Question

lf $f\left(x\right)={\sin }^{-1}\left(\sin x\right)\forall x\in [6\pi ,7\pi ]$, then

• A. $f$ is continuous and differentiable everywhere
• B. $f$ is differentiable at $x=\frac{13\pi }{2}$
• C. $f$ is not differentiable at $x=\frac{13\pi }{2}$
• D. $f$ is not continuous at $x=\frac{13\pi }{2}$

Answer 1

The correct option is C $f$ is not differentiable at $x=\frac{13\pi }{2}$

$f\left(x\right)=\{\begin{array}{c}x-6\pi ;6\pi \le x<\frac{13\pi }{2}\\ \frac{\pi }{2};x=\frac{13\pi }{2}\\ 7\pi -x;\frac{13\pi }{2}<x\le 7\pi \end{array}$
$f^{\prime }\left(x\right)=\{\begin{array}{c}1;6\pi \le x<\frac{13\pi }{2}\\ -1;\frac{13\pi }{2}<x\le 7\pi \end{array}$
For continuity,
Left hand limit $f\left(c^{-}\right)=$ Right hand limit $f\left(c^{+}\right)=f\left(c\right)$
$f\left({\frac{13\pi }{2}}^{-}\right)={lim}_{h\to 0}(\frac{13\pi }{2}-h-6\pi )=\frac{\pi }{2}$
$f\left({\frac{13\pi }{2}}^{+}\right)={lim}_{h\to 0}(7\pi -\frac{13\pi }{2}-h)=\frac{\pi }{2}$
$f\left({\frac{13\pi }{2}}^{+}\right)=\frac{\pi }{2}$
$\therefore f\left(x\right)$ is continuous in $[6\pi ,7\pi ]$
But clearly, $f\left(x\right)$ is not differentiable at $x=\frac{13\pi }{2}$
$\because f^{\prime }\left({\frac{13\pi }{2}}^{-}\right)\ne f^{\prime }\left({\frac{13\pi }{2}}^{+}\right)$