Question

The area of the region
$A=\left\{(x,y):|\cos x-\sin x|\le y\le \sin x,0\le x\le \frac{\pi }{2}\right\}$

• A. $1-\frac{3}{\sqrt{2}}+\frac{4}{\sqrt{5}}$
• B. $\sqrt{5}+2\sqrt{2}-4.5$
• C. $\frac{3}{\sqrt{5}}-\frac{3}{\sqrt{2}}+1$
• D. $\sqrt{5}-2\sqrt{2}+1$

Answer 1

Answer: (D)

$|\cos x-\sin x|\le y\le \sin x$
Intersection point of $\cos x-\sin x=\sin x$
$\Rightarrow \tan x=\frac{1}{2}$
Let $\psi ={\tan }^{-1}\frac{1}{2}$
So, $\tan \psi =\frac{1}{2},\sin \psi =\frac{1}{\sqrt{5}},\cos \psi =\frac{2}{\sqrt{5}}$

Area $=\underset{\psi }{\overset{\pi /2}{\int }}(\sin x-|\cos x-\sin x|)dx$
$=\underset{\psi }{\overset{\pi /4}{\int }}(\sin x-(\cos x-\sin x))dx$
$+\underset{\pi /4}{\overset{\pi /2}{\int }}(\sin x-(\sin x-\cos x))dx$
$=\underset{\psi }{\overset{\pi /4}{\int }}(2\sin x-\cos x)dx+\underset{\pi /4}{\overset{\pi /2}{\int }}\cos xdx$
$=[-2\cos x-\sin x{]}_{\psi }^{\pi /4}+[\sin x{]}_{\pi /4}^{\pi /2}$
$=-\sqrt{2}-\frac{1}{\sqrt{2}}+2\cos \psi +\sin \psi +\left(1-\frac{1}{\sqrt{2}}\right)$
$=-\sqrt{2}-\frac{1}{\sqrt{2}}+2\left(\frac{2}{\sqrt{5}}\right)+\left(\frac{1}{\sqrt{5}}\right)+1-\frac{1}{\sqrt{2}}$
$=\sqrt{5}-2\sqrt{2}+1$