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
Solution
<p>Assume $\sin ^{-1} x=\theta$</p>
<p>$$\begin{aligned}
& \cos (2 \theta)=\frac{1}{9} \\
& \sin \theta= \pm \frac{2}{3}
\end{aligned}$$</p>
<p>as $\mathrm{m}$ and $\mathrm{n}$ are co-prime natural numbers,</p>
<p>$\mathrm{x}=\frac{2}{3}$</p>
<p>i.e. $m=2, n=3$</p>
<p>So, the quadratic equation becomes $2 x^2-3 x+1=0$ whose roots are $\alpha=1, \beta=\frac{1}{2}$</p>
<p>$\left(1, \frac{1}{2}\right)$ lies on $5 x+8 y=9$</p>
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
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