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

$f: R - \left(-\frac{3}{5}\right) \rightarrow R$ is defined by $f(x) = \frac{3x-2}{5x+3}$,then $f \circ f(1)$ is

If $f: A \rightarrow B$ and $g: B \rightarrow C$ are two functions such that $g \circ f: A \rightarrow C$ is a bijection,then which one of the following is always true?

If $f(x)=2^{100} x+1$ and $g(x)=3^{100} x+1$,then the set of real numbers $x$ such that $f(g(x))=x$ is

If $f : R \rightarrow R$ is defined by $f(x) = x^{2} - 3x + 2$,find $f(f(x))$.