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

Let $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Define $f: S \rightarrow S$ as $f(n) = \begin{cases} 2n, & \text{if } n = 1, 2, 3, 4, 5 \\ 2n - 11, & \text{if } n = 6, 7, 8, 9, 10 \end{cases}$. Let $g: S \rightarrow S$ be a function such that $f \circ g(n) = \begin{cases} n + 1, & \text{if } n \text{ is odd} \\ n - 1, & \text{if } n \text{ is even} \end{cases}$. Then $g(10) \cdot (g(1) + g(2) + g(3) + g(4) + g(5))$ is equal to

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

If $f:R \to R$ and $g:R \to R$ are given by $f(x) = |x|$ and $g(x) = |x|$ for each $x \in R$,then $\{ x \in R : g(f(x)) \le f(g(x)) \} = $

If $f(x) = \frac{2x+3}{3x-2}$,$x \neq \frac{2}{3}$,then $(f \circ f)(x)$ is:

If $f(x) = (p - x^n)^{1/n}$,$p > 0$ and $n$ is a positive integer,then $f[f(x)]$ is equal to