Question

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

• A. $\frac{\pi}{8}$
• B. $\frac{\pi}{48}$
• C. $\frac{\pi}{16}$
• D. $\frac{\pi}{12}$

Answer 1

Answer: (D)

$\frac{\pi}{12}$

Answer 2

sin−1sinθ−(2π​−sin−1sinθ)>0
⇒sin−1sinθ>4π​
⇒sinθ>2​1​
So, θ∈(4π​,43π​)
θ∈(4π​,43π​)=(a,b)
⇒b−a=2π​=α−β
⇒β=α−2π​
⇒αx2+βx+sin−1[(x−3)2+1]+cos−1[(x−3)2+1]=0
x=3,9α+3β+2π​+0=0
⇒9α+3(α−2π​)+2π​=0
⇒12α−π=0
α=12π​
So, the correct option is (D) : $\frac{\pi}{12}$