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.

Number of points where the function $f(x) = (x^2 - 1) | x^2 - x - 2 | + \sin(|x|)$ is not differentiable,is

Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be functions satisfying $f(x+y)=f(x)+f(y)+f(x)f(y)$ and $f(x)=x g(x)$ for all $x, y \in R$. If $\lim _{x \rightarrow 0} g(x)=1$,then which of the following statements is/are $TRUE$?
$(A)$ $f$ is differentiable at every $x \in R$
$(B)$ If $g(0)=1$,then $g$ is differentiable at every $x \in R$
$(C)$ The derivative $f^{\prime}(1)$ is equal to $1$
$(D)$ The derivative $f^{\prime}(0)$ is equal to $1$

If $f(x) = \begin{cases} x^2 & \text{if } x \leqslant x_0 \\ ax + b & \text{if } x > x_0 \end{cases}$ is derivable for all $x \in \mathbb{R}$,then the values of $a$ and $b$ are respectively:

The function $f(x)=|x^{2}-2 x-3| \cdot e^{|9 x^{2}-12 x+4|}$ is not differentiable at exactly :

Assertion $(A)$: If $y = f(x) = (|x| - |x - 1|)^2$,then $\left(\frac{dy}{dx}\right)_{x=1} = 1$.
Reason $(R)$: If $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$ exists,then it is called the derivative of $f(x)$ at $x = a$.
Then: