Question

Let \(f(x) = (sin^{-1}x)^2+(cos^{-1}x)^2\) be a real-valued function defined on its domain. Then the sum of the greatest and the least values of \(f(x)\) is

• A. \(\frac{π^2}{8}\)
• B. \(\frac{11π^2}{8}\)
• C. \(\frac{3π^2}{8}\)
• D. \(\frac{7π^2}{8}\)

Hint:

Use \(\cos^{-1}x = \frac{\pi}{2}-\sin^{-1}x\) to get a quadratic in \(t=\sin^{-1}x\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The domain is \(x\in[-1,1]\). Let \(t = \sin^{-1}x\), so \(t\in[-\frac{\pi}{2},\frac{\pi}{2}]\) and \(\cos^{-1}x = \frac{\pi}{2}-t\).

Step 2: Form the quadratic:
\[ f = t^2 + \left(\frac{\pi}{2}-t\right)^2 = 2t^2 - \pi t + \frac{\pi^2}{4} \]

Step 3: Least value:
The parabola opens upwards, with vertex at \(t = \frac{\pi}{4}\), which lies in the interval. \(f_{\min} = 2\cdot\frac{\pi^2}{16} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}\).

Step 4: Greatest value:
The maximum is at an end of the interval, whichever is farther from \(\frac{\pi}{4}\). That is \(t = -\frac{\pi}{2}\).
\(f(-\frac{\pi}{2}) = 2\cdot\frac{\pi^2}{4} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}\). At \(t=\frac{\pi}{2}\), \(f = \frac{\pi^2}{4}\), which is smaller.

Step 5: Sum:
\[ \frac{\pi^2}{8} + \frac{5\pi^2}{4} = \frac{\pi^2 + 10\pi^2}{8} = \frac{11\pi^2}{8} \]

Final Answer:
The sum of the greatest and least values is \(\frac{11\pi^2}{8}\), option (B). \[ \boxed{\frac{11\pi^2}{8}} \]