Question

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

• A. $\frac{11\pi^{2}}{8}$
• B. $\frac{3 \pi^{2}}{8}$
• C. $\frac{3 \pi^{2}}{2}$
• D. $4 \pi^{2}$

Answer 1

Answer: (A)

Given, $2\left(\cos ^{-1} x\right)^{2}-\pi \cos ^{-1} x+\frac{\pi^{2}}{4}$
Let $\cos ^{-1} x=y$
$\therefore 2 y^{2}-\pi y+\frac{\pi^{2}}{4}=2\left[y^{2}-\frac{\pi}{2} y\right]+\frac{\pi^{2}}{4}$
$=2\left[\left(y-\frac{\pi}{4}\right)^{2}-\frac{\pi^{2}}{16}\right]+\frac{\pi^{2}}{4}=2\left(y-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}$
$\quad=2\left(\cos ^{-1} x-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}$
We know that $0 \leq \cos ^{-1} x \leq \pi \Rightarrow -\frac{\pi}{4} \leq \cos ^{-1} x-\frac{\pi}{4} \leq \frac{3 \pi}{4}$
$0 \leq\left(\cos ^{-1} x-\frac{\pi}{4}\right)^{2} \leq \frac{9 \pi^{2}}{16}$
$0 \leq 2\left(\cos ^{-1} x-\frac{\pi}{4}\right)^{2} \leq \frac{9 \pi^{2}}{8}$
$\frac{\pi^{2}}{8} \leq 2\left(\cos ^{-1} x-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8} \leq 5 \frac{\pi^{2}}{4}$
$\therefore $ Required sum $=\frac{\pi^{2}}{8}+\frac{5 \pi^{2}}{4}=\frac{11 \pi^{2}}{8}$