Pure
Mathematics
for
CAPE®
Unit 2
Pure
Mathematics
for
CAPE®
Kenneth
Charles
Sue
Baisden
Cadogan
Chandler
Mahadeo
Deokinandan
Unit 2
3
Great
Clarendon
Oxford
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Acknowledgements
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TechSet
referenced
for
in
all
at
Contents
Introduction
5
1.
16
1.
17
Section
1
Complex
calculus
1.
1
1.2
Complex
numbers
and
numbers
Operations on
The
Integration of
and
40
exponential
logarithmic functions
42
2
1.
18
Partial fractions
1.
19
Applications of
44
6
complex
numbers
1.3
Partial differentiation
partial
fractions
48
8
square
roots of
a
1.20
Integration
using
substitution
1.21
Integration of trigonometric
50
complex
number
10
The Argand diagram
12
functions
1.4
1.5
1.6
Modulus
and
Graphical
argument
1.7
1.22
Integration
by
1.23
Integration of
parts
54
14
representation of
operations on
52
inverse
trigonometric functions
56
1.24
Reduction formulae
58
1.25
The trapezium
62
complex
numbers
16
De
18
Moivre’s theorem
Section
1.8
Complex
numbers
1.9
Differentiation of
and
loci
1
rule
Practice questions
64
22
Section
2
Sequences,
series
and
exponential
approximations
functions
26
2.
1
1.
10
Differentiation of
functions, tangents
normals to
1.
11
1.
12
and
parametric
curves
Implicit functions
2.2
28
30
Inverse trigonometric
functions
1.
13
Differentials of
and divergence of
68
2.3
Number
2.4
Method of differences
2.5
Proving
inverse
and
series
70
74
sequences
series
76
Power
series
and
Maclaurin’s
theorem
78
36
2.7
Second differentials
properties of
34
combinations of
function
1.
15
Convergence
sequences
2.6
Derivatives of
66
32
trigonometric functions
1.
14
Sequences
logarithmic
38
Applications of
theorem
Maclaurin’s
82
3
Contents
2.8
Taylor’s theorem
and
applications
2.9
2.
10
Derivation of the
binomial
theorem for

n

Applications of the
The
n
binomial

Matrix
3.
11
Square
n

unit
3.
12
binomial

3.
13
expansion for
2.
13
Interval
2.
14
Linear
a
root of
an
equation
98
matrices,
matrices
zero
and
matrices,
inverse
144
Determinants,
minors
and
cofactors
92
146
Simplification of
determinants
94
Locating
140
matrices
88

2.
12
multiplication
84
expansion for
2.
11
3.
10
3.
14
150
Multiplicative
inverse of
a
matrix
bisection
152
102
3.
15
interpolation
Systems of
2

2
linear
104
equations
2.
15
Newton–Raphson
2.
16
Using
method
106
3.
16
156
Systems of
3

3
linear
equations
a
given
iteration
3.
17
Section
2
Practice questions
Using
row
3
Counting,
matrices
differential
3.
1
The
principles of
reduction to find
an
112
inverse
Section
160
108
matrix
164
3.
18
Differential
equations
3.
19
Integrating factors
3.20
First order differential
166
and
equations
counting
168
114
equations
3.2
Permutations
3.3
Combinations
170
116
3.21
Differential
equations of the
120
2
d
y
dy
y
____
form
3.4
Sample
3.5
Basic
3.6
Probability that
does
3.7
spaces
an
126
event
happen
Probabilities
3.9
4
more
d x

cy

0
172
d x
3.22
The
particular
integral
1
176
3.23
The
particular
integral
2
178
3.24
The
particular
integral
3
180
3.25
Using
boundary
3.26
Using
substitution
128
conditions
182
130
Probabilities
or
b
involving two
events
3.8
___

2
122
probability
not
a
184
involving two
events
Introduction to
matrices
134
Section
138
Index
3
Practice questions
186
188
Introduction
This
Study
Guide
has
been
developed
exclusively
with
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•
On
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our
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These
Y
ourself
you
that
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syllabus.
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on
assist
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can
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Pure
Mathematics.
5
1
Complex
1.
1
Complex
Learning outcomes
numbers
and
calculus
2
numbers
Imaginary
numbers
2
When

To
define
imaginary
To
define
complex
try
to
solve
the
equation
2
x

1

0,
we
get
x

1
giving
___
x

we
numbers


√
1
numbers
Up
to
this
number
with
point,
we
have
left
whose
square
is
1,
equations
whose
roots
it
at
so
are
the
the
not
statement
equation
real,
we
that
has
no
need
to
there
real
is
no
roots.
introduce
real
T
o
work
another
You need to know
type
of
number
.
___

How
to
solve
a
If
quadratic
equation
we
introduce
the
symbol
i
to
√
represent
1,
we
can
say
that
the
equation
2
x

1

0
has
two
roots,
i
and
i.
___

The
relationship
between
the
i
coefficients
of
a
is
called
an
imaginar y
number ,
quadratic
and
the
roots
of
i
√

1
___
2
equation
where
It
the
follows
that
i

(
2
√
1)

1
equation
This

How
to factorise
a
cubic
is
consistent
product
of
the
expression
of
what
a

1

we
know
quadratic
b
__
2
x
with
roots
about
equation:
the
for
sum
the
of
the
roots
and
the
equation
c
__
0,

0
and

a
1;
and
the
sum
of
the
roots
is
i

(
i)

0
a
2
and
the
product
Any
of
negative
the
roots
number
imaginary
is
i
has
number

(
two
and
i)

i
square
can
be

(
roots,
1)
each
expressed
in

of
1
which
ter ms
___
For
example,
the
square
roots
of
4
are

√
Imaginary
For
the
square
numbers
example,
2i
i
roots
can

be

9i
i

i(
3
49
added,
7i
√
of
√
3
are

√

49
subtracted,
is
an
i.
___
4

√
4
_____
and
of

√
1
___


√
49
multiplied

2i
___

and
√
1

7i
divided.
1)
2
Powers
of
i
2i

7i

14i
10i

5i

2
can
be
example,
1
__
1
i

2
i

i
__
14

1

14
simplified.
3
For

(i
4
)

i

i,
i
2

(i
2
)
2

(
1)

1

2i
and
i
___



i
2
i
i
1
Complex
numbers
2
Consider
The
the
solution
quadratic
of
this
equation
equation
x

___
√
2  4
1
__________

These
numbers
Numbers
A
of
is
the
are
this
complex
a
6
of
the
number
called
the
sum
are
is
real
0
1
2
equation
form


2
roots
5
is

two

√
2 
16
__________
2
The
2x
_____
_______
√
2 
4
20
____________
x
are
of
a
called
of
real
for m
and
1
number
complex
the
part
therefore
ib
a
is

2i
and
and
an
1
2i
imaginary
number
.
numbers

ib
where
called
the
a
and
b
are
imaginary
real.
part.
Section
Like
much
problem,
main
used
mathematics
complex
application
to
that
numbers
of
and
devised
have
complex
understand
is
many
numbers
analyse
to
deal
with
applications
is
in
alternating
signals.
to
√
denote
current
in
Complex
numbers
and
calculus
2
theoretical
in
real
electronics,
___
i
a
1
life.
where
The
they
are
Mathematicians
use
___
1
but
engineers
use
j
to
√
denote
1
because
i
is
used
for
electronics.
Conjugate
complex
numbers
2
The
and
roots
1
These
of
the
x
2x

5

0
were
found
above
to
be
1

2i
2i
two
Any
complex
two
conjugate
We
equation
use
z
to
conjugate,
numbers
complex
complex
represent
denoted
are
numbers
numbers
a
by
called
of
(or
by
the
and
complex
z*
conjugate
for m
each
is
number
,
z),
is
a

the
so
given
complex
ib
a
conjugate
when
by
and
z*
z


a
a
numbers

of

ib
are
the
ib,
other.
its
ib
2
The
solution
of
the
general
quadratic
________
2
√
b 
b
4ac
_______________
x
equation
ax

bx

c

0
is
given
by
________


2a
2
√
4ac
b
__________
b
___

i
2a
2a
________
√
4ac
b
__________
b
___
Using
p
2

and
q

,
2a
Therefore
complex
We
know,
real
and
when
roots,
from
complex
In
can
factor
.
when
the
a
quadratic
those
Unit
coefficients
quadratic
these
roots
can
be
expressed
as
p

iq
2a
roots
1,
be
that
the
roots
a
equation
a
to
cubic
are
pair
of
left-hand
factorised
Therefore
other
are
not
give
with
side
one
equation
real,
real
conjugate
they
of
a
cubic
linear
will
will
coefficients
complex
equation
factor
always
be
a
has
numbers.
and
have
pair
of
with
one
one
real
root
conjugate
numbers.
fact,
expressed
any
as
a
polynomial
product
Therefore
any
of
equation
quadratic
complex
roots
complex
with
factors
will
real
and
be
coefficients
possibly
pairs
of
can
linear
be
factors.
conjugate
numbers.
Example
2
Find
all
the
roots
2
of
the
equation
2
(x
3x

2
(x
3x

2)(x
2)(x

x


2)
2)

0
2

x

2)

(x
2)(x
1)(x

x
______
√
1 
1
8
_____________
2
The
roots
of
(x

x

2)

0
are

2
2
∴
the
roots
of
√
7
___
1
__


i
2
2
2
(x
3x

2)(x
√
7
___
1
__

x

2)

0
are
1,
2,
i
2
√
7
___
1
__
,
2

i
2
2
Exercise 1.1
1
Simplify:
2
5
(a)
5
i
(b)
i
4n
(d)
i
(g)
8i
Find
2n
(c)
i
(a)
x
8n  1
(e)

i
all
the
roots
of
each
equation.
2

5x

2x

8

0
3

0
3
(f )
5i

2i
(b)
x
2i
7
1.2
Operations
Learning outcomes
on
Equality of
T
wo

To
add,
subtract,
multiply
complex
complex
complex
numbers,
a
numbers
numbers

ib
and
c

id
are
equal
if
and
only
if
the
real
and
parts
divide
complex
are
equal
and
the
imaginary
parts
are
equal,
numbers
i.e.
a

ib

c

id
⇔
a

c
and
b

d
You need to know
Addition

The form

The
of
a
complex
The
meaning
complex
of
conjugate
The
relationship
coefficients
equation
product
subtraction of
complex
numbers
real
parts
and
the
imaginary
parts
are
added
and
subtracted
separately.
numbers
For

and
number
of
and
of
its
a
between
example,
(2

3i)

(5
2i)
quadratic
the
sum
roots

(2


7

(2

3
5)

(3i
2i)
5)

(3i
(
the

i
and
and
(2

3i)
(5
Multiplication of
T
wo
complex
expand
(a

numbers
b)(c

2i)
complex
are

2i))
5i
numbers
multiplied
together
in
the
same
way
that
we
d)
2
For
example,
(2

3i)(5
2i)

10
4i


10

11i

16

11i

4

4

13
15i

6(i
)
6
2
and
(2
The
fact
product
that
of
a
(2


pair
3i)(2
3i)
3i)(2
of
3i)
conjugate

6i

13
is

6i
9(i
9
a
complex
particular
numbers
2
This
is
because
Division of
We
can
divide
multiplying
(a
one
This
gives
example,
a

ib)

a
number
and
real
(2

the
by
another
denominator
.
3i)(5
(5
2i)(5


2i)
2i)
2
10

19i

6(i)
_______________

25

19
___
4
___


29
8
(ib)
a
of
real
the
a
fact
that
the
number
.
2

2

i
29
complex
denominator
_______________
2i
case
2


5
is
b
numbers
complex
numerator
2  3i
______
For
ib)(a
complex
the
denominator
.

)
4
by
the
number
by
conjugate
of
the
Section
1
Complex
numbers
and
calculus
2
Example
Find
∴
the
values
of
a
2z

3z*

2(a
2z

3z*

5
Equating
real
5a
and
b


b
where
ib)
2i
and

and

⇔
z
3(a
imaginary
⇒
a

1
2
⇒
b

2
a
ib)
5a
5

ib
parts

ib


such
5a
5
that
2z

3z*

5
2i
ib
2i
gives
Example
One
root
Find
the
of
a
quadratic
equation
with
real
coefficients
is
3
i.
equation.
2
Let
If
the
one
The
equation
root
sum
is
of
be
3
i,
the
ax

the
other
roots
is
bx

c

root
(3
i)

is
0
its
conjugate,
(3

i)

6
i)(3

i)

9
3

i
b
__
∴

6
a
The
product
of
the
roots
is
(3

1

10
c
__
∴

10
a
2
The
equation
is
x
6x

10

0
Example
Find
∴
the
values
of
x
and
y
for
which
(3
2i)(x

iy)

3x

2y
(3
2i)(x

iy)

16

11i
(3
2ix
2i)(x


iy)

16

11i
3iy
⇔
3x

2y
⇔
3x

2y

2ix

3iy
16
[1]

16

and
11i
2x

3y

11
[2]
Equating
Solving
2

[1]
[1]
and

3

[2]
simultaneously
[2]
13 y
⇒
from
[1]
real
and
imaginary
parts
gives

65
y

5
x

2
Exercise 1.2
1
Find,
in
the
form
a

3
ib
Find
the
values
2
i
_____
(a)
(2
4i)(
1

2i)
(b)
(a)
3
One
root
of
coefficients
(x

iy)
x

a
is

2i

5i.


3
quadratic
3
y
for
which
5
12i
iy
(b)
1
i
and
______

2
x
i
4  3i
______
3
_____
(c)
2

of
2
equation
Find
the
with
real
equation.
4
Find
that
2

i
4i
the
z(2
values

i)

of
a
2z*
and

4
b
where
z

a

ib
such
5i
9
1.3
The
square
Learning outcomes
roots
The
If

To find
the
square
roots
of
x
of
square

iy
is
a
a
complex
roots of
square
root
a
of
complex
the
complex
number
number
number
a

ib,
then
a
2
(x
complex

iy)

a

ib
number
2
Expanding
Equating
the
real
left-hand
and
side
imaginary
gives
parts
2
x
gives
y

2ixy
a
pair
of

a

ib
simultaneous
equations
You need to know
which

How
to
multiply
we
can
solve
to
find
values
for
x
and
y
complex
The
equations
are
quadratic,
so
there
will
be
two
values
for
x
and
y.
numbers

The
relationship
roots
and
quadratic
between
coefficients
of
the
Therefore
a
complex
number
has
two
square
roots.
a
equation
Example
Find
the
square
roots
of
3
4i
2
If
x

iy
is
a
square
root
of
3
4i,
then
(x
2
⇒
real
and
imaginary
2
parts
iy)

3
4i

3
4i
2
x
Equating

y

2ixy
gives
2
x
y

3
[1]
2 xy

4
[2]
y


2
__
[2]
⇒
[3]
x
4
__
2
[3]
in
[1]
⇒
x

3
4

0

1)

0
number
,
so
x
2
x
4
⇒
2
x
3x
2
⇒
2
(x
4)(x
2
x
is
a
real

1
does
not
give
a
valid
value
for
x
2
∴
x
from
∴
4
⇒
[3]
the
Note

square
that
number
,
this
z
is
roots
x

2
y

1
of
example
the
or
3
or
4i
shows
other
2
1
are
2
that
if
square
i
z
is
and
one
2

i
square
root
of
a
complex
root.
Exercise 1.3a
1
Find
10
4
square
roots
of
2i
(a)
2
the

3i
(b)
is
one
square
(a)
Find
the
values
(b)
Find
the
other
root
of
a
of
and
square
8
a

b.
root.
6i
ib
(c)
1

2i
√
2
Section
Quadratic
equations
with
complex
1
Complex
numbers
and
calculus
2
coefficients
2
A
quadratic
using
the
equation
for mula,
such
which
as
z

leads
to
(6

i)z
fi nding

10
the

0
can
square
root
be
of
solved
a
complex
number.
2
For
example,
to
solve
z

(6

i)z

10

0
____________
2
(6

i)
√(6


i)
40
________________________
⇒
z

2
_________
√
6
i 
5  12i
___________________
⇒
z

2
_________
If
a

ib
√

imaginary
5
parts

12i ,
then
squaring
both
sides
and
equating
real
and
gives
2
2
a
b
and

5
[1]
2 ab

12
[2]
b

6
__
From
[2]
[3]
a
36
___
2
[3]
in
[1]
⇒
a

5
36

0
4)

0
2
a
4
⇒
2
a

5a
2
⇒
2
(a
∴
a


2
9)(a
and
b

3
or
a

2
and
b

3
_________
√
i.e.
5

12i
z


2
6
i


3i
(2

3i)
_________________
Hence
2
⇒
z

Notice
2
that
numbers.
quadratic
We
can

i
the
or
roots
Roots
are
equation
check
product
of
the



z
the

of
4
this
only
with
equation
pairs
real
answer
Exam tip
2i
of
are
coefficients
to
the
not
conjugate
example
a
pair
of
complex
has
complex
above
conjugate
numbers
complex
when
roots.
using
the
sum
a
Note
that
arithmetic
it
is
very
easy
mistakes
with
complex
your
working.
to
when
numbers,
make
working
so
check
and
roots:
b
__

(
2

i)

(
4
2i)

6
i


a
c
__
and


(
2

i)(
4
2i)

10

a
Exercise 1.3b
1
(a)
Find
the
complex
numbers
u

x

iy,
x,
y

,
where
2
u
(b)

16
Hence

solve
30i
for
z
the
quadratic
equation
2
z

(1

i)z

(4
7i)

0
2
2
Solve
for
z
the
quadratic
equation
z
(3
i)z

(14
5i)

0
11
1.4
The Argand
Learning outcomes
The Argand diagram
A

To
represent
a
diagram
complex
complex
ordered
on
an Argand
number
,
a

ib,
can
be
represented
on
a
diagram
by
using
the
number
pair
( a,
b)
to
represent
a
point
A
on
a
pair
of
perpendicular
axes,
diagram
as
shown
on
the
diagram.
___
Then
the
vector
OA
You need to know

The
difference
represents
the
complex
number
a

ib
iy
between
a
A(a,
b)
b
displacement
position

How
to
vector
and
a
vector
represent
the
sum
This
and
difference
of
is
called
the
vectors
imaginary
axis
geometrically
a
O
This
This
is
called
an
Argand
is
called
the
real
x
axis
diagram
___
Any
complex
point
(x,
number
x

iy
number
5

3i
can
be
represented
by
OP
where
P
is
the
y).
___
The
complex
point
Any
(5,
can
be
represented
by
OA
where
A
is
the
3).
other
represent
vector
5

with
3i,
for
the
same
example
length
and
___
___
DE
or
direction
can
also
be
used
to
BC.
iy
E(
1,
4)
A(5,
3)
z
C(9,
2)
z
z
D(
6,
1)
x
O
B(4,
Therefore
vector
.
It
a
represented
12
complex
can
also
by
be
the
number
can
represented
point
A.
be
by
1)
represented
a
position
by
a
vector
,
displacement
when
it
can
also
be
Section
When
z

5

3i,
the
vector
representing
z
1
arrow
must
be
marked
with
1
Complex
numbers
and
calculus
2
an
1
to
show
its
direction.
iy
A(5,
3)
3
2
z
1
1
x
O
2
1
1
2
3
4
5
1
2
Example
(a)
Illustrate
the
on
complex
an
Argand
numbers
diagram
z

3
the
2i
points
and
z
1
(b)
On
the
same
of
the
diagram
vectors

and
1
B

representing
2i,
respectively.
2
illustrate
z

z
1
terms
A
representing
and
interpret
the
result
in
2
z
and
z
1
2
iy
(a)
3
B(
1,
2)
2
z
2
1
z

z
1
2
C(2,
0)
x
2
1
O
1
z
2
z
1
2
A(3,
(b)
z

z
1

(3
2i)

(
1

2i)

2)
2
2
____
This
is
____
represented
___
___
OC

by
OA

the
point
C
and
the
vector
OC.
___
AC
and
AC
represents
z
2
Therefore
the
vector
representing
z

1
of
the
vectors
representing
z
and
z
is
represented
by
the
sum
2
z
1
2
Exercise 1.4
1
Given
On
2
z
the
Find

3
same
the

2i,
diagram
square
Represent
z
represent
and
roots
its
z
on
represent
of
two
z

an
Argand
diagram.
z*.
2i
square
roots
on
an
Argand
diagram.
13
1.5
Modulus
Learning outcomes
and
The
The

To
define
the
modulus
argument
modulus of
point

To
of
introduce
form
of
a
a
complex
the

ib)
can
be
located
O
and
the
,
angle,
using
___
that
OA
number
the
makes
distance,
with
the
r,
of
A
positive
from
the
x-axis.
number
iy
polar-argument
complex
complex
and
origin
argument
A( a
a
number
A(a,
b)
r
You need to know
θ

How
to
number
represent
on
a
complex
an Argand
x
O
diagram
The
|a
length

of
OA,
r,
is
called
the
modulus
of
a

ib,
which
is
written
as
ib|
_______
∴
|a

ib|

r
2
√

a
2

b
_______
For
example,
The
The
the
modulus
argument of

angle
is
called
a
the
of
3
4i
complex
argument
of
2
√
is
2
3

4

5
number
a

ib
and
is
written
as
arg ( a

ib)
b
__
∴
arg (a

ib)


tan 
where

and





a
T
o
so
find
you
For
are
the
can
argument
see
example,
drawn
in
which
the
the
iy
of
a
complex
quadrant
complex
it
number
,
is
numbers
diagrams
3)
(
4,
4
on

3i,
4

3i,
an
4
iy
3)
Argand
3i
4
3i
iy
4
θ
x
θ
5
3
x
3
3
5
5
θ
θ
4
x
x
4
(
4,
3)
(4,
3
4

3i
is
in
the
first
quadrant,
so
tan



⇒

0.644 rad
4
4

3i
is
in
the
second
quadrant,
3
tan 


so
1
⇒



4
3i
is
in
the
3
tan

4
2.50 rad
4
third
quadrant,
so

is
negative
3
Therefore
tan 

1
⇒




4
3i
is
in
the
fourth
tan 

obtuse.

2.50 rad
4
quadrant,
so

is
negative
3
Therefore
and
3
tan
4

⇒
4
14
diagram
and
4
5
3
it
below.
iy
(4,
draw
in.


0.644 rad
and
acute.
3)
Section
The
polar-argument form of
a
complex
1
Complex
numbers
and
calculus
2
number
iy
P(x,
y)
r
y
In
the
diagram
x
r cos 

Therefore
x
Hence

r(cos 
x

above,

iy
iy
x
and
can
i sin )
y


be
is
iy
is
any
number
r cos 
as

that
the
‘ ’
sign
the
is
polar-argument
form
important:
of
a
complex
3 (cos
i sin
form
but
can
be
)
is
converted
as

cos
(
in
polar-
sin

sin
(
and
)
3
)


i sin
easy

3
(cos
(
convert

)
3
to
examples

)
3
is
not
3
3(cos
It
number
.

3
∴
that

3
cos
see
3


can

3
argument
we
ir sin 

Note
and
i sin )

called
complex
r sin 
written
r(cos 

x
x
O

i sin
(
))
3
between
the
3
two
forms,
as
the
following
worked
show.
Example
Express
1

i
in
the
r(cos 
form

iy
i sin )
______
1
3
___
1

i
⇒
r

√
1

1

√
2,
and
from
the
diagram,


4
3
___
Therefore
1

i
√

2
(cos
3
___

)
i sin
4
4
x
1
Example

__
Express
3
cos
(
(

__

)
i sin
(
))
6
√
3
___

__
cos
(
)


__
and
6
sin
3
cos
(
(
(


form
x

iy

6
2

__
)
the
1
__
)
2

__
∴
in
6
i sin
(
6
))
√
3
3
____
3i
__
2
2

6
Exercise 1.5
1
Find
the
modulus
following
(a)
1
complex
i
and
argument
of
each
numbers.
(b)
of
the
2
Express
the
each
form
x
of


__
4
(a)
2
cos
2i
(e)
i(1
(d)
3

following
numbers
i sin
(b)
)
cos 

i sin 
3
4i
2
___
(c)
i)
complex

__

(
3
(c)
the
iy
3
(cos
2
___

3
)
i sin
3
15
in
1.6
Graphical
complex
Learning outcomes
representation
To
show
a
operations
on
numbers
The
The

of
sum of two
complex
complex
numbers
z
and
z
1
graphical
numbers
iy
are
2
C
z
1
represented
representation
of
products
the
the
Argand
___
sums,
by
differences,
on
vectors
OA
___
and
OB
diagram
B
respectively.
and
z
z

on
of
complex
an Argand
numbers
diagram
Using
z

z
1
is
vector
is
addition
represented
we
by
can
see
____
OC,
2
that
where
OC
z
2
a
2
z
1
quotients
2
diagonal
of
the
parallelogram
A
OACB.
z
1
O
x
You need to know

How
to
number

The
represent
on
a
complex
an Argand
meaning
of
the
diagram
modulus
The difference of two
Using
argument
of
a
complex
The
polar-argument form
complex
numbers
the
same
notation
as
above
iy
and
of
a
1
that
vector
z
z
1
where
number
C
z
number
using

complex
and
subtraction,
is
we
represented
can
___
by
see
B
BA,
2
BA
is
the
other
diagonal
z
z
1
2
z
2
of

How

The
to
add
and
subtract
trigonometric
the
parallelogram
OACB.
z
vectors
2
A
compound
z
1
angle formulae
O
The
product of two
When
z

then
z
z
1
(cos 
r
1
1

r
2
r
2
r
2
Therefore
2
|z
when
when
z
is
z
z
cos 

|
||z
by
an
i sin 

represented

)
cos 
1

sin 
1
arg (z
z
1
an
Argand
)

arg z
2

by
z
arg z
1
,
it
is
enlarged
by
diagram,
a
scale
2
we
2
can
iy
Not
θ

1
drawn
to
z
1
2
θ
2
z
2
z
1
θ
2
θ
1
O
|z
see
|
2

z
factor
x
scale
))
2
))
2
16
cos 
2
multiplied
angle

2
2
and
on
),
2
i(sin 
2
1
rotated
|
i sin 
2
1
|z

2
2
i sin ( 
1
is

sin 
2

2
1
)
(cos 
r
2
sin 


)(cos 
2
2
z
z
1
1
1
that
i sin 
1
1
Therefore

(cos ( 
r
1
and
numbers
2
1
(cos 
r
)
1
(cos 
r
1

i sin 
1
1


complex
x
and
Section
The quotient of two
When
z

1
i sin 

1
(cos 
r
z
(cos 
r
1
complex
i sin 

)
and
z
1
1
Complex
numbers
and
calculus
2
numbers

(cos 
r
2
2
i sin 

2
),
then
2
)
1
1
1
_________________
1
__

z
(cos 
r
2
2
i sin 

2
(cos 
r
)
2
i sin 

)(cos 
i sin 
)
1
1
2
2
______________________________
1
__


(cos 
r
2
i sin 

2
2
(cos 
r
)(cos 
cos 

i sin 
2
sin 
)
2
sin 
)
i(sin 

cos 
cos 
sin 
)
1
2
1
2
1
2
1
2
____________________________________________________
1
__


2
r
2

cos
2


sin
2
cos ( 
r

)
i sin ( 


2
)
1
2
1
2
__________________________
1
__


r
1
2
r
1
__

( cos
(

1
)
i sin ( 

2

1
2
))
r
2
|z
z
⎥
z
1
___
____
⎥
z
|
1
1
___
Therefore

and
|z
2
arg
(
|

)
z
2
arg z

arg z
1
iy
2
Not
2
drawn
to
scale
z
1
__
Therefore
when
is
represented
on
an
Argand
diagram,
we
can
see
z
that
z
2
1
z
2
1
____
when
z
is
divided
by
z
1
,
it
is
enlarged
by
a
scale
factor
and
θ
rotated
2
2
|z
|
2
θ
1
by
an

angle
2
O
θ
x
θ
1
z
2
z
1
2
Example
Find
the
modulus
and
argument
of
(1

i)(1
i
√
3)

__
|1

i|
√

2
and
arg (1

i)

;
4

__
|1
i
√
3|

2
and
arg (1
i
√
3)


3
∴
|(1
and

i)(1
arg (1

i
√
3 )|
i)(1
i

√
3)
|1


i|
arg (1
|1

i
i)

√
3|

√
2
arg (1
i

2
√
3)

2
√
2

__

__
4
3


___
=
12
Exercise 1.6
1
Given
that
z

2

2i
√
3
and
z
1

2
2i,
find
and
argument
of
and
hence
illustrate
z
,
z
1
and
on
z
2
diagram.
2
Using
Hence
3
Given
(a)
an
2
2
2
modulus
z
1
__
z
Argand
the
2
z
1
__
z

r(cos 
find
z

the
1
Express

z

two
i sin ),
square
prove
roots
that
of
z
2

√
3
(cos 2 
r

i sin 2 )
i
i,
in
polar-argument
form.
1
__
(b)
Represent
z,
iz
and
on
an
Argand
diagram.
iz
1
__
(c)
The
points
A,
B
and
C
represent
z,
iz
and
respectively.
iz
___
Find
the
complex
number
represented
by
BC
in
the
form
a

ib
17
1.7
De
Moivre’s
Learning outcomes
De
De
To

state
and
prove
theorem
De
Moivre’s theorem
Moivre’s
theorem
states
that
Moivre’s
theorem
n
(cos 

To
use
De
Moivre’s

To
introduce
cos n

i sin n

for
all
integral
values
of
n
theorem
De
the
i sin )

Moivre’s
theorem
is
important
because
it
links
complex
numbers
and
exponential
trigonometry
form
of
a
complex
number
Proof
You need to know
(T
o
by
remind
induction of
yourself
of
this
method
of
proof
(cos 
by
of
2
i sin )

Moivre’s theorem
method
2
The

De
proof,
go
to
Unit
1,
T
opic
1.6.)
2



cos
sin

cos 2 

i sin 2 

cos n

i sin n
2i sin  cos 

induction

The
meaning
of

The
properties

The
compound
z
n
(cos 
∴
sin  and
of
i sin )

when
n

2
cos 
n
Assume
angle
that
(cos 
i sin )


cos n

i sin n
when
n

k,
and
k
Pythagorean
trig
identities
(cos 
i.e.

i sin )

cos k
i sin k

k  1
(cos 
then

i sin )

(cos k
i sin k)(cos 

cos k cos 

cos (k


i sin )
Did you know?
Abraham
De
was
in
born
Moivre
France,
(1667–1754)
but
England
because
of
He
was
mathematicians
one
who
of
the
the
huge

He
is
of
made
because
he
also
the
mathematics
remembered
of
his
of
of
now
theorem.
contributed
study
a
analytic
1)

if
De
Moivre’s
theorem
is
true
when
n

k
it
is
also
true
when
k

1
contributed
advances
in
have
that
shown
that
De
Moivre’s
theorem
is
true
when
n

2
so
it
is
also
the
true
study
i sin (k
cos k sin )
many
We
to


religious
n
intolerance.
1)
i(sin k cos 

moved
Therefore
to

sin k sin 
when
n

3
time.
It
mainly
follows
that
De
Moivre’s
theorem
is
true
for
all
positive
integer
values
of
n
However,
great
deal
geometry
to
n
Now
consider
(cos 

i sin )
where
n
is
a
positive
integer
.
and
n
probability.
(cos 

i sin )
n

{(cos 

(cos n
i sin )

1
}
1
i sin n)

Using
the
result
above
1
_______________

cos n
i sin n

(cos n
i sin n)
________________________________

(cos n

n)
cos (
i sin n)(cos n

i sin (
i sin n)
n)
_____________________

cos n
2
cos

cos(
n)
and
sin n


sin
18
cos (
n)
n
2

sin(
2
n
n)

i sin (
n)
cos
2
A

sin
A

1
Section
1
Complex
numbers
and
calculus
2
n
i.e.
(cos 
Therefore
and
i sin )

De

Moivre’s
n)
cos (
theorem

is
n)
i sin (
true
for
all
integer
values
of
n,
positive
negative.
Example
Use
De
First
Moivre’s
express
theorem
i
√
1

3
2
and
in
to
find
(1

i
5
√
polar-argument
3)
in
the
form
a

ib
form:

__
|1

i
√
3|

arg (1

i
√
3)

3

__
⇒
1

i
√
3

2

__
cos

(
i sin
)
3

__
5
(
1
∴

i
5
)
√
3

2
3

i sin
)
3
3
5
___

32
5

__
cos
(
(cos
5
___

)
i sin
3
√
3
___
1
__

32
3
(
)
i
2

2
16
16i
√
3
Example
1
___
When
z

1
___

i
1
__
3
use
De
Moivre’s
theorem
to
show
that
z

3
√
is
real
First
and
2
find
express
z
its
in
1
___
z

z
√
2
value.
polar-argument
1
___

i
√
√
2
2

__


__
cos

(
i sin
)
4
z
4

__
3
∴


i sin
)
4
3
___
3
___
(cos

)
i sin
4
1
__
and

__
z

3
cos
i sin
4
(
cos


3
(cos

i sin
(
)
theorem
4
3
___

z
3
___
)
i sin
4

(cos (
4
3
___

Moivre’s
3
___
)
3
___
1
__
3
De
4
4
z
theorem
)
3
___
∴
Moivre’s
3

__

(
z

De
4
3

3

__
cos
(
4

form:
(cos
3
___


i sin

4
(cos
))
3
___
)
i sin
4
(
4
3
___
)
i sin
4
3
___
)
4
4
3
___

2 cos
which
is
real
4
1
___

2

(
)
√
2


√
2
19
Section
1
Complex
numbers
and
calculus
2
Example
Given
that
z
cos 

i sin ,

1
__
z
show
that
1
__

2 cos 

and
z
2i sin 

z
z
From
De
Moivre’s
theorem,
1
z
)

cos (

cos 

)
i sin (
i sin 
1
__
∴
z


(cos 

2 cos 

(cos 

2i sin 

i sin )

(cos 
i sin )
(cos 
i sin )
z
1
__
and
z

i sin )
z
Notice
that
when
1
__
z

cos 
i sin ,

1

z
cos 

i sin 
z
1
__
i.e.

z*
z
The
result
from
the
example
above
can
be
1
___
n
z
extended
2 cos n

and
z


n
results
can
2i sin n
n
z
These
give
1
___
n

to
z
be
used
to
prove
some
trigonometric
identities.
Example
3
Prove
cos 3 
that
Starting
with
z

cos 


4 cos
3 cos 
i sin 

3
1
__
(z
3
__
3
)


z

3z
1
__


3
z
z
1
__
3

(z
z

3
1
__
)

3
(z
Using
z
cos 


i sin 

and
the
2 cos 
and
z
result
above
gives
1
__
3

[1]
z
1
__
z
)

z


2 cos 3 
3
z
z
3
[1]
⇒
(2 cos )
⇒
4 cos

2 cos 3 

cos 3 

6 cos 
3


3 cos 
3
∴
De
20
cos 3 

Moivre’s
4 cos

theorem
3 cos 
can
also
be
used
to
simplify
expressions.
Section
1
Complex
numbers
and
calculus
2
Example
cos 
i sin 
_______________
Simplify
cos 3 
i sin 3 
1
__
Using

z*
z
1
_______________
cos 3 

cos 3 
i sin 3 

i sin 3 
cos 
i sin 
_______________
∴
cos 3 
The
i sin 3 

(cos 
i sin )(cos 3 

cos 3  cos 

cos 2 
i sin 3 )

sin 3  sin 


i(sin 3  cos 
cos 3  sin )
i sin 2 
exponential form of
a
complex
number
i
Euler ’s
for mula
states
that
cos

i sin 


e
i
Therefore
(Euler ’s
De
z
r(cos 

formula
Moivre’s
is
proved
theorem
exponential
form,
for
when
i sin )

can
in
can
T
opic
be
written
as
z

re
2.8.)
sometimes
be
easier
to
apply
using
the

__
i
example
z

4
2e
,

__
3
___
3
i
3
then,
The
the
following
links
an
using
a
laws
use
of
of
combination
indices,
Euler ’s
of
z

8
formula
irrational
(
i
4
e
)

gives
numbers
an
8e
4
interesting
and
an
equation
imaginary
that
number
to
integer
.
i
e

cos 

i
sin 
but
cos 

1
and
sin 

0
i
Therefore
e

1
Exercise 1.7
4
1
Show
2
Use
De
Moivre’s
theorem
to
prove
3
Use
De
Moivre’s
theorem
to
simplify
that
(1

i)
is
real
__
4
(a)
Express
2
√
2
2
Use
z

2
find
its
value.
that
sin 2 
(cos 2 


2 sin  cos 
i sin 2 )(cos 

i sin )
__

2i
i
√
2
__
(b)
and
in
the
form
re
__
√
2

2i
√
2
__
to
find
the
two
square
roots
of
2
√
2
__

2i
√
2
i
in
the
form
re
3
5
Find
the
value
of
(1

i)
3

(1

i)
21
1.8
Complex
Learning outcomes
numbers
To
investigate
the
locus
of
locus
(plural
in
the Argand
by
complex
is
a
set
of
points
that
satisfy
a
given
condition.
For
the
locus
of
points
that
are
at
a
fixed
distance
from
a
fixed
point
diagram
is
defined
loci)
a
example,
point
loci
Loci
A

and
a
circle.
numbers
In
z
an

Argand
x

length
iy
of
4
diagram,
and
we
the
impose
point
the
P( x

iy)
condition
can
| z|
be

anywhere.
4,
then
OP
However
,
is
a
if
fixed
units.
You need to know
iy

How
to
represent
sums
iy
and
locus
differences
on
of
an Argand
complex
of
P
numbers
diagram
P(x

iy)
4

The
meaning
and
a
of
The
properties
a
line
segment
ray
O

perpendicular
of
4
O
x
x
the
bisector
of
a
line
segment

How
to find
the
points
of
Therefore
intersection
of
curves
and
P
is
any
point
on
a
circle
of
radius
4
units
and
centre
O.
lines
Any
equation
of
the
for m
circle
Now
consider
the
equation
of
| z|

radius
|z
z
|
r
r
defines
and

4
where
z
1
A(x

iy
1
the
centre
locus
that
is
a
O.
is
the
fixed
point
1
).
1
iy
P
z
z
1
z
A
z
1
O
x
iy
iy
P(x

iy)
4
z
z
1
A(x

iy
1
)
AP

z
A(x
1
O
x
z
,
so
AP
is
a
fixed

iy
1
)
1
O
length
of
4
units.
x
Therefore
the
locus
of
P
is
1
a
circle
of
radius
4
units
and
centre
z
1
Any
equation
of
the
for m
|z

z
|

a,
where
a
is
a
real
constant,
1
defines
a
locus
that
is
a
circle
of
radius
a
and
centre
z
1
22
Section
When
you
drawing
a
need
to
work
out
the
locus
of
a
point,
it
is
sensible
to
start
1
Complex
numbers
and
calculus
2
by
diagram.
Example
Sketch
on
an
Argand
diagram
the
locus
of
points
such
that
| z
2i|

iy
3
5i
When
we
compare
|z
2i|

3
with
|z
z
|

a
1
we
can
see
that
the
locus
is
a
circle
whose
centre
2i
is
the
point
2i
and
whose
radius
is
3.
O
3
x
1
i
Example
Describe
the
locus
of
the
points
on
the
Argand
diagram
given
by
iy
P
|z
z
|

|z
z
1
z

2
|
where
z
represents
the
point
P( x

iy),
2

2i
and
z
1

4
i
2
z
z
1
z
z
In
the
diagram,
|z
z
|z
z
|

AP
|

BP
z
2
1
A
and
2
z
1
∴
|z
z
|

|z
z
1
A
point
|
⇒
AP

BP
2
that
is
equidistant
from
two
fixed
points
is
on
the
x
2
perpendicular
z
2
bisector
of
Therefore
segment
This
any
is
a
the
the
line
required
between
2
particular
equation
segment
of

locus
2i
is
and
example
the
joining
for m
of
|z
the
the
perpendicular
the
general
z
bisector
bisector
of
the
line
i
|

result,
|z
z
1
the
i
B
points.
perpendicular
4

two
of
points
|
i.e.
defines
a
locus
that
is
2
the
z
line
and
1
segment
between
the
z
2
Example

__
Describe
the
locus
of
points
on
the
Argand
diagram
given
by
arg ( z)
iy

4
z
arg (z)
is
the
Therefore
angle
arg ( z)
that
z
makes
describes
a
ray
with
from
the
the
positive
origin
real
at
axis.
an

__
π
angle
of
to
the
real
axis.
4
4
O
x
23
Section
1
Complex
numbers
and
calculus
2
This
is
another
any
example
equation
from
the
of
of
origin
a
general
the
at
result,
for m
an
of
arg ( z)

angle
the
to
real
i.e.


the
describes
positive
a
ray
direction
axis.
Intersection
T
o
to
find
the
points
Cartesian
often
of
form.
suggest
a
intersection
However
,
simple
of
this
two
is
not
loci,
we
always
can
convert
necessary.
A
the
equations
diagram
will
solution.
Example

__
Find,
in
the
form
a

ib,
the
complex
number
that
satisfies
both
| z|

2
and
arg (z)

3
We
can
see
from
the
diagram
that
A
has
a
modulus
of
2
and
an
iy
π

__
argument
arg(z)
2i
of
A

__
Therefore
A
is

3
3
the
point
2
cos

__

(
3
i sin
)
3
π
i.e.
1

i
√
3
3
2
O
2
2i
Example
Find
the
|z
4|
complex

|z

numbers
2|
and
that
|z
satisfy
2
i|

the
equations
4
iy
z
4
2
i


z

2
i
z
O
2
i

4
x
4
The
locus
of
points
perpendicular
i.e.
x
|z
and
24

that
bisector
of
satisfies
the
line
|z
4|
between

x
|z


4
2|
and
is
x
1
2
this
i|

4
⇒
represents
a
|z
(2
circle,

i)|
centre

2
4

i
and
radius
4.
the

2,
x
Section
T
o
to
find
find
i.e.
the
the
complex
points
where
|1
⇒

|
numbers
on
the
iy
1
that
circle
2
i(y
satisfy
where
i|

4
1)|

4

16
14

0
y

1
x
the

given
equations,
we
1
Complex
numbers
and
calculus
2
need
1,
2
⇒
1

(y
1)
2
⇒
y
2y
___
⇒

√
15
___
Therefore
the
complex
numbers
are
1

i(1
___
√

15 )
and
1

√
i(1
15 )
Region of the Argand diagram
A
locus
can
also
be
a
set
of
points
in
a
region
of
the
Argand
diagram.
iy
z
2
i

4
i
O
x
4
z
For
example,
circle
|z

the
2

set
i|

of
set
of
points
|z
The
set
of
points
that
contained
in

the

satisfying
z
|z

2

2

i|

4
lie
inside
the
4
The
are
points
4
4|

|z
satisfy
shaded

2|
both
region
|z
lie

to
2
the

bounded
right
i|

by
4
the
of
the
and
|z
circle
line

and
2
x


the
1
i|

4
line.
Exercise 1.8
1
2
Sketch
(a)
|z|
(c)
|z
on


Sketch
an
Argand
diagram
the
2
2|
on

an
3
Argand
diagram
the
locus
of
points
(b)
|z

2i|
(d)
|z

2
locus
of


arg z
Find
the
for

2
which

__

(b)
arg z


4
3
which
3
2i|
points

__
(a)
for
6
complex
numbers
satisfied
by

__
(a)
|z|
(b)
|z

5
and
arg z


4
4
Show
|z


3
on
2|

an

i|

2
Argand
and
|z|
diagram

|z
the

set
2i|
of
points
for
which
x

3
and
4
25
1.9
Differentiation
of
exponential functions
x
Learning outcomes
Graphs of
y

The
curves
where
a
a

0
x

To
differentiate
family
of
with
equation
y

a
where
a

0,
are
exponential
exponential
curves.
They
are
investigated
in
Unit
1
T
opic
1.20.
functions
x
y
y

4
x
y

x
3
y

2
You need to know

What
an
exponential function
is
x

The
shape
of
the
curve
y

e
2
dy
d
___

The
meaning
of
y
____
and
x
2
dx

How
to
sums,
differentiate
differences,
(0,
dx
y
1)

1
(
1)
multiples,
products
quotients
of functions

The
chain
rule

The
meaning
and
x
This
of
stationary
diagram
shows
some
members
of
the
family.
Each
of
these
curves
points
has
a
property
drawing
the
gradients
of
that
can
tangents
these
be
at
found
some
by
drawing
points
on
the
accurate
graph,
plots
then
and
by
calculating
the
tangents:
dy
___
the
value
of

y
is
constant.
dx
The
and
table
4,
below
and
the
gives
graph
approximate
shows
a
these
values
values
for
this
plotted
2
3
4
0.7
1.
1
1.4
constant
against
for
the
a

value
2,
of
3
a
dy
___

y
dx
dy

y
dx
1.0
0.5
a
O
1
2
3
4
Base
x
The differential of
The
graph
shows
that
e
there
is
a
number
dy
y

a
,
number
is
and
3
for
which,
___

y

1,
i.e.
y

dx
This
2
dy
___
x
when
between
dx
e.
dy
___
x
Therefore
when
y

e
x

,
e
dx
x
and
when
f( x)

e
x
,
f(x)

e
x
The
function
f( x)
differentiated.
26

e
is
the
only
function
that
is
unchanged
when
it
is
Section
1
Complex
numbers
and
calculus
2
f(x)
The differential of
e
f(x)
e
is
a
composite
function
dy
___
f(x)
When
y

e
dy

dx
we
use
the
chain
rule:
du
___
___
,
so
u

where
du
u

f(x)
⇒
y

e
dx
dy
___
u
⇒

f(x)
e

f(x)

f(x)e
dx
f(x)
i.e.
the
differential
of
f(x)
e
is
f(x)e
Example
Find
the
derivative
of
x
x
(a)
2x  1
3e
2
e
(b)
x
(c)
(3x
e
_____
2)
e
(d)
sin x
d
___
d
___
x
3e
(a)
=
3
dx
x
e
x
=
3e
dx
d
___
2x  1
2x  1
e
(b)

2e
Using
the
result
above
dx
dy
2
(c)
x
(3x
dv
___
___
2)
e
is
a
product
so
we
use

u
dx
(3x
and
v

v
2
where
u

x
dx
2)
e
2
y
du
___

dx
(3x

x

(x

(3x
2)
e
dy
___
2
⇒
(3x
)
2)
(3e
(3x
)

2)
(e
)(2x)
dx
2
(3x

2)
2x)e
x
dy
e
_____
vdu
udv
__________
___
is
(d)
a
quotient
so
we
use

x
with
u

e
and
2
sin x
v

dx
v
sin x
x
e
_____
y

sin x
x
(sin
dy
___
x)(e
x
)
(e
)(cos x)
____________________
⇒

2
dx
sin
x
x
e
(sin x
cos
x)
_______________

2
sin
x
Exercise 1.9
1
Find
the
derivatives
of
the
following
functions.
x
(a)
5e
(c)
e
x
(b)
e
cos x
2x
e
______
sin x
(d)
2
x

4
2
d
Given
y

e
y
dy
____
x
2
sin x
show
that
___

2

2y

0
2
dx
dx
x
3
Find
the
coordinates
determine
its
of
the
stationary
point
on
the
curve
y

e

x
and
nature.
27
1.
10
Differentiation
tangents
Learning outcomes
and
of
logarithmic functions,
normals
The differential of
to
parametric
curves
ln x
y
We

To
differentiate
know
that
y

ln x
⇔
x
functions
d
___
y
Now
To find
the

e
,
gradients
of
y
therefore

dy
normals
to
curves
whose
We
also
know
are

x
dx
___
that

1

From Unit
dx
equations
e
dy
tangents
___
and
e
dx
___
y
e
dy


logarithmic
1 Topic
3.9
dy
parametric
dy
1
__
___
therefore
when
y

ln x,

dx
x
1
__
i.e.
You need to know
when
f( x)

ln x,
f(x)

x

The
meaning
of
ln
x
and
its
The differential of
ln f(x)
x
relationship
to
e
ln f(x)
is
a
composite
function
so
we
use
the
chain
rule:
x

The
differential
of
e
dy
dy
___

The
laws
of
When
logarithms
y



dx

How
to
differentiate
dy
multiples,
differences,
quotients
products
u

f(x)
⇒
y

ln u
dx
f(x)
____


dx
and
where
du
1
__
___
⇒
sums,
du
___
___
ln f(x),
f(x)

u
f(x)
of functions
f(x)
____
i.e.

The
chain

How
the
differential
of
is
f(x)
rule
dy
to
ln f( x)
differentiate
parametric
example,
when
y

2x
______
___
2
For
ln (x

1),

2
x
dx
equations
tangents
and
and
the
meaning
normals

1
of
The
example
simplify
the
below
shows
how
differentiation
of
the
laws
many
log
of
logarithms
can
be
used
functions.
Example
dy
1
___
___
Find
when
y

ln
(
)
√
dx
x
1
1
___
y

ln
(
1

)
ln 1
ln x
2

0
ln x
2
√
x
dy
1
__
___
Therefore

1
__
(

dx
2
1
___
)

x

2x
Exercise 1.10a
Find
the
differential
of
each
function.
3
1
2
ln 2x
3
ln x
ln (sin x)
______
x
______
4
ln
(
5
)
2
x

Differentiation of
We
know
where
t
is
(from
a
dy
28
dx
___
___

dx
Unit
1
parameter
,
dy
___
x ln
√
2
x

1
1

dt
dt
parametric
T
opic
3.9)
that
equations
when
y

f(t)
and
x

g(t),
to
Section
Equations of tangents
When
y

the
g(t),
This
equation
we
can
means
curve
at
that
any
of
use
curve
is
g(t))
as
can
find
the
on
the
the
Complex
numbers
and
calculus
2
normals
given
(f( t),
we
point
a
and
1
parametrically,
coordinates
equation
of
a
of
i.e.
any
x

point
tangent
or
f(t)
on
and
the
normal
curve.
to
the
curve.
1
__
For
example,
when
x

3t
and
y

1
,
the
coordinates
of
any
point
t
1
__
on
the
curve
(3t,
are
)
1
and
the
gradient
at
any
point
on
the
curve
is
t
dy
dy
___
given
dx
___
___
by


1
__
1
___


3

2
dx
Therefore
dt
the
dt
equation
1
__
(1
y
equation
of
of
3t
the
tangent
at
any
point
is
given
by
1
___
)

(x
3t)
Using
y
y

m(x
x
1
2
t
The
2
t
)
1
3t
the
normal
at
any
point
can
also
be
found:
the
gradient
2
of
the
normal
is
3t
so
the
1
__
(1
y
equation
is
given
by
2
)

3t
(x
3t)
t
The
equation
found
by
of
the
tangent
substituting
the
and
value
normal
of
t
at
at
that
a
particular
point
can
be
point.
Example
The
equations
of
a
curve
are
x
equation
of
the
normal
to
this
equation
of
the
normal
at
the

cos 
curve
and
in
y


terms
sin .
.
of
Find
the
find
the
Hence

__
point
where


2
dy
1
cos 
_________
___

cos 
1
_________

sin 
_________
so
sin 
dx
the
gradient
of
the
normal
is
sin 
1
cos 
sin 
_________
The
equation
of
the
normal
is
(
y
sin )

cos )
(x
cos 
1

__
When


,
the
equation
becomes
2

__
y

__
1
(
)

x
⇒
y

x

2
1
2
Exercise 1.10b
1
Find,
in
terms
of
t,
the
equation
of
the
tangent
to
the
curve
1
__
x

t,
y

t
Hence
where
2
Find,
x

find
t
in

the
terms
2 cos ,
Hence
equation
of
the
tangent
at
the
point
on
the
curve
2
find
y
of

the
,
the
equation
of
the
normal
to
the
curve
3 sin 
equation
of
the
normal
at
the
point
on
the
curve

__
where


4
29
1.
11
Implicit functions
Learning outcomes
Implicit functions
2
The

To
describe

To
differentiate
equation
of
some
curves,
such
as
2
y

xy

x
y

2,
are
not
easy
to
implicit functions
express
in
the
form
y

f(x)
implicit functions
A
relationship
implied
by
like
the
this
is
called
an
implicit
function
because
y

f(x)
is
equation.
You need to know
Differentiation of

The
product
rule
and
The
rule for
method
The
we
use
to
differentiate
an
implicit
function
is
to
differentiate
differentiation
term

implicit functions
quotient
chain
by
term
with
respect
to
x
rule
dy
___
The
differential
of
y
with
respect
to
x
is
dx
2
T
o
differentiate
2
y
with
respect
to
x,
we
start
with
g( y)

y
where
y

f(x)
2
Then
g(y)
This
a
is

[f(x)]
composite
substitution
u

function
so
we
use
the
chain
rule
with
the
f(x)
2
Then
g(y)
d
___

u
d
___
2
⇒
(u
)

du
___
2
(u
dx
)

du
dx
du
___

2u

dx
dy
d
___
But
y

u

f(x),
___
2
so
(y
)

2y
dx
This
is
a
particular
dx
example
of
the
general
d
___
to
differentiate
a
dy
d
___
(g(y))

dx
i.e.
result:
___
g(y)
(
dy
function
of
y
with
)

dx
respect
to
x,
differentiate
the
dy
___
function
with
respect
to
y
and
multiply
by
dx
dy
d
___
For
example,
4
(2y
3
)

and
dx
We
can
with
now
respect
dx
differentiate
to
d
___
___
8y
any
(ln y)
dx
expression
dy
1
__
___
y
dx

involving
x
and
y,
term
x
2
T
o
differentiate
such
as
x
y
we
use
dy
d
___
so
terms
2
(x

x

dx
product
rule,
dy
___
2
y)
the
2

2x

y

dx
___
x

2xy
dx
x
__
and
to
differentiate
terms
such
as
we
y
dy
___
y
d
___
so
dx
30
y
x
dx
________
x
__
(

)

2
y
use
the
quotient
rule,
by
term,
Section
1
Complex
numbers
and
calculus
2
Example
dy
___
2
Find
in
terms
of
x
and
y
when
y
2

xy

x
y

2
dx
Differentiating
d
___
)
term
(xy)
dy

y

___
x

dx
dy
2xy

0
dx
2xy
___
(2)
dx
2
x
dx

dy
___

y)
(x
dx
dy
___
d
___
2

dx
2y
gives
d
___

dx
⇒
by
d
___
2
(y
term

y
___________
⇒


2
dx
2y

x

x
Example
2
Find
the
point
gradient
(4,
2)
on
of
the
the
tangent
to
the
curve
y

xy
y

6
at
the
curve.
dy
___
We
need
the
value
of
when
x

4
and
y

2
dx
2
y

xy
y

dy
6
dy
___
dy
___
⇒

y

___
x
2y
dx
dx
dy

0
dx
y
___
___________
∴

dx
2y
x
1
dy
___
When
x

4
and
y

2,

2
dx
Therefore
gradient
at
the
point
(4,
2)
is
2.
Exercise 1.11
1
Differentiate
2
(a)
2x
each
equation
with
respect
to
2

y
x.
y

y
(b)
xe

x

y
2
y
______
2
(c)
x ln (y
)

4

(d)
x

y

1
1
dy
___
2
Find
in
terms
of
x
and
y
when
dx
2
2
(a)
x

(a)
Find
xy

y

6
(b)
2 cos x

3 sin y

4
dy
___
3
in
terms
of
x
and
y
when
sin x

cos y

1
dx

__
(b)
At
the
point
,
(

)
the
gradient
of
the
curve
is
1.
Find,
in
the
4
range
0



,
the
value
of
.
31
1.
12
Inverse
Learning outcomes

To
define
the
trigonometric functions
The
inverse
The
sine
The
graph
sine function
function
is
normally
given
as
f( x)

sin x
for
the
domain
x


inverse
trigonometric functions
of
f( x)

sin x
for
this
domain
is
given
below.
f(x)
You need to know

The
properties
and
1
graphs
of
1
1
sine,
cosine
and
x
π
π
2
2
the
tangent
1
functions

The
meaning
of
domain
and
range
This

The
definition
of
an
function
does
not
have
an
inverse
because
it
is
not
one-to - one.
inverse
However
,
if
we
define
the
function
f
such
that
f( x)

sin x
for
the
domain
function
1
1


x
The
condition for
a function
an
then
the
graph
of
f
is
the
solid
line
in
the
graph
above.
2
to
This
have
,

2

shows
that
f( x)
is
one-to - one
and
so
does
have
an
inverse.
inverse
1
y
The
graph
of
y

f
(x)
is
obtained
by
reflecting
y

f(x)
f(x)
by
in
the
line
y

x
1
1
and
π
the
equation
y

f
(x)
is
obtained
from
y

interchanging
2
x
and
y
1
Therefore
when
y

1

sin x,

x


2
function
the
equation
of
the
inverse
2
is
1
1
1
x
sin y

x,
1

for

y
,

2
i.e.
1

x

1
2
1
so
y
is
the
angle
whose
sine
is
x
1

where

y


2
2
1
The
1
‘angle
whose
sine
is
x’
is
denoted
by
sin
x
(an
alternative
notation
π
2
is
arcsin x).
1
_
Therefore
when
f( x)

sin x,
1
_



x


2
1
f
2
1
( x)

sin
x,
 1

x

1
1
Note
that
1
[
sin
x
is
an
angle
and
that
this
angle
is
in
the
interval
1
,

2
]
2
1
The
angles
in
the
interval
[
1
,
2

]
are
called
the
principal
values
of
2
1
sin
x
1
1
For
example,
(
sin
1
)
is
the
angle
1
is
so
sin
)
inverse
function
whose
sine
2

2
The
The
(

and
2

__
1
__
1
,
2
1

between
2
6
cosine function
f( x)

cos x,
x


is
not
one-to - one
so
it
does
not
have
inverse.
However
,
the
1
so
32
f
exists.
function
f
given
by
f( x)

cos x,
0

x


is
one-to - one
an
Section
Therefore
when
y


y
cos x,
0

x


the
equation
of
the
inverse
function
1
Complex
numbers
and
calculus
2
y
is
π
cos y

x,
for
0
,

i.e.
1

x

1
1
y
so
y
is
the
angle
whose
cosine
is
x
where
0

y


cos
x

1
The
‘angle
notation
whose
is
cosine
is
x’
is
denoted
by
cos
x
(an
alternative
arccos x).
Therefore
when
f(
x)

cos x,
0

x


1
1
f
1
x
π
1
( x)

cos
x,
1

x

1
1
y

cos
x
1
Note
that
cos
x
is
an
angle
and
that
this
angle
is
in
the
interval
[0,
].
1
The
angles
in
the
interval
example,
are
called
the
principal
values
of
cos
x
1
1
For
]
[0,
(
cos
)
is
the
angle
between
0
and

whose
cosine
2
1
is
so
(
cos
2
___
1
1
,
)
2

2
The
3
inverse tangent function
1
When
f(x)

tan x,
x

,
f
does
not
exist,
but
when
y

tan
x
y
1
f(x)

1

tan x,

x
1
,

2
f
does
exist.
2
1
Therefore
when
y

1

tan x,

x
,

2
the
equation
of
the
1
2
π
2
inverse
function
is
1
y
1
tan y

x

y
y
is
the
tan
x
angle
,

2
so

1

for
2
whose
tangent
is
x
where
x


1
x
1
π
π
2
2
1
The
‘angle
whose
tan
is
x’
is
denoted
by
x
tan
(or
arctan x).
1
π
2
1
_
Therefore
when
f( x)

tan x,
1
_



x
1
f
in
the
(
interval
2
1
(x)
1
Angles


2

tan
x,
x


1
,

2
)
are
called
the
principal
values
of
2
1
tan
x
1
1
For
example,
tan
(
1)
is
the
angle
1

between
1,
so
tan
whose
tangent
2

__
1
is

and
2
(
1)


4
1
Note
that
the
range
1
sin
of
tan
x
is
all
real
values
of
x
whereas
the
ranges
of
1
x
and
cos
x
are
each
[
1,
1].
Exercise 1.12
Find
the
principal
value
in
terms
2
tan
1
1
cos
of

of
the
following.
1
(
1)
1
___
1
3
(1)
sin
(
)
√
2
1
4
cos
√
3
___
(
)
1
5
tan
(
√
3)
2
33
1.
13
Differentials
of
inverse
trigonometric
functions
1
Learning outcomes
The derivative of
sin
x
1
Let

To
determine
the
inverse
the
differentials
y

sin
x
⇒
x

sin y
of
trig functions
Differentiating
x

sin y
with
respect
to
y
gives
dx
___

cos y
dy
You need to know
dy
dx
___
___
Therefore
using

1

gives
dx

The
definitions
of
the
inverse
dy
trig
dy
1
_____
___
functions

dx

The
derivatives
of
sin x,
cos x
cos y
and
1
___________
2
_________
tan x

Using
cos
2
y

sin
y  1
2
√1
dy
That
y
dx
___

sin
___

1

dx
dy
1
1
Note
that
when
y

sin
x,
the
range
of
y
1

is

y


2

The
Pythagorean

The
chain
rule
trig
identities
range,
cos y

and
for
this
2
0
2
Therefore
we
only
use
the
dy
sin y

x,
so
square
root
of
1
sin
y
1
________
___
But
positive
______

dx
2
√
1
x
d
___
i.e.
1
________
1
(sin
x)
______

dx
2
√
1

x
1
The derivative of
cos
x
1
We
use
Let
y
the
same
method
to
find
the
derivative
of
cos
x
1

cos
x
Differentiating
⇒
x
x


cos y
cos y
with
respect
to
y
gives
dx
___

sin y
dy
dy
dx
___
___
Therefore
using

1

gives
dx
dy
dy
1
_____
___

2

dx
Using
cos
2
y

sin
y  1
sin y
1
___________
_________


2
√1
cos
y
1
When
y

cos
x,
the
range
of
y
is
0

y


and
for
2
So
we
only
use
the
positive
dy
cos y

x,
so
root
of
1
cos
______


dx
√
1
2
x
d
___
dx
1
________
1
(cos
i.e.
x)


______
√
1
34
y
1
________
___
But
square
2

x
this
range,
sin y

0
Section
1
Complex
numbers
and
calculus
2
1
The derivative of tan
x
1
The
same
method
again
gives
the
derivative
of
tan
x
1
Let
y

tan
x
⇒
Differentiating
dx
___
x
x


tan y
tan y
with
respect
to
y
gives
2

sec
y
dy
dx
___
Therefore
dx
___
using

1

dy
dx
___
1
_____

gives
dy
1
_________
sec
Using
1

dy
tan y

x,
1
 tan
2
y

sec
y
2
y
tan
y
1
______
___
But
2

2
dy
so

2
dx
1

x
d
___
1
______
1
i.e.
(tan
x)

2
dx
1

x
Example
1
Find
the
derivative
of
cos
(3x
2)
1
Let
y

cos
1
(3x
2)
and
u

3x
dy
the
chain
rule
gives
dy

cos
u


3
2
√
1
u
3
______________
___
_____________


2
dx
This
y
_______

dx
⇒
so
1
________
___
Using
2
√1
example
is
a
(3x
2)
particular
case
of
the
general
result,
i.e.
f(x)
d
___
___________
1
(sin
f(x))

__________
2
dx
√1

(f(x))
f(x)
d
___
___________
1
(cos
f(x))


__________
2
dx
√1

(f(x))
f(x)
d
___
__________
1
(tan
f(x))

2
dx
1

(f(x))
Exercise 1.13
Find
the
derivative
of
each
of
the
following.
1
1
cos
2
tan
3
cos
4
sin
5
tan
2x
1
(2x
1

1)
2
(x
)
1
(x

4)
1
x
(1

e
)
35
1.
14
Derivatives
Learning outcomes
of
To find
the
derivative
of
have
found
the
differentials
of
a
variety
of
functions
in
previous
topics
a
and
combination
of functions
Summary of differentials
We

combinations
in
Unit
1.
of functions
The
results
are
Standard
summarised
here.
results
You need to know
d
___
f(x)

The
rules for
f(x)
gf( x)
(gf(x))
differentiating
dx
products
and
quotients
of
n
functions
and
composite
n
1
n
n
1
x
nx
(ax
sin x
cos x
sin f( x)
f(x) cos f(x)
cos x
sin x
cos f(x)
f(x) sin f(x)
tan x
sec
tan f(x)
f(x) sec

na(ax
b)

b)
functions

How
to
use
logarithms
2
x
2
x
f(x)
x
e
e
f(x)
e
f(x)e
f(x)
1
__
ln x
_____
ln f(x)
f(x)
x
f(x)
1
1
sin
_______
1

2

(f(x))
f(x)
____________
1
cos

___________
f(x)

2
√
1

2
1
√
x

1
_______
tan
x
__________
f(x)
2
1
Any
of
these
Y
ou
to
need
be

results
to
able
2
1
x
can
lear n
to
(f(x))
f(x)
1
1
1
√
x
________
x
___________
f(x)
1
_________
1
tan
sin
2
√
cos
____________
1
________
x
f(x)
be
quoted
these
unless
results.
recognise
the
When
function
differentials
you
in
are
asked
integrating
which
the
gives

to
you
any
(f(x))
derive
also
of
them.
need
these
table.
2x
______
For
example,
given
,
you
need
to
recognise
this
as
the
differential
2
x

1
2
of
ln (x

1)
General
results
dy
d
___
___
f(y)

f(y)
dx
dx
dy
dy
___
The
that
to
36

dt
use of
Logarithms
dx
___
___

dx
logarithms
were
contained
differentiate
dt
used
the
in
Unit
unknown
functions
1
to
help
quantity.
where
the
solve
The
equations
same
variable
is
with
technique
contained
in
exponents
can
an
be
used
exponent.
Section
1
Complex
numbers
and
calculus
2
x
For
example,
when
ln y
Then

y
,
taking
logs
of
both
sides
gives
with
respect
to
x
gives
dy
1
__
___
)
y
a
x ln a
differentiating
(


ln a

a
dx
dy
___
⇒
x

y ln
a
ln a
dx
Example
2
Differentiate
2
1
x
sin

1
2
sin
(x
)
2
(x
2
)
1
v
x
is
a
product
so
using
y

1
x
2
sin
(x
2
)
with
u

x
and
2
sin
(x
),
du
___
then
2x
________
dv
___
=
2x
______
and
dx
=
dx
4
√
1
x
3
dy
dv
___
___
∴

u
2x
________
du
___

dx
1
______
v
dx


dx
2x sin
2
(x
)
4
√
1
x
Example
3x
_____
Differentiate
with
respect
to
x
ln 5x
3x
_____
3x
_____
is
a
quotient
so
using
y

with
ln 5x
u

3x
and
v

ln 5x
ln 5x
du
___
then
dv
___

3
and
1
__

dx
dx
dv
___
du
___
v
dy
u
dx
dx
___________
___
∴
x

3 ln 5x
3
__________

2
dx
2
v
(ln 5x)
Example
dy
ye
y cot x
___
(2y  1)
Given

sin x
show
that
_______

dx
1

2y
(2y  1)
ye
is
a
product,
so
we
use
the
product
rule.
dy
(2y  1)
Differentiating
with
respect
to
x
gives
dy
___
e
(2y  1)

dy
y cos x
cos x
_____________
___
⇒

y
_____________

___
2ye

dx
cos x
dx
cot x
_______

(2y  1)
dx
(1

2y)e
(1

2y)
sin x
1

2y
Exercise 1.14
dy
___
1
Find
in
terms
of
x
when
dx
3x
(a)
y

1
x
(b)
y

x tan
1
x
(c)
sin
(xy)

x
2
1
dy
___
2
Find

ln(1

x)
_____________
in
terms
of
x
and
y
when
y

2
dx
1

ln(1

x)
dy
1
3
Given
y

x tan
___
2
x
show
that
x(1

x
)
2

x
2

y(1

x
)
dx
37
1.
15
Second
differentials
Learning outcomes
The
second differential of
with
y
respect to
x
2
d
y
dy
____

To find
and
use
the
second
We
___
met
in
Unit
1.
It
means
the
differential
of
with
respect
to
x,
2
dx
differentials
dx
of functions
dy
d
___
___
(
i.e.
dx
)
dx
dy
You need to know
___
So,
for
example,
2
when

x
,
dx

The
differentials
of
standard
2
y
d
____
2

functions
differential
of
x
with
respect
to
x

2x
2
dx

The
product,
quotient
and
chain
rules
Example
2
d

The
differential
of
f(y)
with
y

e
y
____
3x
If
sin 2x,
find
in
terms
of
x,
simplifying
sin 2x

2e
your
answer
.
2
dx
respect
to
x
x

The
relationship
between
e
dy
and
___
3x
y

e
sin 2x
⇒
3x

3x
3e
cos 2x
Using
the
product
rule
dx
ln x
3x

e
(3 sin 2x

2 cos 2x)
2
d
y
____
∴
3x

3e

e
3x
(3 sin 2x

2 cos 2x)

e
(6 cos 2x
4 sin 2x)
2
dx
3x
When
an
implicit
(5 sin 2x
function
is

12 cos 2x)
differentiated,
we
often
get
terms
such
as
dy
2
___
x
dx
Differentiating
such
terms
will
result
in
a
combination
of
first
and
second
derivatives.
dy
___
2
For
example,
if
y

x
,
then
using
the
product
rule
to
differentiate
dx
2
dy
d
___
with
respect
to
x
gives
y
dy
____
2

x
___

2x
2
dx
dx
dx
Example
2
_______
2
dy
If
y
3x
d
___
2
√


2
show
(
that
y
____
)

y

3
2
dx
dx
_______
1
y

dy
3x
2

2

(3x

2)
2
3x
_________
___
⇒
2
√

1
dx
2
(3x

2)
2
1
2
Now
y

(3x

2)
2
and
we
require
a
relationship
that
involves
1
2
is
sensible
at
this
dy
gives
to

y
for
(3x

2)
2
___
⇒
dx
substitute
dy
3x
___
___
This
stage
y
y

3x
dx
2
dy
___
Differentiating
with
respect
to
x
gives
(
dx
2
2
dy
d
___
i.e.
(
dx
38
y
____
)

y

2
dx
3
dy
d
___
) (
dx
y
____
)

y

2
dx
3
y
so
it
Section
1
Complex
numbers
and
calculus
2
Example
2
_____
The
parametric
equations
of
a
curve
are
y

t
and
x

t

1
2
dy
d
___
(a)
Show
that
y
____
2

x

0
2
dx
dx
2
d
y
____
(b)
Hence
find
in
terms
of
t
2
dx
(a)
We
require
a
eliminating
relationship
t
to
give
a
that
direct
does
2
_____
y

t
and
x
⇒

involve
t,
so
between
we
x
start
and
by
y
2
______

t
not
relationship
x

⇒
1
y

x(y

1)

2
1
dy
___
∴
x

(y

1)

0
dx
2
dy
d
___
y
dy
____
⇒

___
x


0

1)
2
dx
dx
dx
2
dy
d
___
⇒
y
____
2

x

0
2
dx
dx
dy
dy
___
(b)
We
can
use
___
x

(y

0
to
give
in
dx
terms
of
t
dx
2
(t
dy
___
y
i.e.




1)
_______
1
______


dx
2
x
2
dy
d
___
Then
2
y
____
2


0
⇒
(t

1)

2
dx
(
dx
y
____
)
t

1

0
2
dx
3
2
d
d
2
_____
2
x
(t
y
____
⇒

1)
_______

2
dx
2
2
y
d
we
can
dy
d
__
____
Alternatively
2
dx
dt
)

dx
giving
dx
2
2
d
dt
___
___
(

use
y
(t
d
__
____

1)
dx
dx
___
_______
(

2
)
dt

2
dt
2
_______

(
(t

1))

2
(t

1)
3
(t

1)
_______

2
Exercise 1.15
2
d
y

__
____
1
If
tan y

x,
find
the
value
of
when
y

2
dx
4
1
(Hint:
If
e

sin x,
show
y
that
parametric
equations
of
d
y
that
y

tan
x)
(
)

1

0
dx
a
curve
are
x

sin 
and
y

cos 
dy
____
Show
to
2
2
(a)
x
dy
2
The

___

dx
3
tan y
2
2
d
____
y
2
change
y
___

2
dx
(
)

1
of


0
dx
2
d
y
____
(b)
Hence
find
in
terms
2
dx
39
1.
16
Partial
differentiation
Learning outcomes
Functions of two or
Many

To
introduce functions
than
one
of
To
define
depend
on
more
than
one
variable.
more
variable
For

quantities
more variables
and
use
example,
wage
partial
differentiation
costs,
the
produce
If

the
the
to
profit
price
made
the
market,
by
farm
and
a
farmer
produce
several
can
sells
other
depend
for
,
the
on
cost
the
of
weather
,
transporting
variables.
2
z
You need to know

How
to
differentiate
If
The

product,
w

quotient
and
chain
The
to find
then
z
is
a
function
of
two
variables,
x
and
y,
and
we
write
y)
f(x,
y,
farmer
variables
wage
How
y
f(x,
z)
then
w
is
a
function
of
three
variables,
x,
y,
z
Partial differentiation
rules

x
standard
functions

z
a
second
may
want
changes
costs
to
while
know
how
keeping
all
profit
the
changes
others
when
constant,
one
of
such
as
the
when
change.
derivative
This
is
then
the
a
where
partial
partial
differentiation
differential
of
f
is
with
useful:
respect
if
to
f
x
is
a
is
function
found
by
of
x
and
treating
y
y,
as
constant.
f
___
The
partial
differential
of
f
with
respect
to
x
is
written
as
x
f(xy)
______
(The
formal
definition
of
is
x
f(x

h,
y)
f(x,
y)
_________________
lim
where
h

x)
h
h → 0
2
For
example,
if
f( x,
y)

xy

2y
f
___
then
d
___

y
This
is
equivalent
to finding
x
f
___
and
d
___

x

4y
This
is
equivalent
to finding
you
that
get
are
2
(ay  2y
confused
treated
as
when
finding
constant
by
partial
letters
derivatives,
that
look
like
replace
the
variables
constants.
Example
2
If
f(x,
y,
z)

x
2
y

y
z

xyz
find
f
___
f
___
(a)
(b)
x
z
d
___
f
___

(a)
d
___
2
y
x
x


yz

2xy

0

2

d
___
y
y
and
z
as
constants
y
as
constants

d
___
z
2
y
Treating
yz
dz

x
dx
z
40
0
dx
f
___
(b)
)
dy
y
If
(ax  b)
dx
xy

xy
z
dz
Treating
x
and
Section
1
Complex
numbers
and
calculus
2
Example
f
___
x
If
f(x,
y)

xye
find
x
x
T
reating
product
y
as
a
constant,
we
have
to
differentiate
xe
,
so
we
use
the
rule.
y)

d
___
f
___
x
f(x,
xye
⇒

x
(
y
x
x
)
xe

x
y(e

xe
)
dx
Exercise 1.16a
2
1
If
f(x,
y)

f
___
2
x

y
f
___
find
and
x
y
f
___
Hence
show
that
f
___
x

y

x
2
2
Given
f(x,
y,
z)

x
2
y
2f(x,
y)
y

y
f
___
2
z

z
x
f
___
find
and
y
z
f
___
3
Find
when
x
x

y
2
______
f(x,
(a)
y)

x sin (x

y)
f(x,
(b)
y)
f(x,
(c)
x
Second
(x


y)

 2)
ye
y
partial derivative
2

f
___
The

___
symbol
respect
to
x,
f
___
(
means
2
x
x
then
find
the
),
partial
3
Therefore
if
f( x,
y)

x
i.e.
find
derivative

partial
xy
with
f
___
3
y
the
derivative
of
f
with
x
,
then
respect
2

3x
to
x
of
the
result.
3
y

y
x
2

f
___
so

___

2
(3x
3
y

y
)

6xy
2
x
With
partial
differentiation
we
can
also
x
have
a
mixed
second
derivative,
2

___
for
f
___
(
example
y

f
_____
)
is
written
as
x
yx
3
For
example
when
f
___
f( x,
2

3x
y)

x
3
y

xy
,
then
3
y

y
y
constant
x
constant
x
2

f
_____

___
So
f
___
(

yx
y

___
)

x
2
(3x
3
y

y
2
)

3x
2

3y
y
Exercise 1.16b
2
(x
1
(a)
Given
f(x,
y)

2
find
2

f
___
(i)
(ii)
State
whether
2
the
y  x
following
2
Repeat
 xy
true
2

f
_____

f
_____

(ii)
2
x
are
2

f
___

(i)

f
_____
(iv)
x

f
___
2

f
_____
(iii)
2
y
2
2

f
___
2
(b)
 y)
e
2
y
question
yx
1
when
f( x,
y)

 xy
x sin (x

y)
41
1.
17
Integration
of
exponential
and
logarithmic
functions
Learning outcomes
Integration
When

To
integrate
a
as the
function
is
reverse of differentiation
recognised
as
the
differential
of
a
function
it
can
be
exponential
integrated,
functions
and
logarithmic
d
___
functions
so
f(x)

f(x)
∫f(x)
⇔
dx

f(x)

c
dx
Integration of
exponential functions
You need to know
d
___
We
know
x
that
x
e

e
x
,
x
∫e
therefore
dx

e

c
dx

How
to
differentiate
exponential
d
___
functions
and
We
logarithmic
also
know
x
that
d
___
x
ae

ae
and
(ax  b)
(ax  b)
e
dx

ae
dx
functions
d
___
Using

That
integration
is
the
reverse
the
chain
rule,
we
also
f(x)
have
f(x)
e

f(x)
e
dx
of
differentiation
x
x
∫ae
therefore
dx

ae
1
__
(ax  b)

c
∫e
and
dx
(ax  b)

e

c
a

How
to find
a
definite
integral
f(x)

and
its
interpretation
The
laws
of
as
an
∫f(x)e
and
area
f(x)
dx

e
logarithms
d
___
4x
For
example,
to
∫e
find

dx
we
know
that
c
4x
4x
e

4e
dx

The
modulus function
1
4x
∫e
so
dx

4x
e

c
4
Example
2
2
(x
∫
Evaluate
1)
4xe
dx
1
2
2x
is
the
differential
f(x)
∫f(x)e
of
x
1,
so
this
integral
is
of
the
form
f(x)
dx

e

c
2
2
2
2
(x
∫
∴
2
1)
4xe
(x
dx

2
∫
2
1)
2xe
dx

2
[
e
(x
1)
]
3

0
2(e
e
)
1
1
1
3

2(e
1)
Exercise 1.17a
1
Find
2x
(a)
∫2e
(c)
∫x e
(3x
dx
(b)
∫e
(d)
∫(cos
(b)
∫
2)
dx
2
x
sin x
dx
1
x)e
dx
1
2
2
Evaluate
(a)
∫
4x
(x
1
5e
dx
0
(2x
x)
1)e
dx
0
1
__
Integration of
x
1
__
If
we
try
to
integrate
1
(
x
1
__
using
0
x
0
42

c,
which
is
1
______
n
)
x
meaningless.
∫x
dx

n  1
x
n

1

c
we
get
Section
d
___
However
,
we
know
that
that
x

Complex
numbers
and
calculus
2
1
__
ln x

.
dx
provided
1
Now
ln x
is
defined
only
for
x

0,
so
x
0,
1
__
∫
dx

ln x

0,
∫

c
[1]
x
1
__
When
x
1
__
dx

ln x

c
is
not
true,
but
the
function
exists
x
x

for
x
0
Also
the
shaded
part
of
the
graph
shows
that
the
area
represented
y
by
d
1
__
∫
1
__
dx
exists,
so
it
must
be
possible
to
integrate
for
x
c
negative
values
of
x
x
When
x

0,
c
1
__
x

0,
∫

∫
x
[1]
and
[2]
can
d
1
_____
dx
dx
(
be

ln(
x)

c
[2]
O
x
x)
combined
using
| x|
to
give
1
__
∫
dx

ln|x|

c
x
f(x)
_____
Integration of
f(x)
f(x)
d
___
We
know
from
T
opic
1.9
that
____
ln f(x)

,
dx
therefore
f(x)
f(x)
____
∫
dx

ln|f(x)|

c
f(x)
2
_______
For
example,
to
find
∫
dx,
2x

we
see
that
2
is
the
differential
of
2x

1
1
2
_______
∫
So
dx
2x


ln
|2x

1|

c
1
Example
3
3x
______
Evaluate
∫
dx
2
x
2
1
3
3
3x
______
∫
3
__
dx
2x
______
∫

2
x
2
1
2
2
2
dx
and
2x
is
the
differential
of
x
1,
so
2
x
1
3
3
3x
______
∫
2
3
__
dx

2
x
1
[
3
__
2
ln
|x
1|
2
]
2

3
__
ln 8
3
__
ln 3
2

2
8
__
ln
2
3
Exercise 1.17b
1
Find
x
e
______
4x
______
(a)
∫
dx
(b)
2
1
∫
x
dx
x
e
1
1
__
cos x
_____
(c)
∫
1
_____
dx
(d)
sin x
∫
(
dx
Hint:
x ln x
2
1
_____
x
____

x ln x
)
ln x
1
x
e
_______
x
2
________
2
Evaluate
(a)
∫
1
dx
x(x
4)
(b)
∫
0
dx
x
2e
3
43
1.
18
Partial fractions
Learning outcomes
Partial fractions
In

To
decompose
a
this
section
numerators
function
whose
factorises
difference
into
of
we
deal
with
rational
functions,
i.e.
fractions
whose
rational
and
denominators
are
polynomials.
denominator
a
sum
or
We
can
add
or
subtract
fractions
to
give
a
single
fraction,
for
example,
simpler fractions
3
______
3(x
1
______

1)

(x

1)

1
x
4x
2
_____________
_________________

x

1
(x

1)(x
1)
(x

1)(x
1)
4x
2
_____________
You need to know
The
reverse
process,
i.e.
starting
with
a
fraction
such
as
(x

The
meaning
polynomial

The

How
meaning
to
of
the
and
of
express
and
terms
rational function
a
as
improper
the
sum
of
and
a
as
the
sum
into
or
difference
partial
of
two
simpler
1)
fractions,
is
fractions
Fractions
with
linear factors
in the denominator
a
When
polynomial
it
decomposing
1)(x
proper fraction
an
rational fraction
expressing
called

a
fraction
is
proper
(i.e.
the
highest
power
of
x
in
the
numerator
less
than
will
also
the
be
highest
power
of
x
in
the
denominator)
and
The
B
partial
A
______
example,
can
(x
A
the
fractions
proper
.
2x
1
_____________
For
are

1)(x
be
expressed
as
2)
B
______

x

where
1
x
2
numbers.
worked
example
shows
how
the
values
of
A
and
B
can
be
found.
Example
2x
1
_____________
Express
in
(x

1)(x
partial
A
______
B
______

(x
First
over

1)(x
express
a
2)
the
common

1
right-hand
x
side
A(x
2x
1
_____________
2
of
this
identity
as
a
single
fraction
2)

B(x

1)
__________________

(x
This
is

an
1)(x
⇒
2x
These
any
3
using
3

two
value
Choosing
are
1
are
2)
identity:
numerators
and

x
denominator
.
⇒
x
fractions.
2)
2x
1
_____________
to

x

the
the
x
3B

B(x
to
to
1)(x

B
the

same
A

2)
A
B
gives
1
1
______
1
______

x
the
expression,

gives
1
eliminate
⇒
are
same
then
the
1)
eliminate


2)
denominators
stating
2
⇒
1
3A

2x
1
_____________
(x
1)(x
same.
2)
of

choose.
use

since
A(x
ways
we
(x
also
∴
44
is
proper fraction
1
x
2
so
we
can
assign
Section
1
Complex
numbers
and
calculus
2
Example
2
x
2x  1
_____________
Express
in
(x
2)(x

partial
fractions.
3)
2
x
2x  1
_____________
is
(x
2)(x
and
a

improper
so
we
express
it
as
a
sum
of
a
polynomial
3)
proper
fraction.
2
2
2
x
2x  1
_____________
(x
x
2x  1
___________

x
6)
2)(x


7

2
(x
3x
____________________

3)
2
x

x
6
x

x
6
3x  7
__________

1

2
x

x
6
2
(This
can
also
be
done

x
2
x
2x
A(x
B
______

1

by
3)

x

x
B(x
6)
2)
__________________

x
2

dividing
A
______
3x  7
__________
x
by

2
x

3
(x
3x

7
2)(x

3)
6
∴

A(x

3)

B(x
2)
16
__
x

3
⇒
16

5B
⇒
B


5
1
x

2
⇒
1

5A
⇒
A

5
2
x
2x  1
_____________
∴

(x
2)(x

1
1
________
16
________
5(x
5(x

3)
2)

3)
Exercise 1.18a
Express
each
fraction
in
partial
fractions.
6
______________
3x
_____________
1
2
(x

1)(x

(x
2)

1)(2x

1)
2
2x  1
________
x
 1
_____________
3
4
x(x

1)
(x
Fractions
with
a

repeated factor
2)(x

3)
in the denominator
x  1
________
The
fraction
is
a
proper
fraction
and,
by
adjusting
the
2
(x
numerator
,
can

be
2)
expressed
as
the
sum
of
two
fractions
with
numerical
numerators.
x  2  3
__________
x  1
________
i.e.



2
2)
(x

2
2)
(x

2
2)
(x

2)
3
________
1
_______

3
________
x  2
________

2
(x

2
(x
Any
two
fraction
whose
fractions
When
next
there
worked
partial

with
are
2)
(x

2)
denominator
numerical
other
example
factors,
shows
is
a
repeated
factor
can
be
expressed
as
numerators.
adjusting
how
such
the
a
numerator
fraction
can
is
be
not
easy.
The
decomposed
into
fractions.
45
Section
1
Complex
numbers
and
calculus
2
Example
2x
1
_______________
Express
in
partial
fractions.
2
(x
3)
(2x
2x
1
_______________

1)
A
________
3)
C
________
B
_______


2
(x

2
(2x

1)
(x
3)
(x
3)
(2x

1)
2
A(2x

1)

B(x
3)(2x

1)

C(x
3)
______________________________________

2
(x
3)
(2x

1)
2
∴
2x
1

A(2x

1)

B(x
3)(2x

1)

C(x
3)
5
x

3
⇒
5

7A
so
A

7
49
__
1
x


⇒
2

The
the
Choose
so
C


4
value
two
8
__
C
2
of
B
values
an
can
be
49
found
already
easy
by
substituting
any
value
for
value:
we
will
use
x


0
⇒
1

A
3B

9C
⇒
1
5
_________
2x
1
_______________

3B
3)
__
4
so
B

49
49
4
_________
8
__________
49(x
49(2x

2
(x
from
72
__

7
∴
(apart
0
5
x
x
used).
2
(2x

1)
7(x
3)
3)

1)
Exercise 1.18b
Express
each
fraction
in
partial
fractions.
2
3
______________
2
x
 2
______________
1
x
 1
_________
2
3
2
(x
1)
2
(x
Fractions
Fractions

1)
(x
with
with
decomposed
a

1)
x
a quadratic factor
quadratic
into
2)(x
2
partial
factor
in
the
(x
4)
in the denominator
denominator
can
also
be
fractions.
2x
1
______________
For
example
is
a
proper
fraction,
so
its
partial
fractions
2
(x
will
also
be

2)(x

1)
proper
.
2
Therefore
linear
the
partial
fraction
with
denominator
( x
numerator
.
2x
1
______________
Bx  C
_______
A
______
Therefore


2
(x

2)(x
2

1)
x

2
x

1
2
A(x
2x
1
______________

1)

(Bx

C)(x

2)
__________________________
⇒

2
(x

2)(x
2

1)
(x

2)(x

1)
2
⇒
2x
1

A(x

1)


C)(x

so
A

1

0
x

2
gives
5

5A
x

0
gives
1

1

2C
so
C
x

1
gives
1

2

(B)(3)
so
B
2x
1
______________
∴
1
______

x
______

2
(x
46
(Bx

2)(x
2

1)
x

2
x

1

1
2)

1)
will
have
a
Section
1
Complex
numbers
and
calculus
2
Repeated quadratic factors
2
x
3x
________
The
fraction
has
2
(x
numerator
this
a
repeated
quadratic
factor
.
By
rearranging
the
2

1)
can
be
expressed
as
the
sum
of
two
fractions,
2
2
(x
x
3x
________

1)
3x
1
_________________
i.e.

2
2
(x

2
1)
2
(x

1)
3x  1
________
1
______

2
2
x
Any
repeated
with
linear
with
the

1
quadratic

factor
numerators,
repeated
2
(x
one
quadratic
1)
can
with
be
expressed
the
single
as
the
quadratic
sum
of
factor
two
and
factors
the
other
factor
.
Example
3x  6
_______________
Express
in
2
(x
1)(x
partial
2)
3x  6
_______________
1)(x
Bx  C
________
A
______

2
(x
fractions.
2

Dx  E
________


2

2
2)
x
1
2
A(x
2
(x

2)
2
(x

2)
2

2
2)

(Bx

C)(x
1)(x

2)

(Dx

E)(x
1)
__________________________________________________

2
(x
2
∴
3x
There
are
5

6

unknowns
A(x
so
x

1
gives
9

9A
x

0
gives
6

4
⇒
x

1
gives
3
2C

⇒
x

2
gives
12
⇒
x

2
gives
0
⇒
Solving
these
9


6(C
B)
6(C

2B)

18(2B
3x  6
_______________
1)(x

2)

(Dx

E)(x
1)
1
[1]
2(E
(E
6(C
6(2B
C)(x
equations.
B)

36

2
36

5
(Bx
E

3(C
equations
∴
A
2C
E

need
2)
2
2)
so

2

2

we
1)(x


C)
D)
2B)
E
D)


E
2D
C)
(E



24

gives
1
______
x  1
________
x
(x
[2]
2D
3(E
2D)
simultaneously
3
[3]
2D)
12
B

[4]
1,
C

1,
D

3,
E

0
3x
________

2
(x
1)(x
2

2
2)
1
2

2)
(x
2

2)
Exercise 1.18c
Express
each
fraction
in
partial
fractions.
3
1
______________
1
2
x
 x
 2x
_______________
2
(x

2)(x
2

1)
1
_________
3
2
(x

1)(x
2

1)
2
x
2
(x

1)
47
1.
19
Applications
Learning outcomes
The
of
partial fractions
use of
partial fractions to
simplify the
differentiation of fractions

To
use
partial fractions
to
1
_____________
simplify
the
differentiation
of
We
can
use
the
quotient
rule
to
differentiate
but
(x
1)(x

the
1)
fractions
simplification

To
integrate
using
of
the
result
is
complicated.
partial
1
_____________
fractions
When
we
express
as
(x
i.e.
1)(x
1
________
1
________
2(x
2(x
as
,
1)


partial
fractions,
1)
then
we
can
differentiate
two
simpler
1)
You need to know
fractions

The
differentials
of
simple
d
___
∴
functions

How
to
and
of
dx
log functions
decompose
a
and
the
resulting
d
___
1
_____________
(
(x
1)(x

1)
simplification
)
1
________

(
dx
2(x
d
___
rational
into
dx
partial fractions
f(x)
The
integral

of
How
to find
a
definite
1)
d
___
)
1)
(x
)
1
__
(
dx
1
(x


1)
)
2
1
__
1)
2
(x
2
integral

1
(x
2

f(x)

2(x
2
1
__
_____

easier
.
1
________
1)
1
__
(

function
is

1)
2
1
_________
1
_________

2
2(x

The
laws
of

2
1)
2(x
1)
logarithms
2
(x
2
1)
(x

1)
_________________

2
2(x
1)
2
(x

1)
4x
________________


2
2(x
1)
2
(x

1)
2x
_______________


2
(x
1)
2
(x

1)
2x
________


2
2
(x
1)
Exercise 1.19a
Express
each
fraction
in
partial
fractions
and
hence
differentiate
each
fraction.
3x
1
______________
1
(x

2)(2x
1)
5x
___________________
2
(x
1)(x
2)(x
3)
2
3x
x
______________
3
2
(x
The

3)(x

use of
1)
partial fractions
in
integration
1
_____________
The
fraction
is
(x
1)(x

not
recognisable
as
the
differential
of
a
1)
1
_____________
standard
function
so
the
integral
∫
is
(x
we
48
express
the
fraction
in
partial
1)(x
fractions,

1)
not
obvious,
but
if
Section
1
_____________
i.e.
1
________
1
________
2(x
2(x
1
Complex
numbers
and
calculus
2

(x
1)(x

1)
1)
1
_____________
dx
(x
these
1)(x
integrals

is
∫

2(x
2(x
1
________
dx
1)(x
dx
1)

and
each
of
1)
recognisable.
1
_____________
(x
1
________
∫
dx
1)
∫
∴
1)
1
________
∫
then



∫
1
________
∫
dx
1)
2(x
dx
1)
2(x
1
__

1)
1
__

ln |x
1|
ln |x
2

1|

c
2
1
__
x
1
______

ln
⎥
2
⎥
x


c
1
Example
3
2
x
 4x
x
_____________
∫
Find
dx
(x
3
1)(x

4)
2
x
 4x
x
_____________
is
(x
1)(x
into
the

2

improper
fraction,
so
first
divide
the
denominator
numerator:
x
x
an
4)
3x
)
4

1

4x

3x
3
x
2
3
x
x
2
4x
2
x

3x

3x
2
x
4
4
3
2
x
 4x
x
_____________
so
4
_____________

(x
1)(x


1

(x

3
x
4)
x

1
1)(x
4)
4
________
4
________
5(x
5(x

1)

4)
2
 4x
x
x
_____________
∴

∫
4
________
dx
(x
1)(x


∫
(x

1) dx

∫
4)
1
__
1)
x

x

2
dx
5(x
4
__
2

4
________
∫
dx
5(x

4)
4
__
ln
|x
1|
ln
5
|x

4|

c
5
2
4
__
x
__


x

2
x
1
______
ln
5
⎥
⎥
x


c
4
Exercise 1.19b
1
Use
partial
fractions
to
find
the
following
3x  4
________
(a)
∫
2t
____________
dx
x(x

integrals:
(b)
∫
1)
dt
(t

2)(t

2)
2
4x
 3x  2
______________
x
______
(c)
∫
dx
x

(d)
2
∫
dx
(x

1)(2x

3)
0
2
1
________
s
 s
_____________
2
Evaluate
∫
1
ds
2
(s

1)(s

1)
Hint:
(
∫
1

2
(1

x
)
tan
x
)
49
1.20
Integration
Learning outcomes
using
Integration
When

To
use
substitution
substitution
y

using
gh(x)
we
substitution
can
use
the
substitution
u

h(x)
and
the
chain
rule
to find
d
___
integrals
of
some
products
to
find
gh(x)
giving
dx
d
___
du
___
g(u)

g(u)
dx
dx
You need to know
du
___
∫g(u)
∴
dx

g(u)

c
[1]
dx

That
integration
is
the
reverse
of

The
chain
∫g(u)du
Now
differentiation

g(u)

c
[2]
rule
du
___
Comparing
[1]
and
[2]
∫g(u)
gives
dx
∫g(u)du

dx

The
differentials
of
the
standard
du
___
functions
When
we
replace
g( u)
by
f(u)
we
∫f(u)
get
dx

∫f(u)du
dx
du
___
Therefore
(f(u)
integrating
)
with
respect
to
x
is
equivalent
to
dx
integrating
f( u)
with
respect
to
u,
i.e.
du
___
…
dx

…
du
dx
Note
not
that
an
the
relationship
equation
nor
is
it
above
an
is
a
pair
of
equivalent
operations.
It
is
identity.
1
2
For
example,
to
∫x
find
3
(1
x
)
2
3
(1
x
)
2
du
___
Now
2
dx
∫x
⇒
u
3x
∴
substitution
2
∫

x
gives
dx
2
dx

…
du
⇒
…(
3x
) dx

… du
u
2
du
3
3
(
method
one
We
of
3
2
)
(
3
factor
.
1
1
dx
1
when

1
u

This
u
dx
1
2
∫x
2
…
dx
∴
the
du
___
2

using
1
1
2
∫x
3
dx
2
)
u
2

is
substitute


the
u
3
(1
x
)
2

c
9
substitution
function
c
3
is
used
to
differential
for
this
integrate
of
‘inside’
the
a
product
function
of
‘inside’
functions
the
other
function.
3
For
example,
differential
we
of
can
use
it
to
cannot
2 cos x sin
x dx
because
cos x
is
the
sin x
x
We
∫
find
use
it
to
find
∫
e
2
cos
x
x dx
because
e
is
not
the
differential
of
cos x
Example
2
∫
Find
cos x
2 cos x sin
is
the
x dx
differential
of
sin
x
so
we
will
du
___
u

sin
x
⇒
cos x,
∴
…
2 cos x sin

2
x dx

∫
2u
2

du
2
3
u
3
50
dx
dx
2
∫
the
substitution
u

sin x
du
___

dx
∴
use

c

3
sin
3
x

c
… du
⇒
…
cos x dx

… du
Section
Definite
When
do
you
not
use
need
change
After
integration
the
substitution
substitute
limits
some
directly
the
to
to
practice
without
using
back
u

to
a
corresponding
you
having
may
to
find
make
a
Complex
numbers
and
calculus
2
substitution
f(x)
to
evaluate
function
values
that
a
1
of
of
you
x.
a
definite
Y
ou
can
integral,
use
u

f(x)
you
to
u
can
integrate
some
functions
substitution.
Example
3
∫
Evaluate
______
x
2
√
x
1
dx
2
2
2x
is
the
differential
of
x
1,
so
we
will
use
the
substitution
2
u

x
u

x
1
du
___
2
1
⇒

2x
dx
du
___
∴
dx
…

… du
⇒
… 2x dx

… du
dx
3
∴
∫
3
______
x
______
1
2
√
x
1
dx
∫

2x
2
√
x
1
dx
2
2
2
x  3
1
1
∫

u
2
du
2
x  2
When
x

3
2,
u

3
and
when
x

3,
u

8
8
______
1
∴
∫
x
√
1
2
x
1
dx
∫

u
2
du
2
3
2
3
1
8
2
[

2
u
2
]
3
3
3
1

(
8
3
2
3
2
)
3
√
√
16
2
3
3
____________

3
Exercise 1.20
1
Use
the
given
substitution
to
find
_____
(a)
∫
(b)
∫
sin x√cos x
1
__
dx;
u

cos x
2
(ln x)
dx;
u

ln x
x
2
Use
a
suitable
substitution
to
find
______
x
x
√
(a)
∫
e
(b)
∫
cos 2x sin
e

1
dx
2
3
2x dx
Evaluate
0.5
(a)
∫
______
x
√
2
1

x
2
dx
using
the
substitution
u

1

x
0
__
√
3
x
________
(b)
∫
0
______
dx
√
1
2

x
51
1.21
Integration
Learning outcomes
of
Standard trigonometric
From

To
integrate
some
trigonometric functions
the
derivatives
of
the
integrals
standard
trig
functions
we
know
that
trig functions
2
∫
You need to know
cos x dx

sin x

∫
c,
sin x dx

cos x

∫
c,
sec
x dx

tan x

Using
f(x)

The
differentials
of
the
____
standard
∫
dx

ln |f(x)|

c
f(x)
trig functions
d
___

That
gives
f(x)

f(x)
dx
sin x
_____
⇔
∫f(x)

The
laws
of

The
double
dx

f (x)

∫
c
tan x dx
dx

ln|cos x|

c
cos x
logarithms
angle
∫

trig

ln|1|

ln
identities
ln|cos x|

c
ln 1

0
1
_____
⎥

⎥
c

ln|sec x|

c
cos x
and
cos x
_____
∫
cot x dx

∫
dx

ln|sin x|

c
sin x
∫tan
i.e.
Using
x dx
integration

by
ln|sec x|

c
∫cot
and
x dx

ln|sin x|
nx dx
1
__

sin nx

∫sin
c,
nx dx


n
x cos
1
______
x dx


x dx

sin
2
x dx

sin x
and
x

use
the
identities
for
cos 2 x
c,
x

c
1
cos x
1
2
We
c,
n  1
tan


1
1
______
n
x tan
n
powers of

x
1
n  1

n
Even

1
______
n
x sin
∫sec

n  1
cos
n
∫cos
cos nx
n
n
∫sin
and
in
the
forms
cos
x

(1

cos 2x)
2
1
2
and
sin
x

(1
cos 2x)
2
1
2
For
example,
∫
sin
x dx
∫

(1
cos 2x) dx
2
1
1

x
sin 2x
2
Odd
powers of
sin x
use
the
identity
cos
and
sin
c
cos x
2
x

sin
3
∫

4
2
We
x

1.
For
example,
2
x dx

∫
sin x
(sin

∫
sin x
(1

 cos x
x) dx
2
cos
1

2
x) dx
3
cos
3
52
c
substitution
1
__
∫cos
gives

x

c

∫
(sin x
sin x cos
x) dx
c,
c
Section
Multiple
T
o
For
2.6
products
in
Unit
example,
sin 6x

Complex
numbers
and
calculus
2
angles
integrate
(T
opic
1
to
sin 2x
∫
find

such
as
cos 2 x sin 4x
we
can
can
use
use
the
factor
for mulae
1).
2
cos 2x sin 4x dx
sin 4x
cos
we
2x
1
∫
∴
cos 2x sin 4x dx
∫

2
(sin 6x

sin 2x) dx
__
1

1

cos 6x
cos 2x
12
A
variety
The
aim
forms
of
is
trig
to
and/or
functions
convert
to
trigonometric
a
reduce
can
be
integrated
trigonometric
the

c
4
using
integral
trigonometric
to
the
one
function
to
ideas
of
a
the
given
above.
standard
number
of
single
ratios.
Example
5
∫
Find
 d
sin
5
4

sin

sin 
2

sin
2

sin 
(sin
)

sin 
(1
cos

sin 
(1
2 cos
2
2
)
2
5
∫
∴
4


cos
)
2
 d
sin

∫
(sin 
2 sin  cos
2

 cos 

4

1
3

cos
sin 

) d 
cos
5

cos

c
5
3
Example

2
∫
Evaluate
sin 2x sin x dx
0
1
sin 2x sin x


(cos 3x
cos x)
2


2
2
1
∫
∴
sin 2x sin x dx
∫

(cos x
cos 3x) dx
2
0
0

1
1
2
[sin

x
sin 3x
2
1

0
1
(
2
]
3
1
2
(
1))

3
3
Exercise 1.21
Find
5
1
∫
cos 3x dx
3
∫
sec
2
2
∫
sin  cos
4
∫
cos
3
x
tan
3
x dx
 d
2
x
sin
x dx
___________
5
∫
7
∫
sin x
cos 3x dx
6
∫
sin 2x
√
1
2

2 sin
x
dx
cos x
__________
_________
√
1

dx
sin x
Evaluate

__

__
4
8
∫
0
2
(cos 5x cos 3x) dx
9
∫
2
sin
2
 cos
 d
0
53
1.22
Integration
Learning outcomes
by
parts
Integration
by
parts
x
We
To

integrate
functions
a
by
product
cannot
∫
find
of
parts
xe
dx
using
d
___
If
we
start
to
x,
uv
we
of
du
___
with

the
methods

u
and
dx
uv
then
∫

v
Rearranging
product
∫

u
The
and
this
formula
with
dx
gives
du
___
∫
integrals
sides
a
of functions
differentials
both
dx
v
dv
___
dx

uv
∫

u
dx
dx

integrate
dv
___
dx
dx
differentiating
far
.
dx
du
___
The formula for
so
get
You need to know

introduced
dv
___
v
dx
respect
any
dx
of
This
version
of
the
formula
can
be
used
to
integrate
a
product
of
functions
standard functions
du
___
where
v
and
are
the
two
functions;
this
is
called
integrating
by
parts
dx
T
o
use
the
formula,
the
right-hand
side
shows
that
one
of
the
functions
du
___
in
the
product,
,
has
to
be
integrated,
and
the
other
function,
v,
has
to
dx
be
differentiated.
function
whose
integrated,
neither
When
both
differential
then
v
function
is
the
can
be
functions
is
the
other
can
simpler
.
function.
be
integrated,
When
This
only
formula
choose
one
as
function
cannot
be
v
the
can
used
be
when
integrated.
x
So,
to
find
∫
xe
dx,
we
have
two
functions
that
can
be
integrated.
differential
of
x
is
simpler
than
that
of
e
The
du
___
x
,
so
we
choose
v

x
and
x

dx
dv
___
x
This
gives
u

e
and

1
dx
du
___
∫
Then
v
dv
___
dx

∫
uv
u
x
dx
dx
⇒
∫
xe
x
dx

xe
x
∫
(e

1) dx
dx
x

x
xe
e
x

c

e
(x
1)

c
Example
2
Find
∫
x
ln x dx
2
T
o
use
integration
by
du
___
v

ln x
and
parts
x
⇒
dx
v
∫

uv
∫
u
2
ln x dx
u

⇒
∫
see
that
ln x
cannot
be
integrated
but
x
can.
So
we
3
x
3
1
2
dx
we
1
and
x
dv
___
dx
dx
x
1
__

dx
du
___
∫
find
dv
___
2

to
x
ln x dx

1
3
x
ln x
3
dx
1

1
ln x
∫
ln x
3
1
__
3
)
x
3
(
(
)
dx
x
2
x
)
dx
3
1
3
x
(
1
3
x
3

∫
1
3
x

c

9
3
x
(3 ln x
1)

c
9
Example
Find
We
∫
ln x dx
said
ln x
54

1
in

the
previous
ln x,
we
can
example
find
∫
that
ln x dx
ln
x
cannot
be
integrated,
but
using
integration
by
parts
with
choose
e
Section
du
___
T
o
∫
find
(1

ln x) dx,
let
v

ln x
and
dv
___

1
du
___
∫

dx

∫
uv
numbers
and
calculus
2
u

x
x
1
__
u
dx
dx
and
dx
dv
___
v
Complex
1
__
⇒
dx
1
∫
⇒
ln x dx

∫
x ln x
(
x
dx
)
dx

x ln x
x

c

x(ln x
1)

c
x
Example
x
∫
Find
e
sin 3x dx
du
___
x
Using
v

e
dv
___
and

sin 3x
gives
1
e
sin 3x dx


and
u


cos 3x
3
1
x
e
∫
cos 3x
3
1

e
dx
x
∫
1
x

dx

cos 3x

3
3
x
cos 3x
)
e
dx
3
1
x
e
(
x
∫
e
cos 3x dx
x
Using
integration
by
parts
again
on
∫
du
___
x
e
cos 3x dx
with
v

e
and

cos 3x
gives
dx
1
x
∫
e
sin 3x dx


1
x
e
cos 3x
1

3
1
x
e
sin 3x
required
10
__
Y
ou
may
appears
1
x
∫
9
integral
e
sin 3x dx

on
find
it
easier
it
in
1
cos 3x
sides
1
x
e
to
the
3
9
both
sin 3x
of
the
e
sin 3x dx
)
sin 3x dx
equation.
x
e
the
e
x
∫
Collecting
∫
⇒
for mula
for
e
it
__
1
x
cos 3x
3
apply
x
∫
sin 3x
x
e
9
remembering
1
x
e
3
3
9
The
1
(

3
sin 3x dx

on
the
left-hand
side
gives
x
e
(sin 3x
3 cos 3x)

c
10
integration
by
parts
by
for m
d
___
∫
f(x) g(x) dx

(∫
f(x)
)

g(x)

∫ (∫
f(x)

g(x)
)
dx
dx
Example

2
Evaluate
2
∫
x
cos x dx
0
2
Using
integration
We
need
∴
∫
to
use
by
integration
2
x
parts,
∫
x
by
2
cos x dx
parts

again
x
to
∫
sin x
∫
find
2x
2x

sin x dx
sin x dx
2
cos x dx

x
2
(
sin x
2x
cos x
∫
2 cos x dx
)

x
sin x

2x cos x
2 sin x

c


2
∫
x
2
cos x dx

[ x
2

__
2
∴
sin x

2x cos x
2 sin x
]
2
0
0

(
2
)
2
Exercise 1.22
1
2
Find
Evaluate

1
4
3x
(a)
∫
2xe
x
dx
(b)
∫
e
cos x dx
(a)
∫
ln (1

2x) dx
0
(b)
∫
3x sin 3x dx
0

2
2
(c)
∫
x ln x dx
(d)
∫
x
sin 2x dx
(c)
∫
2x
e
cos 2x dx
0
55
1.23
Integration
of
inverse
trigonometric
functions
1
Learning outcomes
The
integral of
sin
x
1

To find
the
inverse
integrals
of
the
T
o
find
∫
sin
x dx
we
use
integration
by
parts
and
the
same
device
1
we
used
to
∫
find
ln x dx,
i.e.
we
∫
using
differentials
of
the
dx
inverse

∫
uv
v

sin
1
x dx
∫
as
(1

u
dx
dx
du
___
1
with
sin
dv
___
v
dx
The
∫
write
du
___
Then,
You need to know

x
and

1

x
dx
trigonometric functions
1
________
dv
___

How
to
integrate
using
______
gives

and
dx
u
2
√
1
x
substitution

How
to
use
integration
by
parts
1
∫
∴
x
________
1
sin
x dx

x sin
______
∫
x
dx
2
√
1

Methods for
functions
integrating
using
x
rational
x
________
partial fractions
______
T
o
find
∫
dx
we
can
use
the
substitution
2
√
1
x
2
u

1
x
⇒
2x dx

du
1
2
x
________
____
______
∴
∫
dx

∫
du
1
2
√
1
x
u
2
1

u
2

c
______

(Y
ou
can
also
find
2
√

1
this
x

integral
c
by
sight.)
______
1
∴
∫
sin
1
x dx

x sin
(
x
2
√
1
x
)

c
______
1
i.e.
∫
sin
1
x dx

x sin
x

2
√
1

x

c

c
1
The
integral of
Using
a
similar
cos
method
x
as
we
1
∫
cos
used
above,
1
x dx

∫

x cos
1

cos
x dx
1
________
1
______
∫ (x
x

2
√
1
x cos
dx
)
x
x
________
1

______
x

∫
dx
2
√
1
x
______
1

x cos
x
√
1
2
x

c
______
1
i.e.
56
that
trigonometric functions
∫
cos
1
x dx

x cos
x

√
1
2

x
sin
x) dx
Section
1
Complex
numbers
and
calculus
2
1
The
integral of tan
x
1
∫
Using
1
tan
x dx
1
∫
tan
∫

1

tan

x tan
and
∫
x
as
of
the
dx
2
1
__
dx
1
__
____
∫
form
x
x
______
∫
gives
f(x)
1
__
dx

gives
x
2
1
parts
dx

x
______
∫
by
2
1
Recognising
integration
x
______
1
x dx
x dx

f(x)
ln f(x)

c
2
2

ln (1

x
)


x tan
c
2
1

x
2
1
∫
∴
tan
1
__
1
x dx
x

2
ln (1

x
)

c
2
These
results
although
it
is
can
be
better
quoted
to
unless
remember
the
their
the
derivation
method
is
rather
asked
than
for,
lear n
integrals.
Example
1
Find
∫
Using
x tan
x dx
integration
by
parts
gives
2
1
__
1
∫
x tan
x dx
2

x
x
_________
1
tan
∫
x
2
2

x
Exam tip
)
2
x
______
1  x
1
__________
Now


1

using
integration
1

x
is
sensible
to
check
differentiating
1
__
dx
2

x
)
dx
1

it
should
by
give
the
∫
1 dx
1
__
x
1
__
x dx


)
x
1
tan
x

c
2
2
x
1
__
1
tan
1
x
(x
2
tan
x)

c
2
1
__

dx
2
1
2
1
integrated.
1
______
(∫
1
__
x tan
you
x
2

function
2
2
1
__
∫
this
answer
x
______
∫


∴
it:
your
2
∫
2(1
parts,
2
x
2
x
_________
∴
by
1
2
x
When
1
______

2
1
dx
2
2(1
2
(x
1
__
1

1)
tan
x
x
2

c
2
Exercise 1.23
1
Find
1
(a)
2
∫
tan
1
3x dx
(b)
∫
sin
∫
√
2
2x dx
(c)
∫
x
1
tan
x dx
Evaluate
1
1
1
2
(a)
∫
0
1
sin
x dx
(b)
0
1
2
cos
x dx
(c)
∫
1
tan
(1

x) dx
0
57
1.24
Reduction formulae
Learning outcomes
Finding
a
reduction formula
If
to
4

To
derive
and
use
reduction
we
need
∫
find
sin
we
then
square
it
to
can
use
the
identity

sin

cos 2)
(1
2
1
4
formulae
1
2
x dx
give

sin
2

2 cos 2 
(1

2).
cos
4
1
2
W
e
can
then
use
the
identity
2
cos

(1
cos 4)

to
give
an
integral
2
cos 2 
involving
of
You need to know
higher
T
o

How
to
use
integration
by
do
powers
this
we
The
integrals
of
of
use
sin
but
this
is
tedious
if
used
to
find
the
integral
x.
integration
by
parts
to
give
a
formula
that
parts
systematically

cos 4 ,
and
reduces
the
power
to
one
that
we
can
easily
integrate.
standard
n
For
example
to
∫
find
functions
calling
this
integral
sin
x dx,
where
n
is
a
positive
integer
,
we
start
by
I
n
n
then,
by
parts,
∫
writing
we
n
sin
x dx
∫
as
sin x
1
sin
x dx
and
using
integration
by
have
n
I

∫
sin

∫
sin x

 cos x
sin

 cos x
sin

 cos x
sin

 cos x
sin
n
x dx
n
1
sin
x dx
n
1
n
cos x
(n
cos x)((n

∫
x

(n
1)
∫
x

(n
1)
(∫
n
cos x
x

(n
1)

x

n
cos x
(n
1)
is
the
Using
A
x
a
where
reduction
power
on

find
the
n
to
for mula
one
function
by
1
or
∫
formula
the
sin

to
by
because
involving
be
can
integral
be
can
used
be
it
reduces
power
integrated,
a
we
have

I
to
found
systematically
58
integral
2.
formula
may
reduce
the
power
to
easily.

x

sin
I
x dx
and
again
on
[1]
4
6
formula
∫
5
5
cos x sin
1

4
an
reduction
I
gives
4
I
n
6
x dx
6
the
the
2.
6
6
Using
2
reduction formula
1
I
2
I
6
T
o
)
n
n
power
reduction
one
a
the
Depending
reduce
2
n
called
)
n
1
______
1
sin
n
involving
x dx
I
n
This
n
sin
1
sin
1
__

x ) dx
∫
x dx
I
n
I
2
sin
2
sin
(I
n
⇒
n
x)
1
sin
n

x) dx
2
sin
x ) dx
2
x sin
((1
n
nI
n
1)(cos
2
cos x sin
2
x
n
⇒
1)
1
n

(
1
n
I
∫
1
n
∴
n
x

3
3
cos x sin
4
x

I
4
[2]
2
using
the
formula
above
Section
1
Complex
numbers
and
calculus
2
2
Now
I

∫

∫
2
sin
x dx
1
(1
cos 2x) dx
2
1

1
x
sin 2x
2
1
From
[2],
I

cos x sin
3
x

4
1
cos x sin
x

c
)
sin 2x

c
16
5
__
5

cos x sin
6
sin 2x
4
3
__
4
1

1
x
2
3
x
8
I
1
(

4
3
[1],
c
3

4
From

4
5
__
3
x
cos x
sin
x

5
__
x
24
6
cos x sin x
16
Using
Note
that
c.
c
when
we
found
I
,
we
introduced
a
constant

c
16
sin 2x
of

2 sin x cos x
integration,
2
As
is
fractions
an
in
unknown
the
constant,
reduction
there
is
no
need
to
multiply
it
by
the
formulae.
Example
n
(a)
Given
I
∫

n
x
x
Hence
∫
find
n
I
(a)

∫

x

x
n
x
n
dx
x

x
x

4
∫
x

x
x
e
nI
n
1
n
∫
1
x
1
nx
x
e
dx
x
e
dx
x
e
4
I
x
e
n
∫
n
nI
n
(b)
I
dx
n
dx
e
n
that
x
e
x
e
show
n
4
(b)
n
e
1
x
e
dx
x
Now
I

∫

xe

e
1
xe
dx
x
x
∫
e
dx
x
(x
1)
n
Using
I

x

nI
2
x
2

x
x
2e
e
(x
I

(x
2x
x
gives


e

x
a
2)

c
(x
2
3x

6x

12x
6)

c
x
e
4I
3

find
c
2
x
formula

3I
3
4
I
1)
x
e
4
also
2
2
x
the

x
e
3
can
n
1
3
case
parts
2I
x
We
by
1

and
integration
x
e
2
Similarly
with
n

Using
x
e
n
I
c
e
4
(x
reduction
is
often
3
4x
2
formula
easier
to
for
24x
a

definite
24)

c
integral,
and
in
this
use.
59
Section
1
Complex
numbers
and
calculus
2
Example
1
(a)
Given
I
2
∫

n
n
______
n
x(x
1)
dx
show
that
I


I
n
n
n
0

1
1
1
(b)
Hence
2
∫
evaluate
5
x(x
1)
dx
0
2
(a)
W
e
need
an
integral
involving
n
(x
1
1)
so
we
start
by
writing
I
n
1
2
∫
2
x(x
n
1)(x
1
1)
dx
0
1
∴
I
2
∫

n
2
x(x
n
1)(x
1
1)
dx
0
We
can
now
express
this
as
two
integrals,
1
I
3
∫

n
i.e.
1
x
2
n
(x
1
1)
2
∫
dx
0
n
x(x
1
1)
dx
0
1
3
∫

x
2
n
(x
1
1)
dx
I
n
1
0
1
∴
I

I
n
3
∫

n
1
x
2
n
(x
1
1)
dx
0
2
If
we
the
2
to
n
(x
be
integration
power
3
x
use
n
2,
by
parts
which
with
we
do
v

not
n
(x
1
1)
want.
we
Therefore
1
2
1)
so
that
there
is
a
term
involving
will
we
n
(x
reduce
rearrange
1
1)
that
integrated.
1
1
3
∫
Now
x
2
n
(x
1
1)
dx

0
2
∫
(x
2
)
du
___
Using
n
( x(x
1)
1
) dx
0
integration
by
parts
with
2

x(x
n
1
1)
gives
dx
2
n
(x
I

I
n
[(x

n
1
)
(
1
__
0
I
n
n
1
__
∴
I

I
n

n

I
1
n
n
n
______
⇒
I


I
n
n
n

1
1
1
(b)
∫
2
x(x
5
1)
dx

I
5
0
1
Now
I

1
2
∫
x(x
1) dx
0
1

3
∫
(x
x) dx
0
4
x
__

1
__

4
60
1
]
4

2
x
__
[
2
1
2
0
n
(x
1)
________
)]
2n

1
1)
________
2
0
∫
0
2x
(
)
2n
dx
can
as
Section
1
Complex
numbers
and
calculus
2
n
______
Using
I


I
n
gives
n
n

1
1
2
I



2
I
1
3
2
1
)(
(

)
3
4
1

6
3
so
I


I
3
2
4
3

1


4
6
1


8
4
and
I


I
4
3
5
4

1



5
8
__
1

10
__
1
5
so
I



5
10
6
__
1


12
1
2
∫
i.e.
__
1
5
x(x
1)
dx


12
0
n
______
Note
that
a
formula
like
I


I
n
a
sequence
giving
values
for
I
,
I
2
Such
a
formula
is
called
a
,
where
n  1
n
,

1
,
…
I
3
I
is
known,
produces
1
4
recur rence
relation .
Exercise 1.24
n
1
If
I
∫

n
cos
x dx
1
__
I

show
n  1
______
n  1
sin x cos
that
x

I
n
,
n

2
n  2
n
n
6
Hence
2
Use
∫
find
the
cos
x dx
reduction
formula
given
in
question
1
to
show
that,

__
2
when
I
n
∫

n
cos
x dx,
0
n  1
______
I

I
n
n  2
n
1
3
If
I

n
∫
______
n
x
√
1

x
dx
show
that
0
2n
_______
I

I
n
,
n

0
n  1
2n

3
1
Hence
find
∫
______
6
x
√
1

x
dx
0
61
1.25
The
trapezium
We
have
rule
covered
a
variety
of
methods
to
integrate
functions.
However
,
Learning outcomes
there

To
derive
and
use
the
are
several
to find
an
trapezium
approximate
the
x-axis
area
and
between
two
a
values
curve,
of
when
that
the
can
integrals
be
function
used
that
to
involved
cannot
give
an
be
found.
But
approximation
cannot
be
there
for
a
are
definite
integrated.
value
We
for
indefinite
methods
integral
rule
many
look
at
one
such
method
here.
the
x
The trapezium
rule
b
You need to know
The
∫
integral
f(x) dx
represents
the
area
between
the
curve
y

f(x),
a
the

How
to find
the
area
of
trapezium
When
the

The
x-axis
meaning
of
the
and
the
ordinates
x

a
and
x

b.
a
a
function
area
into
a
whose
finite
derivative
number
of
is
f( x)
vertical
cannot
strips
as
be
found,
shown
we
in
can
the
divide
diagram.
word
Joining
the
tops
of
the
strips
as
shown
gives
a
set
of
trapezia.
ordinate

The
shape
of
the
graphs
The
of
sum
of
the
areas
of
these
trapezia
then
gives
an
approximate
value
b
simple functions
for
∫
f(x) dx
a

How
to
use
integration
by
parts
y
O
T
aking
same
n
strips
width,
at
d,
x
equal
and
intervals
labelling
the
along
the
vertical
x-axis
sides
so
(i.e.
that
the
each
strip
ordinates)
y
is
,
the
y
0
,
...,
y
1
,
n
1
then
the
area
of
the
first
strip
is
d
(y

y
0
2
),
1
1
the
area
of
the
second
strip
is
d
(y
y

y
1
2
),
and
so
on.
2
y
)
)
y
)

1
the
strips
is
given
by
1
d
(y

1
2
2

0
1
(y
y(d
all
x
1
d
2
n
1
of
1
areas
2
the
1
of
y (d
sum
2
The
O
x
d
1
d
2
1
d
y(d
y(d
d
n
1
0
O

n
y
1
2
y
n
y 
2

y
n
y
y
2

y
1
n
2
y
y
0
n
1
)
1
)
y
y
)

…
1

2
d
(y

n
2
y
2
)
n

1
d
(y

n
2
1
1

d
(y

2y
0
2

2y
1
…

2y
2

n
2y
2

n
y
1
)
n
b
1
_
∴
∫
f(x) dx

d
2
a
62
(y

0
2y

1
2y
…
2

2y

n
2
2y

n
1
y
)
n
y
)
n
Section
This
It
half
the
is
width
of
Note
that
there
Note
also
that
i.e.
the
value
a
is
to
is
called
remember
strip
one
the
of
for mula
easy

(first
more

gets
trapezium
for mula
last
ordinate
approximation
d,
the
this

than
gets
twice
the
better
in
all
Complex
numbers
and
calculus
2
r ule.
words
the
number
when
1
of
the
as
other)
ordinates.
strips.
width
of
the
strips,
smaller
.
Example
6
(a)
Find
an
approximate
value
for
3
∫
x
dx
using
the
trapezium
rule
with
1
five
(b)
intervals.
Use
or
a
an
sketch
to
determine
whether
your
answer
is
an
over-estimate
under-estimate.
6
(a)
3
∫
3
x
dx
represents
the
area
between
y

x
,
the
x-axis
and
the
1
ordinates
There
x
are
intervals

1
five
as
and
units
one
x

6
between
unit
wide,
x

i.e.
d

1
1

to
x

6
so
3
This
gives
six
ordinates:
y

64,
y
4

125,
y
5
The

take
our
five
3

1,
y
0
y
we
1

2

8,
y
2

27,
3
216
6
trapezium
rule
gives
6
∫
1
3
x
dx

( 1

216

2(8

27

64

125) )

332.5
y
2
1
200
(b)
The
the
sketch
area
shows
under
that
that
the
part
area
of
the
of
each
trapezium
is
greater
than
curve.
6
Therefore
332.5
is
an
over-estimate
for
the
value
of
∫
150
3
x
dx
1
6
Alternatively,
∫
6
1
3
x
dx

4
[
]
x
4

323.75
1
100
1
This
is
is
over-estimate.
an
the
exact
value
of
the
area,
confirming
that
332.5
50
x
O
1
2
3
4
5
6
7
Exercise 1.25
1
(a)
Use
the
trapezium
rule
with
five
intervals
to
find
an
approximate
2
1
__
value
∫
for
dx
2
x
1
2
1
__
(b)
Sketch
the
graph
showing
the
area
represented
by
∫
1
2
(a)
Use
the
trapezium
rule
with
five
intervals
to
find
an
dx
2
x
approximate
_
1
_
1
2
value
∫
for

2
(1

x
)
2
dx
0
_
1
_
1
2
(b)
Find
the
exact
value
of
∫

2
(1

x
)
2
dx
0
(c)
Use
your
answers
to
(a)
and
(b)
to
find
an
approximate
value
for
.
63
Section
1
A
quadratic
1
Practice
equation
with
real
questions
coefficients
10
has
Find
the
smallest
positive
value
of
x
for
which
x
one
root
equal
to
3

2i
y

the
W
rite
e
sin x
nature
(a)
the
other
(b)
the
equation.
root
of
the
11
that
value
stationary
and
determine
value.
Determine
on
y

2i
3


ln
2i
(
x

and
y
b,

i(a
write

b)
for
down
real
two
values
of
x,
relationships
y,
a
and
and
whose
nature
of
equation
stationary
is
y,
)

1
a
between
12
x,
curve
2
x
If
the
number
2x
______

3
the
1  2i
______
Simplify
(b)
stationary
equation
points
(a)
of
a
down
2  i
______
2
has
The
parametric
equations
of
a
curve
are
b.
t
x
3
Find
the
values
of
a
and
b
where
z

a
ib
such
(a)
(i)
Find
in

e
terms
,
y
of

t
t
the
equation
of
the
t
that
tangent
2iz

z*(2

i)

2z

2i
(ii)
Hence
the
4
(a)
Find
the
(b)
Hence
square
find
the
roots
roots
of
of
11

the
60i
(b)
equation
Find
(ii)
Use
the
the
(4

i)x

(1

17i)

curve
the
at
point
Cartesian
of
at
the
equation
the
Cartesian
equation

the
find
curve
(i)
2
x
to
the
point
of
the
where
equation
equation
tangent
to
t
of
to
,
t).
tangent

the
(e
to
2
the
find
curve.
the
curve
at
the
0
2
point
5
Y
ou
are
given
(a)
Express
that
z

1

where
x

e
i
y
z
in
the
r(cos 
form
13
i sin )

Given
that
y
2
Find
(b)
the
modulus
and
argument
of
e
sin x
show
that
y
dy
z

e
cos x
_______
___

2
Illustrate
(c)
z
and
dx
2
z
and
z

z
in
an
1

y
Argand
2
d
diagram.
y
dy
____
Hence
find
a
relationship
between
___
,
and
2
dx
___
__
√
15
____
6
Y
ou
are
given
that
z
dx
√
5
___


i
2
2
14
The
parametric
equations
of
a
curve
are
2
(a)
Express
(b)
Hence
z
in
find
the
the
r(cos 
form
two
square
i sin )

roots
of
x

12t
,
y

3t
z.
2
dy
d
___
y
____
Find
and
in
terms
of
t.
2
dx
7
Solve
z
(2
the


simultaneous
dx
equations
(2

i)w

5

i
i)z

3w

3

i
15
Find
the
3x
derivatives
of
x
(a)
x
(b)
sin
e
(c)
x tan
1
8
(a)
Using
the
binomial
theorem
or
(3x
otherwise,

2)
1
(2x)
3
expand
(b)
Hence
(cos 

i sin )
express
2
(i)
cos 3 
in
terms
16
cos 
of
Given
that
z

x
 z
___
2

xy

y
,
find
 z
___
and
 x
(ii)
sin 3 
in
terms
of
2
2
 z
___
Hence
show
that
(
 z
___

)
 x
9
(a)
Describe
(i)
|z

the
4|
locus

|z

of
points
6|
(ii)
satisfied
|z

1|
6
x
__
1
Hence
find
in
the
form
a

ib
the
values
of
Given
that
z

sin
(
y
z
2
which
satisfy
the
simultaneous
equations
 z
___
find
 x
|z
64

4|

|z

6|
and
|z

1|

6
 z
___
,

z
_____
and
 y
(
 y
by

17
(b)
 y
sin 
 x y
)
,
)

5z

3xy
y.
Section
n
18
27
Evaluate
(a)
Given
I
∫

n
x
1
Practice
questions
2x
e
dx
show
that

4
(a)
cos 2x
∫
e
1
__
sin 2x dx
I
n
x

n
__
2x
I
e
n
n
0
2
1
2
1
2x
1
2e
________
(b)
∫
dx
2x
x
0
3
e
(b)
Hence
(a)
If
x
1
______________
19
in
partial

2x
e
dx
fractions.
2
2)(x
x

Express
(x
∫
find
6
1)
28
I
n
∫

n
sec
x dx
show
that
0
dy
x
1
______________
___
Hence
find
when
y
n
2
2
_____________

2
dx
(x
2)(x
I

1)
n
2
______


I
n
for
n
n
1
√
3
n
(n
n

2
2
1
1)
2

3x
6x
2
_______________
20
Express
in
partial
fractions.
6
2
(2x

1)(x
1)
(b)
Hence
4
∫
evaluate
sec
x dx
0
0
2
3x
6x
2
_______________
Hence
evaluate
∫
dx
2
1
(2x

1)(x
1)
29
(a)
Use
the
find
an
trapezium
rule
with
three
intervals
to
1
21
1
______
Find
approximate
value
∫
for
dx
2
1
0

x
3
(a)
∫
sin
x dx
1
1
______
(b)
(b)
∫
sin 5x cos 3x dx
(c)
∫
cos
Find
the
exact
value
∫
of
dx
use
it
to
determine
and
2
1
0
whether

x
your
answer
to
4
x dx
part
is
(a)
an
over-estimate
or
an
under-
estimate.
x
22
(a)
Use
the
substitution
u

e
to
find
x
e
_______
∫
y
30
dx
2x
e

1
1
1.5
x
e
_______
(b)
Hence
or
otherwise
find
∫
0
dx
2x
e

1
1
2
x
23
Use
the
substitution
u

2
to
evaluate
∫
x
2
dx
1
0.5
24
Find
x
(a)
∫
xe
dx
x
O
2
x
(b)
∫
e
(c)
∫
x sec
cos x dx
0.5
2
x dx
(a)
25
The
diagram
shows
Find
the
area
between
the
2
x
______
curve
y

,
the
x-axis
and
the
2
1

x
2x
(a)
∫
e
(b)
∫
2x log
sin x dx
ordinates
26
Show
x

2
and
x

2
x dx
2
that
(b)
2 cos x
____________

sin x
1
the
find
an
Use
the
to
cos x
sin x
____________

cos x
Use
trapezium
rule
approximate
diagram
judge
whether
to
with
value
explain
your
four
for
why
answer
intervals
this
is
it
to
area.
is
an
difficult
over-

cos x

estimate
sin x
or
an
under-estimate.

2
2
Hence
evaluate
0
2
2 cos x
____________
∫
x
______
dx
cos x

(c)
Find
the
exact
value
of
∫
sin x
2
dx
2
1

x
65
2
Sequences,
2.
1
Sequences
Learning outcomes
series
define
a
sequence
To
use
a formula for
the
or
a
recurrence
and
so
an
ordered
list
of
terms.
There
is
a
first
term,
a
second
on.
nth
A
term
is
sequence
term,

approximations
Sequences
A
To

and
relation
sequence
can
have
a
finite
number
of
terms
or
an
infinite
number
of
to
terms.
find
a
specific
term
of
a
sequence
We
denote
the
terms
of
a
sequence
as
u
,
u
1
To

define
arithmetic
nth
geometric
,
…,
u
2
,
…
where
u
n
is
the
n
and
term.
progressions
(The
notation
a
,
a
1
When
u
is
a
,
… ,
a
2
,
…
is
also
used.)
n
function
of
n,
we
can
use
this
to
find
a
specific
term.
n
n
For
example,
if
the
nth
term
of
a
sequence
is
given
by
u

2
1
then
n
Example
The
nth
term
of
a
sequence
we
can
i.e.
the
find
a
specific
term
by
replacing
n
by
the
number
of
that
term,
1
is
first
term
is
given
by
replacing
n
by
1,
so
u

2
1

1
similarly,
1
2
n
 1
_________
2
given
by
u

u
n
Find
the
value

5
2
1

3,
u
2
2
2n

10
2
1

31,
u
5

2
1

1023,
and
so
on.
10
3n
of
u
4
Recurrence
2
 1
4
___________
u

17
___
Sometimes
20
This
the
relations
terms
of
a
sequence
are
related
by
a
recur rence
relation .

4
2
2(4
)
3(4)
is
an
equation
example,
u

2u
n
on
of
its
at
own
least
When
u

n
is
not
one

which
connects
3
or
u
1
the

u
n
enough
to
nth

n
define
a
term
u
1
to
.
n
sequence;
A
previous
terms,
recurrence
for
relation
2
we
need
to
know
the
value
term.
2u
n

n
3,
if
we
know
the
first
term
we
can
generate
the
1
sequence:
if
u

2,
the
recurrence
relation
tells
us
that
each
term
is
twice
the
1
previous
so
u

term
4

plus
3

3,
7,
u
2
So
the
When
sequence
u

u

2
2,

n
the
2u
7,
17,
n
2
u
1

3

17,
and
so
on.
2
37,
we
77,
need
…
to
know
the
first
two
and
of
u
the

two
4,
then
the
previous
recurrence
terms,
so
u
relation

4

tells
2

us
6,
3
So
so
the
Any
the
in
that
u

each
6

is
given
by
u

7
and
u
1
sequence
sequence
one
is
2,
where
above)

3(2

2u
n  1
n
u
is
4,
6,
each
called
a
10,
16,
term
is
26,

1
⇒
u

3(2
)

the
Fibonacci
sum
of
the
two
previous
terms
sequence .
1.
Show
that
n
u

3(2
)

1
n
1
n  1
n

2

3(2
)

1

2(u
1)

1

n
2u
1
n
n
and
u

3(2
)

1
⇒
n
This
66
verifies
u

7
(n

1)
1
that
the
given
formula
for
the

is
10,
…
n  1
)
n
term
4
4
n
sequence
to
on.
Example
A
order
2
sum
and
terms
sequence.
1
the
is
u
n
generate
If

3
nth
term
gives
the
first
term
and
the
recurrence
relation.
(like
Section
2
Sequences,
series
and
approximations
Alternatively
u

2u
n  1
1
⇒
u
n

1

2u
n  1
n
n  1
⇒
3(2
n  1
)

1

1

2u
Using
u

1

3(2
)

1
n
n
n
⇒
6(2
)

2

2u
n
n
⇒
u

3(2
)

1
n
Example
n
The
nth
term
of
a
sequence
is
given
by
u

5

1.
Find
u
n
n
u

5
in
terms
of
u
n  1
n
n  1

1
⇒
u
n

5

1
n  1
n

5(5
)

1
n

5(5

5u

1)
4
4
n
Arithmetic
An
arithmetic
the
previous
For
example,
A
general
(called
The
progression
(AP)
is
a
sequence
where
2,
5,
8,
whose
14,
first
common
recurrence
11,
…
term
is
is
difference )
relation
that
an
a
and
can
gives
arithmetic
can
see
term
be
an
where
the
written
AP
is
progression
u
difference
as

a,
that
the
nth
term
is
given
by
u

a

n
a

u
n
We
each
differs
by
a
constant
from
term.
AP
the
progressions

d,
as
successive
between
a

2d,
terms
successive
a

3d,
differ
terms
by
is
3.
d
…
d
1
(n
1)d
n
Geometric
A
geometric
previous
progressions
progression
(GP)
is
a
sequence
where
each
term
is
a
constant
multiple
of
the
term.
1
For
example,
64,
32,
16,
8,
4,
2,
…
is
a
GP
as
each
term
is
the
previous
term.
2
A
general
GP
whose
first
term
is
a
and
where
each
term
2
(called
The
the
common
recurrence
ratio)
relation
can
that
be
written
gives
a
GP
as
is
a,
u
ar,

ar
see
that
the
nth
term
is
given
by
u

the
previous
term
multiplied
by
r
3
n
n
can
is
ar
,
…
ru
n
We
,
1
1
ar
n
Y
ou
need
to
be
able
to
recognise
an
AP
for mula
or
for
a
GP
the
from
nth
a
recur rence
relation
or
from
a
ter m.
Exercise 2.1
1
State
which
of
the
following
sequences
are
APs
2
A
sequence
is
defined
by
u

10
and
the
1
and
which
are
GPs
and
in
each
case
determine
the
recurrence
relation
u

u
n  1
10th
term.
Find
a
formula
for
u
in
3
n
terms
of
n
n
(a)
5,
3,
1,
1,
…
n
3
The
nth
term
of
a
sequence
is
given
by
u

5
4
n
1
(b)
1,
(c)
1,
(d)
1,
1
,
2
1
,
,
4
1,
…
Find
equation
giving
u
in
terms
of
n  1
8
1,
1,
1,
u
.
n
…
4
1
The
nth
term
of
a
sequence
is
given
by
1
,
2
an
0,
,
2
1,
…
n
u

3

2
n.
Find
the
value
of
the
10th
term.
n
67
2.2
Convergence
Learning outcomes
and
divergence
Convergent
To
of
describe
the
convergent
the
sequence
1,
1
1
behaviour
and
sequences
sequences
1
Consider

of
,
__
1
1
1
2
,
1
4
,
1
8
,
…
16
1
___
divergent
The
nth
term
of
this
sequence
is
given
by
u

1

n
n
2
sequences
1
___
Now
as
n
increases,
→
0,
so
u
To
define
alternate,
oscillating
periodic
1,
i.e.
lim
So
the
said
A
to
terms
be
of
this
(u
)

1
n
n → ∞
2
and
sequences
→
n
n

sequence
converge
to
the
value
1,
and
the
series
is
convergent
sequence
is
convergent
if
the
nth
is
finite
ter m
is
such
that
lim
(u
)

c
n
n → ∞
where
c
a
constant.
You need to know

How
to find
a
limit
of
a function
Divergent
of
n
as
Consider

The

How
sequences
n → ∞
limit
The
to
the
express
an
1,
3,
5,
7,
9,
…
nth
term
as
the
polynomial
sum
and
a
of
this
sequence
is
given
by
u
of
2n
1
increases
so
lim
(u
)

∞.
2n
1
This
and,
as
sequence
n
does
not
n
a
proper fraction

n
improper
increases,
fraction
sequence
theorems
n → ∞
converge,
the
terms
A
diverge
sequence
and
that
the
is
sequence
not
is
said
convergent
is
to
be
divergent.
divergent.
Example
2
4n
1
_____________
Determine
whether
the
sequence
whose
nth
term
is
2
5n
converges
or

2n
1
diverges.
1
___
4
2
2
n
___________
4n
1
_____________

By
dividing
both
numerator
and
2
5n

2n
1
2
__
1
___
2
5

denominator
2
n
by
n
n
1
___
4
2
2
n
___________
1
4n
_____________
∴
lim

lim
4
__

2
n → ∞
5n

2n
n → ∞
1
5
2
__
1
___
n
n
5

2
Therefore
the
sequence
Alternating
When
we
the
have
For
terms
an
An
sequences
in
a
sequence
alter nating
example,
alternating
converges.
1,
1,
alternate
between
positive
and
negative,
sequence.
1,
1,
1,
…
and
0.5,
0.05,
0.005,
0.0005,
…
are
sequences.
alternating
sequence
may
be
convergent
or
divergent.
n  1
The
nth
term
of
1,
1,
1,
1,
1,
…
is
given
by
u

(
1)
and
n
lim
(u
)
does
not
exist,
so
this
sequence
is
divergent.
n
n → ∞
The
nth
term
1
u

n
Note
1
(10
0.5,
n
0.05,
0.005,
0.0005,
…
is
given
by
n  1
)(
1)
and
lim
(u
)

0
so
this
sequence
is
convergent.
n
2
n → ∞
that
alternate
68
of
a
negative
between
number
positive
to
and
a
power
negative
involving
values.
a
multiple
of
n
will
Section
Periodic
When
called
A
the
Periodic
terms
sequences
terms
Examples
(a)
1,
(c)
1,
An
or
in
of
1,
0,
series
and
approximations
a
are
also
not
form
1,
be
2,
a
3,
repeating
1,
2,
3,
alternating,
1,
for
pattern,
2,
3,
the
…
is
example,
1,
sequence
is
periodic.
1,
1,
1,
1,
…
convergent.
oscillating
1,
0,
3,
…
0,
1,
(b)
…
2,
5,
(d)
may
sequence
sequences
sequence
sequences
sequence
oscillating
Oscillating
may
oscillating
1,
2,
sequence
example,
sequences
an
oscillating
an
of
For
sequence
Oscillating
The
Sequences,
sequences
periodic.
periodic
2
are
3,
5,
be
may
not
move
higher
and
lower
values.
are
1,
2,
6,
an
be
between
3,
6,
…
7,
7,
alternating
a
periodic
…
sequence
sequence
as
as
in
in
and
(a)
(a)
and
(d),
(b)
convergent.
u
u
n
u
n
n
n
n
n
An oscillating sequence
Note
that
the
nth
term
of
an
arithmetic
An alternating sequence
A periodic sequence
(also convergent)
(also oscillating)
progression
is
u

a

(n
1)d
so
lim
(u
n
)

∞
n
n → ∞
Therefore
all
arithmetic
progressions
are
divergent.
n
The
nth
term
of
a
geometric
progression
is
u

1
ar
and
lim
(u
n
)
depends
on
the
value
of
r.
n
n → ∞
n
If
1

r

1,
1
r
→
0
as
n
→
∞
so
lim
(u
)

0
and
the
sequence
is
convergent.
n
n → ∞
Example
n  1
1
Determine
whether
the
sequence
whose
nth
term
is
given
by
u

(
5
n
oscillating
lim
(u
)
or

none
5,
so
of
the
)
is
alternating,
periodic,
3
these.
sequence
converges
and
so
is
neither
periodic
nor
oscillating.
n
n → ∞
n  1
1
(
)
n  1
1
alternates
in
sign,
but
⎥(
)
3
⎥
Therefore
the
sequence
is
not
n  1
1

1
therefore
5
(
3
)
is
always
positive.
3
alternating.
Exercise 2.2
1
Determine
whose
nth
which
term
of
is
the
following
given,
sequences,
converges.
2
n  1
_______
(a)
2n
 1
________
2
Determine
sequences
none
of
whether
is
each
alternating,
of
the
following
periodic,
oscillating
or
these.
(b)
2
n
2

1
n

1
(a)
u

1
1,
u

2
1
and
u

u
n  2

n  1
2u
n
3
n
 1
_______
(c)
n
(d)
n
___
1)
(b)
2
n
(

1
u

n
cos n
(c)
u

sin
n
2
69
2.3
Number
Learning outcomes
series
Series
A

To
define

To
introduce

To
use
terms
a
the
of
number
the
sum
a
∑
of
series
series
is
the
sum
of
the
terms
of
a
sequence.
series
notation
the first
to find
n
the
For
example,
When
the
1

terms
2

are
3

real
4

…
numbers
is
a
the
series.
series
is
called
a
number
series
sum
We
use
u
to
denote
a
general
term
of
a
series.
r
to
To

infinity
define
of
the
series
convergence
and
Example
divergence
of
series
1
_____
Find
the
rth
term
of
the
3
_____
2
_____
series


(2)(3)
(3)(4)

…
(4)(5)
You need to know
The
The

meaning
of
an
numerator
of
denominator
is
Therefore

each
the
term
product
is
of
equal
r

1
to
the
and
r
term

number
,
r,
and
the
2
arithmetic
r
____________
progression
and
a
geometric
u
r
(r

1)(r

2)
progression
Check
The

general
arithmetic
term
of
sequence
geometric
to
see
that
the
answer
does
give
the
first
3
terms.
an
and
of
a
sequence
The
sum of the first
The
nth
term
The
sum
of
of
the
the
n terms of
series
first
n
1

terms
2
is

1
a
3

series

2
4


3
…

is
…
n.

(n
1)

n
Exam tip
We
When
you
are finding
a
general
can
T
aking
of
a
sequence
or
a
series,
relationship
between
this
more
briefly
using
∑
to
mean
‘the
sum
of ’.
the
rth
term
as
a
general
term
(i.e.
any
term
between
the
first
and
look for
nth
a
write
term
the
term),
term
r = n
number,
r,
and
the
numbers
in
the
then
term. Common
relationships
r
∑
means
the
sum
of
all
the
values
of
r
from
r

1
to
r

n
are
r = 1
multiples
r, of
r 
k,
multiples
of
r = n
2
r

k,
where
k
is
a
constant.
i.e.
r
∑

1

2

3

…

(n
1)

n
r = 1
r = n
1
_____
Similarly
∑
r

1
r = 1
1
__
means
1
__

The
Any
a

The
1
__

2
3
1
_____

…
n
sum of the first
AP
has
(r

4
the
form
a,

1
n terms of
a

d,
a

an
2d,
arithmetic
a

3d,
2d)

…
progression
where
the
rth
term
1)d
sum
of
(a
(r
the
first
n
terms
is
r = n
∑

1)d)

a

(r

(a

d)

(a

r = 1
r = n
Using
S

n
∑
r = 1
70
(a
1)d)
we
have
…

(a

(n
1)d)
is
Section
S

a

(a

d)

(a
2d)

…

(a

2
(n
Sequences,
1)d)
series
and
approximations
[1]
n
and
S

(a

(writing
(n
1)d)

the
(a
right-hand

(n
2)d)
side

(a
in

reverse
(n
order)
3)d)

…

a
[2]
n
Adding
[1]
and
[2]
gives
2S

n(2a

(n
1)d)
n
Therefore
n
__
S

(2a

(n

1)d
n
2
Y
ou
may
quote
this
for mula
unless
you
are
asked
to
derive
it.
n
__
An
alternative
version
of
the
formula
above
is
S

(a

l)
where
l
is
the
n
2
n
__
last
term.
This
version
is
derived
from
[2]
where
S

(a

a(n
l)d)
n
2
For
a
example,

1
and
d
the

terms
of
the
series
1

3

5

7

…
are
an
AP
,
n
__
The
sum
of
where
2
the
first
n
terms
is
given
by
S

2
(2

2(n
l))

n
n
2
The
sum of the first
n
terms of
2
Any
GP
has
The
sum
of
the
form
the
first
a,
n
ar,
ar
terms

a

ar

ar
ar

ar
geometric
3
,
is
ar
n
,
…
given
2
S
a
,
1
ar
,
ar

ar
…
by
3

progression
n

…

ar

…

ar
1
[1]
n
2
Now
rS

3
n
1
n

ar
[2]
n
n
[1]
[2]
gives
(1
S
r)

a
ar
n
Therefore
n
a(1

r
)
_________
S

n
1
Y
ou
may
also
quote
this
for mula

r
unless
you
are
asked
to
derive
it.
Example
r  m  1
r
1
__
Find
5
∑
(
)
3
r = 1
r  m  1
r
1
__
∑
5
(
)
is
recognised
as
the
sum
of
the
first
m

1
terms
of
Exam tip
a
3
r = 1
5
__
GP
,
with
first
term
1
__
and
common
If
you
do
not
recognise
the form
of
3
3
series,
write
out
the first few
terms:
r  m  1
5
r  m  1
∑
5
(
(
3
1
__
Therefore
)
)
in
this
example
using
3
∑
5
(
)
3
______________
)

r  1
1
3
r
1
m  1
1
(1
r
1
with
3
r = 1
5
__
2
m  1
1
(1
=
a
ratio
(
)
5
)
r

1,
5
3
3
…

…
gives
___

2
3
3,
5
___

2,
3
3
71
Section
2
Sequences,
series
and
approximations
Example
r  n
2
Given
u
∑

n(2
n
)
find
u
r
n
r  1
r  n
∑
r  n
u

u
r

u
1

…

u
2

n
u
1
and
1
u
∑
n
r  1

u
r

u
1

…

u
2
n
1
r  1
r  n
Therefore
u

r  n
u
∑
n
1
u
∑
r
r  1
r
r  1
2

n(2
n
2
)
(n
1)( 2
(n
1)
)
2

1

3n
3n
Exercise 2.3a
1
Find
the
rth
term

(a)

1
_____
Find
the
1
_______
sum
of
(4)(10)
the
first
n


…
(5)(13)
terms
of
the
series
3
__
1
__

0
1
2
3

(3)(7)
1
__
1
1
_______

(2)(4)
2
…
17

(b)
series

10
1
_____
the
4
___

5
2
of
3
___
2
__
1
__
…
2
2
Evaluate
r  5
r
1
__
(a)
∑
3
(
)
2
r  1
r  10
r
1
__
(b)
3
∑
(
)
2
r  1
r  10
r
1
__
Hence
find
3
∑
(
)
2
r  6
r  n
3
n
______
4
Given
u
∑

,
find
u
r
in
terms
of
n
n
3

n
r  1
r  n
r  20
n
______
5
Given
u
∑

,
find
n

1
r
r  10
r  1
The
u
∑
r
sum to
infinity of
a
series
r  n
1
__
The
sum
of
the
first
n
terms
of
a
series
is
given
by
∑
u

1

r
n
r  1
1
__
As
n
→ ∞,
1

→
1
so
the
sum
of
the
terms
of
this
series
converges
to
n
This
is
called
the
sum
to
infinity
of
the
series.
r  n
A
series
is
convergent
when
the
sum
to
infinity
(i.e.
lim
n → ∞
(∑
r  1
is
a
finite
constant.
r  n
r  n
2
If
∑
u

n
2

1,
then
as
n
→ ∞,
n

1
→ ∞
72
not
lim
r
n → ∞
r  1
is
so
(∑
r  1
a
finite
constant
and
the
series
diverges.
u
r
)
u
r
)
)
1.
Section
A
Clearly
series
any
that
series,
does
u

not
u
1
zero
as
n
increases,
continue
to
Therefore
converge

u
2
will

…,
is
called
where
a
the
divergent
terms
do
Sequences,
series
and
approximations
series.
not
approach
3
diverge
as
the
sum
of
the
first
n
terms
will
increase.
a
is
converge
2
necessary
that
Arithmetic
the
(but
nth
not
term
progressions
sufficient)
condition
approaches
and
zero
as
geometric
n
for
a
series
approaches
to
infinity.
progressions
n
__
The
sum
that
this
of
the
first
n
terms
of
an
AP
is
( 2a

(n
1)
d )
and
it
is
clear
2
sum
Therefore
diverges
the
sum
of
as
an
n
→ ∞
AP
always
diverges.
n
a(1
r
)
________
The
sum
of
the
first
n
terms
of
a
GP
is
where
1
and
r
is
the
common
a
is
the
first
term
r
ratio.
n
a(1
r
)
________
Whether
(
lim
)
n → ∞
1
is
a
constant
depends
on
the
value
of
r:
r
n
a(1
|r|

1,
r
r
)
________
n
If
→ ∞
as
n
→ ∞
so
→ ∞
1
and
the
series
diverges.
r
n
a(1
r
)
________
If
r

1,
1
r

0
so
is
1
meaningless.
r
n
a(1
|r|

1,
r
)
a
_____
________
n
If
r
→ 0
as
n
→ ∞
so
→
1
r
and
1
the
series
r
converges.
a
______
Therefore,
provided
that
| r|

1,
the
sum
to
infinity
of
a
GP
is
1

r
Example
1
__
Show
that
the
series
1
___
1
___

sum
to
infinity
of
this
1
__
u
1
___


3
3

5
3
,
u

u
2
(

1
3
is
geometric
and
find
the
3
series.
1
___

1
…
7
3
2
1
___
),
u

u
3
(

2
3
2
)
and
so
on.
Therefore
each
3
1
___
term
is
times
the
previous
term,
so
the
series
is
geometric
with
first
2
3
1
__
term
1
___
and
common
ratio
(1).
2
3
3
1
3
__
3
______
So
the
sum
to
infinity
is

__
1
8
1
2
3
Exercise 2.3b
1
S
is
the
sum
of
the
first
n
terms
if
it
of
a
series.
Determine
whether
the
n
series
is
convergent,
and
is,
give
the
sum
to
infinity
when
2
2n
______
(a)
S
n
______

(b)
S
n
n

(a)
n
Show
that
the
series
1
__

2
(b)
Find
the
(c)
S
the
sum
sum
to
of
the
first

2
n
1
1
__
2
n

n

1
1
___

6
n
1
___

18
terms
of

…
is
geometric.
54
the
series
in
(a)
and
hence
find
infinity.
73
2.4
Method
Learning outcomes
of
differences
Finding the
We
To

use
the
method
of
the
sum
of
found
the first
of
a
number
sum
of
the
first
n
terms of
terms
of
a
a
series
series
whose
terms
are
arithmetic
progression
and
a
series
whose
terms
are
in
geometric
n
progression.
terms
the
n
differences
in
to find
have
sum of the first
There
is
no
general
method
that
will
give
the
sum
of
the
first
series
n
terms
of
any
series,
but
there
are
methods
that
work
for
some
types
of
series.
You need to know
Method of differences
How

to
decompose
function
into
a
rational
This
method
f(r
1)
works
with
a
series
whose
general
term
can
be
expressed
as
partial fractions

f(r),
because
most
of
1
______
Consider
the
the
terms
1
______
series

1

2
cancel
1
______

2

when
3

are
listed.
1
_______

3
they
…

4

r(r

…
1)
1
_______
So
u

and
we
use
partial
fractions
to
express
this
as
two
r
r(r
separate

1)
fractions:
1
A
__
_______

r(r
r

0

⇒
B
_____

⇒
r
1)
A

r
1
1

A(r

1)

Br
r

1
⇒
B

1
1
_____
r
1)
r  n
r

1
r  n
1
__
Hence
u
∑

r
r  1
Exam tip


r(r
We
1
1
and
1
__
_______
∴

now
1
_____
(
∑
)
r
r

1
r  1
list
the
terms
vertically
(this
makes
it
easier
to
see
the
cancel):
r  n
You
need
to
list
enough
terms
at
the
1
__
start
and
at
the
end
so
that
you
∑
can
1
_____
(
1
__
)
r
r


1
1
r  1
clearly
see
the
pattern
of
cancelling.
1
__

3
1
__

4
1
__
1

∴∴
…
…
1
______


1
n
1
1
______

n
1
_______
∴
∑
r  1
74
n
______

r(r

1)
n

1
1
______

n

1
1
n
______

n

1
n

1
terms
that
Section
2
Sequences,
series
and
approximations
Example
1
____________
(a)
Express
in
(r
1)(r
partial
fractions.
2)
r  n
1
____________
(b)
Hence
find
∑
(r
1)(r
2)
r  3
(c)
Deduce
the
1
______
sum
1
______

infinity

1
3


2
4
1

r
1)(r

1
A

1
1
(b)
that
r

A(r
2)

B(r
1)
2
⇒
B

1

r
first

1
_____

2)
the
1
2
1
_____

Note
r
and
____________
1)(r
⇒
1
∴
(r
…

r
2)
⇒
series
B
_____

(r
the
3
A
_____
____________
(a)
of
1
______

2
to
1
term
of
r
this
2
series
is
given
by
r

3,
1
____________
so
is
(r
1)(r
the
(r
2)th
term.
2)
r  n
r  n
1
____________
1
_____

∑
(r
1)(r
2)
∑
r  3
(
1
_____
1
1
__
)

r
r


2

2
1
r  3
1
__


3
1
1


 …


…

n
n
1
______

1

n
1
n
1
______

1
(c)
Sum
to
infinity

lim
(the
sum
of
the
n
2
______

n
terms
up
1
to
n
r

1
n)
n → ∞
1
______

(1
lim
)
n → ∞
n

1
1
Exercise 2.4
1
____________
1
(a)
Express
in
(r

1)(r

partial
fractions.
1)
r  n
1
____________
(b)
Hence
find
∑
(r

1)(r

1)
r  2
(c)
Deduce
the
1
______
sum

1

to
1
______
3
infinity

2

of
the
series
1
______
4

3

…
5
1
_____________
2
Express
in
r(r

1)(r

partial
fractions
and
hence
find
2)
r  n
1
_____________
∑
(
r(r

1)(r

2)
)
r  1
75
2.5
Proving
properties
Learning outcomes
To
use
proof
by
induction
a
properties
of
is
defined
term of
by
a
a
series
sequence
recurrence
relation,
we
may
be
able
to
a
formula
for
the
nth
term
that
works
for
the
first
few
terms,
but
sequences
we
and
sequence
nth
and
to
deduce
prove
sequences
A formula for the
When

of
need
to
prove
that
it
works
for
all
the
terms.
We
can
do
this
using
series
proof
by
induction.
Example
You need to know
A
sequence
of
positive
integers,
{U
},
is
defined
by
U
n

How
to
use
proof
by
3U
induction

2U
n  1

1
and
1
1
n
n
2
__
Prove
by
mathematical
induction
that
U

3
n
(
)
1
3
n
2
__
Let
P
be
the
statement
U
n

3
n
(
)
1
3
1
2
__
Now
P
is
U
1

3
1
(
)
1

1,
which
is
true.
3
k
2
__
Assume
that
P
is
true
when
n

k,
i.e.
that
U
n

3
k
(
)
1
3
Using
the
recurrence
relation
gives
k
2
__
3U

2
k  1
(3(
)
1
)
1
3
k
2
__
⇒
U
3
Therefore
if
P
is
P
is
true
1
__
)
1
true,
2
__
)

3
3
3
P
k
As
k  1
2
__
(3(

k  1
is
also
(
)
1
3
true.
k  1
when
k

1,
then
it
is
true
when
k

2,
3,
4,
…,
n
k
n
2
__
Therefore
U

3
n
(
)
1
is
true
for
all
n


3
A formula for the
sum of the first
It
to
is
not
series.
terms,
this
always
We
may
but
using
we
possible
be
able
need
proof
by
to
to
find
the
deduce
prove
that
a
sum
it
works
for
the
first
that
for
all
n
terms
works
the
for
a
of
the
terms.
series
a
given
first
We
few
can
do
induction.
2
Consider
,
of
formula
terms of
n
example,
the
series
2
1

2
2

3
2

4
2

…

n

1,
5,
14,
…
n
2
Now
when
r

1,
2,
3,
4,
…
∑
r
gives
the
sequence
30,
…
r  1
n
n
__
2
From
this
we
may
be
able
to
deduce
that
∑
r  1
true
for
n

1,
2,
3,
4
Example
n
n
__
2
Prove
by
mathematical
induction
that
∑
r

(n
6
r  1
n
n
__
2
Let
P
n
be
the
statement
∑
r  1
76
r

(n
6

1)(2n

1)

1)(2n

1)
for
all
n


r

(n
6

1)(2n

1)
is
Section
1
__
When
n

1,
P
2
Sequences,
series
and
approximations
2

(2)(3)

1

1
,
i.e.
P
1
is
true.
1
6
k
k
__
2
Assume
that
P
is
true
when
n

k,
i.e.
P
n

r
∑
k

(k

1)(2k

1)
[1]
6
r  1
k  1
k
__
2
then
adding
the
next
term
of
the
series
gives
P

r
∑
k  1

2
(k

1)(2k

1)

(k

1)
6
r  1
We
now
aim
to
simplify
the
right-hand
k
__
side
so

1)(2k

1)

(k

it
becomes
[1]
with

1)
(k

1)
(
6
k
(k
k
__
2
(k
that


1
replacing
k
1)
_______
(2k

1)

(k

1)
)

(k(2k
6
(k


1)

6(k

1))
6
1)
_______
2

(2k

7k

6)
6
(k

1)
_______

(k

2)(2k

3)
6
(k

1)
_______

[(k

1)

1][2(k

1)

1]
6
Therefore
if
P
is
true,
P
k
As
P
is
true
is
also
true.
k  1
when
k

1,
then
it
is
true
when
k

2,
3,
4,
…,
n
k
n
n
__
2
Therefore
r
∑

(n

1)(2n

1)
is
true
for
all
n


6
r  1
There
are
These
are:
some
number
series
whose
sums
are
worth
remembering.
n
n
__
the
sum
of
the
first
n
natural
numbers:
∑
r

(n

1)
2
r  1
(This
is
the
sum
of
the
terms
of
an
AP
so
can
be
verified
using
the
Example
formula
derived
in
T
opic
2.3.)
n
Find
the
sum
of
the
squares
of
the
first
n
natural
numbers:
n
r

1)
(n

1)(2n

1)
n
n
6
2
r  1
∑
proved

n
__
2
is
r(2r
r  1
∑
(This
∑
r(2r

1)

∑
(2r

r  1
r  1
n
the
r)
above.)
sum
of
the
cubes
of
the
first
n
natural
n
2
numbers:

2
n
∑
r

∑
r
2
n
___
3
∑
r

r  1
2
(n

r  1
1)
4
n
__
r  1

2
(n
(

1)(2n

1)
6
(This
can
be
proved
by
induction
and
is
part
of
question
1
in
n
__
Exercise
2.5

below.)
(n
(

1)
)
2
n
__
These
results
can
be
used
to
find
the
sums
of
series
whose
general

(n

1)(4n

5)
6
2
term
is
the
sum
or
difference
of
ar,
ar
3
and/or
ar
Exercise 2.5
1
Prove
by
induction
that
n
n
2
1
n
1
______
_______
(a)
∑
r(r
(b)
n
1)
r  2
2
n
___
3

∑
r

2
(n

1)
4
r  1
(a)
Find
the
(b)
Prove
by
series
is
rth
term
of
induction
the
that
series
the
1(4)
sum
of

2(7)
the

first
3(10)
n

terms
4(13)
of
this
2
n(n

1)
77
)
2.6
Power
series
Learning outcomes
Power
A

To
define
a
power

To
introduce
and
series
Maclaurin’s
series
whose
terms
involve
increasing
is
called
a
power
For
example
power
derive
a
power
and
to
integral
powers
of
a
series
2
theorem
decreasing
the factorial
notation
To
or
series
variable

theorem
use
2

3x

3
4x

n
5x

….
and
n – 1
x
x
n

2
x
…
are
series.
Maclaurin’s
expand functions
as
series
The factorial
There
1

2
are
notation
several

3
is
a

4
occasions

5

…
when

40
products
such
as
such
as
occur
.
You need to know
There

How
to
differentiate
simple
We
shorthand
denote
1

2

notation
3
by
3!
of
all
for
products
(called
3
these.
factorial).
functions
6!

How
to
differentiate
products
of
means
means
the
the
product
product
of
all
the
the
integers
integers
from
from
1
1
to
to
n
6
inclusive
and
n!
inclusive,
functions
i.e.

The
values
multiples
of
of
the
trig
n!

(1)(2)(3)
…
(n

2)(n

1)(n)
ratios for
(including fractional)
Example

20!
_____
Evaluate
17!3!
20!
of
is
the
the
product
integers
of
from
the
1
integers
to
17,
so
from
we
1
can
to
20
cancel
and
this
17!
is
the
product
product.
18  19  20
_____________
20!
_____
∴


17!3!
3

2

3

19

20

1140
1
Exercise 2.6a
Evaluate
9!
____
5!
__
1
2
4!
3
5!
3  4
______
4
5
3!
5!
3!6!
Maclaurin’s theorem
If
we
assume
ascending
term,
that
a
powers
function
of
x
and
of
that
x,
f(x),
this
can
series
be
can
expanded
be
as
a
of
term
by
then
2
f(x)

a

a
0
where
a
,
0
a
,
x

a
1
a
1
Substituting
,
…
x
3

a
2
are
x
4

a
3
x
r

…

4
a
x

…
constants.
2
0
for
Differentiating
x
in
[1]
gives
f(0)

[1]
with
respect
to
x
a
,
i.e.
a

a

2a
1
Substituting
0
for
x

2
x
in
3a
x
3

4a
3
[2]
gives

f(0)
0
gives
2
f(x)
[1]
r
0
x
4

4
f (0)

5a
x

…
[2]
5
a
,
1
78
series
differentiated
i.e.
a

1
f(0)
Section
Differentiating
[2]
with
respect
to
x

2a

(2)(3)a
2
x

Sequences,
series
and
approximations
gives
2
f (x)
2
(3)(4)a
3
3
x

(4)(5)a
4
x

…
[3]
5
f (0)
_____
Substituting
0
for
x
in
[3]
gives
f (0)

2a
,
i.e.
a
2

2
2!
Differentiating
[3]
with
respect
to
x
gives
2
f (x)

(2)(3)a

(2)(3)(4)a
3
x

(3)(4)(5)a
4
x

…
[4]
5
f (0)
_____
Substituting
0
for
x
in
[4]
gives
f (0)

(2)(3)a
,
i.e.
a
3

3
3!
After
differentiating
r
times
we
get
r
f
(x)

(2)(3)(4)…(r
1)(r)a

(2)(3)…(r

1)a
r
x

…
r1
r
f
0
for
x
gives
f
(0)
_____
r
Substituting
(0)

r!a
i.e.
a
r

r
r!
Substituting
these
values
for
a
,
…
in
[1]
gives
1
r
f (0)
f (0)
_____
f(x)

f(0)

f(0)x
x
x
n
f
(0)
_____
3

2!
∞
f
_____
2


…

r
x
3!

…
r!
n
(0)x
_______

∑
n!
n  0
This
The
for
series
all
is
can
values
converge
Some
to
which
found
n.
For
converge
range
the
be
of
Maclaurin’s
if
the
the
theorem
nth
series
and
derivative
expansion
you
of
to
f( x)
need
to
exists
equal
f( x),
lear n
when
the
x
it.

series
0
must
f( x).
series
limited
called
of
series
to
values
f( x)
of
x.
converges
for
In
is
all
values
the
of
x
following
given
but
and
some
examples,
without
converge
the
values
for
of
a
x
for
proof.
Example
x
Use
Maclaurin’s
theorem
to
find
the
power
series
expansion
of
f( x)

e
r
f (0)
f(0)
_____
Using
f(x)

f(0)

f(0)x


2!
f(x)

e
f(x)

e
f (x)

e
f(x)

e

e
(0)
____
3
x

…

3!
x
gives
f
_____
2
x
r
x

…
r!
0
so
f(0)

e
so
f(0)

e
so
f (0)

e
so
f(0)

e

e
x

1

1

1

1

1
0
x
0
x
0
…
r
f
x
(x)
r
so
f
0
(0)
2
e
3
x
__
x
Therefore

1

x

This
series
converges
for
4
x
__

2!
all
r
x
__

3!
x
__

4!
values
of
…


…
r!
x
79
Section
2
Sequences,
series
and
approximations
Example
Use
Maclaurin’s
theorem
to
find
the
power
series
expansion
of
f( x)

cos x
r
f (0)
f(0)
_____
Using
f(x)

f(0)

f(0)x
x
(0)
____
3

x
2!
gives
f
_____
2


…

r
x
3!

…
r!
f(x)

cos x
so
f(0)

cos 0

1
f(x)

sin x
so
f(0)

sin 0

0
f (x)

cos x
so
f (0)

cos 0

1
f(x)

sin x
so
f(0)

sin 0

0
f (x)

cos x
so
f (0)

cos 0

1

…
3
2
cos x

1

the
can
see
series
that
values
involves

cycle
only

3!
from
even
x
__
_____
(0)x
2!
We
4
(0)x
x
__
Therefore
1
to
powers
4!
0
of
to
x.
1
to
0
to
Therefore
1
again
the
and
general
so
term
2r
x
____
has
the
form

;
when
r
is
odd
the
term
is
negative,
and
when
r
is
(2r)!
r
even
the
term
is
positive.
We
2
cos x

series
using
1)
,
of
f( x)
x
___
r
…

(
1)

4!
for
(
2r

converges
this
x
__
1
2!
This
show
4
x
__
i.e.
can
…
2r!
all
values
of
x
Example
Use
Maclaurin’s
theorem
to
find
the
series
expansion

ln (1
x)
r
f (0)
f(0)
_____
Using
f(x)

f(0)

f(0)x


f(x)

f(x)

ln (1

x)
(0)
____
3
x
2!
gives
f
_____
2
x

…

r
x
3!

…
r!
so
f(0)

ln 1
so
f (0)

1
so
f (0)

1
so
f(0)

2
so
f (0)

2

0
1
_____
1
x
1
_______
f (x)


2
(1
x)
2
_______
f(x)


3
(1
x)
2  3
_______
f (x)



3

3!
4
(1
x)
2
ln (1

x)

0

x
2x
____
3!x
____
3!
4!


2!
term
has
the
4
3
x
__
Therefore
…
and
the
general
form
r
(r
r
1)!x
x
__
_________



r!
This
term
is
r
positive
when
r
is
negative
and
vice-versa,
which
we
r  1
show
using
(
1)
2
3
x
__
⇒
ln (1

x)

x
2
This
Note
f(x)
80
series
that

ln x
it
converges
is
not
because
for

r
x
__
r  1

3
possible
f(0)
4
x
__

1
to
ln 0
…

(
4

use
and
x
x
__
1)

…
r

1
Maclaurin’s
ln 0
is
theorem
undefined.
to
expand
can
Section
Standard
These
are
the
series
you
are
1

x
2
1)

of
x
…
for
all
values
of
x
for
all
values
of
x
x
________
r
…

(
1)

5!
(2r
3

r
x
__
x
__
r  1

3
…

(
1)

4
2
…
for
1

x

1
for
1

x

1
r
3
4
x
__
r
x
__
…
3
x
__
r  1
x
2
…
1)!
4
x
__


values
2r  1
x
__
x) 
all
2r!

2
ln (1
(
5
2
for
2r

x
__
x
__
…
x
___
r

x

r!
4!
x
x) 

4!
…
3!

…
4
3
ln (1
approximations
r

3!

x
__

and
x
__

x
__
1
2!
sin x
series
know:
x
__

x
__

to
4
x
__

2!
cos x
expected
3
x
__
x

Sequences,
expansions
2
e
2
(
1)

4
…
r
Example
x
Expand
e
3
sin 2x
as
a
power
series
as
far
as
the
term
in
x
x
Using
the
standard
expansions
for
3
e
and
sin x
as
far
as
the
term
in
x
gives
3
2
3
x
__
x
e

1

x

(2x)
x
__
_____


2!
…
and
2
e
(1


x
the


…
) (2x

3!
and
…
any
terms
series
found
so
2x

far
powers
of
x
greater
than
3
gives
3
2
2x
x
__
2
2x

…

2x

2x

3
The
with
)
involving
3

x
3
ignoring
x
__
sin 2x
Replacing
4x
____

brackets
x
e
…
3
x
__

2!
Multiplying
(2x)
3
x
__
sin 2x

3!
x
∴
sin 2 x
3!
…
3
have
been
infinite,
but
some
series
terminate.
4
For
example,
f(x)

using
Maclaurin’s
theorem
to
expand
(1

x)
gives
r
f (0)
f(0)
_____
f(0)

f(0)x
f
_____
2


x
2!
…
f(x)

(1
so
f(0)

1,
All
further

x)
(1

differentials
f(x)

4(1
f (0)

4,
are
0,
so
12
___
x)

1

4x
the
x
4
(1

x)

1

4x

x)
series

…
6x
4x
f (x)

12(1
f (0)

12,

x)
,
f(x)

24(1
f (0)

24,

x),
f (x)

24
f (0)

24
terminates.

4
x
4!
3

2
,
24
___
3
x
3!
2


24
___
2

2!
i.e.
r
x
r!
3
,
4
∴

3!
4

(0)
____
3

x
4

x
n
Note
that
there
are
easier
ways
to
expand
functions
of
the
form
(1

x)
which
we
will
look
at
later
in
this
section.
and
u
Exercise 2.6b
Use
Maclaurin’s
theorem
to
expand
as
the
each
of
the
(b)
W
rite
down
u
and
u
n
where
u
n1
n
are
n1
4
following
functions
the
of
range
values
as
of
far
x
for
term
which
in
they
x
are
and
give
the
valid.
nth
and
(n1)th
terms
respectively
of
this
series.
Hence
find
a
recurrence
relation
between
u
n
x
1
f(x)

2
e
tan 2x
and
u
n1
x
3
5
ln (1
(a)
Use
4
3x)
Maclaurin’s
e
theorem
1
(1
x)
to
2

1

x

x
(c)
cos x
show
3

x
x
the
recurrence
series
is
geometric
series
converges
relation
and
to
hence
that
4

Use
show
verify
that
that
the
the
1
to
(1
x)
,
stating
the
range
5

x

…
of
values
for
which
this
is
true.
81
2.7
Applications
Learning outcomes
of
Maclaurin’s
theorem
Euler’s formula
i

To
prove

To
use
We
introduced
We
can
the
formula
e
cos 

i sin 

in
T
opic
1.7.
Euler’s formula
Maclaurin’s
theorem
expand further functions
to
now
use
the
and find
Maclaurin
2
4

__
cos 

expansions

cos
sin 
and
to
prove
it:
2r

__
1
of

___
r

…

(
1)

…
approximations
2!
4!
3
5

__
sin 
and
2r  1

__


2r!
3!

________
r


…
(
1)

5!
(2r

…
1)!
You need to know
2
3

__
cos 
∴

How
to
evaluate
powers
of
i sin 


1
4
i
___
i

5

__
i
___

2!
i

3!

4!
…
5!
___
(i.e.
√
1 )
x
Now

The
standard
Maclaurin

The
values
the
replacing
x
i
with
in
the
expansion
2
3
(i)
trig

1
i


____

3!

4!
3

__

1
meaning
of
a
4
i
___
i

…
5!
5

__
i
___

2!
The
(i)
____

(including fractional)


5
(i)
____

2
of
gives
4
(i)
____
e
ratios for
2!
multiples
e
series
i
of
of
3!


4!
…
5!
quadratic
cos 

i sin 

function
Expanding
a
composite function
sin x
T
o
expand
and
By
a
replace
function
x
with
terminating
approximation
sin x.
the
for
such
We
series
the
as
we
f( x)
can

then
can
x
e
find
we
start
replace
a
sin
with
x
by
polynomial
the
its
that
expansion
series
is
of
e
expansion.
an
function.
Example
sin x
Find
a
quadratic
function
that
is
an
2

1

sin x
for
e
3
(sin x)
(sin x)
_______
sin x
e
approximation

_______


2!
…
3!
3
x
__
Now
sin x

x

…
so
we
can
replace
sin
x
with
its
expansion.
3!
2
3
e

1
(x


…
)
of
find
x
a
quadratic
greater
than
we
2
…

can
ignore
x

all
And
(x
the
x
higher
powers
term).
3
3
x
__
3

…
)

x

higher
powers
of
3!
so
82
we
can
ignore
3!
______________

terms
3!
ignore
this
term
and
further
)
…
3!
4
we
…

containing
4
)
3!
(and

2x
____
2

(x
2.
x
__
(x
)
2!
function
3
Now
…

3!
T
o
x
__

3!
______________
x
__
sin x
∴
3
3
x
__
(x
3
terms.
x
of
x
powers
Section
2
Sequences,
series
and
approximations
2
x
__
sin x
∴
e

1

x


…
2!
2
x
__
sin x
⇒
e

1

x

2
Using
a
series
expansion of
approximate value of
By
expanding
a
function
as
a function to find
an
a function
a
Maclaurin
series
of
ascending
powers
of
x,
Did you know?
we
are
first
expressing
few
terms
of
the
function
the
as
an
polynomial
to
infinite
find
an
polynomial.
We
approximate
can
value
use
for
the
the
The
function.
By
adding
more
terms
we
can
improve
on
the
approximation
summation
goes
give
a
value
to
as
great
a
degree
of
accuracy
as
we
choose,
provided
back
to
series
converges
For
example,
for
the
value
of
x
we
infinite
series
the Ancient Greeks.
that
Archimedes
the
of
to
used
the
summation
use.
of
an
infinite
series
to find
the
area

__
we
can
find
an
approximate
value
for
cos
by
using
the
under
an
arc
of
a
parabola.
He
also
4
used
Maclaurin
expansion
of
cos
x
which
converges
for
all
values
of
a
series
value for
first
three
terms
of
the
series,
i.e.
cos
x

1

(
cos

)
(
1
…
2!

)
4
_____

4
…
4!
4

__
4
_____

__
gives
.
x
__
2!
2

__
accurate
4
x
__
the
a fairly
x
2
Using
to find
4!
0.707429…

__
The
calculator
gives
cos

0.707106…
4
so
the
approximation
Adding
more
terms
is
correct
will
to
improve
3
decimal
the
places.
approximation.
6

__
(
)
4
_____
Adding
the
next
term
in
the
series,
i.e.
gives
6!
2

__
(
cos

this
agrees
(
)
4
_____
4!
6!
2!

6

__
)
4
_____

4
and
(
1
4

__
)
4
_____

__
0.707102…
with
the
calculator
value
to
5
decimal
places.
Exercise 2.7
2
1
Expand
ln (1

2x
)
as
a
series
of
ascending
powers
of
x
as
far
as
and
4
including
the
expansion
2
Use
the
values
is
first
term
in
x
.
Give
the
range
of
values
of
x
for
which
the
valid.
four
terms
of
a
Maclaurin
series
to
find
approximate
for:
2
(a)
e
(b)
ln 1.1
(i.e.
1

0.1)

__
(c)
sin
3
x
3
W
rite
down
(a)
By
(b)
Find
the
first
substituting
the
hence
value
1
of
estimate
five
for
the
the
terms
x,
in
find
sixth
the
an
approximate
term
accuracy
of
Maclaurin
of
the
your
series
value
expansion
expansion
for
of
e
e.
when
x

1
and
approximation.
83
2.8
Taylor’s
Learning outcomes
theorem
To
derive
and
applications
Taylor’s theorem
We

and
have
seen
that
we
cannot
expand
ln
x
using
Maclaurin’s
theorem.
use Taylor’s
This
problem
and
others
where
the
Maclaurin
series
does
not
give
a
valid
theorem
expansion
gives
an
can
sometimes
expansion
i.e.
f(x)

a

a
0
The
meaning
of
a
The
to
ascending
(x
–
a)

a
1
values
of
,
a
using
powers
(x
–
a)
a
of
one
differentiate
an
(x
a
(x
–
we
used
to
,
a
1
find
4
a)

,
…
can
be
found
Maclaurin
using
series

f(a)

f(a)(x

a)

a
method
f(a)(x

Did you know?
the
f
r
(a)(x

a)


r!
n
(a)
n
(x
∑

a)
n!
n  0
This
are
named
and Taylor
Brook Taylor,
both
early
several
to
__________
…
3!
f
However,
…
r
a)

_____
the

similar
___________

=
in
a)
3
a)

∞
the UK
–
giving
2!
Maclaurin
(x
4
2
after
a
2
the
f (a)(x
Colin
which
a),
3
____________
f(x)
series
series
implicit
function
Maclaurin
T
aylor
3

2
a
0
How
in
by
differential
equation

f( x)
overcome
2
You need to know

of
be
18th
is
called
Taylor ’s
theorem
and
you
need
to
lear n
it.
after
series
working
in
Y
ou
can
any
expansion
assume
that
you
this
are
series
asked
converges
to
for
values
of
x
close
to
a
for
find.
century.
Maclaurin
Example
series
and Taylor’s
theorem
were
Find
discovered
some
decades
by James Gregory,
a
the
first
four
terms
in
the
T
aylor
expansion
of
ln ( x)
about
Scottish
2
f (a)(x
3
a)
f(a)(x
___________
mathematician.
Using
f(x)

f(a)

f(a)(x
a)


3!
gives
f(x)

f(x)

ln (
x)
so
f( a)

so
f (a)

1
__
ln a
1
__
x
a
1
__
f (x)

1
__

so
f (a)


2
2
x
a
2
__
f(x)
2
__

so
f (a)


3
3
x
a
2
(x
x
a
______
∴
ln x

ln a
3
a)
(x
________

a)
________


2
a
2a
…
3
3a
Exercise 2.8a
Find
the
first
three
terms
in
the
T
aylor
expansion
about
a
of
x
1
tan x
2
sin x
a)
____________

2!
84
a
earlier
3
e
cos x
…
…
Section
Using Taylor
A
T
aylor
is
not
series
valid
or
series to find
can
the
sometimes
series
be
Sequences,
series
and
approximations
approximations
used
converges
2
when
either
the
Maclaurin
series
slowly.
Example
Find
the
first
three
terms
of
the
expansion
of
sin
x
as
a
series
of

__
ascending
powers
of
x
(
)
.
4
Hence
find
an
approximate
value
of
sin 46°
given
that
1°

0.017 rad.

__
Using
T
aylor ’s
theorem
with
a

gives
4
2

__
x
(

__
sin x


__
sin


__
cos
x
(
)(
4
)
4
_________


√
2
…
2!
2

__
x
(
4
______
√

)
4

__

sin
(
4
x
1
___

)
4
)
4
_________

__
2
2
...
√
2

__
Now
sin 46°

sin (45°

1°)

sin

(
0.017
)
4

__
so
when
x

46°,
i.e.

__

(
0.017
rad,
)
x
(

)
4
0.017
4
2
sin 46°
______


√
2
Therefore
(sin 46°
Note

that
2
2
0.707106…
sin 46°

0.7193
we
________

√

(0.017)
0.017
1
___
∴
2
0.0120208…
0.000102…

0.719025…
0.7190
correct
could

…
√
use
to
the
4
decimal
Maclaurin
places)
series
to
find
an
approximate

__
value
for
sin

(
0.017
)
but
the
terms
decrease
in
value
more
slowly
4

__
the
(
third
term
of
the
expansion
of
sin

(
0.017
)
is
0.00277…
so
)
we
4
would
need
more
Using Taylor
terms
to
give
a
series to find
reasonable
polynomial
solution of differential
equations
There
equations
y

that
T
o
are
f(x),
is
do
some
but
an
differential
we
can
sometimes
approximation
this
we
need
to
for
know
use
f( x)
a
approximation.
for
pair
that
cannot
T
aylor ’s
values
of
approximations for the
be
solved
theorem
of
x
close
corresponding
to
to
to
find
a
a
give
polynomial
given
values
of
x
value.
and
y
2
dy
d
___
for
an
equation
involving
y
____
and,
for
an
equation
involving
,
2
dx
dx
dy
___
corresponding
values
of
x,
y
and
.
dx
These
are
called
Then,
stopping
approximate
value
such
of
as
x.
the
the
the
How
the
initial
series
solution
good
number
the
of
conditions .
after
for
a
given
values
of
number
x
approximation
terms
included,
close
is
of
to
terms,
a
depends
how
close
we
where
x
on
is
a
can
is
several
to
a,
often
the
initial
things
and
so
on.
85
Section
2
Sequences,
series
and
approximations
2
dy
d
___
Differential
equations
are
usually
given
in
terms
of
x,
y,
y
____
,
,
…,
so
2
dx
we
use
T
aylor ’s
theorem
in
the
form
2
2
dy
d
___
y

y
(

a
(x
y
____
)
(x
a)
(

a
dx
2!
dy
y
We
initial
,
…
means
the
value
of
a)
_______
)

a
dx
y,
,
a
…
3!
…
when
x

a,
where
a
is
dx
value
illustrate
3
dy
)
dx
the
(x
y
___
(
,
a
d
____
(

a
dx
3
3
a)
_______
)
2
___
where
dx
of
this
x
with
a
simple
first
example.
Example
2
Find
a
T
aylor
series
polynomial
up
to
and
including
the
term
in
x
to
dy
___
approximate
the
solution
of

xy
for
values
of
x
close
to
0,
given
dx
that
y

1
when
x

0
2
We
stop
i.e.
y
the
series
after
the
term
containing
y
(

a
d
(x
y
____
)
(x
a)
a)
________

2
a
dx
,
2
2
dy
___

x
dx
2!
2
d
term
involving
y
____
2
The
x
involves
so
we
differentiate
the
given
2
dx
2
d
y
____
differential
equation
to
give
an
equation
containing
2
dx
2
dy
d
___
y
dy
____

xy
⇒
___


y
2
dx
a
is
the
and
y
dx
initial

1
value
when
x
of

dx
x,
0
so
so
a
y


0,
1
a
2
dy
d
___
(
∴
y
____
)

(0)(1)

0
(
⇒
a
dx
2
)

(0)(0)

y

(

y
a
d
(x
y
____
)
(x
a)

a
dx

1
2
2
dy
___
∴
1
a
dx
(
2
a)
_______
)
a
dx
2!
2
x
__
gives
y

1

2
1
2
x
The
T
ry
differential
to
judge
the
2
1
x
__
1
equation
in
accuracy
the
of
example
the
above
approximate
has
an
exact
solution
by
solution,
finding
y

e
values
of
2
x

and
e
2
when
x

0.01,
0.1,
0.2
2
Example
The
the
displacement,
differential
s
metres,
of
a
particle
at
time
t
seconds
is
given
by
equation
2
d
s
___
ds
___

2

sin s

0
2
dt
dt
ds
___
When
t

0,
s

0
and

0.5
dt
Find
a
T
aylor
series
approximation
3
and
86
including
the
term
in
t
for
s
in
ascending
powers
of
t
up
2
to
Section
2
Sequences,
series
and
approximations
3
d
s
___
3
For
the
term
in
t
we
need
the
value
of
so
we
differentiate
the
3
dt
given
differential
equation.
2
d
s
___
ds
___

2

sin s

0
[1]
2
dt
dt
3
2
s
d
___
d
s
___
⇒


use
(cos s)

0
[2]
2
dt
We
ds
___
2
3
dt
the
dt
T
aylor
series
in
the
form
2
2
ds
___
s

s
(

a
the
initial
(t
a)

(
a
dt
Using
(t
d
s
___
)
we
have
a
(
2!

(t
d
s
___

a
dt
values
3
3
a)
_______
)
2
3
dt
a)
_______
)

a
…
3!
0
ds
___
so
s

0
(
and
a
)
Substituting
these

0.5
a
dt
values
in
[1]
gives
2
d
s
___
(
2
)

2(0.5)

1

sin 0

0
a
dt
2
d
s
___
(
⇒
2
)
a
dt
Substituting
these
values
in
[2]
gives
3
d
s
___
(
3
)

2(
1)

(cos 0)(0.5)

0
a
dt
3
d
s
___
(
⇒
3
3
__
)

a
dt
2
2
3
3
__
t
__
Therefore
s

0

(0.5)t

(–1)

(
2
2
⇒
s
t
__
t
__
2
2
2
6
3
t
__


Remember
small
t
__
)
(i.e.
4
that
this
close
to
approximation
is
only
reasonable
when
t
is
very
zero).
Exercise 2.8b
1
Find
a
T
aylor
series
approximation
to
y
in
ascending
powers
of
x
up
to
3
and
including
the
term
in
x
when
x
is
close
to
zero,
given
that
dy
___
x

2xy

e
dx
and
that
Hence
2
Use
a
y

find
1
an
T
aylor
when

0
approximate
series
approximately
x
value
expansion
equal
to
y
to
of
y
find
when
a
x
cubic

0.1
function
that
is
when
2
2
d
y
dy
____
___

2
dx
(
)
dx
dy
___
given
y

1
and

2
when
x

1
dx
Hence
find
an
approximate
value
of
y
when
x

0.9
87
2.9
Derivation
for
n

Learning outcomes
of
To
introduce
Pascal’s
binomial
theorem

Binomials
A

the
binomial
is
an
expression
with
two
terms,
for
example,
2

x,
triangle
2
3x

To
derive
the
binomial
n

2y,
s
5t
theorem
In
for

this
topic
we
investigate
how
to
expand
powers
of
binomials
as
a
series.

n

To
introduce
and
use
the
C
r
Pascal’s triangle
notation
5
We
can
but
a
expand,
quicker
for
example,
method
is
to
(a
use

b)
,
by
Pascal’s
multiplying
out
the
brackets,
triangle.
You need to know
First
look
at
these
expansions:
1

Maclaurin’s
theorem
(a

b)
(a

b)
=
a
=
a
2

The
use
of factorial
+
b
2
2
+
2ab
+
b
notation
3
(a

b)
(a

b)
3
=
a
=
a
4
Notice
2
+
3a
+
4a
4
that
2
b
+
3ab
b
+
6a
3
the
2
powers
of
3
+
b
2
3
b
a
+
4
4ab
and
b
+
form
b
a
pattern.
4
From
then
the
the
power
and
b
Now
power
of
is
expansion
b
4.
look
a
just
array
(a

is
by
a
the
1.
4
b)
decreases
increases
There
at
triangular
of
of
In
similar
you
by
all
1
can
in
the
of
each
in
the
that
the
first
succeeding
terms,
pattern
coefficients
see
the
the
sum
other
terms.
of
term
term
the
is
a
while
powers
and
the
of
a
expansions.
W
riting
these
in
a
gives:
1
1
1
1
2
1
1
This
Each
array
row
is
called
starts
two
numbers
row
are
Y
ou
can
in
the
symmetric
now
write
ends
row
with
the
as
1
6
triangle
1
above
about
down
3
4
Pascal’s
and
1
3
and
it,
as
many
and
it
each
also
other
shown.
middle
1
4
of
rows
the
as
has
a
pattern:
number
Also,
the
is
row.
you
need.
6
For
example,
to
expand
(a

b)
,
go
as
far
as
row
6:
1
1
1
1
1
1
1
88
3
4
5
6
1
2
6
10
15
1
3
1
4
10
20
1
5
15
1
6
the
sum
numbers
1
in
of
the
each
Section
Using
the
what
array
we
know
The
the
pattern
of
the
powers
and
using
row
six
Sequences,
series
and
approximations
of
gives
6
(a
about
2

6
b)

5
a

6a
4
b

15a
2
3
b

binomial theorem for
20a

x
3
2
b

15a
4
5
b

6
6ab

b

n
We
can
will
use
clearly
Pascal’s
be
a
triangle
to
expand
time-consuming
(a

activity
b)
for
for
any
values
of
n
n

,
but
greater
this
than
5.
n
However
,
we
can
use
Maclaurin’s
expansion
of
(1

x)
to
get
a
general
n
form
for
the
expansion
of
(a

b)
for
any
n


n
Using
f(x)

(1

x)
,
n  1
f(x)

n(x

1)
,
f (x)

n(n

1)(x
f(x)

n(n

1)(n
(x)

n(n

1)…(n

r
(x)

n(n

1)…(n

(n
n  2

1)
,
n  3

2)(x

1)
,
…,
r
f
r

1)(x

1)
,
…,
n
f
so
all
0
further
∴
f(0)
derivatives

1,
f(0)

n(n

of
f( x)
n,

are
f (0)
1))(x
zero

n(n

and


1)…(n

r
n(n
(1
n(n

series
f(0)

1)…(1)

n!
terminates.
n(n

1)(n

2),
…,
n
(0)
+


1),
x)

1
+
nx
…,
f
1)

(0)
n(n
________
n
⇒

the
1),
r
f
1)


1)(n


2)
_______________
2
x

3
x
2!
n(n
n!

…
3!
1)…(n

r

1)
____________________
r

x
n

…

x
r!
n!
___
(Note
that

1
)
n!
n
This
but
expansion
before
we
can
do
be
adapted
that,
we
will
to
give
the
introduce
a
expansion
simpler
of
( a

notation
b)
for
,
the
2
coeffi cients
of
x,
x
,
…
.
(These
coeffi cients
are
called
the
binomial
coeffi cients .)
n
The
notation
C
r
n(n
coefficient
of
x

1)(n

2)
_______________
3
The
in
the
expansion
above
is
,
which
3!
n!
_________
we
can
write
using
only
factorials
as
(n
n(n

1)…(n

r


3)!3!
1)
n!
_________
____________________
Similarly,
can
be
written
as
r!
,
(n

which
we
r)!r!
n
denote
by
C
,
i.e.
r
n!
_________
n
C

r
(n
4!
C

r)!r!
4  3  2  1
______________
_________
4
Therefore



6
2
(4

2
2)!2!
8!
C

2
8  7  6
__________
8!
_______
_________
8
and




56
3
(8

3)!3!
5!

3!
6
89
Section
2
Sequences,
series
and
approximations
n
Now
r
C
is
the
coefficient
of
n
x
in
the
expansion
of
(1

n
x)
so
C
r
is
the
n
n
coefficient
of
x
n!
__________
n
,
which
we
know
is
1,
but
C

n!
____

.
n
(n
T
o
make
this
equal
to
1,
we
define
0!
0!
as

n)!n!
0!n!
1
1
Example
n
Show
that
n
C

n
C
r
r
n!
___________________
n
C
n!
n!
_________

n
_________

n


C
r
r
(n
(n
r))!(n
r)!
r!(n
r)!
(n
r)!r!
Example
n
Find
a
relationship
between
n
and
r
given
that
n
C
1

C
r
r
1
n!
_________
n
C

and
r
(n
r)!r!
(n
n
1)!
(n
______________________
1
C
1)!
______________

r

1
(n
1
r

(n
n!
_________
1)!(r
1)!
(n
r)!(r
1)!
1)!
______________
∴

(n
r)!r!
Now
n!

(n
r)!(r
n(n
1)!
1)!
and
r!

r(r
n(n
1)!
1)!
(n
_______________
Cancelling
gives
1)!
______________
i.e.

(n
r)!
r(r
1)!
(n
r)!
(r
1)!
n
__

1

r
r
⇒
n
Exercise 2.9a
n
1
Find
the
value
of
n
when
n
C
1

C
8
7
n
2
Find
the
value
of
n
when
5(
n  1
C
)

4(
C
3
)
3
n  1
3
Find
a
relationship
between
n
and
r
given
that
n
C

r
C
r  1
n
The
expansion of
(a

b)
for
n


n
We
can
now
n
write
the
n
C

expansion
n
C
0
x

of
2
C
1
x
(1

n

…

x)
as
r
C
2
x
n

…

n
C
r
x
n
n
n
Then
(a

b)
b
__
n

a
(1
b
__
)

and
replacing
x
by
a
in
the
expansion
above
a
gives
r
2
n
(a

b)
n

a
n
(
b
__
n
C

0
C
1
(
a
90
b
__
n
)

C
2
(
a

…

n
b
__
n
)
C
r
(
a
b
__
n
)

…

C
n
(
)
a
)
Section
n
Multiplying
through
by
a
n
and
noting
that
n

n
b)

a
n

n  1
C
a
n
b

n  2
C
1
Y
ou
need
a

C
lear n
this,
n(n
n

n
b)

but

n

na

you
in
…
b
may
the

n(n

1)…(n

r
r
n
b

…

b
it
easier
to
for m
n(n
n  2
a


1)(n

2)
_______________
2
b
n  3
a

3
b
3!

1)
_____________________
…
approximations
gives
a
2!

1
n  r
C
find
1)
________
n  1
a
and
r
remember
(a

2
to
series
n
2
b
Sequences,
n
C
0
(a
2
n  r

a
r
n  1
b

…

nab
n

b
r!
Either
of
these
triangle,
power
i.e.
of
a
increases
forms
the
decreases
by
confirms
sum
of
the
by
1
in
each
have
shown
that
of
a
and
b
succeeding
we
in
made
each
term
from
term
while
is
the
Pascal’s
n
and
power

expand
replace
n
(1

C
so
the
coefficients
are
symmetric
about
r
with

3
x)
in
10,
a
10
(1
ascending
with
1
and
10
x)

powers
b
with

10(1)
x
x
Knowing
four
the
terms,
1


properties
of
as
the
term
in
x
we
give
8
(1)

far
10  9  8
___________
2
x

1
7
(1)
3

2

3
x


120x

expansion,
…
we
can
also
write
down
the
8
120x

9
45x

10x
expand
x
(1

6
x)
in
descending
powers
8
can
either
and
b
write
with
1
to
(1


x)
as
(x
8
x)

of
x
as
far
as
the
term
in
x
,
we
8

1)
(1)

,
then
replace
n
with
8,
a
with
x
give
8
(1
last
10

8
T
o
…
1
3
45x
the
as
i.e.
7

10x
x
to

2

of
10  9
_______
9
1(1)
2
…
b
centre.
10
T
o
the
of
n
C
n  r
the
observations
1.
n
We
the
powers
1(x)
8  7
______
7

8(x)
6
(x)
2
(1)
…
2
8

7
x

6
8x

28x

…
8
or
we
can
property,
expand
(1

x)
in
ascending
8
(1
then
Note

2
x)

reverse
that
1
to
the

x)

8x
give

term
8
is
r
C

(x)
in
look
at
further
…


the
use
of
7

the
symmetry
(1)
8x
8

x
x.
expansion
r
C
28x
powers
8
r
We
and
6
28x
descending
general
8
(1
powers
i.e.
of
r
x
r
expansions
using
this
work
in
T
opic
2.10.
as
far
the
Exercise 2.9b
7
1
Expand
2
Find
(1

2x)
3
in
ascending
powers
of
x
as
term
in
x
4
the
coefficient
of
the
term
in
x
.
5
in
the
expansion
of
(3

x)
91
2.
10
Applications
for n

To
apply
the
the
binomial
n
The
binomial
expansions of



x)

1

nx
n
x)
and
1)
(1
n(n
________
n
n
(1
n(n
expansion
(1
for
expansion

Learning outcomes

of
x)
1)(n
2)
_______________
2

x
3

n
x
2



…

x
3!
and
n(n
x)

1
nx
The
expansion
of
(a  b)
x

These
are
the
recognise
most
the
3
n
x
straightforward
left-hand
side
2
The
2)

…

(
1)
n
x
3!
binomial
when
you
see
expansions
it.
For
and
example,
you
you
need
should

recognise

1)(n

for
to
n
n(n
_______________
2

2
n

1)
________
n
(1
You need to know
meaning
of
1

3x

3x
3

3
x
as
the
expansion
of
(1

x)
compound
interest
Compound

The
sum
of
the first
n
terms
of
interest
problems
a
r
____
geometric
Suppose
progression
$A
is
deposited
in
an
account
that
pays
interest
of
of
$A
100
(where
r
account
is
the
each
withdrawals
rate
year
are
%
on
per
the
made,
annum
(pa))
anniversary
at
the
end
of
and
of
the
the
year
1,
interest
deposit.
the
is
credited
Then,
amount
in
if
to
the
no
the
r
____
account
is
$A
1

(
)
100
at
the
end
of
year
r
____
$A
1
(

the
end
1
(
is
r
____
of
$A
1

(
year
2
3,
amount
$A
1

(
)
100
is
r
____
2
r
____

)
of
100

100
the
$A
1

(
2
r
____
)
100
of

amount
r
____
r
____
$A
the

)
100
at
2,
3
r
____

)
100
$A
1

(
100
)
100
n
r
____
By
deduction,
the
amount
at
the
end
of
year
n
is
$A
1

(
)
100
This
is
formula
added
For
to
example,
compound
is
the
used
to
capital
if
$10 000
interest,
calculate
each
is
then
compound
deposited
the

0.02)
in
amount
4
$10 000(1
interest
(where
the
interest
year).
an
in
account
the
paying
account
after
2%
4
pa
years
is
4

$10 000(1.02)
Example
(a)
Rachel
Her
has
initial
a
pension
pension
that
was
each
$3000
year
increases
when
she
by
retired.
3%
of
What
its
value
was
her
the
previous
pension
at
year
.
the
end
of
the
8th
year
retirement?
(b)
How
much
(a)
Pension
in
total
did
Rachel
receive
in
pension
payments
for
the
first
8
years
of
her
retirement?
8
at
the
end
of
the
8th
year

$3000(1.03)

$3800
(to
the
nearest
2
(b)
T
otal
The
pension
paid
expression
in
for
the
first
brackets
is
8
years
the
is
sum
of
$3000(1
the
first

8
1.03
terms

of
1.03
a
GP
,
$)
3

1.03
with
a
8


...
1

and
1.03
r

)
1.03
8
1(1
1.03
8
)
1.03
1
_________
____________

total
paid

$3000


1
92
1.03
$3000


0.03
$26 677
(to
the
nearest
$)
of
her
Section
Expansions
The
examples
using the
that
follow
2
Sequences,
series
and
approximations
binomial theorem
illustrate
some
problems
involving
expansions.
Example
3
Find
the
terms
up
to
and
including
x
the
expansion
of
(1

(1
2x)
6
1
4
in
x
)
2

6
1
4
(1
2
(1
2x)
x
)

( 1

4(2x)


6  5
______
1
3
6(2x)
4(2x)

… )
2
(1
6
(
x
)
2
1
(

2
x
6  5  4
__________
3
1
)
(
2
2!
x
)

…
2
3!
)
3
There
2

(1

8x

15
___
3
24x

32x

…
)(1
3x

is
no
5
__
2
x

1
3x

5
__
2
x
8x
beyond

…

…

…

30x

…
24
expansion
Exam tip
x
72x
systematic
brackets
32x

…

…
second
so
15
___
1
__
2

either
when
you
expand
3

5x
in
)
3

in x
2
Be
2
1
term
3
24x

the
2
2

go
3
x
4
to
3
x
4
15
___
need
x
12
this:
bracket
on. Then
multiply
by
add
1,
the
then
the
by
8x
and
results.
3
x
4
like
2
Example
9
1
__
Find
the
term
independent
of
x
in
the
expansion
2
(
of
)
2x
x
9
1
__
The
general
term
in
the
expansion
of
2
(
)
2x
is
x
9
r
2r
1
__
9
C
r
(
2
)
(
2x
r
9
)

This
x
____
r
C
(
2)
(
r
x
9
r
)
x
term
is
independent
of
x
when
2r

9
r,
i.e.
when
9!
9
Therefore
the
term
independent
of
x
is
(
2)

(

3
8)
_________
3
C
r


672
3
3!6!
Example
8
Use
the
binomial
expansion
of
8
(1
(1
2
2x)

1
8(2x)

0.98

1
2(0.01)

1
0.16
28(4x
8
2x)
to
find
the
3
)
56(8x
value
of
0.98
4
)

70(16x
correct
to
3
decimal
places.
5
)
56(32x
)

…
So
substituting
0.01 for
x
gives
8
0.98

0.0112
0.000448
We

stop
decimal

0.851
correct
to
3
0.0000112
here
place
as
0.0000001792
the first
so
will
significant figure
not
alter
the
4th
of
...
the
decimal
next
term
will
be
in
the
7th
or
8th
place
d.p.
Exercise 2.10
3
1
Find
the
coefficient
of
x
2
in
the
expansion
of
(1

x
6
(Hint:
treat
it
as
(1

X)
x
6
)
7
3
Find
the
coefficient
then
substitute
x
x
x
in
the
expansion
8
2
and
of
for
X.)
2
of
1
__
(x
)
x
6
2
Find
the
real
part
of
(1

2i)
6
(Hint:
even
expand
powers
(1
of

x)
,
replacing
x
by
2i
and
only
consider
i.)
93
2.
11
The
binomial
Learning outcomes
The
expansion for
binomial theorem for
n

n



n
Using

To
derive
the
expansion
Maclaurin’s
theorem
to
expand
(1

x)
gives
of
n
(1

x)
when
n
is
a fraction
or
a
n(n
(1
integer


1)
n(n
________
n
negative
x)

1

nx

1)(n
x
To
apply
the
binomial
3
x
3!
theorem
n(n

1)…(n

r

1)
_____________________
to
2)

2!


_______________
2

problems

…
r

x

…
r!
Now
r
there
is
is
a
positive
no
value
integer
,
of
r
for
but
when
which
(n
n
r
is

a
negative
1)
is
zero.
integer
In
this
or
a
case
fraction,
the
series
You need to know
does
not
terminate.
n

Maclaurin’s

Factorial
theorem
The
series
expansion
only
of
(1

when
n
x)
1
converges

x

to
(1

x)
1
notation
n
Note

The
meaning
of
a
that
the
ter m
in
x
is
the
(n

1)th
ter m,
not
the
nth
term.
convergent
1
1
series
For
example,
to
expand
(1

x)
2
we
substitute
for
n
giving
2

How
to
express
a
rational
1
in
1
partial fractions

x)
2

1

x
(
)
x
1

__
1
x
(
)
3
x

…
3!
2
x
2


2!
1
1
3
1

2
2
2
_________________
2

2

1
)
2
2
_________
1
(1
(

1
function
3

x
8

…
for
1

x

16
n
Note
that
is
valid
is
infinite
There
when
for
all
is
a
values
and
are
n
positive
of
x,
converges
other
but
only
integer
,
when
n
when
the
is
| x|
series
not

a
(1

x)
positive
terminates
integer
,
the
and
series
1.
differences:
n
we
cannot
use
C
for
the
coefficients
r
n
and
we
cannot
use
the
form
of
the
expansion
for
( a

b)
n
T
o
expand
(a

b)
when
n

,
we
take
a
outside
the
bracket
to
give
n
b
__
n
a
(1
)
a
________
1
________
2
1
2
For
example,
to
√(2
expand
x
x
__
2
),
we
express
√(2
x
)
as
2
2
(1
2
2
x
__
1
then
replacing
n
by
and
x
(
by
2
)
we
have
2
1
1
2
1
2
(1
2
)

2
(
2
x
__
1
1

(
)
2
2
(
)
)
2!
4
√
x
2
_____
√
x
2
_____
4
32
2
…
2
x
__
This
expansion
is
valid
when
1


1
2
2
2
x
__
i.e.
when
0

x
__

1
cannot
2
2
2
⇒
94
x

2
⇒
√
2

x

√
2
be
negative
2
2
x
__
(

2
2
(
2
2
_________
√

1

2
1
x
__
2
)
2

…
2
)
)
1
Section
2
Sequences,
series
and
approximations
Example
1
Expand
(1

x)
as
a
series
of
ascending
powers
of
x
up
to
and
4
including
the
term
in
x
.
n
Give
the
Using
term
the
in
x
binomial
theorem,
1
(1

x)
(
1)(
2)
(
_________

1
x
1)(
x
x
The
of
x

pattern
is
3
x
even

now
and
x

the
2)(
3)(
4)
4
x

…
4!
clear
,
1
…
i.e.
when
the
the
term
in
x
coefficients
powers
n
Therefore
1)(

4
x
is
(
_________________
3
x
3!
2
1
3)

2!

2)(
_____________
2

n
is
(
of
x
are
are
1
when
the
power
odd.
n
1)
x
1
The
series
above,
expansion
of
(1

x)
is
similar
to
the
series
in
the
example
i.e.
(1)(2)

(1)(2)(3)
_________
1
(1
x)

1

(x)
_____________
2

(x
)
3

(x
2!
)
3!
(1)(2)(3)(4)
_________________
4

(x)

…
4!
2

1

x

3
x

x
4

x

…
1
The
series
expansions
remembering
and
of
may
(1
be

asked
1
(1

x)
(1

x)
2

1

x

x

1

x

x
1
Note
that
x

x
2
both
right-hand
3

of
side
these
and
series
1
and
for,
(1
unless

x

x
are
x)
are
worth
derivation
is
i.e.
n

…

(1)

…

x
4
the

their
4
3
finding
x)
quoted
n
x

…
1

x

1
1

x

1
n
geometric,
sum
to

so
infinity
…
starting
of
a
GP
with
the
verifies
these
expansions.
Exercise 2.11a
_
1
1
Expand
(1

x)
2
as
a
series
of
ascending
powers
of
x
as
far
as
the
5
term
in
x
Give
the
.
range
of
values
for
which
the
expansion
is
valid.
_
1
2
(a)
Expand
(1

3x)
2
as
a
series
of
ascending
powers
of
x
as
far
as
the
5
term
in
x
.
n
(b)
Find
the
term
expansion
is
in
x
and
give
the
range
of
values
for
which
n
3
Find
the
the
term
range
of
in
the
valid.
x
values
2
in
of
the
x
binomial
for
which
expansion
the
of
expansion
(1
is

2x)
and
give
valid.
95
Section
2
Sequences,
series
and
approximations
Applications of the
We
can
apply
the
express
them
as
binomial theorem
binomial
theorem
to
a
variety
of
functions
if
we
can
binomials.
1
T
o
use
the
binomial
theorem
to
expand
a
function
such
as
f( x)

(x
1)
1
1
__
1
we
write
it
as
(1
x
)
.
We
can
then
expand

)
the
function
as
a
series
x
of
descending
powers
of
x,
2
1
__
i.e.
f(x)
1
__
(1

)
x
1

x
1
__
(

x
(

1
__
)
x
…

1
__

1
__


2
x
2

3
1
__

x
…
3
x
x
3
x

x

…
1
__
This
series
is
valid
for
1


1,
i.e.
for
x

1
or
x

1
x
Example
n
Find
of
x
the
for
coefficient
which
the
of
x
2
in
the
expansion
is
expansion
of
(3
2x)
in
ascending
powers
of
x
and
give
the
range
of
values
valid.
2
2
(3
2x)
2x
___
2

(1
3
)
3
1
__
(
2x
___
(1


(
(
2)
9
2)(
2
3)
3
)
2!
2x
___
(1

2
9
You
need
to
(

3
(
3
write
down

4
(
3
sufficient
so
2)(
3)(
4)(
4
5)
2x
___
_________________
)
(

3
)
4!

…
3
4
2x
___
)

(
5
)
3
terms
(
2x
___
3!
2x
___
)
3
4)
(
3
2x
___
)
3)(

3
2
1
__

2)(
_____________
(

(
2x
___
_________
)

…
)
3
that
the
pattern
of
the
coefficients
is
clear.
n
n
1
__
n
From
this
we
can
see
that
the
coefficient
of
x
is
2
2
___
(n

(n
n
expansion
is
valid
for
1


1

x

2
The
the
2
binomial
factors
in
sum
the
or
3
3
__
⇒
3
1)
n  2
3
3
__
2x
___
The

__________

1)
9
can
be
used
denominator
theorem
by
using
difference
of
simpler
to
expand
partial
rational
fractions
functions
to
express
with
them
as
functions.
Example
1
______________
(a)
Express
f(x)

in
partial
fractions.
2
(1
(b)
Hence
of
find
the
ascending
the

x
first
four
powers
expansion
is
)(1
of
x)
terms
x,
in
stating
the
the
expansion
range
of
Find
the
coefficient
1
______________
of
x
C
______
Ax  B
_______

(a)

2
(1

x
2
)(1
x)
1

x
1
x
2
⇒
1
1
∴
C


(Ax

B)(1
1
,
B

2
x)
A

x  1
_________

96
1
________

2
x

2
1
______________

C(1
1
and
2
⇒
(1

2
)(1
x)
2(1

x
)
2(1
x)
x
f( x)
values
valid.
r
(c)
of
)
of
as
x
a
series
for
which
)
Section
1
f(x)
(b)
2

(x

1)(1

x
1
1
)
Sequences,
series
and
approximations
1

(1
2
2
x)
2
1
2

(x

1)(1
4
x

1
6
x
x

…)

2
(1
2

x

3
x

4
x

x
…)
2
1
3

(x
5
x

1
7
x
x

…)

2
(1
2
x
4

6
x
x

…)
2
1
2

(1

x

3
x

4
x

2
x

…)
The
terms
in
3
x
and
x
2
cancel
1
2

(1

x
3
x
4
x

so
we
have
to
5
x

x
…)
add
2
another
term
to
the
1
1
2

(1

x

3
x

4
x

x
5

x

expansion
…)
of
(1
x)
2
4

1

x

The
series
is
The
coefficient
5
x

valid
x

for
…
1

x

1
r
(c)
The
binomial
some
of
x
theorem
irrational
is
can
1.
also
be
used
to
find
approximate
values
for
numbers.
Example
1
(a)
Expand
(1
(b)
Substitute
x)
3
2
as
far
as
the
term
in
x
1
Hence
0.02
find
accuracy
an
of
for
x
in
(1
approximate
your
1
(
(1
x)
 1
expansion.
and
x
(
)
state
the
)(
degree
of
3
)(
2
x)
2

1
 …
3!
__
1
2
x
3
x
2
x
8
0.02
3
1
)(
2
2!
1
Substituting
2
1
2
)(
x 
1
(b)
its
√
for
2
2

and
value
1
)(
2
1
(a)
2
answer
.
1
2
x)
for
…
16
x
gives
1
(0.98)
This
2

1
expansion
0.01
is valid
0.000 05
because
x

0.000 000 5
0.02
is
within
the
…
range
1
 x

1
____
98
____
⇒

√
0.989 949 5…
This
is
correct
to
7 d.p.

10
as
the
100
9
next
term
is
5
7
___
√
⇒
2

0.989 949 5…
√
⇒
2

1.41421
correct
to
5 d.p.
10
Exercise 2.11b
_
1
1
Expand
(x
including

the
expansion
is
2)
2
as
a
fourth
series
term.
of
descending
Give
the
range
powers
of
of
values
x
of
as
x
far
for
as
and
which
the
valid.
1
______________
2
Express
in
(1

x)(1

partial
fractions.
Hence
expand
3x)
1
______________
as
(1

x)(1

a
series
of
ascending
powers
of
x
as
far
as
and
3x)
4
including
the
expansion
is
term
in
x
.
Give
the
range
of
values
of
x
for
which
the
valid.
1
_______
3
Use
the
expansion
of
______
√
1
___
√
10
correct
to
4

with
x

0.1
to
find
the
value
of
x
d.p.
97
2.
12
Locating
Learning outcomes
a
root
The
To
introduce
value
the
To
theorem
use
the
theorem
The
intermediate
to
equation
a
function
f( x)
that
is
continuous
between
x

a
and
x

b
intermediate
c

an
intermediate value theorem
Consider

of
locate
a
diagram
lies
shows
between
a
that
and
if
f( c)
is
a
value
of
f( x)
between
f(a)
and
f(b),
then
b
value
root
of
an
y

f(x)
equation
f(a)
f(c)
You need to know

The
meaning
of
a
continuous
f(b)
function

How
to
sketch
graphs
of
simple
functions
O
There
may
diagram
be
more
below
a
than
one
c
value
of
x
b
x
between
x

a
and
x

b,
as
the
shows.
y

f(x)
f(a)
f(c)
f(b)
O
a
c
c
1
However
,
a
value
This
is
of
if
x
f( x)
is
not
between
illustrated
y
in

continuous
x

the
a
and
next
x
between

x
b
2
a
and
b
then
there
b
diagram.
f(x)
f(a)
f(c)
f(b)
O
98
x
may
not
be
Section
The
intermediate
Provided
there
Locating
It
is
not
If
is
in
the
a
we
an
f( x)
to
can
states
is
one
has
find
a
an
the
Sequences,
series
and
approximations
that:
continuous
least
root of
value
given
between
of
x
value
x

between
between
a
x
and

f( a)
a
x

and
and
b,
x

b
f(b).
equation
exact
sometimes
use
roots
the
of
some
equations.
intermediate
theorem
to
locate
a
interval.
equation
between
f( x)
at
which
continuous
The
be
possible
However
,
root
that
must
for
theorem
2
x

f( x)
in
a
and
intermediate
Therefore
f(x)

0
this
has
a
root
interval,
x

value
changes
between
then
the
x

curve
a
and
y

f( x)

x
f(x)

b,
and
crosses
if
the
f(x)
x-axis
b
theorem
sign
tells
us
between
x
that

a
and
x
0

is
between
b,
f(a)
and
f(b).
i.e.
y
f(b)
O
a
x
b
f(a)
If
if
one
then
The
first
where
f(x)
root
f(a)
step
is
of
is
continuous
the
and
to
equation
f(b)
are
roughly
between
f( x)

0
opposite
locate
the
x

lies
in
a
sign,
roots
and
x
between
of
i.e.
an

b
and
x

a
f( a)

f(b)
equation
and
x

using

b
0
a
sketch
possible.
x
For
example,
the
equation
e
2x
2

0
has
roots
where
the
graphs
of
x
y

e
and
y

2x

2
intersect.
y
6
4
2
O
3
2
1
1
2
3
x
2
From
and
x
the

sketch,
we
can
see
that
there
appears
to
be
a
root
between
x

1
2
x
We
can
finding
test
f(1)
this
and
f(1)

e
f(2)

e
by
using
f( x)

e

As
f(1)
and
between
1

f(2)
and

2
(which
is
continuous)
and
f(2).
4

0
2
and
2x
2

6
are

0
opposite
e
in
sign,

i.e.
7
.3…
f(1)

f(2)

0,
there
is
a
root
2.
99
Section
2
Sequences,
series
and
approximations
Example
3
Use
one
a
sketch
real
Find
to
show
consecutive
3
roots
of
y

The
as
T
o
x
→
2
x
2x

x

1

0
has
only
integers
between
which
this
root
lies.
2x

x

1

0
are
the
values
of
x
where
where
y

the
graph
2
2x
curve
x
equation
2
x
3
of
the
root.
two
The
that
is
a

x

cubic
1
intersects
which
the
crosses
x-axis.
the
y-axis
1,
and
y
→
∞
∞
locate
the
curve
we
will
find
the
turning
points:
dy
___
2

3x
4x

1
dx
2
⇒
3x
4x
⇒
(3x

1
1)(x

1)
0

0
1
⇒
x

and
x

1
3
__
4
1
When
x

,
y

3
The
curve
1
and
when
x

1,
y

1
27
crosses
the
x-axis
once
so
there
is
only
one
real
root.
y
1
O
x
1
1
3
3
Alternatively,
3
x
the
roots
of
2
x
2x

x

2x
x

0
are
where
1
3
A
1
2

sketch
also
of
shows
the
that
curves
there
y
is

x
only
2
and
one
y

real
2x
x
1
(
(2x
root.
y
4
2
O
2
1
2
4
100
2
x
1)(x
1))
Section
From
either
sketch
it
3
Using
f(
f(
1)

1)
and
1
(It
f(x)
and
is

3
that
the
root
is
between
1
and
Sequences,
series
and
approximations
0.
2
x
2x
and
f(0)
appears
2

f(0)
are

x

1,
1
opposite
in
sign,
therefore
the
root
lies
between
0.
likely
that
this
root
is
nearer
0
than
1,
as
f(0)
is
nearer
zero
1
than
is
f(
1).
We
can
test
this
by
finding
the
sign
of
f
(
).
2
1
If
f(
1
)

0,
the
root
lies
between
and
2
0.)
2
Exercise 2.12
1
__
1
Draw
a
sketch
to
show
that
the
equation
ln
x

has
one
real
root.
x
Hence
find
equation
two
consecutive
x ln x
1
0
integers
between
which
the
root
x
2
Show,
using
only
one
Find
two
a
sketch
of
the
lies.
or
otherwise,
that
the
equation
e
2

x
1
has
root.
consecutive
integers
between
which
this
root
of
the
equation
lies.
3
The
diagram
shows
a
sketch
1
y

tan
of
the
curve
2
x
ln (1

x
)
y
1
0.5
x
1
0.5
1
1.5
2
(a)
V
erify
that
zero
is
one
1
tan
(b)
Use
the
the
root
of
the
equation
2
x

ln (1

x
intermediate
equation
lies
)

0
value
between
1
theorem
and
to
show
that
another
root
of
1.5
101
2.
13
Interval
bisection
Learning outcomes
Numerical
Numerical

To
use
interval
bisection
to find
of
degree
an
of
equation
to
a
methods
use
solving
repeated
equations
applications
of
a
method
to
successively
a
improve
root
methods for
on
an
approximation
for
a
root
of
an
equation.
specified
accuracy
Interval
In
the
last
bisection
topic,
we
method
showed
how
to
locate
a
root
between
successive
You need to know
integers.
value

How
to
use
sketch
graphs
of
The
the
inter val
root
to
bisection
any
degree
method
of
refines
this
approach
to
give
the
accuracy.
and
x
the
to
intermediate
locate
a
root
value
of
an
theorem
Consider
an
the
equation
e
2x
2

0
equation
W
e
within
again
have
shown
in
T
opic
2.12
that
this
equation
has
a
root
between
1
and
2
interval
and
So
that
if
f(1)
this

root
0
is
and

we
f(2)

know
0
that
1



2
y
f(2)
O
x
0.5
f(1)
We
then
bisect
the
interval
to
give
x

1.5
and
find
the
sign
of
f(1.5):
1.5
f(1.5)
1.5



e
5


0.5…
Therefore
f(1.5)

0
and
f(2)

0
so
2
y
f(2)
O
W
e
then
bisect
x
f(1.5)
0.5
the
interval
again
to
give
x

1.75
and
find
the
sign
of
f(1.75):
1.75
f(1.75)

e
Therefore
5.5
f(1.5)


0
0.25…
and
f

(1.75)
0

0
so
1.5



1.75
y
f(1.75)
O
Bisecting
f(1.625)
102
the

interval
0.17…
x
f(1.5)
0.5
1.5

0
to
1.75
gives
x

1.625
and
Section
Therefore
f(1.625)

0
and
f(1.75)

0
so
1.625



2
Sequences,
series
and
approximations
1.75
y
f(1.625)
f(1.75)
O
x
0.5
We
are
have
a
narrowing
value
There
is
no
need
Bisecting
the
f(1.6875)

Therefore
Bisecting
The
last
get
an
1.7
0.005
This
method
finding
to
feed
until
The
of
In
a
desired
the



less
0
1
bisection
root
we
need
lies,
but
we
still
do
not
continue.
to
keep
track
of
the
sign
of
f( x).
1.6875
that
f(1.625)

0
1.65625
f(1.6875)
0.05,
2
so

0
we
an
a
is
with
a
better
accuracy
method
places,
iterative
starts
give
can
now
say
that
the
root
of
the
place.
decimal
of
to
of

decimal
to
step
degree

just
the
so
1.6875
than
to
x
and
equation
next
we
know
gives
example
an
we
x
which
place,
1.6875
correct
an
of
the
interval
the
is
root
into
the
is
in
decimal
gives
again

1
and

correct
answer
than
0
0.072…
1.65625
is

interval
diagrams,
again
interval

the
even
draw


interval
equation
T
o
0.03…
the
Therefore
to
to
interval
1.625
f(1.65625)
down
correct
is
we
method .
first
the
An
iterative
approximation.
to
interval
approximation
obtained.
slow
need
Each
converge
step
(i.e.
is
to
is
then
then
called
get
be
less
method
and
This
to
that
repeated
an
close
for
uses
iteration.
to
the
value
root).
example
above
it
took
five
iterations
to
get
an
answer
correct
to
1
decimal
place.
However
,
it
conditions
not
In
does
for
continuous
the
next
bisection
have
the
or
topic
the
advantage
intermediate
there
we
is
look
more
at
an
that
value
than
the
method
theorem
are
one
in
iteration
root
method
will
not
the
only
met,
initial
that
fail
i.e.
if
if
the
the
function
is
interval.
improves
on
the
interval
method.
Exercise 2.13
3
1
(a)
Find
the
sketch
(b)
Use
stationary
the
your
points
on
the
curve
y

x

3x

4
and
hence
curve.
sketch
to
find
consecutive
intervals
in
which
the
root
of
the
3
equation
(c)
Use
1
the
x

3x
interval
decimal

4
lies.
bisection
method
to
find
this
root
correct
to
place.
103
2.
14
Linear
interpolation
Learning outcomes
Linear
Linear

To
use
linear
interpolation
interpolation
interpolation
proportion
find
a
root
of
an
equation
is
similar
to
the
interval
bisection
method
but
uses
to
to
to
find
the
next
value
in
the
interval
rather
than
taking
the
a
mid-point.
specified
degree
of
accuracy
Consider
between
an
x

equation
a
and
x
f( x)


0
which
has
a

root
which
we
know
lies
b
You need to know

How
the
to
to
locate
within

use
sketch
intermediate
The
a
an
root
graphs
value
of
an
and
f(b)
theorem
equation
interval
properties
of
x
similar
triangles
f(a)
Did you know?
The
There
is
similar
years
evidence
to
this
ago.
that
was
a
used
method
over
2000
x

line
b
diagram
be
joining
cuts
Therefore
The
a
at
of
c.
on
the
interval
is
curve
Assuming
interval
method
joining
pair
points
the
the
bisection
line
forms
that
than
this
the
x-axis
shows
smaller
interval
the
between
from
likely
y
that
x
to

a

f(x)
where
f( a)

and
x
to

a
the
converge
0
and
x

interval
more
x

f(b)
c
a

is
and
0,
the
likely
bisection
quickly
than
to
point.
the
method.
the
points
similar
on
the
triangles.
curve
y

Therefore
f(x)
the
where
point
x
x


a
c
divides
and
x

b
the
1
line
between
x

a
and
x
c

b
:
|f(b)|,
i.e.
the
ratio
|f(a)|
a
1
______
|f(a)|
in
______

b
c
|f(b)|
1
a|f(b)|

b|f(a)|
________________
⇒
c

1
|f(a)|
where
c
is
the
first

|f(b)|
approximation
for

1
Y
ou
need
to
lear n
this.
x
Consider
W
e
have
and
that
again
shown
f(1)

the
in
0
equation
T
opic
and
e
2.12
f(2)

that
2x
this
2

0
equation
has
a
root
between
1
and
0
y
f(2)
O
x
0.5
f(1)
104
2
Section
Working
f(1)

with
the
first
1.2817…
four
and
decimal
f(2)

places
throughout
2
Sequences,
series
and
approximations
gives
1.3890…
(1)(1.3890)

(2)(1.2817)
_______________________
Therefore
c


1.4799
1
1.2817

1.3890
This
f(c
)

0.5672…

0

so
is
in
the
interval
is
the
1st
1.4799
to

approximation for
2
1
(1.4799)(1.3890)

(2)(0.5672)
____________________________
Repeating
the
process:
c


1.6307
2
0.5672

1.3890
2nd
f(c
)

0.1539…

0

so
is
in
the
interval
1.6307
to
approximation
2
2
(1.6307)(1.3890)

(2)(0.1539)
____________________________
Repeating
again:
c


1.6675
3
0.1539

1.3890
3rd
f(c
)

0.0360…

0

so
is
in
the
interval
1.6675
to
approximation
2
3
(1.6675)(1.3890)

(2)(0.0360)
____________________________
And
again:
c


1.6759
4
0.0360

1.3890
4th
f(1.6759)
see
if

the
f(1.675)

have


Therefore
We
1
the
If
rate
of
shape
the
deal
1.68
rate
the
c
this
value
using
is
is
small
correct
to
of

2

five
interval
enough
2
to
to
decimal
0.022…
decimal
correct
with
to
so
be
worth
checking
to
places:
1.675



1.685
places.
2
decimal
iterations
bisection
to
places
give
(T
opic
a
in
four
value
2.13).
correct
Therefore
the
quicker
.
curve
not
of
in
changes
is
is
f(1.685)
correct
convergence
of
if
and
compares
place
gradient
and
the
This
convergence
The
0.011…

decimal
and
approximation
found
iterations.
to
0.0081…
4th
approximation
a
very
linear
the
interpolation
initial
depends
on
interval.
great
close
1
,
to
is
If
the
rate
of
a
convergence
c
slow.
the
gradient
much
and
if
does
c
is
not
close
b
x
b
x
change
to
,
1
the
As
rate
with
of
convergence
interval
is
fast.
bisection,
this
a
method
fails
if
the
function
is
not
c
continuous
root
in
the
or
has
initial
more
than
one
interval.
Exercise 2.14
1
(a)
(b)
Show
that
and

Use
x
the
equation
ln
x
x

twice
to
2

0
has
a
root
between
x

3
3.5
linear
interpolation
get
an
approximate
value
for
to
places.
this
root.
(c)
Show
that
the
approximation
is
correct
3
decimal
105
2.
15
Newton–Raphson
Learning outcomes
Newton–Raphson
The
To

use
the
method
an
any
method
uses
a
linear
approximation
for
a
function.
approximate
If
to
Newton–Raphson
method
Newton–Raphson
to find
value for
method
the
root
degree
of
an
equation
the
equation
x-axis
where
x
f( x)

0
has
a

root
then
the
curve
y

f(x)
cuts
the


specified
If
c
is
an
approximate
value
,
of
then
the
tangent
to
the
curve
at
the
1
To

give
a
geometric
point
A
where
x

c
cuts
the
x-axis
at
a
point
where
x

c
1
interpretation
of
the
2
method
y
You need to know
c
f(x)
c
α
1

2
x
How

to
which
How

locate
a
root
to find
tangent
to
a
an
of
interval
an
the
in
equation
equation
curve
at
a
lies
of
a
A
given
In
most
cases,
c
will
be
closer

to
than
is
c
2
point
approximation
The
coordinates
of
A
are
(c
,
f(c
1
Therefore
the
.
Therefore
c
1
is
a
better
2

to
equation
of
))
and
the
gradient
of
the
curve
at
A
is
f (c
1
the
tangent
is
y
f(c
)

f(c
1
)(x
c
1
)
1
Did you know?
f(c
)
1
_____
This
tangent
cuts
the
x-axis
where
y

0
⇒
x

c
1
f(c
The
by
method
Sir
Isaac
was first
Newton
it
is
)
1
published
and
often
Therefore
if
c
is
an
approximation
for
a
root
of
an
1
called
simply
Newton’s
method.
f(c
)
1
_____
However,
it
was
simplified
equation
by Joseph
f(x)

0
then
c

c
2

1
f(c
)
1
Raphson
a few
years
later.
Neither
of
is
these
this
early
methods
was first
used
introduced
Simpson. The
version
we
calculus
better
approximation.
–
Y
ou
by Thomas
use
a
also
need
to
today
lear n
this.
x
Using
this
as
first
method
to
find
the
root
of
e
2x
2

0
and
using
c

1
was
by
published
the
French
nearly
a
century
later
x
the
approximation,
we
have
f (x)

e
2
mathematician Joseph
2
e
6
______
Lagrange.
Therefore
c

2

2
1.74224...
2
e
2
1.74224
e
5.48449…
___________________
and
c

1.74224…
3

1.68142…

1.67835…
1.74224
e
2
1.68142
e
5.36284…
___________________
and
c

1.68142…
4
1.68142
e
so

We
is
have
correct
If
probably
we
already
to
do
equal
2
tested
decimal
another
to
1.68
this
in
correct
T
opic
2
to
2
2.14,
decimal
so
we
places.
know
that


places.
iteration,
we
get
1.67835
e
5.3567…
_________________
c

1.67835…

5
106
we
can
see
1.67834…
1.67835
e
so
that

is
likely
to
).
1
be
2
1.6783
to
4
decimal
places.
1.68
2
Section
We
can
check
f(1.67825)
1.67825
The
the
a


the
1.67835
Raphson
root
of
example
However
,
an

are
1.6783
is
f(1.67835)
so

=
method
equation
above
there
first

0.0003…,

Newton
finding
as


whether
correct

is
the
4
decimal
0.00001…
1.6783
because
to
correct
best
to
it
Sequences,
series
and
approximations
places:
Therefore
4
method
when
2
decimal
places.
considered
works
it
so
converges
far
for
rapidly,
shows.
factors
that
approximation,
c
,
cause
the
is
far
too
method
from
to
fail:

1
y
c
c
2

f(x)
α
1
x
A

the
gradient
of
the
curve
at
the
point
where
x

c
is
too
small
1
y
c

f(x)
c
α
1
2
x
A

the
gradient
of
the
curve
increases
rapidly
y
c

f(x)
c
α
1
2
x
A
Exercise 2.15
2
1
(a)
Use
a
sketch
to
show
that
the
equation
x

ln (x

2)
has
two
roots.
(b)
Use
the
Newton Raphson
approximate
(c)
State,
with
value
of
reasons,
the
the
method
larger
three
times
to
find
an
root.
accuracy
of
your
approximation.
107
2.
16
Using
a
given
Learning outcomes
Iteration
As

To
use
a
given
approximate
iteration
value
of
iteration
a
to find
root
of
we
have
seen
with
linear
interpolation
and
Newton–Raphson,
an
iteration
produces
iteration
formula)
a
sequence
of
values
by
using
a
formula
(called
an
an
of
the
form
equation
x

f(x
n1
T
aking
x
as
the
first
value,
then
x
1
)
n

f(x
2
)
1
You need to know
x

f(x
3

How
to
value
use
the
x
intermediate

f(x
4
theorem
)
2
)
3
x

f(x
5
)
and
so
on.
4
1
For
example,
when
x

(x
n  1

1)
2
and
x
n

2
1
1
x

1)
2

(2


(1.732…
1.732…
2
1
x

1)
2

1.652…

1.628…

1.621…
3
1
x

(1.652…

1)
2
4
1
x

(1.628…

1)
2
and
so
on.
5
This
and
is
the
we
now
finding
The
n
same
a
as
look
root
of
sequence
increases,
of
x
at
an
a
recurrence
the
convergence
of
used
such
a
to
generate
sequence
a
in
sequence
the
context
generated
closer
and
above
closer
to
n
converge
x
,
i.e.
to
x
n  1
a
value
→
.
,
This
because
value,
when
x

x
n
,
i.e.

when
(


1)
2
.
n
Therefore

is
a
root
of
the
n  1
1
equation
Not
all

(


iterations
1)
give
2
values
that
converge.
_______
x
For
example,
using
the
iteration
formula,
x

n
√
e

2,
and
taking
n  1
x

2
gives
1
______
x

2
√
e

2

3.064…
2
__________
x

3.064…
√
e

2

4.839…
3
__________
x

√
4.839…
e

2

11.32…
4
This
sequence
(rapidly
in
Using
We
have
this
an
values
diverges
because
the
values
seen
are
increasing
case).
iteration formula to find
that
approximation
108
of
to
we
a
can
root,
use
,
of
an
iteration
an
equation
a
root
formula
f( x)

0
to
find
a
good
as
,
1
is
of
equation.
values
gets
formula
Section
When
make
f(x)
the

0
can
iteration
x

g(x
n  1
The
roots

x
The
and
written
in
the
form
x

g(x)
we
can
use
this
Sequences,
series
and
approximations
to
formula
)
n
of
intersection
y
be
2
the
of
the
diagram
equation
the
x

g(x)
are
the
values
of
x
at
the
points
of
line
curve
y
shows

g(x)
how
this
iteration
works.
y
y
D{x
,
g(x
2

x
E
)}
2
A{x
C
,
g(x
1
)}
1
B
y

g(x)
O
α
x
x
2
Using
x
as
the
first
x
3
approximation
to
x
1
the
,
root
then
in
the
diagram
1
A
is
the
point
on
y

g(x)
where
x

x
so
y

g(x
1
B
is
the
point
where
x


and
y

)
1
g(x
)
1
C
is
the
point
on
the
line
y

x
where
x

x
and
y

g(x
2
Now
x
will
be
closer

to
than
is
x
2
curve
that,
near
the
root,
less

steep
than
the
gradient
of
the
line
y

x,
i.e.
provided
x
is
a
better
approximation
to

than
is
2
is
on
the
line
y

x,
that
|g ( x)|
therefore
x


g(x
2
We
can
repeat
this
process
to
get
x
x
3,
The
near
The
rate
the
of
x
,
1
provided
C
that
1
Therefore
Now
the
1
is
|g(x)|
provided
)
1
convergence
of
this
,
1
)
1
…
.
4
sequence
depends
on
the
value
of
g (x)
root.
smaller
|g (x)|
is,
the
more
rapid
is
the
y
rate
of
convergence.
y
O
O
α
x
x
1
α
x
x
1
109
Section
2
Sequences,
series
and
approximations
The
sequence
diverges
(i.e.
fails
to
find
a
y
if
|g (x)|
1
y
O
O
α
α
x
will
use
this
method
x
to
try
x
1
x
1
We
root)
and
find
the
roots
of
the
equation
x  1
e
x
3

0
x  1
The
one
graph
of
near
3
y

e
and
x
the
other
3
shows
near
that
the
equation
has
two
roots,
0.
y
2
1
O
4
3
2
x
1
1
1
2
x  1
Rearranging
iteration
the
equation
formula
x

as
x

e
3
and
changing
this
to
the
gives
1
n
x

e
3
n  1
T
aking
x

3

e
gives
1
2
x
3

2.8646…
2
1.8646…
x

e
3

2.8450…
3

2.8419…
3

2.8414…
3
1.8450…
x

e
4
1.8419…
x

e
5
so
this
iteration
is
converging.
x  1
Using
f(x)

e
x
3,
5
f(
2.8415)

7.9


2.841
therefore
Now
taking
x


0
10
as
4

0
correct
the
and
to
first
4
f(
2.8405)
significant
approximation

7.6

10
to
the
other
root
1
1
x

e
3

0.2817…
2
x

1
0.2817…
1
0.9490…
e
3

0.9490…
3

1.9477…
3
x

e
4
This
110
sequence
is
diverging
so
it
fails
to
find

0,
figures.
the
root
near
zero.
gives
Section
We
could
predict
that
this
will
happen
by
looking
at
the
gradient
2
Sequences,
series
and
approximations
function
x  1
of
e
3:
d
___
x  1
x  1
(e
3)

x  1
e
and
e

1
for
x

1,
dx
i.e.
|g(x)|

1
for
values
of
x
near
x

0
Example

__
(a)
Show
that
the
(b)
T
aking
0.75
x
cos x
equation
cos
x
x

0
has
a
root
between
0
and
3

n1
as
a
first
three
approximation
times
to
find
an
to
this
root,
use
approximation
the
to
iteration
this
root.
n
(c)
Hence
(a)
f(x)
show
that
the
root
is
0.74
correct
to
2
decimal
places.

__

cos x
x
so
f(0)

1
and
f
(
)

0.5
1.04…

0.5…
3

__
f(0)

0
and
f
(

)
0
therefore
cos x
x

0
has
a
root
between
3
Exam tip

__
0
and
3
Iterations
(b)
Using
x

0.75
and
x
1
x

cos x
n1

cos 0.75

are
easy
to
do
on
most
gives
n
scientific
calculators:
of
press
enter
the
value
0.73168…
2
x
and
EXE
(or
ENTER). Then
1
x

cos 0.73168…

0.74404…
enter
3
x
the formula for
g(x
)
using ANS
n

cos 0.74404…

for
0.73573…
each
value
of
x. Then
press
EXE
4
and
Therefore
0.73573…
is
an
approximate
value
of
the
continue
to
press
EXE for
each
root.
iteration.
(c)
f(0.735)

cos 0.735
0.735

0.0068…
f(0.745)

cos 0.745
0.745

0.0099…
Therefore
correct
Note
to
that
the
2
,
root,
decimal
x
is
lies
between
0.735

0

and
0
0.745,
so


0.74
places.
measured
in
radians,
so
the
root
has
a
is
0.74 rad.
Exercise 2.16
3
(a)
Show
(b)
Use
that
the
0.5
as
equation
a
first
x
5x
3
approximation

for
0
this
root
root
between
and
the
1
and
0.
iteration
3
x
3
n
______
given
by
x

n  1
5
Use
six
iterations
down
5
decimal
(c)
Show
that
(d)
The
your
to
find
places
root
is
a
for
better
each
correct
approximation
for
the
root,
writing
iteration.
to
3
decimal
places.
3
equation
Explain
why
x
the
5x
3
iteration

0
also
formula
has
given
a
root
will
near
fail
to
x

find
2
this
root.
111
Section
2
Practice
questions
1
_____________
1
A
sequence
is
given
by
u

8
and
u
1

u
n

2
9
Express
(a)
in
(r
Show
that
the
sequence
is
an
partial
fractions.
n  1

1)r(r

1)
arithmetic
r  n
progression
and
write
down
the
1
_____________
common
Hence
(b)
find
∑
(r

1)r(r

1)
difference.
r  2
10
The
nth
term
of
a
sequence,
u
,
is
given
by
n
2
The
first
three
terms
in
a
sequence
are
2
u

n

n
n
a
__
,
a
and
ab
respectively,
b

0
b
Prove
by
mathematical
induction
_
1
(a)
Show
that
the
terms
are
in
geometric
the
first
n
terms
is
given
that
the
sum
of
2
by
n(n

1)
3
progression.
(b)
The
first
three
Find
term
terms
the
is
is
2
and
the
product
of
the
11
of
is
interest
216.
values
$2000
a
and
b
and
the
fifth
of
term.
each
Show
invested
at
5%
year
that
in
per
$500
the
an
account
annum
is
paid
withdrawn
amount
$A
in
that
accrues
yearly.
from
the
At
the
the
end
account.
account
after
n
3
The
nth
term
of
a
sequence,
u
,
is
given
by
n
n
years
is
given
by
n
u

2(3
n
)

4
A
n
Show
that
u

2000 (5
n

3u
n  1


4(1.05)
)
8
n
12
The
rth
term
of
a
series,
u
,
is
given
by
r
4
The
nth
term
of
a
sequence,
u
,
is
given
by
u
n

(2r

1)(r
2)
r
r  n
2
2n
 n
________
u

Find
n
2
4n

1
u
∑
r
r  1
Show
that
value
to
the
sequence
which
it
converges
and
give
the
13
converges.
(a)
Use
Maclaurin’s
terms
in
the
theorem
expansion
to
find
the
first
two
of
x
5
Determine
whether
the
sequence
whose
nth
term
f(x)

e
sin x
n
___
is
n sin
is
alternating,
periodic
or
oscillating.
as
a
series
of
ascending
powers
of
x.
2
(b)
Use
your
series
to
find
an
approximate
value
r  n

__
n  1
_______
6
Given
∑
u

find:
for
e
6
.
r
2n

1
r  2
r  2n
14
(a)
u
(b)
(a)
Use
Maclaurin’s
theorem
to
find
the
first
u
∑
n
r
four
terms
in
the
expansion
of
r  n
______
(c)
the
sum
to
infinity
of
the
1  x
______
series.
ln
,
√
1
7
(a)
Show
3
_
that
_
1


2
(b)
Find
in
the
as
series

sum
1

x

1
x
series
of
ascending
values
of
x
for
powers
which
the
of
x,
stating
expansion
is
…
valid.
of
progression.
the
first
n
(b)
terms
of
the
Use
for
series
your
series
to
find
an
approximate
value
ln 3.
(a).
15
(c)
a
the
geometric
the
of
18
6
in
terms
__
1

2
are
the
_
1

State
with
a
reason
whether
the
series
(a)
Find
the
first
three
terms
of
the
expansion
of
is

__
cos x
convergent.
as
ascending
powers
of
x
(

)
3
(b)
Hence
find
an
approximate
value
of
cos 61°
1
____________
8
(a)
Express
in
(r

1)(r

partial
fractions.
given
that
1°

0.017 rad.
3)
r  n
16
1
____________
(b)
Hence
find
∑
(a)
Expand
tan x
as
a
series
of
ascending
powers
2
(r

1)(r

3)
of
(x

a)
as
far
as
the
term
in
(x

a)
r  1

__
(b)
(c)
Deduce
the
sum
to
infinity
of
the
Use
a
1
______

2
112

4

5
an

__

4

6
find
…
a
quadratic
approximate
1
______

3
to
function
that
3
gives
1
______

series
x
is
close
to
3
value
for
tan
x
when
Section
n
17
Prove
(a)
that

(b)
C
n
a
Use
relationship
between
n
and
r
the
n

bisection
root
lies
method
between
x
twice

to
1.75

and
2
C
r
r
1
26
expansion
(1
of
(a)
Sketch
the
graphs
of
9
1
the
this
n  1
C
In
interval
that
when
x
18
questions
r
show
Find
Practice
n
C
r
(b)
2
x
)
in
ascending
2
y
powers
of
x

x
1
and
y

ln (x

2)
find:
Use
(a)
the
first
four
(b)
the
coefficient
your
sketch
to
show
that
the
equation
terms
1
x

ln (x

2)

0
has
only
one
positive
7
of
x
root,
the
(c)
general

term.
(b)
Use
two
the
intermediate
consecutive
value
integers
theorem
to
find
between
which
twice
find

lies.
2
19
Find
the
terms
up
to
and
including
x
in
the
(c)
expansion
Use
linear
interpolation
to
an
of
approximate
4
(1
2x)
value
.
for
Give
your
answer
6
(1

4x)
correct
to
3
significant
figures.
5
20
Find
the
real
part
of
(1
2i)
3
27
(a)
Show
has
21
Find
the
term
independent
of
x
in
the
a
that
root
the
equation
between
expansion
3
curve
y

the
2
x
4x
turning

5

points
0
on
the
2
x
4x

5
9
2
__
2
of
(x
(b)
)
Use
the
intermediate
value
theorem
to
find
x
consecutive
Find
the
coefficient
of
x
between
which
this
root
lies.
4
22
integers
in
the
expansion
of
(c)
1
________
Use
the
this
root
Newton–Raphson
method
to
find
_______
√
2
to
2
decimal
places.
3x
3
1
____________________
23
correct
28
Express
(a)
The
diagram
shows
the
curve
y

x
6x

4
2
(x

1)(x

2)(x

1)
y
in
partial
fractions.
10
1
____________________
Hence
(b)
expand
as
a
2
(x
series
of

ascending
1)(x

powers
2)(x
of
x

up
1)
to
8
and
3
including
of
values
the
for
term
in
which
x
the
,
and
give
expansion
the
is
range
valid.
4
2
2
24
Expand
(1

x

2x
1
)
in
ascending
the
term
powers
of
x
3
up
to
and
including
in
x
O
4
x
2
2
25
The
diagram
x
y

e
shows
the
graphs
of
the
curves
x  2
______
1
and
y

3
(a)
x
Confirm
has
one
that
root
the
equation
equal
to
x

6x

4

0
2.
y
(b)
Using
1
as
a
first
approximation
to
the
other
4
positive
of
the
root,
show
that
an
iteration
form
2
x

g(x
n  1
converges
O
4
2
2
4
formula
to
correct
to
2
T
aking
the
the
value
decimal
)
n
of
this
root
and
find
it
places.
x
(c)
3
and
negative
2,
show
root
that
as
the
lying
between
same
iteration
2
formula
(d)
Use
fails
another
to
converge
numerical
to
this
method
root.
to
find
this
4
root
(a)
V
erify
that
one
x
solution
between
the
equation
29
Use
a
to
numerical
2

to
places.
solve
the
equation
x
x
x
decimal
method
1
x e
lies
of
correct
1
2
and

x
e
0

2
giving
the
roots

correct
3x
to
3
decimal
places.
113
3
Counting,
3.
1
The
matrices
principles
Learning outcomes
To
introduce
and
use
equations
counting
answer
any
question
starting
‘How
many
…
?’,
we
need
an
efficient
the
method
fundamental
differential
Counting
T
o

of
and
counting
of
counting.
principle
When
the
entities
to
be
correspondence
with
For
count
example,
time
counting
However
,
For
of
can
1,
there
example,
choice
We
to
3
2,
are
how
main
illustrate
counted
the
the
3,
…
in
you
a
3,
in
…
you
where
meals
desserts
different
placed
2,
box,
situations
2
be
1,
,
a
one-to - one
counting
can
take
them
them
is
out
easy.
one
at
a
go.
different
courses,
the
balls
as
many
many
can
numerals,
meals
this
are
and
2
that
is
not
possible
drinks
can
be
possible.
when
on
a
there
is
a
menu?
chosen
using
a
diagram:
drink
1
dessert
1
drink
2
main
1
drink
1
dessert
2
drink
2
drink
1
dessert
1
drink
2
main
2
drink
1
dessert
2
drink
2
drink
1
dessert
1
drink
2
main
3
drink
1
dessert
2
drink
2
For
of
each
of
Therefore
For
each
Now
different
T
aking
of
the
are
these
there
has
a
a
ways
just
of
three
3
3

are
3

2
of
choosing
2
of
this
first
an
different
ways
ways
multiple
can
the
2

choice
choosing
four
ways
a
main
course
there
are
two
ways
dessert.
consider
which
ways
a
there
of
Therefore
of
the
choosing

of
there
2
are
a
questions:
answer
and
meals
answers
for
for
of
course
and
choosing
a
a
dessert.
drink.
possible.
with
possible
be
each
main
ways
examination
different
examination
two
two
different
choice
four
choosing
30
questions,
answers.
In
how
each
many
answered?
question
of
these
question
2.
1
there
four
This
can
gives
are
be
4
four
different
paired

4
with
different
2
ways
of
answering
the
first
two
questions,
i.e.
4
different
ways.
30
Repeating
114
this
argument
for
all
30
questions
gives
4
different
ways.
one
Section
Question
Question
Question
1
1
1
4

These
if
two
there
4
examples
are
n
then
illustrate
ways
of
the

of
doing
doing
number
of
is
known
as
matrices
and
differential
equations
Question

…
…

4
that
task,
another
different
n
This
Counting,
30
4
one
yet
…
3

m
m
ways
task
ways

l
…
of

doing
so
doing
another,
l
ways
on,
all
the
tasks
is
…
fundamental
the
of
and
counting
principle
Example
Three
How
ordinary
many
There
are
six
Therefore
six-sided
different
ways
there
in
are
dice,
one
outcomes
which
6

6
each

red,
are
6

one
blue
and
one
green,
are
rolled
and
a
coin
is
tossed.
there?
dice
can
2
432

land
and
two
different
ways
in
which
the
coin
can
land.
outcomes.
Example
A
company
selling
products
is
the
first
The
next
two
characters
are
letters
of
The
next
two
characters
are
one
the
The
final
character
many
There
and
are
fifth
is
different
9
choices
characters
Therefore
there
one
a
letter
codes
for
9

There
are
counting,
many
and
other
we
look
25

of
the
first

21
9 922 500
some
a
to
the
for
digits
0
when

the
10
we
in
to
on
each
item.
not
10
including
including
and
choices
last

not
vowels.
9.
digits
21
different
them
code
9.
alphabet,
where
of
six-character
alphabet,
character
,
choices
21
1
possible
situations
at
digits
of
are
the
and
are
of
uses
The
How
character
software
the
letters
for
the
letter
can
next
O.
be
repeated?
two
characters,
10
choices
for
the
fourth
character
.

25
codes.
need
the
an
next
efficient
few
method
of
topics.
Exercise 3.1
1
There
be
are
used
one
the
of
to
three
four
print
different
make
a
poster
different
on
the
colours
and
colours
of
there
that
paper
is
can
a
that
choice
be
used
poster
.
can
of
for
2
The
number
followed
one
are
digit.
0
to
include
How
many
there?
different
colour
combinations
by
plate
two
The
9,
the
the
on
a
letters
first
two
letters
car
of
digit
is
letters
I
or
O
consists
the
1
of
to
9,
the
and
of
three
alphabet,
the
next
alphabet
the
last
digits
followed
two
do
digit
by
digits
not
is
1
or
0.
are
How
many
when
digits
different
and
number
letters
can
plates
be
are
possible
repeated?
115
3.2
Permutations
Learning outcomes
Permutations
A

To
define
a
permutation
per mutation
is
an
ordered
arrangement
of
a
number
of
objects.
and
n
introduce
the
notation
P
r

To find
a
variety
of
types
For
of
of
example,
arranging
if
four
them
books,
is
A,
B,
A,
C,
B,
D.
C
and
D,
Another
are
is
placed
B,
D,
A,
on
a
shelf,
one
way
C.
permutation
A
B
C
D
B
D
A
C
or
You need to know

Factorial
notation

The fundamental
Each
principle
of
these
arrangements
is
called
a
permutation
of
the
books
and
each
of
arrangement
is
a
different
permutation.
counting
The
For
number
the
leaves
ways
of
books,
3
of
first
the
three
are
the
the
in
There
four
the
and
book
books.
the
different
for
first
third
of
4
choices
is
number
choices
of
for
different
the
left-hand
next
book,
so
second
book.
There
the
is
row,
only
books
is
giving
one
4

book
3

2
there
4

are
are
3

left,
so


1
arrangements.
4
now
2
the
book.

3
only
ways
This
different
2
of
ways
arranging
number
of
4!
general
the
For
number
example,
ordinary
The
in
there
selecting
permutations
In
permutations
different
choosing
the
of
a
the
pack
is
examples
line.
In
per mutations
number
of
of
n
permutations
different
of
the
52
objects
playing
is
n!
cards
from
an
52!
of
the
arrangements
of
the
books
next
of
and
example
some
of
the
cards
we
n
are
look
at
straightforward
the
number
of
arrangements
different
objects.
Example
How
many
integers
There
2,
are
second
The
be
3,
5

5,
ways
6
of
and
there
three-digit
if
digit
choosing
3
are
each
5
ways

4
numbers
of

the
can
only
first

60
be
be
the
made
used
integer
,
choosing
3
can
4
different
the
once ?
ways
third
using
of
choosing
the
integer
.
three-digit
numbers
that
made.
example
is
permutations
n
4,
integer
Therefore
can
different
(n
1)
an
of

r
illustration
objects
(n
of
from
2)

…
in
factorial
n
(n
a
general
different
r

case:
the
objects
number
of
is
1)
n!
_______
This
can
be
written
notation
n
as
and
is
denoted
by
P
r
(n
i.e.
the
number
of
per mutations
r)!
of
n
n
different
objects
is
P

r
objects
from
n!
________
r
(n
In
the
placed
116
next
on
examples
them.
we
look
at
arrangements

that
r)!
have
conditions
Section
3
Counting,
matrices
and
differential
equations
Example
How
be
many
The
number
Starting
The
So
next
there
When
in
5
n
chairs
on
5
if
for

even
so
right-hand
5
can

3
be

objects
five
a
even
the
end
any
75
5!
are
people,
circular
chair
1,
4
numbers
last
of
of
digit
the
the
can
is
be
made
restricted
number
,
5
different
the
arranged
A,
B,
table,
choices
different
includes
using
given
there
to
are
2,
3
the
integers
2,
3,
4,
5,
6
if
the
digits
can
for
any
have
number
chairs
is
round
a
or
6.
different
digits
three-digit
even
numbers
that
can
ways
five
C,
D
then
for
to
in
and
there
chair
be
a
2,
seated.
arrangements
circle,
E,
are
there
to
is
no
first
one
the
of
same
ways
five
these
five
that
can
be
used.
be
made.
last
on
seating
the
and
each
either
the
five
E
D
1
1
1
are
5
and
2
number
of
the
side.
ways
4
people
2
5
2
3
in
4
C
3
C
4
B
3
B
C
B
1
1
A
can
will
Therefore
of
5
This
person
people
object.
A
sit
shown
arrangements,
times
people
of
times
circular
or
D
clockwise
of
4
digits.
diagrams:
moved
still
be
digits
round
giving
the
Now
be
three-digit
arrangements
example,
number
in
two
are
to
the
different
choices
so
has
from
Circular
For
different
repeated?
5
the
2
5
2
numbered
seating
them
4
table.
3
A
4
E
3
E
D
5!
__
Therefore
there
are

(5
1)!
ways
of
arranging
five
different
objects
in
a
circle.
5
In
general
there
are
(n

1)!
ways
of
ar ranging
n
different
objects
in
a
circle
and
n!
_________
ways
r(n
Now

consider
of
ar ranging
r
objects
from
n
different
objects
in
a
circle.
r)!
the
number
of
arrangements
of
five
different
beads
on
a
circular
ring.
5!
__
The
different
arrangements
include
these
two:
A
5
A
ring
can
be
turned
over
,
so
these
two
arrangements
are
the
A
same.
E
Therefore
the
number
of
different
arrangements
in
a
ring
is
half
of
different
arrangements
in
a
circle.
So
there
are
ways
In
of
arranging
the
five
beads
on
a
D

2

B
E
the
5!
______
number
B
C
C
D
12
5
ring.
general
when
n
different
objects
n!
___
tur ned
over
there
are
are
ar ranged
in
a
ring
that
can
be
1
__

2n
(n

1)!
different
ways
of
doing
this.
2
Exercise 3.2a
1
In
how
word
2
How
the
(a)
many
PAGES
many
digits
the
different
be
5,
6
number
can
the
letters
in
(b)
the
and
is
numbers
7
odd
can
be
made
from
3
In
how
from
if
and
the
number
used
arranged?
three-digit
3,
ways
each
digit
can
be
used
more
many
five
is
even
than
different
different
and
each
digit
can
be
once?
ways
beads
be
can
three
threaded
beads
on
a
ring?
once
117
Section
3
Counting,
matrices
and
differential
equations
Permutations
Consider
word
the
when
number
of
not
all the objects
different
ways
of
arranging
are different
the
letters
in
the
LOOK.
There
are
two
letters
O
in
this
word.
If
we
label
them
as
O
and
O
K
4!
1
the
number
of
different
arrangements
of
the
letters
L
O
1
But
L
this
O
O
1
so
number
and
L
O
2
the
This
K
includes
the
O
2
arrangement
means
that
L
the
two
then
2
O
is
2
arrangements
K
1
O
O
K
appears
twice
in
the
number
of
arrangements
argument
to
the
of
4!
the
number
.
letters
L
O
O
K
is
4!
__

12
2!
Applying
we
have
the
the
two
letters
same
Cs,
C
U
1
But
10!
two
R
1
includes
arranging
the
Therefore
R
1
and
I
C
2
the
two
the
Rs
of
L
U
so
M
in
the
is
the
word
number
CURRICULUM,
of
arrangements
of
10!
3
of
the
letters
Us,
2
ways
and
number
U
2
2!
Rs
three
arranging
3!
ways
of
arrangements
the
two
Cs,
arranging
of
the
the
the
letters
2!
ways
three
in
of
Us.
CURRICULUM
10!
______
is

151 200
2!2!3!
In
general
the
number
of
per mutations
of
n
objects
when
p
are
the
n!
____
same
and
q
are
the
same
is
p!q!
Permutations
together or
T
o
find
when
object,
T
H
This
is
the
the
R
(EE),
total
where
the
Es
find
the
that
of
are
the
are
have to
be
kept
apart
permutations
kept
is
number
Es
some objects
together
,
number
which
means
the
kept
number
two
i.e.
when
of
of
we
the
can
letters
in
consider
permutations
of
the
the
the
word
two
four
THREE
Es
as
one
objects
4!
number
of
of
permutations
permutations
of
the
where
letters
the
minus
two
the
Es
are
apart
number
together
,
5!
__
i.e.
4!

36
2!
Independent
T
wo
on
tasks
the
For
are
independent
execution
example,
permutations
the
letters
followed
letters
of
of
the
number
by
any
when
other
of
four
the
execution
of
one
task
has
no
effect
task.
different
digits
is
number
the
plates
number
of
with
any
two
permutations
of
two
26
the
alphabet,
P
,
and
the
number
of
permutations
of
four
2
10
digits,
P
.
These
two
permutations
have
no
effect
on
each
other
,
so
4
the
permutations
are
independent.
Using
the
fundamental
26
counting,
the
number
of
different
number
plates
is
principle

P
2
Therefore
doing
118
both
when
is
the
two
tasks
product
are
of
independent,
the
number
of
the
ways
of
10
P
number
of
doing
4
of
ways
each
of
task.
Section
Mutually
T
wo
For
tasks
exclusive
are
example,
number
Using
a
the
mutually
it
is
digits
1,
exclusive
either
2,
3,
Counting,
matrices
and
differential
equations
permutations
impossible
number
3
4,
to
make
has
5
when
two
a
they
two -digit
digits
without
cannot
or
it
number
has
repeating
both
a
three
be
executed.
and
a
three-digit
digits,
not
both.
digit,
5!
__
the
number
of
permutations
giving
a
two -digit
number
is

20
3!
5!
__
the
number
of
ways
of
making
a
three-digit
number
is

60
2!
and
these
two
three-digit
number
permutations
numbers
or
a
When
of
so
three-digit
two
are
number
tasks
doing
cover
there
are
either
all
20
the

from
mutually
one
number
task
of
1,

2,
80
3,
the
of
4,
5
of
is
a
sum
of
of
and
two -digit
repeating
number
the
each
numbers
making
without
the
other
doing
two -digit
ways
exclusive,
or
ways
different
60
a
digit.
ways
the
task.
Example
A
number
The
repetition
The
3
in.
So
The
are
5
digits
3

is
letters
3
4
are

and

4
4

3
letters
two

cases
2160
4
9
consists
from
of
the
at
least
letters
A,
three
letters
B,
D,
C,
E,
together
F
and
followed
the
digits
by
are
at
least
selected
one
digit.
without
inclusive.
letter
A.
and
available

5
of

2
Find
the
digits
4
from
the
of
arranging

4
or
letters

9
5
number
4
of
letters
the
remaining
two
permutations

and
digits
8
4320
permutations
and
and
1
the
digit

of
number
is
considered

it
different
and
1
number
plates
possible.
digit:
of
are
letters
the
letters
and
(B,
for
letters.
C,
each
There
independent,
so
D,
of
are
the
E,
F)
these
9

8
number
and
2
there
digits
are
3
from
positions
permutations
of
the
of
permutations
the
of
9
digits.
that
A
can
be
digits.
3
letters
and
digit
of

to
letters
ways
3
5
1
number
1
the
on
repetition
digits
letters
permutations
The
4320
2
characters
without
include
and
there
4
The
five
either
2
digits
of
be
are
2
5
the
must
letters
There
with
chosen
from
must
There

are
letters
There

plate
letters
5
are

4
letters
of

including
permutations
3

mutually
4

9

exclusive,
A
of
(using
one
similar
digit
is
9.
reasoning
Therefore
to
the
the
first
case)
number
of
is
permutations
2160
so
the
number
of
different
number
plates
is
6480
Exercise 3.2b
1
Find
the
number
PROBABILITY
2
(a)
the
Bs
(b)
the
Is
(c)
the
Bs
Three
three
In
3
A
In
are
many
code
how
is
together
the
letters
in
the
word
choice
ways
made
and
the
Is
questions
are
apart.
each
have
one
correct
answer
and
answers.
correct
many
of
together
multiple
one
arrangements
which
apart
incorrect
how
least
are
are
of
in
can
from
of
these
questions
be
answered
so
that
there
is
at
5,
6.
answer?
three
these
digits
codes
are
selected
the
from
digits
in
the
digits
ascending
1,
2,
order
3,
of
4,
size?
119
3.3
Combinations
Learning outcomes
Combinations
We

To
define

To find

To
a
have
shelf
a
seen
that
the
number
of
different
arrangements
of
4
books
on
a
combination
variety
of
A
distinguish
is
between
and
but
there
is
only
one
set,
or
combination,
of
books.
combination
a
is
a
group
of
objects
when
the
order
of
the
objects
in
the
a
group
permutation
4!,
combinations
does
not
matter
.
combination
Suppose
from
8
we
want
different
to
find
how
many
groups
of
5
books
can
be
selected
books.
8
There
You need to know
How
to find

What
a
P
different
arrangements
of
5
books
selected
from
the
5
8

are
of
permutation
independent
books,
the
5
but
this
books
number
selected
of
arrangements
among
the
5!
arrangements
themselves.
different
permutations
includes
permutations
are

What
mutually
permutations
exclusive
are
A
B
C
D
E
B
D
E
C
A
n

The
meaning
of
C
r
same
Therefore
8
different
the
number
books
of
combination
different
combinations
of
5
books
selected
from
is
8
P
8!
_________
5
___

5!
5!(8
5)!
8!
_________
8
Now

C
so
we
can
denote
the
number
of
combinations
of
5
5!(8
5)!
8
objects
chosen
from
8
different
objects
by
C
5
The
same
the
argument
number
of
applies
to
different
the
general
combinations
n
different
given
by
of
r
objects
selected
from
objects
n!
_________
n
is
case:
C

r
r!(n
For
example,
group
of
10
the
number
people
of
different
ways
of

r)!
selecting
8
people
from
is
10!
___________
10
C


45
8
8!(10
8)!
Example
In
how
There
many
must
ways
be
4
can
a
set
students
of
in
8
students
each
be
divided
into
two
equal
groups?
group.
8
The
number
of
ways
of
selecting
4
students
from
8
is
C
and
this
leaves
the
remaining
4
other
group.
Labelling
120
the
students
A,
B,
C,
D,
E,
F
,
G,
H,
one
selection
is
the
group
(A,
B,
C,
D).
students
as
the
a
Section
This
gives
(A,
B,
C,
D)
and
(E,
F
,
G,
H)
as
the
two
groups.
But
3
Counting,
(E,
F
,
G,
H)
matrices
is
one
of
and
the
8
in
the
differential
selections
equations
included
8
C
selections
and
this
gives
(E,
F
,
G,
H)
and
(A,
B,
C,
D)
as
the
two
groups.
So
C
4
number
gives
twice
the
4
of
divisions
Therefore
the
into
number
two
of
equal
ways
the
groups.
students
can
be
divided
into
two
equal
groups
is
8
C
8!
___________
4
____


2
2

4!

35
4!
Exercise 3.3a
1
A
box
holds
green,
a
black
large
and
number
brown
of
balls.
red,
How
blue,
four
balls
can
be
made
they
are
all
different
Distinguishing
Problems
Y
ou
need
whether
For
of
or
not
not
a
an
the
the
the
hand
of
different
pack
cards
is
a
numbers
permutation
as
of
number
ordinary
the
permutations
include
problem
order
of
of
and
the
be
the
same
colour?
In
how
many
two
ways
groups
of
can
6
10
children
children
and
4
be
divided
children?
in
use
hands
of
is
order
a
the
combinations
or
combination.
context
to
decide
matters.
cards
made
digits
and
permutation
and
selection
playing
can
of
words
different
52
than
the
carefully
any
group
order
are
colours
between
usually
read
example,
from
as
do
to
only
if
into
(a)
balls
selections
2
of
two
(b)
yellow,
many
cards
a
does
from
that
number
4
not
of
number
5
of
can
matter
.
dealt
The
different
does
be
combinations
number
digits
is
a
matter
.
Example
How
if
many
either
Either
different
Martha
or
Martha
is
ways
Sergio
can
but
selected
4
students
not
and
both
Sergio
be
selected
must
is
be
not,
from
10
students
selected?
which
leaves
3
students
8
to
be
selected
from
the
remaining
8
students,
giving
C
3
or
different
groups,
Sergio
selected
3
is
students
to
be
and
Martha
selected
is
from
not,
the
which
again
remaining
8
leaves
students,
8
giving
C
different
groups.
3
These
are
mutually
exclusive
combinations
so
the
number
of
different
8
groups
is
2
C

112
3
Exercise 3.3b
1
Find
the
placed
number
in
a
line
of
so
ways
that
in
which
Alice,
10
Grace
girls
and
can
Maria
be
B
are
separated.
2
The
diagram
horizontal
Starting
each
at
shows
a
grid
of
8
vertical
lines
and
7
lines.
A,
and
intersection,
either
how
moving
many
left
routes
or
are
up
at
there
to
A
get
to
B?
121
3.4
Sample
Learning outcomes
spaces
Sample
When

To
define
a
sample

To
introduce
a
different
sample
perform
a
task,
one
of
the
items
that
results
from
the
task
is
space
called
drawing
we
spaces
ways
an
outcome
of
For
space
D,
example,
if
the
task
is
one
possible
outcome
the
possible
outcomes
All
choosing
is
two
letters
from
the
letters
A,
B,
C,
AB.
of
a
task
is
called
a
sample
space
You need to know
Tables

How
to
draw Venn
diagrams
When
good
For
are
a
task
example,
chosen
The
side
This
dice
table
is
are
B,
all
also
but
this
all
and
drawing
the
the
possible
possible
up
a
outcomes
A
B
C
D
A
AA
AB
AC
AD
B
BA
BB
BC
BD
C
CA
CB
CC
CD
D
DA
DB
DC
DD
letter
are
the
shows,
a
two -way
outcomes
score
when
listed
table
when
for
along
can
a
be
coin
example,
greater
the
than
top
filled
is
and
in
tossed
that
there
4
5
6
H
H1
H2
H3
H4
H5
H6
T
T1
T2
T3
T4
T5
T6
a
Mild
a
sample
occurs
shows
in
a
letters
the
drug
space
several
for
a
left-
six-sided
two
outcomes
contains
only
a
few
different
times.
outcomes
trial
a
the
outcomes.
4.
3
table
two
the
and
are
2
when
is
listed.
down
with
1
outcome
table
are
D.
outcomes
occurred
improvement
items,
all
and
Then
the
useful
each
outcome
of
table
head
two
that
lists
C
table.
The
a
example,
each
table
A,
choices
the
gives
in
outcomes
Big
of
just
ensuring
this
rolled.
resulting
T
ables
for
from
possible
hand
For
involves
method
and
new
improvement
the
number
treatment
No
of
for
times
migraine.
improvement
Gender
No
Some
No
Some
No
Some
side-effects
side-effects
side-effects
side-effects
side-effects
side-effects
Male
25
3
65
2
38
1
Female
21
5
97
1
14
0
From
a
122
big
this
we
can
read,
improvement
is
for
54
example,
in
total.
that
the
number
of
outcomes
giving
Section
3
Counting,
matrices
and
differential
equations
Tree diagrams
T
ables
task
are
not
suitable
involving
letters,
more
chosen
when
than
from
the
there
two
are
items,
letters
A,
B,
several
for
C,
different
example
D,
are
outcomes
when
three
arranged
in
a
for
a
different
line.
We
4
know
that
there
are
P

24
different
outcomes,
but
not
what
these
3
outcomes
are.
We
Start
by
can
drawing
branches
4
to
different
for
each
the
these
4
show
the
each
end
3
of
the
second
of
drawing
a
At
for
the
the
different
the
the
diagram :
the
list
end
branch
along
choices for
letter
tree
Repeat
2
different
choices
on
branch
by
branch,
repeat for
letter,
letter
outcomes
At
the
choices
the first
writing
find
the
the
of
read
path
to
outcome
third
letter
C
ABC
ABD
D
B
B
ACB
C
ACD
D
B
D
ADB
ADC
C
C
BAC
A
BAD
D
A
A
BCA
C
BCD
D
B
A
D
BDA
BDC
C
B
CAB
CAD
D
A
C
A
CBA
B
CBD
D
A
D
CDA
D
CDB
B
B
DAB
DAC
C
A
A
DBA
B
DBC
C
A
C
DCA
DCB
B
The
list
at
the
right-hand
end
of
the
diagram
is
the
sample
space
for
this
task.
Venn diagrams
When
a
task
diagram
to
Consider
,
entered
for
for
involves
illustrate
overlapping
and
example,
CAPE
find
this
the
outcomes,
numbers
information
for
can
the
50
sometimes
different
students
use
a
V
enn
outcomes.
from
a
college
examinations:

10
students
entered
for
physics,

12
students
entered
for
chemistry,

5
students
we
of
entered
for
both
P
C
physics
and
chemistry.
123
Section
3
Counting,
matrices
and
differential
This
but
for
equations
does
not
for
We
enter
in
the
the
physics
the
5,
numbers
or
and
this
so
the
on
numbers
which
of
students
vice-versa.
chemistry,
represent
represent
7
and
for
can
give
chemistry,
physics
entered
we
not
there
number
a
V
enn
entered
leaves
5
non- overlapping
It
is
in
entered
both
of
tell
for
for
and
how
for
C.
circle
physics
many
entered
the
numbers
subjects.
overlapping
non- overlapping
the
for
between
both
using
P
entered
us
overlap
diagram
the
part
does
an
who
In
the
part
of
circles
to
overlap
the
circle
region,
P
and
C.
5
33
Therefore
leaving
box
17
33
students
who
outside
were
the
were
entered
entered
for
for
physics
neither
or
subject.
chemistry
This
or
number
both,
goes
in
the
circles.
Example
Of
the
100
45
were
entered
for
mathematics
25
were
entered
for
geography
18
were
entered
for
mathematics
36
were
entered
for
none
(a)
Draw
(b)
Find
and
M
V
enn
the
needs
G
and
for
E
a
for
all
V
enn
a
school
of
of
of
to
just
entered
students
and
CSEC
the
G
examinations:
geography
were
entered
for
economics
15
were
entered
for
mathematics
15
were
entered
for
economics
and
and
economics
geography
subjects.
this
information.
who
(ii)
subjects
with
label
M
and
show
these
for
30
these
to
diagram
geography
overlap,
in
diagram
number
entered
(i)
This
a
students
three
entered
overlapping
circles
containing
overlap,
just
E
12
and

for
circles.
the
G
x
mathematics
We
will
numbers
overlap
15
use
for
and
but
M
each
where
for
not
mathematics,
subject.
all
economics
three
We
need
nor
E
geography.
for
economics
regions
where
just
overlap.
x
x
x
18
x
15
x
x
8
36
(a)
We
do
three
not
Looking
region
G
124
know
circles
at
many
students
entered
for
all
three
subjects
so
we
put
x
in
the
region
where
all
overlap.
the
where
overlap.
how
G
numbers
M
and
G
for
mathematics
overlap
already
and
geography,
contains
x,
so
we
that
know
leaves
that
18
18
x
students
for
the
entered
region
both.
where
just
The
M
and
Section
Now
who
We
M
know
does
The
(b)
looking
enter
sum
45
Reading
in
of
are
Therefore
(ii)
numbers
subjects
is
overlap
numbers
There
the
three
that
not
The
(i)
at
all
the
the
100
12
total
either
the
of
the
15
x
number
numbers
students
mathematics
the
remaining
students
from
for
leaves
all
the
total,
were
diagram,
can
so
be
economics,
region
that
15
where
leaves
filled
regions
therefore
entered
24
M,
and
the
Counting,
enter
just
(45
M
matrices
for
and
(33
x))
both.
E
and
differential
T
aking
out
the
equations
number
overlap.

12

x
in
the
region
where
circles.
regions
in
in
in
other
for
3
for
students
is
88
all
in
88


x
three
entered
using
for
reasoning,
x

of
similar
100
⇒
these
x

12
subjects.
mathematics
but
not
economics
nor
geography.
Exercise 3.4
1
2
An
ordinary
(a)
How
(b)
Draw
The
many
table
Age of
six-sided
a
different
table
shows
the
4
the
outcomes
35
15
Under
4160
5
8
56
How
many
people
up
different
(b)
How
(a)
(b)
the
coins
tree
many
100
are
to
tossed
diagram
to
outcomes
customers
at
a
bought
sweet
28
bought
tomatoes
56
bought
carrots
15
bought
sweet
12
bought
tomatoes
16
bought
sweet
12
did
Draw
a
V
enn
buy
over
age
at
show
result
37
not
the
of
the
all
in
market
10
60
rolled.
the
age
of
cars
owned.
over
60
1825
2640
4160
55
60
32
over
60
65
own
60
a
old?
car
that
is
10
years
old
or
less?
time.
outcomes.
least
stall
over
20
years
same
the
at
both
5 years
2640
(b)
into
30
1825
car
are
5 10 years
40
a
dice
10 years
4160
owned
a
investigation
2640
people
blue
there?
1825
many
Draw
an
six-sided
space.
Over
How
Of
of
ordinary
are
sample
car
(a)
an
outcomes
(a)
Four
and
car
Age of owner
3
dice
showing
Age of owner
Age of
red
two
selling
heads?
vegetables:
potatoes
potatoes
and
diagram
tomatoes
carrots
potatoes
sweet
and
and
carrots
potatoes,
to
show
tomatoes
this
or
carrots.
information.
(i)
How
many
people
bought
sweet
(ii)
How
many
people
bought
just
potatoes,
tomatoes
and
carrots?
carrots?
125
3.5
Basic
probability
Learning outcomes
Terminology
Probability

To
introduce
the
i.e.
used

To
in
gives
a
measure
for
how
likely
it
is
that
an
event
will
happen,
terminology
probability
gives
a
measure
of
predictability.
probability
define
and
use
Up
basic
to
we
probability
C
now
call
and
An
we
tasks
D
is
event
example,
have
called
is
talked
about
experiments .
an
the
an
For
tasks,
but
example,
in
the
context
choosing
three
of
probability
letters
from
A,
B,
experiment.
outcome
outcome
or
a
ABC
group
is
an
of
outcomes
event
when
from
an
choosing
experiment.
three
letters
For
from
You need to know
A,
B,
C
letter

How
to find
permutations
and
D.
An
event
can
also
be
all
the
outcomes
containing
the
A.
and
combinations
When

How
to
read
tables
and Venn
of
diagrams
one
random
any
letter
when
is
the
other
letter
.
coins
or
selected
from
selection
In
this
of
A,
any
case,
we
B,
C
and
one
letter
can
say
D,
is
then
as
that
the
likely
each
selection
as
the
outcome
is
selection
is
equally
likely

What
outcomes
spaces
mean
and
sample
When
fair
up
or
dice
unbiased
and
if
the
if
dice
are
the
are
involved
coins
in
are
equally
experiments,
equally
likely
to
likely
show
they
to
any
are
land
one
described
heads
of
the
up
or
as
tails
possible
scores.
Definition of
When
an
all
event
the
A
probability
outcomes
is
likely
to
the
of
an
experiment
happen
number
is
denoted
of
equally
are
by
known
P( A)
likely
the
and
is
probability
given
outcomes
that
by
giving
A
____________________________________________
P(A)

the
Depending
can
the
be
any
on
the
nature
number
denominator
total
from
(all
of
number
the
zero
equally
event
(no
When
P(A)
certain

to
example,
are
likely.
T
o
the
find
6’
2
of
an
probability
number
is
event
given
when
equally
the
the
A
likely
of
the
equally
equally
as
ordinary
that
the
likely
likely
P(5
or
6)

126
numerator
giving
outcomes
0

outcomes
A)
giving
P(A)
and

of
to
this
the
fraction
number
in
A).
1
when
P( A)
P(score
is
7)

or
fair
decimals
dice
score
is
will
outcomes
is
or

1,
the
event
rolled,
be
6
percentages.
each
greater
and
the
of
than
event
outcomes.
1
__

6
Also
the
impossible
fractions
2
__
Therefore
is
A,
likely
happen.
Probabilities
For
0,
equally
outcomes
Therefore
is
of
0
3
and
P(score
is
6
or
less)

1
the
4,
‘a
6
we
scores
know
score
of
is
that
5
or
Section
3
Counting,
matrices
and
differential
equations
Example
The
table
gives
a
breakdown
of
car
theft
in
an
Age of
Cost of
a
than
$10 000
$10 000$30 000
More
Find
than
the
$30 000
probability
(a)
of
a
car
up
(b)
of
a
car
older
(a)
There
are
Thefts
stolen
for
car
the
year
2012.
in years
replacement
Less than
Less
island
of
to
3
that
years
than
377
cars
a
one
to
13
Older than
22
30
60
78
56
84
14
25
8
randomly
selected
theft
3
was
old
thefts
up
1
year
listed
3
and
in
years
costing
the
old
table
are
$10 000
and
listed
any
in
or
more
one
the
of
first
to
replace?
these
two
is
equally
columns:
likely
there
to
225
be
of
selected.
these.
225
____
∴
P(theft
was
a
car
up
to
3
years
old)


0.597
(3 s.f.)
377
(b)
Let
B
be
These
the
are
event
listed
‘thefts
in
the
of
cars
lower
older
two
than
one
right-hand
year
and
columns
costing
and
rows
$10 000
of
the
or
more
to
replace’.
table:
there
are
173.
173
____
∴
P(B)


0.459
(3 s.f.)
377
Example
T
wo
and
cards
5
are
green
drawn
cards.
at
random
Find
the
from
a
probability
pack
that
of
20
both
cards
cards
containing
are
5
red
cards,
5
blue
cards,
5
yellow
cards
red.
5
The
number
of
combinations
of
two
red
cards
is
20
C
and
the
number
of
combinations
of
any
two
cards
is
C
2
Therefore
the
probability
that
two
red
cards
are
drawn
2
is
5
C
5  4
________
2
____


0.0526
(3 s.f.)
20
C
20

19
2
Exercise 3.5
1
T
wo
1,
digits
2,
3,
What
4,
is
are
5,
the
selected
6,
7
to
at
make
probability
random
a
that
from
two -digit
this
the
digits
number
.
3
A
box
mixed
is
(b)
contains
even
two
odd
colours
the
letters
tree
different
A,
B,
C,
diagram
probability
letters,
D,
in
that
are
T
opic
the
a
different
white
patterned
tiles
of
include
blue,
include
red
65
53
and
tiles
background.
tiles
blue,
include
include
18
red,
39
yellow,
tiles
tiles
20
include
tiles
red
and
digits?
tiles
Three
on
pattern,
yellow,
2
200
number
In
(a)
contains
chosen
at
arranged
3.4
or
letters
random
in
a
line.
otherwise,
A
and
B
are
from
next
the
(a)
include
include
all
three
blue
and
yellow
and
10
colours.
Draw
a
V
enn
diagram
to
show
this
information.
the
to
tiles
the
Using
find
25
each
(b)
One
tile
is
selected
at
random.
Find
the
other
.
probability
(i)
only
(ii)
red
that
the
pattern
on
it
contains
red
and
blue
but
not
yellow.
127
3.6
Probability
Learning outcomes
that
The
If

Finding
event
the
does
probability
not
that
A
an
event
probability that
is
an
event,
then
When
an
ordinary
the
an
event
does
not
event does
‘not
A’
is
happen
not
denoted
by
happen
A.
an
happen
S

{1,
2,
3,
S
contains
4,
6
5,
six-sided
dice
is
rolled,
the
sample
space
is
the
set
6}.
elements
so
n(S)

6
You need to know
This

Simple
set
What

How

The
a
space
contains
every
possible
outcome
so
it
is
exhaustive
notation
If

sample
sample
space
is
the
dice
event
of
is
unbiased,
scoring
1
or
the
outcomes
are
all
equally
likely,
so
if
A
is
the
2,
2
__
to
use
basic
probability
then
the
number
of
ways
in
which
A
can
occur
is
2,
so
P( A)

6
meaning
of
mutually
Now
the
number
of
ways
in
which
A
cannot
occur
is
6
2,
exclusive
6
2
______
Therefore
P( A)

In
general
is
x
and
if
the
the
6
__
2
__
6
6


6
number
sample
of
space
equally
is
cannot
n(S)
x
P(A)

n(S)


n(S)
example,
the
is
ways
an
number
n(S)

event
of
A
ways
can
in
happen
which
1
P(A)
probability
P(A)
that

an
1

P(A)
unbiased
dice
shows
6
when
6
is
rolled
1
is
therefore
the
probability
that
the
dice
does
not
show
1

6
6
and
if
the
probability
probability
In
is
5
1
,
simple
chosen
that
it
cases,
at
that
will
P( A)
random
it
not
can
from
A
x,
n(S)
i.e.
For
the
x
____
____

P(A)
likely
then
happen
n(S)
________
∴
S,
1
will
rain
be
rain
tomorrow
found
the
tomorrow
is
directly,
letters
A,
B,
is
67%,
100%
for
C,
then
67%
example
D,
the

6
the
33%
when
one
probability
letter
that
it
is
3
not
the
letter
D
is
4
In
other
cases
it
may
be
easier
to
find
P( A)
first.
Example
A
two -digit
integers
chosen
not
It
a
is
number
,
random
more
than
multiple
easier
many
A
at
are
to
of
greater
from
once.
than
the
What
is
0
is
to
the
9
made
by
probability
that
find
how
many
numbers
are
multiples
is
divisible
by
5
if
it
ends
in
0
or
number
of
permutations
of
two
digits
ending
The
number
of
permutations
of
two
digits
is
A
is
digit
the
two
can
be
number
is
of
5
than
0
or
how
5.
The
if
A
not.
number
So
choosing
inclusive.
5?
A
‘the
integer
is
9  2
_______
P(A)

a

10
number
multiple
cannot
of
9
start

in
is
with
0
but
it
∴
P(A)
9
can
5’
1
__
,
5
5

1
P(A)

2
10
1
__

9
128
zero,
digits

1
4
__

5
5
end
with
0
Section
3
Counting,
matrices
and
differential
equations
Example
This
who
V
enn
were
entered
One
of
diagram
entered
for
T
opic
CSEC
mathematics,
these
probability
(from
for
100
that
mathematics,
shows,
economics
students
they
3.4)
examinations,
were
economics
is
chosen
entered
or
and
at
for
out
the
of
100
students
who
were
geography.
random.
at
the
numbers
least
What
one
of
is
the
the
subjects
geography?
24
12
3
12
4
36
It
is
at
easier
least
is
to
one
find
of
the
the
G
probability
subjects.
The
that
a
V
enn
student
diagram
was
not
shows
entered
that
this
for
number
36.
T
aking
A
subjects
student
or
as
the
event
‘a
mathematics,
is
not
entered
student
is
economics
for
any
of
entered
or
the
for
at
geography ’,
subjects
least
one
A
the
is
of
the
event
mathematics,
‘a
economics
geography ’.
36
____
Then
P(A)


0.36
100
Using
∴
P(A)

1
P(A)
P(A)

0.64
gives
0.36

1
P(A)
Exercise 3.6
1
T
wo
pens
What
2
This
new
is
are
the
table
chosen
at
probability
from
treatment
T
opic
for
random
that
3.4
at
from
least
shows
a
box
one
the
containing
pen
is
outcomes
6
red,
4
blue
and
8
black
pens.
blue?
for
272
people
taking
part
in
a
drug
trial
for
a
migraine.
Big
improvement
Mild
improvement
No
improvement
Gender
No
Some
No
Some
No
Some
side-effects
side-effects
side-effects
side-effects
side-effects
side-effects
Male
25
3
65
2
38
1
Female
21
5
97
1
14
0
Find
the
probability
that
one
person,
chosen
at
random,
had
no
side-effects.
129
3.7
Probabilities
Learning outcomes
Mutually
T
wo
To find

the
involving
probability
events
two
exclusive
are
mutually
events
events
exclusive
when
they
cannot
both
occur
that
simultaneously.
events
A
and
B
will
both
occur
For
To find

either
B
will
the
probability
event
A
will
example,
occur
or
mutually
exclusive
However
,
choosing
If
A
number
and
occur
You need to know
B
and

The
to
use
set
meaning
do
notation
of
union
not
are
are
the
common.
How
an
even
because
a
number
number
and
choosing
cannot
be
both
an
odd
even
number
and
are
odd.
event
occur
even

choosing
that
a
number
not
mutually
set
They
of
overlap.
be
The
in
is
S
is
multiple
B
the
can
in
a
sample
of
because
events,
which
represented
set
a
exclusive
exclusive
ways
can
that
mutually
the
set
occur
V
enn
3
6,
of
and
for
choosing
example,
ways
will
in
have
diagram
as
any
is
which
no
both.
A
can
members
two
circles
in
that
space.
and
S
intersection

Basic

About
of
sets
A
B
probability
independent
permutations
and
combinations
If
there
occur
are
in
p
n
equally
ways
and
likely
B
can
outcomes
occur
in
q
in
the
sample
p

q
p
______
then
the
probability
of
A
or
B
occurring
probability
that
Therefore
A
or
B
when
will
A
occur
and
P(A

B
B)
is
are

of
which
A
can

mutually

__


n
denoted
P(A)
q
__
is
n
The
space
ways,
by
P(A)

P(B)
n
P( A

exclusive
B).
events
P(B)
Example
3
1
The
probability
that
a
girl
walks
to
school
is
and
the
probability
that
she
takes
a
bus
to
school
is
What
is
P(she
goes
W
alking
the
probability
to
and
school
taking
by
a
that
she
another
bus
are
goes
does
walk
or
take
a
mutually
bus)
school
method)


4
by

P(she

1
another
does
P(she
exclusive,
not
P(she
goes
to
school
by
another
nor
walk
or
take
take
a
a
bus)
bus)
therefore

5
20
method)

3
__
1

20
20
Independent
T
wo
has
For
are
events
no
are
effect
example,
The
number
The
number
a
head
events
independent
on
whether
rolling
independent
show
130
walk
does
17
__
∴
method?
17
__
3
1
P(she
to
.
5
4
of
of
is
an
or
when
not
the
ordinary
one
fair
whether
event
six-sided
or
not
it
occurs,
occurs.
dice
and
tossing
a
fair
coin
experiments.
ways
the
dice
can
land
ways
the
dice
can
show
1
event,
other

1
and
the
five
coin
can
uppermost
land
and
is
the
6

coin
2
can
Section
1  1
______
Therefore
P(5
and
H)
6
If
there
in
p
are
n
1
__
outcomes

in

2
an
Counting,
matrices
and
differential
equations
1
__


3

6
P(5)

P(H)
2
experiment
in
which
an
event
A
can
occur
ways,
p
__
then
P(A)

n
If
there
are
m
outcomes
in
an
independent
experiment
in
which
event
B
q
___
can
occur
in
q
ways,
then
P( B)

m
As
the
experiments
There
are
A
B
and
n

can
m
are
independent,
outcomes
occur
in
p
p

and
q
P(A
and
B)
P(A
and
B)
is

m
denoted
Therefore
first
and
second

n
by
are
independent.
experiment
in
which
___


B
q
__

n
the
and
ways,
p
_______
so
for
q
A
P(A)

P(B)
m
P( A

when
A
P(A
B).
and

B)
B

are
independent
P(A)

events
P(B)
Example
Three
is
ordinary
twice
as
six-sided
likely
The
equally
dice
shows
to
likely
dice
show
6
are
as
outcomes
rolled
any
together
.
other
from
the
T
wo
number
.
biased
of
Find
dice
are
the
the
1,
dice
2,
3,
2
6
is
6
,
6
a
are
and
three
∴
way
P(6
The
probability
dice,
so
we
in

which
6

6
b
one
)

of
P(6
c
that
Probability that
mutually
Suppose
6
the
multiple
and
We
12
can
one
and
multiple
is
6.
Therefore
dice
dice
the
is
will
biased
show
probability
so
that
it
6.
a
fair
dice
shows
6
that
this
is
the
fair
dice
a
and
b
and
the
biased
dice
c
and
events
of
showing
a
6
as,
)
we
of
event
of
are
dice

P(6
)

has
P(6
b
1

6
or
effect
on
the
way
the
others
land,
so
the
events
are
independent.
)
c

252
7
B
no
___
2
2

A
lands
occurs
when
A
and
B
are
not
exclusive
that
inclusive
A
call
a
6
a
will
the
1
If
6,
third
three
6

a
5,
the
all
c
a
or
4,
but
that
6
b
The
fair
1
.
7
There
are
probability
3’
number
want
to
is
chosen
find
the
at
random
probability
from
that
the
the
integers
number
is
1
to
12
even
3.
‘choosing
then
both
illustrate
A
even
the
an
and
B
and
even
are
number ’
not
multiples
sample
space
in
and
mutually
of
a
is
the
event
‘choosing
because
3.
V
enn
A
1
B
exclusive
diagram.
B
3
4
6
2
9
5
12
8
10
11
7
A
and
B
131
Section
3
Counting,
matrices
and
differential
This
A

equations
diagram
B,
is
because
not
this
shows
that
(number
includes
the
of
number
elements
the
of
in
elements
A)
6,
elements

12,
in
(number
i.e.
in
A
either
of

A
or
B,
elements
B,
i.e.
in
in
B)
twice.
6
___
The
number
of
ways
in
which
A
can
occur
is
6
so
P( A)

12
4
___
The
number
of
ways
in
which
B
can
occur
is
4
so
P( B)

12
2
___
The
number
of
ways
in
which
A
and
The
number
of
ways
in
which
A
or
B
can
occur
is
2,
so
P( A

B)

12
6  4
2
__________
so
P(A

B)
6
___


any
(the

two
events
number
(number
can
This
ways
can
is
of
of
ways
ways
which
occur)
12
and
in
in
occur)
because
in
A
which
(number
can
when
P(A
that
described
part
of
A
of
4
___
2
___
12
12
A
A

are
or
not
B
can
of
in
can
in
8
P(A)


6

4
2,
P(B)
P(A

B)
exclusive:
occur)

(number
which
which
includes
is
mutually
occur)
ways
ways
occur)
occur
A
A
can
(number
of
and
of
B
occur)
ways
ways
can

in
in
which
the
by
set
the
that
of
set
A

and
B)
B

elements
{ A
excludes
{A
(number
which
A
of
and
∩

{A
B}

are
P(A)
in
and
not

the
mutually
P(B)
part

of
similarly
exclusive
P(A
B
for

for
table
from
T
opic
3.4
shows
the
outcomes
that
the
events
excludes
set
of
{A

elements
B}
in
B}
{A
∩
B
A
∩
for
272
people
taking
part
in
a
drug
trial
for
B}
a
new
improvement
Mild
improvement
No
treatment
improvement
Gender
No
Some
No
Some
No
Some
side-effects
side-effects
side-effects
side-effects
side-effects
side-effects
Male
25
3
65
2
38
1
Female
21
5
97
1
14
0
Find
132
the
probability
that
one
person,
chosen
at
random,
is
male
or
has
side-effects.
is
the
B}
migraine.
Big
B
B)
Example
This
B
occur).
twice.
Therefore
Notice
that
which
(number
B
B
can

12
For
B
Section
Being
male
and
P(male
The
the
number
number
having
or
of
of
side-effects
side-effects)
{males}
{male
is
and

are
not
P(male)
134,
the

mutually
P(side-effects)
number
side-effects}
is
of
P(male
or
no
side-effects)
Counting,
is
matrices
and
differential
equations
Therefore
P(male
{side-effects}
and
side-effects)
12,
6.
134
____
Therefore
exclusive.
3

12
____
6
____
272
272

272
140
____


0.515
(3 s.f.)
272
Example
T
wo
fair
least
At
least
P(at
normal
one
6
or
one
least

at
6
dice
5.
and
one
P(at
six-sided
least
6
or
least
one
are
at
least
one
5
at
least
one
5)
one
6)

P(at
rolled
are
together
.
not
least
Find
mutually
one
5)
least
one
6)

1
P(no
6)

P(at
least
one
5)

1
P(no
5)

one
6
and
at
least
one
5)
least
one
6
or
at
least
one
at
least
one
5)
4
___

5)
and
36

36
22
4
_______
P(at
6

36
∴
therefore
11
___
1
2  2
______
least
at
36
36
P(at
rolling

5  5
______
least
one
of
11
___
1
36
P(at
probability
exclusive,
5  5
______
P(at
the

1
__

36
2
Exercise 3.7
1
T
wo
boys,
sided
A
and
B
play
a
game
that
involves
rolling
The
first
that
B
(Hint:
person
wins
For
on
B
to
to
roll
his
a
first
A
ordinary
six-
win
on
six
wins.
A
goes
first.
Find
and
B
are
the
probability
turn.
his
first
turn,
A
must
lose
on
_
2
2
an
dice.
independent
events.
P( A)

his
first
turn.)
_
1
,
P(B)

and
5
4
__
1
P(A

B)

10
Find
(a)
P(A

B)
P(A
(b)

B)
3
_
3
A
tennis
player
A
has
a
probability
of
of
winning
a
set
against
player
5
B.
The
Find
4
A
the
and
P(A
first

B
to
probability
are
B)
Explain
player

two
win
that
events
2
sets
when
such
out
A
that
of
plays
P( A)
3
wins
B,

A
the
wins
0.3,
match.
the
P(B)

match.
0.5
and
0.6
why
A
and
B
are
neither
mutually
exclusive
nor
independent.
133
3.8
Probabilities
Learning outcomes
Events that
If

To
use
tree
diagrams
to
a
involving
two
Finding
the
probability
is
removed
not
at
or
more
events
independent
random
from
a
pack
of
4
cards
numbered
1,
2,
3
and
not
put
back,
and
then
a
second
card
is
chosen,
the
options
for
events
the

card
are
two
solve
2,
problems
involving
that
number
on
the
second
card
depend
on
which
number
was
removed
an
fi rst.
event
occurs
event
has
given
that
another
Therefore
already
these
two
events
are
not
independent
as
the
ways
in
which
the
occurred
second
event
number
The
was
can
occur
removed
number
of
have
been
reduced
by
one
and
depend
on
which
first.
equally
likely
outcomes
is
therefore
4

3

12
You need to know
We

How
to
use
basic
on

How
to
draw

How
to find
a
can
show
the
probabilities
tree
a
tree
each
card
and
the
different
outcomes
diagram:
diagram
First
mutually
for
probability
probabilities
exclusive
card
Second
card
of
events
2
P(2)

P(3)
(1,
2)
1
(1,
3)
(2,
1)
(2,
2)
(2,
3)
3
(3,
1)
2
(3,
2)

3
1
P(1)
3

4
1
P(1)

3
1
P(2)

2
1
P(2)

3
1
P(3)

3
1
P(3)

1
4
P(1)

P(2)

3
Notice
which
that
are
outcomes
For
this
(there
example,
second
tree
mutually
card
are
there
is
2,
diagram
shows
exclusive,
the
it
seven
does
not
two
(1,
2
ways
)
and
in
(1,
1
which
2
P(1st
card
is
1
the
the
tree
and
2nd
card
is
2)

card
equally
likely
first
card
is
1
and
the
1

6
diagram,
1
P(1st
outcomes,
the
2
12
Using
all
)
__
2
so
different
show
12).
are
i.e.
but
is
1)

P(2nd
card
is
2)

2

4
1

3
6
Similarly,
1
P(1st
card
is
2
and
2nd
card
is
1)


6
134
P(1st
card
is
2)

P(2nd
card
is
1)
Section
So,
to
find
multiply
There
(1,
2)
and
the
are
and
These
are
the
the
probability
of
probabilities
two
(2,
outcomes
an
on
outcome
the
where
path
the
shown
leading
two
cards
on
to
a
that
are
1
tree
diagram,
and
2
in
any
mutually
is
2,
exclusive,
we
add
so
the
to
find
the
probabilities
probability
of
P(the
cards
removed
find
the
probability
add
the
probabilities
T
ree
diagrams
This
diagram
which
are
and
differential
equations
we
order
,
i.e.
that
one
are
numbered
can
be
shows
fair
and
Biased
of
two
of
or
each
extended
the
one
more
to
biased
coin
Fair
is
1
1
and
2)

1
outcomes
shown
1

on

3
6
a
tree
diagram,
outcome.
cover
probabilities
is
card
each.
6
we
matrices
outcome.
1
T
o
Counting,
1).
other
Therefore
3
so
more
when
that
a
than
three
head
is
coin
two
events.
coins
twice
Fair
are
as
tossed,
likely
as
two
a
of
tail.
coin
1
P(H)

2
(H,
H,
H)
1
(H,
H,
T)
2
(H,
T
,
H)
1
(H,
T
,
T)
1
P(H)

2
P(T)

2
2
P(H)
1

P(H)
3

1
P(T)

2
P(T)

2
1
P(H)

2
(T
,
H,
H)
1
(T
,
H,
T)
2
(T
,
T
,
H)
1
(T
,
T
,
T)
1
P(H)

2
1
P(T)
P(T)


2
3
1
P(H)

1
P(T)

2
P(T)

2
T
o
find
the
probability
probabilities
For
and
along
example,
the
to
other
the
find
two
of
the
coins
_
2
giving
P(H,
T
,
T)

of
find
the
of

that
we
outcomes,
we
multiply
the
outcome.
the
biased
follow
the
coin
path
shows
leading
a
to
head,
(H,
T
,
T)
_
1

2
of
tails,
_
1
the
that
probability
show

probability
one
giving
_
1
3
T
o
any
path
2
more
6
than
one
outcome
we
add
the
probabilities
each.
For
example,
heads
and
a
to
find
tail,
we
the
add
probability
the
that
the
probabilities
of
three
the
_
1
(H,
H,
T),
(H,
T
,
H)
and
(T
,
H,
H)
giving
_
1

6
coins
__
1

6
land
showing
two
events
5
__

12
12
135
Section
3
Counting,
matrices
and
differential
There
only
equations
are
problems
some
simplified
of
the
tree
involving
possible
two
events
outcomes.
In
when
cases
we
like
are
interested
these
we
can
in
draw
a
diagram.
Example
Three
three
Find
We
ordinary
times
the
are
sixes
as
six-sided
likely
probability
interested
or
not
in
as
dice
any
that
are
other
exactly
whether
rolled
together
.
T
wo
of
the
dice
are
fair
and
one
is
biased
so
that
a
six
is
score.
two
sixes
sixes
are
are
rolled
rolled.
or
not,
so
we
need
only
show
probabilities
and
outcomes
and
2
for
sixes.
Biased
dice
Fair
dice
Fair
dice
1
P(6)

6
1
P(6)

6
5
P(6)
(6,
6,
6)
(6,
6,

6
3
P(6)
1

P(6)
8

6
6)
5
P(6)

6
5
P(6)

6
1
P(6)

6
(6,
6,
6)
1
P(6)

6
5
5
P(6)
P(6)


6
8
1
P(6)

6
5
P(6)

6
5
P(6)

6
3
Therefore
P(two
6s)

(
5
1

8
3
)

6

5
(

5
1
)

8
6
6

1
(

8
6
35
___
1
)

6


0.122
288
6
Exercise 3.8a
1
A
pack
card
is
second
of
ten
card
probability
two
cards
removed
cards
is
are
numbered
random
then
that
is
at
the
and
removed
sum
of
at
the
1
not
to
10.
random.
numbers
2
One
replaced.
Find
on
A
bag
pen
A
a
the
red
pen
and
red
at
is
pens
random
removed.
one
blue
pen
and
Find
are
blue
not
the
pens.
One
replaced,
then
probability
that
removed.
3.
Conditional
We
a
refer
card
and
we
2,
to
is
not
is
P(2nd
put
card
to
back,
find
that
an
card
In
the
probability
removed
want
given
This
2|1st
general
then
outlined
from
a
of
card
is
that
P(A|B)
is
at
pack
card
the
the
of
is
start
4
of
cards
this
removed
second
card
topic,
numbered
at
namely,
1,
random.
removed
is
2,
2,
probability
and
we
write
it
number
as
1)
means
B
the
has
probability
already
that
occur red.
A
occurs
3
Suppose
1.
conditional
card
a
second
probability
first
example
is
random
and
the
the
situation
at
that
136
3
removed
second
one
the
contains
is
given
Section
If
the
card
first
card
is
numbered
numbered
2,
1,
there
are
2
out
of
3
ways
of
3
Counting,
removing
matrices
and
differential
equations
a
so
2
P(2nd
card
is
2|1st
card
is
1)

3
1
Now
P(1st
card
is
1

2nd
card
is
2)

2
P(1st
card
is
1

2nd
card
is
1

and
4
P(1st
card
is
1)

3
4
2)
____________________________
∴

P(1st
In
card
is
general
if
P(A
P(2nd
card
is
2|1st
card
is
One
of
1)
1)
A

and
B)
B

are
P(A)
two

events
then
P(B|A)
Example
One
coin
coins
coin
the
is
is
selected
biased
fair
.
biased
The
it
is
two
shows
If
so
at
that
the
random
a
coin
head
shows
from
is
a
two
twice
head,
as
coins
likely
what
is
and
as
the
a
tossed.
tail
and
the
probability
the
other
that
it
is
coin?
events
a
we
are
interested
in
are
the
choice
of
coin
and
whether
head.
Choice
of
coin
Shows
a
head
2
P(H)

(biased,
3
H)
1
P(biased)

2
1
P(H)

(not
2
biased,
H)
1
P(not
biased)

2
If
B
is
the
selection
probability
Using
that
P(H

the
B)
of
the
coin

P(H)
1
P(H
and

B)

P( H)
1
2


then
is
a
tossing
head
from
is
the
a
head,
tossed
tree
then
is
the
P( B|H).
diagram
1
3
1

3

7
__
1

2
7
__
1
P(B|H)
H
that

3

and
given
P(B|H),
2

coin
biased

2
2
∴
biased
is

2
12
4

3
12
7
Exercise 3.8b
1
A
A
bag
contains
second
balls.
removed
If
the
came
bag
One
ball
out
white
contains
bag
at
3
is
of
black,
the
1
chosen
random
is
balls
white
at
from
what
second
and
black
ball
random
that
is
2
and
and
equipment,
balls.
4
operator ’s
black
one
ball
is
bag.
the
the
that
probability
is
0.001,
A
telephone
cell
phone
call
goes
from
phone
and
making
the
cell
the
phone
call,
the
receiving
is
0.01
the
that
the
probability
outgoing
that
the
phone
is
network
faulty
is
faulty
it
and
the
probability
that
the
receiving
bag?
phone
2
cell
call.
The
probability
the
network
one
through
cell
three
phone
sets
of
to
another
independent
Find
it
is
is
the
at
faulty
is
0.005
probability
least
partly
a
that
fault
if
a
of
call
the
fails
to
connect,
network.
137
3.9
Introduction
Learning outcomes
To
define
a

To
add

To
multiply
matrix
is
subtract
a
matrix
by
a
of
elements
expressions)
in
rows
and
columns
of
are
enclosed
in
brackets,
2
1
0
3
4
2
6
4
5
0
(
matrix
The
size
is
denoted
of
columns,
meaning
that
scalar
You need to know
The
array
(numbers
or
for
example
matrices
A

an
matrix
algebraic
and
matrices
Matrices
A

to
a
in
by
matrix
that
)
A,
is
B,
etc.
defined
order
.
For
by
the
number
of
rows
and
the
number
of
example,
commutative
2
1
0
3
4
has
operations
(
)
2
When
The
a
6
matrix
position
which
4
row
has
of
and
a
5
m
0
rows
and
particular
column
(in
n
rows
a
2
columns
element
that
2
called
is
order)
and

it
5
is
is
in.
columns
called
identified
it
5
and
is
matrix.
by
For
an
m
suffixes
example,

n
to
show
a
matrix.
means
the
21
element
in
the
second
row
and
first
column
and
a
means
the
element
in
ij
the
ith
row
and
jth
column.
In
the
example
above,
a

2
21
A
matrix
with
are
just
with
one
denoted
just
row
by
a,
one
is
b,
column
called
a
is
row
called
a
vector.
column
Column
vector
and
a
vectors
and
row
matrix
vectors
etc.
6
For
example,
a

2
(
and
)
b
(

8
5
10
)
1
A
matrix
matrix,
T
wo
with
e.g.
the
same
3
8
0
9
(
number
of
and
columns
is
called
matrices
A
and
B
are
element
A

in
B
equal
when
B,
i.e.
a

⇔
b
ij
3
example,
8
(
for
all
element
values
8
(
9
each
of
in
i
A
and
is
equal
j
ij
3

)
0
3
but
)
0
8
(
9

)
0
9
3
6
1
9
(
)
Example
x
Find
x
and
y
given
that
0
a

ij
b
for
and
can
be
corresponding
For
138
values

)
of
i
)
4
and
0
(
1
j,
x

3
and
y
y

4
ij
Addition
Matrices
all
3
(
1
As
example,
a
square
)
corresponding
For
rows
subtraction of
added
when
they
are
matrices
the
same
size
elements.
(
2
3
4
0
6
1
)

(
4
0
2
7
5
5
6
)

(
3
2
7
11
6
)
by
adding
to
the
Section
The
matrices
added,
The
so
size,
it
1
8
2
(
their
addition
same
2
and
)
sum
of
real
has
no
(
1
2
6
4
1
can
be
corresponding
For
example,
Subtraction
the
same
of
is
commutative,
subtracted
when

that
B
they
2
3
4
0
6
1
real
it
(
)
numbers
follows
4
0
2
7
5
5
and
cannot
differential
equations
be
so
if
A
and
B
are
two
matrices
of
the

are
B
the

A
same
size,
by
subtracting
is
not
are
the

)
(
2
3
6
7
1
4
commutative,
)
so
if
A
and
B
are
two
matrices
of
that
A
Matrices
sizes
and
elements.
(
size,
different
matrices
that
A
Matrices
are
)
Counting,
meaning.
numbers
follows
4
3
same
size
are

B
said

to
B
be

A
confor mable
for
addition
and
subtraction
Multiplication of
The
For
elements
example,
in
the
a
matrix
A
matrix
when
A

(
are
2
3
4
0
6
1
by
a
scalar

each
times
the
6
)
,
3A

(
corresponding
elements
in
A
9
12
0
18
3
)
Example
5
Given
A

4
A

B
(a)
A

B
and
)
2
(a)
1
(
5
B

2
0
2
4
6
1
(
0
)
,
find
2A
(b)
5

4
2
)
2
5
3

1
(

2
(
4
6

8
2
4
10
0
10
8
2
)
1
2A
)
1
10
(
3
(
6
2A
(b)
)
0
4
0
B
B

(
)
1
4

,
10
0
12
8
0
0
16
1
(
therefore
2
0
2
4
6
1
(
)
)
Exercise 3.9
4
5
1
x
2
2x
x
5x
2y
x
y
xy
y
3y
2x
2y
2
A

2
0
3
6
(
)
,
B

3
2
3
1
(
)
,
C

y
xy
3x
2x
(
x
2
)
,
D

(
)
2
1
Use
(a)
2
If
the
B
c

is
given
matrices
A
an
to
find:
(b)
element
in
C,
3A
write

2B
down
(c)
the
ij
3
Explain
xy
element
D
c

C
.
31
why
A

C
has
no
meaning.
139
3.
10
Matrix
multiplication
Learning outcomes
The
product of
Provided

To
define
the
product
of
a
and
a
column
a
row
row vector
vector
and
a
and
column
a
column vector
vector
each
have
the

b
same
row
number
vector
that
a
of
elements,
vector
b
11

To
define
the
product
of
a
square
b
21

To
and
define
a
the
column
the
vector
product
of
product
(a
a
11
…
a
12
)
1n
is
…
matrix
(
two
defined
as
a
b
11
)
a
11
12
…

a
21
b
1n
n1
b
n1
matrices
3
For
example,
(2
1
1)
4
(
The
meanings
row
matrix,
The
notation for
m
of
column

n
matrix,
The
product of

(1)(4)

(
1)(
2)
a
matrix
and
a
the
elements
that
a
matrix
has
the
same
number
of
of
columns
a
…
as
the
1n
b
11
…
a
21
The
double
number
a
matrix
a

12
column vector
11
a

matrices
Provided

(2)(3)
2
You need to know


)
angle
trig
identities
of
elements
in
a
column
vector
,
the
product
(
2n
…
…
…
)(
…
)
b
n1
a
…
a
m1
is
defined
as
top
row
the
product
so
of
the
the
of
column
matrix
the
vector
and
the
second
whose
column
row
of
the
top
mn
element
vector
,
matrix
is
whose
and
the
the
product
second
of
the
element
column
vector
,
is
and
on,
a
…
a
a
11
b
11
1n

a
11
b
12

…

a
21
b
1n
n1
b
11
a
…
a
21
i.e.
(
…
a
2n
…
b
21
)(
…
…
)

b

a
11
b
22

…

a
21
(
b
2n
n1
)
…
n1
a
…
a
m1
For
a
mn
b
m1

a
11
b
m2

…
 a
21
b
mn
n1
example,
1
(
5
2
7)
1
5
2
7
1
6
2
(
2
(
)
5
)
2
(
)

(
5
1
(
1
6
2)
2
(
)
)
5
(5)(
1)

1)(
1)

(2)(2)

(7)(
5)
6)(2)

(2)(
5)

(
)
(
(
36

(
)
21
This
definition
vector
only
elements
bA
For
is
140
in
of
the
the
is
of
is
precise:
columns
in
Ab
A
exists
is
equal
when
to
b
the
is
a
column
number
of
b.
product
an
product
number
meaningless
any
result
if
(m
of

and
an
1)
bA
(m
is

column
said
n)
to
be
matrix
vector
.
non-confor mable
by
an
(n

1)
column
vector
,
the
Section
3
Counting,
matrices
and
differential
equations
Example
2
x
1
y
1
Evaluate
(
0
2
)
(
)
2
z
x
2

2x
1

2y
1
1
(
y
0
)
(
)

(
2
z
)
2z
Exercise 3.10a
Evaluate
these
products.
x
4
4
1
(5
2)
(
2
)
(2
5
1)
2
0
(
3
)
(3
1
4)
y
( )
3
5
4
2
1
3
1
1
5
)
2
1
(
2
0
The
For
product of two
two
matrices
number
of
matrices
A
and
columns
that
3
1
4
2
2
0
1
5
1
x
3
4
)(
(
z
in
satisfy
the
is
this
(
6
)
(
2
3
y
)( )
z
matrices
B,
A
)
product
equal
to
condition
AB
the
are
exists
number
called
provided
of
rows
that
in
confor mable
B.
the
T
wo
for
multiplication .
AB
is
then
defined
as
the
matrix
C
where
the
element
c
is
the
product
of
ij
the
ith
row
a
of
A
…
and
the
a
11
jth
b
1n
column
…
of
B,
i.e.
b
11
1m
c
…
c
11
a
…
a
21
(
b
2n
…
…
…
)(
…
1n
b
21
2m
…
…
)
…

…
(
c
…
c
…
c
m1
a
…
a
m1
b
mn
…
)
ij
nm
b
n1
nm
b
1j
c
is
the
product
of
the
ith
row
and
jth
column,
i.e.
c
ij

(a
ij
…
a
i1
)
in
…
where
(
)
b
For
nj
example,
3
1
(3
3
1
1
1
(
1
3
2
1
1
2
1)
(
(3
)
1)
1
)(
)

(
1)
(
2
1
3
1)
three
matrices
Now
1
columns
are
not
consider
and
2
the
(
2
(1
)
1)
(
)
)
2
1
)
3
1
1
1
)(
1
has
2
)
2
8
(
2
(
(
3
(1
)
10
1)
1
1

2
(3
)
1
(1
However
,
(
2
)
second
is
0
meaningless
matrix
has
only
because
two
the
rows,
first
i.e.
matrix
the
conformable.
the
matrices
A
and
B
where
A

2
4
1
1
0
1
(
)
and
B

(
2
1
0
2
1
)
1
141
Section
3
Counting,
matrices
and
differential
The
equations
product
rows,
AB

2
4
1
1
0
1
)
(
1
0
2
1
1

(4)(0)

((1)(2)

(0)(0)

three
columns
and
B
has
three
)
(1)(
1) )
((2)(1)

(4)(
2)

1)(
1) )
((1)(1)

(0)(
2)

(1)(1) )
)
3
5
3
0
(
1)(1) )
(
)
BA
also
exists

(
2
1
0
2
1

because
B
(

(
example
has
two
columns
and
A
has
two

(0)(2)

1)(2)

1
0
1
)
(1)(1)
(2)(4)

2)(1)
(0)(4)

1)(4)

(
(1)(1)
8
1
2
0
2
1
4
2
illustrates
in
A
1
5
so
order
4
1
When
that
2
(
)
(2)(2)
(
The
has
so
BA
say
A
2
((2)(2)
(
product
This
because
(

rows,
exists
(

The
AB
so
AB
which
When
A
is
following
BA
both
key
exist,
multiplication
the
matrices
premultiplies
an
m
(2)(1)

2)(0)
(0)(1)

1)(1)

(
(1)(0)
(
(1)(
1)
2)(
1)
(1)(
1)
(
B

the
and
n
are
for
of
in
is
general
not
we
and
AB
is
say
B
m
AB

BA
commutative.
multiplied
BA
matrix
size
points.
is

matters,
that
an
n
A

so
for
AB
that
A
means
p
matrix,
p
AA
Example
4
Given
A

1
(
)
3
and
I

1
0
0
1
(
6
show
)
that
0
9
2
(a)
AI

IA
(b)
A
A
9I

(
)
27
4
(a)
AI

IA

142

0
(

1
4
IA
1
(
A)
4
1
3
6
(
)
1
)(
0
AI
)
0
6
1
∴
0
)(
3
and
1
1
(
)
3
6

4
1
3
6
(
)
we
postmultiplies
2
Note
)
)
the
and
matrix
(
(1)(0)
18
B.
Section
4
1
4
1
13
10
30
33
3
Counting,
matrices
and
differential
equations
2
(b)
A

(
)(
3

)
6
3
(
6
13
10
)
4
9
1
9
2
A
A

(
)
30
(

)
33
3
9
(
)
6
27
9
9
27
0
0
9
2
∴
A
A
9I

(
(
)
27
0
27
0

)
(
)
9
27
18
9
2
∴
A
A
9I

(
)
27
18
Example
2
Given
A

1
)
3
show
that

1
(

4
)
0
7
(
1

2
A(BC)

)




18
21
34
42
(
2
0
2
3
(
)
)
3
10
12
2
3
(
)
10
12

)
2
2
18
21
34
42
(
3
)
A(BC)
general
if
the
products
three
matrices
(AB)C
i.e.
C
,
0
)(
(AB)C
In
14
)
1
(
3
Therefore
3
3
2
∴
7
2
)
1
and
1
2
0
)(
0
)
(
2
14
2
4
(
4
(
)(
3
BC

1
2


A(BC)
1
2
(AB)C
B
0
)(
3
∴
,
2
(AB)C
2
AB
1
(
matrix

can
A,
be
B
found,
and
then
for
C
A(BC)
multiplication
is
associative.
Exercise 3.10b
1
Evaluate
1
(a)
0
(
3
1
0
) (
2
4
5
(c)
2
2
6
)
1
4
1
1
5
3
4
2
5
2
1
1
(
3
(b)
)
2
(
1
2
4
5
1
3
0
2
1
2
(
1
0
1
2
4
1
)(
)
1
1
2
1
2
2
Given
A

(
)
4
and
B

1

__
cos
)
(
)
show
that
(A

B)
2

A
2

B

__
sin
2
2
1
0
0
1
2
3
Given
A

(

__
sin

__
cos
2
)
show
that
A

(
)
2
143
3.
11
Square
unit
matrices,
matrices
Learning outcomes
To
define

To
determine
for
zero
addition
and
unit
identity
and
and
Square
A

zero
square
matrices,
inverse
matrices
matrices
matrix
has
an
equal
number
of
rows
and
columns.
matrices
matrices
multiplication
For
a
3
example

3
2
4
1
3
(
is
)
square
a
2

2
square
matrix
and
(
1
2
1
2
0
3
1
4
2
is
)
matrix.
You need to know
Unit

How
to find
the
product
of
matrices

What
is
meant
A
by
the
matrices
two
concept
unit
matrix
is
diagonal
(that
elements
are
a
is
square
top
left
matrix
to
such
bottom
that
right)
the
are
elements
all
1
and
in
all
the
the
leading
other
of
zero.
identity
For
A
example,
unit
All
is
0
1
and
)
is
denoted
(
by
1
0
0
0
1
0
0
0
1
elements
not
in
a
zero
necessarily
Multiplication
of
a
matrix
square,
matrix
by
for
a
matrix,
for
are
zero
c
0
0
unlike
when

AB
0,
real
)
d
numbers
neither
2
example,
A
nor
0
1
f
(
0
0
0
real
numbers,
ab
0
ac
1
However
,
given
A

1

1
1
(
1
1
0
1
and
AC

AB


0
1
AC
but
1
1
1

0
0
0
0
(
of
a
suitable
0
0
0
0
(


0

A
or

0
1
1
1
1
(

0
⇒
a

0
or

b
0

or
B
)
,
0
c
0
(

C

1
0
1
1
0
(
)
)
and
AB

B

AC
C

is
a
denoted
zero
size
will
)
)
0
B

)
is
)
)
0
i.e.
144
1
(
)
A
0
0
1
1
0
1
)(
1
,
matrix
0
0
0
1
)
1
(
1
so
)
a
ab
be
0
1
0
)(
⇒
(
1
AB

zero
0
(
2
AB

may

)
i.e.
where
B
1
)(
(
0
For
matrices.
0
e
0
However
,
A
matrix
example,
b
zero.
example,
(
For
unit
I
a
zero
are
)
matrices
the
and
0
(
matrix
Zero
1
A

0
or
B

C
b

0,
by
0
matrix.
give
a
Section
Identity
Under
as
it
from
The
is
no
single
matrix
subtraction
any
will
for
of
real
number
identity
that
matrix
c
e
0

)
b
differential
equations
matrix
leave
it
that
that
addition
numbers,
is
added
can
be
matrix
and
the
to
or
identity
number
subtracted
added
to
or
is
0
from.
subtracted
unchanged.
subtraction
is
a
zero
matrix
of
the
d
0

)
0
f
a
0
(
0
c
e
a
(
but
)
0
b
d
c
e
(

)
f
b
d
0
0
0
0
(
)
f
meaningless.
Under
leaves
multiplication
unchanged
Again,
there
is
postmultiply
The
For
a
a
on
square
c
(
1
b
d
a
d
g
b
e
h
c
f
i
Therefore,
0
0
0
1
0
0
0
1
a
unit
Now
We
consider
can
1
the
a
c
0
1
and
a
we
c
)
b
)
In
d
(

0
0
f
general,
(
0
1
0
0
0
1
for
an
m
premultiplicative
postmultiplicative
)
a
by
can
unit
a
is
premultiply
matrix
b
e
h
c
f
i
A

the
size
a
c
e
b
d
f
(
c
e
b
d
f

)
and
(
For
1
as
it
and/or
its
size
a
c
b
d
)
1
0
0
0
1
0
0
0
1
as
a
example,
(
d
and
)
same
)(
square
a
d
g
b
e
h
c
f
i
matrix
is
)

(
both
a
d
g
b
e
h
c
f
i
)
a
identity.
)
unit
a
size.
c
b
1
g
the
same
) (
d
of
matrix
1
0
0
1
(
)
giving
)
postmultiply
1
is
0
(
a
(
f
that
postmultiplicative
A
e
(
b
d
can
A
e
)(
a
number
1.
unchanged.
the
0
matrix
premultiply
0
(
and
is
1
and

)
it
identity
by
multiplies.
identity
matrix
premultiplicative
leave
d
1
matrix
multiplication
)
b
the
multiplied
c
(
1
)(
and
it
the
a

)
0
for
matrix
0
numbers,
identity
matrix
matrix,
) (
real
number
single
matrix
the
of
any
no
any
identity
depends
a
and
example,
a
(
matrices
size.
(
is
and
unchanged
identity
same
For
addition
any
Counting,
matrices
leaves
There
3
by
the
a
c
e
b
d
f
unit
matrix
(
1
0
0
0
1
0
0
0
1
)
giving

(

n
)
matrix,
identity
and
an
an
m
n


n
m
unit
unit
matrix
matrix
is
is
a
identity.
Exercise 3.11
___
i
1
Given
A

0
(
0
)
where
i

√
1
4
show
that
A

I
i
145
3.
12
Determinants,
minors
and
cofactors
Learning outcomes
Determinant of

To
define

To
calculate
The
the
determinant
n

n
To
define
matrix
deter minant
of
a
square
matrix
is
a
real
number
that
matrix.
Only
square
matrices
have
…
…
a
singular
1n
matrix
a
…
…
a
21
The
determinant
of
a
matrix
A
is
…
You need to know
a
…
…

The
a
meaning
the
…
…
in
a
square
a
⎥
leading
nn
2n
…
a
…
…
…
…
…
⎥
a
n1
diagonal
…
by
a
21

matrix
of
…
denoted
)
…
1n
a
identifying
|A|
in
…
a
11
notation for
…
n1
a
elements
2n

(
The
associated
a
11

is
determinants.
matrix
a

that
of
with
an
a
determinants
nn
matrix
|A|
is
also
written
as
det
A
The determinant of
a
The
matrix
i.e.
determinant
as
the
minus
value
the
of
of
the
the
product
of
2

product
the
2
if
A

b
c
d
of
is
)
the
elements
2
example,
a
(
Therefore
For
matrix
defined
elements
in
the
a
b
c
d
⎥

⎥
1
(
)
3
then
|A|

in
other
ad
the
the
value
leading
of
ad
bc,
diagonal
diagonal.

bc
2
1
3
5
⎥
5
as
⎥

(2)(5)
(
1)(3)

13
Exercise 3.12a
Find
the
determinant
of
each
matrix.
1
3
1
6
2
(
2
)
1
2
(
3
)
5
2
2
2
(
3
5
10
__
6
9
2x
)
4
y
(
1
2
x
)
x
Cofactors
The
determinant
of
an
of
a
n

n
matrix
is
based
on
extracting
smaller
determinants.
The
determinant
extracted
These
and
146
2
from

2
column
the
3
3


3
a
matrix
is
based
on
2

2
determinants
determinant.
determinants
through
3
are
found
particular
from
entry
are
the
elements
crossed
out.
left
when
the
row
Section
For
example,
through
the
when
we
element
cross
8
out
the
elements
are
This
with
the
determinant
Each
sign
left
minor
in
depends
a
2
is

signs
The
are
determinant
the
position

These
2
called
the
⎥
minor

minor
has
of
so
for
the
example,





an
cofactor
the
of
8
a
of
2,
element
sign
is
determinant
each
element
1
of
in
for
example,
and
a
3
the
1
5
9
or
,
of
8.
associated
which
it
is
with
it.
This
minor
.
of
that
with
its
sign
is
called
the
element.
with
the
element
8
in
⎥
2
1
3
6
4
8
5
9
7
⎥
3
are

a
3
3
⎥
2
1
3
6
4
8
,
⎥
⎥
9

2
6
4
5
9
⎥
7
respectively.
⎥
matrix
is
and
and
⎥
5
3

8
⎥
7
matrix
row
6
8
⎥
defined
its
as
6
8
⎥

⎥
9
the
sum
of
the
products
cofactor
.
1
8
⎥

⎥
7
5
3
6
4
5
9
⎥
7
⎥
7
(2)(
44)
(1)(2)

(3)(34)

12
general
a
a
11
a
12
13
a
a
22
a
a
21
⎥
a
⎥
not

sensible
3
a
11
⎥
a
23
a
a
32
⎥
a
21

a
12
⎥
a
33
a
23
⎥
a
31
a
21

a
13
⎥
a
33
22
⎥
a
31
32
33
the
to
try
and
remember
this
as
a
general
method,
just
process.
determinants
3

a
32
remember
The
23
a
31
is
a
22
a
of

element
⎥

It
9
the
associated
2
first
5
In
column
⎥
of
4
So,
equations
is
The determinant of
of
5
together
9
The
1
element
4
cofactors
differential
⎥
⎥
The
2
⎥
sign,
the
cofactor
So,
and
and


of
row
matrices
⎥
determinant
on
the
Counting,
in
⎥
we
from
3
of
larger
determinant,
i.e.
matrices
by
the
are
sum
found
of
the
by
extending
products
of
the
the
definition
elements
in
Exam tip
the
first
For
a
in
4
turn

4
are
involved
usually
row
in
and
their
cofactors.
determinant,
broken
finding
done
using
down
the
to
2
cofactors

2
a
determinant
a
specialist
are
3

3
determinants.
larger
than
calculator
or
3
determinants,
However
,

3
is
software.
the
tedious
and
these
It
is
easy
calculation
finding
and
not
is
it
in
be
a
to
3
make

3
tempted
one
mistakes
when
determinant,
to
try
and
so
do
calculate
step.
147
Section
3
Counting,
matrices
and
differential
equations
Example
Evaluate
⎥
1
2
0
3
1
1
2
0
5
1
2
0
3
1
1
⎥
1

⎥
2
(
1)
⎥
0
3
1
⎥
(2)
⎥
0
1
⎥
⎥
5
2

(0)
3
1
2
0
⎥
⎥
5
5

(
1)(5)

21
(2)(
13)
Example
Solve
the
equation
Expanding
⎥
the
⎥
2
x
1
x
3
1
4
1
1
⎥
determinant
2
x
1
x
3
1
4
1
1
⎥

2(
3

4


24
gives

1)
x(
x

4)

1(x
12)
2
x
4x

x
12
2

x
3x
16
2
∴
x
⇒
x
⇒
(x
3x
16

24
3x
40

0
5)

0
x

8
2
∴
An
x

8)(x

5
or
application of
3

3 determinants
in
coordinate
geometry
y
C(x
,
3
y
)
3
B(x
,
2
A(x
,
1
O
The
area
(area
148
of
the
SACT)

)
2
)
1
S
triangle
(area
y
y
ABC
TCBU)
T
in
the
diagram
(area
SABU)
U
is
x
Section
_
1
_
1

(y

y
1
2
)(x
3

x
3
)
Counting,
matrices
and
differential
equations
_
1

(y
1
3

y
2
2
)(x
3

x
2
)

3
(y

y
1
2
)(x
2

2
x
)
1
_
1

(x
W
riting
y
2
2
this

x
3
y
3

x
2
y
1

x
3
y
3

x
1
y
1

x
2
y
2
)
1
as
_
1
((1)(x

shows
y
2
2
that
it
is

x
3
y
3
the
)
(1)(x
2
y
1
expansion
of

x
3
the
1
y
3
)
(1)(x
1
y
1

x
2
) )
y
2
1
determinant
1
1
1
__
x
⎥
2
x
1
y
the
the
area
of
points
a
(x
y
2
3
triangle
,
y
1
),
(x
1
,
whose
y
2
1
⎥
3
y
1
Hence
x
2
)
vertices
and
(x
2
1
,
y
3
)
are
at
is
3
1
1
__
⎥
2
x
x
1
y
zero,
if
the
points
A,
B
and
C
are
⎥
3
y
1
Also,
x
2
y
2
3
collinear
,
the
area
of
triangle
ABC
is
so
the
condition
for
three
points
(x
,
y
1
to
be
1
⎥
1
x
x
1
y
(x
,
y
2
)
and
(x
2
,
y
3
)
3
is
1
x
2
y
1
),
1
collinear
⎥

0
3
y
2
3
Exercise 3.12b
1
Calculate
(a)
⎥
each
determinant.
2
1
7
0
1
2
1
2
3
(b)
⎥
⎥
1
4
3
0
5
2
0
3
1
(c)
⎥
1
⎥
1
0
1
3
2
0
1
2
4
1
⎥
1
2
2
Expand
and
simplify
the
determinant
⎥
cos 
cos
sin 
sin

1

1
⎥
2
a
b
2
3
Show
that
⎥
2
a
b
3
Determine
(a)
(0,
6),
(b)
(0,
1),
b
whether
(1,
(1,
2
c
3
a
4
c
3),
0),
(1,
⎥

abc(a

b)(b

c)(c

a)
3
c
the
(3,
following
points
are
collinear
.
3)
1)
149
3.
13
Simplification
Learning outcomes
To
simplify
determinants
Simplification of determinants
When

of
the
elements
in
a
determinant
are
large
numbers
or
complicated
determinants
algebraic
expressions,
determinant.
used
to
it
However
,
reduce
is
easy
there
elements
to
to
are
make
mistakes
properties
more
of
manageable
when
evaluating
determinants
that
the
can
be
quantities.
You need to know
a
a
1


How
The
to find
a
meaning
3
of

a
3
determinant
T
ransposing
the
rows
and
columns
of
| A|

b
⎥
a
2
b
1
a
3
b
2
b
1
gives
⎥
3
a
⎥
c
1
b
2
1
c
2
2
cofactor
c
c
1
c
2
a
3
b
3
c
3
T
which
Now
we
denote
|A|

a
by
(b
1
|A
|.
c
2
b
3
c
3
)
a
2
(b
2
c
1
b
3
c
3
)

a
1
(b
3
c
1
b
2
c
2
)
1
T

a
(b
1
c
2
Therefore
b
3
the
The
any
following
A
rows
property
properties
deter minant
example,
⎥
The
value
is
of
be
be
the
3
4
0
2
6
(1)
a
added
a
3
c
3
)
deter minant
for
are
rows
proved
expanded
be
(a
1
is
b
2
not
also
a
a
3
b
3
changed
valid
method
using
respective
c
)
example,
6
1
2
⎥

deter minant
⎥
0

is
It
follows
from
simplification
2
3
5
1
2
1
equal
A
to
⎥
3
this
will
|
any
for
columns.
similar
row
or
to
that
above.
column
cofactors.
found
from
0

the
first
column
giving
10
unchanged
subtracted
2
|A
when
from
any
when
other
0
0
3
1
2
1
any
row
row
(or
(or
column)
column).
Subtracting
For

2
transposed.
is
using

2
2
2
or
c
2
columns
can
⎥
⎥
to
a
proved
can
1
(a
1
of
1
0
b
and
can
and
For
)
2
value
the
Hence
c
3
a
⎥
2
property
give

2
that
row
or
if
⎥
3
two
of
the
third
the first
row
2
rows
column
row from
or
columns
zeroes,
so
are
the
the
same,
determinant
is
zero.
matrix
whose
The
of
determinant
value
any
of
row
a
(or
is
zero
deter minant
column)
other
2
1
4
1
2
2
2
3
1
is
is
is
called
singular
unchanged
added
row
a
(or
to
or
when
matrix
a
subtracted
multiple
from
any
column).
2
1
0
1
2
0
2
3
3
Adding twice the first
For
example,
The
aim
possible
risk
of
make
150
⎥
when
in
one
simplifying
row
mistakes,
mistakes
⎥
or
but
careful
adding
⎥
determinant
column
be
when
a

to
make
that
and
is
the
you
⎥
do
to
get
column to the third column
as
many
evaluation
not
subtracting
overdo
zeroes
easier
it.
multiples
of
It
and
is
also
rows
as
with
or
less
easy
to
columns.
3
⎥
Section
3
Counting,
matrices
and
differential
equations
Example
1
Solve
the
equation
Subtracting
column
top
the
x
2x
2x
1

column
the
1
x
1
1

⎥
x
from
elements

0
1
the
second
containing
x
and
and
from
gives
the
a
third
zero
in
the
row.
1
⎥
2
x
x
2x
2x
Expanding

x
1

1

⎥
x
0
x
1
1
2x
1
⎥
⎥
2x
1
x
(x
1)
1
0
1
x

0

0
1
2
⇒

1
⎥
1
⎥
gives
x
1
x
1
⎥
1
determinant
⎥
1
1
1
1
the
1
(1)
x
⎥
first
simplifies
2
(
2
x
x
⇒
x (x
2x)

0
4)

0
x

0
⇒
⇒
x
or
x


4x

0
4
Example
2
1

x
1

y
1

z
x
1
y
1
z
1
2
Given
f(x,
y,
z)

⎥
⎥
show
that
(x
y)
is
a
factor
of
the
2
function
f.
Subtracting
the
top
row
from
the
second
and
third
row
gives
2
1

x
x
2
f(x,
y,
z)

⎥
2
y
x
2
x
using
the
y,
z)

x
0
z
x
0
third
2
f(x,
y
⎥
2
z
Expanding
1
column
and
its
cofactors
2
(1)((y
x
2
)(z
x)
(y

x)((y
(y

x)(z
2

x
Therefore
(x
y)
is
x
a
) )
2
(z
x
) )
2
y)((z
(x
x)
2
x)(z
2
gives
)
factor
(y
of

the
x)(z
x) )
function
f.
Exercise 3.13
1
Evaluate
(a)
⎥
1
6
10
2
8
16
1
8
14
(b)
⎥
⎥
100
200
100
20
18
16
21
36
14
⎥
2
x
1
x
2
2
Express
⎥
x
1
x
3
x
⎥
as
a
product
of
factors.
3
1
x
151
3.
14
Multiplicative
Learning outcomes
The
If

To find
the
multiplicative
for
inverse
meaning of
a
matrix
A,
a
a
matrix
multiplicative
matrix
B
exists
so
that
inverse of
AB

I,
B
is
a
called
the
inverse
of
A
matrix
1
B
is
denoted
1
AA
You need to know

This
by
A
1
,
so
AA
1

I,
and
we
will
show
that
when
any
is
A
A

How
to

What

How
multiply
similar
to
a
multiplicative
inverse
for
real
numbers.
real
number
by
its
reciprocal
gives
1,
e.g.
2

unit
matrix
1

1,
so
evaluate
a
The
meaning

The
effect
of
of
future
a
singular
by
a
in
2,
and
vice-versa.
we
will
call
a
multiplicative
inverse
of
a
matrix
simply
an
multiplying
a
inverse of
a
2

2
matrix
scalar
to find
element
of
matrix
a
How
inverse
matrix
The
matrix
the
determinant
inverse

is
2
is
In
to
Multiplying
matrices
multiplicative
a
exists,
I
2

A
1
1

matrix
inverse
multiplicative
of
a
a
of
the
a
cofactor
of
an
If
A

b
(
d
then
)
c
d
a
b
postmultiplying
A
by
the
matrix
b
(
gives
)
c
a
matrix
(
d
b
) (
c
ad

)
d
c
0
(ad
d
A
by
d
a
b
c
ad
bc

1
0
0
1
(
)
gives
ad

bc
0
(
)
d
0

Now

bc
a
b
)
c
a
bc)
)
) (
ad
b
(
c
(
0
)
a

Premultiplying
bc
(
(ad
ad
bc)

bc
1
0
0
1
(
)
|A|
d
Therefore
both
premultiplying
and
postmultiplying
A
by
b
(
so
both
premultiplying
and
postmultiplying
A
by
|A|
a
d
1
____
|A|I
b
(
)
c
gives
gives
)
c
a
I
a
b
when
A

(
)
c
,
A
d
b
c
a
1
____
1
Therefore

|A|
(
d
)
1
If
|A|

0,
A
does
not
d
Notice
that
the
matrix
transposing
signs
152
of
the
the
in
is
)
elements
elements
the
and
A
is
a
singular
matrix.
b
(
c
by
exist
obtained
from
the
matrix
a
in
the
other
leading
diagonal
diagonal.
and
a
b
c
d
(
)
changing
the
Section
3
Counting,
matrices
and
differential
equations
Example
2
Find
the
inverse
of
(a)
First
the
transpose
sign
Then
of
the
the
2
1
5
3
⎥

3
5
(
)
3
elements
other
⎥
(b)
)
5
(a)
2
1
(
in
elements
the
to
leading
give
4
diagonal
3
1
5
2
(
and
change
and
change
)
1
1
2
therefore
1
(
5
First
(b)
the
Then
of
the
the
2
5
3
4
⎥

1
5
2
)
elements
other
⎥
3
(
3
transpose
sign

)
in
elements
the
to
leading
give
diagonal
4
5
3
2
(
)
7
1
2
therefore
5
1
__
(

)
3

Note
by
that
the
each
original
of
(
these
5
3
2
)
4

4
(
7
4
5
7
7
3
2
7
7
answers
)
can
be
checked
by
multiplying
it
matrix.
Exercise 3.14a
Find,
when
4
1
it
exists,
2
)
1
inverse
4
2
(
The
the
each
2
3
)
4
a
3

3
of
the
matrix
A

sin 
cos 
cos 
sin 
(
4
)
4
6
)
matrix
a
1
inverse
3
(
1
a
The
matrix.
2
(
1
inverse of
of
(
b
a
2
b
1
c
b
2
c
1
a
3
3
found
by
first
transposing
c
2
b
1
is
)
3
c
1
1
T
the
rows
and
columns
to
give
(
a
b
2
a
b
3
then
replacing
each
element
of
A
c
2
2
This
)
by
1
A
and
so
on,
1
this
gives
with
the
its
cofactor
.
Denoting
B
matrix
(
A
B
| A|,
the
cofactor
1
C
2
B
3
by
A
C
1
2
A
dividing
by
3
1
a
denoted
c
3
A
of
is
2
)
and
finally
C
3
3
i.e.
153
Section
3
Counting,
matrices
and
differential
equations
a
a
1
when
A

b
(
a
2
3
b
1
b
2
c
c
1
)
3
c
2
3
A
B
1

|A|
(
A
B
2
,
A
,
…
are
the
2
B
3
1
C
2
A
A
1
1
____
1
A
where
C
1
)
C
3
3
cofactors
of
a
2
,
a
1
,
…
2
1
If
|A|

0,
A
does
not
exist
and
A
is
a
singular
4
1
0
1
2
1
3
2
1
matrix.
T
For
example,
to
find
the
inverse
4
1
3
1
2
2
0
1
1
of
A

(
)
,
first
find
A
T
A

(
)
T
Next
replace
each
2

2
⎥
1
3
1
1
⎥
|A|:

1
4
3
4
1
0
1
0
1
⎥
⎥
3

⎥
1
0
1
2
1
3
2
1
⎥

⎥
4
1
1
2
⎥
2
4
⎥
)
(
1
1
4
4
4
4
11
9
)
⎥
4
1
0
4
4
0
3
2
1
the

4
⎥
third
row from
the
second
row
20
4
1
1
4
4
4
4
11
9
1
___
1

20
4
⎥
0

A
2
1
Subtracting
Therefore
1
⎥
0
1
⎥
cofactor:
⎥
⎥
2
its
2
4
⎥
find
by
⎥
3
2
Then

⎥
1
A
⎥
1
⎥

in
1
⎥
1
(
element
(
1
0
)
4
1
1
4
4
4
4
11
9
1
___
Check:
1
(
2
3
1
2
)

20
(
1
20
20
20
20
20
20
20
20
20
)
1
___

20

Note
step
154
that
at
a
(
it
(
1
0
0
0
1
0
0
0
1
is
time.
very
)
)
easy
to
make
mistakes
using
this
process,
so
take
one
Section
3
Counting,
matrices
and
differential
equations
Example
Show
T
o
it
find
is
we
⎥
that
the
⎥

Row
that
the
1
2
3
2
1
4
and
3
This
is
the
any
(
7
1
2
3
2
1
4
matrix
with
gives
A,
does
not
does
)
we
evaluating
⎥
the
2
1
4
1
2
3
2
1
4
same,
⎥
Therefore
Properties of
square

1
not
need
this
to
have
an
evaluate
determinant
inverse.
⎥
⎥
A
and
.
if
Therefore
it
is
zero
exist.
⎥
are
zeroes.
A
inverse
7
2
of
start
1
row
1
of
to
3
1
Rows
row
⎥
matrix
inverse
sensible
know
A
the
3
property
we
so
⎥
A
inverse
⎥
taking

0,
one
hence
from
A
does
the
other
not
have
will
an
give
a
inverse.
matrices
proved
for
2

2
matrices
and
it
is
true
for
all
matrices:
1
1
AA

A
A
1
It
follows
from
this
that
as
A
is
the
inverse
1
(A
1
Now
postmultiplying
Matrix
multiplication
AB
is
by
B
1

the
inverse
of
AB
is
B
i.e.
A
1
gives
(AB)(B
1
A
)
therefore
1
)
,
1
A
1
Hence

1
A
A
1
)
associative,
( AB)(B
of
A(BB
1
)A
1

AIA

I
1
A
,
i.e.
1
(AB)
1

B
1
A
Exercise 3.14b
1
Find,
(a)
(b)
A
(
when

it
exists,
the
1
1
0
1
3
0
2
1
1
(
5
1
9
3
1
5
2
0
4
inverse
of
the
(c)
)
(d)
)
matrices
B

1
0
4
0
3
2
1
5
0
(
4
1
3
3
5
2
1
2
1
(
)
)
1
2
V
erify
3
A,
B
that,
and
C
for
are
the
matrices
non-singular
1
Prove
that
A
(ABC)
1

C
3
and

1
B
3
B
in
question
1,
( AB)
1

B
1
A
matrices.
1
A
155
3.
15
Systems
Learning outcomes
of
To
a
pair
in

investigate
of
two
To
the
consistency
simultaneous

2
linear
Systems of
A

2
system
of
2

linear
2
equations
equations
equations
is
a
set
of
linear
equations
containing
the
of
same
variables.
A
of
equations
unknowns
use
matrices
simultaneous
to
solve
equations
a
pair
in
of
of
set
two
linear
linear
equations
with
two
variables
is
called
a
2

2
system
equations,
two
a
x

a
1
unknowns
y

a
2
3
e.g.
b
x

b
1

To
define
the
equivalent
meaning
systems
of
y

b
2
3
of
equations
Consistency of
a
a
system of
x

a
x

b
1
The
y

equations
a
2
3
equations
can
be
represented
by
two
lines
in
the
You need to know
b
1
y

b
2
3
xy-plane.

How
to
represent
equation
line
in
in
the
two
a
linear
unknowns
as
xy-plane
a
These
case
of

How
to
multiply
x
How
2

to find
2
the
and
y
may
is
intersect,
only
that
one
set
satisfies
in
of
the
which
y
values
equations,
matrices
i.e.

lines
there
inverse
of
there
is
a
unique
a
solution.
x

a
1
y

3
matrix
x
O
b
x

b
1
The
lines
case
there
may
is
be
no
parallel,
solution,
in
which
y
they
may
there
solutions,
is
be
an
the
same
infinite
3y

3y
line,
in
number
which
system
system
infinite

1
y
of
e.g.
an
A
2
x
2x
3y

1
4x
6y

2
x
O
A
3
y
O
case
b
e.g.
2x
Or

2
2x
156
a
2
a
of
of
equations
infinite
equations
number
of
that
number
that
of
does
solutions
is
has
either
solutions
not
not
have
is
a
unique
called
either
consistent.
a
solution
or
consistent
unique
solution
or
an
Section
Matrix
Counting,
matrices
and
differential
equations
representation
a
x

a
1
The
3
y

a
2
3
equations
can
b
x

b
1
y

be
represented
by
a
single
matrix
b
2
3
equation.
a
a
x
a
1
Since
(
)(
b

)
b
1
x

a
1
2
(
)
b
y
2
y
2
x

b
1
,
y
2
a
a
1
we
can
express
the
equations
in
the
form
)(
(
b
b
1
a
A

3
)

(
)
b
y
2
3
)
b
1
,
the
matrix
equation
can
be
written
as
A
1
that
)

(
)
b
3
1
A
exists,
premultiplying
each
side
by
A
gives
a
x
3
1
A
(
y
2
provided
3
2
(
b
then,
a
x
a
1
Using
a
x
2
1
A
(

)
A
(
)
,
b
y
3
a
x
3
1
i.e.
(

)
A
(
)
b
y
3
1
The
i.e.
For
equations
provided
that
example,
2
have
x
1
if
A

provided
that
A
exists,
0
2x

3y

5
4x

3y

1
can
be
written
as
5

)
y
2
Then

solution
equations
)(
4
unique
| A|
the
3
(
a
(
)
1
3
(
)
4
,
|A|

14
so
the
equations
have
a
unique
1
solution.
A

1
1
___
14
(
y
)
2
2

4
1
y
)

14
3
(
5
) (
4
2
)
1
5
) (
4
1
1
___
) (
3
(
14
x
3
1
1
___
)
2
) (
x
I
4
3
(
4
⇒
3
(
14
so
1
1
___
1
Now
2
)
1
_
1
x
22
1
___
⇒
(
y
)

(
14
7
)

22
(
__
11
)

7
_
1
Therefore
x

__
11
and
y


7
The
i.e.
advantage
requires
car r y
it
use
using
matrices
thought,
so
is
that
computers
the
can
process
easily
be
is
mechanistic,
programmed
to
out.
However
,
to
no
7
of
the
to
solve
basic
a
system
method
of
of
2

2
equations
elimination
or
by
hand,
it
is
often
simpler
substitution.
157
Section
3
Counting,
matrices
and
differential
equations
Equivalent
systems
2x
The
equations
several
For
ways
example
have
the
Any
The
aim
solution
so
the
the
pair
give
[1]
algebraic
sets
For
to
same
equations
T
wo
given
a
[2]
as
[1]
the
same
[1]

1
[2]
of
can
4x

3y
pair
of
equations.
2x
3y

5
6x
2y

4
and
first
[1]
an
the
same
equations
equations
equivalent
set
[1]
and
[2],
14x

7
14x

y
3[2]

of
[1]
have
and
are

the
2
matrix
easier
to
equations
x
3
(
4
2
equations
will
give
another
set
of
are
called
equations
equivalent
is
to
make
systems
the
gives
the
1
see
(
that
by

2,
solution
the

3y
5

as
the
original
14x

2
14x

y
1
row
(
1
)
1
y
by

2

)
)(
first
1
x
0
14
replacing
3y
(
1
y
same

and
14
and
)
14 x
solve.
3
(3

row
2

row
1)
of
both
5
and
)
4
[2]
of
5

)
)(
(
in
1
4x
can
and
solution
2x
we
these
combined
easier
.
Comparing
i.e.
be
set.
equations
equations
of
5
solution.
with
producing

gives
the
combination
equations
in
and
3y
i.e.
different
solution
with
of

above,
the
column
vector
(
)
1
we
can
obtain
the
second
matrix
1
equation.
By
placing
the
column
augmented
Operating
the
the
rows
vector
,
so
Calling
on
of
matrix
this
the
top
row
the
matrix
2
3
5
4
1
1
to
get
a
third
column,
we
get
)
matrix
matrix,
an
of
in
(
augmented
square
producing
the
vector
we
equivalent
the
ensures
also
do
system
augmented
that
with
of
whatever
the
rows
of
we
the
do
with
column
equations.
matrix
r
and
the
second
row
1
r
,
2
then
3r

r
2
on
the
first
row
gives
1
14
0
2
4
1
1
(
)
1
1
0
7
r
÷
14
gives
(
1
)
4
1
1
0
1
1
7
r
4r
2
(
gives
1
)
__
11
0
1
7
1
1
1
x
0
x
7
This
gives
the
matrix
equation
)(
(
0
1
)

(
__
11
y
so
y


and
7
158
1
x

7
)
(
)

(
__
11
y
7
__
11
7
⇒
7
)
Section
This
to
method
get
This
the
as
an
worth
of
using
next
give
solving
square
method
quick
the
of
matrix
solving
simple
topic
easier
the
to
solve
time
a
the
pair
of
systems
than
2
is
called
a
0
0
b
(
form
algebraic
solution
spending
in
equations

of
3
practising
2
linear

3
equations
However
,
linear
algebraic
the
reduction .
Counting,
The
aim
matrices
and
differential
equations
is
)
elimination.
purely
row
3
method
is
it
equations,
methods.
on
clearly
is
when
it
Therefore
simpler
2
not
extended

2
as
in
does
it
is
equations.
Example
3x
Use
the
row
reduction
method
to
solve
the

2x
3
Expressing
the
equations
in
matrix
form
gives
3r
the

augmented
r
1
2r
2
⇒
)(
10
33
(

)
33y
x
4
)
10
2
3
4
11
30
10
0
33
24
4
)(
3
2
)

(
)
4
y
gives
)
)
10
)

(
)
24
y
(
)
24
10
__
So
3
0
2

10
11x
⇒
2
11
x
30
(
2
(
1
11
∴
1
(
⇒
2
11r
3
(
matrix

x
1
(
2
Using
y
3y
equations

8
__
and
y


11
11
Exercise 3.15
1
Determine
which
of
the
following
systems
of
equations
are
consistent.
3x

2y

5
4x

2y

1
(a)
2
Express
Hence
3
6x

4y

2
3x

2y

1
(b)
Solve
the
the
5x

3y

8
2x

2y

4
equations
solve
the
as
equations
using
6x

y

8
2x

y

0
equations
augmented
x

y

1
x

y

1
(c)
row
a
matrix
equation.
reduction.
using
row
reduction
on
an
matrix.
159
3.
16
Systems
Learning outcomes
of
3
To
investigate
the
set
of
a
3

3
three
system
of
of
a
represent
equations
single

To
in
a
reduce
a
set
of
three
matrix

linear
linear
x

a
x

b
1
To
3
3
equations
equations
equations
with
three
variables
is
called
a
3

3
equations,
linear
equations

linear
consistency
system
of
3
Systems of
A


three
linear
unknowns
as
b
e.g.
1
c
a
y

a
y

b
y

c
2
2
x

c
1
z

a
z

b
3
4
3
2
4
z

c
3
4
equation
matrix
to
row
Consistency of
a
system of
equations
echelon form
a
x

a
1

To
solve
a
3

3
system
of
The
equations
using
row
reduction
equations
b
x

b
1
a
z

a
z

b
3
y

b
4
3
4
can
be
represented
by
three
planes.
of
x

c
1
augmented

2
c
an
y
2
linear
y

c
2
z

c
3
4
matrix
If
the
planes
intersect
in
only
intersect
in
a
one
point,
the
equations
have
a
unique
solution.
You need to know

That
a
linear
unknowns
a
plane
in
equation
can
be
three
in
three
represented
as
dimensions
If

How
3

to find
3
the
determinant
of
a
the
planes
equations
are
multiples
of
common
each
line,
other),
or
are
there
is
identical
an
(i.e.
infinite
the
set
of
solutions.
matrix
In
both
Any
the
these
other
cases
they
of
of
the
represent
equations
three
are
planes
not
is
consistent.
will
not
give
any
solution
and
consistent.
representation
a
x

a
1
The
system
configuration
equations
Matrix
the
equations
b
x

b
1
c
equation

c
a
2
b
(
b
2
c
3
c
1

a
y

b
z

b
y

c
4
3
4
z

can
be
expressed
as
the
single
matrix
c
3
4
a
4
y
)( )
c
2
z
x
3
b
1
a
3
2
a
1

2
x
1
a
y
2
3

(
b
4
)
as
each
row
of
the
matrix
multiplied
by
c
z
4
x
the
column
y
vector
( )
gives
the
left-hand
side
of
each
equation.
z
a
a
1
a
2
3
1
If
A

(
b
b
1
c
b
2
c
1
3
)
then,
provided
that
c
2
3
1
side
160
of
the
matrix
equation
by
A
gives
A
exists,
premultiplying
each
Section
a
x
a
x
4
1
A
y

( )
b
(
)
4
,
i.e.
I
y
( )
c
z

A
matrices
and
differential
equations
a
x
4
1
A
Counting,
4
1
A
3
1
(
b
4
y
⇒
)
c
z
4

( )
A
(
b
4
)
c
z
4
4
1
(If
A
does
not
exist,
the
system
4x
For
example,
the
equations
of

x

2y
2y
3x
(
x
4
1
0
1
2
1
3
2
1
4
equations
2y

A


y
)( )
z

(This
was
z
4
in
20
be
expressed
as
1
2
1
and
)
4
1
1
4
4
4
4
11
9
4
1
1
4
4
4
4
11
9
1
4
1
1
3
4
4
4
4
11
9
1
___
A

20
(
1
)
3.14)
4
1
1
x
0
1
___
∴
can
0
T
opic
1

2
1
2
found

( )
z
2
3
consistent.)
3
1
1
(
not
3

1
Then
is
3
1
___
4
(
4
4
4
11
1
)(
2
3
9
1
2
y
)( )
1

20
(
z
x
2
)( )
1
___
⇒
I
y
( )

20
(
z
2
)( )
1
__
11
11
x
20
1
___
⇒
y
( )

20
4
(
16
)

( )
5
19
__
19
z
20
__
11
∴
x

Again,
as
y

3

We
3
with
of
the
just
as
First
we
we

20
solving
aid
that
equivalent
equations
z
being
matrix,
know
,
5
advantage
without
19
__
4
,
20
of
is
of
the
when
at
2
linear
the
equations,
programmable.
software
equations
equations.
method
solving
form
of
of
2
row

an
to
This
2
this
method
However
,
means
time-consuming
combining
using
look

appropriate
which
system
did
2
easily
and
finding
prone
eliminate
means
using
we
reduction
of
to
a
the
an
the
method
inverse
of
a
mistakes.
variable
can
has
this
solve
produces
3

3
augmented
an
linear
matrix,
systems.
augmented
matrix
that
we
need
to
achieve.
Row
The
echelon form of
leading
elements
in
a
a
row
matrix
of
a
matrix
are
the
elements
reading
from
Did you know?
left
A
to
right
matrix
than
the
matters;
is
along
in
row
the
row
above
other
row.
echelon
it.
(It
elements
is
for m
the
can
be
when
number
each
of
row
has
leading
more
zeroes
leading
in
a
row
zeroes
Echelon
is
the Greek
word for
ladder.
that
zero.)
161
Section
3
Counting,
matrices
and
differential
These
These
equations
matrices
are
matrices
Reduced
are
row
A
each
example,
echelon
Using
(
row
not:
(
echelon
form:
1
2
1
0
0
3
0
0
2
)
,
(
(
matrix
has
the
is
in
reduced
more
first
leading
non-zero
1
4
0
2
0
1
3
1
0
0
0
1
1
0
3
5
0
0
2
0
0
5
0
0
2
1
0
2
3
0
a
row
than
element
(
)
,
(
0
4
0
0
0
0
0
1
0
0
0
0
)
)
matrix
echelon
zeroes
and
)
2
0
echelon form of
row
and
For
in
1
in
for m
the
each
when
row
row
0
1
3
2
0
0
0
1
0
0
0
0
)
above
is
are
it
1.
in
reduced
row
form.
reduced
row
reduction to
solve
systems of
3 
3
linear
equations
Using
i.e.
(
the
4
1
0
1
2
1
3
2
1
(
We
on
page
x
1
2
1
2
3
2
1
1
tempted
to
use
,
the
augmented
matrix
is
)
combinations
important
again,
1
3
use
2
( )
z
0
is

y
)( )
1
It
161
3
4
now
form.
equations
that
columns
of
rows
you
as
use
this
to
change
this
combinations
will
not
give
an
to
of
reduced
rows;
do
equivalent
row
not
echelon
be
system
of
equations.
Using
r
,
r
1
leading
Adding
and
to
3r
r
r
now
to
gives
2
a
4r
zero
from
2
162
denote
the
gives
4r
want
subtracting
in
3
1
We
to
the
rows
of
a
matrix,
we
want
zeroes
3
2
Adding
r
2
elements
in
second
(
(
1
0
3
1
2
1
2
3
4
4
7
4
1
0
3
0
9
4
11
0
4
4
7
second
gives
3
third
4
the
9r
and
(
rows.
)
)
element
in
the
4
1
0
3
0
9
4
11
0
0
20
19
)
third
row:
in
the
Section
Next,
divide
each
row
the
value
of
the
first
non-zero
element
in
Counting,
that
matrices
and
differential
equations
row:
3
1
1
0
4
(
by
3
0
4
4
__
11
9
9
1
)
19
__
0
0
1
20
3
1
x

y

4
This
augmented
matrix
gives
the
equivalent
y
system

[1]
4
__
11
4
z

9
[2]
9
19
__
z

[3]
20
which
can

) (
(
easily
19
___
4
__
y
be
9
solved
using
220
76
_________
11
___
)

⇒
20
substitution,
y

i.e.
144
____
in
substituting
the
value
of
y
into
180
[1]
5
gives
x
3
__

11
___
the
solution
is
x

y

⇒
x

4
20
19
___
4
__
,
,
20
11
___

5
Therefore
gives

180
1
__
then
[2]
4
__

9
[3]
z

5
20
Example
4x
Use
reduced
Starting
3r
row
with
5r
2
reduction
the
⇒
(
3
to
augmented
solve
the
matrix:
equations
(
4
1
5
8
0
1
14
2
3
4
1
27
4
1
5
8
0
1
14
9
0
0
255
255
)
2y
5x

7y
3x

4y
4
1
5
8
5
7
3
42
3
4
1
27
;
4r

5z

8
3z

42
2z

27
)
3r
3

⇒
1
(
4
1
5
8
0
1
14
9
0
19
11
84
1
5
4
5
1
r
19r
3
⇒
(
2
This
gives
the
y
equivalent
;
r
1
14z

9
⇒
y

5

2
⇒
x

2
÷
4,
r
1
equations

÷
255
⇒
3
(starting
with
the
last
(
2
0
1
14
9
0
0
1
1
)
row)
5
y

4
Therefore
of
z
1
x
set
)
)
z
4
x

2,
y

5,
z

1
Exercise 3.16
Use
of
1
row
reduction
of
an
augmented
matrix
to
solve
the
following
systems
equations.
2x

y

3z

8
4x

2y

z
2x

3y

4z

13

x
5
2

2y

4z

0
3x

y

2z

7
5x

y

4z

3
163
3.
17
Using
row
reduction
to find
an
inverse
matrix
Learning outcomes
Finding the
row

To find
the
multiplicative
multiplicative
inverse of
a
matrix
using
row
matrix
using
reduction
inverse
3x
of
a

2y

x

4
2z

2
z

2
reduction
Consider
the
system
of
equations
x
3y
2x
3

3y
2
x
1
4
You need to know
These

How
to
linear

How
represent
equations
to
reduce
a
a
in
system
can
be
represented
by
(
of
1
3
2
2
3
1

y
)( )
)
2
z
matrix form
matrix
2
(
to

(
1
0
0
0
1
0
0
0
1
4
2
)(
)
[1]
echelon form
We
know
that
operating
on
the
rows
produces
an
equivalent
2
system
of
equations.
T
o
find
the
matrix.
So
inverse
of
the
left-hand
matrix
we
want
to
reduce
it
to
a
unit
if
3
2
x
1
4
1
A

(
1
3
2
2
3
1
)
,
we
reduce
the
system
to
I
y
( )

A
(
2
)
2
z
4
Any
row
operation
right-hand
side
of
either
on
I
or
on
2
(
gives
)
the
same
result
on
the
[1],
2
e.g.
r

r
1
on
I
⇒
(
2
1
1
0
0
1
0
0
0
1
4
and
r

1
r
on
2
2
(
⇒
)
(
2
6
4
2
)(
)

2
(
2
1
0
0
0
1
0
0
0
1
)
2
6
2
)(
6
)

(
2
2
)
2
1
If
we
just
want
augmented
(
We
calculate
A
,
2
1
1
0
0
1
3
2
0
1
0
2
3
1
0
0
1
now
work

r
1
r

2
we
can
operate
just
on
A
and
matrix:
3
r
164
to
on
the
⇒
3
2r
⇒
3
(
(
rows
to
)
reduce
the
left-hand
5
1
0
1
0
1
1
3
2
0
1
0
2
3
1
0
0
1
5
1
0
1
0
1
5
9
0
0
1
2
2
3
1
0
0
1
)
;
)
;
side
to
I
I
using
the
Section
9r
r
1
⇒
40
0
0
9
0
7
5
9
0
0
1
2
2
3
1
0
0
1
40
0
0
9
1
7
(
2
1
r
r
2
⇒
0
(
1
8
r
3
8
0
40
0
0
9
⇒
r
8
1
0
3
0
0
0
9
0
9
8
8
8
9
__
__
1
13
__
20
20
20
9
1
7
0
9
9
9
8
8
8
9
__
39
__
33
__
40
40
40
0
__
1
7
__
40
40
0
2
___
,
3
___
_
1
,
⇒
9
0
(
3
1
0
_
1
0
have
now
0
reduced
the
system
0
0
x
3
__
13
__
0
1
0

y
)( )
0
1
7
__
40
_
1
__
1
7
__
40
40
_
1
1
A

_
1
40
(

8
3
__
8
13
__
method
of
advantages
over
arithmetic,
and
check
give
that
your
9
1
7
)

40
(
5
5
5
3
13
11
)
__
11
40
40
finding
the
so
)
2
40

40
2
1
___
_
1

8
40
__
11


40
40
to
)(
8
13
__
40
__
11
_
1

8
z
9
__
[1]
40

(
)
8
4
8
3
__
0
in
__
1
_
1
This
)


40
∴
)

8
1
9
__
1
(
;
_
1

8
40
We
;
r
r
40
equations

40
1
___
differential
3
9
__
r
7
9
and
;
(
1
1
9
matrices
;
2
0
1
Counting,
1
40
⇒
0
)
0
9
(
3
9
8
3
0
3r
9
2
1
20
9
0
9
__
1
r
)
3
an
inverse
method
mistakes
calculated
of
using
are
a
matrix
cofactors
less
inverse
likely.
by
row
because
However
,
multiplied
by
the
reduction
it
has
simplifies
it
is
the
sensible
original
to
matrix
does
I.
Using
row
singular
as
reduction
it
will
is
also
produce
a
a
quicker
row
of
method
zeroes,
for
showing
proving
that
that
| A|

a
matrix
is
0
Exercise 3.17
1
2
Use
Find
(a)
(
row
the
reduction
inverse
of
4
1
1
2
0
2
1
2
1
)
to
show
each
of
(b)
that
the
(
the
matrix
following
2
1
0
5
5
1
1
2
1
(
1
3
1
3
7
2
2
4
1
)
is
singular
.
matrices.
)
(c)
(
2
1
4
2
1
5
1
2
4
)
165
3.
18
Differential
Learning outcomes
Differential
A

To
explain
equations
differential
differential
equations
equation
connects
an
unknown
function
and
its
equations
derivatives.
as

mathematical
To formulate
equations
of
models
The
differential
in
the form
order
the
of
a
differential
equation
is
the
highest
derivative
contained
equation.
dy
___
dy

ky 
f(x)
where
k
is
___
a
For
2
example,
xy

x
is
a
first
order
differential
equation
and
dx
dx
function
of
x
or
a
constant
2
d
y
dy
____
___

y
xy

0
is
a
second
order
differential
equation.
2
dx
dx
You need to know
Models

The
meaning
second
of
a first
and
Unlike
derivative
some
many
time
equations

The
basic facts
about
topics
after
in
their
comes
mathematics
development,
directly
from
the
which
the
find
real-world
formulation
need
to
describe
of
applications
differential
real-world
phenomena
integration
mathematically.

The
derivatives
of
standard
functions
A
differential
equation
phenomenon.

How
to
differentiate
a
product
It
is
is
used
a
mathematical
to
predict
description
results
and
it
is
of
a
called
real-world
a
mathematical
of
model
functions

How
to
differentiate
implicit
How
functions
to
good
the
measured
There
are
model
results
many
is
depends
from
the
well-known
For
example,
Newton’s
laws
the
relationship
between
the
body.
are
on
how
real-world
equations
of
the
that
motion
forces
closely
the
results
it
predicts
are
phenomenon.
are
acting
are
a
extremely
set
on
a
of
good
equations
body
and
the
models.
that
describe
motion
of
Did you know?
needed
Newton’s
in
the
Isaac
name
study
of
Newton
keeps
These
to
place
accurate
satellites
in
enough
mathematics.
(16431727)
mathematician.
arguably
world
the
has
greatest
known.
used
to
determine
the
forces
orbit.
Sir
He
was
was
a
scientist
equations
also
the
The
solution
variables
a
first
will
so
include
the
There
and
In
Unit
an
1
we
any
solving
unknown
will
we
them
look
the
more
at
how
two
be
of
When
equation
the
integration
the
connecting
differential
operation
differential
operations
are
the
equation
so
equation
needed
to
is
it
is
solve
a
it
constants.
different
types
of
differential
equation
solved.
the
of
an
If
one
unknown
solution
In
types
a
involves
integration
number
cannot
variables.
two
it
gives
involved.
constant.
two
involve
covered
equation
derivatives
enormous
of
of
differential
equation,
separable
solution
First,
one
solution
is
a
equation,
order
many
with
of
without
order
second
166
be
appearing
Solution of differential
prolific
to
of
a
first
remaining
differential
particular
type
of
order
topics
differential
in
Unit
2
equation
we
look
at
the
equation.
differential
equation
can
arise.
Section
Formulation of differential
3
Counting,
matrices
and
differential
equations
equations of the form
dy
___

ky

f(x)
where
k

h(x)
dx
d
___
We
know
that,
if
u
and
v
are
functions
of
x,
dv
___
then
(uv)

u
also
know
(yg(x))

g(x)

yg(x)
dx
dy
d
___
example,
dx
___
that
dx
for
v
dx
dy
d
___
We
du
___

dx
___
(y sin x)

sin x

dx
y
cos
x
dx
dy
___
Therefore,
given
the
differential
equation
sin
x

y cos x

2x,
we
can
dx
recognise
equation
the
by
left-hand
side
integrating
as
both
the
differential
of
y sin x,
and
so
solve
the
sides.
dy
___
Hence
sin x

y cos x

2x
y sin x

x
dx
2
⇒
This
type
of
differential

equation
A
is
where
called
A
an
is
an
exact
unknown
constant.
differential
equation
dy
___
If
the
equation
sin x

y cos x

2x
is
divided

y cot x

2x cosec x
by
sin x
(sin x

0)
it
dx
becomes
dy
___
dx
dy
___
so
it
is
of
the
form
ky

f(x)
where
k

h(x)
but
the
left-hand
dx
side
is
not
now
differential
equations
The
of
that
solution
derivative
integration
have
of
integration
boundar y
the
equations
a
is
we
of
not
this
been
differential
called
need
to
the
of
a
product.
form,
Before
solve
a
few
finding
exact
ways
of
solving
differential
simplified.
equation
general
know
we
containing
solution .
initial
values
T
o
of
x
unknown
evaluate
and
y.
constants
constants
These
are
of
called
conditions .
Example
dy
1
__
___
Find
the
general
solution
of
the
differential
equation
x

dx
The
left-hand
therefore
the
side
is
general
the
derivative
solution
is
xy
of

y

x
xy,
ln |x|

A
Exercise 3.18
Find
the
general
solution
of
each
of
the
following
dy
x
1
___
e
x

ye
2

dx
t
equations.
2x
2
___
x

2xy

cos x
dx
dv
___
3
differential
dy
2

v

t
dt
167
3.
19
Integrating factors
Learning outcomes
Integrating factors
We

To
solve
differential
equations
know
that
if
the
left-hand
side
of
a
differential
recognise
that
this
equation
has
the
form
of
dy
___
dy
___
the form
g(x)

yg(x)

f(x)

yg(x)
we
is
the
derivative
of
the
product
dx
dx
yg(x).
using
an
integrating factor
However
,
function
You need to know
This
when
of
x
this
that
function
is
is
not
will
the
give
called
an
case,
the
it
is
possible
derivative
integrating
of
the
factor ,
to
multiply
product
and
we
by
a
yg(x).
will
denote
it
by
I
dy

The formula for
differentiating
a
___
Consider
an
equation
of
the
form

Gy

F
where
both
G
and
F
are
dx
product
functions

The
chain

How
Now
to
solve
equation
of
x
rule
a
with
multiply
the
equation
by
I
where
I
is
a
function
of
x,
differential
dy
separable variables
___
i.e.
I

(y)(GI)

FI
[1]
dx

The
integrals
of
standard
dy
d
___
functions
We
want
to
find
I
such
___
that
(Iy)

I
dx

(y)(GI)
dx
dv
___
Comparing
the
left-hand
side
of
[1]
with
u
du
___

v
dx
dy
dv
___
u

du
___
___

I,
dx
and
v

the
y,

dx
chain
rule,
GI
dx
dI
___
dI
___
Using
gives
dx
i.e.

dx
du
___

dI
___
,
du
gives
dx

1

GI

GI
dx
dI
___
Now

GI
is
a
first
order
differential
equation
whose
variables
are
dx
separable.
1
__
∫
Therefore
dI

∫
ln I

∫
G dx
I
⇒
G dx
∫
⇒
∫G
So
I

I

G dx
e
dy
dx
___
e
is
an
integrating
factor
for
the
expression

Gy
dx
Assuming
equations
∫
that
you
G dx
meet
in
dy
can
the
be
found
(it
cannot
examination),
always,
but
can
for
any
then
dy
___
___

Gy

F
⇒
dx
I

(y)(GI)

FI
dx
dy
___
⇒
∫ (I

(y)(GI)
)
dx

∫
IF dx
dx
⇒
Both
be
168
I
and
found
F
for
are
any
Iy

∫
IF dx
functions
equations
of
x,
you
so
the
meet
integral
in
the
on
the
right-hand
examination.
side
can
Section
3
Counting,
matrices
and
differential
equations
Example
Find
the
dy
2y
___
___
general
of
the
differential
dy
2

dx
solution
x
2
__
___
x
e
⇒

x
y
(
2
)
dx

x
dy
2y
___
___
dx
x
2
equation

x
x
e
x
e
x
2
∫
∴
the
integrating
factor
I
is
e
dx
2 ln x
x

e
2
ln(x

Multiplying
1
__
both
dy
2y
___
___
sides
2y
___
dx
x
2

2

x
x
x
2
e
by
x
gives
x

2
dy
___
of
)
e
e
3
x
dx
x
y
__
x
⇒

∫

e
2
e
dx
x
x

2
⇒
y

x
A
x
2
e

Ax
Example
dv
___
Find
the
solution
of
the
differential
equation
2
t
v

t
t
e

0
dt
given
that
v

0
First,
rearrange
when
t

1
dv
___
the
equation
so
that
it
is
in
the
form

vg(t)

f(t)
dt
dv
___
2
t
v

t
dv
___
t
e

0
⇒
dt
1
__

v
(
t
)
dt

te
t
1
∫
∴
I

dt
1
__
ln t
t
e

e

t
1
__
dv
___
t
dt
1
__
∴
v
(
2
t
)
1
__
⇒
v
(

e
t
t
)
∫

e
dt
t
t
⇒
v

te

At
1
v

0
when
t

1
t
∴
v

gives
0

e
1

A
so
A

e
1
te
te
Exercise 3.19
1
Find
the
general
solution
of
dy
dy
___
___

(a)
3y

x
(b)
sin x
dx

y
cos x

1
dx
dy
___
2
Find
the
solution
of

xy

x


given
y

0
cos 
given
when
x

0
dx
dy
___
3
Solve
the
equation


__
2

d
y
y

0
when


2
169
3.20
First
order
Learning outcomes
differential
Summary of
equations
solutions of first order differential
equations

To
summarise
methods for
dy
___
solving first
order
differential
There
are
many
differential
equations
involving
that
cannot
be
dx
equations
solved
to
methods
give
that
be
solved.
so
you
All
need
a
direct
can
of
to
be
relationship
used
these
know
for
methods
all
of
between
some
first
rely
these
on
from
x
and
order
y.
recognising
Unit
1
The
table
differential
as
the
well
shows
equations
standard
as
from
that
can
integrals,
Unit
2.
You need to know

The
integrals
of
standard
Form of
equation
Method of
solution
functions
dy
Recognise
___


How
to
differentiate
the function
of
which
the
differential
f(x)
implicit
dx
is
f(x)
functions

How
to
use
or
use
a
substitution
to
or
use
partial fractions
or
use
integration
simplify
f(x)
integrating factors
by
(for
parts
a
rational function)
(for
a
product
of
functions).
Exam tip
dy
___

g(y)f(x)
Separate the variables to
Any
equation
examination
you
can
are
be
given
in
dy
1
____
___
g(y)
dx
give
dx

f(x)

∫
an
1
____
solved.
∫
then
dy
f(x) dx
g(y)
dy
Recognise
___
g(x)

yg(x)

the
left-hand
side
as
the
differential
of
f(x)
dx
yg(x)
giving
yg(x)
(You
may

∫
need
f(x) dx
to
rearrange
the
equation
to
give
this form.)
This form
dy
When
___

yg(x)

is
called
an
exact
rearrangement
does
differential
not
give
an
equation.
exact
f(x)
dx
differential
equation,
multiply
throughout
by
integrating factor
∫
I
This

then
g(x) dx
e
gives
an
exact
differential
equation.
Example
Solve
the
following
differential
equations.
dy
(1

x
dy
___
2
(a)
)
2x
(b)
(1

x
dy
(1

x
)
170

2y
dy
2
___

dx
)
dx
___
2
(c)
___
2

dx
2xy

0
(d)
x
x
2

dx
2x
y

4xe
the
Section
3
Counting,
matrices
and
differential
equations
dy
___
2
(a)
(1

x
)

2x
dx
This
equation
contains
no
term
involving
y,
so
first
rearrange
it
to
dy
___
isolate
.
dx
dy
2x
______
___
2

⇒
y

ln |1
x
|

c
2
dx
1

x
dy
___
2
(b)
(1

x
)

2y
dx
The
variables
dy
1
__
___
y
dx
in
this
equation
can
be
separated
to
give
2
______

2
1

x
1
__
∫
⇒
2
______
dy
∫

1
dx
⇒
ln |y|

2 tan
x

c
2
y
1

x
dy
___
2
(c)
(1

x
)

2xy

0
dx
2
2x
is
the
differential
equation
of
(1

x)
so
this
is
an
exact
differential
giving
2
y(1
(The

x
)
variables
recognition
neater
of

in
c
this
an
equation
differential
can
equation
be
gives
separated
a
quicker
but
and
solution.)
dy
2
___
(d)
differential
exact
x
2
x

2x
y

4xe
dx
The
left-hand
separated
but
side
we
is
not
can
exact
and
rearrange
the
the
variables
equation
to
cannot
the
be
form
dy
___

yg(x)

2xy

f(x)
and
use
an
integrating
factor:
dx
dy
2
___
x

4e
dx
∫2x
then
I
2
x
∴

dx
2
x
e
⇒
dy
I

e
2
___
2
x
e

2xe
x
y

4
⇒
e
y

4x

c
dx
Exercise 3.20
Solve
each
of
the
following
differential
equations.
dy
dy
___
1
sin x
2x
______
___

y cos x

tan x
2
x

2
dx
dx
dy
x
x
___
3

dx

dy
___
3
1
y

3x
4

y

0
dx
171
3.21
Differential
equations
of
the form
2
d
y
dy
____
a
___

b

cy

0
2
dx
dx
Learning outcomes
Formulation of
When

To
solve
equations
of
an
equation
y
a
second order differential

f(x)
contains
two
unknown
equation
constants,
A
and
B,
the form
2
dy
2
d
y
___
dy
____
a
___

b
differentiating

cy

twice
gives
y
____

f(x)
and

f (x)
2
0
dx
2
dx
d
dx
dx
where
a,
b,
c  
These
two
eliminate
We
now
and
look
constants
You need to know
equations,
A
A
at
B,
together
giving
three
and
B,
a
types
all
of
with
second
of
the
which
the
original
order
equation
give
rise
equation,
differential
y
to

an
f(x)
can
be
used
containing
equation
of
the
unknown
form
2
d
y
dy
____
___
a

The
relationship
between

the
b

roots
and
the
coefficients
of
cy

0
where
a,
b
and
c
are
real
constants.
2
dx
dx
a
2
d
y
dy
____
quadratic
equation
A
differential
equation
of
the
form
a
___

b

cy

0
is
called
2
dx

The
meaning
of
dx
conjugate
linear
complex
second
order
differential
equation
numbers
 x
The
equation
y

Ae
 x

Be
3x
Consider
the
equation
y

Ae
4x

Be
dy
___
3x

3Ae
4x

4x
4Be

3y

Be
dx
2
d
y
dy
____
dy
___


dy
___
4x
3
4Be

3
___

(
4
2
dx
dx
dx
3y
)
dx
2
d
y
dy
____
___
⇒
(3

4)

(3)(4)
y

0
2
dx
and
dx
the
coefficients
of
this
equation
are
the
roots
of
the
quadratic
2
equation
u
Therefore
7u

working
12

0.
backwards,
by
using
the
coefficients
of
2
d
y
dy
____
___
7

12y

0
to
give
the
quadratic
equation
2
dx
dx
2
u
7u

12

0
⇒
(u
3)(u
4)

0
then
the
3x
equation
give
the
general
solution
as
y

Ae
Be
x
Now
consider
the
general
case,
i.e.
y

Ae
x

Be
dy
___
x

x
Ae

Be
dx
2
d
y
____
and
x
2


Ae
x
2


Be
2
dx
Eliminating
A
and
B
from
these
two
2
d
y
dy
____
___
(

)

2
dx
172
dx
y

0
equations
roots
4x

gives
to
equation.
of
this
a
Section
Then
the
coefficients
of
this
equation
give
the
quadratic
3
Counting,
matrices
and
differential
equations
equation
2
(
u
whose
roots
)u


are



0

and
2
In
general,
the
quadratic
equation
au

bu

c

0,
formed
from
the
2
d
y
dy
____
differential
equation
___
a

b

cy

0
is
called
the
auxiliar y
2
dx
dx
equation
2
When
the

roots
auxiliary
,
and
the
equation
general
au

bu
solution
of

c
the

0
has
real
differential
distinct
equation
2
d
y
dy
____
___
a

b

cy

0
can
be
quoted
as
2
dx
dx
x
y

x
Ae

Be
Example
Find
the
general
solution
of
the
differential
equation
2
d
y
dy
____
___
2
3

y

0
2
dx
dx
2
The
auxiliary
equation
is
2u
3u

1

0
1
2
2u
3u

1

0
⇒
(2u
1)(u
1)

0
⇒
u

or
u

1
2
2
d
y
dy
____
Therefore
the
general
solution
of
___
2
3

y

0
2
dx
dx
1
x
is
y

Ae
x
2

Be
 x
The
equation
y

(A

Bx)e
3x
Consider
the
equation
y

(A

Bx)e
dy
___
3x

3(A

Bx)e
3x

3x
Be

3y

Be
dx
2
d
y
dy
____
dy
___

3
dy
___
3x

3Be

3
___

3
2
dx
dx
(
dx

3y
)
dx
2
dy
y
d
___
____
⇒

(3

3)

(3

3)
y

0
2
dx
dx
2
This
and
time
it
has
the
a
auxiliary
repeated
quadratic
root
of
equation
is
u

(3

3)u

(3

3)

0
3.
 x
Now
consider
the
general
case,
i.e.
y

(A

Bx)e
dy
___
 x

y

Be
dx
2
d
y
dy
____
dy
___


Be
dy
___
 x



___


2
dx
dx
dx
(

y
)
dx
2
d
y
dy
____
⇒
___

2
2


y

0
2
dx
dx
2
d
y

(A

Bx)e
dy
y
___
____
x
i.e.
⇒

2
2


y

0
2
dx
dx
173
Section
3
Counting,
matrices
and
differential
equations
2
When
the
auxiliary
,
root
the
equation
general
au
solution

bu
of
the
0
can

c

0
has
differential
a
repeated
equation
2
d
y
dy
____
___
a

b

cy

be
quoted
as
2
dx
dx
 x
y

(A

Bx)e
Example
Find
the
general
solution
of
the
differential
equation
2
d
y
dy
____
___
4

12

9y

0
2
dx
dx
2
The
auxiliary
equation
is
4u

12u

9

0

0
2
⇒
(2u

3)
3
This
equation
has
a
repeated
root
equal
to
2
2
y
d
dy
____
so
the
general
solution
of
4
___

12

9y

0
2
dx
dx
3
x
is
y

(A

2
Bx)e
 x
The
equation
y

(A cos  x
e
B sin  x)

2x
Consider
the
equation
y

e
(A cos 3x

B sin 3x)
dy
___
2x

2e

2y

2
2x
(A cos 3x

B sin 3x)

e
(
3A sin 3x

3B cos 3x)
dx
2x

e
(
3A sin 3x

3B cos 3x)
2
d
y
dy
____
___
2x

2e

2
2x
(
3A sin 3x

3B cos 3x)

e
(
9A cos 3x
9B sin 3x)
2
dx
dx
dy
dy
___

___
2
(
dx
2y
)
9y
dx
2
d
y
dy
____
___
i.e.
4

13y

0
2
dx
dx
2
The
auxiliary
equation
is
u
4u

13

0
and
the
roots
of
this
_____
√
4 
36
__________
equation
are
the
conjugate
complex
numbers

2

3i
2
 x
Now
consider
the
general
case
y

(A cos x
e
B sin x)

dy
___
 x

e

y


 x
(A cos x
B sin x)


e
 A sin x
(

 B cos x)
dx
 x

e
 A sin x
(
 B cos x)

2
d
y
dy
____
dy
___
___


2
dx
(
dx
y
2
)

y
dx
2
d
y
dy
____
___
2
⇒
2

(
2


)y

0
2
dx
dx
2
The
auxiliary
equation
is
u
2
2 u

(
2


)y
________________
2
2

2
√4
4(
2


)
______________________
equation
are

2
174


i

0
and
the
roots
of
this
Section
3
Counting,
matrices
and
differential
equations
2
When
the

roots
auxiliary
i ,

the
equation
general
au

solution
bu
of

the
c

0
has
complex
differential
equation
2
d
y
dy
____
a
___

b

cy

0
can
be
quoted
as
2
dx
dx
x
y
Note
that

i.e.
then

y
if
the
roots

of
e
(A
the
 x
cos
auxiliary

B
sin
equation
 x)
are
purely
imaginary,
0,
A cos x


B sin x
Example
Find
the
general
solution
of
the
differential
equation
2
d
y
dy
____
___

4

5y

0
2
dx
dx
The
auxiliary
equation
is
________
√
4 
16
20
_______________
2
u

4u

5

0
⇒
u

2

2

i
2x
Therefore
the
general
solution
is
y

e
(A cos x

B sin x)
Summary
The
general
solution
of
the
differential
equation
2
d
y
dy
____
a
___

b

cy

0
2
dx
depends
on
the
nature
of
the
dx
roots
of
the
auxiliary
equation
2
au
If
the
roots

are
and


bu

c

0
then
x

when

and

when



when

and

are
real
and
distinct,
y

Ae
y

(A
y

e
x

Be
x
,

are
complex

Bx)e
conjugate
x
(A cos  x

B sin  x)
numbers,
Exercise 3.21
Find
the
general
solution
of
each
differential
2
d
y
dy
____
1
d
___

7
y
12y

0
2
___

2

5y

0
9y

0
2
dx
dx
dx
2
dx
2
y
d
____
3
y
____

4y

4
0
2

4y

6

0
2
dx
dx
2
d
dy
____

2
d
equation.
2
2
y
dy
____
5
d
___

7
y
dx
dy
____

0
6
2
___

2
dx
dx
dx
175
3.22
The
particular
Learning outcomes
The
integral
particular
1
integral
when
f(x)
is
a
polynomial or
a
constant

To
solve
differential
equations
of
2
d
y
dy
____
the form
Consider
the
differential
___
equation

2
4

3y

6x

x

2
2
dx
dx
2
d
y
dy
____
a
___

b
 cy  f(x)
2
dx
The
dx
polynomial
on
the
bx

right-hand
side
suggests
that
a
polynomial
of
the
2
form
where
f(x)
is
a
y

ax

c
might
be
a
solution
of
the
differential
equation.
polynomial
We
call
this
a
trial
We
can
test
this
solution
by
differentiation,
to
see
if
values
of
a,
b
and
c
exist
so
2
that
y

ax

bx

c
is
a
solution,
You need to know
2
dy
How
of
to find
the
the
general
differential
y

d
___
2
i.e.

ax

bx

c
⇒
y
____

2ax

b
and

2a
2
solution
dx
dx
equation
Substituting
into
the
left-hand
side
of
the
differential
equation
gives
2
d
y
dy
____
a
___

b

cy

0
2
dx
2
dx
2a

4(2ax

b)

2
3(ax

bx

c)

6x

4b

3c)


5

2
⇒
3ax
x

2
2

Comparing
(8a

3b)x

coefficients
(2a
gives
a

2,
b
6x

and
c
x


2
6
2
Therefore
y

2x
5x

6
is
a
solution
of
the
equation
2
d
y
dy
____
___

2
4

3y


6x
x

2
2
dx
dx
2
However
,
not
y
contain

2x
any
5x

constants
6
cannot
of
be
the
full
solution
because
it
does
integration.
2
The
function
2x
5x

6
is
called
the
particular
integral
2
d
y
dy
____
The
general
solution of
a
___

b

cy

2
d x
where
We
have
f(x)
is
found
a
a
d x
polynomial
solution
of
the
equation
2
d
y
dy
____
___

2
4

3y

6x

x

2
2
dx
dx
and
we
can
differential
find
the
general
solution
by
first
solving
the
simpler
equation
2
d
y
dy
____
___

4

3y

0
2
dx
dx
The
auxiliary
equation
is
2
u

4u

3

0
⇒
3x
∴
y

Ae
Ae

and
Ae
we
Be
to
x
can
Be
d
y
show
the
2x
that
particular
5x
this
is

the
dy
___

4
176
2

2
dx
dx
1)

0
⇒
u

3y
integral
gives
2

2
____

x


3)(u
Be
3x
y

x

3x
Adding
(u

6x

x

2:
6
general
solution
of
3
or
1
f(x)
Section
3x
y
x

Ae


3Ae
3
Counting,
matrices
and
differential
equations
2
Be

2x
5x

6
2
dy
d
___
3x
y
dy
____
x
Be

4x
___
5

4
2

3y

6x

x

2
2
dx
}
2
d
dx
y
____
3x

dx
x
9Ae

Be

4
2
dx
3x
x
Ae

the
Be
is
general
called
solution
complementary
the
of
complementar y
the
function
given
and
function ,
differential
the
particular
and
equation
by
we
have
adding
found
the
integral.
2
d
y
dy
____
For
any
differential
equation
of
the
for m
___
a

b

cy

f(x)
2
dx
where
y

f(x)
is
a
polynomial,
(complementary
the
general
function)

dx
solution
(particular
is
given
by
integral)
2
d
y
dy
____
where
the
complementary
function
is
the
solution
of
a
___

b

cy

0
2
dx
and
the
particular
as
f(x)
and
integral
whose
is
a
general
coefficients
substitution
into
can
the
be
given
polynomial
found
by
of
the
dx
same
differentiation
differential
order
and
equation.
Example
2
d
y
dy
____
Find
the
general
solution
of
the
equation
___

2

5y

7x
1
2
dx
First
find
the
particular
integral:
try
y

ax
dx

b
2
dy
d
___
____

⇒
y
a
and

0
2
dx
dx
Substituting
2a

5(ax
into

b)
the

given
7x
differential
equation
7
Comparing
coefficients
gives
a

19
__
and
b


5
7
Therefore
y

find
the
25
19
__
x
is
5
Next
gives
1
the
particular
integral.
25
complementary
function:
2
the
auxiliary
equation
is
u

2u
5

0
_______
√
2 
4
20
______________
⇒
u


1

2i
2
So
the
complementary
function
is
x
y

e
Therefore
(A cos 2x
the
general

B sin 2x)
solution
is
7
19
__
x
y

e
(A cos 2x

B sin 2x)

x
5
25
Exercise 3.22
Find
the
general
solution
of
each
differential
2
d
equation.
2
y
dy
____
1
d
___

y
____

y

1

x
2
2

2
dx
9y

x

2
2
dx
dx
177
3.23
The
particular
Learning outcomes
The
integral
particular
2
integral
when
f(x)
is
a trigonometric
function

To
solve
differential
equations
of
2
d
y
dy
____
the form
Consider
the
differential
equation
___

4

3y

2 cos x
3 sin x
2
dx
dx
2
d
y
dy
____
a
___

b
 cy  f(x)
2
dx
The
dx
y
where
f(x)
is
a

function
p cos x

on
the
right-hand
q sin x
might
be
side
a
suggests
solution
of
that
the
a
function
differential
of
the
form
equation.
trigonometric
Using
this
as
a
trial
solution,
we
can
differentiate
it
to
find
out
if
values
function
of
p
and
q
y
You need to know
exist
so
that

p cos x


p sin x

p cos x
y

p cos x

q sin x
is
a
solution.
q sin x
dy
___

q cos x
dx

How
to find
the
general
solution
2
d
y
____
of
the
differential
equation
q sin x
2
dx
2
d
y
dy
____
a
___

b

cy

0
2
dx
dx
Substituting
⇒
these
expressions
(
p cos x
q sin x)

2 cos x
3 sin x
(2p
p


4q) cos x
2q



(
4(
4p
into
the
p sin x


2q) sin x
given
q cos x)

2q

y



we
so
f(x)
The
y
p
__
1

and
q

5
10
is
a
solution
of
the
given
differential
equation
__
1
cos x
is
p cos vx
use
⇒
sin x

sin x
5
When
3 sin x
10
4
y
q sin x)
__
1
cos x
5
and

gives
3
4
∴
3(p cos x
2 cos x
4
}


equation
1
⇒
4p
differential

is
the
particular
integral.
10
any

combination
q sin vx
p cos 4x
general

as
the
of
cos
trial
vx
and/or
solution.
sin
For
vx
we
example,
use
if
f( x)

3 sin 4x,
q sin 4x
solution
when
2
d
y
dy
____
a
___

b

cy

a trigonometric function of
x
2
d x
d x
2
d
y
dy
____
For
the
differential
equation
___

4

3y

2 cos x
3 sin x
the
2
dx
dx
2
d
y
dy
____
complementary
function
is
the
general
solution
of
___

4

3y
2
dx
The
auxiliary
equation
dx
is
2
u

4u

3

0
⇒
(u

3)(u
1)

0
⇒
u

3
3x
Therefore
The
the
general
complementary
solution
3x
y

Ae
of
the
Be

is
Ae
differential
__
1
cos x
5
178
given
4
x

function

sin x
10
or
1
x

Be
equation
is
therefore

0
Section
3
Counting,
matrices
and
differential
equations
2
d
y
dy
____
For
any
differential
equation
of
the
for m
___
a

b

cy

f(x)
2
dx
where
f(x)
is
a
combination
the
y

(complementary
where
the
of
general
sines
and
solution
function)
complementary
dx
cosines
is
given

of
the
(particular
function
is
same
angle,
by
the
integral)
solution
of
2
d
y
dy
____
___
a

b

cy

0
and
the
particular
integral
is
p
cos
ux

q
sin
ux
2
dx
dx
and
where
p
and
and
q
are
constants
substitution
into
which
the
can
given
be
found
differential
by
differentiation
equation.
Example
Find
the
general
solution
of
the
differential
equation
2
d
y
dy
____
___
6

9y

4 sin 2x
2
dx
dx
Using
y

p cos 2x

q sin 2x
as
the
trial
solution
gives
dy
___

2p sin 2x


4p cos 2x
2q cos 2x
dx
2
y
d
____
4q sin 2x
2
dx
Substituting
(
into
4p cos 2x

⇒
the
given
4q sin 2x)
differential
6(
equation
2p sin 2x

gives
2q cos 2x)

9(p cos 2x

q sin 2x)
4 sin 2x
(5p
12q) cos 2x
Equating
coefficients
5p
12q

0

4
of

5q
(12p

cos 2 x
5q) sin 2x
and
sin 2x
⇒
p


4 sin 2x
gives
20
____
48
____
}
12p

,
q

169
169
1
____
∴
the
particular
integral
is
(48 cos 2x

20 sin 2x)
169
The
complementary
function
comes
from
the
general
solution
of
2
d
y
dy
____
___
6

9y

0
2
dx
dx
2
The
auxiliary
equation
is
u
6u

9

0
⇒
u

3
(repeated)
3x
so
the
complementary
Therefore
the
general

(A

Bx)e
solution
is
of
(A
the

Bx)e
given
differential
equation
is
1
____
3x
y
function

(48 cos 2x

20 sin 2x)
169
Exercise 3.23
Find
the
general
solution
of
each
differential
equation.
2
d
y
2
dy
____
1
d
x
____
___

3

2y

5 cos x
2

2
dx
3x

cos 2 

2 sin 2 
2
dx
d
179
3.24
The
particular
Learning outcomes
The
integral
particular
3
integral
when
f(x)
is
an
exponential
function

To
solve
differential
equations
of
2
d
y
dy
____
the form
Consider
the
differential
___
equation

x
4

3y

e
2
dx
dx
2
d
y
dy
____
a
___

b
 cy  f(x)
2
dx
The
dx
function
on
the
right-hand
side
suggests
that
a
function
of
the
form
x
y
where
f(x)
is
an

pe
might
be
a
solution
of
the
differential
equation.
Using
this
as
a
exponential
trial
solution
gives
function
2
dy

d
___
x
y
pe
⇒
y
____
x

pe
x
and

pe
2
dx
dx
x
You need to know
Substituting
into
the
given
differential
equation
gives
1
x
8 pe

e
⇒
p

8
2
d

How
to find
the
general
solution
1
Therefore
y
e
is
a
solution
the
differential
dy
____
a
and
___

b
4
x

3y

e
dx
equation
1
y

2
dx
2
d
___
of
8
of
dy
____
x

y
the
particular
integral
x
is
e
8

cy

0
2
dx
dx
The
auxiliary
equation
is
2
u

4u

3

0
⇒
(u

1)(u

3)

0
⇒
u

1
3y

e
u

1
3x
Therefore
Hence
the
the
complementary
general
solution
function
is
or
3
x
Ae

Be
of
2
y
d
dy
____
___

x
4

3y

3x
e
is
y

1
x
Ae

Be
e
8
dx
The failure
x

2
dx
case
2
d
y
dy
____
Now
consider
the
differential
___
equation
4
x

2
dx
The
auxiliary
equation
dx
is
2
u
4u

3

0
⇒
(u
1)(u
3)

3x
so
the
complementary
function
is
Ae
0
⇒
or
u

3
x

Be
x
If
we
use
y

pe
3x
y

the
Ae
as
a
x

Be
left-hand
trial
solution
x

3x
pe
side
of

Ae
the
we
get
x

given
Ce
which
is
only
the
general
solution
equation.
x
This
means
we
cannot
use
y

pe
when
the
complementary
function
x
already
includes
a
multiple
of
e
x
Instead
we
use
y

pxe
as
a
trial
solution,
giving
2
dy

d
___
x
y
⇒
pxe
x

pxe
pe
y
____
x

and
x

pxe
x

2pe


2
dx
Substituting
into
x
p(xe
the
dx
given
x

2e
differential
x
)
4p(xe
equation
x

e
x
)

3pxe
gives
1
x

e
⇒
p
2
1
Therefore
y


x
xe
is
a
solution,
2
differential
equation
3x
y

Ae
x

is
1
Be
180
x
xe
2
so
the
general
solution
of
the
given
of
Section
3
Counting,
matrices
and
differential
equations
2
d
y
dy
____
For
any
differential
equation
of
the
for m
___
a

b

cy

f(x)
2
dx
dx
x
where
is
f(x)
given
e

by
y
where
,


where

and
are
(complementary
the
complementary
constants,
function)
function

is
the
general
(particular
the
solution
integral)
solution
of
2
d
y
dy
____
___
a

b

cy

0
and
the
particular
integral
depends
on
the
2
dx
dx
powers
of
e
in
the
complementary
function:
x
use

pe
when
the
complementary
function
does
NOT
x
contain
e
x
use

pxe
when
the
complementary
function
DOES
contain
x
e
x
2
use

px
e
when
x
the
complementary
function
contains
both
x
e
and
and
xe
where
p
substitution
can
be
into
found
the
by
given
differentiation
differential
and
equation.
Example
Find
the
general
solution
of
the
differential
equation
2
d
y
dy
____
___
6
3x

9y

e
2
dx
dx
First
find
the
complementary
function.
2
The
auxiliary
equation
is
u
6u

9

0
⇒
u

3
3x
Therefore
the
complementary
3x
This
contains
function
is
(A
3x
e
and

2
xe
so
we
use
y

px
Bx)e
3x
e
as
a
trial
solution.
dy
2
y

px
___
3x
e
2
⇒

3px
3x
e
3x

2pxe
dx
2
d
y
____
2
⇒

9px
3x
3x
e

12pxe
3x

2pe
2
dx
Substitution
2
(9px
into
the
3x
e
given
equation
3x

gives
3x
12pxe

2pe
2
)
6(3px
3x
3x
e

2pxe
2
)

9px
9y

e
3x
e
3x

e
1
⇒
p

2
2
d
1
Therefore
y

2
x
is
a
solution
the
dy
of
___
6
3x

2
dx
1
So
y
____
3x
e
2
particular
integral
is
2
x
dx
3x
e
2
1
3x
and
the
general
solution
is
y

(A

Bx)e

2
x
3x
e
2
Exercise 3.24
Find
the
general
solution
of
each
differential
2
d
equation.
2
y
dy
____
1
d
___

2
y

e
y
dx
dy
____
2x

2
___

2
2
x

5y

4e
2
dx
dx
dx
181
3.25
Using
boundary
Learning outcomes
conditions
Summary of
differential

To
summarise
solution
of
the
general
solutions of
second order
linear
equations
general
differential
equations
2
d
of
y
dy
____
the form
a
___

b

cy

0
2
dx
2
d
y
dx
dy
____
a
___

b
 cy  f(x)
2
dx
dx
2
Auxiliary

To find
the
solution
equation
au

bu

c

0
General
solution
of
x
real
and
distinct
roots,


and
x
y

Ae

Be
y

(A  Bx)e
y

e
2
d
y
dy
____
a
___

b
 cy  f(x)
2
dx
x
dx
given
repeated
boundary
root,

conditions
x
complex
conjugate
roots,

i

(A cos x

B sin x)
2
You need to know
d
y
dy
____
a
___

b

cy

f(x)
2
dx

How
to
differentiate
functions
and
dx
standard
products
of
General solution:
y

complementary function
functions

particular
d
y
dy
____
the

The
integral where
2
meaning
of
the
complementary function
is the
solution of
___
a

b

cy

f(x)
2
auxiliary
dx
dx
equation
and
the
form

How
from
to find
a
trial
a
particular
of
particular
integral
comes from
a
trial
solution
that
depends
on
the
f(x)
integral
solution
f(x)
polynomial
or
Trial
constant
polynomial
e.g.
f(x)

of
3x
same

2,
solution
order
y

as
px
f(x),

q
trigonometric:
u cos  x
v sin  x
}
u cos  x

y

p cos x
y

pe

q sin x
v sin  x
x
exponential:
x
ue
x
when
e
is
not
part
of
the
complementary function
x
y

pxe
x
when
e
is
part
of
the
complementary
function
x
2
y

px
e
x
when
e
x
and
xe
are
part
of
the
complementary function
When
trial
f(x)
is
not
one
of
the
forms
given
in
the
table,
you
will
be
given
a
solution.
Boundary
When
we
are
conditions
given
boundary
conditions,
i.e.
corresponding
values
of
x,
y
dy
___
and
possibly
,
we
can
use
these
in
the
dx
equation
182
to
find
the
particular
solution.
general
solution
of
a
differential
Section
3
Counting,
matrices
and
differential
equations
Example
2
d
v
____
Solve
the

__
equation

4v

8t
given
that
v

0
when
t

0
and
when
t

2
dt
4
2
d
v
____
2

4v

8t
gives
the
auxiliary
equation
u

4

0
⇒
u
A cos 2t

B sin 2t


2i
2
dt
Therefore
the
complementary
function
is
2
dv
___
Using
v

pt

q
as
a
trial
solution
d
v
____
gives

p
and

0
2
dt
Substituting
Therefore
into
the
the
given
particular
integral
⇒
When
v

0
and
t

When
v

0
and
t

equation
0:
is
gives
dt
4 pt

4q


2t
so
p
v


2
and
q

0
2t
v

A cos 2t
0

A
0

B

__

B sin 2t

__
:
8t

__

4
⇒
B


__

∴
2
2t
sin 2t
2
2
Example
2
d
y
dy
____
Solve
the
___
x
equation

y

5e
sin x
given
that
y

0
and

2
when
x

0
2
dx
dx
x
Use
y

pe
x
cos x

qe
sin x
as
a
trial
solution.
2
The
auxiliary
equation
is
u

1
⇒
u

i
so
the
complementary
function
is
A cos x

B sin x
dy
x
y

pe
___
x
cos x

qe
sin x
x
⇒

x
e
(p cos x

q sin x)

e
(
p sin x

q cos x)
dx
x

e
x
(p

q) cos x

e
(
p

q) sin x
2
d
y
____
and
x
x

e
(p

q) cos x

2qe
cos x
into
the

e
x
(
p

q) sin x
e
x
(p

q) sin x

e
(
p

q) cos x
2
dx
x
x
2pe
sin x
x
Substituting
given
equation
gives
e
x
so
the
particular
integral
is
2e
y

A cos x

B sin x
2e

2q) cos x

(
2p
x

e
x

0,
y

0
⇒
A

q) sin x )

5e
sin x
⇒
p

2,
q

1
sin x
x
cos x

e
sin x
x
When

x
cos
x
∴
x
((p
2,
so
y

2(1
e
x
) cos x

(B

e
) sin x
dy
___
⇒
x

2(1
e
x
) sin x
2e
x
cos x

(B

e
x
) cos x
e
sin x
dx
dy
___
When
x

0,
x

2
⇒
B

3,
∴
y

2(1
e
x
) cos x

(3

e
) sin x
dx
Exercise 3.25
2
d
y
dy
____
1
Solve
the
equation
dy
___

___

2y

10 sin x
given
y

0
and

1
2
dx
when
x

dx
dx
0
2
d
y
dy
____
2
Solve
the
equation
___

5
x

6y

(4x

3)e
using
2
dx
dx
dy
___
x
y

(px

q)e
as
a
trial
solution
and
given
that
y

0
and

0
dx
when
x

0
183
3.26
Using
substitution
Learning outcomes
Substitution
We

To
use
substitution
to
reduce
have
seen
in
Unit
1
that
we
can
sometimes
use
a
substitution
to
find
a
dy
differential
equation
to
a form
___
in
y
when

f(x)
dx
which
it
can
be
solved
It
is
also
order
sometimes
differential
possible
equation
to
to
a
use
a
form
substitution
that
can
be
to
reduce
a
second
solved.
2
d
You need to know
the
differential
equation
dy
____
2
Consider
y
___
x

2x
12y

6
2
dx

How
to find
the
general
solution
We
of
an
equation
of
dx
the form
know
that
when
the
left-hand
side
is
a
second
order
linear
equation,
x
the
solution
often
involves
e
u
,
so
we
will
try
the
substitution
x

e
2
d
y
dy
____
a
___

b
u
 cy  f(x)
When
x

e
,
using
the
chain
rule
gives
2
dx
dx
dy
dy
___
The
chain
dy

relationship
du
dx
___
dy
1
__
___

differentiate
implicit
dy
use
an

How
to
integrate
x
du
___
x

[1]
dx
to

e
dy
___
⇒
How
___

u
functions

dy
1
__

du
to
dx
___
du
dy
How
du

dx


dx
1
___
___
The
1
___
___

rule
dx

dy
du
___
___


du
integrating factor
Differentiating
by
[1]
with
parts
respect
to
x
gives
2
y
d
dy
____
x
dy
d
___
___

___
(

2
dx
dx
dx
)
du
2
d
2
y
d
du
___
____


y

2
du
2
d
2
⇒
1
__
____

2
dx
du
x
2
y
dy
____
x
d
___

x
y
____

[2]
2
2
dx
dx
du
2
d
2
Expressing
the
given
equation
as
y
dy
____
dy
___
x

x
___

x
12y

6
2
dx
2
d
can
now
dx
2
y
d
____
we
dx
2
substitute
for
y
dy
____
x
dy
___

2
dy
___
x
and
___
for
x
giving
2
du
dx
dx
du
dx
2
d
y
dy
____
___

12y

6
2
du
The
du
left-hand
side
is
now
linear
and
second
order
,
so
the
equation
can
be
solved.
Substitution
can
equations
first
to
also
be
order
used
to
transform
equations.
This
some
usually
second
makes
order
the
differential
integration
easier
.
2
d
y
dy
____
Consider
the
equation
___

2

4x
2
dx
There
is
no
term
involving
dx
y
in
this
equation,
so
we
can
reduce
it
to
a
2
dy
first
order
equation
with
the
substitution
u

d
du
___
___
so
that
y
____

2
dx
dx
dx
du
___
The
given
equation
then
becomes

2u
dx
∫2
using
the
integrating
2x
Therefore
factor
du
___
e
2x
184
ue
2ue
2x


2x

dx
⇒
I
∫4xe
dx
e
2x

4xe
dx
2x

e

4x
which
can
be
solved
Section
Using
integration
2x
by
parts
2x
ue

2xe

2xe

2x
3
Counting,
matrices
and
differential
equations
gives
2x
∫2e
2x
dx
2x
e

A
2x
⇒
u
Substituting
1
back

for
Ae
u
gives
another
first
order
differential
equation:
dy
___
2x

2x
1

Ae
dx
1
2
Integrating
again
gives
y

x
x
2x
Ae

B
2
Example
dy
___
Use
the
substitution
u

to
find
the
general
solution
of
the
dx
differential
equation
2
2
d
y
dy
____
___

(
2
2
dx
)

0
dx
2
dy
d
du
___
___
u

y
____
⇒

2
dx
dx
dx
2
2
d
dy
y
___
____

∴
(
2
2
)

0
dx
dx
du
___
⇒
2

2u

0
dx
This
equation
1
___
can
be
integrated
du
___
i.e.
by
separating
the
variables,
1
__

2
⇒

2x

A
2
u
dx
u
1
_______
∴
u

2x
dy

A
1
_______
___
so

dx
2x

A
1
⇒
y

ln|2x

A|

B
substitution
u
2
Exercise 3.26
dy
___
1
Use
the

to
find
the
general
solution
of
the
dx
2
2
d
y
dy
____
equation
___

x
2
dx
(
)

0
dx
dy
___
Given
that
y

0
and

1
when
x

0,
find
y
in
terms
of
x.
dx
u
2
Use
the
substitution
x

e
to
show
that
the
differential
2
d
2
equation
2
y
dy
____
x
d
___

x
y
____

y

0
can
be
expressed
as

2
dx
y

0.
2
dx
du
2
d
2
Hence
find
the
general
solution
of
y
dy
____
x
___

x

y

0
2
dx
dx
185
Section
1
A
a
sim
card
unique
of
one
3
Practice
manufacturer
registration
digit
including
chosen
the
code.
from
letters
marks
O
1
This
to
and
6,
I,
questions
each
sim
code
consists
two
two
with
letters
digits
3
are
batsmen
5
are
bowlers
8
are
neither
and
and
wicket
wicket
keepers,
keepers,
not
batsmen
nor
bowlers
nor
chosen
wicketkeepers.
from
0
to
9
inclusive
and
ending
with
two
letters,
(a)
again
not
including
O
and
I.
All
digits
and
Draw
a
V
enn
diagram
to
show
this
letters
information.
can
be
repeated.
(b)
The
manufacturer
has
made
20 000 000
sim
One
member
random.
How
many
more
sim
cards
can
be
made
before
format
for
the
codes
needs
to
be
Four-digit
numbers
1,
7
are
made
from
the
4,
6,
and
9.
Each
digit
is
used
only
many
different
even
chosen
at
that
chosen
is
not
a
batsman,
nor
a
the
bowler
,
a
wicket
keeper
.
A
numbers
bag
contains
5
red
discs,
8
blue
discs
and
once.
6
How
(a)
is
probability
digits
8
2,
club
the
introduced?
nor
2
the
Calculate
a
person
new
of
cards.
can
white
discs.
be
One
disc
is
removed
at
random
and
not
replaced,
made?
then
How
(b)
many
different
numbers
can
be
a
second
Calculate
that
are
greater
than
Three
the
coins
are
number
of
tossed
ways
simultaneously.
they
can
land
so
at
(b)
4
Three
on
least
one
least
two
cubical
the
coin
lands
coins
dice
uppermost
land
are
with
with
thrown
faces
is
removed
probability
are
a
head
a
head
and
added
the
to
are
the
same
3
5
1
A

uppermost
(
6
0
4
1
4
3
two
discs
colour
.
1
0
4
and
)
B

2
2
0
3
1
2
(
)
uppermost.
numbers
(a)
Find
A

2B
(b)
Determine
x
and
y
if
form
A
the
that
number
the
less
(a)
score
than
of
ways
in
which
they
can
One
cubical
3B

x
12
(
8
is
6
greater
(b)
dice

land
than

y
13
is
biased
so
that
when
xy
4
1
9
it
)
3
10.
10
5
random.
score.
Find
so
at
the
Calculate
0
the
that
that
9
at
(a)
the
4200?
removed
3
disc
made
A

(2
1
4)
and
B

is
0
( )
1
thrown,
it
is
twice
as
likely
to
show
a
six
on
its
Show
uppermost
cubical
face
dice
is
as
any
other
score.
A
that
| AB|

10
but
of
these
thrown.
what
is
If
a
the

0
unbiased.
cos 
One
|BA|
second
dice
six
is
chosen
shows
probability
on
at
the
that
random
and
uppermost
the
biased
then
Given
that
A

(
)
sin 
face,
dice
sin 
2
11
show
that
A
cos 
was
1
x
2
1
0
4
2
1
1
chosen?
12
6
The
an
cards
numbered
ordinary
pack
smaller
pack.
smaller
pack.
of
Three
2
52
to
9
are
playing
cards
are
withdrawn
cards
drawn
to
form
from
Given
A

(
a
(a)
Find
the
(b)
When
value
same
the
probability
that
they
all
x

1
show
A
the
27
are
batsmen,
27
are
bowlers,
16
are
wicket
8
are
186
50
members
of
a
cricket
batsmen
and
bowlers,
which
find
the
value
1
(
)

(
13
)
4
club,
13
keepers,
for
the
number
.
Of
x
6
y
7
of
|A|

0
this
1
Calculate
)
from
Find
the
⎥
value
of
a
given
1
1
1
a
2a
3a
2
a
⎥
2
a
2

1
a

that
4
of
y
for
which

I
Section
14
A

(
1
1
4
0
1
1
2
0
1
and
)
T
Determine
(a)
A
B

that
1
1
2
0
1
1
2
0
Find
(b)
)
x

the
solution
given
Practice
that
y

questions
1
when
0
T
22
B
T
Show
(b)
(
0
3
T
( AB)

A
Find
the
general
solution
of
each
differential
equation.
T
B
2
d
y
dy
____
___

(a)
8

12y

0

16y

0
2
15
Determine
(a)
equations
which
are
of
the
following
sets
dx
of
dx
2
consistent:
d
2
y
2x
(i)

y

4

8
y

2x
2x

y

4

4


(c)
2
9y

0
2
dx
(ii)
y
____
___

0
d
dy
____
(b)
dx
dx
0
2
d
y
____
x

y

2

0
23
For
the
differential
equation

16y

3x

1
2
dx
(iii)
2x

y

3y

6x
4

0
find
12
the
(a)
(b)
Express
(i)
the
set
of
equations
that
has
the
(b)
unique
solution
as
a
matrix
row
reduction
to
solve
integral
complementary
function
equation.
the
(c)
Use
(ii)
particular
a
the
general
solution.
matrix
equation.
24
1
2
2
2
3
1
Given
that
integral
of
y

the
a cos 2x

differential
b sin 2x
is
a
particular
equation
2
d
y
dy
____
16
Show
that
the
matrix
A

(
4
can
be
reduced
to
(
2
1
2
2
0
1
3
0
0
)
___

2

3y

10 cos 2x,
find
2
dx
dx
1
(a)
the
values
(b)
the
general
)
of
the
constants
solution
of
the
a
and
b
differential
equation.
11
2
Hence
find
d
|A|.
y
dy
____
25
Given
___
that

2x
4

4y

e
2
dx
17
Use
row
reduction
to
solve
the
equations
find
(a)
2x

y

x

2y
2z

2
z

6
3x

4y

2z

7
the
explain
not
a
find
(c)
why
Express
the
system
of

the
ae
where
particular
particular
solution
integral
of
the
a
is
a
constant
is
integral
and
hence
differential
give
the
equation
find
the
particular
solution
given
that
dy

y

3z

4
3x

y

2z

2
4y

6z

1
___
y

function
equations
(d)
2x
2x
y
suitable
general
18
complementary
2x
(b)

dx
as
a
matrix

1
and

0
when
x

x

0
dx
equation.
u
Hence
show
that
the
system
is
not
26
consistent.
Use
the
substitution
e
to
show
that
the
2
d
2
differential
equation
y
dy
____
___
x

2x

3y

0
2
3
2
1
1
1
2
4
3
2
2
6
5
dx
dx
2
d
y
dy
____
19
Given
A

(
0
1
and
)
B

(
1
0
3
reduces
)
___
to


3y

0
2
du
du
1
Hence
find
the
general
solution
of
the
differential
2
find
the
matrix
C
that
satisfies
the
equation
d
2
equation
1
AC
20
(a)

A
x
dy
___

2x

3y

0
2
B
Find
y
____
dx
the
general
solution
of
the
dx
differential
dy
___
27
dy
2
equation
___
x

2xy

Use
the
substitution
u

to
reduce
cos x
dx
Find
the
particular
second
2
d
y
dy
____
(b)
the
dx
solution
given
that
when
order
differential
equation
___

x

3x

0
2
dx
dx

__
x

,
y

0
to
a
first
order
differential
equation.
2
Hence
21
(a)
Find
the
integrating
factor
for
solving
find
the
general
solution
of
the
equation
the
2
d
dy
___
differential
equation
2

x
2
y

y
dy
____
___

x
x

3x

0
2
dx
dx
dx
187
Index
conjugate
A
addition
of
complex
addition
of
matrices
addition
of
vectors
alternating
angle

and
17
finding
a
of
62–3,
of
auxiliary
complex
first
n
matrix
graphs
square
22–5
numbers
67,
terms
14–15,
69,
70–1,
158–9,
equations
17
sum
73
172–6,
bias,
dice
135,
131,
137
coefficients
binomial
expansion
binomial
theorem
applications
derivation
binomials
94–7
96–7
of
88–92
interval
28,
136–7
complex
roots
of
equations
67,
176–7,
integration
sequences
convergent
series
converging
iteration
odd
11,
174
system
156,
160,
68,
69
79–80,
84–5,
94
110
determinants
powers
in
148–9
52
1
bisector
,
perpendicular
boundary
brackets,
23,
conditions
167,
expanding
cos
24
x
32–3
derivative
182–3
principle
93
counting
cubic
C
calculators,
car
theft
cards
127,
certainty
136–7
116,
121
28,
30,
35,
42,
146–7,
115,
column
122,
vectors
combinations
difference
150,
138,
126
gradient
curves,
tangent
De
Moivre,
De
Moivre’s
difference
ratio
complementary
106
to
26,
106
140,
158,
18
theorem
into
integral
derivation
137
of
18–21
partial
51,
44,
fractions
121
45,
46
binomial
of
combinations
of
trig
of
functions
inverse
67
theorem
88–92
functions
36–7
34,
52
35
determinants
function
coefficients
complex
conjugate
complex
numbers
of
in
177–83
quadratic
roots
182
6–11,
12
equations
11
coordinate
expansion
of
matrices
minor
in
geometry
7
argument
14–15,
diagrams,
17
dice
146–9
146–7
Argand
115,
148–9
148
simplification
8
applications
44,
59–60
derivatives
160
67
complex
of
7
of
Abraham
denominator
153–4
135,
permutations
common
addition
curves,
definite
120–1
common
188
50,
117
126,
from
35
33
114–15
equations
decomposing
arrangements
cofactors
34,
values
D
22–5
circular
coins
134,
126
rule
circles
111
127
playing
chain
scientific
of
122,
161
182
72–3,
geometry,
even,
7,
182
59
convergent
x,
30
92
probability
conjugate
cos
102–3
27,
82–3
numbers
coordinate
88
bisection,
182
complex
of
93,
175,
conjugate
92–7
89,
8
interest
constants
89
19
10–11
of
functions
consistency
136
binomial
15,
16
compound
12
form
17
expanding
178–83
24
8–9
of
roots
composite
161–4
17,
16
of
of
complex
73
16–17
8
on
subtraction
B
coins
14,
roots
conditional
bias,
of
of
quotient
83
12–17,
progressions
augmented
of
curve
diagrams
arithmetic
sum
product
21
22–5
polar-argument
under
argument
loci
operations
83
33
Argand
8
form
representation
modulus
85–7
32
arctan
axes
graphical
69
174
16
of
exponential
68,
11,
of
multiplication
Archimedes
area
division
33
arcsin
7,
difference
16
138–9
16,
approximations,
8,
16
sequences
14–15,
arccos
numbers
150–1
12–17,
126,
128,
22–5
131,
133
46
Index
difference
of
differences,
differential
exact
order
170,
case
of
16
equivalent
74–5
Euler ’s
166–7
166,
solutions
185
182–4
166–85
approximate
general
trial
170–1,
172,
176,
happening
not
independent
not
mutually
probability
169,
or
174–85
differentials
134–6
exclusive
two
131–3
130–3
134–7
sample
expanding
130
128–9
with
more
exhaustive
178–83
157–8
130–1
exclusive
not
85–7
167,
equations
82
126
mutually
168,
166,
of
21,
independent
166
order
systems
formula
events
171
180–1
order
second
numbers
of
equations
167,
failure
first
complex
method
space
brackets
127
93
expanding
composite
expansion
of
f(x)
functions
82–3
n
of
e
of
e
of
inverse
27
(a

b)
90–1
x
26–7
second
expansion,
trigonometric
function
34
38–9
summary
36
differentiation
of
exponential
of
fractions
of
implicit
of
logarithmic
of
parametric
partial
as
by
functions
of
vectors
series
28
of
factorial
42
fair
12
122,
69
first
8
82,
180
128,
(z)
the

auxiliary
cubic
172–6,
second
trial
first
of
a
terms
168,
170–1,
185
76–7
sequence
76
108
58–61
terms
71
166,
168,
166,
solutions
172,
176,
systems
170–1,
of
185
182–4
178,
48
94
partial
44–9,
proper
44,
with
179–83
of
45
negative
166–7
order
equivalent
n
of
improper
order
166,
82
differentiation
178–83
7
first
equations
fractions
24
differential
66
108–10
term
sum

126
72
21,
reduction
130
equations
arg
dice
recurrence
126,
116
168–9
differential
iteration
x
likely
78,
sequence
Euler ’s
132–3
E
81,
and
order
for
e
notation
constant
nth
equally
42–3
formulae
numbers
129,
182
26–7
integrating
coins
finite
72–3
complex
180–1,
of
of
21
F
Fibonacci
68,
numbers
26
38
78–9
sequences
trial
function
differentiation
28
integration
divergent
drug
exponential
factors,
divergent
division
curves
integration
30–1,
equations
term
displacement
complex
exponential
48
functions
81
126
exponential
26–7
40–1
reverse
term
functions
standard
experiments
74–5,
45,
quadratic
96
46
factors
46–7
functions
157–8
x
linear
,
systems
matrix
matrix
of
157,
156,
normals
29
parametric
28
represented
roots
of
6,
row
of
|z|

| zz
r
of
82
expansion
trigonometric,
160–1
two
or
of
83
integration
more
fundamental
by
7,
three
9,
10,
29,
planes
160
107
reduction
variables
counting
of
52–3
40
principle
114–15
methods
158–9
106
G
general
47
22
|  a
157,
of
numerical
tangents
e
7
simultaneous
solving,

series
158
representation
polynomial
f(x)
160–3
terms
geometric
102–3
sum
to
infinity
sum
of
first
gradient
of
graphical
22–3
70,
graphs
74
progressions
a
n
terms
curve
67,
69,
73,
95
73
71–2
106
representation,
complex
numbers
16–17
26
1
equivalent
operations
50
axes,
real
and
imaginary
12
189
Index
M
I
identities
identity
Maclaurin,
44
matrices
imaginary
axes
imaginary
numbers
implicit
of
graphs
30–1
30–1,
38
independent
permutations
infinite
sum
81,
to
integral,
identity
factors
of
reverse
partial
59
of
differentiation
fractions
in
of
exponential
of
logarithmic
of
trigonometric
inverse
by
parts
using
of
functions
of
value
linear
inverse
cosine
inverse
matrices
reduction
sine
function
inverse
tangent
inverse
trigonometric
iteration
iteration
to
33
function
find
a
103,
154,
158
140–3,
142,
of
of
complex
of
145,
145,
152,
152,
160
157,
160–1
74–5
146–7
numbers
complex
inverse
154
157,
equations
determinants
modulus
144
142
142,
differences
166
158
154
representation
models
140
153
157,
commutative
in
vectors
164
143,
multiplicative
97
108–10
root
145,
145
of
164–5
138–9
144,
multiplication
32–5
108–111
formula
of
144,
minor
164–5
56–7
numbers
152,
zero
method
139
152–4,
161–3
144–6,
unit
matrix
32
function
integration
150,
premultiplying
164–5
scalars
column
138,
141
141–2
postmultiplying
152–4,
inverse
irrational
by
164–5
inverse
form
associative
153–4
161–4
141
140,
multiplication
32–3
152,
with
echelon
non
166
153–4
equations
98–9
106,
reduction
matrix
104–5
function
159,
matrix
102–3
multiplicative
row
50–2
42–3
84,
146–9
two
subtraction
52–3
theorem
bisection
product
square
185
x
of
singular
42–3
43
56–7
interpolation,
using
row
product
row
functions
substitution
intermediate
42
48–9
functions
54–60,
reciprocal
interval
of
multiplicative
constant
76
166
139,
multiplication
168–9
integration
as
158,
non-conformable
176–83
94
138–9
152,
using
94
89,
144–5
inverse
50
particular
integrating
of
determinants
85
negative
18,
conformable
94
84,
138–45
addition
118
73
function
integers,
models
augmented
conditions
‘inside’
induction
mathematical
21
series
infinity,
initial
of
78–81,
82–3
mathematical
matrices
130–1
85
theorem
mathematicians
126
events
law
84
83,
applications
independent
indices,
Colin
series
Maclaurin’s
12
6
functions
differentiation
impossibility
Maclaurin
144–5
of
14,
17,
numbers
matrices
mutually
exclusive
events
mutually
exclusive
outcome
mutually
exclusive
permutations
24
8
152–4,
164–5
130
134–5
119
108–111
N
L
law
of
indices
law
of
logarithms
leading
line
21
28
diagonals
segments
144,
146,
n
→
n
terms,
n!
152–3
∞
72–3
of
first
70–2
,
74–5
116
natural
23
68,
sum
numbers
77
n
linear
equations
C
notation
89–90
r
second
order
differential
systems
of
2
×
2
systems
of
3
×
3
linear
loci
loci
interpolation
166,
fractions
negative
integers
160–3
negative
numbers
104–5
24
Sir
Newton’s
functions
differentiation
integration
190
Newton,
94
94
6
Isaac
106,
Newton–Raphson
intersection
logarithms,
182–4
negative
22–5
logarithmic
172,
156
of
of
law
of
normals,
28,
43
28
36–7
nth
term
of
a
laws
of
motion
equations
66,
67,
sequence,
approaching
166
(Newton’s)
of
69,
29
70,
formula
zero
method
166
73
94
76
106–7
Index
number
plates
115,
number
series
70–3
numbers,
imaginary
numbers,
irrational
numbers,
real
numbers,
negative
118,
proof
119
by
induction
properties
of
18–20,
sequences
and
76
series,
proving
76–7
6
97
Q
8
quadratic
with
numerators
numerical
44,
equations
6,
7,
9,
10,
172–3
6
complex
coefficients
11
45
methods
to
solve
equations
quadratic
factors
quadratic
formula
7,
46–7
102–3
questions
and
7
answers
64–5,
186–7,
112–13
O
operations
orders
of
on
complex
differential
numbers
sequences
outcome
115,
122,
of
complex
quotient
rule
numbers
17
16–17
equations
oscillating
quotient
166,
170–2,
30,
48
185
69
123,
126,
128,
130
R
mutually
exclusive
134–5
radians
overlapping
111
123–5
radius
22
random
selection
126,
128,
129
P
Raphson,
parametric
curves
parametric
equations,
differentiation
of
fractions
44–9,
of
decomposing
74–5,
functions
23–4
96
real
axes
real
numbers
of
graphs
into
integration
particular
one
12
48–9
44,
8
46
reciprocals
in
44
28
rays
applications
106
28–9
rational
partial
Joseph
152
48–9
integral
182,
recurrence
formula
recurrence
relations
108
183
61,
reduction
two
formulae
76
58–61
178–9
repeated
three
Pascal’s
roots
173–4,
triangle
88–9,
of
equations
6,
pensions
sequences
10–11,
116–19,
arrangements
from
98–101,
iteration
combinations
mutually
real
6
116,
polar-argument,
polynomial
vectors
series
15,
142,
counting
vectors
145,
152,
scale
154
145,
152,
157,
160
114–15
136–7
126
happening
128–9
two
events
130–3
with
two
or
events
more
terminology
product
of
product
rule
factor
progressions,
30,
progressions,
n
order
38,
geometric
to
infinity
sum
of
first
n
second
partial
67,
69,
70–1,
61,
68,
69
68,
69
divergent
68,
Fibonacci
66
term,
67,
69,
73,
sum
182–4
69,
110
66
formula
76
69
69
to
72–3,
79–80,
84–5,
94
72–3
expansion
infinite
71–2
172,
70
divergent
95
166,
41
66–74
convergent
73
73
130
equations
derivative
convergent
series
127,
111
alternating
periodic
16
41
73
terms
122–5,
127
differential
generating
numbers
37,
terms
sum
161–3
164–5
16–17
oscillating
arithmetic
first
161,
140
calculators
second
nth
134–7
126
complex
138,
spaces
sequences
not
158–9,
exhaustive
with
of
row
sample
182
126–7
definition
sum
reduction
scientific
142,
conditional
182
matrices
S
176–7,
probability
basic
row
of
19
78–81
premultiplying
principles,
numbers
7
12–13
postmultiplying
form
23
complex
44,
echelon
121
equations
polynomials
row
119
bisector
cards
173–4,
118
exclusive
perpendicular
107
108–111
121
repeated
independent
104–5,
formula
117
no
difference
events
108–111
182
128
using
circular
175,
69
locating
permutations
position
9
106–7,
92
complex
playing
7,
91
approximations
periodic
182
180–1
roots
power
66–7,
176–7
83,
81,
94,
95
94
infinity
72–3
191
Index
Maclaurin
sum
of
83,
first
number
n
84,
terminating
simultaneous
x,
even,
odd
solving
row
square
67
terms,
general
70,
74
intermediate
94
10,
powers
inverse
47
trapezium
52
tree
32
150,
trial
152,
153
Moivre’s
rule
114,
solutions
triangles,
methods
102–3
method
138,
using
154,
area
inverse
158
turning
184–5
134–6
148–9
of
178–9,
52–3,
32–5,
points
two -way
98–9
178–83
function
56–7
56–7
differentials
81
18–21
value
123,
176,
integration
158–9
144–6,
94
62–3
diagrams
trigonometric
expansions
of
34
100
table
122
of
matrices
complex
66,
theorem,
reduction
subtraction
finding
81,
matrices
substitution,
81,
terms,
De
matrices
standard
84–7
series
theorem,
equations
numerical
theorem
terminating
85–7
equations
function,
singular
74–5
78–81
T
aylor ’s
sine
T
aylor ’s
70–2,
70–3
power
sin
85
terms
138–9
U
numbers
8
u
66
n
vectors
sum
16
unbiased
of
unit
()
to
coins
dice
144,
126
145,
164
70
infinity
cubes
of
squares
first
n
72–3,
the
of
first
the
terms,
complex
summary
75
n
first
n
numbers
natural
formula
numbers
of
V
natural
numbers
77
77
variables,
vectors
76–7
8,
differentials
function
column
36
position
two -way
122
40
16
138,
displacement
row
of
12–17
addition
16
T
table,
and
matrices
140,
158,
160
12
12–13
138,
140
subtraction
16
1
tan
x,
tangent
principle
values
function,
tangents
to
a
33
inverse
curve
26,
V
enn
diagrams
123–5,
33
106
X
n
tangents,
equations
tangents,
gradients
T
aylor
,
Brook
T
aylor ’s
192
series
of
of
29,
26,
84
84,
106
x
94
31
Z
85–7
zero
matrices
144,
145
129–32
182
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