Let x = m/n (m, n are co-prime natural numbers) be a solution of the equation cos(2sin^-1 x) = 1/9

Question

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 \(\mathrm{mx}^{2}-\mathrm{nx}-\mathrm{m}+\mathrm{n}=0\). 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) \(5 x+8 y=9\)

Answer 1

Correct option is (4) \(5 x+8 y=9\)

Assume \(\sin ^{-1} \mathrm{x}=\theta\)

\( \cos (2 \theta)=\frac{1}{9}\)

\(\sin \theta= \pm \frac{2}{3}\)

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 \(5 x+8 y=9\).