If $\alpha \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \beta$, then
(a) $\alpha=-\frac{\pi}{2}, \beta=\frac{\pi}{2}$
(b) $\alpha=0, \beta=\pi$
(c) $\alpha=-\frac{\pi}{2}, \beta=\frac{3 \pi}{2}$
(d) $\alpha=0, \beta=2 \pi$
$2 \sin ^{-1} x+\cos ^{-1} x$
$=\sin ^{-1} x+\sin ^{-1} x+\cos ^{-1} x$
$=\sin ^{-1} x+\frac{\pi}{2}$ $\left(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)$
Now,
$-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$
$\Rightarrow-\frac{\pi}{2}+\frac{\pi}{2} \leq \sin ^{-1} x+\frac{\pi}{2} \leq \frac{\pi}{2}+\frac{\pi}{2}$
$\Rightarrow 0 \leq \sin ^{-1} x+\frac{\pi}{2} \leq \pi$
$\therefore 0 \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \pi$
Comparing with $\alpha \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \beta$, we get
$\alpha=0, \beta=\pi$
Hence, the correct answer is option (b).