Question

If $A$ and $B$ are two sets, then $(A\cap B)\text{'}$ is equal to

• A. $A\text{'}\cap B\text{'}$
• B. $A\text{'}UB\text{'}$
• C. $A\cap B$
• D. $AUB$

Answer 1

The correct option is B

$A\text{'}UB\text{'}$

Explanation for the correct option:

Find the value of $(A\cap B)\text{'}$:

As $\left(A\cup B\right)=$ either in $A$or in $B$

$\Rightarrow$$\left(A\cup B\right)\text{'}=$ neither in $A$ nor in $B$

Also, $A\text{'}=$ Not in $A$

$B\text{'}=$ Not in $B$

Now, $A\text{'}\cap B\text{'}=$ Not in $A$ and not in $B$

$$\begin{gathered} LetP=(AUB)\text{'}andQ=A\text{'}\cap B\text{'} \\ LetxbeanarbitraryelementofPthenx\in P \\ \Rightarrow x\in (AUB)\text{'} \\ \Rightarrow x\notin (AUB) \\ \Rightarrow x\notin Aandx\notin B \\ \Rightarrow x\in A\text{'}andx\in B\text{'} \\ \Rightarrow x\in A\text{'}\cap B\text{'} \\ \Rightarrow x\in Q \\ Therefore,P\subset Q\dots \dots \dots \dots \dots ..\left(i\right) \\ Again,letybeanarbitraryelementofQtheny\in Q \\ \Rightarrow y\in A\text{'}\cap B\text{'}\Rightarrow \\ y\in A\text{'}andy\in B\text{'} \\ \Rightarrow y\notin Aandy\notin B \\ \Rightarrow y\notin (AUB) \\ \Rightarrow y\in (AUB)\text{'} \\ \Rightarrow y\in P \\ Therefore,Q\subset P\dots \dots \dots \dots \dots ..\left(ii\right) \\ Nowcombine\left(i\right)and\left(ii\right)weget;P=Qi.e.(AUB)\text{'}=A\text{'}\cap B\text{'} \end{gathered}$$

$\mathbf{\therefore }\mathbf{(}\mathit{A}\mathbf{}\mathbf{\cap }\mathbf{}\mathit{B}\mathbf{)}\mathbf{\text{'}}\mathbf{}\mathbf{=}\mathbf{}\mathit{A}\mathbf{\text{'}}\mathbf{}\mathbf{\cup }\mathbf{}\mathit{B}\mathbf{\text{'}}$

Hence, Option ‘B’ is Correct.