Question

Let $A$ and $B$ be two sets then ${(A\cup B)}^{\prime }\cup (A^{\prime }\cap B)$ is equal to

• A. $A^{\prime }$
• B. $A$
• C. $B^{\prime }$
• D. $A\cap A^{\prime }$

Answer 1

The correct option is A $A^{\prime }$

Let $S=(A\cup B{)}^{\prime }\cup (A^{\prime }\cap B)$
According to De Morgan' Law, we have
$S=(A^{\prime }\cap B^{\prime })\cup (A^{\prime }\cap B)$
$\Rightarrow S=(A^{\prime }\cup A^{\prime })\cap (A^{\prime }\cup B)\cap (B^{\prime }\cup A^{\prime })\cap (B^{\prime }\cup B)$
$\Rightarrow S=A^{\prime }\cap \{A^{\prime }\cup (B\cap B^{\prime }\left)\right\}\cap \bigcup$
$\Rightarrow S=A^{\prime }\cap (A^{\prime }\cup \varphi )\cap \bigcup$
$S=A^{\prime }\cap (A^{\prime }\cup \varphi )\cap \bigcup$
$\therefore S=A^{\prime }\cap A^{\prime }\cap \bigcup =A^{\prime }\cap \bigcup =A^{\prime }$