NCERT Solutions Class 11 Maths
The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 2 Relations and Functions Exercise 2.1 has been provided here to help the students in solving the questions from this exercise.
Chapter 2 (Relations and Functions)
| Exercise – 2.1 |
1. If 
, 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 
, so the corresponding elements should also be equal.
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
2. If set A has 3 elements and 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, the number of elements in (A × B) will be 9.
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 × 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)}
4. State whether each of the following statements is 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)}
Solution – The given statement is False.
The correct statement is: If P = {m, n} and Q = {n, m}, then P × Q = { (m, m), n), (n, m), (n, n) }
(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.
Solution – The given statement is true.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ
Solution – The given statement is true.
5. If A = {–1, 1}, find A × A × A.
Solution – The 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)}
6. If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Solution – Given,
A × 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}
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 × 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)}
Since, all the elements of set A x C are the elements of set B × D
Thus, A × C is a subset of B × D
Hence, verified.
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 is n(A × B) = 4
We know that,
If C is a set with n(C) = m, then n[P(C)] = 2m.
Thus, the set A × B has 24 = 16 subsets.
And these subsets are as given below:
∅, {(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 – Given, 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
So, clearly, x, y, and z are the elements of A; and
1 and 2 are the elements of B.
As n(A) = 3 and n(B) = 2, it is clear that set A = {x, y, z} and set 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 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)
