Let ( are co-prime natural numbers) be a solution of the equation and let be the roots of the equation . Then the point lies on the line

Assume as and are co-prime natural numbers, i.e. So, the quadratic equation becomes whose roots are lies on

JEE Main 2020 Inverse Trigonometric Functions: Let ( are co-prime natural numbers) be a solution of the equation and let be the roots of the equati...

The correct answer is A. Assume as and are co-prime natural numbers, i.e. So, the quadratic equation becomes whose roots are lies on

Let ( are co-prime natural numbers) be a solution of the equ... | JEE Main 2020 Inverse Trigonometric Functions

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Inverse Trigonometric Functions

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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

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Solution

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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