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

For a suitably chosen real constant $a$,let a function $f: R-\{-a\} \rightarrow R$ be defined by $f(x)=\frac{a-x}{a+x}$. Further suppose that for any real number $x \neq-a$ and $f(x) \neq-a$,$(f \circ f)(x)=x$. Then $f\left(-\frac{1}{5}\right)$ is equal to

If $f: R \rightarrow R$ is defined by $f(x) = (3 - x^3)^{\frac{1}{3}}$,then $f \circ (f \circ f)(x) = $ . . . . . . .

Let $Q$ be the set of all rational numbers in $[0,1]$ and $f:[0,1] \rightarrow [0,1]$ be defined by $f(x) = \begin{cases} x & \text{for } x \in Q \\ 1-x & \text{for } x \notin Q \end{cases}$. Then, the set $S = \{x \in [0,1] : (f \circ f)(x) = x\}$ is equal to

If $R$ is a relation $  < $ from $A = \{1, 2, 3, 4\}$ to $B = \{1, 3, 5\}$ i.e.,$(a, b) \in R \iff a < b$,then $R \circ R^{-1}$ is

If $g(x)=x^2+x-2$ and $\frac{1}{2}(g \circ f)(x)=2 x^2-5 x+2$,then one such function $f(x)=$