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CBSE Class 11 Maths Revision Notes Chapter 2 Relations and Functions

Relations and Functions explains how elements of two sets can be connected through ordered pairs.

For CBSE Class 11 Maths 2026–27, this chapter covers Cartesian product, relations, domain, range, codomain and functions.

Relations and Functions is built directly on Chapter 1 Sets. Once you understand sets, the next step is to understand how elements from two sets can be paired and connected.

Think of a simple classroom example. One set has students. Another set has their roll numbers. When each student is connected to a roll number, a relation is formed. If each student has exactly one roll number, the relation becomes a function.

Use these CBSE Class 11 Maths Revision Notes Chapter 2 to revise Cartesian product of sets, ordered pairs, relation, domain, range, codomain, function, image, preimage, standard functions and algebra of real functions.

Key Takeaways

Ordered pair: An ordered pair is written as (a, b), where order matters.
Cartesian product: A × B is the set of all ordered pairs formed from A and B.
Relation: A relation from A to B is a subset of A × B.
Function: A function is a relation in which every element of the domain has exactly one image.

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Access Class 11 Maths Relations and Functions Notes in 30 Minutes

These notes are arranged for 30-minute revision, so you can revise the chapter in the same order you study it in class.

Start with ordered pairs and Cartesian product. Then revise relations, domain, range and codomain. After that, understand when a relation becomes a function. Finally, revise standard functions and algebra of real functions.

The chapter is easier when you first understand ordered pairs, then relations, and then functions.

Topic	What You Revise
Ordered pairs	Meaning and equality of ordered pairs
Cartesian product	Formation of A × B
Relation	Subset of Cartesian product
Domain	First elements of ordered pairs
Range	Second elements of ordered pairs
Codomain	Target set of the relation or function
Function	Relation with exactly one image for each input
Image and preimage	Output and related input
Standard functions	Identity, constant, polynomial and other functions
Algebra of real functions	Addition, subtraction, multiplication and quotient

Why Do Students Study Relations and Functions in Class 11 Maths?

Relations and Functions help you understand how one quantity depends on another.

You already see this idea in real life:

Situation	Relation or Function Idea
Student and roll number	Each student is linked to a roll number
Person and date of birth	Each person has one date of birth
Product and price	Each product may have a price
Input and output in a machine	One input gives an output
x and y in a graph	x-values connect to y-values

This chapter becomes important for later topics such as inverse trigonometry, limits, continuity, differentiation, integration and coordinate geometry.

For students using Class 11 Maths Chapter 2 Notes for revision, the main aim should be to understand the difference between a relation and a function.

Ordered Pairs

An ordered pair is a pair of elements written in a fixed order.

It is written as:

(a, b)

Here, a is the first element and b is the second element.

Order matters in an ordered pair.

So,

(a, b) ≠ (b, a), unless a = b.

Example:

(2, 5) and (5, 2) are different ordered pairs.

Equality of Ordered Pairs

Two ordered pairs are equal only when their corresponding elements are equal.

If:

(a, b) = (c, d)

Then:

a = c and b = d

Example:

If (x + 1, y - 2) = (3, 1), then:

x + 1 = 3, so x = 2
y - 2 = 1, so y = 3

Cartesian Product of Sets

Once ordered pairs are understood, the next step is to form all possible ordered pairs from two sets.

If A and B are two non-empty sets, then the Cartesian product A × B is the set of all ordered pairs (a, b), where a ∈ A and b ∈ B.

A × B = {(a, b) : a ∈ A, b ∈ B}

Example:

A = {1, 2}
B = {3, 4}

Then:

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

Here, every element of A is paired with every element of B.

Number of Elements in Cartesian Product

If n(A) = p and n(B) = q, then:

n(A × B) = pq

Example:

If A has 3 elements and B has 2 elements, then:

n(A × B) = 3 × 2 = 6

Sets	Number of Elements
n(A) = 2, n(B) = 3	n(A × B) = 6
n(A) = 4, n(B) = 5	n(A × B) = 20
n(A) = p, n(B) = q	n(A × B) = pq

Important Points on Cartesian Product

Point	Explanation
Order matters	A × B is generally not equal to B × A
Ordered pairs are used	Elements are written as (a, b)
Null set rule	If A = φ or B = φ, then A × B = φ
Number of pairs	n(A × B) = n(A) × n(B)
Repeated pairing	Every element of A pairs with every element of B

Example:

A = {a, b}
B = {1, 2}

A × B = {(a, 1), (a, 2), (b, 1), (b, 2)}

B × A = {(1, a), (1, b), (2, a), (2, b)}

So, A × B ≠ B × A.

Cartesian Product of Three Sets

The Cartesian product of three sets contains ordered triplets.

A × B × C = {(a, b, c) : a ∈ A, b ∈ B, c ∈ C}

Example:

If A = {1, 2}, then:

A × A × A = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}

These are ordered triplets.

Relation in Class 11 Maths

After forming ordered pairs, we can select some pairs based on a rule. This selected set of ordered pairs is called a relation.

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

R ⊆ A × B

Example:

A = {1, 2, 3}
B = {2, 3, 4}

Let R = {(1, 2), (2, 3), (3, 4)}

Here, R is a relation from A to B because every ordered pair in R belongs to A × B.

Relation as a Real-Life Idea

Think of:

A = {students}
B = {subjects}

If a student studies a subject, we can write the pair as:

(student, subject)

The set of all such pairs forms a relation.

A relation connects elements of one set with elements of another set according to a rule.

This is why Relations and Functions Class 11 Notes usually begin with Cartesian product before moving to functions.

Domain, Range and Codomain of a Relation

Domain, range and codomain help describe a relation properly.

Term	Meaning
Domain	Set of all first elements of ordered pairs
Range	Set of all second elements of ordered pairs
Codomain	The complete second set B

If R is a relation from A to B, then:

Domain of R = {x : (x, y) ∈ R}

Range of R = {y : (x, y) ∈ R}

Codomain = B

Also:

Range ⊆ Codomain

Example of Domain, Range and Codomain

Let:

A = {1, 2, 3, 4, 5}
B = {2, 3, 4, 5, 6}

R = {(1, 2), (2, 3), (3, 4)}

Then:

Domain = {1, 2, 3}
Range = {2, 3, 4}
Codomain = {2, 3, 4, 5, 6}

Notice that codomain can contain elements that are not used in the range.

Representation of a Relation

A relation can be represented in different ways.

Method	Meaning
Roster form	Listing ordered pairs
Set-builder form	Writing relation using a rule
Arrow diagram	Showing mapping using arrows

Example:

R = {(1, 2), (2, 3), (3, 4)}

This is roster form.

Set-builder form:

R = {(x, y) : y = x + 1, x ∈ A, y ∈ B}

Arrow diagrams help students see which element of one set is connected to which element of another set.

Number of Relations from A to B

The total number of relations from A to B depends on the number of subsets of A × B.

If n(A) = p and n(B) = q, then:

n(A × B) = pq

Total number of relations from A to B = 2^(pq)

Example:

If A = {1, 2} and B = {3, 4}, then:

n(A) = 2 and n(B) = 2

n(A × B) = 2 × 2 = 4

Total number of relations = 2^4 = 16

Inverse Relation

The inverse of a relation is formed by reversing the ordered pairs.

If R is a relation from A to B, then inverse relation R^-1 is a relation from B to A.

If:

R = {(a, b) : a ∈ A, b ∈ B}

Then:

R^-1 = {(b, a) : (a, b) ∈ R}

Example:

R = {(1, 2), (3, 4)}

Then:

R^-1 = {(2, 1), (4, 3)}

Function in Class 11 Maths

A function is a special type of relation.

A relation f from set A to set B is called a function if every element of A has one and only one image in B.

It is written as:

f : A → B

This means f is a function from A to B.

Class 11 Maths Relations and Functions Notes often test this idea through ordered pairs, arrow diagrams and input-output examples.

Simple Way to Understand a Function

Think of a function as a machine.

You give one input. The machine gives one output.

For a relation to be a function:

Every input must be used.
Every input must have exactly one output.
One input cannot have two different outputs.

Function vs Relation

Every function is a relation, but every relation is not a function.

Basis	Relation	Function
Meaning	Any subset of A × B	Special relation from A to B
Input rule	An input may have no image or many images	Every input has exactly one image
Output	One input can be linked to more than one output	One input cannot have two outputs
Example	{(1, 2), (1, 3)}	{(1, 2), (2, 3)}

Example:

R = {(1, 2), (1, 3)}

This is not a function because input 1 has two images.

R = {(1, 2), (2, 3), (3, 4)}

This is a function because each input has exactly one image.

Domain, Codomain and Range of a Function

For a function f : A → B:

Term	Meaning
Domain	Set A
Codomain	Set B
Range	Set of actual images in B

If f : A → B and f(x) = y, then y is the image of x.

Also, x is the preimage of y.

Image and Preimage

If f(x) = y, then:

y is called the image of x.
x is called the preimage of y.

Example:

Let f(x) = 2x + 1.

For x = 3:

f(3) = 2(3) + 1 = 7

Here, 7 is the image of 3, and 3 is the preimage of 7.

How to Check Whether a Relation is a Function

A relation is a function if every first element has only one second element.

Relation	Function or Not	Reason
{(1, 2), (2, 3), (3, 4)}	Function	Each input has one image
{(1, 2), (1, 3), (2, 4)}	Not a function	Input 1 has two images
{(2, 5), (3, 5), (4, 5)}	Function	Different inputs may have same image
{(1, 3), (2, 4), (3, 5)}	Function	Each input has one image

Different inputs can have the same output. But one input cannot have two different outputs.

Real Valued Function and Real Function

A function whose range is R or a subset of R is called a real valued function.

If its domain is also R or a subset of R, it is called a real function.

Example:

f : R → R defined by f(x) = x^2 is a real function.

Here, domain is R and range is {x : x ≥ 0}.

Standard Real Functions

Class 11 Mathematics Revision Notes Chapter 2 should include the standard real functions because they appear in examples, graphs and domain-range questions.

Function	Formula	Domain	Range
Identity function	f(x) = x	R	R
Constant function	f(x) = c	R	{c}
Polynomial function	f(x) = a0 + a1x + ... + anx^n	R	Depends on function
Modulus function	f(x) =	x
Signum function	f(x) = sgn(x)	R	{-1, 0, 1}
Greatest integer function	f(x) = [x]	R	Z
Exponential function	f(x) = a^x	R	(0, ∞)
Logarithmic function	f(x) = loga x	(0, ∞)	R

Identity Function

The identity function is defined as:

f : R → R

f(x) = x for every x ∈ R

In this function, every input gives the same output.

Examples:

f(2) = 2
f(-5) = -5
f(0) = 0

The graph of the identity function is a straight line passing through the origin.

Constant Function

A constant function is defined as:

f : R → R

f(x) = c for every x ∈ R

Here, c is a constant.

Example:

f(x) = 3

For every value of x, the output is 3.

The graph of a constant function is a line parallel to the x-axis.

Polynomial Function

A function f : R → R is called a polynomial function if it is of the form:

f(x) = a0 + a1x + a2x^2 + ... + anx^n

Here, n is a non-negative integer and coefficients are real numbers.

Examples:

f(x) = x^2 + 3x + 2

g(x) = x^3 - x + 5

These are polynomial functions.

Modulus Function

The modulus function is defined as:

f(x) = |x|

It can be written as:

f(x) = x, if x ≥ 0
f(x) = -x, if x < 0

Examples:

|5| = 5
|-5| = 5
|0| = 0

The modulus function always gives a non-negative value.

Signum Function

The signum function is defined as:

f(x) = 1, if x > 0
f(x) = 0, if x = 0
f(x) = -1, if x < 0

It is written as:

f(x) = sgn(x)

This function tells whether a number is positive, negative or zero.

Greatest Integer Function

The greatest integer function gives the greatest integer less than or equal to x.

It is written as:

f(x) = [x]

Examples:

[2.4] = 2
[5] = 5
[-1.2] = -2

This function is also called the floor function.

Exponential Function

An exponential function is written as:

f(x) = a^x

where a > 0 and a ≠ 1.

Domain = R
Range = (0, ∞)

Examples:

f(x) = 2^x
g(x) = 3^x

Exponential functions are used in growth and decay problems.

Logarithmic Function

A logarithmic function is written as:

f(x) = loga x

where a > 0 and a ≠ 1.

Domain = (0, ∞)
Range = R

Examples:

f(x) = log2 x
g(x) = log10 x

Logarithmic functions are closely related to exponential functions.

Algebra of Real Functions

Algebra of real functions means performing operations on functions.

If f and g are real functions, then we can define:

Operation	Formula
Addition	(f + g)(x) = f(x) + g(x)
Subtraction	(f - g)(x) = f(x) - g(x)
Multiplication	(fg)(x) = f(x)g(x)
Division	(f/g)(x) = f(x)/g(x), where g(x) ≠ 0
Scalar multiplication	(αf)(x) = αf(x)

These operations are performed point-wise.

Addition of Two Real Functions

If f and g are real functions, then:

(f + g)(x) = f(x) + g(x)

Example:

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

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

= 2x + 1

Subtraction of Two Real Functions

If f and g are real functions, then:

(f - g)(x) = f(x) - g(x)

Example:

f(x) = x + 5
g(x) = 2x

(f - g)(x) = x + 5 - 2x

= 5 - x

Multiplication of Two Real Functions

If f and g are real functions, then:

(fg)(x) = f(x)g(x)

Example:

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

(fg)(x) = x(x + 1)

= x^2 + x

Quotient of Two Real Functions

If f and g are real functions, then:

(f/g)(x) = f(x)/g(x)

This is defined only when:

g(x) ≠ 0

Example:

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

(f/g)(x) = (x + 1)/(x - 2), where x ≠ 2

Even and Odd Functions

Some functions show symmetry.

Type	Condition	Graph Symmetry
Even function	f(-x) = f(x)	Symmetric about y-axis
Odd function	f(-x) = -f(x)	Symmetric about origin

Example of even function:

f(x) = x^2

Since f(-x) = (-x)^2 = x^2 = f(x), it is even.

Example of odd function:

f(x) = x^3

Since f(-x) = (-x)^3 = -x^3 = -f(x), it is odd.

Periodic Function

A function f(x) is periodic if there is a positive real number T such that:

f(x + T) = f(x)

The least positive value of T is called the fundamental period.

Examples:

sin x and cos x are periodic functions.

A periodic function repeats its values after a fixed interval.

Quick Highlights of Class 11 Maths Chapter 2 Notes

Topic	Quick Revision Point
Ordered pair	Pair written in fixed order
Cartesian product	Set of all ordered pairs
Relation	Subset of Cartesian product
Domain	Set of first elements
Range	Set of second elements actually used
Codomain	Complete target set
Function	Relation with exactly one image for every input
Image	Output of an element
Preimage	Input related to an output
Identity function	f(x) = x
Constant function	f(x) = c
Modulus function	f(x) =
Signum function	Gives -1, 0 or 1
Greatest integer function	Greatest integer ≤ x
Algebra of real functions	Operations on functions

Important Terms from CBSE Class 11 Maths Revision Notes Chapter 2

The terms below cover the main definitions students need while revising this chapter.

Term	Meaning
Ordered pair	Pair of elements written in a fixed order
Cartesian product	Set of all ordered pairs from two sets
Relation	Subset of Cartesian product
Domain	Set of all first elements of ordered pairs
Range	Set of all second elements used in relation or function
Codomain	Complete second set
Inverse relation	Relation formed by reversing ordered pairs
Function	Relation where each input has exactly one image
Image	Output of an element under a function
Preimage	Input corresponding to an image
Real function	Function with domain and range as R or subsets of R
Identity function	Function defined by f(x) = x
Constant function	Function defined by f(x) = c
Polynomial function	Function expressed using powers of x
Modulus function	Function defined by f(x) =
Signum function	Function giving sign of x
Greatest integer function	Function giving greatest integer less than or equal to x
Algebra of real functions	Addition, subtraction, multiplication and division of functions

Useful Links for Class 11 Maths

Section	Useful Links
Syllabus	CBSE Class 11 Maths Syllabus
Revision Notes	CBSE Class 11 Maths Revision Notes
Maths Notes	CBSE Class 11 Maths Revision Notes Chapter 1
Maths Notes	CBSE Class 11 Maths Revision Notes Chapter 2
NCERT Solutions	NCERT Solutions Class 11 Maths
Sample Papers	CBSE Sample Papers for Class 11 Maths
Important Questions	Important Questions Class 11 Maths
NCERT Books	NCERT Books for Class 11 Maths

Q.1 If P = {2, 3}, form the set P × P × P.

Ans

P × P × P = {(2, 2, 2), (2, 2, 3), (2, 3, 2), (2, 3, 3), (3, 2, 2), (3, 2, 3), (3, 3, 2), (3, 3, 3)}

Q.2 In given figure write the relation in

(i) Set–builder from (ii) Roster form
Also write the domain and the range.

Ans

(i) Set builder form R = {(x, y): y = 2x + 1 ∀ x ∈ A}
(ii) Roster form {(1, 3), (2, 5),(3, 7)}
Domain = {1, 2, 3} and Range = {3, 5, 7}

Q.3 a) If (x – 2, y + 3) = (4, 1), find the values of x and y.
b) If set A = {2, 3} and B = {3, 4}, find the number of elements in A × B.

Ans

a) Since ordered pairs are equal, so
x – 2 = 4 and y + 3 = 1
x = 4 + 2 = 6 and y = 1 – 3 = –2

b) n(A × B) = n(A) × n(B) = 2 × 2 = 4 elements

Q.4 Find the range:

a) f = {(x, x²): x ∈ R)} be a function from R into R.
b) A = {9, 10, 11, 13, 17, 18} and f : A → N defined by f(x) = highest prime factor of x.

Ans

a) Range : set of all positive real numbers.
b) Factors of 9 = 3 × 3 ⇒ highest prime factor = 3
Factors of 10 = 2 × 5 ⇒ highest prime factor = 5
Factors of 11 = 1 × 11 ⇒ highest prime factor = 11
Factors of 13 = 1 × 13 ⇒ highest prime factor = 13
Factors of 18 = 2 × 3 × 3 ⇒ highest prime factor = 3
∴ range = {3, 5, 11, 13}

Q.5 Let R be a relation from N to N defined by R = {(x, y): x, y ∈ N and y = x²}.
Are the following statements true? Give reason also.

(i) (x, x) ∈ R ∀ x ∈ N
(ii) (a, b) ∈ R ⇒ (b, a) ∈ R, ∀ a, b ∈ N
(iii) (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ N
(iv) Is the relation a function?

Ans

i) No, as only (1, 1) ∈ R, (2, 2), (3, 3), …, ∉ R.
ii) No, (2, 4) ∈ R but (4, 2) ∉ R.
iii) No, as (2, 4) ∈ R and (4, 16) ∈ R but (2, 16) ∉ R.
iv) Yes, as all elements in the domain have a unique image in the co-domain.

Q.6 Find the domain of the function

$f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - 7 x + 12} .$

Ans

Since, x² – 7x +12 = (x – 4)(x – 3), the function is defined for all real numbers
except x = 4 and x = 3. Hence, the domain of f is R – {3, 4}.

Q.7 A function f is defined by f(x) = x² – 5. Find the value of

(i) f(0).
(ii) f(–3).

$(iii) f (\frac{2}{3}) .$

Ans

(i) f(x) = x² – 5 ⇒ f(0) = 0² – 5 = –5
(ii) f(–3) = (–3)² – 5 = 9 – 5 = 4

$(iii) f (\frac{2}{3}) = {(\frac{2}{3})}^{2} - 5 = \frac{4}{9} - 5 = \frac{4 - 45}{9} = \frac{- 41}{9}$

Q.8 Let A = {2, 3, 4}, B = {4, 5}, C = {5, 6, 7}. Find (A × B) ∩ (A × C).

Ans

A × B = {(2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5)}
A × C = {(2, 5), (2, 6), (2, 7), (3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7)}
(A × B) ∩ (A × C) = {(2, 5), (3, 5), (4, 5)}

Q.9 Let A = {1, 2, 6, 8} and let R be a relation on A defined by
{(a, b): a, b ∈ A, b is exactly divisible by a}

a) Write R in roster form.
b) Find the domain of R.
c) Find the range of R.

Ans

a) R = {(1, 1), (1, 2), (1, 6), (1, 8), (2, 2), (2, 6), (2, 8), (6, 6), (8, 8)}
b) Domain of R = {1, 2, 6, 8}
c) Range of R = {1, 2, 6, 8}

Q.10 Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

a) Is f a relation from A to B? Why?
b) Is f a function from A to B? Why?

Ans

a) Yes, f is a relation because every element of f is a subset of A × B.
b) No, because the element 2 of A has two image in B, i.e., 9 and 11.
The mapping is not unique.

Q.11 Let f, g: R → R defined by f(x) = x – 1, g(x) = 2x – 5. Find

a) f – g.
b) f/g.

Ans

a) (f – g)x = f(x) – g(x)
= x – 1 – (2x – 5)
= x – 1 –2x + 5
= 4 – x

b) (f/g)x = f(x)/g(x) = (x – 1)/(2x – 5), where x ≠ 5/2.

Q.12

$If f (x) = x^{2}, find \frac{f (2.1) - f (1)}{2.1 - 1}$

Ans

$f (x) = x^{2} ∴ f (2, 1) = {(2.1)}^{2} = 4.41 f (1) = 1^{2} = 1 On putting these values, we get \frac{f (2.1) - f (1)}{2.1 - 1} = \frac{4.41 - 1}{1.1} = \frac{341}{110}$

Q.13 Determine the domain and range of the relation R defined by R = {(x, x²): x ∈ (1, 2, 3, 4)}.

Ans

R = {(1, 1), (2, 4), (3, 9), (4, 16)}
Since domain is the set of first components in R and range is the set of second components in R .
Therefore,
Domain = {1, 2, 3, 4}
Range = {1, 4, 9, 16}

Q.14 If A = {–1, 0}, find A × A × A.

Ans

A × A × A = {(–1, –1, –1), (–1, –1, 0), (–1, 0, –1), (–1, 0, 0), (0, –1, –1), (0, –1, 0), (0, 0, –1), (0, 0, 0)}

Q.15 Let A = {1, 2, 3} and B = {1, 4}. Find A × B and B × A.

Ans

A × B = (1, 1), (1, 4), (2, 1), (2, 4), (3, 1), (3, 4)
B × A = (1, 1), (1, 2), (1, 3), (4, 1), (4, 2), (4, 3)

Q.16 Under which condition a relation from A to B is said to be a function?

Ans

If every element of A has one and only one Image in B, then f is a function from A to B.

Q.17

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

Ans

$We know that all the values of x for which function is defined is called domain. ∴ For function to be defined, x^{3} - 6 x^{2} + 11 x - 6 \geq 0 \dots (1) Now, for getting its factor put x = - 1, 1, 0 \dots we find that x = 1, satisfies (1) {Hence}_{r} (x - 1) is one of its factors Now, on dividing (1) by (x - 1), we get x^{2} - 5 x + 6, which is a quadratic poynomial On splitting the middle term, we get (x - 2) (x - 3) Finally, we get x^{3} - 6 x^{2} + 11 x - 6 = (x - 1) (x - 2) (x - 3) eqn (1) becomes (x - 1) (x - 2) (x - 3) \geq 0 Using number line, we get$

$Expression is +ve in the interval [1, 2] and [3, + \infty), Hence, the domain of the given function is [1, 2] \cup [3, + \infty) .$

Q.18 Define a relation.

Ans

Let A and B be two sets. Then a relation from A to B is a subset of A × B, i.e., R ⊆ A × B.

Q.19 If R is a relation defined by R = {(x, 2x): x ∈ (1, 2, 3, 4)}, find the range and domain of R.

Ans

Range of R = (2, 4, 6, 8)
and Domain of R = (1, 2, 3, 4)

Q.20 If A = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Define a relation on a set A by R = {(x, y): 3x – y = 0, where x, y ∈ A}. Show this relationship using arrow diagram.

Ans

R = {(1, 3), (2, 6), (3, 9)}

Q.21 If n(A) = 3 and n(B) = 3, then find n(A × B).

Ans

n(A × B) = n(A) × n(B) = 9

Q.22

$If a relation f is defined as f (x) = 2 x + 3, then find the value of f (f (\frac{3}{2}))$

Ans

$f (x) = 2 x + 3 f (\frac{3}{2}) = 2 \times \frac{3}{2} + 3 = 6 f (f (\frac{3}{2})) = f (6) = 2 \times 6 + 3 = 15$

Q.23 Let f, g: R → R defined by f(x) = 6x and g(x) = 3x. Find f – g.

Ans

(f – g)(x) = f(x) – g(x) = 6x – 3x = 3x

Q.24 Let A = {a, b, c, d}, B = {e, f, g, h, i, j} and f = {(a, e), (b, f), (c, g), (d, h), (a, i)}. Check whether f is a function from A to B or not.

Ans

No. f is not a function from A to B since element ‘a’ of set A has more than one image in set B.

Q.25

$Let f and g be real function defined by f (x) = \sqrt{x + 2} and g (x) = \sqrt{4 - x^{2}} . Then find \frac{f}{g} and ff .$

Ans

$a) \frac{f}{g} = \frac{f (x)}{g (x)} = \frac{\sqrt{x + 2}}{\sqrt{4 - x^{2}}} = \frac{1}{\sqrt{2 - x}} b) ff = f (x) f (x) = [f (x)]^{2} = {(\sqrt{x + 2})}^{2} = x + 2$

Q.26 If P = {a, b} and Q = {2, 3}, find P × Q.

Ans

P × Q = {(a, 2), (a, 3), (b, 2), (b, 3)}

Q.27 Given that f: X → Y, the image of x is f(x). Write the preimage of y.

Ans

Preimage of y is f^–1(y) = {x: f(x) = y}, or all x whose image is y.

Q.28 Define the signum function.

Ans

$The function f : R \to R defined by f (x) = \{\begin{cases} 1 & if x > 0 \\ 0 & if x = 0 \\ - 1 & if x < 0 \end{cases} is called signum function .$

Q.29

$Find the domain of f (x) = \sqrt{4 - x^{2}} .$

Ans

f(x) is defined for 4 – x²≥ 0
or x²– 4 ≤ 0
or (x – 2)(x + 2) ≤ 0
Therefore, Domain of function = {x: – 2 ≤ x ≤ 2}

Q.30 The cartesian product of A × A has 9 elements among which (–1, 0) and (0, 1) are found. Find the set A and the remaining elements of A × A.

Ans

A = {–1, 0, 1}
The remaining elements of A × A are (–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), (1, 1).

Q.31 Find the range of function, f(x) = |x|.

Ans

Since function, f(x) = |x| is defined for all the real values of x ∈ R and value of y = f(x) = |x| is always a positive real number ∀ x ∈ R.
Hence, its range is the set of all positive real values, i.e.,
[0, ∞).

Q.32

$Find the domain of the function f (x) = \frac{x}{x - 1}$

Ans

$We know that all values of x for which function is defined is called domain. ∵ f (x) = \frac{x}{x - 1} For function to be defined, denominator should be non-zero. Here, function is not defined for x = 1 . Hence, its domain is all the real values of x except 1, i.e., R - {1} .$

Q.33

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

Ans

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

Q.34 Find the domain of function, f(x) = |X|.

Ans

Since function, f(x) = |X| is defined for all values of x ∈ R.
Hence, its domain is the set of all real values of x, i.e., R or (– ∞, ∞).

Q.35 Let f = {(1,1), (2,3), (0, –1), (–1, –3)} be a linear function from Z into Z. Find f(x).

Ans

Since f is a linear function,
Therefore, let us suppose f(x) = mx + c… (1)
Also, since (1, 1), (0, – 1) ∈ R,
⇒ f (1) = m + c = 1 …(2)
and f (0) = c = –1.
On putting the value of c in (2), we get
m – 1 = 1
⇒ m = 2
On putting the value of c and m in (1), we get
f(x) = 2x – 1.

Q.36 If (x + y, x – y) = (6, 2), find the values of x and y.

Ans

$\begin{array}{l} Since the ordered pairs are equal, therefore, \\ the corresponding elements are also equal. \\ x + y = 6 \dots \dots . (1) \\ x - y = 2 \dots \dots \dots (2) \\ On adding equations (1) and (2), we get \\ x + y = 6 \\ \underline{x - y = 2} \\ 2 x = 8 \\ X = 4 \\ On putting the value of x in (1), we get \\ 4 + y = 6 \\ or y = 2 \\ Hence, x = 4, y = 2 \end{array}$

Q.37

$Find the domain of function f (x) = \frac{x}{\sqrt{49 - x^{2}}} .$

Ans

$We know that all the values of x for which function is defined is called domain. f (x) = \frac{x}{\sqrt{49 - x^{2}}} For function to be defined, 49 - x^{2} > 0 \Rightarrow x^{2} - 49 < 0 \Rightarrow x^{2} - 7^{2} < 0 \Rightarrow (x - 7) (x + 7) < 0 \Rightarrow x > - 7 and x < 7 \Rightarrow - 7 < x < 7 \Rightarrow x \in (- 7, 7) Hence, the domain of the given function is (- 7, 7) .$

Q.38

$Find the domain of the function f (x) = \frac{{3x}^{2} +7}{\sqrt{x^{2} - 5x+6}} .$

Ans

$We know that all the values of x for which function is defined is called domain. ∵ f (x) = \frac{3 x^{2} + 7}{\sqrt{x^{2} - 5 x + 6}} For function to be defined, x^{2} - 5 x + 6 > 0 \Rightarrow (x - 2) (x - 3) > 0 \Rightarrow (x - 2) > 0 and (x - 3) > 0 \Rightarrow x > 2 and x > 3 \Rightarrow x > 3 \dots (1) \Rightarrow Also, (x - 2) < 0 and (x - 3) < 0 \Rightarrow x < 2 and x < 3 \Rightarrow x < 2 … (2) From (1) and, (2) get - \infty < x < 2 and 3 < x < \infty Hence, the domain of the given function is (- \infty, 2) \cup (3, \infty)$

Q.39

$Find the range of the function f (x) = \frac{1 - x^{2}}{1 + x^{2}}$

Ans

$Since, the value of y for which domain (x) is defined is called range, Let y = \frac{1 - x^{2}}{1 + x^{2}} \Rightarrow y + {yx}^{2} = 1 - x^{2} \Rightarrow x^{2} + x^{2} y = 1 - y \Rightarrow (1 + y) x^{2} = 1 - y \Rightarrow x^{2} = \frac{1 - y}{1 + y} \Rightarrow x = \sqrt{\frac{1 - y}{1 + y}} For domain (x) to be defined, \frac{1 - y}{1 + y} \geq 0 On using number line, we get$

$- 1 < y \leq 1 [\begin{matrix} ∵ denominator \neq 0 \\ \Rightarrow (1 + y) \neq 0 \\ \Rightarrow y \neq - 1 \end{matrix}] \Rightarrow y \in (- 1, 1] Hence, range is (- 1, 1] .$

Q.40

$Find the range of the function f (x) = \frac{x}{1 - x} .$

Ans

$Since, the value of y for which domain (x) is defined is called range, Let y = \frac{X}{1 - x} \Rightarrow x = y - yx \Rightarrow x + xy = y \Rightarrow (1 + y) x = y \Rightarrow X = \frac{y}{1 + y} For domain (x) to be defined, denominator \neq 0 \Rightarrow (1 + y) \neq 0 \Rightarrow y \neq - 1 Hence, the range is set of all real numbers R except - 1_{1} i.e., R - {- 1} .$

Q.41

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

Ans

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

Q.42 If Set A has m elements and Set B has n elements, find the number of relations defined from Set A to Set B.

Ans

Since the number of elements in set A = m
Number of elements in set B = n
∴ Number of elements in set (A × B) = mn
The number of all possible subsets of Set (A × B) = 2^mn
⇒ The number of relations defined from Set A to B = 2^mn

Q.43 If Set A has m elements, find the number of all possible subsets of Set (A × A).

Ans

$Since, the number of elements in set A = m \Rightarrow Number of elements in set (A \times A) = m \cdot m = m^{2} The number of all possible subsets of set (A \times A) = 2^{m^{2}}$

Q.44 If Set A has m elements and Set B has n elements, find the elements in Set (AXB).

Ans

Since the number of elements in set A = m
Number of elements in set B = n
∴ Number of elements in set (AXB) = mn

Q.45

$\begin{array}{l} Express the following functions as a sets of ordered \\ pairs and determine their ranges \\ (a) f : A \to R, f (x) {= x}^{2} +1, where A = \{-1,0,2,4\} \\ (b) g : A \to N, g (x) = 2x, where A = \{x:x \in N,x \leq 10\} \end{array}$

Ans

$(a) We have, f (- 1) = (- 1)^{2} + 1 = 2, f (0) = 0^{2} + 1, f (2) = 2^{2} + 1 = and f (4) = 4^{2} + 1 = 17 ∴ f = {x, f (x)) : x \in A} = {(- 1, 2), (0, 1), (2, 5), (4, 17)} Hence, Range of (f) = {2, 1, 5, 17} (b) We have, A = {1, 2, 3, \dots, 10} . Therefore, g (1) = 2 \times 1 = 2, g (2) = 2 \times 2 = 4, g (3) = 2 \times 3 = 6, g (4) = 2 \times 4 = 8, g (5) = 2 \times 5 = 10, g (6) = 2 \times 6 = 12, g (7) = 2 \times 7 = 14, g (8) = 2 \times 8 = 16, g (9) = 2 \times 9 = 18, g (10) = 2 \times 10 = 20 ∴ g = {x, g (x)) : x \in A} = {(1, 2), (2, 4), (3, 6), \dots, (10, 20)} Range of g = g (A) = {g (x) : x \in A} = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}$

Q.46 Let f and g be real function defined by f(x) = 4x + 1 and g(x) = 6x – 7.

(a) For what real numbers x, f(x) = g(x)?
(b) For what real numbers x, f(x) < g(x)

Ans

$We have f (x) = 4 x + 1 and g (x) = 6 x - 7 \sqrt{b^{2} - 4 ac} Now, (a) For f (x) = g (x), we must have, 4 x + 1 = 6 x - 7 Solving above equation for x, we get, 4 x + 1 = 6 x - 7 \Rightarrow 6 x - 4 x = 7 + 1 \Rightarrow 2 x = 8 \Rightarrow x = \frac{8}{2} = 4 ∴ x = 4 (b) Now, For f (x) < g (x), we must have, 4 x + 1 < 6 x - 7 Solving above inequality for x, we get 4 x + 1 < 6 x - 7 \Rightarrow 2 x > 8 \Rightarrow x > 4 Hence, for x = 4, f (x) = g (x) and, for x > 4, f (x) < g (x)$

Q.47

$\begin{array}{l} If X = \{1,3,5\}, Y = \{a,b\}, then represent the following products by arrow diagrams: \\ (i) X×Y \\ (ii) Y×X \\ (iii) X×X \\ (iv) Y×Y \end{array}$

Ans

$(i) We have, X \times Y = {1, 3, 5} \times {a, b} = {(1, a), (1, b), (3, a), (3, b), (5, a), (5, b)} Following arrow diagram represent X \times Y . (ii) We have, Y \times X = {a, b} \times {1, 3, 5} = {(a, 1), (a, 3), (a, 5), (b, 1), (b, 3), (b, 5)} Following arrow diagram represent Y \times X . (iii) We have, X \times X = {1, 3, 5} \times {1, 3, 5} = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)} Following is the arrow diagram represent X \times X . (iii) We have, Y \times Y = {a, b} \times {a, b} = {(a, a), (a, b), (b, a), (b, b)} Following is the arrow diagram represent Y \times Y .$

Q.48

$A relation R is defined on the set Z of integers as follows: lim_{x \to \infty} (x, y) \in R \Leftrightarrow x^{2} + y^{2} = 100 {Express R and R}^{- 1} as the sets of ordered pairs and hence find their respective domains.$

Ans

$We have, (x, y) \in R \Leftrightarrow x^{2} + y^{2} = 100 \Rightarrow y = \pm \sqrt{100 - x^{2}} We observe that x = 0 \Rightarrow y = \pm 10 ∴ (0, 10) \in R and (0, - 10) \in R x = \pm 6 \Rightarrow y = \sqrt{100 - 36} = \pm 8 ∴ (6, 8) \in R, (- 6, 8) \in R, (6, - 8) \in R and (- 6, - 8) \in R x = \pm 8 \Rightarrow y = \sqrt{100 - 64} = \pm 6 ∴ (8, 6) \in R, (- 8, 6) \in R, (8, - 6) \in R and (- 8, - 6) \in R x = \pm 10 \Rightarrow y = \sqrt{100 - 100} = 0 (10, 0) \in R and (- 10, 0) \in R We also notice that for any other integral value of x, y is not an integer ∴ R = {(0, 10), (0, - 10), (6, 8), (- 6, 8), (6, - 8), (- 6, - 8), (8, 6) (- 8, 6), (8, - 6), (- 8, - 6), (10, 0), (- 10, 0)} R^{- 1} = {(10, 0), (- 10, 0), (8, 6), (8, - 6), (- 8, 6), (- 8, - 6), (6, 8) (6, - 8), (- 6, 8), (- 6, - 8), (0, 10), (0, - 10)} Clearly domain (R) = {0, 6, - 6, 8, - 8, 10, - 10} = domain (R^{- 1})$

Q.49

$\begin{array}{l} If A = {x : x < 6, x \in N}, B = \{x : x \leq 4, x \in W\} . Find (A \cup B) \times (A \cap B) \\ where W is the set of whole numbers, and N is the set of natural numbers. \end{array}$

Ans

$It is given that A = {x : x < 4, x \in N} where N is the set of natural numbers. Thus, we have A = {1, 2, 3} Similarly, we are given that B = {x : x \leq 3, x \in W} where W is the set of whole numbers. Thus, we have, B = {0, 1, 2, 3} Now, A \cup B = B = {0, 1, 2, 3}, and A \cap B = A = {1, 2, 3} For finding (A \cup B) \times (A \cap B) we will make following table :$

(A ∪ B) ↓ (A ∩ → B)	1	2	3
0	(0,1)	(0,2)	(0,3)
1	(1,1)	(1,2)	(1,3)
2	(2,1)	(2,2)	(2,3)
3	(3,1)	(3,2)	(3,3)

$Thus, (A \cup B) \times (A \cap B) = {(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3) (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)},$

Q.50

$If 4f (x) + 6f (\frac{1}{x}) = \frac{1}{x} - 5, then find f (x) .$

Ans

$We have been given, 4 f (x) + 6 f (\frac{1}{x}) = \frac{1}{x} - 5 . . . (1) Replace x by \frac{1}{x} in the above expression 4 f (\frac{1}{x}) + 6 f (x) = x - 5 . . . (2) On adding equationsa (1) and (2) 10 f (x) + 10 f (\frac{1}{x}) = x + \frac{1}{x} - 10 f (x) + f (\frac{1}{x}) = (\frac{1}{10}) (x + \frac{1}{x}) - 1 . . . (3) Now, subtracting (1) from (2) 2 f (x) - 2 f (\frac{1}{x}) = x - \frac{1}{x} . . . (4) Now, multiply equation (3) by (2) 2 f (x) + 2 f (\frac{1}{x}) = (\frac{1}{5}) (x + \frac{1}{x}) - 2 . . . (5) Adding equationa (4) nd equation (5) 4 f (x) = \frac{x}{5} + \frac{1}{5 x} + x - \frac{1}{x} - 2 = \frac{x}{5} + x + \frac{1}{5 x} - \frac{1}{x} - 2 = \frac{6 x}{5} - \frac{4}{5 x} - 2 4 f (x) = \frac{6 x}{5} - \frac{4}{5 x} - 2 \Rightarrow f (x) = \frac{3 x}{10} - \frac{1}{5 x} - \frac{1}{2}$

Q.51 There are 4 elements in a set A and 5 elements in the set B, then there will be ___ elements in A × B.

Ans

It is given that n(A) = 4
and n(B) = 5
Therefore, n(A × B) = 4 × 5
= 20
There are 4 elements in a set A and 5 elements in the set B, then there will be 20 elements in A × B.

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FAQs (Frequently Asked Questions)

A relation from set A to set B is a subset of A × B. It connects the first element and second element of ordered pairs using a rule.

A function is a relation in which every element of the domain has one and only one image in the codomain. One input cannot have two different outputs.

A relation can connect an input to one, many or no outputs. A function must connect every input to exactly one output. So, every function is a relation, but every relation is not a function.

The domain is the set of all input values. The range is the set of all actual output values obtained from the function. In f : A → B, A is the domain and range is a subset of B.

The Cartesian product A × B is the set of all ordered pairs (a, b), where a ∈ A and b ∈ B. If n(A) = p and n(B) = q, then n(A × B) = pq.

CBSE Class 11 Maths Revision Notes Chapter 2 Relations and Functions

Key Takeaways

Access Class 11 Maths Relations and Functions Notes in 30 Minutes

Why Do Students Study Relations and Functions in Class 11 Maths?

Ordered Pairs

Equality of Ordered Pairs

Cartesian Product of Sets

Number of Elements in Cartesian Product

Important Points on Cartesian Product

Cartesian Product of Three Sets

Relation in Class 11 Maths

Relation as a Real-Life Idea

Domain, Range and Codomain of a Relation

Example of Domain, Range and Codomain

Representation of a Relation

Number of Relations from A to B

Inverse Relation

Function in Class 11 Maths

Simple Way to Understand a Function

Function vs Relation

Domain, Codomain and Range of a Function

Image and Preimage

How to Check Whether a Relation is a Function

Real Valued Function and Real Function

Standard Real Functions

Identity Function

Constant Function

Polynomial Function

Modulus Function

Signum Function

Greatest Integer Function

Exponential Function

Logarithmic Function

Algebra of Real Functions

Addition of Two Real Functions

Subtraction of Two Real Functions

Multiplication of Two Real Functions

Quotient of Two Real Functions

Even and Odd Functions

Periodic Function

Quick Highlights of Class 11 Maths Chapter 2 Notes

Important Terms from CBSE Class 11 Maths Revision Notes Chapter 2

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1. What is relation in Class 11 Maths?

2. What is function in Class 11 Maths?

3. What is the difference between relation and function?

4. What is domain and range of a function?

5. What is Cartesian product of two sets?

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