Question

Let the maximum value of \( (\sin^{-1}x)^2 + (\cos^{-1}x)^2 \) for \( x \in \left[ -\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}} \right] \) be \( \frac{m}{n}\pi^2 \), where \( \gcd(m, n) = 1 \). Then \( m + n \) is equal to _________.

Hint:

For any quadratic function \( f(t) \) on an interval \( [a, b] \), the extrema occur either at the vertex or at the endpoints. If the vertex is at one endpoint, the maximum must be at the other endpoint.

Answer 1

Correct Answer: 65

To solve the problem, we need to find the maximum value of the expression \( f(x) = (\sin^{-1}x)^2 + (\cos^{-1}x)^2 \) for \( x \in \left[ -\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}} \right] \). We know that for any \( x \) in the domain of the inverse trigonometric functions, \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \). Therefore, let's express \( f(x) \) in terms of a single variable: \[ f(x) = (\sin^{-1}x)^2 + \left(\frac{\pi}{2} - \sin^{-1}x\right)^2 \]. This simplifies to: \[ f(x) = (\sin^{-1}x)^2 + \left(\frac{\pi}{2} - \sin^{-1}x\right)^2 = (\sin^{-1}x)^2 + \left(\frac{\pi^2}{4} - \pi\sin^{-1}x + (\sin^{-1}x)^2\right). \] Thus, \[ f(x) = 2(\sin^{-1}x)^2 - \pi\sin^{-1}x + \frac{\pi^2}{4}. \] To find the maximum value, we can take the derivative and set it to zero: \[ f'(x) = 2\sin^{-1}x \cdot \frac{1}{\sqrt{1-x^2}} - \pi \cdot \frac{1}{\sqrt{1-x^2}} = \frac{2\sin^{-1}x - \pi}{\sqrt{1-x^2}}. \] Setting \( f'(x) = 0 \) gives \( 2\sin^{-1}x = \pi \), so \( \sin^{-1}x = \frac{\pi}{2} \), which is impossible within the given domain for \( x \). Therefore, we evaluate \( f(x) \) at the endpoints of the interval: \[ f\left(-\frac{\sqrt{3}}{2}\right) = 2 \left(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)^2 - \pi\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) + \frac{\pi^2}{4} = 2\left(-\frac{\pi}{3}\right)^2 + \frac{\pi^2}{3} + \frac{\pi^2}{4}, \] \[ f\left(\frac{1}{\sqrt{2}}\right) = 2\left(\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)^2 - \pi\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) + \frac{\pi^2}{4} = 2\left(\frac{\pi}{4}\right)^2 - \pi\left(\frac{\pi}{4}\right) + \frac{\pi^2}{4} = \frac{\pi^2}{8}. \] The expression at \( x = \frac{1}{\sqrt{2}} \) simplifies to: \[ \frac{\pi^2}{8}, \] and at \( x = -\frac{\sqrt{3}}{2} \), we get: \[ \frac{2\pi^2}{9} + \frac{\pi^2}{3} + \frac{\pi^2}{4} = \frac{7\pi^2}{36}. \] Comparing \(\frac{\pi^2}{8}\) and \(\frac{7\pi^2}{36}\), \(\frac{7}{36} < \frac{1}{8}\), so the maximum is \(\frac{\pi^2}{8}\), which is of form \(\frac{m}{n}\pi^2\) where \( \gcd(m, n) = 1 \). Here, \( m = 1 \), \( n = 8 \), thus \( m+n = 9 \). Therefore, the answer is confirmed to be within the range and is \( 9 \).