If $f: A \rightarrow B$ and $g: B \rightarrow C$ are functions such that $g \circ f: A \rightarrow C$ is onto,then a necessary condition is:

If $f(x)$ and $g(x)$ are two functions such that $g(x) = x - \frac{1}{x}$ and $(f \circ g)(x) = x^3 - \frac{1}{x^3}$,then $f'(1)$ is equal to:

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

Show that if $f: R - \{\frac{7}{5}\} \rightarrow R - \{\frac{3}{5}\}$ is defined by $f(x) = \frac{3x+4}{5x-7}$ and $g: R - \{\frac{3}{5}\} \rightarrow R - \{\frac{7}{5}\}$ is defined by $g(x) = \frac{7x+4}{5x-3}$,then $f \circ g = I_{A}$ and $g \circ f = I_{B}$,where $A = R - \{\frac{3}{5}\}$,$B = R - \{\frac{7}{5}\}$; $I_{A}(x) = x, \forall x \in A$,$I_{B}(x) = x, \forall x \in B$ are called identity functions on sets $A$ and $B$,respectively.

Let $f(x) = \sin^{-1} x$ and $g(x) = \frac{x^2 - x - 2}{2x^2 - x - 6}$. If $g(2) = \lim_{x \to 2} g(x)$,then the domain of the function $f \circ g$ is .... .

If $f(x) = \sin^2 x$ and the composite function $g(f(x)) = |\sin x|$,then the function $g(x)$ is equal to