# Class 11, Relations and Functions | NCERT Solutions 11 Maths Exercise 2.1 | Class XI CBSE Maths Guide

## CBSE Mathematics Guide - Class XI NCERT Mathematics Solutions

### Chapter 2, relations and functions

#### Cbse Ncert Solutions Class 11 Maths Exercise 2.1

(Cbse Class 11 Ncert Math Textbook Page 33, 34)
Question 1:

Solution:

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:  It is given that set A has 3 elements while the elements of set B are 3, 4, and 5.
i.e., 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
Hence, 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.
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)}, and
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.
Solution:
(i)  False.
Since, 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.
Solution:
For any non-empty set A,
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)}

#### CBSE Class 11 NCERT Mathematics Solutions Chapter 2, Relations and Functions Exercise 2.1

Question 6: If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Solution: Given that A × B = {(a, x), (a, y), (b, x), (b, y)}
For two non-empty sets P and Q is defined as P × Q = {(p, q): p P, q Q}
Since, 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
Solution:
(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
From the given elements we have, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
L.H.S = A × (B ∩ C) = A × Φ = Φ
From the given elements we have,
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
Thus, L.H.S = R.H.S verified.
(ii) To verify: A × C is a subset of B × D
From the given elements we can write,
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)}
We can observe that all the elements of set A × C are also present in the elements of set B × D.
Therefore, A × C is a subset of B × D, verified.

Question 8: Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.
(Hint: If C is a set with n(C) = m, then n[P(C)] = 2m. Try yourself)

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 that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.
A and B are Sets of the first and second elements respectively of the ordered pair elements of A × B
As x, y, and z are the elements of A; and 1 and 2 are the elements of B, so
n(A) = 3 and n(B) = 2.
Therefore, it is clear that A = {x, y, z} and 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 n(A) = p and n(B) = q, then n(A × B) = pq.
So, n(A × A) = n(A) × n(A)
Given that the Cartesian product n(A × A) has 9 elements so,
n(A × A) = 9
Or, n(A) × n(A) = 9
Or, 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}. Hence –1, 0, and 1 are elements of A.

Since n(A) = 3, then A = {–1, 0, 1} and the remaining elements of set A × A are

(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1).