Let K be the set of all real values of x wher | vedclass

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(A) The function is $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$.We check for differentiability at $x = 0$ and $x = \pi$.Case $1$: At $x = 0$.For $x

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$\phi$ (empty set)

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(A) The function is $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$.We check for differentiability at $x = 0$ and $x = \pi$.Case $1$: At $x = 0$.For $x > 0$,$f(x) = \sin x - x + 2(x - \pi) \cos x$. Then $f'(x) = \cos x - 1 + 2 \cos x - 2(x - \pi) \sin x$. Thus,$f'(0^+) = 1 - 1 + 2 - 0 = 2$.For $x Since $f'(0^+) = f'(0^-) = 2$,the function is differentiable at $x = 0$.Case $2$: At $x = \pi$.For $x > 0$,$f(x) = \sin x - x + 2(x - \pi) \cos x$. $f'(x) = \cos x - 1 + 2 \cos x - 2(x - \pi) \sin x$. $f'(\pi^+) = \cos \pi - 1 + 2 \cos \pi - 0 = -1 - 1 - 2 = -4$.For $x Thus,the function is differentiable for all real $x$. The set $K$ is $\phi$.

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

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