The set of points where the function $f(x)=|x-1| e^{x}$ is differentiable,is

If $f(x)=|x-3|$,then $f^{\prime}(3)$ is

If $f$ is a real-valued differentiable function satisfying $|f(x) - f(y)| \le (x - y)^2$ for all $x, y \in R$ and $f(0) = 0$,then $f(1)$ is equal to:

Let $f: R \rightarrow R$ be a function defined by $f(x) = \max \{x, x^2\}$. Let $S$ denote the set of all points in $R$ where $f$ is not differentiable. Then $S$ 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 $f(x) = \begin{cases} x^{\alpha} \sin \left( \frac{1}{x} \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$; Which of the following is true?