If $f(x) = \begin{cases} \sin x, & x \neq n\pi, n \in \mathbb{Z} \\ 0, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \neq 0, 2 \\ 4, & x = 0 \\ 5, & x = 2 \end{cases}$,then $\lim_{x \to 0} g(f(x)) = $

If $f: R \rightarrow R$ is defined by $f(x)=e^{x}$ and $g: R \rightarrow R$ is defined by $g(x)=x^{2}$,then the mapping $(g \circ f): R \rightarrow R$ is defined by $(g \circ f)(x) = g(f(x))$ for all $x \in R$. Which of the following is true?

Find the function $g(t)$ if $f(t)=3t-2$ and $(g \circ f)^{-1}(t)=t-2$.

If $f:R \to R$ and $g:R \to R$ are given by $f(x) = |x|$ and $g(x) = |x|$ for each $x \in R$,then $\{ x \in R : g(f(x)) \le f(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 $(g \circ f)(x) = 1$

Let $f(x) = e^x$ and $g(x) = x^2$. Then,the number of solutions of $f(g(x)) = g(f(x))$ is equal to: