Question

If $\mathit{A}\mathbf{=}\mathbf{\{}\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}\mathbf{\}}$, $\mathit{B}\mathbf{=}\mathbf{\{}\mathit{b}\mathbf{,}\mathit{c}\mathbf{,}\mathit{d}\mathbf{\}}$ and$\mathit{C}\mathbf{=}\mathbf{\{}\mathit{a}\mathbf{,}\mathit{d}\mathbf{,}\mathit{c}\mathbf{\}}$ then $\mathbf{(}\mathit{A}\mathbf{-}\mathit{B}\mathbf{)}\mathbf{}\mathbf{\times }\mathbf{(}\mathit{B}\mathbf{\cap }\mathit{C}\mathbf{)}$ is equal to

• A. $\mathbf{\left\{}\mathbf{\right(}\mathit{a}\mathbf{,}\mathit{c}\mathbf{)}\mathbf{,}\mathbf{(}\mathit{a}\mathbf{,}\mathit{d}\mathbf{\left)}\mathbf{\right\}}\mathbf{}$
• B. $\mathbf{\left\{}\mathbf{\right(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}\mathbf{,}\mathbf{(}\mathit{c}\mathbf{,}\mathit{d}\mathbf{\left)}\mathbf{\right\}}$
• C. $\mathbf{\left\{}\mathbf{\right(}\mathit{c}\mathbf{,}\mathit{a}\mathbf{)}\mathbf{,}\mathbf{(}\mathit{d}\mathbf{,}\mathit{a}\mathbf{\left)}\mathbf{\right\}}$
• D. $\mathbf{\left\{}\mathbf{\right(}\mathit{a}\mathbf{,}\mathit{c}\mathbf{)}\mathbf{,}\mathbf{(}\mathit{a}\mathbf{,}\mathit{d}\mathbf{)}\mathbf{,}\mathbf{(}\mathit{b}\mathbf{,}\mathit{d}\mathbf{\left)}\mathbf{\right\}}$

Answer 1

The correct option is A

$\mathbf{\left\{}\mathbf{\right(}\mathit{a}\mathbf{,}\mathit{c}\mathbf{)}\mathbf{,}\mathbf{(}\mathit{a}\mathbf{,}\mathit{d}\mathbf{\left)}\mathbf{\right\}}\mathbf{}$

Explanation for correct option:

$$\begin{gathered} \mathit{A}\mathbf{=}\mathbf{\{}\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}\mathbf{\}} \\ \mathit{B}\mathbf{=}\mathbf{\{}\mathit{b}\mathbf{,}\mathit{c}\mathbf{,}\mathit{d}\mathbf{\}} \\ \mathit{C}\mathbf{=}\mathbf{\{}\mathit{a}\mathbf{,}\mathit{d}\mathbf{,}\mathit{c}\mathbf{\}} \end{gathered}$$

To find $\mathbf{(}\mathit{A}\mathbf{-}\mathit{B}\mathbf{)}\mathbf{\times }\mathbf{(}\mathit{B}\mathbf{\cap }\mathit{C}\mathbf{)}$

$A-B=\{a,b,c\}-\{b,c,d\}=\mathbf{\left\{}\mathit{a}\mathbf{\right\}}$

$B\cap C=\{b,c,d\}\cap \{a,d,c\}=\mathbf{\{}\mathit{c}\mathbf{,}\mathit{d}\mathbf{\}}$

So,

$\mathbf{(}\mathit{A}\mathbf{-}\mathit{B}\mathbf{)}\mathbf{\times }\mathbf{(}\mathit{B}\mathbf{\cap }\mathit{C})\mathbf{=}\mathbf{\left\{}\mathit{a}\mathbf{\right\}}\mathbf{\times }\mathbf{\{}\mathit{c}\mathbf{,}\mathit{d}\mathbf{\}}$

$\mathbf{=}\mathbf{\left\{}\mathbf{\right(}\mathbf{a}\mathbf{,}\mathbf{c}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{a}\mathbf{,}\mathbf{d}\mathbf{)}\mathbf{\}}$

Hence, the correct option is (A).