NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1.
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 11 |
| Subject | Maths |
| Chapter | Chapter 2 |
| Chapter Name | Relations and Functions |
| Exercise | Ex 2.1 |
| Number of Questions Solved | 10 |
| Category | NCERT Solutions |
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1
Ex 2.1 Class 11 Maths Question 1.
If \(\left( \frac { x }{ 3 } +1,y-\frac { 2 }{ 3 } \right) =\left( \frac { 5 }{ 3 } ,\frac { 1 }{ 3 } \right) \), find the values of x and y.
Solution.
Since the ordered pairs are equal. So, the corresponding elements are equal
∴ \(\frac { x }{ 3 } +1=\frac { 5 }{ 3 } \) and \(y-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
⇒ \(\frac { x }{ 3 } =\frac { 5 }{ 3 } -1\) and \(y=\frac { 1 }{ 3 } +\frac { 2 }{ 3 } \) ⇒ x = 2 and y = 1.
Ex 2.1 Class 11 Maths Question 2.
If the set A has 3 elements and the set B {3, 4, 5}, then find the number of elements in (A x B).
Solution.
According to question, n(A) = 3 and n(B) = 3.
∴ n(A x B) = n(A) x n(B) = 3 x 3 = 9
∴ There are total 9 elements in (A x B).
Ex 2.1 Class 11 Maths Question 3.
If G = {7, 8} and H = {5, 4, 2}, find G x H and H x G.
Solution.
We have G = {7, 8} and H = {5, 4, 2} Then, by the definition of the cartesian product, we have
G x H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H x G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}.
Ex 2.1 Class 11 Maths 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 x Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then Ax 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 x (B ∩φ) = φ
Solution.
(i) False, if P = {m, n} and Q = {n, m}
Then P x Q = {(m, n), (m, m), (n, n), (n, m)}.
(ii) True, by the definition of cartesian product.
(iii) True, We have A = {1, 2} and B = {3, 4}
Now, B ∩ φ = φ ∴ A x (B ∩ φ) = A x φ = φ.
Ex 2.1 Class 11 Maths Question 5.
If A = {-1, 1},find A x A x A.
Solution.
A = {-1, 1}
Then, A x A = {-1, 1} x {-1, 1} = {(-1, -1), (-1,1),(1,-1), (1,1)}
A x A x A = ((-1,-1),(-1,1),(1,-1),(1,1)} x {-1,1}
= {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1,1), (1, -1, -1), (1, -1,1), (1,1,-1), (1,1,1)}
Ex 2.1 Class 11 Maths Question 6.
If A x 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)}
If {p, q) ∈ A x B, then p ∈ A and q ∈ B
∴ A = {a, b} and B = {x, y}.
Ex 2.1 Class 11 Maths Question 7.
Let A = {1, 2}, B = (1, 2, 3, 4), C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A x (B ∩ C = (A x B) ∩ (AxC)
(ii) A x C is a subset of B x D.
Solution.
Given, A = {1, 2}, B ={1, 2, 3, 4}, C = {5, 6}, D = (5, 6, 7, 8}
Ex 2.1 Class 11 Maths Question 8.
Let A = {1, 2} and B = {3, 4}. Write 4 x B. How many subsets will 4 x B have? List them.
Solution.
Given, A = {1, 2} and B = {3, 4}
Then, A x B = {(1, 3), (1,4), (2, 3), (2, 4)}
i. e., A x B has 4 elements. So, it has 24 i.e. 16 subsets.
The subsets of A x B are as follows :
φ, {(1, 3)1, ((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, 4), (2, 3), (2,4)},{(1, 3), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2,4)}.
Ex 2.1 Class 11 Maths 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 x B, find A and B, where x, y and z are distinct elements.
Solution.
Given, n(A) = 3 and n(B) = 2
Now (x, 1) ∈ A x B ⇒ x ∈ A and 1 ∈ B,
(y, 2) ∈ A x B ⇒ y ∈ A and 2 ∈ B
(z, 1) ∈ A x B ⇒z ∈ A and 1 ∈ B
∴ x, y, z ∈ A and 1, 2 ∈ B
Hence, A = {x, y, z} and B = {1, 2}.
Ex 2.1 Class 11 Maths Question 10.
The Cartesian product 4×4 has 9 elements among which are found (-1, 0) and (0, 1). Find the set 4 and the remaining elements of 4 x 4.
Solution.
Since, we have n(A x A) = 9
⇒ n(A) x n(A) = 9 [ ∵ n (A x B) = n(A) x n(B)]
⇒ (n(A))2 = 9 ⇒ n(A) = 3
Also, given (-1, 0) ∈ A x A ⇒ -1, 0 ∈ A ,
and (0,1) ∈ A x A ⇒ 0, 1 ∈ A
∴ -1, 0,1 ∈ A
Hence, A = {-1, 0, 1} (∵ n(A) = 3)
and the remaining elements of A x A are (-1, -1), (-1,1), (0, -1), (0,0), (1, -1), (1,0), (1,1).
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