If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=x-[x]$ and $g(x)=[x]$ for $x \in R$, where $[x]$ is the greatest integer not exceeding $x$, then for every $x \in R, f(g(x))$ is equal to

Show that if $f: R - \{\frac{7}{5}\} \rightarrow R - \{\frac{3}{5}\}$ is defined by $f(x) = \frac{3x+4}{5x-7}$ and $g: R - \{\frac{3}{5}\} \rightarrow R - \{\frac{7}{5}\}$ is defined by $g(x) = \frac{7x+4}{5x-3}$,then $f \circ g = I_{A}$ and $g \circ f = I_{B}$,where $A = R - \{\frac{3}{5}\}$,$B = R - \{\frac{7}{5}\}$; $I_{A}(x) = x, \forall x \in A$,$I_{B}(x) = x, \forall x \in B$ are called identity functions on sets $A$ and $B$,respectively.

If $R$ is a relation from a set $A$ to a set $B$ and $S$ is a relation from $B$ to a set $C$,then the relation $S \circ R$ is:

If $f(x) = \frac{1}{1 - x}$,then the derivative of the composite function $f[f\{ f(x)\} ]$ 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

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