$$\mathop {\lim }\limits_{x \to 0} \frac{x \tan 4 x-2 x \tan 2 x}{(1-\cos 4 x)^2}= $$

Assertion (A) $f(x)=|x-a|+|x-b|$, is continuous on $\mathbf{R}$

Reason (R) $\frac{|x-\alpha|}{x-\alpha}$ is continuous at $x \in \mathbf{R}-\{\alpha\}$.

The correct option among the following is

A function $y=f(x)$ with $f(-1)=-249$ has no maximum and has only one minimum at $x=5$ with $f(5)=75$. Which one of the following is true?

$$ \mathop {\lim }\limits_{x \to 0} \frac{1-\cos \left(x^2+\pi(x+2)\right)}{x^2}= $$