If $f(x) = [x]$,where $[x]$ is the greatest integer not greater than $x$,then $f^{\prime}(1^{+}) = \dots$.

The set of all points,where the derivative of the function $f(x) = \frac{x}{1+|x|}$ exists,is

Assertion $(A)$: $f(x) = |x|$ is differentiable at $x = a \neq 0$ and continuous but not differentiable at $x = 0$.
Reason $(R)$: If a function is differentiable at a point,then it is continuous at the point. But the converse is not true.

Let $f(x) = \begin{cases} 1, & \forall x < 0 \\ 1 + \sin x, & \forall 0 \le x \le \pi/2 \end{cases}$,then what is the value of $f'(x)$ at $x = 0$?

If $f(x) = \begin{cases} \frac{1}{2}(b^2 - a^2), & 0 \leq x \leq a \\ \frac{1}{2}b^2 - \frac{x^2}{6} - \frac{a^3}{3x}, & a < x \leq b \\ \frac{1}{3}\left(\frac{b^3 - a^3}{x}\right), & x > b \end{cases}$,then which of the following is true?

The function $f(x)=|x^{2}-2 x-3| \cdot e^{|9 x^{2}-12 x+4|}$ is not differentiable at exactly :