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)) \} = $

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 $f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions defined by $f(x)=2x-3$ and $g(x)=5x^2-2$,then the least value of the function $(g \circ f)(x)$ is

If $f(x) = \frac{4x+7}{7x-4}$,then the value of $f\{f[f(2)]\} =$ ?

If $f(x) = \frac{3x - 4}{2x - 3}$, then $f(f(f(x)))$ will be

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 .... .