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(B) Given $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$.Since $|x|$ is not differentiable at $x = 0$,we check the differentiability of $f(x)$ at $x =

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an empty set

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(B) Given $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$.Since $|x|$ is not differentiable at $x = 0$,we check the differentiability of $f(x)$ at $x = 0$.For $x \ge 0$,$f(x) = \sin x - x + 2(x - \pi) \cos x$.For $x Now,we find the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$ at $x = 0$.$f'(x)$ for $x > 0$ is $\cos x - 1 + 2 \cos x - 2(x - \pi) \sin x = 3 \cos x - 1 - 2(x - \pi) \sin x$.$f'(0^+) = 3(1) - 1 - 2(-\pi)(0) = 2$.$f'(x)$ for $x $f'(0^-) = 1 + 1 - 2(-\pi)(0) = 2$.Since $f'(0^+) = f'(0^-) = 2$,the function is differentiable at $x = 0$.Since the function is composed of differentiable functions everywhere else,$f(x)$ is differentiable for all $x \in \mathbb{R}$.Therefore,the set $K$ of points where $f(x)$ is not differentiable is an empty set.

Let $K$ be the set of all real values of $x$,where the function $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$ is not differentiable. Then the set $K$ is

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