Let $f(x) = \begin{cases} x + 1, & \text{when } x < 2 \\ 2x - 1, & \text{when } x \ge 2 \end{cases}$,then $f'(2) = $

Prove that the function $f$ given by $f(x) = |x - 1|, x \in R$ is not differentiable at $x = 1$.

The number of points in the interval $(0,2)$ at which $f(x)=|x-0.5|+|x-1|+\tan x$ is not 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 $x=1$

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 $f'(0)$:

Suppose $f: R \rightarrow R$ is given by $f(x) = \begin{cases} 1, & \text{if } x=1 \\ e^{(x^{10}-1)} + (x-1)^2 \sin \frac{1}{x-1}, & \text{if } x \neq 1 \end{cases}$. Then: