MathematicsMedium66×since 2002Q4119Let x=mnx=\frac{m}{n}x=nm (m,nm, nm,n are co-prime natural numbers) be a solution of the equation cos(2sin−1x)=19\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}cos(2sin−1x)=91 and let α,β(α>β)\alpha, \beta(\alpha >\beta)α,β(α>β) be the roots of the equation mx2−nx−m+n=0m x^2-n x-m+ n=0mx2−nx−m+n=0. Then the point (α,β)(\alpha, \beta)(α,β) lies on the lineA3x−2y=−23 x-2 y=-23x−2y=−2B3x+2y=23 x+2 y=23x+2y=2C5x+8y=95 x+8 y=95x+8y=9D5x−8y=−95 x-8 y=-95x−8y=−9Check answerSkip