The function $f(x) = e^{-|x|}$ is

Let $f(x) = |x|$. Then which of the following is true?

Let the functions $f, g$ and $h$ be defined as follows:
$f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right) & \text{for } -1 \le x \le 1, x \ne 0 \\ 0 & \text{for } x = 0 \end{cases}$
$g(x) = \begin{cases} x^2 \sin \left( \frac{1}{x} \right) & \text{for } -1 \le x \le 1, x \ne 0 \\ 0 & \text{for } x = 0 \end{cases}$
$h(x) = |x|^3$ for $-1 \le x \le 1$.
Which of these functions are differentiable at $x = 0$?

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 $x=1$

Assertion $(A)$: If $f(x)$ is not continuous at $x=a$,then it is not differentiable at $x=a$.
Reason $(R)$: If $f(x)$ is differentiable at a point,then it is continuous at that point.

The total number of points of non-differentiability of $f(x) = \min \{ |\sin x|, |\cos x|, \frac{1}{4} \}$ in $(0, 2\pi)$ is