Let $(\mathrm{a}, \mathrm{b}) \subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi)$, holds. If $\alpha \mathrm{x}^{2}+\beta \mathrm{x}+\sin ^{-1}\left(\mathrm{x}^{2}-6 \mathrm{x}+10\right)+\cos ^{-1}\left(\mathrm{x}^{2}-6 \mathrm{x}+10\right)=0$ and $\alpha-\beta=\mathrm{b}-\mathrm{a}$, then $\alpha$ is equal to :

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Inverse Trigonometric Functions

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 :

Answer: (D)

The inequality

sin^{- 1} (sin θ) > cos^{- 1} (sin θ)

means

sin^{- 1} (sin θ) > \frac{π}{4}

, hence

sin θ > \frac{1}{2}

and

θ \in (\frac{π}{4}, \frac{3 π}{4})

, so

b - a = \frac{π}{2} = α - β .

Since

x^{2} - 6 x + 10 = (x - 3)^{2} + 1 \geq 1

, both inverse functions are defined only at

x = 3

, where

sin^{- 1} 1 + cos^{- 1} 1 = \frac{π}{2} .

Then

9 α + 3 β + \frac{π}{2} = 0

; using

β = α - \frac{π}{2}

gives

12 α = π

, so

α = \frac{π}{12} .

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Single correct option