Class 12 chapter 1
Relations and functions
Basic concepts of topic relations and functions class XII chapter 1 of mathematics. Equivalence relations, different types of functions, composition and inverse of functions.
In class 11 we have studied about
Cartesian product of two sets, relations, functions, domain, range and
codomains. Now in this chapter we have studied about the different types of
relations, different types of functions, composition of functions and
invertible functions.
Relations
A relation R from set A to set B is the subset of Cartesian product A x B.
In Cartesian product A x B number of relations equal to the number of subsets.
Domain
Set of all first elements of the ordered pairs in the relation R is called its domain.
Range
Set of all second elements of the ordered pairs in the relation R is called its range.
In any ordered pair second element is also called the image of first element.
Types of relations
i) Empty Relation:
A relation R in set A is called empty
relation, if no element of A is related to any element of A, i.e., R = Ï† ⊂ A x A
Example:
Let A = {1, 2, 3} and R be a relation on the set A defined as R = {(a, b) : a + b = 10; a, b ∈ A}
Here R is empty relation. R = Î¦
ii) Identity Relation
A relation R in set A is called identity relation, if every element of A is related to itself, i.e., R = (a, a) ⊂ A x A
A relation R is an identity relation in set A if for all a ∈ A, (a, a) ∈ R.
iii) Universal Relation :
A relation R in a set A is called universal relation, if each
element of A is related to every element of A, i.e., R = A x A
iv) Reflexive relation
A relation R in a set A is called reflexive, if (a, a) ∈ R, for every a ∈ A
Number of reflexive relations
The number of reflexive relations on a set with n elements is presented by N = 2^{n(n1)} where capital N is the number of reflexive relations and small n is the number of
elements in the set.
Example
If n(A) = 5, the number of reflexive relations in a set A = 2^{5(51)} = 2^{5x4} = 2^{20}.
If n(A) = 3, the number of reflexive relations in a set A = 2^{3(31)} = 2^{3x2} = 2^{6}
= 64.
v) Symmetric Relation
A relation R in a set A is called symmetric, if (a, b) ∈R implies that (b, a) ∈ R, for all a, b ∈ A.
Number of Symmetric Relations If n(A) = 3, then number symmetric relations are given by
vi) Transitive Relation
A relation R in a set A is called transitive, if (a, b) ∈R and (b, c) ∈R, implies that (a, c) ∈ R, for all a, b, c_{ }∈ A
Number of Transitive Relations
To find the number of Transitive Relations, there is no implicit formula yet discovered.
However number of Transitive Relations for the set having number of elements from 0 to 6 are given below.
Number
of elements

Number of Transitive relations

0

1

1

2

2

13

3

171

4

3994

5

154303

6

9415189

vii) Equivalence relation:
A relation R in set A is said to be equivalence relation if it
is reflexive, symmetric and transitive
relation.
Viii) Equivalence Class:
Let R be an equivalence relation on a set A and let a ∈ A. Then we define equivalence class of a as
[a] ={b ∈ A : b is related to a} = {b ∈ A : (b, a) ∈ R}
Note : Both empty relation and universal relations are sometimes called equivalence relation of trivial relation.
Functions
Functions : Functions are those relations (or subsets) of cartesian product in which no two ordered pair have same first element.
If any two ordered pair have same first element, then the relation is not a function.
Number of elements
If n(A) = n and n(B) = m, then number of functions from A to B = m^{n}
If n(A) = 3 and n(B) = 4, then number of functions from A to B = 4^{3 }= 64
If n(A) = 4 and n(B) = 3, then number of functions from A to B = 3^{4 }= 81
There are mainly three types of function
1) OneOne functions (or Injective functions)
2) Many one Functions
3) Onto functions (Surjective Functions)
4) Bijective Functions
OneOne functions (or injective
functions):
A function f : A →B is said to be one –one or injective function if
different elements of set A has different images in set B
Mathematically:
Case 1: For all a, b ∈ A, we have: If f(a) = f(b) ⇒ a_{ }= b
Case 2 : For a, b ∈ A, such that : If a ≠ b ⇒ f(a) ≠ f(b)
In the above figure different elements of set A have different
images in set B. So this function is called oneone function.
Number of Elements in OneOne functions
Case I If n(A) ≤ n(B)
Let set A has n elements and the set B has m elements (n ≤ m). Then the number of injective functions that can be defined from set A to set B is given by
If a function has 'n' elements and is defined from A to A then number of injective relations from A to A = ^{n}P_{n} = n!
If n > m, then Then the number of injective functions that can be defined from set A to set B is = 0
Example
Set A has 3 elements and the set B has 4 elements. Then the number of injective functions that can be defined from set A to set B is = ^{4}P_{3} = 4! = 24.
Many One functions:
If two or more elements of set A have same image in set B then it is called many one function.
If a function is not oneone then it is
called many one.
Here two different elements a and b
of set A has same image in set B. So it is not an oneone function. It is only
a many one function.
Onto function ( or surjective
function)
A function f : A →B is said to be onto (or
surjective) function, if every element of set B is the image of some element in
set A.
Let y ∈ B, then for every element
y ∈ B, ∃ (there exist) an element x ∈ A
such that f(x) = y.
Other definitions of onto functions are as follows
A function f : A →B is said to be onto (or surjective) function, if every element of set B has pre  image in set A.
If range of the function is equal to the codomain(or set B), then it is called onto function.
In the above figure every element of set B is the image of some
element in set A so it is a onto function. It is neither many one nor oneone.
Number of onto functions
Case I: If n(A) ≥ n(B) or m ≥ n
If n(A) = m, and n(B) = n, then number of onto functions from A to B is
Example 1
If n(A) = 6, n(B) = 3, then find the total number of onto functions
Solution
Number of onto functions from A to B are given by
= 729  3 ✕ 64 + 3 ✕1
= 540
Example 2:
If n(A) = 6, n(B) = 4, then find the total number of onto functions
Solution
Number of onto functions from A to B are given by
= 4096  4 ✕ 729 + 6 ✕64  4
= 4096  2916 + 384  4
= 1560
Case II : If n(A) < n(B) or m < n
If number of elements in set A is less than the number of elements in set B then there is no onto function possible.
Bijective Function
The functions which are both oneone (or injective) as well as onto (or surjective) are called Bijective Functions
Number of Bijective Functions
If n(A) = n(B) = n, then number of bijective functions from A to B is given by n!
Example
If n(A) = 3, then number of bijective functions from set A to A are 3! = 6
If n(A) ≠ n(B), then no bijective function is possible
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Exercise 1.2
Q No. 9
\[f:N\rightarrow N,\; \; f(n)=\left\{\begin{matrix} \frac{n+1}{2}, if\; \; n\; \; is\; \; odd\\ \frac{n}{2}, if\; \; n \; \; is\; \; even \end{matrix}\right.\]
State whether the function f is bijective. Justify your answer.
OneOne
\[Let\; (1,2)\epsilon N\; then\]\[f(1)=\frac{n+1}{2}=\frac{1+1}{2}=\frac{2}{2}=1\]\[f(2)=\frac{n}{2}=\frac{2}{2}=1\]\[f(1)=f(2)\: but\: \: 1\neq 2\]⇒ f is not one one function.
Onto
Case 1: If n = odd = 2q+1=4q+1\[f(n)=\frac{n+1}{2}\]\[f(4q+1)=\frac{4q+1+1}{2}=\frac{4q+2}{2}=2q+1\]⇒ 2q + 1 is also an odd function ∈ N
Case 2: If n = even = 2q = 4q, \[f(n)=\frac{n}{2}=\frac{4q}{2}=2q\]⇒ 2q is also an even function ∈ N
From case 1 and case 2 we conclude that :
*For all odd numbers in the domain we have odd numbers in the codomain .
*For all even numbers in the domain we have even numbers in the codomain .
⇒ R_{f} = Odd natural number and Even natural numbers.
⇒ R_{f} = All Natural Numbers = Codomain
⇒ f(x) is an onto function.
Now f(x) is an onto function but not oneone.
So f(x) is not a bijective function.
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Deleted Topics fro CBSE Syllabus
Composition of two functions
Let f : A →B and g
: B →C be two functions.
Then the composition of f and g, denoted by gof, is defined as the function gof
: A →C given by
gof(x) = g[f(x)], ∀ (for all) x ∈ A
Composition of two functions exists only if codomain of first function
is equal to the domain of second function
In the above figure function f is
defined on the element of set A (i.e.
x), function g is defined on the elements of set B (i.e f(x) ) so
composition from set A to set C becomes gof or g[f(x)]
Bijective functions:
The functions which are oneone and
onto both are called bijective functions.
Invertible functions and existence of inverse
If a function f is oneone and onto
then it is an invertible function.
If a function is an invertible
function then inverse of f exist and is denoted by f^{1}.
If f is invertible the f must be
oneone and onto, conversely if f is oneone and onto then f must be invertible. If f is invertible
then f^{1} exists.
Let f(x) = y is an invertible
function then f^{1} exists and we can write f^{1}(y) = x
Example : f(x) = { (a, 1), (b, 2),
(c, 3), (d, 4)}, this can be written as
f(a) = 1, f(b) = 2, f(c) = 3,
f(d) = 4
If f is invertible then we can write
f^{1}(1) = a, f^{1}(2) = b, f^{1}(3) = c, f^{1}(4) = d
If f and g are two invertible
functions then their composition gof is also invertible.
Give an example of a relation which is
i) Symmetric but neither reflexive nor transitive
Ans R = { (1,2), (2,1)}
ii) Transitive but neither reflexive nor symmetric
Ans: R = { (1,1), (2,2), (1,2), (2,1), (3,4)}
iii) Reflexive and symmetric but not transitive.
Ans: R = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}
iv) Reflexive and transitive but not symmetric.
Ans: R = {(1,1), (2,2), (1,2), (2,1), (2,3), (3,3)}
v) Symmetric and transitive but not reflexive
{(5,6), (6,5),(5,5) }
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Very nice and helpful thanks
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