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

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Aicharya · Accepted Answer

Given, $2\left(\cos ^{-1} x\right)^2-\pi \cos ^{-1} x+\frac{\pi^2}{4}$ Let $\cos ^{-1} x=y$ $\begin{aligned} \therefore \quad & 2 y^2-\pi y+\frac{\pi^2}{4}=2\left[y^2-\frac{\pi}{2} y\right]+\frac{\pi^2}{4} \ & =2\left[\left(y-\frac{\pi}{4}\right)^2-\frac{\pi^2}{16}\right]+\frac{\pi^2}{4}=2\left(y-\frac{\pi}{4}\right)^2+\frac{\pi^2}{8} \ & =2\left(\cos ^{-1} x-\frac{\pi}{4}\right)^2+\frac{\pi^2}{8} \end{aligned}$ We know that, $\begin{aligned} 0 & \leq \cos ^{-1} x \leq \pi \Rightarrow-\frac{\pi}{4} \leq \cos ^{-1} x-\frac{\pi}{4} \leq \frac{3 \pi}{4} \ 0 & \leq\left(\cos ^{-1} x-\frac{\pi}{4}\right)^2 \leq \frac{9 \pi^2}{16} \ 0 & \leq 2\left(\cos ^{-1} x-\frac{\pi}{4}\right)^2 \leq \frac{9 \pi^2}{8} \ \frac{\pi^2}{8} & \leq 2\left(\cos ^{-1} x-\frac{\pi}{4}\right)^2+\frac{\pi^2}{8} \leq 5 \frac{\pi^2}{4} \ \therefore & \text { Required sum }=\frac{\pi^2}{8}+\frac{5 \pi^2}{4}=\frac{11 \pi^2}{8} \end{aligned}$ There is no option match.

The sum of the maximum and the minimum values of \(2\left(\cos ^{-1} x\right)^2-\pi \cos ^{-1}…

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