The sum of the maximum and the minimum values | vedclass

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(B) Let $f(x) = 2(\cos ^{-1} x)^2-\pi \cos ^{-1} x+\frac{\pi^2}{4}$.Let $y = \cos ^{-1} x$. Since $x \in [-1, 1]$,we have $y \in [0, \pi]$.Then $f(y)

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $f(x) = 2(\cos ^{-1} x)^2-\pi \cos ^{-1} x+\frac{\pi^2}{4}$.Let $y = \cos ^{-1} x$. Since $x \in [-1, 1]$,we have $y \in [0, \pi]$.Then $f(y) = 2y^2 - \pi y + \frac{\pi^2}{4}$.Completing the square: $f(y) = 2(y^2 - \frac{\pi}{2}y) + \frac{\pi^2}{4} = 2(y - \frac{\pi}{4})^2 - \frac{\pi^2}{8} + \frac{\pi^2}{4} = 2(y - \frac{\pi}{4})^2 + \frac{\pi^2}{8}$.For $y \in [0, \pi]$,the minimum value occurs at $y = \frac{\pi}{4}$,which is $f(\frac{\pi}{4}) = \frac{\pi^2}{8}$.The maximum value occurs at the boundary $y = \pi$ (since $|\pi - \frac{\pi}{4}| > |0 - \frac{\pi}{4}|$).$f(\pi) = 2(\pi - \frac{\pi}{4})^2 + \frac{\pi^2}{8} = 2(\frac{3\pi}{4})^2 + \frac{\pi^2}{8} = 2(\frac{9\pi^2}{16}) + \frac{\pi^2}{8} = \frac{9\pi^2}{8} + \frac{\pi^2}{8} = \frac{10\pi^2}{8} = \frac{5\pi^2}{4}$.The sum of the maximum and minimum values is $\frac{5\pi^2}{4} + \frac{\pi^2}{8} = \frac{10\pi^2 + \pi^2}{8} = \frac{11\pi^2}{8}$.

The sum of the maximum and the minimum values of $2(\cos ^{-1} x)^2-\pi \cos ^{-1} x+\frac{\pi^2}{4}$ is

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