Question

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 \(\frac{\beta}{\alpha}\) is:

• A. \(10 \)
• B. \(4 \)
• C. \(6 \)
• D. \(2 \)
• E. \(12 \)

Hint:

Check endpoints of range when dealing with squared expressions.

Answer 1

The Correct Option is A

Concept: The range of inverse sine function is: \[ \sin^{-1}x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] For squared expressions, minimum occurs at zero and maximum at endpoints.

Step 1: Let $t = \sin^{-1}x$.
\[ t \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]

Step 2: Transform expression.
\[ t - \frac{\pi}{4} \in \left[-\frac{3\pi}{4}, \frac{\pi}{4}\right] \]

Step 3: Minimum value.
\[ \alpha = \frac{\pi^2}{8} \]

Step 4: Maximum value.
\[ \beta = \frac{\pi^2}{8} + 2\left(\frac{3\pi}{4}\right)^2 = \frac{10\pi^2}{8} \]

Step 5: Ratio.
\[ \frac{\beta}{\alpha} = 10 \]