Question

If \(0\leq x\leq 1\) and \((sin^{-1}x)^3+(cos^{-1}x)^3 = aπ^3\) then

• A. \(a\geq \frac{1}{32}\)
• B. \(a\geq \frac{1}{16}\)
• C. \(a\leq \frac{1}{32}\)
• D. \(a\leq \frac{1}{16}\)

Hint:

Use s + c = pi/2 and find the minimum of the cubic expression.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Let \(s = \sin^{-1}x\) and \(c = \cos^{-1}x\) with \(s + c = \frac\pi2\). For \(0 \leq x \leq 1\), \(s \in [0, \frac\pi2]\).

Step 2: Key Formula or Approach:
\(s^3 + c^3 = (s + c)^3 - 3sc(s + c) = \frac{\pi^3}{8} - \frac{3\pi}{2}s\left(\frac\pi2 - s\right)\).

Step 3: Detailed Explanation:
The product \(s\left(\frac\pi2 - s\right)\) lies between \(0\) and its maximum \(\frac{\pi^2}{16}\) (at \(s = \frac\pi4\)).
So \(s^3 + c^3\) lies between \(\frac{\pi^3}{8} - \frac{3\pi}{2}\cdot\frac{\pi^2}{16} = \frac{\pi^3}{32}\) and \(\frac{\pi^3}{8}\).
Thus \(a\pi^3 \geq \frac{\pi^3}{32}\), giving \(a \geq \frac{1}{32}\) (and \(a \leq \frac18\)).
Option B, \(a \geq \frac1{16}\), fails at \(x = \frac{1}{\sqrt2}\) where \(a = \frac{1}{32}\).

Final Answer:
\(a \geq \frac{1}{32}\), option (A). \[ \boxed{a \geq \frac{1}{32}} \]