Question

$$\begin{gathered} \mathrm{If}\alpha \le 2{\sin }^{-1}x+{\cos }^{-1}x\le \beta ,\mathrm{then} \\ \left(a\right)\alpha =-\frac{\mathrm{\pi }}{2},\beta =\frac{\mathrm{\pi }}{2}\left(b\right)\alpha =0,\beta =\pi \left(c\right)\alpha =-\frac{\mathrm{\pi }}{2},\beta =\frac{3\mathrm{\pi }}{2}\left(d\right)\alpha =0,\beta =2\pi \end{gathered}$$

Answer 1

$2{\sin }^{-1}x+{\cos }^{-1}x$

$={\sin }^{-1}x+{\sin }^{-1}x+{\cos }^{-1}x$

$={\sin }^{-1}x+\frac{\mathrm{\pi }}{2}\left({\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}\right)$

Now,

$-\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}x\le \frac{\mathrm{\pi }}{2}$

$\Rightarrow -\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}x+\frac{\mathrm{\pi }}{2}\le \frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}$

$\Rightarrow 0\le {\sin }^{-1}x+\frac{\mathrm{\pi }}{2}\le \mathrm{\pi }$

$\therefore 0\le 2{\sin }^{-1}x+{\cos }^{-1}x\le \mathrm{\pi }$

Comparing with $\alpha \le 2{\sin }^{-1}x+{\cos }^{-1}x\le \beta$, we get

$\alpha =0,\beta =\mathrm{\pi }$

Hence, the correct answer is option (b).