NCERT Solutions for Class 12 Mathematics Chapter 1

NCERT solutions for Class 12 Mathematics Chapter 1 Relations and Functions – complete answers & explanations

Class 12 Mathematics Chapter 1 Relations and Functions is an important topic that builds the foundation for higher-level concepts in algebra and calculus. This chapter helps students understand how elements are related and how functions behave under different conditions. It introduces key ideas like reflexive, symmetric, and transitive relations, along with one-one and onto functions. These concepts are essential for exams and help improve logical reasoning skills. This blog provides clear and reliable NCERT solutions that help students understand the correct approach to answering questions. Students can strengthen their preparation by carefully going through each solution and practicing regularly. Download the worksheet and practice alongside solutions for better clarity. Book a free trial now to get expert guidance.

What this NCERT chapter covers?

1. Understanding the concept of relations and how elements of sets are connected 
2. Types of relations such as reflexive, symmetric, and transitive 
3. Identification of equivalence relations and their properties 
4. Real-life examples of relations like parity, similarity, and parallel lines 
5. Introduction to functions and mapping between sets 
6. Concept of one-one (injective) functions 
7. Concept of onto (surjective) functions 
8. Understanding bijective functions and their importance 
9. Invertible functions and how inverses are calculated 
10. Composition of functions and their properties 
11. Graphical and logical interpretation of functions 
12. Application of relations and functions in problem-solving 

How to use these NCERT solutions?

1. First attempt all questions from the worksheet on your own to build understanding 
2. After solving, use the solutions to check your answers step-by-step 
3. Focus on understanding where mistakes occur and correct them 
4. Follow the exact order of questions to stay aligned with the worksheet 
5. Revise important concepts like reflexive, symmetric, and transitive relations 
6. Practice function-based questions multiple times for better clarity 
7. Parents and teachers can guide students by discussing each solution 
8. Use the solutions to improve answer presentation for exams 
9. Re-attempt difficult questions after reviewing the answers 
10. Regular practice will help build confidence and accuracy 

Important tips & tricks for students

1. Always check all three properties before classifying a relation 
2. Do not assume symmetry or transitivity without proper verification 
3. Carefully observe function behavior before deciding one-one or onto 
4. Avoid skipping steps while solving function-based problems 
5. Practice identifying equivalence relations through examples 
6. Learn how to test invertibility correctly 
7. Double-check calculations in composition of functions 
8. Read each question carefully before answering 
9. Practice similar problems to strengthen concepts 
10. Keep formulas and definitions clear for better performance 

NCERT solutions – complete answer key

Exercise No. 1.1

(i) R: x < y 
Reflexive: x < x is false ⇒ Not reflexive 
Symmetric: x < y does not imply y < x ⇒ Not symmetric 
Transitive: x < y, y < z ⇒ x < z , but since other properties fail 
⇒ Not transitive overall classification 

(ii) R: x − y = 1 
Reflexive: x − x = 0 ≠ 1 ⇒ Not reflexive 
Symmetric: x − y = 1 ⇒ y − x = −1 ≠ 1 ⇒ Not symmetric 
Transitive: x − y = 1, y − z = 1 ⇒ x − z = 2 ≠ 1
⇒ transitive 

(iii) R: x divides y 
Reflexive: x|x ⇒ true 
Symmetric: 2|4 but 4∤2 ⇒ Not symmetric 
Transitive: x|y, y|z ⇒ x|z ⇒ true

(iv) R: x − y divisible by 3 
Reflexive: x − x = 0 divisible by 3 ⇒ true 
Symmetric: if x−y divisible by 3 ⇒ y−x also ⇒ true 
Transitive: x−y and y−z divisible ⇒ x−z divisible ⇒ true 
⇒ Equivalence relation 

(v)a. Reflexive: For every element a, (a, a) belongs to R ⇒ TRUE 
Symmetric: If (a, b) ∈ R
⇒ (b, a) also belongs to R ⇒ TRUE 
Transitive: If (a, b) and (b, c) ∈ R, then (a, c) should be in R 
⇒ Final: Reflexive, Symmetric, Transitive 
⇒ Equivalence Relation 

b. Reflexive: (a, a) ∈ R for all a ⇒ TRUE 
Symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R ⇒ TRUE 
Transitive: (a, b) and (b, c) ⇒ (a, c) also belongs ⇒ TRUE 
⇒ Final: Reflexive, Symmetric, Transitive 
⇒ Equivalence Relation 

c. Reflexive: (a, a) does NOT satisfy relation ⇒ FALSE 
Symmetric: (a, b) ∈ R does NOT imply (b, a) ∈ R ⇒ FALSE 
Transitive: Chain condition fails ⇒ FALSE 
⇒ Final: Not reflexive, not symmetric, not transitive

d. Reflexive: Fails for some element ⇒ FALSE 
Symmetric: Does not hold in reverse ⇒ FALSE 
Transitive: Chain condition 
⇒ Final: Not reflexive, not symmetric, but transitive 

e. Reflexive: (a, a) not always in R ⇒ FALSE 
Symmetric: Reverse condition fails ⇒ FALSE 
Transitive: Chain condition fails ⇒ FALSE 
⇒ Final: Not reflexive, not symmetric, not transitive 

R: a ≤ a² 
Reflexive: For a=½, ½ ≤ ¼ false ⇒ Not reflexive 
Symmetric: a ≤ b² does not imply b ≤ a² ⇒ Not symmetric 
Transitive: a ≤ b² and b ≤ c² does not guarantee a ≤ c²
⇒ Not transitive 

3 Relation fails basic checks for reflexivity, symmetry, and
transitivity 
⇒ Not reflexive, not symmetric, not transitive 

R: a ≤ b 
Reflexive: a ≤ a ⇒ true 
Symmetric: a ≤ b does not imply b ≤ a ⇒ false 
Transitive: a ≤ b and b ≤ c ⇒ a ≤ c ⇒ true 

Reflexive: relation holds for same element (fails)
Symmetric: fails (direction matters) 
Transitive: fails (chain may break) 

R: |a−b| = k 
Reflexive: |a−a|=0 ≠ k ⇒ Not reflexive 
Symmetric: |a−b| = |b−a| ⇒ true 
Transitive: may fail ⇒ Not transitive 

Reflexive: Every element is related to itself 
⇒ relation holds for (a, a).
Symmetric: If (a, b) ∈ R, then (b, a) also belongs to R ⇒
symmetric property holds.
Transitive: If (a, b) and (b, c) ∈ R, then (a, c) ∈ R ⇒
transitive holds.
⇒ Since all three properties are satisfied, it is an
equivalence relation.

Same parity relation 
Reflexive: number has same parity as itself 
Symmetric: if a same parity as b ⇒ b same as a 
Transitive: a,b same parity and b,c same ⇒ a,c same 
⇒ Equivalence relation 
Equivalence classes: 
{1,3,5} (odd), {2,4} (even) 

(i) Relation is based on modulo (a ≡ b). 
Reflexive: a−a=0 ⇒ true 
Symmetric & Transitive also hold ⇒ equivalence relation 
⇒ Elements related to 1: {1, 5, 9}

(ii) Relation is equality. 
Only identical elements are related 
⇒ all three properties hold 
⇒ Elements related to 1: {1}

(i) Missing (a,a) ⇒ not reflexive 
(a,b) ⇒ (b,a) present ⇒ symmetric 
Chain fails ⇒ not transitive 

(ii) Not reflexive, not symmetric 
(a,b), (b,c) ⇒ (a,c) present ⇒ transitive 

(iii) All (a,a) present ⇒ reflexive 
(a,b) and (b,a) ⇒ symmetric 
Chain works ⇒ transitive 
⇒ equivalence relation 

(iv) Reflexive (all (a,a) present) 
Not symmetric (missing reverse pair) 
Transitive holds 

(v) Not reflexive 
Symmetric (pairs reversed present) 
Not transitive

Similarity of triangles 
Reflexive: triangle similar to itself 
Symmetric: if A~B ⇒ B~A 
Transitive: A~B, B~C ⇒ A~C 
⇒ Equivalence 

Check condition a = b − 2 
For (6,8): 6 = 8 − 2 ⇒ true 
Hence, pair satisfies relation 
⇒ Correct answer: (C) 

R: distance from origin same 
Reflexive: same distance 
Symmetric: interchangeable 
Transitive: equal distances chain 
⇒ Equivalence 
Set: circle centered at origin 

Relation: triangles having same area 
Reflexive: every triangle has same area as itself 
Symmetric: if A has same area as B ⇒ B same as A 
Transitive: A=B and B=C ⇒ A=C 
⇒ Equivalence relation 

Relation: parallel lines 
Reflexive: every line is parallel to itself 
Symmetric: if l₁ ∥ l₂ ⇒ l₂ ∥ l₁ 
Transitive: l₁ ∥ l₂ and l₂ ∥ l₃ ⇒ l₁ ∥ l₃ 
The set of all lines y = 2x + c, c ∈ R
⇒ Equivalence relation 

Check options for properties 
Only option (B) satisfies reflexive and transitive but NOT
symmetric 
⇒ Correct answer: (B) 

Exercise No. 1.2

f(x)=1/x 
One-one: If f(x₁)=f(x₂), then 1/x₁ = 1/x₂ ⇒ x₁ = x₂, so
injective. 
Onto: For any y ≠ 0, we can find x = 1/y, so every value is
covered. 
⇒ Hence, bijective (but not defined for 0 or natural
numbers fully). (No)

2 (i) Increasing function 
⇒ different inputs give
different outputs
⇒ one-one, but range is
limited 
⇒ not onto. 

(ii) Same output for x and −x
⇒ not one-one, and
negative values missing
⇒ not onto. 

(iii) Constant-type 
⇒ many inputs give same
output 
⇒ not one-one, also not
onto. 

Greatest integer function gives same value for many
inputs (e.g., 1.2 and 1.8). 
So not one-one. Also, it only gives integer outputs, not all
real numbers. 
⇒ Not onto.

3

f(x)=x² 
f(2)=f(−2) ⇒ same output ⇒ not one-one. 
Also, output is always ≥0, so negative numbers are not
covered. 
⇒ Not onto.

4

(v) Linear with full domain
⇒ covers all values
⇒ one-one and onto.

(iv) Linear but restricted 
⇒ one-one, but does not cover
full codomain 
⇒ not onto. 

There exist different inputs giving same output ⇒ not
one-one. 
Also, some values in codomain are not achieved. 
⇒ Neither one-one nor onto.

5

Different inputs map to different outputs ⇒ one-one. 
Every value in codomain has a pre-image ⇒ onto. 
⇒ Bijective.(Yes)

Function satisfies both conditions: 
Unique outputs for each input (one-one) and covers
full codomain. 
⇒ (No)

f(a,b)=(b,a) just swaps values. 
Each pair has a unique image and can be reversed
back. 
⇒ One-one and onto ⇒ bijective.

Each input produces a unique output. 
No two different inputs give the same value. 
⇒ Function is one-one (injective).

6

(i) Since it is a linear function (f(x)=ax+b, a≠0),
different inputs give different outputs ⇒ one-one. 
Also, for every y, x=(y−b)/a exists ⇒ onto. 
⇒ Hence, it is bijective.

(ii) There exist different inputs giving the same output
⇒ not one-one. 
Also, some values in codomain are not obtained ⇒
not onto. 
⇒ Hence, it is neither one-one nor onto.

f(x)=x⁴ 
f(2)=f(−2) ⇒ not one-one. 
Range is only non-negative numbers ⇒ not onto. 
⇒ Correct option: (D)

A function is invertible only if it is one-one. 
This means different inputs must give different
outputs (no repetition). 
If f(x₁)=f(x₂) ⇒ x₁=x₂, then inverse exists. Otherwise,
inverse is not possible.

f(x)=3x 
If 3x₁=3x₂ ⇒ x₁=x₂ ⇒ one-one. 
For any y, x=y/3 exists ⇒ onto. 
⇒ Correct option: (A)

Miscellaneous Exercise

1

First check if function is one-one (important). 
Then write y = f(x) and solve for x. 
Replace y with x to get inverse. 
Verify using f(f⁻¹(x)) = x. (B)

2

To find inverse: let y = f(x). 
Solve this equation to express x in terms of y. 
Then replace y by x to get f⁻¹(x). 
Finally, verify: f(f⁻¹(x)) = x.

3 For (f∘g)(x), first apply g(x). 
Then put that result into f(x), i.e., f(g(x)). 
Simplify step-by-step carefully to avoid mistakes.

4

Find (f∘g)(x) and (g∘f)(x) separately. 
Compare both results. 
If they are equal, functions are commutative;
otherwise not (usually not equal).

5

Compute both compositions: f(g(x)) and g(f(x)). 
Simplify both expressions fully. 
If results differ, then f∘g ≠ g∘f.

6

For inverse of composite function: 
(f∘g)⁻¹ = g⁻¹∘f⁻¹ 
Order reverses while taking inverse. 
Apply inverse of outer function first, then inner. (A)

Why NCERT solutions help students?

NCERT solutions help students build strong conceptual understanding and prepare effectively for exams. They provide clarity in answering questions correctly and help students follow the expected answering pattern. With regular practice, students gain confidence and improve accuracy, making it easier to score well in Mathematics.

Help your child build strong Mathematics fundamentals with expert-guided learning support.
Book a free trial!