If $f(x) = |x - 3|,$ then $f$ is

Assertion $(A)$: If $f(x)$ is not continuous at $x=a$,then it is not differentiable at $x=a$.
Reason $(R)$: If $f(x)$ is differentiable at a point,then it is continuous at that point.

If $f(x + y) = f(x) + f(y) + |x|y + xy^2$,$\forall x, y \in R$ and $f'(0) = 0$,then

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 the function $g(x) = \begin{cases} k\sqrt{x+1}, & 0 \le x \le 3 \\ mx + 2, & 3 < x \le 5 \end{cases}$ is differentiable,then the value of $k+m$ is:

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