Question:

Step 1: Identify potential points of non-differentiability. The function is given by $f(x) = \sin|x| - |x| + 2(x - \pi)\cos|x|$. The absolute value function $|x|$ is not differentiable at $x=0$. Consequently, terms involving $|x|$, such as $\sin|x|$ and $|x|$, are potentially not differentiable at $x=0$. The term $\cos|x|$ is differentiable everywhere because $\cos|x| = \cos x$. The term $2(x-\pi)$ is a polynomial and thus differentiable everywhere. Products and sums of differentiable functions are differentiable. Therefore, the only point where $f(x)$ might not be differentiable is $x=0$. Step 2: Analyze the function for $x > 0$. For $x > 0$, $|x| = x$. The function becomes $f(x) = \sin x - x + 2(x - \pi)\cos x$. To find the right-hand derivative at $x=0$, we compute $f'(x)$ for $x > 0$: $$f'(x) = \frac{d}{dx}(\sin x - x + 2(x - \pi)\cos x)$$ $$f'(x) = \cos x - 1 + 2\left[\frac{d}{dx}((x - \pi)\cos x)\right]$$ $$f'(x) = \cos x - 1 + 2[\cos x + (x - \pi)(-\sin x)]$$ $$f'(x) = \cos x - 1 + 2\cos x - 2(x - \pi)\sin x$$ $$f'(x) = 3\cos x - 1 - 2(x - \pi)\sin x$$ The right-hand derivative at $x=0$ is: $$f'(0^+) = \lim_{x \to 0^+} (3\cos x - 1 - 2(x - \pi)\sin x)$$ $$f'(0^+) = 3\cos(0) - 1 - 2(0 - \pi)\sin(0)$$ $$f'(0^+) = 3(1) - 1 - 0 = 2$$ Step 3: Analyze the function for $x < 0$. For $x < 0$, $|x| = -x$. The function becomes $f(x) = \sin(-x) - (-x) + 2(x - \pi)\cos(-x)$. This simplifies to $f(x) = -\sin x + x + 2(x - \pi)\cos x$. To find the left-hand derivative at $x=0$, we compute $f'(x)$ for $x < 0$: $$f'(x) = \frac{d}{dx}(-\sin x + x + 2(x - \pi)\cos x)$$ $$f'(x) = -\cos x + 1 + 2\left[\frac{d}{dx}((x - \pi)\cos x)\right]$$ $$f'(x) = -\cos x + 1 + 2[\cos x + (x - \pi)(-\sin x)]$$ $$f'(x) = -\cos x + 1 + 2\cos x - 2(x - \pi)\sin x$$ $$f'(x) = \cos x + 1 - 2(x - \pi)\sin x$$ The left-hand derivative at $x=0$ is: $$f'(0^-) = \lim_{x \to 0^-} (\cos x + 1 - 2(x - \pi)\sin x)$$ $$f'(0^-) = \cos(0) + 1 - 2(0 - \pi)\sin(0)$$ $$f'(0^-) = 1 + 1 - 0 = 2$$ Step 4: Compare the one-sided derivatives at $x=0$. Since $f'(0^+) = 2$ and $f'(0^-) = 2$, the left-hand derivative and the right-hand derivative at $x=0$ are equal. Therefore, $f(x)$ is differentiable at $x=0$. Step 5: Conclusion. The function $f(x)$ is differentiable for all $x \neq 0$ because it is a composition, sum, and product of differentiable functions in these intervals. As shown in Step 4, $f(x)$ is also differentiable at $x=0$. Thus, the function $f(x)$ is differentiable for all real values of $x$. The set $K$ of all real values of $x$ where $f(x)$ is not differentiable is the empty set. $$K = \emptyset$$