If $f(x) = \frac{1+x}{1-x}$ where $x \neq 1$,then $f(x) \cdot f(y) = $ . . . . . . .

If $f(x)=3x+6$,$g(x)=4x+k$ and $f \circ g(x)=g \circ f(x)$,then $k =$

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x)=\begin{cases} \log _e x & , x>0 \\ e^{-x} & , x \leq 0 \end{cases}$ and $g(x)=\begin{cases} x & , x \geq 0 \\ e^{x} & , x < 0 \end{cases}$. Then $gof: R \to R$ is . . . .

Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on the set $A = \{1, 2, 3\}$. Then $R \circ S = $

$[x]$ represents the greatest integer function. Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -3, & x < 0 \\ 0, & x = 0 \\ 5, & x > 0 \end{cases}$. Then $f(g(x))$ is:

Consider functions $f: A \rightarrow B$ and $g: B \rightarrow C$ $(A, B, C \subseteq \mathbb{R})$ such that $(g \circ f)^{-1}$ exists. Then: