Question

The area bounded by the curve $y=\cos x$, the line joining $(-\pi /4,\cos (-\pi /4))$ and $(0,2)$ and the line joining $(\pi /4,\cos (\pi /4))$ and $(0,2)$ is

• A. $\left(\frac{4+\sqrt{2}}{8}\right)\pi -\sqrt{2}$
• B. $\left(\frac{4+\sqrt{2}}{8}\right)\pi +\sqrt{2}$
• C. $\left(\frac{4+\sqrt{2}}{4}\right)\pi -\sqrt{2}$
• D. $\left(\frac{4+\sqrt{2}}{4}\right)\pi +\sqrt{2}$

Answer 1

Answer: (A)

Given, $y=\cos x$
Equation of line joining
$\left(\frac{-\pi }{4},\cos \left(\frac{-\pi }{4}\right)\right)$ and $\left(0,2\right)$ is
$y-2=\frac{2-1\sqrt{2}}{0+\pi 4}\left(x-0\right)$
$\Rightarrow y-2=\left(8-2\sqrt{2}\right)x$
$\Rightarrow y=\frac{\left(8-2\sqrt{2}\right)}{\pi }x+2$
Equation of line joining $\left(\frac{\pi }{4},\cos \frac{\pi }{4}\right)$ and $\left(0,2\right)$ is
$y-2=\frac{2-\frac{1}{\sqrt{2}}}{0-\frac{\pi }{4}}\left(x-0\right)$
$\Rightarrow y=\left(\frac{-8+2\sqrt{2}}{\pi }\right)x+2$
Graph of given curve, $y=\cos x$, and line are

Area of shaded region $=2$ area of curve $APCA$
$=2\underset{0}{\overset{\pi /4}{\int }}\left[\frac{\left(-8+2\sqrt{2}\right)}{\pi }x+2-\cos x\right]dx$
$=2{\left[\frac{-8x^2}{2\pi }+\frac{2\sqrt{2}{x}^2}{2\pi }+2x-\sin x\right]}_0^{\pi /4}$
$=2\left[\frac{-4}{\pi }{\left(\frac{\pi }{4}\right)}^2+\frac{\sqrt{2}}{\pi }{\left(\frac{\pi }{4}\right)}^2+\frac{2\pi }{4}-\frac{1}{\sqrt{2}}\right]$
$=2\left[\frac{\pi }{4}\left(\frac{-4}{4}+\frac{\sqrt{2}}{4}+2\right)-\frac{1}{\sqrt{2}}\right]$
$=\left(\frac{4+\sqrt{2}}{8}\right)\pi -\sqrt{2}$