Let $S = \{1, 2, 3, 4, 5, 6, 7\}$. Then the number of possible functions $f: S \rightarrow S$ such that $f(m \cdot n) = f(m) \cdot f(n)$ for every $m, n \in S$ and $m \cdot n \in S$ is equal to $......$

Consider the following statements:
$(i)$ $A$ relation is a special case of a function.
$(ii)$ $A$ function is a special case of a relation.
$(iii)$ Both relation and function are the same.

If the number of elements in the sets $G$ and $A$ are $3$ and $4$ respectively,then match the items of List-$I$ with those of List-$II$.

List-$I$	List-$II$
$A$. The number of non-bijective functions from $G \times G$ to $G$	$I$. $24$
$B$. The number of bijective functions from $A$ to $A$	$II$. $0$
$C$. The number of functions from $G$ to $G \times A$	$III$. $1728$
$D$. The number of surjective functions from $A$ to $A \times A$	$IV$. $12$
	$V$. $19683$

If $f(x)=2\{x\}+5x$,where $\{x\}$ is the fractional part function,then $f(-1.4)$ is

Let $f:(-1,1) \rightarrow \mathbb{R}$ be such that $f(\cos 4 \theta) = \frac{2}{2-\sec^2 \theta}$ for $\theta \in \left(0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then the value$(s)$ of $f\left(\frac{1}{3}\right)$ is (are):

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