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

Let $f(x)=\log (\sin x), 0 < x < \pi$ and $g(x)=\sin ^{-1}(e^{-x}), x \geq 0$. If $\alpha$ is a positive real number such that $a=(f \circ g)^{\prime}(\alpha)$ and $b=(f \circ g)(\alpha)$,then

Let $f: \{1,3,4\} \rightarrow \{1,2,5\}$ and $g: \{1,2,5\} \rightarrow \{1,3\}$ be given by $f = \{(1,2), (3,5), (4,1)\}$ and $g = \{(1,3), (2,3), (5,1)\}$. Write down $g \circ f$.

If $f(x) = \frac{4x+3}{6x-4}$,$x \neq \frac{2}{3}$ and $(f \circ f)(x) = g(x)$,where $g: R - \{\frac{2}{3}\} \rightarrow R - \{\frac{2}{3}\}$,then $(g \circ g \circ g)(4)$ is equal to

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = x^{2} - 3x + 4$ and $g(x) = 2x + 1$,then the value of $x$ for which $f(x) = (f \circ g)(x)$ is

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