Our ncert solutions for Exercise 2.1 Class 11 maths 2. Relations and Functions - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 2.1 Class 11 maths 2. Relations and Functions - ncert solutions - Toppers Study

Topper Study classes prepares ncert solutions on practical base problems and comes out with the best result that helps the students and teachers as well as tutors and so many ecademic coaching classes that they need in practical life. Our ncert solutions for Exercise 2.1 Class 11 maths 2. Relations and Functions - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

## Exercise 2.1 Class 11 maths 2. Relations and Functions - ncert solutions - Toppers Study

CBSE board students who preparing for **class 11 ncert solutions maths and Mathematics** solved exercise **chapter 2. Relations and Functions** available and this helps in upcoming exams
2023-2024.

### You can Find Mathematics solution Class 11 Chapter 2. Relations and Functions

- All Chapter review quick revision notes for chapter 2. Relations and Functions Class 11
- NCERT Solutions And Textual questions Answers Class 11 Mathematics
- Extra NCERT Book questions Answers Class 11 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

NCERT Solutions do not only help you to cover your syllabus but also will give to textual support in exams 2023-2024 to complete **exercise 2.1 maths class 11 chapter 2** in english medium. So revise and practice these all cbse study materials like class 11 maths chapter 2.1 in english ncert book. Also ensure to repractice all syllabus within time or before board exams for ncert class 11 maths ex 2.1 in english.

See all solutions for class 11 maths chapter 2 exercise 2 in english medium solved questions with answers.

### Exercise 2.1 class 11 Mathematics Chapter 2. Relations and Functions

Sure! The following topics will be covered in this article

- Exercise 2.1 Class 11 Maths 2. Relations And Functions - Ncert Solutions - Toppers Study
- Class 11 Ncert Solutions
- Solution Chapter 2. Relations And Functions Class 11
- Solutions Class 11
- Chapter 2. Relations And Functions Exercise 2.1 Class 11

## 2. Relations and Functions

### | Exercise 2.1 |

## Exercise 2.1 Class 11 maths 2. Relations and Functions - ncert solutions - Toppers Study

**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)] = 2^{m}.

Therefore, the set A × B has 2^{4} = 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).

##### Other Pages of this Chapter: 2. Relations and Functions

### Select Your CBSE Classes

Important Study materials for classes 06, 07, 08,09,10, 11 and 12. Like CBSE Notes, Notes for Science, Notes for maths, Notes for Social Science, Notes for Accountancy, Notes for Economics, Notes for political Science, Noes for History, Notes For Bussiness Study, Physical Educations, Sample Papers, Test Papers, Mock Test Papers, Support Materials and Books.

*Mathematics Class - 11th*

NCERT Maths book for CBSE Students.

books

## Study Materials List:

##### Solutions ⇒ Class 11th ⇒ Mathematics

## Topper's Study

New Books