Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f(x)=\sin |x|-|x|+2(x-\pi )\cos |x|$ 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 $x=0$ $f^{\prime }\left(0^{\ast }\right)\mathrm{R}\cdot \mathrm{HD}=3-1=2$ $f^{\prime }(0)\mathrm{L}.H.D.=1+1=2$ $\mathrm{L}.\mathrm{HD}=\mathrm{R}.H.D$ Then, function $f(x)$ is differentiable for $x=0.$ So it is differentiable everywhere Hence, $k=\varphi$