Class 11 Maths Chapter 2 Relations And Functions Exercise 2.1 Solutions in English Medium

Free NCERT Solutions for Class 11 Maths Chapter 2 Relations And Functions Exercise 2.1 prepared by expert Mathematics teacher as per CBSE (NCERT) books guidelines.

Free NCERT Solutions for Class 11 Math’s Chapter 2 Relations And Functions Exercise 2.1, Exercise 2.2, Exercise 2.3 and Miscellaneous Exercise in English Medium for CBSE.

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Class 11 Maths Chapter 2 Relations And Functions Exercise 2.1 Solutions in English Medium

EXERCISE : 2.1

Exercise 2.1 Class 11 Maths Question no 1

Solution :-

Since, the ordered pairs are equal the corresponding elements are equal.

Therefore,

Solving we yet x = 2 and y = 1.

Exercise 2.1 Class 11 Maths Question no 2

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 :-

Number of elements in set A = 3 and Number of elements in set B = 3

∴ Number of elements in A × B = 3 × 3 = 9

Exercise 2.1 Class 11 Maths Question no 3

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}

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

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

Exercise 2.1 Class 11 Maths Question no 4

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 ∩ ϕ) = ϕ

Solution:-

(i) Here P = {m, n} and Q = { n, m}

Number of elements in set P = 2 and Number of elements in set Q = 2

∵ Number of elements in P × Q = 2 × 2 = 4

But P × Q = {(m,n),(),}and here number of elements in P × Q = 2

Therefore, statement is false.

(ii) True

(iii) True

Exercise 2.1 Class 11 Maths Question no 5

5. If A = {–1, 1}, find A × A × A.

Solution :-

It is known that for any non-empty set A, A × A × A is defined as

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)}

Exercise 2.1 Class 11 Maths Question no 6

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

Solution :-

It is given that A × B = {(a, x), (a, y), (b, x), (b, y)}

We know that the Cartesian product of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q}

∴ 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}

Exercise 2.1 Class 11 Maths Question no 7

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)

We have

B ∩ C = {1, 2, 3, 4} ∩ {5,6} = ϕ

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) }

∴ R.H.S = (A × B) ∩ (A × C) = ϕ

∴ L.H.S = R.H.S

Hence, A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To 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)}

We can observe that all elements of set A × C are the elements of set B × D.

Therefore, A × C is a subset of B × D .

Exercise 2.1 Class 11 Maths Question no 8

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

Solution :-

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

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

⇒ n(A × B) = 4

We know that if C is a set with n(C) = m, then n[P(C)] = 2^m .

Therefore, the set A × B has 2⁴ = 16 subsets. These are 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)}

Exercise 2.1 Class 11 Maths Question no 9

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 :-

We are provided with the fact that n(A)=3 and n(B)=2 ; and(x,1),(y,2),(z,1)

are in A×B

We also know that, A

is the set of all the first elements and B

is the set of all the second elements.

So, we can conclude, A

having elements x, y, z and B having elements 1,2

Thus, we get, n(A)=3,n(B)=2

So, A={x, y, z},B={1,2}

Exercise 2.1 Class 11 Maths Question no 10

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 are provided with, n(A×A)=9

We also know that, if n(A)=a, n(B)=b, then n(A×B)=a, b

As it is given that, n(A×A)=9

It can be written as,

n(A)×n(A) =9

⇒ n(A) = 3

And it is also given that (−1,0), (0,1)

are the two elements of A×A

Again, the fact is also known that, A×A={(a, a): a ∈A}. And also −1,0,1 are the elements of A

Also, n(A)=3

⇒ A={−1,0,1}

So, (−1,−1),(−1,1),(0,−1),(0,0),(1,−1),(1,0),(1,1)

are the remaining elements of A×A .

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CLASS 11 English MEDIUM NCERT SOLUTIONS. CLASS 11 MATHEMATICS SOLUTIONS IN ENGLISH MEDIUM. CLASS 11 MATHEMATICS RELATIONS AND FUNCTIONS CHAPTER SOLUTIONS IN ENGLISH MEDIUM. CLASS 11 MATHS RELATIONS AND FUNCTIONS EXERCISE 2.1 SOLUTIONS IN ENGLISH MEDIUM.

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Class 11 Maths Chapter 2 Relations And Functions Exercise 2.1 Solutions in English Medium