Question

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

• 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)

$|\cos x-\sin x| \leq y \leq \sin x$
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 $ =\int\limits_\psi^{\pi / 2}(\sin x-|\cos x-\sin x|) d x$
$=\int\limits_\psi^{\pi / 4}(\sin x-(\cos x-\sin x)) d x $
$ +\int\limits_{\pi / 4}^{\pi / 2}(\sin x-(\sin x-\cos x)) d x $
$=\int\limits_\psi^{\pi / 4}(2 \sin x-\cos x) d x+\int\limits_{\pi / 4}^{\pi / 2} \cos x dx $
$=[-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}} $
$= \sqrt{5}-2 \sqrt{2}+1$