If $A$ and $B$ are two sets then $(A -B) \cup (B -A) \cup (A \cap B)$ is equal to
$A \cup B$
$A \cap B$
$A$
$B'$
(a) From Venn-Euler's diagram,
$\therefore (A – B)\, \cup (B – A) \cup (A \cap B) = A \cup B$.
Let $P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\}$ and $Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\}$ be two sets. Then
If ${N_a} = \{ an:n \in N\} ,$ then ${N_3} \cap {N_4} = $
If $A=\{3,6,9,12,15,18,21\}, B=\{4,8,12,16,20\},$ $C=\{2,4,6,8,10,12,14,16\}, D=\{5,10,15,20\} ;$ find
$A-B$
If $A$ and $B$ are disjoint, then $n(A \cup B)$ is equal to
Show that $A \cap B=A \cap C$ need not imply $B = C$
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