Discussion
If \(\alpha\) and \(\beta\) are respectively the minimum and maximum values of \(\frac{\pi^{2}}{8} + 2\left(\sin^{-1}x - \frac{\pi}{4}\right)^{2}\) , then \(\frac{\beta}{\alpha} =\)
Answer: A
\(\frac{\beta}{\alpha} = 10\).
Let \(t = \sin^{-1}x\), which ranges over \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). The expression \(\frac{\pi^{2}}{8} + 2\left(t - \frac{\pi}{4}\right)^{2}\) is smallest when \(t = \frac{\pi}{4}\) and largest at the end of the interval farthest from \(\frac{\pi}{4}\), which is \(t = -\frac{\pi}{2}\).
\[ \begin{aligned} \alpha &= \frac{\pi^{2}}{8} \\ \beta &= \frac{\pi^{2}}{8} + 2\left(\frac{3\pi}{4}\right)^{2} = \frac{\pi^{2}}{8} + \frac{9\pi^{2}}{8} \\ &= \frac{10\pi^{2}}{8} \end{aligned} \]
So \(\frac{\beta}{\alpha} = 10\). Using \(t = \frac{\pi}{2}\) for the maximum gives \(\beta = \frac{2\pi^{2}}{8}\) and a ratio of 2, the trap of choosing the wrong endpoint.
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