Question

The range of ${\sin }^{-1}\left(2^x\right)$ is

• A. $\left(\frac{-\pi }{2},\frac{\pi }{2}\right]$
• B. $(0,\frac{\pi }{2}]$
• C. $(0,\pi )$
• D. $\left(0,\pi \right]$

Answer 1

The correct option is B $(0,\frac{\pi }{2}]$

$\mathrm{We}\mathrm{know}\mathrm{that}{2}^x>0{\sin }^{-1}\left(2^x\right)\mathrm{in}\mathrm{order}\mathrm{to}\mathrm{be}\mathrm{defined},\mathrm{we}\mathrm{should}\mathrm{have}{2}^x\le 1.\mathrm{Therefore},0<2^x\le 1\Rightarrow 0<{\sin }^{-1}\left(2^x\right)\le \frac{\pi }{2}$