Let (a , b) $\in$ (0,2 $π$ ) be the largest interval for which $\sin^{- 1} (\sin θ) - \cos^{- 1} (\sin θ) > 0, θ \in (0, 2 π)$ holds. If $α x^{2} + β x + \sin^{- 1} (x^{2} - 8 x + 17) + \cos^{- 1} (x^{2} - 8 x + 17) = 0$ and $α - β = b - a$ then $α$ is equal to.

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$\frac{π}{12}$

$\frac{3 π}{20}$

$\frac{3 π}{40}$

$\frac{π}{40}$

answer is D.

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Detailed Solution

$\sin^{- 1} (\sin θ) - \cos^{- 1} (\sin θ) > 0 \sin^{- 1} (\sin θ) - \frac{π}{2} + \sin^{- 1} (\sin θ) > 0 \sin^{- 1} (\sin θ) > \frac{π}{4} \sin θ > \frac{1}{\sqrt{2}} \Rightarrow θ \in (\frac{π}{4}, \frac{3 π}{4}) = (a, b) a = \frac{π}{4}, b = \frac{3 π}{4} b - a = \frac{π}{2} = α - β β = α - \frac{π}{2} \to (1)$

$α x^{2} + β x + \sin^{- 1} ({(x - 4)}^{2} + 1) + \cos^{- 1} ({(x - 4)}^{2} + 1) = 0$ this is possible only if $x = 4$

$16 α + 4 β + \frac{π}{2} + 0 = 0 16 α + 4 β + \frac{π}{2} = 0 16 α + 4 (α - \frac{π}{2}) + \frac{π}{2} = 0 20 α - 2 π + \frac{π}{2} = 0 20 α = \frac{3 π}{2} α = \frac{3 π}{40}$