QJEEM23Mad9ff: JEE Main 2023 Mathematics - Class 12 JEEM23Mad9ff Solution: The area of the region is

$|\cos x-\sin x| \leq y \leq \sin x$

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 $= \int\limits_{{{\tan }^{ - 1}}{1 \over 2}}^{{\pi \over 4}} {(\sin x - (\cos x - \sin x))dx + \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {(\sin x - (\sin x - \cos x))dx} }$

$=\int\limits_{{{\tan }^{ - 1}}{1 \over 2}}^{\pi / 4}(2 \sin x-\cos x) d x+\int\limits_{\pi / 4}^{\pi / 2} \cos x d x$

$=[-2 \cos x-\sin x]_{{{\tan }^{ - 1}}{1 \over 2}}^{\pi / 4}+[\sin x]_{\pi / 4}^{\pi / 2}$

$=-\sqrt{2}-\frac{1}{\sqrt{2}}+2 \cos ({{{\tan }^{ - 1}}{1 \over 2}}) +\sin ({{{\tan }^{ - 1}}{1 \over 2}})+\left(1-\frac{1}{\sqrt{2}}\right)$

$=-\sqrt{2}-\frac{1}{\sqrt{2}}+2\left(\frac{2}{\sqrt{5}}\right)+\left(\frac{1}{\sqrt{5}}\right)+1-\frac{1}{\sqrt{2}}$

$= \sqrt 5 - 2\sqrt 2 + 1$