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

#### Page No 33:

#### Question 1:

If, find the values of

*x*and*y*.#### Answer:

It is given that.

Since the ordered pairs are equal, the corresponding elements will also be equal.

Therefore,

and.

∴

*x*= 2 and*y*= 1#### Question 2:

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

#### Answer:

It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.

⇒ Number of elements in set B = 3

Number of elements in (A × B)

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

= 3 × 3 = 9

Thus, the number of elements in (A × B) is 9.

#### Question 3:

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

#### Answer:

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)}

#### Question 4:

State whether each of the following statement 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.

#### Answer:

(i) False

If P = {

*m*,*n*} and Q = {*n*,*m*}, then
P × Q = {(

*m*,*m*), (*m*,*n*), (*n**,**m*), (*n*,*n*)}
(ii) True

(iii) True

#### Question 5:

If A = {–1, 1}, find A × A × A.

#### Answer:

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}
It is 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)}

#### Question 6:

If A × B = {(

*a*,*x*), (*a*,*y*), (*b*,*x*), (*b*,*y*)}. Find A and B.#### Answer:

It is given that A × B = {(

*a*,*x*), (*a,**y*), (*b*,*x*), (*b*,*y*)}
We know that the Cartesian product of two non-empty sets P and Q is defined as P × Q = {(

*p*,*q*):*p*∈ P,*q*∈ Q}
∴ A is the set of all first elements and B is the set of all second elements.

Thus, A = {

*a*,*b*} and B = {*x*,*y*}#### Question 7:

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

#### Answer:

(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

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

B × 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)}

We can observe that all the elements of set A × C are the elements of set B × D.

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

#### Question 8:

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

#### Answer:

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)}

#### 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 × B, find A and B, where*x*,*y*and*z*are distinct elements.#### Answer:

It is 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, it is clear that A = {*x*,*y*,*z*} and B = {1, 2}.#### Page No 34:

#### Question 10:

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.

#### Answer:

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)

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