1. Tardigrade
  2. Question
  3. Mathematics
  4. Let (a, b) ⊂(0,2 π) be the largest interval for which sin -1( sin θ)- cos -1( sin θ)>0, θ ∈(0,2 π), holds. If α x2+β x+ sin -1(x2-6 x+10)+ cos -1(x2-6 x+10)=0 and α-β=b-a, then α is equal to :

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 :

JEE MainJEE Main 2023Trigonometric Functions

Solution:

$ \sin ^{-1} \sin \theta-\left(\frac{\pi}{2}-\sin ^{-1} \sin \theta\right)>0 $
$ \Rightarrow \sin ^{-1} \sin \theta>\frac{\pi}{4}$
$ \Rightarrow \sin \theta>\frac{1}{\sqrt{2}} $
$ \text { So, } \theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right) $
$ \theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)=(a, b) $
$ \Rightarrow b-a=\frac{\pi}{2}=\alpha-\beta $
$ \Rightarrow \beta=\alpha-\frac{\pi}{2}$
$ \Rightarrow \alpha x^2+\beta x+\sin ^{-1}\left[(x-3)^2+1\right]+\cos ^{-1}\left[(x-3)^2+1\right]=0 $
$ x=3,9 \alpha+3 \beta+\frac{\pi}{2}+0=0$
$ \Rightarrow 9 \alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0 $
$ \Rightarrow 12 \alpha-\pi=0 $
$ \alpha=\frac{\pi}{12}$