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

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Aicharya · Accepted Answer

$\begin{aligned} \mathrm{f}(x) & =\sin |x|-|x|+2(x-\pi) \cos |x| \ 	herefore \quad \mathrm{f}(x) & =\sin x-x+2(x-\pi) \cos x, x \geq 0 \ & =-\sin x+x+2(x-\pi) \cos x, x \lt 0 \end{aligned}$ Now, $\begin{aligned} \mathrm{f}^{\prime}(x) & =\cos x-1-2(x-\pi) \sin x+2 \cos x, x \geq 0 \ & =-\cos x+1-2(x+\pi) \sin x+2 \cos x, x \lt 0 \ \mathrm{f}\left(0^{+}ight) & =\cos 0-1-2(0-\pi) \sin 0+2 \cos 0=2 \ \mathrm{f}\left(0^{-}ight) & =-\cos 0+1-2(0-\pi) \sin 0+2 \cos 0=2 \ 	herefore \quad \mathrm{f}^{\prime}\left(0^{+}ight) & =\mathrm{f}^{\prime}\left(0^{-}ight)=2 \end{aligned}$ $	herefore \quad \mathrm{f}(x)$ is differentiable at $x=0$ $	herefore \quad \mathrm{f}(x)$ is differentiable everywhere. $	herefore \quad \mathrm{k}=\phi$

Let K be the set of all real values of $x$, where the function $\mathrm{f}(x)=\sin |x|-|x|+2(x-\pi) \cos…

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