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

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Solution

Q.
Let $x = \frac{m}{n}$ $(m, n$ are co-prime natural numbers $)$ be a solution of the equation $\cos (2 \sin^{- 1} x) = \frac{1}{9}$ and let $α, β (α > β)$ be the roots of the equation $m x^{2} - n x - m + n = 0$ . Then the point $(α, β)$ lies on the line [2024]

1 $3 x - 2 y = - 2$

2 $5 x - 8 y = - 9$

3 $3 x + 2 y = 2$

4 $5 x + 8 y = 9$

Ans.
(4)

We have, $\cos (2 \sin^{- 1} x) = \frac{1}{9}$

Put $\sin^{- 1} x = θ$

$\cos 2 θ = \frac{1}{9} \Rightarrow 1 - 2 \sin^{2} θ = \frac{1}{9}$

$\Rightarrow \sin θ = \frac{2}{3}$ $(∵ x > 0 and \sin θ lies in the first quadrant)$

$\Rightarrow x = \frac{2}{3} = \frac{m}{n} .$ So, $m = 2$ and $n = 3$

The given equation becomes, $2 x^{2} - 3 x - 2 + 3 = 0$

$\Rightarrow 2 x^{2} - 3 x + 1 = 0 \Rightarrow 2 x^{2} - 2 x - x + 1 = 0$

$\Rightarrow (2 x - 1) (x - 1) = 0$ $\Rightarrow x = \frac{1}{2} or x = 1$

$\Rightarrow α = 1 and β = \frac{1}{2}, which lies on the line 5 x + 8 y = 9$

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