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Let \mathrm{(a, b) \subset(0,2 \pi)} be the largest interval for which \mathrm{\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi)}, holds. If \mathrm{\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 :

 

Option: 1

\frac{\pi}{16}
 


Option: 2

\frac{\pi}{48}
 


Option: 3

\frac{\pi}{12}
 


Option: 4

\frac{\pi}{8}


Answers (1)

best_answer

\mathrm{\mathrm{x}^2-6 \mathrm{x}+10=(\mathrm{x}-3)^2+1 \geq 1}
So, \mathrm{x=3} is the only element in the Domain
So, \mathrm{\alpha x^2+\beta x+\sin ^{-1}\left(x^2-6 x+10\right)+\cos ^{-1}\left(x^2-6 x+10\right)=0}
\mathrm{ 9 \alpha+3 \beta+\frac{\pi}{2}=0 }
\mathrm{ \sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0 }
\mathrm{\sin ^{-1}(\sin \theta)-\left(\frac{\pi}{2}-\sin ^{-1}(\sin \theta)\right)>0 }
\mathrm{2 \sin ^{-1}(\sin \theta)>\frac{\pi}{2} }
\mathrm{ \sin ^{-1}(\sin \theta)>\frac{\pi}{4} }

So, \mathrm{\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)}
\mathrm{ \alpha-\beta=\frac{3 \pi}{4}-\frac{\pi}{4}=\frac{\pi}{2} }..............(1)
\mathrm{ \quad 9 \alpha+3 \beta=\frac{-\pi}{2} }
\mathrm{ 3 \alpha+\beta=\frac{-\pi}{6} }..................(2)
Adding (1) & (2)
\mathrm{ 4 \alpha=\frac{\pi}{2}-\frac{\pi}{6}=\frac{2 \pi}{6} }
\mathrm{ \alpha=\frac{\pi}{12} }
 

Posted by

Divya Prakash Singh

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