JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 4)
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 :
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 :
$\frac{\pi}{16}$
$\frac{\pi}{48}$
$\frac{\pi}{8}$
$\frac{\pi}{12}$
Explanation
$\sin ^{-1} \sin \theta-\left(\frac{\pi}{2}-\sin ^{-1} \sin \theta\right)>0$
$\Rightarrow \sin ^{-1} \sin \theta>\frac{\pi}{4}$
$\Rightarrow \sin \theta>\frac{1}{\sqrt{2}}$
So, $\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$
$\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)=(\mathrm{a}, \mathrm{b})$
$b-a=\frac{\pi}{2}=\alpha-\beta$
$\Rightarrow \beta=\alpha-\frac{\pi}{2}$
$\Rightarrow \alpha x^{2}+\beta \mathrm{x}+\sin ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]+\cos ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]=0$
$x=3,9 \alpha+3 \beta+\frac{\pi}{2}+0=0$
$$ \begin{aligned} & \Rightarrow 9 \alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0 \\\\ & \Rightarrow 12 \alpha-\pi=0 \\\\ & \alpha=\frac{\pi}{12} \end{aligned} $$
$\Rightarrow \sin ^{-1} \sin \theta>\frac{\pi}{4}$
$\Rightarrow \sin \theta>\frac{1}{\sqrt{2}}$
So, $\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$
$\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)=(\mathrm{a}, \mathrm{b})$
$b-a=\frac{\pi}{2}=\alpha-\beta$
$\Rightarrow \beta=\alpha-\frac{\pi}{2}$
$\Rightarrow \alpha x^{2}+\beta \mathrm{x}+\sin ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]+\cos ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]=0$
$x=3,9 \alpha+3 \beta+\frac{\pi}{2}+0=0$
$$ \begin{aligned} & \Rightarrow 9 \alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0 \\\\ & \Rightarrow 12 \alpha-\pi=0 \\\\ & \alpha=\frac{\pi}{12} \end{aligned} $$
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