Suppose that $g(x) = 1 + \sqrt{x}$ and $f(g(x)) = 3 + 2\sqrt{x} + x$,then $f(x)$ is

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions defined by $f(x)=2x-3$ and $g(x)=x^{3}+5$,then $(fog)^{-1}(x) = $

$[x]$ 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:

If $g(x)=1+\sqrt{x}$ and $f(g(x))=3+2 \sqrt{x}+x$,then $f(f(x))$ is

Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ and the composed mapping $g \circ f: S \rightarrow U$ be defined. If $g \circ f$ is an injective mapping, then:

Two functions $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined as follows: $f(x) = \begin{cases} 0, & x \text{ is rational} \\ 1, & x \text{ is irrational} \end{cases}$ and $g(x) = \begin{cases} -1, & x \text{ is rational} \\ 0, & x \text{ is irrational} \end{cases}$. Then, $(f \circ g)(\pi) + (g \circ f)(e)$ is equal to: