Let $x=\frac{m}{n}\left(\mathrm{~m}, n\right.$ 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 $\mathrm{mx}^2-\mathrm{nx}-\mathrm{m}+\mathrm{n}=0$. Then the point $(\alpha, \beta)$ lies on the line
Select the correct option:
A
$3 x+2 y=2$
B
$5 x-8 y=-9$
C
$3 x-2 y=-2$
D
$5 x+8 y=9$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Assume $\sin ^{-1} \mathrm{x}=\theta$
$$
\begin{aligned}
& \cos (2 \theta)=\frac{1}{9} \\
& \sin \theta= \pm \frac{2}{3}
\end{aligned}
$$
as m and n are co-prime natural numbers,
$$
x=\frac{2}{3}
$$
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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