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: Plan:
Complete the square in $\sin^{-1}x$.

Step 2: Steps:
Write $f = 2\left(t - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{8}$. The square is least (zero) at $t = \frac{\pi}{4}$, giving $f_{\min} = \frac{\pi^2}{8}$.
The square is largest when $t$ is farthest from $\frac{\pi}{4}$ in $[-\frac{\pi}{2},\frac{\pi}{2}]$, which is $t=-\frac{\pi}{2}$, where $\left(t-\frac{\pi}{4}\right)^2 = \frac{9\pi^2}{16}$. Then $f_{\max} = 2\cdot\frac{9\pi^2}{16} + \frac{\pi^2}{8} = \frac{9\pi^2}{8}+\frac{\pi^2}{8} = \frac{5\pi^2}{4}$.
Sum $= \frac{\pi^2}{8} + \frac{10\pi^2}{8} = \frac{11\pi^2}{8}$.

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