Let K be the set of all real values

Solution:

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

There are two cases,

Case (1), $x>0$

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

$f^{\prime}(x)=\cos x-1+2(1-0) \cos x-2 \sin (x-\pi)$

$f^{\prime}(x)=3 \cos x-2(x-\pi) \sin x-1$

Then, function $f(x)$ is differentiable for all $x>0$

Case (2) $x<0$

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

$f^{\prime}(x)=-\cos x+1-2(x-\pi) \sin x+2 \cos x$

$f^{\prime}(x)=\cos x+1-2(x-\pi) \sin x$

Then, function $f(x)$ is differentiable for all $x<0$

Now check for $x=0$

$f^{\prime}\left(0^{+}\right)$R.H.D. $=3-1=2$

$f^{\prime}\left(0^{-}\right)$L.H.D. $=1+1=2$

L.H.D. $=$ R.H.D.

Then, function $f(x)$ is differentiable for $x=0$. So it is differentiable everywhere

Hence, $k=\phi$