The function $f(x) = x^2 \sin \frac{1}{x}$ for $x \ne 0$ and $f(0) = 0$ at $x = 0$ is:

Which one of the following functions is continuous everywhere in its domain but has at least one point where it is not differentiable?

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 $f(x) = |x-3| + |x+5|$ and $A = \{a \in \mathbb{R} \mid \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \text{ exists} \}$. Then the number of real numbers which are in $(-\infty, -3) \cup (5, \infty)$ but not in $A$ is

Let $f(x) = x|x|$ and $g(x) = \sin x$.
Statement-$1$: $gof$ is differentiable at $x=0$ and its derivative is continuous at that point.
Statement-$2$: $gof$ is twice differentiable at $x=0$.

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