If the sets $A$ and $B$ are defined as $A = \{ (x,\,y):y = {1 \over x},\,0 \ne x \in R\} $ $B = \{ (x,y):y = - x,x \in R\} $, then
$A \cap B = A$
$A \cap B = B$
$A \cap B = \phi $
None of these
If ${N_a} = \{ an:n \in N\} ,$ then ${N_3} \cap {N_4} = $
If $n(A) = 3$, $n(B) = 6$ and $A \subseteq B$. Then the number of elements in $A \cup B$ is equal to
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 $A, B, C$ be three sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then
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
$B-C$