Inverse Trigonometric Functions Question 2
Question 2 - 2024 (29 Jan Shift 2)
Let $x=\frac{m}{n}\left(m, n\right.$ 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 $mx^{2}-nx-m+n=0$. Then the point $(\alpha, \beta)$ lies on the line
(1) $3 x+2 y=2$
(2) $5 x-8 y=-9$
(3) $3 x-2 y=-2$
(4) $5 x+8 y=9$
Show Answer
Answer (4)
Solution
Assume $\sin ^{-1} x=\theta$
$\cos (2 \theta)=\frac{1}{9}$
$\sin \theta= \pm \frac{2}{3}$
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 $5 x+8 y=9$
