Let $R$ be the set of real numbers and the mapping $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x) = 5 - x^2$ and $g(x) = 3x - 4$, then the value of $(f \circ g)(-1)$ is

Let $R$ be the set of real numbers and the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x) = x^{2} + 2x - 3$ and $g(x) = x + 1$. Then, the value of $x$ for which $f(g(x)) = g(f(x))$ is

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 $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = \begin{cases} x+2, & x>0 \\ 2-x, & x \leq 0 \end{cases}$ and $g(x) = \begin{cases} x^2-2x-2, & 1 \leq x < 2 \\ x-7, & x \geq 2 \\ x+5, & x < 1 \end{cases}$, then $\lim _{x \rightarrow 0} g(f(x))$

For a suitably chosen real constant $a$,let the function $f: R-\{-a\} \rightarrow R$ be defined by $f(x)=\frac{a-x}{a+x}$. Further,suppose that for any real number $x \neq-a$ and $f(x) \neq-a$,$(f \circ f)(x)=x$. Then,$f\left(-\frac{1}{2}\right)$ is equal to

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: