JEE Main 2023 — Mathematics Calculus

Given,$|\mathrm{cos}x-\mathrm{sin}x|\leq y\leq \mathrm{sin}x,0\leqslant x\leqslant \frac{\pi }{2}$∣cosx−sinx∣≤y≤sinx,0⩽x⩽2π​

For intersection point:

$$\mathrm{cos}x-\mathrm{sin}x=\mathrm{sin}x$$cosx−sinx=sinx

$$\Rightarrow \mathrm{tan}x=\frac{1}{2}$$⇒tanx=21​

Let $\psi ={\mathrm{tan}}^{-1}(\frac{1}{2})$ψ=tan−1(21​), so $\mathrm{tan}\psi =\frac{1}{2} \mathrm{sin}\psi =\frac{1}{\sqrt{5}} \mathrm{cos}\psi =\frac{2}{\sqrt{5}}$tanψ=21​sinψ=5​1​cosψ=5​2​

Required area is $\displaystyle ={\int }_{\psi }^{\pi /2}(\mathrm{sin}x-|\mathrm{cos}x-\mathrm{sin}x|)dx$=∫ψπ/2​(sinx−∣cosx−sinx∣)dx

$$={\int }_{\psi }^{\pi /4}(\mathrm{sin}x-(\mathrm{cos}x-\mathrm{sin}x))dx+{\int }_{\pi /4}^{\pi /2}(\mathrm{sin}x+(\mathrm{cos}x-\mathrm{sin}x))dx$$=∫ψπ/4​(sinx−(cosx−sinx))dx+∫π/4π/2​(sinx+(cosx−sinx))dx

$$={\int }_{y}^{\pi /4}(2\mathrm{sin}x-\mathrm{cos}x)dx+{\int }_{\pi /4}^{\pi /2}(\mathrm{cos}x)dx$$=∫yπ/4​(2sinx−cosx)dx+∫π/4π/2​(cosx)dx

$$={[-2\mathrm{cos}x-\mathrm{sin}x]}_{\psi }^{\pi /4}+{[\mathrm{sin}x]}_{\pi /4}^{\pi /2}$$=[−2cosx−sinx]ψπ/4​+[sinx]π/4π/2​

$$=-\sqrt{2}-\frac{1}{\sqrt{2}}+2\mathrm{cos}\psi +\mathrm{sin}\psi +(1-\frac{1}{\sqrt{2}})$$=−2​−2​1​+2cosψ+sinψ+(1−2​1​)

$$=1-\sqrt{2}-\frac{2}{\sqrt{2}}+2(\frac{2}{\sqrt{5}})+(\frac{1}{\sqrt{5}})$$=1−2​−2​2​+2(5​2​)+(5​1​)

$$=1-2\sqrt{2}+\frac{5}{\sqrt{5}}$$=1−22​+5​5​

$=(\sqrt{5}-2\sqrt{2}+1)$=(5​−22​+1) sq. units