Question

If $\alpha \leq 2 \sin^{-1}x+\cos^{-1}x \leq \beta$, then

• A. $\alpha=\frac{-\pi}{2} \,\,\, \beta=\frac{\pi}{2}$
• B. $\alpha=\frac{-\pi}{2} \,\,\, \beta=\frac{3 \pi}{2}$
• C. $\alpha=0 \,\,\, \beta=\pi$
• D. $\alpha=0\,\,\, \beta=2 \pi$

Answer 1

Answer: (C)

We know that , $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$
$\Rightarrow 0 \leq \frac{\pi}{2}+\sin ^{-1} x \leq \pi$
$\Rightarrow 0 \leq \sin ^{-1} x+\cos ^{-1} x+\sin ^{-1} x \leq \pi$
$\Rightarrow 0 \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \pi$
On comparing. we get
$\alpha=0$ and $\beta=\pi$