NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Relations and Functions |

Exercise |
Ex 2.1 |

Number of Questions Solved |
10 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1

**Ex 2.1 Class 11 Maths Question 1.**

If \(\left( \frac { x }{ 3 } +1,y-\frac { 2 }{ 3 } \right) =\left( \frac { 5 }{ 3 } ,\frac { 1 }{ 3 } \right) \), find the values of x and y.

**Solution.**

Since the ordered pairs are equal. So, the corresponding elements are equal

∴ \(\frac { x }{ 3 } +1=\frac { 5 }{ 3 } \) and \(y-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)

⇒ \(\frac { x }{ 3 } =\frac { 5 }{ 3 } -1\) and \(y=\frac { 1 }{ 3 } +\frac { 2 }{ 3 } \) ⇒ x = 2 and y = 1.

**Ex 2.1 Class 11 Maths Question 2.**

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

**Solution.**

According to question, n(A) = 3 and n(B) = 3.

∴ n(A x B) = n(A) x n(B) = 3 x 3 = 9

∴ There are total 9 elements in (A x B).

**Ex 2.1 Class 11 Maths Question 3.**

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

**Solution.**

We have G = {7, 8} and H = {5, 4, 2} Then, by the definition of the cartesian product, we have

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

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

**Ex 2.1 Class 11 Maths Question 4.**

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 x Q = {(m, n), (n, m)}.

(ii) If A and B are non-empty sets, then Ax 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 x (B ∩φ) = φ

**Solution.**

**(i)** False, if P = {m, n} and Q = {n, m}

Then P x Q = {(m, n), (m, m), (n, n), (n, m)}.

**(ii)** True, by the definition of cartesian product.

**(iii)** True, We have A = {1, 2} and B = {3, 4}

Now, B ∩ φ = φ ∴ A x (B ∩ φ) = A x φ = φ.

**Ex 2.1 Class 11 Maths Question 5.**

If A = {-1, 1},find A x A x A.

**Solution.**

A = {-1, 1}

Then, A x A = {-1, 1} x {-1, 1} = {(-1, -1), (-1,1),(1,-1), (1,1)}

A x A x A = ((-1,-1),(-1,1),(1,-1),(1,1)} x {-1,1}

= {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1,1), (1, -1, -1), (1, -1,1), (1,1,-1), (1,1,1)}

**Ex 2.1 Class 11 Maths Question 6.**

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

**Solution.**

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

If {p, q) ∈ A x B, then p ∈ A and q ∈ B

∴ A = {a, b} and B = {x, y}.

**Ex 2.1 Class 11 Maths Question 7.**

Let A = {1, 2}, B = (1, 2, 3, 4), C = {5, 6} and D = {5, 6, 7, 8}. Verify that

(i) A x (B ∩ C = (A x B) ∩ (AxC)

(ii) A x C is a subset of B x D.

**Solution.**

Given, A = {1, 2}, B ={1, 2, 3, 4}, C = {5, 6}, D = (5, 6, 7, 8}

**Ex 2.1 Class 11 Maths Question 8.**

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

**Solution.**

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

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

i. e., A x B has 4 elements. So, it has 2^{4} i.e. 16 subsets.

The subsets of A x B are as follows :

φ, {(1, 3)1, ((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, 4), (2, 3), (2,4)},{(1, 3), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2,4)}.

**Ex 2.1 Class 11 Maths Question 9.**

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 x B, find A and B, where x, y and z are distinct elements.

**Solution.**

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

Now (x, 1) ∈ A x B ⇒ x ∈ A and 1 ∈ B,

(y, 2) ∈ A x B ⇒ y ∈ A and 2 ∈ B

(z, 1) ∈ A x B ⇒z ∈ A and 1 ∈ B

∴ x, y, z ∈ A and 1, 2 ∈ B

Hence, A = {x, y, z} and B = {1, 2}.

**Ex 2.1 Class 11 Maths Question 10.**

The Cartesian product 4×4 has 9 elements among which are found (-1, 0) and (0, 1). Find the set 4 and the remaining elements of 4 x 4.

**Solution.**

Since, we have n(A x A) = 9

⇒ n(A) x n(A) = 9 [ ∵ n (A x B) = n(A) x n(B)]

⇒ (n(A))^{2} = 9 ⇒ n(A) = 3

Also, given (-1, 0) ∈ A x A ⇒ -1, 0 ∈ A ,

and (0,1) ∈ A x A ⇒ 0, 1 ∈ A

∴ -1, 0,1 ∈ A

Hence, A = {-1, 0, 1} (∵ n(A) = 3)

and the remaining elements of A x A are (-1, -1), (-1,1), (0, -1), (0,0), (1, -1), (1,0), (1,1).

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