The set of points where $f(x) = \frac{4x}{5 + 6|x|}$ is differentiable is:

Let $f$ be a differentiable function from $R$ to $R$ such that $|f(x) - f(y)| \le 2|x - y|^{\frac{3}{2}}$ for all $x, y \in R$. If $f(0) = 1$,then $\int_{0}^{1} f^2(x) dx$ is equal to

$f(x) = \begin{cases} 4, & -\infty < x < -\sqrt{5} \\ x^2-1, & -\sqrt{5} \leq x \leq \sqrt{5} \\ 4, & \sqrt{5} < x < \infty \end{cases}$
If $k$ is the number of points where $f(x)$ is not differentiable,then $k-2=$

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

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 $f(x) = \begin{cases} |4x^2 - 8x + 5|, & \text{if } 8x^2 - 6x + 1 \geq 0 \\ [4x^2 - 8x + 5], & \text{if } 8x^2 - 6x + 1 < 0 \end{cases}$,where $[\alpha]$ denotes the greatest integer less than or equal to $\alpha$. Then the number of points in $\mathbb{R}$ where $f$ is not differentiable is $.......$