JEE MAIN - Mathematics (2024 - 29th January Evening Shift - No. 19)
Let $$x=\frac{m}{n}$$ ($$m, n$$ are co-prime natural numbers) be a solution of the equation $$\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$$ and let $$\alpha, \beta(\alpha >\beta)$$ be the roots of the equation $$m x^2-n x-m+ n=0$$. Then the point $$(\alpha, \beta)$$ lies on the line
$$3 x-2 y=-2$$
$$3 x+2 y=2$$
$$5 x+8 y=9$$
$$5 x-8 y=-9$$
Explanation
Assume $$\sin ^{-1} x=\theta$$
$$\begin{aligned} & \cos (2 \theta)=\frac{1}{9} \\ & \sin \theta= \pm \frac{2}{3} \end{aligned}$$
as $$\mathrm{m}$$ and $$\mathrm{n}$$ are co-prime natural numbers,
$$\mathrm{x}=\frac{2}{3}$$
i.e. $$m=2, n=3$$
So, the quadratic equation becomes $$2 x^2-3 x+1=0$$ whose roots are $$\alpha=1, \beta=\frac{1}{2}$$
$$\left(1, \frac{1}{2}\right)$$ lies on $$5 x+8 y=9$$
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