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 :

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