## NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Exercise 2.1

Chapter 2 Relations and Functions Exercise 2.1 NCERT Solutions for Class 11 Maths is very helpful in improving problem solving skills and increase your efficiency. These will useful if you want to score better marks in your examination. These Class 11 Maths NCERT Solutions are prepared by Studyrankers experts that are detailed so you can frame your own answers for your notebook.

1. If (x/3 + 1, y â€“ 2/3) = (5/3 , 1/3), find the values of x and y.

Answer

Here, (x/3 + 1, y â€“ 2/3) = (5/3, 1/3)

x/3 + 1 = 5/3 and y â€“ 2/3 = 1/3

x/3 = 5/3 â€“ 1 and y = 1/3 + 2/3

x/3 = (5 â€“ 3)/3 and y = (1 + 2)/3

x/3 = 2/3 and y = 1

x = 2, y = 1

hence, x = 2, y = 1

2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of element in (A Ã— B)

Answer

We have n(A) = 3, n(B) = 3

Thus n(AÃ—B) = n(A). n(B) = 3.3 = 9

Hence, number of elements in A Ã— B is 9

3. If G = {7, 8}, H = {5, 4, 2}, find G Ã— H and H Ã— G.

Answer

We have

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

And H Ã— G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}.

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 Ã— Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A Ã— B is a non-empty set of ordered pairs (x, y) such that x âˆˆ B and y âˆˆ A.
(iii) If A = {1, 2}, B = {3, 4}, then A Ã— (B âˆ© Ð¤) =Ð¤.

Answer

(i) False. Since, n (P) = 2, n (Q) = 2, therefore n (P Ã— Q) = 4. Hence, P Ã— Q = {(m, n), (m, m), (n, n), (n, m)}. Thus, the correct statement is if P = {m, n} and Q = {n, m}, then P Ã— Q = {(m, n), (m, m), (n, n), (n, m)}.

(ii) False. The Correct statement is if A and B are non-empty sets, then A Ã— B is nonempty set of ordered pairs (x, y) such that x âˆˆ A and y âˆˆ B.

(iii) True.

5. If A = {-1, 1}, find A Ã— A Ã— A.

Answer

We have, A Ã— A Ã— A = {(â€“1, â€“1, â€“1), (â€“1, â€“1, 1), (â€“ 1, 1, 1), (1, â€“ 1, 1), (1, 1, â€“ 1), (1, â€“ 1, â€“1), (â€“1, 1, â€“1), (1, 1, 1)}.

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

Answer

A = set of first distinct elements = {a, b}

B = set of second distinct elements = {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 Ã— (B âˆ© C) = (A Ã— B) âˆ© (A Ã— C).

(ii) A Ã— C is a a subset of B Ã— D.

Answer

Here, A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
(i) B âˆ© C = {1, 2, 3, 4} âˆ© {5, 6} = Ð¤
A Ã— (Bâˆ© C) = A Ã— Ð¤ = Ð¤ ...(i)
Now, A Ã— B = {1, 2} Ã— {1, 2, 3, 4}
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A Ã— C = {1, 2} Ã— {5, 6}
= {(1, 5), (1, 6), (2, 5), (2, 6)}
Hence (A Ã— B) âˆ© (A Ã— C) = Ð¤ ...(ii)
From (i) and (ii), we have
A Ã— (B âˆ© C) = (A Ã— B) âˆ© (A Ã— C).

(ii) A Ã— C = {1, 2} Ã— {5, 6}
= {(1, 5), (1, 6), (2, 5), (2, 6)}
B Ã— D = {1, 2, 3, 4} Ã— {5, 6, 7, 8}
= {(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)}.
Clearly, A Ã— C âŠ† B Ã— D
because every element of set (A Ã— C) is in set (B Ã— D).

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

Answer

Here, A = {1, 2}, B = {3, 4}
A Ã— B = (1, 2) Ã— (3, 4)
= {(1, 3), (1, 4), (2, 3), (2, 4)}
No. of subset of a set having n element = 2n
Hence, No. of subset of A Ã— B = 24 = 16

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 Ã— B, find A and B where x, y and z are distinct elements.

Answer

Given n (A) = 3, n (B) = 2
Therefore, n(A Ã— B) = n (A). n (B) = 3 Ã— 2 = 6
If (x, 1), (y, 2), (z, 1) are in A Ã— B then
A = {x, y, z} and B = {1, 2}.

10. The cartesian product A Ã— A has 9 elements among which are found (â€“ 1, 0) and (0, 1). Find the set A and the remaining elements of A Ã— A.

Answer

Clearly, set A has 3 elements since (â€“ 1, 0) âˆˆ A Ã— A
and (0, 1) âˆˆ A Ã— A. Therefore, (â€“ 1, 0) âˆˆ A Ã— A
â‡’ â€“ 1, 0 âˆˆ A and (0, 1) âˆˆ A Ã— A â‡’ 0, 1 âˆˆA.
Therefore, â€“ 1, 0, 1 âˆˆ A.
Hence, A = {â€“ 1, 0, 1}
Thus remaining elements of
A Ã— A = (â€“ 1, â€“ 1), (â€“ 1, 1), (0, â€“1), (0, 0), (1, â€“ 1), (1, 0), (1, 1).
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