If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=x^3-x$ and $g(x)=\sin 2x$,then the values of $x \in (0, 2\pi)$ that satisfy $f(g(x)) > 0$ lie in the interval

If $g(x) = x^2 + x - 2$ and $(g \circ f)(x) = 2x^2 - 5x + 2$, then $f(x)$ is equal to

Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ be functions such that $g \circ f: S \rightarrow U$ is surjective. Then,

If $f(x)=2^{100} x+1$ and $g(x)=3^{100} x+1$, then the set of real numbers $x$ such that $f(g(x))=x$ is

Define the functions $f, g$ and $h$ from $R$ to $R$ such that $f(x) = x^2 - 1, g(x) = \sqrt{x^2 + 1}$ and $h(x) = \begin{cases} 0, & x \leq 0 \\ x, & x \geq 0 \end{cases}$ Consider the following statements:

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 $(g \circ f)(x) = 1$