JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
Let \(\mathrm{x}=\frac{\mathrm{m}}{\mathrm{n}}\) ( \(\mathrm{m}, \mathrm{n}\) 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
- A \(3 x+2 y=2\)
- B \(5 x-8 y=-9\)
- C \(3 x-2 y=-2\)
- D \(5 x+8 y=9\)
Answer & Solution
Correct Answer
(D) \(5 x+8 y=9\)
Step-by-step Solution
Detailed explanation
Assume \(\sin ^{-1} x=\theta\) \( \cos (2 \theta)=\frac{1}{9} \) \( \sin \theta= \pm \frac{2}{3}\) as \(\mathrm{m}\) and \(\mathrm{n}\) are co-prime natural numbers, \(\mathrm{x}=\frac{2}{3}\) i.e. \(m=2, n=3\) So, the quadratic equation becomes \(2 x^2-3 x+1=\) 0 whose roots…
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