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

\(\mathrm{f}(x) =\sin |x|-|x|+2(x-\pi) \cos |x| \)
\( \therefore \quad \mathrm{f}(x) =\sin x-x+2(x-\pi) \cos x, x \geq 0 \)
\( =-\sin x+x+2(x-\pi) \cos x, x \lt 0\)
Now,
\(\mathrm{f}^{\prime}(x) =\cos x-1-2(x-\pi) \sin x+2 \cos x, x \geq 0 \)
\( =-\cos x+1-2(x+\pi) \sin x+2 \cos x, x \lt 0 \)
\( \mathrm{f}\left(0^{+}\right) =\cos 0-1-2(0-\pi) \sin 0+2 \cos 0=2 \)
\( \mathrm{f}\left(0^{-}\right) =-\cos 0+1-2(0-\pi) \sin 0+2 \cos 0=2 \)
\( \therefore \quad \mathrm{f}^{\prime}\left(0^{+}\right) =\mathrm{f}^{\prime}\left(0^{-}\right)=2\)
\(\therefore \mathrm{f}(x)\) is differentiable at \(x=0\)
\(\therefore \mathrm{f}(x)\) is differentiable everywhere.
\(\therefore\mathrm{k}=\phi\)