Class 11 NCERT Solutions – Chapter 2 Relation And Functions – Exercise 2.1

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

Solution:

We know that,

If two ordered pairs are equal, then their corresponding first elements and second elements are equal.

We are given that the pairs (x/3 + 1, y – 2/3) and (5/3, 1/3) are equal, so the corresponding elements should also be equal.

So we have, (x/3 + 1) = 5/3 and (y – 2/3) = 1/3

On solving both the equations, we get –

x/3 + 1 = 5/3     and    y – 2/3 = 1/3

x/3 = 5/3 – 1    and    y = 1/3 + 2/3

x/3 = 2/3    and    y = 3/3

x = 2    and    y = 1

Therefore, 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).

Solution:

Given, number of elements of set A = 3

The elements of set B are 3, 4, and 5.

So, number of elements of set B = 3

Then, number of elements in A×B = (Number of elements in A) × (Number of elements in B)

                                                       = 3 × 3 = 9

Therefore, 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.

Solution:

Given, G = {7, 8} and H = {5, 4, 2}

The cartesian product of two non-empty sets P × Q is the set of all ordered pairs of elements from P and Q, i.e.,

P × Q = {(p, q) : p ∈ P, q ∈ Q}

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

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

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 × 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 ∩ ∅) = ∅.

Solution:

(i) The given statement is False. 

    The correct statement is: If P = {m, n} and Q = {n, m}, then P x Q = { (m, m), n), (n, m), (n, n) } 

(ii) The given statement is true.

(iii) The given statement is true.

Question 5. If A = {–1, 1}, find A × A × A.

Solution:

A × A × A for a non-empty set A is given by –

A × A × A = {(a, b, c) : a, b, c ∈ A}, where (a, b, c) is called an ordered triplet

Here, given A = {–1, 1},

So, 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.

Solution:

Given, 

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

Since, the cartesian product of two non-empty sets P × Q is given by –

P × Q = {(p, q) : p ∈ P, q ∈ Q}

So, A is the set of all first elements and B is the set of all second elements.

Therefore, 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. 

Solution:

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

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

     Since B ∩ C= {1,2, 3,4} ∩ {5,6} = ∅

     Thus, L.H.S.= A × (B ∩ C) = A × ∅ = ∅

     Now, 

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

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

     Thus, R.H.S. = (A × B) ∩ (A × C) = ∅

     Therefore, L.H.S. = R.H.S 

     Hence, verified.

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

     Here,

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

     B x 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)}

     Since, all the elements of set A x C are the elements of set B x D

     Thus, A x C is a subset of B × D

     Hence, verified.

Question 8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

Solution:

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

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

Number of elements in A × B = n(A × B) = 4 

We know that,

For a set S with n(S) = m, number of subsets of S is given by n[P(S)] = 2^m. 

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

These subsets are: ∅, {(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.

Solution:

Given, 

n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.

We know that,

A is the set of first elements and B is the set of second elements of the ordered pair elements of A x B.

So, the elements of A are x, y, z and the elements of B are 1, 2

As, n(A) = 3 and n(B) = 2, it is clear that set A = {x, y, z} and set B = {1, 2}. 

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.

Solution:

We know that,

If there are p elements in A and q elements in B, then there will be pq elements in A × B

, i.e., if n(A) = p and n(B) = q, then n(A × B) = pq

Given, n(A × A) = 9

So, n(A) × n(A) = 9

Thus, n(A) = 3

Also given that, the ordered pairs (-1, 0) and (0, 1) are two of the nine elements of A × A.

And, we know A × A = {(a, a): a ∈ A}. 

So, -1, 0, and 1 should be the elements of A. 

As n(A) = 3, clearly A= {-1, 0, 1}.

Hence, the remaining elements of set A × A are as follows: (-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), and (1, 1) 

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