Let x = frac{m}{n} (m, n are co-prime natural | vedclass

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(D) Let $\sin^{-1} x = 	heta$. Then $x = \sin 	heta$. Given $\cos(2	heta) = \frac{1}{9}$. Using the identity $\cos(2	heta) = 1 - 2\sin^2 	heta$,w

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$5x + 8y = 9$

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(D) Let $\sin^{-1} x = 	heta$. Then $x = \sin 	heta$. Given $\cos(2	heta) = \frac{1}{9}$. Using the identity $\cos(2	heta) = 1 - 2\sin^2 	heta$,we have $1 - 2x^2 = \frac{1}{9}$. $2x^2 = 1 - \frac{1}{9} = \frac{8}{9} \implies x^2 = \frac{4}{9} \implies x = \frac{2}{3}$ (since $x$ must be positive for $\sin^{-1} x$ to be defined in the context of natural numbers $m, n$). Thus,$m = 2$ and $n = 3$. The quadratic equation is $2x^2 - 3x - 2 + 3 = 0$,which simplifies to $2x^2 - 3x + 1 = 0$. Factoring the equation: $(2x - 1)(x - 1) = 0$. The roots are $x = 1$ and $x = \frac{1}{2}$. Given $\alpha > \beta$,we have $\alpha = 1$ and $\beta = \frac{1}{2}$. Checking the point $(1, \frac{1}{2})$ in the options: For $5x + 8y = 9$: $5(1) + 8(\frac{1}{2}) = 5 + 4 = 9$. Thus,the point lies on the line $5x + 8y = 9$.

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 $\alpha, \beta$ $(\alpha > \beta)$ be the roots of the equation $mx^2 - nx - m + n = 0$. Then the point $(\alpha, \beta)$ lies on the line

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