Question

The function $f\left(x\right)={\sin }^{-1}\left(\cos x\right)$ is

• A. continuous at x = 0.
• B. not continuous at x = 0.
• C. differentiable at x = 0.
• D. not differentiable at x = 0.

Answer 1

Answer: (A), (D)

The correct options are
A continuous at x = 0.
D not differentiable at x = 0.

We have,
$f\left(x\right)={\sin }^{-1}\left(\cos x\right)\Rightarrow f\left(x\right)=\frac{\pi }{2}-{\cos }^{-1}\left(\cos x\right)\Rightarrow f\left(x\right)=\left(\begin{array}{c}\frac{\pi }{2}-x,\mathrm{if}0\le x\le \pi \\ \frac{\pi }{2}-\left(-x\right),\mathrm{if}-\pi \le x<0\end{array}f\right(0^{+})=f(0^{-})=f(0)\therefore f(x)\mathrm{is}\mathrm{continuous}x=0.\Rightarrow f\text{'}(x)=(\begin{array}{c}-1,\mathrm{if}0\le x\le \pi \\ 1,\mathrm{if}-\pi \le x<0\end{array}f\text{'}\left(0^{+}\right)=-1\&f\text{'}\left(0^{-}\right)=1\therefore f\left(x\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{at}x=0.$