Let $S$ be the set of points where the function $f(x) = |2 - |x - 3||, x \in R,$ is not differentiable. Then $\sum_{x \in S} f(f(x))$ is equal to

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 $f'(0)$:

If $f(x)=|x|+|sin x|$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,then its left hand derivative at $x=0$ is

Assertion $(A)$: If $y = f(x) = (|x| - |x - 1|)^2$,then $\left(\frac{dy}{dx}\right)_{x=1} = 1$.
Reason $(R)$: If $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$ exists,then it is called the derivative of $f(x)$ at $x = a$.
Then:

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 $f(x) = x(\sqrt{x} - \sqrt{x + 1}),$ then