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NCERT Exemplar · Q41

Q.If A={1,3,5,7,9,11,13,15,17}A = \{1, 3, 5, 7, 9, 11, 13, 15, 17\}, B={2,4,…,18}B = \{2, 4, \ldots, 18\} and N\mathbb{N} the set of natural numbers is the universal set, then A′∪((A∪B)∩B′)A' \cup ((A \cup B) \cap B') is
(A) ϕ\phi
(B) N\mathbb{N}
(C) AA
(D) BB

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Simplifying A′∪((A∪B)∩B′)A' \cup ((A \cup B) \cap B') gives A′∪A=NA' \cup A = \mathbb{N}, the universal set — option (B).

Here A={1,3,5,…,17}A = \{1,3,5,\dots,17\} (odd numbers) and B={2,4,…,18}B = \{2,4,\dots,18\} (even numbers), so AA and BB are disjoint, and the universal set is N\mathbb{N}.

Simplify the inner bracket using the distributive law:

(A∪B)∩B′=(A∩B′)∪(B∩B′)=(A∩B′)∪ϕ=A∩B′(A \cup B) \cap B' = (A \cap B') \cup (B \cap B') = (A \cap B') \cup \phi = A \cap B' …

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