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 \( m x^{2}-\frac{9}{n} x-m+n=0 \). Then the point \( (\alpha, \beta) \) lies on the line [Jan. 29, 2024(II)] (a) \( 5 x-8 y=-9 \) (c) \( 5 x+8 y=9 \) (b) \( 3 x-2 y=-2 \) (d) \( 3 x+2 y=2 \)

Answer 1

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

Assume \(\sin^{-1}x = \theta\)

\(\cos(2\theta) = \frac 19\)

\(\sin \theta = \pm \frac 23\)

as m and n are co-prime natural numbers,

\(x = \frac 23\)

i.e. m = 2, n = 3

So, the quadratic equation becomes \(2x^2-3x+1= 0\) whose roots are \(\alpha = 1, \beta = \frac 12 (1, \frac 12)\) lies on \(5x + 8y = 9\).