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Sets, Relations and Functions

MODULE - IV
Functions

15

Notes

SETS, RELATIONS AND FUNCTIONS

Let us consider the following situation :
One day Mrs
...
Mehta went to the market
...
Mehta purchased the following objects/
items
...
Where as Mrs
...

"Lady fingers, Potatoes and Tomatoes"
...
What can you say about the
collection of students who speak the truth ? Is it well defined? Perhaps not
...
For a collection to be a set it is necessary that it
should be well defined
...
D) to define a set
...
Now-a-days set theory has become
basic to most of the concepts in Mathematics
...
The
concept of relation has been developed in mathematical form
...
Function is a special type of relation
...
In this lesson we shall discuss
some basic definitions and operations involving sets, Cartesian product of two sets, relation
between two sets, the conditions under when a relation becomes a function, different types of
function and their properties
...


Notes

EXPECTED BACKGROUND KNOWLEDGE
l

2

Number systems, concept of ordered pairs
...
1 SOME STANDARD NOTATIONS
Before defining different terms of this lesson let us consider the following examples:
(i) collection of tall students in your school
...

(ii) collection of honest persons in your (ii) collection of those people in your colony
who have never been found involved in any
colony
...


Notes

(iii) collection of interesting books in your (iii) collection of Mathematics books in your
school library
...

(iv) collection of intelligent students in your (iv) collection of those students in your school
who have secured more than 80% marks in
school
...

In all collections written on left hand side of the vertical line the term tallness, interesting, honesty,
intelligence are not well defined
...
Hence
those collections can not be considered as sets
...
'
'mathematics books' 'never been found involved in theft case,' ' marks more than 80%' are well
defined properties
...

If a collection is a set then each object of this collection is said to be an element of this set
...

For example, A = Toy elephant, packet of sweets, magazines
...

MATHEMATICS

3

Sets, Relations and Functions

MODULE - IV For example N is the set of natural numbers and we know that 2 is a natural number but −2 is
Functions not a natural number
...

15
...

15
...
1 (i) Roster method (Tabular form)
Notes
In this method a set is represented by listing all its elements, separating these by commas and
enclosing these in curly bracket
...

then A={1, 2, 3, 4, 5, 6}, is in the Roster form
...
e
...
For example if A be the set of letters used in the word mathematics, then
A = {m, a, t, h, e, i, c, s}
15
...
2 Set-builder form
In this form elements of the set are not listed but these are represented by some common
property
...

then

A = {x : x ∈ N and 1 ≤ x < 7}

Note : Symbol ':' read as 'such that'
Example: 15
...
2

Write the following in Roster form
...


D = {2,3}

15
...
3
...
A is said to be an infinite set and B is said to be is finite set
...


Notes

15
...
2 Empty (Null) Set : Consider the following sets
...
Therefore this
set consists of no element
...
Such a set is said to be a null (empty) set
...

15
...
3 Singleton Set : Consider the following set :
A = {x : x is an even prime number}
As there is only one even prime number namely 2, so set A will have only one element
...
Here A = { 2}
...

15
...
4 Equal and equivalent sets : Consider the following examples
...


In example (i) Sets A and B have the same elements
...
In example (ii) set D and E have the same number of elements but elements
are different
...

Two sets A and B are said to be equivalent sets if they have same number of elements but they
are said to be equal if they have not only the same number of elements but elements are also the
same
...
3
...

For example,sets A= { 1,3,5} and B = { 2,4,6 } are disjoint sets
...
3

Given that

A = { 2, 4} and B = { x : x is a solution of x 2 + 6 x + 8 =0 }
Are A and B disjoint sets ?
Solution : If we solve x 2 + 6 x + 8 = 0 ,we get
x = − 4, − 2
...

Example 15
...

Each set is having five elements but elements are different
A ≠ B but A ≈ B
...
5 Which of the following sets
A = {x : x is a point on a line}
B = {y : y ∈ N and y ≤ 50}
are finite or infinite ?
Solution : As the number of points on a line is uncountable (cannot be counted) so A is an
infinite set while the number of natural numbers upto fifty can be counted so B is a finite set
...
6

Which of the following sets

A = { x : x i s irrational and x 2 − 1 = 0 }
...
If we solve
x 2 − 1 = 0 we get x = ±1
...
Therefore A is an empty set
...
B is not an empty set as it has five elements
...
7 Which of the following sets are singleton ?
A = { x : x ∈ Z and x − 2 = 0 }

B = { y : y ∈ R and y 2 − 2 = 0 }
...
∴ A ={2}
...

⇒
B is a set of those real numbers which are solutions of y 2 − 2 = 0 or y = ± 2
∴
6

B=

{−

2, 2 } Thus, B is not a singleton set
...
1
1
...

(ii) The collection of natural numbers upto fifty
...

(iv) The collection of fat students of your school
...


Notes

If A = {1 , 2 , 3
...
A
(ii)
4
...

3
...


(ii)

B=

{x : x ∈R

and x 2 − 1 = 0 }
...

(iv)
D = {x : x is a prime number and exact divisor of 60}
...
Write each of the following sets are in the set builder form ?
(i)
A = {2, 4, 6, 8, 10} (ii) B = {3, 6, 9,
...
Which of the following sets are finite and which are infinite ?
(i) Set of lines which are parallel to a given line
...

(iii) Set of Natural numbers less than or equal to fifty
...

6
...


(ii)

B = { x : x ∈ Z and x is a solution of x − 3 = 0}
...

(iv) D = {x : x is a student of your school studying in both the classes XI and XII }
7
...

(i)
(ii)

A = {a}, B = {x : x is an even prime number}
...


(iii)

A = {x : x is a solution of x 2 − 5x + 6 = 0 },B = {2 , 3}
...
4 SUB- SET
Let set A be a set containing all students of your school and B be a set containing all students of
class XII of the school
...
Such a
set B is said to be subset of the set A
...
}
E = {
...
}

Clearly each element of set D is an element of set E also ∴ D ⊆ E
If A and B are any two sets such that each element of the set A is an element
of the set B also, then A is said to be a subset of B
...
e
...

(ii) Null set has no element so the condition of becoming a subset is automatically satisfied
...

(iii) If A ⊆ B and B ⊆ A then A = B
...
i
...
A ⊂ B or B ⊃ A
...
8 If A = {x : x is a prime number less than 5} and
B = {y : y is an even prime number} then is B a proper subset of A ?
Solution : It is given that
A = {2, 3 }, B = {2}
...

Example 15
...

is A ⊆ B o r B ⊆ A ?
Solution : Here 1∈ A but1∉ B ⇒ A ⊆ B
...

/

Hence neither A is a subset of B nor B is a subset of A
...
10 If A = {a, e, i, o, u}
B = {e, i, o, u, a }
Is A ⊆ B o r B ⊆ A or both ?
8

MATHEMATICS

Sets, Relations and Functions
Solution : Here in the given sets each element of set A is an element of set B also
∴

A ⊆B

MODULE - IV
Functions

...
∴ B ⊆ A
...
5 POWER SET
Let

A = {a, b}

Subset of A are φ , {a}, {b} and {a, b}
...

Notation : Power set of a set A is denoted by P(A)
...

Example 15
...


(ii)

B = {y : y ∈ N and1 ≤ y ≤ 3}
...

∴

P (B) = { φ , {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} }
...
6 UNIVERSAL SET
Consider the following sets
...

MATHEMATICS

9

Sets, Relations and Functions

MODULE - IV A can be considered as the universal set for this particular example
...

In a particular problem a set U is said to be a universal set if all the sets in that problem are
subsets of U
...

(ii) A set which is a universal set for one problem may not be a universal set for another
problem
...
12 Which of the following set can be considered as a universal set ?
X = {x : x is a real number}
Y = {y : y is a negative integer}
Z = {z : z is a natural number}
Solution : As it is clear that both sets Y and Z are subset of X
...


15
...
According to him universal set is represented by the interior of a rectangle and
other sets are represented by interior of circles
...
15
...


15
...

A new set having those elements which are in A but not B is said to be the difference of sets A
and B and it is denoted by A − B
...

∴
10

B − A = {6}
MATHEMATICS

Sets, Relations and Functions

MODULE - IV
Functions

In general, if A and B are two sets then
A − B = { x : x ∈ A and x ∉ B }
B − A = { x : x ∈ B and x ∉ A }
Difference of two sets can be represented using Venn diagram as :

Notes

or

Fig
...
2

Fig
...
3

15
...
COMPLEMENT OF A SET
Let X denote the universal set and Y, Z its sub set where
X = {x : x is any member of the family}
Y = {x : x is a male member of the family}
Z = {x : x is a female member of the family}
X − Y is a set having female members of the family
...

X − Y is said to be the complement of Y and is usally denoted by Y' or Y c
...

If U is the universal set and A is its subset then the complement of A is a set of those elements
which are in U which are not in A
...

A' = U − A = {x : x ∈ U and x ∉ A}
The complement of a set can be represented using Venn diagram as :

Fig
...
4

Remarks
(i) Difference of two sets can be found even if none is a subset of the other but complement
of a set can be found only when the set is a subset of some universal set
...


MATHEMATICS

(iii)

Uc = φ
...
13 Given that
A = {x : x is a even natural number less than or equal to 10}
and
B = {x : x is an odd natural number less than or equal to 10}
Find (i) A − B
(ii) B− C
(iii) is A − B=B − A ?
Solution : It is given that
Notes
A = {2, 4, 6, 8, 10}, B = {1, 3, 5, 7, 9}
Therefore,
(i)
A − B ={2, 4, 6, 8, 10}
(ii)
B − A ={1, 3, 5, 7, 9}
(iii)
Clearly from (i) and (ii) A − B ≠ B − A
...
14 Let U be the universal set and A its subset where
U={ x : x ∈ N and x ≤ 10 }
Find

A = {y : y is a prime number less than 10}
(i) A c
(ii) Represent A c in Venn diagram
...

A = { 2, 3, 5, 7}
(i) A c = U − A = {1, 4, 6, 8, 9, 10}
(ii)

Fig
...
5

CHECK YOUR PROGRESS 15
...


Insert the appropriate symbol in the blank spaces, given that
A={1, 3, 5, 7, 9}
(i) φ
...
A

2
...
A
(iv) 10
...


Let A = { φ , {1} ,{2}, {1,2}}
Which of the following is true or false ?

12

MATHEMATICS

Sets, Relations and Functions
(i) {1,2} ⊂ A (ii) φ ∈ A
...


5
...


Which of the following statements are true or false ?
(i) Set of all boys, is contained in the set of all students of your school
...

(iii) Set of all rectangles, is contained in the set of all quadrilaterals
...

If A = { 1, 2, 3, 4, 5 } , B= {5, 6, 7} find (i) A− B (ii) B− A
...


MODULE - IV
Functions

Notes

15
...
INTERSECTION OF SETS
Consider the sets
A = {1, 2, 3, 4} and B = { 2, 4, 6}
It is clear, that there are some elements which are common to both the sets A and B
...

Here A ∩ B = {2, 4 }
If A and B are two sets then the set of those elements which belong to both the sets is said to be
the intersection of A and B
...

A ∩ B = {x : x ∈ A and x ∈ B}
A ∩ B can be represented using Venn diagram as :

Fig
...
6

Remarks
If A ∩ B = φ then A and B are said to be disjoint sets
...
15
...
15 Given that
A = {x : x is a king out of 52 playing cards}
and
B = { y : y is a spade out of 52 playing cards}
Find (i) A ∩ B (ii) Represent A ∩ B by using Venn diagram
...
The set B has 13 elements as there are 13 spade cards but out of these 13 spade
cards there is one king also
...
∴ A ∩ B =
{ King of spade}
...
15
...
11 UNION OF SETS
Consider the following examples :
(i)
A is a set having all players of Indian men cricket team and B is a set having all players of
Indian women cricket team
...
Union of these two sets is
a set having all players of both teams and it is denoted by A ∪ B
...
Suppose three players are common to both the teams then union
of D and E is a set of all players of both the teams but three common players to be
written once only
...

In set builder form :
A ∪ B = {x : x ∈ A or x∈ B}
OR
A ∪ B = {x : x ∈ A − B or x ∈ B − A or x ∈ A ∩ B }
A ∪ B can be represented using Venn diagram as :

Fig
...
9
14

Fig
...
10
MATHEMATICS

Sets, Relations and Functions
n(A ∪ B) = n(A −B) +n(B − +
A) n(A ∩
...

Example 15
...

Solution : We have,
A={1, 2, 3, 4, 5} B = {2, 3, 5, 7}
...

(ii)

Fig
...
11

CHECK YOUR PROGRESS 15
...


Which of the following pairs of sets are disjoint and which are not ?
(i) {x : x is an even natural number}, {y : y is an odd natural number}
(ii) {x : x is a prime number and divisor of 12}, { y : y ∈ N and 3 ≤ y ≤ 5 }

2
...


Given that A = {1, 2, 3, 4, 5}, B={5, 6, 7, 8, 9, 10}
find (i) A ∪ B (ii) A ∩ B
...


If A ={ x : x ∈ N }, B={ y : y∈ z and −10 ≤ y ≤ 0 } find A ∪ B and write your
answer in the Roster form as well as set-builder form
...


6
...

Let U = {1, 2, 3,
...


(i) ( A ∪ B ) ' (ii) ( A ∩ B ) ' (iii) (B')' (iv) ( B − A )'
...


Find

MATHEMATICS

15

Sets, Relations and Functions

MODULE - IV
Functions

(i) A ∩ B when B ⊂ A
(ii) A ∩ B when A and B are disjoint sets
...

8
...


Notes

9
...


(iii) A ∪ B when A and B are neither subsets of each other nor disjoint sets
...

(ii) A − B and B − A when A and B are disjoint sets
...


15
...


Set of all ordered pairs of elements of A and B
is

{(1,3), (1,4), (1,5), (2,3), (2,4), (2,5)}

This set is denoted by A × B and is called the cartesian product of sets A and B
...
e
...

In the present example, it is given by
B×A = {(3, 1),(3, 2),(4, 1),(4, 2),(5, 1),(5, 2)}
Clearly A×B ≠ B×A
...

Example 15
...

Find

(i) A×B(ii) B×A
(v) (A ∩ B) × C

Solution :

(iv) (A ∩ C) × B

(vi) A × (B − C)
...

(ii) B×A = {(d, a),(d, b), (d, c), (e, a) (e, b),(e, c)}
...

MATHEMATICS

Sets, Relations and Functions
∴

A × ( B ∪ C ) ={(a, a),(a, d),(a, e),(b, a),(b, d),(b, e), (c, a),(c, d),(c, e)
...


∴

( A ∩ C )×B={(a, d), (a, e)}

(v)

A ∩ B = φ , c={a,d}, ∴ A ∩ B × c = φ

(vi)

A = {a,b,c}, B − C = {e}
...


Notes

15
...
If we define a relation R " is a brother of" between the elements of A and
B then clearly
...

After omiting R between two names these can be written in the form of ordered pairs as :
(Mohan, Rita), (Sohan, Rita), (David, Marry), (Karima, Fatima)
...
e
...

If

(i) R = φ , R is called a void relation
...

(iii) If R is a relation defined from A to A, it is called a relation defined on A
...


15
...
1 Domain and Range of a Relation
If R is a relation between two sets then the set of its first elements (components) of all the
ordered pairs of R is called Domain and set of 2nd elements of all the ordered pairs of R is
called range, of the given relation
...

Domain = {Mohan, Sohan, David, Karim}
Range = {Rita, Marry, Fatima}
Example 15
...

R is a relation from A to B defined by

MATHEMATICS

17

Sets, Relations and Functions

MODULE - IV
Functions find

R = {(a, b) : a ∈ A, b ∈ B and a is divisible by b}

(i) R in the roster form (ii) Domain of R (iii) Range of R
(iv) Repersent R diagramatically
...
15
...
19 If R is a relation 'is greater than' from A to B, where
A= {1, 2, 3, 4, 5} and B = {1,2,6}
...
(ii) Domain of R (iii) Range of R
...
4
1
...


2
...


If U is a universal set and A, B are its subsets
...

A = {1,3,5}, B = {x : x is a prime number} find A' × B'
If A = {4, 6, 8, 10}, B = {2, 3, 4, 5}
R is a relation defined from A to B where
R= {(a, b) : a ∈ A, b ∈ B and a is a multiple of b}

4
...

If R be a relation from N to N defined by
R= {(x,y) : 4x + y = 12, x , y ∈ N }

5
...

If R be a relation on N defined by
R={ (x,x 2 ) : x is a prime number less than 15}

18

MATHEMATICS

Sets, Relations and Functions
Find (i) R in the Roster form (ii) Domain of R (iii) Range of R
6
...


15
...
So we observe that in a function no two ordered
pairs have the same first element
...
15
...
e
...
Thus here:
(i)
the set B will be termed as co-domain and
(ii)
the set {1, 2, 3, 5} is called the range
...

Symbolically, this function can be written as
f:A → B

or

f
A  B
→

Example 15
...
Write their
domain and range
...


MATHEMATICS

19

Sets, Relations and Functions

MODULE - IV Solution :
Functions (a)
It is a function
...
Because Ist two ordered pairs have same first elements
...

Domain= { a , b , c , d } ≠ A ,

(d)

Range = { −2,7, −6,1}

Range = { b , c}

It is a function
...

Domain = {1,2,3,4,5 } ≠ A , Range = { −1, −2, −3, −4, −5 }
(f) It is a function
...

First two ordered pairs have same first component and last two ordered pairs have also same
first component
...
21 State whether each of the following relations represent a function or not
...
15
...
15
...
15
...
15
...


(b)

f is not a function because the element c of A does not have a unique image in B
...


(d)

MODULE - IV
Functions

f is a function because every element in A has a unique image in B
...
22 Which of the following relations from R →
...

∴

It is a function
...


∴

For any real value of x we get more than one real value of y
...


(c)

y = 2x 2 + 1

∴

For any real value of x, we will get a unique real value of y
...


CHECK YOUR PROGRESS 15
...


Which of the following relations are functions from A to B ?
(a) { (1, −2),(3,7),(4, −6),(8,11) } ,

A = {1,3,4,8} , B = { −2,7, −6,11}

(b) { (1,0),(1, −1),(2,3),(4,10) } ,

A = {1,2,4} , B = {1,0, −1,3,10 }

(c) { (a,2),(b,3),(c,2),(d,3) } ,

A = { a , b , c , d} , B = { 2,3}

(d) { (1,1),(1,2),(2,3),( −3,4) } ,

A = {1,2, −3 } , B = {1, 2 , 3 , 4}

1 
 1   1 

(e)   2,  ,  3,  ,
...
,
2 3
11

}

(f) { ( 1,1 ) , ( −1,1 ) , ( 2,4 ) , ( −2,4 ) } , A = { 0,1, −1,2, −2 } , B = {1, 4 }
2
...
15
...
15
...
15
...
15
...


Which of the following relations defined from R → R are functions ?
(a) y = 2x + 1 (b) y > x + 3 (c) y < 3x + 1 (d) y = x2 + 1

4
...


{(

{ ( Deepak,16 ) , ( Sandeep,28 ) , ( Rajan,24 ) }

)(

5, −1 ,

1
 
,  −2,
2
 

3,5

)}

1 
 
,  −1,  
2 
 

Write domain and range for each of the following mappings :
(a)

Fig
...
22

22

)(

2,2 ,

(b)

Fig
...
23

MATHEMATICS

Sets, Relations and Functions
(c)

MODULE - IV
Functions

(d)

Notes

Fig
...
24

Fig
...
25

(e)

Fig
...
26

15
...
1 Some More Examples on Domain and Range
Let us consider some functions which are only defined for a certain subset of the set of real
numbers
...
23 Find the domain of each of the following functions :
(a) y =

1
x

(b) y =

1
x−2

(c) y =

1
(x + 2)(x − 3)

Solution : The function y = 1 can be described by the following set of ordered pairs
...
,  −2, −  , ( −1, −1 ) , ( 1,1 )  2,  ,
...
e
...

0
∴

Domain = R − { 0 } [Set of all real numbers except 0]

Note : Here range = R − { 0 }
(b)

x can take all real values except 2 because the corresponding image, i
...
,

∴

not exist
...
24 Find domain of each of the following functions :
(a) y = + x − 2

(b) y = +

( 2 − x)(4 + x )

Solution :(a) Consider the function y = + x − 2
In order to have real values of y, we must have
Notes

( x − 2) ≥ 0

i
...


x≥2

∴

Domain of the function will be all real numbers ≥ 2
...

⇒
But, x cannot take any real value which is greater than or equal to 2 and less than or equal to
−4
...
25 For the function
f ( x ) = y = 2x + 1, find the range when domain = { −3, −2, −1,0,1,2,3}
...

24

MATHEMATICS

Sets, Relations and Functions
i
...
,
∴

MODULE - IV
Functions

{ ( −3, −5 ) , ( −2, −3 ) , ( −1, −1 ) , ( 0,1 ) ( 1, 3 ) , ( 2,5 ) ( 3,7 ) }
Range = { −5, −3, −1,1,3,5,7}

Example 15
...


0≤x≤ 4

Solution : Here

Notes

or

0 + 3 ≤ x + 3 ≤ 4 +3

or

3 ≤ f (x )≤ 7

∴

Range = { f ( x ) : 3 ≤ f ( x ) ≤ 7 }

Example 15
...


Solution : Given − 3 ≤ x ≤ 3
or
∴

0 ≤ f (x) ≤ 9

0 ≤ x 2 ≤ 9 or

Range = { f ( x ) : 0 ≤ f ( x ) ≤ 9 }

CHECK YOUR PROGRESS 15
...


Find the domain of each of the following functions x ∈ R :
(a)

(i)

y = 2x

(b)

(i) y =

1
3x − 1

(iii) y =

(ii)

(ii) y =
1

( x − 3)( x − 5 )
6 −x

(i) y =

(iv) y =

(iii)

y = x 2 +5

1

( 4x + 1 ) ( x − 5 )
1

(3 − x)(x − 5 )

3x +5

(iii) y =

(d)

(i) y =
(iii) y =

(c)

y = 9x + 3

(ii) y =

7 +x

( 3 − x )( x −5 )

(ii) y =

( x −3 )( x +5 )

1

(iv) y =

1

(3 + x)( 7 +x )

( x − 3) ( 7 + x )

2
...

x ∈ {1, 5, 7, − 1, − 2 }
x ∈ { −3, 2, 4, 0 }

(iii) f ( x ) = x 2 − x + 2 ,
MATHEMATICS

(i) f ( x ) = 3x +10 ,
(ii) f ( x ) = 2x 2 +1 ,

(a)

x ∈ {1, 2, 3, 4, 5 }
25

Sets, Relations and Functions

MODULE - IV
Functions

(i) f ( x ) = x −2 , 0 ≤ x ≤ 4

(ii) f ( x ) = 3x + 4 , −1 ≤ x ≤ 2

(c)

(i) f ( x ) = x 2 , −5 ≤ x ≤ 5

(ii) f ( x ) = 2x , −3 ≤ x ≤ 3

(iii) f ( x ) = x 2 +1 , −2 ≤ x ≤ 2

(iv) f ( x ) =

(i) f ( x ) = x +5 , x ∈ R

(ii) f ( x ) = 2x −3 , x ∈ R

(iii) f ( x ) = x 3 , x ∈ R

Notes

(b)

(iv) f ( x ) =

1
, { x : x < 0}
x

(vi) f ( x ) =

1
, { x : x ≤ 0}
3x − 2

(viii) f ( x ) =

x
, { x : x ≠ −5 }
x +5

(d)

(v) f ( x ) =
(vii) f ( x ) =

1
, { x : x ≤ 1}
x−2
2
, { x : x > 0}
x

x , 0 ≤ x ≤ 25

15
...
If every element of the set B is the image of at least one element
of the set A i
...
if there is no unpaired element in the set B then we say that the function f maps
the set A onto the set B
...

Functions for which each element of the set A is mapped to a different element of the set B are
said to be one-to-one
...
15
...
Such
a type of function is said to be many-to-one
...
15
...

Notes

Fig
...
29

Fig
...
30

Fig
...
31

Fig
...
32

Fig
...
29 shows a one-to-one function mapping { A , B , C} into{1, 2 , 3 , 4}
...
15
...


Fig
...
31shows a many-to-one function mapping { A , B , C} into{1, 2 , 3 , 4}
...
15
...


Function shown in Fig
...
30 is also a bijective Function
...
The following
figure illustrates this fact
...
15
...
28 Without using graph prove that the function
F : R → R defiend by f ( x ) = 4 +3x is one-to-one
...


Example 15
...
It has no real value of x1 and x 2
...


Again let y = ( x )
We have
∴

( x 2 − x1 ) ( x12 + x1x 2 + x 22 ) = 0

where y ∈ codomain, x ∈ domain
...


Thus F is onto function
...

∴
Example 15
...

Solution : We have F ( x1 ) = F ( x 2 ) ∀ x1 , x 2 ∈
domain giving

x12 + 3 = x 2 2 +3

⇒ 12
x

or

x12 − x2 2 = 0 ⇒ x1 = x2

or

= 22
x

F is not one-one function
...


⇒
⇒

∀ y <3

∴
28

y = x2 + 3

⇒

F is not an onto finction
...

no

MATHEMATICS

Sets, Relations and Functions

15
...
For example, consider y = x 2
...
15
...

Now consider the equation x 2 + y2 = 25
x 2 + y2 = 25
x 0

0 3

3 4

4 5 − 5 − 3 − 3 −4 −4

y 5 −5 4 −4 3 −3 0

0

4 −4

3 −3

Fig
...
35
MATHEMATICS

29

Sets, Relations and Functions

MODULE - IV This graph represents a circle
...


CHECK YOUR PROGRESS 15
...


(i) Does the graph represent a function?

Fig
...
36

(ii) Does the graph represent a function ?

Fig
...
37

2
...
15
...

(c)
3
...


f ( x ) = 115x + 49

(b)

f (x ) = x

Which of the following functions are one-to-one functions ?
(a)

f : { 7,8,9 } → {10 } defined as f ( x ) = 10

(c)

f : I → R defined as f ( x ) = x 3

(d)

f : R → R defined as f ( x ) = 2 + x 4

(d)
5
...
15
...


f : N → N defined as f ( x ) = 5x + 7

Draw the graph of each of the following functions :
(a) y = 3x 2

(b) y = − x 2

(c) y = x 2 − 2

(d) y = 5 − x 2 (e) y = 2x 2 + 1
7
...
15
...
15
...
15
...
15
...
15
...


15
...
17
...

For function to be increasing on an interval (a,b)
x1 < x 2

⇒ F ( x1 ) < F ( x2

a,
) ∀ x1 x 2 (∈ b )

and for function to be decreasing on (a,b)
x1 < x 2

⇒ F ( x1 ) > F ( x2

) ∀ x1 x 2 ∈a, b )
(

A function may not be monotonic on the whole domain but it can be on different intervals of the
domain
...

Now

∀ x1 , x 2 ∈ 0, ∞
[
]
x1 < x 2 ⇒ F ( x1 ) < F ( x 2 )

⇒
32

F is a Monotonic Function on [ 0, ∞ ]
...


Notes

Again consider the function F : R → R defined by f ( x ) = x 3
...
e
...


15
...
2 Even Function
A function is said to be an even function if for each x of domain
F( −x) = F(x)

For example, each of the following is an even function
...
15
...

Observation
Graph is symmetrical about y-axis
...
17
...
15
...
15
...

15
...
4 Greatest Integer Function (Step Function)
f ( x ) = [ x ] which is the greatest integer less than or equal to x
...
Its graph is in the form of steps, as
shown in Fig
...
47
...

Range of the step function is the set of integers
...
15
...
17
...

For example,
(i)
(ii)
34

f ( x ) = 3x 2 − 4x − 2
f ( x ) = x 3 − 5x 2 − x + 5
MATHEMATICS

Sets, Relations and Functions
(iii)

MODULE - IV
Functions

f(x) = 3

are all polynomial functions
...

15
...
6 Rational Function
Function of the type f ( x ) =

Notes

g(x )
, where h ( x ) ≠ 0 and g ( x ) and h ( x ) are polynomial
h (x )

functions are called rational functions
...

15
...
7 Reciprocal Function
Functions of the type y = 1 , x ≠ 0 is called a reciprocal function
...
17
...
In
fact
e =1 +

1 1
+
1 2

1
+
3

1
+
n


...

+


...
e
...
This number e is a transcendental irrational
number and its value lies between 2 an 3
...


xn
+
n


...
Thus,

ex = 1 +

x x2
+
1
2

x3
+
3


...

+


...

We easily see that we would get (1) by putting x = 1 in (2)
...

The graph of the exponential function
y = ex
MATHEMATICS

35

Sets, Relations and Functions

MODULE - IV is obtained by considering the following important facts :
Functions (i) As x increases, the y values are increasing very rapidly, whereas as x decreases, the y
values are getting closer and closer to zero
...

The y intercept is 1, since e 0 = 1 and e ≠ 0
...
15
...

−3

x
y = ex

−2

−1

0

1

2

3

0
...
13

0
...
00

2
...
38

20
...
15
...


For example, we may take a = 2 or a = 3 and get
the graphs of the functions
y = 2x (See Fig
...
49)
and

y = 3x (See Fig
...
50)
Fig
...
49

36

MATHEMATICS

Sets, Relations and Functions

MODULE - IV
Functions

Notes

Fig
...
50

Fig
...
51

15
...
9 Logarithmic Functions
Consider now the function
y = ex
We write it equivalently as


...
(4)

is the inverse function of y = e x
The base of the logarithm is not written if it is e
and so log e x is usually written as log x
...
r
...
the line
y= x
The graph of the function y = log x can be
obtained from that of y = e x by reflecting it in
the line y = x
...
15
...


CHECK YOUR PROGRESS 15
...


Tick mark the correct statement
...


(iii)

f ( x ) = x1 / 2 − x3 + x5 is a polynomial function
...


Function f ( x ) = 2x 4 + 7x 2 + 9x is an even function
...

3+ x

5
is a constant function
...

(vii)
Greatest integer function is neither even nor odd
...

4
...

Specify the following functions as polynomial function, rational function, reciprocal function
or constant function
...

9

15
...

Graphically one can represent this as given below :
z = y + 1,

Fig
...
53

The composition, say, gof of function g and f is defined as function g of function f
...


= 3 ( 3x +1 ) +
1

= 9x +3 + =
1 9x

4
+

( gog )( x ) = g ( g ( x ) ) = g ( x 2 +2 ) [ Read as function of function g ]
= ( x2 + 2 )

2

+2

= x 4 +4x 2 +
4

+
2

= x 4 + 4x 2 +6

Example 15
...


fog ( x ) = f ( g ( x ) )

Solution :

= f ( x 2 +2 )
=

x2 +2 +
1

=

x 2 +3

( gof ) ( x ) = g ( f ( x ) )
= g( x +1)
= ( x +1)

2

+2

= x +1 +2
= x + 3
...
32 If f ( x ) = x 3, f : R → R
g(x ) =
40

1
, g : R − { 0} → R − {0}
x
MATHEMATICS

Sets, Relations and Functions

MODULE - IV
Functions

Find fog and gof
...
31 and Example 15
...


CHECK YOUR PROGRESS 15
...


Find fog, gof, fof and gog for the following functions :
f ( x ) = x 2 + 2, g ( x ) = 1 −

2
...

1−x

For each of the following functions write fog, gof, fof and gog
...


Let

4
...
Verify that fog ≠ gof
...

6
...
Show that fog = gof
...

x
Find (a) fog
(b) goh
(c) foh
(d) hog
(e) fogoh


 Hint :


MATHEMATICS

1
( fogoh ) ( x ) = f ( g ( h ( x ) ) ) = f  g    
   
x
 

 

41

Sets, Relations and Functions

MODULE - IV
Functions

15
...
15
...
Now let us find the inverse of this relation
...
55

Clearly this relation does not represent a function
...
15
...
Now let us find the inverse of this relation, which is
represented pictorially as

Fig
...
57
42

MATHEMATICS

Sets, Relations and Functions

MODULE - IV
Functions

This represents a function
...
15
...
Now find the inverse of the relation
...
15
...
Also note that the elements of B does not have a unique image
...
15
...

Find the inverse of the relation
...
15
...
From the above relations we see that we may or may not get a relation as a function when
we find the inverse of a relation (function)
...
e
...


Notes

CHECK YOUR PROGRESS 15
...


2
...
Write the
domain and range of its inverse function
...

To represent a set in Roster form all elements are to be written but in set builder form a
set is represented by the common property
...

If each element of set A is an element of set B also then A is called sub set of B
...

Complement of a set A is a set of those elements which are in the universal set but not in
A
...
e
...

Union of two sets is a set of those elements which belong to either of the two sets
...
It is denoted by A × B
...
e
...

Relation is a sub set of A×B where A and B are sets
...
e

---