If $f(x) = \begin{cases} x^2 \left| \cos \frac{\pi}{x} \right|, & x \neq 0 \\ 0, & x = 0 \end{cases}$,then at $x = 2$,$f(x)$ is

The function $f(x) = |\cos x|$ is

If $f(x) = \begin{cases} \tan^{-1} x, & \text{when } |x| \leq 1 \\ \frac{1}{2}(|x|-1), & \text{when } |x| > 1 \end{cases}$,then the domain of $\frac{d}{dx} f(x)$ is

Let $f(x) = \begin{cases} a \sin(x + b) & x \ge 0 \\ 6x^7 - x + 1 & x < 0 \end{cases}$ be differentiable for all real $x$. If $a \in \mathbb{R}$ and $b \in [0, 2\pi]$,then the number of ordered pairs $(a, b)$ is:

Let $[x]$ denote the greatest integer function and $f(x) = \begin{cases} 4x^2 + [2x]x, & \text{if } x \in [-\frac{1}{2}, 0) \\ ax^2 - bx, & \text{if } x \in [0, \frac{1}{2}) \end{cases}$. 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?