1
Working
with
functions
HSC Outcomes
A student
develops understanding and fluency in mathematics through exploring and connecting mathematical concepts, choosing
and applying mathematical techniques to solve problems, and communicating their thinking and reasoning coherently
and clearly
applies algebraic techniques and the laws of indices and surds to manipulate expressions and solve problems
uses functions and relations to model, analyse and solve problems
In this chapter,
1.1 Real numbers ................................................... 6
1.2 Quadratic equations ......................................... 9
1.3 Properties of real functions ............................ 11
1.4 Basic real function graphs ............................. 15
1.5 More real function graphs .............................. 20
1.6 Piecewise-defined functions .......................... 26
1.7 Region and Inequalities ................................. 28
1.8 Review Exercise 1 ......................................... 30
Solutions ........................................................... 33
Mathematics Advanced Years 11 + 12
6
One of the most important aspects of any science is the establishment of relationships among
variables. For example, a medical researcher might investigate the effect of a new drug on blood
pressure; an educational psychologist might study the relationship between teaching time and learning
time. This chapter deals with common functions defined in terms of real numbers.
1.1 Real numbers
1.1.1 The real number line
Real numbers are a mathematical abstraction commonly used when modelling real phenomena. Real
numbers include all rational numbers, i.e., numbers in the form
, where and
p
pq
q
are integers,
0,q ≠
and irrational numbers such as
2, 3,
π
, etc. A real number may be represented by a point on the real
number line.
Example 1.1
Plot
2
in its exact place on the real number line.
By Pythagoras theorem
22
2 11= +
, we can plot
2
this way: First draw an isosceles right-angled
triangle OAB, where O and A are points 0 and 1 respectively. The arc with centre O, radius OB would
cut the number line at
2
.
1.1.2 Understanding basic formulae
Three basic and very important formulae are the identities
22 2
() 2a b a ab b±=± +
and the Pythagoras
theorem
2 22
cab= +
, where c is the hypotenuse of the right-angled triangle of sides a, b and c.
One way to understand them is using geometric means.
Referring to the diagram in the left, clearly the area of the largest square of side
(a + b) is equal to the sum of the areas of two smaller squares of side a and side b
and two rectangles of sides a and b.
Thus,
2 22
() 2a b a b ab+ =++
Referring to the diagram in the left, the area of the largest square of side (a + b)
is equal to the sum of the areas of the central square of side c and 4 triangles of
legs a and b, noting the diagram is not to scale.
Thus,
22
1
() 4
2
a b c ab+ = +×
22 2
222
22
.
a b ab c ab
abc
++ =+
∴+=
−1
0 1 2
O A
B
a
a b
b
b
c
c
a
a b
b
a
b
a
c
c
Chapter 1: Working with functions
7
1.1.3 Index laws
( )
.
.
.
m n mn
m
mn
n
n
m mn
aaa
a
a
a
aa
+
−
×=
=
=
Example 1.2
Simplify
(a)
2
3
3
6
−
−
. (b)
43
36
9
−
×
. (c)
1
2
4
n
n+
. (d)
( )
2
2
2
4
n
n
.
(a)
2 3 33
3
32 2
3 6 32
3 2 3 8 24
63 3
−
−
×
= = =× =×=
. (b)
43 4 4
233 53 3
36 3 3 1 1
9 332323224
−
×
= = = =
×× × ×
.
(c)
( )
1
1 22 2
2
22 2 2 1
4 2 2 4 42
2
nn n n
n
n nn n
+
++
= = = =
××
. (d)
( )
2
2 22
2
2
2
24
4
4 44
n
nn
nn
n nn
−
= = =
.
1.1.3 Quadratic surds
Example 1.3
(a) Simplify
(i)
8 18+
. (ii)
1
52+
.
(b) Prove that
41
2 5 9 45
−
+−
is a rational number.
(a) (i)
8 18 2 2 3 2 5 2+=+=
.
(ii)
( )( )
1 52 52
52
54
52
52 52
−−
= = = −
−
+
+−
.
(b)
( )
( )( )
( )( )
42 5
4 1 9 45
2 5 9 45
2 5 2 5 9 45 9 45
−
+
−= −
+−
+− − +
( )
42 5
9 45
4 5 81 80
−
+
= −
−−
( ) ( )
42 5 9 4 5=− − −+
17, which is a rational number.= −
Exercise 1.1
1 Plot
3 and 5
in their exact positions on the real number line.
2 Use a geometric means to prove that
(a)
2 22
() 2a b a b ab− =+−
. (b)
22
( )( )a b a ba b−=+ −
.
Mathematics Advanced Years 11 + 12
8
3 Prove the Pythagoras theorem by using similar triangles.
1
4 Simplify the following. Leave your answers in positive index form or surd form.
(a)
33
xx
−
×
. (b)
3
3
x
x
−
. (c)
( )
2
34
xx×
. (d)
( )
2
3
3
x
.
(e)
( )
1
8
2
a
a
−
.
(f)
( )
3
4
9
3
a
a
−
.
(g)
( )
( )
1
4
1
2
2
16
4
a
a
−
−
. (h)
( )
( )
3
2
4
2
6
3
81
64
a
a
−
.
(i)
2
8
n
n
. (j)
1
1
2
4
n
n
+
−
. (k)
12
2
29
6
nn
n
−−
+
×
. (l)
2
12
12
43
n
nn
−
−+
×
.
(m)
1
1
24
42
n
n
+
−
×
×
. (n)
2
2
23
6
nn
n
−
−
×
. (o)
( )
1
2
1
4 18
6
nn
n
−
+
×
. (p)
4
2
10 25
5
n
n
−
+
×
.
5 By raising the following to a common power, arrange the following in ascending order.
(a)
6
4
2, 3, 6
. (b)
35
2,3,5
. (c)
22
11
33
22
25,34××
. (d)
432
555
544
,,
333
.
6 Simplify, where possible.
(a)
8 50+
. (b)
90 40−
. (c)
75 27−
. (d)
18 12−
.
(e)
12 48 98+−
. (f)
98 200 20+−
. (g)
3 5 10 20++
. (h)
2 3 3 2 108+−
.
(i)
( )
3 26−
. (j)
( )
38 6−
. (k)
( )
2 6 2 10+
. (l)
( )
5 15 5−
.
(m)
10 40
2
+
. (n)
24 54
3
+
. (o)
4 12
2
+
. (p)
3 27
6
−
.
(q)
( )
2
23 2+
. (r)
( )
2
5 22−
. (s)
( )
2
36 2+
. (t)
( )
2
22 33−
.
7 Express with rational denominators, then simplify, where possible.
(a)
11
32
+
. (b)
11
52
−
. (c)
11
25 32
+
. (d)
22
35
+
.
(e)
3
32+
. (f)
5
51+
. (g)
6
62+
. (h)
23
52−
.
(i)
1
32 1+
. (j)
8
23 2+
. (k)
31
21
+
−
. (l)
31
31
+
−
.
(m)
11
21 31
+
++
. (n)
12
2 31 2
+
++
. (o)
15
35 35
−
+−
. (p)
31
31 21
+
+−
.
(q)
3 2 31
2 32 3
+−
−
−+
. (r)
( )
2
1
65−
. (s)
( )
2
5
52−
. (t)
( )
2
21
21
+
−
.
8 Prove that the following are rational numbers.
1
Professor Elisha Scott Loomis (1852‒1940) presents 367 proofs of the Pythagoras theorem in his book, The
Pythagorean Proposition, published in 1940, republished in 1968 by the National Council of Teachers of
Mathematics in the United States of America.
Chapter 1: Working with functions
9
(a)
21
3 2 22 1
+
+−
. (b)
2 11
3 36 3
+
−+
. (c)
36
3 2 31
+
+−
. (d)
14
4 5 57
−
++
.
(e)
1 32
, where
32
aa
a
−
+=
+
. (f)
2
2
1
, where 1 2aa
a
+=−
.
9 Simplify, leaving your answers in factored form.
(a)
33
22
2
a b ab
a ab b
−
++
. (b)
2
2
23 1
xx
xx x
−
+− −
. (c)
32
23
1mm m
x x xx
++
÷
−+
. (d)
3
22
1
2
mm m
xxxx
−+
÷
+ −−
.
10 Evaluate the following.
(a)
11
, where
1
21
x
x
x
−
=
+
−
. (b)
2
2
11
, where
1
12
x
x
x
−
=
+
+
.
1.2 Quadratic equations
Example 1.4
Solve the following quadratic equations by using all available methods, where possible: factorising,
quadratic formula and completing the square.
(a)
2
2 30xx− −=
. (b)
2
4 30xx− −=
. (c)
2
60xx−−=
.
(a)
2
2 30xx− −=
.
2
2
by factorising.
Using
.
the quadratic form
2 3 ( 3)( 1) 0,
1 or 3.
4
, where 1, 2, 3,
2
2 4 4(1)(
l
3)
24
1 or 3
2
u
2
a,
xx x x
x
b b ac
x ab c
a
x
− −= − + =
∴=−
−± −
= ==−=−
±− −
±
= = = −
( )
( )
2
2
2 22
2
2 3.
2 1 3 1, since 2 .
1 4.
1 2.
2 1.
1 or 3.
By completing the square,
xx
x x x ax a x a
x
x
x
x
−=
− +=+ − + = −
−=
−=±
=±+
= −
(b)
2
4 30xx− −=
.
Impossible to factorise.
4 16 12
2
2 7, which approximately are 0.65 or 4.65.
Using the quadratic formula,
x
±+
=
=±−
Mathematics Advanced Years 11 + 12
10
2
.
By completing the squa
4
re,
3xx−=
2
2
4 4 3 4.
( 2) 7.
2 7.
2 7.
xx
x
x
x
− +=+
−=
−=±
= ±
2
2
2
2
a
(
by factorising,
using the quadr tic
6 ( 3)( 2),
( 3)( 2) 0.
2 or 3.
1 1 4( 6)
,
2
1 25
2
15
2
2 or 3, as
formula,
By completing the squar
above.
6.
11 1
6.
24 4
1
,
c)
2
0
e
6.
xx x x
xx
x
x
xx
xx
xx
x
−−= − +
− +=
∴=−
± −−
=
±
=
±
=
= −
−=
− +=+
−
−−=
2
25
.
4
15
.
22
2 or 3, as above.
x
x
=
−=±
= −
In the quadratic formula
2
2
4
, the value of 4 , denoted by
2
b b ac
x b ac
a
−± −
= −∆
, is called the
discriminant, because it “discriminates” different types of roots (also called solutions or zeros) of the
quadratic equation, without the need to solve it.
2
If 0 then the quadratic equation has 2 distinct real roots.
Further, if is a perfect square, then the equation has 2 distinct rational roots.
If 0 then the quadratic equation has 1 repeated real
∆>
∆
∆= root (equal roots).
If 0 then the quadratic equation has no real roots (or 2 complex roots).∆<
2
These terms are often used interchangeable, but there is a subtle distinction. Solution is used for an equation
f(x) = 0 (i.e., the value of x in an equation that makes the statement true), and root is also the solution of an
equation, but typically used for polynomials, while zero is used for a function f(x). Zero is used for the values of
x that makes the function equal to zero.
Therefore, finding the zeros of a function f(x) is equivalent to finding the solutions of the equation f(x) = 0.
Chapter 1: Working with functions
11
Example 1.5
Find k if the equation
2
(2 3) 4 4 0k x kx+ − +=
has equal roots.
The equation has equal roots if
2
( 4 ) 4(2 3)(4) 0kk∆= − − + =
.
2
2
16 16(2 3) 0
2 30
( 3)( 1) 0
1 or 3.
kk
kk
kk
k
− +=
− −=
− +=
∴=−
Exercise 1.2
1 Solve the following quadratic equations by using all available methods, where possible: (
α
)
factorising, (
β
) quadratic formula and (
γ
) completing the square.
(a)
2
2 30xx+ −=
. (b)
2
6 90xx+ +=
. (c)
2
6 50xx− +=
. (d)
2
3 12 36 0xx− −=
.
(e)
2
3 4 60xx+ −=
. (f)
2
3 10xx+ −=
. (g)
2
40xx+−=
. (h)
2
3 2 40xx+ −=
.
(i)
2
3 5 20xx− +=
. (j)
2
53 2 0xx+− =
. (k)
2
11
60
32
xx+ −=
.
2 If the quadratic equations
22
( ) and ( ) ( 3) 1fx x x fx ax bx c= − = + + +−
have the same graph, find
the values of a, b and c.
3 Determine the nature of the roots (i.e., state whether there are none, one or two roots, real, rational
or complex) of the following equations.
(a)
2
5 6 0.xx+ −=
(b)
2
12 4 3 0.xx+− =
(c)
2
5 3 2 0.xx− +=
(d)
2
2 5 7 0.xx− + −=
(e)
2
6 9 0.xx+ +=
(f)
2
9 6 1 0.xx+ +=
(g)
2
3 8 4 0.xx− +=
(h)
2
4 6 12 0.xx+−=
4
Find the values of k for which the roots of the equation
2
60x xk+ +=
are
(i) equal, (ii) real, (iii) real and distinct, (iv) complex.
5 Repeat question 4 for the following equations.
(a)
2
( 3) 4 4 0kxx+ − +=
. (b)
2
( 5) 4 0k x xk− + −=
.
(c)
2
3 20kx x k+− −=
. (d)
2
( 3) 3 0x k xk+ − −+=
6
Prove that
2
10x kx k− + −=
, where k is a rational number, always has rational roots.
7 Prove that
2
2 (2 3 ) 3 0px p q x q+ + +=
, where p and q are rational numbers, always has rational
roots. What can you conclude if 2p = 3q?
1.3 Properties of real functions
1.3.1 Definition of a function
A function is a rule that assigns each value of one variable to one value of another variable, such that
the value of the second variable is uniquely determined by the value of the first.
For a function f, this means that to each element x in a given set of real numbers, there is assigned
exactly one real number y. This y is called the value of the function f at x, and the dependence of y on f
and x is explicitly denoted by y = f(x).
Mathematics Advanced Years 11 + 12
12
The set of real numbers x on which f is defined is called the domain of f, while the set of values f(x)
obtained as x varies over the domain of f is called the range or image of f.
The variable x is called the
independent variable since it may be chosen freely within the domain of f,
while
()y fx=
is called the dependent variable since its value depends on the value chosen for x.
Example 1.6
Which of the following is a function?
(a)
{( , )} {(1,5),(4,5),(3,3)}xy =
. (b)
{( , )} {(1,5),(1, 2),(3,3)}xy =
.
(c) (d)
2 3
1 2
3 4
4 1 4 1
As a function is defined such that each element of x can only have one real value of y, ∴
(a) and (d) are
functions, and
(b) and (c) are not.
The pictorial representation of a function is extremely useful, as is the idea that algebraic and
geometrical descriptions of functions are both helpful in understanding and learning about their
properties. The function
()y fx=
may be represented pictorially by its graph, which is the set of points
(x, f(x)) for each x in the domain of f, indicated with respect to Cartesian coordinate axes.
The graph of
()y fx=
is called a curve and the part of the curve lying between 2 points is called an
arc.
If to each value of x there is a unique corresponding value of y, their relation is called a function of y in
terms of x.
When a function is represented in the Cartesian graph, a line drawn parallel to the y-axis through any
point of the domain must meet the curve only once. This is known as the Vertical Line test.
It is also common to refer to ‘the function f(x)’ where f(x) is prescribed but no domain is given. In such
cases, the understanding required to be developed is that the domain of f is the set of real numbers for
which the expression f(x) defines a real number.
Example 1.7
(a) Find the domain and range of the following curves. Which curves are functions?
(i) (ii) (iii)
y
x
−1 1
2
y
x
−1 1
2
y
x
−2
2
−2
Chapter 1: Working with functions
13
(b) If
2
( ) 5 and ( ) 1f x x gt t=+=+
, find
(i)
(3)f
. (ii)
( 3)fa+
. (iii)
( )
()f gt
.
(a) (i)
Domain: All real x. Range: y ≤ 2.
3
It passes the vertical line test, ∴It is a function.
(ii) Domain: −1 ≤ x ≤ 1. Range: −2 < y ≤ 2.
It fails the vertical line test, ∴It is not a function.
(iii) Domain: All real x ≠ 0. Range: y = {−2, 2}.
It passes the vertical line test, ∴It is a function.
(b)(i)
2
(3) 3 5f = +
. (ii)
2
( 3) ( 3) 5fa a+=+ +
(iii)
( ) ( )
2
() () 5f gt gt= +
95
14.
= +
=
2
2
6 95
6 14.
aa
aa
= + ++
=++
2
2
( 1) 5
2 6.
t
tt
=++
=++
1.3.2 Symmetry properties (Odd and even)
A function
()fx
is even if
( ) ()f x fx−=
for all values of x in the domain. Its graph is symmetric with
respect to reflection in the y-axis, i.e., it has line symmetry about the y-axis.
A function
()fx
is odd if
( ) ()f x fx−=−
for all values of x in the domain. Its graph is symmetric with
respect to reflection in the point (0,0), i.e., it has point symmetry about the origin.
Example 1.8
Determine which of the following functions are odd, even or neither.
(a)
3
()fx x=
. (b)
2
() 4fx x= +
. (c)
3
() 4fx x x= −
. (d)
() 3fx x= −
.
(a) If
3 33 3
( ) then ( ) ( ) ( ), is odd.fx x f x x x fx x= −=− =−=− ∴
(b) If
2 22 2
( ) 4 then ( ) ( ) 4 4 ( ), 4 is even.fx x f x x x fx x= + − =− += += ∴ +
(c) If
( )
3 3 33 3
( ) 4 then ( ) ( ) 4( ) 4 4 ( ), 4 is odd.fxxx fx x xxx xx fxxx= − −=− −−=−+ =− − =− ∴−
(d) If
( ) 3 then ( ) 3 3 ( ) nor ( ), 3 is neither.fx x f x x x fx fx x= − − =−− = + ≠ − ∴ −
3
The domain and range can be expressed in interval notation.
For example,
(
]
( )
[ ] [
)
2 , 2 , 1 1, , 1 2 1, 2 , or 1 2 1, 2 .yy yy xx xx≤⇔∈−∞ >⇔∈ +∞ −≤≤⇔∈− ≤<⇔∈
y
x
y
x
y
x
−2 2
y
x
3
4
3
Mathematics Advanced Years 11 + 12
14
Exercise 1.3
1 Which of the following graphs are not functions? State the domain and range for the following.
(a) (b) (c) (d)
(e) (f) (g) (h)
2 If
2
( ) 3, ( ) 2 1f x x gt t=−=+
find
(a)
(2)f
. (b)
(2)g
. (c)
( 1)fa+
. (d)
( )
2ga+
.
(e)
( (2))fg
. (f)
( ( ))f gt
. (g)
( ( ))gfx
. (h)
( ( 1))gfa+
.
3 If
2
1
( ) , ( ) 2 and ( )f x x xgt t hu
u
=+=+ =
find
(a)
(2)f
. (b)
(2)g
. (c)
(2)h
. (d)
( )
2ha+
.
(e)
( 2)ga−
. (f)
( 1)fa+
. (g)
( ( ))f gt
. (h)
( ( ))gfx
.
(i)
(())f hu
. (j)
( ( ))hfx
. (k)
( ( ))hgt
. (l)
(())g hu
.
(m)
( ( 2))f ga−
. (n)
( ( 1))gha+
. (o)
( ( 1))gfa+
. (p)
( ( ( )))f hga
.
4 Find
( ( )) and ( ( ))fgx gfx
for the following functions
( ) and ( )f x gx
.
(a)
() , () 4f x xgx x= = +
. (b)
2
() , () 1fxxgxx= = −
. (c)
() 2, () 1f x x gx x=−=+
.
(d)
2
() 1, () 2 1f x x gx x=−=−
. (e)
1
() 2 3, ()
2
f x x gx
x
=+=
+
. (f)
2
1
() 1, ()
2
f x x gx
x
=+=
+
.
5 True or false, given that
( ) and ( )f x gx
are any two real functions? If true, prove it. If false, give
an example to justify your answer.
(a)
() ()fg gf=
. (b)
2
()ff f=
. (c)
()fg f g= ×
. (d)
( 1) ( ) (1).fg fg f+= +
6 The graphs of eight functions are shown below.
Copy the following graphs into your book and complete them so that they are
(i) odd functions, (ii) even functions.
(a) (b) (c) (d)
y
x
y
x
y
x
−2 2
y
x
−2
2
−2
−1
−1 1
2
y
x
y
x
y
x
−2
y
x
−1
4
−4
1
−1
−1 1
−1 1
y
x
y
x
y
x
y
x
Chapter 1: Working with functions
15
(e) (f) (g) (h)
7
Classify the following functions
()fx
as odd, even or neither.
(a)
( 1)xx−
. (b)
3
xx−
. (c)
3
5x +
. (d)
4
10x +
.
(e)
( 1)( 3)xx+−
. (f)
12 x−
. (g)
22
( 1)(3 )xx+−
. (h)
2
5 x−
.
(i)
3
5
x
x
+
. (j)
2
2
x
x +
. (k)
2
2x
x
+
. (l)
12x
x
+
.
8 Algebraically prove the following.
(a) If
f
and
g
are two even functions then
,, ,
f
f gf gf g
g
⋅+−
are even functions.
(b) If
f
and
g
are two odd functions then
and fg fg+−
are odd while
and
f
fg
g
⋅
are even.
(c) If
f
is an odd function and
g
is an even function then
and
f
fg
g
⋅
are odd while
and fg fg+−
are not odd neither even.
(d) If
543 2
Ax Bx Cx Dx Ex F+ + + ++
is an odd function then
0BDF= = =
.
(e) If
543 2
Ax Bx Cx Dx Ex F+ + + ++
is an even function then
0AC E= = =
.
(f) If
()fx
is an odd function and
(0)f
is defined then
(0) 0f =
.
1.4 Basic real function graphs
1.4.1 Linear functions
Linear functions or straight line graphs can be in one of these forms:
- General form:
0, where 0.ax by c a+ += >
- Gradient form:
y mx b= +
, where m is the gradient and b is the y-intercept.
- Special lines: x = a (parallel to the y-axis) or y = b (parallel to the x-axis).
Example 1.9
Sketch the following straight lines, showing any axis intercepts:
(a)
4 3 12 0xy+−=
. (b)
24yx= −
. (c)
2, 1xy= = −
.
(a)
Let 0, 4.xy= =
(b)
Let 0, 4.xy= = −
(c)
Let 0, 3.yx= =
Let 0, 2.yx= =
y
x
2
y
x
−1
2
−4
y
x
3
4
y
x
y
x
y
x
y
x
Mathematics Advanced Years 11 + 12
16
Example 1.10
Find the equations of the following straight lines.
(a) It passes through (2,3) with gradient −3.
(b) It passes through (−1,3) and (2,4).
(c) It passes through (5,−2) and inclines an angle of 30°.
(d) It is the perpendicular bisector of AB, where A(5,2) and B(1,0)
The gradient of the line joining
( ) ( )
21
11 2 2
21
, and , is
yy
xy xy
xx
−
−
.
Two lines are parallel if they have the same gradient.
Two lines are perpendicular if the product of their gradients equals to −1.
If a line inclines an angle α with the positive x-axis, then its gradient is
tan
α
.
The equation of a line passing through
( )
11
,xy
with gradient m is
( )
11
y y mx x−= −
.
(a)
3 3( 2).yx−=− −
(b)
43 1
21 3
m
−
= =
+
.
3 3 6.
3 9.
yx
yx
−=− +
=−+
1
3 ( 1).
3
3 9 1.
yx
yx
−= +
−=+
3 10 0.xy−+=
(c)
1
tan 30
3
m = °=
. (d)
12
20 2 1
, 2.
51 4 2
AB
mm m
−
= = ==∴=−
−
1
2 ( 5).
3
3 2 3 5.
3 2 3 5 0.
yx
yx
xy
+= −
+=−
− − −=
Midpoint of has coordinates :
51 20
3, 1.
22
Equation of the perpendicular of :
1 2( 3).
AB
xy
AB
yx
++
= = = =
∴
−=− −
1 2 6.
2 7.
yx
yx
−=− +
=−+
1.4.2 Quadratic functions
Quadratic functions are functions of the standard form
2
, where 0y ax bx c a= ++ ≠
. Their graphs are
in the shape of a parabola. They have line symmetry about their axes of symmetry (called the axes).
The parabola is upward if a > 0, and downward if a < 0.
If the parabola meets the x-axis, its equation can be expressed in factor form
( )( )y ax p x q=−−
,
where (p,0) and (q,0) are the x-intercepts. If difficult to find its factors, the x-intercepts can be found
by the quadratic formula
2
4
2
b b ac
x
a
−± −
=
.
The standard form
2
y ax bx c= ++
can be changed to the vertex form
( )
2
y ax h k= −+
by completing
the square, as shown below.
Chapter 1: Working with functions
17
2
22
2
2
2
2
44
4
.
24
b
y ax x c
a
bb b
ax x c
aa a
b b ac
ax
aa
= ++
= ++ −+
−
=+−
2
4
This form shows that it has axis (of symmetry) , and vertex , .
2 24
b b b ac
x
a aa
−
=− −−
Example 1.11
Sketch the following curves, showing the coordinates of the vertex and any axis intercepts.
(a)
2
yx=
. (b)
2
( 2)yx= −
. (c)
2
4yx= −
. (d)
2
1
( 2) 1
2
yx= ++
.
(a) (b) (c) (d)
Example 1.12
Sketch the following curves, showing the coordinates of the vertex and any axis intercepts.
(a)
2
6y xx=+−
, by factoring. (b)
2
43yx x=−−
, by quadratic formula.
(c)
2
23yx x=++
, by completing the square.
(a)
Let 0, 6.xy= =
The -intercept is (0,6)y∴
.
2
6 (3 )(2 ).
Let 0, 2, 3.
The -intercepts are ( 2,0) and (3,0).
xx x x
yx
x
+− = − +
= = −
∴−
2 3 1 1 1 25 1 25
Vertex: , 6 . , .
2 2 2 4 4 24
xy
−+
= = =+−= ∴
Since the coefficient of
2
x
is negative, this is a
maximum point.
(b)
Let 0, 3xy= = −
.
The -intercept is (0, 3)y∴−
.
2
Let 0, 4 3 0.yxx= − −=
4 16 12
2 7 0.65, 4.65.
2
The -intercepts are (2 7,0) and (2 7, 0).
x
x
±+
= =±−
∴ −+
( ) ( )
27 27
4
Axis of symmetry: 2
4
Alternatively,
.
22
2.
22
−
=−=−
+ +−
= = = =
b
x
a
x
y
x
(2,4)
y
x
2
y
x
−2 2
y
x
4
−2
4
3
1
Mathematics Advanced Years 11 + 12
18
( )
Vertex: 2, 4 8 3 7. 2, 7 .xy= = −−=−∴ −
Since the coefficient of
2
x
is positive, this is a minimum
point.
(c)
Let 0, 3xy= =
.
The -intercept is (0,3)y∴
.
2
Let 0, 2 3 0.yxx= + +=
4 12 8 0, No -intercepts.x∆= − =− < ∴
22
2
By completing the square: 2 3 2 1 2
( 1) 2.
xx xx
x
++=+++
=++
This is a translation of the graph of
2
yx=
to the left 1 unit and up 2 units.
Vertex: ( 1, 2).∴−
Since the coefficient of
2
x
is positive, this is a minimum point.
(a) (b) (c)
1.4.3 Reciprocals of linear functions
Reciprocals of linear functions or hyperbolic graphs have a point symmetry which coincides with the
point of intersection of the two asymptotes.
To find the equations of the asymptotes of the hyperbola
b
ya
xc
= +
−
.
Let , 0, . is the vertical asymptote.
Let , 0, . is the horizontal asymptote.
y xc x c xc
b
x y a ya
xc
→∞ − → ∴ → ∴ =
→∞ → → ∴ =
−
Example 1.13
Sketch the following curves, showing the equations of any asymptotes and the coordinates of any axis
intercepts.
(a)
1
y
x
=
. (b)
1
2
y
x
=
−
. (c)
1
2y
x
= −
. (d)
1
1
2
y
x
= +
+
.
(a) (b) (c) (d)
y
x
y
x
−3
−1
−2 3
y
x
3
2
6
−0.65 4.65
(2,
−
7)
y
x
(1,1)
y
x
2
y
x
y
x
1.5
1
−3 −2
2
Chapter 1: Working with functions
19
Exercise 1.4
1 Sketch the following straight lines, showing any axis intercepts.
(a) 2x + 3y − 6 = 0. (b) 2x − 3y = 12. (c) 3x + 2y = 12. (d) 3x − 2y + 12 = 0.
(e) 3x + y + 6 = 0. (f) 3x + y = 6. (g) x − 2y − 2 = 0. (h) x − 5y + 5 = 0.
(i) y = 3x + 2. (j) y = 5 + 2x. (k) y = 5 − x. (l) y = 2 − 3x.
(m)
1
45
xy
+=
. (n)
3
32
xy
−=
. (o)
34
2
x
y
−
=
. (p)
83
4
x
y
−
=
.
(q) y = 4. (r) x = −2. (s)
12
0
35
xy+−
+=
. (t)
12
1
32
xy+−
+=
.
2 State the equations of the following straight lines.
(a) It passes through (1,2) and (−3,4). (b) It passes through (1,0) and (3,−1).
(c) It passes through (1,3) and (−2,−6).
(d) It passes through (−1,2) and is parallel with the x-axis.
(e) It passes through (3,1) and is parallel with the y-axis.
(f) It passes through (2,4) with gradient 3.
(g) It passes through (−2,1) with gradient −2.
(h) It passes through (2,−1) with inclination angle of 60°.
(i) It passes through (0,2) with inclination angle 135°.
(j) It passes through (−1,2) with inclination angle of 150°.
(k) It passes through (3,−1) and is parallel to the line y = 3x + 5.
(l) It passes through (2,4) and is perpendicular to the line x − 3y + 5 = 0.
(m) It passes through A(5,1) and is perpendicular to AB, where B(−3,−1).
(n) It is the perpendicular bisector of CD, where C(0,5) and D(−2,3).
(o) It passes through (1,−2) and the point of intersection of
3 2 4 and 2 12xy xy+= −=
.
3 Sketch the following parabolas, showing the coordinates of the vertices and any axis intercepts.
(a)
2
1
, where , 1 and 2
2
y cx c= = −
. Describe in words the effect of the parameter c on the graphs.
(b)
2
, where 1 and 2yx c c=+=−
. Describe in words the effect of the parameter c on the graphs.
(c)
2
( ) , where 1 and 2y xc c=−=−
. Describe in words the effect of the parameter c on the graphs.
(d)
2
( 2) 1yx=−+
. (e)
2
( 1) 4=+−yx
. (f)
2
4 ( 2)yx=−−
. (g)
2
2( 1)yx= −
.
(h)
2
1
( 4) 2
2
= −−yx
. (i)
2
4 2( 1)yx=−−
. (j)
2
2( 1) 4yx= ++
. (k)
2
1
8 ( 2)
2
=−−yx
.
(l)
( 2)y xx= −
. (m)
2( 2)( 6)yx x=+−
. (n)
2
23yx x=+−
. (o)
2
6 16yx x=+−
.
(p)
2
64 2y xx=+−
. (q)
2
54y xx=+−
. (r)
2
1
62
2
y xx=+−
. (s)
2
1
4
2
y xx= +−
.
(t)
2
24yx x=++
. (u)
2
6 10yx x=−+
. (v)
2
45yx x=++
. (w)
2
3 12 36yx x=++
.
(x)
2
3 46yx x= +−
. (y)
2
31yx x=+−
. (z)
2
4yx x= +−
. (aa)
2
3 24yx x= +−
.
(ab)
2
3 65yx x= −+
. (ac)
2
31yx x=++
. (ad)
( 2)( 3) 1=+ ++yx x
(ae)
( 2)( 4) 6.yx x=− +−
4 Sketch the following hyperbolas, showing the equations of the asymptotes and the coordinates of
any axis intercepts.
(a)
10xy = −
. (b)
( 1) 5xy+=
. (c)
( 2) 10xy−=−
. (d)
( 2) 5xy−=
.
(e)
3
5
y
x
=
−
. (f)
2
2
3
y
x
= +
−
. (g)
1
4
2
y
x
= −
+
. (h)
1
2
1
y
x
= −
+
.
Mathematics Advanced Years 11 + 12
20
(i)
26
3
x
y
x
+
=
−
. (j)
63
21
x
y
x
+
=
−
. (k)
21
2
x
y
x
+
=
−
. (l)
41
23
−
=
+
x
y
x
.
1.5 More real function graphs
1.5.1 Square root graphs
The curve
yx=
is the positive half of the ‘sideway’ parabola
2
yx=
.
Example 1.14
Sketch the following curves, showing any axis intercepts.
(a)
yx=
. (b)
2yx= −
. (c)
2yx= −
. (d)
2yx=−+
.
(a) (b) (c) (d)
1.5.2 Cubic graphs
Example 1.15
Sketch the following curves, showing any axis intercepts.
(a)
3
yx=
. (b)
2
( 3)y xx= −
. (c)
2
( 1)( 2)yx x=+−
. (d)
( 1)( 1)(2 )yx x x=+−−
(a) (b) (c) (d)
1.5.3 Circles and semi-circles
The equation of the circle of centre (0,0) and radius r is found by the Pythagoras theorem:
2 22
xyr+=
.
y
x
−r r
r
−r
r
x
y
y
x
(1,1)
y
x
2
y
x
2
y
x
−2
x
y
x
(2,8)
y
−2
−1 1 2
y
x
−1 2
y
x
3
4
Chapter 1: Working with functions
21
Example 1.16
Sketch the following curves, showing the coordinates of the centre and any axis intercepts.
(a)
22
4xy+=
. (b)
22
( 1) 1xy−+=
. (c)
22
( 1) 1xy++ =
. (d)
2
3yx= −
.
(a) (b) (c) (d)
1.5.4 Absolute value graphs
Geometrically,
x
, where x is a real number, is the distance from the origin to x on the number line.
Algebraically,
, if 0
, if 0
xx
x
xx
≥
=
−<
∴
( ), if ( ) 0
()
( ), if ( ) 0
fx fx
fx
fx fx
≥
=
−<
Example 1.17
(a) Sketch the following curves, showing any axis intercepts.
(i)
yx=
. (ii)
1yx= −
. (iii)
2
4yx= −
. (iv)
1
1
y
x
=
−
.
(b) Solve both algebraically and graphically the equation
2 13x −=
.
(a)
(i) (ii) (iii) (iv)
(b) By algebraic method,
1
2 1 3, if
2
2 13
1
(2 1) 3, if
2
21 3
2 31
4 or 2.
2 or 1.
xx
x
xx
x
x
x
−= ≥
−=⇔
− −= <
∴ −=±
=±+
= −
∴= −
y
x
−2 2
y
x
1 2
y
x
−1
−2
y
x
−
2
−2
y
x
(1,1)
y
x
1
y
x
−2 2
y
x
1
4
1
1
Mathematics Advanced Years 11 + 12
22
By graphical method,
11
The graph 2 1 has 2 branches, 2 1 for and 2 1 for .
22
Solving 2 1 3 gives 1 and solving 2 1 3 gives 2.
1 or 2.
y x y x x yx x
xx xx
x
= − =−+ < = − ≥
− += =− −= =
∴=−
1.5.5 Transformation of graphs
Earlier, in section 1.4.2, we learnt how to reflect and translate quadratic functions. We now apply these
techniques to other functions and also learn about dilation.
Example 1.18
Given . It has a maximum at (−2,27) and a minimum at (1,0). It meets the y-
axis at (0,7). The graph of is shown below.
Sketch the following graphs, showing stationary points and axes intercepts where possible.
(a)
()yfx= −
. (b)
()y fx−=
. (c)
( )
3y fx= −
. (d)
( )
1y fx= +
.
(e)
( )
1y fx−=
. (f)
2
x
yf
=
. (g)
( )
2
y
fx=
. (h)
( )
1
3
y
fx= +
.
(a)
changes signs of the -axis, ( )x x yfx− ∴= −
is the reflection of
()y fx=
about the y-axis.
2
( ) ( 1) (2 7)fx x x=−+
()y fx=
x
y
1
(−1,3) (2,3)
y
x
− −2 1
27
7
y
27
−1 2
x
7
Chapter 1: Working with functions
23
(b)
changes signs of the -axis, ( ) or ( )y y y fx y fx− ∴− = =−
is the reflection of
()y fx=
about the
x-axis.
(c) The graph of
( ) ( )
3 shifts the graph of y fx y fx=−=
to the right 3 units. Therefore, the x-
intercepts are
71
3 and 1 3 4
22
xx=−+=− =+=
and the stationary points are (1,27) and (4,0).
( )
y fx a= −
is called horizontal translation, it shifts the graph of
()y fx=
to the right if a > 0 or to
the left if a < 0.
()y a fx−=
is called vertical translation, it shifts the graph of
()y fx=
up if a > 0 or down if a < 0.
(d)
The graph of
( ) ( )
1 shifts the graph of y fx y fx=+=
to the left 1 unit. Therefore, the x-intercepts
are
79
1 and 1 1 0
22
xx=− −=− =−=
and the stationary points are (‒3,27) and (0,0).
(e)
As parts (c) and (d) are horizontal translations,
1 ()y fx−=
is a vertical translation. The stationary
points are (‒2,28) and (1,1). The y-intercept is (0,8).
−7
y
x
− −2 1
y
x
4
(1,27)
( )
2, 27−−
y
x
(‒3,27)
Mathematics Advanced Years 11 + 12
24
(f) Considering the x-intercepts,
7
7 and 1 2,
22 2 2
xx x
x x yf
=−⇔=− =⇔=∴=
spreads the graph
of
()y fx=
twice across both sides of the y-axis. Therefore, the x-intercepts are (‒7,0) and (2,0) and
the stationary points are (‒4,27) and (2,0).
( )
For 0, 1, k k y f kx>≠ =
is horizontal dilation, while
()ky f x=
is vertical dilation of the function
()y fx=
. The graph is expanded if 0 < k < 1, and compressed if k > 1.
(g) As part (f) is a horizontal dilation, part (g) is a vertical dilation of factor
1
2
k =
. Therefore, the y-
intercept is (0,14), and the stationary point is (‒2, 54) and (1,0).
(h)
( )
1
3
y
fx= +
combines a horizontal translation, shifting to the left 1 unit, and a vertical dilation,
expanding the curve three times.
The stationary points
( ) ( )
are 3,81 and 0,0 .−
9
The -intercepts are at
2
xx= −
and the origin.
y
x
−2 1
28
8
(1,1)
y
x
−7 ‒6 −4 ‒2 2
27
7
y
x
− −2 1
54
14
1
Further
work with
functions
HSC Outcomes
A student
develops understanding and fluency in mathematics through exploring and connecting mathematical concepts, choosing
and applying mathematical techniques to solve problems, and communicating their thinking and reasoning coherently
and clearly
solves problems involving inequalities, functions and their inverses, graphical relationships between functions, and
parametric equations
In this chapter,
1.1 Further transformation of graphs ..................... 6
1.2 Rational functions .......................................... 10
1.3 Using graphs .................................................. 14
1.4 Parametric equations ...................................... 18
1.5 Inverse functions ............................................ 20
1.6 Review Exercise 1 ......................................... 23
Solutions ........................................................... 25
Mathematics Extension 1 Years 11 + 12
6
This chapter delves deeper into functions and requires prior knowledge of the following chapters from
the Advanced course: Chapter 1: Functions, Chapter 2: Trigonometric functions, Chapter 3:
Introductory Calculus, Chapter 4: Exponential and Logarithmic functions, Chapter 6: Mathematical
Modelling, and Chapter 8: Differential Calculus.
1.1 Further transformation of graphs
Example 1.1
Given
2
( ) ( 3)f x xx= −
.
(a) Sketch the curve, showing all x-intercepts and stationary points.
(b) Without using Calculus, sketch the following graphs, showing stationary points, axes of
asymptotes and axes intercepts, where possible.
(i)
()y fx=
. (ii)
( )
y fx=
. (iii)
()y x fx= +
. (iv)
1
()
y
fx
=
.
(a)
2
( 3)y xx= −
.
( ) ( )
2
3 22
2
0 gives ( 3) 0, , 3.
-intercepts 0, 3
3 , 3 3 3( 1).
0 gives 1, 1.
Stationary points 1,2 and 1, 2 .
y xx x
xx
yx x y x x
y xx
= − =∴=±
= ±
′
= − ∴ = −= −
′
= =∴=±
−
(b)
(i)
()y fx=
( ), if ( ) 0
()
( ), if ( ) 0
fx fx
fx
fx fx
≥
=
−<
‒1 1
2
‒2
y
x
‒1 1
2
y
x
Chapter 1: Further work with functions
7
(ii)
( )
y fx=
.
, if 0, we retain this part of ( )
, if 0, we flip the part where 0 about the -axis
x x fx
x
xx x y
≥∴
=
− <∴ ≥
(iii)
()y x fx= +
.
For addition and subtraction of two curves:
1) They must have the same domain. For example, if
( ) exists for 1 2, and ( ) exists for f x x gx−≤ ≤
02x≤≤
then the domain of
() () is 0 2f x gx x+ ≤≤
.
2) To add or subtract two graphs, add or subtract their y-values for each x-value. Key points to
consider are where the graphs intersect the axes or each other.
‒1 1
‒2
y
x
‒1 1
‒2
y
x
(1,1)
(1,‒1)
(2,2)
(2,4)
‒1 1
‒2
y
x
Mathematics Extension 1 Years 11 + 12
8
(iv)
1
()
y
fx
=
.
To graph reciprocal functions:
Key points to consider are special points such as stationary points, zeros and asymptotes.
1) Points
( )
1
, ( ) become ,
()
xfx x
fx
. It also means that the reciprocal graph will have the same sign
(positive or negative) as the original function for the same values of x.
2) Vertical asymptotes: As
1
,
0
→∞
the zeros (where the original function equals zero) of the original
function become vertical asymptotes in the reciprocal graph.
3) Horizontal asymptotes:
(i) If
1
( ) as then 0fx x→∞ →∞ →
∞
: the x-axis becomes the horizontal asymptotes.
(ii) If
1
( ) as then 0fx x a→∞ → →
∞
: the reciprocal curve is not defined at x = a. Use open
circles for these points.
(iii) If
1
( ) as then fx a x y
a
→ →∞ =
is the new horizontal asymptote.
Exercise 1.1
1 Given the graph of as shown below. Sketch the following graphs.
(i)
()y fx=
. (ii)
()yfx= −
. (iii)
( )
y fx=
. (iv)
1
()
y
fx
=
.
()y fx=
x
−
3
−
2
−
1 1 2 3 4 5
y
2
−1
2
‒2
‒1 1
y
x
Chapter 1: Further work with functions
9
2 Repeat question 1 for the following curves
()y fx=
.
(a) (b) (c)
(d) (e) (f)
(g) (h)
3 Given
2
() 2fx x x= −
. Express the following in algebraic forms and sketch the curves.
(a)
()y fx=
. (b)
()yfx= −
. (c)
( )
y fx=
. (d)
1
()
y
fx
=
.
4
Repeat question 3 for
2
() 2 3fx x x=−−
.
5
Given
( ) and ( )f x gx
, sketch the following graphs
() () and () ()f x gx f x gx+−
. Verify your
answers by proving it algebraically.
(a)
2
() 4, ()fxx xgxx=−=
. (b)
3
() , () 2fxxgxx= = −
.
(c)
1
() , ()f x xgx
x
= =
. (d)
2
2
1
() , ()f x x gx
x
= =
.
6 Sketch the following graphs.
(a)
3
logyx=
. (b)
3
logyx=
. (c)
3
logyx x= +
. (d) .
(e)
3
x
y =
. (f)
3
x
yx= +
. (g)
3
x
yx= −
. (h)
1
31
x
y =
−
.
7 Sketch the graphs of the following reciprocal trigonometric functions, showing any turning points,
axes intercepts and asymptotes.
(a) . (b) . (c) .
3
1
log
y
x
=
sec , 2 2yx x
ππ
= − ≤≤
cosec , 2 2y xx
ππ
= − <<
cot ,yx x
ππ
= −<<
2 3 −1 2
(2,1) 1
−1
x
y
x
y
x
y
x
y
−2 2 4 4 −1 2
−2
2
x
y
(2,2)
x
y
(−1,2) (3,2)
(1,−1)
1
x
−2
y
2
1
x
−2
y
2
‒2
Mathematics Extension 1 Years 11 + 12
10
1.2 Rational functions
Example 1.2
Sketch the following curves, showing the asymptotes and any stationary points. Do not use Calculus.
(a)
1
( 3)( 1)
y
xx
=
−+
. (b)
9( 3)
( 2)( 1)
x
y
xx
−
=
−+
.
(a)
This is the reciprocal of the parabola
( ) ( 3)( 1)fx x x=−+
.
( ) ( 3)( 1)y fx x x= =−+
. y =
11
( ) ( 3)( 1)fx x x
=
−+
.
x-intercept: (3,0) and (−1,0). Vertical asymptotes: x = 3, −1.
y-intercept: (0,−3). y-intercept:
1
0,
3
−
.
As
2
,x yx→ ±∞ → +∞
. As
2
1
,0xy
x
+
→ ±∞ →
.
Axis of symmetry:
3 ( 1)
2
x
+−
=
= 1.
Minimum point: (1,−4). Maximum point:
1
1,
4
−
.
In the above diagram,
( 3)( 1)yx x=−+
is drawn in red, and
1
( 3)( 1)
y
xx
=
−+
is drawn in blue.
(b) Using the “guide graph” method.
The guide graph is the polynomial formed by the product of all the non-positive factors of the given
function. Notably, positive factors, such as
22
1, 2 4,x xx+ −+
can be disregarded. If the guide graph is
above the x-axis, draw the curve above the x-axis, and if the guide graph is below the x-axis draw the
curve below the x-axis.
If the coordinates of the turning points are required, use Calculus.
The following notes explain how the curve is drawn.
y
x
−1 3
Chapter 1: Further work with functions
11
Step 1: Finding the asymptotes:
9( 3) 9
.
( 2)( 1)
x
y
xx x
−
=
−+
As , 0xy
+
→ +∞ →
, and as
9
,0xy
x
−
→ −∞ →
. Draw the asymptote y = 0, noting that as
x → +∞
,
the curve approaches the asymptote from above, and as
x → −∞
, the curve approaches the asymptote
from below.
As
2 or 1,xy→ − →∞
. Draw the asymptotes x = 2, x = −1.
Step 2: Sketch the guide graph
( ) ( 3)( 2)( 1)fx x x x=−−+
(shown in pink), noting that 9 is always
positive, so ignore it.
1
Now we can draw the required curve:
For x < −1: The guide graph is below the x-axis, so the curve must be drawn below the x-axis. There
are two asymptotes in this region so draw a curve to approach both asymptotes;
For −1 < x < 2: The guide graph is above the x-axis, so the curve must be drawn above the x-axis.
There are two asymptotes in this region so draw a curve to approach both asymptotes. Also, note the y-
intercept (0, 13.5).
For 2 <
x ≤
3, The guide graph is below the x-axis, so the curve must be drawn below the x-axis. Note
the x-intercept (3,0), so draw a curve from (3,0) to approach the asymptote x = 2;
For
x ≥
3: The guide graph is above the x-axis, so the curve must be drawn above the x-axis. Draw a
curve from (3,0) up to a maximum value then bend it down to approach the asymptote y = 0.
The curve is now complete. You can safely remove the guide graph or label it.
How to find the asymptotes.
(a) Vertical asymptotes: Let the denominator equal zero to find vertical asymptotes.
(b) Horizontal asymptotes:
(i) If the degree of the numerator is less than the degree of the denominator, as
,0xy→∞ →
.
(ii) If the degree of the numerator is equal to the degree of the denominator, as
,xy→∞ →
a
constant C, found by dividing the two leading coefficients. If it is necessary to investigate the
behaviour of the curve for large values of x, use long division. See example 1.3.
1
Advantages of the guide graphs:
(1) You immediately know where the curve is positive (above the x-axis) or negative (below the x-axis).
(2) Vertical asymptotes and x-intercepts are easily noticed.
(3) In combination with the asymptotes the curve can be drawn with high accuracy in the least amount of time.
−1 2 3
y
x
13.5
in blue
The guide graph
in red
Mathematics Extension 1 Years 11 + 12
12
22
22 2222
If the coordinates of the stationary points are required, we may use Calculus.
9[( 2) (2 1)( 3)] 9( 6 5) 9( 1)( 5)
( 2) ( 2) ( 2)
xx xx xx xx
y
xx xx xx
−−−−− −−+ −−−
′
= = =
−− −− −−
y
′
= 0 gives x = 1 or 5, ∴Stationary points (1,9) and (5,1).
Determining the nature of the stationary points:
2
As
22
( 2)xx−−
≥ 0 for all
1, 2x ≠−
, the sign of
y
′
is determined by the
concave down parabola −9(x − 1)(x − 5) which cuts the x-axis at 1 and 5
(see inset at the right), therefore, (1,9) is minimum and (5,1) is maximum.
Example 1.3
Given
2
2 ( 2)
1
xx
y
x
−
=
+
.
(a) Investigate the behaviour of the curve for large values of x.
(b) Sketch the curve, showing the main features but do not use Calculus.
(a) Using long division,
22
22 2 2
2 ( 2) 2 4 2( 1) 4 2 4 2
2
11 1 1
xx x x x x x
xx x x
− − +− − +
= = = −
++ + +
.
For large values of x,
2
42 4
approximately equals to
1
x
xx
+
+
. ∴As
4
,2 2xy
x
−
→ +∞ − →
, i.e., the curve
approaches the horizontal asymptote y = 2 from below. As
4
,2 2xy
x
+
→ −∞ − →
, i.e., the curve
approaches the horizontal asymptote y = 2 from above.
As
2
10x +>
, there is no vertical asymptote.
(b) The following notes explain how the curve is drawn.
To draw this curve without using Calculus, we use the guide graph x(x − 2), since 2 and
2
1x +
are
always positive, so they have no effect on the signs of the curve.
For x < 0: the guide graph is above the x-axis so the curve must be drawn above the x-axis. Draw a
curve from the x-intercept (0,0) to approach the asymptote y = 2. Note that the curve must approach
this asymptote from above, i.e., it must go through the asymptote before going down to approach it.
For 0 < x < 2: the guide graph is below the x-axis so the curve must be drawn below the x-axis. Draw a
curve below the x-axis joining the two x-intercepts (0,0) and (2,0).
2
There are three methods to determine the nature of stationary points:
Method 1: Use the signs of
y
′′
, if positive it’s a minimum point, if negative it’s a maximum point. However, if
y
′′
= 0, it’s inconclusive, use one of the other two methods.
Method 2: Use the table of the signs of
y
′
at two sides of each turning point, if the gradient goes from positive to
negative it’s a maximum point, if the gradient goes from negative to positive it’s a minimum point, if the
gradient does not change its sign it’s a horizontal point of inflexion.
Method 3: Use the graph of
,y
′
ignoring positive factors, as we are interested in its signs only. If the graph is
above the x-axis,
0y
′
>
, and if the graph is below the x-axis,
0y
′
<
.
+ +
− −
1 5
Chapter 1: Further work with functions
13
For x > 2: draw a curve from (2,0) to approach the asymptote y = 2. Note that the curve must approach
this asymptote from below.
Exercise 1.2
1 (i) Sketch the graph of
( 2)( 6)y xx x=−+
, showing the x-intercepts only. Hence, state the values of
x for which the curve is positive or negative.
(ii) Sketch the following curves. You are not required to find the co-ordinates of any turning points.
(a)
1
( 2)( 6)
y
xx x
=
−+
. (b)
2
( 6)
x
y
xx
−
=
+
. (c)
( 2)( 6)
x
y
xx
=
−+
. (d)
6
( 2)
x
y
xx
+
=
−
.
2 Sketch the following curves. Do not use Calculus.
(a)
( 2)
( 1)( 3)
xx
y
xx
−
=
+−
. (b)
( 1)
( 2)( 3)
xx
y
xx
+
=
−−
. (c)
( 3)
( 1)( 2)
xx
y
xx
−
=
+−
. (d)
( 1)( 3)
( 2)
xx
y
xx
+−
=
−
.
3 Sketch the following curves. Do not use Calculus.
(a)
3
16
4
y
xx
=
−
. (b)
1
( 1)( 3)
y
xx
=
++
. (c)
2
1
( 3)
y
xx
=
−
. (d)
2
45
1
x
y
x
+
=
−
.
(e)
2
( 1)( 4)
x
y
xx
+
=
−+
. (f)
92
( 4)
x
y
xx
−
=
−
. (g)
2
( 1)
( 1)( 3)
x
y
xx
+
=
−−
. (h)
( 1)( 4)
( 1)( 3)
xx
y
xx
+−
=
−+
.
(i)
2
2
1
1
x
y
x
−
=
+
. (j)
2
2
2
4
x
y
x
=
+
. (k)
2
2
( 1)
( 1)( 3)
xx
y
xx
+
=
++
. (l)
3
3
1
x
y
x
=
−
.
4 Given
2
()
4
x
fx
x
=
−
.
(a) Show that
( ) 0 for all in the domain of ( )f x x fx
′
>
.
(b) Sketch the graph of
()y fx=
.
5 Consider the function
42
4
()
1
xx
fx
x
+
=
+
.
(a) Show that
()fx
is an even function. What does it mean geometrically?
(b) Find the equations of any asymptotes.
(c) Find the x-coordinates of all stationary points for the graph
()y fx=
.
(d) Sketch the graph of
()y fx=
. You are not required to find the y-coordinates of any stationary
points, nor the coordinates of any points of inflexion.
2
2
y
x
Guide graph
( 2)y xx= −
2
( 2)
1
xx
y
x
−
=
+
Mathematics Extension 1 Years 11 + 12
14
1.3 Using graphs
Graphs can be used with supporting algebraic working to solve a variety of practical problems, e.g.,
finding approximate solutions of equations that are impossible to solve algebraically or solving
inequalities of high degrees.
Example 1.4
(a) Determine the number of solutions of the equation
3
2 40xx+ −=
.
(b) Hence, find two consecutive integers between which each of the solutions lies.
(a)
33
2 40 2 4xx x x+ −=⇔ =− +
.
We now sketch two curves
3
yx=
and y = −2x + 4. These two curves intersect at only one point, thus,
there exists only one root of the equation
3
24xx+−
= 0.
(b) When
3
1, (1) 1 2 1 4 1 0xf= = + ×− =−<
.
When
3
2, (2) 2 2 2 4 8 0xf= = +×−=>
.
∴Thus, there exists a solution in the interval [1,2] (i.e., between x = 1 and x = 2).
Note: If the question asks for solutions correct to 1 decimal place, both curves must be drawn to scale,
e.g., you may need to show (2,8) on the curve
3
yx=
. The answer, correct to 1 decimal place, is 1.2.
Example 1.5
Solve the following inequalities. Show the results on the number line.
(a)
2 13x −≤
. (b)
24 1xx−> +
. (c)
2
60xx−−≥
. (d)
3
1
2x
≥
−
.
(a) By algebraic method,
11
2 1 2 1, if , or 2 1, if .
22
11
If , 2 1 3. If , 2 1 3.
22
2 4. or 2 2.
2. 1.
11
21
22
x xx xx
xx x x
xx
xx
xx
−= − ≥ − + <
∴ ≥ −≤ < − +≤
≤ −≤
≤ ≥−
∴≤≤ −≤<
∴Conclusion: −1 ≤ x ≤ 2.
−1 2
x
y
4
2
Chapter 1: Further work with functions
15
By graphical method,
11
The graph 2 1 has 2 branches, 2 1 for and 2 1 for .
22
Solving 2 1 3 gives 1 and solving 2 1 3 gives 2.
Points of intersection ( 1,3) and (2,3).
2 1 3 for 1 2.
y x y x x yx x
xx xx
xx
= − =−+ < = − ≥
− += =− −= =
∴−
∴ − ≤ −≤ ≤
(b) By critical values method,
Solving
2 4 0 and 0xx−= =
gives two critical values: x = 2 and 0.
For
, 2 4 2 4 d0 an xxxx x< −=−+ =−
:
2 4 1.
3.
3.
xx
x
x
− + >− +
− >−
<
Combining
3 and gives 0.0x x x<<<
For
, 2 402 2 4 and x x x xx−=− +≤ =≤
:
2 4 1.
3 3.
1.
xx
x
x
− +>+
− >−
<
Combining
1 and gives 02 0 1.xxx < ≤<
≤≤
For
, 2424 d2 an xx xx x−= −> =
:
2 4 1.
5.
xx
x
−>+
>
Combining
5 and gives 5.2x x x>>>
Conclusion:
1 or 5.xx<>
By graphical method,
The graphs of 1 (in red) and 2 4 (in blue) are shown.yx y x=+=−
0 2
0 1 5
x
y
1
(−1,3) (2,3)
x
y
1
4
2
A
B
Mathematics Extension 1 Years 11 + 12
16
Clearly, the points of intersection are A, where
1 and 2 4yx y x=+ =−+
meet, and B, where
1 and 2 4yx y x=+=−
meet, and
2 4 1 for or
AB
x x xx xx−> + < >
.
Solving 2 4 1 gives 1 and solving 2 4 1 gives 5.
Points of intersection (1, 2) and (5,6).
2 4 1 for 1 or 5.
xxx xxx
A
x x xx
−+=+= −=+=
∴
∴ −> + < >
(c) Sketch the graph of
2
6 ( 3)( 2)yx x x x= −−= − +
, showing only the x-intercepts.
From the graph,
2
6 0 for 2 or 3xx x x− − ≥ ≤− ≥
.
(d)
3
1
2x
≥
−
.
( )
2
2
Multiplying both sides by ( 2) gives
3( 2) ( 2) .
( 2) 3 ( 2) 0.
( 2)(5 ) 0.
( 2)( 5) 0.
From the graph of ( 2)( 5), ( 2)( 5) 0 for 2 5.
x
xx
xx
xx
xx
yxx xx x
−
−≥−
− −− ≥
− −≥
− −≤
=−− −−≤ ≤≤
3
But 2, 1 for 2 5.
2
xx
x
≠∴ ≥ <≤
−
Example 1.6
Two tangents are drawn from a point to a circle. If the area enclosed by the tangents and the radii
drawn to the points of contact is equal to the remaining area of the circle, prove that tan
α
=
π
−
α
,
where
α
is the angle subtended at the circumference by the minor arc.
Hence, find the value of
α
to the nearest degree.
−2 3
−2 3
x
2 5
2 5
x
A O C T
T
B
D
Chapter 1: Further work with functions
17
Let the radius of the circle be r and ∠BAD =
α
.
∠OBT = 90° (the tangent is perpendicular to the radius at the point of contact)
∠BOD = 2
α
(angle at the centre is twice as the angle on the circumference)
∠BOC =
α
(= half of ∠BOD, due to symmetry)
tanα =
BT
OB
, ∴BT = OB tan
α
= r tan
α
.
∴Area of ∆OBT =
2
11
. tan
22
OB BT r
α
=
.
∴Area of quadrilateral OBTD =
2
tanr
α
(1)
Area of sector OBAD = Area of the circle − Area of sector OBCD
=
22
1
2
(2 )rr
πα
−
=
2
()r
πα
−
(2)
(1) = (2) gives
tan
απα
= −
.
As this equation cannot be solved algebraically, we use the graphs of
tan and ,yy
α πα
= = −
0
απ
<<
. Note: To find the y-intercept of the line (
y
π
=
), you must find y = 1
tan 1
4
π
=
first.
Given
0.8 and
42
ππ
1.6, the point of intersection of the two curves has its x-coordinate of about
1.1,∴
α
≈ 1.1 radians ≈ 63°.
Exercise 1.3
1 Determine the number of solutions of the equation
4
2 52xx−+
= 0, hence, find two consecutive
integers between which each of the solutions lies.
2 Use graphical means to find the solutions of the following equations correct to 1 decimal place.
(a)
3
1xx−+
= 0. (b)
x
ex+
= 0. (c)
tan 0xx−=
. (d)
3cos 2 4 tanxx−
= 0.
3 Find the values of k for which the equation
32
30x xk+ −=
has two distinct solutions.
4 Find the values of k for which the line
y kx=
meets the curve
3
21yx= +
at 2 distinct points.
5 Find the values of k for which the equation
x
e kx=
has 2 distinct solutions.
π
π
≈ 3.1
y
α
1.1
1
Mathematics Extension 1 Years 11 + 12
18
6 Solve for x, using both algebraic and graphical methods. Show the results on the number line.
(a)
2 31x −>
. (b)
3 15x −<
. (c)
83 4x−≤
. (d)
34 2x− ≤−
.
7 Solve for x, using both critical values and graphical methods. Show the results on the number line.
(a)
36 4xx−> +
. (b)
24 2xx+> −
. (c)
3 14 1xx++≥ −
. (d)
1 12xx++ −≤
8 Solve for x.
(a)
( 2)( 1) 0xx− +≥
. (b)
3 ( 4) 0xx−<
. (c)
2
4 12 0xx−−>
. (d)
2
3(6 ) 0xx−− ≤
.
(e)
( 1) 2xx+>
. (f)
( 2)( 4) 7xx− +<
. (g)
2
12 xx+≥
. (h)
(2 1) 1xx−≤
.
(i)
(3 2 ) 2xx− ≥−
. (j)
2( 2) 5 7x xx− <−
. (k)
2 ( 3) 3xx x− ≥−
. (l)
2
5 (1 ) (1 ) 1xx x− ≤− +
.
(m)
( 2)( 1) 0xx x+ −>
. (n)
(2 )(2 ) 0xx x− +≥
. (o)
2
( 1)( 2) 0xx+ +>
. (p)
32
20x xx− +>
.
(q) . (r) . (s) . (t)
2
0
1
x
x
x
−<
+
.
(u) . (v)
5
23
x
x
x
≤
+
. (w)
11
52xx
≥
+
. (x)
1
21
x
xx
≤
−
.
9 (a) Find the size of the angle subtended by a segment of area of 18 cm
2
in a circle of radius 4 cm.
(b) Find the size of the angle subtended by a chord that divides the area of a circle in the ratio of 1:
3.
1.4 Parametric equations
We have defined a curve so far by an equation in terms of two variables x and y. However, sometimes
it is more convenient to define a curve by two equations, where each equation is expressed in terms of
a common third variable, called a parameter.
For example, for the parabola
2
4ay x=
we may define
2
2x at
y at
=
=
Notes:
(i) The relation of x and y may be obtained by eliminating t:
From x = 2at we have
2
x
t
a
=
.
Substituting it to
2
y at=
gives
22
2
2
, i.e., 4
44
xx
y a ay x
aa
= = =
.
(ii) For every value of t, there is only one value of x and only one value of y. Similarly, every point on
the curve is satisfied by only one value of t.
Example 1.7
(a) Which curve is represented by these parametric equations
2
2,xt yt=−=
?
(b) Which point on the curve corresponds to t = 0?
(c) If t is defined for non-negative real numbers, what is the domain of the curve?
(a)
22
Substituting 2 into gives ( 2) .tx yt y x=+= =+
∴The curve is the parabola
2
( 2)yx= +
.
1
2
1x
≥
−
2
3
x
x
≤
+
3
3
3
x
x
+
≥
−
2
1
23
x
x
<
+
Chapter 1: Further work with functions
19
(b) When t = 0, x = −2, y = 0. ∴Point (−2,0) corresponds to t = 0.
(c) If
0 then 2 0, 2tx x≥ + ≥ ∴ ≥−
: The parabola is defined for x ≥ −2 only.
Example 1.8
(a) When a particle moves in a circle of radius r, centre (a,b), with period T (i.e., time to complete a
revolution), prove that its parametric equations are
22
cos , sinx a r ty b r t
TT
ππ
=+=+
.
(b) Hence, find the parametric equations of a gymnast who cycles a stationary bike with a 23 cm
flywheel radius at 80 rpm (rpm stands for revolutions per minute).
( ) ( )
22
22
22
2
(
h
22
cos ,sin .
Using cos sin 1 gives 1.
The equation of t e circle is .
60 3 2 2 8
Given 23 cm, frequency 80 r d
a
pm, period secon s, = =
3
80 4
)
(b)
4
xa yb
tt
Tr Tr
xa yb
xx
rr
xa yb r
rT
T
ππ
πππ
−−
= =
−−
+= + =
∴ − +− =
= =∴== ∴
.
3
88
23cos , 23sin , where is in cm, and is in seconds.
33
x ty t x t
ππ
∴= =
Exercise 1.4
1 By eliminating t find the Cartesian equations of the following curves.
(a)
,
c
x ct y
t
= =
. (b)
4, 2 1xt y t=+=−
. (c)
2
2,xt yt=+=
. (d)
2
2, 3xt y t=+=
.
(e)
12
,
11
t
xy
tt
= =
++
. (f)
22
12
,
11
t
xy
tt
= =
++
(g)
2
22
12
,
11
tt
xy
tt
−
= =
++
(h)
2
21
,
11
tt
xy
tt
+
= =
++
.
(i)
cos , sin .x a ty a t= =
(j)
cos , sin .x a ty b t
αβ
=+=+
(k)
sec , tan .x a ty b t= =
(l)
cos , sec .x a ty b t= =
(m)
( ) ( )
cos 1 tan , sin 1 tanxttytt=+=+
.
2 (a) Find the equation of a parabola which is defined parametrically as
2
2, 3x ty t= =
.
(b) Find the coordinates of the point corresponding to t = 2.
(c) If t is defined to be non-negative, find the domain of the function.
3
( ) ( )
22
2 , and 2 ,P ap ap Q aq aq
are two points such that their parameters are the roots of the
equation
2
2 3 20tt+ −=
.
(a) Find the coordinates of the points P and Q.
(b) Show that the line PQ passes through the point (0,4).
4 P and Q are two points of parametric equations
( ) ( )
22
2 , and 2 ,ap ap aq aq
such that OP is
perpendicular to OQ. O is the origin.
Mathematics Extension 1 Years 11 + 12
20
(a) Find the equation of the line PQ.
(b) Show that the line PQ passes through a fixed point.
5 A record player spins a 15 cm radius vinyl disc at 45 revolutions per minute. Find the parametric
equations of a point on the rim of the disc.
6 Kathy starts running clockwise from point A(50,0) at 0.25 rpm around a circle centred at the
origin, radius 50 m. Find her parametric equations.
1.5 Inverse functions
The inverse of a function y =
()fx
is defined by interchanging x and y,∴ x =
()fy
.
For example, the inverse of {(1,3), (2,5), (4,6), (5,6)} is {(3,1), (5,2), (6,4), (6,5)}.
The domain of x
=
()fy
is the range of y =
()fx
.
The range of x =
()fy
is the domain of y =
()fx
.
The graph of x
=
()fy
is the reflection of y =
()fx
about the line y = x.
However, the inverse of a function is not necessarily a function.
• If there exists a
one-to-one correspondence between x and y in the function y
=
()fx
then its inverse
is a function.
3
• If there exists a many-to-one correspondence between x and y in the function y
=
()fx
then its
inverse is not a function. In this case, we need to reduce its domain to where
()fx
is one-to-one.
If the inverse is a function, it may be denoted by y =
1
()fx
−
.
Example 1.9
For each of the following functions:
(i)
( ): 2 4fx y x= −
. (ii)
2
( ): 4 5fx y x x=−+
.
(a) Explain whether the inverse function exists. If it does, find it. If it does not, find the largest domain
of f(x) that contains x = 4 so that the inverse is a function then find the inverse function.
(b) State the domain and the range of the inverse function.
(c) Sketch the curve of the original function (on the restricted domain, if necessary) and its inverse
function.
(d) Find
( )
1
( ) , where is a real number.f fa a
−
(e) Find x for which
1
() ()f x fx
−
=
.
(i)
(a)
The inverse function exists because there is a one-to-one correspondence in
24yx= −
, i.e., for
every value of x there exists one value of y, and for every value of y there exists one value of x.
3
A relation can be one-to-one (e.g., the straight line y = 2x + 1), one-to-many (e.g., the sideway parabola
2
yx=
), many-to-one (e.g., the parabola
2
yx=
) or many-to-many (e.g., the circle
22
4xy+=
).
A function must satisfy a vertical line test, so it can only be one-to-one or many-to-one. The inverse of a many-
to-one function, which becomes one-to-many, is therefore not a function.
Chapter 1: Further work with functions
21
1
: 24
: 2 4.
2 4.
4
.
2
fy x
f xy
yx
x
y
−
= −
= −
= +
+
=
(b)
1
:f
−
Domain = Real, Range = Real.
(c) See diagram.
(d)
11
( ( )) ( )f fx f y x
−−
= =
for all x.
∴
1
( ( ))f fa a
−
=
.
(e) The two curves
1
( ) and ( )fx f x
−
meet each other on the line y = x.
Solving 2x − 4 = x gives x = 4.
∴Point of intersection (4, 4).
(ii)
(a)
The inverse is NOT a function because it fails the horizontal line test.
4
If we define the domain of
()fx
to be {x: x ≥ 2} then upon this domain there exists a one-to-one
correspondence between x and y in y =
()fx
, hence, its inverse is a function.
5
2
12
2
: 4 5, Domain: 2, Range: 1
: 4 5, Domain: 1, Range: 2.
4 5 0.
4 16 4(5 )
.
2
fyx x x y
f xy y x y
yy x
x
y
−
=−+ ≥ ≥
=−+ ≥ ≥
− +−=
± −−
=
4
There is a many-to-one correspondence in
2
45yx x=−+
, i.e., there exist more than one value of x for a value
of y, e.g., let y = 5 we obtain x = 0 and 4 (i.e., a horizontal line cuts the curve at two points. Therefore, it’s
acceptable to declare that if a function f(x) fails the horizontal line test, then its inverse is not a function.
5
To prove a function is one-to-one in an interval [a,b], prove that it is either monotonic decreasing or increasing
in the interval.
y
x
−4
2
2
−4
Mathematics Extension 1 Years 11 + 12
22
4 44
2
2 1.
Since 2, we take 2 1.
x
y
x
y yx
±−
=
=±−
≥ =+−
(b)
1
:f
−
Domain = {x: x ≥ 1}, Range = {y: y ≥ 2}.
6
(c) See diagram.
(d)
If
1
2, ( ( ))a f fa a
−
≥=
, but if
2a <
, i.e., a does not belong to the restricted domain of
()fx
, so
we have to find the corresponding value of a in the restricted domain. Let this value be b, where
() ()fa fb=
.
As the parabola is symmetrical about x = 2, using the midpoint formula,
2, 4
2
ab
ba
+
=∴=−
.
( ) ( )
11
( ) ( ) 4 if 2ffa ffbb aa
−−
= ==−<
.
∴
( )
1
( ) if 2 or 4 if 2f fa aa aa
−
=≥ −<
.
(e) The two curves
1
( ) and ( )fx f x
−
meet each other on the line y = x.
22
Solving 4 5 gives 5 5 0.
5 25 20 5 5
.
22
xx x xx
x
− += − +=
±− ±
∴= =
55
Take because 2.
2
xx
+
= ≥
7
5555
Point of intersection , .
22
++
∴
6
The domain of
1
()fx
−
is the range of
()fx
, and the range of
1
()fx
−
is the restricted domain of
()fx
.
7
The point of intersection of two curves must satisfy the domains of both curves. Here, x ≥ 2 for f(x) and x ≥ 1
for
1
()fx
−
, ∴x ≥ 2.
y
x
(2,1)
(1,2)
Chapter 1: Further work with functions
23
Exercise 1.5
1 Given
()fx
: y =
1x +
.
(a) By finding y ′ show that the function is one-to-one.
(b) Explain why the inverse is a function and find the inverse function
1
()fx
−
.
(c) Find the exact coordinates of any points of intersection of the two curves, if possible.
(d) Find
( )
1
()f fa
−
, where a belongs to the domain of
()fx
.
(e) Sketch both curves in the same graph.
2 Repeat question 1 for the following functions.
(i)
38yx=−+
. (ii)
6yx= −
. (iii)
log
e
yx=
. (iv)
3
x
y
x
=
−
.
(v)
1
2
1
y
x
= −
+
. (vi)
1
x
ye= −
. (vii)
ln(4 )yx= −
. (viii)
3
x
x
e
y
e
=
+
.
3
Given
()fx
: y =
2
2xx+
.
(a) Explain why the inverse is NOT a function and find the largest domain of
()fx
that contains
the point where x = 1 for which the inverse function
1
()fx
−
exists.
(b) Find the inverse function
1
()fx
−
.
(c) Find the coordinates of any points of intersection of the two curves, if possible.
(d) Find
( )
1
()f fa
−
if a belongs to the domain of
()fx
, but not the restricted domain of
()fx
.
(e) Sketch both functions y =
()fx
(on the restricted domain) and y =
1
()fx
−
in the same graph.
4
Repeat question 3 for the following functions.
(i)
2
4yx x= −
. (ii)
2
22yx x=+−
. (iii)
2
( 2)yx=−−
. (iv)
42
8 ,2 2yx x x= − −≤ ≤
.
(v)
2
2
1
y
x
=
+
. (vi)
2
8
9
y
x
=
−
. (vii)
2
ln(4 )yx= −
. (viii)
xx
ye e
−
= +
.
1.6 Review Exercise 1
1 The graph of the curve
()y fx=
is shown below. Sketch the following curves.
(a)
()y fx=
. (b)
()yfx= −
. (c)
( )
y fx=
. (d)
1
()
y
fx
=
.
1 2
y
x
Mathematics Extension 1 Years 11 + 12
24
2 Repeat question 1 for the following graphs.
(α) (β)
3 Given
2
() 2fx x x=−−
.
On separate diagrams, sketch the following graphs without using Calculus.
(a)
()y fx=
. (b)
()yfx= −
. (c)
( )
y fx=
. (d)
1
()
y
fx
=
.
4
Given
( ) and ( )f x gx
, sketch the following graphs
() () and () ()f x gx f x gx+−
. Verify your
answers by proving it algebraically.
(a)
2
() 4, () 2fxx gxx=−=+
. (b)
1, for 3 0
2 4, for 0
( ) , ( ) 1 , for 0 1
1, for 0
1, for 1
x
xx
f x gx x x
xx
xx
−≤<
+<
= = − ≤<
+≥
−≥
.
5 Sketch the following curves, showing axes intercepts and asymptotes. Do not use Calculus.
(a)
2
1
x
y
x
=
−
. (b)
1
( 1)
x
y
xx
−
=
+
. (c)
2
2
1
x
y
x
=
−
. (d)
2
2
( 1)x
y
x
−
=
.
(e)
5
( 2)( 1)
x
y
xx
−
=
−+
. (f)
2
( 1)(5 )
x
y
xx
−
=
+−
. (g)
2
2
2 43
1
xx
y
x
++
=
−
. (h)
2
(2 )( 1)
1
xx
y
x
−+
=
+
.
6
Consider the function
2
( ) (3 )fx x x= −
.
(a) Find
()fx
′
and the coordinates of any turning points.
(b) Hence, sketch the graph of the curve
( ),y fx=
showing clearly the coordinates of any turning
points and axes intercepts.
(c) On separate diagrams, without working, sketch the graphs of the following curves, showing
clearly the coordinates of any turning points and axes intercepts.
(i)
(3 )yfx=
. (ii)
( 3)y fx= +
. (iii)
(3 )yf x= −
. (iv)
( )
y fx=
. (v)
1
()
y
fx
=
.
7 Determine the number of solutions of the equation
4
1xx+−
= 0, hence, find two consecutive
integers between which each of the solutions lies.
8 Determine the values of k for which
x
e kx
−
=
has exactly (i) 1 solution, (ii) 2 solutions.
9 Solve the following inequalities, using a graphical means.
(a)
3 24x −>
. (b)
2 15x +<
. (c)
26 3xx−> +
. (d)
2 1 2 12xx++ −>
.
(e)
–2( + 7m – 2) + 21m > 0. (f) (2m + 1)(m + 2) < – 3.
(g) . (h) . (i) . (j) .
2
m
2
( 1)m +
1
1
t
t
−
≤
−
10 1
1
41
t
t
t
+
≤+
−
1
1
t
t
t
−
<
+
6
32
1
t
t
t
−≤
+
1 3
y = 1
1
y
x
(1,‒2)
x
‒2
y
2
y = ‒1
1
The nature
of proof
HSC Outcomes
A student
develops understanding and fluency in mathematics through exploring and connecting mathematical concepts, choosing
and applying mathematical techniques to solve problems, and communicating their thinking and reasoning coherently
and clearly
selects and applies the language, notation and methods of proof to prove results
In this chapter,
1.1 The nature of proof .......................................... 6
1.2 Inequality proofs ............................................ 11
1.3 Induction proofs ............................................. 14
1.4 Review Exercise 1 ......................................... 19
Solutions ........................................................... 23
Mathematics Extension 2 Year 12
6
1.1 The nature of proof
Mathematicians accept nothing as true without proof. This is because a mathematical statement is
inherently either true or false, but never both. Therefore, to prove the truth of a statement means
providing a logical reason ‒ or a finite sequence of logical reasons ‒ to support its validity. In this
chapter, we’ll first explore the language of proof before diving into six basic proof techniques: Direct,
Contrapositive, Contradiction, Counterexample, Inequality, and Mathematical Induction proofs.
1.1.1 The language and notation of proof
A good proof must use not only logical reasons correctly, but also the language correctly.
(a) Conventions and Context:
It uses conventions. For example, given two points A and B,
AB
is the vector from A to B (not the
other way around),
AB
m
is the gradient of the line AB
1
, but AB could refer to the line AB, the
interval AB, or the distance AB. To differentiate these usages, one must consider the context of the
sentence. For example:
(i) if you are asked to prove that
AB CD⊥
, you must understand that AB and CD refer to two lines,
and the intervals AB and CD may not intersect.
(ii) if you are asked to find the ratio of AC:AB when C is a point that divides AB internally in the
ratio 1:2, then both AC and AB refer to distances.
(b) Symbols:
The following symbols are commonly used:
(i) In set theory,
(belong to), (subset), or (union), or (intersection)∈ ⊂ ∪∨ ∩∧
.
For example, when tossing a die, if P is the set of odd numbers and Q is the set of numbers higher
than 3, then
3 ,3 ,{3} ,{3} , or {1,3,4,5,6}, and or {5}.P Q P QP Q PQ P Q PQ∈ ∉ ⊂ ⊄ ∪ ∨= ∩ ∧=
(ii) In statements,
or PP¬
(not P),
or ⇒→
(implies), and
⇔
(equivalent).
The negation of a statement P is denoted by
or PP¬
(read: “not P”). We’ll use
P¬
in this book.
For example, if P is the statement that a = b, then
P¬
is the statement that
ab≠
; Or if P is the set
of odd numbers then
P¬
is the set of even numbers.
In Mathematics, negate statements include the following:
- the negation of a negation:
( )
PP¬¬ =
.
- the negation of an implication:
( ) ( )
PQ P Q¬ ⇒ = ∧¬
(read: “P and not Q”).
- the negation of a set:
( ) ( ) ( )
PQ P Q¬ ∧ =¬ ∨¬
(read: not the set of P and Q = not P or not Q);
( ) ( ) ( )
PQ P Q¬ ∨ =¬ ∧¬
(read: “not the set of P or Q = not P and not Q).
The symbol
or ⇒→
, as in
or PQPQ⇒→
, is short for ‘if P then Q’. We’ll use
⇒
in this
book. These two statements are equivalent:
( )
and PQ PQ⇒ ¬∨
(read: not P or Q).
2
1
Why is m the symbol for gradient? No one knows, although in CRC Concise Encyclopedia of Mathematics,
Eric Weisstein claimed that it was first used in 1844 by British mathematician Matthew O’Brien. Some of you
may like this explanation of the historian Howard Eves: “because the word slope starts with the letter m”
(Mathematical circles revisited, 1971, page 142).
2
The statement
PQ⇒
(if P then Q) means that if P is true, Q must also be true. The statement
PQ¬∨
(not P or
Q) means that either P is false or Q is true, or both. For example, let P be the statement “It is raining”, and Q be
the statement “The ground is wet”. These two statements (a)
PQ⇒
, which means if it is raining then the ground
is wet, and (b)
PQ¬∨
, which means it is not raining or the ground is wet, are equivalent. This might seem odd,
Chapter 1: The nature of proof
7
The symbol
⇔
(equivalent), for example,
log
x
e
ye x y= ⇔=
, refers to the equivalence of two
statements P and Q, as both
PQ⇒
(P implies Q) and
( implies ), denoted by Q PQ P P Q⇒⇔
,
read as P if and only if Q, commonly abbreviated ‘P iff Q’.
(iii) Other symbols.
-
or ≡≅
(congruent), e.g.,
ABC DEF∆ ≡∆
if they both have sides 3, 4, and 5 units long.
- ||| or
(similar), e.g.,
ABC∆
|||
DEF∆
if their corresponding angles are equal.
-
∴
(therefore), e.g.,
2 1 5, 2xx+=∴=
.
-
(since), e.g.,
6 2 3, ABC>+∆
with side lengths 2, 3 and 6 units cannot be formed.
-
∀
(for all), e.g., “
2
2 , 5,
n
nn n> ∀≥ ∈
” means
2
2 for all values of 5, where
n
n nn>≥
is an
integer.
-
∃
(there exists), e.g., “
2
:2
n
nn∃∈ =
” means there exists a positive integer n such that
2
2
n
n=
.
Students should understand that
= the set of natural numbers
3
,
= the set of integers
4
,
= the
set of rational numbers,
= the set of real numbers, and
= the set of complex numbers.
(c) Specialized vocabulary:
It uses words that may not have the same meaning as in our commonly used language.
For example, you may have asked “What is the combination of this safe?”, when you actually
meant to refer to the permutation of the safe.
5
Can you answer this question? When a die is tossed, what is the probability that the upward face
shows numbers ‘1’ and ‘2’?
The answer is 0, as it cannot show both numbers simultaneously. If the above question changes
“and” to “or”, then the answer would be one-third.
(d) The converse:
The converse of ‘if A then B’ is ‘if B then A’, where A and B are statements.
The converse of a statement is not always true. For example, “if it rained last night then the grass is
wet today” does not imply that “if the grass is wet today then it rained last night”.
1.1.2 Direct proof
Example 1.1
If n is an odd integer, prove that
2
n
is also an odd integer.
but the statement
PQ⇒
only guarantees what happens if it rains. Since it’s not raining, the condition for the
“then” part isn’t met, so the implication itself isn’t violated. The ground could be wet for other reasons, like
sprinklers.
3
Is 0 a natural number? Natural numbers are counting numbers, beginning with 1. So, 0 is not a natural number.
4
Why is
used for integers?
stands for the German word zahlen, which means numbers (which literally
means all numbers, not only integers). Apparently, as ‘I’ is not a good choice for integers (in English, ‘I’ is a
pronoun), ‘Z’ and the next letter ‘J’ have been chosen.
was first used in Grundlagen der Analysis, 1930, by Edmund Landau. We did not know when J was first used,
but J was mentioned in Survey of Modern Algebra, 1953, by Garrett Birkhoff and Saunders MacLane.
In this book,
is used, instead of J, which has been used in my other books. Further, these symbols
,,,
…
are all written in double-struck font.
5
A combination of n different numbers has n! different permutations. For example, a combination of 3 numbers
{1, 2, 3} has these 3! = 6 permutations: 123, 132, 213, 231, 312, and 321.
Mathematics Extension 2 Year 12
8
Let
2 1, where , 1nk k k= + ∈ ≠±
.
2 22
(21)4 41.n k kk= += ++
22 2
As 4 and 4 are even, 4 4 1 is odd, If is an odd integer, then k k kk n n++ ∴
is also an odd integer.
6
1.1.3 Proof by contradiction.
Example 1.2
Prove that
2
is an irrational number.
Assume that
2
is rational. Let
2
p
q
=
, where p and q are positive integers,
0q ≠
, and have no
common factor.
2
2
22
2
2222
2
2 , on squaring both sides.
2 . (1)
is even, is even. Let 2 , where , 0.
Substituting to (1), 4 2 2 .
is even, is even.
p
q
pq
p p pk k k
kqqk
qq
=
=
∴ ∴ = ∈≠
= ⇔=
∴∴
But p and q assumingly do not have any common factors, they cannot both be even. ∴The
assumption that
2
is rational is incorrect. ∴
2
is irrational.
1.1.4 Proof by contrapositive
The contrapositive of “if A then B” is “if not B then not A”, where A and B are statements.
A statement and its contrapositive are equivalent.
For example, “if there is a blackout then the light must be off” is the same as “if the light is on then
there is no blackout”. Note: It does not necessarily mean that if there is no blackout then the light is
on. Use this method if a direct proof is not easy.
Example 1.3
If
2
n
is an odd integer prove that n is odd.
Assume n is not an odd integer, i.e., an even integer, then since the product of 2 even numbers is an
even number, ∴
2
n
is even, i.e.,
2
n
is not odd.
Therefore, by contrapositive, if
2
n
is an odd integer then n is odd.
6
The converse of this statement is also true, but a direct proof is tricky. It’s easy to prove it by contrapositive.
Refer to example 1.3.
However, below is the direct proof:
2
2
Let 2 1, , 0.
1 2.
( 1)( 1) 2
As ( 1) and ( 1) are two integers different by 2, so, if their product is even then both ( 1) and ( 1) are
even. Let 1 2 and 1 2 , where and are s
n kk k
nk
nn k
nn nn
n mn p mp
+
=+∈ ≠
−=
− +=
−+ −+
−= +=
2
ome integers.
2 1 2 1, which is an odd integer. If is an odd integer, then is an odd integer.nm p n n∴= += − ∴
Chapter 1: The nature of proof
9
1.1.5 Proof by counterexamples
Example 1.4
Prove that the statement “
2 1, for 2,
n
nn− ∀≥ ∈
, is a prime number” is not true.
Since the statement refers to all values of n such that
2,nn≥∈
, we only need to find just one
counter-example to declare that the statement is not true.
By trial and error, when
11
11, 2 1 2047 23 89n = −= = ×
.
Therefore,
2 1, for 2,
n
nn− ∀≥ ∈
, is not a prime number.
Exercise 1.1
1 Fill in the blanks with the most correct symbols, using these symbols
, , and∀∃⇒ ⇔
. Briefly
explain your choice.
(a) He scored 3 out of 10 he failed the test. (b) even numbers are divisible by 4.
(c)
2
2 30xx− +≥
x ∈
. (d)
2
x
y =
2
logxy=
.
2 True or false? Justify your answers.
(a)
1
sin siny xx y
−
= ⇔=
. (b)
sin cos 1,xx x< ∀∈
. (c)
22
A B AB= ⇒=
.
(d)
// , 3 3AB CD AB CD=⇒=
. (e)
22
:x x y xy∃∈ + = +
. (f)
( ) ( )
AB CA BC∩ ∪=∩ ∪
.
(g) Lines
and , , , , 0, are iff 0 and // iff .Ax By C Dx Ey F A B D E AD BE AE BD+= += ≠ ⊥ + = =
3 Directly prove the following.
(a) Prove that the angle bisectors of the base angles in an isosceles triangle are equal.
(b) Prove that the square of any integer either is divisible by 3 or gives a remainder of 1 when it is
divided by 3. Hint: Any integer is either divisible by 3 or not.
(c) If
,xy∈
, prove that
xy x y+≤+
algebraically and interpret it geometrically. Consider 4
cases (i)
0, 0xy≥≥
, (ii) x < 0, y < 0, (iii) x > 0, y < 0 and
0xy+≥
(iv) x > 0, y < 0 and
0xy+<
(d)
22
, prove that sec cosec 4
θ θθ
∀∈ + ≥
. (e) Prove that
1
2 1 3, 2,
n
nn
n
< + < ∀≥ ∈
.
Hint for part (e): Use
( )
23
( 1) ( 1)( 2)
1 1 ...
2! 3!
n
n
nn x nn n x
x nx x
− −−
+ =+ + + ++
.
4 Prove by contradiction.
(a) Prove that
3
is irrational. (b) Prove that
6
is irrational.
(c) Prove that
2
log 5
is irrational. (d) Prove that
3
log 5
is irrational.
(e) If
ABAB∪=∩
, where A and B are two non-empty sets, prove that A = B.
(f) In question 3, part (a), we asked you to directly prove that the angle bisectors of the base angles
in an isosceles triangle are equal. But it is not easy to prove the converse by direct method. Prove
by contradiction that if two angle bisectors of a triangle are equal, the triangle is isosceles.
5 Prove by contrapositive.
(a) Suppose
3
, if xx∈
is even, prove that x is even.
(b)
Suppose
,xy∈
, if xy and x − y are even, prove that x and y are both even.
(c) Suppose
, , if ( )xy x y∈−
is even, prove that x and y have the same parity.
Mathematics Extension 2 Year 12
10
(d) Suppose
2
, if 3 1x xx∈ ++
is even, prove that x is odd.
(e) Suppose
, if 2 1
n
n ∈−
is a prime number then n is a prime number.
6 Prove that the following statements are not true by counterexamples.
(a)
3
,3
n
nn∀∈ >
. (b)
22
, 0, ab a b a b∀> +=+
.
(c) A quadrilateral is formed by joining any 4 points in a plane.
(d) The point of inflexion of a curve
()y fx=
corresponds to the point where
() 0fx
′′
=
.
7 Find the mistakes in the following “proofs”.
(a) Question: Prove that 1 = 2.
Let ,
2 , on adding to each side
2 2 2 , on subtracting 2 to each side
2( )
1 2, on dividing by
xy
xy y y
xyxyx x
yx yx
yx
=
+=
+− = −
−= −
= −
(b) Question: Prove that 2 = 3.
22
2 22
4 10 9 15
25 25 25
4 10 9 15 , on adding to each side
44 4
55
2 3 , using 2 ( )
22
55
2 3 , on taking square roots
22
23
a ab b a b
−=−
−+ =−+
− =− − +=−
−=−
∴=
(c) Question: Prove that 1 = 2 = 3 = 4 = …
( )
( )
22
22
22
2
22
2
22
2
2
2
2 1 ( 1) .
11 11
2 1 2( 1) ( 1) 2( 1) .
22 22
1
LHS 2 1 2 3 1
2
1
2
2
11
2
22
1
.
2
1
RHS ( 1)
2
xx x
xx xx x x xx x
xx xx x
x xxx
x xx x
xx
xx
+ += +
++− + + ++ =+− + +++
= + +− − −+ +
= − −+ +
=− + ++
= −+
= +− +
( ) ( )
2
.
11
LHS RHS, ( 1) .
22
1.
Thus, when 1 we have 1 2, when 2 we have 2 3, and so on, 1 2 3 4 ...
xx x x
xx
xx
= ∴−+ =+−+
∴=+
= = = = ∴= = = =
Chapter 1: The nature of proof
11
1.2 Inequality proofs
1.2.1 Inequality by Calculus
Example 1.5
Prove that
ln 1
. Hence, deduce that
ex
x
xe
xe
≤≤
.
ln
Let ( )
x
fx
x
=
.
22
1
ln
1 ln
() .
xx
x
x
fx
xx
×−
−
′
= =
2
4 43
1
2 (1 ln )
2 2 ln 2 ln 3
() .
xx x
x x xx x
x
fx
x xx
−× − −
−− + −
′′
= = =
( )
3
( ) 0 when ln 1, .
11
When , ( ) 0 : , is a maximum point.
fx x x e
x ef e e
ee
′
= =∴=
−
′′
= = <
ln 1
.
ln , on multiplying by , noting 0.
.
.
x
e
ex
x
xe
x
x xx
e
xe
xe
∴≤
∴≤ >
∴≤
∴≤
1.2.2 Inequality by Algebra
Example 1.6
(a) (i) Prove that
222
a b c ab ac bc++≥ ++
, if a, b, c are any real numbers.
(ii) Hence, prove that
222 2
3( ) ( )a b c abc+ + ≥ ++
.
(iii) By multiplying both sides of the result in part (i) by (a + b + c), prove that
333
3a b c abc++≥
.
(b) (i) If a, b are any positive real numbers, prove that
2a b ab+≥
.
(ii) Hence, if a, b, c, d are positive real numbers, prove that
(
α
)
4
4
a b c d abcd+++ ≥
. (
β
)
3
3a b c abc++≥
.
(a) (i) Since
2 22
( ) 0, 2 0a b a b ab− ≥ +− ≥
,∴
22
2a b ab+≥
.
Similarly,
22
2a c ac+≥
and
22
2b c bc+≥
.
222
2( ) 2( )a b c ab ac bc∴ ++ ≥ ++
, on adding the above lines.
∴
222
a b c ab ac bc++≥ ++
. (1)
(ii) Adding
222
abc++
to both sides of (1) gives
222 222
3( ) 2( ).abc abc abacbc++ ≥+++ ++
222 2
3( ) ( ) .a b c abc∴ + + ≥ ++
Mathematics Extension 2 Year 12
12
(iii)
( )
( ) ( )( )
222
.a b c abc abacbcabc+ + ++ ≥ + + ++
7
333222222222222
333
3.
3.
a b c ab ac ba bc ca cb ab ac ba bc ca cb abc
a b c abc
++++++++≥++++++
∴++≥
(b) (i) If a, b > 0, since
2
( ) 0, 2 0a b a b ab− ≥ +− ≥
, ∴
2a b ab+≥
. (2)
(ii) (
α
) Similarly, if
, 0, 2c d c d cd> +≥
. (3)
∴
( )
( )( )2a b c d ab cd+++ ≥ +
, on adding (2) and (3).
∴
44
4 since 2 2a b c d abcd ab cd ab cd abcd+++ ≥ + ≥ =
.
(
β
) Let
3abc d++=
then
4
44a b c d d abcd+++ = ≥
, from the result of part (
α
).
4
d abcd∴≥
.
∴
3
d abc≥
.
∴
3
d abc≥
.
3
But , 3
3
abc
d a b c abc
++
= ∴++≥
.
The results in (
α
) and (
β
) are special cases where n = 4 and 3 of the Cauchy’s inequality. Generally, it
is expressed as: The arithmetic mean of n positive numbers is always greater than or equal to their
geometric mean, i.e.,
12
12
n
n
n
aa a
aa a
n
+++
≥
. Do not use it without proof.
Exercise 1.2
1 Show by Calculus that
ln , 0x xx> ∀>
.
2 If
() 1
n
x
x
fx e
n
= −
.
(a) Find the coordinates of any stationary points of f(x).
(b) Sketch the curve.
(c) Deduce that
1 for
n
x
x
e xn
n
−
≤− <
.
(d) Show that
1
1
nn
nn
e
nn
+
≤≤
−
.
7
In Chapter 2, Section 2.4, of our book ‘Mathematics Extension 1’, we employed the following brilliant
notations by Euler:
2 2 22
,, ,
α α β γ α α β γ αβ αβ αγ βγ α αβγ
∑=++∑ = + + ∑ = + + ∏=
.
Using these notations, these three lines can be written as:
( )
( ) ( )( )
2
32 2
3
.
3.
3.
a a ab a
a ab ab a
aa
∑ ∑ ≥∑ ∑
∑ +∑ ≥∑ + ∏
∴∑ ≥ ∏
Clearly, it’s more convenient than writing all of them out.
Chapter 1: The nature of proof
13
3 (a) Show that
sin xx<
for
0x >
and deduce that
0
sin
lim 1
x
x
x
→
=
. Hence, sketch the curve
sin x
y
x
=
.
Do not find any stationary points.
(b) Show that
3
sin 0
6
x
xx−+ >
for
0x >
and deduce that
2
sin
1,
6
xx
x
x
≥ − ∀∈
.
4 If a, b are two positive real numbers, prove that
(a)
2
ab
ab
+
≥
. (b)
2
ab
ba
+≥
. (c)
11
() 4ab
ab
+ +≥
. (d)
2
22
11
() 8ab
ab
+ +≥
.
5 Suppose
, and ab c
are the sides of a triangle, explain why
ab c−<
, hence, prove that
(a)
( )
222
2a b c ab ac bc++< ++
. (b)
( )
2
( )4a b c ab ac bc++ < + +
.
6 (a) If
, , 0 and (1 )(1 )(1 ) 8, prove that 1x y z x y z xyz> + + += ≤
.
(b) If
111
, , 0 and 1, prove that 1 1 1 64xyz x y z
xyz
> ++= + + + ≥
. You may use the fact
that
4
4 , ,,, 0a b c d abcd a b c d+++ ≥ ∀ >
, proven in example 1.6.
7 If a, b, c are positive real numbers prove that
(a)
( )
222
0a b c ab ac bc++− ++ ≥
. (b)
( )( )( ) 8a b a c b c abc+ + +≥
.
(c)
333
3a b c abc++≥
, by expanding
( )
222
()a b c a b c ab ac bc++ + + − − −
.
(d)
( )
22 22 2 2
0a b b c c a abc a b c+ + − ++ ≥
. (e)
444
()a b c abc a b c+ + ≥ ++
.
(f)
2
( ) 3( )a b c ab ac bc++ ≥ + +
. (g)
111
6abc
abc
+++ + + ≥
.
(h)
111
() 9abc
abc
++ + + ≥
. (i)
2
222
111
( ) 27abc
abc
++ + + ≥
.
(j)
3
2
abc
bc ca ab
++≥
++ +
. (k)
222
2
a b c abc
bc ac ab
++
++≥
+++
.
8 If a, b, c are positive real numbers prove that
(a)
333
2( ) ( ) ( ) ( ) 6a b c ab a b ac a c bc b c abc++ ≥ ++ ++ +≥
.
(b)
333 222
3( ) ( )( ) 9a b c a b c a b c abc+ + ≥ ++ + + ≥
.
9 If a, b, c are the sides of a triangle, assuming a ≥ b ≥ c, prove that
(a)
()()()a b c a c b b c a abc+− +− +− ≤
.
(b)
222
3()()()0abc a b c a b a c b c a b c− +− − +− − +− ≥
.
When does the equality hold true in each of the above parts?
10 Factorise
(a)
333
3a b c abc++−
(b)
3 333
()abc a b c++ − − −
(c)
222
()()( )abc bc a ca b−+ −+ −
.
11 (a) If
22
1xy+=
prove that
2xy+≤
.
(b) If
2 2 22
1x yab+=+=
prove that
1ax by+≤
.
12 Let
,,abc
+
∈
.
(a) If
222
a b c ab ac bc++=++
, prove that
abc= =
.
(b) Prove that
333
3a b c abc++=
if and only if
abc= =
or
0abc++=
.
Mathematics Extension 2 Year 12
14
13 (a) Prove that
1
( 1) , , : 1
2
y
y xy x xy x y
+
≤ −+ ≤ ∀ ∈ ≤≤
.
(b) Hence, prove that
( )
( )
1
!,
2
n
n
n
nn n
+
≤ ≤ ∀∈
.
1.3 Induction proofs
In Chapter 6 of our book ‘Mathematics Extension 1’, we covered two types of Mathematical
Induction: the series type and the divisibility type. In this chapter, we will learn the following types:
the product type, the inequality type, the strong induction type, and miscellaneous types.
1.3.1 The Product type
Example 1.7
Prove by Mathematical Induction that
22
2
1 11 1 1
1 1 1 1 , 2,
49 2
n
k
n
nn
k nn
=
+
− = − − − = ∀≥ ∈
∏
.
Let n = 2, LHS =
13
1
44
−=
. RHS =
21 3
22 4
+
=
×
. ∴It’s true for n = 2.
Assume
22
2
1 11 1 1
:1 1 1 1
49 2
n
k
n
n
k nn
=
+
∃∈ − = − − − =
∏
(Read: Assume there exists a natural
number n such that …).
Required to prove that
( )( )
1
2 22
2
1 11 1 1 2
1 11 1 1
4 9 ( 1) 2( 1)
n
k
n
k nn n
+
=
+
−=− − − − =
++
∏
.
( )
( )
( )
( )
( )
( )
2
2
2
2
2
2
11
LHS 1 , from the assumption,
2 ( 1)
1 ( 1) 1
2 ( 1)
12
2 ( 1)
1 ( 2)
2 ( 1)
n
nn
nn
nn
n nn
nn
n nn
nn
+
= −
+
+ +−
=
+
++
=
+
++
=
+
2
RHS.
2( 1)
n
n
+
= =
+
∴The statement holds true for n + 1 if it holds true for some integer n.
∴By the principle of Mathematical Induction, it’s true for all integers n ≥ 2.
1.3.2 The Inequality type
Example 1.8
Prove that
2
2 for all integers 5
n
nn>≥
.
Let n = 5, LHS =
52
2 32, RHS 5 25, LHS RHS= ==∴>
,∴It is true for n = 5.
Assume
2
: 2
n
nn∃∈ >
.
Chapter 1: The nature of proof
15
Required to prove that
12
2 ( 1)
n
n
+
>+
.
LHS =
1
2 22
nn+
= ×
2
2 22 2
2 , by assumption.
We now need to prove that 2 ( 1) 2 1, which is equivalent to 2 1, for 5.
n
nn n n n n n
>
>+ =+ + > + ≥
2
22
2
M
.
ethod By Calculus, let ( ) 2 1 for 5, .
( ) 2 2 8 0 for 5, ( ) is increasing for 5
But (5) 25 11 14,
( ) 14 0.
2 1 0 or 2 1 for 5.
By2 graphs, the 2 cM urves
1:
ethod :
fx x x x x
f x x x fx x
f
fx
xx xx x
yx
=−− ≥ ∈
′
= −≥> ≥∴ ≥
= −=
∴ ≥>
∴ − −> > + ≥
=
and 2 1 meet at 1 2.
The points of intersection approximate ( 0.4,0.17) and (2.4,5.8).
yx x=+=±
−
22
22
Since 2 1 for 2.4, 2 1 for 5.
2 ( 1) .
xxxxxx
nn
≥+ ≥ >+ ≥
∴ >+
∴The statement holds true for n + 1 if it holds true for some integer n.
∴By the principle of Mathematical Induction, it’s true for all integers n ≥ 5.
1.3.3 The Strong Induction type
Example 1.9
Consider the Fibonacci sequence: 1, 1, 2, 3, 5, 8, … which is defined as
12 1 1
1 and
nnn
TT T T T
+−
= = = +
.
Prove by Mathematical Induction that
7
4
n
n
T
<
for all integers n ≥ 1.
Let n = 1,
1
7
1
4
T = <
; and let
2
2
2
7 49
2, 1
4 16
nT= =<=
.
∴It is true for both n = 1 and 2.
8
Assume that
1
1
77
and
44
nn
nn
TT
−
−
<<
for some integer n.
Required to prove that
1
1
7
4
n
n
T
+
+
<
.
8
Why is it “strong”? It’s called “strong” because the inductive hypothesis is stronger. In this example, you prove
the truth for P(n + 1) by assuming the truth for P(n ‒ 1) and P(n), after proving the truth for both P(1) and P(2).
(2.4,5.8)
y
x
Mathematics Extension 2 Year 12
16
11
LHS , by definition of the sequence,
n nn
T TT
+−
= = +
1
1
1
1
1
12
1
77
, by assumptions,
44
77
1
44
7 11
44
7 44
4 16
7 49
4 16
77
44
7
.
4
nn
n
n
n
n
n
n
−
−
−
−
−
−
+
<+
= +
=
=
<
=
=
∴The statement holds true for n + 1 if it holds true for some integers n and n – 1.
∴By the principle of Mathematical Induction, it’s true for all integers n ≥ 1.
1.3.4 Miscellaneous types
Example 1.10
In a room of n people, if everyone has to shake hands with each other once, prove by Mathematical
Induction that the number of handshakes is
( 1)
2
nn−
.
If there are two people, obviously there is only one hand-shake.
Substituting n = 2 into the formula gives
2( 2 1)
1
2
−
=
. ∴Therefore, the formula holds true for n = 2.
Assume that there are
( 1)
2
nn−
handshakes amongst n people.
Required to prove there are
( 1)( )
2
nn+
handshakes if there are n + 1 people.
Now, suppose a new guest arrives. They must shake hands with everyone already in the room; thus,
they must shake n hands.
∴The total number of hand-shakes =
( 1)
2
nn
n
−
+
2
2
2
( 1)( )
, as required.
2
nnn
nn
−+
=
+
=
∴By the principle of Mathematical Induction, it’s true for all integers n ≥ 2.
Chapter 1: The nature of proof
17
Exercise 1.3
1 Prove by Mathematical Induction for all integers n ≥ 1, unless stated otherwise.
(a)
3
2 222 2 2
1 1 1 1 ... 1
3 4 5 ( 1)
n
k
k n nn
=
−=−−− −=
−
∏
for all integers n ≥ 3.
(b)
2
1
(2 1)(2 3) 2 3
(2 1) 3(2 1)
n
k
kk n
kn
=
−+ +
=
++
∏
. (c)
1
2 ( 1)( 2)
1
2
n
k
nn
k
=
++
+=
∏
.
(d)
2
22
1
( 1) 3
n
k
n
kk n
=
+
−=
+
∏
. (e)
2
3
21
1
(2 1)
n
k
k nn
=
−=
−
∏
.
(f)
( )
( 1)( 2)...(2 ) 2 1 3 5 ...(2 1) for all integers 2
n
nn n n n+ + = ××× − ≥
.
2 Prove by Mathematical Induction for all integers n ≥ 1, unless stated otherwise.
(a)
321
n
n≥+
. (b)
1
!2
n
n
−
≥
. (Note:
! 1 2 3 ...nn=××× ×
)
(c)
2 ( 1)
n
nn>+
for n ≥ 5. (d)
3 ( 1)( 2)
n
nn n>++
for n ≥ 5.
(e)
3
3
n
n>
for n ≥ 4. (f)
2
32
n
n>
for n ≥ 2.
(g)
32
23
n
n
>
for n ≥ 2. (h)
2
2
53
n
n
>
for n ≥ 4.
3 (a) Prove by Mathematical Induction that
12 1 12 2
sin( ) sin sin sin
n nn
xx x a xa x a x+++ = + ++
,
where
12
,,,
n
aa a
are real numbers such that
1
i
a ≤
for
1, 2, ,in=
.
(b) Hence, deduce that
sin sin for 0nx n x x
π
≤ ≤≤
.
4 (a) Prove that
43
1 2 3 , for 1
6
n
n nn
+
+ + ++ ≤ ≥
, by Mathematical Induction.
(b) Prove that
2
123
3
n
nn≤ + + ++
by using a graphical means.
(c) Hence, estimate
1 2 3 100+ + ++
to the nearest hundred.
5 (a) Prove by Mathematical Induction that
22 2
11 1 1
1 2 , for all 1
23
n
nn
+ + + + ≤− ≥
.
(b) Prove that
2 22 2
11 1 1 1 13 1
, hence, deduce that 1 , for all 1
1 23 2 1
n
n nn n n
>− + + + + >− ≥
++
.
(c) Show that
22 2
11 1
1.49 1 1.99
2 3 99
<+ + + + <
.
6 (a) Let n be a positive integer and
12
, ,...,
n
aa a
be n positive real numbers. Prove by Mathematical
Induction that
( )
2
12
12
11 1
... ...
n
n
aa a n
aa a
+ ++ + ++ ≥
for all integers
1n ≥
.
(b) Deduce that
(i)
22
sec cosec 4xx+≥
. (ii)
2 22 2
sec cosec cot 9cosx xx x+ +≥
.
7 Let
1 1 11
... , where 1
1 2 2 12
n
tn
nn n n
= + ++ + ≥
++ −
.
(a) Graphically show that
1
ln 2
2
nn
tt
n
< <+
.
Mathematics Extension 2 Year 12
18
(b) Let
111 1 1
1 ...
234 2 12
n
s
nn
=−+−++ −
−
. Prove by Mathematical Induction that
nn
st=
.
(c) Hence, find to 3 decimal places, the value of
111 1 1
1 ...
2 3 4 9999 10 000
−+−++ −
.
8 A sequence
{}
n
U
is given by
1
( 1)( 2)nn++
for all counting numbers n ≥ 1.
(a) Prove by Mathematical Induction that
1
1
( 1)( 2) 2( 2)
n
k
n
kk n
=
=
++ +
∑
.
(b) Evaluate
1
1
lim
( 1)( 2)
n
n
k
kk
→∞
=
++
∑
. Let this result be S.
(c) Evaluate
1
( 1)( 2)
dx
xx
∞
++
∫
. Let this result be I.
(d) Find the appropriate area under the curve
1
( 1)( 2)
y
xx
=
++
for x ≥ 1 by the trapezoidal rule
using equal strips of unit width. Let the result be T.
(e) Using a graphical means, explain the relations of S, I and T.
9 A sequence
n
T
is defined by
12 1 1
5, 7 and 3 2 .
n nn
T T T TT
+−
= = = −
Prove by Mathematical Induction for all integers n ≥ 1 that
32
n
n
T = +
.
10 The Fibonacci sequence
n
T
is defined by
12 2 1
1 and .
n nn
TT T T T
++
= = = +
(a) Prove by Mathematical Induction for all n ≥ 1 that
, where and , ,
5
nn
n
T
αβ
α βα β
−
= >
are
the roots of the equation
2
10xx− −=
.
(b) Prove that
3n
T
is always even for all integers n ≥ 1.
(c) Assume the ratio
1n
n
T
T
+
approaches a limit as
n →∞
, show that
1
15
2
n
n
T
T
+
+
→
.
11 A sequence
n
S
is defined by
12 1 2
1, 2 and ( 1) , for 2
nn n
S S SS n S n
−−
= = = +− ≥
.
(a) Find
34
and SS
.
(b) Prove that
( 1)x x xx+> +
.
(c) Prove by Mathematical Induction that
! for all integers 1
n
Sn n≥≥
.
12 (a) Prove that if
22
( ) then ( ( ))
1 12
xx
fx f fx
xx
= =
++
.
(b) Prove by Mathematical Induction that
2
times
( (... ( )...)
1
n
x
f f fx
nx
=
+
for all integers n ≥ 2.
13 (a) If a and b are two positive numbers, explain why
( )( ) 0
nn
a ba b− −≥
.
(b) Hence, prove by Mathematical Induction that
22
n
nn
a b ab++
≥
for all positive integers n ≥ 1.
14 Prove by Induction that the sum of exterior angles of a n-sided convex polygon is 360°, n ≥ 3.
Chapter 1: The nature of proof
19
15 Given n points in a plane, no three points are collinear, prove by Induction that the number of lines
joining any two points is
( 1)
2
nn−
for all integers n ≥ 2.
16 Prove by Mathematical Induction that the number of diagonals in a n-sided regular convex
polygon is
( 3)
2
nn−
for all integers n ≥ 3.
17 Prove by Mathematical Induction these binomial theorems:
(a)
( )
0
, 1
n
n
n nr r
r
r
a b Ca b n
−
=
+= ≥
∑
. (b)
0
1
1
, 1 and 2
(1 )
k
n
k
nk
x xn
k
x
∞
=
+−
= <≥
−
∑
.
1.4 Review Exercise 1
1 (a) Directly prove that the general solution of
is
x
dy
y y Ae
dx
= =
, where A is a constant.
(b) Prove by contradiction that given three non-collinear points in a plane, there exists one and
only one circle through these three points.
(c) Prove by contrapositive that if (
33
mn+
) is odd then (m + n) is odd.
(d) Prove by counterexample that if m and n are irrational numbers then (m + n) is irrational.
2 Prove by Mathematical Induction that
1111 1
...
1234 2 12
n
n
+++++ >
−
for all integers
2n ≥
.
3 (a) Using Mathematical Induction show that for each positive integer n there are unique positive
integers
n
p
and
n
q
such that
(1 2 ) 2
n
nn
pq+=+
.
(b) Hence, show that
22
2 ( 1)
n
nn
pq−=−
.
4 (a) Show that for k ≥ 0,
2 3 2 ( 1)( 2)k kk+> + +
.
(b) Hence, prove by Induction that for n ≥ 1,
11 1
1 2 11
23
n
n
+ + + + > +−
.
5 It is given that a, b, c and d are positive numbers. Prove the following inequalities.
(a)
22
2a b ab+≥
. (b)
2
( ) 4( )a b c d ab ac bd cd+++ ≥ + + +
.
(c)
( )
44 44 44 222 2 2 2
ab ac bc abc a b c+ + ≥ ++
. (d)
222 2
3( ) ( )a b c abc+ + ≥ ++
.
(e)
222
222
a b c abc
b c a bca
+ + ≥++
. (f)
222 2 2 2
( )( ) 9abca b c
−−−
++ + + ≥
.
6 Prove the following inequalities. Do not use Mathematical Induction.
(a)
1.3.5...(2 1)
n
nn−≤
.
(b)
(
)
11 1
1 ... 2 1 1
23
n
n
+ + + + > +−
.
(c)
0 ( )( )( ) , if , ,a b c a c b b c a abc a b c< +− +− +− ≤
are the sides of a triangle.
7 Let n be a positive integer and x be any positive approximation to
n
. Choose y so that xy = n.
(a) Prove that
2
xy
n
+
≥
.
Mathematics Extension 2 Year 12
20
(b) Suppose that
xn>
. Show that
2
xy+
is a closer approximation to
n
than x is.
(c) Suppose
xn<
. How large must x be in terms of n for
2
xy+
to be a closer approximation to
n
than x is?
8 The general term of a sequence is defined by
1
n
un n= +−
for all positive integers n.
(a) Show that
1nn
uu
+
<
, and that
lim( 1 ) 0
n
nn
→∞
+− =
.
(b) Show that
1
(1 )
11
n
k
n
kk
n
=
+− =
++
∑
.
(c) If
12
,,,
n
aa a
form an AP prove that
12 23 1 1
11 1 1
nn n
n
aa aa a a aa
−
−
+ ++ =
++ ++
.
9 (a) Prove by Mathematical Induction that
22 2
11 1 1
1 2 , for all 1
23
n
nn
+ + + + ≤− ≥
.
(b) Prove that
2
11 1
1n nn
>−
+
, hence, prove that
22 2
11 13 1
1 , for all 1
23 2 1
n
nn
+ + + + >− ≥
+
.
(c) Show that
22 2
11 1
1.49 1 1.99
2 3 99
<+ + + + <
.
10 (a) By decomposing
2
( 1)( 2)xx x++
into partial fractions, evaluate S =
2
2
( 1)( 2)
n
k
kk k
=
++
∑
.
(b) Evaluate I =
1
2
( 1)( 2)
n
dx
xx x++
∫
.
(c) Find the appropriate area under the curve
2
( 1)( 2)
y
xx x
=
++
for
1 xn≤≤
by the trapezoidal
rule using equal strips of unit width. Let the result be T.
(d) Explain why S < I < T.
11 Given
222 2
( 1)(2 1)
123
6
nn n
n
++
++++=
. Evaluate the following series.
(a)
222 2
2 4 6 (2 )n++++
. (b)
222 2
1 3 5 (2 1)n++++ −
.
(c)
222 2
2 5 8 (3 1)n++++ −
. (d)
222 2
1 4 7 (3 2)n++++ −
.
12 In the following series, each term is written in a pair of brackets. Write an expression in terms of
n for the n
th
term, hence, evaluate the sum of the first n terms of each of the following series.
(a) (1) + (2 + 3) + (4 + 5 + 6) + … (Hint: You are given that
22
3
1
( 1)
4
n
k
nn
k
=
+
=
∑
)
(b) (1) + (2 + 4) + (8 + 16 + 32) + …
(c) (1) + (2 + 3 + 4) + (5 + 6 + 7 + 8 + 9) + …
13 (a) Prove by Calculus that
2
1
( ) , , has a minimum when
1
n nt
fs t s s
s ts n
=+> =
−+
.
(b) Suppose
0
i
x >
for
1, 2, ,in=
, where
2n ≥
. Using the result in part (a), show by
Mathematical Induction that
2
12
11 1
n
n
xx x s
+ ++ ≥
, where s denotes
12 n
xx x+++
.
Chapter 1: The nature of proof
21
14 It is given that a, b and c are three positive real numbers. Prove that
(a)
222
abc
abc
bca
+ + ≥++
.
(b)
333
abc
abc
bc ac ab
+ + ≥++
.
(c)
22 22 22
2( )
ab ac bc
abc
cba
+++
+ + ≥ ++
.
(d)
222 2 2 2
2
a b c ab bc ca
ab bc ac
++
≥++
+++
.
(e)
111
( 1)( 1)( 1) 8, given that 1abc
abc
− − −≥ ++=
.
15 (a) Let a, b, c, and d be four positive real numbers such that a + b + c + d = 4. Prove that
222 2
1111
2
1111abcd
+++ ≥
++++
.
(b) What is the minimum value of
222
111
if 3
111
abc
abc
+ + ++=
+++
?
16 It is given that a and b are two positive real numbers. Let their QM (quadratic mean) =
22
2
ab+
,
AM (arithmetic mean) =
2
ab+
, GM (geometric mean) =
ab
, and HM (harmonic mean) =
2
11
ab
+
.
9
Prove that QM
≥
AM
≥
GM
≥
HM.
17 It is given that a, b and c are three positive real numbers. Let their QM (quadratic mean) =
222
2
abc++
, AM (arithmetic mean) =
3
abc++
, GM (geometric mean) =
3
abc
, and HM
(harmonic mean) =
3
111
abc
++
.
Prove that QM
≥
AM
≥
GM
≥
HM.
18 (a) Prove that
2
ln( 1) ,
2
x
xx x
+
+ > − ∀∈
.
(b) Prove that
tan , 0,
2
xxx
π
> ∀∈
.
(c) Prove that
3
tan , 0,
32
x
xx x
π
> + ∀∈
.
9
The arithmetic mean is the average used when the data is additive in nature (e.g., the average of a class’s test
scores); the geometric mean is the average when the data points are linked by multiplication (e.g., the average of
a set of discount rates); the harmonic mean is the reciprocal of the arithmetic mean of the reciprocals of the data
points (e.g., the average speed of a set of speeds); the quadratic mean (also known as ‘root mean square’) of a set
of numbers is the square root of the arithmetic mean of the squares of the set. It’s useful when the set has both
positive and negative numbers, as you’d get zero with a simple average (e.g., the effective value of AC voltage).
Mathematics Extension 2 Year 12
22
19 (a) Prove by Mathematical Induction for n positive numbers,
1n ≥
, their arithmetic mean is
always greater than or equal to their geometric mean, i.e.,
12
12
n
n
n
aa a
aa a
n
+++
≥
.
(b) Show that if the term
n
a
is allowed to take any positive real number, while the other terms are
kept fixed, then the ratio of the arithmetic mean to the geometric mean is minimum when
n
a
is
the arithmetic mean of
12 1
,,,
n
aa a
−
.
20 (a) Determine the range and the coordinates of the turning points of
( ) cos(sin )fx x=
and
( ) sin(cos )gx x=
, hence, sketch the curves
( ) and ( )y f x y gx= =
.
(b) Algebraically, prove that
cos(sin ) sin(cos ) for all x xx>∈
.
21 Let a, b, c be positive real numbers such that
1ab ac bc++=
.
(a) Prove that we can let
tan , tan , tan , where 2 , 2 , and 2abc
α β γ αβ γ
= = =
are angles of a
triangle.
(b) Prove that
222
222
222 111
3
111 222
abc abc
abc abc
−−−
++≥ ++
−−−
.
22 The following are a few famous inequality theorems.
(a) Nesbitt’s inequality:
3
, , , prove that
2
abc
abc
bc ac ab
+
∀∈ + + ≥
+++
.
(b) Bernoulli’s inequality:
1, , if all
ii
xi x∀ >− ∈
have the same sign, prove that
1 2 12
(1 )(1 )...(1 ) 1 ...
nn
x x x xx x+ + + ≥+ + + +
.
Hence, derive its corollary:
For 1, (1 ) 1
n
x x nx>− + ≥ +
.
(c) Cauchy-Schwarz inequality: Prove that
22
1 2 12
11
, ,..., , , ,..., ,
nn
n n ii
ii
aa a bb b a b
= =
∀∈
∑∑
2
1
n
ii
i
ab
=
≥
∑
. (Note: We will prove this inequality for n = 2 by using vector in the next chapter.)
Hence,
(i)
,, ,abc
+
∀∈
prove that
22 22 22
ab ac bc
abc
ab ac bc
+++
+ + ≥++
+++
.
(ii)
222
3
, , , where 1, prove that
2
abc
a b c ab ac bc
bc ac ab
+
∀ ∈ ++= + + ≥
+++
.
(d) Jensen’s inequality:
( )
12
Let ( ) be a convex function on the interval , , and , ,...,
n
f x ab
αα α
be
real numbers such that
12
... 1,
n
αα α
+ ++ =
prove by Mathematical Induction that
( )
11
nn
ii i i
ii
f x fx
αα
= =
≤
∑∑
.
Hence, if
,,
αβγ
are angles of a triangle, prove that
(i)
tan tan tan 3
222
αβγ
++≥
.
(ii)
cot cot cot 3
αβγ
++≥
.
(iii)
33
sin sin sin
2
αβγ
++≤
.
(iv)
tan tan tan 3 3, iff the triangle is acute.
αβγ
≥
Chapter 1: The nature of proof
23
Solutions
Exercise 1.1
1
(a)
⇒
, he failed the test because he scored 3/10.
(b)
∃
, only some are divisible by 4.
(c)
2
, because 2 2 0xx∀ − +>
always.
(d)
⇔
, because it is how logarithm is defined.
2
(a)
false, because it works only for
22
x
ππ
− ≤≤
.
(b) true, because
11
sin cos sin 2 1
22
xx x= ≤<
.
(c) false, missing
AB= −
.
(d) false, line CD // line AB. The length of CD is
unknown.
(e) true, when x = 0.
(f) false, as seen in the Venn diagram below.
LHS =
RHS =
(g) true, as the 2 lines are perpendicular iff
( )
( )
1 0,
AD
AD BE
BE
− − =−⇔ + =
and parallel iff
AD
AE BD
BE
−=−⇔ =
.
3
(a)
Data:
is isosceles with .
bisects , bisects .
Aim:
Prove that .
ABC ACB ABC
BN B CM C
BN CM
∆ ∠=∠
∠∠
=
Consider and ,
is common
(given. Let them be )
(both )
2
(AAS)
(corresponding sides of congruent
triangles)
BNC CMB
BC
ACB ABC
NBC MCB
BNC CMB
BN CM
θ
θ
∆∆
∠=∠
∠=∠ =
∴∆ ≡ ∆
∴=
(b)
If 3 , nk=
( )
( )
22
2
22 2
2
22 2
2
9,
is divisible by 3.
If 3 1,
9 6 1 3 3 2 1,
gives a remainder of 1 when divided by 3.
If 3 2,
9 12 4 3 3 4 1 1,
gives a remainder of 1 when divided by 3.
nk
n
nk
nkk kk
n
nk
nk k kk
n
=
∴
= +
= + += + +
∴
= +
= + += + + +
∴
∴The square of any integer either is divisible by 3
or gives a remainder of 1 when it is divided by 3.
(c) Case 1: If
0, 0xy≥≥
.
,,x xy yxy xy= = +=+
.
.
Case 2: If 0, 0.
,, .
xy x y
xy
x xy yx y x y
∴+ = +
<<
=− =− + =−−
A B C
A B C
A
M N
B C
Mathematics Extension 2 Year 12
24
.xy x y∴+ = +
Case 3: If 0, 0 and 0.
,, ,
.
x y xy
x xy yxy xyxy x y
xy x y
> < +≥
= =− + =+≤−= +
∴+ ≤ +
Case 4: If 0, 0 and 0.
,, ,
.
,, .
x y xy
x xy yxy xyxy x y
xy x y
xy x y x y
> < +<
= =− + =−− < − = +
∴+ < +
∴∀ ∈ + ≤ +
Note: Due to symmetry, we do not need to
consider x < 0, y > 0.
If
and xy
represent the lengths of two sides of
a triangle, then
xy+
represents the length of the
third side. This is the triangular inequality: A side
of a triangle is less than the sum of the other two
sides. The equal sign occurs when the triangle is
flat.
(d) LHS =
22
11
cos sin
θθ
+
22
22
sin cos
cos sin
θθ
θθ
+
=
2
2
1
4, as sin 2 1
1
sin 2
2
θ
θ
=≥≤
= RHS.
(e)
23
1 ( 1) ( 1)( 2)
1 1 ...
2! 3!
n
n nn nn n
n nn n
− −−
+ =++ + +
23
23
23
2
2
1
( 1) ( 1)( 2) 1
2 ... . (1)
2! 3!
1
Since line (1) 2, 1 2.
1
(1) 2 ...
2! 3!
11 1
2 ... , since 2,
12 123
11 1
2 ...
22 2
11
2 ... (to infinity)
22
n
n
n
n
n
n
n
nn nn n
nn n
n
nn
nn n
n
n
+
− −−
=+ + ++
>+>
<+ + ++
=+ + ++ ≥
× ××
<++ ++
<++ +
1
2
2 , using the limiting sum of this GP
1
1
2
2 1 3.
<+
−
= +=
1
21 3
n
n
∴< + <
.
4
(a)
Assume that
3
is rational. Let
3
p
q
=
,
where p and q are integers,
0q ≠
, and have no
common factors.
2
2
3.
p
q
=
22
3.pq=
(1)
2
3 is a factor of p∴
. But 3 is a prime number and
p is an integer,
∴3 is also a factor of p.
22
Let 3 , where is an integer.
Substituting to (1),
9 3.
pk k
kq
=
=
22
3,qk∴=
2
3 is a factor of .q∴
But 3 is a prime number and
q is an integer, ∴3 is also a factor of q.
But p and q assumingly do not have any common
factors, they cannot both have 3 as factors.
∴The assumption that
3
is rational is incorrect.
∴
3
is irrational.
This method can be applied to prove the
irrationality of the square root of any prime
numbers higher than 3.
(b) Assume that
6
is rational. Let
6
p
q
=
,
where p and q are integers,
0q ≠
, and have no
common factors.
2
2
6.
p
q
=
22
6.pq=
(2)
