pure-mathematics-2-3-pdf

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4 a 2i + 6j − 4k = 4 + 36 + 16 = 56 = 2 14 BOA = cos−1 1 54  = 88.7° correct to
34
b OA = 1 + 4 + 25 = 30,
1 decimal place.
OB = 9 + 16 + 1 = 26 and 30 + 26 = 56
c 1 × 30 × 26 = 195 3 6a + (−2)(4) + (5)(−2) = 0 a = 3

2 4 a 5k2 − 3(k + 2) − (7k + 9) = 0
5 16 + (q − 2)2 = 22 and so q = 2 ± 6. 5k2 − 10k − 15 = 0 k2 − 2k − 3 = 0
(k + 1)(k − 3) = 0 → k = −1 or k = 3
6 a 25 cm

b ON = 9.6i + 20j + 4.2k  10   2

Review 7 4 + 25 + a2 = 1 + (1 + a)2 + (−3)2 so a = 9. b OP =  −3  , OQ =  4 
 23   −1 
8 a = λOQ and using the y-component, λ = 1.
OP 4 OP ⋅ OQ = 10(2) + (−3)(4) + (23)(−1) = −15

Hence, −6k = 1 (2k + 13), k = − 1 and OP = 102 + (−3)2 + 232 = 638,
42
OQ = 22 + 42 + (−1)2 = 21
checking 8(1 + k) = 1 (−32k) gives k = − 1. → θ = cos−1 −15 
42 638 = 97.4°

 3 21
 
b OP =  −2  = 3i − 2j + 4k and
4
5 NP = 2j + 3k and MP = −3i − 2j + k and so
 12 
  NP ⋅ MP = 2(−2) + 3(1) = −1, NP = 13,
OQ =  −8  = 12i − 8j + 16.k MP = 14
16
339

c PQ = 9i − 6j + 12k and NPM = cos−1 −1  = 94.2509 … = 94.3°
13 14
Review
PQ = 92 + (−6)2 + 122 = 3 29 6 a ⋅ j = (4)(0) + (−8)(1) + (1)(0) = −8

 0 a = 42 + (−8)2 + 12 = 81 = 9, j = 1

9 Home is the null displacement  0  . Total θ = cos−1  −8  = 152.733 … = 152.7° correct to
 13   0   9 

vector sum is   so to get home the 1 decimal place.
 
8 7 a ⋅ b is a scalar and the dot product is a product of
0 two vectors.

 −13 
 
displacement is  −8  . The distance home is 8 a OM = 2i + 4j + 4k, NG = −4i + 3j + 4k
0

10 169 + 64 = 153 cm , correct to the nearest cm. b OM ⋅ NG = 2(−4) + 4(3) + 4(4) = 20,

OM = 6, NG = 41, cos−1  20  = 58.6°
 6 41 
Exercise 9C correct to 1 decimal place.

Review 1 a a ⋅ b = 0, a ⊥ b 9 AM = −77i + 30j + 36k and

b θ = cos−1 13 12 19  = 81.4° DB = 772 + 362 = 85 so
2
= + 1 = + 1
c e ⋅ f = 0, e ⊥ f AN 60 j 5 BD 60 j 5 (−77i + 36k)

2 OB ⋅ OA = (−5)(1) + (0)(7) + (3)(2) = 1 = − 77 i + 60j + 36 k
OA = 34, OB = 54 55

AM ⋅ AN = −77  − 77  + 30(60) + 36  36  = 3245
 5  5

AM = 25 13

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AN = 3889  0   1  t 
     
MAN = cos−1 25 3245  = 54.7° correct to OA + tAB =  4  + t  −3  =  4 − 3t  and so
13 3889 −2 8 −2 + 8t

1 decimal place. x=t

 −3  y = 4 − 3t

10 a AN =  1.5  , AN = 3 14 z = −2 + 8t
 4.5  2
b Oxy plane → z = 0, −2 + 8t = 0, t = 1
cos−1  4.5  4
 3 × 1.5 14 
= 74.5° correct to x 1, y 4− 3 13 →  1 , 13 , 0 
4 4 4  4 4 
→ = = =

Review 1 decimal place. 6 a r = (µ + 4)i + (µ − 7)j + (3µ)k

 −3  → r = 4i − 7j + µ(i + j + 3k) → direction is
 
b MN =  0  i + j + 3k which is not a scalar multiple of
4.5 6i + j + 3k, so the lines are not parallel.
 3
 
c PN ⋅ MN = 0, PN =  1.5  , b cos−1  16  = 44.7° correct to 1 decimal
 4.5 − 11 46
p   0

3(−3) + 4.5(4.5 − p) = 0, p = 2.5, OP =   place.
 
0  x   5   4
2.5  y   −3   
  =   +  
7 a  z  2 t −1
−3
Exercise 9D

1 a r = −j + 5k + λ(2i + 6j − k)  1  4

340 b r = λ(7i − j − k) b BA ⋅ dL2 =  4  ⋅  −1 
 0   −3 
c r = 7i + 2j − 3k + λ(3i − 4k)

Review2 a x = 2λ b x = 7λ = 1(4) + (4)(−1) + (0)(−3) = 0

y = −1 + 6λ y = −λ  −3   0   −3 

z =5−λ z = −λ 8 a AB =  −2  so OB + tAB =  −1  +t  −2 
 −3   2   −3 
c x = 7 + 3λ

y=2  –3t + 3   –3t + 2 

z = −3 − 4λ b ON =  –2t + 1  , and then CN =  –2t − 1 
 –3t + 5   –3t + 2 
3 Direction of line through 9i + 2j – 5k and
i + 7j + k is, for example, 8i – 5j – 6k. The

direction of this line is a scalar multiple of Since CN is perpendicular to L,
16i − 10j − 12k and so the lines are parallel.
( −3t + 2 ) ( −3 ) + ( −2t − 1) ( −2 ) + ( −3t + 2 ) ( −3 ) = 0

4 a x =2+t b x = 2t 22t − 10 = 0, so t = 5 ,
y = 13 + t y = 10 + 5t 11
z = 1−t z=0
 3  5  −3  1  18 
Review c x = 1 + 2t ON =  1  + 11  −2  = 11  1 
y = −3 + 3t  5   −3   30 
z = 4t

 1
 
5 a AB =  −3  so
8

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Exercise 9E b cos−1  19 41  = 62.3923 … ° = 62.4°
41
1 a Skew
b Parallel correct to 1 decimal place
c Intersecting (8, 5, 15)
d Intersecting (5, − 1, 3) 5 a 8 − 4 + 5p = 0 → p = − 4
5
2 p = −80, P(5, − 3, 16)
b i r = −3i + j + 5k + λ(7i − j − k)

ii Proof

3 a 4i − 3k, − 8i + 4j, 12j + 5k 6 a6
b r = −4i + 6j − 6k + λ(−2i − 14j + 2k)
Review b 55.8°, 72.3°, 51.9° c −5i − j − 5k
d 50.6°
c 13, 4 14, 3 17

4 a r = 3i + 7j + 9k + λ(4i + 4j + 5k)  −2   2 

b Proof 7 a AB ⋅ CB =  −4  ⋅  −4 
 6   −2 
5 a AB = −2i + 2j + 4k
= (−2)(2) + (−4)(−4) + (6)(−2) = 0
b e.g. r = i + 5k + λ(−2i + 2j + 4k)
 −10   −2 
cos−1 2 10      ,
c 6 14 = 56.938 … ° = 56.9° correct b AD =  20  and BC =  4 AD = 5BC .
10 2
to 1 decimal place
The lines AD and BC are parallel.
d (0, 1, 7)
 −1 
  341
 
End-of-chapter review exercise 9 c OE = 12 r = −i + 12 j + 4k + λ(i − 10 j + 3k)
4
Review
1 a cos−1  2 40 62  = 47.2466 … ° = 47.2° 8 a 36.3°correct to 1 decimal place.
14
b Point of intersection is (4, 0, 1)
correct to 1 decimal place

b r = 2i + 3j + 7k + λ(2i − 5j − 13k) c Foot of perpendicular is N(3, 2, 4) and

2 a Not perpendicular as EN = −2i − j = 22 + 12 = 5

OA ⋅ OB = (−2)(1) + (0)(−1) + (6)(4) = 22 ≠ 0  −9.5   −4.5 
   
 3 9 a PQ =  4  PS =  −6 
2.5 −7.5
bi AB =  −1 
 −2 
b R(−5, 0, −1)

ii r = −2i + 6k + λ(3i − j − 2k) c Proof and side length = 7.5 2

c (4, –2, 2) d T (2, 1, 1.5)

3 a AH = −9i + 15j + 12k, NH = 2.5i + 15j + 6k e i e.g. r = v + λ(t − v)

Review b cos−1  274.5  = 37.6695 … ° = 37.7°  5   −3 
 1069 × 15     
2 2 r =  17.5  + λ  −16.5 
−13.5  15 

correct to 1 decimal place ii Proof
iii Right, squared-based pyramid
c e.g. r = 9i + λ(−9i + 15j + 12k)

4 a n=7

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10 a When λ = −1 the position vector given is P. c s3 = sin(t + 5) + C
3
b 185
c 143.0° correct to 1 decimal place d ln V = 2t + c or V = 0
d Foot of perpendicular is (−3, 29, − 4) so e ln y + 1 = ln x + C or y = −1
f − cos y = (1 − x) sin x − cos x + C
perpendicular distance is 67.
2 a y = 5etan x
11 a r = 7i + j + 6k + λ ( 3i + 4j − 5k )
5
b P(4, −3, 11) and r = −5j + 7k + µ(4i + 2j + 4k)
b y2 − 2 y 2 = x3 − x − 54
c PQ = 6 25 3 5
12 a p = 2, q = −1
Review 3 y = ln(ex + 1)

b 7i + 4j + 8k 4 a dy = 5y3x
dx
cos−1 51  = 17.8584 … ° = 17.9°
c 29 99 b y2 = 1
6 − 5x2
correct to 1 decimal place.
1 + 1
5 a 3(2 − x) 3(x + 1)

10 Differential equations x = 2(e3t − 1) ii x → 2
e3t + 2
Prerequisite knowledge bi

1 w = 0.02r3 6 Proof

x 7 x = 2

342 2 y= 60e 4 6− 1 sin t

x e4

1 + 3e 4 8 yey+1 − ey+1 = x2 − ln x + C
2
Review3 a 1 ln 3x − 1 + c
3 tan 1 tan−1 x  π 
2 2  4
b −ln cos x + c 9 y = +

c −xe−x − e−x + c 10 ln 1 − 2y = 1 + B
x2
2 ( )+c
4 7 ln x − 2 − ln 3x + 1 11 x = et + 1 sin 2t
2

5a y 12 a 1 et2 + C
2

b x3 = 1 et2 + 5t2 − 1
32 26

13 y = 2 3
− x2

Review 1 x 14 k = ln 7, 1 ln(3x2 + x4 ) = t ln 7 + ln 2
O 2

b k = 12 15 a ln ln x + C

b y = B ln x where B = eC

Exercise 10A 16 48 hours (to the nearest hour)

1 a y = x4 − 6x + C Exercise 10B
4
1 a dh = −kh2 where k > 0
b A = πr2 + C dt

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b dn = kn where k > 0 9 Approximately 37 or 74%
dt 50

c dv = −kv(v + 1) where k > 0 10 dL = k where k is a positive constant;
dt dt
L = 0.3t + 20
d dV = −kV where k > 0
dt 11 a 7.47 minutes correct to 3 significant figures
b 24 °C
e dC = kC 3 where k > 0
dt 12 a dn = k n where k is a positive constant;
dt
f dv = k where k > 0 2 n = kt + C
dt
t b9
c t = 11 ⇒ n = 12100 and t = 12 ⇒ n = 14 400
Review e2
13 a 0.458 hours correct to 3 significant figures
2 a dx = −k x where k > 0 b Unlimited growth, unrealistic
dt

b t = 600 − 240 x

c 176 seconds correct to 3 significant figures

3 a dA = kA where k is a positive constant
dt

b Proof End-of-chapter review exercise 10

c $334 correct to 3 significant figures − ln 5 1 x 1 
4 4 2 2 
4 a dx = k(100 − x) where k is a positive 1 y = − e2x + e2x − x2e2x
dt
constant t 343

2 a x = 2500e 2

b x = 100 − 75e−kt t
c 86.9 °C
Review 4 + e2

d 100 °C b As t → ∞, x → 2500

5 a Proof 3 y4 = 17e4x − 1

b dh = dV × dh = −36 4 a dy = kx y ; 24 y = 93 − 5x2
dt dt dV πh2 dx

c t = π 1125 − h3  b5
36 3 5 a 10(ln 10 − x − ln 5 − x ) + C

d 85.9 seconds correct to 3 significant figures

6 a dP = kP where k is a positive constant b i Proof
dt
t
b Proof
c 5.42 minutes correct to 3 significant figures ii x = 10(e10 − 1)
7 a dr = 3.92
t
dt r
Review b r = 3 ( 5.88t − 1.568 )2 2e10 − 1
c V = 10.3 cm3 correct to 3 significant figures
8 At about 6.34 pm iii 5 grams

6 y5 = xex − ex + 1025; y = 4.06 correct to
3 significant figures

7 a 1 y4 − 1 + C
2

b 1 y4 − 1 = 2x + ln x + 2 − 2
2

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Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

8 a dh = k(8 − h) where k > 0; 3A
dt

h = 8 −  1 ln 0.8  t
5 
7.5e

b As t → ∞, h → 8

9 y = − 1 ln(2 − e3x ) BC
3
Review 4 a − π , 3π
10 a dx = −kxt 44
dt
b 2.29 radians correct to 3 significant figures
b ln x = − kt2 + C
2 5 a5

c 47.4 seconds correct to 3 significant figures b 0.927 radians correct to 3 significant figures or
53.1° correct to 1 decimal place
11 a 1 ( ln P − ln 5 − P )+C
5 c −2i + 11j

b P = 15e5t
2 + 3e5t

c As t → ∞, P → 5

12 73.4 million correct to 3 significant figures

13 a − 1 cos x2 + c Exercise 11A b 2i
2 3
1 a 12i
344 b y2 = − 1 cos x2 + 1 c (3 10 )i d 13i
22
2 a8
14 y = ln(1 − 3 ln cos x )
3
Review b (9 + 2 2 )i

c 29 d5
6
15 a dx = k(2000 − x)x, t = 15 ln 3x 
dt 2 2000 − x  3 a 8i i7
5 b 2

b 30.3 hours correct to 3 significant figures c 1i
2
y = x2 − 4
16 x2 + 4

17 ln x + 2 = θ − 1 sin 4θ + ln 2 , x = 1.09 correct Exercise 11B b 4+i
28 d − 1 − 13 i
1 a 6 − 5i
to 3 significant figures 55
c 11 + 7i b −2 ± i
d 3±i 6
Review11 Complex numbers 2 a −1 ± ( 2 3 )i f − 5 ± 7i

Prerequisite knowledge c 1 ± 3i 44
2 2 b x = 1, y = 3
1 a 3a − 2bx
b 2a2 − abx − 3b2x2 e − 4 ± 14 i b 41
2 a −2 2 b −1 33
c 2− 3
3 a x = 3, y = −1
c x = 1, y = 2

4 a 5 + 5i

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c 40 − 42i d 28 − 96i e (41, −1.79) f  2, − 2π 
e 5 − 2i f −1 + i g (3, 0.730) 3
g 5−i h 1 − 5i
h (25, −2.86)
5 a Proof 22
b z = 1 − 2 6i i  2k, − π 
6 a z2 + 49 = 0  4 
c z2 − 4z + 13 = 0 55
b z2 − 2z + 26 = 0 4 a 10 ( cos(1.89) + i sin(1.89) )
7 x = 1, y = 3 d z2 + 5z + 14 = 0 10 ( cos(0.322) + i sin(0.322) )
22 10 ( cos (−1.25) + i sin(−1.25) )

8 1 + 2i b AC is a straight line, midpoint O as z1 = z3 .
9 z2 − 10z + 28 = 0
Review 10 3.2 − 2.4i amps Angle AOB = π − tan−1(3) − tan−1  1 = π .
 3 2

Triangles AOB and AOC are isosceles, since

z1 = z2 and z2 = z3 . QED.

5 a 3 + 33 i b 1.91 + 4.62i
2 2

Exercise 11C c − 1 d 32 − 3 2 i
2 2 2
All angles are given in radians correct to 3 significant
figures where rounded. 6 z2 = 73 , arg z2 = 3.02
97

1a Im(z) iπ b 2 e − iπ 345
2 4
C B 7 a 5e 6

1 2 e− 5πi
12
c5
Review –5 –4 –3 –2 –1 O 12345
–1 Re(z) 8 a = 5 3 , b = 5
A 2 2
–2

9 a r2 b cos 2θ + i sin 2θ
b Proof
b (−u)* = −5 − 2i 10 a Proof
2a
Im(z)
5 P Exercise 11D
4
3 All angles are given in radians correct to 3 significant
2 figures where rounded.
1 1 a1

b z2 = i, z3 = −1 and real

Review 2 a z = 7, z = 4 + i 3, z = 4 − i 3

–7 –6 –5 –4 –3 –2 –1 O 1 Re(z) b z = − 11 , z = −25 + i 3, z = −25 − i 3
–1 8 16 16

Q

b −3 + 2i 3 6 (cos1.78 + i sin1.78), −
3 a (13, 2.75) 2

c (17, 1.08) b  5, π  + ), 6 (cos(−1.78) + i sin(−1.78))
2 2

d (61, − 0.181) 4 z = 3, z = 5i, z = −5i

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Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity

5 x = 8, y = 3 b Im(z) Re(z)
6 z = −0.5, z = 3 + i, z = 3 − i
7 z = 3i, z = −3i, z = 1 + 2i, z = 1 − 2i O
8 a −5 + i, 5 − i
(0, −3)
b −3 − i 2, 3 + i 2

c 2 −i 3, − 2+i 3 − π3–
22
d 4 − 3i, − 4 + 3i

Review e 1+ i 5, −1 − i 5 c Im(z)

f 2 e π i , 2 e −3π i 7
4 4 6
5
22 4
3
9 a z=2−i b p = 37, q = −40 2
1
c Im(z)

2

1B –7 –6 –5 –4 –3 –2 –1 O 1 2 3 Re(z)
–1
C
O 1 2 3 4 5 6 7 8 9 Re(z) –2
–1 A

–2

Review34610 a = 16, b = −1, z = 4i, z = −4i, z = 1 + 15 i, z d Im(z)
2 2
z 5
i, z = 1 − 15 i 4
22 3
2
Exercise 11E 1

1 a A half-line from (2, − 3) at an angle of π –1 O 1 2 3 4 5 6 7 Re(z)
12 –1

radians.

b The region to the right of the perpendicular –2
bisector of the points (0, 6) and (10, 0)
–3

c A circle, centre (−6, 1), radius 7. 3 Im(z) 1±i√15
d A half-line from (0, 0) at an angle of 5π 5

12 4
radians.
3

Review 2 a Im(z) 2

7 1
6
5 –5 –4 –3 –2 –1 O 1 2 3 4 5 Re(z)
4 –1
3 –2
2
1 –3

–1 O 1 2 3 4 5 6 7 Re(z) –4
–1
–5

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4 Im(z) y = –4 8 Im(z) z = 12 + 5i
Re(z)
O 15
10
5

–15 –10 –5 O 5 10 15 Re(z)
–5
(0, –4)
–10

–15

(0, –8)

5 Im(z) 9 Im(z) z = 1 + 2i

(x – 3)2 + (y + 6)2 = 9

Review –3 –2 –1 O 1 2 3 4 5 6 7 Re(z)
–1
π4–
–2
(–1, 0) O Re(z)

–3

–4

–5 10 a (x − 5)2 + ( y − 5)2 = 25
b Im(z)
–6

–7

–8 347
–9
–10

Review 6 Im(z) No

5π O Re(z) O Re(z)
6
c Least arg z = 0 , greatest arg z = π
(–4, –2) 2

(0, –5) End-of-chapter review exercise 11

7 Im(z) least = 6√5 All angles are given in radians correct to 3 significant
greatest = 10√5 figures where rounded.
20 1 a −0.1 − 1.7i
Review
10 b w = 1 − 5i, w = 1 + 5i
c Im(z)
O 10 Re(z)
7
6
5
4
3
2
1

–4 –3 –2 –1 O 1 2 3 Re(z)

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2 a zz* = k2 + 36, z = (k2 − 36) − 12ki 6 a  8, − π  ( )− π i
z* k2 + 36 6
b 2 2e 4

( ) ( )b= − 7π i u − 7π i 7 a w = −1 + i
uw 8e 12 = 2e 12
,
w
b i Im(z)
3a Isosceles triangle
Im(z)

2P

1

–1 O 1 Re(z) π–3
–1 (3, 3)
Review
–2
Q

b u= 2  cos 3 π  i sin 3   O Re(z)
w 4  4 
+ π

4 a x = 1, y = 3 ii z = 4 + ( 3 + 3 ) i
2
8 a x = −3, y = 2 or x = 3, y = − 2
b Im(z) Right-angled

C5 B

4 b i Proof

3 ii z1 = 3, z2 = − 1 + i, z3 =− 1 − i
2 2 2 2
2
348 9 a z2 = 3 + 2i

1

Review –3 –2 –1 O 12 3 b z = − 4 , z = 1 − 3 i, z = 1 + 3 i
–1 Re(z) 3 62 62
–2
–3 A 10 a i f (−3) = 0
–4
–5 ii z = 1, z = −3, z = − 1 + 15 i, z
44
c i 21 − 20 i
29 29 z i, z = − 1 − 15 i
44
ii cos( −0.761) + i sin( −0.761)
b i Im(z)
5 a z1 = −2 3 + i, z2 = −2 3 − i
(–1, √3)
b Im(z)

A1

O Re(z)

Review –2√3 O Re(z) ii Min. arg z = π
B –1 2

Reflection in the real axis iii Max. arg z = π + π + π = 5π
c z1 = z2 = 13, arg z1 = 2.86, arg z2 = −2.86 266 6

11 a i z2 + 3z + 4 = 0

ii z1 = 2

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b i z = −1, z = 1 + i 3 , z = 1 − i 3 16 i 3 + 1 i ii − 3 + 1 i
22 22 22

ii Im(z) Cross-topic review exercise 4

1B  −5   6 

1 a r =  0  + λ  7 
 3   −1 
C
–1 O 1 2 Re(z) b i m = −2

–1 A ii Proof

Review 2 y2 = 4(x2 − 1)

3 a p = −2, q = −25

Equilateral b i 14 (−10i + j − 5k)
42
12 a Proof
ii Angle POQ = 90°; 63 5
b z = 1, arg  z  = −0.841
z* z*
4 i z = 2, arg z = π or 30° or 0.524 radians
c 3z2 − 4z + 3 = 0 6

13 a k = −1 b arg z = 0.862 ii a 3 3 + i

14 i 7 − 2i b 3+i
22
349
1
ii 6.69e 4 πi Im(z) iii Im(z)

Review 6 3 B(1, √3)
×
5 y=x 2
×A(√3, 1)
1

4 O ×
–1
3 P y = 1 x + 2 12345
√3 Re(z)

2 5 x = 2 sin 2θ − 1

1

O 3 6 Re(z)  4
–1  
6 a OS =  −3 
–2 0

15 i u = −2 − 2i, v = 1 + 2i b 33, 26

ii Im(z) c 65.8°, 114.2° correct to 1 decimal place

Review 2 7 i − 2 + 11i
π 55
4
ii Im(z)
1
4
–2 –1 O 1 3π 3 Re(z) 3
4 2×
2 1

–1

–2 –3 –2 –1 O 1 2
Re(z)

Least z−w = 3 sin  π  − 1 = 32 −1
 4  2

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8 i u = 8 or 2 2, arg u = π or 45° Least z = 3 sin π = 32
4 4 2

ii Im(z) Im(z)

3 O Re(z)

2

1× 3
–π
–1–1O × 4
12345

Re(z)

iii z = 8 − 1 = 7 iii Circle, centre 3 − 3i, radius 1.

Review Im(z) Im(z)
3
–1–1O
2 1 –2 1234
1 √8 |z| –3 Re(z)
–4
–1–1O 1 2 3 45 ×
Re(z)

9 i Proof Greatest  z = 32 + (−3)2 + 1 = 3 2 + 1
ii − ln(20 − x) = 0.05t − ln 20
Im(z)

iii 7.9 –1–1O 1234
–2 Re(z)
iv t becomes very large, x approaches 20 –3
–4 ×
350 10 a r = −2i + j − k + µ(i + 3k) b 265
5

Review11 i ln R = ln x − 0.57x + 3.80 R = xe−0.57x+3.80 17 a m = −1, m = −4
b m = −3
ii R = 1 e−1+3.80 = 28.850...
0.57 c AB and CD do not intersect.

πi 18 i Proof
ii −6 ø p ø 2
12 i 9e 3
Im(z)
ii reiθ = π i or reiθ = π i− πi = − 5 πi 4

3e 6 3e 6 3e 6

13 y = 4(2 + e3x )2

14 a  −7  b 50.1°

 3 
 −13 

15 ln y = ln x − x2 + 1 + ln 2 and y = 2xe−x2 +1. 3–π –π
4

16 i u = 3 4
ii Half-line from (0, − 3)
–4 O 4
Im(z)
2 Re(z)

Review 1

–1–1O 123456
–2 π– Re(z)
–3 4

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iii w* = 2 − 4i 26 i The roots are: −3 + i 2 and 3 − i 2.

Im(z) ii w − (1 + 2i) = 1: circle, centre (1, 2) radius 1
6

4 ×× arg(z − 1) = 3 π: half-line, above and to the left
4
2 of the point (1, 0)

–4 –2 O× 2 4 6 8 10 Re(z) Im(z)
–2 3

2×

Review –4 1

–6 –2 –1 O –3 π
–1 4
× 2 3
1 Re(z)

z−5 =5

19 y = 3e16x2 − 1 b Least w − z = 2 − 1
3 − e16x2

20 i Proof 27 a Proof

5 5 5 55 b m = −4

ii 2 h2 = − 1 H 2t + 2 H2 t = 60H 2 − 60h 2

5 150 5 5 c 73.2°

H2 1
2
iii t = 49.3933... = 49.4 1800e t 351

21 a r = 2i + 5j + 7k + λ(7i − 6j − 9k) 28 i N = 1
b 127.0° 2
5 + e t

Review ii 1800

22 i Im(z) 29 i The roots are: − 3 + i 2 and − 3 + i 2 .
2 ×C

B ii Circle, centre (0, 3), radius 2. Greatest
1× value of arg z = 2.3005... = 2.30 correct to
3 significant figures.

–1 O× 1 2 3 4 Im(z)
Re(z) 6

–1 ×A

OABC is a parallelogram. 4
×
ii u* = 0.8 + 0.6i iii Proof
u 2

23 i Second root is 1 − ( 2 )i .

ii Other two roots are −1 ± i . –4 –2 O 24
Re(z)
Review
24 a h = 3 30 i dV = 80 − kV
dt
b 10 (j + 3k)
10 ii kn+1 = 4 − 4e−15kn ; 0.14
25
c 94.9° iii V = 540 cm3 correct to 2 significant figures, V

d r = 5i + 4j + λ(5i − 3k) approaches the value given by 80 = 570 cm3
correct to 2 significant figures. k
25 y3 = 2xe3x − 2 e3x + 10 , y = 2.437... = 2.44
3 33
correct to 3 significant figures

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Pure Mathematics 2 3 a = 7, b = −12
Practice exam-style paper
4 −101.9°, 78.1°

1 Proof 5 2 11

2 x ø 1 or x ù1 6 a x 5 2 + 2x − 1
5 − x2 + 5

3 1.46 b − 27 − 17x − 117x2
10 20 200
4 a p = −17 b 71
7 a dy = sin 2x + 2x cos 2x
Review5 a 1 → 1.2003, 1.1794, 1.1895, 1.1849, 1.1871, 1 dx →
.1871, 1.1861, (1.1865, 1.863), ... → 1.19
bπ
b Answer x = 8x2 obtains sec x = 8 4

3 sec x 3 8 a u = 2 + i, w = 3 + 9i

cos x = 3 → 1.186 399 6 b A(1, − 1), B(1, 1), C(0, − 2); BC = 10
8
9 a p = ±3
6 a 1 (2t + 1)(1 − 2e2t ) b y = 2x − 1
2 b i r = 5i + 2k + λ(i − 6j + k)
ii Proof
7 a 10 sin(θ + 36.87)° b 7.56°, 98.7°
10 a Proof
c 13
b Proof
8 a Proof b Proof c 0.1 → 0.077 19, 0.071 46, 0.069 98, 0.069 59, 0
→
352 , 0.069 49, 0.069 46 → 0.069 
d 90.5 °C
ReviewPure Mathematics 3
Practice exam-style paper

1 2.652

2 Proof

Review

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A Direction of a vector: this is represented by the arrow on
the vector and can be, for example, the angle of the path
Absolute value: the non-negative value of a number that takes you from one point to another along the length
without regard to its sign of the vector
Algebraic improper fraction: the algebraic fraction P(x) ,
Dividend: the quantity being divided by another quantity
Q(x)
Review where P(x) and Q(x) are polynomials in x, is said to be Division algorithm for polynomials:
an improper fraction if the degree of P(x) ù the degree dividend = divisor × quotient + remainder
of Q(x)
Divisor: the quantity by which another quantity is to be
Arbitrary constant: a constant that may be assumed to be divided
any value

Argument of a complex number: the argument of a Double angle formulae: sin 2A ≡ 2 sin A cos A
complex number x + iy is the direction of the position
 cos 2A ≡ cos2 A − sin2 A
 x 
vector  y  tan 2A ≡ 2 tan A
1 − tan2 A
E

C Explicit functions: functions of the form y = f(x) are

Complex number: a number that can be written in the form called explicit functions as y is given explicitly in terms
x + iy where x and y are real (this form of a complex
number is called the Cartesian form) of x

Compound angle formulae: F 353
sin(A + B) ≡ sin A cos B + cos A sin B
Factor: a portion of a quantity that, when multiplied by
Review sin(A − B) ≡ sin A cos B − cos A sin B other factors, gives the entire quantity

cos(A + B) ≡ cos A cos B − sin A sin B Factor theorem: if for a polynomial P(x), P(c) = 0 then
x − c is a factor of P(x)

cos(A − B) ≡ cos A cos B + sin A sin B G

tan( A + B) ≡ tan A + tan B General binomial theorem:
1 − tan A tan B
n(n − 1) n(n − 1)(n − 2)
tan A − tan B (1 + x)n = 1+ nx + 2! x2 + 3! x3 + …
tan( A B) 1 + tan A tan B
− ≡

where n is rational and x , 1

Converging: the sequence of values generated by an General solution: a solution for a differential equation that
iterative formula is convergent or said to be converging to works for any value of the constant C
a root if the values in the sequence approach the actual
value of the root I

Cosecant: cosec θ = 1 Imaginary number: any multiple of the unit imaginary
sinθ number i where i2 is defined to be −1

Cotangent: cot θ = 1  = cosθ  Implicit function: when a function is given as an equation
tanθ    sinθ  connecting x and y, where y is not the subject

Review D ∫ ∫Integration by parts: u dv dx = uv − v du dx
dx dx
Decompose: split a single algebraic fraction into two or
more partial fractions Integration by substitution: can be considered as the
reverse process of differentiation by the chain rule
Degree: the highest power of x in the polynomial is called
Iteration: an iteration is one trial in a process that is
the degree of the polynomial. repeated such as when using an iterative formula (see
For example, 5x3 − 6x2 + 2x − 6 is a polynomial of iterative process)
degree 3.

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Iterative formula: this type of formula is one that is used Particular solution: a solution for a differential equation
repeatedly. The output from each stage is used as the input that works for a specific value of C

for the next stage. It is commonly used in numerical Polynomial: an expression of the form
methods. anxn + an−1xn−1 + an−2xn−2 + … +a2x2 + a1x1 + a0

Iterative process: a process for finding a particular result Position vector: a means of locating a point in space
by repeating trials of operations. Each repeat trial is called relative to an origin
an iteration and each iteration should produce a value
closer to the result being found. Principal argument: the principal argument, θ , of a
complex number is an angle such that −π , θ ø π.
L Sometimes it might be more convenient to give θ as an
angle such that 0 ø θ , 2π. (Usually the argument is
Laws of logarithms: given in radians.)

ReviewMultiplication law Division law Power law d dv du
dx dx dx
loga (xy) = loga  x  = loga (x)m = Product rule: ( uv ) = u +v
loga x + loga y  y  m loga x
Q
loga x − loga y
Quotient: a result obtained by dividing one quantity by
Locus: a locus is a path traced out by a point as it moves
following a particular rule. The rule is expressed as an another
inequality or an equation.
d  u = v  du  −  u  dv
Logarithm: the power to which a base needs to be raised to dx  v dx dx
produce a given value Quotient rule:
v2

R

M Real number: the set of all rational and irrational
numbers
Magnitude of a vector: the length or size of the vector
Remainder: the amount left over after a division
354 Modulus: the magnitude of the number without a sign
attached. The modulus of a number is also called the Remainder theorem: if a polynomial P(x) is divided by
x − c, the remainder is P(c)
absolute value.
Resultant vector: the combination of two or more single
ReviewModulus of a complex number: the modulus of the vectors
complex number x + iy is the magnitude of the position

vector  x  S
 y 
Scalar: a quantity (often a number) that is used to scale a

vector 1
cosθ
N Secant: sec θ =

Natural exponential function: the function y = ex

ReviewNatural logarithms: logarithms to the base of e. ln x is T
used to represent loge x
Trapezium rule: this numerical method involves splitting
P the area under the curve y = f(x) between x = a and
x = b into equal width strips
Parallel vectors: two vectors are parallel when one is a
scalar multiple of the other U

Parameter: sometimes variables x and y are given as a Unit vector: a vector with magnitude or length one
function of a third variable t. The variable t is called a
parameter and the two equations are called the parametric V
equations of the curve.
Vector addition: the process of finding the sum of two or

Partial fraction: each of two or more fractions into which more vectors
a more complex fraction can be decomposed as a sum Vector subtraction: the addition of the negative of a vector

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Review Index degree of a polynomial 11
differential equations 244–5
absolute value see modulus function 355
area under a curve forming an equation from a problem 252–5
Review general solution 246
trapezium rule 126–8 modelling growth 259–60
Review see also integration Newton’s law of heating or cooling 251–2
Argand diagrams 273 particular solution 246
argument of a complex number 274 separating the variables 245–9
differentiation
binomial expansion 181 chain rule 88
of (a + x)n where n is not a positive integer 177–8 derivative of ef(x) 88–90
of (1 + x)n where n is not a positive integer 174–6 derivative of ex 87–8
and partial fractions 179–80 derivative of ln f(x) 92–3
derivative of ln x 91–2
calculus see differential equations; differentiation; derivative of tan−1 x 186–7
integration derivatives of trigonometric functions 95–8
implicit 99–101
chain rule 88 parametric 103–5
change of sign method, numerical solutions 136 product rule 81–3
circles, loci 289–92 quotient rule 84–6
cobweb pattern, iterative processes 142 displacement (translation) vectors 211–12
complex conjugate pairs 284 addition and subtraction 213–15
complex conjugates 270 components of 212
complex numbers (z) 267, 269–70 magnitude of 215–16
notation 212, 213
argument of 274 unit vectors 217–18
calculating with 271, 280–1 dividend 12
Cartesian form 273 division law, logarithms 33
cube roots of one 287 divisor 12
equal numbers 270 dot product see scalar product
loci 288–94
modulus of 273 e (Euler’s number) 42
polar forms 277 derivative of ef(x) 88–90
roots of polynomials 283–5 derivative of ex 87–8
square roots of 286 integration of exponential functions 112–13
complex plane
Argand diagrams 273 equations
exponential form 277–81 exponential 38–9
modulus-argument form 273–7 of the form a sinθ + b cosθ = c 68–70
components of a displacement 212 involving the modulus function 3–6
compound angle formulae 58–60 logarithmic 35–6
cosecant (cosec) 53–6 parametric 103, 232–4
graph of 54 quadratic 283
cosine (cos) see also differential equations; numerical solutions of
compound angle formulae 58–60 equations
derivative of cos x 96
derivative of cos(ax + b) 97 Euler, Leonhard 42, 277
double angle formula 61, 62–4 explicit functions 99
graph of 55 exponential equations 38–9
integration 118–19, 122 exponential form of the complex plane 277–81
cotangent (cot) 53–6 exponential growth 259–60
graph of 55 exponential inequalities 40–1
cube roots of one (unity) 287
cubic equations 284–5

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factor theorem 14–17 subscript notation 142–3
trigonometric equations 149–52
factors 12 using a calculator 143
first order differential equations 244 worked examples 143–5
Fourier, Joseph 70
fractals 267 loci 288
circles with centre (0, 0) 289–90
Reviewgeneral binomial theorem 174 circles with centre (a, b) 290–2
general solution of a differential equation 244, 246 half-lines and part-lines 292–3
golden ratio (1:φ ) 153 perpendicular bisectors 293–4
golden spiral 153
graph sketching 135–7 logarithmic equations 35–6
graphs logarithms (logs) 26

of the modulus function 7 to base 10 26–9
of reciprocal trigonometric functions 54–6 to base a 30–2
transforming a relationship to linear form 44–6 laws of 33–4
growth models 259–60 natural 42

i 267–9 derivative of ln f(x) 92–3
derivative of ln x 91–2
imaginary numbers 266–9 solving exponential equations 38–9
solving exponential inequalities 40–1
implicit differentiation 99–101 transforming a relationship to linear form 44–6
logistic growth models 259–60
implicit functions 99
magnitude (modulus) of a vector 215–16
improper algebraic fractions 166–7 modulus function (absolute value) 2–3

partial fractions 171–2 graphs of y = f(x) 7
solving equations 3–6
indices, relationship to logarithms solving inequalities 8–10
modulus-argument form of the complex plane 273–7
in base 10 26–9 multiplication law, logarithms 33

in other bases 30–2 Napier, John 42
356 inequalities natural exponential function 42
natural logarithms 42
exponential 40–1
derivative of ln f(x) 92–3
Review involving the modulus function 8–10 derivative of ln x 91–2
Newton, Isaac 175
integration law of heating or cooling 251–2
numerical solutions of equations 135
of 1 115–17 iterative processes 140–5
ax + b starting points 135–7

of 1 115–17 parallel lines, vector equations 236
x parallel vectors 213
1
of x2 + a2 187–8 scalar product 227
parametric differentiation 103–5
of exponential functions 112–13 parametric equations 103

Review of k f ′(x) 189–90 of a line 232–4
f(x) partial fractions 181

solving differential equations 245–9 application of 172–3
trapezium rule approximation 126–8 and binomial expansions 179–80
of trigonometric functions 118–19, 121–4 Heaviside’s cover-up method 174
use of partial fractions 194–6
using different techniques 201–2
integration by parts 197–200
integration by substitution 191–3
intersecting lines, vector equations 236–7
iterations 141
iterative formulae 142
iterative processes 140–1
cobweb pattern 142
staircase pattern 145

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Review improper 171–2 skew lines, vector equations 237–8
with a quadratic factor in the denominator that cannot be square roots of a complex number 286
staircase pattern, iterative processes 145 357
factorised 171 subscript notation, iterative processes 142–3
Review with repeated linear factor in the denominator 169–70 substitution, integration by 191–3
use in integration 194–6
where the denominator has distinct linear factors 168–9 tangent (tan)
particular solution of a differential equation 244, 246 compound angle formulae 58–60
perpendicular bisectors 293–4 derivative of tan x 96
perpendicular vectors, scalar product 227 derivative of tan−1 x 186–7
polar coordinates 277 derivative of tan(ax + b) 97
polar forms of a complex number 277 double angle formula 62–3
polynomials 11 graph of 55
complex roots 283–5 integration 118–19, 123
division algorithm 13
division of 11–13 telescoping series 172–3
factor theorem 14–17 translation vectors see displacement vectors
remainder theorem 18–20 trapezium rule 126–8
population growth models 259–60 trigonometry
position vectors 220
Cartesian components 220–3 compound angle formulae 58–60
power law 33, 46 derivatives of trigonometric functions 95–8
principal argument of a complex number 274 double angle formulae 61–4
problem solving, forming differential equations 252–5 equations of the form a sinθ + b cosθ = c
product rule 81–3
68–70
quadratic equations, complex roots 283 integration of trigonometric functions 118–19, 121–4
quartic equations 285 proving identities 66
quotient 12 reciprocal functions 53–6
quotient rule 84–6 using iterative processes 149–52

unit vectors 217–18

Review reciprocal trigonometric ratios 53–6 vector equation of a line 231–2
remainder theorem 18–20 intersection of two lines 236–7
resultant vectors 213–15 parametric form 232–4
roots of polynomials 283–5 skew lines 237–8

scalar product (dot product) 225–6 vector spaces 225
for component form 227–9 vectors
parallel vectors 227
perpendicular vectors 227 addition and subtraction 213–15
angle between 225
secant (sec) 53–6 displacement (translation) vectors 211–18
graph of 55 magnitude of 215–16
integration 118–19, 123–4 multiplication by a scalar 213
notation 212, 213
separating the variables, differential equations 245–9 parallel 213, 227
sine (sin) perpendicular 227
position vectors 220–3
compound angle formulae 58–60 scalar product 225–9
derivative of sin x 95–6 unit vectors 217–18
derivative of sin(ax + b) 97 uses of 211
double angle formula 61, 62–4
graph of 54
integration 118–19, 122, 124

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