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