NCERT Solutions for Class 12 Mathematics Chapter 1 Relations and Functions

This worksheet focuses on Class 12 Mathematics Chapter 1 Relations and Functions and helps students understand important concepts related to relations and functions in a clear and structured way. The chapter explains how elements are connected through relations and how functions behave under different conditions. It is an important part of Mathematics as it builds the base for higher-level topics and improves logical thinking. This worksheet provides complete and accurate NCERT Solutions, helping students check their answers and understand the correct approach step by step. It is useful for both practice and revision, making learning more effective and confidence-building.

Chapter summary: 

This chapter is concept-based. It focuses on mathematical understanding of relations and functions. Students learn how to identify different types of relations such as reflexive, symmetric, and transitive. They also understand equivalence relations and their real-life applications like parity, similarity of triangles, and parallel lines. The chapter further introduces functions and explains different types such as one-one, onto, and bijective functions. The main theme of the chapter is logical reasoning and understanding mathematical relationships between elements.

What this NCERT chapter covers?

• Understanding relations between elements of sets 
• Identifying reflexive, symmetric, and transitive properties 
• Learning about equivalence relations and their conditions 
• Studying examples like same parity, similarity, and parallel lines 
• Introduction to functions and mapping concepts 
• Understanding one-one (injective) functions 
• Understanding onto (surjective) functions 
• Learning about bijective functions 
• Concept of invertible functions 
• Composition of functions and their properties 

How to use these NCERT solutions?

• Students should first try solving each question from the worksheet on their own 
• After attempting, they can use these NCERT Solutions to check answers 
• Solutions follow the exact NCERT order and structure for easy understanding 
• Parents and teachers can guide students by explaining each step 
• These solutions help in revising important concepts clearly 
• Students can reattempt difficult questions after reviewing the answers 

Student tips & learning tricks

• Always check all properties before classifying a relation 
• Do not assume results without proper verification 
• Carefully observe function behavior before deciding its type 
• Practice identifying one-one and onto functions regularly 
• Avoid skipping steps while solving problems 
• Revise definitions and properties for better accuracy 
• Focus on understanding concepts rather than memorising 
• Double-check answers to avoid small mistakes 
Why NCERT solutions are important?

NCERT Solutions are important because they help students understand the correct method of solving questions as per exam expectations. They strengthen basic concepts and improve accuracy in answering. Regular practice with these solutions builds confidence and helps students perform better in assessments.

Complete answer key – NCERT solutions

Exercise No. 1.1

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 

2 
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 

4 
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) 

5 
R: |a−b| = k 
Reflexive: |a−a|=0 ≠ k ⇒ Not reflexive 
Symmetric: |a−b| = |b−a| ⇒ true 
Transitive: may fail ⇒ Not transitive 

6 
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.

7 
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) 

8 
(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}

9 
(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

10 
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

1 
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. 

3 
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.

4 
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.

5 
(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.

6 
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).

7 
(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.

8 
f(x)=x⁴ 
f(2)=f(−2) ⇒ not one-one. 
Range is only non-negative numbers ⇒ not onto. 
⇒ Correct option: (D)

9 
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.

10 
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)

7 
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.

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