If $f(x) = \frac{2x+3}{3x-2}$,$x \neq \frac{2}{3}$,then $(f \circ f)(x)$ is:

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

Let $f(x) = \sin \left(\frac{\pi}{6} \sin \left(\frac{\pi}{2} \sin x\right)\right)$ for all $x \in R$ and $g(x) = \frac{\pi}{2} \sin x$ for all $x \in R$. Let $(f \circ g)(x)$ denote $f(g(x))$ and $(g \circ f)(x)$ denote $g(f(x))$. Then which of the following is (are) true?
$(A)$ Range of $f$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(B)$ Range of $f \circ g$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(C)$ $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)} = \frac{\pi}{6}$
$(D)$ There is an $x \in R$ such that $(g \circ f)(x) = 1$

If $f(x) = x^2 + 1$,then $fof(x)$ is equal to

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) = $

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