Let ( a , b ) ( 0 , 2 ) be the largest interval for which sin - 1 ( sin ) - cos - 1 ( sin ) 0 , ( 0 , 2 ) , holds. If x 2 + x + sin - 1 ( x 2 - 6 x + 10 ) + cos - 1 ( x 2 - 6 x + 10 ) = 0 and = b - a , then is equal to [2023]

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Solution

Q.
Let $(a, b) \subset (0, 2 π)$ be the largest interval for which $\sin^{- 1} (\sin θ) - \cos^{- 1} (\sin θ) > 0,$ $θ \in (0, 2 π),$ holds.

If $α x^{2} + β x + \sin^{- 1} (x^{2} - 6 x + 10) + \cos^{- 1} (x^{2} - 6 x + 10) = 0$ and $α - β = b - a,$ then $α$ is equal to [2023]

1 $\frac{π}{8}$

2 $\frac{π}{48}$

3 $\frac{π}{16}$

4 $\frac{π}{12}$

Ans.
(4)

We know that $\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$

$\sin^{- 1} (\sin θ) - (\frac{π}{2} - \sin^{- 1} (\sin θ)) > 0;$ This equation holds true if, $\sin^{- 1} (\sin θ) > \frac{π}{4} \Rightarrow \sin θ > \frac{1}{\sqrt{2}}$ So, $θ \in (\frac{π}{4}, \frac{3 π}{4})$

$θ \in (\frac{π}{4}, \frac{3 π}{4}) = (a, b) \Rightarrow b - a = \frac{π}{2} = α - β [Given]$

$\Rightarrow β = α - \frac{π}{2}$

$Now, α x^{2} + β x + \sin^{- 1} [{(x - 3)}^{2} + 1] + \cos^{- 1} [{(x - 3)}^{2} + 1] = 0$

$\sin^{- 1} [{(x - 3)}^{2} + 1] + \cos^{- 1} [{(x - 3)}^{2} + 1] = \frac{π}{2} if x = 3$

$Put x = 3, 9 α + 3 β + \frac{π}{2} = 0; Put β = α - \frac{π}{2}$

$\Rightarrow 9 α + 3 (α - \frac{π}{2}) + \frac{π}{2} = 0$

$\Rightarrow 12 α - π = 0 \Rightarrow α = \frac{π}{12}$

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