If $f(x) = \frac{x}{\sqrt{1 + x^2}}$,then $(fofof)(x) = $

If $f(x) = \begin{cases} \sin x, & x \neq n\pi, n \in \mathbb{Z} \\ 0, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \neq 0, 2 \\ 4, & x = 0 \\ 5, & x = 2 \end{cases}$,then $\lim_{x \to 0} g(f(x)) = $

Let $f(x) = \begin{cases} x, & x < 0 \\ 1 + x^2, & x \geq 0 \end{cases}$ and $g(x) = 1 + x - [x]$,then the range of $f(g(x))$ is (where $[.]$ denotes the greatest integer function).

If $f: R \to R$ is defined by $f(x) = (x + 1)^2$ and $g: R \to R$ is defined by $g(x) = x^2 + 1$,then $(fog)(-3)$ equals:

If $f: R \rightarrow R$ and $g: R^{+} \rightarrow R$ are such that $g\{f(x)\}=|\sin x|$ and $f\{g(x)\}=(\sin \sqrt{x})^2$,then a possible choice for $f$ and $g$ is

If $f(x) = 3x + 10$ and $g(x) = x^2 - 1$,then $(fog)^{-1}(x) = $