If $f(x)=|x|+|\sin x|$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,then its left hand derivative at $x=0$ 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 $x=1$

If $f(x) = \begin{cases} x^2 \left| \cos \frac{\pi}{x} \right|, & x \neq 0 \\ 0, & x = 0 \end{cases}$, then at $x = 2$, $f(x)$ is

Let $f(x)=a_0+a_1|x|+a_2|x|^2+a_3|x|^3$, where $a_0, a_1, a_2, a_3$ are real constants. Then $f(x)$ is differentiable at $x=0$ if and only if:

If $x + |y| = 2y,$ then $y$ as a function of $x,$ at $x = 0$ is

At $x=1$,the function $f(x)=\begin{cases} x^{3}-1, & 1 < x < \infty \\ x-1, & -\infty < x \leq 1 \end{cases}$ is