Question

# Number of points , where $f(x) = \cos \left| x \right| + \left| {\sin x} \right|$ is not differentiable in x$\in \left[ {0,4\pi } \right],{\text{ is:}}$A. 2B. 3C. 4D. 5

We have been given that x$\in \left[ {0,4\pi } \right]$ for which $\left| x \right| = x$
$f(x) = \cos \left| x \right| + \left| {\sin x} \right|$
$f(x) = \cos x + \sin x{\text{ }}0 \leqslant x < \pi \\ {\text{ }} = \cos x - \sin x{\text{ }}\pi \leqslant x < 2\pi \\ {\text{ }} = \cos x + \sin x{\text{ 2}}\pi \leqslant x < 3\pi \\ {\text{ }} = \cos x - \sin x{\text{ 3}}\pi \leqslant x < 4\pi \\ {\text{Therefore, }} \\ f'(x) = - \sin x + \cos x{\text{ }}0 \leqslant x < \pi \\ {\text{ }} = - \sin x - \cos x{\text{ }}\pi \leqslant x < 2\pi \\ {\text{ }} = - \sin x + \cos x{\text{ 2}}\pi \leqslant x < 3\pi \\ {\text{ }} = - \sin x - \cos x{\text{ 3}}\pi \leqslant x < 4\pi \\ {\text{Now at }}x = \pi \\ f'(\pi ) = - \sin \pi + \cos \pi = - 1{\text{ if x < }}\pi \\ f'(\pi ) = - \sin \pi - \cos \pi = 1{\text{ if x}} > \pi \\ {\text{ }}\therefore {\text{ Function is not differentiable at }}x = \pi \\ {\text{Similarly, we can find that the function is not differentiable at }}x = 2\pi ,3\pi \\ \\ {\text{ }} \\$
Also, any function is not differentiable at the end points for a given closed interval. So the function is not differentiable at x=0, $4\pi$