Pure
Mathematics
for
CAPE®
Unit 1
Pure
Mathematics
for
CAPE®
Kenneth
Charles
Sue
Baisden
Cadogan
Chandler
Mahadeo
Deokinandan
Unit 1
3
Great
Clarendon
Oxford
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information
work.
in
Tech-Set
referenced
for
in
all
at
Contents
Introduction
Section
1.
1
1
5
Basic
algebra
Terminology
and
1.20
Exponential
and
logarithmic
functions
56
1.21
Modulus functions
58
1.22
Modulus
and functions
principles
equations
and
6
inequalities
1.2
Binary operations
8
Section
1.3
Surds
1.4
Logic
60
1
Practice questions
62
12
Section
and truth tables
2
Trigonometry,
geometry
14
and vectors
1.5
Direct
proof
18
2.
1
1.6
1.7
Proof
by
induction
20
Remainder theorem
and factor
theorem
1.9
1.
10
1.
11
Factors of
Quadratic
Curve
and tangent
functions
64
2.2
Reciprocal trig functions
2.3
Pythagorean
2.4
Compound
2.5
Double
2.6
Factor formulae
68
a

and
b
,
n
cubic

6
24
equations
1.
13
Inequalities
angle formulae
curves
angle
identities
76
expressions
36
– quadratic
and
expressions
curves
2.8
Intersection of
lines
42
1.
15
Functions
44
1.
16
Types of function
48
The
a
38
and
78
32
1.
14
expression
cos


b
sin
Trigonometric

80
identities
and
equations
2.9
General
82
solution of trig
equations
2.
10
Coordinate
straight
1.
17
Inverse function
Logarithms
1.
19
Exponential
equations
52
and
86
geometry
and
lines
90
50
2.
11
1.
18
72
26
2.7
Rational
70
30
Transformation of
1.
12
identities
n
sketching
rational
cosine
22
n
1.8
Sine,
logarithmic
and the
equation of
a
circle
2.
12
54
Loci
92
Equations of tangents
normals to
circles
and
94
3
Contents
2.
13
Parametric
equations
2.
14
Conic
2.
15
The
parabola
100
2.
16
The
ellipse
102
2.
17
Coordinates
sections
in
3-D
96
3.
10
Rates of
change
98
3.
11
Increasing
functions
146
3.
12
Stationary values
148
3.
13
Determining the
stationary
Unit vectors
2.
19
Scalar
2.20
Equations of
2.21
Pairs of
and
problems
product
3.
1
a
line
3.
14
Curve
lines
3.
15
Tangents
3.
16
Integration
3.
17
Integration of
2
Practice questions
Functions
–
continuity
160
sums
and
differences of functions
162
3.
18
Integration
164
3.
19
Calculus
substitution
and the
area
under
curve
166
3.20
Definite
integration
3.21
Area
under
3.22
Area
below the
168
124
a
curve
170
126
3.4
Gradient of
130
3.5
Differentiation from first
a
using
and
128
curve
principles
x-axis
area between two
and
curves
3.23
Volumes of
revolution
3.24
More volumes of
3.25
Forming differential
revolution
General differentiation
rule
and quotient
equations
134
174
176
rule
136
3.26
180
Solving differential
equations
chain
172
132
3.8
The
3.9
Parametric
rule
and
differentiation
182
138
Section
4
158
1
notation
Product
normals
122
Limit theorems
3.7
and
118
3.3
3.6
154
116
Calculus
Limit
sketching
112
discontinuity
3.2
150
108
a
3
points
110
Planes
Section
nature of
104
2.
18
Section
and decreasing
and
vectors
2.22
144
3
Practice questions
186
general
142
Index
188
Introduction
This
Study
Guide
has
been
developed
exclusively
with
the
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Examinations
candidates,
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additional
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resource
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Advanced
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Proficiency
Examination
It
prepared
(CAPE
)
programme.
®
has
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teaching
and
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a
team
examination.
with
The
expertise
contents
are
in
the
CAPE
designed
to
syllabus,
support
learning
®
by
providing
tools
and
the
and
requirements
for
full
Inside
features
guidance
this
activities
•
On
Y
our
answer
and
•
of
you
that
it
is
in
an
an
Do
examiner
will
CAPE
to
master
to
which
the
where
your
includes
concepts
syllabus
format!
electronic
techniques:
candidate
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short
answers
could
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skill
level
questions.
designed
questions
study
to
Mathematics
key
examination-style
your
specifically
Pure
the
examination
example
show
build
refer
examination
sample
with
to
and
examination
examination
inside
in
you
CD
good
provide
questions,
are
best
remember
interactive
answering
sections
for
requirements
developing
activities
your
easier
and
guide
so
to
provide
helpful
that
experience
feedback
you
can
will
revise
areas.
Answers
so
to
course
activities
in
achieve
syllabus.
type
multiple-choice
problem
This
These
Y
ourself
you
make
activities
from
confidence
refer
•
the
essay
feedback
T
est
the
you
Marks
and
help
Guide
assist
improved.
and
of
on
Study
to
to
included
are
you
unique
examination
included
can
check
on
combination
practice
the
your
will
of
CD
own
focused
provide
for
exercises
work
as
you
syllabus
you
with
and
practice
questions,
proceed.
content
and
invaluable
interactive
support
to
help
you
®
reach
your
full
potential
in
CAPE
Pure
Mathematics.
5
1
Basic
1.
1
Terminology
algebra
Learning outcomes
To
use
words
and
and
principles
Language of
The

and functions
language
of
mathematics
mathematics
is
a
combination
of
words
and
symbols
where
symbols
each
symbol
is
a
shorthand
form
for
a
word
or
phrase.
When
the
words
and
correctly
symbols
are
properly
constructed
Many
the
of
used
correctly
words
a
piece
sentences
used
have
in
of
mathematical
the
precise
same
way
as
reasoning
a
mathematical
piece
of
can
be
read
in
prose.
definitions.
For
example,
You need to know
the
word
when

How
to
solve
simultaneous
a
pair
of
‘bearing’
used
has
many
mathematically
meanings
it
means
when
the
used
in
direction
of
everyday
one
language,
point
from
but
another
.
linear
equations
in
two
Y
ou
need
unknowns
to
be
able
cor rect
to
present
mathematical
your
solutions
language
and
using
clear
and
symbols.
n

The
meaning
of
x
Symbols
A
used for operators
mathematical
Y
ou
are
already
describe

operator
familiar
is
a
with
rule
for
several
combining
operators
or
and
changing
the
quantities.
symbols
used
to
them.
means
‘plus’
or
‘and’
or
‘together
with’
or
‘followed
by ’,
depending
on
context.
For
example,
2
a
means
For
For
‘is
not
use
‘take
2
a

plus
5
or
together
2
and
with
b
5,
or
a
followed
by
b
away ’.
means
and
2
also
minus
have
5
or
2
familiar
take
away
5.
meanings.
comparison
symbol
used
for
comparing
two
quantities
is

and
it
to’.

6
‘is
means
symbol
x
is
symbols
greater
is
than
used
equal
are
or
to

is
to
6.
which
equal
mean
means
to’.
‘not’,
A
for
‘is
greater
forward
than’
slash
example,
,
and
across
which

a
means
to’.
expressions,
comparison
between
A
or
familiar
equal
Terms,
T
o
x
means
comparison
‘is

equal
other
which
means
used for
example,
Some
means
5
commonest
means
5
b
2
operators
Symbols
The

‘minus’
example,
The

terms,
equations
symbols
correctly,
expressions,
mathematical
expression
you
equations
is
a
group
and
need
and
of
identities
to
recognise
the
difference
identities.
numbers
and/or
variables
2
5x
_______
(for
example,
x)
and
operators.
For
example,
2 x,
3
2y
and
are
3
2x
expressions.
The
parts
example,
6
of
3
an
and
expression
2y
are
separated
terms
in
the
by

or
expression
are
3
called
2y
terms.
For
Section
An
equation
For
is
example,
This
a
2x
3
statement
Some
is
equations
example,
example
x

of
Therefore
x
an
we
Symbols
statement

7
true
are

is
identity
can
is
a
only
true
2x
saying
for
and
x
any
used for
for
we

two
statement
when
true
write
that
x
x

use

that
reads
are
‘2 x
equal
3
is
in
Basic
algebra
and functions
value.
equal
to
7’.
5
value
any
quantities
1
that
value
the
the
of
variable
x.
symbol
This

to
can
take.
equation
mean
‘is
is
For
an
identical
to’.
2x
linking
statements
2
When
that
one
is
words
Some
statement,
logically
or
such
as
connected,
x
for

4,
is
example
followed
x

2,
by
another
they
statement
should
be
linked
by
symbols.
examples
of
words
that
2
can
be
used
are:
2
‘x

4
therefore
x

2’,
‘x

4
implies
2
that
x

2’
2
‘x

4
so
it
follows

4
hence
that
x

2’
‘x

4
gives
x

2’
2
‘x
The
symbols
‘therefore’
For
or
example,
x


and
is
⇒
can
‘hence’
and
2x

Setting out
It
2’
1
a
be
the
5
used
x
or


to
set
out
also
important
of
following
example
simultaneous

3x
[1]

3
⇒
6x
[2]

2
⇒
6x
[4]
⇒
[3]
The
solution
Notice
briefly
the
1
that
is

the
what
we
1

5
that’
⇒
x

or

means
‘gives’.
3
3y
of

[1]
4y
10
[2]
9y
3
[3]
8y
20
[4]
17
y

1
y
in
2
[1]
and
equations
are
way

doing
y
gives

are
to
2x
or
to
problems
using
cor rect
words.
explain
the
steps
you
take
reasoning.
1


you
your
a
2x
17y
for
x
where
‘implies
2x
solutions
3y

Substituting
shows
equations:
2x
statements,
means
symbols
that
and
The
3
your
linking
is
link
⇒
solution
important
It
to
symbol

explaining
1
and
3(
3x
1)

the
solution
4y
1

⇒
of
the
pair
10
x

2
1
numbered.
combine
This
them
in
gives
order
a
way
to
of
explaining
eliminate
one
of
variables.
Exercise 1.1
In
each
and
question,
write
down
a
explain
correct
the
incorrect
use
of
symbols
2
Find
the
value
sin
1
Solve
the
equation
3x
1

1

5

3x

6

x

2
A
A°

0.5
given
that
sin
A°

0.5
A

30°
5

3x
of
solution.
7
1.2
Binary
operations
Learning outcomes
Binary operations
A

To
know
perform
the
meaning
binary
of
binary
To
of
be
able
to
identity,
use
the
closure,
commutativity,
concepts
inverse,
example,
addition,
these
we
can
rule
for
combine
subtraction,
operations,
combining
two
members
of
a
the
real
set.
two
members
multiplication
that
is
4

2

6,
or
of
division.
4
2

2,
set
We
4

of
know
2

8
numbers
the
and
rules
4

by
for
2

2
can
define
other
operations.
For
example,
for
a
and
b,
where
a,
b
are
addition,
members
multiplication
binary
a
associativity,
We
distributivity,
is
operations
For

operation
and
and
other
simple
2a
of
the
set
of
real
numbers,
,
then
a
and
b
are
combined
to
give
b
operations
We
write
Then,
this
for
briefly
example,
as
3
a
*
*
7
b


2
2a

b
3
7

1
You need to know
Properties of operations

The
meaning
numbers,
of
the
set
of
real

An
operation,
*,
as
For
example,
is
b
commutative
*
addition
a
for
on
the
a
Multiplication
on
the
set
subtraction
is

of
a
However
,
any
two
set
b

real

not
of
b
b

also
not
b
by
is

on
two
example,
because
e.g.
(2
(a

3)
Addition
(a

e.g.
b)
(2
(a

b)
set
of
(b
3)

4

2
is

a
4

also
(3
is

a

3)

4


not
(2

8
set
4
because
commutative
because

in
general,
3

7
in

3
any
*
general,
a
first
a
result
commutative
three
two
that
(b
*
members
members
can
the
be
second
c)
of
real
numbers
a,
b,

c
or
is
is
associative

b,
is
c
also

associative
because

associative
a,

5
b,
c

because,
whereas
because,
a,
b,
c

in
general,

2
in
(3
4)


(
general,
3
whereas
6
2

1

3


same
4)
c),
2
e.g.
c
b
is
the
set.

also
7
first,
c),
a,
c),



the
4)
associative
(b
*
b
the
numbers
4
7
gives
of
a
when
the

c),
1
not
c
on
(b

real
(b


(3

subtraction

a


3)
Division
2
a
c
(2


b)
e.g.
the

c
c
However
,
(a
4

b)
b
because,
b


either
*
multiplication
b)

on



a,

members
(a
For
3
b
*
b
is
because,
associative
operating
a,
a,
7
commutative
operation
a,

3
e.g.
An

b
a
combined

a
numbers
commutative
e.g.
is
real
numbers
a
Division
when
members
2

(3

4)

2

8

4
3
1)

3
Section
An
operation,
*,
three
For
example,
members
of
distributive
of
multiplication

the
is
over
set
another
a
*
(b
◊
distributive
operation,
c)
over

(a
*
b)
addition
◊,
◊
when
(a
and
*
for
Basic
algebra
and functions
any
c)
subtraction
on
because
a
but
is
members
1

(b
multiplication

is
c)

not
ab

ac
and
distributive
a

over
(b
c)

division
ab
ac
because
ab
___
a

(b

c)

c
b
__
whereas
(a

b)

(a

c)

c
so
a

(b

c)

(a

b)

(a

c)
unless
a

1
Example
An
operation
Determine
(a)
x
*
2x
y


*
is
defined
whether
the
2x
and
2y


2y
2y

2x
for
all
real
operation
y
*
x

because
numbers
*
2y
is:

the
(a)
x
and
y
as
x
commutative
*
y
(b)

2x

2y
associative.
2x
addition
of
real
numbers
is
commutative.

(b)
the
operation
T
aking
(x
x
*
*
y)
(y
x,
*
*
y
and
z
z

(2x
z)

x
Therefore
(x
*
*
is
as

*
*
three
2y)
(2y
y)
commutative.

z
*
real
numbers,
z

2(2x
2z)

2x

x
*
(y
*


z)
2y)
4y
so


2z

4x

4y

2z
4z
the
operation
is
not
associative.
Example
(a)
For
two
real
numbers,
2
x
*
y

x
Determine
(b)
For
two
Determine
(a)
For
any
*
y)
whether
real
*

(b)
(y
(x
For
*
*
z
z)
y)
any
the
x
operation
and
the
y,
the
operation
numbers,

(x

*
z
y
*
(x

x
*
real
y
*
z

◊
(y
*

so
◊
x
y)
z)
◊
the
*
(y
*
is
given
by
associative.
operation
◊
is

x(y

(y
*
z
)
z)

so

◊
is
given
distributive
by
over
x
◊
y

xy
the
x
z)

(x
*
◊
z)
z)


(x
operation
xy
◊
◊
*
y)
x(y
is
xz
*
not
y

x,
2
2
)


x
◊
z
2
(x

z
2
)
operation
y
and
z
not
2
)

xy
2
y
is
2

associative.
z,
2

(x
z,
2
the
numbers,
*
and
2
2
(x
is
2
(x
2
2
x
operation
2
)
2
x
three
x,
2

2
x
y,
*.
three
*
the
numbers,
2
(x
and
y
whether
real
operation
x
2

x
2

xz
2
z
z)
distributive
over
the
operation
*.
9
Section
1
Basic
algebra
and functions
Closed
A
set
is
sets
closed
under
set,
For
example,
any
two
the
set
integers,
However
,

is
a
a
not
an
*
b
of
operation
gives
integers,
and
b,
closed
a
*
when
another

,
b
under
is
is
for
any
member
closed
also
an
division
of
under
two
the
members
of
the
set.
addition
because
for
integer
.
because
a

b
does
not
always
3
give
an
integer
,
for
example
3

4

,
which
is
not
an
integer
.
4
Identity
If
a
such
is
any
that
member
under
an
identity
For
example,
member
0
is
of
a
set
member
the
identity
Also
1
there
no
is
member
b
is

set
one
b
*
a
under
members
of
member

the

a
b
then
b
of
the
is
set,
called
the
operation.
under
addition
as,
for
any
is
real
the
no
identity
number
identity
b
for

a
for
such

a

0

members
that
members
a
of
b

a
of


under
b
under
subtraction
because
a
multiplication,
as
for
any
a,
However
,
is
*
the
for
1
there
there
a
a,
However
,
is
*,
of
0
there
and
operation,
there
no
is
real
no
identity
number
b

a
for

a

1

members
such
that
a
has
a

b
a
of


b
under

division
because
a
Inverse
Any
if
there
member
is
another
a
of
set
member
of
gives
Clearly,
a
member
can
have
For
example,
then
a
since
Also,
is
a
as

1
as
the
a
is
0
is
the
inverse

the
a

identity
of
(
any
a)
identity

for
the
inverse
of
a
only
there
members
one
members
since
important
of
of


under
under
a


1,
exception
to
this:
1
__
is
0
10
there
is
an
with
identity
addition,
a,
a
is
if
0
a
but
operation
combined
operation.
1
__
is
an
when
identity.
member
1
__
then
under
which
inverse
the
for
inverse
set,
the
an
under
an
the
meaningless.
multiplication,
a
Section
1
Basic
algebra
and functions
Example
An
operation,
*,
is
defined
for
all
real
numbers
(x

x
and
y
as
y)
_______
x
*
y

2
(a)
Show
that
the
(b)
Show
that
x
(a)
When
set
has
is
closed
no
under
inverse
the
under
operation
the
(x

*.
operation
*.
y)
_______
x
and
y
are
real
numbers,
is
also
a
real
number
.
2
Therefore
(b)
For
x
to
the
have
i.e.
one
i.e.
such
set
an
closed
inverse,
member
,
(x
is
b,

of

under
there
such
the
operation
needs
that
x
to
*
be
b

an
*.
identity
member
,
x,
b)
_______
that

x
2
Solving
Now
x
this
is
any
therefore
As
there
for
gives
member
there
is
b
no
is
no
b
of

x
,
so
identity
identity,
x
b
is
not
member
has
no
a
of
single
the
member
and
set.
inverse.
Exercise 1.2
1
Determine
set
2
of
The
real
whether
addition
is
distributive
over
multiplication
on
the
numbers.
operation
*
is
given
by
2
x
for
all
real
values
Determine
3
x
whether
(a)
commutative
(b)
associative
(c)
distributive
The
of
operation
the
over
*
is
and
*
y

x
y
y.
operation
*
is:
addition.
given
by
___
x
for
4
all
positive
real
(a)
Show
that
(b)
W
rite
down
(c)
Determine
The
for
the
operation
x,
y

numbers
operation
the
identity
whether
~
is
by
y

√xy
including
*
is
0.
closed.
member
.
each
given
*
member
x
~
y

has
the
an
inverse.
difference
between
x
and
y
.
(a)
Determine
whether
(b)
Show
that
the
(c)
Show
that
each

identity
is
closed
member
member
is
its
under
is
this
operation.
0.
own
inverse.
11
1.3
Surds
Learning outcomes
Surds
The

To
perform
operations
square
roots
of
most
positive
integers
and
fractions
cannot
be
involving
expressed
exactly
as
either
a
fraction
or
as
a
terminating
decimal,
i.e.
they
surds
are
A
not
rational
number
exactly
numbers.
such
when
√
as
left
2
is
an
√
as
2.
In
irrational
this
form
number
it
is
and
called
a
can
only
be
expressed
surd.
You need to know
Note

The
meaning
of
a
√
that
2
means
the
positive
square
root
of
2.
rational
number
Simplifying
Many
surds
surds
can
be
simplified.
______
___
For
example,
√
18
√

9

2
√

9
√

2

3
√
2
______
And
In
√
8

both
√
2
√

cases,
3
When
4
√
2
a

is
2
the
Operations on
expression
i.e.
(3
2
such

2
√
simplest
calculation
answer
An
√

in
2
2
possible
involves
the
√


√
2
surd
surds,
simplest
3
form.
you
possible
should
surd
give
your
for m.
surds
as
√
(3
2)
√
(2
3)
can
be
expanded,
______
When
For
(5
√
√
2 )(2
the
3)
same
surd

6
3
occurs
√
3
in
2
each
√
2
√

6
bracket
(
the
√
3 )(3

2
√
3)


particular
,
rational
(5
√

3

can
be
√
2

3)
simplified.
example,
2
For
2
expansion
15
6
√
3

10
√
3
12
(2
In
√
3

expressions
4
of
√
3

2
√
3

4
√
9

12)
√
3
the
form
√
(a
b )(a
√

b)
simplify
to
a
single
number
.
example,
2
√
3 )(5

2
2
√
3)

5
(2
2
√
3)
((2

25
12

√
3)
2

4
√
3
√
3

4
√
9

4

3)
13
Example
Simplify
(2
√
√
(2
5 )(3
5 )(3

2
√
5)


2
√
6
5)
3
√
5

4
√
5
10
___
(2

12
4

√
5
√
5

√
5

2
√
25

10)
Section
1
Basic
algebra
and functions
Rationalising the denominator
When
the
a
fraction
has
a
surd
in
the
denominator
,
it
can
be
transferred
to
numerator
.
When
the
bottom,
For
denominator
by
that
surd
is
will
a
single
change
surd,
the
multiplying
denominator
the
into
fraction,
a
top
rational
and
number
.
example,
√
2 
3
_______
√
2 
3
_______
√
5
___


√
√
5
√
5
5
___
√
√
2
5

15
__________

5
When
top
the
and
denominator
bottom,
by
is
of
√
a
b
the
will
form
a
change
√

b,
the
multiplying
denominator
the
into
fraction,
a
rational
number
.
For
a
denominator
of
the
form
√
a
b
multiply
top
and
bottom
by
a

√
b
Example
√
2
1
___________
Rationalise
the
denominator
and
simplify
√
3(
This
Do
fraction
not
start
has
attempt
with
a
to
single
rationalising
√
√
2
3(
1
___________
surd
rationalise
the
and
single
√
2
3(
√
2

bracket
both
at
2
in

the
the
3)
denominator
.
same
time.
We
will
surd.
1)
√
√
6
________________
3
_________

√
a
them
√

√
3)
3
√

3(
√
2

3)
3(
√
2

3)
___
√
√
6
√
3
2
_________

√
12
3
_______
3(
√
2

√
3)
2
3
3(
√
√
√
2
3
4
6
 3
3
_________________
to
do
written
some
of
3
√
6

3
√
3
√
2
2
9)
√
√
4
6
5
3
__________

3(2
have
6


We
√
______________________

down
these
9)
every
steps
in
21
step
in
your
this
example,
but
you
should
be
able
head.
Exercise 1.3
Expand
and
simplify
__
1
(3

2
√
Rationalise
3 )(
__
√
the
3
when
possible.
__

√
2)
__
2
denominator
of
(
__
__
2
√
2
√

each
3
5)
surd
and
(1
simplify

(
√
3
when
__

√
2
2 ))
possible.
__
1
2
___
4

√
2
_______
__
__
7
√
2
1

√
2
__
__
2
√
√
3
2
____
__________
__
5
__
8
√
3
2
__
√
3

5
√
5
__
√
8
1
_______
__
6
3

√
2
____________
9
__
√
2(
__
√
3
__

√
2)
13
1.4
Logic
and
Learning outcomes
truth
Propositions
A

To
identify
simple
sentence
To
establish
compound
the
truth
value
statements

To
went
to
school
today ’
is
a
closed
sentence,
but
went
to
who
school
could
be
today ’
any
is
not
closed
because
it
contains
the
variable
female.
Closed
p,
sentences
q,
are
called
statements
or
propositions
and
are
denoted
etc.
tables
state
the
A
converse,
contrapositive
a
‘Sonia
of
using
by
truth
as
propositions
‘she’,

such
and
‘She
compound
tables
conditional
and
inverse
proposition
is
either
true
or
false.
of
(implication)
Negation
statement

To
determine
statements
whether
are
two
The
proposition
‘It
is
not
raining’
contradicts
the
proposition
‘It
is
raining’.
logically
‘It
is
not
raining’
is
called
the
negation
of
‘It
is
raining’.
equivalent
If
You need to know
p
is
The
meaning
distributive
binary
of
and
proposition
‘It
is
raining’,
the
‘It
is
raining’,
if
~p
is
true.
negation
of
p
is
denoted
by
~ p.
Truth tables
For

the
the
proposition
p:
p
is
true
then
~p
is
false.
commutative,
associative for
But
if
p
is
false,
then
operations
We
can
We
use
show
1
to
this
logic
represent
in
table
true
and
form
0
to
(called
a
represent
truth
table).
false.
p
~p
1
0
0
1
Did you know?
The
George
a
Boole
system
(1815–1864)
using
values
0
numbers
in
each
column
are
called
the
tr uth
values
invented
and
1
and
Conjunction
truth
tables
system
is
algebra.
to formalise
now
known
as
logic. This
Boolean
The
is
statements
raining
Using
We
p
put
for
the
can
can
is
symbol
true
or
possible
and
across,
(If
it
construct
be
all
p
and
p:
q
we
either
in
p
or
is
raining’
cold’.

a
to
This
mean
truth
false,
q
first
can
two
complete
q
is
false,
is
and
for
also
of
q:
called
‘and’
table
combinations
the
can
‘It
p
be
1
we
third
then
p
a
is
true
or
this
false.
and
Then,
column
q
can
be
of
combined
two
as
propositions.
conjunction
as
p

0
(false)
reading
for
must
We
as
statements
‘It
is
raining’
propositions
written
Using
14
as
the
‘It
p:
and
is
‘It
is
and/or
the

is
word
raining
symbol
raining’
‘it
to
or
‘and’
it
and
cold’.
is
mean
q:
This
is
be
p

q
false.)
q
p
1
1
1
1
0
0
0
1
0
0
0
0
‘It
is
is
cold’
called
implied
so
it
a
can
be
combined
disjunction
would
of
normally
two
be
cold.’
‘
or ’
we
q
p
Disjunction
The
‘It
q
(true)
and
cold’
conjunction
write

columns.
the
‘It
write
this
disjunction
as
p

q

q
Section
We
p
can
can
be
before,
p
and
we
p
q
q
true
we
can
or
construct
in
or
put
all
the
first
both
‘it
is
are
Using
The
the
raining’
the
q
table
can
possible
two
the
Conditional
If
truth
false,
complete
or
a
also
third
then
symbol
logic,
example,
The
and
in
then
‘it
→
p
p
for
q
The
is
to
is
p
the
is
q
‘5
q
mean
is
one
is
1
and
reading
p

must
called
‘If
the
q.
be
0
for
across
(If
either
true.)
...
a
then
conditional
...’
hypothesis
tr ue
p
a
except
a
false
p
q
p
1
1
1
1
0
1
0
1
1
0
0
0
q

we
write
and
the
statement
p
→
q
proposition
q
is
called
prime
when
a
tr ue
hypothesis
inverse
is
not
the
of
p
→
is
of
q
is
the
a
is
prime
of
→
‘It
conclusion.
number ’
such
that
of
p
q
q
is
of
→
and
q
is
‘6
is
a
of
is
‘5
cold’
~p
is
of
→
→
‘5
number ’
‘It
prime
number ’
p
→
and
q:
q
p
is
true
p
q
p
1
1
1
1
0
0
0
1
1
0
0
1
→
q
p
a
number ’
is
inverse
prime
inverse
q
converse
a
p
→
leads
false.
combination
converse
contrapositive
example
number ’
is
the
prime
→
‘It
‘5
is
is
number ’
a
prime
raining’
is
→
‘6
is
a
prime
number ’.
‘It
is
raining’ →
‘It
is
cold’.
~q
is
→
a
prime
‘6
is
raining’ →
‘6
is
not
bi-conditional
statement
reads
‘if
p
p
it
is
is
number ’ →
not
‘It
is
a
prime
cold’
is
‘6
is
a
prime
number ’
number ’.
‘It
is
not
raining’ →
‘It
is
q
and
it
is
its
is
then
as
~q
of
→
‘5
~p
is
number ’
if
q
‘It
is
is
the
then
raining
it
is
a
prime
→
‘5
is
number ’
not
conjunction
converse
bi-conditional
simply
q
statement
then
‘If
→
a
→
prime
‘6
is
a
prime
number ’.
statements
with
a
p
prime
q
raining
written
a
→
example,
raining’
of
contrapositive
Bi-conditional
‘If
of
As
cold’.
The
For
is
is
for
the
‘6
example
‘5
and
A
q
false.
false.
example,
The
is
→
table
converse
Also
For
or
cold’
called
→
p
p
only
number ’
not
if
logic
truth
false
is
or
Then
for
and functions
q
true
column
algebra
statements
proposition
then
For

combinations
to
For
be
columns.
true,
p
Basic
conclusion
In
For
for
1
q
→
p,
that
of
is
the
(p
conditional
→
q)

(q
→
p).
This
p’.
then
it
is
cold’
and
‘If
it
is
cold
then
it
is
statement.
cold’
and
raining’
if
‘If
it
and
is
cold
only
if
then
‘It
is
it
is
raining’
can
be
cold’.’
15
Section
1
Basic
algebra
and functions
Using
‘It
is
We
the
cold’
can
Start
symbol
and
with
the
→
q,
then
q
→
p.
Lastly,
the
third
a
truth
add
a
add
fourth
table
can
to
q)
mean

a
(q
truth
table
for
the
columns.
now
be
and
p)
table
column
of
‘if
→
for
column
conjunction
and
This
→
construct
p
for
(p
⇔
only
can
for
p
be
⇔
if ’
we
can
written
as
write
p
⇔
‘It
is
raining’
⇔
(q
p)
q
q
p
q
1
1
1
1
1
1
0
0
1
0
0
1
1
0
0
0
0
1
1
1
written
as
a
p
→
simpler
q
q
truth
→
p
table
(p
for
→
a
q)

→
bi-conditional
statement.
p
q
1
1
1
The
1
0
0
p
0
1
0
0
0
1
⇔
Compound
A
compound
combination
A
q
p
q
shows
are
both
that
true
p
⇔
or
q
is
both
true
only
when
false.
statements
two
bi-conditional
compound
and
statement
of
table
or
combines
more
statement,
of
(p
the
→
two
or
more
symbols
q)

(q
~,
→
p),
propositions
,
is
,
an
→,
using
a
←
example
of
a
statement.
Example
Let
p:
p,
q
‘Students
Express
in
and
the
symbolic
r
be
play
the
propositions:
soccer ’,
compound
q:
‘Students
statement
‘Students
play
soccer
or
basketball’
‘Students
play
soccer
or
basketball
‘and
cricket’,
play
r:
‘Students
soccer
or
play
basketball’.
basketball
but
not
both
and
students
play
cricket’
form.
‘Students
Adding
play
students
play
cricket’
is
p
but
to

r.
not
this
‘Students
both’
gives
The
way
(p
truth
to
is
( p


r)
table
the
do
not
r)


play
~(p
~(p
for
a

both

r)
and
basketball’
is
~( p

r).
r).

q
compound
bi-conditional
soccer
table
statement
can
be
constructed
in
a
similar
above.
Example
Construct
16
a
truth
table
for
the
compound
statement
p
q
~q
1
1
0
0
1
1
0
1
1
1
Then
0
1
0
0
0
to
0
0
1
0
0
~q

p
p

(~q

p

(~q

p)
p)
Always
start
add
build
with
p
columns
up
statement.
the
and
in
q.
stages
compound
Section
1
Basic
algebra
and functions
Equivalence
T
wo
statements
same,
that
is
are
in
logically
the
equivalent
completed
truth
when
tables
their
the
truth
final
values
columns
are
are
the
identical.
Example
Determine
whether
the
statements
p

q
and
~p
→
q
are
logically
equivalent.
We
construct
a
truth
table
for
q
~p
1
1
0
1
1
The
1
0
0
0
1
~p
→
0
1
1
0
1
the
statements
0
0
1
0
0
Identity
This
law
q
q
truth
q
values
are
equivalent.
not
for
the
are
W
e
p

q
same.
and
Therefore
logically
write
p

q
not

~p
→
q
law
states
Algebra of
The

→
statement:
p
p
~p
each
symbols
that
p

p
and
p

p
are
both
equivalent
to
p
propositions

and

are
called
logical
connectors.
Example
These
connectors
are
commutative,
that
is
p

q

q

p
and
p

q

q

p
Use
They
are
also
associative,
that
They
are
also
distributive
over
for
p
p
(p

each
q)

other
r

and
p

over
(q

the
r)
conditional
→,
example


(q

(q
The
r)
→
These

r)
(p

(p
properties
can

also
q)

properties
compound
It
is

q)
can
can
(p
→
be
also

(p
r)

and
r)
proved
be
and
using
used
p
to

p
(q

(q
truth
prove

→
r)

r)
(p

(p

q)


q)
(p
→

(p
algebra
show

(p

q)

p
p

(p

q)

(p
using
the
using
the
r)

to
p

that
(p


p)

distributive

p

(p
q)

distributive
(p

q)
law
q)
law
r)
tables.
the
equivalence
between
two
statements.
be
shown
that
p
→
q

~q
→
~p
Exercise 1.4
In
this
exercise,
1
W
rite
2
(a)
(b)
down
p,
the
Construct
State,
q
a
with
and
r
are
propositions.
contrapositive
of
truth
p
a
table
reason,
for
~p
→
whether
p

~q
→
q
and
~q
p

and
~q
p

~q
are
logically
equivalent.
3
p:
‘It
is
Using
‘The
raining’,
logic
sun
is
q:
‘It
symbols,
shining
is
cold’,
write
and
it
r:
‘The
down
is
cold
in
sun
terms
and
it
is
is
of
shining’.
p,
not
q
and
r
the
statement:
raining’.
17
1.5
Direct
proof
Learning outcomes
Direct
proof
Mathematics

To
construct
simple
is
the
Mathematicians
specifically
direct
Proof
by
the
use
of
numbers,
shapes,
space
and
change.
look
for
patterns
and
formulate
conjectures.
They
then
proofs
try

study
proofs,
of
to
prove
the
truth,
or
otherwise,
of
conjectures
by
proof
that
is
built
counter
up
from
axioms.
The
axioms
are
the
basic
rules
or
definitions,
and
all
examples
other
true
facts
can
be
inferences
conditional
rules
are
from
→).
the
derived
those
(We
moves
from
can
rules
use
that
these
are
(an
the
by
deduction,
inference
game
allowed
of
for
is
chess
each
that
the
as
same
an
piece,
is
by
as
using
the
analogy
and
–
logic
the
games
are
basic
built
You need to know
up

The
basic
rules
of
from
How
to
solve
a
n
i.e.
quadratic
example,
x

a
equation
moves.)
logic
For

these
by factorisation
or
x
by
x


x
b

x
a
x
x

is

defined
….
x,
to
mean
and
from
n
lots
this
of
x
multiplied
definition
we
together
,
can
deduce
that
b
the formula
Example

How
to find
the
area
of
a
triangle
Prove
that

8

8
20

8
Using
⇒
4x

28
Adding
⇒
x

7
Dividing
5)

8
Starting
if
4( x
with
4( x
⇒
4( x
This
is
with
p
(Note
T
opic
is
an
example
then
that
1.4
know
also
this
that
converse
in
this
of
‘
A
polygon
the
if
we
the
polygon
four
a
converse
s
⇒
an

→
8
or
a
q
⇒
an
each
x
is
⇒
side
by

7,
⇒
a
⇒
p
⇒
is
then
that
a
keeps
4
keeps
the
equality
the
equality
true
true
if
i.e.
true
a
x
q.
the
x

prove
We

know
p
⇒
q,
start
equal
needs
from
question.)
contrapositive
7
7
then
⇒
has
is
four
a
4(x
4(x
implication
polygon
implication
to
another
polygon
four
i.e.
q
true
true
of
has
side
law
7
p
say
sides
rhombus
of
to
each
deduction,
so
is
also
square
equal
distributive
20

q
converse
is
the
by
q,
7
implication
true
p

x
proof
can
5)
but
because
Therefore
is
x
⇒
⇒
of
p
that
4( x
has
r
proof
logic
case,
example,
‘
A
a
then
direct
⇒
Therefore
true
but
is
p
whether
from
true.
of
deduce
The
For
5)
4x

W
e
5)
5)
is
not


8
is
→
sides’
is
not
~p
8
also
always
equal
square’
~q
5)
true
is
true
true
sides.
to
be
proved
to
be
true.
Example
Prove
that
the
sum
of
the
interior
angles
of
any
triangle
is
C
180°.
D
ABC
is
DCA
ECB
DCA
any



triangle.
is
parallel
to
AB
CAB
CBA
ACB
⇒
CAB
⇒
the
18
DE

sum

ECB
ACB
of
the


180°
ECB
interior

Alternate
angles
are
equal
Alternate
angles
are
equal
Supplementary
angles
180°
angles
of
any
triangle
is
180°.
E
Section
Use of
As
well
also
counter
as
important
A
it
being
important
for
statement
disproves
it.
to
can
necessary
This
shown
is
to
that
converse
be
called
a
prove
of
a
to
be
a
that
statement
true
a
is
statement
false.
This
is
example,
a

0
⇒
true,
is
algebra
and functions
it
is
particularly
implication.
false
counter
if
we
can
find
just
one
example
that
example
2
For
Basic
examples
prove
the
1
2
a

0
is
true,
but
the
converse
a

0
⇒
a

0
3

is
false.
2
We
can
For
example,
prove
odd
use
this
a

the
using
9
⇒
a

3
statement
the
or
3
‘all
counter
as
a
prime
example
counter
numbers
‘2
is
a
example
are
prime
odd’
because
is
not
number
true.
and
2
is
W
e
not
0.
can
an
number ’.
Example
Use
a
counter
example
to
prove
that
the
converse
of
the
true
2
statement:
The
‘n
converse
is
an
of
integer ’
the
given
⇒
‘n
is
an
statement
integer ’
is
false.
is
2
‘n
(
is
an
integer ’
⇒
‘n
is
an
2
√
2)

2
is
an
integer
integer ’.
√
but
2
is
not
an
integer
.
2
Therefore
‘n
is
an
integer ’
⇒
‘n
is
an
integer ’
is
false.
Exercise 1.5
2
1
Prove
2
Find
3
(a)
that
if
x
3x

2

0
then
x

1
or
x

⇒
a
2
2
a
counter
example
to
show
that
a

b
2

b
is
not
true.
2
(b)
Prove
that
(Start
with
Use
in
a
(a)
‘n
n
is
an
odd

2k
1
counter
is
integer
where
example
to
⇒
k
is
show
n
is
any
that
an
odd
integer ’.
integer
.)
the
converse
2
4
5
(a)
Prove
that
‘x
(b)
Prove
that
the
In
the
Prove
is
diagram,
that
twice
the
the
of
the

4c’
statement
false.
2

D
c

0
converse
of
the
is
area
area
of
bx

the
of
has
triangle
roots
statement
midpoint
triangle
equal
of
in
⇒
(a)
b
is
also
true.
C
AB .
ABC
ADC.
D
19
1.6
Proof
by
Learning outcomes
induction
Proof
by
induction
2
Consider

Establish
the
simple
principle
of
proofs
by
these
results:
1

2
1

2,
2

2
2

6,
3

3
2

12,
4

4

20
using
mathematical
In
every
case,
the
right-hand
side
is
a
multiple
of
2.
induction
This
suggests
that
the
proposition
2
‘for
is
any
true
positive
but
it
integer
does
not
n,
n
prove

n
is
a
multiple
of
2’
it.
You need to know

The
set
of
denoted

Any
positive
by
even
integers
We
can
W
e
call
We
start
2k
number
and
any
written
using
a
method
called
mathematical
induction
2
is
can
be
the
proposition
odd
as
number
2k 
with
the
p(n)
and
proposition,
written
rephrase
it
as
‘n

n

2m
for
n,
m
∈
’.
1 for k
that
2
n

k,
‘k

k
is
a
multiple
of
2,
k
∈
’.
[1]
can
The
be
it

when
as
prove
∈
next
step
is
to
replace
k
by
k

1
(i.e.
by
the
next
consecutive

integer)

A
natural
number
is
a
member
of
2
the
set
1,
2,
3,
4,
...
⇒
(k


k
1)

(k

1)
2

1

k

2k

2(k

1
2

(k

k
)

From
1)
[1]
which
this
is
a
is
also
a
multiple
of
multiple
of
2.
2
2
Therefore
of
we
have
shown
that
if
for
any
integer
,
n,
n

n
is
a
multiple
2
2
then
(n
1)

(n

1)
is
also
a
multiple
of
2.
a
multiple
of
2.
[2]
2
We
know
that
p(1)is
true,
i.e.
1

1
is
Therefore
[2]
shows
that
p(1)
⇒
p(2)
so
p(2)
is
true,
then
[2]
shows
that
p(2)
⇒
p(3)
so
p(3)
is
true,
[2]
shows
that
p(3)
⇒
p(4)
so
p(4)
is
true,
again
again
This
process
can
be
continued
indefinitely,
i.e.
for
all
…
positive
integers.
2
Therefore
multiple
An
of
you
fall,
one
Proof
to
20
by
cover
for
on
can
domino
have
proved
that,
for
any
positive
integer
n,
‘n

n
is
a
2’.
analogy
standing
If
we
proof
show
fall,
that
then
domino
induction
pushing
pushing
after
induction
any
by
is
a
row
of
evenly
spaced
dominoes
end.
the
can
positive
be
over
over
any
the
domino
first
will
domino
make
will
the
make
next
the
whole
row
other
.
used
integer
to
from
prove
a
many
result
results
proved
for
that
a
are
generalised
particular
integer
.
Section
The
1
proof
Let
has
p(n)
Prove
three
be
a
2
Prove
3
Combine
that
that
p(k
Basic
algebra
and functions
steps:
proposition
directly
convenient
distinct
1
involving
1)
is
n,
true.
then
(Note
assume
that
k
is
that
an
p(k)
is
arbitrary
true.
and
integer
.)
p(1)
is
steps
true.
1
and
2
to
prove
that
p(2),
p(3),
p(4),
...
are
true.
Example
n
Prove
by
induction
that
10
1
is
a
multiple
of
9,
n
∈

k
Assume
that
p(k)
is
10
k
1,
1

9m,
k
Replace
k
by

giving

where
k
and
m
1
are
natural
numbers.
k
10
1

10

10
1
k

10

10
10

9
k

10(10
1)

9

10(9m)

9
which
is
a
k

k
Therefore
multiple
when
of
10
1
is
a
multiple
of
9
then
multiple
is
the
10
1
is
also
a
proposition
from
multiple
a
[1]
1
10
1
is
a
multiple
of
9,
and
10
1

2
therefore
9.
9.
1
p(1)
of
1
of
[1]
9,
9,
3
10
1
is
a
multiple
of
9,
from
[1]
again
10
1
is
...
n
Therefore
10
1
is
a
multiple
of
9,
for
all
n
∈

Example
2
Prove
(The
by
induction
second
2(3)
1,
...
odd
the
that
the
number
nth
odd
is
sum
of
the
2(2)
1,
number
is
first
the
n
odd
third
2n
numbers
odd
number
is
n
is
1)
2
Let
p(n)
be
the
proposition
that
1

3

5

1)

...

(2n
1)

n
2
Assume
Then
that
adding
1

the
3

5

next
odd


...

(2k
number
to
both
k
sides
gives
2
1

3

5
...
(2k
1)

(2k

1)

2
k

2k

1

(k

1)
2
Therefore
when
1

3

5

...

(2k
1)

k
then
2
1

3

5

...

(2k

1)

(k

1)
2
i.e.
when
the
sum
of
the
first
k
odd
numbers
is
k
,
the
sum
of
the
first
2
(k

1)
odd
numbers
is
(k

1)
2
Now
so
p(1):
1

1
2
,
so
the
sum
of
the
n
odd
first
two
odd
numbers
is
2
,
and
on.
2
Therefore
the
sum
of
the
first
numbers
is
n
Exercise 1.6
Prove
by
induction
that:
3
1
n

n
is
a
multiple
of
6
for
all
positive
integral
values
of
n
_
1
2
the
sum
of
the
first
n
natural
numbers
is
n(n

1)
2
21
1.7
Remainder
Learning outcomes
theorem
To
apply

To
use
the
remainder
general
for m
n
factors
and
to
theorem
evaluate
of
a
polynomial
expression
is
theorem
a
the factor
theorem
Polynomials
The

and factor
unknown
n
x

a
n
to find
where
n
is
a
positive
1
2
x
n

....

a
1
x

a
2
integer,
a
,
a
n
,
n
...
a
1
,
x

a
1
a
1
are
0
real
numbers
and
a
0

0
n
coefficients
The
order
5
x
T
wo
The
meaning
of
the
notation
the function
2
f(x)
where
a
a
polynomial
has
order
of
is
How
For
x
to
expand
polynomials
power
of
expressions
example,

0
(the
x
are
has
identical
equal
How

(ax

c)(a
to factorise
a
power
of
x.
For
example,
when
they
have
the
same
order
and
when
coefficients.
2
x
5
5x
order

must
2
be

the
ax
4

2
bx
same)

and
b
cx


1,
c
dx


e
5,
if
d
and

only
0,
e
if

2
of
(coefficients
the form
highest
a
a

the
polynomials
4
value
is
5.
f(a)
each
for

Identical
You need to know

of
2
x
must
be
equal).
polynomial)
quadratic
The
remainder theorem
expression
When
17
is
divided
17
___
i.e.

5

3
5
is
by
which
called
the
f(x)

remainder
.
can
7x
The

Substituting

be
written

be
written
as
5,
remainder

2.
2,
as
17

5

3
In
this
form,
2
is
divided
between
by
these
x
2,
we
quantities
get
can
a
quotient
be
and
written
as
2
x
2
6x
relationship
7x
for
x

6

(quotient)(x
eliminates
the
term
2)

remainder
containing
the
quotient,
giving
remainder
.
3
f(2)
divided
This
can
2
x
3
f(x)
Now
result
quotient.
3
f(2)
the
3
When
a
3
2
__
by
is
when

an
a
2
2
7(2
x
2,
3
)
the
6

14,
remainder
illustration
polynomial

of
f( x)
the
is
so
is
f( x)

2
x
7x

6x
2
is
14.
general
divided
when
case:
by
( ax
b)
then
b
__
f(x)

(quotient)(ax
b)

remainder
⇒
f
(
)

remainder
a
This
result
is
called
the
remainder
theorem
and
can
be
summarised
as:
b
__
when
a
polynomial
f( x)
is
divided
by
( ax

b),
the
remainder
is
f
(
)
a
Example
Find
the
3
2x
remainder
Example
when
2

7x
3
is
divided
by
3
When
2x
2
x
ax
1.
and
3
Let
f(x)

x
3

2
2x

7x
when
Find
3
the

b
is
divided
by
x
1,
the
ax
values
of

a
x

and
b
is
divided
by
x
3,
b
1
2x
1

0
when
x

Using
,
the
remainder
theorem
gives
2
3
therefore
is
when
divided
by
2x
2x
1
2

7x
a

1

b

4
and
27
9a

3
3
1
i.e.
b
a

and
b
9a
2
[1]
1
the
remainder
is
f(
).
2
1
f(
1
)

2
2(
)
2
1
3

7(
the
22

14
[2]
3

1
[1]
[2]
⇒
8a

16
⇒
a
in
[1]
⇒
b

2
2
Substituting


2
)
remainder
is
1.
remainder
is
4
2
x
2
for
a

4

b

16,
the
remainder
is
16.
Section
1
Basic
algebra
and functions
The factor theorem
When
the
(x
a)
is
remainder
a
is
factor
zero
of
⇒
the
f(a)
This
i.e.
For
if,
for
example,
4
a
when
3
x
polynomial
x

polynomial

f( x),
0
is
the
f( x),
factor
f(a)

0
theorem,
then
x

a
is
a
factor
of
f( x).
3,
2
3x
3x

11x
6

81
81
4
Therefore
x
3
is
a
factor
of
27

3
x
33
6

0
2
3x
3x

11x
6
Example
3
Given
5
that
when
a
__
2
ax

3x
divided
by
b
x

has
2,
a
factor
find
the
2x
of
and
a
leaves
and
a
remainder
b
3
__
1

b
8

0
⇒
a
8b

6
[1]
Using
the factor
theorem
with
x

2
4
8a

12
b

5
⇒
8a
b

17
Using
8
1
values

[1]

[2]
Substituting
The
factor
⇒
1
for
65b

b
[1]
theorem
in
can
65
be
⇒
gives
used
to
b

1
a

2
find
[2]
the
remainder
factors
of
theorem
with
x

2
polynomials.
Example
3
Factorise
2
x
x
x
2.
3
If

x
3
is
a
factor
of
2
x
x
x
2
x


x
c

2
2,

so
)(x
(x
possible
3
T
ry


1:


(1)
1:
(


2:
3


then
of
c)

are
1
1)
(1)
(
2

0
1)
(
1)
2

0
(2)
(2)
2

0
so
x
so
x
1
is
not
a
factor
.

1
is
not
a
factor
.
therefore
2
x
2.
2
(2)
x
and
2
2
x
bx
values
(1)
3
T
ry

2
3
T
ry
2
2
x
2  (x
2)(x
3
 bx  c)  x
x
2
is
a
factor
.
2
 (b
2)x
 (c
2b)x
2c
2
Comparing
so
b

1
and
3

coefficients
2

of
2c
x
so
and
c

2
x
the
constant
gives
1

b
2
1
2
x
x
2

(x
2)(x

x

1)
Exercise 1.7
3
1
Given
that
the
values
f(x)

(x
of

a
3
2
5x
and
(x

2)
are
factors
of
x
2

ax

bx

6,
find
b.
2

remainder
Given
1)
and
that
is
px

x

is
a
q.
When
f(x)
is
divided
by
x

2,
the
3.
(x

1)
factor
of
f( x),
find
p
and
q.
23
n
1.8
Factors
of
n
a

2
extract
for
all factors
positive
integers
n

of
a
n 
n

6
2
a
Factors of
n
To
,
2
Learning outcomes

b
b

2
a
b
2
is
the
difference
between
two
squares,
so
a

b
2
b

(a

0
b)(a

b)
b
6
3
3
a
Factors of
b

3
From
the
factor
theorem,
when
a

b,
3
a
3
b
3
b
You need to know
3
Therefore

How
to factorise
quadratic
(Y
ou
can
a
b
verify
is
a
this
factor
by
of
3
a
3
b
expanding
⇒
the
3
a
2
b

right-hand
(a
2
b)(a

ab

b
)
side.)
expressions
2
2
(a

How
to
expand
expressions
3
as
(x

1)
,
2
(2x

1)
(3x

ab

b
)
cannot
be
factorised.
such
3

3
Therefore
4)
3
For
example,
8

2
b

(a

b
is
2
i.e.
ab

b
)
2
2

(x
2)(x

2x

2)
4
a
b

(a
the
2
difference
2
)
(b
2
between
2
)
2

(a
b

(a
of
the
two
2
squares,
2
)(a

b
)
2
using
the
2
factors
b)(a

2
b)(a
difference

b
)
between
two
squares
twice.
2
(a

b
)
cannot
be
factorised.
4
For
example,
Therefore
a
4
4
x
16

5
a
Factors of
4

2
b

(a

b)(a
the
2
5
theorem,
b
can
4
(a
b
when
is
a
factor
(a
verify
3

a
a
(x


a
this
by
2

a
2)(x



b
b,

32
b

6
6
a
6
b
3

(a
a
3
(b
a
3
b

the
ab
)
has
no
linear
b
a

(a

b)(a
(Y
ou
can
a

,
2
which
3

(a

(a
b,
verify
b
b


ab
0
)
side.)
3

a
5
x
2
b

a
4

(x
is
3
b
the
2)(x
2
b
3
3

difference
3
)(a
2x
4

b
)
2

4x

8x

16)
b)(a
a
b
b
ab
is
a
this
factor
by

3

of
b
a
b
3
)(a
squares.
3

b
)
3

b

3

expanding
two
)
2

3

between
3

3
Therefore
b
factors.
4

3
when
5
b
4

right-hand
2
Now
5

2
)
6
a
b
6
2
)
4)
b

6
Therefore

5
x
Factors of
b
5
a
2
5
a
5
example,

4
ab
5
Therefore
For
2)(x
b
2
expanding
3
b
)
5
a
3
b)(a
2
b
of
4

b
5
5
a
(Y
ou
a
2

2

5
Therefore
b)(a
b

factor

4
x
5
From
24

4
a
⇒
2
b)(a
3
x
4
4
3

3
x
Factors of
a
b
the
0
3
⇒
a
3

b
right-hand
2

(a

side.)
b)(a
2
ab

b
)
Section
2
Neither
2
(a

ab

b
2
)
6
Therefore
nor


6
For
example,
These
in
(a

6
x
results
expressed
ab
64
can
one

b
)
can
be
of
be


2
b)(a

ab


2)(x
to

(x
2)(x
factorise
forms
b
2
)(a
and functions
2

ab
2
2
used
the
b)(a
6
x
algebra
factorised.
2
b
Basic
2
(a
6
a
1
given
any

b
)
2

2x
polynomial
4)(x
that
can
2x
4)
be
above.
Example
3
Factorise
8x
27
completely.
3
3
8x
27
can
be
3
Using
written
as
3
a
(3)
2
b
and
3
(2 x)

(a
replacing
2
b)(a
a
by
2x

ab
and

b
3
b
by
)
3
gives
2
8x
27

(2x
3){(2x)

(2x
3)(4x
2

(2x)(3)

(3)
}
2

6x

9)
Example
4
Show
that
(x

4
Using

(a
(x
b)(a
expanding
4
the

a
by



x
(a

1
two
b
b

by
a
x

bracket



(x

1
x

(x

1)
){(x

1)

b
1)

x
)
2

x(x
2
x(x

gives
3

3
(x
3
ab
3
x
x
gives
2
b
2
1)
)
2
b)(a
and

4
1)
x(x
2
b)(a
3
b
4
3
1)
last
4
a


2
b
and
(x
3
x
4
a
Replacing
4
1)

2

1)


1)


1)
x
2
1)

x
3
(x

1)


(x

1)
x
}
3
(x
x
Example
3
(a)
Show
that
(b)
Hence
(a)
7x
2
7x
3
3x
3x
1

3
8x
(x
3
or
3
otherwise
factorise
2
3
3x
3x
1 

8x
3x
3
1 
8x
2
7x
(x
3x
3
Using
and
3
a

3x

replacing
3
a
(a
by
b)(a
2x
and

1)
(x

1)
(x

1)
3x

1)
3
2

b
ab
by

(x
b

)
1)
3
8x

3
(x
2
b
1
2
3
3x
1
3x
3
3
(b)
3x
2
x
3
8x
3x
3
8x

2
7x
gives
2

{2x
(x

1)}{(4x
2

2x(x

1)
}
2

3

(x
1)(7x

4x
1)
2
7x
3x
2
3x
1

(x
1)(7x

4x
1)
Exercise 1.8
3
1
Factorise
2
Show
8x
3

1

(x
completely.
4
that
x
(Hint:
( 1)
4

2)

1)
2

8(x

1)(x

2x

2)
25
1.9
Quadratic
Learning outcomes
and
Polynomial
A

To
investigate
roots
of
a
the
nature
quadratic
cubic
of
polynomial
equations
equations
equation
equation
n
x

a
n
the
relationship
sum
and
form
1
2
x
n

....

a
1
x

a
2
x

a
1

0
0
between
The
the
the
n
a
and
has
the
product
of
roots
of
a
polynomial
equation
are
the
values
of
x
that
satisfy
the
these
equation.
roots
and
the
coefficients
of
2
ax

bx

c

0
The
order
Some,

To
use
the
relationship
or
of
the
all,
of
polynomial
these
roots
gives
may
the
not
number
be
real.
of
For
of
the
equation.
example,
roots
the
quadratic
between
2
equation
the
sum
of
the
roots,
of
the
roots,

the
(Y
ou
will
the
wise
product
and
3
a
the
of
the

2

0
has
two
roots,
although
neither
of
them
roots
coefficients
discover
the
nature
of
these
roots
if
you
study
are
Pure
sum
Mathematics
of
x
the
real.
product
x
Unit
2.)
pair-
of
2

bx

cx

d

0
The
nature of the
roots of
a quadratic
equation
2
The
general
The
values
form
of
You need to know
x
of
a
that
quadratic
satisfy
this
equation
is
ax
equation
are
given
√
to
expand
brackets
(ax

of
b)(a
c

0.
by
2

and
it
is
the
value
of
b
4ac
that
determines
the
2a
the
nature
form

2
b 
b
4ac
_______________
How
bx
________
x


of
these
polynomial)
roots.
2
Note
that
b
4ac
is
called
the
discriminant
________
2
When
b
4ac
therefore
the

0,
roots

are
2
√
b
4ac
real
and
has
two
real
and
different
values,
different.
________
2
When
b
4ac

0,

2
√
b
4ac
b
___
∴
x



0
and
x


2a
so
is
there
said
is
to

0
b
___
0
2a
only
have
one
a
value
of
repeated
x
that
satisfies
the
equation
and
the
equation
root
________
2
When
no
b
real
The
4ac

0,

√
2
b
4ac
has
no
real
value,
so
the
equation
has
roots.
relationship
equation
between the
and the
coefficients of
a quadratic
roots
2
The
If

the
general
and

form
are
equation
of
the
can
a
quadratic
roots
be
of
this
expressed
equation
is
ax

bx

c

0
)

0


0
[1]
equation,
as
)(x
(x
2
⇒
[1]
and
[2]
are
the
identical
b
__
2
x
equation,

c
__
x
so
we

say
)x

that
2
(

)x
coefficients
of
x

a
(
x

x


a
2
([1]
is
divided
Comparing
by
a,
so
that
coefficients
of
the
this
identity
shows
b
__





a
c
__
and


a
26
are
that
equal.)
[2]
Section
the
sum
of
the
roots
of
the
equation
ax
Basic
algebra
and functions
b
__
2
i.e.
1

bx

c

0
is

and
a
c
__
the
product
of
the
roots
is
.
This
is
tr ue
whether
or
not
the
roots
a
are
real.
Example
Determine
(a)
the
nature
of
the
roots
of
the
equation
2
3x
2x

2

0
2
If
(b)


and
are
the
roots
of
the
equation
1
__
equation
whose
roots
2
2x

2

0,
find
the
1
__
are
and


2
3x
(a)
3x
2x

2

0
so
‘b
4ac’

4
4(6)

20
2
Therefore
3x
2x

2

0
has
no
2
3x
(b)
real
roots.
2
__
2x

2

0

gives


2
__


and

3
3
1
__
For
the
equation
whose
roots
1
__
are
and
,

1
__
the
sum
of
the
roots

1
__
is


______






2
3


1
2
3
1
__
and
the
product
of
the
roots
is
1
__
1
___





1
__

3
__

2
2
3
3
2
Therefore
the
required
equation
is
x
x


0,
2
2
i.e.
2x
2x

3

0
Exercise 1.9a
2
1
One
root
Find
of
is
the
roots
of
the
equation
3x
x

0
c

0
is

and
the
other
2
the
value
of
c
2
2
The
Find
Cubic
The
roots
the
of
the
equation
equation
whose
x

roots
3x
are

5



2

are
and

and


2
equations
formula
for
Did you know?
solving
a
general
quadratic
equation
was
known
to
the
It
ancient
Greeks.
However
,
the
search
for
a
general
solution
for
the
is
thought
(1501–1576)
equation
continued
until
a
method
was
developed
during
the
general
the first
method
of
to
publish
solution for
the
Italy.
cubic
This
and
that
method
is
you
does
difficult
are
reasons
If
was
Renaissance
a
in
that Girolamo Cardano
cubic
not
it
are
‘general
to
real,
not
lead
work
and
interested
of
a
in
formula,
with.
such
included
solution
to
It
also
but
it
relies
numbers
are
is
on
not
not
at
all
working
covered
easy
with
in
to
equation.
remember
numbers
Unit
1.
For
these
here.
finding
cubic
this
formula,
search
on
the
internet
for
equations’.
27
Section
1
Basic
algebra
and functions
The
roots of
a
cubic
equation
3
The
general
The
order
If
these
form
of
this
roots
are
of
a
cubic
equation
,

is

and
equation
three,
then
By
expanding
form,
the
we
can
this
get
coefficients
form
a
of
of
the
the
the
general

bx
it
equation
)(x
equation
relationship
2
ax
therefore
)(x
(x
is

)
and
between

has
can

cx
three
be
d

0
roots.
written
as
0
comparing
the

roots
of
a
with
cubic
the
general
equation
and
form.
2
)(x
(x
)(x
)


(x

x
)(x
(
)x

3
Dividing
the
general
3
)

2
(
form
of

the

)x

cubic
(

equation

by
2
(
x




a
(



)x




b
__
x
)x

c
__
2
x

a
3
Therefore
,
if


and
are
the
roots
of
ax


gives
3
)x

d
__
x

a
a
2

bx

cx

d

0,
then
b
__







a
c
__






a
d
__



a
3
For
is
example,
2,
roots
the
is
the
sum
product
of
of
the
the
roots
roots
of
the
pair-wise
equation
is
5,
and
x
2

the
2x

product
5x
of
7

0
the
7.
Example
3
T
wo
of
Find
If

the
the
is
roots
values
the
third
of
of
the
p
equation
and
root,
[2]
gives

therefore
and
from

2

px

2x

q

0
are
1
and
2.
q
then
2
x

1


p
Sum
2

2
Pair-wise
2

q
Product


of
the
product
of
[1]
roots
the
of
the
roots
[3]
roots
4,
from
[3],
[1],
q

p

5
8
Example
3
The
Find
The

equation
a
ax
2

relationship
sum
of
the
3


b
__

cx
between
roots
a
28
bx
is
⇒
a,
(


b,
p)

d


c

b
___
3a
0
has
and


roots

d
(

p)

3
p,

and
[2]


p
Section

is
a
root
of
the
equation,
so
x
3
i.e.
a


b
___
(
b



)
c


the
algebra
and functions
equation,
⇒
3
ab

9a

0
3
bc
2
2ab
d
2
3ab
3
⇒

)
3a
3
27 a
b
___
(
3a
3
by
b
___
(
3a
Multiplying
satisfies
Basic
2

)


1

9a
27a
d

0
3
bc

27a
d


0
0
Example
3
The
roots
of
the
equation
2
2x
x

3x
1
__
Find
the
equation
whose
roots
are
1
__
,
the
given
are
,

and

1
__
and

From
1


equation
1
__






2
3
__






2
1
__


2
For
the
required
equation,
the
1
__
sum
1
__
the

1
__



of

product
of
the
roots
pair-wise
1
___
1
___



_____________

3
is

1
___



is



The
roots




__________






1
1
____
The
product
of
the
roots
is

2
is
x

3
Therefore
the
required
equation
2
3x

x
2

0

1

Exercise 1.9b
3
1
T
wo
of
the
roots
of
the
equation
2x
2

px

qx
0
are
_
1
and
1.
2
Find
the
values
of
p
and
q.
3
2
The
Find
roots
the
of
the
equation
equation
whose
2
x

roots
2x
The
Find
roots
a
of
the
equation
relationship
2x
between
p

are
3
3

5x


1,

1


1
0
are
and
,



and
.
1
2

x
and

px

q

0
are
,

and



q.
29
1.
10
Curve
sketching
Learning outcomes
Straight
The
To

revise
basic
curve
equation
of
any
straight
line
can
be
written
as
y

mx

c
where
m
techniques for
is
simple
lines
the
gradient
of
the
line
and
c
is
the
intercept
on
the
y-axis.
sketching
T
o
sketch
points
You need to know
The
on
the
graph
the
line.
most
of
a
straight
straightforward
the
axes.
For
example,
line,
points
to
you
find
need
are
the
those
coordinates
where
the
of
line
two
crosses
2
How

to
express
ax

bx

c in
2
the form
a(x

p)

to
sketch
the
line
2x
3y

9

0,
first
find
where
the
line
q
1
crosses
the
axes:
when
x

0,
y

3
and
when
y

0,
x

4
,
so
draw
2
1
the
line
through
(0,
3)
and
(
4
,
0).
2
y
5
Curves
4
A
2x
3y

9

2
on
1
4
3
2
the
the
of
a
curve
1
1
2
3
4
coordinate
curve
accurate
x
O
5
sketch
should
show
the
shape
of
the
curve
and
its
position
0
such
plot,
as,
so
axes.
for
It
should
example,
these
also
where
features
will
in
show
the
any
significant
curve
many
turns.
cases
be
A
features
sketch
is
of
not
an
approximate.
5
1
2
Parabolas
3
2
A
curve
whose
equation
has
the
form
y

ax

bx

c
has
a
characteristic
4
shape
called
a
parabola.
5
When
y
has
a
a

a
minimum
y
value.
In
a

0,
maximum
value.
both
where
has
cases,
the
the
curve
curve
turns,
is
as
symmetrical
shown
in
the
about
the
line
through
the
point
diagrams.
2
T
o
sketch
either
the
find
graph
the
of
the
curve
coordinates
of
whose
the
equation
points
is
y
where

the
ax

curve
bx

c,
you
can
crosses
2
the
axes
(this
symmetry
to
is
easy
find
if
the
ax

bx

coordinates
c
of
factorises)
the
turning
2
or
express
ax
curve
bx
of
crosses

the
the
c
in
the
turning
y-axis
at
form
point
the
a(x
p
together
point
(0,
use
)

q
with
to
find
the
fact
the
that
the
c).
y
Example
5
2
Sketch
then
point
2

coordinates
and
the
curve
whose
equation
is
y

2x
3x

1
4
2
y

2x
3x

1

(2x
1)(x
1)
3
1
The
curve
crosses
the
y-axis
at
(0,
1)
and
crosses
the
x-axis
at
(
2
,
0)
and
(1,
0).
3
2
(
Therefore
the
curve
is
symmetrical
about
x
1
,
4
3

)
8
1

4
1
(halfway
between
x

and
x

1)
2
1
1
1
1
has
a
minimum
value
where
x

4
30
3
of
2(
4
3
2
)
3(
4
2
1
)

1

1

8
1
1
1
2
2
2
3
y
2
2
x
Section
1
Basic
algebra
and functions
y
Example
2
2
Sketch
the
curve
y

x

1

(x
x
1
1
1
2
y

(x
x)
(
3
2
curve
and
y
has
crosses
a
the
y-axis
maximum

)
4
4
value
at
(0,
x
O
2
The
3
,
2
)
2
2
4
1)
where
2
3
1
x

of
2
4
4
Cubic
curves
3
A
curve
whose
equation
is
y

2
ax

bx

cx

d
has
a
characteristic
shape.
a

0
a

0
y
8
6
or
The
curve
is
easy
or
to
sketch
when
the
cubic
expression
factorises.
2
For
example,
the
graph
of
y

(x

1)(x
2)(x
3)
crosses
the
x-axis
at
1,
0),
(2,
0),
(3,
x
O
2
(
0).
2
3
When
the
shows
brackets
that
a

1
are
and
expanded,
d

6,
so
and
the
comparing
curve
with
crosses
the
ax
2

y-axis
bx
at

(0,
cx

d
4
6).
1
__
The
curve
whose
equation
is
y

x
1
We
know
that
is
meaningless,
so
there
is
no
point
on
the
curve
where
y
0
x

in
0.
the
When
first
x

and
0,
y
third

0
and
when
x

0,
y

0
so
the
curve
exists
only
6
quadrants.
4
1
We
also
know
that
as
x
increases
the
as
x
approaches
for
positive
curve
gets
values,
closer
to
as
x
the
gets
larger
,
x-axis.
x
Using
gets
smaller
,
similar
i.e.
2
reasoning,
x
For
negative
zero
values
from
of
x,
positive
as
x
values,
approaches
y
6
increases.
zero,
y
decreases,
and
as
4
2
4
6
x
4
approaches
,
y
increases.
The
curve
gets
closer
and
closer
to
the
axes
6
but
never
Any
line
crosses
crosses
that
is
a
them.
curve
called
an
gets
closer
and
closer
to
but
never
asymptote
1
__
y

0
and
x

0
are
asymptotes
to
the
curve
y

x
The
curve
is
symmetric
about
the
line
y

x
Exercise 1.10
1
Draw
sketches
of
the
graphs
whose
equations
are
given.
Mark
2
On
the
graphs
all
significant
points
on
the
same
whose
set
of
axes,
equations
draw
sketches
of
the
are
2
__
curves.
y

and
2y
3x

6

0
x
2
(a)
y

x
5x

6
(c)
y

x(x
1)(x
3)
2
(b)
y

3x
x

1
31
1.
11
Transformation
Learning outcomes
To
understand
how
curves
knowledge
and
to
to
use
sketch
the
curve
whose
equation
is
y

f(x)
and
the
curve
whose
are
equation
transformed
curves
Translations
Consider

of
is
y

f(x)

2
this
curves
y
y

f(x)

2
You need to know
3
2

The
meaning
of
translation
and
y

How
to
sketch

f(x)
1
reflection
graphs
of
simple
x
O
equations
Comparing
of
x,
the
Therefore
points
y

on
f(x)
y

value
for
y
by
For
is
f(x),
units
any
the
with
f( x)
equal

2
f(x)
of

in
function
the
f,
c
curve

x,
of
we
y

units
on
f(x)
that
than

2
equation
y
a
particular
value

2
are
of
2
value
f( x).
units
translation
of
above
the
curve
y-axis.
equation
is
y
equation
the
is
a
the
whose
to
for
the
f(x)
is
of
whose
cur ve
parallel
whose
y

direction
cur ve
the
see
greater
points
curve
the
2,
units
positive
translation
consider
f(x)
2
of
the
the

is
values
i.e.
by
Now
y
2

is
f(x)
y


c
f(x)
y-axis.

f(x
2)
y
y
y

f(x)


f(x)
4
2
4
y

f(x
2)
x
O
Comparing
y
same
the
in
when

f(x)
with
value
of
y
x

in
f(x
2),
f(x
2)
is
2
see
that
units
the
values
greater
than
of
the
y
are
the
value
of
x
f(x).
Therefore
right
of
units
curve
for
equal
points
of
Using
the
curve
similar
y

For
f(x)
any
on
the
y

f(x),

4
f(x)
units
function
When
of
values
y
reasoning,
by
translation
c
of

x-axis
f,
the
0,
y,
i.e.
in
the
in
the
points
the
the
y

translation
when
c

0,
by
is
the
of
f(x
2)
f(x
2)
direction

of
whose
f(x)


f(x
direction

y
y
positive
cur ve
y
on
curve
curve
the
cur ve
the
and
of
direction
32
we
4)
the
is
of
a
units
in
the
the
units
is
y
is
to
the
by
x-axis.
of
the
x-axis.

negative
x-axis.
2
translation
translation
parallel
translation
the
a
negative
equation
c
are
is
f(x
to

the
c)
is
a
x-axis.
direction
in
the
positive
2
Section
1
Basic
algebra
and functions
Reflections
y
Consider
y

the
equation
is
y
Comparing
y

f(x),
given
x,
value
a
point
reflection
point
y
any
function
of
Consider
the
curve
y
whose
f(x)

we
Therefore
for

f(x)
see
of
x,
with
that
f(x)
for

a
f(x)
x
O
So,
curve
f(x)
the

f,
in
y
the
y

the

same
value
f(x)
is
the
of
the
x-axis
of
f(x)
f(x)
the
cur ve
whose
on
for
on
cur ve
y

equation
y
f(x)
is
y

in

f(x)
the
f(
is
the
reflection
x-axis.
x)
y
y

f(x)
y

f(
x)
x
O
Comparing
when
the
Therefore
y
f(x)
of
points
symmetrical
So

values
are
with
about
for
with
x
any
y
the
the

x),
same
we
in
see
sign,
that
i.e.
the
f( a)
y-coordinates
on

values
f(
the
(
of
y
are
equal
a))
curves
are
y-axis.
function
of
f(
opposite
the
f,
the
cur ve
cur ve
y

y
f(x)

in
f(
x)
the
is
the
reflection
y-axis.
y
Example
6
1
______
Sketch
the
curve
whose
equation
is
y

x

2
1
y
1
__
Start
with
the
curve
y

whose
shape
and

x
4
position
x
is
known.
1
y

x
1
__
If
f(x)

1
______
then
x
is
x
f(x
2).
2
1
______
So
the
curve
y

1
__
is
x
a
translation
2
of
y

by
2
units
x
6
in
the
positive
2
2
direction
of
the
4
O
x
x-axis.
6
33
Section
1
Basic
algebra
and functions
Example
2
Sketch
the
curve
y

2
(x

5)
y
2
y

2
(x

5)
2
x
8
6
4
2
2
2
4
2
y

(x

2
5)
y

x
6
2
Start
with
y

x
whose
shape
and
position
is
2
Then
y

direction
(x
of

the
known.
2
5)
is
a
translation
negative
of
y

x
by
of

5
units
in
2
Therefore
parallel
y
to

the
2
(x

positive
the
x-axis.
2
5)
is
a
translation
y
(x

5)
by
2
units
y-axis.
Exercise 1.11a
1
Sketch
each
of
the
following
curves
whose
2
(a)
y

x
4
y
(c)

(x
y

1
______

2
y
(d)

3
x
2
On
the
x
same
set
of
axes,
sketch
the
1
curves
3
(a)
y

y

x
(x
(c)
y



equations
are
(d)
y

1
(x

2)
3
2)
One-way
stretches
Consider
curve
the
whose
3
3
(b)
are
1)
1
__
(b)
equations
3
whose
(x
equation
is
y


2)
af(x)
y
y

af(x)
y

f(x)
ay
y
x
O
Comparing
points
y-coordinate
of
the
So
of
34
the
on
y

point
the
cur ve
f(x)
on
y
cur ve
y

and

y
f(x)
y

af(x)

af(x)
is
af(x)
parallel
a
is
with
times
a
to
the
the
same
one-way
the
x-coordinate,
y-coordinate
y-axis
on
stretch
by
a
factor
a
y

the
f(x)
Section
Consider
the
curve
whose
equation
is
y

1
Basic
algebra
and functions
f(ax)
y
y

f(ax)
y

f(x)
1
x
a
x
x
O
Comparing
points
on
y

f(x)
and
y

f(ax)
with
the
same
y-coordinate,
1
__
the
x-coordinate
of
the
point
on
y

f(ax)
is
times
the
x-coordinate
a
on
y

f(x)
So
the
cur ve
y

f(ax)
is
a
one-way
stretch
1
__
of
the
cur ve
y

f(x)
parallel
to
the
x-axis
by
a
factor
a
Example
2
On
the
same
set
of
axes
sketch
the
curves
y

x
2
,
y

2x
and
2
y

(2x)
for
values
of
x
from
3
to
3.
y
2
y

(2x)
30
2
y

2x
25
20
15
2
y

x
10
5
x
3
2
1
O
1
2
3
2
Start
with
y

x
2
Then
double
the
y-coordinate
of
points
on
y

x
2
to
give
y
2
Halve
the
x-coordinate
of
points
on
y

x

2x
2
to
give
y

(2x)
Exercise 1.11b
On
of
the
x
same
from
3
set
to
of
axes
1
__
(a)
y

the
graphs
2
__
(b)
x
sketch
of
the
curves
given
for
values
3.
y

1
___
(c)
x
y

2x
35
1.
12
Rational
Learning outcomes
expressions
Rational
An

To
express
an
improper
expressions
expression
polynomials
expression
as
polynomial
the
and
a
sum
where
both
the
numerator
and
denominator
are
rational
of
proper
is
called
a
rational
expression
a
rational
1
__
For
example,
3x
______
x
_____________
,
,
are
rational
expressions.
2
expression
x
These
the
(x
1)(x
expressions
numerator
is
are
less
2)
all
x
proper
than
the

1
rational
order
of
expressions
the
because
the
order
of
denominator
.
You need to know
When

The
meaning
of
the
order
of
the
order
of
the
denominator
,
For
example,

How
to
sketch
is
is
greater
called
than
or
equal
to
the
order
of
improper
.
2
x
_______
x  1
_______
and
2x
The factor
numerator
expression
a
polynomial

the
the

are
1
2x
improper
1
theorem
use
transformations
to
curves
Expressing
an
polynomial
There
form
The
are
two
where
first
improper fraction
and
a
methods
the
a
sum of
a
proper fraction
we
remaining
method
as
can
use
to
fraction
involves
is
express
an
improper
fraction
in
a
proper
.
rearranging
the
numerator
so
that
we
can
cancel.
2x  3
_______
For
example,
in
the
case
of
we
x
that
x
1
is
part
the
2(x
2x  3
_______
1)
rearrange
the
numerator
so
numerator
,

2

3
________________
i.e.

x
We
of
can
1
1
can
x
now
2(x
express
1)

the
1
right-hand
side
as
the
sum
of
two
fractions,
5
______
________
i.e.

x
1
x
1
5
______
We
can
now
cancel
(x
1)
in
the
first
fraction
to
give
2

x
2x  3
_______

5
______

x
The
1
2

1
x
second
1
method
2
involves
Start
by
dividing
dividing
x
the
into
numerator
2x.
It
goes
by
2
the
denominator
.
times.
_______
x
1
)
2x

2x
Multiply
3
2
2
is
the
x

1
by
quotient
2
then
and
5
is
subtract
the
this
from
2x

remainder
.
5
2x  3
_______
Then
This
5
______

x
2

second
x
method
is
5
__
12
___
(in
1
the
same
way
as
1

1

7
useful
when
the
denominator
)
7
is
quadratic.
3
x
2x  5
___________
For
example,
to
express
in
2
x
proper
,
36
we
divide
by
the

4x

5
denominator
.
a
form
where
the
fraction
is
3
Section
1
Basic
algebra
and functions
2
x
There
4
_________________
2
x

4x
)
5 x

3
is
no
x
term
in
the
2

0x

4x
3
2x

numerator
5

term.
5x
It
2

4x

4x
so
we
add
zero
for
this
2
2
x
7x

Start
goes
x
by
dividing
3
x
into
x
.
times.
5
2
2
16x

Multiply
x
subtract.
Bring

4x

5
by
x
then
20
2
9x

25
down
5.
Divide
x
2
into
4x
until
no
,
and
more
repeat
the
division
is
process
possible.
3
9x
x
2x  5
___________

x
This
x
4
25

2

4x

second
polynomial,
is

___________

2
not
5
x
method
we
of
need

4x
division
to
find
the
is

5
also
other
useful
factor
when,
and
given
finding
one
it
by
factor
of
a
inspection
straightforward.
Example
4
Given
that
2x
1
is
a
factor
of
3
2x
2
x

6x
x
1,
find
the
cubic
factor
.
3
x

3x

1
______________________
2x
)
1
2x
4
3
x
4
2

6x

x
1
3
2x
x
2
0

6x
x

1
2
6x
3x
2x

1
2x

1
3
Therefore
cubic
x

3x

1
is
the
factor
.
Example
x
______
Sketch
the
curve
whose
equation
is
y

x
1
x
______
x
1  1
__________

y
x
1
______

1
x
1

1
x
1
4
1
______

y

1

x
1
1
__
2
Start
with
y

1
______
,
then
y

x
x
1
1
__
is
the
translation
of
y

by
x
x
O
2
4
1
unit
in
positive
the
direction
of
the
x-axis.
1
______
So
y

1

is
x
the
1
1
______
4
translation
of
y

by
x
parallel
to
the
1
unit
1
positive
y-axis.
Exercise 1.12
1
Express
each
expression
in
a
form
where
3
6x
_______
2x
_______
(a)
the
fraction
is
proper
.
2
x
 x
 3
____________
(b)
(c)
2
2x

1
2x

1
x

4
2x
_______
2
Sketch
the
curve
whose
equation
is
y

2x
4
3
Show
that
x

2
is
a
factor
4
Hence
factorise
x
of
3

x
x
3

x

1
2

2x

3x

6
2

2x

3x

6
37
1.
13
Inequalities
–
quadratic
and
rational
expressions
Learning outcomes
Quadratic
A

To
revise
quadratic
quick
To
solve
rational
sketch
is
the
easiest
way
to
solve
an
inequality
such
as
inequalities
(x

inequalities
inequalities

1)(x
2)

0
involving
y
expressions
10
8
6
You need to know
4

How
to
sketch
a
curve
whose
2
equation
has
the form
2
y


The
ax


c
conditions for
equation
real
bx
to
have
a
roots
or
to
complete
4
no
The
roots
How
2
quadratic
real
the

x
O
2
the
sketch
x-axis,
of
the
i.e.
(x
curve

y

1)(x
(x
2)


1)(x
0,
2)
when
shows
1

x
that

the
curve
is
2
square
The
We
of
inequality
know
and

we
When
1)

(x
(x
and
x

1

x
x
Therefore
2,
(x
1)(x
solved
2)
2)

is
0
algebraically.
when
positive
x
or

1
and
negative
x

2,
depends
on
the
signs
2).
these
both

is
2,
signs
brackets
(x
1)
is
for
x
are

1,
negative,
positive
and
1

x
so
(x

(x
2)

2
and
1)(x
is
x
2)

is
negative,
2
positive.
so
negative.
both

be
1)(x
(x
2)

also


1,
1)(x
When
can
investigate
When
(x
that
whether
(x
So
brackets
1)(x
2)

are
0
positive,
when
1
so

(x
x


1)(x
2)
is
positive.
2
Example
2
Find
the
values
of
a
for
which
x

ax

a

0
for
x


2
The
curve
y

x

ax

a
is
a
parabola
with
a
minimum
value.
2
a
__
2
Completing
the
square
gives
x

ax

a

(x

)
2
2
a
__
x
(

)

0
for
all
values
of
x,
2
2
a
__
2
so
for
x

ax

a

0,
a

0
0
⇒
4
2
a
__
Now,
a
2

4
38
below
0
⇒
4a
a

a(a
4)

0
a
__
2

a
4
Section
1
Basic
algebra
and functions
y
A
sketch
of
y

a(a
x
O
2
4)
shows
that
a(a
4)

0
when
0

a

4
2
Therefore
x
Rational
An


a

0
when
0

a

4
expressions
expression
polynomials
For
ax
where
is
both
called
a
the
numerator
rational
and
denominator
are
expression.
example,
1
__
3x
______
x
____________
,
,
are
rational
expressions.
2
x
(x
The

1)(x
2)
x

1
range of values that
a
rational function
can take
1
__
The
graph
of
y

(see
T
opic
1.10),
shows
that
0
is
the
only
value
that
x
y
cannot
take.
1
__
We
can
show
this
algebraically:
y

1
__
⇒
x

x
when
y

0,
so
1
__
there
is
no
1
__
i.e.

0
or
of
x
x
is
undefined
for
which
y

0,
1
__

x
value
and
y
0
but

x
0
x
3x
______
Now
x
can
take
all
real
values
when
y

2
x
T
o
find
the
quadratic
values
that
equation
in
y
can
have,
we

1
rearrange
the
equation
to
give
a
x,
2
i.e.
yx

3x

y

0
2
For
x
to
be

4y
real,
this
equation
has
to
have
real
roots,
so
‘ b

4ac’

0,
2
i.e.
9
y
0
9
__
2
⇒


4
3
__
⇒
y

3
__

and
y

2
2
3
__
⇒

3
__

y

2
We
can
2
use
this
information,
together
with
the
following
observations,
3x
______
to
sketch
the
graph
of
y

2
x

1
39
Section
1
Basic
algebra
and functions

y

0
when
x

0

y

0
when
x

0

when

as
x
approaches

as
x
→
x

0,
,
y
y

0
very
→
large
values
(we
write
this
as
x
→
),
y
→
0
0
y
3
2
1
20
15
10
x
O
5
5
10
15
20
1
2
3
Solving
inequalities
It
to
involving
rational
expressions
3x
______
is
easy
see
the
values
of
x
for
which

0,
but
it
is
not
so
2
x

1
x
_____________
obvious
for
the
expression
(x

1)(x
2)
x
_____________
The
values
of
x
for
which

(x
x
1
The
x

and
x
value
1
of
and

1)(x
0
depend
on
the
signs
of
x
the

expression
2,
so
we
is
need
zero
to
when
x
investigate

0
the
and
sign
undefined
of
the
when
expression
when
x
The

easiest
0,
way
0
to
do
x
x
x
1
x
2

x

this

is
1,
to
0
1
use
0

a

x

2
and
x

2
table.
x

1
1

x

2
x

















2
x
_____________
(x
1)(x
2)
x
_____________
Now
we
can
see
that

(x

1)(x
0
when
0

x

1
and
x

2)
Example
x
_______
Solve
the
inequality
inequality
right-hand
40
is
side
easier
is
1
__

2x
An
x,
2)
2
zero.

to
1
solve
x
if
it
is
first
rearranged
so
that
the
2
Section
x
_______
1
Basic
algebra
and functions
1
__

2x

1
x
_______
x
1
__
⇒
2x

1

0

0
x
2
(x
1)
_________
⇒
x(2x
The
1)
numerator
is
positive
for
all
x


so
the
significant
values
of
1
x
are
0
and
2
1
We
need
to
investigate
the
ranges
x

0,
0

x
1

and
x

2
2
1
_
x

0
0

x
1
_

x

2
2
2
(x
1)
x
2x
1












2
(x
1)
_________
x(2x
1)
1
__
x
_______
1

Therefore
2x
for
1
0

x

2
x
Example
2
x
2x  k
___________
Find
the
values
of
k
for
which
can
x
for
all
x


take
all
real
values
1

2
x
2x  k
___________
Let
y

x
Rearranging

as
1
a
quadratic
equation
in
x
gives
2

x
x(2

y)

(k
y)

0
2
For
x
to
be
real,
(2

y)

4(k
y)

(4k
4)
2
⇒
y

8y
2
Completing
the
square
gives
(y

4)
16

4k
4
2
The
minimum
values
provided
value
of
that
(y
16


4)
16
4k
4,
is
16,
i.e.
k

so
y
can
take
all
real
3
Exercise 1.13
2
1
Find
the
values
of
x
for
which
x
2
Find
the
values
of
k
for
which
(kx)
3
Find
the
set

4

2x

1
2

(3k

2)x

4

0
for
x

.
x  1
________
of
values
of
x
for
which

x(x

0
2)
2x
______
4
Find
the
range
of
values
of
y
for
which
y

2
1

x
x  k
______
5
Find
the
minimum
value
of
k
for
which

1
for
all
x

.
2
x

1
41
1.
14
Intersection
Learning outcomes
of
curves
investigate
the
intersection
curve
and
a
on
the
shape
of
a
curve,
a
line
may
intersect
the
curve
at
of
several
a
lines
Intersection
Depending
To

and
points,
it
may
touch
the
curve
at
one
of
these
points,
or
it
may
line
not
intersect
For
example,
may
touch
the
a
curve
line
the
at
may
parabola
any
point.
intersect
at
one
a
parabola
point
(in
in
two
which
distinct
case
it
is
points,
called
a
or
it
tangent
You need to know
to
How

to
linear,
sketch
the
quadratic
graphs
and
the
parabola),
or
it
may
not
intersect
the
parabola.
of
cubic
functions
How

to
solve
a
simultaneous
one
is
linear
pair
of
equations
and
the
where
other
is
quadratic
T
o
The

conditions for
a
find
curve
equation
roots,
or
to
a
have
two
repeated
the
points
of
intersection,
we
need
to
solve
the
equation
of
the
quadratic
and
the
equation
of
the
line
simultaneously.
distinct
root,
or
For
no
example,
to
find
the
points
of
intersection
of
the
line
with
equation
2
real
y
roots

3x
5
equations
How

to
to find
use
the
the factor
roots
of
a
with
the
curve
with
equation
y

x
2x
simultaneously.

1,
we
solve
the
y
theorem
A
cubic
rough
idea
equation
of
above
which
exists
sketch
The
sketch
is
of
of
in
these
the
this
curves
three
case.
gives
cases
an
illustrated
However
,
this
inconclusive.
nature
of
the
solution
will
tell
us
if
this
x
O
line
intersects,
y
x
touches
or
misses
the
1
curve.
2

2x

1
[1]
y

3x
5
[2]
2
[2]
in
[1]
There
two
are
3x
two
distinct
From
or
⇒
x
[3],

3
5
real

2
x
and
2x

distinct
1
⇒
x
⇒
(x
roots
so
5x

6
2)(x
the

3)
line
0

0
[3]
intersects
the
curve
in
points.
the
and
coordinates
y

4,
i.e.
of
(2,
these
1)
and
points
(3,
are
x

2
and
(from
[2])
y

1,
4).
y
Example
2
______
Prove
that
the
line
y

3
2x
and
the
curve
y

do
x
not
intersect.
3
2
______
Solving
y

3
2x
and
y

simultaneously
x
y

3
2x
gives
3
[1]
2
______
y

[2]
x
x
3
2
______
Substituting
[2]
in
[1]
gives
x

3
2x

(3
3
⇒
2
2x)(x
3)
2
⇒
2x
9x

11

0
2
‘b
4ac’
is
81
88
which
is
less
than
zero,
so
there
are
no
real
The
sketch
shows
that
the
line
and
2
values
of
x
Therefore
42
for
the
which
line
2x
and
the
9x

curve
11
do

0
not
intersect.
curve
a
do
proof.
not
(A
intersect
sketch
is
but
also
this
is
not
unreliable.)
Section
1
Basic
algebra
and functions
Example
(a)
Find
the
condition
on
m
and
c
for
which
the
line
y

mx

c
is
a
2
tangent
(b)
Hence
to
the
find
curve
the
whose
equation
equation
of
the
line
is
y
with

3x
2x
gradient
2
1
that
is
a
2
tangent
to
the
curve
whose
equation
is
y

3x
2x
1
2
(a)
Solving
y

3x
2x
1
and
y

mx

c
simultaneously
gives
2
mx

c

3x
1)

0
2x
1
2
⇒
3x
For
x(m
the
line

to
2)
(c
touch

the
curve,
this
equation
must
have
a
repeated
2
root,
i.e.
‘b
4ac’

0,
2
so
(m

2)

12(c

1)
1
(b)
When
m

2,
16

12(c

1)
⇒
c

3

the
equation
of
the
line
with
gradient
2
that
is
a
tangent
to
2
y

3x
2x
1
is
1
y

2x

3
⇒
3y
6x
1

0
Example
Show
that
the
3
y

x
The
7x
values
roots
line
y

5x
4

of

of
the
intersects
the
curve
10x
x
at
5
once
which
the
and
intersects
4
5x
once.
the
curve
3

10x
of
f(x)
5

3
factors
the
it
line
2
7x
Possible
touches
are
given
by
the
equation
3
x
Using
0
2
factor

2
x
7x

15x

1),
9

0
2
x
7x
theorem,
⇒

when
x
15x

9
1,
f(x)
are

(x
0,

(x
(x
1)

9),
is
a
(x

3)
factor
.
2
So
f(x)

(x
1)(x

(x
1)(x
6x

9)
2
3)
3

the
equation
repeated
Therefore
y

7x

15x
9

0
has
one
single
root
and
one
root.
the
3
2
x
line
y

5x
4

0
intersects
the
curve
2
x
7x

10x
5
once
and
touches
it
once.
Exercise 1.14
1
Find
xy

the
4
value

of
k
for
which
the
line
y

kx

2
touches
3
2
Find
the
where
3
y
nature

Determine
0.
of
the
Hence
whether
points
sketch
the
on
the
line
y
the
curve
y

curve
the
curve
whose
x
2

5x

8x

4
curve.

x

5
intersects,
2
intersect
the
0
equation
is
x
touches
or
does
not
2

2y

7
43
1.
15
Functions
Learning outcomes
Mappings
2
When

To
define
mathematically
the
pressed,
terms: function,
domain,
number
2
is
entered
in
a
calculator
and
then
the
x
button
is
the
the
display
shows
the
number
4.
range,
2
is
mapped
to
4,
which
is
denoted
by
2

4
composite functions
Under
this
rule,
which
is
squaring
the
2
x

To
use
the fact
that
a function
input
may
be
defined
as
a
set
3

9,
25

625,
of
0.2
ordered
number
,
pairs

0.04,
number)

2

(the
4
and
square
(any
of
real
that
number).
2
This
is
denoted
This
mapping
by
x
can

be
x
,
for
x


represented
You need to know
2
graphically
against

How
to
sketch
curves
by
values
are
of

ax
square

bx
x.
The
of
graph,
x
of
what
happens
and
when
our
we
the form
2
y
of
values
whose
knowledge
equations
plotting

a
number
,
show
that
one
input
x
O
c
number
But
the
gives
just
mapping
one
that
output
maps
a
number
.
number
to
√x
its
square
root,
output
only
greater
than
numbers
This
x
The
when
do
or
for
gives

2,
input
to
have
real
is
(negative
square
written
roots).
as
x
O

that
output
a
number
zero
real
be
gives
representation
shows
two

the
can
x
graphical
mapping
4
equal
not
mapping
√x ,

e.g.
one
of
input
this
value
values.
Functions
2
For
the
mapping
x

x
,
for
x

,
one
input
number
gives
one
output
number
.
The
mapping
The
one
word

√x
function
output
A
x
is
gives
used
two
for
outputs
any
for
mapping
every
where
one
one
input
input
number
.
value
gives
value.
function
is
a
r ule
that
number
for
maps
a
each
defined
single
set
of
number
input
to
another
single
numbers.
2
Using
f
for
function
and
the
symbol
:
to
mean
‘such
that’,
we
write
f :
x

x
2
for
x
The
not


to
mapping
satisfy
have
function
44
they
do
The
set
x
this
Domain
We
mean

and
unless
of
is
the
√x
function
for
x

0,
that
x

maps

is
x
to
not
x
a
for
all
real
function
values
because
of
it
range
give
that
some
real
inputs
for
we
can
use
particular
numbers
a
as
function
any
real
numbers
number
have
to
as
be
an
input
excluded
for
a
because
output.
is
called
the
domain
of
the
function.
x
does
condition.
assumed
not
f
Section
The
domain
is
The
domain
does
wide,
or
fully,
If
as
the
the
also
not
domain
is
as
must
not
the
have
restricted,
domain
numbers
called
be
Basic
algebra
and functions
pre-image.
to
we
1
contain
choose
all
to
possible
make
it.
inputs;
Hence
to
it
can
define
be
a
as
function
stated.
stated,
we
assume
that
it
is
the
set
of
all
real
().
2
The
mapping
domain
we
x

x
choose.

3
can
Some
be
used
examples,
to
define
together
a
function
with
their
2
1
f : x

f
over
graphs,
any
are
given.
2
x

3
for
x

2

f : x
f(x)

x

3
for
x


x

0
f(x)
The
is
point
on
included,
solid
point
circle.
x
curve,
open

0
and
the
and
For
curve
we
the
would
we
where
denote
domain
not
indicate
be
x
this
x

part
this
by

by
0,
of
0
a
the
the
using
an
circle.
3
3
x
O
x
O
2
3
f : x

This
x

time
3
for
the
x

1,
2,
graphical
3,
4
representation
is
four
discrete
points.
y
20
16
12
8
4
x
O
These
1
2
three
3
4
examples
are
not
the
same
function
–
each
is
a
different
function.
For
each
The
is
set
also
The
domain,
of
there
output
called
the
notation

x
For
the
in
range
is
A
corresponding
numbers
is
called
the
set
of
output
range
of
the
values
of
a
numbers.
function.
The
range
f( x)
represents
the
output
function,
so
for
2
for
x,
f(x)
function
given
a
image.
2
f : x
is
2,
the
the
function

defined
range
set
can
x
of
is
in
numbers
be
1
also
above,
f( x)
4,
represented

7,
the
3
range
and
12,
for
is
the
f( x)

3,
function
for
the
defined
function
in
3,
the
19.
pictorially.
2
For
example,
This
first
f : x
function
number

can
in
x
also
the

3
be
pair
is
for
x

1,
2,
represented
the
value
of
3,
by
a
x,
4
can
set
and
of
be
illustrated
ordered
the
second
pairs,
1
4
2
7
3
12
4
19
domain
range
as:
where
number
is
the
the
2
value
by
of
the
f(x).
set
Therefore
{(1,
4),
(2,
f : x
7),
(3,

x
12),

(4,
3
for
x

1,
2,
3,
4
can
be
represented
19)}.
45
Section
1
Basic
algebra
and functions
Example
The
diagram
B{a,
b,
c,
d,
shows
Give
(b)
Construct
(a)
In
two
A,
(b)
For
f
2
to
doing
4
reasons
a
of
members
of
the
set
A{1,
2,
3,
4,
5}
to
the
1
a
2
b
3
c
4
d
5
e
set
why
this
function,
f,
mapping
that
maps
is
A
not
to
a
B,
function.
giving
your
answer
as
a
set
of
pairs.
maps
be
a
this
maps
mapping
e}.
(a)
ordered
a
to
different
function,
but
to
two
e
the
(or
every
simplest
d).
f

{(1,
members
member
is
to
b),
of
of
A
change
(2,
d),
B
and
must
the
(3,
(d
c),
map
two
(4,
e).
In
to
A,
just
mappings
e),
(5,
4
does
one
in
not
map
member
A
so
that
of
2
to
B.
any
member
There
maps
to
are
either
of
B.
several
d
(or
ways
e),
and
of
then
a)}
Example
2
The
function,
f,
is
defined
by
and
(a)
(a)
Find
f(4)
and
f(
4)
For
x

0,
f(x)

x
For
x

0,
f(x)

x
f( x)

x
f( x)

x
Sketch
(b)

f(4)

f(
the

for
for
x
x


graph
0,
0,
of
x
f.


Give
(c)
the
range
of
f.
4
f(x)
2
(b)
T
o
sketch
and
that
the
curves
part
graph
in
of
the
the
2
of
a
4)
y

(
function,
xy-plane.
line

x
So
we
which
4)
we

can
can
16
use
what
interpret
corresponds
we
f( x)
to

to
f(x)

x
negative
for
positive
2
and
know
x
about
x
lines

0,
as
values
of
x,
2
for
x
values

of
0
as
the
part
of
the
curve
y

x
that
corresponds
x
x
O
(c)
The
range
of
f
is
f( x)

0
Composite functions
1
__
2
T
wo
functions
f
and
g
are
given
by
f( x)

x
,
x


and
g(x)
and
g(x)

,
x
x

1
0,
x

They
.
can
These
be
two
added
functions
or
f(x)

g(x)

be
combined
in
several
ways.
subtracted,
1
__
2
i.e.
can
x

,
x

0,
x


x

0,
x


x
1
__
2
and
f(x)
g(x)

x
,
x
2
They
can
be
multiplied
2
i.e.
f(x)g(x)

x
or
divided,
1
__


x,
x

0,

0,
x


x
2
f(x)
x
__
____
and
3


g(x)
x
,
x
x


1
__
x
3
The
output
f
g
f
i.e.
of
be
made
input
or

of
g,
2
x

the
1
__
2
x
can
g[f(x)]

g(x
),
x

0,
x


2
x
1
__
Therefore
the
function
f :
x

,
x

is
a
0,
x


is
obtained
by
taking
the
2
x
function
as
g
of
the
function
f.
This
composite
gf(x).
2
For
f(x)
gf(x)

x
means
,
x

the

and
function
2
i.e.
46
gf(x)

g(x
g(x)
g

of
3x
the

3x
1,
x

x

function
2
)
1,


f( x),
function
and
is
written
Section
fg(x)
means
the
function
f
of
the
function
1
Basic
algebra
and functions
g( x),
2
i.e.
fg(x)
This

f(3x
shows
composite
For
any
the
input
1)
that
the
function
composite
domain
values
of

(3x
1)
composite
x


function
fg( x)
is
not
the
same
as
the
gf( x).
function
of
,
g.
gf( x),
Therefore
f(x)
the
is
the
range
range
of
f
of
must
f
and
be
this
range
included
in
gives
the
g.
Example
f,
g
and
h
are
functions
given
by
2
f(x)

x
,
x
as

(a)
Find
a
(b)
Calculate
,
g(x)

function
the
2x
of
value

x:
of:
1,
x

fg
(i)
,

1
x,
x


ghf.
(ii)
gf(3)
(i)
h(x)
hfg(3)
(ii)
gg(3).
(iii)
2
(a)
fg(x)
(i)

f(2x

1)

(2x

2
ghf(x)
(ii)

gh(x
1)
,
x


2
)

g(1
x

2(1
)
2
x
)

1
2

3
2
(b)
gf(x)
(i)

g(x
2x
,
x


2
)

2
2x

1,

gf(3)

2(3)

1

19
2
hfg(x)
(ii)

hf(2x

1)

h((2x

1)
2
)

1
(2x

1)
,
2

hfg(3)
gg(x)
(iii)



1
g(2x
gg(3)

(7)


1)

4(3)

3
48
2(2x


1)

1

4x

3
15
Example
1
_______
f
and
g
are
functions
of
x
such
that
f( x)

and
2x
Find

gf(x)

1
_______
T
o
x
1
g(x).
change
1
_______
to
2x

x,
we
need
to
first
take
the
reciprocal
of
1
,
2x

1
1
__
so
let
h(x)

,
then
hf(x)

2x
1
x
1
T
o
change
2x
1
to
x,
we
need
to
halve
2x
1
and
then
add
,
2
1
so
let
j(x)

1
x

,
2
1
___
jh(x)

1
then
jhf(x)

2
1
(2x
1)

2

x
2
1
__


2x
g(x)
2
Exercise 1.15
1
The
and
function
f(x)

(a)
Find
(b)
Sketch
x
the
f
is
for
given
x

value
the
0,
of
by
x
f(5),
graph
of
f( x)


f( 3)
the
x
for
x

0,
x


.
and
f(0).
function.
2
2
The
(a)
functions
(i)
(ii)
(b)
(i)
(ii)
Find
the
Sketch
Find
f
g
are
function
the
the
Sketch
and
curve
curve
given
whose
function
the
given
given
whose
by
by
f( x)

,
x


and
g(x)

2

x
fg( x).
equation
by
x
is
y

fg(x)
is
y

gf(x)
gf( x).
equation
3
_______
3
f
and
g
are
functions
such
that
f( x)

and
2x

gf(x)

x.
Find
g(x).
1
47
1.
16
Types
of function
Learning outcomes
Codomain
The

To
define
mathematically
codomain
of
a
function
is
the
possible
values
that
can
come
out
of
a
the
function.
terms:
one-to-one function
(injective function),
function
onto
This
(surjective function),
many-to-one,
onto function
one-to-one
will
include
and
(bijective function)
know
include
other
all
the
function
or
codomain

To
prove
whether
simple function
is
or
not
a
the
values
values
one
the
we
actual
as
that
The
will
have
values
values
well.
come
not
that
that
seen
might
come
codomain
out
of
a
before).
come
(i.e.
useful
function
We
out
out
is
of
can
a
the
range)
when
(e.g.
then
a
we
but
do
may
not
complicated
choose
as
a
function.
given
one-to-one
or
2
For
the
function
f : x

x
for
x

1,
2,
3,
we
can
onto
1
choose
the
from
to
codomain
to
be
the
set
of
integers
2
1
10
1
inclusive.
3
4
5
You need to know
2
6
7

The
meaning
and
range
of function,
domain
8
3
9
10
domain
codomain
One-to-one functions
A
function
different
to - one
A
is
one-to -one
member
of
the
when
each
codomain.
member
The
of
the
function
f
domain
defined
maps
above
to
is
a
a
one-
function.
one-to - one
function
is
also
called
an
injective
function
Onto functions
A
function
by
a
is
onto
member
without
a
of
when
the
every
domain,
matching
member
i.e.
member
of
no
the
of
the
members
codomain
of
the
is
mapped
codomain
are
to
left
domain.
2
For
x
example,

2,
1,
when
0,
1,
g
2
is
given
and
the
as
g( x)

codomain
x
for
is
2
0
the
1
set
{0,
1,
member
image
4},
of
in
the
the
the
diagram
codomain
shows
has
at
that
every
least
0
one
1
domain.
1
Therefore
An
onto
g
is
an
onto
function
is
function.
also
called
a
2
4
domain
codomain
surjective
function.
2
The
function
some
f : x
members

of
x
the
for
x

1,
codomain
2,
3
given
do
not
onto
the
above
have
an
is
not
image
surjective
in
the
because
domain.
2
However
,
because
number
,
not
one
48
h : x
every
and

real
for
of
x

number
,
every
one-to - one
member
x
positive
because
the

when
real
more
it
positive
squared,
number
than
codomain,
is
e.g.
one
has
a
real
real
member
both
2
and
numbers
maps
of
2
to
a
square
the
map
is
root.
domain
to
surjective
positive
4.
real
But
h
maps
is
to
Section
1
Basic
algebra
and functions
Bijective functions
A
function
This
is
a
member
comes
For

of
the
from
just
to
one
is
both
one-to - one
where
each
codomain
one
example,
maps
T
o
that
function
member
f : x
one

2x
for
member
member
of
and
of
).
of
where
the
x
and
of
is
the
each
called
a
domain
member
of
bijective
maps
the
to
function.
one
codomain
domain.

)
onto
member

onto

and
onto
(each
Therefore
f
is
a
is
one-to - one
member
bijective
(each
of

member
comes
of
from
function.
summarise:
injective
but
Every
domain
not
surjective
member
maps
member
of
to
the
of
a
surjective
the
Some
different
domain
codomain.
not
map
member
and
but
injective
members
of
the
every
codomain
to
of
the
Every
same
different
of
mapped
member
domain
codomain
member
is
bijective
the
and
member
codomain
is
of
the
to
member
codomain
to.
of
maps
a
of
the
every
the
mapped
to.
Example
2
Determine
the
whether
codomain

is
the
function
injective,
given
surjective
by
or
f( x)

x
2,
x


onto
neither
.
2
x
2
are

not
When
the
2
for
mapped
x

3
domain
Therefore
f
all
to.
and
maps
is
x
x
to
not

,
therefore
Therefore

3,
one
f(x)
f
is

some
not
7,
member
members
therefore
of
of
the
codomain
surjective.
the
more
than
one
member
of
codomain.
injective.
2
So
f(x)

x
2,
x


onto
the
codomain

is
neither
injective
nor
surjective.
Exercise 1.16
1
2
Let
A
f
{(1,


{1,
2),
Show
that
Let
A

f
{(1,

f
is
{1,
(a)
Show
(b)
Is
f
2,
(2,
1),
4}
and
(3,
1),
the
(4,
one-to - one
0,
(0,
that
an
3,
4),
f
onto
1,
0),
is
2}
not
and
and
(1,
function
1),
f : A

A
be
given
by
3)}
onto.
the
(2,
function
f : A

A
be
given
by
4)}
one-to - one.
function?
Give
a
reason
for
your
answer
.
49
1.
17
Inverse function
Learning outcomes
Inverse functions
f

To
define
term
mathematically
is
the
function
where
f( x)

2x
for
x

2,
3,
4
the
2
4
3
6
4
8
2
4
3
6
4
8
inverse function
The
domain
The
mapping
of
the
range
{2,
3,
can
can
4}
be
be
maps
to
the
reversed,
mapped
i.e.
back
range
each
to
{4,
6,
8}.
member
the
domain.
You need to know
We
can
express
this
reverse
mapping
as
1
x

The
meaning

What

The
a
of
shape
of
is
a
x
a function
curve
when
for
x

4,
6,
8
2
one-to-one function
equation

is
its
y 
a
quadratic
of x or
a
cubic function
This
It
is
is
a
function
called
the
in
its
inverse
own
right.
function
of
f
where
f( x)

2x
1
Denoting
function
the
inverse
we
write
f
function
of
f
by
6,
8
f
,
1
1
(x)

x
for
x

4,
2
of x
Notice
that
the
range
of
Not
every
function
has
an
the
of
the
inverse
function
is
the
domain
function.
inverse.
2
Consider
for
the
example,
function
both
f( x)
map
to

4.
x
for
x
When

this
,
which
mapping
is
is
such
that
reversed,
2
and
each
2,
value
of
2
x
maps
this
is
not
domain
i.e.
to
two
a
function.
maps
only
values
to
a
x,
for
Hence
different
one-to - one
A
of
only
f
has
Y
ou
can
When
tell
any
whether
line
a
have
an
and
function
parallel
to
the
of
an
the
onto
f( x)
is
x-axis
to
both
where
2
each
codomain
and
2,
member
have
an
of
and
the
inverse,
inverse.
inverse
an
maps
functions
member
functions
function
example 4
only
if
f
is
a
one-to - one
function.
one-to - one
will
cut
the
from
the
graph
graph
only
of
once,
y
f

f(x)
is
one-to - one.
y
y
x
O
one-to - one
The
The
line
The
graph of
diagram
y

A
50
shows
of
a
coordinates
gives
the
curve
and
that
is
its
one-to - one
inverse
obtained
by
reflecting
y

f(x)
in
the
x
reflection
whose
not
a function
x
O
the
point
are
A( a,
( b,
coordinates
of
a),
A
b)
i.e.
on
the
curve
y
interchanging

f(x)
the
x-
is
the
and
point
A
y-coordinates
of
Section
Therefore
we
can
interchanging
Now
the
x
obtain
and
y
coordinates
the
in
of
equation
the
A
of
equation
on
y

f(x)
y
the

can
reflected
curve
1
Basic
algebra
by
y
f(x)
be
written
as
[ a,
f(a)].
Therefore
A(a,
the
coordinates
the
reflected
of
A'
curve
on
is
the
such
reflected
that
the
curve
output
and functions
are
of
f
[f( a),
is
a],
i.e.
mapped
the
to
equation
the
input
of
b)
of
f.
x
O
A(b,
Hence
if
the
equation
of
the
reflected
cur ve
can
be
a)
written
1
in
the
for m
y

g(x),
then
g
is
the
inverse
of
f,
i.e.
g

f
y
Any
curve
whose
equation
can
be
2
y
written
be
in
the
reflected
However
,
form
in
the
this
y

line
f(x)
y
reflected


x
can
x
curve
may
not
2
y
have
an
equation
that
can
be

x
written
1
in
the
form
y

f
(x)
2
The
diagram
shows
the
curve
y
the
line


x
x
O
and
The
its
reflection
equation
of
in
the
image
y
x
curve
is
√x
is
2
x

y
⇒
y
a
function.
 √x

and
x

not
y
2
y
(We
on
x
can
the
y
this
reflected
maps
case
see
to
two
cannot
function
of
from
the
curve,
values
be
diagram
one
of
value
y.
written
So
as
in

x
,
x

of
this
a
x.)
y
However
,
0
as,
if
we
change
the
domain


0
to
2
give
x

so
the
,
function
then
does
f
have
is
f( x)
a
an

x
,
x

one-to - one
0,
function
inverse.
x
O
Example
1
Find
y

f
(4)
5x
when
f(x)

5x
1,
x


1
1
For
the
reflected
curve
x

5y
1
⇒
y

1
(x

1)
and
(x
5
1
1
function,
so
f
(x)

(x

1),

f

1)
is
a
5
1
1
(4)

5
(4

1)

1
5
Exercise 1.17
2
1
A
function
f
is
defined
by
f :
x

(3

x)
,
x

3,
x


1
Define
2
The
x

f
(x)
fully.
functions
f
1
Find:
3
(a)
and
g
are
given
by
f( x)

2x,
x


and
g(x)

2
x,

(a)
Show
g
1
f
that
1
(x)
f( x)

(b)
(x
–
(gf)
1)(x
(x)
–
2)(x
–
3),
x


does
not
have
an
inverse.
1
(b)
Redefine
f(x)
with
a
different
domain,
so
that
f
(x)
exists.
51
1.
18
Logarithms
Learning outcomes
Indices
Logarithms

To
use
the
laws
of
logarithms
depend
on
the
laws
of
indices,
so
here
is
a
reminder
of
these
to
laws.
simplify
expressions
p

a

a

(a
q

a

a
p
p

a

a
q
p

p
q
q
3
For
example,
x
For
example,
x
For
example,
(x
q

x

x
3
pq
)
4

x

x
4
3
a
3

3
4
3
)



4
7

x

x
4
1
4
x
12

x
You need to know
1
0
1
__
n
a


1,
n
n
a

,
a
√

a
n
a

The
2,
value
3,
of
simple
powers
of
5
Logarithms
2
We
can
read
the
This
this
The
statement
base
10
relationship
2
In
the
raised
can
be
the
power
to
form
the
power
is
relationship
2
or
the
which
called
can
is
2
to

100
then
the

to
the
a
as
power
rearranged
is
whole
10
gives
give
base
the
10
100.
same
must
information,
be
raised
to
i.e.
give
100.
logarithm
be
abbreviated
logarithm
log
2
to
the
to
base
read
10
of
100
100
10
3
In
the
same
way,
2

8
⇒
3

log

81
⇒
4

log
8
2
4
and
3
81
3
2
Similarly,
log
25

2
⇒
25

5
⇒
3

9
5
1
1
and
log
3

9
The
base
of
a
2
2
logarithm
can
be
any
positive
number
,
so
c
b

a
⇔
log
b

c,
a

0
a
The
symbol
⇔
means
that
each
of
these
facts
implies
the
other
.
0
Also,
as
a

1
this
means
log
1

0,
a
i.e.
the
The
logarithm
power
of
a
4
16,
to
4
any
base
number
is
zero
always
gives
a
positive
result,
__
1
2

1
positive
2
e.g.
of

,
…
16
c
This
means
that,
if
log
b

c,
i.e.
b

a
,
then
b
must
be
positive.
So
logs
a
of
positive
but
the
logarithm
Natural
There
numbers
is
of
a
negative
number
does
not
exist.
logarithms
an
irrational
mathematics.
This
exist,
constant
It
is
was
number
denoted
first
by
that
e
named
appears
and
e
by
is
in
equal
Euler
several
to
who
different
areas
2.71828…
showed
that
as
x
1
__
x
→
,
(x
)

→
e
x
1
__
Newton
discovered
that
the
sum
1

1
______

1
1
__________

1

2
1

2

3
1
______________


1
52

2

3

4
…
→
e
as
more
and
more
terms
are
added.
of
Section
When
e
is
used
logarithms
as
and
the
are
base
for
denoted
logarithms
by
ln
they
are
called
1
Basic
algebra
and functions
natural
x
y
ln x
means
log
x
so
ln x

y
⇔
e

x
e
Logarithms
by
lg x
or
with
log x,
a
base
i.e.
if
of
the
10
are
base
is
called
not
common
given,
it
is
logarithms
taken
to
be
and
10.
denoted
So
log
Did you know?
x
y
means
log
x
and
log x

y
⇔
10

x
Natural
10
logarithms
Napierian
invented
Evaluating
scientific
base
e
or
the
calculator
can
be
used
to
find
the
values
of
logarithms
with
10.
‘ln’
–
in
called
Napier
1614
he
logarithms
tables
Use
button
to
evaluate
natural
logarithms
and
the
‘log’
explanatory
related
an
to
internet
‘Napier’s
Use
also
logarithms. John
logarithms
published
A
are
text
natural
search
and
logarithms.
to
look
up
bones’.
button
x
to
evaluate
button)
is
Laws of
common
used
to
logarithms.
evaluate
The
powers
of
e
button
(usually
above
the
‘ln’
e.
logarithms
x
Given
x

log
b
and
y

log
a
x
Now
bc

Therefore
log
(a
bc
c
then
a
y

b
and
a

c
a
y
)(a

x
)
x
⇒

y
bc

i.e.

y
a
log
a
bc

log
a
b

log
a
Example
c
a
2
Express
This
is
the
first
law
of
logarithms
and,
as
a
can
represent
any
base,
the
law
is
applies
used
for
to
all
the
the
logarithm
of
logarithms
any
product
the
formula
in
provided
that
the
same
√
log pq
r
in
terms
of
this
simplest
possible
logarithms.
base
2
√
log pq
r
2

Using
x
and
y
again,
a
law
for
the
log
of
a
fraction
can
be
log p

log q

log
√
r
found.
1
x
b
__
a
__

b
__

⇒
x

log p

2 log q
y

log r
2
a
y
c
a
c
b
__
Therefore
log
a
(
b
__
)

x
y
i.e.
log
a
(
c
)

log
b
log
a
c
a
Express
n
A
third
law
allows
Example
c
us
to
deal
with
an
expression
of
the
type
log
b
a
3 log p

n log q
4 log r
x
n
Using
x

log
b
x
⇒
n
a

n
b
i.e.
a

b
a
as
x
__
Therefore
a
single
logarithm.
n

log
b
⇒
x

n log
a
b
i.e.
log
a
b

n log
a
b
a
3 log p

n log q
4 log r
n
3
These
are
the
most
important
laws
of
logarithms.
Because
they
are

log p

log
n

any
base
we
do
not
include
a
base,
but
in
each
of
these
laws
4
log r
true
3
for
log q
p
every
n
q
____
4
logarithm
must
be
to
the
same
r
base.
b
__
log bc

log b

log c,
log
n

log b
log c,
log b

n log b
c
Exercise 1.18
1
Find
(a)
the
log
value
of:
16
(b)
log
2
2
2
(c)
log
4
Express
in
terms
of
the
8
4
simplest
possible
logarithms:
p
__
(a)
x
______
2
log
(b)
ln 5x
(c)
log p√q
(d)
ln
q
3
Express
as
(a)
log p
(b)
ln 3
x
a
single
log q

1
logarithm:
(c)
2 log p

5 log q
1

ln x
(d)
2 ln x
ln (x
1)
4
53
1.
19
Exponential
Learning outcomes
and
logarithmic
Exponential
equations
equations
x
An

To
solve
logarithmic
exponential
exponential
equations
the
base
of
has
x
as
part
of
the
index,
for
example,
2
3

8
including
When
changing
equation
and
a
logarithm
you
solution
need
is
to
solve
2
For
an
exponential
equation,
first
look
to
see
if
the
obvious.
example,
for
x
3
5
2
Therefore

125,

5
x
notice
that
125

5
3
5
so
2
x

3
⇒
x

1
You need to know
When

The
laws

How
to
of
logarithms
simplify
the
change
solution
the
index
is
to
logarithms
x
For
example,
for
not
a
obvious,
taking
logarithms
of
both
sides
can
factor
.
3
3

8,
taking
logs
of
both
sides
gives
log 8
_____
(x
3) log 3

log 8
so
x
3

log 3
Therefore
x
3

1.892...
⇒
x

4.89
log 8
that
is
NOT
equal
to
log
.
log 3
well
have
used
s.f.)
8
__
_____
Note
(3
Note
also
that
we
could
equally
3
natural
logarithms.
Example
x
Solve
The
the
equation
left-hand
x
2
side

of
2(2
this
)

3
equation
cannot
be
simplified
so
taking
logs
x
will
not
help,
but
using
x
Let
y

2
y

x
,
then
2
will.
2
__
x
2

2(2
) 
3
⇒
y

2

3
⇒
y
3y

2

0
y

(y
2)(y
1)

0
⇒
y
x
So
2

1
or
y

2
x

1
⇒
x

0
or
2

2
⇒
x

1
Example
x
Solve
the
x
4(3
equation
4(3
x
)(5
x
)

7
⇒
(3
x
)(5
logs
gives

7
x
)(5
)

x
T
aking
)
ln (3
1.75
x
)(5
)

ln 1.75
0.207
(3
s.f.)
⇒
x ln 3

x ln 5

ln 1.75
ln 1.75
__________

x


ln 3

ln 5
Logarithmic
A
logarithmic
x,
T
o
for
equation
example,
solve
a
equations
ln ( x
contains
2)
logarithmic

1
the
logarithms
of
expressions
containing
ln x
equation,
again
look
to
see
if
the
solution
is
obvious.
2
For
example,
for
log
(2x
1)

2 log
2
x,
we
can
write
2 log
2
then
log
(2x
1)

log
2
x
x
as
log
2
2
x
2
2
⇒
2x
⇒
x
1

x
2
2
When
single
54
the
solution
logarithm
is
and
not
obvious,
then
remove
express
the
the
2x

1

logarithmic
logarithm.
0
⇒
terms
x

as
1
a
,
Section
For
example,
for
3 log
x

log
2
16
1,
collecting
the
logarithmic
1
Basic
algebra
and functions
terms
2
3
x
___
on
one
side
and
expressing
as
a
single
term
gives
log

1,
then
2
16
3
x
___
removing
the
log
gives
1
__

16
2
3
Therefore
x

8
⇒
x

2
Example
Solve
the
ln x
2
equation

ln (x

ln
x
1)
2

⇒
ln x
⇒
ln
ln (x
1)
ln (x

1)

2
x
______

x

x
______
2
⇒

x

2
1
e
1
2
2
So
x(1
e
e
______
2
)

e
⇒
x


1.16
(3
s.f.)
2
1
e
Changing the
base of
When
logarithmic
the
bases
simplified
change
If
x

to
the
log
a
single
base
c
of
of
and
logarithm
terms
logarithmic
the
we
a
are
term.
different,
T
o
do
they
that,
we
cannot
need
to
be
be
able
to
logarithm.
want
to
change
the
base
of
the
logarithm
to
b,
then
a
x
x

log
c
⇒
c

a
a
log
c
b
_____
T
aking
logarithms
to
the
base
b
gives
log
c

x
log
b
a
⇒
x

b
log
a
b
log
c
ln c
____
b
______
i.e.
c
log

and
in
particular
log
a
c

a
log
a
ln a
b
The
base
of
an
exponential
expression
can
x
T
o
express
change
x
a
as
a
power
of
e,
then
using
x
in
a
similar
way.
p
a

e
gives
x ln a

p,
therefore
x ln a
a

e
Example
Solve
the
equation
3 log
x

1

2 log
3
First
change
the
base
of
x
9
log
x
to
3.
9
log
x
log
3
_____
log
x

x
3
_____

,
9
log
9
2
3

3 log
x

1

2 log
3
x
⇒
3 log
9
x

1

log
3
x
3
1
1
2 log
x

1
so
log
3
x

⇒
3
x

3
2

1.73
(3
s.f.)
2
Exercise 1.19
2x
1
Solve
the
2
Solve
the
ln(x
3

Solve
1)
the
equation

(4
ln 2

equation
x

2
)(5
simultaneous

1
1
)


x
6
equations
ln y
and
ln(x
log
x
log

2

2
2y


1)

0
2
x
3
4
Given
that
ln y

3,
find
the
value
of
x
given
that
ln x

4 log
5

8
y
55
1.20
Exponential
Learning outcomes
and
logarithmic functions
Exponential functions
x
The

To
define
exponential
mapping
x

2
2
is
such
that
2

1
2
2

4,
2

2

,
and
and
any
4
real
number
maps
to
a
single
real
number
.
logarithmic functions
x
Therefore
x

2
,
x


is
a
function.
x
But
is
x

not
a
(
2)
x
has
a
real
value
only
when
x
is
an
integer
,
so
(
2)
,
x


function.
You need to know
x
However
,

The
the
definition
meaning
of
of
a function
a
for
any
value
of
a

0,
x
a


,
x


is
a
function.
and
x
one-to-one
The
function
f( x)

a
,
x
is
called
an
exponential
function.
function
x

How
to find
an
For
inverse function
is

The
the
meaning
laws
of
of
logarithms
all
f(x)
values

of
a

0,
x
a

0,
therefore
the
range
of
f( x)

a
,
x


0
and
logarithms
X

The
meaning
of
The
natural
y
curve
a

logarithms
x
The
family
of
curves
whose
equations
are
y

a
go
through
a
point
that
0
is
common
point
(0,
to
all
of
them:
when
x

0,
a

1,
i.e.
they
all
go
through
the
1).
x
y
x
y

When
a

1,
y

1
When
a

1,
and
x

1
3
10
x
y

x
2

0,
a
increases
as
x
increases
x
2
1
y

(
2
(e.g.
)
2
3
,
2
10
,
...,
2
,
...)
8
x
and
when
x

0,
a
decreases
2
(e.g.
2
as
x
decreases
3
,
2
10
,
...,
2
6
,
...),
but
never
reaches
0,
x
i.e.
as
x
→
,
a
→
0
x
When
a

1,
the
opposite
happens:
y

(
(e.g.
x
(1.5)
as
x
→
,
2
1
4
)

4,
...
→
0
(
2
a
increases
5
1
)
32...)
2
x
as
x
→
,
2
x
This
4
graph
shows
the
curve
y

a
for
some
different
x
O
2
a
2
values
of
x
X
4
The
f(x)
inverse of the function
a

x
The
2
function
f( x)

a
is
x
If
y

a
a
one-to - one
function,
so
it
has
x
where
f(x)

a

x
an
inverse.
1
,
we
obtain
the
graph
of
y

f
(x)
by
reflecting
x
y

We
a
in
can
the
line
obtain
y
the
equation
of
this
reflected
curve
1
y,
y
so
the
equation
of
y

f
by
interchanging
x
and
y
(x)
is
given
by
x

a
.
T
aking
logarithms
to
x
y

a
base
a,
we
get
log
x

y,
i.e.
y

log
a
Therefore
when
x
a
x
f( x)

a
1
,
f
(x)

log
x
a
The
function
log
x
has
domain
x

0.
a
y

log
x
a
(The
range
of
a
function
is
the
domain
of
the
inverse
function.)
1
The
function
f : x

log
x,
x

0,
x


is
called
a
logarithmic
a
O
1
x
x
function
and
it
is
the
inverse
of
f :
x

a
,
x


x
The
graph
shows
the
curves
with
equations
y

a
and
y

log
x
a
56
Section
1
Basic
algebra
and functions
X
e
The functions
When
a

ln x
and
e,
x
f(x)
f(x)

The

e
,
ln x,
x
x



is
0,
logarithmic
x
called


the
is
function
exponential
called
is
function
the
and
the
inverse
vice
function
logarithmic
of
the
and
function.
exponential
versa.
x
The
graph
shows
the
curves
y

e
and
y

ln x
y
10
x
y

e
8
6
4
y

lnx
2
4
x
O
2
2
4
6
8
10
2
4
x
Note
and
that
the
the
x-axis
y-axis
is
an
is
an
asymptote
asymptote
to
to
the
the
curve
curve
y

y

ln x
variations
of
the
e
x
These
sketches
show
some
simple
graph
of
y

e
Example
y
y
y
3x
Given
O
O
x
x
find
(0,
f(x)

e

1,
x
∈
,
1
f
(x).
1)
(0,
1)
3x
When
x
y

e
y
x
y

y

e

e

e
interchanging
(0,
1,
x
x
and
y
gives
3y
1)
x

e

1
3y
O
⇒
e

x
⇒
3y

ln (x
1
x
1)
1
1
Therefore
f
(x)

ln (x
1),
3
x

1,
x


Exercise 1.20
1
On
the
same
set
of
axes,
x
y
2

On
y

1

the
1
e
sketch
the
x
,
y
same
ln x,

1
set
y
e
of

,
y
axes,
ln x

1
Given
f(x)

1
e
4
Given
f(x)

1
2
of
graphs
of
e
sketch
1,
y

the
ln (x
x
3
graphs
x
1)
1
,
x

,
find
f
(4).
1
ln
x,
find
f
(2).
57
1.21
Modulus functions
Learning outcomes
The
x
modulus of
y
The

To
define
the
modulus
its
x
is
written
as
| x|
and
it
modulus function
means
and
of
the
positive
value
of
x
whether
or
properties
not
x
itself
e.g.
|2|
Hence

is
2
the
positive
and
|
graph
or
2|
of
y
negative,


2
|x|
can
be
found
y

O
x
O
x
x
You need to know
from
part

How
to
sketch

Algebraic
simple
the
of
graph
the
of
graph
y

for
x
by
changing
which
y
is
the
negative
curves
to
methods for
the
equivalent
positive
values,
i.e.
by
y
solving
reflecting
the
part
of
the
graph
where
y
is
inequalities
negative
Hence
in
we
the
x-axis.
define
the
function
f :
x

|x|,
⎫
|x|

x
for
|x|

x

y
0
⎬
as
The
|x|
for
x

0
always
positive


Now
the
for
positive
any
|x|

x
|y|

y
for
two
or
zero,
x

for
y
square
real
0
so
0
then
x
root
of
numbers
and

we
|x|

x
x
x
can
,
x
|x|


y
|y|
so
|x|


y
so
x
i.e.
and
for
2
If
write
|x|

__
2
√
x
y,
x

0,
and
|y|

y
y
conversely
if
x
| x|

|y|
⇔
x
2
y
follows
2
⇒
x
⇒
(x
that
| x|

|y|
⇔
x

y)(x
y)

2
that
x
The
can
,
then
y

y
illustrate
2
y
when

y
| x|
last
this

y
y

x

y
x

y
y)



y)



2

i.e.
We
y
0:
(x
shows
2

0

(x
table
and
2
y
x
The
0
2

2


|y|,
It
2
y
2
and
2
x
for
2

i.e.
Now
|x|
| x|
2
|x|



properties of
is
x
on
property
a
is


|x
x

|y|

y|
y,
⇔


y
|x|


x

y
|y|
diagram.
x
The
y
modulus
representing
x
y
x
x
x



a
number
number
is
equal
from
to
zero,
the
distance
shown
here
of
as
the
the
point
vertical
line.
y
y
The
diagram
when
y
of
that
but
x
and
when
shows
y
one
are
is
that
both
positive
positive
and
or
the
both
other
negative,
|x

y|

|x|

|y|
is
negative,
| x

y|

|x|

|y|,
so
| x

y|

|x|

|y|
0
58
Section
The
The
the
C
modulus of
graph
curve
for
For
of
C
any
which
f( x)
example,
curve
W
e
y

then
to
(x
sketch
in
y
negative.
1)(x
reflect
whose
equation
is
y

equation

f(x),
The
by
is
y

remaining
|(x
|f(x)|
reflecting
1)(x
in
can
the
sections
2)|
the
and functions
we
are
start
be
found
x-axis
by
not
the
from
parts
of
changed.
sketching
the
x-axis
the
part
of
this
curve
which
is
below
the
x-axis.
y
y
y

(x
1)(x
any
function
f,
the
mapping
x

1
|f(x)|
1)(x
2)|
is
2
also
a
function.
y
the
graph
of
1

|(x
x
O
Example
Sketch

2)
x
O
y
algebra
2)
y
For
Basic
a function
curve
with
1
|1
1
x|
and
y
write

1
x
2
2
1
the
equations
y
in

(1
x)
2
1
non-modulus
each
part
on
form
the
of
sketch.
x
O
Start
with
a
sketch
2
of
1
y

1
x,
then
reflect
the
2
part
the
below
the
x-axis
in
x-axis.
Example
2
Sketch
the
graph
of
y

2

|x
4|
2
Start
with
positive
y

|x
direction
4|,
of
the
then
translate
the
curve
by
2
units
in
the
y-axis.
y
y
10
10
8
8
6
4
2
2
O
x
4
2
4
4
2
2
x
O
2
4
2
4
Exercise 1.21
Sketch
the
graphs
of:
1
______
1
y

1

|x

1|
2
y

|
x

2
|
3
y

|(x

1)(x

1)(x

2)|
59
1.22
Modulus
Learning outcomes
equations
To
solve
modulus
inequalities
Intersection
T
o

and
equations
find
the
points
of
intersection
between
two
graphs,
we
need
to
solve
and
the
equations
of
the
graphs
simultaneously.
When
those
equations
inequalities
involve
a
modulus,
modulus
For
a
sketch
helps
to
identify
example,
x
equations
in
non-
form.
to
find
the
y
values
y
You need to know
of
those
where
the
graph

1
2x

(1
2x)
y
of
y
y

How
to
sketch
the
graph

|x

intersects

(x

the
y

|1
2x|,
we
draw
|f(x)|
a

2|
of
graph
y

How
to
solve
linear
sketch
and
on
it
y
the

(x

2)
and
equations
quadratic
write
equations
of
each
section
in
0.5
2
non-modulus
form.
O

How

The
to
solve
x
inequalities
There
properties
of
modulus
are
two
intersection,
functions
(1
2x)
points
one

x
of
where

2
⇒
x

3
1
and
one
where
1
2x

x

2
⇒
x


3
2
Alternatively,
using
the
property
that
| x|

|y|
⇔
2
|x

2|

|1
2x|
⇒
(x
2


2)
2
x

y
:
2

(1
2x)
2
x

4x

4

4x

3

0
4x

1
2
⇒
3x
8x
⇒
(3x

1)(x
3)

0
1

x


or
x

3
3
Check:
1
when
x


2
,
|x

2|

when
It
is
x

3,
We
|x
can
2|
that

the
can
5
an
2x|

1

or
g(x)
by
and
|1
of
2x|
x
found
give
such
f(x)
as
the


5,
so
x



this
of
|2 x
1|
and

is
a
solution
3
x
x
3x
3
is
also
method
that
modulus
following
g(x)
so
using
values
involving
using
then
1
,
3
values
equation
above,
|f(x)|
|1
sometimes
equations
solve
illustrated
when

squaring
Solving
and
3
essential
because
2
1
3
are
a
are
not
solution.
checked
solutions.
signs
by
sketching
graphs
as
fact:
f(x)

g(x)
Example
Solve
the
equation
|2 x
1|

3x
1
2x
1

3x
gives
x

1
and
(2x
1)

3x
gives
x

5
Check:
when
x

when
x


1,
|2x
1|

3
and
3x
3
1
,
|2x
1|
5

3x

5
x

is
5
60
the
3,
so
only
solution.
x

1
is
not
a
1
,
5
1
Therefore

3
and
so
x

is
5
a
solution.
solution
2)
Section
Example
1
Basic
algebra
y
and functions
y

x

2
1
______
Solve
the
equation
x

2

|
x

3
|
1
______
From
the
sketch,
x

2

|
x

3
|
where
1
______
x

2
2

⇒
x

x
x
7

0
3
1
⇒
x

3.19
(3
y
s.f.)

1

y
x
x
and
sketch
shows
that
we
only
want
the
positive
3
x
3
O
(the

3
root)
where
1
______
x

2

2

⇒
x

⇒
Solving
Simple
inequalities
inequalities
can
be
involving
solved
x
x
5

0
3
from
a
x

2.79
or
modulus
sketch
of
1.79
(3
s.f.)
signs
the
graphs.
y
y
y

(x

x
a
a)
b
a
a
For
a
example,
b

x

Otherwise
the
a
the
sketch

b
shows
O
that
b
the

x
a
inequality
| x
a|

b
is
true
for
b
method
used
is
the
same
as
for
equations.
y
y
Example

3
x
y
y
Solve
the
inequality
From
the
sketch,
|3
x|

x|


(3
x)
x
|x|
y
|3

|x|
where
3
x


x
x,
3
i.e.
where
x

2
3

|3
x|

|x|
for
x

x
O
2
3
2
Exercise 1.22
1
Solve
(a)
these
|3x

equations:
2|

5
2
Solve
(a)
the
|2x
following

1|

|1
inequalities:

x|
1
______
(b)
|2

x|

|x|
(b)
|
x

|


x
1
|x

1|
x
(c)
|e

2|

1
(c)
|ln x|

1
61
Section
1
Practice
questions
3
1
f(n)

x
n

10
2n
Solve
(a)
the
equation
e
x

4e

3

0
2
(a)
Show
that
(b)
Hence
f( k

1)

f(k)

3(k

k

1)
Find
(b)
log
by
3
prove
for
all
by
induction
positive
that
integer
f( n)
values
is
of
the
x

values
2 log
2
2
divisible
of
x

for
which
1
x
n.
11
(x

1)
4
x
and
(x
3


2)
are
factors
of
2
px

qx

16x

12
n
2
f(n)

9

(a)
Show
(b)
Hence
1
that
f(k
prove

by
1)

9f(k)
induction

(a)
Find
the
(b)
Hence
that
f( n)
is
solve
4
8
for
all
positive
integer
of
p
and
q.
values
of
the
equation
divisible
x
by
values
8
3

px
2

qx

16x

12

0
n.
2
x
______
12
Given
that
y

for
x
3
3
Prove
by
induction
that
n

n
is
a
multiple
of
the
for
all
positive
integer
values
of

all
real
values
of
x,
find
1
6
range
of
values
of
y.
n.
___
√
18
 1
________
13
4
p
and
q
are
Construct
__
Simplify
(a)
propositions.
a
truth
table
√
2
to
show
the
truth

3
3
values
Given
(b)
that
x

5
,
find
the
value
of
log
5.
x
of
~p
→
Hence
q
and
p

determine
equivalent
q
whether
~p
→
q
and
p

q
are
14
Find
the
range
of
values
of
x
for
y

which
x  1
___________
statements.

0.
2
x
5
p
and
q
are

5x

6
2x
______
propositions.
15
Sketch
(a)
the
graph
of
.
x
(a)
W
rite
(b)
Use
down
the
contrapositive
of
~p
→
On
(b)
the
algebra
of
propositions
to
show

1
~q
a
separate
diagram
sketch
the
graph
of
that
2x
______
~p
→
(p

~q)

q
→
y
p

|
x

1
|
.
2x
______
Solve
(c)
6
Sketch
(a)
the
graphs
the
equation
x

|
of
x

1
|
.
2
__
y

x

1
and
y

|
x
16
|
The
the
same
diagram.
Show
the
*
y

the
Find
(b)
points
the
where
range
of
the
graphs
values
of
x
7

Prove
1
by
2


x

x

for
binary
…

operation
2
x
*
y
n

(n

1)(2n


all
State
y

.
whether
the
operation
is:
closed
(b)
associative
(c)
distributive
(d)
such
that
x
over
has
multiplication
an
inverse.
2
17
(a)
*
is
defined
Sketch
Solve
the
the
graph
of
inequality
y

|ln
2

x|
|1


x
|
2
by
2
x
x
x,
1)

y
and
y
18
for
for
reasons
(a)
(b)
The
y
which
6
8

with
n
__
2
3
by
that
2
2
defined
|
induction
2
1
|
is
intersect.
2
__
x
*
2
coordinates
Explain
of
operation
2
x
on
binary
with

Given
f(x)

2

ln x,
sketch
the
graph
of:
.
reasons
whether
the
operation
*
(a)
y

f(x)
(b)
y

f
is:
1
(a)
closed
(x)
commutative.
(b)
x
19
3
9
f(x)

When
2x
f(x)
is
5x

px
divided

the

1
6
2x

5(3

1
)
by
(x

2)
the
remainder
2
f(x)

(x

1)
for
x


and
10.
1
_______
_
1
One
root
of
the
equation
f( x)

0
is
g(x)
.
2
(a)
Find
(b)
Factorise
real
62
equation
q.
20
is
Solve
2

the
values
roots
f( x)
of
of
and
the
p
and
hence
state
f( x)
the

for
(x
q.
equation

number
0
of

(a)
Find
(b)
Explain
not.
x

0,
x

.
1)
fg(x)
and
why
gf(x).
g( x)
has
an
inverse
but
f( x)
does
Section
1
Practice
questions
1
Find
(c)
g
f(x).
Define
(d)
a
28
domain
for
the
function
2
h(x)
21
The
f


(x

function
{(2,
3),
f
(3,
(a)
Show
(b)
Suggest
is
f
that
given
2),
that
so
(4,
is
h
where
f(x)
h
function
f
is
defined
by
⎧
1
1)
The
x

2,
x

2
4

x,
x

4
⎨

exists.
⎩
for
(a)
Find
(b)
Explain
x

.
ff(2).
by
4),
onto
(1,
why
f
does
not
have
an
inverse.
2)}
but
not
one-to - one.
ln x
29
a
change
to
one
of
the
ordered
Show
that
Hence
of
f
to
give
a
function
g
such
that
g
is
e

find,
in
terms
(ln 6
and
of
e,
the
value
of
both

2 ln 3

ln 13)
e
onto
x.
pairs
.
one-to - one.
1
Using
(c)
your
ordered
definition
of
g,
give
g
as
a
set
of
pairs.
30
Find
the
range
of
f
where
f
is
defined
by
2
x
______
f(x)
22
Find
the
the
line
relationship
x

ay

b
between

0
is
a
a
and
tangent
b
such
to
the

for
that
x
curve
Hence

all
x

.
1
sketch
the
curve
whose
equation
is
2
y

2x
Hence

x
find
tangent

4
the
with
coordinates
gradient
1
of
the
touches
point
the
where
y

f(x)
p
and
the
curve.
31
Draw
23
Find
the
maximum
value
x
______
of
k
for
q
are
a
propositions.
truth
table
for
~p

q.
which
2

k

(x

1)
2
x

1
32
for
all
real
values
of
Give
(a)
x.
a
counter
example
following
statement
The
of
sum
any
is
two
to
show
that
the
false.
prime
numbers
is
an
is
an
2
24
Given
that
4
x

3
x

1
is
a
factor
of
even
2
3x

3x

3x

4,
2
Prove
(b)
find
the
number
.
other
factor
and
hence
even
4
3
x

that
if
n
is
any
the

x

3x

2
same
diagram,
sketch
the

x
2
,
y


x
(x

1)
33
2

__
√y
x
__
Simplify


√y
_______
__

√y
x

√y
2
,
y

x

1,
34
for

_______
following
x
2
y
n
completely.
curves:
(a)

2
3x
__
On
n
integer
.
x
25
integer
,
factorise
Find
the
(x
2)
conditions
satisfied
by
a
and
b
such
2
y
(b)

for
ln x,
0

y
x


that
2
1

ln x,
y

1


is
a
factor
of
(x

a)(x
2

3b

2b
)
ln x,
2
4
x
y
(c)
26
The

e
roots
3
x
,
y

3e
of
the

6x
3x
,
y

e
,
for
2

x

2
35
Solve
36
The
the
inequality
|x

16|

1.
equation
function
f
is
defined
by
2
2x

5x

3

0
2
x
are
,

and
f(x)
.

and
27
the


equation
whose
1.
Factorise
roots
are


1,


2,
x

1
x

1
{
x
Find


2,
1
(a)
Sketch
(b)
Find
the
graph
f(0)
(i)
of
y

f(x).
f(2)
(ii)
completely:
2
(c)
g(x)

x

3x

10
4
(a)
81x

16
Find
3
(b)
(a

b)
the
points
of
intersection
of
the
curves
3

b
y

f(x)
and
y

g(x).
63
2
Trigonometry,
2.
1
Sine,
cosine
Learning outcomes
and
To
revise
circular
To
revise
the
line
rotates
an
from
angle
its
initial
position
OP
sine,
cosine
about
the
fixed
point
O
0
measure
to

a
and vectors
tangent functions
The definition of
When

geometry
any
other
position
OP
,
the
amount
of
rotation
is
measured
by
the
angle
and
between
OP
and
OP
.
0
tangent functions
When
the
rotation
rotation
is
is
anticlockwise,
clockwise,
i.e.
a
the
negative
angle
angle
is
the
angle
is
positive,
and
when
the
negative,
represents
a
clockwise
rotation
You need to know
The

The
of
sine,
an
cosine
angle
in
a
and
rotation
of
OP
can
be
more
than
one
revolution.
tangent
right-angled
P
P
triangle

The
exact
cosine
values
and
of
tangent
the
of
sine,
30°,
45°
45°
and
O
60°
P
O
P
0

How
of
to
use
curves
to
O
P
0
0
transformations
help
with
curve
sketching
P
Q
The
r
radian
Degrees
r
and
(sometimes
revolutions
called
are
circular
two
units
measure)
used
is
to
measure
another
unit
angle.
used
to
The
radian
measure
angles.
1 rad
O
r
P
One
radian
by
The
an
is
arc
circumference
the
angle
equal
of
a
in
circle
subtended
length
is
2 r
to
so
at
the
the
the
centre
radius
of
number
the
of
of
a
circle
circle.
radians
in
one
2 r
____
revolution
is
2

Therefore
2  rad

360°.
r
The
diagrams
show
some
other
angles
measured
in
radians.
3
π
π
2π
π
2
6
π
π
2
4
π
3
y
The
P(x,
sine,
cosine
and tangent functions
y)
When
OP
is
coordinates
defined
1
drawn
of
P
on
are
( x,
x-
and
y),
y-axes,
then
the
where
sine,
OP
cosine

1
and
as:
y
y

,
1
θ
O
64
x
x
y
x
__
__
sin 
cos 

__
and
1
tan 

for
x



unit
and
tangent
the
functions
are
Section
The
2
Trigonometry,
geometry
and
vectors
sine function

__

Measuring
in
radians,
for
0



,
OP
is
in
the
first
quadrant,
y
is
2

__
positive
and
increases
in
value
from
0
to
1

as
increases
from
0
to
2
sin 
Therefore
increases
from
0
to
1

__
For



,
OP
is
in
the
second
quadrant,
y
is
positive
and
decreases
2

__
in
value
from
1
to
0

as
increases
from
.
to
2
sin 
Therefore
decreases
from
1
to
0.
3
___

For



,
OP
is
in
the
third
quadrant,
y
is
negative
and
decreases
2
3
___
in
value
from
0
to
1
as

increases
from

to
2
sin 
Therefore
decreases
from
0
to
1.
3
___
For



2 ,
OP
is
in
the
fourth
quadrant,
y
is
negative
and
2
3
___
increases
in
value
from
1
to
0
as

increases
from
to
2 .
2
sin 
Therefore
increases
from
–1
to
0.
y
P(x,
y)
y
y
y
P(x,
y)
θ
θ
1
1
O
y
x
x
O
y
y
y
θ
θ
1
x
O
O
P(x,
This
and
OP
As
shows
sin 
moves
OP
from
in
round
rotates
1
sin 
that
varies
to
0,
is
positive
value
the
quadrants
clockwise,
and
so
for
between
sin
on.

The

0
<
1
and
<
1.

1
x
and
The
negative
pattern
for

repeats
y)
P(x,



itself
y)
2
as
again.
decreases
graph
sin
of
from
f( )

0
to
sin 
1,
then
shows
increases
these
properties:
θ
1
π
O
π
π
3
π
θ
1
This
The
graph
also
shows
that,
for
any
angle
,
sin (
)

sin 
cosine function
x
__
For
any
value
of
,
cos 

1
P(x,
y)
P(x,
y)
1
1
x
θ
θ
x
θ
θ
x
x
1
P(x,
y)
1
P(x,
y)
65
Section
2
Trigonometry,
geometry
and
vectors
As
OP
moves
decreases
increases
The
round
again
to
again
graph
of
1,
to
y
the
1.
then
So
cos 

quadrants,
cos
cos
increases
,
looks
like
to
sin ,
like
cos

decreases
0
and
varies
in
from
the
in
1
to
fourth
value
0,
and
then
quadrant
between
1
and
1.
this:
θ
1
α
3π
2π
α
α
O
π
π
2π
3π
θ
4π
1
The
curve
)
cos (
is
symmetrical
about
the
vertical
axis,
showing
that
cos 

Comparing
the
curves
for

cos
sin ,
and
we
see
that
when
we
translate

__
the
cosine
curve
by
in
the
positive
direction
of
the
horizontal
axis,
2

__
we
get
the
sine
curve,
i.e.
sin


cos

(
)
2
The tangent function
y
__
For
P(x,
any
value
,
of
tan 

x
y)
As
OP
rotates
through
the
first
quadrant,
x
decreases
from
1
to
0
and
1
y
y
__
y
increases
from
0
to
1.
Therefore
the
fraction
increases
from
0
to
very
x

__
large
O
values,
and

as
→
x
,
tan 
→

2
y
__
The
behaviour
of
in
the
other
quadrants
shows
that
in
the
second
x
quadrant
third
fourth
The
tan 
is
quadrant
quadrant
cycle
Therefore
then
the
negative
tan 
is
tan 
is
repeats
graph
of
and
increases
positive
and
negative
and
f( )
2π

tan 
looks
The
cosine
and

sin ,

sine,


1

For
f( )

is
continuous
repeats
every
cos ,


is
repeats
in
the
,

and
to
in
the
0
this:

(i.e.
1
continuous
every


2π
θ
and tangent functions
functions
show
that:
has
pattern
1
sin 
has
no
breaks)
and
a
that
2  rad

cos (
cos 
like
tangent
sin 
)
sin (
sin 
to
π
cosine
f( )
0,
0
from
π
the
to
θ
sine,
For
66
increases
Properties of the
of

from
itself.
tan
graphs
from
increases
cos 
)
(i.e.
2  rad


1
cos 
has
no
breaks)
and
has
a
pattern
that
Section
For
f( )

tan ,



the
range
)
tan (
is

is
undefined

when
has
a
pattern
that
geometry
and
vectors
tan 

...
repeats
3
___

__
,
,
2
and
Trigonometry,
unlimited

__
tan 
2
,
2
…
2
 rad.
every
Example
5
___
Find
exact
values
for:
5
___
sin
(a)
(a)
From
the
sin


From
(b)
sin

3
3
√
3
___

__
the
sketch,
5
___


2
the

tan
3

__

2

tan

3

√
3
3
f(θ)
f(θ)
1
4
0.5
2
π
π
3
3
π
π
sketch,
5
___
1
__
cos
3
f(θ)
From
(c)

__
cos
3
tan
(c)
3
sketch,
5
___
5
___
cos
(b)
3
π
3π
θ
π
π
π
θ
2π
3
2
2
2
2
1
4
π
3π
θ
π
2
2
0.5
1
π
π
3
2
2
π
π
O
π
3
2
3
π
π
O
π
2
Example
Sketch
the
maximum
graphs
value
of
of
the
the
following
functions
for
values

of
in

__
(a)
f()
sin 2 

(b)
the
range
0


2 .

In
each
case
give
the
function.
f()

3 cos 
2
f()
(c)


__
(
2 sin
)
2
2

__
(a)
Comparing
y

sin 2
with
(b)
Start
with
y

cos ,
then
(c)
Start
with
y

sin
(this
is
sin 
2
y

f(ax)
shows
y

sin 2
that
reflect
the
curve
in
the
stretched
is
compressed
by
horizontal
axis,
stretch
to
a
factor
of
2
horizontal
sin 2
parallel
axis.
goes
to
the
curve
Therefore
through
two
to
cycles
by
the
a
factor
vertical
translate
the
of
axis
3
the
every
one
cycle
sin 
of
up
the
vertical
by
translate
by
direction
of
2
of
axis)
2
parallel
then

in
the
the
positive
horizontal
axis,
units
lastly
stretch
by
a
factor
to
the
vertical
1

sin 2θ
f(θ)
1

2

3
f(θ)
cos θ
2π
4π
5π
3
3
3
θ
π
(
1
θ
π)
2
1
2
3
2 sin
2
0.5
π

2
4
O
2
axis.
f(θ)
f(θ)
f(θ)
of
axis.
parallel
f(θ)
factor
then
and
for
a
horizontal
parallel
and
curve
by
this
O
O
π
π
3
0.5
3
π
2π
3
3
π
4π
5π
3
3
2π
θ
π
θ
3
1
2
1
2
4
The
maximum
value
is
1.
The
maximum
value
is
The
5.
maximum
value
is
2.
Exercise 2.1
7
___
5
___
1
Find
exact
values
for:
(a)
cos
(
)
(b)
tan
(
2
Sketch
the
graphs
of
the
following
9
___
)
4
sin
(c)
(
)
4
functions
for
2
values

of
in
the
range
0



2

__
(a)
f()

tan 2 
(b)
f()

2

cos 
(c)
f()

sin

(

)
2
3
W
rite
down
the
maximum
and
minimum
values
of
each
function.
_
1
(a)
5 cos (2 

)
(b)
5

cos 2 

(c)
2 sin 
2
67
2.2
Reciprocal
Learning outcomes

To
define
and
use
trig functions
The
the
reciprocal trigonometric functions
reciprocal
The
reciprocals
of
the
three
main
trigonometric
(trig)
functions
have
their
trigonometric functions
own
names:
1
_____
1
_____

cosec 
You need to know
where
cosec
secant

The
the
properties
sine,
and
cosine
graphs
and
and
is
is
sec 
abbreviation
the
of
abbreviation
tan 
cosecant,
for
cot 

cos 
the
cot
1
_____

sin 
sec
is
the
abbreviation
for
cotangent.
of
tangent
The
cosecant function
functions
The
graph
of
f( )

cosec 
is
given
below.
f(θ)
1
2π
π
θ
3
f(θ)

sin
θ
1
The

graph
the
shows
cosec
multiple
that:
function
of

(this
is
not
is
to
continuous;
be
expected
it
is
undefined
because
these
when
are
the

is
any
values
of

1
__
where
sin 

0
and
is
undefined)
0

the
and
The
The
cosec
function
takes
all
real
values
except
for
similar
to
values
between
1
1.
secant function
graph
of
f( )

sec ,
shown
below,
is
the
graph
for
cosec

θ)
f(θ)

cos
θ
1
O
2π
2π
θ
3π
1
The
properties
function:
68
of
the
secant
function
are
similar
to
those
of
the
cosecant
Section
2
Trigonometry,
geometry
and
vectors

__

it
is
not

it
takes
continuous;
it
is
undefined

when
is
any
odd
multiple
of
2
The
The
all
real
values
except
for
values
between
1
and
1.
cotangent function
graph
of
f( )
cot 

is
given
below.
f(θ)
f(θ)

tan
θ
1
2π
The
properties
tangent
of
π
the
cotangent
function
θ
π
are
similar
to
those
of
the
function:

it
is
undefined

it
takes
From
the
curve
for
all
real
graph

when
is
any
multiple

of
values.
we
can
see
that
the
curve
for
cot

is
the
reflection
of
the

__
tan 
in
the
x-axis,
and
it
is
translated
by
in
the
direction
of
2
the
negative
x-axis.

__
cot 
Therefore


tan (


__
)

)
tan (
2
2
cos
Example
θ
1
Find,
in
radians,
the
smallest
value

of
for
which
sec 

2
1
2
1
_____
cos 

1
__


sec 
2
O
π
2π
3
3
the
sketch,
the
required
value
of

4π
5π
3
3
π
2
___
From
2π
θ
is
1
3
2
1
cos
θ
Example
7
___
Find
the
exact
value
of
1
cosec
6
1
2
7
___
cosec
1
______
7

6
7
__
6
sin
O
6
π
7
___
From
the
sketch,
sin

6
θ
1
1
__

2
2
7
___
cosec

2
1
6
Exercise 2.2
7
___

__
1
Find
the
exact
value
of:
(a)
cot
2
Find,
in
radians,
the
smallest
value
5
___
sec
(b)
4
(c)
cosec
4
of
x
for
3
which:
2
___
(a)
cot x

1
(b)
sec x


__
(c)
cosec x

1
√
3
69
2.3
Pythagorean
Learning outcomes
The
of

To
derive
and
Pythagorean
use
identities
relationship
any
between the
sine,
cosine
and tangent
angle
the
y
identities
P(x,
y)
You need to know
x
O


The
definitions
of
and
tangent functions
The
definitions
of
the
the
sine,
cosine
reciprocal
trig functions
y

Pythagoras’
For
any
,
angle
we
know
that
sin 

,
cos 

,
OP

How
to
solve
a
quadratic
y
sin 
_____
But
equation
cos 
__
tan 

OP
x
y
x
____
____

y
x
____
____
theorem
__

OP
OP
x
sin 
_____
i.e.
for
all
values
,
of
tan 

[1]
cos 
Using
the
reciprocal
functions,
this
identity
can
be
written
as
cos 
_____
cot 

sin 
The
Pythagorean
identities
y
x
O
P(x,
y)
2
For
any
angle
,
a
right-angled
triangle
y
2
x
____
2
Dividing
by
OP
gives
(
can
be
(

)
2
cos
2

y
2

OP
2



sin
identities

1
[2]
2

(cos
these
x
1
2
use
which
OP
i.e.
can
in
____

)
OP
We
drawn
2
to
means
(cos )
produce
,
etc.)
further
identities.
2
2
Dividing
2

cos

sin

______
2

sin

1
by
cos

gives
1

1
______

2
cos
2

sin 
_____
then
using
tan 

cos
2

gives
1

tan
2


sec

[3]
cos 
2
2
Dividing
cos
2


sin
cos

______
2


1
by
sin

gives
1
_____

1

2
sin
2

cos 
_____
then
using
cot 

sin

2
gives
cot
2


1

cosec
sin 
These
70
identities
can
also
be
used
to
solve
some
trig
equations.

[4]
Section
2
Trigonometry,
geometry
and
vectors
Example
2
Solve
the
equation
2 cos
x
sin x
2
Using
[2]
gives

1
for
2
cos
x

1
sin
0
x  2

2
x
so
2 cos
x
sin x

1
becomes
2
2(1
sin
x)
sin x

1
f(x)
2
⇒

2 sin
x
This
a
is

sin x
1
quadratic
(2 sin x

0
1
equation
1)(sin x

1
in

sin x
0
x
O
π
1
⇒
sin x 
or
sin x

1
2
1
5
___

__
⇒
x

,
,
6
These
identities
3
___
6
can
2
be
used
to
prove
the
validity
of
some
other
identities.
Example
Prove
This
that
(1
identity
cos A)(1
has yet to
be

sec A)
proved
working. Work
separately on the
LHS
cos A)(1

(1


so do
sin A tan A
not
left-hand
sec A)
use the
side
complete
(LHS)
identity
and the

1
cos A

sec A

1
cos A

sec A
in your
right-hand
side
(RHS).
cos A sec A
cos A
_____
cos A

sec A
cos A
2
2
1
cos
A
__________

sin
A
______

(1
These
cos A)(1
identities
can

sec A)
also
be
sin A tan A

sin A tan A
to

sin A
cos A

used
sin A
_____

cos A

eliminate
cos A
RHS
trig
ratios
from
a
set
of
equations.
Example
Eliminate

from
the
equations
x

3 sin 
and
y

2 sec 
x
__
x

3 sin 
sin 
⇒
2
__

and
y

2 sec 
⇒
cos 

3
2
x
__

(
)
2
__

(
3
y
y
2
2
)

1
Using
cos
2


sin


1
Exercise 2.3
2
1
Solve
2
Simplify
the
equation
sin x
_____
sec
2
x

tan
x

3
for
0
 x

2
3cos x
______

cos x
3
Eliminate
4
Prove
the

from
sin x
the
identity
equations
tan
A

cot A
x


4 tan 
and
y

2 cos 
sec A cosec A
71
2.4
Compound
Learning outcomes
angle formulae
The
identity
T
rigonometric

To
derive
sin (A
tan (A


and
B),
use
cos (A

B)
You need to know

Pythagoras’

The
are
not
cos A cos B

sin A sin B
distributive,
i.e.
cos (A
We
can
B)
is
NOT
equal
to
cos
A
cos B
and
below,
The
functions

identities for
B)

cos (A  B)
derive
which
We
find
the
two
the
the
correct
shows
length
a
of
identity
circle
PQ
of
for
radius
using
two
cos ( A
1
B)
unit
using
centre
different
the
diagram
O.
methods
and
then
equate
results.
cosine formula
y
theorem
P
properties
of
the
trigonometric functions
Q
S
A
B
N
Using
the
cosine
2
PQ
From
formula
2

1

2
the
in
OPQ
x
M
O
gives
2

1
2
cos (A
2 cos (A
B)
B)
diagram
OM

cos B
and
ON 
cos A
so
QS

(
QM

sin B
and
PN
sin A
so
PS

(sin A
Using
Pythagoras’
theorem
in
2
PQ

PQS
(
cos A


cos
A

2
2(cos A cos B

(sin A
cos
sin B)
sin B)
2

cos B)
2
cos B)
2

gives
2

cos A
2
B
2 cos A cos B


sin
2
A

sin
B
2 sin A sin B
sin A sin B)
2
Equating
the
two
2
⇒
The
angles
identity
72
is
expressions
2cos (A
cos (A
A
and
true
B)
B
for
B)

can
all
for

PQ
2
2(cos A cos B
cos A cos B
be
any
angles.
size
gives


sin A sin B)
sin A sin B
and
the
proof
is
similar
.
Therefore
the
Section
Compound
The
we
identity
can
use
Replacing
derived
it
B
angle
to
by
B
in
Trigonometry,
geometry
and
vectors
identities
above
derive
2
is
one
of
the
compound
angle
identities
and
others.
[1],
cos ( A

B)

cos A cos B
cos ( A

B)

cos A cos (B)

cos A cos B

sin A sin B


[1]
sin A sin (B)
sin A sin B
[2]

__
Replacing
A
by

A
in
[1]
gives
2

__
cos

__

(
(A

B)
)

cos
⇒
Replacing
B
by

B
in
[3],
[3]
cos B

sin
B)

sin A cos B
sin ( A

B)

sin A cos (B)

B)
gives

(
A
)
sin B
2

sin (A
by
)
sin ( A
sin A cos B


cos A sin B

[3]
sin A cos (B)
cos A sin B
[4]
sin A cos B  cos A sin B
______________________
__________
[1]
A
2

Dividing

__

(
2

cos (A

B)
cos A cos B

sin A cos B
__________
sin A sin B
cos A sin B
__________

cos A cos B
cos A cos B
______________________
tan ( A

B)

cos A cos B
__________
sin A sin B
__________

cos A cos B
cos A cos B
tan A  tan B
______________
⇒

[5]
1
Replacing
B
by
B
in
[5]

tan A tan B
gives
tan A  tan B
______________
tan ( A

B)

[6]
1
Y
ou
need
to
lear n
these
recognise
Collecting
the
identities
together
identities
either
and

be
tan A tan B
able
to
side.
gives
sin ( A

B)

sin A cos B

cos A sin B
sin ( A

B)

sin A cos B

cos A sin B
cos ( A

B)

cos A cos B

sin A sin B
cos ( A

B)

cos A cos B

sin A sin B
tan ( A

B)

tan A  tan B
_______________
1

tan A tan B
tan A  tan B
_______________
tan ( A

B)

1

tan A tan B
73
Section
2
Trigonometry,
geometry
and
vectors
Using
These
compound
identities
angle
can
be
identities
used
to
prove
further
identities.
Example
Prove
that
LHS
cos 

)



sin A cos 

2 sin A cos 
)
sin (A
cos A sin 



2 sin A
sin A cos 
cos A sin 
1
Therefore
These
sin ( A
LHS
identities

2 sin A
can
be

used
to
RHS
solve
equations.
Exam tip
The
range
is
given
in
radians
so
the
Example

__
answer
must
be
given
in
radians.
Solve
the
equation

cos


sin
(
for
)
values
of

in
the
range
3
0





__
cos 

sin

(
)
3

__
sin 


__
cos 
cos
sin
3
3
√
3
___
1
sin 

cos 
2
2
√
3
___
(1

1
)

cos 
⇒
(2
⇒
tan 
2
√
3)

cos 
2
identities
can

sin 
√

3
  1.31 rad

The

sin 

2
(3
be
s.f.)
used
to
find
exact
values
of
some
trig
ratios.
Example
Find
the
sin 15°
exact

value
sin (45°
of
sin 15°.
30°)

sin 45° cos 30°
√
3
___
1
___


√
2
2
√
3
____
1
____

√
2
2
√
2
√
√
6
2
________

4
They
can
be
used
to
simplify
expressions.
Example
Simplify
This
is
cos 
74
the
cos 
RHS
cos 2 
cos 2 
of

[2]

sin 
with A
sin 


sin 2  
sin 2 
and
B

2
cos ( 

cos (

cos 
1
__

2
√
2
cos 45° sin 30°
1
___
2 )
)
2
Section
These
identities
can
also
be
used
to
eliminate
an
angle
from
two
2
Trigonometry,
geometry
and
vectors
equations.
Example

__
 from
Eliminate
the
equations
x

sin
 
(
)
and
y

cos 
4

__
x

sin
 
(
)
4

__
sin 


__
cos

 sin
cos
4
1
___
cos 

√
√
1
___
1
___
2
2
sin 


√
y
√
2
x(
4
1
___
sin 

2
√
2
y)
sin 

2
Then,
using
(x
2

2y
gives
2
y)

2
2x
  1
cos
2
√
2
⇒
2
 
sin
2xy
y
√
2

1

1
Exercise 2.4
1
2
Find
the
exact
value
(a)
sin 75°
(b)
sin 50° cos 40°
Prove
of:

cos 50° sin 40°
that:
cot A cot B  1
______________
(a)
cot (A

B)

cot A
(b)
3
sin (60°

)


cot B
sin (120°

)
Simplify:
(a)
sin 
cos 3 

cos 
sin 3 
tan P  tan 3P
_______________
(b)
1
4
Solve

tan P tan 3P
the
equations
for
0





__
(a)
cos

(

)

sin 

sin 
3

__
(b)
cos

(

)
4
5
Eliminate

from
x


the
equations

__

4 tan
(
)
and
y

2 tan 
4
3
_
6
Given
that
sin 

,
find
the
exact
value
of:
5

__
(a)
sin

(

)
4

__
(b)
tan

(

)
4
75
2.5
Double
Learning outcomes
angle
Double
The

To
derive
and
use
ratios
of
angle
compound
identities
angle
identities
are
true
for
any
two
angles,
A
and
B,
so
identities for
they
trig
identities
multiple
can
be
used
for
two
equal
angles,
i.e.
when
B  A
angles
Replacing
B
by
A
in
the
compound
sin 2A

angle
formulae
gives
2 sin A cos A
You need to know
2

The
compound
angle
cos 2A

tan 2A

2
cos
A

sin
A
identities
2tan A
__________
2
1
The
identity
for
cos 2 A
2
cos
can
be
2
A
sin
A
sin

tan
expressed
A
in
2
A

cos
A

(1
other
forms
2
A
(1
because
2
cos
A)

2 cos
A
1
and
2
cos
2
2
2
sin
A)
2
sin
Therefore
A

2
cos
1
2 sin
A
2
A

sin
A
2
cos 2A

2cos
{
A

1
}
2
1
The
last
two
identities
above
can
be
2sin
A
rearranged
as
1
_
2
cos

A

(1

cos 2A)
(1

cos 2A)
2
1
_
2
sin
A

2
Y
ou
also
need
alter native
some
As
with
The
of
the
the
to
lear n
identities
most
the
useful
previous
following
all
double
involving
for
they
illustrate
can
some
identities
The
simplifying
identities,
examples
angle
cos 2 A.
in
trigonometric
be
of
including
identities
used
their
in
a
this
all
set
the
are
expressions.
variety
of
problems.
uses.
Example
3
Given
that

is
an
acute
angle
and
that
sin


,
find
the
value
of
:
5
(a)
cos 3 
(b)
tan 2 
3
Given

is
acute
sin 
and

(a)
cos 3 

cos (2 

)

3
4
,
then
cos 

5
and
cos 2 
cos 
2
(2 cos
sin 2 

[2(
1) cos 
16
___
2 sin
1
]

2

5
2
2 tan 
_________


tan


9
__
1
16
76
24
___
4
______

2
1
sin 
compound
cos 
4
__

25
3
tan 2 

9
___
4
__
)
25
(b)
a
2

7
44
____

5

4
Using

tan 
5

125
angle
identity
Section
2
Trigonometry,
geometry
and
vectors
Example
Solve
the
equation
cos 2 x
cos 2 x


3 sin x
3 sin x


2
for
Using
the
0

x


2
2
2
⇒
(1
2 sin
⇒
2 sin
⇒
(2 sin x
x)

3 sin x

so
2
that
terms
identity
the
in
cos 2x
equation

1
contains
2 sin
x
only
sin x
2
x
3 sin x

1
1)(sin x

1)
sin x


or
1,
so
x

0
5
___

__
1
⇒
0
,
2
,
6

6
Example
3
Prove
LHS
the

identity
sin 3x
sin 3 x  3 sin x

sin (2x


sin 2x cos x

2 sin x
4 sin
x
x)

cos 2x sin x
2
cos x cos x

(1
2 sin
x) sin x
2

2 sin x(1
sin
3
x)

sin x
2 sin
x
3

3 sin x
4 sin
x

RHS
3
Therefore
sin 3 x  3 sin x
The
Note
as
that
well
as
we
a
identity
used
a
4 sin
for
sin 3 x
compound
double
angle

the
x
is
angle
identity
in
worth
remembering.
identity
this
and
a
Pythagorean
identity
proof.
Example
Eliminate
from
equations
x

1
_____
y
cos 2 
and
y

cosec 
2

and
x

1
2

sin
sin 
2
__

x

1
2
⇒
y
(1
x)

2
2
y
Exercise 2.5
12
___
1
Given
that
cos 

and
that

is
acute,
find
the
exact
value
of:
13
(a)
sin 2 
(b)
cos

__

__
(Hint:
use
a
double
angle
identity
with
A
2
)

2
2
1  tan
A
__________
2
Prove
the
identity
cos 2 A

2
1
3
Solve
4
Eliminate
the
equation
t
from
tan
the

tan
x tan 2x  2
equations
x
A
for

0

cos 2t
x

and

y

sec 4t
77
2.6
Factor formulae
Learning outcomes
The factor formulae
The

To
derive
and
use
identities
convert
formulae
in
this
set
are
called
the
factor
formulae
because
they
the factor
expressions
such
as
sin
A

sin B
into
a
product.
identities
Starting
with
the
sin A cos B

compound
angle
formulae
cos A sin B

sin (A
cos A sin B

sin (A

B)
You need to know
sin A cos B

The
properties
of
the
B)
trig
adding
gives
2 sin A cos B

sin (A

B)

sin (A

B)
[1]
2 cos A sin B

sin (A

B)

sin (A

B)
[2]
functions

The
compound
angle
subtracting
gives
Now,
the
identities
using
other
cos A cos B
cos A cos B
adding

cos (A
sin A sin B

cos (A
2 cos A cos B
gives
right-hand
following
compound
sin A sin B
gives
subtracting
The

two

2 sin A sin B
side
of
each
of
angle

formulae
B)
B)
cos (A

these

cos (A
B)


B)
formulae
cos (A

can

cos (A
be
B)

[3]
B)
simplified
[4]
by
the
substitutions.
1
P

A

B
⎫
A
⎧

(P

Q)
2
⇒
⎬
⎨

⎩
1
Q

A
B
B

(P
Q)
2
1
_
Then
sin P
sin P


sin Q

sin Q
2 sin

1
_
(P

Q) cos
2
2
1
_
1
_
2 cos
(P

Q) sin
2

cos Q


Q)
[5]
(P

Q)
[6]
2
1
_
cos P
(P
2 cos
1
_
(P

Q) cos
2
(P
1
_
cos P

cos Q

2 sin
For
may
find
it
easier
to
remember
Q)
[7]
1
_
(P

Q) sin
2
Y
ou

2
(P

Q)
[8]
2
the
last
group
using
words.
example,
sum
The
first
of
sines
group,

twice
[1]–[4],
is
sine
used
(half
sum)
when
we

cosine
want
to
(half
express
difference)
a
product
as
a
Exam tip
sum
Identities
[4]
and
[8]
involve
a
or
difference.
For
example,
using
[1]
1
minus
sin 5x
cos 3x

(sin 4x

sin x)
2
sign,
so
take
care
when
using
these.
The
second
difference
sin 5x
As
well
as
identities,
calculus
78
group,
as

a
sin 3x
being
the
[5]–[8],
product.

problems
used
when
example,
we
using
want
to
express
a
sum
or
[5]
2 sin 4x cos x
useful
factor
is
For
for
solving
formulae
which
we
are
will
some
trig
equations
particularly
consider
in
useful
Section
and
for
3.
proving
certain
some
types
of
Section
2
Trigonometry,
geometry
and
vectors
Example
cos 2x
cos 2y
______________
Prove
that

sin 2x
cos 2x

2sin (x
cos 2y
______________
LHS
tan(y
x)
sin 2y

y) sin (x
y)
______________________


sin 2x

Using
sin 2y
2sin (x

y) cos (x
sin (x
[8]
and
[5]
y)
y)
___________

cos (x
y)
sin (y
x)
Using
__________
sin A

 sin (
A)


cos 2x
cos (y
x)
tan (y
x)
and

cos A

cos (
A)
RHS
cos 2y
______________


sin 2x

tan(y
x)
sin 2y
Example
Solve
the
equation
cos 2x

⇒
2 cos 3x
⇒
cos 3x

either
cos 4x
cos x

cos 2 x
0
or



cos 4x
Using
cos x

i.e.
in
3
of
x
times
⇒
3x
between
the
0

3
___
,
x

,
x


__
x

This
ax
The
,
,
2
6
example
cos ax
of
5
___

__
,
6

in
b.
a
T
o
is
true
2
3x
between
0
and
6,
6
3
___
,
11
____
,
6
2
6
3
___
or
2
6
a
special
values
that
of
11
____
2
includes
find
11
____
,
2
,
6
times
same
3
___
,
7
___
,
2
2

2

9
___
,
2
need values
5
___

__

__
⇒
,
2
we
,
6
7
___
x
x

__
cos x  0

7
___
5
___
,
2
2,
and
range for
⇒
or
0
0
2
values
for
[7]

__
For
0
0
0
cos 3 x  0

of
x
case
in
a
of
an
given
equation
range,
we
of
the
need
form
to
find
values
range.
for
equations
of
the
form
sin
ax  b
and
tan ax  b
Exercise 2.6

__
1
Simplify
sin
x
(


__
)

sin
x
(
3
2
Prove
3
Solve
the
the
identity
)
3
sin
equation

A

sin 3 x
sin 2A


sin 4x
sin 3A


sin 2x
sin 2A (1
for
0

x


2 cos A)

79
2.7
The
expression
Learning outcomes
The
The

To
express
a cos 

b sin 

You need to know
 b sin 
can
be
reduced
to
a
single
term
such
as
)
can
find
i.e.
of
values
angle
for
r

and
by
expanding
r cos ( 
)
using
a
identity,
r cos ( 
if
r (cos 
then
properties
b sin 

)
compound
The
a cos 
expression
sin 
b
as
We

a cos 
expression
r cos ( 
r sin/cos (
cos  
a
cos 
sin 

)
sin )

a cos 
 b sin 

a cos 
 b sin 
the
Comparing
the
coefficients
of
cos

sin 
and
gives
trigonometric functions
r cos 

The
compound

a
[1]
r sin 
and

b
[2]
angle formulae
b
__
[2]

[1]
tan 
⇒

a
2
and
[1]
2

2
[2]
⇒
r
2
2
  sin
(cos
2
)  a
2

b
_______
r
⇒
However
,
always
For
it
use
is
not
the
example,
expansion
of
r (cos 
Comparing
the

sensible
to
express
r cos ( 
cos 
2
a
method
to
2
√

4 cos
sin 

to
Using
rely
work
),
2
b

sin )
of
the
the


sin
formulae

for
1
r
and
.
Y
ou
should
result.
 3 sin 
which
coefficients
on
out
2

cos
as
r cos ( 
),
start
with
the
gives

4 cos 
cos 
 3 sin 
and sin 
gives r cos 

4
and
r sin 

3
r sin 

4
_______
3
Then
tan 

and
r
2
√

2
3

4

5
4
3
4 cos 
Therefore
 3 sin 

5 cos ( 
)
where
tan  
4
We
can
This
also
time
express
start
r (sin 
Comparing
with
cos 
the
the
cos 


4 cos
 3 sin 
expansion
sin )
coefficients
of

as
of
r sin ( 
r sin ( 
4 cos 
cos 
 )
 ),
which
gives
 3 sin 
and sin 
gives r cos 

3
and
_______
4
Then
tan  
and
r

2
√
4
2

3

5
3
4
Therefore
4 cos 
 3 sin 

5 sin ( 
 )
where
tan 

3
Exercise 2.7a
1
Express
2
Express
Finding
a cos 
find
a cos 
the
80

12 sin 
the
the

form:
(b)
24 sin 
in
the
)
and
r sin (
 )
form:
(b)
maximum

in
)
7 cos 
r cos (
(a)
T
o
5 cos 
r cos (
(a)
r sin (
)
minimum values of
b sin 
maximum
 b sin 
maximum
we
and
and
express
minimum
it
as
minimum
a
values
single
values.
sine
of
or
an
expression
cosine.
We
of
can
the
then
form
‘read’
Section
For
example,
3 sin x
to
find
2 cos x,
the
we
maximum
can
express
r (sin x cos 
Comparing
coefficients
r
of
2
√

sin

x
and
minimum
x
values
2 cos x

as
3 sin x
cos x
Trigonometry,
geometry
and
vectors
of
):
r sin (x
2 cos x
gives
cos 
r

3
and
r
sin 

2
___
2
2
2
3 sin
cos x sin )
______
Therefore
and
2
3
√

13
tan 
and

3
___
So
3 sin x
The
2 cos x
maximum
√

value
13
of
)
sin (x
the
sine
of
We
an
do
angle
not
is
1
need
and
to
evaluate
the

minimum
value
___
is
1,
therefore
the
maximum
value
of
3 sin
x
2 cos x
√
is
13
and
the
___
minimum
T
o
find
occur
,
value
the
we
value
do
√
is
13 .
of
need
x
to
at
which
the
maximum
and
minimum
values
:
evaluate
2
tan 


⇒

0.588 rad
Make
sure
your
calculator
is
in
radian
mode
3

__
)
sin (x
is
maximum
when
)
(x

and
minimum
when
2
3
___
)
(x

,
i.e.
when
x


2
The
equations
For
we
a cos 
of
the
example,
can
a cos 
to
express
 b sin 
solve
cos x
cos

r
as
x
√
3
b sin 
a
single
x



0.588 rad
√
3
sin x
sin x as


c

or
cosine
makes
solving
straightforward.
1,
for
0

x
2 ,

),
r cos (x
√
cos x 
c

sine
 b sin 
r sin x sin 
3
sin x

__
2
⇒
and
2

a cos 
form
r cos x cos  
i.e.
0.588 rad
2
equation
Expressing
3
___

__

4
tan 
and

√
3
so


3

__

cos x
√
3

sin x

1
becomes
2 cos
x
(

)
1
3

__
⇒
cos
x
(
1
__

)
3
2
f(x)
1
2
π
O
π
2π
4π
5π
3
3
π
3
x
2π
3
1
2
1

__
T
o
find
values
of
x
in
the
range
0
to
2
we
need
to
find
values
of
x 
3

__
in
the
range


__
to
2

,
3

__
cos
x
(

3

__
1
__

)
⇒
3
x

2

__


3
5
___

__
,
3
,
3
3
2
___

x

0,
,
2
3
Exercise 2.7b
1
Express
Hence
and
3 cos x
find
the
the
values

4 sin x
in
maximum
of
x
at
the
and
which
form
r cos (x
minimum
they
occur

).
values
in
the
of
3 cos
range
0

x
x


4 sin x
2
__
2
Solve
the
equation
cos
x

sin x

√
2
for
0

x

2
81
2.8
Trigonometric
Learning outcomes
The
In

To
prove

To
solve
trigonometric
important
previous
in
a
given
equations
identities
we
have
introduced
some
trigonometric
at
a
time.
equations
We
range
pages
and
identities
group
trigonometric
identities
now
collect
them
together:
sin 
_____
tan 

cos 
2
cos
2


sin


1


sec
1

cosec
You need to know
2
1

The
properties
of

the
tan

2
2

cot
trigonometric
2


ratios
sin (A

sin (A
cos (A

cos (A
B)

sin A cos B

B)

sin A cos B
cos A sin B
B) 
cos A cos B
sin A sin B
B) 
cos A cos B

cos A sin B
sin A sin B
tan A  tan B
______________
tan(A

B)

1
tan A tan B
tan A
tan B
______________
tan(A
B)

1
sin 2A

tan 2A


tan A tan B
2 sin A cos A
2 tan A
__________
2
1
tan
A
2
cos
2
A
sin
A
2
cos 2A

{
2 cos
A
}
1
2
1
A
1
2
cos
2 sin
A

(1

cos 2A)
2
1
2
sin
A

(1
cos 2A)
2
2 sin A cos B

sin (A

B)
2 cos A sin B

sin (A

B)
2 cos A cos B

cos (A

B)
2 sin A sin B

cos (A

B)


sin (A
B)
sin (A
B)
cos (A
B)
cos (A
B)
1
sin P

sin Q

2 sin
1
(P  Q) cos
2
sin P
sin Q

2 cos
(P  Q) sin
cos Q

2 cos
(P
Q)
2
1

Q)
1
2
cos P
(P
2
1
1
(P  Q) cos
2
(P
1
cos P
cos Q

2 sin
1
(P  Q) sin
2
82
Q)
2
(P
2
Q)
identities
one
Section
Proving trigonometric
T
o
prove
The
an
identity,
examples
with
a
mixed
and
specific
set
identities.

work
with

convert

when
it
is
the
in
to
the
identities.
Some
In
work
on
previous
Exercise
guidelines
more
Trigonometry,
geometry
and
vectors
identities
sensible
exercises
of
2
on
complicated
one
side
pages
2.8a
where
to
at
have
you
start
a
all
are
time.
been
asked
associated
to
prove
are:
side
Exam tip
all
ratios
to
sine
and
cosine
Remember
a
multiple
angle
is
involved,
start
with
that
and
break
it
guidelines,
to
ratios
of
a
single
angle
using
the
compound
angle
formulae
multiples
multiples

use
the
vice
The
factor
sin 3 A),
and
the
double
angle
formulae
for
these
points
practice
will
are
only
help
strategies
that
work for
you
you.
even
sin 2 A)
formulae
to
express
a
product
of
ratios
as
a
sum,
or
versa.
following
both
(e.g.
(e.g.
and
for
develop
odd
that
down
sides,
example
but
illustrates
that
it
may
be
necessary
to
work
on
separately.
Example
Prove
that
There
the
are

multiple
single
RHS:
sin 3 A
angle
A
(sin 2A
(sin 2A
angles
and
sin A)(1
on
both

sides,
2 cos A)
so
we
will
start
with
the
RHS
and
express
that
in
terms
of
ratios
of
simplify.
sin A)(1

2 cos A)

(2 sin A cos A

sin A(2 cos A

sin A(4 cos
sin A)(1
1)(1


2 cos A)
2 cos A)
2
We
now
LHS:
turn
sin 3A
to
the
LHS
and

sin 2A cos A

2 sin A cos
express

that
in
A
terms
1)
of
ratios
of
the
This form for

sin A (2 cos
2

sin A (4 cos
A
cos 2A
is
chosen
because
we
have
the factor
2
A
A
1)
sin A
sin A (2 cos
angle
cos 2A sin A
2

single
and
we
want
the
other factor
to
involve
cos A
2
A

2 cos
A
1)
2
Therefore
sin 3 A

A
(sin 2A
1)

RHS
sin A)(1

2 cos A)
Exercise 2.8a
Prove
the
following
identities.
1  cos A
_________
2
1
(cot A

cosec A)
1
2
cot 2A

cosec 2A


cosec 
______________
2
8

sec


cosec 
cos A

sin 
cot A
cot A cot B  1
______________
9
3

1

cot(A

B)

sin A
_________
cos A
_________
cot A

1
tan A

sin A

sin( 

)
10
2
___

sin(
3


sin 4 
)
sin 5 
sin 6 

3
3
1  cos 2A
__________
5
cot B
sin 4 
______
sin 3   sin 5 
______________

__
4

cos A
cot A

11
cos 3A
12
tan(A

4 cos
A

3 cos A
tan A
sin 2A
sin B
_______________
cos 2A  cos 2B
_______________
6

cos 2B

cot(A

B)cot(A


B)

tan A
cos A cos(A
cos x

cos y

B)
_
1
____________
cos 4x

cos 2A
4
7

B)
8 cos
13
2
x

8 cos
x

1

tan
(x

y)
2
sin x

sin y
83
Section
2
Trigonometry,
geometry
and
vectors
Solving trigonometric
Any
of
a cos x
the

trig
identities,
b sin x

equations
together
r sin/cos (x
with
),

the
can
be
in
a
given
range
transformation
used
to
help
solve
a
trig
equation.
There
are
whether
an
two
the
identity.
equation
The
tables
together
equation
If
to
general
a
this
form
that
with
exhaustive,
Equation
an
will
not
that
follow
nor
Equations
is
approaches
factorise,
possible
involves
list
most
appropriate
are
they
to
solving
either
(and
only
of
the
method
it
containing one
its
often
one
trig
for
given
is
try
of
them.
or
to
one
categories
solving
First
form
not),
ratio
common
see
by
using
reduce
the
angle.
of
The
trig
lists
equations,
are
not
angle only
Method
1
a cos x

b sin x

0
Divide
2
a cos x

b sin x

c
T
ransform
3
a cos
2

in
equations.
infallible.
category
x
trig
b sin x

Use
c
the
by
the
cos x,
the
provided
LHS
to
Pythagorean
LHS
in
terms
of
that
cos x
identities
ratio
0
)
r cos (x
one

to
express
only
2
a sin
x

b cos x

c
b sec x

c
2
a tan
4
x
a cos x
a sin x


Equations
Equation
1
a cos x

b tan x
b tan x


Multiply
0
multiples of one
category
2
cos ax


b cos 2x

Use
c
b cos 2x

The
methods
equation,
may
of
a
be
nor
able
trig
Practice
that
cos x

angle
double
angle
formulae
containing
trig
to
ratios
reduce
of
to
only
Solve
for
then
ax
in
a
divide
times
by
x
the
required
range
a
c
listed
do
to
do
they
be
not
give
always
simplified
Sometimes
the
lead
only
to
quickly
you
the
way
when
may
of
solving
quickest
part
need
to
of
a
particular
solution.
it
is
classify
An
equation
recognised
each
side
as
of
help
you
recognise
the
best
way
to
tackle
any
part
an
independently.
will
0
c
c
identity.
equation
84

the
equation
and
sin ax
provided
Method
an
a sin x
cos x,
0
containing

by
equation.
Section
2
Trigonometry,
geometry
and
vectors
Example
Solve
the
sin 2  cos 
equation

cos 2  sin 

tan 3 
for
values

of
in

__
the
range
0



2
sin 2  cos 
cos 2  sin 

⇒
sin 3 

⇒
sin 3 

⇒
sin 3 
⇒
sin 3 
(1
⇒
sin 3 

sin 3 

The
LHS
is
the
expansion
of
sin (2

)
tan 3 
sin 3 
______
cos 3 
sin 3  sec 3 
sec 3 )
0

sec 3 
or

0
0

1

__
For
values

of
in
the
range
0



,
we
need
to
find
values
of
3
2
3
___
in
the
range
0



,
2
sin 3 

0
3
⇒

0

or
and
sec 3 

1
⇒
3

0

__



0,
3
Example
Solve
the
cos 4 
Using
This
cos 4 
equation
cos 2 

the
factor
cos 3 
formula
cos 3 
gives

⇒
2 cos 3 
cos 
⇒
cos 3  (2 cos 
⇒
cos 3 
as

a

on
factor
cos 3 

cos 2 


1)

0
cos 



0
for
0




0
the
of
cos 3 

first
the
two
terms
gives
a
factor
cos 3 
LHS.
0
1

0
or
2
3
___

__
3


,
5
___

__
,
2
so
2


2

__
,
6
5
___
,
2
6
2
___
or


3

__
i.e.



__
,
2
5
___
2
___
,
6
,
3
6
Exercise 2.8b
Give
answers
Solve
range
the
0
that
are
following



not
exact
equations
correct
for

values
to
of
three

in
significant
the
figures.
Solve
range
the
0
following



equations
for
values
of

in
the
2
2
1
cos 2  sin 
2
sin

sin 2  cos 

1
7
2 sec
8
4 sin
2


tan 


2 cos
2


2 cos 

1
5 sin 

12 cos 

9
13
sin 3 
5
cos
6
sin 5 

sin 
2

0
2


4 cos

cos 
10
tan 2 
11
sin 3 
12
4 sin 



1
1
_________

1
4
3
2
1
_________
3


1

cot 2 
4
cos 

0
2



2 sin
sin 



2
cos 3 

0


sin 

cos 

cos 3 
sec 
85
2.9
General
Learning outcomes

To find
the
general
solution
solution
General
trig
equations
solutions of trigonometric
equations
of
The
trig
of
equation
cos x

1
has
a
finite
number
of
solutions
in
a
given
range
equations
of
values
All
of
x
possible
has
one
values
solution
of
in
cos
this
x
occur
range,
in
the
namely

range
x


x


and
cos x

1
0
You need to know
However
,
In

The
properties
of
fact
cos x
cos x


1
1
for
for
an
every
infinite
number
multiple
(both
of
values
positive
of
and
x.
negative)
of
2 
the
trigonometric functions
cos

The
main
x
trigonometric
identities
1
6π
4π
O
2π
2π
4π
6π
x
1
Therefore
the
x
when
solutions
2n

This
The
is
for
of
n
called

the
range

1
of
values
can
be
of
x
is
written
unrestricted,
as

the
general
Within
the
cos x
general
solution
solution of

range

x

cos x
,
of
the

the
c,
equation.
|c|

equation
1
cos x

c
has,
in
general,
two
solutions.
cos
x
1
c
6π
4π
O
2π
2π
4π
6π
x
1
If
one
The
2 .
solution
graph
86
f( x)
Therefore
multiples
We
of
can
of
is
we
2
write
x  , then

to
this
cos x
can

get
and
has
the
to
solution
as
the
a
other
pattern
general
solution
that
is
x
repeats
solution
of


every
the
interval
equation
by

x

2n

, where
n
is
an
of
adding
integer
.
Section
Therefore
the
general
x
where

is
a
solution
2n


solution
of

in
the
for
the
n
equation

cos
x

c
2
Trigonometry,
geometry
and
vectors
is


range

x


Example
1
__
Find
the
general
solution
of
the
equation
cos
x

2
1
__
cos x

2

__
⇒
x


in
the

range

x


3
Therefore
the
general
solution
is

__
x
2n


3
The
All
general
possible
the
solution of
values
equation
sin x
of

sin
c
x
has
sin x
occur
two

in
c,
|c|
the
range
solutions
sin

in
1
0

this
x

2
and,
in
general,
range.
x
1
c
π
O
π
π
x
π
1
If
the
The
smaller
graph
Therefore
of
2
We
is
to
can
an
of
solution
f( x)
we

can
and
write

sin x
get
the

to
this
is
x  , then
has
a
the
pattern
general
other
that
solution
solution
repeats
of
sin
is
every
x  c
by


interval
adding
of
2
multiples

solution
as
x

2n   and
x
2n  

 where
n
integer
.
Therefore
x
where


the
2n
is
general

the

solution
and
x
smallest

of
(2n
the

solution
equation
1)
in

the

for
range
sin
n
0


x

c
is

2

x
Example
√
2
___
Find
the
general
solution
of
the
equation
sin
x

2
√
2
___
sin x

2

__
⇒
x

as
the
smallest
solution
for
values
of
x
in
the
range
4
0

x

2

__
Therefore
the
general
solution
is
x

2n


__
or
4
x

(2n

1)
4
87
Section
2
Trigonometry,
geometry
and
vectors
The
solution of tan x
general

c

__
All
possible
values
of
tan
x
occur
in
the
range

__

x

and
2
equation
tan x

c
has
one
solution
tan
in
this
the
2
range.
x
10
5
3π
2π
π
π
O
2π
3π
x
5
10
The
graph
of
f( x)

tan
x
has
a
pattern
that
repeats
every
interval

__
if
x
 is

the
solution
of
tan
x

c
in
the
range
x

2
solution
x
can
be
written
n  ,

n
Therefore

,
the
,
so
general
2
as

the
general
x
solution

of
n   ,
the
n

equation
tan
x
is
the
solution
in
the

c
is


__

where
of

__

range

__


x

2
2
Example
Find
the
general
solution
of
the
equation
tan
x 
1

__
tan x


1
⇒
x



__
for
values
of
x
in
the
range

__

4
x

2
2

__
Therefore
the
general
solution
is
x

n
4
When
a
multiple
multiple
angle.
angle
Then
is
use
involved,
that
to
first
find
find
the
the
general
general
solution
solution
for
the
for
angle.
Example
1
__
Find
the
general
solution
of
the
equation
sin 2 x

2
1
__
When
sin 2x

,
the
solution
for
2x
in
2
5
___

__
2x

and
2x

6
6
Therefore
the
general
solution
for

__
2x


2n
and
6
5
___
2x


2n
6
giving
the
general
solution
x


12
88
for
x as
5
___

___
n
and
x 

12
n
2x
is
the
range
0

2x

2
the
single
is
Section
2
Trigonometry,
geometry
and
vectors
Example
Find
the
general
solution
of
the
sin x
sin x
⇒
∴

2 sin x cos 2x
sin x (1
Either

2 cos 2x)
sin x

Then
0

0

0
⇒
the
x


general
equation
2 sin x cos 2x
0,
,
solution
cos 2 x


in
the
0
interval
0

x

2
x  n
is
2
___
1
__
Or
2
or

⇒
2x


in
2
the

interval

x


3

__
2
___
Then
the
general
solution
is
2x

2n 
⇒
x

n

3
3

__
i.e.
x
n

or
x

n

3
Example
Find
the
general
solution
of
the
equation
x
__
x
__
cos
sin
2
x
__
x
__
cos


)
where
r

√
2
and
tan 

1
2
sin

2
1
2
1
___

__
x
__
cos

(
x
__
cos
⇒
r cos
2
x
__

1
x
__
sin
2

2

(
2
)

4
√
2
For
values
of
x
in
the
range
need
values
of
(

2

__
x
__

(

__
)
2


4
⇒
x

Therefore

x

,

__
)
from
4

__

2

__
to
4

__

2
4

__
or
4



__
x
__
we

or
the
4
0
general
solution
is
is
x

n
Exercise 2.9
Find
the
general
solutions
of
the
equations.
_
1
1
sin x

1
5
cos 3x

2
2
2
cos x

1
6
sin
3
tan x

1
7
tan 2x
8
sin 3x
x

sin x

0
__

√
3
__
4
√
2
cos x
1

sin x

0
89
2.
10
Coordinate
Learning outcomes
geometry
The
gradient of
a
and
straight
straight
lines
line
y
The

To
revise
the
gradient
of
gradient
of
a
straight
line
is
given
a
A(x
by
straight
finding
the
increase
in
y
divided
by
,
y
1
)
1
line
the
increase
in
x
when
moving
from
y

To find
the
angle
between
two
B(x
one
straight
point
on
the
line
to
another
point
,
y
2
the
line.
x
x
1
To find
from
a
2
lines
on

y
1
)
2
the
distance
straight
of
a
2
point
line
x
O
You need to know
y
y
1
2
_______
In
the
diagram,
the
gradient
of
the
line
is
.
x

The
Pythagorean

The
compound

The
exterior
of
This
is
also
the
value
x
1
identities
2
tan 
angle formulae
Therefore
angle
property
of
the
gradient
of
a
straight
line
is
equal
to
the
tangent
of
the
a
angle
that
the
line
makes
with
the
positive
direction
of
the
x-axis.
triangle
When
two
Therefore
lines
if
are
the
perpendicular
,
equation
of
a
the
product
of
their
is
y

mx

c,
form
y


line
L
gradients
any
line
is
1.
perpendicular
1
___
to
L
will
have
an
equation
of
the
x

k
m
The
The
angle
lines
L
between two
and
L
1
and
m
have
lines
gradients
y
m
2
L
1
respectively,
1
where
2
L
2
m

tan 
1
and
m
1
The
angle

tan 
2

2
between
the
lines

is

1
tan 
Therefore
tan( 


1
α
2
)
2
θ
tan 
tan 
1
2
1
2
_______________

x
O
1

tan 
tan 
1
Therefore
the
angle,
,
2
between
lines
with
gradients
m
and
1
m

is
given
by
tan 

1

m
m
1
2
Example
L
is
the
line
y

3x
5
and
L
1
is
m
3

the
angle
and
m
1
between

2,
so
(
2)
the
line
2x

y

1

0.
the
lines?
angle
between
the
lines
is
given


1
is
the
3
___


4

2)
obtuse

__

4
(3)(
3
___
5
__
___________
tan 
This
the
2
3
90
is
2
What
angle
⇒

5
between

4
the
lines;
the
acute
angle
is
m
2
m
1
2
__________
by
Section
The distance of
The
In
is
distance
the
of
a
diagram,
the
length
point from
point
the
of
a
the
from
distance
line
a
line
of
the
is
a
Trigonometry,
geometry
and
vectors
line
the
point
2
perpendicular
A( a, b)
from
distance.
the
line
y  mx  c
y
AN.
A(a,
b)
B((b
B
is
a
point
on
the
line
such
that
AB
is
parallel
to
the
x-axis,
so
b
is
c)/m,
θ
the
d
y-coordinate
of
B.
N
b
c
______
Therefore
from
the
equation
of
the
line
the
x-coordinate
of
B
is
m
b
c
______
This
gives
the
length
of
AB
as
b
c
ma
___________
|
a
|
|

θ
|
m
m
x
O
AN

d
AB sin 

y
m
________
tan 
But

m,
2
_______
sin 
therefore

Using

1
cot

mx

c
2


cosec

2
√
m

1
b
c
ma
___________
_______
Hence
d

|
|
2
√
m
Note
the
that
right
AB.
The
the
also
1
diagram
line,
is
point
then
we
needed
x-coordinates
that
sin ( 
But
the
the
modulus
between
Note
in
of

if

is
is
on
A
and
the
the
the
write
because
of
obtuse,
)  sin , so
A
would
the
left
a
of
the
(b
length
line.
c)/m
of
AB
is
If
for
A
were
the
the
on
length of
difference
B.
angle
result
is
ABN
valid



when
the
line
has
a
negative
gradient.
The
distance
of
the
point
( a,
b)
from
the
line
y

mx
 c
b  c  ma
___________
_______
is
given
|
by
|
2
√
m

1
Example
Find
the
distance
of
the
point
(2,
3)
from
the
line
2
Rearranging
the
equation
of
the
line
as
y

x
⇒
a

2,
b

3,
m
the

distance
and
of

4

0
3
4
c

3
Therefore
3y

3
2
2x
4
3
(2,
3)
from
2x
3y

4

0
is
given
by
___
4
4
3
√
17
13
______
3
3
___________
________
|
2
√(
|
2
)


1
13
3
Exercise 2.10
1
Find
y
2
A

is
the
3
4
A

acute
2
the
line
Find
y
the
3x
point
that
the
3
Find
2x
(2,
5)
bisects
distance
between

of
3y
and
1
B
is
the

lines
whose
equations
are
0
the
point
(
5,
the
angle
AOB,
where
the
point
(1,
from
5)
2).
O
the
is
Find
the
line
the
equation
of
origin.
whose
equation
is
x
point
point
angle
and
P( a,
(1,
a
b)
is
equidistant
from
the
line
2y

5x
1
and
from
the
1).
relationship
between
a
and
b
91
b)
2.
11
Loci
and
Learning outcomes
the
equation
To
define
the
meaning
of
loci
context
of
the
To find
a
given
a
condition
restricted
the Cartesian
equation
of
To find
on
the
possible
positions
of
a
point
P
,
then
P
to
a
particular
set
of
points.
This
set
of
points
is
called
the
of
When
locus
P
is
the Cartesian
equation
the
this
point
( x, y),
condition
is
the
relationship
called
between
x
and
y
that
the
y
of
Cartesian
a
placed
P
.
satisfies

is
xy-plane
locus

circle
in
is
the
a
Loci
When

of
equation
of
the
locus
of
P
.
circle
P(x,
For
example,
condition
on
is
the
if
A
is
the
P( x, y)
is
point
that
(1,
the
4)
and
gradient
y

y)
the
of
AP
4
______
You need to know
2,
then
gradient
of
AP
is
x


How
to find
the
the
Cartesian
equation
of
the
1
A(1,
locus
4)
of
distance
x
O
y

4
______
between
two
points
P
is
x

How
to find
the
equation
of

How
2

2x
1
a
⇒
straight

y
2
line
to form
a
perfect
square
Example
A
point
the
P(x,
point
Find
the
y)
(1,
is
2)
the
as
it
equation
same
is
of
distance
from
the
the
locus
line
of
y
from
x

3.
P
.
P(x,
The
x

distance
3
The
is
3
of
P
from
the
line
P
from
the
point
y)
x
distance
of
(1,
2)
is
_________________
2
√(x


the
x
2
1)

(y
2)
equation
of
the
locus
of
P
is
given
_________________
2
by
3
x
√(x


1)
2
⇒
(3
⇒
9
⇒
y
x)

(y
2)
2

(x

1)
2
6x
2

x
2

(y

2)
2

2
x
2x

1

of
the
y
4y

4
2
4y

4x
4

0
Exercise 2.11a
Find
the
Cartesian
following
equation
of
1
P
is
the
same
distance
from
the
point
2
P
is
the
same
distance
from
the
points
3
P
is
twice
distance
from
the
line
the
The Cartesian
A
circle
point.
92
locus
is
P( x, y)
when
P
satisfies
the
conditions.
the
locus
equation of
of
points
that
a
are
y
(0,
4)
(1,

and
2)
5
the
and
as
it
(
is
line
2,
x

6
4).
from
the
point
(2,
circle
a
constant
distance
from
a
fixed
0).
Section
If
P(x,
y)
is
any
point
on
the
circle
of
radius
r
and
centre
C( p, q),
2
then
Trigonometry,
geometry
and
vectors
y
_________________
2
CP

r
and
CP

√(x
2
p)

(y
q)
P(x,
2

(x
2
p)

(y
q)
2

r
C(p,
i.e.
the
Cartesian
equation
of
a
circle,
centre
2
radius
r
is
(x

2
p)

(y

( p,
q)
the
equation
of
2
(x
a
circle,
centre
2
2)

(y
(
1))
(2,
2
q)

2

9
⇒
1)
r
and
radius
centre
at
the
(
1,
circle
3)
whose
and
its

equation
radius
equation
where
f, g
of
and
the
c
form
are

2
2gx 
y

can
y
4x
is
(x


2y
4

2gx
be


(y
3)

2fy

rearranged
c

8
has
its

[1]
0
2fy


g


g
2
f

c
2
g)
and

[2]
[1]
as
2
Comparing

2
y
(x
x
0
2
1)
2
g
⇒
2
by
8.
2

given
2
x
constants,
x
is
√
is
2
An
3,
x
2
x
2
Conversely
q)
and
O
Therefore
y)
r
f

f
2
(y  f)
shows
2

2
c
[2]
that
2

y

2gx

2fy

c

0
is
the
equation
of
a
circle
___________
2
with
centre
( g,
f)
and
2
√g
radius

f

c
___________
2
provided
2
√g
that

f

2
Notice
xy
that
the
coefficients
of
c
is
a
real
number.
2
x
and
y
are
equal
and
that
there
is
no
term.
Example
Find
the
2
x
centre
and
radius
of
the
y

2x
4y

2
Rearranging
(x

1)
(
1,
2)
4

whose
equation
is
0
2
x

y
2

2x
4y

(y
and
2)
the

1
radius

as
4
4
4
comparing


1
0
as
gives
the
centre
as
the
point
1.
2
2

2
Alternatively
x
circle
2

2
x


0
y

2x
4y

4

0
with
2

y

2gx

2fy

c
gives
g

1,
f

2
and
c

4
__________
2
Then
using
(
g,
f)
and
√g
2

f
c,
gives
the
centre
as
the
point
__________
(
1,
2)
and
the
radius
as
√
1

4
4

1
Exercise 2.11b
1
Find
the
radius
2
Find
the
2
x
3
is
of
the
circle
whose
centre
y
6x
the
2
is
(3,
2)
and
whose

and
radius
centre
2y
6
and

of
the
circle
whose
equation
is
the
circle
whose
equation
is
0
radius
of
2

(Hint:
2y

6x
divide

the
4y

1
equation

0
by
2.)
2
4
centre
5.
2

Find
2x
equation
Explain
why
equation
of
the
a
equation
x
2

y

2x
y

6

0
cannot
be
the
circle.
93
2.
12
Equations of tangents
Learning outcomes
The
equation of
When

To find
the
equations
of
normals
to
know
To find
be
a
the
condition for
tangent
a tangent to
coordinates
of
the
a
circle
centre
and
at
a
the
circles
given
radius
of
a
point
circle,
can
use
the
fact
that
the
tangent
at
a
given
point
on
the
circle
is
circles
perpendicular

the
normals to
tangents
we
and
we
and
to
a
a
line
to
the
radius
through
the
point
of
contact.
to
circle
Example
Find
You need to know
the
point
equation
A(3,
11)
of
the
on
the

10y
tangent
at
the
circle
C(2,
2

How
to find
the
centre
x
and
5)
2

y
4x
8

0
A(3,
radius from
First
equation
of
a
rearrange
the
equation
2
The
basic facts
geometry

How
of
to find
a
about
of
Therefore
points
a
line
2)
2

(y

5)

8

4

25
the
circle
the
intersection
as
circle
(x

11)
the Cartesian
of
and
The
a
the
gradient
centre
of
AC
of

the
circle
is
the
point
C(2,
5)
6
curve
1

the
gradient
of
the
tangent
at
A
is
6

How
to find
the
distance
of
a
1
So
point from
a
the
equation
of
the
tangent
at
A
is
y
(
11)

(x
3)
6
line
⇒
x
The
The
normal to
normal
through
the
to
a
a
of
69

0
curve
curve
point
6y
is
the
line
perpendicular
to
the
tangent
to
a
curve
contact.
normal
tangent
In
the
case
through
The
of
the
a
circle,
point
of
the
condition for
There
are
two
normal
is
the
line
containing
the
radius
contact.
methods
a
line to
for
be
determining
a tangent to
whether
a
line
is
a
a
circle
tangent
to
a
circle.
The
first
distance
the
94
the
uses
the
centre
fact
of
the
that
a
circle
line
to
is
the
a
tangent
line
is
to
equal
a
circle
to
the
if
the
radius
of
circle.
The
the
method
from
second
method
equations
of
the
uses
line
the
and
fact
the
that
there
circle
are
will
be
solved
a
repeated
root
simultaneously.
when
Section
2
Trigonometry,
geometry
and
vectors
Example
(a)
Find
the
values
of
c
2
to
(b)
the
Find
the
circle
the

y
equation

which
of
2x
the
the
8

diameter
2
x

∴
the
x
of
y
the
2
y
First
For
line

c

0
is
a
tangent
0
circle
that
is
parallel
to
tangent.
2
(a)
for
2
x

2x
point
8
(

1,
0
0)
⇒
is
(x
the

2
1)
centre

of
y

the
9
circle
and
the
radius
is
3.
method
the
circle,
line
the
x
y

distance
c
of

0
the
(i.e.
line
y  x  c)
from
(
1,
to be
0)
is
a
tangent
to
the
3.
b
c
ma
___________
_______
∴
using
d

|
|,
2
√
m
m

1

where
d

0
c
(1)(
|
Second
method
Solving
x

⇒
3


⇒
1
y

c

0
and
(x
(x

0
and
to
be


1)

y
(x  c)

a
2x(1

tangent
c)

this
c
8
0’

9

9

0
equation
2

1

3
√
2
simultaneously
gives
2
2
4ac

2
2x
line
c
2
1)
2
⇒
‘b

2
2
i.e.
b
√
2
the
1,
1
c
______
|
√
1
For

1)
______

a
gives
_______________
3
3,
1
⇒
4(1

must
have
equal
roots,
2
c)
8c

64

0

0

1
2
⇒
c
2c
17
___
√
2 
72
________
⇒
c


3
√
2
2
(b)
The
i.e.
diameter
through
The
i.e.
1,
diameter
Therefore
y

x
the

goes
(
is
through
the
centre
of
the
circle,
0).
parallel
equation
to
of
the
the
tangents
diameter
so
is
its
gradient
(y
0)

is
1.
1(x
(
1)),
1
Exercise 2.12
2
1
Find
at
2
the
the
equations
points
Determine
2
x
on
of
the
whether
the
tangents
circle
the
where
line
3x
to
y

the

circle
(a)
y

4x
Explain
2
x
y

6y

11

0

8y

4y

5

0
is
a
tangent
to
the
circle
0
why
the
circles

y

4x

12y

(b)
Find
the
coordinates
(c)
Find
the
equation
Find
(x
1)
2

(y

2)

9
and
2
the
4

2

2
3
2
x
1
point
the
of
of
36
of

the
the
0
touch.
point
of
common
contact
normal
to
condition
to
the
the
the
two
circles.
circles
through
contact.
that
m
and
c
satisfy
if
the
line
2
tangent
of
circle
whose
equation
is
x
y

mx

c
is
a
2

y

6x

5

0
95
2.
13
Parametric
Learning outcomes
equations
The definition of
When

To
define

To find
a
a
the Cartesian
equation
relationship
between
x
and
y
is
difficult
to
work
with,
it
is
easier
given
in
to
express
each
of
x
and
x

y
in
terms
of
a
third
variable.
This
of
variable
curve
parameter
parameter
often
a
direct
a
is
called
a
parameter
parametric form
2
For
example,
equations
A
point
in
are
P(x,
the
equations
called
y)
is
the
on
t
,
parametric
the
curve
given
y

t
1,
equations
by
these
t
is
of
the
the
parameter
.
The
curve.
equations
if
and
only
if
the
You need to know
2
coordinates

How
to find
the
centre
of
a
circle from
P
are
(t
,
t
1).
and
By
radius
of
giving
t
any
value
we
choose,
we
get
a
pair
of
corresponding
values
of
its
x
and
y.
For
example,
when
t

2,
x

4
and
y

1.
Therefore
(4,
1)
is
a
equation
point

The
Pythagorean
and
the
trig
identities
angle
identities
By
double
on
this
giving
t
curve.
several
other
values
we
can
plot
points
and
draw
the
curve.
t
3
2
1
0
1
2
3
x
9
4
1
0
1
4
9
y
4
3
2
1
0
1
2
y
4
2
x
O
2
4
6
8
10
2
4
6
The
relationship
Cartesian
The
between
parametric
equations
and
equations
Cartesian
equation
of
a
curve
can
be
found
by
eliminating
the
parameter
.
2
In
the
case
of
the
equations
x

t
,
y

t
1,
eliminating
t
gives
2
x

(y
When
,
the
For
x
96


1)
the
trig
parametric
identities
example,
2 sin 
a
[1]
curve
and
equations
are
useful
has
y
these

involve
to
help
trigonometric
eliminate
parametric
3 cos 2 
[2]

equations:
ratios
of
an
angle
Section
2
Trigonometry,
geometry
and
vectors
2
The
cos 2 
identity
equation
of
the

1

2 sin
can
be
sin 
⇒
find
the
Cartesian
y
__

and
[2]
cos 2 
⇒

2
3
y
2
x
__
__
Therefore
to
curve.
x
__
[1]
used

1
2
(
3
)
2
2
⇒
It
2y
is
not
always
equations.
to

6
3x
easy
to
However
,
parametric
convert
the
a
Cartesian
Cartesian
equation
equation
of
a
to
circle
parametric
can
be
2
For
example,
radius
3
and
converted
equations.
the
circle
centre
whose
(2,
equation
is
(x
2
2)

(y
1)

9
has
1).
y
4
P(x,
y)
3
2
θ
C(2,
N
1)
x
O
2
6
2
From
the
diagram,
CN

3 cos 
x

2
PN

3 sin 
y

1
therefore
therefore
Hence,
of
the
x

2
circle,

3 cos 

where


and
is
y
the
3 cos 
3 sin 

1

3 sin 
are
the
parametric
equations
parameter
.
Exercise 2.13
1
Find
the
Cartesian
are
given
(a)
x
equation
of
the
following
curves
whose
equations
parametrically.
2

t

1,
y

(b)
x

t
t
1
_____


2
1
_____
,
y

2
1
(c)
2
x

Show

t
sec ,
that
t
y
the

3 tan 
curve
x
represents
Give
the
a
whose

3(1
parametric

sin )
equations
and
y

are
3 cos 
circle.
coordinates
of
the
centre
and
the
radius
of
the
circle.
97
2.
14
Conic
sections
Learning outcomes
Conic
There

To
define
the
conic
is
sections
a
set
of
curves
called
conic
sections
that
come
from
the
sections
intersection
of
a
plane
and
a
right
circular
cone.
You need to know

The
definition
of
a
right
circular
When
the
plane
is
perpendicular
to
the
axis
of
cone
the

The
meaning
of
the
axis
of
not
a
cone,
the
always
curve
is
a
considered
circle.
to
be
(The
one
of
circle
the
is
conic
sections.)
cone
Did you know?
This
group
among
the
of
curves
earliest
is
believed
whose
to
be
properties
When
were
the
plane
inclination
They
were
studied
extensively
ancient Greeks. Apollonius
wrote
conic
was
an
eight-volume
sections
translated
Khayyam
last
is
at
an
angle
less
than
the
investigated.
near
around
by
the
the
curve
is
the
an
slant
of
the
curved
surface,
ellipse
Perga
treatise
300
into Arabic
the
of
of
on
BC. This
by Omar
beginning
of
the
millennium.
When
the
surface,
plane
the
is
curve
parallel
is
a
to
the
slant
When
plane
of
the
the
is
the
at
cone
an
curved
formed
98
of
curved
parabola
called
is
double-ended
angle
greater
surface,
a
a
and
than
two -part
hyperbola
the
the
slant
curve
is
Section
Like
many
purely
many
mathematical
academic
such
activity
activities,
with
mathematical
no
conic
interest
activities,
they
sections
in
their
end
up
were
investigated
applications.
having
wide
And
as
2
Trigonometry,
geometry
and
vectors
a
like
scientific
uses.
It
was
in
planets
round
the
flight
One
of
the
of
a
in
in
a
same
Sun
most
time
ball
path
from
the
a
about
property
axis
is
to
that
shape
another
scientific
to
its
fact,
found
whose
thrown
important
In
discovered
our
that
moon
the
and
orbits
of
satellites
the
move
orbits.
Galileo
a
Kepler
ellipses.
telescopes
parabola
to
are
that
elliptical
follows
giant
comes
parallel
century
in
cricket
the
rotating
the
Earth
vertical
mirrors
This
17th
round
the
Around
to
the
focus
an
is
object
a
player
is
a
applications
light.
(A
projected
parabola.
For
at
an
angle
example,
the
parabola.
is
the
parabolic
use
of
surface
parabolic
is
made
by
axis.)
of
the
reflected
parabola
onto
one
that
point
means
(called
all
the
light
coming
focus).
Focus
The
Hubble
mirror
In
the
space
which
next
parabola
topics
and
telescope
collects
we
currently
orbiting
the
Earth
has
a
parabolic
starlight.
use
coordinate
geometry
to
find
equations
for
the
ellipse.
99
2.
15
The
parabola
Learning outcomes
The
We

To
define
a
parabola
as
a
parabola
are
already
familiar
with
the
shape
of
a
parabola
–
the
graph
of
locus
2
y

To find
and
the
the Cartesian
the
parametric

ax

bx  c
is
a
parabola.
equation
equations
One
of
of
that
parabola
focus)
We
the
any
and
can
properties
point
a
now
standard
on
a
fixed
use
of
a
parabola
parabola
straight
this
is
line
property
discovered
equidistant
(called
to
the
derive
by
from
the
a
ancient
fixed
point
Greeks
(called
is
the
directrix).
the
Cartesian
equation
of
the
parabola.
You need to know
The

The
meaning
of
a
by

How
to find
simplest
equation
is
y
obtained
parameter
taking
and
the Cartesian
the
the
fixed
fixed
line
point
as
x

as
A( a,
0)
a,
P(x,
equation
of
a
then
locus
P(x, y)
is
such
that
PA

y)
PN
N

How
to find
between

The
the
two
distance
points
condition for
tangent
to
a
a
line
to
be
a
curve
a
2
2
PA

(x
2
a)

y
(x
A(a,
2
and
PN
2

O
a)

(x

a)

(x

a)

4ax
2

y
⇒
PN
x
0)
2

(x

a)
2
2
⇒
y
2
y

whose
4ax
is
vertex
the
is
at
Cartesian
the
and
origin,
whose
equation
whose
focus
is
of
line
the
the
of
standard
symmetry
point
( a,
parabola
is
the
x-axis
0)
2
For
example,
point
(2,
the
equation
y

8x
gives
a
parabola
whose
focus
is
at
the
0).
Example
y
2
Find
the
focus
and
vertex
of
the
parabola
(y
1)

8(x
2)
and
10
hence
sketch
the
curve.
2
Comparing
(y
1)

8(x
2)
5
2
with
Y

4aX
whose
focus
is
at
X

a,
Y

0
and
vertex
is
at
focus
X

0,
gives
Y

y

when
Y

0
O
1,
Y

X
0,

y
x

2,
1,
a

when
2
X

0,
x

2
5
Therefore
The
line
When
100
X
the
of

vertex
is
symmetry
2,
x

4,
at
is
so
the
Y

the
point
0,
i.e.
focus
is
(2,
y
1).

the
10
1,
point
(4,
1).
x
6
8
10
Section
Parametric
The
equations for the
parametric
equations
of
standard
the
standard
2
Trigonometry,
geometry
and
vectors
parabola
parabola
are
2
x
Therefore
the
coordinates

of
at
any
and
point
y

on
2at
the
parabola
can
be
written
as
2
(at
,
2at).
Using
that
parametric
apply
to
coordinates
any
point
on
means
the
that
we
can
find
general
properties
parabola.
Example
Find
the
equation
of
a
chord
that
passes
through
the
focus
of
the
2
parabola
x

at
and
y

2at
from
any
point
on
the
parabola.
y
2
P(at
O
A
chord
of
a
curve
is
(a,
a
line
,
2at)
x
0)
joining
any
two
points
on
the
curve.
2
P(at
,
2at)
is
any
The
equation
point
on
the
parabola.
2
of
the
line
through
( a,
0)
and
(at
,
2at)
2at
_______
is
y

(x
a)
2
at
a
2
⇒
y(t
1)

2t(x
a)
Exam tip
Example
Find
the
value
of
a
for
which
the
line
y

2x
1
is
a
tangent
to
the
This
problem
highlights
the
2
parabola
y

4ax
importance
Solving
the
equation
of
2
(2x
For
line
and
the
parabola
simultaneously
solutions
gives
are
checking
valid
in
that
the
context
of
2
1)
the
the
of

line
4ax
to
⇒
be
a
4x
4x(1
tangent
to

the
a)
1

curve,
0
this
the
equation
must
problem.
have
2
equal
roots,
i.e.
16(1 
a)

16
⇒
1

a

1
⇒
a

2
2
(a

0
which
is
is
not
not
a
a
valid
solution
to
the
problem
because
it
gives
y

0,
parabola.)
Exercise 2.15
2
1
Find
the
focus
and
vertex
of
the
parabola
2
(a)
(y

2)
given
by:

16x
(b)
x

3t
3
The
y
2
,
y


parametric
16t.
Find
the
coordinates
of
the
points
of
equations
in
terms
of
of
a
p,
the
curve
are
x
equation

of
8t
,
the
6t
chord
2
Find,
intersection
joining
the
points
on
the
curve
where
t

2
of
and
t

p
2
the
line
y

x

6
and
the
parabola
x

4t
,
y

2t
101
2.
16
The
ellipse
Learning outcomes
The
One

To
define

To find
the
ellipse
as
a
of
properties
of
an
ellipse
discovered
by
the
ancient
Greeks
is
when
a
point
is
constrained
so
that
its
distance
from
a
fixed
point
and
and
parametric
the
locus
that
the Cartesian
ellipse
equations
of
a
fixed
straight
line
are
in
a
constant
ratio
which
is
less
than
1,
the
an
locus
is
an
ellipse.
ellipse
The
the
of
You need to know

The
meaning

How
to find
of
a
position
fixed
the
of
line.
ellipse
the
ellipse
These
are
depends
eccentricity
of
the
(Notice
that
when
We
find
the
also
on

value
and
1,
on
called
the
ellipse
e
depends
the
is
the
the
of
position
focus
the
the
constant
denoted
definition
and
by
of
the
fixed
directrix.
ratio;
this
point
The
is
and
shape
called
the
e
gives
a
parabola.)
parameter
the Cartesian
will
simplest
Cartesian
equation
for
an
ellipse.
This
is
when
a
__
equation
of
a
locus
the
point
(ae,
0)
is
the
focus
and
the
line
x

is
the
directrix.
e

How
to
solve
quadratic
y
2
a
__
2
PN
inequalities

2
x
(
and
)
2
PS

(ae
2
x)

y
e
2
Now
ePN

PS
so
e
2
2
PN

PS
P(x,
e
y)
2
a
__
2

2
x
(

(x

a
by
b
)
2
ae)
N
y
e
2
⇒
x
2
(1
e
2
)

2
y
2
(1
e
)
2
2
y
x
2
Replacing
a
2
(1
e
__
2
)
gives

2
shape
of
the
curve
can
be
x
0)
1
2
a
The
S(ae,
O
__

a
b
x
deduced
from
the
equation
as

e
follows:
2
2
2
y
x
__
2
__
2

1
⇒
x

2
2
(b
2
a
y
a
__

y
2
)
and
x

0
2
b
b
b
2
so
2
b
y

Therefore
0
b
⇒

y
(b

y)(b
b,
and

y)

0
as
_______
a
__
x


2
a
a
O
x
2
√b
y
,
b
the
curve
is
symmetrical
about
the
b
y-axis
2
2
y
x
__
__
Solving

for
1
for
y
,
gives
similar
2
a
results
2

2
b
x,
i.e.
a

x

y

b
a
and
the
curve
is
symmetrical
about
the
x-axis.
Did you know?
Also
A
property
of
the
ellipse
is
that
The
sum
of
the
distances
between
and
point
on
between
the
each focus
ellipse
is
x

0,
and
when
y

0,
curve
is
symmetrical
and
symmetrical
foci
and
use
this
to
draw
an
a
about
both
axes
and
so
we
can
see
y


it
has
a
x

e
e
ellipse.
B
The
the
line
focus
focus
A
(
ae,
0)
O
is
is
axis
called
and
its
2a
A
x
(ae
The
is
2b
directrix
line
called
axis
B
directrix
AA'
major
length
102
that
directrices.
a
can

a
constant.
x
You
x
the
two
foci
when
the
and
through
the
its
BB'
minor
length
is
Section
2
Trigonometry,
geometry
and
vectors
2
2
y
x
___
___


2
1
is
the
Cartesian
equation
of
an
ellipse
2
a
b
2
where
a
major

b
axis

of
with
b
foci
length
2

at
2a
a
2
(1
( ae,
and

0)
e
),
and
minor
(ae,
axis
of
0),
length
2b
Example
2
2
y
x
__
Find
the
eccentricity
of
the
ellipse
__


9
Comparing
and
b

the
e
2
b

a
the
standard
equation
gives
a

3
e
2
)
gives
so
e

9(1
e
)

9
3
Parametric
equations of
The
parametric
x
Therefore
4
√
5
___

(a cos ,
with
2
(1
5
2
⇒
equation
2
2
Using
given
1
4
the

coordinates
an
equations
a cos 
of
ellipse
any
and
point
y
of
on
an
ellipse
are
b sin 

the
ellipse
can
be
written
as
b sin ).
Example
3
___

__
Find
the
length
of
the
chord
joining
the
points
A

where

and
B
where


3
x

a cos 
and
y

__
The
coordinates
of
A

where


__
are
a cos
of
B

where

2

ellipse
,
i.e.
(
cos
)
,
2
2
√
a
2
____
3
___
,
),
b sin
4
i.e.
(
4
√
b
2
____
)
,
2
2
2
a
__
2
the
√
3
b
____
a
__
)
3
3
___
(a
are
4
AB
b sin
3
3
___
coordinates

__
,
(
3
The
on
4
b sin 

(
b
___
2
√
2

1)

4
(
√
2
√
2
3)
4
________________________
1
⇒
AB
2
√a

(3

2
2
√
2)

b
(5
2
√
6)
2
Exercise 2.16
1
Find
the
lengths
of
the
major
and
minor
axes
of
the
ellipse
2
2
y
x
___
__

25

Hence
2
(a)
1
9
sketch
W
rite
the
down
ellipse.
the
coordinates
of
F
and
1
x
(b)
4 cos ,
Find,
in
y

terms
F
,
the
of
,
the
lengths
of
F
P
and
F
1
(4 cos ,
(c)
Hence
two
foci
of the
ellipse
2
3 sin 
P
where
P
is
the
point
2
3 sin ).
show
that
the
sum
of
the
lengths
F
P
,
1
F
P
2
and
F
F
1
is
2
constant.
103
2.
17
Coordinates
Learning outcomes
A,
To
define
a

To
define
three-dimensional
B
AB,
and
BC
are
and
introduce
the
unit
three
points
in
A
space.
the
line
AC
(the
are
displacements.
magnitude
segment
AB,
of
i.e.
AB
2 m)
Each
is
the
and
has
a
length
2 m
a
vectors
definite
j,
C
magnitude
of
i,
and vectors
vector
coordinates
To
3-D
Vectors


in
direction
in
space.
k
___
›
C
The

To find
the

To find
a
magnitude
of
a
displacement
from
A
to
B
(written
as
AB)
vector
followed
by
the
displacement
from
B
to
C
2 m
is
B
vector
sum
equivalent
to
the
displacement
___
___
write
There
this
are
as
AB
many

to
C.
›
BC
other
A
___
›
›
We
from

AC
quantities
that
are
defined
by
magnitude
and
You need to know
direction

Pythagoras’
and
can
be
represented
by
vectors.
theorem
A
vector
is
a
quantity
specific
A
scalar
be
as
A
of
quantity
represented
the
length
vector
the
can
line
represents
is
by
of
be
a
a
one
real
piece
is
of
the
fully
string
direction
by
does
a
the
both
in
defined
Length,
for
not
straight
magnitude
of
has
direction
number
.
represented
represents
the
that
which
magnitude
by
magnitude
example,
depend
line
and
and
a
space.
the
on
is
its
segment
a
alone
scalar
can
direction.
where
direction
and
quantity,
of
the
the
line
length
segment
vector
.
B
B
B
→
a
→
BA
A
A
A
___
›
The
line
vector
and
can
the
be
arrow
denoted
shows
by
the
AB,
where
direction,
___
A
i.e.
and
B
from
are
A
the
to
B.
end
A
points
vector
in
of
the
the
›
opposite
for
direction
example,
is
denoted
by
BA.
The
vector
can
also
be
denoted
by,
a
Properties of vectors
The
the
T
wo
are
If
magnitude
line
of
a
vector
representing
vectors
are
a
is
written
as
| a|
or
a,
so
|a|
is
the
length
of
a
equal
if
their
magnitudes
are
equal
and
their
directions
equal.
two
vectors
magnitude
then
b

a
but
and
b
have
opposite
the
same
directions,
a
ta
a
If
t
is
a
positive
real
number
,
then
ta
is
b
in
the
same
magnitude
104
direction
t|a|.
as
a
but
of
a
Section
2
Trigonometry,
geometry
and
vectors
Addition of vectors
If
the
sides
represents
AB
the
and
BC
vector
represent
sum
a

the
vectors
a
and
b,
the
third
side,
AC,
b
b
C
C
B
B
b
a
a


a
a
b
D
A
D
A
b
Notice
that
diagram)
and
b
follow
each
whereas
a
a

in
b
is
other
the
round
opposite
the
sense
triangle
(clockwise
(anticlockwise
in
in
the
this
case).
The
order
above
in
show,
which
i.e.
a
a

and
b

b
b
are

added
does
not
matter
,
as
the
a
This
rule
can
be
extended
diagrams
a
to
cover
the

b

c

d
A
E
addition
of
as
many
vectors
as
we
wish
a
to
add.
In
the
the
diagram
vector
notice
the
sum
that
a,
a
b,
side

c
b
AE

and
c
d
represents

d.
B
d
Again
follow
each
b
other
round
the
pentagon
in
the
D
same
c
sense,
but
opposite
Note
a
that,
equally

b

c
although
represent
vector
called
a
d
is
usually
this
in
the
C
some
in
appears
three
to
be
two -dimensional,
it
can
dimensions.
and displacement vectors
has
no
displacement
However
,
diagram
vectors
Position vectors
A

sense.
particular
position
in
space.
Such
a
vector
is
vector
vectors
___
represent
the
specific
position
of
a
point,
for
›
example
of
the
the
point
vector
A
OA,
relative
where
to
O
is
a
fixed
origin,
represents
the
position
O.
___
›
OA
is
called
represented
the
by
any
Coordinates
T
o
locate
we
start
a
Any
by
giving
other
its
other
vector
line
of
of
A.
the
It
is
same
unique
length
and
and
cannot
be
direction.
in three dimensions
point
from
O.
position
a
in
three
fixed
point
can
distances
y
dimensions
origin,
be
the
point
located
from
O
in
y
each
of
three
directions.
mutually
Therefore
perpendicular
we
need
three
x
coordinates
to
locate
a
point
in
3-D.
P(x,
y,
z)
O
We
use
the
familiar
x-
and
x
y-axes,
z
together
Then
(x,
y,
with
any
z)
a
point
relative
third
has
to
axis
O z.
coordinates
the
origin
O.
z
105
Section
2
Trigonometry,
geometry
and
vectors
y
Cartesian
unit vectors
A
(0,
1,
unit
vector
has
a
magnitude
of
one
unit.
0)
j
(1,
i
0,
i
is
a
unit
vector
in
the
direction
of
Ox
j
is
a
unit
vector
in
the
direction
of
Oy
0)
k
x
(0,
0,
1)
k
is
a
unit
vector
Therefore
the
in
the
position
direction
vector
,
of
Oz
relative
to
O,

4j
of
any
point
P
can
be
given
in
z
terms
y
For
of
i,
j
and
example,
k
the
point
P
distant
3
units
from
O
in
the
direction
of
Ox
4
units
from
O
in
the
direction
of
Oy
5
units
from
O
in
the
direction
of
Oz
5k
4j
3i
___
P(3,
4,
5)
›
x
O
has
coordinates
(3,
4,
5)
and
OP

3i
___
›
This
can
also
be
written
as

5k
3
OP

(
4
)
5
___
›
z
When
P
is
position
y
Then
r
any
point
vector

xi

of
yj
with
coordinates
( x,
y,
z),
then
r

OP
is
the
P
.

zk
x
or
r

(
y
)
z
Displacement
P(x,
y,
vectors
are
also
given
in
the
same
way.
For
example,
z)
r
the
vector
2i
3j

2k
can
represent
the
position
vector
of
the
point
x
O
P(2,
and
we
3,
2)
but
direction
can
it
as
assume
can
OP
.
that
equally
Unless
it
is
a
represent
we
are
any
told
vector
that
displacement
a
of
the
vector
is
a
same
magnitude
position
vector
.
z
Addition
y
3i
and
subtraction of vectors
in
i,
j,
k form
k
V
ectors
2j
in
i,
subtracting
j,
k
the
form
can
be
coefficients
added
of
i,
j,
and
and
subtracted
k
by
adding
or
separately.
5j
For
example,
when
r

2i

5j
k
and
r
1

3i
2j

3)k

3k
2
3k
then
r

r
1
r


(2

3)i

5i

3j

(2

i

(5
2)j

(
1
2
r
1
2

2k
x
2i
and
r
r
1
z
y
The
The
P
is

7j
magnitude of
magnitude
the
Using
P
3)i

(5
(
2))j

(
1
3)k
2
point
of
(4,
a
3,
Pythagoras’
4k
a vector

4i

3j
in

i,
2k
j,
k form
is
the
length
of
OP
where
2).
theorem
twice
gives
4
2
OB
2

OA

OB
2

2
AB

4
2

2
x
O
3
2
2
OP
2
2

BP
2

(4
2

____________
B

OP

√
2
4
2

3
2
2
)

3
2

4
2

3
2

2
___
2

2

√
29
z
____________
2
For
106
any
vector
r

xi

yj

zk,
|r|

√x
2

y
2

z
vector
,
Section
2
Trigonometry,
geometry
and
vectors
Parallel vectors
T
wo
vectors
if
v
and
v
1
For
example
and
3i

3i
2j

k
are
parallel
when
v
2
2j
is
k
also

tv
1
is
parallel
parallel
to
to
6i

3i
4j
2j
where
t


2
2k

k
(t
(t


2)
1)
Equal vectors
V
ectors
v

a
1
only
if
a
i

b
1

a
1
j

c
1
and
2
k
and
v
1
b

b
1

a
2
and
c
2
i

b
2

j

c
2
k
are
equal
if
and
2
c
1
2
Example
Determine
whether
the
vector
2i
i

6j
k
is
parallel
to:
1
i
(a)
2j
k
(b)

3j
k
2
2i
(a)
6j
not
k
is
not
a
multiple
of
i
2j
k
,
so
these
vectors
are
parallel.
1
2i
(b)
6j
k

2(
i

3j

k)
so
these
vectors
are
parallel.
2
Example
___
›
A
is
the
point
(
1,
3,
2)
and
___
B
the
point
(3,
0,
1).
Find
| AB|.
___
›
›
|OA|
___
i

___
›
AB
is
3j
A(
AO
1,
and
| OB|

3i
k
___
›

2k
›

3,
OB
2)
When
in
drawing
three
axes,
as
they
However,
B(3,
0,
1)
this
a
diagram
dimensions,
gives
draw
the
include
reference
showing
not
complicate
always
a
do
the
points
the
diagram.
origin
as
point.
O
Remember
,
the
same
and
sense
vectors
be
in
to
the
be
added
opposite
must
sense
go
to
round
their
the
diagram
in
the
sum.
___
›
AB

(

4i
___
i

3j
3j

2k)

(3i
k)
k
____________
___
›

2
√
|AB| 
4
2

2
3

1

√
26
Exercise 2.17
1
P
is
the
point
(1,
4,
2).
___
›
Give
2
a

| OP|
3i

(a)
3a
(b)
a
in
5j

i,
j,
k
2k.
W
rite
1
3
a

2
(
and
and
find
down
the
the
length
of
OP
.
vectors:
2
)
and
b

0
A
form
1
(
)
are
the
position
vectors
of
the
points
4
B.
___
›
Show
the
vectors
a
and
b
on
a
diagram,
and
find
the
magnitude
of
BA.
107
2.
18
Unit vectors
Learning outcomes
To find
a
unit
vector
problems
Unit vector
A

and
parallel
to
unit
vector
parallel to
has
a
a
magnitude
given vector
of
one
unit.
a
___
›
given
vector
The
vector
a

2i
–
6j

____________

To
solve
problems
in
three
|a|

2
√
2
6

3k
is
represented
by
OA
___
2
2

3
√

49

7
dimensions
O
7
units
You need to know
a

How
in
to
three
add
and
subtract
vectors
dimensions
a
A

How
to find
the
magnitude
of
a
1
unit
vector
1

The
properties
of
cubes
and
right
Therefore
the
unit
vector
parallel
to
a
is
the
magnitude
of
a,
7
prisms
1
i.e.
the
unit
vector
parallel
to
a
is
(2i
–
6j

3k)
and
is
denoted
by
â
7
A
unit
vector
in
the
direction
of
v
is
denoted
by
v̂
v
____
and
is
given
by
|v|
Example
8
Find
a
unit
vector
in
the
direction
of
v

(
1
)
4
____________
8
___
1
__
|v|

√
64

1

16

√
81

9

v̂

(
9
1
)
4
Exercise 2.18a
1
Find
a
unit
vector
in
the
direction
of
the
vector
i

2j

2k
3
2
The
position
vectors
of
the
points
A
and
B
are
(
2
1
)
and
4
(
2
respectively.
)
0
___
›
Find
a
unit
Solving
T
o
solve
Mark
the
to
problem
origin
diagram.
you
length
can
Mark
add
and
in
but
what
displacement
108
parallel
to
AB.
problems
represent
that
in
a
the
vector
three
do
all
not
the
need
what
you
direction
to
to
information
you
vector
.
dimensions,
attempt
to
find.
need
to
another
it
on
When
find.
line
helps
draw
the
the
a
to
be
as
diagram
diagram
Remember
can
draw
axes
is
a
clear
these
and
draw
given,
that
any
represented
diagram.
clutter
by
lines
copy
line
the
it
so
equal
same
Section
2
Trigonometry,
geometry
and
vectors
Example
OABCDEFG
is
a
cube
of
side
4
units.
O
is
the
origin
and
the
unit
M
D
G
vectors
M
is
i,
the
j,
and
k
are
midpoint
parallel
of
the
to
edge
OA,
OE
and
OC
respectively.
DG.
E
F
Find,
in
i,
j,
k
form,
___
vectors
___
›
____
›
OA
(a)
the
›
OG
(b)
OM
(c)
4
___
›
OA
(a)

4i
____
Each
___
›
OG
(b)
edge

of
the

is
4
units
long
›
›
OA
cube
___
___
›
AB

BG

4i

4k

4j
C
B
k

____
____


4j

j
4k
____
›
›
OM
(c)
4i
›
OG

GM
O
____
____
›
____
›
A
i
___
›
›
1
GM

GD
and
GD

OA

–4i
2
____
›
1

OM

(4i

4j

4k)
–
(4i)
2

2i

4j

4k
Example
The
and
position
4i
–
3j
–
vectors
2k
of
the
points
A
and
B
are
2i

5j
–
3k
respectively.
___
›
Find
the
vector
The
vector
___
of
of
magnitude
magnitude
5
5
units
units
in
in
the
the
direction
of
AB.
B
direction
›
of
AB
is
five ___
times
the
unit
vector
in
the
›
direction
___
of
AB.
___
›
AB
A
___
›
›

OB
OA

(4i

2i
___
–
–
3j
8j
–

2k)
–
(2i

5j
–
3k)
k
___
›
O
√
|AB|

69

the
___
›
unit
vector
in
the
direction
of
AB
is
5
____
1
____
___
(2i
–
8j

k)
so
the
required
vector
is
√
___
(2i
–
8j

k)
√
69
69
Exercise 2.18b
1
The
___
position
vectors
of ___
points
›
OA
A
and
B
2
are
OABCDE
–i

2j

k
and
OB

i
–
bj

k
OA
OD
Find,
in

The
respectively.
(a)
is
a
right
triangular
prism
with
›

terms
___of
b,
the
unit
vector
in
2
units,
unit
and
OD
vectors
OC
i,

j,
2
units
and
k
and
are
AB

parallel
4
to
units.
OA,
respectively.
the
›
E
direction
of
Find
AB.
the
unit
___
›
(b)
Given
that
| AB|

2,
find
the
value
of
b
vector
___
parallel
D
›
to
DB.
2
j
C
k
B
4
O
2
A
i
109
2.
19
Scalar
product
Learning outcomes
The
angle
between two vectors
π
Y
ou

To
define
the
scalar
product
can
obtuse
two
either
the
acute
angle
or
α
α
the
angle
for
the
angle
between
two
lines,
vectors
i.e.

use
of
To find
the
angle
between
either


or

two
However
,
vectors
their
the
angle
directions
between
when
they
two
both
vectors
is
converge
defined
or
both
as
the
angle
between
diverge.
a
f
You need to know
c
θ

The Cartesian form
of
a
vector

How
product
θ
e
two
to
expand
the
θ
d
b
of
brackets
(a)
The
The
the
(b)
scalar
scalar
angle
product
product
between
a
.
(c)
b

of
a
two
and
vectors
b
ab cos 
and
is
a

where
and
b
denoted
is
the
is
defined
by
a
.
angle
b,
as
ab cos 
where

is
i.e.
between
a
and
b
Parallel vectors
When
a
and
either
b
a
are
.
b
parallel,

then
ab cos 0
or
a
.
b

a
ab cos 
a
b
a
b
π
Now
cos 0

1
for
and
for
and
cos 

1,
therefore
parallel
vectors
in
the
same
parallel
vectors
in
opposite
direction
a
directions
a
.
b
.
b


ab
ab
2
In
the
For
special
the
unit
case
when
vectors,
i
.
i

i,
j
j
.
a

and
j

b,
a
.
b

a
.
a

a
k,
k
.
k

1
Perpendicular vectors

__
When
a
and
b
are
perpendicular
,
then
a
.
b

ab cos
2

__
but
cos

0,
a
therefore
2
for
In
perpendicular
particular
,
for
the
vectors
unit
a
and
vectors
i,
j
b,
a
and
.
b
k,

0
π
2
i
110
.
j

i
.
k

j
.
k

0
b
Section
The
scalar
When
a

product of vectors
x
i

y
1
a
.
b

(x
i
y
1
involve
Now
i
the
.
i

z
a
j
.
.
j
b
i
k
x
.
example

x
j
2j
y

.
and
vectors
i

y
j

z
2

z
k,
2
k)
2
j,
i
.
involving
k
and
perpendicular

z
2
j
.
i
k
.
i,
j
that
.
j
and
are
all
k
.
k,
zero
as
they
vectors.
z
z
1
k)
.
k)
2
(x
1

j
terms
i
geometry
1,
y
1
y
y
2
of
1
(3 i

gives


2

i
2
k
x
1
For
b
2
product


(x
and
(x
involving
1
i.e.
k
brackets
terms
scalar

z
k)
Trigonometry,
in Cartesian form
1

1
these
with
therefore
j
1
Expanding
together
j
1

2
i

y
2
(2i

j

z
2
5j
k)

x
2
2k)

x
1
(3)(2)


y
2
(
y
1
2)(5)

z
2

z
1
(1)(
2
2)

6
Example
2
Find
the
value
of
a
for
which
the
vectors
1
(
2
1
(
a
)
are
2
2
a
(
)
)

4
a
2
2
1
The

and
1
perpendicular
.
(
2
)
vectors
a

are
perpendicular
when
4
a
2

0
2
Example
___
›
The
position
vectors
of
points
A
and
B
are
OA

i

2j

3k
___
›
and
OB

i
3j

2k,
respectively.
___
___
›
Find
the
angle
___
between
___
___
›

√
14
and
___
| OB|
√

14 ,
›
OA
OB

1
6
___

6

1
___
›
›
|OA|

|OB|
×
cos AOB
1
cos AOB

___
___
›
›

1
1
___
__________
so
OB.
___
›
∴
and
›
|OA|
___
›
OA

14
|OA|| OB|
⇒
AOB

1.50 rad
Exercise 2.19
1
a

4i
Find
2
a
Show
2i

j

.
3j
b
and
that


i

5k
and
the
7j
b

angle

3k
2i

2j
between
is

a
4k
and
perpendicular
b.
to
both
i

j

2k
and
3k
111
2.20
Equations
Learning outcomes
of
To
define
and
the
line
Straight
A

a
straight
lines
line
is
in three dimensions
uniquely
located
in
space
if:
vector, Cartesian
parametric
equations
of
a

it
line
is
parallel
through

it
a
passes
to
a
fixed
given
point,
through
two
vector
,
i.e.
it
has
a
known
direction,
and
passes
or
fixed
points.
You need to know
The vector

The
difference
position
vector
and
line
line
L
passes
through
the
point
A
whose
position
vector
is
a
and
is
a
parallel
displacement
a
between
The
a
equation of
to
the
vector
b
vector
y

How
to
add
and
subtract

The Cartesian form
of
a
vectors
vector
in
L
three
dimensions
b

The
scalar
product
of
two
vectors
A
a
O
x
P(x,
r
y,
z)
z
P(x,
y,
z)
is
any
point
on
the
line.
___
___
›
If
r
is
any
the
real
position
___
›
i.e.
r
and
OP

a
for

any
OA
is
different
For
is
112

›
OP
,
then
AP

tb,
where
t
can
take
b
value

a
the
of

a
,
b
is
This
although
example,
line.
to
the
3i
the
line
k
this
is
b
equation
called
the
position
values.
parallel
the
r
AP
,
and
unique
i.e.
›

where
a
P
,
___
›

r
Now
of
value.
___
Now
vector
position
is
a
vector
means
line
is
i
vector
vector
a
of
parallel
of
point
any
the
on
equation
vector
that
point
a
point
to
on
vector
the
the
the
of
line.
the
on
line
the
line
line.
line
so
equation
it
of
can
a
have
line
is
many
not
unique.
whose
and
the
gives
vector
2j

k
equation
is
the
is
r

position
i
2j
vector

of
k
a

(3i
point
on
k)
Section
2
Trigonometry,
geometry
and
vectors
Example
W
rite
(a)
down
vector
r

5i
j

Determine
(b)
the
position
equation
vector
of
two
points
on
the
line
whose
is
2k
(3i

whether

the
4j
6k)
vector
6i
8j

12k
is
parallel
to
the
line.
(c)
Show
(a)
Comparing
gives
that
a


Giving
gives
r
the
8i
shows
6i
(c)
8j

The
vector
to
3i
of
3i

4j
2i
Parametric
y,
z)
is
then
line
the
P(x,

2j
many
(like
The
is
perpendicular

one
4j
8i
point
on
on

3j
4k
with
the
the
are
(3i

4j
parallel
to
the
2k
is
2(3i
4j
6k
6k
z)


the
is
3j
yj




give
and
5k


4j
is
6k),
parallel
and
3j
3k
on
zk

is
the

of
3
2i

3k)
to
3j
the
with
if
point
xi
yj

line,
3k
is
12
the
so
zero,

3i
2j

5k
k



parametric
the
line
example,
and
the
1
vectors
of
r

a
b

8j

12k
is
the
scalar
then
the
vectors
any
This
of
vector
means
any
the
line
not
line
r
zk

x
i

x

x

of
z
y
j
a,


of
a
equations

4k
is

bj

ck
is
(8i

4j


can
i
any

4j

6k)
6k)

k
on
j
line
of

in
z
k
terms
so
it
can
equations
line
same
(ai

of
line.
line
the
the

in
the
the
although
1
z
the
parametric
found
y
6
on
point
the
be


equations
unique
x
5
is
way,
bj
of
a
unique.

i.e.
if
ck)
1
(ai


bj

ck)
1
y

y

b,
z

z
1
equations
ai
0
line.
5k
point
that
1

the
(8i
parametric
are

line
a

c
1
where
(x
y
vector
x
1,
parallel
to
)
is
a
point
on
1
the
line.
equations
x
7j

gives
of
the
to
2j

coordinates

18
1,
the

taking
line
4 ,
y
position
on


equations
any
6
1

b

line.
the

r
and
8 ,
equation)
Therefore
2i
line.
a
position
6i
2
values.
vector
line
j
called
the

a
1
are

so
therefore
1
For
r
the
line.
line,
6k)
perpendicular
3i
i,

the
are
is
different
then
are
6k)
point
to
line.
coefficients
parametric
y,

6k

point

equations
3i
is
another
and
j
4j

any
parameter
have
2k
(2i
x
Now
3k
(3i

this
equations of
xi
Equating
the

line.
4j

6k)
Therefore
These

the

2k
so
gives
5i


3j
perpendicular
.
(3i
P(x,


4k
the
12k
parallel
are
If
r
3i
also
product
j
on
that
2k
3j
5i
points
j

value

2i
5i
j
any

Comparing
(b)

5i
Therefore
two
r
vector

of
a
1
line
parallel
to

2 ,
y
where
the

3
( 1,

3,
7 ,
1)
z
is

a
1

point
4
on
the
line
and
line.
113
Section
2
Trigonometry,
geometry
and
vectors
Example
Find
and
the
parametric
B(0,
1,
equations
of
the
line
through
the
points
A(1,
1,
4)
2).
A
B
O
___
___
›
AB
___
›
is
parallel
to
the
line
and
AB
___
›

›
OB
OA
___
›
Therefore
Using
x

AB
A(1,
x
1,
a,


j
4)
as
y

a


1,
b
the

(i
point
b,


on
z
y
2,

the

the

c

z
4k)
line
and
c


i
2k
comparing
gives
x

1,
with
y
1
equations
2 ,
1
2j

1,
z
1

4
1
2
z

equations of
with
j
1
parametric
,
1
Cartesian
Starting
2k
1
Therefore
x
a
y
1
and

parametric
of
line
are
2
4
a
the
line
equations
of
a
line,
x

x

a,
1
y

y

b,
z

z
1
c

and
solving
each
equations
of
a
line,
x

y
x

y
the
Cartesian
(x
Using the
of
the
equation
on
ai
y
a
the
bj

x
1,
ck
forms
line
line
and
and
)
a
a
described
in
a
each
of
vector
)
c
is
a
point
on
the
line
1
is
equations of
three
of

z
(
b
1,
and

1
______

a
where
z
1
______

point
gives
i.e.
1
______
Any

for
1
vector
parallel
to
the
line.
line
above
them,
that
is
can
you
be
used
can
parallel
to
to
‘read’
the
describe
the
the
coordinates
of
a
line.
Example
State
whether
r

2i
3j
r

(3
)i
T
o
direction
first
r 
line
2i
r  (3
i
114
this
j

but
the
(i

of

j
the
and
4)k
lines
)j  (2
i
4k)
We
j 
to
equations
(2
line.
parallel
(

equation
(3

j
)j
2k 
is
with
(i
each
the
)i
4k

whether
of
3j 
line
2k
lines
(3
determine
the
so

the
are
can
the
4k)
line
parallel
vector
⇒
(
we
this
second
is
parallel.
parallel
‘read’
4)k
4k)
are
to
r  3i
i
therefore

j
the
need
from
needs
the
to
the
find
a
rearranging
vector
(i
j 
3j  2k  (
4k)
lines
vector
equation
are
parallel.
of
in
the
first.
4k)
i  j
4k),
Section
2
Trigonometry,
geometry
and
vectors
Example
The
point
A(3,
2,
C
2)
is
the
and
midpoint
B(2,
1,
5).
D
of
the
line
is
the
point
segment
(1,
joining
4,
1).
A
D
Find
the
Cartesian
equations
of
the
line
through
C
and
D.
C
T
o
find
the
Cartesian
equations
of
the
line
through
C
and
D,
B
we
need
a
vector
parallel
to
CD.
___
___
›
C
is
the
point
___
with
___
›
___
›
AC

vector
›
OA

AC
O
___
›
1
and
position
›
1
AB

(OB
2
OA)
2
1

{(2i

j

5k)
(3i

2j
2k)}
2
1

(
i
j

7k)
2
___
›
1

OC

(3i

2j
2k)

(
i
j

7k)
2
1

(5i

3j

3k)
2
___
____
›
___
›
CD

OD

(i
›
OC
1
1
4j

k)
(5i

3j

3k)

Using
D(1,
equations
1)
as
a
3i
11j
k)
point
on
the
line
and
3i

11j
,
z

k
as
a
vector
parallel
to
CD
gives
the
Cartesian
as
y
x
1
______
4,
(
2
2

4
z
1
______
______


3
11
1
Example
The
line
L
has
equations
x

3 ,
2
y

1


2
5
and
the
1
line
L
has
equations
x

1

4,
y

3,
4
z

1
.

2
Find
the
angle
between
L
and
L
1
L
is
parallel
to
the
2
vector
3i

j
2k
and
L
1
4i
If
is
parallel
to
the
vector
2
3j

is

the
k
angle
between
L
and
L
1
(
3i

j
2k)
(4i
3j
,

k)

|
12
3
2

(
√
9

the
scalar
j
2k|
product,
1

3i

|4i
3j

k| cos 
___________
__________
⇒
using
2
4 )(
√
16

9

1)
cos 
17
________
cos 
⇒

√
14


 2.67 rad
This
the
___
___

is
√
26
the
acute
obtuse
angle
is
angle
between
the
lines;
0.471 rad
Exercise 2.20
1
A
is
the
point
parametric
(1,
2,
2)
equations
and
for
B
the
is
the
line
point
2
2
The
line
l
has
equation
r
1

1
(
5
r

(
1
that
l
and
1
0,
5).
and
Find
vector
and
B.
)

2
(
)
and
the
line
l
has
2
1
3
)


(
a
)
8
1
Given
A
4

5
equation
(2,
through
l
are
perpendicular
,
find
the
value
of
a.
2
115
2.21
Learning
Pairs
of
outcomes
lines
Pairs of
T
wo

To
in
determine
three
whether
dimensions
intersecting
or
two
are
lines
lines
in
in
space
space
may
be
parallel
or
not
parallel,
in
which
case
they
may
lines
intersect
or
they
may
A
non-parallel
not.
parallel,
skew
pair
of
lines
that
do
not
intersect
are
called
skew
You need to know

The
different forms for
the
O
equations
of
a
line
in
three
dimensions

The
condition for
vectors
to
be
Parallel
parallel
lines
Intersecting
lines
Skew
lines
separated
in
space

How
to
solve
simultaneous
a
pair
of
equations
Parallel
It
is
easy
vectors
For
r
lines
to
tell
that
are
example,

2i
k
parallel
i.e.
2i
j
parallel
the

are
whether
k
j
and
Non-parallel
lines
each
are
line
parallel
from
because
their
you
can
‘read’
the
equations.
lines
(2i
because

two
to

the
k)
and
vectors
4i
2j

r

j

parallel
2k
(
2k
to
(4i

the
2(2i
j
2j

2k)
lines,

k)),
are
parallel.
lines
r

2
a

μb
2
2
P
r

a
1
r

λb
1
1
1
r

r
1
2
O
r
2
T
wo
lines
whose
vector
equations
are
r

a
1
intersect
if
values
of

and

can
be

b
1
found
for
and
1
which
r

2
r

1
If
no
The
such
values
parametric
determining
116
can
be
found
equations
whether
two
of
then
lines
lines
the
are
lines
easiest
intersect
or
are
to
are
a

2
r
2
skew.
work
skew
with
when
b
2
Section
2
Trigonometry,
geometry
and
vectors
Example
Show
x

of
If
that
2
their
the
the
2,

y
point
lines
lines

of
4
x

1
,

z
,


6
y

Solving
When

When


then
lines
values
equating
of
equations
3,

Therefore
the
the
these
x
2,

3
and


find
and
the
coordinates
intersection.
intersect
equating
z
intersect
the
1
and
,
1
3

x
these

y
2,
values
intersect
at
gives

gives
2,

y

y


of
the
2
1

3
and
2
z
and
and
point

(
values

z
give
2,

2
[1]


4


[2]


2
(first
0
line)
(second
the
2,
gives
2
0

x

and

of

same
line)
point
on
each
line,
so
0).
Example
Determine
whether
the
lines
r

i

k

(i

3j
k)
and
1
r

2i

3j

k
(4i

j
5k)
are
parallel,
intersect
or
are
skew.
2
The
are
lines
not
W
riting
x
are
not
parallel
the

1
equations
,

in
y
1
i

parametric

2
4,

3 ,

z

,
3
z
2
Equating
x
and
x
1
Equating
and
4i
j
4k
y
and
[1]
1


1
5
2
gives
1
gives
3



2
4

[1]
y


3
[2]
2
and
these

2
1
[2]
values

gives
z

0

and
1
values
are

and
z
1
These
k
1
y
2
Solving
3j
form:
1
x
With
because
parallel.


0
1
2
not
equal,
therefore
the
lines
do
not
intersect
and
are
skew.
Exercise 2.21
Show
that
the
lines
r

(
1
)


2
2
T
wo
r
lines
which
intersect

2i

9j

13k

ai

7j

2k

(i
5
2
2
1
and
)
r
2j

(
1
0
2
have

3
(
1
)


0
(
)
are
skew.
2
equations

3k)
and
1
r

(i

2j

3k)
2
Find:
(a)
the
value
(b)
the
position
of
(c)
the
angle
a
vector
between
of
the
the
point
of
intersection
lines.
117
2.22
Planes
Learning outcomes
Defining
There

To
determine
Cartesian
the
vector
equation
of
a
are
several
one
one
(b)
You need to know
and
add
and
subtract
one
only
only
three
to
define
and
only
plane
three
one
therefore
a
unique
plane,
for
example:
plane
two
one
can
be
given
can
given
plane
drawn
points
be
through
specify
drawn
to
intersecting
can
be
drawn
a
three
contain
lines
non-collinear
unique
plane
two
specify
a
perpendicular
intersecting
unique
to
a
plane
given
vectors
direction
in
one
therefore
and
lines,
(c)
to
ways
plane
points,
How
plane
and
(a)

a
at
a
given
distance
from
the
origin,
therefore
the
normal
to
dimensions
a

The
scalar
product

The
equations
of
two
plane
and
the
distance
of
the
plane
from
be
drawn
the
origin
specify
a
unique
vectors
plane
of
a
line
in
three
one
(d)
and
only
one
plane
can
through
a
given
point
and
dimensions
perpendicular

The
meaning
of
a
unit
a
vector
normal
to
to
the
a
given
plane
direction,
specify
a
therefore
unique
a
point
on
the
plane
and
plane.
normal
(A
nor mal
therefore
to
a
The vector
We
of
use
a
plane
is
perpendicular
the
any
to
line
any
equation of
definition
of
a
perpendicular
line
a
plane
in
the
to
the
plane.
A
normal
plane
given
in
(d)
to
derive
the
vector
plane.
N
A
P
r
a
n
r
a
O
___
›
A
is
a
The
point
vector
on
n
is
the
plane
and
perpendicular
OA
to

a
the
plane.
___
›
P
is
AP
any
is
a
point
line
on
in
perpendicular
the
the
to
plane
plane,
and
and
OP
is

r
therefore
n
___
›
AP

r
a,
therefore
the
scalar
i.e.
This
118
is
called
the
vector
is
plane.)
product
(r

a)
equation
of
a
.
of
n
r

plane.
a
0
and
n
is
zero,
equation
Section
The
vector
equation
of
a
plane
can
be
written
in
another
2
Trigonometry,
geometry
and
vectors
form.
N
P
a
A
r
n
a
r
θ
O
In
If
the
diagram,
ON
But
r
.
ON

d,
then
in
n̂

OP cos 
Therefore
r
perpendicular
This
Therefore
an
is
the
distance
triangle
.

n̂
to
OPN,
of
d
the
plane
from
the
origin.
OP cos 

d

d
the
is
the
unit
vector
vector
equation
can
equation
of
be
n̂
equation
and
multiplied
the
for m
r
.
of
distant
n
by

a
d
any
D
plane
from
that
the
is
origin.
scalar.
represents
a
plane
D
__
perpendicular
to
n
and
distant
from
the
origin.
n̂
Example
The
vector
equation
Find
the
distance
2i
j
2k


Dividing
form
r
.
is
a
both
n̂

of
of
the
vector
sides
a
plane
plane
is
r
.
from
(2i
the
perpendicular
by
|2 i

j



2k)

12.
origin.
to
2k|
j
the
plane.
converts
the
equation
to
the
d
1
|2i

j

2k|

3,
so
the
equation
becomes
r
.
(2i

j

2k)

4
3
Therefore
the
plane
is
4
units
from
the
origin.
Example
(a)
Find
the
equation
of
the
plane
that
is
4
vector
(
4
and
)
contains
the
point
5
(
1
(b)
Find
(a)
Using
r
.
(
the
form
4
2
4
5
)
1
(
of
the
(r
a)
.
plane
n

.
(
4
the
)
from
0
⇒
the
r
.
origin.
n
a
.
n

0
gives
4
(
)
3
4
)

0
1
4
r
to
3
distance
the
perpendicular
2
4
)
25

0,
i.e.
r
1
.
(
4
)

25
1
25
________________
25
____
_______________
(b)
The
distance
of
the
plane
from
the
origin


2
√4
2

4
2

(
1)
___
√
33
119
Section
2
Trigonometry,
geometry
and
vectors
Example
3
Show
that
the
line
whose
equation
is
r

(
1
2
)


2
(
is
)
parallel
to
5
2
8
the
plane
r
.
(
2

)
5
4
If
the
line
normal
is
to
parallel
the
to
the
plane,
then
it
is
perpendicular
to
the
plane.
8
2
The
line
is
parallel
to
2
(
and
)
the
plane
is
perpendicular
to
(
2
(
(
)
2
5

)
The
point
unit
vector

0,
therefore
the
line
is
parallel
to
the
plane.
4
The Cartesian
n̂
4
8
2
Now
)
2
5
li
P( x,
y,
equation of
z)
is
any
point
on
a
plane
the
plane
that
is
perpendicular
to
the
where

mj

nk
N
P(x,
y,
z)
a
r
n

li

mj

nk
O
Using
r
.
n̂

(xi
⇒
lx
d


gives
yj

my
zk)

(li
nz


mj

nk)

d
d
Therefore
lx

my
where
and
li
Multiplying
the

d
mj
this
Cartesian

is

the
nk
ai
the

bj
plane
d
is
the
Cartesian
distance
is
equation
ax
Then

a
unit
by
a
of
the
vector
constant
equation
plane
from
of
the
a
plane
origin
perpendicular
to
gives
general
the
more
the
plane.
form
of
equation,
i.e.
of
nz

ck
from
is

a
the
by

vector
origin
is
cz

D
perpendicular
given
to
the
plane
and
the
distance
by
D
_____________
____________
√
2
a
For
example,
3i
from
2j

the
6k
the
is
origin
2

equation
b
2

3x
2y
perpendicular
is
given
c
to

the
6z
21
___
____________

3
120
21
and
by
21
_____________
√

plane
2
2

2
2

6

7
3
represents
the
a
distance
plane
of
where
the
plane
Section
2
Trigonometry,
geometry
and
vectors
Example
(a)
Find
the
A(1,
(b)
Hence
(a)
The
(1,
Cartesian
5,
2),
B(1,
write
down
Cartesian
5,
2)
equation
1,
1)
a
equation
satisfies
the
1,
1)
satisfies
therefore
These
This
also
three
is
all
the
[1]
using

b
b
are
the
gives
gives
6b
[4]
gives
b
c


ax
the

contains
by
the
points
D
of

to
cz
the

plane.
D
plane,
[1]
the
plane,
D
4c
any
is

[2]

enough
D
to
[3]
find
multiple
a,
of
b
and
ax

c
in
terms
by

cz

of
D
D.
is
plane.
4b
[3]
of
2c
that
4).
perpendicular
plane

because
of
is
equation
5a
need
a
plane
1,
that
5b
the
equations
we
equation
[2]
5[2]
C
of

a
similarly,
the
equation
a
also
of
C(5,
vector
therefore
(1,
and
3c

c


0
[4]
4D
[5]
6D
___
3[5]


11
3D
___
8D
___
Then
[4]
gives
c

and
[2]
gives
a


11
Therefore
the
equation
of
11
the
plane
is
ai
3D
_____
6D
___
x
y
11

bj

ck
8D
___


z
11

D
11
A
⇒
3x

6y

8z

11
B
C
(b)
ai

bj

ck
Therefore
is
3i
perpendicular

6j

8k
is
to
the
plane
ax
perpendicular

to
by
the

cz

D
plane.
Exercise 2.22
1
Find,
the
in
point
Hence
2
3
the
the
(1,
1)
(a)
5,
and
that

.
3
(3i
Show
in
the

that
Hence
is
is
j

the
find

and
the
D,
is
the
equation
line
L
2k)

point
equation
perpendicular
distance
of
with
parallel
the
plane
n
perpendicular
the
2
.
1)
Cartesian
z

r
down
Show
r
(b)
(2,
write
Find
0
form
to
of
the
the
to
the
plane
P
x

plane
vector
from
that
vector
equations
the
to
the
plane
plane
the
of
i
the
that
3i
2
2j

whose

,

j

k.
origin.
contains

contains
the
point
k.
y
vector

1

,
equation
is
6
(0,
8,
equation
1)
of
is
the
contained
line
in
parallel
the
to
L
plane.
that
is
contained
P
.
121
Section
1
Give
exact
2
Practice
values
questions
13
for:
The
equation
2
7
___
4
___
(
cos
(a)
x
)
sin
(c)
(
y
7
___
(
(
cosec
(d)
the

a
circle
6y

3
is

0
equation
gradient
is
of
the
diameter
of
this
circle
1.
)
3
14
Give
the
whose
5
___
)
4
2
2x
4
Find
tan

)
3
(b)
of
2

minimum
value
of
3
The
focus
of
a
parabola
directrix
is
Find
equation
the
the
line
y
of
is

the
point
(2,
4)
and
the
8
the
parabola.
__________

__
2sin( 

)
3
15
The
(a)
equations
of
two
circles
are
5
__
3
Given
sin A

and
that
A
is
2
acute,
(x
13
find
the
value

the
2


cot

x

y

4x

find
the
2

c

0
value
of
c
such
Find
the
coordinates
of
that
the
the
circles
point
of
touch.
the
contact
circles.
Find
the
solution
of
the
equation
the
of
the
of
contact.
point
common
tangent
equation
2



cot

4
16
values

of

between
Prove
(a)
.
and
that
tangent
to
the
line
the
circle
2
(x
Prove
6y
the
through
5
9

(c)
for

4 cot 2 
_______
2

sin 2 
tan
1)
identity
tan
Hence

Find
of
(b)
(y
2
cot A
(b)
(b)
Prove
(a)

2
and
4
1)
of:
sin 2A
(a)
2

3x

2y

12

0
is
a
2
1)

(y

2)

13
that
Find
(b)
cos A cos 5A

sin A sin 5A

the
coordinates
of
the
point
of
contact
cos 6A
of
the
line
and
the
circle.
__
6
Express
cos
x

√
3
sin x
in
the
form
r cos (x

).
17
Find
the
equations
of
2
to
Hence
find
the
maximum
and
minimum
the
circle
the
tangents
from
the
origin
2
x

y

8x

6y

16

0
values
__
of
cos x
√

3
sin x
and
the
values
of
x
at
which
18
they
occur
in
the
range
0

x

Show
(a)
that
the
line
y

2x

2
is
a
tangent
to
2
the
parabola
whose
parametric
equations
are
2
x
7
Show


5 cos )
Find
(b)
2
(2 sin 

,
y

8t
all
values
of
the
coordinates
of
the
point
of
the
line
and
parabola.
.
2
19
8
Prove
the
9
2
2x cot
Solve
the
The
equation
2
x

2x tan
x

(a)
Sketch
(b)
Find
the
sin 2 


sin 

(a)
Find
the
0
y

x
of

between
0
and
cos 

4x
2

9y
general
cos 3 

solution
of
the
of
the
foci.
points
and
of
intersection
the
of
the
y

2y

25

the
circle
The
of
the
locus
of
the
point
P( x,
at
equations
each
of
of
the
these
tangents
points.
line
l
has
vector
equation
2
y)
r
distance
of
P
from
the
line
y

4
is

twice
(
2
)


(
4
the
distance
between
P
and
the
point
(2,
0
)
1
1).
and
the
line
1
has
vector
equation
2
3
2
12
Find
the
Cartesian
parametric
x

122
2

equation
equations
cos 2 
and
y
of
the
are

5
curve
whose
r

(
1
)


(
1

line
0
cos 2 
equation
the
the
circle
1
when

ellipse.
coordinates
5
1
Find
36
2

Find
equation
21
11

2 .
(b)
the
the

2
Find
is
equation
x
10
ellipse
4cos 2x
20
values
an
2
sin
sin 3 
for
of
identity
2
sin
contact
29
of
for
4t
that
sin 
where
1
)
2

and

are
scalar
parameters.
to
the
0
Section
Show
(a)
that
position
the
lines
vector
of
intersect
their
and
point
of
give
the
28
Show
that
the
2
the
acute
angle
between
whose
Practice
vector
questions
equation
is
intersection.
r
Find
(b)
line
2
the
two

(
lines.
1
1
)


(
1
5
4
whose
equation
is
)
parallel
to
the
plane
1
22
The
(a)
position
r

2i

3j
r

5i

k,
r

i
vectors

of
three
points
are
j

the
Find
(a)
the
equation
of
the
plane
these
find

)
8
the
equations
three
the
point
(2,
1,
of
5)
the
that
is
which
to
the
plane
whose
equation
is
points.
3x
Hence
5
parametric
through
perpendicular
(b)
(
k
vector
contains
.
1
line
Find
r
2k,
29

is
distance
of
the
plane
from

3y

z

9
the
Find
(b)
the
coordinates
of
the
point
of
origin.
intersection
23
The
equations
of
two
lines

5
,

y

2
3 ,

z

1



4
2,

y

3
6,

Show
z

1
that
2)
Show
(b)
Show
that
the
lines
are
the
vector
2i

j

k
is
these
as
if
P( x,
it
Find
P(a
cos 
is
y)
is
from
the
twice
the
centre
,
sin 
b
1
perpendicular
two
to
the
plane.
as
far
origin
from
then
P
the
lies
point
on
and
radius
of
the
a
circle.
parallel.
31
that
and
2

circle.
(a)
line
and
(4,
x
this
are:
30
x
of
)
and
cos 
Q(a
1
points
on
the
,
sin 
b
2
)
are
2
ellipse
lines.
2
2
y
x
__
___


2
The
Cartesian
equation
of
a
line
l
1
2
a
24
b
is
1
y
x  1
______

2
5
the
Find
the
(b)
Deduce
equation
of
the
chord
PQ.

2
and
(a)
z  3
______
______

2
Cartesian
equation
of
a
line
l
y
x  1
______

3
the
z  1
______
______

that
the
equation
of
the
tangent
to
is
2
ay
ellipse
sin 
at

the
point
cos 
bx

cos ,
(a
b
sin )
is
ab

2
3
a
2
Given
that
l
and
l
1
value
25
The
3x
of
Cartesian

y
are
perpendicular
,
find
the
32
2
P(5t
,
10t)

2z
equation

of
a
plane
is
(a)
Find
(b)
M
7
any
point
Find
the
distance
of
this
plane
from
the
on
Cartesian
is
the
midpoint
Find
the
Cartesian
as
(a)
is
a
curve.
a.
t
equation
of
the
of
line
equation
the
curve.
OP
.
of
the
locus
of
M
varies.
the
origin.
33
Show
(b)
that
the
line
whose
vector
(2i
2j
equation
(a)
Prove
that
is
2
cos
r

lies
i

in

the
Hence
from
2j



(b)
down
the
The
position
distance
of
the
Find
the

2j
relative

to
k)
a
the
hence
vectors
and
fixed
acute
find
general
line
cos
of
points
A
and
B
x

1
solution
of
the
equation
2
x

5
sin
x

2
Show
that
the
(3i

origin
6j

2k)
angle
respectively
r

O.
the
area
of
4
OA
triangle
and
whose
vector
equation
is
(
1
)

t
8
between
line
are
is
Find
sin
origin.
0
(2i

2
34
26
2
x
3k)
plane.
write
the
4k
OB
(
0
)
2
contained
in
the
plane
whose
and
2
OAB.
equation
is
r
.
0
(
)

8
1
27
Show
r
.
(i
and
that

r
.
Hence
5j
the

(2i
find
planes
2k)

the

10j
whose
equations
are
10

4k)
distance

8
are
between
parallel.
the
two
planes.
123
3
Calculus
3.
1
Functions
Learning outcomes
1
–
continuity
To
investigate
the
meaning
of
the
and
discontinuity
using
that
functions .
we
For
have
a
drawn
function
so
to
far
be
in
this
unit
continuous,
have
there
involved
must
be
no
of
breaks
functions
graphs
of
continuous
continuity
discontinuity
Continuous functions
Most

and
in
its
graph
and
no
points
at
which
it
is
undefined.
graphs
2
For
example,
graph
x
in
of
its
y

f( x)
f(x)

x
has
for
no
x


breaks
is
a
and
continuous
f( x)
has
a
function
real
value
because
for
every
the
value
of
domain.
You need to know
y

The
meaning
of
a function

The
meaning
of
the
domain
10
of
a
function

The
shapes
5
of
graphs
of
simple
functions
O
3
2
x
1
1
2
3
Discontinuity
The
graph
below
illustrates
the
graph
of
a
function
over
the
domain
x

.
y
10
5
discontinuity
x
O
4
4
6
8
5
10
There
point
is
a
clear
missing
break
from
in
the
the
graph
graph
where
where
x

x

2.
2.
There
These
is
also
breaks
a
are
called
y
discontinuities .
The
function
Some
is
functions
continuous
are
for
all
continuous
other
even
values
though
of
the
x
graphs
have
breaks
in
them.
O
x
1
__
For
example
the
function
f( x)

,
x

0,
x


is
continuous
because
x
1
__
although
the
graph
of
y

has
x
is
124
not
in
the
domain
of
f( x).
a
break
in
it
where
x

0,
this
value
of
x
Section
3
Calculus
1
Example
(a)
Sketch
the
x
f(x)
graph

State,
its
the
x

4
4,
x

4
{

,
x
(b)
of
4,
with
a
reason,
function
x

given
by

whether
the
function
is
continuous
over
all
of
domain.
f(x)
(a)
10
5
4
x
O
2
2
4
6
8
5
(b)
There
is
so
curve
the
a
break
is
where
not
x

4
and
continuous
x
over

4
all
is
the
included
in
the
domain,
domain.
Example
(a)
Sketch
the
graph
of
the
function
given
by
2
x
f(x)
,
State

0
x

0
{

,
x,
(b)
x
whether
x
the


function
is
continuous.
f(x)
(a)
5
x
O
(b)
There
are
in
nature
the
missing
is
no
5
breaks
of
the
because
in
this
graph
f( x)
is
graph
where
defined
and
x

when
although
0,
x
there

0.
is
there
not
is
a
a
change
point
Therefore
the
function
continuous.
Exercise 3.1
Sketch
each
of
the
following
functions
and
state
whether
the
function
is
continuous.
x
1
f(x)

x,
x

{2,
4,
4
6}
f(x)


x
x
1
______
2
f(x)

,
x

x
f(x)

1,


x

0
x

0
,

1,
5
f(x)


x


9,
x

4
x

4
1,
x

4
x

5
{
x
{
x
x
1
x,
3

1,
{

9,
,
x


,
x



125
3.2
Limit
notation
Learning outcomes
Limits
1
__
In

To
introduce
basic
concepts
T
opic
1.10
we
looked
at
the
behaviour
of
as
the
value
of
x
gets
large.
of
x
1
__
limits
This
table
of
values
shows
that
approaches
zero
as
x
approaches
x

To
of
investigate
f(x)
above
as
x
and
the
behaviour
approaches
infinity.
a from
below
x
5
10
100
1000
10 000...
0.2
0.1
0.01
0.001
0.000 1...
1
__
x
You need to know
1
__
We

The
meaning

The
shape
of
write
this
as
→
0
as
x
→
,
and
we
say
that
the
limiting
1
__
1
__
of
the
curve
y
value
x
continuity
of
1
__
as
x
→

is
0.
The
notation
for
this
statement
is
lim
x→
x


0
x
x
1
__

How
to
sketch
simple functions
If
we
now
look
at
as
x
→
0,
there
are
two
cases
to
consider
because
x
x
can
approach
x
zero
from
approaches
0
positive
from
20
As
x
→
0
from
x
→
10
from
(
approaches
0
5
100),
10
0
from
1
__
(
1000)…
),
→
(
100),
1
__
(
1000)…
(
),
0.001
→

x
1
__
x→0

x
0.01
lim
above
1
_______
10),
0.1
Therefore
values.
20
0.001
1
______
(
negative
below
1
_____
(i.e.
from
1
______
10),
0.01
0
x
5
1
_____
(
0.1
As
below
or
above
1
____
(i.e.
values
1
__
)
does
not
have
a
unique
value
so
lim
x
(
For
lim [f(x)]
to
exist
and
equal
k
)
does
not
exist.
x
x→0
then
x→
→a
as
x
→
a
and
as
x
from
above
}
The
limit
of
f( x)
as
x
→
a
→
from
a
from
above
f(x)
→
k
below
is
written
as
lim
f(x)
and
the

x→a
of
f(x)
as
x
→
a
from
below
is
written
as
lim
f(x)
x→a
Limits
and discontinuity
We
can
now
For
f(x)
to
be
define
a
discontinuity
continuous
where
x
in

terms
a,
lim

x→a
126
of
f(x)
limits.

lim
x→a
f(x)

f(a)
limit
Section
3
Calculus
1
2
x
For
example
the
function
f( x)
,
sketched
Using
the
in
T
opic

0

0
,
x,
is
x
x
{

x

,
3.1.
condition
for
continuity
at
x

0,
2
lim
x

0,
lim
x

0
and

x→0
f(0)
x→0

0,
therefore
However
applying
x
f(x)


4,
x

4
4,
x

4
,
3.1,
gives
is
this
{
x
T
opic
f( x)
condition
x
lim (x
continuous


,
4)
to
where

8
at
x
the
x
0
function

and

4,
which
lim (x
is
4)

at
x
also
sketched
in
0

x→4
x→4
They
are
not
equal
so
there
is
a
discontinuity

4
Example
2x
The
function
f
is
defined
as
f( x)

{
Find
lim
x

1
x

1
,
x
(a)
1,
x


2
,
f(x)

x→1
(b)
Find
lim
f(x)
x→1
(c)
Hence
First
show
sketch
that
the
f( x)
is
graph
continuous
y

at
x

1
f(x)
y
5
x
O
2
(a)
From
2
the
graph,
4
lim
f(x)

lim

(2x
1)

1

x→1
x→1
2
(b)
From
the
graph,
lim
f(x)

x→1
lim
x

1
x→1
2
(c)
f(1)

lim
1

f(x)

1
lim
f(x)

f(1)

x→1
x→1
Therefore
f(x)
is
continuous
at
x

1
Exercise 3.2
(a)
Sketch
the
graph
of
the
function
given
by
2
f(x)
(b)

Find
x
,
2

x
{

1
,
x


for
values
of
3

x

3
2
lim
x
,
x

1
f(x)

x→1
(c)
Find
lim
f(x)

x→1
(d)
Hence
show
that
f( x)
is
continuous
at
x

1
127
3.3
Limit
theorems
Learning outcomes

To
list
and
use
the
The
limit theorems
1
lim
If
f(x)

F
then
lim
kf(x)

kF
where
k
is
a
constant.
limit
x→a
x→a
theorems
2
For
example,
lim
2
x

4,
therefore
lim
x→2
sin 
3x

3

4

12
x→2
_____

To find
and
use
lim
→0

2
If
lim
f(x)

F
and
if
lim
x→a
g(x)

G
then
lim [f(x)
x→a

g(x)]

FG

g(x)]

F
x→a
2
For
example,
lim
You need to know
x

4
and
x→2
lim
(x

1)

3,
x→2
2
therefore

The
notation for
limits

The
trigonometric
double
How
to factorise
x
(x

If
lim
f(x)

F
and
if

lim
x→a
4

3
g(x)

G
The factor
12
then
lim [f(x)

G
x→a
2
For
y
example,
lim
x

4
and
x→2


x→a
3

1)]
angle
and factor formulae
3

[x
x→2
3
formulae
lim
theorem
lim
(x

1)

3,
x→2
2
therefore
lim
[x

x

1]

4

3

7
x→2
f(x)
F
__
____
4
If
lim
f(x)

F
and
if
x→a
lim
g(x)

G
then,
provided
G

0,
lim
x→a
x→a
(
g(x)
)

G
2
For
example,
lim
x

4
and
x→2
lim
(x

1)

3,
x→2
2
x
______
therefore
4
__
lim

x
x→2

1
3
0
__
Now
is
meaningless,
but
these
theorems
can
be
used
to
find
the
limit
0
0
__
of
a
function
which
appears
to
be
.
The
example
below
illustrates
this.
0
Example
2
x
9
____________
Find
(
lim
)
2
x
x→3
7x
2

(x
x
9
____________
12
3)(x

3)
x  3
______
_____________


provided
that
x

3
2
x
7x
The
as
x

limit
gets
12
as
(x
x
We
4)
approaches
closer
towards.
3)(x
and
do
closer
not
want
3
x
means
to
3
the
to
4
we
see
value
want
what
when
the
values
value
x

3,
they
so
of
are
we
the
function
tending
can
say
that
2
x
9
____________
lim
x→3
(
x  3
______
)
2
x
7x


12
lim
x→3
lim
(
)
x
(x
4

3)
6
___
x→3
__________


lim
(x
4)

6
1
x→3
sin x
_____
The
limit of
as
x → 0
x
This
is
an
directly,
origin.
128
so
important
we
start
limit
with
and
the
the
limit
graphs
of
y
theorems

sin x
do
and
y
not

help
x
to
close
find
to
it
the
Section
3
Calculus
1
y
x
0.6
x
0.4
0.2
π
π
6
6
x
0.2
0.4
0.6
sin x
_____
When
x

0,
0
__
0
__

x

__
and
is
0
meaningless.
0

__
For

x

,
6
sin x
and
x
are
nearly
equal,
and
as
x
approaches
0
from
6
sin x
_____
above,

1
x
sin x
_____
Also
as
x
approaches
0
from
below,

1
x
sin x
_____
Therefore
lim

Note
that
this
result
is
valid
only
1
x
x→0
when
x
is
measured
in
radians.
Example
sin 2 
______
Find
lim

→0
sin 2 
______
2 sin  cos 
___________
sin 
_____



sin 2 
______

lim

sin 
_____


→0
lim


→0
lim
lim

1
2 sin 2 
_______


2

2
sin 2 
______
lim

→0
(2 cos )
→0
sin 2 
______
Alternatively,
2cos 



2lim
2
→0
→0

2
2
Exercise 3.3
2
x
 5x  6
___________
1
Find
the
limit
of
as
x

x
→
2
2
2
 1
x
_______________
2
Find
the
limit
of
3
x
as
x
→

→
0
1
2

x

x

1
sin 3   sin 
_____________
3
Find
the
limit
of
as
2
sin x
_____
4
Use
the
results
above
to
show
that
the
function
f( x)

has
a
x
discontinuity
where
x

0
129
3.4
Gradient
Learning outcomes
of
a
Tangents,
The

Find
the
point
on
gradient
the
of
curve
a
curve
at
line
chords
joining
two
and
points
normals
on
a
curve
is
a
chord
a
curve
A
line
that
touches
a
curve
point
at
a
point
is
a
tangent
to
the
curve.
of
contact
You need to know
A

The
a

definition
straight
The
of
the
gradient
chor
d
of
B
line
concept
of
a
limit
The
gradient of
The
a
curve
gradient
of
gradient
We
can
curve
find
at
another
a
the
point
point
B
by
on
a
of
gradient
A
at
a
cur ve
the
of
given
at
a
point
point
tangent
to
A
the
is
defined
cur ve
at
as
the
A.
a
taking
the
curve
B
close
As
B
to
A
A.
moves
gradient
closer
of
to
tangent
closer
the
the
at
to
chord
gradient
A,
the
gets
of
the
A.
limit
(gradient
of
chord
AB)

gradient
of
tangent
at
A
B→A
2
We
can
use
this
to
find
the
gradient
of
the
curve
y

x
at
the
point
1
where
x

2
1
A
is
the
point
on
the
curve
where
x

,
2
1
B
is
a
point
on
the
curve
whose
x-coordinate
is
a
bit
larger
than
2
h
We
denote
the
‘bit
larger ’
by
is
sometimes
used
as
an
alternative
x
notation for
the
‘bit
larger’
1
So
B
is
the
point
on
the
curve
where
x


x
2
y

is
not
a variable
–
it
is
a
prefix
8
that
means
‘a
small
increase
in’
6
the
1
B(
4
1

δx
2
1
A(
1
,
2
2

δx)
)
value
follows
the
y
variable
means
that
a
small
2
2
increase
in
the
means
small
value
of
y,
t
)
2
2
value
x
O
3
130
of
it. So
2
1
1
2
3
a
of
t,
and
increase
so
on.
in
the
Section
1
The
coordinates
of
A
are
(
of
B
are
)
1
(

x,
2
(
2

x)
)
2
1
1
2
(
1
4
1
coordinates
Calculus
1
,
2
The
3

x)
2
4
____________
The
gradient
of
AB
is
1
1

x
2
2
1
1
2

x

(x)
4
4
_________________


1

x
x

gradient
of
tangent
at
A

limit
(gradient
of
chord
AB)
B→A

limit
(1

x)

1
x→0
1

at
the
point
where
x

2
,
the
gradient
of
y

x
is
1.
2
2
We
can
apply
the
same
process
to
a
variable
point
on
y

x
2
A
B
is
is
any
the
point
point
x-coordinate
on
on
of
the
the
curve
curve
so
its
coordinates
whose
can
x-coordinate
is
be
a
denoted
little
by
larger
( x,
than
x
).
the
A,
2
i.e.
x

x.
Hence
B
has
coordinates
2
(x

(x

x,
(x

x)
)
2
x)
x
_____________
The
gradient
of
AB
y
is
(x

x)
x
2
x
2

2xx

2
(x)
x
_____________________

x
2
B(x

δx,
(x

δx)
2
2xx

(x)
x(2x
____________

x)
___________
2


A(x,
x

2x

x
)
x
x
O
The
gradient
of
the
tangent
at
A


lim
(gradient
lim
(2x

of
x)
x
AB)

2x
x→0
We
can
now
use
this
result
to
find
the
gradient
at
any
particular
point
2
on
y
For
2

x
example,

3

6,
at
and
the
point
where
x
where

4
x

the
3,
the
gradient
gradient
is
2

of
4
the

curve
is
8
Exercise 3.4
1
Use
the
method
above
to
find
the
gradient
at
any
point
on
the
on
the
3
curve
Use
(a)
2

your
x
Use
y

the
x
result
to
1
find
(b)
method
above
the
x

to
gradient
on
the
curve
where:
5
find
the
gradient
at
any
point
curve
2
y

x
Use

your
2x
result
to
find
the
gradient
at
the
points
where:
1
(a)
x

0
(b)
x


2
131
)
3.5
Differentiation from first
Learning outcomes
The
In

Find
the
gradient function
of
principles
gradient function
the
previous
topic,
we
found
that
the
gradient
of
any
point
on
the
a
2
curve
curve
by
y

x
is
given
by
2x
differentiation from first
2
principles
Now
2x
is
a
function
and
it
is
called
the
gradient
function
of
x
.
2
Because
the
2x
is
derived
from
x
,
2x
is
often
called
the
derived
function
or
derivative
You need to know
dy
___
The
gradient
function
of
a
general
curve
y

f(x)
is
denoted
by
f(x)
or
by
dx

The
concept

How
to
of
a
limit
manipulate
algebraic
f(x)
fractions
limit
of
sin 

The
as

The factor formulae

→
0
x
f(x
x,


δx,
f(x
δx)

δx)
f(x)
f(x))
δx
x
O
dy
___
2
So
for
y

x
,
we
write
f(x)

2x
or

2x
dx
f(x

x)
f(x)
______________
For
any
curve
y

f(x),
gradient
of
AB

x
f(x

x)
f(x)
______________
and
the
gradient
at
A

lim
x
x→0
f(x

x)

f(x)
______________
i.e.
f (x)

lim
x
x→0
We
it
can
is
use
called
this
general
formula
differentiation
from
to
find
first
the
gradient
of
any
function,
principles
Example
1
__
Differentiate
from
first
principles.
2
x
1
__
f(x)

,
1
________
so
f(x

x)

2
2
x
(x

x)
2
1
________
f(x

x)
−
f(x)
2
x
1
__

(x

2
(x

2xx
2
x
(x)
_____________

2
x)
2
x)
_____________

x
2
(x

2
x)
x
2
(x

2
f(x

x)
f(x)
2xx
______________

f(x)

lim
x→0
2x  x
__________

lim
when
y

(x

x)
dy
1
__

x
2
__
___
,


2
x
132
3
dx
x
__

2
4
x
x(x
2
2x
____

2
x→0
lim
2
x
x→0
(x)
_____________


3
x

x)
2
x
x)
and
Section
3
Calculus
1
Example
Differentiate
f(x)
f(x


sin x,
x)
−
sin x
so
f(x
f(x)

from

first
x)

sin (x

principles.
sin (x
x)

−
x)
sin x
Using
1

2 cos (x


x
2
1
2 cos (x
1
x) sin
1
x
cos (x
2
2
____________________

f(x)

lim

1
x) sin
x
2
2
__________________

x
x→0
a factor formula
1
x) sin
2
lim
1
x→0
x
2
1
Now
as
x
→
0,
cos (x

x)
→
cos x
and,
provided
that
x
is
in
2
1
sin
x
2
_______
radians,
→
1
1
x
2

f(x)

cos x
dy
___
i.e.
if
y

sin x,
then

cos x
dx
Example
__
√
Differentiate
the
curve
f(x)

at
xl
the
from
point
first
principles
where
x

and
hence
find
the
gradient
of
4
__
1
√
xl

x
2
1
1
f(x

x)
f(x)

(x

x)
2
x
2
1
1
((x

x)
2
x
2
1
1
)((x

x)
2

x
2
)
_____________________________

1
1
((x
x


x
x)
2

x
2
)
x
_______________
2

Using
((x

x)
1
1
2
2

x

(a
b)(a
b)

2
a
b
)
x
_______________

1
1
((x

x)
2

x
2
)
x
1
_________________

f(x)

lim
x→0
x((x

dy
1
if
y

x
2
,
dy

4,
x
2
)
x→0
(x

x)
2


x
x
2
1
2
2x
2
2
x
2
2
1
1
__
___
x

1
1
__
___

1
1
1

dx
When
2
lim
1
__
___
i.e.
x)
1
_____________

1
1

dx
1
__
(4)
2
2

4
Exercise 3.5
Differentiate
the
following
functions
from
first
principles.
1
1
f(x)

kx
2
f(x)

cos x
3
f(x)

5x
where
k


4
f(x)
5
(x
6
f(x)

x
2

2x

1)
3

sin 2x
133
3.6
General
Learning outcomes
differentiation
Differentiation of
a
constant
y
The

To
derive
rules for
equation
y

c,
where
c
is
a
constant,
differentiating
represents
a
straight
line
parallel
to
the
simple functions
y
c

c
x-axis.
Therefore
the
gradient
of
the
line
is
zero,
dy
i.e.
when
y

x
O
___
You need to know
c,

0
dx
dy

The
equations
of
straight
___
lines
For
example,
when
y

5,

0
dx
dy
___

The
meaning
of
dx
y
Differentiation of
ax

y
The
equation
represents
a
y

line
ax,
where
through
a
the
is
a
constant,
origin
y
with
gradient

a
a
dy
___
Therefore
when
y

ax,

a
dx
x
O
dy
___
For
example,
when
y

4x,

4
dx
n
y
Differentiation of
The
table
shows
some
of

the
x
results
from
T
opic
3.5:
1
3
2
x
y

x
x
1
2
2
x
x
f(x)
1
dy
1
__
2
___
3x
2x
2
2
x
3
x
2x
2
dx
These
that
results
power
suggest
and
that
reduce
to
the
differentiate
power
by
a
power
of
x,
we
multiply
by
1,
dy
___
n
i.e.
when
y

x
,
n

1
nx
for
all
values
of
n
dx
dy
example,
when
y

x
dy
___
10
For
,
9

10x
___
4
and
when
y

x
,
dx
5

4x
dx
n
Differentiation of
y

ax
3
The
result
from
Exercise
3.5,
question
3,
shows
that
when
y

dy
___
2

15x
2

5
×
3x
dx
This
is
a
particular
example
of
the
general
result,
dy
___
n
i.e.
when
y

ax
n

,
1
anx
where
a
is
a
constant.
dx
dy
1
example,
when
y

4x
2
,
134
1

dx
3
1
___
For
4
×
1
x
2
2

2x
2
5x
,
Section
y
Differentiation of
The
result
from
Exercise
f(x )

3.5,
Calculus
1
g( x )

question
3
5,
shows
that
when
dy
___
2
y

x
2x

1,

2x
2
dx
This
is
dy
the
d
___
___

same
2
(x
dx
as
differentiating
d
___
)
(2x)
dx
each
term
separately,
i.e.
d
___

dx
(1)
dx
d
d
___
The
notation
___
f(x)
means
the
differential
of f(x)
with
respect
to
x,
i.e.
dx
This
result
is
f(x)
 f(x)
dx
true
for
the
differential
of
the
sum
or
difference
of
any
functions,
dy
d
___
___
i.e
when
y

f(x)

g(x),

dx
T
wo
other
results
from
T
opic
3.5
are
d
___
f(x)

dx
g(x)
dx
important:
dy
dy
___
when
y

sin x,
___

cos x
and
when
y

cos x,

dx
All
these
We
have
results
used
are
the
important
letters
sin x
dx
y
and
and
x
for
you
the
need
to
remember
variables,
but
any
them.
letters
can
be
ds
___
used,
for
example
when
s

2t

cos t,

2
sin t
dt
(Letters
s
and
concerned
t
are
with
often
used
for
displacement
and
time
in
problems
movement.)
Example
3
dy
6x
5x  2
_____________
___
Find
when
y
Exam tip

dx
3x
d
___
3
f(x)
3
6x
5x  2
_____________
6x
____
5x
___

2
___
3x
3x
f(x)
d
___

3x
3x
dx
dx
____
(
g(x)
______
)

d
___
g(x)
5
2

2x
2

3
dx
1
x
3
dy
5
2

y

2x
2

___
1
x
3
3
so
2

dx
4x
2
x
3
Exercise 3.6
dy
___
Find
when
y
is:
Exam tip
dx
4
1
5
5x
4
3 cos x
d
d
___
f(x)
dx
d
___

g(x)

f(x)
dx
___

g(x)
dx
3
2x
3
________
1
__
2
6
2
x
x
3
3
4x
4
(3x
2x

4)(2x
5

1)
7
x(2
8
5 sin x
√
x)
4 cos x
135
3.7
Product
Learning outcomes
rule
The
If

To
for
derive
and
use
y
differentiating
derive
uv,
where
a
product
and
use
if
x

f(x)
and
v

product of functions
g(x)
is
a
small
increase
increases
in
in
y,
the
u
value
and
v,
of
x,
and
y,
u
and
v
are
the
then
a formula for
y
a
u
a
of
corresponding
differentiating
rule
rule for differentiating

and
To
quotient
a formula
functions

and
quotient

y

(u

uv

u)(v

v)
of

u(v)

v(u)

(u)(v)
functions
As
y

uv,
this
y
simplifies

u(v)

to
v(u)

(u)(v)
You need to know
Dividing

How
to
differentiate
sums
by
x
gives
y
and

How
to
of
powers
differentiate
of
sin x
u
x
x
and
y
lim

The
limit
x→0

dy
u
x
dv
___
v
___
,
x
v
___

x
___

lim
dx
x→0

x
du
___
u
___
,
lim
dx

x→0
and
x
lim u
dx

0
x→0
theorems
dy
y
___
Trigonometric

lim
dx
identities
dv
___
___
Therefore

v
x
___
Now
cos x
u
___
v
___
___

differences
(
)

u
x
x→0
(uv)

u
dx
Y
ou
may
T
o
find
it
easier
the
to
a
first
,
dx
du
___

v
dx
differentiate
multiply
v
dv
___
d
___
i.e.
du
___

dx
dx
remember
product
of
function
this
r ule
in
words:
functions,
by
the
differential
of
the
second
function
then
of
add
the
the
second
function
multiplied
2
For
by
the
differential
first.
example,
if
y

x
2
cos x,
then
using
u

x
and
v

cos x
dy
___
2

−x
2
sin x

(cos x)(2x)

2x cos x
−
x
sin x
dx
Exercise 3.7a
dy
___
Use
the
product
rule
to
find
when
y
is:
dx
1
3
(a)
x
(b)
sin x cos x
sin x
2
3
(d)
x
(e)
(x
(3x
4)
2
1
__
(c)
sin x
x
136

1) cos x
gives
Section
The
rule for differentiating
3
Calculus
1
a quotient of functions
u
__
If
y

,
where
u

f(x)
and
v

g(x)
v
and
if
x
is
a
small
corresponding
increase
increases
in
in
y,
the
u
value
and
of
x,
and
y,
u
and
v
are
the
v,
u  u
_______
then
y

y

v
u
__
y


v
u  u
_______
,

y
u
__


v
v

v
v
vu  uv
__________

2
v
Dividing
by
x

vv
gives
u
___
v
___
v
y

u
x
x
__________
___

2
x
v
y

,
x
dv
___
v
___
___
lim
x→0
vv
dy
___
Now

lim
dx

x
x→0
lim
x→0
du
___
y
___
Therefore

lim
dx
(
)

0
2
dy
dv
___

u
dx
dx
___________
___


v
du
___
y
v
u
v
if
lim
x→0

u
__
i.e.
and
dx
dx
dx
__________
x
x→0

x
dv
___
v
dy
___
du
___
u
___
,
dx
then

2
v
Y
ou
need
to
important.
computer
dx
remember
One
way
monitor
–
to
this
v
r ule.
remember
‘visual
The
it
display
is
order
by
unit’
in
the
using
or
numerator
the
VDU.
old
So,
word
VDU
sin x
_____
For
example,
if
y

is
for
a
comes
first.
2
,
then
using
u

sin x
and
v

x
gives
2
x
2
dy
x
cos x  2x sin x
________________
___
x cos x  2 sin x
_______________


4
dx
3
x
x
2
Alternatively,
writing
y
as
a
quotient,
i.e.
y

x
sin x,
dy
___
product
rule
gives
3

2x

x
using
the
x cos x  2 sin x
_______________
2
sin x
and
cos x

3
dx
The
disadvantage
simplification
of
of
the
x
writing
result
a
is
quotient
often
as
a
product
is
that
the
complicated.
Exercise 3.7b
dy
___
Use
the
quotient
rule
to
find
when
y
is:
dx
sin x
_____
(a)
tan x
(
x
______
)
(d)
2
cos x
x

1
2
x
______
(b)
(e)
x

cot x
1
cos x
_____
(c)
__
√
x
137
3.8
The
chain
Learning outcomes
rule
Differentiating
When

To
derive
the
chain
rule
and
it
to
differentiate
is
a
composite
If
write
y

x
is
a
small
increase
y
y

The
meaning
of
a
x
→
0,y
and
composite
u
the
in
also
dy
y
y
of
lim
(
where
u

f(x),
we
x
and
y
and
u
are
the
then
zero.
y
x→0

lim
x
(
x
y
u
___
___
(
lim
)
dy
product


du
dx
dy
dy
___
du
___
___
i.e


dx
is
known
composite
as
the
function
chain
gf( x)
by
)
x
u→0
du
___
___

of functions
This
(
lim
u
u→0
a
)

u
x→0
simple
differentiate
quotient
u
___
___
)
functions
and
of
u,
approach

to
value
and
___

dx
differentials
How
gf(x)
x
___



u
Therefore
The
y
theorems
function

when
u
___
___

x
As
in
increases
___
You need to know
limit
i.e.
g(u).
corresponding
The
function,
composite
functions

composite function
to
can
use
y
a
du
r ule
and
making
dx
can
the
be
used
to
differentiate
substitution
u

6
For
example,
when
y
dy
___

(2x

3)
6
,
using
u

2x

1
gives
y

u
du
___
5
then

a
f(x)
6u
and

du
2
dx
dy
dy
___
dy
du
___
___

___

dx
du
5
gives

dx
5

6u
2

12u
dx
dy
___
Substituting
2x

1
for
u
gives
5

12(2x

1)
dx
Y
ou
f(x)
will
not
when
y
usually

be
given
the
substitution,
so
you
need
gf(x)
Example
_______
Find
f(x)
when
f(x)

2
√
2x
5
_______
1
Let
y

2
√
2
2x
5
then
if
dy
dy
4x
dx

du
dy
u
2

dx

2

2
4x
√
u
2x
___
u
f(x)

2x
_________
_______
√
2x
138
2
−
5

1
____
√

y

2
1
____
___
dx
5,
1

du
gives
−
1
__
and
du
___
___

2x
___

dx
___

dy
du
___
Therefore
u
√
u
√
u

u
2
to
recognise
Section
3
Calculus
1
Example

__
Differentiate
sin
3

(
with
)
respect
to

4

When
is
the
variable,
we
replace
x

__
Let
y

sin
3

(
with

__
and
)
u

3

4
dy

with
d
y

sin u
dy
du
___

d
then
4
d
___
du
___
___
Using

___

3
and

d
du
cos u
gives
du
dy
___

(cos u)(3)

3 cos
d

__
3
(

)
4
After
a
bit
of
practice
the
substitution
For
example,
with
mentally
you
could
simpler
and
go
functions,
write
straight
down
you
the
should
differential
be
able
to
make
directly.
from
dy
___
3
y

2(3x
1)
to
2

(2)(3)(3)(3x

18(3x
1)
dx
2
1)
Example
dy
2
_______
___
Find
when
y

dx
We
(x
could
use
the
3)
quotient
rule
for
this,
but
by
writing
the
equation
as
1
y

2(x
3)
y

2(x
3)
,
we
can
use
the
chain
rule:
i.e.
dy
___
1
⇒
2

(2)(


1)(x
3)
dx
dy
2
________
___
⇒
2
dx
(x
3)
Exercise 3.8a
1
Differentiate
each
function
with
respect
to
x.
4
(a)
(3x

3
4)
(d)
(2

(c)
sin 2x
x
1
_____
3
(b)
cos
x)
(e)
sin x
4
2
Differentiate
3
Find
cos

dy
respect
to
.
1
________
___
______
when
dx
with
y

√
x
2

1
139
Section
3
Calculus
1
T
o
differentiate
the
more
complicated
functions,
it
is
sensible
to
write
down
substitutions.
Example
dy
___

__
2
Find
given
y

(3x
2)
sin
2x
(
)
dx
This
3
is
the
product
finding
the
differential
product
of
two
of
composite
each
functions.
composite
t

(3x
−
and
by
apply
the
u

3x
2
so
t

u
du
___

2u
and

du
3
dx
dt
___
dt
___

then
du
___

dx
du
dx

6u

6(3x
−
2)

__
s
start
then
2
2)
dt
___
⇒
Let
will
and
rule.
2
Let
We
function

sin
2x
(

__
and
)
u

2x
(
du
___
so
)
3
s

sin u
3
ds
___
⇒

2
and

dx
cos u
du
ds
___
then
ds
___

dx
du
___

du
dx

(cos u)(2)

2 cos

__
2x
(
)
3
Now
using
dy
the
ds
___
___

t
dx
product
rule
gives
dt
___

s
dx
dx

__
2

(3x
2)
×
2 cos
2x
(

__
)

sin
2x
(
)
3
×
6(3x

__

2(3x
−
2)
(3x
−
[
2)
cos
2x
(
We
of
can
extend
three
where
the
functions,
u

dy
f(x)
and
dy
___
chain
i.e.

3 sin
2x
(
)]
then

dv
3
rule
rule
where
dv
___
___

dx
chain
to
y
using
cover

functions
hgf(x),
the
by
extended
du
___

du
dx
__________
2
For
example,
when
y

√cos(x

1) ,
1
2
then
140
u

x

1
and
v

2)

__
)
3
Extending the
−
3
cos u
so
y

v
2
that
using
y
version
are

of
a
h(v)
the
composite
and
v
chain

g(u)
rule,
i.e.
Section
dy
3
Calculus
1
1
___
1
then
2
v


(
sin u)

2x
2
dx
1
1

2
cos
u

(
sin u)

2x
2
2
x sin(x

1)
____________
__________


2
√cos(x

1)
Example
1
____________
Differentiate
2
sin
(3x
1)
1
____________
First
we
can
write
2
as
sin
(3x
1)
2
sin
(3x
1)
2
Now
sin
where
(3x
f(x)

1)
(3x
−

hgf(x)
1),
2
g(x)
so

we
sin x
need
and
two
h(x)

x
substitutions.
2
Let
u
y


sin
3x
and
v
(3x
1),
1

sin u
2
then
y

v
dy
dy
___
then
dv
___
___

dx

dv
du
___

du
dx
dy
___
gives
3

2v
×
cos u

6(sin u)


×
3
dx
3
cos u
6 cos (3x
1)
____________
3
sin
(3x
1)
Exercise 3.8b
______
4
1
Differentiate
2
Find
(2 x

1)
√
2
x

dy
1
with

__
___
when
y

cos
2
(

to
x.
2
)
d
respect
sin

4
5
3
Find
4
Find
f(x)
when
f(x)

sin [(x

4)
]
_____________
dy
___
4
when
y

√1

(3x

1)
dx
141
3.9
Parametric
and
general
differentiation
dy
d x
___
Learning outcomes
___
The formula

1

d x
dy
This
is
a
formula
we
dy
need
to
be
able
to
differentiate
a
parametric
___

To find
when
the
equation
of
equation
dx
a
curve
is
given
parametrically
dy
y
___
When
y

f(x),
x
___
___
then

dx
lim
x→0

dy
y
→
0
as
x
→
0

1

lim
dx
(
y→0
meaning
of
parametric
y
)
y
dx
___
)

1

dy
dx
___
___
The

x
___
so
dy

1
(
x→0
___
But
You need to know
lim
x
i.e.

1

dy
dx
equations

How
to
differentiate
simple
Parametric differentiation
functions
dy
___
When

The
product
rule,
quotient
the
chain
The
limit
of
a
curve
is
given
parametrically,
we
find
in
dx
of
the
parameter
.
rule
dy
dy
___

can
rule
terms
and
equation
theorems
If
x

f(t)
and
y

g(t),
then
the
chain
rule

dx
dt
___
Using
the
formula
above,
dy
dy
___
1

becomes
dx

dt
,
dt
____

dx
f(t)
1
_____
2
For
dx
___
___

dx
g(t)
dy
___
then
dy
___

dt
dx
dt
dy
dt
___
___

dx

dt
dx
___

dx
so
dt
___
___
gives
example,
when
x

t
and
y

then
t
dy
dx
___
2t
1
1
_______
___


and


Using
the
quotient
rule
2
dt
dt
dy
dy
___


1)
dx
___
___

(t

dx
dt
dt
1
_______



2t
2
(t

1)
1
_________


2
2t(t

1)
Exercise 3.9a
dy
___
Find
in
terms
of
the
parameter
when:
dx
2
(a)
y

t
1
__
3
,
x

1
(c)
t
x

2
,
y

(1

t)
t
(b)
y

cos ,
x

2 sin 
General differentiation
In
the
previous
functions
are
142
and
topics
derived
summarised
in
we
have
rules
the
for
table.
given
the
differentials
combinations
of
of
various
functions.
types
These
of
results
Section
3
Calculus
1
dy
___
y
General
results
dx
d
___
c
(a
constant)
0
(f(x)

g(x))

f(x)

g(x)
dx
n
d
___
n1
x
nx
(af(x))

af(x)
dx
dy
n
dy
___
n1
ax
du
___
___
anx


dx
du
dx
dy
dx
___
___
sin x
cos x

1

dx
dy
d
___
cos x
dv
___
(uv)
sin x

du
___
u
dx

v
dx
dx
du
___
v
d
___
2
tan x
sec
x
(
dx
Any
of
asked
The
these
can
be
used
u

)
2
v
v
directly
unless
their
derivation
is
for
.
next
applying
some
results
dv
___

dx
dx
__________
u
__
exercise
the
gives
most
functions
practice
direct
may
fall
in
method
into
recognising
to
more
find
its
than
the
type
of
differential.
one
category,
function
Remember
so
two
rules
and
that
may
3x
________
be
needed.
For
_______
example,
√
4x
a
composite
some
function.
expressions

is
Remember
before
a
quotient
and
the
denominator
is
1
also
that
differentiating
you
may
be
able
to
simplify
them.
Exercise 3.9b
dy
___
1
Find
when
y
is
equal
dx
to:
______
2
√
1  x
________
2
(a)
sin (x

1)
1  x
______
3
(b)
(1
(c)

x
) sin x
(d)
2
x
2
Differentiate
each
of
the
1
following
3x
________
3
_______
(a)
√
4x

functions
(e)
(4

x
with

respect
5
x
to
x.
2
)
(i)
x
( j)
(
cos 2x
1
2
cos x
_____
x  2
___________
(b)
x  1
______
(f )
2
sin x
x
1
_____

4x

4
x
)

1
x  2
___________
(c)
(g)
2
cos x
x

4x

4
sin x
_____
(d)
(1

2x) tan x
(h)
x
dy
___
3
Find
in
terms
of
the
parameter
when:
dx
2
(a)
x

(b)
x

t

1,
y


2
(c)
x

cos ,
y

3 tan 
t  1
_____
1
_____
,
t
y

2
1

t
t
143
3.
10
Rates
Learning outcomes

To
solve
rates
of
of
change
Rate of
increase
change
The
problems
gradient
increase
in
of
a
straight
line,
y

mx

c,
is
calculated
from
y
____________
from
increase
in
Therefore
increase
one
point
on
the
line
to
another
point
on
the
line.
x
the
in
x,
gradient
i.e.
the
measures
rate
of
the
rate
increase
of
at
y
which
with
y
increases
respect
to
per
unit
x
You need to know
dy
___
gives

The
chain
the
gradient
of
the
tangent
at
a
point
on
the
curve
y

f(x),
rule
dx

How
to
differentiate
dy
simple
___
so
is
a
measure
of
the
rate
at
which
y
is
increasing
with
respect
dx
functions
to
x
For
at
that
point
example,
at
on
the
the
curve.
point
y
where
dy
___
2
x

2
on
the
curve
y

x
8
,

2x

4
dx
So
where
x

2,
y
is
increasing
at
the
6
rate
in
of
4
units
for
every
unit
increase
x.
4
Note
that
because
varies
this
the
as
x
is
rate
only
at
true
which
where
y
x

2
changes
varies.
2
dy
___
At
the
point
where
x

2,

4
dx
The
negative
decreasing
in
sign
at
4
shows
units
per
that
y
is
unit
increase
4
x
O
2
2
4
x
Connected
The
chain
variables
variable
rule
y
u
rates of
is
and
and
useful
x,
we
we
change
when
know
want
to
we
the
find
know
rate
the
of
the
relationship
change
rate
of
of
y
change
with
of
x
between
respect
with
the
to
a
respect
to
u
Example
1
__
The
equation
of
a
curve
is
y

4
x
A
point
P
is
increasing
Find
the
moving
at
rate
the
at
along
constant
which
the
the
rate
curve
of
so
0.01
that
units
x-coordinate
is
dy
rate
at
which
y
is
increasing
is
the

4
,
dy

dt
1
__
___
therefore

2
x
dx
x
dx
___
The
rate
of
change
of
x
with
respect
to
t
is
dt
144
is
second.
when
x

1
___
so
time.
1
__
y
y-coordinate
dy
,
dt
is
per
increasing
___
The
the
0.01
where
t
seconds
Section
dy
dy
___
dt
___
___
Using

dt
___

dt

dx

dx
Calculus
1
100
____
1
__
gives
0.01

2
dx
3
2
x
x
2
dx
___
Now
dt
___
using

dx
___
1/
dt
x
____
gives

dx
dt
100
dx
___
When
x

1,

0.01
dt
Therefore
when
x

1,
x
is
increasing
at
the
rate
of
0.01
units
per
second.
Example
3
Air
is
per
second.
Find
leaking
the
rate
out
of
a
spherical
of
change
of
a
of
the
balloon
radius
at
the
when
constant
the
volume
rate
of
of
the
0.3 cm
balloon
3
is
36 cm
4
(The
volume
sphere
is
V
3
r

)
Give
your
answer
correct
to
one
3
significant
figure.
3
___
When
V

36,
r

3
√

dr
___
We
require
when
the
volume
is
decreasing
at
the
rate
of
dt
dV
___
3
0.3 cm
per
second,
i.e.
when

0.3
dt
4
From
V
dV
___
3
r

,
dV
___

dt
___

dr

0.3
×
dt
___
⇒
dr
dt
___

dt
___
2
4 r
gives
dt
dr
___
Then
4 r
dr
dV
___
Using
2

3
40
__

dr
2
r

3
dr
3
______
1/


2
dt
3
___
When
r
40r
dr

1
______
dr
___
,


1
3
√
The
radius
dt

is

0.006
3
120
decreasing
at
the
rate
of
0.006 cm
per
second.
Exercise 3.10
1
The
area
of
a
circular
oil
slick
on
a
lake
is
increasing
at
the
rate
of
2
2 m
2
per
second.
Find
the
rate
3 m.
Give
The
of
your
equation
change
answer
of
of
the
radius
correct
to
of
three
y

2 sin .
increasing
at
the
a
curve
is
the
slick
when
significant
A
point
is
the
radius
is
figures.
moving
along
the
_
1
curve
so
that

is
constant
rate

of
radians
per
4
7
_
second.
Find
the
rate
of
change
of
y
when



6
3
A
right
circular
contains
grain,
cone
has
which
is
its
axis
pouring
vertical
out
of
and
a
its
hole
in
vertex
the
downwards.
vertex
at
50 cm
It
rate
_
1
3
of
the
per
second.
The
semi-vertical
angle
of
the
cone
.
is
Find
6
the
of
rate
the
of
change
circular
of
the
surface
of
height
the
of
grain
grain
is
in
the
cone
when
the
radius
2 m.
145
3.
11
Increasing
Learning outcomes
and
decreasing functions
Increasing
and decreasing functions
dy

To
determine
whether
___
a function
The
value
of
at
any
point
on
a
curve
whose
equation
is
y

f(x)
dx
is
increasing
or
decreasing
measures
the
rate
at
which
y
measures
the
rate
at
which
the
to
is
increasing
function
as
f( x)
x
is
increases,
increasing
to find
the
rate
of
change
How
to
of
The
y
a function

differentiate
respect
3
Consider
,
How
f (x)
with
x
You need to know

i.e.

for
graph
f(x)
example,
shows
and
y

the
the
function
relationship
given
by
between
f( x)
the

x
3x

2
curves
f(x)
y
simple
2
f(x)

3x
3
functions

How
to
differentiate
products,
5
quotients
and
composite
functions
3
f(x)

How
to
rational
solve
quadratic

x
3x

2
and
inequalities
x
O
3
2
5
This
the
graph
x-axis
above
the
shows
(i.e.
that
f(x)
x-axis

(i.e.
f( x)
0),
f(x)

Therefore
and
So,
to
to
determine
determine
decreasing
f(x)

a
f (x)
is
as
x
increases
increasing
as
x

0
0
when
when
function
is
is
positive
f(x)
f(x)
is
the
range
of
values
is
of
x
f(x)
increasing
or
or
decreasing,
negative.
for
which
the
function

3x
2x

4
decreasing.
2
f(x)
⇒

3x
2x
f(x)

6x
f(x)

0

4
2
1
when
6x

2,
i.e.
when
x

3
1
Therefore
f(x)
is
decreasing
for
values
of
x
less
than
3
146
is
below
f (x)
is
increasing
2
is
f (x)
where
decreasing.
Example
Find
where
increases
0).
f(x)
f(x)
whether
whether
is
and
we
need
Section
3
Calculus
1
Example
2x
___________
Determine
the
range
of
values
of
x
for
which
f( x)

is
increasing.
2
x

5x

4
2x
___________
f(x)

2
x

5x

4
2
2(x

5x

4)
2x(2x

5)
__________________________
⇒
f(x)

2
(x
2

5x

4)
2
2x
 8
_____________

2
2
(x

5x
2(2

4)
x)(2

x)
_______________

2
[(x
f(x)
is
zero
x

4
x

4,

1)(x
when
and
x
4


x

1
x


4)]
2
so
and
we
2,
x

2
need
to
investigate
2

x

and
f(x)
1,
1
is
undefined
the

x
sign

2
of
when
f (x)
and
x
when

2
2
Note
The
that
table
[(x

1)(x
shows

the
x
2
2

4)]
sign


of
0
for
f (x)
4
all
for
4

values
the
x

of
x
different
2
ranges
2

x
of

x
1
1

x

2
x

x





x















2
2
[(x

1)(x
2(2

x)(2
4)]

x)
_______________
2
[(x

1)(x

4)]
Therefore
f(x)
hence
is
f(x)

0
for
increasing
2

for
x
2

1

x
and

for
1
1
and
for

x
1


2,
x

2
Exercise 3.11
1
Find
the
range
of
values
of
x
for
which
the
3
f(x)
is
2

x
function
given
by
given
by
2

4x

4x
increasing.
Find
the
range
of
values
of
x
for
which
the
function
2x
________
f(x)

2
(1
is

x)
decreasing.
147
3.
12
Stationary values
Learning outcomes

To
define
and find
Stationary values
stationary
We
have
seen
in
T
opic
3.11
that
when
f (x)
is
positive,
f( x)
is
increasing
values
as
x
increases,
and
that
when
f (x)
is
negative,
then
f( x)
is
decreasing
as
x
increases.
You need to know
There
may
f(x)
neither
The

How
to
differentiate
is
value
be
of
How
to
f( x)
at
nor
such
Therefore
differentiate
and
f( x)
is
neither
negative
a
point
is
but
is
called
increasing
equal
a
to
nor
decreasing
so
zero.
stationar y
value
f(x)

0
⇒
f(x)
has
a
stationary
value.
products,
Consider
quotients
where
positive
simple
functions

points
the
graph
of
y

f(x).
y
composite
At
the
points
A,
B
y,
is
and
C,
f( x),
functions
and
therefore
increasing
Therefore
nor
the
neither
A
decreasing.
values
of
y
at
C
these
points
are
stationary
values,
B
dy
___
i.e.

0
⇒
y
has
a
dx
stationary
The
value.
points
stationar y
At
on
the
the
graph
where
y
has
a
stationary
value
are
called
points
stationary
i.e.
x
O
points,
tangents
are
the
gradients
parallel
to
of
the
the
tangents
to
the
curve
are
zero,
x-axis.
y
or
f(x)
has
a
stationary
value
dy
___
Therefore
at
a
stationary
point
,
{
or
f(x),

0
dx
the
tangent
parallel
to
to
the
the
cur ve
is
axis
Example
Find
the
stationary
3
f(x)

x
f(x)

x
3
At
values
of
the
function
given
by
2
2x

x
1

x
1
2
2x
stationary
2
values,
⇒
f (x)

f(x)

3x
4x

1
0,
2
⇒
⇒
3x
4x
(3x

1)(x
1

0
1)

0
x

1
1
⇒
x

or
3
1
When
x

,
f(x)
3
__
1
2
27
9

23
__
1

1


3
and
when
x
27
23
__
Therefore
the
stationary
values
of
f( x)
are
1
and
27
148

1,
f(x)

1
Section
3
Calculus
1
Example
1
___________
Show
that
the
curve
whose
equation
is
y

has
no
2
x
stationary
1
________

1
2
________
___

⇒

2
2
x

dy
1
___________
y
2x
points.
2x

(x
1
3
1)
dx
(x
1)
there
are
dy
___
At
stationary
values,

0,
but
no
values
of
x
for
dx
2
________
which

0,
3
(x
1)
1
___________
therefore
the
curve
y

has
no
stationary
values.
2
x
2x

1
Example
x
________
Show
that
the
curve
whose
equation
is
y

has
only
one
2
(x
stationary
point
and
x  1
________
___

⇒


2
(x
1)
it.
dy
x
________
y
find
3
1)
dx
(x
1)
dy
___
The
curve
has
a
stationary
point
where

0,
dx
x  1
________
i.e.
where

0
3
(x
1)
x  1
________
There
is
only
one
value
of
x
for
which

0,
i.e.
x

1
3
(x
Therefore
the
curve
has
only
one
1)
stationary
point.
1
When
x

1,
y


1
,
so
the
stationary
point
is
the
point
(
1,
4
).
4
Example
2
The
curve
Find
the
y

ax
values

of
a
bx

and
8
has
a
stationary
value
of
5
when
x

1.
b
dy
___
2
y

ax

bx

8
⇒

2ax

b
dx
y
has
a
stationary

5
and
2a
Solving

a

[1]

b
b

and
value

8
of
5
⇒
a
when

b
x


1,
3
[1]
0
[2]
[2]
simultaneously
gives
a

3
and
b

6
Exercise 3.12
2
1
Find
the
stationary
value
2
Find
the
stationary
value
3
by
f(x)

x
of
of
the
function
given
the
function
given
by
f( x)

x

5x

1
2

6x

12x

2
2
x
______
3
Find
the
stationary
points
on
the
curve
y

x
4
Show
that
there
are
no
stationary
points
on

the
1
curve
2x
___________
y

2
x

5x

4
149
3.
13
Determining the
Learning outcomes
Turning
A

To
define
maximum
curve
points
stationary
points
points

f(x)
can
have
several
stationary
points
and
the
shape
of
the
and
curve
minimum
y
nature of
and
points
close
to
one
of
these
points
belongs
to
three
different
types.
of
inflexion
y

To
distinguish
stationary
between
points
A

To
of
define
the
second
differential
a function
C
You need to know
B

How
to
differentiate
simple
x
O
functions

How
to find

How
to
stationary
determine
a function
is
values
whether
increasing
Moving
along
the
curve
in
the
positive
direction
of
the
x-axis:
or
1
Near
the
point
A,
the
gradient
of
the
curve
goes
from
positive
through
decreasing
zero
to
negative.
A
The
2
Near
zero
value
the
to
of
point
y
is
at
B,
3
At
a
C
value
the
point
here,
A
the
is
a
maximum
called
gradient
the
of
tur ning
maximum
the
curve
point.
value
goes
of
from
y
(or
of
negative
f(x)).
through
positive.
B
The
called
of
y
is
at
gradient
on
but
the
the
called
A
is
zero
curve
sense
is
a
called
but
moves
of
minimum
the
the
minimum
gradient
through
curvature
tur ning
C.
does
does
The
point.
value
not
curve
change
of
(or
change
does
from
y
of
sign
not
f(x)).
as
turn
clockwise
to
anticlockwise.
A
point
to
There
are
changes,
the
Note
and
at
that
least
a
cur ve
a
other
at
a
A
point
turning
the
terms
which
(and
in
and
of
the
vice
the
B
is
is
diagram
and
one
is
where
between
not
changes
called
a
the
B
from
point
sense
and
necessarily
of
C.
zero.
clockwise
inflexion.
of
curvature
Therefore
However
the
zero.
maximum
refer
cur vature
versa)
inflexion
point
They
values.
at
points
between
value.
stationary
150
two
one
gradient
gradient
on
anticlockwise
only
and
to
minimum
the
do
behaviour
not
of
a
mean
greatest
function
close
value
to
its
Section
Distinguishing
There
are
three
between
ways
of
3
Calculus
1
stationary values
distinguishing
between
stationary
values.
y
A
A
A
1
2
C
C
2
B
B
2
1
C
1
B
x
O
First
Look
For
method
at
A
the
(a
points
on
maximum
either
side
of,
value):
y
and
at
close
A
to,
the

y
at
A

y
at
A
stationary
values.
1
y
at
A
2
For
B
(a
minimum
value):
y
at
B

y
at
B

y
at
B
1
y
at
B
2
For
C
(a
point
of
inflexion):
y
at
C

y
at
C

y
at
C
1
y
at
C
2
We
can
summarise
these
observations
in
Maximum
V
alues
either
the
of
Both
y
side
a
table:
Minimum
smaller
Both
Inflexion
larger
One
one
of
smaller
and
larger
stationary
value
Second
For
this
method
method
stationary
we
look
at
the
gradient
on
either
dy
A
(a
of,
and
close
to,
maximum
value):
___
at

A
0,
___
at
A

0,
at
dy
(a
minimum
value):
dx
dy
dy
___
at
B

0,
___
at
B

0,
at
B
1
dx
(a
point
of
inflexion):
dy
___
at
C

0,
___
at
C

0,
at
1
results
are
summarised
in
Maximum
the
C

0
2
dx
These
0
dx
dy
dy
___
C

2
dx
For
0
2
dx
___
B

A
1
dx
For
the
dy
dy
___
For
side
values.
dx
dx
table:
Minimum
Inflexion
dy
___
Sign

of
0
0


0

or
0
dx
either
and
side,
at,
stationary
a
value
151
Section
3
Calculus
1
Third
For
A
method
(a
maximum
value):
As
a
point
moves
along
the
curve
dy
___
through
A,
then
as
x
goes
increases
dx
from
positive
to
negative
dy
___
so
decreases
as
x
increases.
dx
dy
___
Therefore
is
a
decreasing
function
dx
at
For
B
(a
minimum
value):
A.
As
a
point
moves
along
the
curve
dy
___
through
B,
then
as
x
increases
goes
dx
from
negative
to
positive
dy
___
so
increases
as
x
increases.
dx
dy
___
Therefore
is
an
increasing
function
dx
at
For
C
(a
point
of
inflexion):
A.
As
a
point
moves
along
the
curve
dy
___
through
C,
then
as
x
increases
goes
dx
from
again
positive
to
to
zero
positive
then
increases
values.
dy
___
Therefore
itself
has
a
stationary
dx
value
at
C.
Second derivative
dy
dy
___
The
rate
of
change
___
of
is
the
derivative
of
with
dx
respect
to
x.
This
is
dx
2
d
y
____
called
the
second
derivative
of
y
and
is
denoted
by
or
when
y

f(x),
2
dx
by
f(x).
3
For
example,
when
y

x
2
2x
,
dy
___
2

3x
4x
and
dx
2
d
y
d
___
____

2
(3x
4x)
2
dx
dx

6x
4
Returning
points,
to
we
the
can
third
now
method
express
for
the
distinguishing
observations
between
above
in
derivative.
2
dy
d
___
At
A,
y
____
is
a
decreasing
function
therefore

0
2
dx
dx
2
dy
d
___
At
B,
y
____
is
an
increasing
function
therefore

2
dx
dx
2
dy
d
___
At
C,
y
____
has
a
stationary
value,
therefore

2
dx
152
dx
0
0
stationary
terms
of
the
second
Section
3
Calculus
1
2
d
y
____
This
is
the
easiest
method
to
use.
However
can
also
be
zero
at
a
2
dx
2
d
y
____
maximum
or
minimum
value,
so
when

0
either
the
first
or
second
2
dx
method
These
has
to
results
be
used
are
to
distinguish
summarised
in
the
between
stationary
values
of
y
table:
Maximum
Minimum
2
d
y
____
Sign
of
at
Negative
a
Positive
2
dx
(or
stationary
zero)
(or
zero)
value
Example
4
Find
the
stationary
distinguish
points
between
on
the
curve
y

3
3x
4x

2
and
them.
dy
4
y

___
3
3x
4x

2
⇒
3

2
12x
12x
dx
dy
___
At
stationary
points,
3

0
⇒
12x
⇒
x
⇒
x
2
12x

0
dx
2
When
x

Therefore
0,
y
(0,

2)
2
and
and
(1,
when
1)
are
x

(x

1,
1)
0
y
stationary
or


0
x

1
1
points.
2
d
y
____
2

36x
−
24x
2
dx
2
d
y
____
When
x

1,

36
24

0
2
dx

(1,
1)
is
a
minimum
point.
2
y
d
____
When
x


0,
0
which
is
inconclusive,
so
we
look
at
the
signs
2
dx
dy
___
of
each
side
of
x

0
dx
dy
___
1
When
x


1
,
2

dy
x

,
Therefore
(0,
12
___
8
4
point
of

dx
1)
is
12(
a
2
)

0
2
12
___

2
1
3
)
2
___
1
When
12(
dx
0
inflexion.
Exercise 3.13
Find
the
them
stationary
when
3
(a)
x
f( x)
points
on
the
2

3x
curve
3

y

f(x)
and
distinguish
between
is:
3x

1
(b)
x
2

2x
4

x

2
(c)
(x

2)
153
3.
14
Curve
sketching
Learning outcomes
Features to
T
o

To
use
a
variety
of
techniques
sketch
curve
whose
when
shape
is
sketching
unknown,
we
curves
can
often
find
several
to
features
sketch
a
look for
by
observation
and
by
calculation
from
the
equation
of
the
curve.
curves
The
main
features

where
the

stationary

vertical

the
range

the
behaviour

where
to
curve
look
for
crosses
are:
the
axes
You need to know

How
to find
stationary
and
distinguish

The
meaning

How
to find

How
to
points
between
of
an
points
and
horizontal
asymptotes
them
of
values
of
y
asymptote
of
y
as
x
→

∞
limits
solve
rational
the
function
is
increasing
or
decreasing.
inequalities
Not

How
to
convert
an
rational function
improper
to
a
proper
A
all
of
picture
diagram
these
of
as
features
the
you
curve
find
need
can
be
to
be
built
considered
up
by
for
marking
a
particular
these
curve.
features
on
them.
rational function

How
that
to find
a
the
range
of
rational function
values
can
take
Example
x
3
______

How
to
determine
whether
Sketch
the
curve
whose
equation
is
y

x
a function
is
increasing
2
or
decreasing
First
find
where
the
curve
crosses
the
axes:
1
when
x

0,
y

1
and
when
y

0,
x

3
2
1
Therefore
the
curve
goes
through
the
points
(0,
1
)
and
(3,
0).
2
Now
look
for
asymptotes:
1
______
as
x
→

∞,
1
→
x
The
value
of
y
is
1
so
the
line
y

1
is
an
asymptote.
2
undefined
when
x

2
so
the
line
x

2
asymptote
1
______
and
lim

x→2
These
(1
1
______
)
x
findings
are

∞
and
lim
2
shown
(1
x→2
on
the
)
x

∞
2
diagram.
y
6
4
2
x
O
4
2
4
2
4
6
154
is
an
a
Section
Next
investigate
stationary
3
Calculus
1
points:
x
3
______
y

x
2
1
______

1
x
2
1

1
(x
dy
2)
1
________
___
⇒

2
dx
(x
2)
no
values
dy
___
There
are
of
x
for
which

0
so
there
are
no
stationary
dx
values.
dy
___
Also

0
for
all
values
of
x
except
x

2,
therefore
y
is
increasing
as
dx
x
increases.
Check
the
range
of
2y
x
3
______
y
⇒
We
x
now
y:
3

so
2
therefore
of
_______

x
values
y
the
curve
have
x
has
real
values
except
when
y

1,
1
does
enough
not
cross
the
information
to
line
y
sketch

1
the
curve.
y
6
4
2
O
4
x
2
2
4
6
Example
2
(x
1)
_________
Sketch
the
curve
y

x(2x
Check
when
for
x
intercepts

0,

0,
y
is
on
1)
the
undefined.
axes:
Therefore
the
curve
does
not
cross
therefore
the
curve
the
y-axis.
when
y
touches
(This
the
must
x

1.
x-axis
be
a
This
at
(1,
is
a
repeated
root,
0).
stationary
point.)
155
Section
3
Calculus
1
Check
for
asymptotes:
1
y
is
undefined
when
x

0
and
x

,
2
1
therefore
the
lines
x

0
and
x

are
vertical
asymptotes.
2
2
(x
1)
________
y

2
2x
x
2
x
2x  1
___________

2
2x
x
1
__
3x
2
_________
2
4x
1
__
3x
2
__________
2
2x(2x

2
2x

1)
3x
2
__________
1
__
As
x
→

∞,
(
2
2x(2x
as
x
→
∞,
1
__
)
→
from
(
2
2x(2x
below
2
3x
2
__________
1
__
and
1)
1
__
→
)
1)
from
above.
2
1
__
Therefore
y

is
a
horizontal
asymptote.
2
1
__
When
y

,
2
1
__
1
__
3x
2
__________
2
2x(2x

2
3x
2
__________
2
__
⇒

1)
2x(2x
0
⇒
x

1)
3
1
__
Therefore
the
curve
crosses
the
line
y
2
__

where
x

2
3
y
6
4
2
4
x
O
2
2
4
2
4
6
This
diagram
shows
the
features
found
crosses
the
asymptote
so
far
.
1
As
the
curve
y

only
once,
and
does
not
2
cross
the
vertical
minimum
asymptotes,
it
is
clear
that
the
point
(1,
0)
is
point.
1
We
can
also
deduce
that
y

when
x

0
2
1
and
y

0
when
x

2
We
now
check
for
other
stationary
points.
2
(x
2
1)
2(x
dy
_________
y

___
⇒
2
1)(2x
x)
(x
2
x(2x
1)
dx
x
(x
1)(3x
2
(2x
1)
______________

2
x
dy
___
1

dx
156
0
when
x

1)
1
(4x
1)
_________________________________

and
3
2
(2x
1)
1)
a
Section
3
Calculus
1
1
Therefore
y
has
stationary
points
where
x

1
and
where
x

3
2
2
(
)
3
________
1
When
x

1,
y

0
and
when
x

,
y

3

1
(
We
will
use
determine
(The
the
values
point
and
values
their
)
x
either
side
of
3
the
stationary
points
to
nature.
chosen
the
of
4
1
)(
3
must
curve
must
be
close
be
to
the
value
continuous
of
between
1
1
5
__
3
4
3
12
4
x
at
the
those
x
stationary
values.)
1
y

4
4

4

2
0
0

0
1
(
Therefore
,
4
)
is
a
maximum
point
and
(1,
0)
is
a
minimum
point.
3
We
of
x
could
for
also
check
which
information
y
to
is
the
range
increasing
sketch
the
of
values
and
that
decreasing,
y
can
but
take
we
and
now
the
have
values
enough
curve.
y
6
2
O
4
2
x
2
4
2
4
6
Exercise 3.14
Sketch
the
graphs
of
the
following
curves.
1
______
1
y

2
x
x
______
2
y

2

x
2
x
________
3
y

2
(2
x)
157
3.
15
Tangents
Learning outcomes
and
normals
Equations of tangents
and
normals
dy
___

To find
the
equations
of
tangents
We
know
that
represents
the
gradient
function
of
a
curve
whose
dx
and
normals
to
curves
including
equation
curves
given
whose
equations
is
y

f(x)
are
We
parametrically
can
the
therefore
tangent
When
x

to
a,
find
the
y

the
curve,
f(a)
gradient
at
and
any
the
of
the
given
curve,
point
gradient
at
on
that
and
the
hence
the
gradient
of
curve.
point
is
f (a).
You need to know
Therefore
(a,

How
to
differentiate
How
to
y
use
quotient
to
the
rule
product
and
the
is
equation
given
of
the
tangent
to
the
curve
y

f(x)
at
the
point
by
simple
functions

f(a))
the
rule,
chain
the
rule
f(a)
The
the

f(a)(x
nor mal
point
of
to
a)
a
curve
contact
of
is
a
the
line
perpendicular
tangent.
differentiate functions
Therefore
to
at
the
the
tangent
point
and
( a,
through
f(a))
the
1
____
gradient
of
the
normal
is
,
and
the
equation
of
the
normal
to
the
f(a)

The
relationship
between
the
curve
gradients

How
of
to find
perpendicular
the
equation
at
the
point
y

f(x)
at
the
point
( a,
f(a))
is
given
by
lines
of
a
1
____
y
f(a)


(x
a)
f(a)
straight
line
______
dy
1
________
___
For
example,
when
y
√

x

1,
dy
x

3,
2
√
x

1
1
__
___
When
______

dx

,
dx
so
y
the
y

14
4x
4
5
gradient
of
the
curve
at
the
4y

x

5
4
1
__
point
where
x

3
is
y
3
4
x

1
2
x
O
2
1
1
When
x

3,
y

2,
therefore
the
line
through
(3,
2)
with
gradient
is
a
4
tangent
to
the
curve
at
the
point
(3,
2).
1
The
equation
of
this
tangent
is
y
2

(x
3)
4
⇒
The
is
equation
y
⇒
of
the
2

4(x
y

14
normal
When
the
the
curve
dy
158
of
a
and
curve
expressed

dt

5
point
normals for
curves
whose
parametric
g(t)
dx
___
___

dx
be
x
4x
equation
can
dy
___
by
are
this

3)
Equations of tangents
equations
at
4y
____

dt
f(t)
as
is
given
(f( t),
as
g(t))
x
and

f(t)
the
and
y

gradient
g(t),
a
point
function
is
on
given
Section
Therefore
the
equation
of
a
tangent
to
the
curve
can
be
given
3
Calculus
1
as
g(t)
____
y
g(t)

(x
f(t))
f(t)
Then
any
particular
value
of
t
gives
the
equation
of
the
tangent
at
that
point.
Similarly,
the
equation
of
a
normal
to
the
curve
can
be
given
as
f(t)
____
y
g(t)


(x
f(t))
g(t)
Example
Find
the
equation
of
the
tangent
and
the
normal
to
the
curve

__
x

3 cos ,
y
4 sin ,

at
the
point
where


6
f( )
f()
⇒
so
the
3 cos 

3 sin 

equation
g( )
and
of
a
4 sin 

g()
and
tangent

to
4 cos 
this
curve
is
4 cos 
______
4 sin 
y


3 cos )
(x
3 sin 
(This
is
curve
at
a
general
any
equation;
point
on
the
it
gives
the
equation
of
a
tangent
to
the
curve.)

__
Therefore
the
equation
of
the
tangent
at
the
point
where


is
6
given
by
√
3
__
4(
)
2
4(
y
√
)

3(
(x

2
√
3
__
_____
1
√
4
3
___
))
⇒
y
2

3
3
___
(x

2
1
3(
)
3
2
)
2
The
equation
of
a
normal
to
this
curve
is
given
by
3 sin 
______
4 sin 
y

3 cos )
(x
4 cos 
(Again
any
this
point
is
on
a
general
the
equation
giving
the
equation
of
a
normal
at
curve.)

__
So
the
equation
of
the
normal
at
the
point

where

is
given
by
6
1
3(
)
2
y
4(
√
3
__
_____
1
)
3(
(x

2
4(
√
3
3
___
3
___
))
⇒
y
2
2
√
3
__
(x

√
4
)
2
3
)
2
Exercise 3.15
1
Find
the
equation
of
the
tangent
and
normal
to
the
curve
whose
2
_______
equation
is
y

at
(x
2
Find
the

equations
the
point
where
x

7
3)
of
the
tangent
and
y
1
normal
to
the
curve
whose
1
_____
equations
are
x

,
1


t

t
159
3.
16
Integration
Learning outcomes
Reversing differentiation
2
When

To
define
reverse
integration
of
as
x
is
differentiation
apply
basic
principles
with
respect
to
x
the
derivative
is
2x
and
Therefore
to
differentiated
the
of
when
the
derivative
of
an
unknown
function
is
2 x
then
the
2
unknown
function
could
be
x
integration
This
process
operation
of
of
finding
a
function
differentiating,
is
from
called
its
derivative,
which
reverses
the
integration
You need to know
The

The
differentials
of
constant of
integration
simple
2
functions
We
know
2
x
that
2x
is
the
derivative
of
x
,
but
it
is
also
the
derivative
of
2

5,
x
3
2
In
fact,
2x
Therefore
is
the
the
derivative
result
of
of
x

integrating
c,
where
2x
is
not
c
is
any
constant.
a
unique
function
but
is
of
2
the
c
form
is
x
called

c
the
where
c
constant
is
of
any
constant.
integration
2
x

∫2x
c
is
called
the
integral
of
2x
with
respect
to
x
and
is

where
x
∫…

function
means
any
f(x)
‘the
function
we
integral
reverses
of
the
…
with
process
respect
of
to
x’.
differentiating,
f(x) dx

f(x)

∫3x
differentiating
2
x

x
any
2
with
respect
3
dx
for
c
3
example,
so
have
∫
so
as
c
dx
Integrating
For
written
2
dx

c
and
it
follows
∫x
that
to
x
gives
1
2
dx

3x
3
x

c
3
1
(We
do
not
need
to
write
c
in
the
second
form,
as
c
represents
any
3
constant
in
either
expression.)
n
Now
we
know
1
______
n
∫
x
that
dx

n

the

Therefore
For
rule
can
example,

dx

1
n
is
(n

1)x
to
integrate
a
power
be
used
∫6x
to
integrate
6
__
3
dx
by

x
∫3x
of
the
any
3
__
4

c


3x
increase
that
power
by
1
and
power.
power
of
x
except
1.
4
x

c,
2
0
dx
x,
new

c
and
∫4x
4
___
3
dx

x
2
160
so
c
4
∫3

x
1
divide
This
of
1
x
n
derivative
2
2

c

2x

c
Section
3
Calculus
1
Exercise 3.16a
Find
the
(a)
∫5x
(b)
∫4x
following
integrals.
dx
(c)
∫6
(d)
∫4x
7
dx
Families of
dx
∫x
(f )
∫5x
3
dx
2
(e)
dx
6
dx
curves
dy
___
When

2x,
then
y

∫2x
2
dx

x

c
dx
2
Therefore
Each
The
the
value
graph
of
equation
c
gives
shows
a
y

x

different
some
c
represents
member
members
of
this
of
a
the
family
of
curves.
family.
family.
y
15
10
5
T
o
find
the
equation
the
curve.
For
example,
if
of
a
y

∫3x
y

x
x
O
4
particular
member
,
we
need
to
know
a
point
on
2
dx
3
then
If
c
we
also
and
know
hence
the
that

(2,
c
5)
equation
is
of
a
point
the
on
the
particular
curve
we
can
find
the
value
of
curve.
3
When
x

2
and
y

3,
y

x

c
⇒
3

8

c
⇒
c

5
3
Therefore
the
equation
of
the
curve
is
y

x

5
Exercise 3.16b
1
Find
the
which
2
Find
y
the
which
y
equation

∫6x
∫9x
the
curve
that
goes
through
(1,
5)
the
curve
that
goes
through
( 1,
and
for
dx
equation

of
2
of
2)
and
for
4
dx
161
3.
17
Integration
of
sums
and
differences
of
functions
Learning outcomes
Integration of
a
sum or difference of functions
dy

To
define
the
integral
of
___
sums
When
y

f(x)

g(x)
we
know
that

f(x)

g(x)
dx
and
differences
of functions
Therefore
For
You need to know
it
follows
example,
∫(cos
∫(f(x)
that
x

sin x) dx

g(x) ) dx

sin x

f(x)
cos x


g(x)

c

c
c
dy

The
differentials
of
___
simple
When
y

f(x)
g(x)
we
know
that

f(x)
g(x)
dx
functions

The
the

The
meaning
reverse
of
of
meaning
integration
as
Therefore
of
the
constant
of
How
to
solve
simultaneous
For
follows
example,
The
effect
curve
by
a
on
g(x) ) dx

f(x)
g(x)
a
pair
2
∫(2x
sin x) dx
2x) dx

x
x

x

cos x

c
and

c
of
equations
Integration of

∫(f(x)
that
2
∫(1
integration

it
differentiation
the
equation
of
translation parallel
a
multiple of
a function
a
to
dy
___
When
y

af(x)
we
know
that
the y-axis

af(x)
dx
∫af(x)
Therefore
For
example,
∫6x

af(x)
2
dx

c
∫3(2x

2
3
) dx
T
o
∫
(f(x)
Note

g(x) ) dx
∫(f(x)
that

/
∫
f(x) dx
g(x)) dx
2x

c
summarise,

is

∫
g(x) dx
not
equal
and
to
∫
(af(x) ) dx
∫f(x)
dx
Example
∫(x
Find
∫(x
3
4 cos x) dx
3
4 cos x) dx

∫x
1

3
4∫ cos x dx
dx
4
x
4 sin x

c
4
Exercise 3.17a
Find
162
the
following
integrals:
2
(a)
∫9x
(b)
∫x(3x
(c)
∫(5
dx
3
sin x
4) dx
(Hint:
6 cos x) dx
Multiply
out
the
bracket.)
/

a∫ f(x) dx
∫g(x)
dx
Section
3
Calculus
1
Example
2
d
y
___
The
equation
of
a
curve
is
such
that

6x
14.
T
wo
points
on
the
curve
have
coordinates
(1,
2)
and
2
dx
(
1,
12).
Find
the
equation
of
the
curve.
2
d
y
___
is
the
second
derivative
of
y
with
respect
to
x,
so
we
need
to
integrate
twice.
2
dx
dy
___
The
first
integral
gives
dx
2
d
y
dy
___
6x
14
y
∫(6x

14) dx
⇒
∫(3x

can
to
integrate
now
When
x
When
x
Solving
again
to
find
an
expression
3
[1]
use



1,
y
1,
y
and
c) dx
the
2,

[2]
y

y

14x

c
of
⇒
7x
the

points
2

6
12

8
simultaneously
for
y
so
we
need
to
introduce
another
unknown
constant.
2
x
⇒
12
3
Therefore
⇒
coordinates

3x
dx
2
14x
2

dx
need
We
___
⇒
dx
We
dy
___

2
gives
c
cx
on



d
the
curve
c

d,
i.e.
c

d,
i.e.
4
and
d

to
c
find

d
c
the

d
values
4

4
of
c
and
d
[1]
[2]
0
2
x
7x

4x
Example
dy
___
The
equations
of
a
family
of
curves
is
given
by

4x
3
dx
Sketch
the
graphs
of
two
members
of
this
y
family.
10
dy
8
___

4x
3
⇒
y

2
∫(4x
3) dx

2x
3x

c
dx
6
We
can
The
use
any
simplest
value
is
c

of
c
to
get
one
member
of
the
family.
4
0
2
This
gives
y

2x
3x

x(2x
3)
which
is
a
parabola
that
passes
2
3
through
O
and
(
,
0
)
2
Then
any
other
value
of
c
translates
the
curve
2
The
sketch
shows
y

a
curve
by
c
units
up
the
y-axis.
2
1
O
x
2
2x
3x
and
y

2x
3x

2
Exercise 3.17b
dy
___
1
The
equation
of
is
such
that

3 cos x
and
the
curve
dx

__
passes
through
the
point
,
(
4
)
.
2
Find
2
A
the
curve
equation
passes
of
the
through
curve.
the
points
(0,
1)
and
(1,
6).
The
equation
of
2
d
y
___
the
curve
is
such
that

6x
2
dx
Find
the
equation
of
the
curve.
163
3.
18
Integration
Learning outcomes
using
Integration
When

To
use
substitution
to
substitution
we
use
using
the
substitution
chain
rule
to
differentiate
a
composite
function
the
integrate
3
result
is
often
a
product
of
functions.
For
example,
when
y

4
(x
4x)
,
functions
3
then
with
u

4
x
4x
so
that
y

u
,
the
chain
rule
gives
dy
___
3

4u
2
2
(3x
4)

3
4(3x
3
4)(x
4x)
dx
You need to know
In

The
chain
rule
general,
dy
dy
___
The
differentials
of
simple
u

dx

f(x)
and
du
___
___


if

g(u)
then
using
the
chain
rule
gives
du
___

du
y
g(u)
dx
dx
functions
Therefore,

The
integrals
of
using
integration
as
the
reverse
of
differentiation
simple functions
du
___
∫g(u)
dx

g(u)

c
[1]
du

g(u)

c
[2]
dx
∫g(u)
Now
Comparing
[1]
and
[2]
gives
∫g(u)
Replacing
g(u)
by
f(u)
du

g(u)

dx

∫f(u)
dx

… du
c
gives
du
___
∫f(u)
du
dx
du
___
Therefore
…
dx
du
___
This
means
that
((a
integrating
function
of
)
u)
with
respect
to
x
is
dx
equivalent
to
integrating
(the
same
function
of
u)
with
respect
to
u.
du
___
This
means
that
the
relationship
...
dx

… du
is
not
an
equation
dx
an
identity
For
–
example,
it
is
to
a
pair
of
∫3x
find
equivalent
2
3
(x
operations.
4

4)
dx
we
can
use
the
substitution
3
u

x
This

4
gives
∫3x
2
3
(x
du
___
Then
as
4

4)
3x
,
...
dx
∴
∫3x
2
dx
∫3x

∫u
4
u
dx
2
dx

… du
becomes
4
1

... 3 x
dx

2
du
Since
5
u

c
5
1

3
(x
5
164
2
dx
4
u

du
___
2

dx
5

4)

c
...
3x
dx

...
du
... du
or
Section
3
Calculus
1
Example
Use
the
substitution
u
sin x
u

sin x
to
2
∫cos
find
x sin
x dx
du
___

⇒

cos x,
dx
du
___
∴
...
dx

… du
⇒
...
cos x dx

...
du
dx
Therefore
∫cos
2
x sin
x dx
∫u

2
du
1

3
u

c
3
1

3
sin
x

c
3
Example
______
Use
the
substitution
u

√
1

______
x
to
(
√
x
1
∫
find

)
x
dx
______
1
du
___
u

√
1

x
⇒
1

(1

2
2
x)
and
u
1

x
2
dx
1
du
___
∴
...
1
dx

… du
⇒
…
(1

2
x)
dx

...
du
2
dx
______
⇒
...dx

...2
√
1

x
du

…2u du
______
Hence
(
√
x
1
∫

x
)
dx

∫(u
2

2
1)(u)(2u) du
2
3
u
4
2
2u
) du

c
3
5
3
2

∫(2u
3
u
5

2
(1

x)
2
(1
5

x)
2

c
3
3
__
2

(1

x)
2
(3(1

x)
5)

c
15
________
__
2
3
2)√(1
(3x


x)

c
15
Exercise 3.18
1
Use
the
substitution
u

2x
2
Use
the
substitution
u

x
3
Use
the
substitution
u

x
4
Use
the
substitution
u

tan 
to
find
∫cos
2

1
4
to
to
to
find
find
find
2x dx
∫6x
∫(x
∫sec
2
(x
3

1)
dx
6

1)(x
2
4)
du
2

tan
 d
d
___
(Hint:
2
(tan
)

sec
)
d
165
3.
19
Calculus
Learning outcomes
and
examine
the
area
under
a
curve
Calculus
The
To

the
process
topics
in
Section
3
of
this
book
all
come
under
the
umbrella
name
of
of
calculus
finding
the
area
under
a
curve
Calculus
studies
You need to know
Our
The
The

meaning
of
definition
limits
of
from
differentiation
in
The
We
many
mathematics,
development
over
many
of
limits,
change
would
that
place.
electronic
look
involve
It
exploration
has
of
derivatives
when
the
very
rate
different
change
and
applications
space
devices.
to
the
Apart
and
of
in
to
integrals;
change
without
predict
almost
the
of
obvious
basically
calculus.
what
will
everything
development
from
it
varies.
the
It
we
tiny
is
used
happen
do
to
when
today,
microchips
scientific
uses,
calculus
principles
Did you know?
with
world
takes
the
covers
As
study
things
situations
change
found
from first
the
modern
model

is
how
branches
of
the
up
of
build
calculus
centuries,
but
to
the full
range
area
will
x-
disciplines
look
the
and
of
under
now
Consider
the
occurred
a
at
area
y-axes
a
A
in
as
economics
and
graphic
design.
curve
how
and
such
the
the
the
fundamental
diagram
vertical
below
line
theorem
enclosed
through
a
of
by
value
calculus
the
of
is
curve
derived.
y

f(x),
x
the
y
defining
step
Newton
in
was
made
by
Sir
Isaac
y
about
Gottfried Wilhelm
independently
but
about
8
the
after
contribution
that
to
symbols
introduce
f(x)
Leibniz
made
years

1665–1667
.
same
step,
Newton. The
Leibniz
made
that
was
are
dy
___
still
used
today
(including
and
∫
).
dx
A
Leibniz
is
considered
the founder
of
by
modern
many
to
be
mathematics.
x
O
We
can
strips
find
as
an
shown
approximate
in
the
next
value
for
this
area
by
dividing
it
diagram.
y
y
(x,

y)
y
δx
x
O
166
x
f(x)
into
vertical
Section
The
area
of
The
area
is
a
typical
then
strip
of
height
approximately
y
the
and
sum
xa
We
write
this
as
A

This
of
y x
∑
between
approximation
x
of
all
x
the
is
Calculus
1
y x
strips.
xa
y x
where
∑
x0
values
width
3
y x
means
the
sum
of
all
x0

0
and
improves
x
as

x
a
gets
smaller
,
so
we
can
write
xa
A

lim
x→0
y x
(∑
)
x0
Considered
putting
We
If
now
A
is
x,
x,
look
the
vertical
this
way,
together
at
area
line
gives
(i.e.
a
area
different
enclosed
through
a
the
under
integrating)
approach
by
the
a
the
point
corresponding
curve
different
to
curve
( x,
small
y)
y
is
the
finding

on
process
that
involves
elements.
f(x),
the
increase
A.
the
x-
curve,
in
A,
and
then
y-axes
a
small
and
the
increase
in
A
y
y
(x,

f(x)
y)
δA
A
y
δx
x
x
O
A
is
approximately
width
i.e.
equal
to
the
area
of
the
rectangle
of
height
y
and
x,
A

y x
A
___
∴

y
and
this
approximation
gets
better
as
x
→
0
x
dA
___
A
___
Now
lim
x→0

x
dx
dA
___
hence

y
dx
This
the
gives
the
connection
differential
of
A
with
between
respect
to
finding
x,
i.e.
A
A
as
is
a
the
summation
reverse
of
a
process
and
differential,
hence
A
This
is
called
the
fundamental

∫
y dx
theorem
of
calculus .
167
3.20
Definite
Learning outcomes
integration
Definite
We

To
define
and
compute
know
integration
that
the
area
between
a
curve
y

f(x),
the
x-
and
y-axes
and
the
definite
line
integration
through
a
value
of
x
is
given
∫
by
y dx,
i.e.
by
∫
f(x) dx
T
o
find
this
area
up
to
the
line
x

b,
we
find
∫
f(x) dx
and
substitute
b
T
o
find
this
area
up
to
the
line
x

a,
we
find
∫
f(x) dx
and
substitute
a
for
x
for
x
You need to know

How
to
integrate
simple
Then
the
x
is

b
area
the
between
difference
the
curve,
between
the
these
x-axis
two
and
the
lines
x

a
and
calculations.
functions
y

How
to
use
substitution
to
y

f(x)
integrate functions
A
a
O
2
For
example,
when
y

x
x
b

∫
2,
y dx
∫(x

1

2

2) dx
3
x

2x

c
3
1
Then
the
area
up
to
x

1
is
1
3
(1)

2(1)

c

2
3
1
and
the
area
up
to
x

2
is

c

c
3
2
3
(2)

2(2)

c

6
3
3
2
Therefore
and
x

the
2
between
1

c
the
curve
y

x

2,
the
x-axis
and
x

1
is
2
(6
area
)
1
(2
3

c
)

4
3
.
Notice
that
the
unknown
constant
disappears.
3
b
This
process
is
called
definite
integration .
It
is
written
∫
as
f(x) dx
and
it
a
means
the
integrand
definite
We
can
value
when
of
x
the

integrand
a.
The
when
values
a
x
and

b
b
minus
are
called
the
value
the
of
limits
the
of
the
integral
write
the
example
above
more
precisely
as
2
(x
2
1
2
∫

2) dx
[

8
3
x

2x
3
]

(
1

4
)
(
3
1
1

2
)

4
3
3
1
The
unknown
because
it
substituted
The
constant
cancels
and
second
out.
the
value
is
of
integration
The
order
then
square
in
which
does
brackets
they
subtracted
are
from
not
need
show
to
the
the
to
be
substituted
first
included
values
of
x
(top
to
one
be
first).
value.
b
∫
f(x) dx
can
be
found
only
when
f( x)
is
continuous
from
x

a
to
x

b
x

a
1
1
__
For
example
∫
(
1
168
1
__
)
x
dx
cannot
be
found
because
is
x
undefined
when
0
Section
3
Calculus
1
Example

2
∫
Evaluate
sin x dx
0


2
2
∫
sin x dx

[
cos x]
0
0

__

 cos
(
cos 0)
2

0
(
1)

1
Exercise 3.20a
Evaluate
the
following
definite
integrals:
2
1
3
∫
(a)
2
4x
dx
∫
(c)
(x
1
4
∫
(b)
3) dx
∫
(d)
general
need
x
(x
5) dx
1
Definite
not
dx
3
(2x
2
In
1)
1
to
when
we
substitute
need
given
integration
to
as
do
the
use
a
back
this
for
limits
to
substitution
substitution
to
a
using
give
the
definite
find
the
u

answer
integral
g(x)
in
to
find
terms
because
corresponding
of
we
an
x.
can
values
of
integral,
we
However
,
we
use
the
do
values
of
u
Example
2
3
Use
the
substitution
u

2
x

2
to
evaluate
∫
x
3
(x
4

2)
dx
0
3
u

x
2

{
and
2
⇒
… du

x

0
⇒
u

2
x

2
⇒
u

10
2
∫
dx
10
2
∴
… 3x
x
3
(x
1
4

2)
dx

∫
0
4
u
du
3
2
10
__
1

[
5
]
u
15
2
100 000
______
32
__
15
15

99 968
_____

15
Exercise 2.20b
3
6
1
Use
the
substitution
u

x

1
to
evaluate
∫
7(x

1)
dx
1

__
4
2
2
Use
the
substitution
u

sin x
to
evaluate
∫
cos
x(1

sin x)
dx
0
______
1
2
3
Use
the
substitution
u
2

1

x
to
evaluate
∫
x
√
1
2

x
dx
0
169
3.21
Area
under
Learning outcomes
a
curve
Calculating the
From
To

calculate
the
area
under
T
opics
3.19
area
and
3.20
under
we
a
curve
y
now
a
know
that
the
area
between
the
curve
curve
6
y

and
f(x),
x
the

b
x-axis
shown
and
in
the
the
lines
x
diagram

y
a

f(x)
is
5
b
the
You need to know
value
∫
of
f(x) dx
4
a
2
How

to
integrate
simple
functions
For
example,
the
x-axis
the
and
x
area

1
between
and
x

y
2

x
3
,
is
2
2
How

to
evaluate
a
definite
∫
by
2
1
2
given
x
dx

[
8
3
x
]
3
7
1


3
1
3
3
b
1
integral
1

f(x)dx
a
How

to
sketch
curves
x
O
a
b
1
Example
Find
the
area
The
curve
is
The
sketch
enclosed
a
by
parabola
shows
the
the
curve
which
area
cuts
y

the
required.
(It
(1
x)(2
x-axis
is
at

x
always
x)

and
2
the
and
sensible
to
x-axis.
x

y
1
draw
a
sketch.)
1
A

∫
(1
x)(2
(2
x

x) dx
2
1
1
2
=
∫
x
) dx

[2x
1
1
2
x
3
]
x
2
3
2
2

1
1
2
3
(2
8
(
)
4
2
)

3
1

4
2
3
2
1
x
O
1
The
area
is
4
square
units.
2
Exercise 3.21a
Find
the
area
lines
given.
1
y

4x
2
y

x
3
y

√
enclosed
by
each
of
the
following
curves,
the
following
curves
the
x-axis
and
3
,
x

0,
x

2
2
Find

x,
the
1,
x
area

170
y

1
5
y

x(1

1,
1,
x

enclosed
2
4
x
x
x)
x

1
4
by
each
of
and
the
x-axis.
the
Section
The
area
between
a
curve
and the
3
Calculus
1
y-axis
2
The
the
We
diagram
line
y
can
First
This

shows
the
area
between
two
different
the
curve
y

x

1,
the
y-axis
and
10.
find
this
area
in
ways.
method
uses
the
bounding
fact
that
rectangle
the
minus
area
the
required
area
is
equal
between
the
to
the
curve
area
and
of
the
the
y
x-axis.
10
When
y

10,
x

3
so
the
area
shown
is
equal
to
8
(area
of
the
rectangle
bounded
by
x
0,
y

0,
x

3
and
y

10)
6
(area
between
the
curve,
the
x-axis,
x

0
and
x

3)
3
4
2
Therefore
the
area
required

∫
30
(x

1) dx
0
2
3
1

[
30
3
x

x
]
3
0
x
O
1
2
3
4
3
4
27
__

{(
30

3
)
(0)
}

30
12

18
3
The
area
is
Second
This
square
units.
method
uses
Consider
The
18
area
a
a
of
direct
approach.
horizontal
this
strip
strip
is
y
of
length
x
and
approximately
x y
width
y
10
x
8
δy
y10
The
area
required

lim
y→0
(
x y
∑
y
6
)
y1
4
10
We
know
that
this
limit
is
equal
to
∫
x dy
1
2
10
Now
∫
x dy
means
integrate
x
with
respect
to
y
so
we
need
to
find
x
in
terms
x
O
1
of
1
2
y
______
2
From
y

x

1,
x

√y
1
______
10
9
3
1
9
2
so
the
required
area
is
∫
(√y
1
) dy
∫
u
2
du

[
3
u
2
]

18
square
units
Using
the
substitution
u  y
1
0
0
1
Exercise 3.21b
2
1
Find
the
y-axis
area
and
enclosed
the
line
y

by
the
curve
3
y

√
x,
the
2
Find
the
y-axis
y
area
and
enclosed
the
line
y
by
the
curve
y

x
,
the
4
y
10
4
8
6
2
4
2
O
x
2
4
6
8
10
O
x
2
171
3.22
Area
two
below
the
x-axis
The
area
between
below the
To
calculate
below
To

the
areas
of
The
areas
between
a
curve
and the
x-axis that
is
axis
curves
x-axis
calculate
area
curves
Learning outcomes

and
definite
y
integral
2
between
2
1
3
∫
curves
x
dx
4
[

]
x
4
1
1
3
1

4

3
4
This
can
be
4
interpreted
as
the
area
You need to know
3
between
the
curve
y

x
,
the
x-axis
How

to
integrate
How
The
to find
the
lines
x
area
below
definite
and
above
1
and
x

1
2
1
2
2
a
integral
1
the
∫
x-axis
1
1
3
curve

simple
functions

the
x
O
2
and
x
dx
4
[

x
]
4
2
2
3
1
How

to
sketch
curves

4

3
4
How

to find
the
points
If
intersection
of
two
4
of
we
look
at
the
diagram,
the
y
area
curves
3
between
and
is
the
the
curve
lines
equal
to
x
the

y

2
area
x
,
and
the
x
found
x-axis

1
above,
3
i.e.
3
square
units.
4
(The
curve
about
has
rotational
symmetry
O.)
x
O
2
1
3
But
the
∫
integral
x
dx
is
negative.
2
This
is
gives
because
the
the
length
of
value
a
of
y
vertical
that
strip
is
b
negative,
so
∫
y dx
will
be
negative
a
when
This
and
y
is
negative
means
the
you
x-axis
for
need
when
a
to

x
be
part

b
careful
of
the
when
area
is
finding
below
the
the
area
between
a
curve
x-axis.
Example
Find
the
First
draw
is
below
area
a
the
Therefore
between
sketch:
x-axis
we
need
the
The
and
to
curve
sketch
the
find
y
3x(x
shows
area
each

that
between
area
2),
x
the

2
the
x-axis
area
and
and
between
x

4
is
the
x

lines
0
above
and
the
x

0
x

2
and
x

4
y
x-axis.
separately.
2
2
A
∫

(3x
6x) dx

[
3
2
2
x
]
3x

(8
12)
0

4
0
0
4
2
B

∫
(3x
6x) dx

[
3
2
x
3x
B
4
]

(64
48)
(8
12)

20
2
2
O
x
1
Therefore
the
area
required

20

4

24
square
2
units.
A
4
2
Note
that
∫
0
172
(3x
6x) dx
gives
the
value
of
the
area
of
B
minus
the
area
of
A
3
4
5
Section
3
Calculus
1
Exercise 3.22a
1
(a)
Sketch
(b)
Find
the
the
curve
area
y

x(x
enclosed
1)(x
between
2).
this
curve
and
the
x-axis.
2
2
(a)
Sketch
(b)
Find
Area
When
a
curve
area
y
you
need
an
which
idea
are
to
This
find
will
where
shown
x
the
area
any
points
the
1.
by
this
curve
and
the
x-axis.
curves
show
the
in

enclosed
between two
diagram.
give
the
the
next
between
areas
of
two
that
curves,
may
intersection
be
are.
you
should
below
the
There
are
draw
x-axis
two
and
methods,
example.
Example
2
Find
the
area
between
the
curves
y

x
and
y

3x(2
x)
y
4
2
The
curves
intersect
where
x

3x(2
6x

0
3)

0
x)
3
2
⇒
4x
2
1
⇒
x(2x
⇒
x

0
or
x

1
1
2
x
1
First
1
2
3
method
1
The
area
between
y

3x(2
x),
the
x-axis
and
x

1
2
1
1
1
1
2
2
2
is
given
by
∫
(6x
3x
) dx

[
2
3
3x
x
]
27
___
27
___
4
8
0
area

between
y

x
8
1
2
The
27
___

0
0
,
the
x-axis
and
x

1
is
given
by
2
1
1
1
1
2
1
2
∫
x
dx

[
9
__
2
3
x
3
]
Therefore
Second
the
9
__

0
0

8
0
area
8
between
the
curves
27
___
9
__
8
8
is
9
__

square
units
4
method
2
A
vertical
strip
in
the
area
has
length
(y

6x
3x
2
)
(y

x
2
)

6x
4x
1
1
1
1
2
2
Therefore
the
area
between
the
curves
is
∫
(6x
4x
2
) dx
4
[3x

0
2
3
x
3
]
0
27
___
9
__
4
2

9
__
0

square
units
4
Exercise 3.22b
2
1
Find
the
area
between
2
Find
the
area
enclosed
by
the
y-axis
enclosed
by
the
curve
the
curves
y

2

x
and
y

2

y

x
x
3
and
the
curves
and
3
y
3

16
Find
y


the
x
area
y

(x

2)(x

2)
and
the
line
3x
173
3.23
Volumes
Learning outcomes

To
calculate
when
round
part
the
the
of
a
of
revolution
Volume of
revolution
volume formed
curve
is
rotated
x-axis
When
an
object
formed
area
is
rotated
about
a
solid
of
straight
line,
the
three-dimensional
You need to know

How
to
integrate
simple
The
volume
is
called
formed
is
a
called
a
revolution
volume
of
revolution
functions
The

How
to
evaluate
a
line
about
which
rotation
takes
place
is
an
axis
of
symmetry
of
definite
the
solid
of
revolution.
integral
All

The formula for
the volume
of
cross-sections
of
the
solid
that
are
perpendicular
to
the
axis
of
a
rotation
are
circular.
cylinder
The
is
diagram
rotated
shows
the
completely
solid
about
of
the
revolution
formed
y

shaded
area
f(x)
x
O
calculate
the
y
y
T
o
when
x-axis.
the
perpendicular
volume
to
the
of
axis
this
of
x
O
solid
we
divide
it
into
‘slices’
rotation.
y
x
O
When
the
When
one
to
this
first,
The
174
cut
the
cuts
cut
is
are
is
V,
of
together
,
through
and
volume
volume
close
of
this
then
this
slice
the
point
another
slice
is
a
each
is
slice
P( x,
cut
is
y)
is
and
made
approximately
small
approximately
increase,
the
at
a
V,
a
volume
distance
of
a
cylinder
.
the
x
cylinder
.
of
the
volume
solid
from
V
up
the
Section
This
‘cylinder ’
has
radius
y
and
depth
3
Calculus
1
x
y
P(x,
y)
P(x,
y)
y
y
x
O
δx
δx
2
Therefore
V

 y

 y
V
___
x
2
⇒
x
This
approximation
V
___
i.e.

x
→
⇒
0,
2
 y

dx
V
the
rotated
 y
x
Therefore
as
dV
___
2
lim
x→0
When
improves
=
area
∫
2
 y
dx
between
completely
a
cur ve
about
the
y

x-axis
f(x),
the
the
x-axis,
volume
of
x

the
a
and
solid
x

b
for med
is
is
b
given
by
the
definite
integral
V
2
∫

 y
dx
a
Example
___
Find
the
the
volume
x-axis
and
generated
the
line
x
when

2
is
the
area
rotated
between
completely
the
curve
about
the
y
y
√

3x ,
x-axis.
4
3
When
we
integrate
with
respect
to
x,
the
limits
of
the
integration
must
2
be
values
of
x
1
2
___
2
2
2
V

∫
 y
dx


∫
0
(√
)
3x
dx
O
1
0

∫
2
3
2
3

x
1
1
2
3x dx


[
2
x
]
2

 (6
0)

6
2
0
0
3
Therefore
the
volume
Note
values
required
is
6
cubic
units.
4
that
for
volumes
of
revolution
are
usually
given
in
terms
of

Exercise 3.23
2
1
The
area
enclosed
completely
about
by
the
the
curve
x-axis.
y
Find

4
the

x
and
volume
the
of
x-axis
the
solid
is
rotated
generated.
2
2
The
x

area
1
volume
enclosed
and
of
x
the

1
by
is
solid
the
curve
rotated
y

x
,
the
completely
x-axis
about
and
the
the
x-axis.
lines
Find
the
generated.
175
3.24
More
volumes
Learning outcomes
Rotation
When
To find

the
volume
a
section
rotated
about
of
a
the
an
revolution
about the
area
is
rotated
y-axis
about
the
y-axis,
the
volume
formed
is
calculated
generated
in
when
of
curve
a
similar
way
to
rotation
about
the
x-axis.
is
y-axis
y
To find

when
curves
or
the
the
the
is
volume
area
generated
between
rotated
about
two
the
x-axis
y-axis
x
P(x,
δy
y)
You need to know
O
How

to
integrate
x
simple
functions

How
to
sketch

How
to find
curves
T
wo
intersection
the
of
points
two
of
slices
cylinder
parallel
of
radius
to
x
the
and
x-axis,
depth
a
distance
y
apart,
form
an
approximate
y
curves
2
The
The formula for

the volume
of
volume
of
this
cylinder
is
 x
y.
a
V
___
cone
∴
dV
___
2

 x
⇒
2

y
 x
⇒
V
∫

2
x
y
dy
Therefore
between
the
y

volume
a
and
y
generated

b
is
when
rotated
the
part
of
completely
the
cur ve
about
the
y

y-axis
f(x)
is
b
2
given
by
V

∫
 x
dy
a
Note
to
y,
that
so
must
∫…dy
the
be
means
function
values
of
to
that
be
the
integration
integrated
must
has
be
in
to
be
done
terms
of
y
with
and
respect
the
limits
y
Example
2
The
region
Find
the
defined
volume
of
by
the
the
inequalities
solid
y

x

2,
2

y

3
is
rotated
about
the
generated.
y-axis.
y
y
2
The
equation
of
the
curve
is
y

x

2
3
3
3
2
2
V
∫

 x
y
2
dy
with
y

x


x

2
2,
2
2
2
i.e.
x

y
2,
gives
3
O
3
1
V


∫
(y
2) dy

2
[
y
2y
2
]
2
2
9


{(
1
6
)
(2
4)
}
2

=
2
1
Therefore
the
volume
generated

is
2
176
cubic
units.
x
O
x
Section
3
Calculus
1
Exercise 3.24a
2
1
Find
the
the
volume
x-axis
is
generated
rotated
when
completely
the
area
about
enclosed
the
by
y

9
y

1
x
and
y-axis.
3
2
The
region
enclosed
completely
about
Rotation of
When
an
formed
area
has
subtracting
volume
i.e.
if
a
between
the

f(x)
is

two
,
the
Find
curves
We
formed
rotation
the
x
y-axis
the
and
volume
between two
section.
volume
by
y
y-axis.
area
hollow
formed
y
an
by
the
of
by
the
equation
of
is
rotated
can
find
outer
the
line
about
an
axis,
volume
of
the
of
inner
the
this
solid
solid
curve
by
from
the
outer
curve
and
y

g(x)
is
the
2
of
the
inner
b
curve,
then
the
volume
between
them
is
given
by
b
2
∫
rotated
curve,
1
equation

is
curves
the
rotation
the
generated.
y
2
∫

dx
1
a
y
dx
where
a
and
b
are
the
values
of
x
where
the
2
a
curves
intersect.
y
y
O
O
An
x
x
alternative
method
may
simplify
the
working.
This
involves
just
one
integral.
A
slice
(An
If,
through
annulus
for
a
is
value
the
the
of
x,
solid
area
y
is
gives
a
shape
between
the
two
whose
cross-section
concentric
y-coordinate
of
a
is
an
annulus.
circles.)
point
on
one
of
the
curves
1
and
y
the
corresponding
point
on
the
other
curve,
where
2
>
y
1
2
the
y
area
of
the
cross-section
is
,
y
then
2
1
y
2
(y
y
1
)
2
2
x
2
The
volume
of
a
slice
of
thickness
x
is
then
2
(y
y
1
2
Therefore
V
=
2
 (y
V
___
2

) x
y
1
⇒
2
)
y
1
⇒
V
This
2
 (y
)
y
1
2
dx
=
is
2

2
x
∴
δx
dV
___
2
 (y
) x
2
∫ (y
2
2
) x
y
1
useful
2
when
the
equations
of
the
curves
are
similar
.
For
example,
2
1
__
when
y

1
__

1
and
1
y

x
Each
,
2
then
problem
A
sketch
y
1

2
x
should
of
the
be
(
1
__

1
)
(
x
assessed
cur ves
2
1
__
2
y
2
to
deter mine
involved
will
help
2
__
)
x
the
you
best
do


1
x
method.
this.
177
Example
Find
the
volume
of
the
solid
formed
when
the
area
enclosed
by
the
______
curve
y
about
First

the
find
√
x
1
and
the
line
2y

x
1
is
rotated
completely
x-axis.
where
the
curve
and
line
intersect.
______
1
√
x
1

(x
1)
2
1
⇒
x
1
2

(x
1)
4
2
⇒
(x
1)
⇒
(x
1)(x
1
⇒
(x
1)(x
5)

⇒
x

5
When

1
x

Therefore
Next
4(x
or
1,
x
y
the
sketch

1)
4)
0

0
0
0
and
graphs
the

when
intersect
x

at
5,
(1,
y
0)

2
and
(5,
2)
y
graphs.
3
The
hollow
rotating
about
the
the
the
line
x-axis
volume
found
section
of
by
completely
is
the
without
formed
a
cone,
cone
and
can
2
be
integration.
1
x
O
1
1
2
3
4
5
1
This
cone
has
base
radius
2
units
and
height
4
units.
16
__
Therefore
the
volume
of
the
hollow
cone

is
3
1
(using
the
formula
V
2
 r

h
)
3
______
The
volume
between
given
x
generated

1
and
x
when

5
is
the
section
rotated
of
the
completely
curve
y
about
the
by
5
5
1
V


∫
(x
1) dx


[
2
x
x
]
2
1
1
25
__


1
{(
5
)
2
8
)}

8
8

3
1
2
16
__
Now
(

=
3
8
Therefore
the
required
volume

is
3
178
cubic
units.

√
x
1
x-axis
is
6
Section
3
Calculus
1
Example
The
by
diagram
the
line
y
shows

x
the

1,
area
the
y
enclosed
curve
6
______
y
√

x
1,
the
x-
and
y-axes
and
the
5
line
x

4
4
Find
the
volume
generated
when
this
3
area
is
rotated
completely
about
the
2
x-axis.
1
This
volume
needs
to
be
calculated
in
x
O
1
two
are
by
separate
calculations
different
the
for
rotation
volume
the
of
volume
the
generated
as
by
line
the
the
limits
1
3
1
generated
and
the
rotation
of
the
curve.
The
x

volume
0
and
x
generated

4
is
when
rotated
4
section
the
of
the
x-axis
line
y

x

1
between
is
4
2
∫

the
about
(x

1)
2
dx

∫

0
(x

2x

1) dx
0
4
1
__
3
[


x
2

x

x
]
3
0
1
__

3
(

(4)
124 
_____
2

(4)

4
0
)

3
3
______
The
volume
between
x
generated

1
and
x
when

4
______
4
is
the
section
rotated
of
about
the
the
curve
x-axis
y
√

x
1
is
4
4
1
__
2

∫
(√
x
1
)
dx


∫
(x
1) dx

[

1
__


[(
the
124 
_____
9
___
3
2
volume
of
]
1
1
__
2
(4)
4
)
(
2
Therefore
x
2
1
1
2
x
9
___
2
(1)
1
)]

2
the
solid
formed
2
is
221 
_____
=
cubic
units.
6
Exercise 3.24b
2
1
Find
and
the
the
volume
line
y

generated
3x
is
when
rotated
the
area
completely
between
about
2
2
3
The
area

between
and
x
2
(a)
Draw
is
a
the
rotated
sketch
curves
about
y
the
showing

x
line
the
curve
y

x
x-axis.
2
,
y
x-axis.
the
the
y

x
Find

x

the

1
1
and
the
volume
and
the
lines
x

0
generated.
curve
3
y
(b)

The
x

(x

area
1
is
1)
between
enclosed
rotated
by
x

the
1
and
line
and
completely
about
x

the
the
2
curve,
x-axis.
between
Find
the
x

0
and
volume
generated.
179
3.25
Forming
Learning outcomes
differential
Differential
equations
equations
2
dy
d
___
Any

To
use
differential
equation
with
terms
involving
,
model
situations
involving
and
so
on,
is
called
a
2
dx
to
y
____
,
equations
differential
dx
equation
change
dy
___
An
equation
involving
only
terms
in
is
called
a
first
order
differential
dx
equation.
dy
___
You need to know
For
example
x

y

2
is
a
first
order
differential
equation.
dx
2

The
meaning
of
rates
of
d
change
y
____
If
an
equation
involves
,
it
is
called
a
second
order
differential
2
dx

The
relationship
between
equation.
quantities
that
are
proportional
Rates of
increase
dy
___
W
e
know
that
represents
the
rate
at
which
y
increases
with
respect
to
x
dx
When
the
varying
value
of
a
quantity
P
depends
on
the
change
in
another
dP
___
quantity
Q,
then
the
rate
of
increase
of
P
with
respect
to
Q
is
dQ
Such
changes
expands
and
the
occur
when
it
is
frequently
heated.
temperature
is
T,
in
everyday
When
then
the
the
life,
volume
rate
at
for
of
a
which
example,
quantity
the
liquid
of
volume
liquid
of
the
is
V
liquid
dV
___
increases
with
respect
to
changing
temperature
can
be
modelled
by
dT
Another
the
example
number
,
n,
is
of
the
profit,
books
sold
P,
made
(among
by
a
other
bookseller
.
factors).
So
This
the
depends
rate
at
on
which
dP
___
profit
increases
as
n
changes
can
be
modelled
as
dn
Formation of differential
The
If
motion
you
are
velocity
of
and
If
velocity
an
particle
with
with
object
is
to
through
a
to
modelled
you
where
respect
respect
falls
often
mechanics
acceleration,
displacement
of
a
studying
equations
will
be
velocity
time
and
by
a
differential
familiar
is
the
with
rate
of
acceleration
is
equation.
displacement,
change
the
of
rate
of
change
time.
medium
that
causes
its
velocity
v
to
decrease
dv
___
with
respect
to
time
at
a
rate
that
is
proportional
to
its
velocity,
then
dt
dv
___
measures
the
rate
of
increase
with
respect
to
time,
so
is
negative.
dt
dv
___
As
is
proportional
to
v,
we
can
model
this
movement
with
the
dt
dv
___
differential
equation

dt
180
kv
where
k
is
a
constant
of
proportionality.
Section
3
Calculus
1
ds
___
As
v
is
the
rate
of
change
of
the
displacement,
s,

v,
we
can
also
dt
2
d
s
___
model
this
movement
with
the
equation

kv
2
dt
Note
we
that
when
assume
we
that
are
the
told
change
the
is
rate
with
at
which
respect
to
a
quantity
time
is
unless
changing,
we
are
told
otherwise.
Example
Form
a
differential
W
ater
is
depth
of
in
the
leaking
water
equation
from
is
a
to
model
cylindrical
decreasing
is
the
tank
following
such
proportional
that
to
information.
the
the
rate
at
which
volume
of
water
the
left
tank.
dh
___
The
rate
at
which
the
depth,
h,
is
changing
is
dt
2
The
volume
of
water
in
the
tank
 r
is
h
dh
___
is
negative
as
h
is
decreasing,
dt
dh
___
2
∴
 r
∝
h
dt

and
r
are
both
constants
so
we
can
write
this
equation
as
dh
___

kh
dt
Exercise 3.25
Form
1
a
differential
When
bacteria
number
that
2
3
A
of
its
to
s.
moving
C
grown
is
in
a
displacement,
rate
at
increases
Grain
is
volume,
volume
which
at
difference
4
are
cells
to
model
in
a
the
culture,
proportional
to
following
the
the
rate
data.
of
number
increase
of
cells
of
the
present
at
time.
body
The
of
equation
a
rate
being
of
s,
cereal
its
drained
grain
grain
from
with
between
V,
of
a
straight
in
line
a
fixed
crop
respect
final
a
remaining
in
the
is
time
H,
hopper
.
hopper
so
point
grows
to
height,
from
the
moves
is
that
is
such
The
rate
inversely
which
and
the
its
that
is
change
proportional
its
height,
proportional
present
rate
inversely
of
of
h cm,
to
the
height.
change
of
proportional
the
to
the
hopper
.
181
3.26
Solving
Learning outcomes
differential
Solving differential
Solving

To
solve
differential
equations
a
differential
equations
equation
means
finding
a
direct
relationship
equations
between
the
variables.
dy
___
For
example,
the
general
solution
of
the
differential
equation

2x
dx
You need to know
2
is

How
to
integrate
simple
T
o
functions

How
of
to find
x
the
constant
integration from
given
information

How
to
use
y

x
get
and

a
unique
solution,
we
need
a
pair
of
corresponding
values
of
y
Differential
solution
equations
will
additional
substitution
c
involve
pieces
of
often
two
involve
unknown
information
to
a
constant
constants.
get
a
of
In
unique
proportionality,
this
case
we
so
need
the
two
solution.
to
ds
___
For
example,
given
the
differential
equation

integrate
kt,
and
that
s

1
when
dt
t

0
and
that
s

ds
___
6
when
1

kt
⇒
s
t

10,
then
2

kt

c
2
dt
s

1
when
t

0
gives
s

6
when
t

10
1

c
__
1
gives
6

50k

1
⇒
k

10
__
1
∴
s

2
t

1
10
We
also
second
need
two
additional
derivative
integrations
are
involved
needed,
pieces
in
each
the
of
of
information
differential
which
will
when
equation.
introduce
a
there
In
this
is
a
case
two
constant.
2
d
y
____
For
example,
if
2

3x
then
integrating
once
gives
2
dx
dy
___
3

x

c
dx
Integrating
1
y

again
gives
4
x

cx

k
that
y
4
If
we
this
If
know
gives
we
also
5

know

4
5

that
when
2c

y

1
c

k
x

2,
k
[1]
when
x

1,
1
we
have
1


[2]
4
1
Solving
[1]
and
[2]
simultaneously
gives
c

1
and
k

4
2
2
d
y
____
Therefore
the
solution
of
the
equation
2

3x
is
2
dx
1
y

1
4
x

4
1
x

4
2
4
⇒

T
o
solve
do
not
to
182
4y
the
a
x

x

2
differential
need
to
equation,
understand
differential
you
every
equation.
need
detail
to
know
about
the
how
to
integrate
situation
that
it.
gives
Y
ou
rise
Section
3
Calculus
1
Example
A
body
moves
so
that
at
time
t
seconds
its
displacement,
s
metres
2
d
s
___
from
a
fixed
point
O
is
modelled
by
ds
___
√

t.
When
t

0,

5
and
2
dt
s

2.
value
Find
of
2
s
the
direct
predicted
relationship
by
ds
___
t
model
between
when
t
s

and
t.
Hence
find
the
4
3
1
d
s
___

this
dt
2
⇒
2

t
2
dt
2

c
3
dt
ds
___
When
t

0,

5,
∴
c

5
dt
3
ds
___
t
t

0,
s

2

5
⇒
s

3
dt
When
5
__
4
2

t
2

5t

k
15
2,
∴
k

2
5
__
4
Hence
s

t
2

5t

2
15
128
___
When
t

4,
s
8
__


22

30
15
15
8
__
The
model
predicts
that
s

30
when
t

4
15
Exercise 3.26a
1
The
rate
of
change
of
a
quantity
dr
___
with
respect
2.
Find
r

to
is
given
by

__
3 sin .

When


,
d
2
r
r

in
terms
of
.
3
The
variation
of
a
quantity
P
with
respect
to
r
is
modelled
by
the
2
d
P
____
differential
2
equation

12r

6r.
It
is
known
that
when
model
predict
that
the
r

1,
2
dr
dP
___
P

6
and

1.
What
does
this
value
of
dr
P
will
be
when
Integration
Many
r
by
differential

3?
separating the variables
equations
used
to
model
situations
are
of
the
form
dy
___

f(y).
We
cannot
integrate
f( y)
with
respect
to
x,
so
we
need
to
change
dx
the
form
of
the
differential
equation.
du
___
We
know
from
T
opic
3.18
that
∫
f(u)
dx
=
∫
f(u) du,
where
u
is
a
dx
function
of
x.
dy
___
Therefore
∫
f(y)
dx
=
∫
f(y) dy
dx
dy
___
This
means
that
integrating
((a
function
of
)
y)
with
respect
to
x,
is
dx
equivalent
to
integrating
(the
same
dy
___
Now
we
can
write
=
f(y)
dx
then
∫(
function
___
f(y)
dx
y)
with
respect
to
y.
dy
1
____
___
f(y)
dx
as
=
dy
1
____
of
1
1
____
)
dx
=
∫
1dx
becomes
∫(
f(y)
)
dy
=
∫
1 dx
183
This
is
called
integration
have
effectively
by
separating
the
variables
dy
in
going
from

f(y)
∫(
to
dx
is
to
gather
containing
all
x
the
on
terms
the
containing
other
side,
i.e.
y
on
we
what
we
1
____
___
done
because
one
have
side
f(y)
and
)
all
‘separated’
dy
=
the
the
∫
1 dx
terms
numerator
dy
___
and
denominator
of
dx
dr
___
For
example,
=
,
dt
∫
r dr
∫
=
2 dt
then
multiplying
by
r
gives
r
r
1
so
dr
___
2
__
given
⇒
=
2
dt
2
r

2t

c
2
Example
The
is
atoms
in
modelled
at
any
a
as
given
radioactive
inversely
time,
t,
material
are
proportional
measured
in
to
days.
disintegrating
the
number
Initially
of
there
at
a
rate
atoms
are
N
that
present
atoms
present.
(a)
Form
and
solve
a
differential
equation
to
represent
this
information.
(b)
Half
the
model
mass
disintegrates
predicts
that
it
will
in
200
take
days.
for
Find
how
three-quarters
long
of
the
the
mass
to
disintegrate.
(a)
If
n
is
rate
the
of
number
change
dn
___
of
n
of
atoms
is
=
⇒
dt
n

n
∫
Therefore
at
any
given
time,
then
the
dn
___
k
__
∴
present
negative,
k
where
k
is
a
constant.
dt
n dn
1
⇒

 ∫ k dt

kt
2
n

c
2
1
Initially,
i.e.
when
t

0,
n

N,
∴
2
N
=
c
2
1
hence
1
2
n

kt
2

N
2
2
1
(b)
When
t

200,
n

N
2
1
2
1
(
∴
2
N
200k
1

200k


2
N
2
3
⇒
)
2
3
____
2
N
so
k

8
2
N
1600
1
i.e.
2
n


3
____
1
2
N
t

2
2
N
2
1600
3
When
1
of
the
mass
has
disintegrated,
n

4
1
2
2
1
(
⇒
N
4
N
)
3
____

4

1
2
N
t

__
1
2
N
⇒
2
1600
3
____


32
⇒
t

1
t

2
1600
250
3
The
model
predicts
that
it
will
take
250
days
for
of
4
disintegrate.
184
the
mass
to
Section
3
Calculus
1
Example
The
rate
of
modelled
people
being
already
person
(a)
increase
as
was
Form
in
the
number
,
proportional
infected.
Nine
to
n,
the
people
of
people
square
were
infected
root
of
infected
the
5
by
a
virus
number
days
after
is
of
the
first
infected.
and
solve
a
differential
equation
to
represent
this
information.
(b)
How
will
many
take
days,
for
to
100
the
nearest
people
to
be
day,
does
the
model
predict
that
it
infected?
1
dn
___
(a)
When
n
people
are
infected,

kn
2
where
k
is
a
constant.
dt
1
∫
∴
When
(This
1
2
n
t
dn

is
first
∴
c

0,
not
the
∫

n
k dt

2n
2

kt

c
1
given
person
⇒
explicitly
is
but
the
days
are
counted
from
when
infected.)
2
4
When
t

5,
n

9
⇒
6

5k

2
⇒
k

5
1
4
∴
2n
2

t

2
5
4
(b)
When
n

100,
20

t

2
5
45
__
⇒
t

2
The
model
100
people
predicts
to
be
it
will
take
approximately
23
days
for
infected.
Exercise 3.26b
1
1
The
velocity,
v m s
,
of
a
ball
rolling
along
the
ground
is
such
that,
_
1

t
seconds
t

a
2
0
and
direct
W
ater
it
that
v
is

dripping
a
dripping,
2
at
when
a
damp
radius
which
v
t
of
the
tap
dv
3,
between
from
circular
the
rate
∫
started,
relationship
forming
The
after
2
v
on
patch.
the
k dt.
solve
and
to
a
the
that
v
differential
concrete
hours
patch
r cm,
Given

5
when
equation
to
give
t.
T
wo
damp
radius,
∫

of
was
the
surface
after
the
where
tap
it
is
started
20 cm.
damp
patch
is
increasing
is
1
__
modelled
as
being
proportional
to
.
r
(a)
Form
and
number
(b)
How
for
3
Grain
long,
the
is
conical
solve
of
differential
equation
elapsed
after
to
nearest
hour
,
the
radius
pouring
pile
a
hours
of
the
from
whose
a
damp
h
is
on
tap
does
patch
hopper
height
the
to
to
giving
starts
the
model
reach
a
barn
increasing
at
r
to
a
in
terms
of
t,
the
drip.
predict
it
will
take
1 m?
floor
rate
where
that
is
it
forms
a
inversely
3
proportional
doubles
height
after
has
to
a
h
.
The
time
grown
to
T.
initial
Find,
height
in
of
terms
the
of
T,
pile
the
is
2 m
time
and
after
the
height
which
its
6 m.
185
Section
1
3
Practice
questions
9
Find:
A
spherical
balloon
is
losing
air
at
the
rate
of
3
h
___________
(a)
_____
lim
√
h→0

0.5 cm
per
second.
__
2
Find
√

2
the
balloon
x  2
___________
rate
of
when
change
the
of
radius
the
is
radius
of
the
20 cm.
lim
(b)
2
x
x→2

2x

_
4
8
(The
volume
of
a
sphere
of
radius
r
3
r
is
.
)
3
2
(a)
The
function
f
is
given
by
10
Find
the
range
of
values
of
x
for
which
the
4x
________
2
x
f(x)
{


1,
x

function
2,
2x,
x

∈
lim
f(x)
and
by
f( x)

Show
increasing.
x)
that
2
f(x).
x
_________

x→2
x→2

2,
lim

is
(2

11
Find
given
2
x
y

2
(3x
Hence
explain
continuous
(b)
Repeat
at
whether
x

or
not
f( x)
has
2
one
stationary
value
Find
the
stationary
3x
{

1,
x

3
x

3
y
x

1,
∈

3x

the
curve
2
4x

6x

12x

5

and
from
first
principles:
13
distinguish
The
curve
value
sin 2x
__
(b)
it.
2
x
Differentiate
(a)
find
on
3
between
3
3
and
points
4

3)
for
(a)
12
f(x)

is
of
when
x
4
y


2
ax
when
them.

x

b
has
x

0
and
a
of
a,
b
and
c.
a
maximum
minimum
value
c
2.
√
x
Find
the
values
dy
___
4
Find
when:
dx
14
(a)
y

(2x

1)(3x
(b)
y

4 sin x
(c)
y


Sketch
the
curve
whose
2)
equation
is
4x
________
y

2
(2

2
x
 2x  1
___________
x

Given
that
y

(Y
ou
can
Find
the
use
your
results
equation
of
x sin x
through
d
find
the
point
the
,
(
10.)
curve
which
goes
1
)
and
for
which
2
y
and
.
y
2

∫(5
cos
) d 
dx
16
Find
question
____
dx
6
from

__
2
dy
___
x)
1
15
5

3 cos x
f(x)
when
f(x)
A
curve
passes
through
the
points
(0,
1)
and
is:
(1,
1).
The
equation
of
the
curve
is
such
that
x
______
(a)
2
d
2
x

y
____
1

______
4

6x
2
dx
(b)
x
√
x

1
Find
the
equation
of
the
curve.
x
_____
(c)
sin x
______
(d)
√
17
2
x

(a)
Use
∫

__
(e)
cos
2x
(

2x

(

sin 2x
u

x
to
find
cos 2x sin
2x dx
2
3
(b)
Use
∫
___
in
u
2
)
dy
Find
substitution

__
sin
)
3
7
the
1
terms
of
t
the
substitution
2
6x(x

1
to
find
4

1)
dx
when:
dx
18
2
(a)
y

2t,
x

(b)
y

3 cos t,
t

Evaluate:
2t

4
x

4

(a)
5 sin t
∫
(3x

4) dx
(b)
∫
2
8
The
equation
of
a
curve
2 cos  d

__
2
is
2
19
(a)
Use
the
substitution
u

x

1
to
evaluate
3
y

3(x

5)
(
∫
A
point
is
increasing
moving
along
the
at
the
constant
Find
the
rate
curve
rate
______
2
of
so
that
0.2 cm
x
is
per
x
√
2
x

1
)
dx
1
(b)
Use
the
substitution
u

__
second.
of
change
of
y
when
x
is
4
3
∫
1.5 cm.
186
0
cos 
(sin 

1)
d

sin 
to
evaluate
Section
20
Find
(a)
the
area
enclosed
by
the
The
(b)
curve
initial
radius
of
3
the
Practice
balloon
questions
was
10 cm.
2
y

4
x
and
the
T
en
x-axis.
radius,
3
Find
(b)
the
the
area
y-axis
enclosed
and
the
by
lines
the
y

curve
1
and
y
y


x
seconds
after
r cm,
of
air
the
started
balloon
to
escape,
was
,
How
2
long
balloon
will
is
it
be
before
the
radius
Sketch
(a)
the
curve
(The
of
the
2 cm?
4
21
the
5 cm.
volume
of
a
sphere
3
 r
is
.)
3
y
Find
(b)
the

area
(x

1)(x
enclosed
1)(x
by
this

2)
curve
and
the
28
Given
that
x-axis.
y

3 cos 2x,
2
d
y
____
22
Find
the
area
enclosed
by
the
find
curves
in
terms
of
y
2
dx
2
y

2
x

1
and
y

5
x
__
44
2
23
Find
the
volume
generated
when
the
29
area
The
(
point
)
,
3
is
a
point
of
inflexion
on
the
9
3
enclosed
x-axis
by
and
the
the
curve
line
x
y


1
x
is

1,
the
rotated
y-axis,
curve
the
completely
3
y
about
the
Find
Give
your
answer
in
terms
Find
(a)
the
equation
of
the
tangent
to
the

x

2
at
the
point
on
the
curve

a
sketch
x-
the
and
to
show
curve
the
and
the
equation
area
tangent
point
∫

4
where
x
of
the

1
tangent
to
the
curve
at
dx

10
where
a

0.
Find
the
value
of
a
2
x
a
between
the
bx
1
__
2
Draw
(b)

1
where
30
x
6x
curve
2
y
2
ax

of
the
24

x-axis.
enclosed
in
part
(a)
and
31
Solve
the
differential
equation
y-axes.
2
d
y
____
Find
(c)
the
volume
generated
when
the

area
6x

4

3
2
dx
described
in
part
is
(b)
rotated
completely
dy
___
about:
given
that,
when
x

0,
and
y

9
dx
(i)
the
x-axis
(ii)
the
y-axis.
32
Find
the
between
25
Solve
the
differential
volume
the
generated
when
the
area
curves
equation
2
y

2
x
and
y

8
x
dy
___
2

6y
is
dx
given
that
y

3
when
x

rotated
completely
about
the
x-axis.
1
y
33
26
Solve
the
differential
equation
dy
x
__
___

dx
given
that
y

3
y
when
x

2
x
O
27
Air
is
escaping
from
a
spherical
balloon
2
that
is
proportional
to
V
at
a
rate
3
,
where
V cm
is
the
The
volume
of
the
diagram
shows
the
area
enclosed
by
the
balloon.
2
curve
(a)
Use
the
information
above
to
form
y
where
has
t
elapsed
escape.
equation
seconds
is
from
in
the
terms
time
when
the
of
in
air
V
x

2x

2,
the
x-
and
y-axes
and
the
a
tangent
differential

and
seconds
started
to
the
curve
at
the
point
where
x

2
t
(a)
Find
this
(b)
Find
the
area.
that
to
rotated
volume
generated
completely
about
when
the
this
area
is
x-axis.
187
Index
0
128,
circumference
129
64
0
clockwise
rotation
codomain
A
acceleration
a cos 
acute
angles
addition
algebra
9,
110,
115
propositions
a
lines
triangle
between
annulus
110,
of
a
curve
between
two
under
curves
below
x-axis
and
curves
compound
angle
identities
compound
statements
associative
operations
31,
of
axes,
and
57,
axioms
(z)
154,
minor
symmetry
third
8,
98,
sections
proportionality
10
base
of
52
a
logarithm
changing
bearings
55
6
bi-conditional
bijective
binary
Boole,
52
statements
functions
George
14
algebra
14
168
128,
124,
180,
182,
15
geometry
in
three
graph
70,
90–1
dimensions
105–7,
68
68
82
cosine
formula
cosine
function
cosine
graphs
72
64,
66,
cotangent
function
cotangent
graphs
65–6
67,
86
68,
69
69
examples
(solids)
19
109
cubic
curves
cubic
equations
31
27–9
curves
area
C
c
(constant
of
calculators
calculus
integration)
53,
Cardano,
families
81
of
a
circle
92–3,
of
a
curve
96–7
of
an
of
a
line
of
a
parabola
of
a
plane
102,
114,
form
Cartesian
unit
rule
circles
64,
equation
188
103,
deduction
92–3,
of
106
144,
130
94,
93,
98
97
30–1,
174,
37,
of
175,
133
51
38,
39,
154–7
32–5
176
terminating
decreasing
111
138–41,
101,
decimals,
100
vectors
132,
D
103
115
120–1
Cartesian
chords
cylinders
97
96–7
163
reflected
transformation
92
173
33,
130–1,
of
sketching
31,
161,
of
equations
167
27
equations
ellipse
of
two
of
of
gradient
theorem
Girolamo
Cartesian
between
equations
168
166
fundamental
chain
160,
164
functions
integrals
definite
integration
denominator
169,
13,
rationalising
function
second
12
146
18
definite
derived
184
125
15
cosecant
cubes
8
160,
(k)
functions
function
counter
49
operations
Boolean
15–16
(c)
cosecant
cosine
15
17
integration
coordinates
B
80
134
of
coordinate
174
105
18
base
logical
of
converse
76–7,
98–9
contrapositive
156
78
14
120,
continuous
102
72–5,
178
constants
9
72–5,
15
conjunction
17
164
16
statements
connectors,
170–1
138–41,
formulae
172–3
connectors
major
171
173
166–8,
associative
axes,
y-axis
9
46–7
angle
conic
between
asymptotes
functions
17
8,
compound
cones
98
area
axes
operations
conditional
Perga
53
commutative
conclusion
111
177
Apollonius
logarithms
differentiating
18
50
connectors
17
90
vectors
64
49,
commutative
composite
10
64
between
of
80–1
90,
8,
of
angles
common
180
+b sin 
48,
172,
175
168–9
36,
39
13
(derivative)
152–3
132
108–9
Index
differential
equations
differentiation
of
composite
of
a
functions
constant
parametric
from
first
180–5
132–5
138–41,
164
of
quotients
54
functions
56,
expressions
6,
of
132–3
functions
of
136
functions
factor
formulae
factor
theorem
factorising
137
78–9
23
24–5
n
reversing
table
factors
160
142–3
of
y

ax
of
y

ax
of
y

f(x)
of
y

x
57
36–7
F
principles
products
equations
exponential
134
142
of
exponential
of
families
134
first
n
a
of
order

b
24–5
curves
161,
differential
163
equations
180
n
focus
134

g(x)
135
point
fractions,
99,
100,
improper
102
36
n
direct
134
proof
directrix
18–19
100,
discriminant
of
a
181,
vectors
point
from
connectors
operations
domain
44–5,
dominoes
double
9,
10,
47,
183
105,
distributive
8,
a
derived
106
line
91
17
48,
of
exponential
56,
124
modulus
onto
types
e
(eccentricity
e
(irrational
of
an
ellipse)
number)
52,
102
56
58–9
48,
49,
50
48,
trigonometric,
E
126–7
56–7
subjective
76–7
124–5,
57
146–7
50–1,
logarithmic
50,
identities
132
discontinuity
inverse
36–7
125
146
increasing
9
20
angle
124,
decreasing
distributive
division
46–7
continuous
180,
displacement
49
composite
14–15
displacement
44–7
bijective
102
26
disjunction
distance
functions
of
49,
50
reciprocal
68–9
48–51
fundamental
theorem
of
calculus
167
53
x
e
G
57
ellipses
98,
99,
Cartesian
equal
a
42–3,
a
diameter
102,
logarithmic
normals
of
a
158,
parabola
of
a
quadratic
of
a
of
equivalent
Euler
52
130–1,
a
straight
of
a
tangent
of
a
132,
90,
130–1,
133
144
144
68
sine
158–9
67,
86
69
function
secant
101
142,
line
66,
50–1
68
65,
straight
curve
66,
67,
line
tangent
66,
87
30
67,
88
H
51
h
7,
94,
112–13,
equivalence
182
132–3
of
cosine
116–17
159
26–7
tangents
vector
112–15,
26
simultaneous
of
equations
90–1
71
reflected
roots
90,
120–1
polynomial
curve
cotangent
100,
96–7,
plane
a
cosecant
42–3,
differential
coordinate
graphs
103
54–5
parametric
solution,
function
163
54
30,
of
161,
95
ellipse
line
96–7,
180–5
exponential
134–5
142–3
gradient
97
of
of
a
general
92–3,
curve
of
differentiation
table
27–9
differential
99
general
geometry,
a
an
103
60–1
of
of
102,
92
circle
cubic
of
107
7,
Cartesian
of
equations
vectors
equations
Galileo
102
42–3,
158,
60
159
118–20
(small
Hubble
increase
telescope
hyperbola
in)
130
99
98
hypotheses
15
17
operations
164
I
identity
7,
26
189
Index
compound
double
law
angle
angle
72–5,
image
47,
implications
82–3,
by
improper
fractions
36
130,
131,
functions
136,
137
infinity,
61,
injective
points
of
integers
150,
48,
151,
49,
50,
152,
definite
laws
of
long
172,
a
difference
of
a
multiple
of
a
sum
of
of
functions
160
using
substitution
162
162
164–5
163
power
variables
of
angles
x
of
183–4
160
a
triangle
18
intersection
curves
between
and
two
lines
lines
116,
of
planes
and
inverse
inverse
42–3
graphs
of
connectors
division
functions
irrational
numbers
44,
induction
mathematical
operators
maximum
turning
maximum
values
turning
minimum
values
models
183,
modulus
functions
modulus
signs
modulus
of
of
indices
laws
of
logarithms
e
52,
Gottfried
length
104
(surds)
Newton,
12–13
128,
180,
182,
non-parallel
190
150,
151,
152,
80–1
58
8,
9,
10
to
184
a
a
52,
53
Sir
Isaac
52,
lines
166
116,
117
158
curve
94
of
plane
158,
118,
13,
159
120
36,
39
6
O
obtuse
53
Wilhelm
Omar
166
angles
onto
128–9
30,
113–14,
144
116,
116,
117
117
115,
116–17
90,
Khayyam
one-way
126–7
90,
logarithms
14
functions
functions
operators,
orbits
origin
of
110,
8,
48,
49,
50
164
mathematical
120
49,
34–5
48,
9,
planets
119,
115
98
stretches
operations
intersecting
152,
53
equations
lines
of
151,
99
153,
157
80–1
58–9
negation
one-to - one
of
31,
60–1
53
52
Leibniz,
gradient
8
150,
185
John
99
laws
equations
184,
x
numerator
theorems
30,
(Napierian)
mathematics
limit
56
20–1
6,
points
natural
L
notation
30,
minimum
to
limit
51,
points
Napier
,
proportionality)
Johannes
of
50,
mathematical
normals
language
46,
56
K
of
45,
50
non-parallel
(constant
17
N
98
50–1,
number
Kepler
,
53
52
36–7
multiplication
60
15
irrational
k
52,
numbers
117
cones
10–11,
(Napierian)
negative
reverse
162
functions
functions
separating
53
53
M
rule
twice
56
55
53
of
mapping
of
54–5
56–7
175
168–9
of
52–3,
52,
natural
153
56
160–5
definite
of
function
logical
169,
102
logarithmic
167
160
integration
interior
154,
10
integrals
92,
evaluating
126,
112–13
180
point
common
112
of
equations
176
function
equations
of
120
117
logarithmic
38–41
119,
104
dimensions
logarithms
146–7
approaching
inflexion,
a
120
118,
of
heating
of
base
quadratic
a
loci
36–7
52
inequalities
116
planes
116,
three
liquid
18–19
expressions
indices
in
vector
improper
increasing
skew
96–7
20–1
48
small
to
segments
70–1
proof
45,
increase,
116–17
114,
perpendicular
10–11
trigonometric
induction,
of
parallel
parallel
Pythagorean
of
pairs
80
17
members
by
76–7,
76–7
6,
8
50,
56
153,
156,
157
Index
reciprocal
P
pairs
of
lines
parabolas
30,
Cartesian
tangent
parabolic
98,
99,
to
42,
parallel
lines
parallel
vectors
parameter
100–1,
equations
of
114,
108,
of
an
of
a
line
ellipse
of
a
parabola
perpendicular
vectors
vector
point
to
points
of
120
by
of
the
y-axis
176–7
of
product
scalar
quantities
function
order
graphs
150,
151,
152,
153
sines,
26
36,
106,
109,
112
48,
solid
50
equations
of
propositions
curves
20–1
72,
identities
line
straight
lines
stretches,
105
subjective
70–1
using
quadratic
quotient
u
71
inequalities
62–3,
sum
38–41
122–3,
of
154–7
131,
136,
137
of
15
in
three
148–53,
30,
42–3,
dimensions
155,
90,
156–7
112–15,
116–17
112
34–5
functions
48,
49,
50
7
139,
8,
sines
164–5
140,
9,
164–5,
169,
171
10
78
12–13
6–9,
symmetry
142
15–16
integration
symbols
137,
39,
145
one-way
138,
surds
186–7
21
rule
38,
174–5
equations
subtraction
26–7
questions
quotient
130,
values/points
straight
Q
of
37,
16
substitution
roots
in
volume
stationary
equations
87
117
revolution
conditional
20–1
58
theorem
67,
30–1,
116,
compound
17
Pythagorean
60
statements
140
14–17,
of
Pythagoras’
quadratic
66,
bi-conditional
x
42–3,
78
increase
of
7,
64–5
65,
lines
small
20
47,
skew
180
82
sum
sketching
39
105,
70,
equations
10–11
simultaneous
118–20
152–3
differential
closed
115
68–9
second
18–19
of
177–9
104
derivative
sets,
120–1
110–11,
second
function
induction
algebra
about
sphere,
direct
properties
178
scalar
180
proof,
curves
x-axis
sine
136,
174–9
the
sine
44–5,
rule
between
91
22,
of
64
area
sine
vectors
160
174–5
about
secant
111
109
profit
an
99
integers
product
proof
158–9
120
110,
equations
pre-image
prism
118,
inflexion
polynomials
positive
91,
equations
line
polynomial
position
142,
120
equation
a
volume
of
175
orbits
of
S
lines
planets’
differentiation
revolution,
142
101
perpendicular
Cartesian
22
116–17
distance
118,
21,
26
solid
103
113–14,
68–9
rotation
110
96–7,
perpendicular
64,
theorem
roots
clockwise
equations
functions
50–1
revolution,
96
parametric
planes
repeated
116
107,
33,
remainder
reversing
differentiation
()
170
99
parametric
pi
102,
100
101
mirrors
trigonometric
reflection
116–17
axis
14,
30,
of
16,
31,
52,
167
102
174
R
radians
64,
radius
range
93,
45,
T
129
radioactivity
tangents
184
94
47,
48
rate
of
change
144–5,
rate
of
decrease
180
rate
of
increase
144,
rational
expressions
rational
numbers
rationalising
real
numbers
the
180,
circles
to
curves
to
180
36–7,
10–11,
graph
13
44–5,
49
of
148
94,
158,
130–1,
42,
telescopes
64,
66,
159
144
101
(trigonometric)
function
13
denominator
130,
of
parabolas
tangents
39–41
158
94–5
equations
181
gradient
12,
8–9,
(lines)
to
70,
82,
90
66
67,
88
99
191
Index
terminating
terms
theorem
three
decimals
12
equal
of
calculus
167
straight
lines
transformation
translations
triangles,
parallel
in
105–7,
in
of
32,
interior
34,
18
functions,
trigonometric
identities
tables
values
turning
see
of
unit
reciprocal
82–3,
68–9
vertex
virus
31,
150–3
minimum,
stationary
points
48–51
U
u
(substitution)
unit
vectors
110,
106,
111
109,
112
104
of
106
108–9
180,
181
100
infection
185
volume
maximum,
function
110
121
of
106,
velocity
96–7
14–15
30,
108,
105,
subtraction
14–17
points
also
position
84–9
106
106
plane
properties
angles
trigonometric
a
of
107,
perpendicular
32–5
59
equations
truth
to
112
curves
trigonometric
truth
108–9
105,
107
magnitude
dimensions
coordinates
types
displacement
6
138,
139,
140,
108–9
164–5,
169,
of
a
cone
of
a
cylinder
178
of
revolution
of
a
solid
of
a
sphere
175,
176
174–9
175,
176,
178,
179
145
171
X
x-axis,
rotation
about
178
about
176–7
n
V
x
variables
VDU
7,
182,
18
183–4
137
Y
x
vector
equations
of
a
line
of
a
plane
vectors
192
of

a
y-axis,
56
rotation
118–20
104–9
addition
angle
y
112–13
Z
105,
between
106,
two
107
110,
111
z
(third
axis)
105
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