Chapter 2 of the Class 11 NCERT Mathematics textbook, titled "Relations and Functions," introduces students to the fundamental concepts of relations and functions in mathematics. This chapter explores different types of relations and functions, their properties, and how to represent them. Exercise 2.1 focuses on basic problems related to defining and understanding relations and functions, helping students build a solid foundation in these concepts.
NCERT Solutions for Class 11 - Chapter 2 Relations and Functions - Exercise 2.1
This section provides detailed solutions for Exercise 2.1 from Chapter 2 of the Class 11 NCERT Mathematics textbook. The exercise includes problems in identifying and representing relations and functions, determining the domain and range, and working with function notation. The solutions are presented step-by-step to help students understand the methods for solving these problems and applying the concepts of relations and functions effectively.
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)] = 2m.
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)
Summary
Chapter 2 of the Class 11 NCERT Mathematics textbook, "Relations and Functions," introduces the concepts of relations and functions. Exercise 2.1 focuses on basic problems related to defining, identifying, and representing relations and functions, including determining their domains and ranges. This exercise helps students understand how to work with functions, including the distinction between general relations and functions where each input is uniquely paired with an output.