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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

माना कि \( (sin^{-1}x)^{2} + (cos^{-1}x)^{2} \) का अधिकतम मान \( x\in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}] \) के लिए \( \frac{m}{n}\pi^{2} \) है, जहाँ gcd (m, n) = 1 है। तब \( m+n \) = ...........

  1. A 55
  2. B 65
  3. C 75
  4. D 45
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Answer & Solution

Correct Answer

(B) 65

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Detailed explanation

\( (sin^{-1}x)^{2}+(cos^{-1}x)^{2} \) \(= (sin^{-1}x+cos^{-1}x)^{2}-2~sin^{-1}x~cos^{-1}x \) \( =\frac{\pi^{2}}{4}-2(sin^{-1}x)(\frac{\pi}{2}-sin^{-1}x) \) \( =2(sin^{-1}x-\frac{\pi}{4})^{2}+\frac{\pi^{2}}{8} \) where \( sin^{-1}x\in[\frac{-\pi}{3},\frac{\pi}{4}] \) Then max…
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