Answer

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Hint: Find f(x) and f’(x) and continue with differentiability criteria. Observe the nature of given trigonometric functions.

Complete step-by-step answer:

We have been given that x$ \in \left[ {0,4\pi } \right]$ for which $\left| x \right| = x$

$f(x) = \cos \left| x \right| + \left| {\sin x} \right|$

Now,

$

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, $4\pi$

Therefore the answer is 5.

Note: One must remember the range and nature of trigonometric functions in order to solve such similar problems.

Complete step-by-step answer:

We have been given that x$ \in \left[ {0,4\pi } \right]$ for which $\left| x \right| = x$

$f(x) = \cos \left| x \right| + \left| {\sin x} \right|$

Now,

$

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, $4\pi$

Therefore the answer is 5.

Note: One must remember the range and nature of trigonometric functions in order to solve such similar problems.

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