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CBSE Class 12 Maths Revision Notes Chapter 1 Relations and Functions

Relations and Functions explains how elements of sets are connected, mapped and operated through relations, functions and binary operations. For CBSE Class 12 Maths, this chapter builds the base for checking relation types, function types, composition and inverse functions.

Relations and Functions begins with the idea that two objects or quantities are related if there is a recognisable connection between them. In Mathematics, a relation from set A to set B is defined as a subset of A × B. A function is a special type of relation where every element of the domain has a unique image.

Use these CBSE Class 12 Maths Revision Notes Chapter 1 for the 2026–27 academic year to revise definitions, conditions, examples and quick checks. Start with relations, then move to functions, composition, inverse functions and binary operations.

Key Takeaways

Relation: A relation from A to B is any subset of A × B.
Function: A function maps every element of the domain to exactly one element in the co-domain.
Equivalence relation: A relation is an equivalence relation only when it is reflexive, symmetric and transitive.
Invertible function: A function is invertible when it is one-one and onto.

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Access Maths Notes for Chapter 1 Relations and Functions Class 12

Chapter 1 connects set theory with functions and operations. A quick revision should cover definitions first, then conditions and examples.

Revision Area	What to Revise
Relations	Meaning, notation, empty relation, universal relation
Properties of Relations	Reflexive, symmetric, transitive
Equivalence Relation	Equivalence classes and partitions
Functions	Domain, co-domain, range
Types of Functions	One-one, onto, bijective
Composition and Inverse	gof, fog, invertible functions
Binary Operations	Closure, commutativity, associativity, identity and inverse

Relations and Functions Class 12 Notes: Core Definitions

Relations and functions are based on ordered pairs. Ordered pairs show how one element is connected with another.

Relation

A relation R from set A to set B is a subset of A × B.

If (a, b) ∈ R, then a is related to b under R. It is written as a R b.

Example:
Let A = {1, 2, 3}.
A relation R on A can be R = {(1, 1), (2, 2), (3, 3)}.

Here, each element is related to itself.

Function

A function f from set A to set B is a special relation in which every element of A has exactly one image in B.

It is written as:

f: A → B

Here, A is the domain and B is the co-domain.

Domain, Co-domain and Range

Term	Meaning
Domain	Set of all input values
Co-domain	Set into which the function maps values
Range	Set of actual output values

For a function f: A → B, the range is always a subset of the co-domain.

Types of Relations in CBSE Class 12 Maths Revision Notes Chapter 1

A relation in a set A is a subset of A × A. NCERT classifies important relations using specific conditions.

Empty Relation and Universal Relation

Type of Relation	Definition	Symbol
Empty Relation	No element of A is related to any element of A	R = φ
Universal Relation	Every element of A is related to every element of A	R = A × A

Empty relation and universal relation are also called trivial relations.

Reflexive, Symmetric and Transitive Relations

Relation Type	Condition	Quick Check
Reflexive Relation	(a, a) ∈ R for every a ∈ A	Every element relates to itself
Symmetric Relation	If (a, b) ∈ R, then (b, a) ∈ R	Pair can be reversed
Transitive Relation	If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R	First connects to third through second

Example:
If A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3)}, then R is reflexive.

It is also symmetric and transitive because all ordered pairs are self-pairs.

Equivalence Relation and Equivalence Classes

A relation R in a set A is an equivalence relation if it is reflexive, symmetric and transitive.

Example:
In the set of integers Z, define R = {(a, b): 2 divides a - b}.

This relation is reflexive, symmetric and transitive. So, it is an equivalence relation.

Equivalence classes:
This relation divides integers into two classes:

Equivalence Class	Elements
[0]	All even integers
[1]	All odd integers

Equivalence classes are mutually disjoint subsets whose union gives the original set.

Types of Functions in Class 12 Maths Chapter 1 Notes

Functions are classified by how elements of the domain map to the co-domain. The main types of functions are one-one, onto and bijective.

One-One or Injective Function

A function f: X → Y is one-one if distinct elements of X have distinct images in Y.

Condition:

f(x1) = f(x2) ⇒ x1 = x2

Example:
f: R → R, f(x) = 2x is one-one.

If 2x1 = 2x2, then x1 = x2.

Onto or Surjective Function

A function f: X → Y is onto if every element of Y is the image of some element of X.

Condition:

For every y ∈ Y, there exists x ∈ X such that f(x) = y.

Quick fact:
f is onto if Range of f = Y.

Bijective Function

A function is bijective if it is both one-one and onto.

Function Type	Main Condition
One-one	Different inputs have different images
Onto	Every co-domain element has a pre-image
Bijective	Function is both one-one and onto

A bijective function is important because it has an inverse function.

Composition of Functions and Invertible Functions

Composition joins two functions in a fixed order. The order matters because gof and fog may give different results.

Composition of Functions

Let f: A → B and g: B → C be two functions.

The composition of f and g is written as gof.

gof: A → C

gof(x) = g(f(x))

This means f acts on x first, then g acts on f(x).

Example:
If f(x) = x + 1 and g(x) = 2x, then:

gof(x) = g(f(x))
= g(x + 1)
= 2(x + 1)
= 2x + 2

fog(x) = f(g(x))
= f(2x)
= 2x + 1

So, gof ≠ fog in general.

Invertible Function

A function f: X → Y is invertible if there exists a function g: Y → X such that:

gof = IX
fog = IY

Here, g is called the inverse of f. It is denoted by f⁻¹.

Important condition:
A function is invertible if and only if it is one-one and onto.

Key Properties

Property	Meaning
gof(x) = g(f(x))	f acts first, then g
fog(x) = f(g(x))	g acts first, then f
gof may not equal fog	Composition is not always commutative
f⁻¹ exists	f must be bijective

Binary Operations in Relations and Functions

A binary operation combines two elements of a set and gives an element of the same set.

Meaning of Binary Operation

A binary operation * on a non-empty set A is a function:

*: A × A → A

For a, b ∈ A, the result a * b must also belong to A.

Example:
Addition is a binary operation on real numbers because a + b ∈ R for all a, b ∈ R.

Properties of Binary Operations

Property	Condition
Closure	a * b ∈ A for all a, b ∈ A
Commutative	a * b = b * a
Associative	a * (b * c) = (a * b) * c
Identity Element	a * e = e * a = a
Inverse Element	a * b = b * a = e

Here, e is the identity element and b is the inverse of a.

Quick Revision Tables for Relations and Functions

These tables help revise Class 12 Mathematics Chapter 1 notes before solving exercises.

Relations Quick Revision Table

Concept	Definition	Exam Check
Relation	Subset of A × B	Write ordered pairs
Empty Relation	R = φ	No pair satisfies the condition
Universal Relation	R = A × A	Every possible pair satisfies the condition
Reflexive	(a, a) ∈ R for all a ∈ A	Check all self-pairs
Symmetric	(a, b) ∈ R ⇒ (b, a) ∈ R	Reverse every pair
Transitive	(a, b), (b, c) ∈ R ⇒ (a, c) ∈ R	Link first to third
Equivalence	Reflexive, symmetric and transitive	All three must hold

Functions Quick Revision Table

Concept	Meaning	Example Check
Function	Every input has one image	No input can have two outputs
Domain	Input set	Values allowed for x
Co-domain	Target set	Given output set
Range	Actual output set	Values obtained by f(x)
One-one	Different inputs, different outputs	f(x1) = f(x2) ⇒ x1 = x2
Onto	Range = co-domain	Every y has a pre-image
Bijective	One-one and onto	Inverse exists

Binary Operations Quick Revision Table

Property	Question to Ask
Closure	Is the result still in the same set?
Commutative	Does changing order give the same result?
Associative	Does grouping change the result?
Identity	Is there an element that leaves every element unchanged?
Inverse	Does every element combine with another to give identity?

Important Terms in Relations and Functions Class 12 Notes

Term	Meaning
Ordered Pair	A pair written as (a, b), where order matters
Cartesian Product	Set of all ordered pairs from two sets
Relation on A	A subset of A × A
Image	Output of an element under a function
Pre-image	Input that gives a particular output
Identity Function	Function that maps every element to itself
Inverse Function	Function that reverses the mapping of a bijective function
Equivalence Class	Set of all elements related to a given element

Useful Links for Class 12 Maths

Section	Useful Links
Syllabus	CBSE Class 12 Maths Syllabus
Revision Notes	CBSE Class 12 Maths Revision Notes
NCERT Solutions	NCERT Solutions Class 12 Maths
Sample Papers	CBSE Sample Papers for Class 12 Maths
Important Questions	Important Questions Class 12 Maths
NCERT Books	NCERT Books for Class 12 Maths
Previous Year Papers	CBSE Maths Question Paper Class 12

Q.1 If f: [1, 2, 3] [a, b, c] and g: [a, b, c] [apple, ball, cat] defined as f(1) = a, f(2) = b, f(3) = c, g(a) = apple, g(b) = ball and g(c) = cat. Show that f, g and gof are invertible. Find out f ^-1, g^-1 and (gof)^-1 and show that (gof)^-1 = f ^-1og^-1.

Ans

$\begin{array}{l} By definition, f and gare bijective functions . \\ Let f^{- 1} : [a, b, c] \to [1, 2, 3] and g^{- 1} : [apple, ball, cat] \to [a, b, c] \\ be defined as \\ f^{- 1} (a) = 1, f^{- 1} (b) = 2, f^{- 1} (c) = 3 \\ g^{- 1} (apple) = a, g^{- 1} (ball) = b, g^{- 1} (cat) = c \\ \Rightarrow f^{- 1} of = I_{(123)} \\ {fof}^{- 1} = I_{(a, b, c)} \\ and g^{- 1} = I_{(a, b, c)} \\ {gog}^{- 1} = I_{(apple, ball, cat)} \\ Now, gof : [1, 2, 3] \to [apple, ball, cat] is given by \\ gof (1) = apple, gof (2) = ball, gof (3) = cat . We can define \\ {gof}^{- 1} [apple, ball, cat] \to [1, 2, 3] by \\ {gof}^{- 1} (apple) = 1 \\ {gof}^{- 1} (ball) = 2 \\ {gof}^{- 1} (cat) = 3 \\ We can see that {gof}^{- 1} o (gof) = I_{\{1, 2, 3\}} and \\ (gof) o {(gof)}^{- 1} = I_{\{apple, ball, cat\}} \\ Thus, f, g and gof are invertible . \\ Now, \\ f^{- 1} {og}^{- 1} (apple) = f^{- 1} (g^{- 1} (apple)) = f^{- 1} (a) = 1 = {gof}^{- 1} (apple) \\ f^{- 1} {og}^{- 1} (ball) = f^{- 1} (g^{- 1} (ball)) = f^{- 1} (b) = 2 = {gof}^{- 1} (ball) \\ f^{- 1} {og}^{- 1} (cat) = f^{- 1} (g^{- 1} (cat)) = f^{- 1} (c) = 3 = {gof}^{- 1} (cat) \\ Hence, {(gof)}^{- 1} = f^{- 1} {og}^{- 1} \end{array}$

Q.2

Ans

$\begin{array}{l} Define a binary operation * on the set \{0, 1, 2, 3, 4, 5\} \\ as a * b = \{\begin{cases} a + b if a + b < 6 \\ a + b - 6 if a + b \geq 6 \end{cases} \\ Show that 0 is the identify for this operation and each \\ element of the set is invertible with 6 - a being the inverse of a . \\ (i) e is the identify element if a * e = e * a = a \\ a * o = a + o = a \\ o * a = o + a = a \\ \Rightarrow a * o = o * a \\ Hence, 0 is the identify of the operation . \\ (ii) b is the inverse of a if \\ a * b = b * a = a \\ Now, a * (6 - a) = a + (6 - a) - 6 = 0 \\ (6 - a) * a = (6 - a) + a - 6 = 0 \\ Hence, each element of the given set is invertible \\ with 6 - a being the inverse of a . \end{array}$

Q.3 Let * be the binary operation on the set Q of rational numbers which is defined as follows

(i) a * b = a – b
(ii) a * b = a² + b²
(iii) a * b = a + ab
Find which of the binary operations are commutative and which are associative.

Ans

(i) Here a * b = a – b
Now b *a = b – a, but a – b ≠ b – a
⇒ a * b ≠ b * a
⇒ * is not commutative

a*(b*c) = a*(b – c) = a – (b – c) = a – b + c
(a*b)*c = (a – b)*c = (a – b) – c
Thus, a*(b*c) ≠ (a * b)*c
⇒ * is not associative

(ii) Here a*b = a² + b²
b*a = b² + a² = a² + b²
a*b = b*a
⇒ * is commutative

a*(b*c) = a*(b² + c²) = a² + (b² + c²)²
(a*b)*c = (a² + b²)*c = (a²+ b²)² + c²
Thus, a*(b*c) ≠ (a*b)*c
⇒ * is not associative

(iii) Here a * b = a + ab
Now b *a = b + ba
⇒ a * b ≠ b * a
⇒ * is not commutative

a*(b*c) = a*(b +bc) = a + a(b + bc) = a + ab + abc
(a*b)*c = (a + ab)*c = a + ab + (a + ab)c = a + ab + ac + abc
Thus, a*(b*c) ≠ (a * b)*c
⇒ * is not associative

Q.4

$\begin{array}{l} If f (x) = \frac{(4 x + 3)}{(6 x - 4)}, show that fof (x) = x_{1} \forall x \neq \frac{2}{3} . \\ What is the inverse of f ? \end{array}$

Ans

$\begin{array}{l} Here, f (x) = \frac{4 x + 3}{6 x - 4}, x \neq \frac{2}{3} \\ (a) fof (x) = f (f (x)) \\ = f (\frac{4 x + 3}{6 x - 4}) \\ = \frac{4 (\frac{4 x + 3}{6 x - 4}) - 3}{6 (\frac{4 x + 3}{6 x - 4}) - 4} \\ = \frac{\frac{4 (4 x + 3) + 3 (6 x - 4)}{6 x - 4}}{\frac{6 (4 x + 3) + 4 (6 x - 4)}{6 x - 4}} \\ = \frac{4 (4 x + 3) + 3 (6 x - 4)}{6 (4 x + 3) - 4 (6 x - 4)} \\ = \frac{16 x + 12 + 18 x - 12}{24 x + 18 - 24 x + 16} = \frac{34 x}{34} = x \end{array}$ $\begin{array}{l} (b) Put f (x) = y \\ \Rightarrow y = \frac{4 x + 3}{6 x - 4} \\ \Rightarrow y (6 x - 4) = 4 x + 3 \\ \Rightarrow 6 xy - 4 y = 4 x + 3 \\ \Rightarrow 6 xy - 4 x = 4 y + 3 \\ \Rightarrow x (6 y - 4) = 4 y + 3 \\ \Rightarrow x = \frac{4 y + 3}{6 y - 4} \end{array}$

Q.5 Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a ∧ b = min{a, b}. Write the composition table of the operation ∧ .

Ans

Operation ∧ table on the set {1, 2, 3, 4, 5} is as follows.

∧	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

Q.6 Let the * binary operation on N be defined by a*b = HCF of a and b. Is * commutative? Is * associative? Does there exist identity for this operation on N?

Ans

Here a*b = HCF of a and b.

(a) We know HCF of a, b = HCF of b, a

⇒ a*b = b*a

Hence, * is commutative.

(b) a*(b*c) = a*(HCF of b, c)

= HCF of a and HCF of b, c

= HCF of a, b and c

Now (a*b)*c = (HCF of a, b)*c

= HCF of a, b and HCF of c

= HCF of a, b and c

⇒ a*(b*c) = (a*b)*c

Hence, * is associative.

i.e., 1*a = a*1 = 1
1 ≠ a

∴ There does not exist any identity for this operation

Q.7 Let f: N → R be a function defined as f(x) = 4x² + 12x + 15. Show that f: N → S where, S is the range of f(x) is invertible. Find the inverse of f.

Ans

$\begin{array}{l} f is invertible if f : N \to S is a bijection . \\ f is one - one : For any x, y \in N \\ f (x) = f (y) \\ \Rightarrow 4 x^{2} + 12 x + 15 = 4 y^{2} + 12 y + 15 \\ \Rightarrow 4 x^{2} + 12 x = 4 y^{2} + 12 y \\ \Rightarrow 4 (x^{2} - y^{2}) + 12 (x - y) = 0 \\ \Rightarrow (x^{2} - y^{2}) + 3 (x - y) = 0 \\ \Rightarrow (x - y) (x + y) + 3 (x - y) = 0 \\ \Rightarrow (x - y) (x + y + 3) = 0 [x + y + 30 for anyx, yN] \\ \Rightarrow x - y = 0 \\ \Rightarrow x = y \\ So, f : N \to S is one - one . \\ Obviously f : N \to S is onto . \\ Hence f : N \to S is invertible . \\ Let f^{- 1} denote the inverse of f . Then, \\ {fof}^{- 1} (x) = x for all x \in S \\ \Rightarrow f (f^{- 1} (x)) = x for all x \in S \\ \Rightarrow 4 {(f^{- 1} (x))}^{2} +12 f^{- 1} (x) +15 = x for all x \in S \\ \Rightarrow 4 {\{f^{- 1} (x)\}}^{2} + 12 f^{- 1} (x) + 15 = 0 \\ \Rightarrow f^{- 1} (x) = \frac{- 12 \pm \sqrt{144 - 16 (15 - x)}}{8} = \frac{- 3 \pm \sqrt{x - 6}}{2} \\ \Rightarrow f^{- 1} (x) = \frac{- 3 + \sqrt{x - 6}}{2} [f^{- 1} (x) \in N, ∴ f^{- 1} (x) > 0] \end{array}$

Q.8

$\begin{array}{l} Let A = {- 1, 0, 1, 2}, B = {- 4, 2, 0, 2} and f, g: A \to B be \\ function defined by f (x) = x^{2} - x, x \in A and g (x) = 2 | x - \frac{1}{2} | - 1, x \in A. \\ Are f and g equal? Justify your answer. \end{array}$

Ans

$\begin{array}{l} At x = -1, \\ f (x) = 1^{2} + 1 = 2 \\ g (x) = 2 | - 1 - \frac{1}{2} | - 1 \\ = 2 \times \frac{3}{2} - 1 = 2 \\ At x = 0 \\ f (0) = 0 \\ g (0) = 2 | \frac{- 1}{2} | - 1 = 2 \times \frac{1}{2} - 1 = 0 \\ At x =1 \\ f (1) = 1^{2} - 1 = 0 \\ g (1) = 2 | 1 - \frac{1}{2} | - 1 = 2 \times \frac{1}{2} - 1 = 0 \\ At x = 2 \\ f (2) = 2^{2} - 2 = 2 \\ g (2) = 2 | 2 - \frac{1}{2} | - 1 = 3 - 1 = 2 \\ ∴ f (a) = g (a) for all a \in A \\ \Rightarrow f and g are equal functions . \end{array}$

Q.9 Let f, g and h be functions from R →R. Then show that
(i) (f + g)oh = foh + goh
(ii) (f.g)oh = (foh).(goh)

Ans

Here f: R →R, g: R →R, h: R →R(i) ((f + g)oh)(x) = (f + g)[h(x)]
= f[h(x)] + g[h(x)]
= foh(x) + goh(x)
⇒ (f + g)oh = foh + goh

(ii) ((f.g)oh)(x) = (f.g)[h(x)]
= f[h(x)].g[h(x)]
=(foh)(x).(goh)(x)
⇒ (f.g)oh = (foh).(goh)

Q.10

$\begin{array}{l} Let A = R – {3} and B =R – {- 1} . Consider the function \\ f: A \to B defined by f (x) = \frac{x - 2}{x - 3} . Is f one-one onto ? \\ Justify your answer. \end{array}$

Ans

$\begin{array}{l} f : A \to B where A = R - {3} and B = R - {1} . f is defined by f (x) = \frac{x - 2}{x - 3} . \\ (a) f (x_{1}) = \frac{x_{1} - 2}{x_{1} - 3} f (x_{2}) = \frac{x_{2} - 2}{x_{2} - 3} \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{x_{1} - 2}{x_{1} - 3} = \frac{x_{2} - 2}{x_{2} - 3} \\ \Rightarrow (x_{1} - 2) (x_{2} - 3) = (x_{2} - 2) (x_{1} - 3) \\ \Rightarrow x_{1} x_{2} - 3 x_{1} - 2 x_{2} + 6 = x_{1} x_{2} - 2 x_{1} - 3 x_{2} + 6 \\ \Rightarrow - 3 x_{1} - 2 x_{2} = - 2 x_{1} - 3 x_{2} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one - one \\ (b) Let y = \frac{x - 2}{x - 3} \\ \Rightarrow xy - 3 y = x - 2 \\ or xy - x = 3 y - 2 \\ or x (y - 1) = 3 y - 2 \\ \Rightarrow x = \frac{3 y - 2}{y - 1} \\ \Rightarrow For every value of y \in B except y =1 there is a pre - image, \\ x = \frac{3 y - 2}{y - 1} \\ \Rightarrow f is onto \\ This f is one - one and onto . \end{array}$

Q.11

$\begin{array}{l} Show that the functionf : \to R given by f (x) = \{\begin{matrix} 1 if x > 0 \\ 0 if x = 0 \\ - 1 if x < 0 \end{matrix}\} \\ is neither one - one nor onto . \end{array}$

Ans

(a) f(1) = f(2) = 1

1 and 2 have the same images.

⇒ f(x₁) = f(x₂) = 1 for x > 0
But x₁ ≠ x₂

Similarly -1 and -2 have the same images.

Hence f is not one-one.

(b) Except -1, 0, 1 no other element of co-domain of f has any pre-image in its domain.

⇒ f is not onto.

Q.12 Check the injectivity and surjectivity of the following:
(i) f: N → N given by f(x) = x²

(ii) f: R → R given by f(x) = x²

Ans

(i) f: N → N given by f(x) = x²
(a) f(x₁) = f(x₂) ⇒ x₁²= x₂²⇒ x₁= x₂

⇒f is one – one, i.e., it is injective

(b) There are many such elements belonging to co-domain that have no pre-image in their domain N e.g. 3 ∈ co domain N. But there is no pre- image in domain of f.

⇒ f is not onto, i.e., not surjective.

(ii) f: R R given by f(x) = x²
(a) f is not one – one because f(-1) = f (1) = 1
-1 and 1 have the same image
⇒ f is not injective.
(b) -2 ∈ co domain R of f. But √-2 does not belong to domain R of f.
⇒ f is not onto, i.e., f is not surjective.

Q.13 Show that the Relation R on the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : |a – b| is a multiple of 4} is an equivalence relation.

Ans

R = {(a, b) : |a – b| is a multiple of 4}, where a, b ∈ A ={x Z : 0 ≤ x ≤ 12} = {0, 1, 2, …, 12}.

Reflexivity: For any a ∈ A, |a – a| = 0, which is a multiple of 4.

⇒ (a, a) ∈ R, for all a ∈ A.

So, R is reflexive.

Symmetry: Let (a, b) ∈ R. Then,

(a, b) ∈ R

⇒ |a – b| is a multiple of 4
⇒ |a – b| = 4k for some k ∈ N
⇒ |b – a| = 4k for some k ∈ N
⇒ (b, a) ∈ R

So, R is symmetric.

Transitive: Let (a, b) ∈ R and (b, c) ∈ R. Then,
⇒ |a – b| is a multiple of 4 and |b – c| is a multiple of 4
⇒ |a – b| = 4k and |b – c| = 4m for some k, m ∈ N
⇒ a – b = ± 4k and b – c = ± 4m
⇒ a – c = ± 4k ± 4m
⇒ a – c is a multiple of 4
⇒ |a – c| is a multiple of 4
⇒ (a, c) ∈ R

So, R is transitive.

Hence, R is an equivalence relation.

Q.14 Determine whether the following relation is reflexive, symmetric and transitive: Relation R on the set N of all natural numbers is defined as
R = {(x, y): y = x + 5 and x < 4}

Ans

R = {(x, y): y = x + 5 and x < 4}
∴ R = {(1, 6), (2, 7), (3, 8)}

Reflexivity: (1, 1), (2, 2) etc. are not in R. So, R is not reflexive.

Symmetry: (1, 6) ∈ R but (6, 1) ∉ R. So, R is not symmetric.

Transitivity: Since (1, 6) ∈ R and there is no ordered pair in R which has 6 as the first element. Same is the case for (2, 7) and (3, 8). So, R is not transitive.

Q.15 Show that 1/x is the inverse of x ≠ 0 for the multiplication operation on R.

Ans

We know that x X (1/x) = 1 (Identity for ‘x’). So, 1/x is the multiplicative inverse of x.

Q.16 Let f = {(3, 1), (2, 3), (1, 2)}, find f ^-1if f is one-one and onto.

Ans f is invertible. So, f ^-1 = {(1, 3), (3, 2), (2, 1)}.

Q.17

$\begin{array}{l} If a function is defined as f (x) = \frac{x}{x - 1}, x \neq 1, then \\ find of f (x) \end{array}$

Ans

$\begin{array}{l} If a function is defined as f (x) = \frac{x}{x - 1}, x \neq 1, then \\ find of f (x) \\ f (x) = \frac{x}{x - 1} \\ \Rightarrow f of (x) = \frac{\frac{x}{x - 1}}{\frac{x}{x - 1} - 1} \\ = \frac{\frac{x}{x - 1}}{\frac{x - x + 1}{x - 1}} \\ = \frac{\frac{x}{x - 1}}{\frac{1}{x - 1}} \\ Now, fof (x) = x \\ \Rightarrow fofof (x) = \frac{x}{x - 1} [same as given function] \\ on repeating the above step, we get fofofof (x) = x . \end{array}$

Q.18 A relation R in a set A is called …, if no element of A is elated to any element of A.

AnsEmpty Relation

Q.19 A relation R in a set A is called …, if each element of A is elated to every element of A.

AnsUniversal Relation

Q.20 Name the relation R in a set A satisfying the condition (a, a) ∈ R, for every a ∈ A.

Ans

The relation R is reflexive.

Q.21 Name the relation R in a set A satisfying the condition (a₁, a₂) ∈ R ⇒ (a₂, a₁) ∈ R for all (a₁, a₂) ∈ A.

Ans

The relation R is symmetric.

Q.22 R is a relation in a set {1, 2, 4} given by R = {(1, 1), (2, 2), (4, 4)}. State whether R is reflexive or symmetric.

Ans

R is reflexive. [ (a, a) ε R, for every a ε {1, 2, 4}]

Q.23 Let f: R R; f(x) = sinx and g: R R; g(x) = x², find fog and gof.

AnsClearly, fog and gof both exist.
Now, (gof)(x) = g(f(x)) = g(sinx) = (sinx)² = sin²x.
And, (fog)(x) = f(g(x)) = f(x²) = sinx².

Q.24 If f is one-one and onto, then what is f ^-1o f ?

Ans

f ^-1of = I_f

Q.25 Let * be a binary operation on set Q of rational numbers as follows:

a*b = (a-b)² . Is this operation commutative?

Ans

a*b = (a-b)²
b*a = (b-a)²
(a-b)²= (b-a)²
Therefore, a*b = b*a
So, the given operation is commutative.

Q.26

$If f (x) = x+8 and g (x) = x-8, \forall x \in R, find fog (8)$

Ans

$\begin{array}{l} f (x) = x + 8, g (x) = x - 8 \\ ∴ fog (x) = f [g (x)] \\ \Rightarrow fog (x) = f (x - 8) \\ \Rightarrow fog (x) = (x - 8) + 8 \\ \Rightarrow fog (x) = x \\ \Rightarrow fog (8) = 8 \end{array}$

Q.27

$If f (x) {= x}^{2} +3 and g (x) = \frac{x}{x+2}, \forall x \in R, find gof (2)$

Ans

$\begin{array}{l} f (x) {= x}^{2} + 3, g (x) = \frac{x}{x+2} \\ ∴ gof (x) = g [f (x)] \\ = g [x^{2} + 3] \\ = \frac{x^{2} + 3}{(x^{2} + 3) + 2} \\ \Rightarrow gof (x) = \frac{x^{2} + 3}{x^{2} + 3 + 2} \\ \Rightarrow gof (x) = \frac{x^{2} + 3}{x^{2} + 5} \\ \Rightarrow gof (2) = \frac{4 + 3}{4 + 5} = \frac{7}{9} \end{array}$

Q.28 Let * be a binary operation defined by a*b=2a+b+ab+1. Find 3*5.

Ans

Given that

a*b = 2a+b+ab+1

⇒ 3*5 = 2×3+5+3×5+1

= 6+5+15+1

= 27

Q.29

$\begin{array}{l} If a function is defined as f (x) = \frac{x}{x - 1}; x \neq 1, then \\ find fof (x) . \end{array}$

Ans

$\begin{array}{l} f (x) = \frac{x}{x - 1} \\ \Rightarrow fof (x) = \frac{\frac{x}{x - 1}}{\frac{x}{x - 1} - 1} \\ = \frac{\frac{x}{x - 1}}{\frac{x - x + 1}{x - 1}} \\ = \frac{\frac{x}{x - 1}}{\frac{1}{x - 1}} = x \\ ∴ fof (x) = x \end{array}$

Q.30

$\begin{array}{l} If a function is defined as f (x) = \frac{x}{\sqrt{1 + x^{2}}}, then \\ find fof (x) . \end{array}$

Ans

$\begin{array}{l} f (x) = \frac{x}{\sqrt{1 + x^{2}}} \\ \Rightarrow fof (x) = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{1 + {(\frac{x}{1 + x^{2}})}^{2}}} \\ = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{1 + \frac{x^{2}}{1 + x^{2}}}} \\ = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{\frac{1 + 2 x^{2}}{1 + x^{2}}}} \\ = \frac{x}{\sqrt{1 + x^{2}}} \times \frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}} \\ = \frac{x}{\sqrt{1 + 2 x^{2}}} \end{array}$

Q.31 Let A = {1, 2, 3, 8}, B = {4, 5, 6} and let f = {(1, 4), (2, 5), (3, 6)} be defined from A to B. Show that f is not a function from A to B.

AnsSince there is no image in set B for every element of set A (that is, element 8 of set A has no image in set B), therefore, f is not a function from set A to set B.

Q.32 Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Ans

Since every element of set A has only one image in set B, that is 1→ 4, 2→ 5, 3→ 6, therefore, f is one-one function.

Q.33 Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 4), (3, 6)} be a function from A to B. Show that f is many one.

Ans

Since elements 1 and 2 have same image 4 in set B, (1→ 4, 2→ 4, 3→ 6 ), therefore, f is many one function.

Q.34 Let A = {1, 2}, B = {4, 5, 6} and let f = {(1, 4), (2, 5), (2, 6)} be a function from A to B. Show that f is not a function.

Ans

Since every element of set A has no unique image in set B, that is, element 2 has two images 5 and 6 in set B (1→ 4, 2→ 5, 2→ 6), therefore, f is not a function.

Q.35 If Set A has 2 elements and Set B has 3 elements, find the number of relations defined from Set A to Set B.

Ans

Since number of elements in set A = 2

Number of elements in set B = 3

∴ Number of elements in set (AXB) = 23=6

The number of all possible subsets of Set (AXB) = 2⁶

⇒ The number of relations defined from Set A to Set B is 2⁶ = 64.

Q.36

$\begin{array}{l} Show that the function f:R \to R defined by f (x) = \frac{2 x - 1}{3}, \\ x \in R is one-one and on to function. Also, find the \\ inverse of the function f. \end{array}$

Ans

$\begin{array}{l} {Let x}_{1,} x_{2} \in R such that f (x_{1}) = f (x_{2}) . \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{2 x_{1} - 1}{3} = \frac{2 x_{1} - 1}{3} \\ \Rightarrow 2 x_{1} - 1 = 2 x_{2} - 1 \\ \Rightarrow 2 x_{1} = 2 x_{1} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one-one. \\ Again, y = \frac{2 x - 1}{3} \\ \Rightarrow 2 x - 1 = 3 y \\ \Rightarrow 2 x = 3 y + 1 \\ \Rightarrow x = \frac{3 y + 1}{2} \\ Now, f (x) = f (\frac{3 y + 1}{2}) \forall x, y \in R \\ = \frac{2 (\frac{3 y + 1}{2}) - 1}{3} \\ = \frac{3 y -}{3} \\ = \frac{3 y}{3} \\ = y \\ ∴ f is onto function. \\ Since f is one-one and onto function, therefore, f is invertible. \\ Again, y = \frac{2 x - 1}{3} \\ \Rightarrow 2 x - 1 = 3 y \\ \Rightarrow 2 x = 3 y + 1 \\ x = \frac{3 y + 1}{2} \\ \Rightarrow f^{- 1} (y) = \frac{3 y + 1}{2} [f (x) = y \Rightarrow x = f^{- 1} (y)] \\ \Rightarrow f^{- 1} (x) = \frac{3 x + 1}{2} [Replacing y by x] \end{array}$

Q.37

$Find the domain of the function f (x) = \sqrt{36 - x^{2} .}$

Ans

$\begin{array}{l} We know that all values of x for which \\ function is defined is called domain. \\ f (x) = \sqrt{36 - x^{2}} \\ For function to be defined \\ 36 - x^{2} \geq 0 \\ \Rightarrow x^{2} - 36 \leq 0 \\ \Rightarrow x^{2} - 6^{2} \leq 0 \\ \Rightarrow (x - 6) (x + 6) \leq 0 \\ \Rightarrow x \geq - 6 and x \leq 6 \\ \Rightarrow - 6 \leq x \leq 6 \\ \Rightarrow x \in [- 6, 6] \\ Hence, domain of the given function is [- 6, 6] . \end{array}$

Q.38

$Find the domain of the function f (x) = \sqrt{x^{2} - 7 x + 10 .}$

Ans

$\begin{array}{l} We know that all values of x for which function \\ is defined is called domain. \\ f (x) = \sqrt{x^{2} - 7 x + 10} \\ For function to be defined, \\ x^{2} - 7 x + 10 \geq 0 \\ \Rightarrow (x - 2) (x - 5) \geq 0 \\ \Rightarrow (x - 2) \geq 0 and (x - 5) \geq 0 \\ \Rightarrow x \geq 2 and x \geq 5 \Rightarrow x \geq 5 . .. (1) \\ \Rightarrow Also (x - 2) \leq 0 and (x - 5) \leq 0 \\ \Rightarrow x \leq 2 and x \leq 5 \Rightarrow x \leq 2 . \dots (2) \\ From (1) and (2) we get, \\ - \infty < \times \leq 2 and 5 \leq \times \infty \\ Hence, domain of the given function is \\ (- \infty, 2] \cup [5, \infty) . \end{array}$

Q.39

$\begin{array}{l} Find the domain of the function \\ f (x) = \sqrt{x^{3} - 6 x^{2} + 11 x} - 6 . \end{array}$

Ans

$\begin{array}{l} We know that all values of x for which function \\ is defined is called domain. \\ ∴ for function to be defined \\ x^{3} - 6 x^{2} + 11 x – 6 \geq . \dots (1) \\ Let P (x) = x^{3} - 6 x^{2} + 11 x - 6 \\ Now for getting its factor, put x = - 1, 1, 0 . \dots \\ {(1)}^{3} - 6 {(1)}^{2} + 11 (1) - 6 \\ = 1 - 6 +11 - 6 \\ = 12 - 12 \\ 0 \\ Hence, (x - 1) is one of the factors of P (x) . \\ Now on dividing P (x) by (x - 1), we get \\ P (x) = (x - 1) (x^{2} - 5 x + 6) \\ P (x) = (x - 1) (x - 2) (x - 3) \end{array}$

$\begin{array}{l} The expression is +ve in the interval \\ [1, 2] and [3, + \infty] . \\ Hence, the domain of the given function is \\ [1, 2] \cup [3, + \infty] . \end{array}$

Q.40

$\begin{array}{l} Show that the function f:R \to R defined by f (x) = \frac{5 x + 2}{3}, \\ x \in R is one-one and on to function. Also find the \\ inverse of the function f. \end{array}$

Ans

$\begin{array}{l} {Let x}_{1,} x_{2} \in R such that f (x_{1}) = f (x_{2}) . \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{5 x_{1} + 2}{3} = \frac{5 x_{2} + 2}{3} \\ \Rightarrow 5 {fx}_{1} + 2 = 5 {fx}_{2} + 2 \\ \Rightarrow 5 {fx}_{1} = 5 {fx}_{2} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one-one \\ Again, y = \frac{fx+2}{3} \\ \Rightarrow 5 x + 2 = 3 y \\ \Rightarrow 5 x = 3 y - 2 \\ \Rightarrow x = \frac{3 y - 2}{5} \end{array}$ $\begin{array}{l} Now, f (x) = f (\frac{3 y - 2}{5}) \forall x, y \in R \\ \Rightarrow = \frac{5 (\frac{3 y - 2}{5}) + 2}{3} \\ = \frac{3 y - 2 + 2}{3} \\ = \frac{3 y}{3} \\ = y \\ ∴ f is onto function. \\ Since f is one-one and onto function. therefore. it is invertible. \\ Again, y = \frac{5 x + 2}{3} \\ \Rightarrow 5 x + 2 = 3 y \\ \Rightarrow 5 x = 3 y - 2 \\ \Rightarrow x = \frac{3 y - 2}{5} \\ \Rightarrow f^{- 1} (y) = \frac{3 y - 2}{5} [f (x) = y \Rightarrow {x = f}^{- 1} (y)] \\ \Rightarrow f^{- 1} (x) = \frac{3 x - 2}{5} \end{array}$

Q.41

$Find the range of the function f (x) = \frac{x^{2} + 2}{x^{2} + 1} \forall x \in R.$

Ans

$\begin{array}{l} Let y = \frac{x^{2} + 2}{x^{2} + 1} \forall \times \in R \\ \Rightarrow {yx}^{2} + y = x^{2} + 2 \\ \Rightarrow (y - 1) x^{2} = 2 - y \\ \Rightarrow x^{2} = \frac{(2 - y)}{y - 1} \\ \Rightarrow x = \sqrt{\frac{(2 - y)}{y - 1}} . \dots\dots. (1) \\ {Since x}^{2} \geq 0 \\ So, \frac{(2 - y)}{y - 1} \geq 0 \\ \Rightarrow \frac{}{} \frac{(y - 2)}{y - 1} \leq 0 \\ Now, using number line, we get \end{array}$

$\begin{array}{l} \Rightarrow 1 \leq y \leq 2 \\ but from (1) we get, y \neq 1 \\ \Rightarrow 1< y \leq 2 \\ \Rightarrow y \in (1, 2) \\ Range of the given function is (1, 2) . \end{array}$

Q.42

$\begin{array}{l} Show that the function f: [- 1, 1] \to R defined \\ by f (x) = \frac{x}{x + 2} is one-one. Find the inverse \\ of the function f. \end{array}$

Ans

$\begin{array}{l} {Let x}_{1}, x_{2} \in [- 1, 1] such that f (x_{1}) = f (x_{2}) . \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{x_{1}}{x_{1} + 2} = \frac{x_{2}}{x_{2} + 2} \\ \Rightarrow x_{1} x_{2} + 2 x_{1} x_{2} + 2 x_{2} \\ \Rightarrow 2 x_{1} = 2 x_{2} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one-one \\ Again, y = \frac{x}{x + 2} \\ \Rightarrow yx + 2 y = x \\ \Rightarrow 2 y = x - xy \\ \Rightarrow x (1 - y) = 2 y \\ \Rightarrow x = \frac{2 y}{1 - y} . \dots.. (1) \\ f (x) = f (\frac{2 y}{1 - y}) \\ = \frac{\frac{2 y}{1 - y}}{\frac{2 y}{1 - y} + 2} \\ = \frac{2 y}{2 y + 2 - 2 y} \\ = \frac{2 y}{2} \\ = y \\ ∴ f (x) = y [so, f (x) is on to] \\ \Rightarrow {x = f}^{- 1} (y) \\ Using this (1) become \\ \Rightarrow f^{- 1} (y) = \frac{2 y}{1 - y} \\ \Rightarrow f^{- 1} (x) = \frac{2 x}{1 - x} [Replacing y by x] \end{array}$

Q.43

$\begin{array}{l} Let f: R \to R be defined as f (x) = 11 \times -9. Find the \\ function g:R \to {such that gof = fog I}_{R} . \end{array}$

Ans

$\begin{array}{l} {We have, gof =I}_{R} \\ \Rightarrow gof (x) = I_{R} (x) for all \times \in R \\ \Rightarrow g [f (x)] = x for all \times \in R [I_{R} (x) = x] \\ \Rightarrow g (11 \times 9) = x for all \times \in R [f (x) = 11 x - 9] \\ \Rightarrow g (y) = x \\ Where y = 11x -9 \\ \Rightarrow 11x = y+9 \\ \Rightarrow x = \frac{y + 9}{11} \\ \Rightarrow g (y) = x \\ \Rightarrow g (y) = \frac{y + 9}{11} \\ \Rightarrow g (x) = \frac{x + 9}{11} \\ Thus, g:R \to R defined by g (x) = \frac{x + 9}{11} is the \\ required function. \end{array}$

Q.44

$\begin{array}{l} Let f : R \to R be defined as f (x) = ax+b. \\ Show that f (x) is one-one. Also find the function \\ g : R \to R such that gof = fog I. \end{array}$

Ans

$\begin{array}{l} Let x 1, x_{2} \in R such that f (x_{1}) = f (x_{2}) . \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow {ax}_{1}_{2} \\ \Rightarrow {ax}_{1} = {ax}_{2} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one - one . \\ We have, gof = I_{R} \\ \Rightarrow gof (x) = I_{R} (x) for all \times \in R \\ \Rightarrow g [f (x)] = x for all x \in R [I_{R} (x) = x] \\ \Rightarrow g (ax + b) = x for all x \in R [f (x) = ax + b] \\ \Rightarrow g (y) = x \\ Where y = ax + b \\ \Rightarrow ax = y - b \\ \Rightarrow x = \frac{y - b}{a} . \dots.. (1) \\ \Rightarrow g (y) = x \\ \Rightarrow g (y) = \frac{y - b}{a} using (1) \\ \Rightarrow g (x) = \frac{x - b}{a} \\ Thus, g : R \to R defined by g (x) = \frac{x - b}{a} is the required function . \end{array}$

Q.45 Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a∧b = min {a, b}. Write the operation table of the operation ∧.

Ans

Since a*b = min {a, b}

⇒1*1 = min {1,1} = 1

⇒ 1*2 = min {1,2} = 1

⇒ 1*3 = min {1,3} = 1

⇒ 1*4 = min {1,4} = 1

⇒ 1*5 = min {1,5} = 1

⇒ 2*1 = min {2,1} = 1

⇒ 2*2 = min {2,2} = 2

⇒ 2*3 = min {2,3} = 2

⇒ 2*4 = min {2,4} = 2

⇒ 2*5 = min {2,5} = 2

⇒3*1 = min {3,1} = 1

⇒3*2 = min {3,2} = 2

⇒3*3 = min {3,3} = 3

⇒3*4 = min {3,4} = 3

⇒3*5 = min {3,5} = 3

⇒ 4*1 = min {4,1} = 1

⇒ 4*2 = min {4,2} = 2

⇒ 4*3 = min {4,3} = 3

⇒ 4*4 = min {4,4} = 4

⇒ 4*5 = min {4,5} = 4

⇒ 5*1 = min {5,1} = 1

⇒ 5*2 = min {5,2} = 2

⇒ 5*3 = min {5,3} = 3

⇒ 5*4 = min {5,4} = 4

⇒ 5*5 = min {5,5} = 5

Thus, operation table of operation (*) calculated above is

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

Q.46 Consider the binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table. Using this table compute

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	1	2	1
3	1	1	3	1	1
4	1	2	1	4	1
5	1	1	1	1	5

(i) (3*4)*5.
(ii) 3*(4*5).
(iii) (2*3)*(4*5).
(iv) if * is commutative.

Ans

We have
(i) (3*4)*5 = 1*5 [Since, 3*4=1] =1
(ii) 3*(4*5) = 3*1 [Since, 4*5=1] =1
(iii) (2*3)*(4*5) = 1*1 [∴2*3=1 and 4*5=1] =1
(iv) Since (2*5)=1 and (5*2)=1
⇒(2*5) = (5*2)
⇒ Operation * is commutative.

Q.47 Show that addition, subtraction and multiplication are binary operations on R, but division is not a binary operation on R. Further, show that division is a binary operation on the set R of non-zero real numbers.

Ans

$\begin{array}{l} + : R \times R \to R is given by (a, b) \to a + b \\ If 2 \in R, 3 \in R then 2+3 = 5 \in R \\ \Rightarrow Operation ‘+’ is binary operation on R \\ - : R \times R \to R is given by (a, b) \to a - b \\ If 2 \in R, 3 \in R then 2 - 3 = - 1 \in R \\ \Rightarrow Operation ‘ - ‘ is binary operation on R \\ \times : R \times R \to R is given by (a,b) \to ab \\ If 2 \in R, O \in R then 2 \times 0 = 0 \in R \\ \Rightarrow Operation ‘ \times ‘ is binary operation on R \\ \div : R \times R \to R is given by (a, b) \to \frac{a}{b} \\ If 2 \in R, 0 \in R then \frac{2}{0} = \infty \in R \\ \Rightarrow Operation ‘ \div ‘ is not a binary operation on R \\ \div : R \times R \to is given by (a, b) \to \frac{a}{b}, where R is the set of \\ non-zero real numbers (given) . \\ If 2 \in R, \in R then \frac{2}{3} \in R . \\ \Rightarrow Operation ‘ \div ‘ is a binary operation on R. \end{array}$

Q.48

$If a function is defined as f (x) = \frac{x}{\sqrt{{1+x}^{2}}}, then find fofofofof (x) .$

Ans

$\begin{array}{l} f (x) = \frac{x}{1 + x^{2}} \\ \Rightarrow fof (x) = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{1 + \frac{x}{1 + x^{2}}}} \\ = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{1 + \frac{x^{2}}{1 + x^{2}}}} \\ = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{\frac{1 + 2 x^{2}}{1 + x^{2}}}} \\ = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{} x \frac{\sqrt{1 + x^{2}}}{\sqrt{1 + 2 x^{2}}} \\ = \frac{x}{\sqrt{1 + 2 x^{2}}} \\ = \frac{x}{\sqrt{1 + x^{2}}} \\ Now, fofof (x) = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{1 + \frac{x^{2}}{1 + 2 x^{2}}}} \\ = \frac{\frac{x}{\sqrt{1 + 2 x^{2}}}}{\sqrt{\frac{1 + 2 x^{2} + x^{2}}{1 + 2 x^{2}}}} = \frac{\frac{x}{\sqrt{1 + 2 x^{2}}}}{\sqrt{\frac{1 + 3 x^{2}}{1 + 2 x^{2}}}} \\ = \frac{x}{\sqrt{1 + x^{2}}} x \frac{\sqrt{1 + 2 x^{2}}}{1 + 3 x^{2}} \\ = \frac{x}{\sqrt{1 + 3 x^{2}}} \\ on observing, we get \\ for fof there is 2, fofof there is 3, fofofof there will be 4 \\ and therefore, fofofofof (x) = \frac{x}{\sqrt{1 + 5 x^{2}}} \end{array}$

Q.49

$\begin{array}{l} Let : A \to B and g:B \to {A be two functions such that fog = I}_{B} . \\ Then, show that f is a surjection and g is an injection. \end{array}$

Ans

$\begin{array}{l} f is surjection: we have to prove here that every element in B \\ has its pre – image in A. \\ Let b be an arbitrary element of B. \\ Since, g:B \to A. Therefore, g (b) \in A \\ Let g (b) = a \\ Now, f (a) = f (g (b)) [a = g (b)] \\ \Rightarrow f (a) = fog (b) \\ \Rightarrow f (a) = I_{B} (b) [fog = I_{B}] \\ \Rightarrow f (a) = b \\ Thus, \forall b \in B there exists a \in A such that f (a) = b. \\ So, f is a surjection. \\ {g is an injection : let x}_{1} {, x}_{2} be any two elements of B that \\ g (x_{1}) = g (x_{2}) . Then, \\ g (x_{1}) = g (x_{2}) \\ \Rightarrow f (g (x_{1})) = f (g (x_{2}^{})) \\ \Rightarrow fog (x_{1}) = fog (x_{2}^{}) \\ \Rightarrow I_{B} (x_{1}) = I_{B} (x_{2}) \\ \Rightarrow x_{1} {= x}_{2} \\ Thus, g (x_{1}) = g (x_{2}) \Rightarrow x_{1} = x_{2} \forall x_{1} {, x}_{2}, \in B. \\ So, g is an injection. \end{array}$

Q.50

$\begin{array}{l} If f, g : R \to R are defined respectively by f (x) = x^{2} + 3 x + 1, \\ g (x) = 2 x - 3, find \\ (i) fog (ii) gof (iii) fof (iv) gog . \end{array}$

Ans

$\begin{array}{l} Since the range of f = doamin of g and range of g = domain of f, \\ therefore, fog, gof, gog and fof exists . \\ (i) For any x \in R, we have \\ fog (x) = f (g (x)) = f (2 x - 3) \\ = {(2 x - 3)}^{2} + 3 (2 x - 3) + 1 \\ = 4 x^{2} - 6 x + 1 \\ Therefore, \\ fog : R \to R is defined by (fog) (x) = 4 x^{2} - 6 x + 1 for all x \in R . \\ (ii) For any x \in R, we have \\ (gof) (x) = g (f (x)) \\ = g (x^{2} + 3 x + 1) \\ = 2 (x^{2} + 3 x + 1) - 3 \\ = 2 x^{2} + 6 x - 1 \\ Therefore, \\ gof : R \to R is defined by (gof) (x) = 2 x^{2} + 6 x - 1 for all x \in R . \\ (ii) For any x \in R_{1} we have \\ (gof) (x) = g (f (x)) \\ = g (x^{2} + 3 x + 1) \\ = 2 (x^{2} + 3 x + 1) - 3 \\ = 2 x^{2} + 6 x - 1 \\ Therefore, \\ gof : R \to R is defined by (gof) (x) = 2 x^{2} + 6 x - 1 for all x \in R . \\ (iii) For any x \in R, we have \\ (fof) (x) = f (f (x)) \\ = f (x^{2} + 3 x + 1) \\ = {(x^{2} + 3 x + 1)}^{2} + 3 (x^{2} + 3 x + 1) + 1 \\ = x^{4} + 9 x^{2} + 1 + 6 x^{3} + 6 x + 2 x^{2} + 3 x^{2} + 9 x + 3 + 1 \\ = x^{4} + 6 x^{3} + 14 x^{2} + 15 x + 5 \\ Therefore, \\ fof : R \to R is defined by \\ (fof) (x) = x^{4} + 6 x^{3} + 14 x^{2} + 15 x + 5 for all x \in R . \\ (iv) for any x \in R, we have \\ (gog) (x) = g (g (x)) \\ = g (2 x - 3) \\ = 2 (2 x - 3) - 3 \\ = 4 x - 9 \\ Therefore, \\ gog : R \to R is defined by (gog) (x) = 4 x - 9 for all x \in R . \end{array}$

Q.51

$\begin{array}{l} Let A = R – {3} and B = R – {1} . Consider the finction f:A \to B \\ defined by f (x) = \frac{x - 2}{x - 3} . Is f one – one and onto? \\ Justify your answer. \end{array}$

Ans

$\begin{array}{l} f : A \to B where A = R – {3} and B = R – {1}, f is defined by \\ f (x) = \frac{x - 2}{x - 3} . \\ (a) f (x_{1}) = \frac{x_{1} - 2}{x_{1} - 3}; f (x_{2}) = \frac{x_{2} - 2}{x_{2} - 3} \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{x_{1} - 2}{x_{1} - 3} = \frac{x_{2} - 2}{x_{2} - 3} \\ \Rightarrow (x_{1} - 2) (x_{2} - 3) = (x_{2} - 2) (x_{1} - 3) \\ \Rightarrow x_{1} x_{2} - 3 x_{1} - 2 x_{2} + 6 = x_{1} x_{2} - 2 x_{1} - 3 x_{2} + 6 \\ \Rightarrow - 3 x_{1} - 2 x_{2} = - 2 x_{1} - 3 x_{2} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one-one. \\ (b) y \in B=R- {1} . Then, y \neq 1 \\ The function f is onto if there exists x \in A such that \\ \Rightarrow f (x) = y \\ \Rightarrow \frac{x - 2}{x - 3} = y \\ \Rightarrow x - 2 = y (x - 3) \\ \Rightarrow x - 2 = yx - 3 y \\ \Rightarrow x - yx = 2 - 3 y \\ \Rightarrow x (1 - y) = 2 - 3 y \\ \Rightarrow x = \frac{2 - 3 y}{1 - y} \in A [y \neq 1] \\ Thus, for any y \in B, there exists \frac{2 - 3 y}{1 - y} \in A such that \\ f (\frac{2 - 3 y}{1 - y}) = \frac{\frac{2 - 3 y}{1 - y} - 2}{\frac{2 - 3 y}{1 - y} - 3} = \frac{\frac{2 - 3 y - 2 (1 - y)}{1 - y}}{\frac{2 - 3 y (1 - y)}{1 - y}} \\ = \frac{2 - 3 y - 2 + 2 y}{2 - 3 y - 3 + 3 y} = \frac{- y}{- 1} \\ f (\frac{2 - 3 y}{1 - y}) = y \\ ∴ f is onto. \\ Hence, function f is one-one and onto. \end{array}$

Q.52

$\begin{array}{l} Three relations R_{1}, R_{2}, R_{3} are defined on set A = (a, b, c) as follows : \\ (i) R_{1} = \{(a, a), (a, b), (a, c), (b, b), (b, c), (c, a) (c, b), (c, c)\} \\ (ii) R_{2} = \{(a, b), (b, a), (a, c) (c, a)\} \\ (iii) R_{3} = \{(a, b), (b, c), (c, a)\} \\ {Find whether each of the relations R}_{1}, R_{2}, R_{3} arereflexive, \\ symmetric and transitive. \end{array}$

Ans

$\begin{array}{l} (i) Reflexive : Clearly, (a, a), (b, b), (c, c) \in R_{1} . \\ So, R_{1} is reflexive on A. \\ Symmetric: We observe that (a, b) \in R_{1} but (b, a) \notin R_{1} . \\ {So, R}_{1} is not symmetric on A. \\ Transitive : We find that (b, c) \in R_{1} and (c, a) \in R_{1} but (b, a) \in R_{1} . \\ S {o, R}_{1} is not transitive on A. \\ (ii) Reflexive: Clearly, (a, a), (b . b), (c, c) \notin R_{2} . \\ So, R_{2} is not a reflexive on A. \\ Symmetric: We find that the ordered pairs obtained by interchanging \\ {the components of ordered pairs in R}_{2} {are also in R}_{2} . \\ So, R_{2} is a symmetric relation on A. \\ Transitive: We find that (a, b) \in R_{2} and (b, a) \in R_{2} . \\ So, R_{2} is not transitive on A. \\ (iii) Reflexive: Since none of (a, a), (b, b), (c, c) is {an element of R}_{3} . \\ {So, R}_{3} is not a reflexive relation on A. \\ Symmetric: We observe that (b, c) \in R_{3} but (c, b) \notin R_{3} . \\ So, R_{3} is not symmetric on A. \\ Transitive:We find that (a, b) \in R_{3} and (b, c) \in R_{3} but (a, c) \notin R_{3} . \\ So, R_{3} is not transitive on A. \end{array}$

Q.53

$\begin{array}{l} Show that f: [01, 1] \to Range f, given by f (x) = \frac{x}{x + 2} is one-one. \\ Find the inverse of the function f. \end{array}$

Ans

$\begin{array}{l} f : [01, 1] \to Range is given as f (x) = \frac{x}{x + 2} . \\ Let f (x) = f (y) . \\ \Rightarrow \frac{x}{x + 2} = \frac{y}{y + 2} \\ \Rightarrow xy + 2 x = xy + 2 y \\ \Rightarrow 2 x = 2 y \\ \Rightarrow x = y \\ ∴ f is a one - one function . \\ It is clear that f : [- 1, 1] \to Range f is onto . \\ ∴ f : [- 1, 1] \to Range f is one - one and onto and therefore, \\ the inverse of the function \\ f : [- 1, 1] \to Range f exists . \\ Let g : Range f \to [- 1, 1] be the inverse of f . \\ Let y be an arbitrary element of range f . \\ Since f : [- 1, 1] \to Range f is onto, we have : \\ y = f (x) for some x \in [- 1, 1] \\ \Rightarrow y = \frac{x}{x + 2} \\ \Rightarrow xy +2 y = x \\ \Rightarrow x (1 - y) = 2 y \\ \Rightarrow x = \frac{2 y}{1 - y}, y \neq 1 \\ Now, let us define g : Range f \to [- 1, 1] as \\ g (y) = \frac{2 y}{1 - y}, y \neq 1 \\ Now, (gof) (x) = g (f (x)) = g (\frac{x}{x + 2}) \\ = \frac{2 (\frac{x}{x + 2})}{1 - \frac{x}{x + 2}} = \frac{2 x}{x + 2 - x} \\ = \frac{2 x}{2} = x \\ (fog) (y) = f (g (y)) = f (\frac{2 y}{1 - y}) \\ = (\frac{\frac{2 y}{1 - y}}{\frac{2 y}{1 - y} + 2}) = \frac{2 y}{2 y + 2 - 2 y} \\ = \frac{2 y}{2} = y \\ ∴ gof = I_{[- 1, 1]} and fog = I_{Range f} \\ ∴ f^{- 1} = g \\ \Rightarrow f^{- 1} (y) = \frac{2 y}{1 - y}, y - \neq 1 \end{array}$

Q.54

$\begin{array}{l} Consider f : {1, 2, 3} \to {a, b, c} given by f (1) = a, f (2) = b \\ and f (3) = {c. Find f}^{- 1} and show that {(f^{- 1})}^{- 1} = f. \end{array}$

Ans

$\begin{array}{l} Function f: {1, 2, 3} \to {a, b, c} is given by f (1) = a, f (2) = b, f (3) = c. \\ If (1) = a, f (2) = b, f (3) = c \\ If we define g : {a, b, c} \to {1, 2, 3} as g (a) = 1, g (b) = 2, g (c) = 3, \\ then we have : \\ (fog) (a) = f (g (a)) = f (1) a \\ (fog) (b) = f (g (b)) = f (2) b \\ (fog) (c) = f (g (c)) = f (3) c \\ And \\ (gof) (1) = g (f (1)) = g (a) 1 \\ (gof) (2) = g (f (2)) = g (a) 2 \\ (gof) (3) = g (f (3)) = g (a) 3 \\ ∴ gof = Ix and fog = Iy, where x = {1, 2, 3} and y = {a, b, c} \\ {Thus , the inverse of exists and f}^{- 1} = g . \\ ∴ f^{-1} : {a, b, c} \to {1, 2, 3} is given by, \\ f^{- 1} (a) = 1, f^{- 1} (b) = 2, f^{- 1} (c) = 3 \\ {Let us now find the inverse of f}^{- 1} i . e ., find the inverse of g. \\ If we define h : {1, 2, 3} \to {a, b, c} as \\ h (1) = a, h (2) = b, h (3) = c, then we have : \\ (goh) (1) = g (h (1)) = g (a) 1 \\ (goh) (2) = g (h (2)) = g (b) 2 \\ (goh) (3) = g (h (3)) = g (c) 3 \\ and \\ (hog) (a) = h (g (a)) = h (1) a \\ (hog) (b) = h (g (b)) = h (2) b \\ (hog) (c) = h (g (c)) = h (3) c \\ ∴ {goh = I}_{x} {and hog = I}_{y}, where x = {1, 2, 3} and y = {a, b, c} . \\ {Thus, the inverse of g exists and g}^{- 1} = h \Rightarrow (f^{- 1}) = h . \\ It can be noted that h = f. \\ Hence, {(f^{- 1})}^{- 1} = f. \end{array}$

Q.55 Let A = {2, 3, 4}, B = {5, 6, 7, 9} and let f = {(1, 4), (3, 6), (4, 7)} be a function from A to B, then function is ______________.

Ans

Since, every element of set A has only one image in set B i.e., 2→5, 3→6 and 4→7. Therefore, f is one-one function.

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FAQs (Frequently Asked Questions)

A function is a special type of relation. In a relation, an element can be related in different ways. In a function, each domain element must have exactly one image. This single-output rule is the main difference.

Check three conditions in order: reflexive, symmetric and transitive. If even one condition fails, the relation is not an equivalence relation. Use ordered pairs carefully because missing reverse pairs or self-pairs often changes the answer.

Co-domain is the set where outputs are expected to lie. Range is the set of outputs actually obtained. For onto functions, range and co-domain are equal. For non-onto functions, range is only a part of the co-domain.

No, gof does not always equal fog. In gof, f acts first and then g acts. In fog, g acts first and then f acts. Since the order changes, the final expressions may also change.

An invertible function must be one-one and onto. One-one ensures no two inputs share the same image. Onto ensures every co-domain element has a pre-image. Together, these conditions allow the reverse mapping to work correctly.

CBSE Class 12 Maths Revision Notes Chapter 1 Relations and Functions

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Access Maths Notes for Chapter 1 Relations and Functions Class 12

Relations and Functions Class 12 Notes: Core Definitions

Relation

Function

Domain, Co-domain and Range

Types of Relations in CBSE Class 12 Maths Revision Notes Chapter 1

Empty Relation and Universal Relation

Reflexive, Symmetric and Transitive Relations

Equivalence Relation and Equivalence Classes

Types of Functions in Class 12 Maths Chapter 1 Notes

One-One or Injective Function

Onto or Surjective Function

Bijective Function

Composition of Functions and Invertible Functions

Composition of Functions

Invertible Function

Key Properties

Binary Operations in Relations and Functions

Meaning of Binary Operation

Properties of Binary Operations

Quick Revision Tables for Relations and Functions

Relations Quick Revision Table

Functions Quick Revision Table

Binary Operations Quick Revision Table

Important Terms in Relations and Functions Class 12 Notes

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CBSE Class 12 Maths Revision Notes Chapter 1 Relations and Functions

Key Takeaways

Access Maths Notes for Chapter 1 Relations and Functions Class 12

Relations and Functions Class 12 Notes: Core Definitions

Relation

Function

Domain, Co-domain and Range

Types of Relations in CBSE Class 12 Maths Revision Notes Chapter 1

Empty Relation and Universal Relation

Reflexive, Symmetric and Transitive Relations

Equivalence Relation and Equivalence Classes

Types of Functions in Class 12 Maths Chapter 1 Notes

One-One or Injective Function

Onto or Surjective Function

Bijective Function

Composition of Functions and Invertible Functions

Composition of Functions

Invertible Function

Key Properties

Binary Operations in Relations and Functions

Meaning of Binary Operation

Properties of Binary Operations

Quick Revision Tables for Relations and Functions

Relations Quick Revision Table

Functions Quick Revision Table

Binary Operations Quick Revision Table

Important Terms in Relations and Functions Class 12 Notes

Useful Links for Class 12 Maths

Please register to view this section

FAQs (Frequently Asked Questions)

1. Why do students confuse relation and function in Chapter 1?

2. How can I quickly check whether a relation is an equivalence relation?

3. Why is range different from co-domain in functions?

4. Does gof always equal fog in composition of functions?

5. Why must an invertible function be bijective?

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