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JEE Main 2024
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Medium

Question

Let (a, b) ⊂(0,2π)\subset(0,2 \pi) be the largest interval for which sin⁡−1(sin⁡θ)−cos⁡−1(sin⁡θ)>0,θ∈(0,2π)\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi), holds. If αx2+βx+sin⁡−1(x2−6x+10)+cos⁡−1(x2−6x+10)=0\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 α−β=b−a\alpha-\beta=b-a, then α\alpha is equal to :

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Solution

sin⁡−1sin⁡θ−(π2−sin⁡−1sin⁡θ)>0\sin ^{-1} \sin \theta-\left(\frac{\pi}{2}-\sin ^{-1} \sin \theta\right)>0 ⇒sin⁡−1sin⁡θ>π4\Rightarrow \sin ^{-1} \sin \theta>\frac{\pi}{4} ⇒sin⁡θ>12\Rightarrow \sin \theta>\frac{1}{\sqrt{2}} So, θ∈(π4,3π4)\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right) θ∈(π4,3π4)=(a,b)\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)=(\mathrm{a}, \mathrm{b}) b−a=π2=α−βb-a=\frac{\pi}{2}=\alpha-\beta ⇒β=α−π2\Rightarrow \beta=\alpha-\frac{\pi}{2} ⇒αx2+βx+sin⁡−1[(x−3)2+1]+cos⁡−1[(x−3)2+1]=0\Rightarrow \alpha x^{2}+\beta \mathrm{x}+\sin ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]+\cos ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]=0 x=3,9α+3β+π2+0=0x=3,9 \alpha+3 \beta+\frac{\pi}{2}+0=0 ⇒9α+3(α−π2)+π2=0⇒12α−π=0α=π12\begin{aligned} & \Rightarrow 9 \alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0 \\\\ & \Rightarrow 12 \alpha-\pi=0 \\\\ & \alpha=\frac{\pi}{12} \end{aligned}

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