Let $\mathrm{x}=\frac{\mathrm{m}}{\mathrm{n}}$ ( 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

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Inverse Trigonometric Functions

Let

x = \frac{m}{n}

( m , n are co-prime natural numbers) be a solution of the equation

cos (2 sin^{- 1} x) = \frac{1}{9}

and let

α, β (α > β)

be the roots of the equation

mx^{2} - nx - m + n = 0

. Then the point

(α, β)

lies on the line

Answer: (D)

Let

θ = sin^{- 1} x

, so

sin θ = x

and

cos (2 θ) = 1 - 2 x^{2}

1 - 2 x^{2} = \frac{1}{9} \Rightarrow x^{2} = \frac{4}{9} \Rightarrow x = \frac{2}{3}

(taking the natural-number ratio,

m = 2, n = 3

, co-prime).

The quadratic

m x^{2} - n x - m + n = 0

becomes

2 x^{2} - 3 x + 1 = 0 \Rightarrow (2 x - 1) (x - 1) = 0 \Rightarrow x = 1

x = \frac{1}{2}

.

Since

α > β

α = 1, β = \frac{1}{2}

, so the point is

(1, \frac{1}{2})

.

Checking

5 x + 8 y = 9

5 (1) + 8 (\frac{1}{2}) = 5 + 4 = 9

. ✓ (the other three given lines fail this point.)

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option