NCERT Class XI Mathematics - Sets - Solutions
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Question : 46
Total: 73
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that
(i) (A ∪ B)′ = A′ ∩ B′ (ii) (A ∩ B)′ = A′ ∪ B′
(i) (A ∪ B)′ = A′ ∩ B′ (ii) (A ∩ B)′ = A′ ∪ B′
Solution:
Here U = {1, 2, 3, 4, 5, 6, 7, 8, 9},
A = {2, 4, 6, 8} and B = {2, 3, 5, 7}
(i) A ∪ B = {2, 4, 6, 8} ∪ {2, 3, 5, 7} = {2, 3, 4, 5, 6, 7, 8}
∴ (A ∪ B)′ = U – (A ∪ B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 4, 5, 6, 7, 8} = {1, 9} ...(1)
A′ = U – A = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8} = {1, 3, 5, 7, 9}
B′ = U – B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 5, 7} = {1, 4, 6, 8, 9}
A′ ∩ B′ = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9} ...(2)
From (1) and (2), we have (A ∪ B)′ = A′ ∩ B′
(ii) A ∩ B = {2, 4, 6, 8} ∩ {2, 3, 5, 7} = {2} (A ∩ B)′ = U – (A ∩ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2} = {1, 3, 4, 5, 6, 7, 8, 9} ...(1)
A′ = U – A = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8} = {1, 3, 5, 7, 9}
B′ = U – B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 5, 7} = {1, 4, 6, 8, 9}
A′ ∪ B′ = {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9} ... (2)
From (1) and (2), we have (A ∩ B)′ = A′ ∪ B′
A = {2, 4, 6, 8} and B = {2, 3, 5, 7}
(i) A ∪ B = {2, 4, 6, 8} ∪ {2, 3, 5, 7} = {2, 3, 4, 5, 6, 7, 8}
∴ (A ∪ B)′ = U – (A ∪ B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 4, 5, 6, 7, 8} = {1, 9} ...(1)
A′ = U – A = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8} = {1, 3, 5, 7, 9}
B′ = U – B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 5, 7} = {1, 4, 6, 8, 9}
A′ ∩ B′ = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9} ...(2)
From (1) and (2), we have (A ∪ B)′ = A′ ∩ B′
(ii) A ∩ B = {2, 4, 6, 8} ∩ {2, 3, 5, 7} = {2} (A ∩ B)′ = U – (A ∩ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2} = {1, 3, 4, 5, 6, 7, 8, 9} ...(1)
A′ = U – A = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8} = {1, 3, 5, 7, 9}
B′ = U – B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 5, 7} = {1, 4, 6, 8, 9}
A′ ∪ B′ = {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9} ... (2)
From (1) and (2), we have (A ∩ B)′ = A′ ∪ B′
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