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Home » NCERT Solutions » Class 11 » Maths » NCERT Solutions for Exercise 2.1, Class 11, Maths

NCERT Solutions for Exercise 2.1, Class 11, Maths

Last Updated on July 11, 2023 By Mrs Shilpi Nagpal

Exercise 2.1 Relations and Functions
Question and Answers
Class 11 – Maths

Class Class 11
Subject Mathematics
Chapter Name Relations and Functions
Chapter No. Chapter 2
Exercise Exercise 2.1
Category Class 11 Maths NCERT Solutions

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

Answer As the ordered pairs are equal, the corresponding elements should also be equal.

Thus, x/3 + 1 = 5/3 and y – 2/3 = 1/3

Solving, we get

x + 3 = 5 and 3y – 2 = 1 [Taking L.C.M. and adding]

x = 2 and 3y = 3

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).

Answer The number of elements in (A×B) = Number of elements in A × Number of elements in B

The number of elements in (A×B) = 3 × 3=9

So, 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.

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

The Cartesian product of two non-empty sets A and B is defined as

A × B = {(a, b) :a ∈ A and b ∈ B

So, the value of G×H will be, 

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

The value of H×G will be,

H × 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)}.

(i) The statement is False.

If P = {m, n} and Q = {n, m}, then

P × Q = {(m, 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.

(ii) True

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.

(iii) True

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

Answer The A × A × A for a non-empty set A is given by

A × A × A = {(a, b, c): a, b, c ∈ A}

Here, it is given 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)}

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

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

The Cartesian product of two non-empty sets A and B is defined as

A × B = {(a ,b) : a ∈ A and b ∈ B}

Hence, 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).

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

Verify : A × (B ∩ C) = (A × B) ∩ (A × C)

Now, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

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

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)}

Thus, R.H.S. = (A × B) ∩ (A × C) = Φ

Therefore, L.H.S. = R.H.S.

Hence verified

(ii) A × C is a subset of B × D.

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

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)}

Now, it’s clearly seen that all the elements of set A × C are the elements of set B × D.

Thus, A × 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.

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

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:

Φ, {(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.

Answer We are given that n(A) = 3 and n(B) = 2 

(x, 1), (y, 2), (z, 1) are in A × B

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}

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.

Answer We are provided with, n ( A × A ) = 9

If n(A) = p and n(B) = q, then n(A × B) = pq.

Also, n(A × A) = n(A) × n(A)

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 in A × A = {(a, a): a ∈ A}

Thus, –1, 0, and 1 have to 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)

Filed Under: Class 11, Maths, NCERT Solutions Tagged With: ex 2.1 class 11, ex 2.1 class 11 maths, Exercise 2.1 Class 11, exercise 2.1 class 11 maths

About Mrs Shilpi Nagpal

Author of this website, Mrs. Shilpi Nagpal is MSc (Hons, Chemistry) and BSc (Hons, Chemistry) from Delhi University, B.Ed. (I. P. University) and has many years of experience in teaching. She has started this educational website with the mindset of spreading free education to everyone.

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