Cambridge International AS A Level Mathematics Pure Mathematics 2 and 3 second edition

? (Page 188) (iii) 4ex3 + c (ii) esin x + c

Yes: Using the chain rule 2 (i) 0.018 (iii) 1 tan2 x + c
2
–1
dy = dy × du (ii) ln 3 (iv) sin x + c
dx du dx
Integrating both sides with 3 (i) 21(e − 1) 3 (i) 1
(ii) 21(e4 − 1) Answers
respect to x, (ii) 1
dy du 16
∫y = ( du × dx )dx
(iii) (21t(oe e4) 1 27.7 (iii) 1
dy + s.f.) − =
du 3
∫= ( )du (iv) e − 1

Activity 8.1 (Page 190) 4 0.490; 0.314 (v) ln 2

2 (x − 2) 5 + 4 (x − 3 +c 5 (i) −(x + 2)e−x 4 (i) tan−1 x + c
5 2 3
2)2

= 2 (x − 2) 3 [3(x − 2) + 10] + c (ii) –(x + 3)e−x + c ( )(ii) 1 tan −1 x +c
15 2 4 4

= 2 (3x + 4)(x − 2) 3 + c (iii) (–2, e2)
15 2
(iv) −e2; max. at x = −2 ( )(iii) 2 tan −1 x c
3 3 +

Exercise 8B (Page 190) (vi) 3 − 4 ( )(iv) 2
e 2
(i) 1 (x3 1)8 c tan −1 2x + c
8
1 + + 6 (i) y

(ii) 1 (x2 + 1)6 + c 1 (v) 3 tan − 1 ⎛ 3x ⎞ + c
6 6 ⎜⎝ 2 ⎠⎟

(iii) 1 (x3 − 2)5 + c 1 (vi) ( )1 tan−1 52x + c
5 2
1 3 10
(iv) 6 (2x2 − + c
5)2

(v) 115(2x + 3 (3x − 1) + c Ox 5 2π
3
1)2

(vi) 2 (x + 9) 1 (x − 18) + c (ii) ln(e 2 + 1) ≈ 1.434 6 (i) x = 0.685
3 2 2

2 (i) 222 000 (iii) ln(e 2 + 1) 1.434 (ii) 8
2 15
(ii) 586 ≈

(iii) 18.1 (iv) The same.The 7 (i) 1 (e 2 − 1)
2
substitution ex = t2
1 (ii) 0.896
3 (i) 22 2 transforms the integral
in part (ii) into that in
(ii) 1 1 part (iii). 9 (ii) 1 π − 1 3
9 6 4

4 y = 2x + 3 + 3 7 (i) 1 10 (a) 3x + 1 tan 2x + c
2
4 ( )(ii) 1
5 3 2 ln p2 + 1 ( )(b) 1
8 3π + ln 4
(iii) 2.53
6 7 11 (i) 1
6 24
3
7 (i) (a) 8 x − 2x 2 +c Exercise 8D (Page 197) (ii) k = 10

(b) 2(1 + x2)23 + c 1 (i) 1 sin 3x + c
3
? (Page 199)
(ii) k = 2, a = 1, b = 2; (ii) cos(1 − x) + c
32.5 Substitution using u = x2 − 1
(iii) − 1 cos4 x + c needs 2x in the numerator. Not
4 a product.
(1 ln x)2
8 + 6 (2 ln x − 1) + c (iv) ln | 2 − cos x | + c

Exercise 8C (Page 192) (v) −ln | cos x | + c

1 (i) ln |x2 + 1| + c (vi) − 1 (cos 2x + 1)3 + c
6

(ii) 1 ln |3x2 + 9x − 1| + c 2 (i) −cos(x2) + c 337
3

Exercise 8E (Page 202) ∫(c) ⇒ x sin x dx (iv) 1 x3 ln 2x − 1 x3 + c
3 9
= −x cos x + sin x + c
3x – 2 2 x 3 (3x c
1– x 15 2
⎪ ⎪1 (i) ln +c (ii) (a) d (x e2x) x ×2e2x e2x 3 (1+ ) − 2) +
dx
= + 115(x − 2)5(5x + 2) + c

1 x –1 4
1– x 2x + 3
⎪ ⎪(ii) + ln +c (b) ⇒ xe2x

Answers x –1 ∫ ∫= 2xe2x dx + e2x dx 5 (i) x ln x − x + c
x2 +1 (ii) x ln 3x − x + c
⎪ ⎪(iii) ln +c ∫⇒ 2xe2x dx (iii) x ln px − x + c

(x − 1)2 ∫= xe2x − e2x dx
2x + 1
⎪ ⎪(iv)ln +c

(v) ln⎪1 x x⎪ − 1 + c ∫(c) ⇒ 2x e2x dx 1 6 x2ex − 2x ex + 2ex + c
– x = xe2x − 2 e2x + c

⎪ ⎪(vi)21ln x +1 +c ? (Page 204) 7 (2 − x)2 sin x − 2(2 − x)cos x
x+3 − 2 sin x + c

x2 +4
x+2
⎪ ⎪(vii) ln +c Each of the integrals in Activity 8.2 8 1 ((x 2 + 1)tan−1 x − x) + c
dv 2
is of the form ∫x dx dx and is
2x + 1
x+2
⎪ ⎪(viii)ln + found by starting with the Exercise 8G (Page 210)

1 c product xv.

2(2x + 1) + Exercise 8F (Page 208) 1 (i) 2 e3 + 1
9 9

2 – x + 1 3, ln ⎜⎝⎛ 2 ⎟⎞⎠ 1 (i) (a) u = x, dv = ex (ii) −2
x2 + – 6 dx
4 x (iii) 2e2

2 1 (b) xex − ex + c (iv) 3 ln 2 − 1
– 2x 1+ x
3 (i) (a) 1 + (ii) (a) u = x, dv = cos 3x (v) π
dx 4
(b) ln )(11 0.318 45 (b) 31x sin 3x +
8 = (vi) 64 ln 4 − 7
3
(ii) (a) 3 + 3x + 9x2 + … 1 cos 3x + c
9

(b) 0.318 00 (iii) (a) u = 2x + 1, 2 (i) (2, 0), (0, 2)

(c) 0.14% dv = cos x (ii)
dx
4 (i) A = 1,B = 2,C = 1, (b) (2x + 1)sin x + y
D = –3
2 cos x + c 2

5 (i) 1 + 1 − 3 (iv) (a) u = x, dv = e−2x y = (2 – x)e–x
2(x + 2(x + dx
1) 3) O2
(b) − 21x e−2x −
6 (i) 2 3 x + 4 4x 41e−2x + c (iii) e–2 + 1
− + x2
x
(v) (a) dv e−x
7 (ii) ln 2 − x − 2 ln x + 1 + 2 u = x, dx =
+ 1 + 2 ln 4x + 3 + c
(b) −xe−x − e−x + c

Activity 8.2 (Page 204) (vi) (a) u = x, dv = sin 2x 3 (i)
d dx
(i) (a) dx (x cos x) y
(b) 1
= −x sin x + cos x − 2 x cos 2x +

1 sin 2x + c y =xsinx
4
(b) ⇒ x cos x
2 (i) 1 x4 ln x 116x4 c
= −xsinx dx + cosx dx 4 − +
∫ ∫ ∫ ∫338
⇒ x sin x dx (ii) x e3x − 13e3x + c O π x
= −x cos x + cos x dx 1
(iii) x sin 2x + 2 cos 2x + c (ii) π

4 5 ln 5 − 4 ∫ x2 x –5 3 dx (v) In this case the numerator
+ 2x – is the differential of the
π 1 denominator and so the
5 2 − 2 3) dx 1 dx integral is the natural
∫ ∫= + − – logarithm of the modulus of
(x (x 1) the denominator.
4 so area 4 square units
6 − 15 = 15 = 2ln| x + 3 | − ln| x − 1 | + c

7 (i) 4π − 3 3 (ii) The derivative of the Since Answers

(ii) 1 (π − 2) expression on the bottom ddx(x2 + sin x) = 2x + cos x,
16 line is 2x + 2, which is twice then
the expression on the top
8 4(ln 4 − 1) ∫ 2x + cosx dx =
9 (ii) line. So the integral is of the x2 + sinx

form

y y=x ∫k f ′(x ) dx = k ln | f(x) | + c. ln|x2 + sinx| + c.
f(x)
y = 2 + e– 1 x (vi) This is a product: sin2 x is a
2 function of sin x, and cos x is
the derivative of sin x, so you
This integral can also be can use the substitution
found using partial fractions, u = sin x.
but using logarithms is
quicker.

3 x +1 dx Using u = sin x
+ 2x –
∫x 2 3

O x ∫= 1 2x + 2 3 dx ∫ ∫cos x sin2 x dx = u2 du
2 x2 + 2x – 1
(iv) 2.31 = 3 u3 + c
1 ln | x2 2x 3 c
1 = 2 + − | + = 1 sin3 x + c
2 3
10 (i) (iii) This is a product of x and ex.
There is no relationship
(ii) π (2 e − 3) Exercise 8H (Page 214)
between one expression and

11 (i) 6e the derivative of the other, so 1 (i) 1 sin(3x − l) + c
you cannot use substitution. 3
As one of the expressions is
(ii) − 3 (ii) (x 2 –1 – 1) + c
4 x, you can use integration +x
by parts.
12 (i) y = x − 1 (iii) −e1−x + c

(ii) 1 π( e2 − 2) ∫ ∫x ex dx = x ex − ex dx (iv) 1 sin 2x + c
4 = x ex − ex + c 2

? (Page 212) (iv) This is also a product, this (v) x ln 2x − x + c
time of x and ex2. ex2 is a
You will return to these integals in function of x2, and 2x is the (vi) –1 1)2 + c
Activity 8.3. derivative of x2, so you can 4(x 2 –
use the substitution u = x2.
(vii) 1 (2x − 3)23 + c
3

Activity 8.3 (Page 214) Using u = x2 (viii) ln ⎪xx – 21⎪ + 1 + c
+ –
(i) This is a quotient.The ∫ ∫x ex2 dx = 1 eu du x 1
derivative of the expression 2
on the bottom is not related (ix) 1 x4 ln x 1 x4 c
to the expression on the = 1 eu + c 4 − 16 +
top, so you cannot use 2
substitution. However, as the = 21ex2 + c (x) ln ⎪2xx––31⎪ + c
expression on the bottom
can be factorised, you can (xi) 1 ex2+ 2x + c
write it as partial fractions. 2

(xii) −ln (sin x + cos x) + c

339

(xiii) −21x2 cos 2x + 21x sin 2x 2 ds = k The model covers the main features
dt s2 of the situation.
+ 41cos 2x + c
3 dh = k ln(H − h) ? (Page 224)
dt
(xiv) −21 cos 2x + 1 cos3
6 2x + c 4 dm = k ln | y | + c1 = 21x2 + c2
dt m
Answers
2 (i) 8 5 dP =k P can be rewritten as
3 dt
1 1
(ii) 3 ln 4 de ln | y | = 2 x2 + (c2 − c1).
dθ
(iii) 48 + 8 ln 4 6 = kθ Exercise 9B (Page 225)

(iv) 2 7 dθ = – (θ – 15) 1 (i) y = 1 x3 + c
3 dt 160 3

(v) 8 ln 2 − 7 dN N (ii) y = sin x + c
3 9 dt 20
2
)((vi) 3 =
2 −1 8 (iii) y = ex + c

(vii) (ln 2)2 9 dv = 4 (iv) y = 2 x 3 + c
dt v 3 2

(viii) − 2 10 dA = 2k π = k′ 2 (i) y = − (x 2 c )
9 dt A A 2+

(ix) 1 (2 2 − 1) dθ s (ii) y2 = 23x3 + c
3 ds 4
= –
(x) 13 11 (iii) y = Aex
4 + 4e2
dV 2V (iv) y = ln | ex + c |
(i) 3 3x 12 dt = – 1125π
3 − + +x (v) y = Ax
2 x 2 2

4 (ii) 2e2 − 10 13 dh = (2 –k h) (vi) y = ( 41x2 + c)2
dt 100
5 (ii) 15 ln 5 − 4 (vii) y= 1
− (sin x + c )

Chapter 9 Investigation (Page 222) (viii) y2 = A(x2 + 1) − 1
H is about (70° N, 35°W) and L
? (Page 217) is about (62° N, 5°W) so they are (ix) y = −ln(c − 1 x2)
separated by 30° in longitude at a 2
Answers depend on the size of the mean latitude of 66°. Reference
cup,the initial temperature of the to the scale shows this to be about (x) y3 = 3 x2 ln x − 3 x2 + c
coffee,whether the cup is insulated 900 nautical miles. 2 4
and whether milk is added,as well as
personal preference. It takes around 1035 Exercise 9C (Page 229)
5–10 minutes for a cup to become
drinkable (around 60°C) and after 996 1 (i) y = 1 x3 − x − 4
about 25 minutes the coffee is 3
probably too cool for most people 957 (ii) y = e31x2
(around 30°C).
isobars O 900 (iii) y = ln(21x2 + 1)
Exercise 9A (Page 220) nautical miles
(iv) y = (2 1 x)
–
(v) y = e21(x2−1) − 1

(vi) y = sec x

1 dv is the rate of change of The mean level is 996 and the 2 (i) θ = 20 − Ae−2t
dt amplitude 39 so a model is (ii) θ = 20 − 15e−2t
(iii) t = 1.01 hours
velocity with respect to time, p= 996 + 39 cos πx
900 3 (i) N = Aet
( )i.e. the acceleration. dp –39π πx (ii) N = 10et
( )The differential equation tells and dx = 900 sin 900

you that the acceleration is or dp = −a sin bx
proportional to the square dx

340 of the velocity. with a = 0.136 and b = 0.0035.

(iii) N tends to ∞, which Investigation (Page 233) Start with the vector
would never be O⎯→Q = a1i + a2 j.
realised because of Using the assumptions in
the combined effects Exercise 9A, question 7, the rate Oy
of food shortage, of cooling is proportional to the
predators and human temperature of the coffee above a1 Answers
controls. the surrounding air.The initial
temperature is 95°C and the
4 ds = 2 ;s = 4t + c cooling rate is 0.5°C s−1. So x a2 Q
dt s
θ = 15 + 80e−16t0.
5 y = 3e16x2 − 1 Length = a 2 + a 2
3 − e16x2 Adding 10% milk at 5°C gives 1 2
θ = 15 + 71e−16t0.
6 (i) θ = A(1 + 3e−kt ) Now look at the triangle OQP.
The final temperature is lower if
(iii) 7A the milk is added at the end. P
3
Chapter 10
( )7 (i) 1 1 e −2t
tan −1 2 − 2 ? (Page 234)

(ii) The value of x tends to To find the distance between the a3
vapour trails you need two pieces
tan −1 1 . of information for each of them:
2 either two points that it goes
through, or else one point and its
(iii) As 1 − 1 e −2t increases direction.All of these need to be in OQ
2 2 three dimensions. However, if you
so does want to find the closest approach
of the aircraft you also need to
( )tan−1 1 − 1 e −2t . know, for each of them, the time at OP2 = OQ2 + QP2
2 2 which it was at a given point on its
trail and the speed at which it was (a12 a ) a
travelling. (This answer assumes = + 2 + 2
⎜⎝⎛1100 h constant speeds and directions.) 2 3
(iii) + h ⎠⎟⎞
8 100 ln − − 20h ? (Page 239) ⇒ OP = a 2 + a 2 + a 2
1 2 3
dN The vector a1i + a2j + a3k is
9 (i) dt = k(N − 150) shown in the diagram. Exercise 10A (Page 239)
1 (i) 3i + 2j
(ii) N = 500e0.08t + 150 z
(ii) 5i − 4j
(iii) No; when t = 15 then a3 P (iii) 3i
N = 1810 or when (iv) −3i − j
N = 1500 then t = 16.4 2 For all question 2:

10 (i) N = 1800e 21t
1
5 + e 2 t

(ii) 1800

11 x = 1 (tan θ + 1)2 − 1 j
8 2 i

12 (i) (i)

( )y =−3 x cos 1 x + 9 sin 1 x + c 2
10 3 10 3

(ii) 203

13 (ii) 2 h 5 = 1 H 52t + 2 H 5 O
5 2 150 5 2 a1

a2 y ( 13, 56.3°)
Q
( )(iii) t= ⎛ h 5 ⎞
60⎜⎜⎝1 − H 2 ⎟⎠⎟

x

341

(ii) (v) 5k Exercise 10B (Page 248)

( 13, −33.7°) (vi) −i − 2j + 3k 1 (i) ⎛6⎞
(iii) (vii) i + 2j − 3k ⎝⎜8⎟⎠
(viii) 4i − 2j + 4k
Answers (4 2, −135°) (ix) 2i − 2k (ii) ⎛1⎞
(iv) (x) −8i + 10j + k ⎜⎝1⎟⎠

( 5, 116.6°) 5 (i) A: 2i + 3j, C: −2i + j (iii) ⎛0⎞
(v) (ii) A⎯→B = −2i + j, ⎜⎝0⎠⎟
C⎯→B = 2i + 3j
(5, −53.1°) (iii) (a) A⎯→B = Ο⎯→C (iv) ⎛ 8⎞
3 (i) 3.74 (b) C⎯→B = Ο⎯→A ⎜⎝−1⎠⎟
(iv) A parallelogram (v) –3j
(ii) 4.47
(iii) 4.90 2 (i) 2i + 3j + k
(iv) 3.32 (ii) i – k
(v) 7
(vi) 2.24 (iii) j – k
4 (i) 2i − 2j
(ii) 2i Activity 10.1 (Page 244) (iv) 3i + 2j – 5k
(iii) −4j (v) –6k
(iv) 4j (i) (a) F
3 (i) (a) b
342 (b) C
(b) a + b
(c) Q
(c) –a + b
(d) T
(ii) (a) 21(a + b)
(e) S
(ii) (a) Ο⎯→F (b) 1 (–a + b)
2
(b) O⎯→Ε, C⎯→F
(c) O⎯→G, →PS,A⎯→F (iii) PQRS is any
(d) B⎯→D
(e) Q⎯→S, P⎯→T parallelogram and

P⎯→M = 21P⎯→R, Q⎯→M = 21Q⎯→S
4 (i) (a) i

? (Page 247) (b) 2i

(c) i − j

OΟ⎯→CC== OΟ⎯→A + s s t A⎯→B (d) −i − 2j
+ (ii) | A⎯→B | = | Β⎯→C | = 2 ,
Ο⎯→A = a and A⎯→B = b − a
| A⎯→B | = | C⎯→B | = 5
OC = a + s (b − a)
s + t
5 (i) −p + q, 21p − 21q,
= a + s s t b − s s t a − 21p, − 21q
+ +

= s + t a − s s t a + s s t b (ii) N⎯→M 1 B⎯→C,
s + t + + 2
=
t s
= s + t a + s + t b N⎯→L = 1 A⎯→C,
2
M⎯→L A⎯→B
= 1
2

⎛2⎞ ? (Page 250) (v) 90°

6 (i) ⎜ 13 ⎟ The cosine rule (vi) 180°
⎜ ⎟ Pythagoras’ theorem
⎜⎝ 3 ⎟⎠ 2 (i) ⎝⎛⎜13⎞⎟⎠, ⎛− 1⎞
13 ? (Page 252) ⎜⎝ 3⎠⎟

(ii) 53i + 4 j Answers
5
(ii) B⎯→A . B⎯→C = 0
⎛ –1 ⎞ ⎛⎜⎝aa12 ⎞ . ⎛⎝⎜bb12 ⎞ = a1b1 + a2b2 (iii) | A⎯→B | = | B⎯→C | = 10
⎜ ⎟ ⎟ ⎟
(iii) ⎜ 2 ⎟ ⎠ ⎠ (iv) (2, 5)
⎝⎜ –1 ⎠⎟
⎛⎜⎝bb12 ⎞ . ⎛⎝⎜aa12 ⎞ = b1a1 + b2a2
2 ⎟ ⎟
⎠ ⎠
(iv) 5 i – 12 j 3 (i) 29.0°
13 13 (ii) 76.2°
(iii) 162.0°
⎛1⎞ These are the same because
⎜ 14 ⎟ ordinary multiplication is
⎜ ⎟ commutative.
⎜ 2 ⎟
7 (i) ⎜ ⎟ ? (Page 253) 4 (i) (0, 4, 3)
⎜ 14 ⎟
3 Consider the triangle OAB with (ii) ⎜⎜⎝⎛−453⎟⎟⎞⎠ , 5 2
angle AOB = θ, as shown in the (iii) 25.1°
⎝ 14 ⎠ diagram. 5 (i) O⎯→Q = 3i + 3j + 6k,

(ii) 23i − 2 j + 1 k O ⎯P→Q = −3i + j + 6k
3 3 (ii) 53.0°

(iii) 3 i − 4 k
5 5

⎛ –2 ⎞ θ
⎜ ⎟ ba
(iv) ⎜ 29 ⎟
⎜ 4 ⎟ BA
⎜ ⎟ b – a = (b1 – a1)i + 6 (i) −2
⎜ 29 ⎟ (b2 – a2)j + (b3 – a3)k
–3 (ii) 40°

⎝ 29 ⎠ (iii) A⎯→B = i − 3j + (p − 2)k;
p = 0.5 or p = 3.5
(v) 5 i − 3 j + 2 k
38 38 38

⎛1 ⎞ cos θ = OA 2 + OB2 – AB2 7 (i) −6, obtuse
(vi) ⎜⎜⎜⎝00⎟⎠⎟⎟ 2 × OA × OB
8 11.74 OA2 = a12 + a22 + a32
9 x = 4 or x = −2 ⎛ 2 ⎞
10 (i) (a) q OB2 = b12 + b22 + b32 (ii) ⎜ 3 ⎟

AB2 = (b1 − aa13))22 + (b2 − a2)2 + ⎜ – 2 ⎟
(b3 − ⎜⎝⎜ 3 ⎟⎟⎠
1
2(a1b1 + a2b2 + a3b3 ) 3
2 | a || b|
(b) −p ⇒ cosθ = 8 (i) 99°

(c) q + p (d) r + p = a.b (ii) 71(2i − 6j + 3k)
(e) p + q + r |a||b| (iii) p = −7 or p = 5

(ii) p + 3 r Exercise 10C (Page 254) 9 (ii) q = 5 or q = −3
5 1 (i) 42.3°
10 (i) P⎯→A = −6i − 8j − 6k,
⎛ 2⎞ (ii) 90° P⎯→N = 6i + 2j − 6k
⎜⎜⎜⎝−63⎟⎟⎟⎠ (iii) 18.4°
12 (i) 1 (iv) 31.0° (ii) 99.1°
7

(ii) m = −2, n = 3, k = −8 343

11 (i) 4i + 4j + 5k, 7.55 m (iii) is parallel to (i) since the (viii) r = ⎛ 3⎞ λ ⎛ –1⎞
⎜⎝ –12⎟⎠ ⎜⎝ 4⎠⎟
(ii) 43.7° (or 0.763 radians) direction vector is the same. +
(iv) is parallel to (ii) since

12 (i) P⎯→R = 2i + 2j + 2k, ⎛–1⎞  = − ⎛ 1⎞ . 3 Note:These answers are not
P⎯→Q = −2i + 2j + 4k ⎜⎝ 2⎠⎟ ⎝⎜–2⎟⎠ unique.
Answers
(ii) 61.9° Exercise 10D (Page 264) ⎛ 2 ⎞ ⎛ 3⎞
(iii) 12.8 units
1 (i) (a) 2i + 8j (i) r = ⎜ 4⎟ + λ ⎜ 6⎟
⎜ –1⎟⎠ ⎜ 4⎟⎠
(b) 68 ⎝ ⎝
(c) 3i + 7j
? (Page 259) (ii) (a) −4i − 3j ⎛ 1 ⎞ ⎛ 1⎞
(b) 5
O⎯→P = O⎯→A + λ(O⎯→B − O⎯→A) (c) 2i + 1.5j (ii) r = ⎜ 0⎟ + λ ⎜ 0⎟
= (1 − λ)O⎯→A + λO⎯→B (iii) (a) 6i + 8j ⎜ –1⎠⎟ ⎜ 0⎟⎠
(b) 10 ⎝ ⎝
(c) i + 3j
(iv) (a) 6i − 8j ⎛ 1⎞ ⎛ 5 ⎞
(b) 10
Activity 10.2 (Page 260) (c) 0 (iii) r = ⎜ 0⎟ + λ ⎜ 3⎟
(v) (a) 5i + 12j ⎜ 4⎠⎟ ⎜ –6⎠⎟
(ii) ⎛−2⎞ , ⎛⎜⎝–50⎞⎠⎟, ⎛ 2⎞ , ⎜⎛⎝13⎠⎞⎟, (b) 13 ⎝ ⎝
⎝⎜−9⎠⎟ ⎜⎝ –1 ⎠⎟ (c) −7.5i − 2j
⎛ 0⎞ ⎛ 2⎞

⎛3 1 ⎟⎞, ⎜⎛⎝⎜43 ⎟⎟⎞⎠ , ⎛ 8⎞ (iv) r = ⎜ 0⎟ + λ ⎜ 1⎟
⎝⎜2 2 ⎠ ⎝⎜⎜11⎟⎟⎠ ⎜ 1⎟⎠ ⎜ 3⎠⎟
⎝ ⎝

⎛ 1⎞

(iv) 0, 1, 1 , 3 (v) r = λ ⎜ 2⎟
2 4 ⎜ 3⎠⎟
⎝
(v) (a) It lies between A and B.

(b) It lies beyond B. 4 (i) Yes, λ = 2
(c) It lies beyond A. (ii) Yes, λ = −1
(iii) No
Activity 10.3 (Page 263) (iv) No
(v) Yes, λ = −5
y 2 Note:These answers are not
unique.

8 (i) r = ⎛ 2⎞ + λ ⎛ 1⎞
6 ⎝⎜ 1 ⎟⎠ ⎜⎝ 2⎠⎟
4 ⎛ −1⎞ ⎛ −1⎞
2 ⎛ 3⎞ –1⎞ (i) λ⎜ 3⎟
(i), (iv) (ii) r = ⎝⎜ 5⎠⎟ + λ ⎛ 1 ⎟⎠ 5 r = ⎜ −2⎟ + ⎜⎝ −3⎠⎟
–2 0 ⎝⎜ ⎝⎜ 1⎟⎠
(iii) –2
2 (iii) r = ⎛ −6⎞ + λ ⎛ 1⎞ ⎛ −1⎞ ⎛ −2⎞
4 ⎝⎜ –6⎠⎟ ⎜⎝ 1⎠⎟ or r = ⎜ −2⎟ + λ ⎜ 6⎟
68 x
(ii) (v) ⎛ 5⎞ 1⎞ ⎝⎜ 1⎠⎟ ⎝⎜ −6⎠⎟
(iv) r ⎜⎝ 3⎠⎟ ⎛ 1⎠⎟
= + λ ⎜⎝ (ii) (–2, 1, –2)

(i) and (iv) are the same since (v) r = λ ⎛2⎞ ⎛ −2⎞ ⎛ 0⎞
putting λ = −1 in (i) gives ⎝⎜ 1⎟⎠ r = ⎜ 1⎟ + λ ⎜ 1⎟
(iii) ⎜⎝ −2⎟⎠ ⎝⎜ 0⎠⎟
⎛ 1⎞ ⎛–1⎞
⎝⎜–3⎠⎟ and ⎛ 1⎞ is parallel (vi) r = λ ⎜⎝ 4 ⎠⎟
⎜⎝ 2⎠⎟

to ⎛ 3⎞ . (vii) r = λ ⎛–1⎞
⎝⎜ 6⎟⎠ ⎜⎝ 4 ⎠⎟

344

Exercise 10E (Page 269) 5 (–2, –6, –1); 30 units ℚ Rational numbers – numbers
6 No which can be expressed exactly as
1 (i) ⎛ 4⎞ (ii) ⎛ 5⎞ a fraction
⎝⎜ 1⎠⎟ ⎝⎜ 5⎟⎠
7 6 units, 9 units, 77 units ℤ Integers – positive or negative
⎛ 12⎞ ⎛ –5⎞ whole numbers, including zero
(iii) ⎜⎝ 17⎟⎠ (iv) ⎜⎝ 6 ⎟⎠ Answers
ℕ Natural numbers – non-
(v) ⎛ 6⎞ Exercise 10F (Page 275) negative whole numbers (although
⎜⎝ 3⎠⎟ 1 53.6° there is some debate amongst
2 81.8° mathematicians as to whether zero
2 (i) Intersect at (3, –2, 5) 3 8.72° should be included!)
(ii) Parallel 4 35.3°
? (Page 281)
(iii) Intersect at (3, 2, –13)

(iv) Intersect at (1, 2, 7)

(v) Skew 5 61.0° Any real number is either rational
6 (i) (a) (–2, 6, 7) or irrational.This means that all
(vi) Intersect at (4, –7, 11) real numbers will either lie inside

(vii) Skew the set of rational numbers, or
inside the set of real numbers
3 (i) 12.8 km (b) 29 units but outside the set of rational
(ii) 20 km h−1, 5 km h−1 (ii) (a) (3, –1, 7)

(iii) After 40 minutes there (b) 17 units numbers.Therefore no separate set
is a collision.
(iii) (a) (2, 7, –3) is needed for irrational numbers.
4 (i) O⎯→L = ⎛ 10 ⎞ ; (b) 7 units The symbol ℚ– is used for irrational
⎜⎝ 4.5⎟⎠ numbers – numbers which cannot

⎛ 7⎞ 7 2 10 units be expressed exactly as a fraction,
⎝⎜ 3.5⎠⎟
O⎯→M = ; 8 35 units such as π.

⎛ 4⎞ Activity 11.1 (Page 281)
⎜⎝ 1⎠⎟
O⎯→N = 9 (i) A(4, 0, 0), F(4, 0, 3)

(ii) 114.1°, 109.5° Real

⎛ 1⎞ 2⎞ (iii) They touch but are not Rational 227 √5
⎝⎜ 0⎠⎟ 1⎟⎠ perpendicular. Integers 109
(ii) AL: r = + λ ⎛ ; Natural −13
⎜⎝
numbers 3.1415
7

BM: r = ⎛ 7⎞ + µ ⎛ 0⎞ ; 10 (ii) 5i + 3j + 4k 0.3 π
⎜⎝ 2 ⎟⎠ ⎝⎜ 1⎟⎠
−√5

⎛ 13⎞ 3⎞ 11 (ii) λ = 3
⎜⎝ 7 ⎟⎠ 2⎟⎠ 8
⎛
CN: r = + ν ⎝⎜ 12 (ii) a = −2, a = 3 Activity 11.2 (Page 281)

(iii) (a) (7, 3) 13 (ii) 79.5° (i) x = 2 Natural number

(b) (7, 3) (or integer)

(iv) The lines AL, BM and (ii) x = 9 Rational number
CN are concurrent. 7
(They are the medians Chapter 11
of the triangle, and (iii) x = ±3 Integers
this result holds for ? (Page 280)
the medians of any (iv) x = −1 Integer
triangle.) ℝ Real numbers – any number
which is not complex (v) x = 0, −7 Integers

345

Activity 11.3 (Page 283) (vii) 3 + 29i Activity 11.5 (Page 287)
(viii) 14 + 5i
z = 3 − 7i (ix) 40 + 42i x 1 iy = p + iq
⇒ z2 − 6z + 58 (x) 100 +
(xi) 43 + 76i ⇒ (p + iq)(x + iy) = 1
= (3 − 7i)2 − 6(3 − 7i) + 58 (xii) −9 + 46i
= 9 − 42i + 49i2 − 18 + 42i + 58 ⇒ px + ipy + iqx + iqy2 = 1
= 9 − 42i − 49 − 18 + 42i + 58
Answers =0 ⇒ (px − qy) + i(py + qx) = 1

px − qy = 1 and py + qx = 0

Solving simultaneously gives

? (Page 283) 2 (i) −1 ± i p = x2 x y2, q = x –y
3 + 2 + y2
i3 = −i, i4 = 1, i5 = i 4 (ii) 1 ± 2i 1 x − iy
All numbers of the form so + iy = x2 + y2
» i4n are equal to 1 (iii) 2 ± 3i x
» i4n+1 are equal to i
» i4n+2 are equal to −1 (iv) −3 ± 5i ? (Page 289)
» i4n+3 are equal to −i.
(v) 1 ± 2i 1 = −i, 1 = −1, 1 =i
Activity 11.4 (Page 284) 2 i i2 i3
(i) (a) 6 (b) 2
(vi) −2 ± 2 i All numbers of the form
(c) 34 (d) 5
(i) 2i » 1 are equal to 1
They are all real. (ii ) 5i and –3i i4n
(ii) z + z∗ = (x + iy) + (x − iy) = 2x (iii) 1 + i and –1 + i
(iv) 2 – 3i and –2 – 3i » 1 are equal to –i
zz∗ = (x + iy)(x − iy) (v) –1 – 4i and 1 – 4i i 4n+1
= x2 − ixy + ixy − i2y2 (vi) –3i and 2i
= x2 + y2 » 1 are equal to –1
i4n+2
These are real for any real
values of x and y. » 1 are equal to i.
i4n+3

Exercise 11B (Page 289)

(i) 2 1 (i) 3 – 1 i
(ii) −4 10 10
(iii) 2 − 3i
(iv) 6 + 4i (ii) 6 + 1 i
(v) 8 + i 37 37
(vi) −4 − 7i
(vii) 0 (iii) – 1 + 3 i
(viii) 0 4 4
(ix) −39
? (Page 285) (x) −46 − 9i (iv) 4 + 11 i
(xi) −46 − 9i 5 10
(xii) 52i
Yes, for example 2 4 , although (v) 5 – 1 i
2 ≠4 and 3 ≠6. 3 6 2 2
=
(vi) 7 − 5i

Exercise 11A (Page 285) (vii) −i

(viii) 11 – 27 i
25 25
1 (i) 14 + 10i
(ii) 5 + 2i (ix) 7 + 32 i
(iii) −3 + 4i 29 29
(iv) −1 + i
(v) 21 (x) –1 – 3 i
(vi) 12 + 21i 2

7 a = 1 or 4, b = −1 or 3 2 (i) a = 5, b = 2

The possible complex (ii) a = 3, b = −7
numbers are 1 + 9i, 1 + i;
16 + 9i, 16 + i (iii) a = 2, b = −3

(iv) a = 4, b = 5

346

(v) a= 5 , b = − 3 Activity 11.7 (Page 293) 4 (i) 5 (ii) 13
4 4
(i) Im z2 , (iii) 65 (iv) 5
(vi) a= 1 , b = 1 5 13
2 2
z2 – z1 (v) 13
5
3 a = 2, b = 2
zw = z w , z = z ,
z1 w w Answers
4 (i) z = 2 − i Re
O w w
(ii) z = 3 + i z1 z1 , z = z
(ii) Im
(iii) z = 11 − 10i
1 1
(iv) z= –35 + 149i (i) z−1 = 2 – 2 i,
34
O –z2 Re | z−1 | = 1
5 0, 2, −1 ± 3 i z1 + (–z2) 2

6 2x z0 = 1, | z0 | = 1
x2 + y2
Exercise 11C (Page 294) z1 = 1 + i, | z1 | = 2
8 (i) a3 − 3ab2 + (3a2b − b3)i 1 Im

(iii) z = 1, − 1 ± 1 3i z2 = 2i, | z2 | = 2
2 2
4i
9 (i) (z − α)(z − β) –5 + i 3 + 2i z3 = −2 + 2i, | z3 | = 2 2
= z2 − (α + β)z + αβ
z4 = −4, | z4 | = 4
(ii) (a) z2 − 14z + 65 = 0 –2 O Re
4 – 3i
(b) 9z2 + 25 = 0 z5 = −4 − 4i, | z5 | = 4 2
(ii) Im
(c) z2 + 4z + 12 = 0 –6 – 5i

(d) z2 − (5 + 3i)z + (i) 13 (ii) 4
4 + 7i = 0 (iii) 26 (iv) 2
(vi) 5
10 (i) 3i and –3i (v) 61 Re
(ii) 2 + i and –2 – i
(iii) 3 + 5i and –3 – 5i 2 Im
(iv) 3 – 4i and –3 + 4i
(v) 5 – 2i and –5 + 2i –z z* (iii) The half-squares
(vi) 2 – 3i and –2 + 3i iz* iz formed are enlarged

O Re tbhyrou2ghan4πd eraocthatetidme.
–iz (iz)*

Activity 11.6 (Page 291) –z* z 6 Half a turn about O

(i) Rotation through 180° about followed by reflection in
the origin
3 Points: the x-axis is the same as
(ii) Reflection in the real axis (i) 10 + 5i reflection in the x-axis
followed by half a turn
? (Page 291)
(ii) 1 + 2i about O.
z and −z∗ (or −z and z∗) are
reflections of each other in the (iii) 11 + 7i ? (Page 294)
imaginary axis. (iv) 9 + 3i
(v) −9 − 3i | z2 − z1 | is the distance between
the points representing z1 and z2
in the Argand diagram.

347

? (Page 296) ? (Page 296)

(i) Im (i) Im

Answers 3 + 4i 3 + 4i
Re
O Re –1 + 2i

(ii) Im O

(ii)

Im

3 + 4i 3 + 4i
Re
O Re –1 + 2i

(iii) O

Im (iii)

Im

3 + 4i 3 + 4i
O Re
–1 + 2i

O Re

Exercise 11D (Page 297)

1 (i) Im

2
O Re

348

(ii) Im (vii) Im

O4 Re O Re Answers
1–i
(iii) Im
(viii) Im

5i –2 O Re

O Re 2 Im

(iv) Im A Re
12 – 5i
–3 + 4i B

O Re

(v) Im | z | is least at A and greatest at B.
| 12 − 5i | = 144 + 25 = 13
O Re At A, | z | = 13 − 7 = 6
6–i At B, | z | = 13 + 7 = 20

(vi) Im 3 (i) O 2 4 6 8 Re

O Re –2 R 5 – 4i

–2 – 4i –4

–6

–8
Im

(ii) 7, 13
4 Not possible

349

5 (i) Im (e) −88.9° (f) −89.7°

−90° Ͻ tan−1 x Ͻ 90°

(ii) – π Ͻ tan−1 xϽ π
2 2

Answers O24 Activity 11.9 (Page 300)

Re π
4
arg(1 + i) = ,

(ii) Im arg(1 − i) = − π ,
4
2i 3π
i arg(−1 + i) = 4 ,
O
arg(−1 − i) = − 3π
4

Re ? (Page 301)

(i) 2(cos (π − α) + i sin (π − α))
(ii) 2(cos (−α) + i sin (−α))

(iii) Im Activity 11.11 (Page 301)

–1 + i πππ
O 1 – i Re 463

tan 1 1 3
3

sin 11 3
22 2
(iv) Im
cos 1 31
2 + 6i 22 2

O Re Exercise 11E (Page 302)

1 (i) r = 8, θ = π
5
1
(ii) r = 4 , θ = 2.3
(iii) r =
–5 – 7i 4, θ = − π
3
? (Page 298)
(iv) r = 3, θ = π – 3

(i) π 2 (i) r = 1, θ = 0,
2 z = 1(cos 0 + i sin 0)
3π
(ii) – 4 (ii) r = 2, θ = π,

(iii) – π z = 2(cos π + i sin π)
4
π
Activity 11.8 (Page 299) (iii) r = 3, θ = 2 ,
(i) (a) 45° (b) 63.4°
z = 3(cos π + i sin π )
(c) 89.4° (d) −63.4° 2 2

350

(iv) r = 4, θ = − π , 6 (i) Im
2 4
π π 3
z = 4(cos (− 2 ) + i sin (− 2 )) 2
1
(v) r= 2 , θ = π ,
4
π π Answers
z= 2 (cos 4 + i sin 4 )

(vi) r=5 2 , θ = − 3π , –3 –2 –1 O 1 2 3 Re
4 –1

z=5 2 (cos (− 3π ) + i sin (− 3π ))
4 4

(vii) r = 2, θ = − π3π3,) –2
z = 2(cos (−
π (ii) Real part = 1
3 4
+ i sin (− ))
7 (i) (a) 2 + i
(viii) r = 12, θ = π , (b) r = 5 , θ = 0.464
6
(ii) −3 + 2i and 3 − 2i
z = 12(cos π + i sin π )
6 6

(ix) r = 5, θ = −0.927, 8 (i) Im
z = 5(cos(−0.927) + i sin(−0.927))
3
(x) r = 13, θ = 2.747,
z = 13(cos 2.747 + i sin 2.747) 2

(xi) r = 65 , θ = 1.052, 1A
z = 65 (cos 1.052 + i sin 1.052)
O C
1 2 3 4 5 Re

(xii) r = 12013, θ = −2.128, –1 B

z = 12013 (cos (−2.128)+ i sin (−2.128) –2

3 (i) z = 2i OACB is a rhombus.

(ii) z= 3 + 3 3 i (ii) 3 + 4 i
2 2 5 5

(iii) z= – 7 3 + 7 i 9 (iii) The locus is a circle, centre 2i, radius 5.
2 2
1 – 1i ? (Page 304)
(iv) z= 22

(v) z= – 5 – 523 i arg(z1 − z2) is the angle between the line joining
2 z1 and z2 and a line parallel to the real axis.
(vi) z = −2.497 − 5.456i
Exercise 11F (Page 306)

4 (i) α − π 1 (i) Im
(ii) −α

(iii) π − α

(iv) π −α
2
π
(v) 2 +α O –π3 Re

5 (ii) Real part = 1
2

351

(ii) Im (iii) The three points are the same

4i distance from the origin and separated by

O Re equal angles of 2π (i.e. 120°).
3
(iii) Im (iv) –(2 + 3) + (2 3 – 1)i

Answers –(2 – 3) – (2 3 + 1)i

4 (ii) −6 ഛ p ഛ 2
(iii) z − 5 = 5

–3 O Re 5 (i) 1 − 3i, −1 − 3i
(ii) Im

2

1

(iv) Im –3 –2 –1 O 1 2 3 Re
–1

–2

(iii) 1− 3i: r = 2, θ = − π
3

O Re −1 − 3i: r = 2,
2π
–1 – 2i θ = − 3

(v) Im (iv) The three points are the same distance

from the origin and separated by equal
2π
angles of 3 (i.e. 120°).

6 (i) (a) 1 + 2i

O π Re (b) – 1 + 21i
3–i 6 2
– 3π
(ii) 4

(vi) Im (iv) OA = BC and OA and BC are parallel

7 (i) u: r = 2 ,θ = – 3 π
4
u2: r = 2, θ = 1 π
2

–5 + 3i π (ii) Im
3

– π 4
4

O Re 3
2 u2

1

–4 –3 –2 –1 O 1 2 3 4 Re
u–1
2 π , 2π –2
3 3

3 (i) r = 1, θ = 2 π –3
3
(ii) wz: modulus R, argument 2 π –4
modulus = R, argument = θ + 3
z = = θ – 23π
352 w :

8 (i) 3 ( )(ix)3 2 cos 7π + i sin 7π
12 12
(ii) 3 at z = 3 − 3 i
2 2 2 3 Exceptions

(iii) 3+ 1 (i) if z = 0 then 1 does not exist
2 z

( )(iii) 1 = arg z Answers
Activity 11.12 (Page 308) if z = real and negative then arg z

(i) Rotation of vector z through + π 4 (i) Enlarge from O ×3
2
(ii) Enlarge from O ×2 and rotate π
(ii) Half turn of vector z + 2

(= two successive π (iii) Complete the parallelogram 3z, 0, 2iz
2
(iv) Reflect in the real axis
rotations: –1 = i × i )

? (Page 309) (v) Find where the circle with centre O
through z meets the positive real axis
3 + i and −3 − i
(vi) Complete the similar
triangles 0, 1, z and 0, z, z2

Exercise 11G (Page 312) 5 3− 1, 3+ 1;
4 4
1 (i) 32(cos0.6 + i sin 0.6)
( )2cos 3π + i sin 3π ,
(ii) 2(cos(−0.2) + i sin(−0.2)) 4 4

( )(iii)12 cos π + i sin π ( )2 cos π + i sin π ; 3+ 1
2 2 3 3 22

( )(iv)3 cos π + i sin π 6 (i) –1
6 6
1+ i
( )(v) 24 cos 5π + i sin 5π (ii) 2
4 4
(iii) –1.209 + 0.698i

( )(vi)6 cos 3π + i sin 3π (iv) –13.129 + 15.201i
4 4

( )2 (i) 6 cos 7π + i sin 7π 7 (i) (a) 10ei (b) 4
12 12

( )(ii)3 π π (c) 6e8i (d) 3e i
2 12 12
cos + i sin (e) 3e3i (f) 4e−i

( ) ( )(iii)2⎣⎢⎡cos– π π ⎤ (ii) (a) 2(cos 3 + i sin 3)
3 12 + i sin – 12 ⎦⎥ × 5(cos(−2) + i sin(−2))
= 10(cos 1 + i sin 1)
( ) ( )(iv)21 ⎡⎢⎣cos π π ⎤
− 4 + i sin − 4 ⎦⎥

( )(v) 9 cos 2π + i sin 2π (b) 8(cos 5 + i sin 5)
3 3 ÷ 2(cos 5 + i sin 5)
=4
32⎡⎢⎣cos 3π 3π ⎤
( ) ( )(vi) 4 – 4 ⎦⎥ (c) 3(cos 7 + i sin 7)
– + i sin × 2(cos 1 + i sin 1)
= 6(cos 8 + i sin 8)
(vii) 432(cos 0 + isin 0)

( )(viii)10cos 3π + i sin 3π
4 4

353

(d) 12(cos 5 + i sin 5) 2 2 – i, –3
÷ 4(cos 4 + i sin 4)
= 3(cos 1 + i sin 1) 3 z = 7, 4 ± 2i

(e) 3(cos 2 + i sin 2) 4 p = 4, q = –10, other roots 1 + i, –6
× (cos 1 + i sin 1)
Answers = 3(cos 3 + i sin 3) 5 z = 3 ± 2i, 2 ± i

(f) 8(cos 3 + i sin 3) 6 z = ±3i, 4 ± 5
÷ 2(cos 4 + i sin 4)
= 4(cos(−1) + i sin(−1)) 7 7, 4 ± 2i

8 (i) −16.3° 8 (i) z = –3

(ii) Im (ii) z = –3, 5 ± 11 i
2 2
5
k 36, other roots are 5 323 i
9 = − 2 ±

10 z3 − z – 6 = 0

−5 O 2.5 5 Re 11 (i) False (ii) True
(iii) True (iv) True

−5 12 a = 2, b = 2, z = −2 ± i,1 ± 2i

Complex numbers common to both loci: 13 (ii) 1 – 2i
(iii) Im
5e ± 1 π i
3

9 (i) − 7 + 23 i
17 17
2
(ii) wz = 17 + 17i O1

Exercise 11H (Page 315) Re

1 4 + 5i is the other root.
The equation is z2 − 8z + 41 = 0

354

Index

A differential equations Fundamental Theorem of Algebra Index
forming from rates of change 313–14
acceleration, definition 218 218–22
addition and subtraction general solution 223 G
particular solution 225–32
of complex numbers 291–3 verifying the solution 228–9 gradient
of fractions 173 of exponential curves 25
of polynomials 2–3 differentiation of a straight line 37
of vectors 241–3 of exponentials 90–3
algebraic fractions 172–4 implicit 184 graphs
alternating current (AC) 78 of implicitly defined functions displacement–time 83
angle 101–8 of exponential functions 25–6
between two lines 272–3 of inverse trigonometrical of logarithms 33
between two vectors 250–8 functions 184–816 of the natural logarithm function
Argand diagrams 291–3, 294–7, 304–7 of natural logarithms 90–3 48
parametric 112–18 of parametric equations 110
B product rule 84–6 of polynomial equations 7
quotient rule 86–7 of trigonometrical functions 58–9
binomial expansion of trigonometrical functions
general 163–71 96–101 growth and decay, exponential 24,
use of partial fractions 180–2 26–7, 49
direction, of a line 261–2
binomial theorem 166 displacement vectors 238–9 I
displacement-time graphs 83
C distance, from a point to a line 273–7 identities see trigonometrical identities
division see multiplication and division identity symbol 199
change-of-sign methods 146–51 dot product 251–3 imaginary number 282
cobweb diagram 154, 155 double-angle formulae 67–71 implicit functions, differentiation
common factors 173
complex conjugates 283–4, 313 E 101–8
complex exponents 310–13 indices, laws 30
complex numbers e, base of natural logarithms 38, 46–7 inequalities, involving the modulus
equation of a line 37
addition and subtraction 291–3 sign 19–21
definition 283 vector form 258–65 integrals
equality 284–6 equations
and equations 313–16 involving exponentials 191–4
geometrical representation 291 and complex numbers 313–16 involving natural logarithms
modulus 293–4 rearranging 152–6
modulus–argument (polar) form error, bounds 149 191–4
estimates, accuracy 135 standard 213
297–304, 307–10 existence theorem 314 integration
multiplication and division experimental data, curve fitting 37–41 by change of variable 186–91
exponential functions by parts 203–12
287–8, 307–10 applications 39–41, 48–52 by substitution 186–91
need for 281–2, 313 and complex numbers 310–13 choice of technique 212–14
notation 282 definition 24, 34, 48–52 of exponentials 90–3
real and imaginary parts 283 differentiation and integration involving the exponential function
square root 288–90
complex plane 291 90–3, 121–2, 191–4 121–2
compound interest 128 graphs 25–6 involving the natural logarithm
compound-angle formulae 61–6, 72–8 growth and decay 24, 26–7, 49
convergence to a limiting value 151 as infinite series 128 function 122–7
cosecant (cosec) 58 of natural logarithms 90–3
cotangent (cot) 58 F numerical 133–40
cubic expressions see polynomials of trigonometrical functions
curves, discontinuous 125 factor theorem 8–10
fixed-point iteration 151–61 129–33, 194–9
D fractions use of partial fractions 199–203
intervals
decimal search 146–8 addition and subtraction 173 bisection 148
Devi, Shakuntala 162 multiplication and division 172 estimation 146–51
simplification 172 notation 145
inverse functions 34, 48
iteration 151, 153, 155

355

Index L parametric equations 108–12 spiral dilatation 309
partial fractions square root
line segment 239
lines types 174–80 of a complex number 288–90
use in integration 199–203 of a negative number 282
angle between 272–3 using with the binomial expansion staircase diagram 154, 155
direction 261–2 stationary points
gradient 37 180–2 of implicit functions 103–6
intersection 265–71 Pascal’s triangle 164 of a parametric curve 115
location on 262–3 point straight line see equation of a line, line
perpendicular distance from 273–7
skew 266, 273 coordinates 237 segment; lines
location, on a line 262–3 perpendicular distance from a line subtraction see addition and
logarithm function 34, 46, 47
inverse 34 273–7 subtraction
logarithmic integrals 123–4 polynomial equations
logarithmic relationships 38–9 T
logarithms 28–33 graphs 7
see also natural logarithms roots 11–12 trapezium rule 133–40
lowest common multiple 173 solution 7–16 trial and improvement 144
trigonometrical equations, general
M with complex numbers 313
polynomials solution 79–80
magnitude trigonometrical functions
of a quantity 17 addition and subtraction 2–3
of a vector 236 definition 1–2 differentiation 96–101
multiplication and division 3–6 graphs 58–9
modelling order 2 integration 129–33, 194–9
use of logarithms 37–41 position, definition 218 inverse, differentiation 184–6
use of trigonometric functions position vector 237–8, 262 reciprocal 58–60
57, 61, 83 principal argument, of a complex trigonometrical identities 59, 61, 67,
of waves 57, 61, 75, 184
number 298 130–1
modulus principal value 69, 79
of a complex number 293–4 product rule 84–6 U
of a vector 236
Q unit vector 235, 245–7
modulus function 17–21
multiplication and division quadratic expressions 1 V
quadratic formula 7, 313
of complex numbers 287–8, quotient 4, 6 vector product 252
307–10 quotient rule 86–7 vectors

of fractions 172 R addition and subtraction 241–3
of logarithms 30 angle between 250–8
of polynomials 3–6 Ramanujan, Srinivasa 1 calculations 240–50
of a vector by a scalar 240–1 rate of change 218 components 235
rational functions see algebraic definition 234
N equal 237
fractions of geometrical figures 244
Napier, John 48 real and imaginary axes 291 joining two points 258–9
natural logarithm function 45–8,122–7 reciprocal trigonometrical functions length 239
natural logarithms magnitude 236
58–60 magnitude—direction (polar)
base 38 remainder theorem 12–13
differentiation and integration resultant 242 form 235
roots multiplication by a scalar 240–1
90–3, 191–4 negative of 241
natural numbers 163 of logarithms 32 notation and terminology 218,
negative number, square root 282 number 75, 145
Newton’s law of cooling 217 of polynomial equations 7–8, 235–6
numbers, types 280, 281–2 perpendicular 251–2
numerical methods 11–12 polar form 235
of trigonometric equations 79–80 in three dimension 236–40,
accuracy 135, 144–6
integration 133–40, 142–3 S 252–3
for polynomial equations 8 in two dimensions 235–6
scalar, definition 234 uses in mathematics 234
P scalar product 251–3 velocity, definition 218
secant (sec) 58
parameter, eliminating 111–12 separation of variables 223–5 W
parametric differentiation 112–18 sets of points, on an Argand diagram
waves, modelling 57, 61, 75, 184
294–7, 304–7
sine function, differentiating 83

356

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