Let $f(x) = \begin{cases} g(x) \cos(\frac{1}{x}) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$ where $g(x)$ is an even function differentiable at $x = 0$,passing through the origin. Then $f'(0)$:

Let $R$ denote the set of all real numbers. Define the function $f: R \rightarrow R$ by $f(x) = \begin{cases} 2-2x^2-x^2 \sin \frac{1}{x} & \text{if } x \neq 0 \\ 2 & \text{if } x=0 \end{cases}$. Then which one of the following statements is True?

If $f(x) = \begin{cases} \tan^{-1} x, & \text{when } |x| \leq 1 \\ \frac{1}{2}(|x|-1), & \text{when } |x| > 1 \end{cases}$,then the domain of $\frac{d}{dx} f(x)$ is

The function $f(x)=|x^{2}-2 x-3| \cdot e^{|9 x^{2}-12 x+4|}$ is not differentiable at exactly :

The set of points where $f(x) = \frac{x}{4+|x|}$ is differentiable is

Let the function $f: R \rightarrow R$ be defined by $f(x)=x-x^2+(x-1) \sin x$ and let $g: R \rightarrow R$ be an arbitrary function. Let $f g: R \rightarrow R$ be the product function defined by $(f g)(x)=f(x) g(x)$. Then which of the following statements is/are $TRUE$?
$(A)$ If $g$ is continuous at $x=1$,then $f g$ is differentiable at $x=1$
$(B)$ If $fg$ is differentiable at $x=1$,then $g$ is continuous at $x=1$
$(C)$ If $g$ is differentiable at $x=1$,then $f g$ is differentiable at $x=1$
$(D)$ If $fg$ is differentiable at $x=1$,then $g$ is differentiable at $x=1$