If $f : R \to R$ and $g : R \to R$ are defined as $f(x) = 2x - |x|$ and $g(x) = 2x + |x|$, then

If $f(x) = \sin^2 x$ and the composite function $g(f(x)) = |\sin x|$,then the function $g(x)$ is equal to

Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be any two functions and $g \circ f: A \rightarrow C$ is one-one,then

Define $f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$. If $f \circ f(x)$ is discontinuous at $a$ and $b$ in $[0, 3]$ and $a < b$, then $2 a + 3 b = $

If $g(x)=x^2+x-2$ and $\frac{1}{2}(g \circ f)(x)=2 x^2-5 x+2$,then one such function $f(x)=$

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