If $f(x)=\sqrt{x}$ $(x \geq 0)$ and $g(x)=1+x^2$, then $(f \circ g)^{\prime}(1)=$

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)) = $

If the function $f(x) = \frac{1}{2} - \tan \left( \frac{\pi x}{2} \right)$ for $-1 < x < 1$ and $g(x) = \sqrt{3 + 4x - 4x^2}$,then the domain of the composite function $(g \circ f)(x)$ is:

Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x \in R$. Then the set of all $x$ satisfying $(f \circ g \circ g \circ f)(x)=(g \circ g \circ f)(x)$,where $(f \circ g)(x)=f(g(x))$,is

Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be any two functions and $g \circ f: A \rightarrow C$ is one-one,then

If $f(x) = \frac{1}{1 - x}$,then the derivative of the composite function $f[f\{ f(x)\} ]$ is equal to