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

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 $f: R - \left\{-\frac{1}{2}\right\} \rightarrow R$ be defined by $f(x) = \frac{x-2}{2x+1}$. If $\alpha$ and $\beta$ satisfy the equation $f(f(x)) = -x$,then $4(\alpha^2 + \beta^2) = $

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

Let $R$ be a relation '$ < $' from $A$ to $B$,where $A = \{1, 2, 3, 4\}$ and $B = \{1, 3, 5\}$ such that $(a, b) \in R \iff a < b$. Then $R \circ R^{-1}$ is:

$f: R \rightarrow R$ 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?