The value of $m$ for which the function $f(x) = \begin{cases} mx^2, & x \le 1 \\ 2x, & x > 1 \end{cases}$ is differentiable at $x = 1$,is

Suppose $f: R \rightarrow R$ is given by $f(x) = \begin{cases} 1, & \text{if } x=1 \\ e^{(x^{10}-1)} + (x-1)^2 \sin \frac{1}{x-1}, & \text{if } x \neq 1 \end{cases}$. Then:

Identify the correct statement,where $[.]$ and $\{.\}$ denote the greatest integer function and fractional part function,respectively.

The total number of points of non-differentiability of $f(x) = \min \{ |\sin x|, |\cos x|, \frac{1}{4} \}$ in $(0, 2\pi)$ is

$f(x) = \begin{cases} [\cos \pi x]; & x \leqslant 1 \\ 2\{x\} - 1; & x > 1 \end{cases}$ Comment on the derivability at $x = 1$,where $[\cdot]$ denotes the greatest integer function and $\{\cdot\}$ denotes the fractional part function.

Let $f : (-1, 1) \to \mathbb{R}$ be a function defined by $f(x) = \min\{-|x|, -\sqrt{1 - x^2}\}$. If $K$ is the set of all points at which $f$ is not differentiable,then $K$ has exactly