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

### Relations and Functions Exercise 2.1 Solutions

**1. If (x/3 + 1, y - 2/3) = (5/3 1/3) , find the values of x and y. **

**Solution**

It is given that (x/3 + 1, y - 2/3) = (5/3, 1/3).

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

Therefore, x/3 + 1= 5/3 and y - 2/3 = 1/3.

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

**Solution**

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.

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

**Solution**

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)}

**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 ∩ Î¦) = Î¦.**

**Solution**

(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

**5. 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}

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)}

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

**Solution**

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*}

**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**

**Solution**

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

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

∴L.H.S. = A × (B ∩ C) = A × Î¦ = Î¦

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) = Î¦

∴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.

**8. 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

Î¦, {(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)}

**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.**

**Solution**

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}.

**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.**

**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)