Relations & Functions Exercise 1.3 NCERT Solutions Class 12 Math Chapter 1 free PDF Download

Exercise 1.3 delves into the advanced concepts of function composition and invertible functions, building upon the foundational knowledge of relations and functions established earlier in the chapter. This exercise is crucial for developing a deeper understanding of how functions interact and transform, paving the way for more complex mathematical analysis. Students will explore the process of combining functions through composition, understand the conditions under which functions can be inverted, and learn to find and work with inverse functions. These skills are fundamental in various branches of mathematics, including calculus, abstract algebra, and complex analysis, and have wide-ranging applications in fields such as physics, engineering, and computer science. Mastering the concepts in this exercise will equip students with powerful tools for solving sophisticated problems and understanding the intricate relationships between mathematical entities.

Class 12 NCERT Solutions – Mathematics Chapter 1 Relations and Functions – Exercise 1.3

Table of Contents

Question 1: Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(i) On Z+, define ∗ by a ∗ b = a – b

Solution: 

If a, b belongs to Z+

a * b = a – b which may not belong to Z+

For eg:  1 – 3 = -2 which doesn’t belongs to Z+ 

Therefore, * is not a Binary Operation on Z+

(ii) On Z+, define * by a * b = ab

Solution: 

If a, b belongs to Z+ 

a * b = ab which belongs to Z+

Therefore, * is Binary Operation on Z+

(iii) On R, define * by a * b = ab²

Solution:

If a, b belongs to R

a * b = ab which belongs to R

Therefore, * is Binary Operation on R

(iv) On Z+, define * by a * b = |a – b|

Solution:

If a, b belongs to Z+

a * b = |a – b| which belongs to Z+

 Therefore, * is Binary Operation on Z+

(v) On Z+, define * by a * b = a

Solution:

If a, b belongs to Z+

a * b = a which belongs to Z+

Therefore, * is Binary Operation on Z+

Question 2: For each binary operation * defined below, determine whether * is binary, commutative or associative.

(i) On Z, define a * b = a – b 

Solution:

a) Binary: 

If a, b belongs to Z

a * b = a – b which belongs to Z

Therefore, * is Binary Operation on Z

b) Commutative: 

If a, b belongs to Z, a * b = b * a 

LHS = a * b = a – b

RHS = b * a = b – a

Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a – b + c

RHS = (a – b) * c = a – b- c

Since, LHS is not equal to RHS

Therefore, * is not Associative

(ii) On Q, define a * b = ab + 1

Solution:

a) Binary:

If a, b belongs to Q, a * b = ab + 1 which belongs to Q

Therefore, * is Binary Operation on Q

b) Commutative: 

If a, b belongs to Q, a * b = b * a 

LHS = a * b = ab + 1

RHS = b * a = ba + 1 = ab + 1

Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:

If a, b, c belongs to Q, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (bc + 1) = abc + a + 1

RHS = (a * b) * c = abc + c + 1

Since, LHS is not equal to RHS

Therefore, * is not Associative

(iii) On Q, define a ∗ b = ab/2

Solution :

a) Binary:

If a, b belongs to Q, a * b = ab/2 which belongs to Q

Therefore, * is Binary Operation on Q

b) Commutative:

If a, b belongs to Q, a * b = b * a

LHS = a * b = ab/2

RHS = b * a = ba/2

Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:

If a, b, c belongs to Q, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (bc/2) = (abc)/2

RHS = (a * b) * c = (ab/2) * c = (abc)/2

Since, LHS is equal to RHS

Therefore, * is Associative

(iv) On Z+, define a * b = 2ab

Solution:

a) Binary:

If a, b belongs to Z+, a * b = 2ab which belongs to Z+

Therefore, * is Binary Operation on Z+

b) Commutative:

If a, b belongs to Z+, a * b = b * a

LHS = a * b = 2ab

RHS = b * a = 2ba = 2ab

Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:

If a, b, c belongs to Z+, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * 2bc = 2a * 2^(bc)

RHS = (a * b) * c = 2ab * c = 22abc

Since, LHS is not equal to RHS

Therefore, * is not Associative

(v) On Z+, define a * b = ab

Solution:

a) Binary:

If a, b belongs to Z+, a * b = ab which belongs to Z+

Therefore, * is Binary Operation on Z+

b) Commutative:

If a, b belongs to Z+, a * b = b * a

LHS = a * b = ab

RHS = b * a = ba

Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:

If a, b, c belongs to Z+, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * bc = ab^c

RHS = (a * b) * c = ab * c = abc

Since, LHS is not equal to RHS

Therefore, * is not Associative

(vi) On R – {– 1}, define a ∗ b = a / (b + 1)

Solution: 

a) Binary:

If a, b belongs to R, a * b = a / (b+1) which belongs to R

Therefore, * is Binary Operation on R

b) Commutative:

If a, b belongs to R, a * b = b * a

LHS = a * b = a / (b + 1)

RHS = b * a = b / (a + 1)

Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:

If a, b, c belongs to A, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * b / (c+1) = a(c+1) / b+c+1

RHS = (a * b) * c = (a / (b+1)) * c = a / (b+1)(c+1)

Since, LHS is not equal to RHS

Therefore, * is not Associative

Question 3. 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 ∧. 

Solution: 

^12345
111111
212222
312333
412344
512345

Question 4: Consider a binary operation ∗ on the set {1, 2, 3, 4, 5} given by the following multiplication table.

(Hint: use the following table) 

*12345
111111
212121
311311
412141
511115

(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)

Solution:

Here, (2 * 3) * 4 = 1 * 4 = 1

2 * (3 * 4) = 2 * 1 = 1

(ii) Is ∗ commutative?

Solution:

The given composition table is symmetrical about the main diagonal of table. Thus, binary operation ‘*’ is commutative.

(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).

Solution:

(2 * 3) * (4 * 5) = 1 * 1 = 1

Question 5: Let ∗′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a ∗′ b = H.C.F. of a and b. Is the operation ∗′ same as the operation ∗ defined in Exercise 4 above? Justify your answer.

Solution:

Let A = {1, 2, 3, 4, 5} and a ∗′ b = HCF of a and b.

*’12345
111111
212121
311311
412141
511115

We see that the operation *’ is the same as the operation * in Exercise 4 above.

Question 6: Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find

(i) 5 ∗ 7, 20 ∗ 16

Solution:

If a, b belongs to N

a * b = LCM of a and b

5 * 7 = 35

20 * 16 = 80

(ii) Is ∗ commutative?

Solution:

If a, b belongs to N

LCM of a * b = ab

LCM of b * a = ba = ab

a*b = b*a

Thus, * binary operation is commutative.

(iii) Is ∗ associative?

Solution:

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

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

Since, a * (b * c) = (a * b) * c

Thus, * binary operation is associative.

(iv) Find the identity of ∗ in N

Solution:

Let ‘e’ is an identity 

a * e = e * a, for a belonging to N

LCM of a * e = a, for a belonging to N

LCM of e * a = a, for a belonging to N

e divides a 

e divides 1

Thus, e = 1

Hence, 1 is an identity element

(v) Which elements of N are invertible for the operation ∗? 

Solution:

a * b = b * a = identity element

LCM of a and b = 1

a = b = 1

only ‘1’ is invertible element in N. 

Question 7: Is ∗ defined on the set {1, 2, 3, 4, 5} by a ∗ b = L.C.M. of a and b a binary operation? Justify your answer. 

Solution:

The operation * on the set {1, 2, 3, 4, 5} is defined as

a * b = L.C.M. of a and b

Let a=3, b=5

3 * 5 = 5 * 3 = L.C.M. of 3 and 5 = 15 which does not belong to the given set

Thus, * is not a Binary Operation.

Question 8: Let ∗ be the binary operation on N defined by a ∗ b = H.C.F. of a and b. Is ∗ commutative? Is ∗ associative? Does there exist identity for this binary operation on N?

Solution:

If a, b belongs to N

LHS = a * b = HCF of a and b

RHS = b * a = HCF of b and a

Since LHS = RHS

Therefore, * is Commutative

Now, If a, b, c belongs to Z, a * (b * c) = (a * b) * c

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

RHS = (a – b) * c = HCF of a, b and c

Since, LHS = RHS

Therefore, * is Associative

Now, 1 * a = a * 1 ≠ a

Thus, there doesn’t exist any identity element.

Question 9: Let ∗ be a binary operation on the set Q of rational numbers as follows:

(i) a ∗ b = a – b 

(ii) a ∗ b = a2 + b2

(iii) a ∗ b = a + ab 

(iv) a ∗ b = (a – b)2

(v) a ∗ b = ab / 4

(vi) a ∗ b = ab2

Find which of the binary operations are commutative and which are associative. 

Solution:

(i) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = a – b

RHS = b * a = b – a

Since, LHS is not equal to RHS

Therefore, * is not Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a – (b – c) = a – b + c

RHS = (a – b) * c = a – b – c

Since, LHS is not equal to RHS

Therefore, * is not Associative

(ii) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = a2 + b2

RHS = b * a = b2 + a2

Since, LHS is equal to RHS

Therefore, * is Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (b2 + c2) = a2 + (b2 + c2)2

RHS = (a * b) * c = (a2 + b2) * c = (a2 + b2)2 + c2 

Since, LHS is not equal to RHS

Therefore, * is not Associative

(iii) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = a + ab

RHS = b * a = b + ba

Since, LHS is not equal to RHS

Therefore, * is not Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (b + bc) = a + a(b + bc)

RHS = (a * b) * c = (a + ab) * c = a + ab + (a + ab)c

Since, LHS is not equal to RHS

Therefore, * is not Associative

(iv) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = (a – b)2

RHS = b * a = (b – a)2

Since, LHS is not equal to RHS

Therefore, * is not Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (b – c)2 = [a – (b – c)2]2 

RHS = (a * b) * c = (a – b)2 * c = [(a – b)2  – c]2

Since, LHS is not equal to RHS

Therefore, * is not Associative

(v) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = ab / 4

RHS = b * a = ba / 4

Since, LHS is equal to RHS

Therefore, * is Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * bc/4 = abc/16

RHS = (a * b) * c = ab/4 * c = abc/16

Since, LHS is equal to RHS

Therefore, * is Associative

(vi) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = ab2

RHS = b * a = ba2

Since, LHS is not equal to RHS

Therefore, * is not Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (bc)2 = a(bc2)2

RHS = (a * b) * c = (ab2) * c = ab2c2

Since, LHS is not equal to RHS

Therefore, * is not Associative

Question 10: Find which of the operations given above has identity

Solution:

An element e ∈ Q will be the identity element for the operation * if

a * e = a = e * a, for a ∈ Q

for (v) a * b = ab/4

Let e be an identity element 

a * e = a = e * a

LHS : ae/4 = a

   => e = 4

RHS : ea/4 = a

  => e = 4

LHS = RHS

Thus, Identity element exists

Other operations doesn’t satisfy the required conditions. 

Hence, other operations doesn’t have identity.

Question 11: Let A = N × N and ∗ be the binary operation on A defined by :

(a, b) ∗ (c, d) = (a + c, b + d)

Show that ∗ is commutative and associative. Find the identity element for ∗ on A, if any. 

Solution:

Given (a, b) * (c, d) = (a+c, b+d) on A

Let (a, b), (c, d), (e,f) be 3 pairs ∈ A

Commutative :

LHS = (a, b) * (c, d) = (a+c, b+d)

RHS = (c, d) * (a, b) = (c+a, d+b) = (a+c, b+d)

Since, LHS is equal to RHS

Therefore, * is Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = (a, b) * [(c, d) * (e, f)] = (a, b) * (c+e, d+f) = (a+c+e, b+d+f)

RHS = [(a, b) * (c, d)] * (e, f) = (a+c, b+d) * (e, f) = (a+c+e, b+d+f)  

Since, LHS is equal to RHS

Therefore, * is Associative

Existence of Identity element:

For a, e ∈ A, a * e = a

(a, b) * (e, e) = (a, b)

(a+e, b+e) = (a, b)

a + e = a    

=> e = 0

b + e = b

=> e = 0

As 0 is not a part of set of natural numbers. So, identity function does not exist.

Question 12: State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation ∗ on a set N, a ∗ a = a ∀ a ∈ N.

(ii) If ∗ is a commutative binary operation on N, then a ∗ (b ∗ c) = (c ∗ b) ∗ a

Solution:

(i) Let * be an operation on N, defined as:

a * b =  a + b ∀ a, b ∈ N

Let us consider b = a = 6, we have:

6 * 6 = 6 + 6 = 12 ≠ 6

Therefore, this statement is false. 

(ii) Since, * is commutative

LHS = a ∗ (b ∗ c) = a * (c * b) = (c * b) * a = RHS

Therefore, this statement is true.

Question 13: Consider a binary operation ∗ on N defined as a ∗ b = a3+ b3. Choose the correct answer.

(A) Is ∗ both associative and commutative?

(B) Is ∗ commutative but not associative?

(C) Is ∗ associative but not commutative?

(D) Is ∗ neither commutative nor associative? 

Solution:

On N, * is defined as a * b = a3 + b3

Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = a3 + b3

RHS = b * a = b3 + a3

Since, LHS is equal to RHS

Therefore, * is Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (b3 + c3) = a3 + (b3 + c3)3

RHS = (a * b) * c = (a3 + b3) * c = (a3 + b3)3 + c3

Since, LHS is not equal to RHS

Therefore, * is not Associative

Thus, Option (B) is correct

Summary

Exercise 1.4 serves as a critical juncture in the study of functions, bridging basic concepts with more advanced applications. Through exploring function composition, students learn to create complex functions from simpler ones, a skill essential in modeling real-world phenomena. The focus on invertible functions introduces the idea of reversibility in mathematical operations, a concept with far-reaching implications in various fields of study. By working through problems involving finding inverses and composing functions, students develop analytical skills and a deeper intuition about functional relationships. This exercise not only reinforces previous learning but also prepares students for more advanced topics in calculus and beyond, such as differential equations and functional analysis. Mastery of these concepts provides a solid foundation for understanding transformation of functions, a key element in many areas of higher mathematics and its applications in science and engineering.

FAQs on Relations and Functions

What is function composition?

Function composition is the operation of combining two functions to create a new function. If f and g are functions, their composition (f ∘ g)(x) is defined as f(g(x)).

When is a function invertible?

A function is invertible if it is both injective (one-to-one) and surjective (onto). In other words, each element in the codomain is paired with exactly one element in the domain.

How do you find the inverse of a function?

To find the inverse of a function f(x), replace f(x) with y, swap x and y, then solve for y. The resulting expression in terms of x is f^(-1)(x).

What is the relationship between (f ∘ g) and (g ∘ f)?

In general, (f ∘ g) is not equal to (g ∘ f). Function composition is not commutative.

Why is function composition important in mathematics?

Function composition is crucial for creating complex functions from simpler ones, modeling intricate relationships, and solving equations involving multiple transformations. It’s a fundamental concept in advanced mathematics and its applications.

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

CBSE Class 12 Maths Syllabus 2025-26 with Marks Distribution

The table below shows the marks weightage along with the number of periods required for teaching. The Maths theory paper is of 80 marks, and the internal assessment is of 20 marks which totally comes out to be 100 marks.

CBSE Class 12 Maths Syllabus And Marks Distribution 2023-24

Max Marks: 80

No.UnitsMarks
I.Relations and Functions08
II.Algebra10
III.Calculus35
IV.Vectors and Three – Dimensional Geometry14
V.Linear Programming05
VI.Probability08
Total Theory80
Internal Assessment20
Grand Total100

Unit-I: Relations and Functions

1. Relations and Functions

Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto functions.

2. Inverse Trigonometric Functions

Definition, range, domain, principal value branch. Graphs of inverse trigonometric functions.

Unit-II: Algebra

1. Matrices

Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix, symmetric and skew symmetric matrices. Operations on matrices: Addition and multiplication and multiplication with a scalar. Simple properties of addition, multiplication and scalar multiplication. Noncommutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries).

2. Determinants

Determinant of a square matrix (up to 3 x 3 matrices), minors, co-factors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.

Unit-III: Calculus

1. Continuity and Differentiability

Continuity and differentiability, derivative of composite functions, chain rule, derivative of inverse trigonometric functions like sin-1 x, cos-1 x and tan-1 x, derivative of implicit functions. Concept of exponential and logarithmic functions.
Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative of functions expressed in parametric forms. Second order derivatives.

2. Applications of Derivatives

Applications of derivatives: rate of change of quantities, increasing/decreasing functions, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as real-life situations).

3. Integrals 

Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them.

Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals.

4. Applications of the Integrals

Applications in finding the area under simple curves, especially lines, circles/ parabolas/ellipses (in standard form only)

5. Differential Equations

Definition, order and degree, general and particular solutions of a differential equation. Solution of differential equations by method of separation of variables, solutions of homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type:

dy/dx + py = q, where p and q are functions of x or constants.

dx/dy + px = q, where p and q are functions of y or constants.

Unit-IV: Vectors and Three-Dimensional Geometry

1. Vectors

Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Definition, Geometrical Interpretation, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors.

2. Three – dimensional Geometry

Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, skew lines, shortest distance between two lines. Angle between two lines.

Unit-V: Linear Programming

1. Linear Programming

Introduction, related terminology such as constraints, objective function, optimization, graphical method of solution for problems in two variables, feasible and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints).

Unit-VI: Probability

1. Probability

Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution, mean of random variable.

Students can go through the CBSE Class 12 Syllabus to get the detailed syllabus of all subjects. Get access to interactive lessons and videos related to Maths and Science with ANAND CLASSES’S App/ Tablet.

Frequently Asked Questions on CBSE Class 12 Maths Syllabus 2025-26

Q1

Is Calculus an important chapter in the CBSE Class 12 Maths Syllabus?

Yes, Calculus is an important chapter in the CBSE Class 12 Maths Syllabus. It is for 35 marks which means that if a student is thorough with this chapter will be able to pass the final exam.

Q2

How many units are discussed in the CBSE Class 12 Maths Syllabus?

In the CBSE Class 12 Maths Syllabus, about 6 units are discussed, which contains a total of 13 chapters.

Q3

How many marks are allotted for internals in the CBSE Class 12 Maths syllabus?

About 20 marks are allotted for internals in the CBSE Class 12 Maths Syllabus. Students can score it with ease through constant practice.