For any two sets A and B- if A - B - A - B - thenA- A - BB- B - AC- A - BD- A - B

The correct option is C A=BGiven that nbsp;(A∪ B)'=A'∪ B' By De Morgan's law, (A∪ B)'=A'∩ B' So, A' ∩ B'=A'∪ B' This is possible only if A=B (∵ P ∪ Q= P ∩ Q ...

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Byju's Answer

Standard XII

Mathematics

De Morgan's Law

For any two s...

Question

For any two sets $A$ and $B,$ if $(A \cup B)^{'} = A^{'} \cup B^{'},$ then

A \subseteq B

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B \subseteq A

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A^{'} = B

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A = B

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Solution

The correct option is D $A = B$
Given that $(A \cup B)^{'} = A^{'} \cup B^{'}$
By De Morgan's law,
$(A \cup B)^{'} = A^{'} \cap B^{'}$
So, $A^{'} \cap B^{'} = A^{'} \cup B^{'}$
This is possible only if $A = B$ $(∵ P \cup Q = P \cap Q \Rightarrow P = Q)$

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For any two sets A and B, if (A∪B)′=A′∪B′, then

The correct option is D A=BGiven that (A∪B)′=A′∪B′ By De Morgan's law, (A∪B)′=A′∩B′ So, A′∩B′=A′∪B′ This is possible only if A=B (∵P∪Q=P∩Q⇒P=Q)

For any two sets $A$ and $B,$ if $(A \cup B)^{'} = A^{'} \cup B^{'},$ then

The correct option is D $A = B$
Given that $(A \cup B)^{'} = A^{'} \cup B^{'}$
By De Morgan's law,
$(A \cup B)^{'} = A^{'} \cap B^{'}$
So, $A^{'} \cap B^{'} = A^{'} \cup B^{'}$
This is possible only if $A = B$ $(∵ P \cup Q = P \cap Q \Rightarrow P = Q)$