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

Step 1: Understanding the Concept:
We need to find the maximum value of a function involving inverse trigonometric squared terms. Using the identity \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \), we can transform the expression into a quadratic in terms of one variable.

Step 2: Key Formula or Approach:
Let \( \sin^{-1}x = t \). Then \( \cos^{-1}x = \frac{\pi}{2} - t \).
For \( x \in \left[ -\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}} \right] \), the range of \( t \) is \( \left[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right), \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \right] = \left[ -\frac{\pi}{3}, \frac{\pi}{4} \right] \).

Step 3: Detailed Explanation:
Let the function be \( f(x) = (\sin^{-1}x)^2 + (\cos^{-1}x)^2 \).
Substituting \( t \):
\[ g(t) = t^2 + \left( \frac{\pi}{2} - t \right)^2 \]
\[ g(t) = t^2 + \frac{\pi^2}{4} - \pi t + t^2 = 2t^2 - \pi t + \frac{\pi^2}{4} \]
This is a quadratic function represented by an upward-opening parabola. The vertex occurs at \( t = -\frac{b}{2a} = \frac{\pi}{4} \).
Since the domain is \( t \in \left[ -\frac{\pi}{3}, \frac{\pi}{4} \right] \), the vertex is one of the endpoints.
To find the maximum value, we check the endpoints:
At \( t = \frac{\pi}{4} \):
\[ g\left(\frac{\pi}{4}\right) = 2\left(\frac{\pi^2}{16}\right) - \pi\left(\frac{\pi}{4}\right) + \frac{\pi^2}{4} = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8} \]
At \( t = -\frac{\pi}{3} \):
\[ g\left(-\frac{\pi}{3}\right) = 2\left(\frac{\pi^2}{9}\right) - \pi\left(-\frac{\pi}{3}\right) + \frac{\pi^2}{4} = \frac{2\pi^2}{9} + \frac{\pi^2}{3} + \frac{\pi^2}{4} \]
\[ g\left(-\frac{\pi}{3}\right) = \frac{8\pi^2 + 12\pi^2 + 9\pi^2}{36} = \frac{29\pi^2}{36} \]
Comparing the values, \( \frac{29}{36} \approx 0.8 \) and \( \frac{1}{8} = 0.125 \). Clearly, the maximum value is \( \frac{29}{36}\pi^2 \).
Thus, \( m = 29 \) and \( n = 36 \).
Since \( \gcd(29, 36) = 1 \), the values are correct.

Step 4: Final Answer:
\( m + n = 29 + 36 = 65 \).