Question

Let $(a, b) \subset(0,2 \pi)$ be the largest interval for which
$\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi)$, holds. If $\alpha x^2+\beta x+\sin ^{-1}\left(x^2-6 x+10\right)+\cos ^{-1}\left(x^2-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to :

• A. $\frac{\pi}{8}$
• B. $\frac{\pi}{48}$
• C. $\frac{\pi}{16}$
• D. $\frac{\pi}{12}$

Answer 1

Answer: (D)

$\sin ^{-1} \sin \theta-\left(\frac{\pi}{2}-\sin ^{-1} \sin \theta\right)>0$
$\Rightarrow \sin ^{-1} \sin \theta>\frac{\pi}{4}$
$\Rightarrow \sin \theta>\frac{1}{\sqrt{2}}$
$\text { So, } \theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$
$\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)=(a, b)$
$\Rightarrow b-a=\frac{\pi}{2}=\alpha-\beta$
$\Rightarrow \beta=\alpha-\frac{\pi}{2}$
$\Rightarrow \alpha x^2+\beta x+\sin ^{-1}\left[(x-3)^2+1\right]+\cos ^{-1}\left[(x-3)^2+1\right]=0$
$x=3,9 \alpha+3 \beta+\frac{\pi}{2}+0=0$
$\Rightarrow 9 \alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0$
$\Rightarrow 12 \alpha-\pi=0$
$\alpha=\frac{\pi}{12}$