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 equal to:

We are given the function

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

The elementary functions $$\sin t,\;\cos t,\;t$$ are differentiable everywhere, but the absolute-value function $$|x|$$ is not differentiable at the single point $$x=0$$. Hence, if $$f(x)$$ fails to be differentiable anywhere, it can only happen at $$x=0$$. Nevertheless, we will verify differentiability explicitly.

First we rewrite $$f(x)$$ without the absolute value by splitting the real line into the two natural regions $$x\ge 0$$ and $$x<0$$. For $$x\ge 0$$ we have $$|x|=x$$, while for $$x<0$$ we have $$|x|=-x$$. Using these facts, we obtain

For $$x\ge 0:$$

$$f(x)=\sin x-\;x\;+\;2\,(x-\pi)\cos x.$$

For $$x<0:$$ since $$\sin(-x)=-\sin x$$ and $$\cos(-x)=\cos x$$, we obtain

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

Thus the piece-wise definition is

$$f(x)= \begin{cases} \sin x-x+2\,(x-\pi)\cos x,& x\ge 0,\\[6pt] -\sin x+x+2\,(x-\pi)\cos x,& x<0. \end{cases}$$

Now we differentiate each piece. We use the standard rules $$\frac{d}{dx}\bigl(\sin x\bigr)=\cos x,$$ $$\frac{d}{dx}(x)=1,$$ $$\frac{d}{dx}\bigl(\cos x\bigr)=-\sin x,$$ and for the product $$2(x-\pi)\cos x$$ we use the product rule $$\frac{d}{dx}\bigl[u(x)v(x)\bigr]=u'(x)v(x)+u(x)v'(x).$$

Right-hand derivative, $$x>0$$ (use the first branch)

$$\begin{aligned} f'(x) &=\frac{d}{dx}\Bigl(\sin x-x+2(x-\pi)\cos x\Bigr)\\[6pt] &=\cos x-1+2\Bigl[\frac{d}{dx}(x-\pi)\cdot\cos x +(x-\pi)\cdot\frac{d}{dx}(\cos x)\Bigr]\\[6pt] &=\cos x-1+2\Bigl[1\cdot\cos x+(x-\pi)(-\,\sin x)\Bigr]\\[6pt] &=\cos x-1+2\cos x-2(x-\pi)\sin x\\[6pt] &=3\cos x-1-2(x-\pi)\sin x. \end{aligned}$$

Left-hand derivative, $$x<0$$ (use the second branch)

$$\begin{aligned} f'(x) &=\frac{d}{dx}\Bigl(-\sin x+x+2(x-\pi)\cos x\Bigr)\\[6pt] &=-\cos x+1+2\Bigl[1\cdot\cos x+(x-\pi)(-\,\sin x)\Bigr]\\[6pt] &=-\cos x+1+2\cos x-2(x-\pi)\sin x\\[6pt] &=\cos x+1-2(x-\pi)\sin x. \end{aligned}$$

Both formulas are valid away from $$x=0$$, so $$f(x)$$ is certainly differentiable for every $$x\ne 0$$. We now test the possible trouble point $$x=0$$ by taking the limits of the two expressions.

Right-hand derivative at $$x=0$$

Substituting $$x=0$$ in $$3\cos x-1-2(x-\pi)\sin x$$ gives

$$ f'_+(0)=3\cos 0-1-2(0-\pi)\sin 0 =3(1)-1-2(-\pi)(0) =2. $$

Left-hand derivative at $$x=0$$

Substituting $$x=0$$ in $$\cos x+1-2(x-\pi)\sin x$$ gives

$$ f'_-(0)=\cos 0+1-2(0-\pi)\sin 0 =1+1-2(-\pi)(0) =2. $$

The two one-sided derivatives are equal and finite, hence

$$f'(0)=2.$$ Therefore $$f(x)$$ is differentiable at $$x=0$$ as well.

Because $$f(x)$$ is differentiable for all $$x\ne 0$$ and we have just proved differentiability at $$x=0$$, the function is differentiable for every real number. Consequently there is no real value of $$x$$ where $$f(x)$$ fails to be differentiable.

So the set $$K$$ of non-differentiability points is the empty set: $$K=\varnothing.$$

Hence, the correct answer is Option A.