Your Complete CBSE Learning Hub

Free NCERT Solutions, Revision Notes & Practice Questions

Notes | Solutions | PYQs | Sample Papers — All in One Place

Get free NCERT solutions, CBSE notes, sample papers and previous year question papers for Class 6 to 12 in Hindi and English medium.

Advertise:

2. Relations and Functions Mathematics Exercise - 2.1 class 11 Maths in English - CBSE Study

2. Relations and Functions Mathematics Class 11 exercise - 2.1 class 11 Maths cbse board school study materials like cbse notes in English medium, all chapters and exercises are covered the ncert latest syllabus 2026 - 27.

• Hi Guest! • LoginRegister

Class 6

NCERT Solutions

Class 7

NCERT Solutions

Class 8

NCERT Solutions

Class 9

NCERT Solutions

Class 10

NCERT Solutions

Class 11

NCERT Solutions

Class 12

NCERT Solutions

Class 6

CBSE Notes

Class 7

CBSE Notes

Class 8

CBSE Notes

Class 9

CBSE Notes

Class 10

CBSE Notes

Class 11

CBSE Notes

Class 12

CBSE Notes

2. Relations and Functions Mathematics Exercise - 2.1 class 11 Maths in English - CBSE Study

NCERT Solutions for Class 11 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 2. Relations and Functions with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 2.1 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 11 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.

Class 11 English Medium Mathematics All Chapters:

2. Relations and Functions

1. Exercise 2.1

Exercise 2.1


Comparing both side as order pairs are equal, so corresponding elements also will be equal,  

Q2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B).

Solution: 

Given : n(A) = 3 and set B = {3, 4, 5}

∴ n(B) = 3 

Number of elements in (A × B)

 = (Number of elements in A) × (Number of elements in B)

= n(A) × n(B) 

= 3 × 3

= 9 

Q3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

Solution: 

Given: G = {7, 8} and H = {5, 4, 2}

We know that the Cartesian product P × Q of two non-empty sets P and Q is defined as P × Q = {(p, q): p∈ P, q ∈ Q}

∴ G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)} 

Q4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.

(i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}.

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.

Solution: 

(i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}.

False

If P = {m, n} and Q = {n, m},

P × Q = {(m, m), (m, n), (n, m), (n, n)}

Solution: 

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.

(ii) True

Solution: 

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.

(iii) True 

Q5. If A = {–1, 1}, find A × A × A.

Solution: 

It is known that for any non-empty set A, A × A × A is defined as;

A × A × A = {(a, b, c): a, b, c ∈ A}

Given that A = {–1, 1}

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

Q6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

Solution: 

The cartesian product of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q} 

Given that: A × B = {(a, x),(a , y), (b, x), (b, y)}

∴ a, b ∈ A abd x, y ∈ B 

So, A = {a, b} and B = {x, y} 

Q7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}.

Verify that:

(i) A × (B ∩ C) = (A × B) ∩ (A × C).

(ii) A × C is a subset of B × D.

Solution: 

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

∴ L.H.S. = A × (B ∩ C) = A × Φ = Φ

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

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

∴ R.H.S. = (A × B) ∩ (A × C) = Φ

∴ L.H.S. = R.H.S

Hence, A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To verify: A × C is a subset of B × D

Solution:

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

A × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

All the elements of set A × C are the elements of set B × D.

Therefore, A × C is a subset of B × D. 

Q8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

Solution: 

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

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

⇒ n(A × B) = 4

We know that if C is a set with n(C) = m,

then n[P(C)] = 2m.

Therefore, the set A × B has 24 = 16 subsets.

These are Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

Q9. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.

Solution: 

Given that n(A) =3 and n(B) = 2;

and (x, 1), (y, 2), (z, 1) are in A × B.

We know that A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B.

∴ x, y, and z are the elements of A; and 1 and 2 are the elements of B.

Since n(A) = 3 and n(B) = 2,

∴ A = {x, y, z} and B = {1, 2}. 

Q10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A.

Solution: 

We know that if n(A) = p and n(B) = q,

then n(A × B) = pq.

∴ n(A × A) = n(A) × n(A)

It is given that n(A × A) = 9

∴ n(A) × n(A) = 9 ⇒ n(A) = 3

The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A×A.

We know that A × A = {(a, a): a ∈ A}.

Therefore, –1, 0, and 1 are elements of A.

Since n(A) = 3,

it is clear that A = {–1, 0, 1}.

The remaining elements of set A × A are (–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1). 

 

Topic Lists:

Disclaimer:

This website's domain name has included word "CBSE" but here we clearly declare that we and our website have neither any relation to CBSE and nor affliated to CBSE organisation.