If $A = \{1, 2, 3, 4, 5\}, B = \{2, 4, 6\}, C = \{3, 4, 6\},$ then $(A \cup B) \cap C$ is

Let $A = \{a, b, c\}$ and $B = \{1, 2, 3, 4\}$. Then the number of elements in the set $C = \{f : A \rightarrow B \mid 2 \in f(A) \text{ and } f \text{ is not one-one}\}$ is

Given $n(U) = 20$,$n(A) = 12$,$n(B) = 9$,$n(A \cap B) = 4$,where $U$ is the universal set,$A$ and $B$ are subsets of $U$,then $n((A \cup B)^C) = $

Range of the function $f(x) = \frac{1}{2 - \sin 3x}$ is

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 $A = \{x : f(x) = 0\}$ and $B = \{x : g(x) = 0\}$,then $A \cap B$ will be