The values of $x$ at which the real-valued function $f(x) = 7|2x + 1| - 19|3x - 5|$ is not differentiable are:

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.

If $y = \sec^{-1} \left( \frac{2x}{1 + x^2} \right) + \sin^{-1} \left( \frac{x - 1}{x + 1} \right)$,then $\frac{dy}{dx}$ is equal to

If $f(x) = \begin{cases} 1, & x < 0 \\ 1 + \sin x, & 0 \le x < \frac{\pi}{2} \end{cases}$,then $f'(0) = $

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

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