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

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=|x|$ and $g(x)=[x-3]$ for $x \in R$, then $\{g(f(x)):-\frac{8}{5} < x < \frac{8}{5}\}$ is equal to

For $x \in R, x \neq 0,$ let $f_0(x) = \frac{1}{1 - x}$ and $f_{n + 1}(x) = f_0(f_n(x)),$ $n = 0, 1, 2, ....$ Then the value of $f_{100}(3) + f_1\left( \frac{2}{3} \right) + f_2\left( \frac{3}{2} \right)$ is equal to

Let $f(x)=\sqrt{x^{2}-3x+2}$ and $g(x)=\sqrt{x}$ be two given functions. If $S$ is the domain of $f \circ g$ and $T$ is the domain of $g \circ f$, then:

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

$f: R \rightarrow R$ and $g:[0, \infty) \rightarrow R$ are defined by $f(x)=x^2$ and $g(x)=\sqrt{x}$. Which one of the following is not true?