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# Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }. Find

(i) A′ (ii) B′ (iii) (A ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′ (vi) (B – C)′

**Solution:**

The given sets are as follows:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9},

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

B = {2, 4, 6, 8} and C = {3, 4, 5, 6}

The complement of a set A is U - A, where U is the universal set.

Thus,

**(i)** A' = U - A = {5, 6, 7, 8, 9}

**(ii)** B' = U - B = {1, 3, 5, 7, 9}

**(iii)** A υ C = {1, 2, 3, 4, 5, 6}

Therefore,

(A υ C)' = U - (A υ C) = {7, 8, 9}

**(iv)** A υ B = {1, 2, 3, 4, 6, 8}

Therefore,

(A υ B)' = U - (A υ B)

= {5, 7, 9}

(v) (A')' = A = {1, 2, 3, 4}

(vi) The difference between two sets B and C is a set denoted by B - C and is obtained by writing the elements of B that are NOT in C in a set.

Thus,

B - C = {2, 8}

Therefore,

(B - C)' = U - (B - C) = {1, 3, 4, 5, 6, 7, 9}

NCERT Solutions Class 11 Maths Chapter 1 Exercise 1.5 Question 1

## Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }. Find (i) A′ (ii) B′ (iii) (A ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′ (vi) (B – C)′

**Summary:**

It is given that U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}. We have found that

(i) A' = {5, 6, 7, 8, 9}

(ii) B' = {1, 3, 5, 7, 9}

(iii) (A υ C)' = {7, 8, 9}

(iv) (A υ C)' = {5, 7, 9}

(v) (A)' = A = {1, 2, 3, 4}

(vi) (B - C)' = {1, 3, 4, 5, 6, 7, 9}

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