For any two non-empty sets A and B-A - BC-AC- B is equal to

The correct option is B ϕAs we know that nbsp; (P∪ Q)^C=P^C∩ Q^C So, (A∪ B)^C∩(A^C∩ B) =A^C∩ B^C∩(B∩ A^C ) =A^C∩(( B^C∩ B)∩ A^C) =A^C ∩ (ϕ∩ A^C) =A^C ∩ϕ=ϕ ...

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Standard XII

Mathematics

De Morgan's Law

For any two n...

Question

For any two non-empty sets $A$ and $B$ , $(A \cup B)^{C} \cap (A^{C} \cap B)$ is equal to

A^{C}

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B^{C}

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ϕ

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A \cup B

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Solution

The correct option is C $ϕ$
As we know that
$(P \cup Q)^{C} = P^{C} \cap Q^{C}$
So, $(A \cup B)^{C} \cap (A^{C} \cap B)$
$= A^{C} \cap B^{C} \cap (B \cap A^{C}) = A^{C} \cap ((B^{C} \cap B) \cap A^{C}) = A^{C} \cap (ϕ \cap A^{C}) = A^{C} \cap ϕ = ϕ$

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For any two non-empty sets A and B, (A∪B)C∩(AC∩B) is equal to

The correct option is C ϕAs we know that (P∪Q)C=PC∩QC So, (A∪B)C∩(AC∩B) =AC∩BC∩(B∩AC)=AC∩((BC∩B)∩AC)=AC∩(ϕ∩AC)=AC∩ϕ=ϕ

For any two non-empty sets $A$ and $B$ , $(A \cup B)^{C} \cap (A^{C} \cap B)$ is equal to

The correct option is C $ϕ$
As we know that
$(P \cup Q)^{C} = P^{C} \cap Q^{C}$
So, $(A \cup B)^{C} \cap (A^{C} \cap B)$
$= A^{C} \cap B^{C} \cap (B \cap A^{C}) = A^{C} \cap ((B^{C} \cap B) \cap A^{C}) = A^{C} \cap (ϕ \cap A^{C}) = A^{C} \cap ϕ = ϕ$