Q70 / 150 · MHT CET 2024 (02 May Shift 1) · Mathematics
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
- A$\{0\}$
- Correct answerBan empty set
- C$\{\pi\}$
- D$\{0, \pi\}$
Solution
$\begin{aligned}
\mathrm{f}(x) & =\sin |x|-|x|+2(x-\pi) \cos |x| \\
\therefore \quad \mathrm{f}(x) & =\sin x-x+2(x-\pi) \cos x, x \geq 0 \\
& =-\sin x+x+2(x-\pi) \cos x, x \lt 0
\end{aligned}$
Now,
$\begin{aligned}
\mathrm{f}^{\prime}(x) & =\cos x-1-2(x-\pi) \sin x+2 \cos x, x \geq 0 \\
& =-\cos x+1-2(x+\pi) \sin x+2 \cos x, x \lt 0 \\
\mathrm{f}\left(0^{+}\right) & =\cos 0-1-2(0-\pi) \sin 0+2 \cos 0=2 \\
\mathrm{f}\left(0^{-}\right) & =-\cos 0+1-2(0-\pi) \sin 0+2 \cos 0=2 \\
\therefore \quad \mathrm{f}^{\prime}\left(0^{+}\right) & =\mathrm{f}^{\prime}\left(0^{-}\right)=2
\end{aligned}$
$\therefore \quad \mathrm{f}(x)$ is differentiable at $x=0$
$\therefore \quad \mathrm{f}(x)$ is differentiable everywhere.
$\therefore \quad \mathrm{k}=\phi$
Hence, the correct answer is option: 'B'.