Let $f(x) = \frac{ax + b}{cx + d}$. Then,$f \circ f(x) = x$ provided that

If $f(x) = (a - x^n)^{1/n},$ where $a > 0$ and $n$ is a positive integer,then $f[f(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 $f(x) = ax + b$ and $g(x) = cx + d$,then $f(g(x)) = g(f(x))$ is equivalent to

If $f(x) = (p - x^n)^{1/n}$,$p > 0$ and $n$ is a positive integer,then $f[f(x)]$ is equal to

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