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(B) Given $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$. Since $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we have $\cos ^{-1} x = \frac{\pi}{2} - \sin ^

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$11 \pi^2$

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(B) Given $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$. Since $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we have $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$. Let $t = \sin ^{-1} x$. Since $x \in [-1, 1]$,$t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Then $f(t) = t^2 + (\frac{\pi}{2} - t)^2 = t^2 + \frac{\pi^2}{4} - \pi t + t^2 = 2t^2 - \pi t + \frac{\pi^2}{4}$. 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 occurs at $t = \frac{\pi}{4}$. $\alpha = f(\frac{\pi}{4}) = 2(\frac{\pi^2}{16}) - \pi(\frac{\pi}{4}) + \frac{\pi^2}{4} = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}$. The maximum value occurs at the endpoints of the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. At $t = -\frac{\pi}{2}$,$f(-\frac{\pi}{2}) = 2(-\frac{\pi}{2})^2 - \pi(-\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}$. At $t = \frac{\pi}{2}$,$f(\frac{\pi}{2}) = 2(\frac{\pi}{2})^2 - \pi(\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}$. Thus,$\beta = \frac{5\pi^2}{4}$. Finally,$8(\alpha + \beta) = 8(\frac{\pi^2}{8} + \frac{5\pi^2}{4}) = 8(\frac{\pi^2 + 10\pi^2}{8}) = 11\pi^2$.

If $\alpha$ and $\beta$ are the least and the greatest values of $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$ for all $x \in [-1, 1]$ respectively,then $8(\alpha+\beta)=$

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