Question Number 63430 by rajesh4661kumar@gamil.com last updated on 04/Jul/19
Answered by MJS last updated on 04/Jul/19
$${y}=\mathrm{sin}^{−\mathrm{1}} \:\sqrt{{x}−\mathrm{1}} \\ $$$$\mathrm{1}\leqslant{x}\leqslant\mathrm{2}\:\Rightarrow\:\mathrm{0}\leqslant\sqrt{{x}−\mathrm{1}}\leqslant\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{0}\leqslant{y}\leqslant\frac{\pi}{\mathrm{2}} \\ $$