If $f(x) = \operatorname{Max} \{3 - x, 3 + x, 6\}$ is not differentiable at $x = a$ and $x = b$,then $|a| + |b| =$

Let the functions $f, g$ and $h$ be defined as follows:
$f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right) & \text{for } -1 \le x \le 1, x \ne 0 \\ 0 & \text{for } x = 0 \end{cases}$
$g(x) = \begin{cases} x^2 \sin \left( \frac{1}{x} \right) & \text{for } -1 \le x \le 1, x \ne 0 \\ 0 & \text{for } x = 0 \end{cases}$
$h(x) = |x|^3$ for $-1 \le x \le 1$.
Which of these functions are differentiable at $x = 0$?

If $f(x) = \begin{cases} \frac{x-1}{2x^2-7x+5}, & \text{for } x \neq 1 \\ -\frac{1}{3}, & \text{for } x=1 \end{cases}$,then $f^{\prime}(1)$ is equal to:

Let the function $f(x) = (x^2 - 1)|x^2 - ax + 2| + \cos|x|$ be non-differentiable at exactly two points $x = \alpha = 2$ and $x = \beta$. Then the distance of the point $(\alpha, \beta)$ from the line $12x + 5y + 10 = 0$ is equal to:

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

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 = $