We need the set of all real $x$ where $f(x)=\sin|x|-|x|+2(x-\pi)\cos|x|$ is not differentiable.

The only possible trouble points come from the presence of $|x|$, i.e. at $x=0$. For $x\neq 0$, both $|x|$ and $\cos|x|,\sin|x|$ are differentiable as compositions of differentiable functions.

So we only need to check differentiability at $x=0$.

1. Rewrite $f(x)$ piecewise

Case 1: $x>0$

Then $|x|=x$, so $f(x)=\sin x-x+2(x-\pi)\cos x.$

Differentiate: $f'(x)=\cos x-1+2\cos x+2(x-\pi)(-\sin x).$ So $f'(x)=3\cos x-1-2(x-\pi)\sin x.$

Hence the right-hand derivative at $0$ is $f'_+(0)=3\cos 0-1-2(0-\pi)\sin 0=3-1=2.$

Case 2: $x<0$

Then $|x|=-x$, so $f(x)=\sin(-x)-(-x)+2(x-\pi)\cos(-x).$ Using $\sin(-x)=-\sin x$ and $\cos(-x)=\cos x$, $f(x)=-\sin x+x+2(x-\pi)\cos x.$

Differentiate: $f'(x)=-\cos x+1+2\cos x+2(x-\pi)(-\sin x).$ Thus $f'(x)=\cos x+1-2(x-\pi)\sin x.$

Hence the left-hand derivative at $0$ is $f'_-(0)=\cos 0+1-2(0-\pi)\sin 0=1+1=2.$

2. Compare left and right derivatives at $x=0$

We get $f'_-(0)=f'_+(0)=2.$ Therefore $f$ is differentiable at $x=0$.

3. Check all other points

For every $x\neq 0$, the function is a combination of differentiable functions, so it is differentiable there as well.

Thus there is no real value of $x$ where $f$ is not differentiable.

Therefore, $K=\varnothing.$

So the correct option is B.

4. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

They agree.