Let K be the set of all real values of x where the function f(x)=sin |x|-|x|+2(x-π) cos |x| is not differentiable. Then the set K is equal to :

$f(x)=\sin |x|-|x|+2(x-\pi) \cos |x|$f(x)=sin∣x∣−∣x∣+2(x−π)cos∣x∣ There are two cases, Case $(1), x>0$(1),x>0 $f(x)=\sin x-x+2(x-\pi) \cos x$f(x)=sinx−x+2(x−π)cosx $f^{\prime}(x)=\cos x-1+2(1-0) \cos x-2 \sin (x-\pi)$f′(x)=cosx−1+2(1−0)cosx−2sin(x−π) $f^{\prime}(x)=3 \cos x-2(x-\pi) \sin x-1$f′(x)=3cosx−2(x−π)sinx−1 Then, function $f(x)$f(x) is differentiable for all $x>0$x>0 Case $(2) x < 0$(2)x<0 $f(x)=-\sin x+x+2(x-\pi) \cos x$f(x)=−sinx+x+2(x−π)cosx $f^{\prime}(x)=-\cos x+1-2(x-\pi) \sin x+2 \cos x$f′(x)=−cosx+1−2(x−π)sinx+2cosx $f^{\prime}(x)=\cos x+1-2(x-\pi) \sin x$f′(x)=cosx+1−2(x−π)sinx Then, function $f(x)$f(x) is differentiable for all $x < 0$x<0 Now check for $x=0$x=0 $f^{\prime}\left(0^{*}\right) \mathrm{R} \cdot \mathrm{HD}=3-1=2$f′(0∗)R⋅HD=3−1=2 $f^{\prime}(0) \mathrm{L} . \mathrm{H.D.}=1+1=2$f′(0)L.H.D.=1+1=2 $\mathrm{L} . \mathrm{HD}=\mathrm{R} . \mathrm{H.D}$L.HD=R.H.D Then, function $f(x)$f(x) is differentiable for $x=0 .$x=0. So it is differentiable everywhere Hence, $k=\phi$k=ϕ