JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

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 :

View answer and explanation

Correct option: D

\frac{\pi}{12}

Option D (

\frac{\pi}{12}

) is correct because by solving the given conditions, we find that

\alpha = \frac{\pi}{12}

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