Let $f(x) = \begin{cases} x, & x < 0 \\ 1 + x^2, & x \geq 0 \end{cases}$ and $g(x) = 1 + x - [x]$,then the range of $f(g(x))$ is (where $[.]$ denotes the greatest integer function).

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 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

If $f(x) = \frac{1+x}{1-x}$ where $x \neq 1$,then $f(x) \cdot f(y) = $ . . . . . . .

$f(x) = (20 - x^4)^{1/4}$ for $0 < x < \sqrt{5}$,then $f(f(1/2))$ is equal to

If $g(x) = x^2 + x - 2$ and $\frac{1}{2}g(f(x)) = 2x^2 - 5x + 2$,then $f(x)$ is