Let f(x) = (sin^{-1} x)^2 + (cos^{-1} x)^2 be | vedclass

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(B) The domain of $f(x)$ is $x \in [-1, 1]$.We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.Substituting this into $f(x)$:$f(x) = (\sin^{-1}

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$\frac{11\pi^2}{8}$

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(B) The domain of $f(x)$ is $x \in [-1, 1]$.We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.Substituting this into $f(x)$:$f(x) = (\sin^{-1} x)^2 + (\frac{\pi}{2} - \sin^{-1} x)^2$.Let $t = \sin^{-1} x$. Since $x \in [-1, 1]$, $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.$f(t) = t^2 + (\frac{\pi}{2} - t)^2 = t^2 + \frac{\pi^2}{4} - \pi t + t^2 = 2t^2 - \pi t + \frac{\pi^2}{4}$.This is a parabola opening upwards. The vertex is at $t = -\frac{-\pi}{2(2)} = \frac{\pi}{4}$.Since $\frac{\pi}{4} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, the minimum value is $f(\frac{\pi}{4}) = 2(\frac{\pi}{4})^2 - \pi(\frac{\pi}{4}) + \frac{\pi^2}{4} = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}$.The maximum value occurs at the endpoints of the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.$f(-\frac{\pi}{2}) = 2(-\frac{\pi}{2})^2 - \pi(-\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}$.$f(\frac{\pi}{2}) = 2(\frac{\pi}{2})^2 - \pi(\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}$.Thus, the greatest value is $\frac{5\pi^2}{4}$ and the least value is $\frac{\pi^2}{8}$.Sum $= \frac{5\pi^2}{4} + \frac{\pi^2}{8} = \frac{10\pi^2 + \pi^2}{8} = \frac{11\pi^2}{8}$.

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

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