If $f(x) = \begin{cases} x[x], & 0 \le x < 2 \\ (x-1)[x], & 2 \le x \le 4 \end{cases}$,where $[.]$ denotes the greatest integer function,then:

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 $f(x) = \begin{cases} x^2 & \text{if } x \leqslant x_0 \\ ax + b & \text{if } x > x_0 \end{cases}$ is derivable for all $x \in \mathbb{R}$,then the values of $a$ and $b$ are respectively:

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

Let $f(x) = \begin{cases} \sin x, & \text{for } x \ge 0 \\ 1 - \cos x, & \text{for } x \le 0 \end{cases}$ and $g(x) = e^x$. Then $(g \circ f)'(0)$ is

If $\alpha \in R - \{-1\}$ and $f(x) = |(|x| + \alpha)(|x| - 1)|$,then the number of points at which $f(x)$ is not differentiable is: