Let $f(x) = \begin{cases} \max \{|x|, x^2\}, & |x| \le 2 \\ 8 - 2|x|, & 2 < |x| \le 4 \end{cases}$. Let $S$ be the set of points in the interval $(-4, 4)$ at which $f$ is not differentiable. Then $S$

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

Let $g: [-2, 2] \rightarrow R$ and $f: [-2, 2] \rightarrow R$ be two functions defined as $g(x) = \begin{cases} -1, & \text{if } -2 \le x < 0 \\ x^2 - 1, & \text{if } 0 \le x \le 2 \end{cases}$ and $f(x) = |g(x)| + g(|x|) + 2$. In the interval $(-2, 2)$, $f$ is not differentiable at $x = $

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

Consider the following statements.
$(a)$ If a function is differentiable at a point $p$ then it is not continuous at $p$.
$(b)$ If a function is not continuous at $x = a$, then it is not differentiable at $x = a$.
$(c)$ If $f(x) = |x|$ then $f(x)$ is not differentiable but continuous on $R$.
$(d)$ If $f(x) = x - [x]$, then $f'(1) = 1$.
Which of the above statements are (is) correct?

Which of the following functions is differentiable at $x = 0$?