Question

Number of points, where $f\left(x\right)=\cos |x|+|\sin x|$ is not differentiable in $x\in [0,4\pi ]$, is:

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is D $5$

For $xϵ[0,4\pi ]$,

$\left|x\right|=x$

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

Now,

$f\left(x\right)=\cos x+\sin x$ $0\le x<\pi$

$=\cos x-\sin x$ $\pi \le x<2\pi$

$=\cos x+\sin x$ $2\pi \le x<3\pi$

$=\cos x-\sin x$ $3\pi \le x<4\pi$

Therefore,

$f^{\prime }\left(x\right)=-\sin x+\cos x$ $0\le x<\pi$

$=-\sin x-\cos x$ $\pi \le x<2\pi$

$=-\sin x+\cos x$ $2\pi \le x<3\pi$

$=-\sin x-\cos x$ $3\pi \le x<4\pi$

At $x=\pi$

$f^{\prime }\left(\pi \right)=-\sin \pi +\cos \pi =-1$ if $x<\pi$

$f^{\prime }\left(\pi \right)=-\sin \pi -\cos \pi =1$ if $x>\pi$

So, function is not differentiable at $x=\pi$

Similarly, we can find that the function is not differentiable at $x=2\pi ,3\pi$

Also, any function is not differentiable at the end-points for a given closed interval. So, function is not differentiable at $x=0,4\pi$

So the answer is $5$