Cambridge International A and AS Level
Mathematics
Pure Mathematics 1 Practice Book
University of Cambridge International Examinations bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.
Answers
1 Algebra
(vi)
x = − 1 or 3
2
(vii) x =
3
2
or −1
(viii) x = − 1 or 2
3
Exercise 1.1
1 (i)
11a − 9b
(ii)
−17cd 3 − 14cd − 15c 4 d 2
(iii)
5a + 3ab
3
2
5
(ix)
x =
or − 3
(x)
x =1
(xi)
x = ±1
3
1
3
or − 3
(iv)
2f g
(xii) x =
(v)
6x
y
(xiii) x = 0 or 2
(vi)
8
x
(xiv) x = − 4 or
3
(vii) 149
2 (i)
60 x
(viii) x + 15
20
2 (i)
3mn2 ( 4 + 3n)
(ii)
( p − 4 )( p + 3)
(iii)
(3q − 1)(q + 2)
(iv)
(t − 2u )( s + p)
3 (i)
(ii)
x =
2
−12
(ii)
(iii)
(iv)
e =
x = ±1
1
2
(ii)
x =
(iii)
x = 4
(iv)
x = 1 or x = 2
(v)
x = ±1
(vi)
x = 4
= −1
6
bd
v − bc
p−n
m2 + w
fu
v =
u−f
k =
p
d
e =
−
3 12π 2w
1 (i)
or 2
2
Exercise 1.2
x = 0 or − 5
(ii)
x = 5 or − 5
(iii)
x = 4 or − 2
(iv)
x = −7 or 2
(v)
x = 8 or − 5
( x − 3)2 − 8
(ii)
( x + 2)2 − 4
(iii)
(x − )
(iv)
( x + 1)2 + 4
2 (i)
3
2
2
−
x = 3±
1
4
8
(ii)
x = 0 or − 4
(iii)
x = 2 or 1
3 (i)
1 (i)
1
2
Exercise 1.3
x = −17
4 x = 168km
5 (i)
2
2( x − 1)2 + 5
(ii)
2 ( x + 3)2 − 7
(iii)
3( x + 2)2 − 16
(iv)
5( x − 4)2 − 8
(v)
4( x + 3)2 − 52
(vi)
9 x −
(
1
3
)
2
−1
1
P1
4 (i)
(ii)
a = 19, b = 4
(v)
a = 2, b = 1
5 a = 15, b = 2, c = 2
6 (i)
y
5
−7 −6 −5 −4 −3 −2 −1 0
−1
−2
−3
−4
−5
4
3
2
Answers
1
−7 −6 −5 −4 −3 −2 −1 0
−1
y
5
4
3
2
1
1
2 3 4 5 6 7x
1 2 3 4 5 6 7x
Vertex(–
1
2
, 1)
−2
−3
(vi)
−4
−5
Vertex (–1, 0)
(ii)
y
5
4
3
y
5
2
4
3
–7 –6 –5 –4 –3 –2 –1 0
–1
2
–2
1
–3
−7 −6 −5 −4 −3 −2 −1 0
−1
1
1 2 3 4 5 6 7x
–4
1 2 3 4 5 6 7x
–5
Vertex (– 2, –1)
−2
−3
(vii)
−4
−5
Vertex (–4, –2)
y
5
4
3
(iii)
y
5
4
3
2
1
2
1
−7 −6 −5 −4 −3 −2 −1 0
−1
1 2 3 4 5 6 7x
−2
−7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 x
−1
−2
−3
−4
−5
Vertex (2, 1)
−3
−4
−5
Vertex (0.5, 2.25)
(viii)
y
5
4
(iv)
3
y
2
5
4
3
2
1
−7 −6 −5 −4 −3 −2 −1 0
−1
−2
−3
−4
−5
1
−7 −6 −5 −4 −3 −2 −1 0
−1
−2
1 2 3 4 5 6 7x
Vertex ( 12 , –4 )
2
1 2 3 4 5 6 7x
−3
−4
−5
Vertex (1, 1)
7 (i)
y = ( x − 4)2 − 2
2
(ii)
y = −( x − 1) + 5
(iii)
y = x ( x + 2) = ( x + 1)2 − 1
(iv)
y = −( x − 3)2 + 4
(iii)
k >
(iv)
k < −1
8 (i)
(ii)
2 or k < − 2
P1
m = −4, n = −12
p = −16
Exercise 1.5
Exercise 1.4
1 (i)
x = 3, y = −1
(ii)
x = −2, y = 5
(ii)
x = 6.14 or −1.14 (3 s.f.)
(iii)
x = 3, y = 12
(iii)
x = 0.180 or −1.85 (3 s.f.)
(iv)
x = 6, y = −3
(iv)
x = −0.354 or 2.35
2 (i)
Exercise 1.6
3 2
(ii)
5 3
(iii)
3 5
Answers
x = 1.79 or −2.79 (3 s.f.)
1 (i)
1 (i)
3 a = 3.
x  −9
(ii)
x < −8
(iii)
x > −2
2 (i)
or
−2 < x
x < 0 or x > 1
y
5
4 Using Δ to be the value of the discriminant:
4
(i)
No solutions
(ii)
No solutions
3
2
(iii) Two solutions
1
(iv)
One solution
(v)
No solutions
(vi)
Two solutions
–3
k = −4
–4
(ii)
k = 4 or − 4
–5
(iii)
k = 2
(iv)
k = 0 or − 4
5 (i)
–5
–4
–3
–2
–1
0
–1
2
3
4
5x
1
2
3
4
5x
–2
(ii)
y
5
0<
–x<
–1
4
1> k
k <1
3
(ii)
k < −6 or k > 6
1
(iii)
k < 0 or k > 4
(iv)
k >
6 (i)
1
2
−36
24
–5
–4
–3
–2
–1 0
–1
–2
k > −3
–3
2
–4
7 (i)
(ii)
− 2 <k <
1
36
2
–5
<k
k >
1
36
3
P1
(iii)
x < –1 or x > 1
y
5
(vi)
4
3
3
2
2
1
1
–4
–3
–2
Answers
–5
(iv)
–3 < x < 3
–4
x
4
0
2
3
4
5x
−2
–2
−3
–3
−4
–4
−5
–5
−6
1
3x
2
y
10
9
8
7
6
5
4
3
2
1
(vii) – 2 x 4
1 2 3 4 5x
y
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
−5−4−3−2−1–10 1 2 3 4 5 x
–2
–3
–4
–5
4
1
−7 −6 −5 −4 −3 −2 −1 0
−1
–1
y
5
4
3
2
1
−5 −4 −3 −2 −1 0
−1
−2
−3
−4
−5
−6
−7
−8
−9
−10
(v)
–1
y
x < –5 or x > 0
−4 −3 −2 −1 0
−1
−2
−3
−4
−5
−6
−7
−8
−9
−10
(viii)
1 2 3 4 5 6 x
x < –7 or x > 2
–10 –8
–6
–4
y
14
12
10
8
6
4
2
–2 –20
–4
–6
–8
–10
–12
–14
–16
–18
–20
–22
–24
2
4
6
8
x
(ix)
x < –5 or x >
3
2
(xii)
y
x < 0 or x > 2
y
4
P1
3
20
2
1
10
−2
−1
0
2
1
3x
−1
–4
0
–2
2
x
Answers
–6
−2
−3
–10
(x)
–
1
2
−4
Stretch and challenge
y
30
<x<3
1 (i)
The equation could be thought of in the form
2
y = − a ( x − b ) + c.
25
20
Since the vertex is at approximately (9, 15) the
equation becomes y = − a ( x − 9) + 15 .
15
2
10
Substituting the point (0, 8) we get:
5
8 = − a ( 0 − 9) + 15
2
–4
0
–2
2
4
6x
−7 = − a × 81
–5
a=
–10
7
81
Equation is y = 15 −
(xi)
−3  x 
1
3
–4
–6
–8
–10
–12
–14
–16
–18
–20
–22
–24
( x − 9)
2
(Answers will vary depending on vertex
chosen.)
y
16
14
12
10
8
6
4
2
0
–6 –5 –4 –3 –2 –1
–2
7
81
(ii)
Many answers are possible.
One approach is to determine the co-ordinates
of the centre of the hoop and ensure these
co-ordinates fit the equation.
1 2 3 x
2 (i)
ax 2 + bx + c = 0 ⇒ x 2 +
c
b
x + = 0
a
a
Given the roots we can write the equation as:
( x − α )( x − β ) =
0
x − α x − β x + αβ = 0
2
x 2 − (α + β ) x + αβ = 0
Equating, we get
− (α + β ) =
b
a
⇒α +β = −
αβ =
b
a
c
a
5
P1
(ii)
4 x 2 + ( k + 2) x + 72 = 0
3 9 x − 3x + 1 − 54 = 0
(3 x )2 − 3 × 3 x − 54 = 0
The roots are α and 2α .
Let t = 3 x
k+2
and α × 2α =
4
Solving the second equation,
α + 2α = −
t 2 − 3t − 54 = 0
72
4
(t
+ 6)(t − 9) = 0
t = −6 or 9
2α 2 = 18
Answers
α2 = 9
α = ±3
Substituting into the first equation we get:
k+2
3+2×3 = −
⇒ k = −38
4
or ( −3) + 2 × ( −3) = −
(iii)
α +β =
4
3
1 1 α +β
+
=
=
α β
αβ
4
3
7
3
4
7
x +
3
7
=
1
=8
t
kt 2 + 1 = 8t
kt +
kt 2 − 8t + 1 = 0
4
7
and 1 =
αβ
3
7
= 0 or 7 x 2 − 4 x + 3 = 0
(iv) α + β = 2 and αβ = 3
(α
x
−x
4 k2 + 2 = 8
k+2
⇒ k = 34
4
So the equation is:
x2 −
x
3 x = −6 has no solutions so 3 = 9 ⇒ x = 2
Let t = 2 x, the equation can then be written as
7
3
and αβ =
3 = −6 or 9
x
+ β ) = α 3 + 3α 2 β + 3αβ 2 + β 3
For a single solution b2 − 4ac = 0
( −8)2 − 4 × k × 1 = 0 ⇒ k = 16
16t 2 − 8t + 1 = 0
( 4t − 1)2
2x =
1
4
=0⇒t =
1
4
⇒ x = −2
3
= α 3 + β 3 + 3αβ (α + β )
Exam focus
1 a = 2, b = 2, c = −20
Rearranging, we get
α 3 + β 3 = (α + β ) − 3αβ (α + β )
3
= 23 − 3(3)(2) = −10
(αβ )
3
= 33 = 27
The equation is x 2 + 10 x + 27 = 0
6
2 a = 4 and b = −2
3 x  −1 or x 
4 (i)
(ii)
1
2
Vertex is (−4, 6)
−6 < x < −2
2 Co-ordinate geometry
3 (i)
(ii)
m=3
(ii)
Exercise 2.3
4
Length 20 ≈ 4.47
1 A:
Mid-point = (1, 0)
y = − x + 1 or x + y = 1
B:
y = 2x − 3
1
x
3
Gradient =
1
2
C:
y =
Length =
116 ≈ 10.8
D:
y = −3
Mid-point = (10, 2)
E:
x = 6
Gradient = −2.5
F:
y = −2x − 2
(iii) Length =
2
80 ≈ 8.94
3
Gradient of line
Mid-point = (−1, 1)
Gradient = −
(iv)
Answers
(ii)
15 y = 12 x + 5
12 x − 15 y + 5 = 0
m=0
(iii) m = − 1
2 (i)
1
2
180 ≈ 13.4
Length =
–1
1
4
–4
−2 1 = − 7
3
7
−3
2
3
3
2
Gradient = 2
0.3 =
3 m=5
3 (i)
n=5
(ii)
x=0
(ii)
x = −4
(iii)
3
10
− 10
3
y = 2x − 7
y = −1x −
2
1
2
y = – 0.5x – 0.5
(iii) x = 1
5 (i)
Gradient of perpendicular
1
3
Mid-point = (−7, −3)
4 (i)
m1 = 26.6°
36.9° (1 d.p.)
4 (i)
Exercise 2.2
(ii)
Gradient: 3, y intercept: −1
(ii)
Gradient: −2, y intercept: 3
(iii)
2 (i)
(ii)
Gradient:
4x + 3y = 1
, y intercept: −2
4x + 3y = –8
Equation: y = − x + 2, Gradient: −1, y intercept: 2
(iii) Equation: y = − 1 x + 9 , Gradient: − 1 , y
2
(iv)
4
2
9
4
Equation: y =
intercept: −4
3
2
x − 4 , Gradient: 3 , y
y
5
4
3
2
1
–7 –6 –5 –4 –3 –2 –1 0
–1
–2
–3
–4
–5
Equation: y = 3 x − 2 , Gradient: 3, y intercept: −2
intercept:
x
3 x − 4 y − 13 = 0
2
(iv)
1 2 3 4 5 6 7
4 x + 3y + 8 = 0
(iii) Gradient: 3 , y intercept: 1
−1
2
2x – y = 7
–2
–3
–4
–5
m3 = 135°
1 (i)
y
5
2x – y = 1
4
3
2
1
–7 –6 –5 –4 –3 –2 –1–10
m2 = 63.4°
(ii)
P1
x + 3y + 6 = 0
Exercise 2.1
1 (i)
3y = − x − 6
5 (i)
2
(ii)
3x – 4y = 13
1 2 3 4 5 6 7x
( 32 , − 4)
x + 2y − 2 = 0
7
y
3
2
1
(iii)
−7 −6 −5 −4 −3 −2 −1 0
−1
−2
−3
−4
−5
−6
−7
−8
−9
−10
−11
−12
−13
Answers
P1
Stretch and challenge
1 The diagram shows the relationship between the
tangent and normal at the point A.
1 2 3 4 5 6 7 8x
4
3
A
1 Area =
8
3
×
–4
–3
–2
(
d =
4 y = −3x +
= 51
−3 − −
3
4
3
2
) (
2
+ 2 − −5
2
)
2
2x 2 + 4x + 2 = 0
= 4.74 (3 s.f.)
x 2 + 2x + 1 = 0
y = 2(−1)2 = 2
x+
31
4
(iii)
y = −4x −
(iv)
D is (−5, 4).
3
So A is (−1, 2).
or 3 x − 4 y + 31 = 0
8
3
or 4 x + 3 y + 8 = 0
1
x+
To find B, solve simultaneously:
2x 2 = 14 x +
( , 6) or (3, 1)
1
2
(4, 1)
7
,
3
−
4
3
9
4
8x 2 = x + 9 ⇒฀8x 2 − x − 9 = 0
) or ( − 1, 2)
(8x − 9)(x + 1) = 0
x=
Substitute x =
9
8
9
8
or −1
into the equation of the curve to
find the y co-ordinate:
3 k > 2 or k < −2
( ) = 2×
So B is ( , ).
y =2
9
8
5 k = 2 or −1
9
8
2
9
8
6 k = −7 or 1
81
64
=
81
32
81
32
Finally, use Pythagoras’ theorem to find the length
0 <k < 4
Point is
9
4
9
4
(0, –2) or (–6, 10)
2 k=8
( 3)
of AB.
1
,
2
8 Area of triangle ABC =
8
1
4
y=
(vi)
(ii)
.
(–1, –3) or (–2, –4)
(
7 (i)
1
4
gradient of the normal is
y = mx + c ⇒฀2฀=฀ 4 ฀×฀−1 + c ⇒฀c =
(iii) (2, 13)
4 k >
The gradient of the tangent to the curve is −4 so the
The equation of the normal is:
Exercise 2.5
(v)
x
x = −1
y =
(iv)
4
(x + 1)(x + 1) = 0
(ii)
(ii)
3
2x 2 = −4x − 2
b = 10
1 (i)
2
4x + y + 2 = 0 ⇒ y = −4x − 2
5 a:b=3:2
6 (i)
1
simultaneously:
3
1
8
8
0
–1
To find where the line meets the curve, solve
16
3
×4=
y = −3 x − 7
(ii)
2
We need to find the co-ordinates of both A and B.
2 y = −2 x − 2
3 (i)
B
1
Exercise 2.4
1
2
y
5
AB =
1
2
× 7 1 × 3 = 11 1 .
2
4
( −1 − 89 ) + (2 − 3281 )
2
= 2.19 (3 s.f.)
2
2
y
x
−
= 4 multiplying every term by ab:
a b
bx − ay = 4ab
bx − 4ab = ay
bx − 4ab
y =
a
b
y = x − 4b
a
−
k 2 − 36
−1=
8
k 2 − 36
−
=
8
k 2 − 36
=
8
k 2 − 36 =
P1
−9
−8
8
64
2
k = 100
k = ±10
From this form the gradient of the line is b ,
a
so b = 1 ⇒ 2b = a
a
2
y
0 2
–14 –12 –10 –8 –6 –4 –2
–2
B1
–4
The y intercept is when x = 0 so
y
0 y
−
= 4 ⇒ − = 4 ⇒ y = −4b, so N is (0, −4b).
a b
b
16a2 + 16b2 =
6
8 10 12 14 x
B2
–8
–10
16a2 + 16b2
x 2 − 4 x + 3 = ( x − 1) ( x − 3) so the x intercepts
4 (i)
are x = 1 or 3.
720
So the line must pass through (1, 0) or (3, 0).
16a2 + 16b2 = 720
y = kx + 1 ⇒ 0 = k × 1 + 1 ⇒ k = −1
16(2b )2 + 16b2 = 720
or y = kx + 1 ⇒ 0 = k × 3 + 1 ⇒ k = − 1
16 × 4b2 + 16b2 = 720
3
64b2 + 16b2 = 720
The x intercepts are x = a or b.
(ii)
80b2 = 720
So the line must pass through (a, 0) or (b, 0).
1
a
1
y = kx + 1 ⇒ 0 = k × b + 1 ⇒ k = −
b
b2 = 9
b = 3
a = 2b = 6
y = kx + 1 ⇒ 0 = k × a + 1 ⇒ k = −
(iii) The x intercepts are x = a or b.
y
16
14
12
10
8
6
4
2
0
−6 −4 −2
−2
4
–6
By Pythagoras’ theorem,
(4a)2 + (4b)2 =
2
A
x 0
x
−
= 4⇒
= 4 ⇒ x = 4a , so M is (4a, 0).
a b
a
MN =
Answers
4
The x intercept is when y = 0 so
So the line must pass through (a, 0) or (b, 0).
c
a
c
y = mx + c ⇒ 0 = k × b + c ⇒ k = −
b
y = mx + c ⇒ 0 = k × a + c ⇒ k = −
2 4 6 8 10 12 14 16 18 20 22 24 26 28 x
−4
−6
−8
−10
−12
−14
−16
−18
5
2 y = kx + 1 ⇒ y =
k
1
x +
2
2
k
1
x +
2
2
4 x 2 + 2 x + 2 = kx + 1
2x 2 + x + 1 =
4 x 2 + (2 − k ) x + 1 = 0
3 Mid-point of A(−6, 1) and B(k, −3) is
( −62+ k , 1 +2−3 ) = ( k 2− 6 , −1) .
For two points of intersection, b2 − 4ac > 0
(2 − k )2 − 4 × 4 × 1 > 0
4 − 4k + k 2 − 16 > 0
4 .
Gradient of AB is 1 − ( −3) =
−6 − k
−6 − k
Gradient of perpendicular is − −6 − k = 6 + k .
4
4
k 2 − 4k − 12 > 0
(k
+ 2) ( k − 6) > 0
k < −2 or k > 6
Equation of perpendicular bisector is
y = mx + c ⇒ −1 =
6+k k −6
×
+c
4
2
2
k − 36
+c
8
2
k − 36
c = −
−1
8
−1 =
Since the y intercept is −9,
Exam focus
1 The co-ordinates of D are (0, 1).
2 (i)
(ii)
The equation of CD is y = − 1 x + 11 .
2
The equation of BD is y = 2x − 14.
(iii) The co-ordinates of D are (10, 6).
3 k = −5
9
P1
3 Sequences and series
Answers
121.5
(ii)
71
(ii)
−73
(iii)
58
15
1260 minutes (3 s.f.)
11 (i)
t = 6.53 minutes
(ii)
−2.15
(iv)
x = −1 or 16
10 (i)
Exercise 3.1
1 (i)
$29900 (3 s.f.)
9
For the A.P., a = 18
12 (i)
u4 = a + (4 − 1)d = 18 + 3d
u6 = a + (6 − 1)d = 18 + 5d
−27
2 (i)
1080
(ii)
First three terms of the G.P. are
18, 18 + 3d, 18 + 5d
(iii) 1591
So r =
84.05
(iv)
(3 s.f.)
18 + 3d
18 + 5d
=
18
18 + 3d
(18 + 3d )2
3 878
= 18 (18 + 5d )
324 + 108 d + 9d = 324 + 90d
2
4 10m + 9n
9d 2 + 18d = 0
9d (d + 2) = 0
d = 0 or − 2
5 78
6 $60
Since d ≠ 0, d = −2
7 −290
8 (i)
(ii)
Solving simultaneously, d = 1.5, a = 23
When d = −2 ,
n = 23
r =
9 −6
18 + 3d
18 + 3 × −2
=
=
18
18
(ii)
54
(iii)
n = 19
12
18
=
Exercise 3.2
256
1 (i)
1
8
(ii)
Exercise 3.3
(iii)
1 × 10
(iv)
519
−9
1 (i)
(3 s.f.)
2 (i)
765
2 (i)
121.5
(ii)
(iii) −213000 (3 s.f.)
−200
(iv)
(ii)
5 u5 = 128 × 0.752 = 72
S∞
227.5
=
= 910.2
1 − 0.75
6 S∞
8
= 12 ⇒
= 12
1− r
−
1
2
10
()
1
3
< x <
8 r =
u6 =
256 x 8 − 3072 x 7 + 16128 x 6 − ...
1
12 60
+ 4 + 2 + ...
x6
x
x
(iii)
2187a7 − 5103a6 b + 5103a5 b2 − ...
(iv)
1
7
21
+ 9 + 4 + ...
x 14
x
x
3 The x2 term is
4 x = ±8
7
1 − 9 x + 27 x 2 − 27 x 3
(ii)
3 Parts (ii) and (iv) have a sum to infinity.
u5 = 8 ×
x 4 + 8 x 3 + 24 x 2 + 32 x + 16
5 −1
=
1
2
2
8
=
=
2
2
2×
8
81
4 The x4 term is
( 2 )6 −1 = 8
()
8
2
6
4
= 13440 x 2
2
( 2x ) ( − 3x )
=
5 The term independent of x is
 15 2
 6  x
( )
6 (i)
(ii)
2
10
(2 x )6 1
x
4
9
 5 
 3 
x
6
= 78 203125
The coeficient is 2160.
The coefficient is 5328.
7 k = ±3
8 b = −1
2
63 4
x
16
2
3
9 (i)
729 + 1458u + 1215u2 + ...
S30 = 450 ⇒
(ii) The coefficient of the x 2 term is −243
The coefficient of the x term is 720
11 (i)
2187 − 10206 x 2 + 20412 x 4
12 (i)
6
a + 2d = 0
6a + 87d = 90
a = −2.4, d = 1.2
4
x + 24 x + 252 x + ...
6
(ii) The coefficient of the x term is −228
5 Value of stamp 1: 55 000, 52 600, …
Value of stamp 2: G.P. with r = 0.96
14 a = 2.5
For Stamp 1,
Stretch and challenge
1 u7 = 400 ⇒ a + 6d = 400
30
2
S30 = 1800 ⇒
u10 = 55000 + (10 − 1) × −2400 = 33 400
For stamp 2:
[2a + (30 − 1) × d ] = 1800
15 [ 2a + 29d ] = 1800 ⇒ 30a + 435d = 1800
2a + 29d = 120
Solving simultaneously,
d = −40 and a = 640
So Anna used Facebook for 640 minutes in week 1 of
a × 0.9610 = 33 400
33 400
a=
= $50 238
0.9610
So the second stamp was bought for $50 238
6 (i)
her plan.
6− r
( )
x 10 + 10ax 9 + 45a2 x 8 + 120a3 x 7 +
r
7
− 3
x
210a4 x 6 + 252a5 x 5 + 210a6 x 4 + 120a7 x 3 +
45a8 x 2 + 10a9 x + a10
Term independent of x is when r = 3
Comparing the coefficients of the terms either
3
 6
( 3 )6 − 3  − 73  = 20k 3 x 9 × −343
 3 kx
 x 
x9
side of the x 3 term:
= −6860k 3
120a7 > 210a6 ⇒ a >
General term for the second expansion is
8−r
( )
Term independent of x is when r = 4
8
kx 4
4
8−4
( )
120a7 > 45a8 ⇒ a <
r
m
x4
4
m
x4
4 16
= 70k x
m4
× 16 = 70k 4 m4
x
7
4
(ii)
< a<
...
100
100
( x )44 ( a)56 +
( x )43 ( a)57
56
57
+
100 57
100 56
a >
a ⇒ a>
57
56
57
44
and
100 57
100 58
a >
a ⇒ a<
57
58
58
43
729
−3
r 5 = −243
r =
5
−243 = −3
100
( x )42 ( a)58 ...
58
We want
ar 5 + ar 6 = 729 ⇒ ar 5 (1 + r ) = 729
ar 5 (1 + r )
=
a(1 + r )
8
3
8
3
98
m4
3 a + ar = −3 ⇒ a(1 + r ) = −3
7
4
The terms we are interested in are:
So 70k 4 m4 = −6860k 3
k = −
10
( x )10 − r ( a)r
r
The general term is
The expansion is:
2 General term for the first expansion is
8
kx 4
r
Answers
A.P. with a = 55 000, d = −2400
13 k = −4
6
kx 3
r
P1
Solving simultaneously:
(ii) The coefficient of the x 4 term is 10 206
8
450
30a + 435d = 450
6a + 87d = 90
2
10
[2a + 29d ] =
30
2
57
44
< a<
58
43
a (1 + ( −3)) = −3 ⇒ −2a = −3 ⇒ a = 1.5
4 u12 = 3 × u6 ⇒ a + 11d = 3 ( a + 5d )
a + 11d = 3a + 15d
2a + 4d = 0
a + 2d = 0
11
P1
The coefficient of the x 3 term is − 241 920.
1 (i)
3
The coefficient of the x term is − 435 456.
(ii)
 12
2  
 4 
 2x
12 − 4
x
3 135a4 = 2160
Answers
a4 = 16
12
4 u21 = − 15
Exam focus
a = 2
2 4
= 495×
8
2
× x 8=
x8
126720
5 n = 29.
6 36
4 Functions
(vi) gf( x ) = 2 x−
Exercise 4.1
(vii) ff( x ) = x
1 (i)
(ii)
f(x) = x  3
f(x) = x
2
(viii) gg( x ) = 2 x 2−
 1
2x
3 (i)
f(x) = 10 x  2
(ii)
fg( x ) = 14
(ii)
4 (i)
2
3x  2
4(3 x  2) 2
=
3x  2
12 x  8 2
=
3x  2
12 x  10
=
3x  2
Is not one-to-one the rest are one-to-one.
Is a function. Is not one-to-one.
Is not a function. Is not one-to-one.
(ii)
5 (i)
(ii)
Is a function. Is not one-to-one.
f( − 2)=
9
11
9
(iii)
x = −
(iv)
k > − 2
4 a = 3, b = 2
x2
f( x  1)=
2)
= 4
(iii) Is a one-to-one function.
(iv)
6
x
Answers
3 (i)
x4
gf( x ) = g(3 x
2 All are functions.
(ii)
P1
2 a = 3, b = − 1 or a = − 3, b = 2
(iii) f(x) = − 2x  1 = 1 −
(iv)
x2
Is not a function. Is not one-to-one.
Is a function. Is not one-to-one.
5 (i)
(ii)
x  − 3 or x − 1
− 5  x −
1
3
(iii) Is a one-to-one function.
(iv)
6 (i)
(ii)
Is a function. Is not one-to-one.
Domain:
x∈ 
Range:
f(x)  − 2
Domain:
x
Range:
f(x)  3
(iii) Domain:
(iv)
1 (i)
x − 1
2
f − 1( x ) =
f(x)
6

4
(–1, –1)
x1
Range:
f(x)  2
Domain:
x∈
Range:
f(x) ∈
–6
–4
–2

8
Domain:
0x4
Range:
0  f(x)  2
Domain:
x∈ 
Range:
f(x)  1
2
(ii)
f − 1( x ) = 5−
x
f(x)
6
4
(2.5, 2.5)
2
f( − 2)=
3
(ii)
g( − 2)=
− 3
(iii) fg( − 2)=
4
(iv) gf( − 2)=
− 8
fg( x ) = x 2
6x
–6
Exercise 4.2
(i)
4
–4
10 Given that x  − 1, the range is f(x)  − 3
(v)
0

Range is f(x)  − 2
1
2
–2
7
9
Exercise 4.3
–6
–4
–2
0
2
4
6x
–2
–4
–6
13
P1
(iii)
f( x ) = 1−
5x = − 5x  1
f(x)
6
(ii)
x − 1 1− x
=
5
− 5
f − 1( x ) =
5
4
f(x)
6
3
2
1
4
Answers
1, 1
6 6
–6 –5 –4 –3 –2 –1 0
–1
2
1
2
3
4
5
6x
–2
–3
–6
–4
–2
0
2
6x
4
–4
–5
–2
–6
–4
f − 1( x ) = 2( x
1) = 2 x
x2
Range f( x ) :
f( x )  − 4
− 1
x − 4
f − 1( x )  2
Domain f ( x ) :
− 1
Range f ( x ) :
–6
(iv)
Domain f( x ) :
2
(iii)
f(x)
6
f(x)
6
5
4
4
3
2
2
1
–6
–4
–2
(–2, –2)
0
2
–6 –5 –4 –3 –2 –1 0
–1
6x
4
–2
1
2
3
4
5
6x
3
4
5
6
7
8x
–2
–3
–4
–4
–5
–6
2 (i)
–6
f(x)
6
Domain f( x ) :
x  –1
Range f( x ) :
f( x )  − 5
− 1
Domain f ( x ) : x  − 5
Range f − 1( x ) :
f − 1( x )  − 1
5
4
3
(iv)
2
f(x)
8
1
–6 –5 –4 –3 –2 –1 0
–1
7
1
2
3
4
5
6x
6
5
–2
4
–3
3
–4
2
–5
1
–6
Domain f( x ) :
x0
Range f( x ) :
f( x )  − 4
–4 –3 –2 –1 0
–1
2
–2
–3
Domain f − 1( x ) : x  − 4
Range f − 1( x ) :
1
–4
f − 1( x )  0
Domain f( x ) :
x  –2
Range f( x ) :
f( x )  − 1
Domain f − 1( x ) : x  − 1
14
Range f − 1( x ) :
f − 1( x )  − 2
3 The largest possible value of k is −
3
2
9x
4x − 1
9
4x − 1
4x − 1
9x
=
×
9
4x − 1
= x
.
g(x)
6
5
4
3
2
1
2
3
4
5
x  2
, x ∈ , x≠
4x − 1
f − 1( x ) =
(ii)
1
–6 –5 –4 –3 –2 –1 0
–1
P1
=
1
.
4
6x
–3
Answers
–2
Stretch and challenge
1 (i) y = af  b( x c )  d
–4
–5
–6
Constant
2
3x − 2
=
x
x
4 j− 1( x ) = 3−
− 1
5 f (x) =
3
Effect on graph
a > 1 Vertical expansion
a
0 <
x  1
− 1
2
2)4 − 1
( x−
x0
c
g( x )  3
d
Domain g− 1( x ) :
x 3
Range g− 1( x ) :
g− 1( x )  0
(ii)
2
The translation of f( x ) = x −
3 x 3 units right
and one up is given by
7 k = π
(ii)
0) or right (c 0)
Shift d units up (d 0) or down
(d 0)
Shift left (c
Domain g( x ) :
Range g( x ) :
8 (i)
1 Vertical compression
b > 1 H orizontal compression
0 < b < 1 Horizontal expansion
b
6 g− 1( x ) =
a<
a < 0 Reflects the graph i n x axis
f( x − 3)
Range is 0  f  10
= ( x−
3) − 3( x−
2
f(x)
1
2
= x −
6 x  9−
= x −
9 x  19
2
3)  1
3 x  9
1
y
5
4
3
2
1
0
−1
1
2
3
4
5
6
7x
−1
x
(iii) 4  x  10
9 (i)
ff( x) = f
x +2
+2
4x − 1
x +2
=
4x − 1
x +2
4
−1
4x − 1
x  2
=
2(4 x − 1)
4x − 1
4( x  2)− (4 x − 1)
4x − 1
−2
2 g(1) = f(0)=
g(2) = fg(1)=
1
f(1) = 3
g(3) = fg(2)=
f(3) = 7
g(4) = fg(3)=
f(7) = 15
We can see the pattern here, every number is one
less than a power of 2 so g(n) = 2n−
1.
h(0) = g(2)= 3
h(1) = gh(0)= g(3) = 7
h(2) = gh(1)=
g(7) = 27−
1 = 127
15
P1
3 f( x  1)
Answers
a=
f( x − 1)=
3x 
1
3x 
1
 3 x− 1=
 3 x−
3x
1
a3 x
=
a= 2
2 (i)
af( x )
x = − 2
(ii)
3 x (3  3− 1)
= 3
3x
3−
1
4 f(2n − 1)=
a(2n − 1)
b = 2an−
a b
f(2n) − 1=
a(2n)  b−
1 = 2an
b− 1
2f(n) − 1=
2( an  b)−
1 = 2an
2b − 1
=
10
3
2an  b is common to all three expression
3 (i)
k = − 3.
g− 1( x ) =
(ii)
x =
4 (i)
1
3
x
7− 3
or x=
g− 1( x ) =
(ii)
2
2 − 7x
2x − 1
(iii) Since b2 − 4ac=
( − 1)2−
4 × 1×
− a,− 1 and b − 1 .
(iv)
Since − 1 is one of the terms, the possible
(v)
f − 1g( x ) =
x 4
2x  7
Point on y = f − 1( x)
Point on y = f( x)
(i)
− 3,− 2, − 1
(0, 2)
(2, 0)
(ii)
− 2,− 1, 0
(1, –1)
(–1, 1)
(iii)
− 1, 0, 1
(–1, 5)
(5, –1)
(i)
a = 3, b=
− 1⇒
f(n) = 3n−
1
a = 2, b=
− 2⇒
f(n) = 2n−
2
(ii)
a = 2, b=
1⇒
(iii)
a = − 1, b = 1⇒
f(n= )
f(x)
8
6
2n  1
f(n) = −
1 = − 3 the
equation has no real solutions.
so the three expressions can be simplified to
consecutive terms are
b= 3
y=x
4
n 1
2
−8
Exam focus
1 The domain of f( x ) is x  − 2
The range of f( x ) is y  − 2
−6
−4
−2
0
−2
−4
−6
− 1
The domain of f ( x ) is x ฀− 2
− 1
The range of f ( x ) is y  − 2
16
−8
2
4
6
8
x
5 Differentiation
Exercise 5.3
1 (i)
Exercise 5.1
y
dy
= 20 x−
dx
1 (i)
3
1
(ii)
dy
= 1−
dx
(iii)
dy
= − 10 x − 3 = −
dx
20 x
3
10
x3
(iv)
dy
6x
=
=
dx
3
2x 2
(v)
y =
4
−
x
dy
=
dx
4x−
(vi)
x
=
4
− 4 x − 2−
(vii) y = 2 x−
dy
=
dx
⇒
(viii) y =
⇒
f ′ ( x ) = − 2x −
2 (i)
(ii)
(iii)
f′ (x) =
1
3 x 2−
f′ (x) =
2x −
π
−
2
π
3
3 f ′ (9) = −
93
2
2
5 3
x
f ′ ( x ) = 6×
(iv)
1
−
3x 2
3
 5=
−
=
− 3x −
4
4
3
4x 4
= −


1
5
,−
3
27
Points are
 , 4
2
=
33 x 2
2
−
x2
−
2
3x 2
−
15
10
5
2
x3
–1
–15
2
π
−
2 π x2
–20
–25
–30
= 10 3 x 2
−5
8
1
= 4
9
9
or (− 1, − 3).

f′  − 
f′
1
2
> 0⇒
1
2
< 0⇒
5 k = 5
1
2
1
2
y
7
12 =
4
2
3
4⇒
Q is 4 1 , 6 3
2
 , 4 is a local minimum
 − ,− 4 is a local maximum
(iii) Point is (1, 2)
(ii) S is (− 1, 0)

0.5
–10
5
(ii)
0
–5
–0.5
3
x
= 3 x−
17
y = 2 x
1.5 x

4
20
1
2 y = 4 x−
4 (i)
1
,−
2
25
6
12
and −
2x
3
1 Equation of the tangent is: y = − 10 x − 8
PR =

1
2
y
30
Exercise 5.2
3 (i)
1.0
(1, − 1) is a local minimum
(ii)
4 Point is (− 5, 18).
5 Points are
3
(0, 0) is a local minimum
− 1
3
 5=
27
x
2
–2
3x
− 2 x− 3=
2
1
4
1
− 3
x
− 2x −

3
2
3
1 −
x
3
dy
=
dx
4
−
x2
=−
1
x 3
1
–1
dy
=
dx
− 3=
0
–1
x
1
2 x 2−
2
=
3x
x
3
1
4
3x =
1
−
x 2
1
4
−
x− 3
⇒
4
1
=
4x3
y =
1
–2
Answers
2
⇒
P1
Stationary points are (0, 0) and (1, − 1).
4

2−x
y  4=
0
2
1
−3 −2 −1 0
1
x = 1 is a local minimum
2
3
4
5
6
7
8x
17
P1
2 0<x<1
3 (i)
3 Since the discriminant is < 0, there are no solutions
2 x − 3 x 2=
when − 1
2 x − 3 x 2=
So − 1
0
0 is always < 0
4 (i)
x
h
y
5
4
x
2
Answers
3
2
h2 
1
0
−1
−1
2x
1
 2x  =
2
x2
x2
4
h2 = x 2−
y = 2x3 – 3x2
−2
3x 2
4
3
x
2
h2 =
−3
h =
−4
−5
(ii)
g(x)
(ii)
4
g(x) = 3x2 – 2x3
2
−2
3 x  2 y=
10 ⇒
y=
A = xy
1
x ×
2
3
x
2
= x
0
−1
Perimeter = 10
1
10 − 3 x
2
10 − 3 x
3 2
+
x
2
4
10 x − 3 x 2
3 2
+
x
2
4
3
3 2
= 5x − x 2 +
x
2
4
3−6 2
= 5x +
x
4
=
2x
−2
−4
(iii) Stationary value when dA = 0
dx
= 2.34 m (3 s.f.)
5 (i)
(ii)
k = 5
The x co-ordinate of the other stationary point
8
=
3
is x = −
(iii)
−
⇒
x=
8
3
= 6×
−
8
3
4 (i)
−

8
3
 5=
− 11<
0
is a maximum
Exercise 5.4
1 Maximum value of P is 9 × 9=
2 Dimensions are 3 m by 4.5 m.
81
V = π r 2h = 5000⇒
h=
5000
π r2
S = 2
π r2  2
π rh
 5000 
= 2
π r2  2
π r
 π r 2 
10 000
= 2
π r2 
r
(ii)
18
x = 2.34 m is a maximum
3
f ′ (1) = 6× 1 5=
11 > 0
⇒ x=
1 is a minimum
f′  −

(iv)
2
2 .
Stationary value when dS = 0
dr
r =
3
10 000
= 9.27cm
4π
(iii) r = 9.27 is a minimum value
2b  2
π r = 400
5 (i)
(ii)
2b = 400−
2π r
b = 200−
π r
dy
=
dx
⇒
400
= 31.8 m (3 s.f.)
4π
b = 200 − π ×
31.8... = 100 m
2
= 2  x
x  3
y =
(vi)
3
3
− 2  x
− 1
− 2
P1
= −
2
2
x  3

r =
(vii) y =
6 Let the length of the end be x
dy
=
dx
⇒
Let the length be y
x
V = x y
= x 2 (120−
4x)
2
= 120 x −
4x
dV
= 240 x−
dx
− 3
5
2
−
3
 5 x
dV
= 0
dx
y = mx
y = 2 x−
0
0
c ⇒
21
3
=
4 × 20=
2
4 g′ ( x ) = 9  x
Since x ≠ 0, x = 20 cm
× 5=
−
25
2  5 x  3
3
2 × 12
c ⇒
2
Hence the function is one-to-one so it has an
inverse.
8 1241.123... = 1240 cm3 (3 s.f.)
The graph of y = g( x ) for x > − 2 and its inverse are
shown on the next page.
9 Dimensions are:
g(x)
length = 1.87 m, width = 7.47 m, height = 3.56 m
3
2
Exercise 5.5
dy
=
dx
4)7 ⇒
7( x − 4)6
dy
=
dx
2)8 ⇒
y = (3 x
dy
=
dx
2 x )9 ⇒
1
8(3 x  2)7×
= 24(3 x
y = (5−
− 21
increasing so there are no stationary points.
7 The largest area is 0.5 units2
y = ( x−
=c
Since g′ ( x ) > 0 for all values of x, the curve is always
40 cm
Volume is 16 000 cm3
(iii)
1
2
−
3 Point is (2, − 2)
x = 0 cm or 20 cm
(ii)
3
2 Equation of the tangent is
240 x − 12 x 2=
12 x (20 − x )=
1 (i)
16
−
1
 2x  3
×− 2=
3
2
−
− 2
12 x 2
Maximum volume when
y = 120−
dy
=
dx
⇒
3
2x 
− 8  1−
2x 
5
= 5  5 x
5x  3
(viii) y =
2
4
= 4  1−
2
2x 
Answers
4x  y = 120 so y = 120 − 4
1−
3
−6
−5
−4
−3
−2
1
2
3 x
−1
2)7
9(5 − 2 x )8×
0
−1
−2
−3
− 2
−4
= − 18(5 − 2 x )8
−5
y =
(iv)
3
dy
=
dx
⇒
4 x = (3
1
2
y =
⇒
3
1−
dy
=
dx
4x 
5 (i)
4
(ii)
3  4x
9x =
1
3
= −
3
− 1
2×
2
=
(v)
1
4 x )2
 1−
6 (i)
9x 
 1−
9x 
− 2
3×
1−
9x 
2
3
2
x < −
5
2
dx 2
− 9
1
2
or x > −
dy
− 2
= − 2  x − 1 
dx
d2 y
(ii)
3
3
1
3
k = −
= 4  x−
1
− 3
2= −
=


2
 2
2
x − 1
4
3
x − 1
maximum point A is (0, − 2),
minimum point B is (2, 6)
19
P1
h
( a + b ) × 200
2
h
h
60 + 60 + 2 ×
=
× 200
2
3
7 0.2cm/s
=
8 0.111m/min (3 s.f.)
= 11.1cm/min (3 s.f.)
9 0.16 units/s
10 (i)
Answers
(ii)
0.810 cm/s (3 s.f.)
247cm2 /s (3 s.f.)
(ii)
= 100h 120 +
2h
3
= 200h 60 +
h
3
0.0818 cm/s
11 0.05 rad/s
12 (i)
Stretch and challenge
Using similar triangles
1 (i)
g′ (x) = k − 2 x so if the gradient of the normal
is 1 , the gradient of the tangent is − 2.
2
k 2 2x 5 22
60 cm
k 2 2 3฀2฀5 22
120 cm
k 2 4 5 22
80 cm
x
k฀5 2
h
30 cm
(ii)
y
The y value when x = 2 is g(2) = 2 3 2 2 22 = 0.
Equation of the normal is y = mx
1
×
2
0 =
y  80
y
=
⇒
60
30
h  y h  80
=
⇒
x
x
y = 80 cm
3h
2  c⇒
=


 30  30
2
Volume = 120 ×
=
(ii)
45 2
h
2
x =
240 = 8 x⇒
3
h
8
1
2
30
× h=
3 2
h
8
 60h
1
x − 1=
2x − x 2
x − 2=
4 x − 2x 2
2
3 2
h
16
=
2 x − 3 x− 2 = 0
(2 x  1)( x− 2) = 0
30h
1
2
x = −
3 2
h + 30h
16
or 2
3600h cm3
g( −
1
)=
2
2× −
1
2
− −
 21  2
= − 1 41

The co-ordinates of the point P are −
h = 10.4 cm or− 170.4 cm
g(x)
3
2
0.983cm/min
1
−2 −1 0
P −1
h
In the right-angled triangle
h
h
tan60˚ =
⇒
3=
⇒
x =
x
x
1
2
4 x
3
−2
−3
60°
B
1
.
2
So the x co-ordinate of the other point is x = −
13 (i)
−4
x
2 (i)
h
3
Length of CD is 60  2 x = 60  2×
x 2  102 so cos α
SP =
PW =
h
3
Volume = area of trapezium × length
=
=
cos β =
20
x−
2
Since h cannot be negative, h = 10.4 cm
(iii)
c = − 1 so y =
c ⇒
To find the other point, solve simultaneously
Area of trapezium cross-section
3
h
8
1
2
(30−
(900−
925−
=
x
x 2  102
x )2  52
60 x  x 2 )
60 x  x
2
30 − x
925 − 60 x
x2
25
1
,−
2

1 41 .
(ii)
Time T =
=
x 2  102

10
925 − 60 x
5
x 2  102
2 925 − 60 x
10
cos α
3 (i)
2
0
−2
−2
5 AE2  30 2 = 50 2 ⇒ AE = 40 km
1
[cos α
10
B
2cosβ ] = 0
−
50 km
= 2cosβ
30 km
V = π r 2h
 h2 
R2 = r 2
2
r =2
⇒
R2 −
D
40 – x
A
x
E
2
h
4
30 km
50 km
 2 h2 
 R − 4  h
V = π
4 x
2
2cosβ ]
−
T ′ = 0 when
4



60) 



P1
Answers
=
f’(x)
x2
− 1
 1 2
( x  100) 2× 2 x 

2
T′ = 1 
1
10
 2 × 1 (925− 60 x  x 2 )− 2× (2 x −

2
2( x − 30)


x

= 1
2
10 
x  100
925 − 60 x x 2 
x
30 − x

− 2
= 1
2
10 
x  100
925 − 60 x x 2
1
[cos α
10
4
x2
C
π h3
= π R h−
4
h2
dV
π
3
= π R2 −
dh
4
dV
3π h2
= 0 whenπ R2 −
= 0
dh
4
2
In triangle BDE, BD2 = x
L = AD
h2
=
4
r 2 = R2−
r =
(ii)
R−2
2
R
3
Vcylinder
=
Vsphere
=
π r h
π R3
4
3
3r 2h
4R 3
3
=
=


3

=
900
= (40−
x )  2( x 2
900) 2
1 2
(x
2

2

2x
x  900
2x
x 2  900
2x
= 0
= 1
2
x 2
900
2
4x = x 
900
2
3 x = 900
4R 3
2R2   2R 
3   3 
4R
3
4R 3
1
3
1
−
 900) 2×
Minimum length is when
2R 

3
4R 3
1
2
2x =
2
3R
900
2x
= − 1
x 2  900
3
=
2
dL
= − 1  2×
dx
R2
2R2
=
3
3
x 2
900 
x)  2 x 
− 1
2
x 2
x) 
= (40−
2
= R2−
x 2  900
BD  CD
= (40−
4π R2
= h2
3π
2R
h=
3
 2R 


3
4
 30 2 ⇒ BD =
Total length of road, L, is:
3π h2
4
π R 2=
2
x 2 = 300
x =
L = (40−
x )  2 x 2
300≈
17.3 km
900
= (40− 17.3...)  2 (17.3...)2
= 91.9615... km
900
= 92 km (3 s.f.)
So ratio is 1: 3 or 3:3
21
P1
Exam focus
1 The points are 1,−
and
16
−2,
3
.
π r 2h=
2 The equation of the tangent is
y = − 3 x  8 or 3 x
y − 8=
Answers
The slope of the normal is
y = mx
c
⇒
5
=
1
3
h=
0
⇒
c
=c
14
3
The equation of the normal is
y =
1
3
14
3
x
or x − 3 y
dx 2
= 12 x 2−
At x = 0,
At x = 2,
d2 y
dx 2
14 = 0
At x = − 2,
minimum.
d2 y
dx 2
= 12×
22 − 16=
32 so (2, − 14) is a
( − 2)2−
16 = 32 so (− 2, − 14) is a
4 The function is increasing for 0 < x < 4.
5 1200 mm2 /minute
6 0.012 units/second
22
200
−
r
200
=

r
200
=

r
2
π
3
r3
1

2
4 π r 2

4π r 2
 3π r 2
3
4π r 2
3π −
3
5π 2
r
3

(iii) Stationary point is when
10 π
r = 0
3
10π
− 200
r3 = 0
3
200
= 19.098...
r3 =
10 π
3
r = 2.67cm (3 s.f.)
−
= 12×
100 −
π r2
=
= − 16 so (0, 2) is a maximum.
dx 2
minimum.
r3
 100 2r 
−
S = 2π r 
  π r 2  2π r 2
3
 π r2
16
d2 y
2
π
3
S = 2π rh  π r 2 
(ii)
3 The stationary points are (0, 2), (2, − 14) and (− 2, − 14).
d2 y
100 −

r 3 = π r 2h 
2
π r3
100
3
=
−
π r2
π r2
= 100
− 2r
3
π r2
1
3
× 1

1 4
π
2 3
Volume = π r 2h 
7 (i)
13
6
200

r2
(iv) r = 2.67 cm is a minimum value.
2
π
3
r 3 = 100
6 Integration
7 Area = 4.5
Exercise 6.1
1 (i)
x4
x2


2
2
c
(ii)
x
3x 2

2
c
(iii)
x 2 − x 5
c
−
(iv)
c
x2 x6


4
3
(v)
−
(vi)
(–2, 5)
1

2x
–5 –4 –3 –2 –1 –1 0 1
–2
–3
–4
–5
c
3x 2  c
93 x 4  c
2
2 (i)
2x 3 − 6 x 5 
3
5
(iii)
4 x  c
(iv)
4 x5  c
5
3 − 
x
1
 c
3x 3
4 y = −
1
 x
x3
2
3
2− 5 −
1
3
Exercise 6.3
x 2  30
3
− 2
x
1
(i)

(ii)
 3 x  1 6
2
3x
+ x
2
x − 2
4
4
 c
(iii) −
 1−
6 x
30
2
10
4
x
 3
(iv) 

1
 c
18
2
4
2
3×2
+2 − 3×1 +1
2
2
 c
 c
2
3
2
4 7
× 1=
2
A=
5.5
2 The area is 1 1 .
3
 2 x − 1 2
3
3 (i)
−
10
2
3
 c
2  1 − x
3
3
 c
1
(ii)
−
(iii)
−
(iv)
2
5  7 x  2
(v)
− 3 5−
(vi)
x
36 3  1 

2
3 Area between the curve and the x axis:
16
=
3
3= 9
2
= 8−2 .5
= 5 .5
2×
6x
10 k = 4
∫ 1 (3x + 1) dx
=
(ii)
5
Area between curve and the y axis:
Exercise 6.2
=
4
1
2
2
7 − 1  = 5
 3
3
3
6 The equation is y = x
A=
3
9 Area between curve and the x axis:
c
= 9×
5 f  x  = 2 x 3−
1 (i)
2
 − 14 2  = 41 2

3
3
Area = 27−
(ii)
(3, 0)
Answers
2

x2
P1
y
10
9
8
7
6
5
4
3
2
1
8
 c
 2 x  1 2
1
6  x − 4
4
 c
Area between the curve and the y axis:
= 4×
2
3
4 − 10 =
4 Area = 2×
5 Area = 10−
5
1
3
5
1
3
= 10
 − 2 =
2
3
12
6 Total area = 5 + 1 = 6
5
 c
2x  c
2
 c
23
P1
Exercise 6.4
Stretch and challenge
1 − 1 − − 1  = 2
36  4  9
1
0
2 x + k − x + 2 dx =
0
2 x + k − x 2 + 4 x + 4 dx =
0
− x 2 − 2 x + k − 4 dx =
∫−2
2 4
∫−2
3 3
∫−2
2
10
3
10
3
10
3
0
 x3

2
− 3 − x + k − 4 x  =

 −2
Answers
Exercise 6.5
4
π
3
(ii)

 −2
0 −  −
3


25.1  3 s.f.
8π =
1 (i)
4.19  3 s.f.
=
−
2 75.4  3 s.f.
V  solid =
3
8
3
+ 4 + 2k − 8 =
20
3
−
1
π
3
× 62×
+ 2k =
2 = 24π
2
10
3
10
3
2 Gradient of the sloping line is m =
4 15π
y = mx
c ⇒
h
× r
R− r
0
=
5 Points of intersection:
2 x  x − 2 =
2
−
0
x  2 x − 3 =
0
1 .5 

x4 − 4x3 +
15
4
x2 d x


 R − r  2 h3 2r (R − r ) h2
+ r 2h 
+
= π 
 h  3
h
2


2
 h (R − r )

= π
+ hr (R − r ) + r 2h 
3



 h R2 − 2Rr + r 2
= π
+ hr R − r + r 2h 
3




πh  2
2
2
2
=
R − 2Rr + r + 3rR − 3r + 3r 
3 
πh  2
=
R + Rr + r 2 
3 


 1.55
5 × 1.53  
4
−
+
1
.
5
 

4
= π  5


 05
5 × 03  
4
−0 +

 −
4  
 5

7 (i)
π
0
24
1
πh
3
=
1
2
R 2 + rR + r 2
A is ( ,10 ), B is (2, 10)
Final volume = 150π − 66 π = 84 π ≈ 264 units
Area =
3
M is (1, 8)
(ii)
dy
h
1 .5
6 a = 8  since a>
2
 R − r  2 y 3 2r R − r y 2

+
+ r2y 
= π 
 h  3
h
2

0
 x5
5x3 
= π
− x4 +
5
4 

0
27
40
hr
R− r
−
2

h R − r
2r (R − r )
= π∫  
y2 +
y + r 2  dy
0  h 
h

2

2
1
= ∫ π  x ( x − 2) −  − x   d x
 2 
0


1 .5  4
1

= π ∫ (x − 4 x 3 + 4 x 2 ) −  x 2   d x
 4 
0 
0
h
∫0 πx
V =
)
1 .5
= π∫
=c
hr
h
x−
R− r
R− r
hr
h
=
x
y 
R− r
R− r
R− r
y  r
x =
h
 R − r  2 2 2r  R − r
y  r2
x2 = 
 y 
h
 h 
3
2
x = 0 or x=
V (s o l i d
c ⇒
y =
1
x
2
− x
2 x − 3 x=
h
R− r
Equation is
V (solid) = 24π − 15 π = 9 π = 28.3 (3 s.f.)
x  x − 2 =
10
3
2k = 10
k =5
27
π
4
=

+ k − 4 × −2  =
 
3
− −2
10
3
k 1
∫k
109
6
2 x 2 dx=
 2x 3 
 3 
k 1
=
109
6
 2  k  1 3 2k 3 
=
–

3
3 
109
6
 2  k 3  3k 2 3k  1 – 2k 3 
 =

3


109
6
3




 =




k


 =










 =
 



 2k 3  6k 2


 =

6k  2 – 2k 3  =

3
 6k 2  6k

3
2  6k 2  6k
2  = 109

6
2 = 109
2
12k  12k
2
12k  12k−
 2k  7  2k−
5 = 0
= 4
7
2
or –
5 − 12 −
 4×
=
2
3
3
x =2
 = r 2=
w /2
−
=
= ∫ 2∫ π
=
= ∫ 2∫ π
R− 2 −
w2
4
8 (i)
w /2
2
units3
=
π x d− yπ − π r w
0
w /2
π

0
1
×
2
Area under line =
2
 R− 2 −  y 2  d− yπ − π r 2w

 
(ii)
2
−
3
=
1
2
1
π
2
Final volume =
w /2
=
1
π
6
=
π wπ 3
−
− π − rπ 2w
12
 
π wπ 3
w2  
= π = Rπ 2w − −
− π − π R2 − −  w
12
4  
 
9 (i)
1
π
3
−
=
1
π
5
1
π
5
4
π
21
or
Final volume =
π wπ 3
π wπ 3
− π − Rπ 2w  
−
12
4
w
π 3
8
π
15
or
=
= π = Rπ 2w −
1
π
6
(b) Final volume =
  R2w w
 3
2
= π= 2
π  2− − 24
 − π − π r w
 
 
1
π=
6
1
π
3
−
or
3

   w  


  2  
 − π  − π r 2w
 R2   w−  −
3 
  2
 

 

= π = Rπ 2w −
1
6
(a) Volume of curve rotated around the x axis:
  2
y 3 
2
= π= 2
π  R − y −  3  − π − π r w
 
 0


π
= π= 2



1
2
1 × 1=
Shaded area = Area under curve − area of Δ
x 2d− y −  V  cylinder
 
∫ − π w / 2π
18
π
5
7 Volume =
R2 − y 2
R⇒ 2 ⇒ = r =2
=V ∫ =
=
3 × 1

1
3
6 Shaded Area = 42 2
R2 ⇒
2
− area of OAFD
5
2
Since x > 0, k=
   w 
Also
  2
5 Area of DFGE = area of OBGE − area of ABGF
Answers
5
2
k =
P1
4 1
105 = 0
35 = 0
3
 c
4  4 x  1
3 2
4 = 109
4k 2  4k−
2
2
4 x  y =
2 −
109
6
(ii)
Volume =
1
π
3
1
π
7
−
=
4
π
21


4π
units3 0.838 units3 to 3 s.f.
15
or
Since the volume does not depend on R, the claim
is true.
Final volume =
3
5
π−
1
4
3
15
π=
π units
Exam focus
1 The curve is y = x 2
x − 4
y
12
10
8
6
4
2
–6
–4
–2
0
–2
2
–4
A(1, –2)
4
x
–6
25
P1
7 Trigonometry
2 (i)
sin x tan x ≡
LHS = sin x×
Exercise 7.1
Answers
5.71cm  3 s.f.
a=
32.65≈
(ii)
b =
7.2225≈
(iii)
c = 8.40 cm
=
(iv)
θ = 40.5°
=
1 (i)
=
2.69 m  3 s.f.
=
=
2 (i)
(ii)
d = 11.1cm
β = 60.9°
θ = 180− 60.9 − 53=
(iii)
x = 17.9 cm
(iv)
θ = 107°
(ii)
(i)
Area = 27.1cm
(ii)
Area = 512 cm
(iii)
Area = 89.1cm2
(iv)
Area = 5.49 m
(ii)
11.2 cm
sin2 x
cos x
1 − cos2 x
cos x
1
cos2 x
−
cos x
cos x
1
− cos x
cos x
RHS


tan x 1 − sin2 x ≡
(iv)
tan2 x − sin2 x≡
tan2 x sin2 x
sin2 x
− sin2 x
cos2 x
sin2 x − sin2 x cos2 x
=
cos2 x
LHS =
=
1
Other angle
the same value
sin 30°
sin 150°
sin 210°
sin 330°
1
2
−
1
2
1
2
cos 300°
1
3
tan 30°
tan 210°
cos 150°
cos 210°
–
tan 120°
tan 300°
– 3
3
2

sin2 x 1 − cos2 x
2

cos x
sin2 x × sin2 x
=
cos2 x
2
sin x
=
× sin2 x
cos2 x
= tan2 x sin2 x
= RHS
Exact value
that has
26
sin x cos x
1  cos x
sin x
≡
sin x
1 − cos x
sin x
1  cos x
RHS =
×
1 − cos x 1  cos x
sin x  1  cos x 
=
1 − cos2 x
sin x  1  cos x 
=
sin2 x
1  cos x
=
sin x
= LHS
Exercise 7.2
cos 60°
sin x
cos x
(iii)
y = 6 cm
Angle
cos x
LHS = tan x× cos2 x
sin x
=
× cos2 x
cos x
= sin x cos x
= RHS
66.1°
3 (Using rounded answers from question 2)
4 (i)
1
−
cos x
3 (i)
sin4 x − cos 4 x≡
sin2 x − cos2 x
LHS =
=
 sin
 sin
2
2
x−
x−
= sin2 x−
= RHS

x   1

cos2 x sin2 x  cos2 x
cos
2
cos2 x
=
=
(ii)
tan x +
(ii)
=
 1  cos x   1 − cos x
 1 − cos x  1 cos x
=
=
=
1  tan2 x
1
≡
1 − tan2 x
2cos2 x − 1
sin2 x
1
cos2 x
LHS =
sin2 x
1−
cos2 x
cos2 x  sin2 x
cos2 x
=
2
cos x − sin2 x
cos2 x
1
2
x
cos
=
2
cos x − sin2 x
cos2 x
1
cos2 x
=
×
2
2
cos x cos x − sin2 x
1
=
cos2 x − sin2 x
1
=
2
cos x − 1− cos2 x

1
2cos2 x − 1
= RHS
=

1
≡
−
sinθ
LHS =
=
=
=
θ


P1
cos x  cos2 x− cos x  cos2 x
1 − cos2 x
2cos2 x
sin2 x
2
sin2 x
cos2 x
2
tan2 x
RHS
1 (i)
(ii)
2 (i)
(ii)
3 (i)
sin x =
15
4
tan2 x = 15
cos x = −
sin2 x =
7
53
4
53
x = 210
° or 330°
(ii)
x = 30
° or 150° or 210° or 330°
(iii)
x = 63.4
° or 243.4°  1d.p.
(iv)
x = 120
°
4 x = 21.1° or 81.1° or 141.1°  1d.p.
y 1
1
 1
 1−

sinθ
cos2 θ
sinθ  1  sinθ

sin θ
2
sinθ  1  sinθ
−
θ
0.25

1x
–1

sinθ   1 sinθ
sinθ  1  sinθ 
1 − sinθ
=
θ
=
θ


cos θ
sinθ
cos θ
 1−

Exercise 7.3
5 (i)
cos θ
tanθ  1  sinθ
θ
Answers
1
1
2

≡
1 − cos x 1  cos x
sin2 x
2
1 − cos2 x
2
=
sin2 x
= RHS
4 (i)

2
tan2 x
cos x
cos x
LHS =
−
1 − cos x 1  cos x
cos x  1  cos x  − cos x 1 − cos x
=
 1 − cos x   1 cos x
=

θ
cos x
cos x
−
≡
1 − cos x 1  cos x
=
(iv)
−

θ

=
cos x
1 − sin x
= RHS
LHS =


1 − sinθ
sinθ
1
sinθ
=
−
sinθ
sinθ
1
=
− 1
sinθ
= RHS
1
cos x
≡
cos x
1 − sin x
sin x
1
LHS =

cos x cos x
sin x  1
=
cos x
sin x  1 cos x
=
×
cos x
cos x
sin x  1 cos x

=
1 − sin2 x
 sin x  1 cos x
=
 1 − sin x  1 sin x
=
(iii)
θ

–1
x = 14.5 ° or 165.5 ° (1 d.p.)
27
P1
y 1
(ii)
x = 90
° , 270° , 60° , 300°
7 (i)
(ii)
x = 120
° , − 120
°
(iii)
x = 180°
8 (i)
–1
4 cos2 x  7sin x−
1x
–0.7

2

4 1 − sin x 
2
4 − 4 sin x
2
Answers
− 4 sin x
7sin x − 2=
7sin x  2=
4 sin x − 7sin x−
 4 sin x
x =134.4 °or 225.6 °
1
4
0
0
0
2= 0
 1  sin x−
sin x = −
y 1
(iii)
7sin x − 2=
2
–1
2= 0
2 = 0
or sin x = 2
Since sin x is never greater than 1, sin x = −
(ii)
–1
θ = 5.5° , 214.5°
1x
Exercise 7.4
1
–1
x =41.2 °or 221.2 °
y 1
(iv)
–1
Degrees
Radians
120°
2π π
3
36°
π
5
330°
11π π
6
150°
5π
6
240°
4π π
3
54°
3π π
10
225°
5π π
4
315°
7π
4
1 x
–2
5
–1
x =203.6 °or 336.4 °
6 (i)
 cos1 θ 
tanθ

2
≡
LHS =
=
=
=
=
1  sinθ
1 − sinθ
sinθ
 cos1 θ  cos
θ
 1 +cossinθ θ 
2
cos2θ
 1 + sinθ  2
 1−
 1−
sin2 θ
θ = 203.6° or 336.4°

 1 + sinθ  2
sinθ   1
1 + sinθ
1 − sinθ
= RHS
28
2
 1 + sinθ  2
=
(ii)
2
sinθ
Degrees
Radians
12°
0.209
10.9°
0.190
145°
2.53
169.0°
2.95
235°
4.10
288.8°
5.04
342.5°
5.98
78.5°
1.37
1
4
3
Angle
Other angle that has
Exact value
the same value
(ii)
sin
π
4
sin
3π
4
1
2
=
2
2
cos
π
6
cos
11π
6
3
2
π
tan
6
7π
tan
6
cos
5π
4
−
1
2
sin
5π
3
sin
4π
3
−
3
2
tan
5π
4
=
π
4=
=

−
−
θ − sinθ =
θ = sinθ 
7 (i)
1
2
cos x =
1
2
(ii)
4
3
tan x =
(v)
1
or tan x=
x = 0.927,−
1
×
2
AC = 8 tanθ
1
2
32
θ
Perimeter =
8π
 8
3
× 82× θ

8 3 ≈ 30.2 cm  3 s.f.
Since triangle OCD is equilateral, angle
OCD = π
3
π = 2π
So angle ACD = π −
3
3
Perimeter = AD  BD  AB
32
π
3
=
− 1
8 tan θ × 8−
= 32tan θ −
= 8×
π 5π
, 0, 2π
,
3 3
x =
AC
⇒
8
tan θ =
= 32  tan θ − θ
8 (i)
or cos x=
2
2
θ = 2.55
=
(ii)
(iv)
36
36
sector OAB
sin x = 3.30 or− 0.303
x = − 0.308 or − 2.83
(iii)
= 36
Shaded area = Area of triangle OAC − area of
1
π 5π
,
6 6
x =
sinθ
×6 × 6
6 Total length is: 49.1cm  3 s.f.

3
or sin x =
2
sin x = −
(ii)
1
×
2
× θ−
18  θ − sinθ  =
tan x = 0 or tan x= 2
− 4.25
=
x = 0,π  , 2π , 1.11,
  
−  =

4 (i)
1 2
6
2
18θ − 18sinθ =
(ii)
P1
Area of shaded region = 15.0 cm2  3 s.f.
Answers
3π
4
Perimeter = 18.2 cm (3 s.f.)
5 (i)
1
3
cos
tan
4 (i)
2
π
3

8  16×
1
π
3
 8 cm
(iii) Shaded area =
64
π
3
− 16 3
2.21, − 0.785, 2.36
Exercise 7.6
sin x =
(vi)
1
2
or sin x=
− 5
1 (i)
π 5π
,
6 6
x =
y
3
2
1
0
Exercise 7.5
1 (i)
180°
270°
360° x
P = 22.4 cm
–2
A = 29.4 cm2
–3
π
2
π
3π
2
2π x
P = 75.2 cm  3 s.f.
(ii)
A = 314 cm2  3 s.f.
2 θ =
90°
–1
12
=
7
(ii)
1
1.71 = 98.2
°
3 Perimeter:
10 π
 10 3=
3
Area: 17.1cm2
y
2
0
–1
27.8 cm  3 s.f.
–2
–3
–4
–5
29
P1
(iii)
(ii)
y
3
2
1
0
90°
180°
270°
0
–1
–2
–3
–4
–5
360° x
–1
–2
Answers
–3
(iv)
y
5
T
5
4
3
2
1
1
t =0.7, 1.3, 2, 2.7, 3.3 and 4 (1 d.p.)
3
t = 12.40 pm, 1.20 pm, 2 pm, 2.40 pm, 3.20 pm,
2
1
4 pm
π
2
0
–1
π
3π
2
2π x
9 (i)
–2
y
4
3
2
1
(ii)
0
90°
180°
270°
10 (i)
–2
y
8
7
6
5
4
3
2
1
y
4
3
2
1
0
−1
−2
−3
−4
π
2
π
f  x  2
 π
1
3
(ii)
f
(iii)
f(x)
5
= 2−
3
π
2
–5
π
2
π
3π
2
2π x
–10
2 A =4, B =1, C =− 2
–15
3 A =5, B =2, C =− 1
4 A =2, B =3, C =− 2
5 A =− 2, B = 1 , C =3
(iv)
2
6 A =2, B =3, C =2
30
f−
1
 x
b = 5.
a = 3.
7 A =2, B =− 1
8 (i) period =
11 (i)
2
3
2π x
π x
π+ x
1
= +
= 1 + x = 1 + 0.318 x
π π
π
π
0
0
–1
3π
2
The line we need to draw is:
y =
360° x
–1
(vi)
4 t
3
(iii) From the graph, the highest points occur at
4
(v)
2
= 2tan−
1
 2 −3 x
π x
(ii)
2 (i) H (m)
π
2
π
cos  − x =
k
x = − k
(iii)
sin x =
1− k 2
(iv)
cos  π 
x = − k
(v)
tan x =
1 − k2
k
(iii)
k
tan  π −
x = k
(v)
tan

1
π
2
1
k
(ii)
x =
π
5π
or π or
3
3
(iii)
x = tan−
(iv)
x = 0.970 or 2.74 or 4.11or 5.88  3 s.f.
− 1
1
2
16 32
,
3
3
2
3
Since t is hours after 9.30 am, refloating can
π
π
5
or
4
4
Period = 40 seconds
occur between 5
1
3
and 10
2
3
hours after 9.30 am,
i.e. between 2.50 pm and 8.10 pm.
4
Sun
Moon
r
Maximum height = 107 m
q
Stretch and challenge
3sin3θ  3=
2cos2 θ3
3sin3θ  3=
2 1 − sin2 θ3
3sin3θ  3=
2sin θ3

−2
θ 
2sin2 3θ  3sin3
 2sin3θ
sin3θ = −
θ 
 1  sin3
1
2
8
1
2
(iii) t = 23 seconds
1
7
 
1
3
π
4π
5
or
3
3
 1 =
6
3cos π t
8
t = 5 , 10
x =
1
5
 = − 1.5
 =−
 = cos  − 
 = 23π , 4π 3
t =
14 (i)
(ii)




1  k2

4
2π
= 4 hours
π
2
10.5 = 9−
3cos π t
8
cos π t
8
π
t
8
π
t
8
k
x =

3
t (hours)
Period =
3
1  k2
x = −
2
(iii) 1 am
1
cos x = −
sin  π 
1
π
H = 2sin t  3
2
(ii)
tan  − x  =
(iv)
15 (i)
2π x
H = 2sin π t + 3
2
A m p li t u d e = 2
Period = 2π : π = 4
2
0
cos  π −
(ii)
3π
2
8
7
6
5
4
3
2
1
–1
–2
–3
–4
(ii)
13 (i)
P1
Answers
12 (i)
θ = − 170° ,− 150° − , 130
° − , °50
− , ° −30 , 10° ,70° ,
90° ,110°
y
8
7
6
5
4
3
2
1
0
–1
–2
–3
–4
–5

2
1= 0
1 = 0
or sin3θ = −
r
1
(i)
A= 2
1
π
2
r 2=

1 2
r 2θ
2
−
1 2
r sin2
θ
2

r 2 2θ − r 2 sin2
θ
π = 4θ − 2sin2
θ
θ − π
2sin2θ = 4
3θ = − 30° ,− 150° − , 390
° − , 510° ,210° ,330
° , − 90
° ,
270° ,− 450°
31
P1
(ii)
4 LHS =
r
=
=
=
q
r
2
=
Answers
r
cos θ = 2 ⇒ cos θ =
r
A= 2
(iii)

1
2
1 2
r 2θ
2
= 2r 2θ−
π
= 2r 2
3
2π 2
r
=
3
 2π
= 
−
 3
 2π
−

 3
% covered =
⇒ θ = 60° =
−
π
3

1 2
r sin2θ
2
− r 2 sin
−
=
2π
3
3 2
r
2
3 2
 r
2 
3 2
 r
2 


Exam focus
The equation is y = 2sin 3 x; i.e. a = 2, b = 3.
c = 2.
(ii)
=
r 2 sin2θ
× 100%
π r2
3
2π
−
3
2 × 100%
=
π
= 39.1%
1 (i)
5 LHS =
=
32

cos2 x  sin2 x
sin x  1 − cos x 
2 − 2cos x
sin x  1 − cos x 
2  1 − cos x 
sin x  1 − cos x 
2
sin x
= RHS
6
x = 15° or 165° or 195
° or 345°
7
x = −
8
sin x = 0 or sin x=
π
x = 0 or x= −
2
7π
11π
π
5
π
or −
or
or
12
12
12
12
9 cos x =
1
2
− 1
or− 2
π
π
,−
3
3
The equation is y = 2 − 4 cos 2x.
11 Perimeter of shaded region
(iii)
 π

tan  − x  =
 2

=
2cos x
10 Area of segment = 5.80 cm2  3 s.f.
cos x =
=
 1−
a = 2.
(ii)
=
cos x   1− cos x  sin2 x
sin x  1 − cos x 
 1−
x =
sin x = sin (π − x)
3 LHS =
1
sin x
cos x
sin x
cos x

cos x
sin x
2
sin x  cos2 x
sin x cos x
1
sin x cos x
RHS
=
b = 4.
2 (i)
sin x

cos x
1−
sin2 θ
1
cos2θ
cos2 θ  sin2θ
cos2 θ
1
cos2 θ
RHS
k.
k2
=
10 π
 10 3
3
12 (i)
1 − k2
k
(ii)
θ =
DE
10
5
3
6.60 (3 s.f.)
(iii) Shaded area = 9.63 cm 2 (3 s.f.)
8 Vectors
1
c
2
Exercise 8.1
 −5
+ 2a = 
 0





P1
1 (i)
c
b
Answers
a
(ii)
2 (i)
 −6 
2a = 

 2 
(ii)
 −5 
−b = 

 0 
4i −5 j
−i +2 j
(iii) g =− 6j
h =4 i −5 j
 1 
c=

2
 −2 
1
g +h =− 6j +4 i −5 j =4 i −11 j
l = 3.61  3 s.f.
3 (i)
l = 2.24  3 s.f.
(ii)
1c
2
→
AB = 6.08  3 s.f.
(iii)
–b
4
y
2a
(iii)
C
2 
a+b =  
 1 
A
x
B
(i)
(ii)
 1 
b − 2c = 

8 
(iii)
(iv)
5 (i)
(ii)
(iii)
 − 3
→
OA = 

 2
→
 − 4
BC = 

 7
→
 − 3
CA = 

 − 2
D is (− 15, 23)
→
OB = 2 m
→
1
n
2
5
m
2
5
n
2
BD = 2 m
3
n
2
→
OE =
33
P1
(iv)
6 (i)
(ii)
→
EC =
1
m
2
2
(ii)
 −0.8 
Unit vector = 

 0.6 


Unit vector = 
 −

4
17
1
17
or





15

2


8

 − 18 


(ii)

3


 − 18 
 13 


Answers
→
8 AB = − i  5 j − 4 k
− 3i  4 k
(v)
5i − 6 j − 4 k
(vi)
2i  6 j
→
→
→
(ii)
12 (i)
(ii)
0
− 6
12




34
 1d.p.
a=
(iii)
→
7 (i)
they are parallel
→
QR is twice the length of PQ
9
19

8
CD = 
4


 − 21 
→
 1
EC =  − 3 


 7
AB = 10
Unit vector = −
→
AF = 4 i  10 j − 3 k
→
BF = − 4i  10 j  3 k
θ = 53.1°  1d.p.
→
OM = 14i
20k
C ′ M = 30i−
20k
→
→
→
MB′ = − 14i  16 j
S is (1, 2, − 3).
→
2
5
→
CB′ = 12i
(ii)
20k
16 j  40k
θ = 47.1°  1d.p.
Stretch and challenge
 a 


 − 1 


1  = 0⇒
 3 


1  b  .
13 G is (8, − 13, − 6)
14 (i)
θ = 147.3ο
(iii)
QR goes in the opposite direction to PQ.
(iv)
 1d.p.
θ = 121.9°  1d.p.
(ii)
→
→




θ = 97.9°  1d.p.
6 (i)
(iii) Since QR = − 2PQ,
→
θ = 169.7ο
(ii)
5 c = −
 0
PQ =  3 


 − 6

QR = 


0.521
− 0.047
0.852
4 θ = 77.5°  1d.p.
10 CD = 9.64 (3 s.f.)
11 (i)
1 (i)
3 (i)
(iii) 5i − 6 j  4 k
(iv)
12k
2 θ = 130.2°  1d.p.
5i  6 j  4 k
(ii)
 5i −
Exercise 8.2
(ii)
3i  4 k
9 (i)
1
13

The unit vector is 


1
or
 4i −− j
17
7 (i)

5 




13
5




=
Unit vector is 1 
0 
0 

13 


12 
 − 12 
 − 13 


− 3n
 c 


1
i
3
−
2
3
j
2
k
3
 a   2
 

 b  . 0
 c   5
 



 = 0⇒


− a
b  3c=
2a  5c=
0
0
(1)
(2)
0
(1) × 2 (3)
2b  11 c = 0
(2)  (3)
−2 a  2 b  6 c
There are infinitely many solutions to this system of
two equations with three unknowns.
Choosing the whole number solutions, b = 11, c = − 2,
 5 


a = 5, the vector is  11  (or any scalar multiple).
 −

 2 
32 =
a2  b2
9 = 16
a2  9
b2 = 16
b=± 4
 a   4
 

 b  . a
 3   3
 



 = 0⇒


4a  ab
1 So the unit vector in the direction of
→
OR is
 6
0
 8
1 
10 
 0.6 




=
0 


 0.8 




2 p= − 6
3 θ = 51.2°  1d.p.
a2  32
42
P1
Answers
a2  b2
2 p = q⇒
Exam focus
9= 0
9
8
If b = 4, 4 a  4 a  9 = 0 ⇒ a = −
If b = − 4, there are no solutions for a.
9
8
So a = −
1202
3 vb =
v a = 140
−
60 
2
402 = 140 m/s
35 = 175 m/s
va = k vb =
 120k
va =
and b = 4
 120k 
 120 




k  − 60  =  − 60k 
 40k 
 40 




2

−
60k 
2
 40k
2
= 175
19600k 2 = 1752
va =
 120
− 60
 40

5

4
k2 =
25
16
k =
5
4
 150 




 =  − 75  = 150i−
 50 




75 j + 50k
35
P1
Past examination questions
(ii)
1 Algebra
1 (i)
(ii)
16 − ( x − 4)2
a = 16, b = −4
Answers
5 (i)
y=

a=
3
, b=
2
3
2
x−

2
6 (i)
−
9
4
1
2
11
2
x
7 (i)
(ii)
or x − 2 y
11 = 0
8 (i)
(ii)
(ii)
C is  13, 12
(iii)
Perimeter = 2 20
2 (i)
2 180 ≈ 35.8
1
2
x
(ii)
(ii)
C is  10, 9
80 x 4  …
Coefficient of x 4 term is 272
(a) d = 3
(a) $6515.58 or $6516
(b) $56 827 or $56 800
Points of intersection are (1.5, 8) and (4, 3)
−
96<
k<
or k <
4 (i)
32  80 x 2
(c) 570
y = − 2 x  29
3 (i)
The coefficient of the x 2 term is 160.
(b) 57
4
Equation of CD is:
(ii)
32  80u 80u 2  …
9 a=5
10 (i)
Equation of BC is:
y=
10836
(b) S∞ = 432
Equation of BC is
y=
$369 000
(ii) (a) x = 96
9
4
2 Co-ordinate geometry
1 (i)
S32 = 3280
(iii) $14300
x < −1or x > 4
(ii)
S10 = 239  3 s.f.
(ii) $3140 000
−2  x  10
2 (i)
4 (i)
4 Functions
96
1 (i)
96
x = 71
2
Equation of CD is:
3
2
y=−
x  24 or 3x
2 y − 48=
0
(ii)
f(x)
4
D is  10, 9
3
5 m < − 10 or m > 2
2
(ii)
6 (i)
M is  5, 2
1
D is  7, − 2
(ii)
7 (i)
(ii)
–4
6:4 = 3:2
–3
–2
k = 8.5
(ii)
2 (i)
(ii)
a=
3
4 x
–3
27
S∞ = 81
$61.50
S∞ = 18
3 Coefficient of x is 1080
36
2
–2
3 Sequences and series
1 (i)
1
–1
Points are  6, 1 and  2, 3
18
=
2
3
0
–1
–4
(iii)
2 (i)
(ii)
3 (i)
(ii)
9
2
x=−
a = 2, b=
x=
or 2
− 3
9
4
The domain of g− 1(x) is x  16 and range is y  4.
g− 1  x =
16−
x 4
4 (i)
(ii)
x < −3 and x > 5
8 The domain of f − 1 is x  3
Range is f  x   − 1 .
9 (i) Range of f is 0  f  4.
(ii) f ( x)
No inverse as f(x) is not one-to-one.
(iii)
P1
gf  x = g  x 2− 2 x = 2  x 2− 2 x  3
= 2 x 2− 4 x  3
2 x 2 − 4 x 3 = 0 has b 2− 4ac = − 8
so has no real solutions.
(iv)
Answers
g(x)
6
5
4
3
2
1
0
–6 –5 –4 –3 –2 –1–10
–2
–3
–4
–5
–6
5 (i)
(ii)

2 x for 0  x  2
x
2
− 2 for 2< x  4

(iii) f -1  x = 
10 (i)
y
5
1
(i)
V =
2 (i)
0
(192r − r )
3
90°
180°
h = 4−
270°
(ii)
(iii) Largest value is when A = 90 ° .
A = 2r×
1
π
2
h

9
4
f does not have an inverse as the function is
not one-to-one.
(iii) x = 25

1
π
2
r2
= 8r−
2r 2 − π r 2 
1
π
2
r2
= 8r−
2r 2 −
1
π
2
r2
(iii)
r = 1.12cm  3 s.f.
(iv)
r = 1.12 cm is a maximum
3 The point is
The domain of f − 1 is 0 < x  2
r2
r− π r 
2
= 2r 4−
 3− x
g− 1  x = sin− 1 
 2 
f( x )  −
π
r
2
r −
360° x
–1

1
,
2

51 .
4
4 y = − 22
f(x)
6
(ii)
1
π
2
(iii) r = 8 cm is a maximum
1
7 (i)
6x
(ii) r = 8 cm
2
(ii)
b= 8
5 Differentiation
3
6 (i)
a= − 2
(ii) fg  x = 22−
Range is 1  f(x)  5.
4
(iv)
x
1 2 3 4 5 6 x
3
5 Angle between the two lines is 8.2° (1 d.p.).
5
6 f ′  x = −
4
12
 2x
 3
2
12
(2 x  3)2
3
Since (2 x  3)>2 0 for all values of x, then −
2
is always negative.
1
Since f (x) is always negative, the curve is always
decreasing.
–1
0
1
2
3
4
5
6x
–1
(iii) x = 1
37
P1
7 f ′  x = 9  3 x
2
2
(ii)
V =
Since (3 x  2)2>
0 for all values of x, f (x) > 0 for all
Answers
8 (i)
x
2
3
24 x
2
7
3
x

y =
(iii)
− 0.018 units/second
9
x= 4
or 2x − 3y 7 = 0
(ii)
3 (i)

 2

(ii)
3
7 (i)
(ii)
5
6
(iv)
0.025 units/second
8 (i)
8
15
Equation of curve is y = x 3−
x <
1
3
y =
8
= 8  3 x
3x  2
or x>
3
8
dy
=
dx
2
2 x 2  x
DC =
4 2−
3
5
When x = 1,
2  c⇒
2
x  9⇒


2  
x  2 y=
3− 6
(iii)
V =
2=
1
∫ 0 5 −
1
7
4
= 36π 

x =
14
=
3
=
42
3
9 (i)
22
3
1
×
2
22
3
× 1=
2
dx
2 x
− 2
dx
1
1

×
− 2
0
− 1

1
= 36π  
  2  5 − 2×
= 36π  1 − 1 
 6 10 
7
4
7
⇒
4
 5 − 62x
∫ 0π

1
1
= 36π 

 2  5 − 2 x  0
2
c =
4
3
0.015 units/second
3
8
=−
dy
=
dx
(ii)
24
c ⇒ = 1 − 3×
−
Area of triangle BDC =
38
2y = −
 5 − 2x


= 36π 
1
−

 3 x  2
24
2 2
x 
9
⇒
2
x 
A = 33.75
Co-ordinates of C is when y = 0
3
8
1
3  3 × 0
Maximum point is  − 1, 32
− 1
3=−
8
0 = −
1
2
= 36π
3×
x
 
 − −
2  
k = 27
Equation of tangent at  2,1 is:
y =−
0
Equation of curve is y = 3 4 x−
1
dy
− 2
= − 8  3 x  2 ×
dx
y = mx
×
(iii) The function is decreasing for − 1 < x < 3
dy
=
dx
At x = 2,
1
3
Equation of normal at P is:
y = −
3
Area = 2
5 (i)
6 (i)
3
Area = 9 1
(iii)
(ii)
2 x
3
y = − 2 x  12
At x = 1=
2
− 1
= 64π  1 
 8
= 8π
The y intercept is 0,2 1
(ii)
4 (i)
 1
Equation is y =
x = 1,
2

1
= 64π   −
  3  3 × 2
= 64π  − 1  1 
 24 6 
6 Integration
2 (i)
∫ 0  3x  2 − 2dx


1
= 64π  −

 3  3 x  2  0
x = 4 is a minimum point
(ii)
dx
2
(ii)
1 (i)
2
 3x  2


= 64π 
− 1

2
 3
2
= 64π
values of x, hence the function is always increasing.
dy
= −
dx
 3x8 2
∫ 0π
4
3
 
1
 − 
1   2  5 − 2×
1
15 
12
π
5
Shaded area = 2×
3
π
2
1−
14
=
9
4
9
= 4.71  3 s.f.
(ii)
V=
(iii)
Anglebetweenthe tangents = 19.4
°


0  
9
k=5
Using Pythagoras′ theorem,
Equation is y = 5 x − x 2 + 5
AB2 = AC 2
a = 1, b = 2
 3 
= 
l
 2 
2
(ii)
A = 2.25
=
l2
(iii)
c =
3 2
l 
4
=
7 2
l
4
10 (i)
(ii)
11 (i)
8 =
6
2 ≈ 1.41
b = −3
a=5
(ii)
(ii)
x = 0.64, 2.50  2 d.p.
(iii)
y
3
π
2
π
3π
2
5 (i)
–3
–4
–5
Area of sector = 68.5cm2  3 s.f.
–7

− 2
θ =
(iii)
AB2 = 82
82 − 2 × 8×
×8
cos
π
3
AB2 = 64
AB = 8 cm
or
–8
–9
Cosine rule:
2
2
2
cos AOB = 20  20 − 32
2 × 20× 20
cos AOB = − 0.28
Since angle AOB = 60 ° , and AO = BO ,
then angle OAB = angle OBA = 60 °
AOB = 1.855  3 s.f.
So triangle OAB is equilateral so AB = 8 cm
or
BC 2 = 82
sin


1
2

16
20
AOB =

1
AOB
2
= sin−
1 16
BC =
20
=
A = 371 cm2
(iii)
Area of cross-section = 502cm2  3 s.f.
3tanθ = 2cosθ
3sinθ
= 2cos θ
cos θ
3sinθ = 2cos2θ

3sinθ = 2−
6 (i)
cos
2π
3
8 3  16

8 3 cm
4 sin4 θ  5=
7cos2θ
4 sin4 θ  5=
7 1 − sin2θ
4 sin θ  5=
7 − 7sin θ


2
4 sin θ  7sin2θ −
2= 0
With x = sin θ we have 4 x 2
2
sin θ
2

(ii)
2sin θ
2
7 (i)
2= 0
(ii)
θ = 30° or 150°
BC = 2×

24
4
2sin θ  3sinθ −
sin30° =
×8
8 3
4
2
cos 30° =
192=
Perimeter = 8
AOB = 1.855  3 s.f.
3sinθ = 2 1−
82 − 2 × 8×
BC 2 = 192
(ii)
4 (i)
1
π
3
(ii)
–6
(ii)
2
3
 2 
− 30
°
x = tan− 1 
 3 
2π x
–2
3 (i)
l
 2 
x  30
°  = tan− 1 
 3 

0
–1
7
3
l
2
°  =
tan  x  30
1
1
l
2
In triangle ABC
tan  x  30°  =
2
2 (i)
7
l =
2
Answers
1 (i)
 l2
7 l2=
4
AB =
7 Trigonometry
P1
BC 2
AC ⇒ AC=
l
DC ⇒ DC=
l
DC = l
cos 30° ×
sin30° ×
0
θ = 30° , 150° , 210° , 330
°
Shaded area = 21.5cm2  3 s.f.
Perimeter = 20.6cm  3 s.f.
3
l
2
l =
l =
7 x − 2=
1
l
2
39
P1
8 (i) y
(ii)
y = 2s i n x
2
14 (i)
1
(ii)
π
0
π
4
−1
3 π
4
2
π x
y= co s 2 x
Answers
−2
(ii)
15 (i)
(ii)
θ = 60° or 300°
BD = 6.66 cm
Shaded area = 10.3cm2  or 9π − 18
2 − 5cos2 x
a = 2, b= − 5
Greatest value is when cos2 x = 0, f  x =
Least value is when cos x = 1, f  x =
2
Two solutions
B
(iii) x = 0.685 or 2.46
16 x = 113.6° or 70.5°
O
q
15 cm
17 (i)
8 cm
A
15 cm
 1
1 
LHS = 
−

tanθ 
 sinθ
 
(ii)
Perimeter = 47.3cm  3 s.f.
(iii)
Shaded area = 50.8 cm2  3 s.f.
11 (i)
cos2 x =
21
25
tan2 x =
4
21
1 − cos θ 2
+
=(

) sinθ


3 − 3cos2 x=
2
0 = 3cos x
(ii)
13 (i)
(ii)
18 (i)
2
(1 − cos θ )
2
(1 − cos θ )(1 − cosθ )
(1 − cos θ )(1+ cosθ )
θ = 64.6° or 295.4°
π
3
= 6×
=3 3
2
3
Distance from E to AB is 10 sin q
Distance from D to AB is 6 sin
10 sinθ = 3 3
8 cos x
sinθ =
8 cos x − 3
Shaded area = 15.3cm2  3 s.f.
2tan θ cosθ = 3
 sin2 θ 
2
 × cos θ =
 cos2 θ 
(ii)
1 (i)
3
3cosθ
2 − 2cos2 θ =
3cosθ
2cos θ  3cosθ −
2= 0
)
 4
→ →
→
MO = MB + BO = 4i − 6k =  0

 −6
→
2sin2 θ
= 3
cos θ
2sin2 θ = 3cosθ

P = 16.20 cm (4 s.f.
8 Vectors
2
2 1 − cos2 θ =
3 3
10
3 3 
θ = sin−1 

 10 
3= 0
0 = 2cos2 θ 
40
(1 − cos θ )
Equating these distances gives:
8 cos x
Perimeter of shaded region = 25.9cm  3 s.f.
2
2
1 − cos θ
1 + cos θ
= RHS
x = 70.5
° or 289.5°

)
=
3cos x  8 cos x−
12 (i)
=
=
2
(ii)
= (1 − cos θ
sin2 θ
3sin x tan x = 8
sin x
3sin x ×
= 8
cos x
3sin2 x
= 8
cos x
3sin2 x = 8 cos x
3 1 − cos2 x =
2
2
 1
cos θ 
= (
=
+
sinθ 
)sinθ
T
θ = tan− 1 15 = 1.08  3 s.f.
tanθ = 15⇒
8
8
So AOB = 2× 1.08 = 2.16  3 s.f.
(ii)
− 3
− 3  f  x  2
9 (i)
10 (i)
2
→
→
→
MC ′ = MO + OC + CC ′
= (4i − 6k )+ 4 j + 12k
 4 
= 4i + 4 j + 6k =  4 


 6 
3cosθ − 2
(ii)
θ = 109.7°(1d.p. )




2 (i)
→
BA = a − b = (3i + 2k )− (2i − 2j + 5k )
4 (i)
(ii)
 −4 
→


The vector must be 5 × AC = 5 2  =


 4
(iii)
p=
= i + 2j − 3k
→
BC = c − b = (2j + 7k )− (2i − 2j + 5k )
= −2i + 4j + 2k
 1   −2 
→ →
 


BA ⋅⋅BC =  2  i  4  = 1 × −2 + 2 × 4 + −3 × 2


 2
 
 −3 
= 0
(ii)
→
AD = d − a = (−2i + 10 j + 7k )− (3i + 2k )
= −5i + 10j + 5k
= 5 (−i + 2j + k )
→
Since both vectors are multiples of
−i + 2j + k they are parallel
→
→
BC : AD = 2 : 5

1
The unit vector is 
6


(ii)
p = 10
(iii)
q = −7 or 5


2  
 
−4  =

4 
 


(ii)
P1
1
2


 4  
 
1
The unit vector is  2 =
6
 
 4  


2
3
1
3
2
3



or 2 i + 1 j + 2 k
 3 3
3



Acute angle is 180° − 105.8° = 74.2°
(iii) Perimeter = 15.4
BC = −2i + 4j + 2k = 2 (−i + 2j + k )
3 (i)
5 (i)
 −20 
 10 




 20 
Answers
→ →
Since BA i BC = 0 the vectors are perpendicular.
θ = 160.5°
1
3
−
2
3
2
3








6 (i)
θ = 66.6°
(ii)
 3 + 2p 


 −2 + p  or (3 + 2p )i + (−2 + p )j + (4 − 3p )k
 4 − 3p 


(iii)
p=
7 (i)
8
14
=
4
7
→
PQ = 3i + 6 j − 3k
→
RQ = −3i + 8 j + 3k
(ii)
θ = 63.2°
41