JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
Let the maximum value of \( (sin^{-1}x)^{2} + (cos^{-1}x)^{2} \) for \( x\in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}] \) be \( \frac{m}{n}\pi^{2} \), where gcd (m, n) = 1. Then \( m+n \) is equal to ........... .
- A 55
- B 65
- C 75
- D 45
Answer & Solution
Correct Answer
(B) 65
Step-by-step Solution
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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