i =
f ( ) (x ) f=
(x ) sin x
80
ii
h ( t ) = sin3 t
cos t ) (3 sin2 t )
(=
( g o f ) ( x ) = cos ( 4x 3 )
=
r ' ( x ) 2 x cos 4 x 3 + x 2 −12 x 2 sin 4 x 3
dy
= 5 cos (5x )
dx
=
h ' (t )
1
sin ( 4 x )
2
dy
=
2 cos ( 4 x )
dx
(
dy 1
x
3=
cos − ( −3 sin (3x ) )
dx 3
3
5
( sin x ) ( − sin x ) − cos x ( cos x )
12 x 2 − sin 4 x 3
=
f=
' ( x ) 4 ( cos x ) + 3 ( − sin x )
( = 3 sec x tan x )
2
f (x) =
sin2 x + cos2 x =
1 ⇒ f '(x) =
0
∴
x − 2 x ln x 1 − 2ln x
=
x4
x3
4 =
f (x)
−1
10 y sin
=
=
(2x ) cos (2x )
= 4 cos x − 3 sin x
2
π
π
sin
) ( − sin (π x ) ) =
x
x
sin2 x
sin2 x + cos2 x
1
=
−
=
−
=
−cosec2 x
sin2 x
sin2 x
Exercise 13A
1
)
1
cos x
=
tan x
sin x
8 =
g (x)
12 x 2 + 7
=
b
(
)
(
7 y cos
=
= cos π x −1
x
20× 4
42
10× 4 + 2 )
f ( ) (x) = f (
(x ) = − sin x
3 cos t sin2 t
© Oxford University Press 2019
Worked solutions
1
Worked solutions
Exercise 13C
Exercise 13B
1
f (π ) = −1
π

1=
f ' ( x ) 3 ( 4 ) cos  4 x −  + 5
6

f '(x) =
− sin x + 2 cos x
π

= 12 cos  4 x −  + 5
6

∴ f ' (π ) =
−2
Therefore at (π , −1) the tangent
has gradient − 2 and the normal
1
has gradient
2
Tangent: y − ( −1) =−2 ( x − π )
2
dy
=
dx
3
dy (1 − cos x ) ( cos x ) − sin x ( sin x )
=
2
dx
(1 − cos x )
⇒y =
−2 x + 2π − 1
Normal: y − ( −=
1)
x
π
⇒y =
−1−
2
2
2
1
(x − π )
2
=
4
π 
f   =1
3
5
cos x − 1
(1 − cos x )
2
= −
3
+2 x2 + x
1
1 − cos x
f ' ( x ) = 2 xe x + x 2 e x + e x
)
2
h ' ( t ) = −2ecos t sin t
dy
6
5e5 x sin (3x ) + 3e5 x cos (3x )
=
dx
π 
Therefore at  ,1  the tangent is
3 
parallel to the x − axis and the normal
is parallel to the y − axis
Tangent: y = 1
π
3
1
π 
f =
 sin  =

4
 
4
b
f ' ( x ) = π cos (π x )
c
π
1
f '  =
2
4
1
=
2
2
2
(
= e5 x 5 sin (3x ) + 3 cos (3x )
f ' ( x ) = −2 sin x
f '(x) =
− 2 =
−2 sin x
1
⇒ sin x =
2
π 3π
⇒ x =,
4 4
)
7
dy
=
− sin x + sin x + x cos x =
x cos x
dx
8
f ' ( x ) = −2 xe x sin e x
( )
2
9 f '(x)
=
11 a
b
2
1
sin (3x ) + 3ln (3x ) cos (3x )
x
10 f ' ( x ) = 3 cos (3x ) ⋅
π 
1
1
∴y − =
x − 
4
2
2
πx
π
1
⇒ y=
+
−
2
2 4 2
4
)
+ 4 x + 1 e4 x
(
π 
f '  = 0
3
3 a
2
= e x x 2 + 2 x + 1= e x ( x + 1)
f ' ( x ) = −6 sin ( 6 x )
Normal: x =
(12x
1
= 3 cot (3x )
sin (3x )
f ' ( x ) = −ecos x sin x
ecos x is always positive, so can just
consider the behaviour of
− sin x
∴ Increasing for π < x < 2π
Decreasing for 0 < x < π
c
f ' ( x ) = 0 ⇒ x = 0, x = π or x = 2π
(
=
f '' ( x ) ecos x sin2 x − cos x
)
f '' ( 0 ) =−e < 0
f '' (=
π ) e−1 > 0
f ''=
(2π ) f '' (0) < 0
© Oxford University Press 2019
2
Worked solutions
d
Alternative method: Since the
exponential function is a
continuous increasing function,
the minima and maxima of cos x
will correspond directly to respective
minima and maxima of ecos x
∴ Local maxima at ( 0, e ) and (2π , e )
(
Local minimum at π , e−1
12 a
)
f ' ( x )= 1 − sin x
Exercise 13D
f '' ( x ) = − cos x
b
1
Concave up when f '' ( x ) > 0
C ' (120 ) = 70
 π 3π 
⇒ x ∈ ,

2 2 
Concave down when f '' ( x ) < 0
π
3π
c =
=
=
f '' ( x ) 0 when
x
or x
2
2
Concavity changes at both of
these values of x, so
c
=
−2 sin2 x − sin2 x
= −3 sin2 x
f ' ( x ) = 0 when
,=
x π
π

∴ ( 0,2 ) ,  , −1  ,
2

c
Look for change in sign of v ( t )
This occurs at t = 1
4 a
(π ,2)
(
)
0.2 10
120 e ( ) − e0
= 12 e2 − 1
10
(
)
b P ' ( t ) 120
=
=
(0.2) e0.2t 24e0.2t
f '' ( x ) = −6 cos 2 x
c
f '' ( x ) = 0 in the interval when
3π
=
, x
4
4
Concavity changes at these
=
x
1−2
= −e−2
e2
 π 
 2π 
 2π 
=
v '
 3 cos 
 + 6 sin 

27
3




 3 
 3 6 3 −3
 1
= 3−  + 6
 =

2
 2
 2 
In the range 0 ≤ x ≤ π ,
2
( )
3 =
v ( t ) d=
' ( t ) 3 cos18t + 6 sin18t
=
−2 sin2 x − 2 sin x cos x
π
e2t
∴ Velocity is − e−2 and speed is e−2
−2 sin2 x + 2 ( − sin x ) ( cos x )
f '(x) =
x 0,
x
=
=
( )
et (1) − t et
et (1 − t ) 1 − t
=
2
et
et
=
b v (2 ) =
π π 
 3π 3π 
,
 ,  and 

2 2
 2 2 
are both points of inflexion
b
This means it costs 70 Euros to produce
the 121st table
2 a =
' (t )
v ( t ) s=
 π   3π

,2π 
⇒ x ∈  0,  U 
 2  2

13 a
C ' ( x=
) x − 50
π
P ' (10 ) = 24e2
At day 10 the number of bacteria
are increasing at a rate of
177 bacteria per day
π 1
 3π 1 
points, so  ,  and 
, 
 4 2
 4 2
are points of inflexion
© Oxford University Press 2019
3
Worked solutions
5 a
P '(x) =
−0.00015x 2 + 12
π
d2 A
=
−128 < 0
At θ =
,
4
dθ 2
So this value of θ gives the
maximum value of the area
P ' (200 ) = 6
The profit gained by selling the
201st unit of the chemical is 6 euros
b
3 a
C=
(x) R (x) − P (x)
vertical and the curved
(
= 10 x − 4 − −0.00005x 3 + 12 x − 200
)
face of the cone be θ , then:
4 2
r
2
tan θ =
=
=
⇒ r = (6 − h)
6 3 6−h
3
= 0.00005x 3 − 2 x + 196
2
c C ' ( x ) 0.00015x 2 − 2
=
=
C ' (200 ) 0.00015
=
(200) − 2 4
2
6 a
b
3.19s (use of GDC)
s (3.18533) − s ( 0 )
3.18533
= −0.408 (3 s.f.)
2

b=
V π=
hr 2 π h  ( 6 − h ) 
3


4

= π h  36 − 12h + h2 
9


4π
2
3
=
36h − 12h + h
9
(
v (t ) =
s ' (t ) =
−9.8t + 15.2
d v ' (t ) = 0 ⇒ t =
15.2
= 1.55
9.8
(3s.f .)
d
)
dV
=0 ⇒ 3h2 − 24h + 36 =0
dh
d2V
4π
16π
= ( −12 ) =
−
<0
dh2
9
3
2
8
∴ h= 2, r=
(6 − 2=) 3
3
At h =
2,
10000
x2
4 a
C ' ( x ) = 0 ⇒ x = 100
20000
x3
⇒ C '' (100 ) > 0 so minimum
C '' ( x ) =
2 a | PQ
=| | SR
=| 8 cos θ
PS | | QR
|=
=| 8 sin θ
∴A=
(8 cos θ ) (8 sin θ )
dP
=
dx
c
3
dP
−2
−1
= 0 ⇒ 2 − 4x 2 = 0 ⇒ x = 2 3 = 4 3
dx
d
Students should verify using their
GDC that this does indeed
d2 A
= −128 sin2θ
dθ 2
2
x
− 4x
give rise to the maximum profit
( ) = 2.38110...
= 32 sin2θ
dA
= 64 cos 2θ
dθ
dA
π
=0⇒θ =
dθ
4
P ( x ) =R ( x ) − C ( x ) =4 x − 2x 2
b
= 64
=
sin θ cos θ 32 (2 sin θ cos θ )
c
(
h = 6 would not make sense,
so consider h = 2 :
Exercise 13E
b
dV
4π
=
36 − 24h + 3h2
dh
9
⇒ h2 − 8h + 12 = ( h − 6 ) ( h − 2 ) = 0
(and changes direction)
C ' ( x )= 1 −
)
d2V
4π
=
( −24 + 6h)
dh2
9
This is the value of t at which
the ball reaches its maximum height
1
)
(
c
so − 0.408 ms−1
c
Let the angle between the downward
P 4
− 13
So maximum profit is $2381.10
5 a
675
Let |AB |=
| CD |=⇒
|BC |=
| AD |=
x
x
 675 
C ( x ) = 10 x + 4 x + 4 (2 ) 

 x 
5400
= 14 x +
x
© Oxford University Press 2019
4
Worked solutions
b
C '(x) =
14 −
(
5400
=
0
x2
)
∴ ∫ 15x 4 sin 3x 5 dx =
∫ sin u du
(
19.64 (2d.p.)
⇒x =
 5400 
Cmax C=
=

 549.91 (2d.p.)
 14 
So minimum cost is $550
Minimum to be verified by use of GDC
3 Let u= 2 x 2 + 3x + 1 ⇒ du=
∴∫
(2 x
4x + 3
2
dx =∫ u −2 du =
−
)
2
+ 3x + 1
1
5∫ sin x dx =
− 5 cos x + C
∫ 5 sin x dx =
2
∫ ( 4 cos x − 2 sin x ) dx
∴ ∫ (2 x + 7 ) e x
2
+7 x
2
+7 x
u
dx =
eu + C
∫ e du =
+C
5 Let =
u x 4 − 3x 2
( 4x
⇒ du =
= 4 sin x + 2 cos x + C
1
sin (7 x ) + C
7
3
( 7 x ) dx
∫ cos=
4
6
(2x ) dx 2 sin (2x ) + C
∫ 6 cos=
= 2∫ u3 d=
u
u4
+C
2
1
∫sin (5x + 3) dx =
− cos (5x + 3) + C
6
∫ (x
1 4
x − 3x 2
2
(
)
6 Let u =
4
4
− 3x 2
)
3
∴∫
dx + ∫ sin (2 x ) dx
7
8
∫ 2π sin (2π x ) dx
x ⇒ du =
=
du
∫ u=
2
(
)
3
( 4x − 2) dx =
2 (2 x − 1) dx
)
8 Let u =
cos x ⇒ du =
− sin x dx
)
+ 4 dx
1 3
u +C
3
1
=
5x 3 + 4 x
3
dx
+C
(
(
2
1
2 x
∴∫ 1
1
∫cos
=
=
udu
sin u + C
2
2
1
=
sin 2 x 2 − 2 x + C
2
)
1 Let u = 5x 3 + 4 x ⇒ du = 15x 2 + 4 dx
2
dx
(2x − 1) cos (2x 2 − 2x ) dx
Exercise 13G
) (15x
3
dx =
2 ∫ e u du =
2eu + C
⇒ du =
2π
=
−
cos (2π x ) + C
2π
=
− cos (2π x ) + C
(
x
)
)
7 Let=
u 2x 2 − 2x
x
x
  dx 2 sin   + C
∫ cos=
2
 
2
∴ ∫ 5x 3 + 4 x
x
= 2e
x4 1
=
− cos (2 x ) + C
4 2
x
e
(
+C
5
+ sin (2 x ) dx
1
8 x 3 − 12 x dx
2
)
− 6 x dx =
)( x
=
5
3
(
∴ ∫ 8 x 3 − 12 x
= 3 sin (2x ) + C
∫x
1
+C
u
1
=
−
+C
2 x 2 + 3x + 1
= ex
= 4∫ cos x dx − 2∫ sin x dx
=
( 4 x + 3 ) dx
4 Let u = x 2 + 7 x ⇒ du = (2 x + 7 ) dx
Exercise 13F
3
)
=
− cos u + C =
− cos 3x 5 + C
+C
u5
∴ ∫ sin x cos4 x dx =
− ∫ u 4 du =
−
+C
5
1
=
− cos5 x + C
5
9 Let u = ln x ⇒ du =
∴∫
1
dx
x
sin (ln x )
dx =
− cos u + C
∫ sin u du =
x
=
− cos (ln x ) + C
2 Let u = 3x 5 ⇒ du = 15x 4 dx
© Oxford University Press 2019
5
Worked solutions
3
5 Let u = x 3 + 1 ⇒ du = 3x 2 dx
3
10 Let u = e x + 5 ⇒ du = 3x 2e x dx
∴ ∫ x 2e x
3
3
e x + 5 dx =
(
2 x3
=
e +5
9
11 f ( x ) = ∫ e
)
3
2
x =1 ⇒ u = 2
2
+C
3
x +1
1
sin x
cos x dx
9
3x 2
∴∫
u
u
sin x
∫ e du = e + C = e + C
2
(
9
π
x = ln π3 ⇒ u=
∴ f ( x ) = esin x + 11
3
π
x = ln π4 ⇒ u=
4x
12 f ( x ) = ∫
dx
2
2 x + e2
∴
ln π3
∫
4
π
x
3
− cos u 
( ) dx =
∫ sin u du =
e sin e
ln π4
Let u = 2 x 2 + e2 ⇒ du = 4 x dx
x
π
3
π
4
π
4
1
1 
2 −1
=
− −
=
2
2
2
∴ f ( x ) =∫ u −1 du =ln u + C
)
= ln 2 x 2 + e2 + C
7 a As ex ≠ 0, consider sin x = 0. x = 0 or
π∴k =
π
f (0) = 2 + C = 5 ⇒ C = 3
(
)
6 Let u = e x ⇒ du = e x dx
f (π ) =1 + C =12 ⇒ C =11
(
−1
dx =
∫ u 2 du

= 2 u=
2 3− 2
 2
1
2
u= sin x ⇒ du= cos x dx
∴ f (x) =
x =2 ⇒ u =9
1
1
2 3
u 2 du = u 2 + C
∫
3
9
)
∴ f ( x=
) ln 2x 2 + e2 + 3
b Using GDC:
π
∫e
x
sin=
x dx
0
Exercise 13H
π
4
sin x=
 04
0
8 Limits of integration are x = 0,
1.27531 and x = 4.06401
2
2
Using GDC:
∫
∫ 2 sin x dx =
π
3
5π
6
+
−2 cos x  π
3
1 a
x = 0 ⇒ u = 0, x = 2 ⇒ u = 6
2
(
∴∫ x + x
0
6
2
)
3
b
3
0
(2x + 1) dx
x =
π
3
π
6
⇒u=
π
c
1
2
2 a
3
2
b
1
2
3
∴ ∫ sin x cos3 x dx =
− ∫ u 3 du
π
3
2
6
1
( −x
3
)
+ 5x 2 − 4 x − sin x dx
v ( t ) =s ' ( t ) =−t 2 + 8t − 12
9
∫ v ( t ) dt =
9
∫ ( −t
2
)
+ 8t − 12 dt
0
9
 u4 
324
=

 4 0
⇒u=
1.27531
 1 3

2
=
−27 m
 − 3 t + 4t − 12t  =

0
4 Let u =
cos x ⇒ du =
− sin x dx
x =
4.06401
0
6
u du
∫=
)
sin x − − x 3 + 5x 2 − 4 x dx
Exercise 13I
(2x + 1) dx
3 Let u = x 2 + x ⇒ du =
∫
(
x=
≈ 11.4

3 1
=
−2  −
−  =3 + 1
 2
2 

=
1.27531
0
5π
6
2
1
=
2
π
dx
∫ cos x =
1
eπ + 1
≈ 12.1
2
c
9
9
0
0
∫ | v (t ) |dt = ∫ | −t
2
+ 8t − 12 |dt = 48.3 m
v ( t=
) s ' (t=) 2t − 6
6
6
0
0
6
2
∫ v (t ) dt =∫ (2t − 6 ) dt =t − 6t 0 =0 m
6
6
0
0
∫ | v (t ) |dt = ∫ | 2t − 6 |dt = 18 m
 u4  2
1 9
1  1
=
−  =
−

=
4  16 16  8
 4  23
© Oxford University Press 2019
6
Worked solutions
3 a
b
v ( 0 ) = −2 ms–1
v=
(t ) s ' =
(t ) 3 (t − 1)
2
3
3
) dt ∫ 3 (t − 1)
∫ v (t=
0
2
π
b v ( t ) =−2 cos t =0 ⇒ t =
2
dt
0
3
= ( t − 1)  = 8 − ( −1) = 9 m

0
c
3
3
0
0
∫ | v (t ) |dt =∫ | 3 (t − 1)
2
4 a Displacement =
Distance =
∫
8
0
∫
14
2
5 a
∫
14
0
v(t ) dt = 22 m
∫
14
2
v(t ) dt = 6 m
∫
14
0
b
v(t ) dt = 10 m
(
s (t )
c=
v(t ) dt = 34 m
is − 3 ms−2
7
∫ | v (t ) |dt = 16.5 m
= −𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡 + 𝐶𝐶
𝑠𝑠(0) = 3𝑒𝑒 ⇒ −𝑒𝑒 + 𝐶𝐶 = 3𝑒𝑒 ⇒ 𝐶𝐶 = 4𝑒𝑒
∴ 𝑠𝑠(𝑡𝑡) = 4𝑒𝑒 − 𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡
=
s (t )
dt
= −5 + C = 0 → C = 5
s(4) = 2sin4 – 5cos4 + 5 = 6.75 m (3 s.f.)
–1
∫ v ( t ) dt =
1 3
t − 16t + C
3
b
4
14.0 m (3s.f.)
∫ | 5 sin t + 2 cos t |dt =
0
5 a i
1 3
t − 16t + 10
3
ii
6
b
2
c i
2
32 m
∫ | t − 16 |dt =
1 a =
v ( t ) s=
' ( t ) e ( cos t − sin t )
v ( t ) = s ' ( t ) = −2 cos t
v ' (1.3) < 0 so slowing down
−6.92 m (use of GDC)
At 6 seconds, the particle
is 6.92m to the left of its
inital position
t
a ( t ) = v ' ( t ) = −2et sin t
−2.37 ms−2 (use of GDC)
3.54s and 5.01s (use of GDC)
Exercise 13J
2 a
(t ) dt ∫ 5 sin t + 2 cos t
∫ v=
= 2 sin t − 5 cos t + C
ii
b
sin t dt
s(0) = 2 sin0 – 5cos0 + C = 0
s ( 0=
) C= 10
c
cos t
∴ 𝑠𝑠(𝑡𝑡) = − � 𝑒𝑒 𝑢𝑢 𝑑𝑑𝑑𝑑 = −𝑒𝑒 𝑢𝑢 + 𝐶𝐶
6 a =
a ( t ) v=
' ( t ) 2t
∴ s (t ) =
v ( t ) dt ∫ e
∫=
4 a Assume initial displacement is 0.
t ∈ ( 0,3) U (5,7 )
s (t ) =
)
Let 𝑢𝑢 = 𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡 ⇒ 𝑑𝑑𝑑𝑑 = − 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 d𝑡𝑡
0
b
a (t ) = v ' (t )
= ecos t cos t − sin2 t
The acceleration is the gradient of
a (2 ) = 4 ms
v ( t ) < 0 whenever sin t < 0
= ecos t cos t + ( − sin t ) ecos t sin t
=
t 3 is − 3 so the acceleration
c
ii
∴ t ∈ (π ,2π )
this graph. The gradient at
b
3 a i v ( t ) 0=
whenever sin t 0
=
∴=
t 0, =
t π, =
t 2π
v(t ) dt = 30 m
c Displacement =
Distance =
0
|dt =9 m
v(t ) dt = 22 m
b Displacement =
Distance =
∫
8
π 
s  = 6 −2 = 4m
2
c
3
d 18.6 m (use of GDC)
6 a
b
−12.8 ms−2 (use of GDC)
t = 0.696, 5.59 (use of GDC)
© Oxford University Press 2019
7
Worked solutions
c
13.2 m (use of GDC)
d
24.8 m (use of GDC)
(
1
∫ (1025t
0
2 33.4 +
∫
3 3800 +
4
∫
10
0
8
0
∫
2
−t
3
5.2te
20
0
(
1152 spectators
) dt =
−0.05t 3 + 2.3
t 

−150 1 −
 dt ≈ 1175 gallons
80 

t
du
∫u=
4
=
e
(
c
h ' ( t ) = 4 cos x sin3 x
d
f ( x ) = ( cos x ) 2
= 2e x
− 12
sin x
=
−
2 cos x
=
4
+5 x
dx =
2∫ eu du =
2eu + C
+C
1
dx
x
cos (ln x )
(
( )
et ( − sin t ) − cos t et
e
2e4 x
1
1
u −1 du =+
dx =
ln u C
e4 x + 5
2∫
2
(
)
3π
4
 − cos x 
∫ sin x dx =
3 a
2t
π
2
sin2 x + 2e
)
1
ln e4 x + 5 + C
2
=
( )
∫ (3x
2
x2 +5 x
⇒=
du 4e4 x =
dx 2 2e4 x dx
et ( sin t + cos t )
sin t + cos t
=
−
=
−
2
t
et
e
2 a
1
( 4x + 10) dx
2
dx =
sin u + C
∫ cos u du =
x
= sin (ln x ) + C
∴∫
2x
∫ (3x
2
4
u du
∫=
=
)(
− 2 x 3 − 2x
−1
3 5
x + sin x + C
5
)
x =2 ⇒ u =4
x =−1 ⇒ u =1
( sin2x + cos 2x )
)
1
2
=
=
2
2
(
2
+ cos x dx =3∫ x 4 dx + ∫ cos x dx
3π
4
π
2
b Let u = x 3 − 2 x ⇒ du = 3x 2 − 2 dx
cos 2 x
3
1
)
3
dx
1 4 4 255
u 
=
4  1
4
π
c
b
1
− cos 4 x + C
∫ sin 4x dx =
4
c
) dx
∫ cos (2x + 3=
sin u + C
h Let =
u e4 x + 5
dy
1
= cos (ln x )
x
dx
= 2e
∴∫
sin x
x
( cos x ) (ln x ) +
2x
du
∫ cos u=
g Let u = ln x ⇒ du =
1
( cos x ) (ln x ) + ( sin x )  x 
 
h f ' ( x ) 2e
=
cos x 3 =
dx
∫ ( 4x + 10) e
1
f '(x) =
( − sin x ) ( cos x )
2
2x
2
Let =
u x 2 + 5x
1
s ' (t ) =
+C
⇒ du = ( 2 x + 5 ) dx =
dy
=
−3 sin (3x − 4 )
dx
g
5
3
1 a =
f ' ( x ) 3 cos x − 4 sin x
f
)
( )
sin ( x ) + C
=
Chapter Review
b
4
Let u = x 3 ⇒ du = 3x 2 dx
∫ 3x
f
=
)
)
+ 5 dx
1 5
u +C
5
1
2 x 3 + 5x
5
=
) dt =
206 cubic feet
20.4e18 dt ≈ 273 billions of barrels
dy
e
=
dx
)(
2
∴ ∫ 6 x 2 + 5 2 x 3 + 5 x dx
Exercise 13K
1.5
(6 x
d Let u = 2 x 3 + 5x ⇒ du =
6
 x + sin x 
∫ (1 + cos x ) dx =
0
=
1
sin (2 x + 3) + C
2
π
6
0
π 1
=
+
6 2
π +3
6
d Let u= 4 x 2 + 1 ⇒ du= 8 x dx
© Oxford University Press 2019
8
Worked solutions
x =1 ⇒ u = 5
x = 0 ⇒ u =1
1
∴ ∫ 8 xe 4 x
0
4 a
2
+1
1
π
 5π

y −
−1
=
 4 x − 
4
2
4



5
5
u
eu  =
dx =
e5 − e
∫ e du =
1
1
The surface area of the box is
b the normal to the graph is
vertical when the tangent is
horizontal
f ′(x) =
0 ⇒ sec2 x =
5
x 2 + 4 xh =
432
432 − x 2
⇒h=
4x
2
b=
V x=
h
c
(
2
)
x 432 − x
1
= 108 x − x 3
4
4
∴ Maximum volume occurs when
and accordingly h 6
=
x 12
=
5 a
A1A1
b Use GDC to find the maximum M1
A1
t = 2.25
c Find intersection of graphs
M1
A1
10 a
b
x2
∫ ( x − sin x ) dx =2
π
+ cos x + C
M1A1A1
x
2

+ cos x 

∫ ( x − sin x ) dx =
2

0
π
0
π2
=− 2
2
( e )′ cos x + e ( cos x )′
x
x
= e x cos x − e x sin x
b
M1A1A1
a(0) = 1
A1
2π
a′(0) = 1
A1
0
y= x + 1
M1A1
∫ | v (t ) |dt = 14.1 m
4.5 +
10
∫ 7t e
2
−1.2t
dt =
12.6 cm (3 s.f.)
0
7=
a g′ ( x ) 2 sin x cos x − 5
b
5 − 1 = 2 M1
d ( t ) = sin2t − sin ( t − 0.24 )
da
11
a
=
dx
t ∈ ( 0.905, 5.38 )
6
1
⇒ tan x =
5
M1A1A1
Using a GDC, plot a ( t ) to
see where it is positive
d
9 a
)
b v ( t ) < 0 when sin t > 0 ⇒ t ∈ ( 0, π )
c
cos2 x =
A1
t = 1.13
a (t ) =
v ' (t ) =
−3ecos t cos t + 3ecos t sin2 t
(
1
5
⇒ x = arccos
5
5
The coordinates of A are


5
5
,5 arccos
− 2 .
 arccos

5
5


v ( t ) = s ' ( t ) = −3ecos t sin t
= 3ecos t sin2 t − cos t
cos2 x =
R1
M1

5
5
f  arccos
=
− 2 A1
 5 arccos

5
5


dV
3
= 108 − x 2
dx
4
dV
= 0 ⇒ x 2 = 144 ⇒ x = 12
dx
d2V
3
= − x so the second
dx 2
2
derivative is negative at x = 12
A1
12 a A1 for shape, A1 for domain, A1 for
scale on axes
A1A1
g′ ( x ) =sin2 x − 5 ≤ 1 − 5 =−4 < 0
M1A1R1
Therefore g is decreasing on all its
domain.
AG
8 a
f ′ ( x )= 5 − sec2 x
M1A1
1
π 
π 
f′  =
5 − sec2   =
4
2
4
4
A1
 π  5π
f=
−1

4
4
A1
b i
© Oxford University Press 2019
Minimum points: (0.785,0.909)
and (2.36,0)
A1A1
Maximum points: (0.304,1) and
(1.27,1)
A1A1
9
Worked solutions
ii
A1
0≤ x ≤1
iii sin1
2.356...
∫
c i
A1
sin (1 + sin2 x ) dx
16 a
A1A1
0 ≤ x ≤ 2π .
π
6
)
A1
−0.369 ≤ y ≤ 1.76
A1A1
iii=
f ′ ( x ) 2 cos 2 x − sin x
A1A1
b
b
∫ ( sin2x + cos x )
−
14 a
π x 
= sin  +  = f (π + x )
2 2
dx =
2.25 M1A1A1
6
s ( 0 ) = 2 mm
ii
′ 15 cos 3t + 2t
v= s=
M1A1
a=
v′ =
−45 sin3t + 2
M1A1
R1A1
a < 0=
⇒ t 0.548,
=
t 2.74
ii
A1
A1
c v = 0 ⇒ t = 0.548, t = 1.50, t = 2.74
M1
15 a i
M1R1
Therefore a = f (π − x ) = f (π + x ) = b
π
b i
π x
f (π − x=
) sin  2 − 2 



 π x 
= sin  π −  −  
 2 2 

π
2
AG
A1
–0.524 (or −
ii
M1
Therefore f ( x ) ≥ 0 for all
ii Using GDC to evaluate the definite
integral
M1
13 a i
x
≤π
2
In the 1st and 2nd quadrants sine is
positive
R1
0
area = 1.76
0 ≤ x ≤ 2π ⇒ 0 ≤
f (0) = 5
A1
f (π ) = 5
A1
c
π − x 
A ( x ) = 2 x sin 

 2 
M1A1
Find maximum point (1.72,2.24)
M1A1
A(1.42,0.652) and B(4.86,0.652)
A1A1
d =
p 2 AB + 2 × 0.6520...
= 8.19 M1A1AG
b A1 for coordinates of A, A1 for
coordinates of B,
A1 for zeros, A1 for shape and
domain
c
b
∫ f ( x ) dx = 2.11 (or 2.07 using 3 s.f. for
a
a and b)
A2
d The graph crosses the x -axis
between a and b
e Either
b
∫ f (x)
dx = 7.39
R1
M1A1
a
Or
1.961...
∫
1.017...
f ( x ) dx +
2.588...
∫
f ( x ) dx =
7.39
1.9601...
M1A1
© Oxford University Press 2019
10
Worked solutions
14
Valid comparisons and informed
decisions: probability distributions
2 a
Skills check
3×3 +5× 4 + 7×5 + 9×6 + 6×7 + 2×8
3+5+7+9+6+2
1 a
=
176
= 5.5
32
3 × 10 + 10 × 12 + 15 × 15 + 9 × 17 + 2 × 20
3 + 10 + 15 + 9 + 2
b
=
568
≈ 14.6
39
2 a 5.5
3 a
c
=
55
= 1.719
32
c
x=
217
= 8.68
25
Exercise 14A
b
8
  = 56
5
b
x=
149
= 2.98
50
b
P(S > 2)
=
P(S =
3) + P(S =
6) + P(S =
10)
1 1 1 1
= + + =
6 6 6 2
4 a The sum of the probabilities in a
probability distribution should be 1, so
1 1
1= + +c+c
3 3
2
1
=
+ 2c
3
2
1− =
2c
3
1
= 2c
3
1
÷2 =
c
3
1
c =
6
b
c
12 9
7
+
=
36 36 12
3 a
9
3
6
  (0.3) (0.7) = 0.267
6
x=
b
P(T > 4) = P(T = 5) + P(T = 6)
b 14.6
6
  = 15
2
4 a
1 a
b
P(1 < X < 4) = P( X = 2) + P( X = 3)
=
1 1 1
+ =
3 6 2
5 The sum of the probabilities must be equal
to 1, so
d
1 = 13 c + 23 c + 33 c
1= c + 8c + 27c
1 = 36c
1
c =
36
© Oxford University Press 2019
Worked solutions
1
Worked solutions
6 As the sum of the probabilities must be 1,
we have
2
2
=
1 2k + 4k + 6k + k
b
2
= b=
+ b2 2b2
P(sum
= 9)
= P(4 and 5)
= b2
=
1 3k + 10k 2
5) b2
P(sum
= 10)
= P(5 and=
0 = 10k 2 + 3k − 1
0=
(2k + 1)(5k − 1)
So the possible values for
k
are k = −
P(sum > 7) = P(sum = 8) + P(sum = 9)
10)
+ P(sum =
1
2
1
but as a probability must be
5
1
is the only possible
non-negative, k =
5
P(sum
=
> 7) 2b2 =
+ b2 + b2 4b2
25
=
144
and k =
9 a
7 Using the fact that the sum of the
probabilities must be equal to 1,
1
2
1 2 1 1
⋅ + ⋅
3 3 3 6
5
=
18
=
3
1
1
1
1
=
1 k  +k  +k  +k 
3
3
3
3
k k
k
1= k + + +
3 9 27
40
k
1=
27
27
k =
40
b
8 a As the sum of the probabilities must
equal 1,
P( X ≥ 2)
2) + P( X =
3) + P( X =
4)
P( X =
=
5)
+ P( X =
= a+b+b+b
Exercise 14B
1
a + b + b + b= 3(a + a)
a + 3b =
6a
5
a
3
5 
1 = 3a + 3 ×  a 
3 
1 3a + 5a
=
1 = 8a
1
a=
8
5 1
b=
×
3 8
5
b=
24
1
1
1
1
× 1 + × 4 + × 9 + × 16
6
6
6
6
1
1
91
+ × 25 + × 36 =
6
6
6
≈ 15.2
2
E( X )
= 1×
3=
b 6a − a
3b = 5a
b=
E( X )
=
P( X < 2) =P( X =0) + P( X =1) =a + a
1= a+a+a+b+b+b
=
1 3a + 3b
P( X ≥ 2)
= 3P( X < 2)
P(C = 3)
= P=
( A 1 and
=
B 2)
+ P=
( A 2 and=
B 1)
solution
0
= 8)
= P(3 and 5) + P(4 and 4)
P(sum
=
3
16
3
1
1
1
1
1
1
+ 2 × + 3 × + 5 × + 8 × + 13 ×
6
6
6
6
6
6
E( X )
1
2
3
4
5
+2×
+ 3×
+ 4×
+5×
36
36
36
36
36
6
7
8
+6×
+7×
+8×
36
36
36
17
=
3
= 1×
4 a Using the fact that the sum of the
probabilities must equal 1,
1 =k + 2k + 3k + 4k + 5k + 4k
+ 3k + 2k + k
1 = 25k
1
k =
25
© Oxford University Press 2019
2
Worked solutions
b
E( X )
P(R = 3)
= P(blue then blue then red)
8 8 2
16
=
×
×
=
10 10 10 125
1
2
3
4
5
+2×
+ 3×
+ 4×
+5×
25
25
25
25
25
4
3
2
1
+6×
+7×
+8×
+9×
25
25
25
25
=5
= 1×
5 a
b
c For the first red to be drawn on the r th
try, there are r − 1 blues picked first so
P(=
R r=
) P(r blues then a red)
0.2 ≤ k ≤ 1
E( X )
n
2
 8 
=
 ×
10
 10 
= 1 × 0.2 + 2 × (1 − k )
+ 3 × (1 − (1 − k ) − 0.2)
= 1.6 + k
d 1
6 a If you pick the first red ball on the r th
try, that means you have picked r − 1
blue balls
P(R=1) =
1
5
P(R=2) =
8
45
P(R=3) =
7
45
8 a As the sum of the probabilities must
x P=
(Z 0)
equal 1, let=
1 =x + 0.2 + 0.05 + 0.001 + 0.0001
1= x + 0.2511
1 − 0.2511 =
x
x = 0.7489
b
= 0 × 0.7489 + 2 × 0.2 + 20 × 0.05
+ 200 × 0.001 + 1000 × 0.0001
= 1.7
so the expected winnings on a ticket
are $1.70
2
P(R=4) =
15
c You expect to lose $0.30 per ticket
1
P(R=5) =
9
b
P(R=6) =
4
45
P(R=7) =
1
15
P(R=8) =
2
45
Exercise 14C
1
 1
X ~ B  4, 
 2
a
b
E(R)
b
c
2
8
2
8
8
2
+
×
+
×
×
10 10 10 10 10 10
61
=
125
=
P( X < 1)= P( X= 0)
4
P( X ≤ 1) = P( X = 0) + P( X = 1)
1
1
5
= +
=
4 16 16
d
P(=
R 2)
= P(blue then red)
2
P(R > 3) =
P(R =
1) + P(R =
2) + P(R =
3)
1 1
4
1
× =
=
2 8 16 4
0
c 1
8
2
4
×
=
10 10 25
3
 4  1   1 
1
1
=      = 1×1×
=
16 16
0  2   2 
1
8
7
2
1
+2×
+ 3×
+ 4×
+5×
5
45
45
15
9
4
1
2
1
+6×
+7×
+8×
+9×
45
15
45
45
11
=
3
=
1
 4  1   1 
P( X
= 1=
)     
1  2   2 
= 4×
= 1×
7 a
E (Z )
P( X ≥ 1) =1 − P( X < 1) =1 −
1
15
=
16 16
 1
X ~ B  6, 
 3
a
2
4
0
6
6  1   2 
= 2)
=      = 0.329
P( X
2  3   3 
1
5
6  1   2 
6  1   2 
b =
P( X < 2)       +      
0  3   3 
1  3   3 
= 0.351
© Oxford University Press 2019
3
Worked solutions
c
b If 13 are not faulty then 3 are faulty,
16 
)   (0.01)3(0.99)13
P( X
= 3=
3
P( X ≤ 2)= P( X < 2) + P( X= 2)
= 0.329 + 0.351 = 0.680
d
P( X ≥ 2) =1 − P( X < 2)
= 0.000491
1 − 0.351 =
0.649
=
3
c
 2
X ~ B  8, 
 7
1 − P( X =
0) − P( X =
1)
=
5
a
16 
= 1 −   (0.01)0 (0.99)16
0
16 
−   (0.01)1(0.99)15
1
3
8  2   5 
= 5=
P( X
)     
5  7   7 
= 0.0389 (3 s.f.)
b
P( X < 5)
= 0.0109
0
8
1
7
8  2   5 
8  2   5 
=      +     
7
7
0
    
1  7   7 
2
6
4
4
3
4
8  2   5 
+     
 4  7   7 
= 0.952 (3 s.f.)
b
5
P( X ≥ 1) =1 − P( X =0)
0
8
8  2   5 
= 1 −       = 0.932
0  7   7 
 1
R ~ B  4,  , then
 4
8
5
3
P( X
= 5)
=   ( 0.55) ( 0.45)= 0.257
5
 
8
8
=
P( X ≤ 3)   (0.55)0 (0.45)8 +   (0.55)1(0.45)7
0
 
1
8
8
+   (0.55)2 (0.45)6 +   (0.55)3(0.45)5
2
3
= 0.260
3
= 0.224
b
7
6
5
P( X
= 1=
)   (0.15)1(0.85)=
0.399
1
 
X ~ B(15, 0.05)
a i
16 
P( X
= 0)
=   (0.01)0 (0.99)16 = 0.851
0
15 
P( X
= 3)
=   (0.05)3(0.95)12
3
= 0.0307
ii
X ~ B(16,0.01)
a
P( X > 1) =
1 − P( X =
0) − P( X =
1)
6
6
=
1 −   (0.15)0 (0.85)6 −   (0.15)1(0.85)5
0
 
1
X ~ B ( 8, 0.55)
b If he misses at least 5 times then he
hits at most 3 times,
X ~ B(6, 0.15)
a
So the most likely number of times the red
face shows is 1
a
X ~ B(5, 0.4) ,
P( X ≤ 3)
=
P( X =
0) + P( X =
1) + P( X =
2)
+ P( X =
3)
= 0.913
6
2
P( X ≥ 3) =1 − P( X < 3)
=
1 − P( X =
0) − P( X =−
1) P( X =
2)
5
5
=   (0.4)0 (0.6)5 +   (0.4)1(0.6)4
0
 
1
5
5
+   (0.4)2 (0.6)3 +   (0.4)3(0.6)2
2
 
3
Exercise 14D
1
10 
5
P( X
= 5)
=   (0.25)5(0.75)=
0.0584
5
 
10 
10 
= 1 −   (0.25)0 (0.75)10 +   (0.25)1(0.75)9
0
1
10 
+   (0.25)2 (0.75)8 = 0.474
2
P( X > 5) =1 − P( X =5) − P( X < 5)
1 − 0.038857.. − 0.9524..
=
= 0.00870 (3 s.f.)
d
X ~ B(10, 0.25)
a
5
8  2   5 
8  2   5 
+      +     
2  7   7 
3  7   7 
c
P( X ≥ 2) =1 − P( X < 2)
15 
P( X
= 0)
=   (0.05)0 (0.95)15
0
= 0.463
iii P( X ≥ 2) =1 − P( X < 2)
=
1 − P( X =
0) − P( X =
1)
© Oxford University Press 2019
4
Worked solutions
15 
= 1 −   (0.05)0 (0.95)15
0
15 
−   (0.05)1(0.95)14
1
0.5n < 1 − 0.99
n log0.5 < log0.01
log 0.01
n>
log0.5
7
n > 6.644 ⇒ n =
= 0.171
b i
(P( X = 0))2 = 0.215
ii
Exercise 14F
(P( X ≥ 2))2 =
0.0292
0.158
iii 2 × P( X = 0) × P( X ≥ 2) =
1 a
1
1

20
X ~ B  40,  , E( X ) = np = 40 × =
2
2


b
1 20
1

X ~ B  40,  , E( X ) = np = 40 × =
6
3
6

c
1
1

10
X ~ B  40,  , E( X ) = np = 40 × =
4
4

Exercise 14E
1
0.0256= P( X < 1)= P( X= 0)
 n
0
(0.4)n 1=
=   (0.6)
=
× 1 × (0.4)n 0.4n
0
log 0.0256 = n log0.4
log 0.0256
n=
log0.4
n=4
2
2
E( X=
) np
= 0.4=
n 10 ⇒ n =
3 a
X ~ B(n, 0.01) ,
 n
0
n
0.5 < P( X =
0) =
  (0.01) (0.99)
0
 
E( X
=
) np
= 15 × 0.25 = 3.75
c
P( X ≥ 10)
=P( X =10) + P( X =11) + P( X =12)
+ P( X =13) + P( X =14) + P( X =15)
 15 
15 
=   (0.25)10 (0.75)5 +   (0.25)11(0.75)4
10 
 11 
 15 
15 
+   (0.25)12 (0.75)3 +   (0.25)13(0.75)2
12 
 13 
log0.5 < n log0.99
log0.5
n<
log0.99
68
n < 68.968 ⇒ n =
 15 
15 
+   (0.25)14 (0.75)1 +   (0.25)15(0.75)0
1
4
 
 15 
= 0.000 795
3
 n
0.25 > P( X < 1) = P( X = 0) =   (0.2)0 (0.8)n
0
4 a
log0.25 > n log0.8
b
n>
X ~ B(n, 0.3) ,
0.95 < P( X ≥
=
1) 1 − P=
( X 0)
 n
= 1 −   (0.3)0 (0.7)n= 1 − 0.7n
0
0.7n < 1 − 0.95
n log0.7 < log0.05
n>
log 0.05
log0.7
n > 8.399 ⇒ n =
9
5
X ~ B(n, 0.5) ,
0.99 ≤ 𝑃𝑃(𝑋𝑋 ≥ 1) = 1 − 𝑃𝑃(𝑋𝑋 = 0)
 n
= 1 −   (0.5)0 (0.5)n= 1 − 0.5n
0
P(girl) =
0 × 13 + 1 × 34 + 2 × 40 + 3 × 13
300
153
= = 0.51
300
= 1 × 1 × (0.8)n = 0.8n
4
X ~ B(15,0.25)
b
= 1 × 1 × (0.99)n = 0.99n
log0.25
log0.8
n > 6.213 ⇒ n =
7
10
= 25
0.4
300 × 0.51 × 0.51 × 0.49 =
38.2
Exercise 14G
1
E( X=
) 12
= np and Var( X )= 3= np(1 − p)
Solving these simultaneously gives
12(1 − p) = 3
3
= 1− p
12
3
p= 1 −
12
3
p=
4
3
n× =
12
4
4
n = 12 ×
3
n = 16
© Oxford University Press 2019
5
Worked solutions
2 a
b
c
X ~ B(20,0.2)
= P(−0.4 < Z < 0.4) = 0.311
E( X ) = np = 20 × 0.2 =
4 and
Var( X )= np(1 − p)= 20 × 0.2 × 0.8 =
3.2
b
= P(Z > 1.24) + P(Z < −1.24)
= 0.215
Exercise 14I
P( X ≥ 10) =−
1 P( X < 10)
1
=
1 − P( X =
0) − P( X =−
1) P( X =
2)
− P( X =
3) − P( X =
4) − P( X =
5)
− P( X =
6) − P( X =
7) − P( X =
8)
− P( X =
9)
X ~ N(14,52 )
a
 20 
 20 
= 1 −   (0.2)0 (0.8)20 −   (0.2)1(0.8)19
 0 
 1 
 20 
 20 
2
18
−   (0.2) (0.8) −   (0.2)3(0.8)17
 2 
 3 
b
 20 
 20 
−   (0.2)4 (0.8)16 −   (0.2)5(0.8)15
4
 
 5 
 20 
 20 
−   (0.2)6 (0.8)14 −   (0.2)7 (0.8)13
 6 
 7 
 20 
 20 
8
12
−   (0.2) (0.8) −   (0.2)9 (0.8)11
 9 
 8 
c
12 − 14 
 9 − 14
P(9 ≤ =
X < 12) P 
≤Z<

5
5


= P(−1 ≤ Z < −0.4) =0.186
0.5
d As the mean is 14, P( X < 14) =
2
X ~ N(48,81)
a
Var(
=
X ) np(1 −=
p) 12 × p(1 −=
p) 1.92 , so

52 − 48 
P( X < 52) = P  Z <

81 

=
P(Z < 0.4444) =
0.672
1.92
p(1 − p) =
12
0 = p2 − p + 0.16
0=
( p − 0.8)( p − 0.2)
b

42 − 48 
P( X ≥ =
42) P  Z >

81 

= P(Z > −0.6667)
= 0.748
Which gives us that p = 0.2 or p = 0.8 .
c
Exercise 14H
 37 − 48
47 − 48 
P(37=
< X < 47) P 
<Z<

81 
 81
= P(−1.2222 < Z < −0.1111)= 0.345
P(−2 < Z < −1) + P(1 < Z < 2)
3
= 0.1359 + 0.1359 = 0.272
X ~ N(3.15, 0.022 )
a
P(−1.5 < z < −0.5) + P(0.5 < Z < 1.5)
= 0.2417 + 0.2417 = 0.483
2 a
9 − 14 

P( X > 9)= P  Z >

5 

= P(Z > −=
1) 0.841
3 We know that
b
16 − 14 

P( X < 16) = P  Z <

5 

= P(Z < 0.4) = 0.655
= 0.00259
1 a
P(| Z |> 1.24)
3.2 − 3.15 

P( X < 3.2) = P  Z <

0.02 

= P(Z < 2.5) = 0.994
P(Z > 1) =
0.159
b
P( X ≥ 3.11)
b
P(Z > 2.4) =
0.0082
c
P(Z < −1) =0.159
d
P(Z < −1.75) =0.0401
3.11 − 3.15 

= P Z ≥

0.02


= P(Z > −=
2) 0.977
3 a
P(Z < 0.65) =
0.7422
c
b
P(Z > 0.72) =
0.2358
c
P(Z ≥ 1.8) =
0.0359
3.15 − 3.15 
 3.1 − 3.15
= P
<Z<

0.02
 0.02

d
P(Z ≤ −0.28) =0.3897
= P(−2.5 < Z < 0) = 0.494
4 a
P(0.2 < Z < 1.2) =
0.3057
b
P(−2 < Z ≤ 0.3) = 0.5952
c
P(−1.3 ≤ X ≤ −0.3) =
0.2853
5 a
P(| Z |< 0.4)
P(3.1 < X < 3.15)
Exercise 14J
1
X ~ N(100,202 )
a
130 − 100 

P( X < 130) =
P Z <

20


© Oxford University Press 2019
6
Worked solutions
P(Z < a) − 0.8413 =
0.12
= P(Z < 1.5) = 0.933
b
P(Z < a)= 0.12 + 0.8413
P(Z < a) =
0.9613
∴a =
1.77
P( X > 90)
90 − 100 

= P Z >

20


= P(Z > −0.5)
= 0.691
c
b
= P(Z < 1.6) − P(Z < a)
P(80 < X < 125)
0.9452 − P(Z < a) =
0.787
125 − 100 
 80 − 100
= P
<Z<

20
 20

P(Z < a) =
0.9452 - 0.787
P(Z < a) =
0.1582
∴ a =−1.00
= P(−1 < Z < 1.25) = 0.736
2
c
X ~ N(4, 0.252 ) ,
0.3821 − P(Z < a) =
0.182
P(Z <=
a) 0.3821 − 0.182
0.2001
P(Z < a) =
∴ a =−0.841
= P(−2 < Z < 2) = 0.9545
, now Y ~ B(500,0.9545) and
E(Y
=
) np
= 500 × 0.9545
= 477.25 so one
would expect 477 to be accepted on
average
3 a
20 − 14 

P( X > 20) = P  Z >

4


b
= P(Z > 1.5) = 0.0668
b
10 − 14 

P( X < 10) = P  Z <

4 

5
1
0.1096 = 0.0548 , so we look for a
2
such that
P(Z < a) =1 − 0.0548 =0.9452 ,
∴a =
1.60
= P(Z < −=
1) 0.1587
= 15.87%
4
1
(1 − 0.3) =
0.35 , so we look for a
2
such that P(Z < a) =1 − 0.35 =0.65 ,
∴a =
0.385
X ~ N(14, 42 )
a
P(a < Z < −0.3)
= 0.182
= P(Z < −0.3) − P(Z < a)
4.5 − 4 
 3.5 − 4
<Z <
P(3.5 < X < 4.5) =
P

0.25 
 0.25
3
P(a < Z < 1.6) =
0.787
4 a
P(Z < a=
) 0.95 ∴ =
a 1.64
b
P(Z > a)= 0.2 ∴ a= 0.842
X ~ N(551.3,15) ,
550 − 551.3 

P( X > 550) = P  Z >

15


Exercise 14L
=
P(Z > -0.08667)=0.5345 ≈ 53.5%
1
X ~ N(500,20)
a
b
a − 5.5
= 0.722
0.2
⇒=
a 0.722 × 0.2 + 5.5 ⇒=
a 5.64
gives that
475 − 500 

P( X < 475) = P  Z <

20


2
= P(Z < −1.25)
= 0.106
a − 5.5 

0.235= P( X > a)= P  Z >
 , this
0.2 

M ~ N(420,102 )
a The first quartile equates to
a − 420 

0.25 = P(M < a) = P  Z <
 , so
10 

P(3 packets less than 475 g)
(P( X < 475))3 =
0.10563 =
0.00118
=
a − 420
=
−0.674 ⇒ a =
−0.674 × 10 + 420
10
⇒a=
413
Exercise 14K
1 a
P(Z <=
a) 0.922 ∴=
a 1.42
b
P(Z >=
a) 0.342 ∴
=
a 0.407
c
P(Z >=
a) 0.005 ∴
=
a 2.58
2 a
b The 90th percentile is
a − 420 

=
0.9 P(M < a) = P  Z <
 , giving
10 

a − 420
= 1.282
us
10
⇒=
a 1.282 × 10 + 420 ⇒=
a 433
P(1 < Z < a) =
0.12
= P(Z < a) − P(Z < 1)
3
X ~ N(502,1.62 )
a
P( X < 500)
© Oxford University Press 2019
7
Worked solutions
2
500 − 502 

= P Z <

1.6


= P(Z < −1.25)
= 0.106
b
20.5 − µ 

=
0.9 P( X < 20.5) = P  Z <
 , so
4


20.5 − µ
= 1.282 ⇒ µ= 20.5 − 1.282 × 4
4
⇒ µ =
15.4
P(500 < X < 505)
505 − 502 
 500 − 502
= P
<Z <

1.6
1.6


= P(−1.25 < z < 1.875)
= 0.864
= 86.4%
c
3
0.95 = P(b < X < a) = P(−a < X < a) , so
, so
∴ a' =
1.96 , so
a = 1.6a '+ 502 = 1.6 × 1.96 + 502 =
505.1
41.82 − µ
= −1.90 ⇒ −41.82 = 1.90σ − µ ,
σ
P(520 < X < 570)
570 − 550 
 520 − 550
= P
<Z<

25
25


solving these simultaneously gives
σ = 4.23 and µ = 49.9
4
a − 550 

0.9 P( X < a) = P  Z <
b =

25 

a − 550
= 1.282 ⇒=
a 1.282 × 25 + 550
25
⇒a=
582
89 − µ
= 1.282 ⇒ 89
= 1.282σ + µ and
σ
94 − µ
= 1.645 ⇒ 94
= 1.645σ + µ , solving
σ
X ~ N(55,152 )
a
d − 55 

0=
.95 P( X < d ) = P  Z <
 , giving
15 

d − 55
= 1.645
15
these simultaneously gives σ = 13.8 and
µ = 71.3
5
145 − 136
145 − 136
σ
= 1.175 ⇒=
σ
1.175
⇒σ =
7.66 cm
6
500 − µ
= −2.326
20
⇒ µ= 500 + 2.326 × 20 ⇒ µ= 546.5 g
X ~ N(30, σ 2 ) ,
40 − 30 

0.115
= P( X > 40) = P  Z >
 , so
σ


40 − 30
40 − 30
= 1.2 ⇒ σ =
⇒ σ = 8.33
σ
1.2
X ~ N(µ , 202 ) ,
500 − µ 

0.01 = P( X < 500) = P  Z <
 , so
20 

Exercise 14M
1
X ~ N(136, σ 2 ) ,
145 − 136 

0.12 = P( X > 145) = P  Z >
 , so
σ


⇒=
d 1.645 × 15 + 55 ⇒=
d 79.7
f − 55 

0.1 P( X < f ) = P  Z <
b =
 , giving
15 

f − 55
= −1.282
15
⇒f =
−1.282 × 15 + 55 ⇒ f =
35.8
X ~ N( µ , σ 2 ) ,
89 − µ 

0.90 = P( X < 89) = P  Z <
and
σ 

94 − µ 

0.95 = P( X < 94) = P  Z <
, so
σ 

= P(−1.2 < Z < 0.8) = 0.673
5
58.39 − µ
=2.02 ⇒ 58.39 =2.02σ + µ
σ
and
X ~ N(550,252 )
a
X ~ N( µ , σ 2 ) ,
58.39 − µ 

0.0217 =
P( X > 58.39) =
P Z >

σ


and
41.82 − µ 

0.0287 =
P( X < 41.82) =
P Z <

σ


1
(1 − 0.95) =
0.025 , so we look for a '
2
1 − 0.025 =
0.975 ,
such that P(Z < a ') =
4
X ~ N(µ , 42 ) ,
7
X ~ N(0.85, σ 2 )
a
1.1 − 0.85 

0=
.74 P( X < 1.1) = P  Z >
,
σ


so
1.1 − 0.85
= 0.643
σ
=
⇒σ
© Oxford University Press 2019
1.1 − 0.85
=
⇒ σ 0.389 kg
0.643
8
Worked solutions
b
Chapter review
1 − 0.85 

P( X > 1) = P  Z >

0.389 

= 0.3497
= 35%
= P(Z > 0.386)
8
1 a 0.3 + 1 + 2 + 0.1 + 0.1 =
1
k
X ~ N(µ ,72 ) ,
0.5 +
68 − µ 

0.025
= P( X > 68) = P  Z >
 , so
7 

68 − µ
= 1.96 ⇒ µ = 68 − 1.96 × 7
7
54.3 cm
⇒ µ =
9
3
= 0.5
k
3
k
=
= 6
0.5
1
2
+0×
6
6
+ 1 × 0.1 + 2 × 0.1
6 1
1
2
=−
− +0+
+
10 6
10 10
3 1
9
5
14
7
=
−
− =
−
−
=
−
=
−
10 6
30 30
30
15
= −2 × 0.3 + −1 ×
3 − 2.9 

0.35 = P( X > 3) = P  Z >
, so
σ 

3 − 2.9
3 − 2.9
= 0.385 ⇒=
σ
σ
0.385
⇒σ =
0.260 m
10 X ~ N(µ , σ 2 )
108 − µ 

0.30 = P( X < 108) = P  Z <

σ


and
154 − µ 

0.20 = P( X > 154) = P  Z >
 , so
σ


2 a 1 = c(6 − 1) + 2c(6 − 2) + 3c(6 − 3)
+ 4c(6 − 4) + 5c(6 − 5)
1 = 5c + 8c + 9c + 8c + 5c
1 = 35c
108 − µ
=
−0.524 ⇒ 108 =
−0.524σ + µ
σ
and
154 − µ
= 0.842 ⇒ 154= 0.842σ + µ ,
σ
solving these simultaneously gives
σ = 33.7 and µ = 125.66
b
3
1
=
k
b E( X )
X ~ N(2.9, σ 2 ) ,
a
k
117 − 125.66 

P( X > 117) = P  Z >

33.7


= P(Z > −0.257)
= 0.601
= 60.1% ,
so this is consistent with the normal
distribution
11 X ~ N(µ , σ 2 ) ,
495 − µ 

0.95 =P( X > 495) =P  Z >
 and
σ


490 − µ 

0.99 =P( X > 490) =P  Z >
 , so
σ


495 − µ
=
−1.645 ⇒ 495 =
−1.645σ + µ
σ
and
490 − µ
=
−2.326 ⇒ 490 =
−2.326σ + µ ,
σ
solving these simultaneously gives
σ = 7.34 and µ = 507.1
c =
1
35
b E( X )
5
8
9
+2×
+3×
35
35
35
8
5
+ 4×
+5×
35
35
5 16 27 32 25
=
+
+
+
+
35 35 35 35 35
105
=
35
=3
=
1×
3 Find the value of x:
1 1 1
+ + +x =
1
4 4 8
5
+x =
1
8
3
x =
8
P(total 6) = P(2, 4) + P(3,3) + P(4,2)
1 3 1 1 3 1
× + × + ×
4 8 8 8 8 4
3
1
3
=
+
+
32 64 32
6
1
6
=
+
+
64 64 64
13
=
64
=
© Oxford University Press 2019
9
Worked solutions
4 a 2,4,6,8,12,16
c i
b 1 ,2 ,1 ,2 ,1 ,1
E( X ) =−5 ×
40 19
21
7
=
−
+
=
−
=
−
27 27
27
9
8 8 8 8 8 8
c
E( X )
Expected loss of $0.78
ii Expected loss of $7
1
2
1
2
+ 4× +6× +8×
8
8
8
8
1
1
+ 12 × + 16 ×
8
8
2 8 6 16 12 16
= + + +
+
+
8 8 8
8
8
8
60
=
8
= 7.5
=2×
9 a
X ~ B(8, 0.3)
8
5
P( X
= 3)
=   0.330.7=
0.254
3
b P( X ≥ 3) =
0.448
10 X = no. of sixes when 6 dices are thrown
 1
X ~ B  6, 
 6
P( X
= 3)
= 0.0536
Y = no. of times three sixes are seen
d E(Money per week)
6
2 30 20 50
+ 10 × =
+
=
8
8
8
8
8
= £6.25
E(Money in 10 weeks)
=5 ×
Y ~ B (5, 0.0536 )
P(Y
= 2)
= 0.0243
10 £6.25 =
£62.50
=×
11 a i
1
5 X ~ B  5, 
 3
X ~ B(10, 0.2)
P( X
= 4)
= 0.0881
3
2
5  1   2 
40
P( X
= 3)
=      =
3
3
243
3




 
ii P( X > 5) =
0.00637
b E( X ) =np =10 × 0.2 =2
6 P( X =0) =0.9 × 0.9 =0.81
P( X =
1) =
2 × 0.1 × 0.9 =
0.18
P( X =2) =0.1 × 0.1 =0.01
E( X ) = 0 × 0.81 + 1 × 0.18 + 2 × 0.01
=
0 + 0.18 + 0.02 =
0.2
7 a P( X < 65) = P( X > a)
By symmetry a =75 + (75 − 65) =85
b P(65 < X < a) =
0.954
P( X < a) − P( X < 65) =
0.954
P( X < a) − 0.023 =
0.954
P( X < a) =
0.977
∴a =
85
P( X > 85) =
0.023
8 a P( X = 1) =
x
P(X = x)
-5
8
27
19
27
P(Y > 1) =1 − P(Y =0) =1 − 0.8n
1 − 0.8n > 0.95
0.05 > 0.8n
log0.05 > n log0.8
log0.05
<n
log0.8
13.4 < n
∴n =
14
By symmetry:
P(Z < a) =1 −
0.935
∴a =
2
1 2 1 2 1
+ × +  ×
3 3 3 3 3
b
c Y ~ B(n, 0.2)
12 P(−a < Z < a) =
0.85
0.85 − 0.5
=0.825
2
13 a P( X < 80) =
0.85
1 2
4
9
6
4
19
= + +
= +
+
=
3 9 27 27 27 27 27
1
8
19
+1×
27
27
80 − 71 

P Z <
0.85
=
σ


80 − 71
= 1.036
σ
9
σ
= 1.036
=
σ
9
= 8.68
1.036
65 − 71

b P  Z >
=
=
) 0.755
 P ( Z > −0.69
8.68 

© Oxford University Press 2019
10
Worked solutions
14 P  Z <

⇒
0.999 (3 s.f.)
b P ( X ≤ 1) =
30 − µ 
0.15
=
σ 
30 − µ
σ
c P ( X = 1 X ≤ 1) =
=
−1.036
σ
⇒ µ = 50 − 1.282σ
∴ 30 + 1.036σ = 50 − 1.282σ
20
⇒ 2.318σ =
np = 3 and npq = 1.2
Solve simultaneously
q = 0.4 ⇒ p= 0.6
A1A1
M1
A1
n=5
A1
21 X : N (50.1, 0.4
0.775 (3 s.f.)
b P ( 49.5 < X < 50.5) =
35 − µ 
0.2
=
2 
c P ( X > 49 X < 49.5)
35 − µ
0.841..
=
2
1.683..
⇒ 35 − µ =
=
5−µ

0.754
⇒ PZ <
=
3 

P( X
= 5)
= 0.328
c Y ~ B(5, 0.2)
5−µ
= 0.6871...
=
⇒ µ 2.94
3
P( X ≥ 2) =
0.263
16 a 0,1,2
A1
10 10 25
×
=
P ( X =2) =
18 18 81
0
1
2
16
81
40
81
25
81
M1A1
0.116
b P ( 4 < X < 5) =
23 a i
17 a 0.2 + k + 0.25 + k − 0.05 + 0.3 =
1
b E(X )
18 a
b
=
0 × 0.2 + 1 × 0.4 + 2 × 0.1 + 3 × 0.3 =
1.5
M1A1
0.05 + 0.22 + 0.27 + a + b =
1
M1A1
⇒ a+b =
0.46
E ( X ) = 2.46
⇒ 0 × 0.05 + 1 × 0.22 + 2 × 0.27 + 3a + 4b
M1A1
= 2.46
3a + 4b =
1.7
Solve simultaneously
a+b =
0.46 and 3a + 4b =
1.7 M1
A1A1
a = 0.14 , b = 0.32
19 a
ii
M1A1A1
X : B (10, 0.005)
M1
P (X
= 1=
) 0.0478 (3 s.f.)
A1
b
M1
M1A1
M1A1
Let X be the number of correct
answers in the 12 questions
answered at random.
X : B (12, 0.5)
M1
A2
⇒k =
0.15
M1A1A1
0.754
a P ( X < 5) =
b X ~ B(5, 0.8)
x
P (X = x)
P ( 49 < X < 49.5)
= 0.955
P ( X < 49.5)
M1A1
22 X : N ( µ,52 )
⇒ µ = 35 − 1.683.. = 33.3
c
)
0.0668 (3 s.f.) M1A1
a P ( X < 49.5) =
⇒
b
M1A1A1
2
σ = 8.63
50 − 1.282(8.63) =
38.9
µ =

P ( X ≤ 1)
20 Let X : B ( n, p ) .
50 − µ 

0.10
P Z >
=
σ 

50 − µ
1.282
⇒
=
15 a P  Z >
P ( X = 1)
= 0.0478
⇒ µ = 30 + 1.036σ
M1A1
12 
12
P (X
= 2=
)   (0.5)= 0.0161
2
M1A1
12 
12
P=
=
)   (0.5=
) 0.000244
( X 12
12 
A1
E ( X ) =12 × 0.5 × 0.5
M1
= 3 correct answers
3 correct random answers = 6 marks
A1
9 incorrect random answers =-9 marks
A1
8 answers known =16 marks A1
If the student answers all the question
the expected number of marks is 13
marks which is 3 less than the total
marks if he just answers the questions
he knows the correct answer. R1
© Oxford University Press 2019
11
Worked solutions
24 a i
(
W : N µ, σ 2
)
82 − µ
P (W < 65) =
0.27
65 − µ 

⇒ PZ <
=
0.27
σ 

M1
P (W > 96 ) =
0.25
= 1.28... ,
σ
40 − µ
= −0.841...
σ
Solve simultaneously
µ = 56.6 and σ = 19.8
M1
M1
A1A1
96 − µ 

⇒ PZ <
=
0.75
σ 

65 − µ
= −0.6128...
σ
96 − µ
= 0.6744...
σ
ii Solve simultaneously
65 − µ
= −0.6128... , §
σ
96 − µ
= 0.6744...
σ
µ = 79.8 and σ = 24.1
0.20
b P (W > 100 ) =
(
c Let Y : N 80.5,10.12
d
e
A1A1
)
P (75 < Y < 85) =
0.379 .
M1A1
630P (Y > 85) =
207
M1A1
630 × 80.5 + 370m
= 79.7573...
1000
M1
25 a i
(
A1
2
T : N 45, 9
)
P (T ≥ 55) =
0.133
ii
M1A1
P (T > 65)
P (T ≥ 65 T > 55) =
P (T ≥ 55)
=
c
M1
M1A1
m = 78.5 kg
b
A1A1
0.01313...
= 0.0986
0.13326...
M1A1A1
M1A1
(0.133...) = 0.00237
N : B (50, 0.133...)
50 × 0.133... =
6.66 M1A1
i E (N ) =
ii P ( N ≥ 5) =1 − P ( N ≤ 4 ) =0.814
3
M1M1A1
0.1
26 P ( X > 82 ) =
⇒ P ( X < 82 ) =
0.9
82 − µ 

⇒ PZ <
=
0.9
σ 

M1A1
P ( X < 40 ) =
0.2
40 − µ 

⇒ PZ <
=
0.2
σ 

A1
© Oxford University Press 2019
12