JEE Main 2022 — Mathematics Calculus

Given differential equation $\frac{dy}{dx}=\frac{1}{1+\mathrm{sin}2x}$dxdy​=1+sin2x1​

$$\int dy=\int \frac{dx}{{(\mathrm{sin}x+\mathrm{cos}x)}^{2}}$$∫dy=∫(sinx+cosx)2dx​

$$\int dy=\int \frac{{\mathrm{sec}}^{2}x}{{(1+\mathrm{tan}x)}^{2}}$$∫dy=∫(1+tanx)2sec2x​

$$y=-\frac{1}{1+\mathrm{tan}x}+C$$y=−1+tanx1​+C

So $y(\frac{\pi }{4})=\frac{1}{2}$y(4π​)=21​

So $\frac{1}{2}=-\frac{1}{2}+C$21​=−21​+C$\Rightarrow C=1$⇒C=1

So $y=\frac{-1}{1+\mathrm{tan}x}+1$y=1+tanx−1​+1

$$y=\frac{-1+1+\mathrm{tan}x}{1+\mathrm{tan}x}$$y=1+tanx−1+1+tanx​

$$y=\frac{\mathrm{tan}x}{1+\mathrm{tan}x}$$y=1+tanxtanx​

Solving this equation with $y=\sqrt{2}\mathrm{sin}x$y=2​sinx

$$\frac{\mathrm{tan}x}{1+\mathrm{tan}x}=\sqrt{2}\mathrm{sin}x$$1+tanxtanx​=2​sinx

i.e. $\mathrm{sin}x=0,\frac{1}{\sqrt{2}}=\mathrm{sin}x+\mathrm{cos}x$sinx=0,2​1​=sinx+cosx

$$x=\pi$$x=π

or $\frac{1}{2}=\mathrm{sin}(x+\frac{\pi }{4})$21​=sin(x+4π​)

$$\Rightarrow \mathrm{sin}\frac{\pi }{6}=\mathrm{sin}(x+\frac{\pi }{4})$$⇒sin6π​=sin(x+4π​)

$$x+\frac{\pi }{4}=\pi -\frac{\pi }{6},2\pi +\frac{\pi }{6}$$x+4π​=π−6π​,2π+6π​

$$x=\frac{5\pi }{6}-\frac{\pi }{4},x=\frac{13\pi }{6}-\frac{\pi }{4}$$x=65π​−4π​,x=613π​−4π​

$$x=\frac{7\pi }{12},x=\frac{23\pi }{12}$$x=127π​,x=1223π​

Hence sum of abscissas of all the points of intersection

$$=\pi +\frac{7\pi }{12}+\frac{23\pi }{12}$$=π+127π​+1223π​

$$=\frac{12\pi +7\pi +23}{12}=\frac{42\pi }{12}$$=1212π+7π+23​=1242π​

$$\frac{k\pi }{12}=\frac{42\pi }{12}\Rightarrow k=42$$12kπ​=1242π​⇒k=42