MCQMediumJEE Main 2024 · 29 January, Shift 2Inverse Trigonometric Functions

Mathematics Question from JEE Main 2024 · 29 January, Shift 2

Let x=mnx = \frac{m}{n} (m,nm, n are co-prime natural numbers) be a solution of the equation cos(2sin1x)=19\cos(2\sin^{-1} x) = \frac{1}{9}, and let α,β\alpha, \beta (α>β\alpha > \beta) be the roots of the equation mx2nxm+n=0mx^2 - nx - m + n = 0. Then the point (α,β)(\alpha, \beta) lies on the line:

  • A

    3x+2y=23x + 2y = 2

  • B

    5x8y=95x - 8y = -9

  • C

    3x2y=23x - 2y = -2

  • D

    5x+8y=95x + 8y = 9

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