[B' ∪ (B' A)]' = [B' ∪ (B' ∩A')]' (∵ A B= A ∩B') =[(B' ∪B') ∩ (B' ∪A'))]' = [B' ∩ (A ∩ B)']' = (B')' ∪ ((A ∩ B)')' (∵(A ∩ B)' = A' ∪B') ...

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Byju's Answer

Standard XII

Mathematics

Complement

[B' ∪ B'-A]' ...

Question

${[(B^{'} \cup (B^{'} - A))]}^{'}$ =___

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A - B

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B - A

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Solution

The correct option is B
B

${[B^{'} \cup (B^{'} - A)]}^{'} = {[B^{'} \cup (B^{'} \cap A^{'})]}^{'} (∵ A - B = A \cap B^{'}) = {[(B^{'} \cup B^{'}) \cap (B^{'} \cup A^{'}))]}^{'} = {[B^{'} \cap (A \cap B)^{'}]}^{'} = {(B^{'})}^{'} \cup ((A \cap B)^{'})^{'} (∵ {(A \cap B)}^{'} = A^{'} \cup B^{'}) = B \cup (B \cap A) = B$

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[(B′∪(B′−A))]′ =___

The correct option is B B [B′∪(B′−A)]′=[B′∪(B′∩A′)]′ (∵A−B=A∩B′)=[(B′∪B′)∩(B′∪A′))]′=[B′∩(A∩B)′]′=(B′)′∪((A∩B)′)′ (∵(A∩B)′=A′∪B′)=B∪(B∩A)=B

${[(B^{'} \cup (B^{'} - A))]}^{'}$ =___