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Question:

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) $\alpha=0, \beta=2 \pi$

Solution:

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

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

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

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).

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