JEE Main 2022 — Mathematics Calculus

Given differential equation, $\mathrm{sin}(2{x}^{2})\mathrm{ln}(\mathrm{tan}{x}^{2})dy+(4xy-4\sqrt{2}x\mathrm{sin}({x}^{2}-\frac{\pi }{4}))dx=0$sin(2x2)ln(tanx2)dy+(4xy−42​xsin(x2−4π​))dx=0

$$\Rightarrow \mathrm{ln}(\mathrm{tan}{x}^{2})dy+\frac{4xydx}{\mathrm{sin}(2{x}^{2})}-\frac{4\sqrt{2}x\mathrm{sin}({x}^{2}-\frac{\pi }{4})}{\mathrm{sin}(2{x}^{2})}dx=0$$⇒ln(tanx2)dy+sin(2x2)4xydx​−sin(2x2)42​xsin(x2−4π​)​dx=0

$$\Rightarrow d(y\cdot \mathrm{ln}(\mathrm{tan}{x}^{2}))-4\sqrt{2}x\frac{(\mathrm{sin}{x}^{2}-\mathrm{cos}{x}^{2})}{\sqrt{2}(2\mathrm{sin}{x}^{2}\mathrm{cos}{x}^{2})}dx=0$$⇒d(y⋅ln(tanx2))−42​x2​(2sinx2cosx2)(sinx2−cosx2)​dx=0

$$\Rightarrow d(y\mathrm{ln}(\mathrm{tan}{x}^{2}))-\frac{4x(\mathrm{sin}{x}^{2}-\mathrm{cos}{x}^{2})}{{(\mathrm{sin}{x}^{2}+{\mathrm{cos}}^{2}x)}^{2}-1}dx=0$$⇒d(yln(tanx2))−(sinx2+cos2x)2−14x(sinx2−cosx2)​dx=0

Now integrating both side,

$$\Rightarrow \int d(y\mathrm{ln}(\mathrm{tan}{x}^{2}))-\int \frac{4x(\mathrm{sin}{x}^{2}-\mathrm{cos}{x}^{2})}{{(\mathrm{sin}{x}^{2}+{\mathrm{cos}}^{2}x)}^{2}-1}dx=\int 0$$⇒∫d(yln(tanx2))−∫(sinx2+cos2x)2−14x(sinx2−cosx2)​dx=∫0

Now let $\mathrm{sin}{x}^{2}+{\mathrm{cos}}^{2}x=t\Rightarrow -2x(\mathrm{sin}{x}^{2}-{\mathrm{cos}}^{2}x)=dt$sinx2+cos2x=t⇒−2x(sinx2−cos2x)=dt

$$\Rightarrow \int d(y\mathrm{ln}(\mathrm{tan}{x}^{2}))+2\int \frac{dt}{{t}^{2}-1}=\int 0$$⇒∫d(yln(tanx2))+2∫t2−1dt​=∫0

$$\Rightarrow y\mathrm{ln}(\mathrm{tan}{x}^{2})+2\cdot \frac{1}{2}\mathrm{ln}|\frac{t-1}{t+1}|=c$$⇒yln(tanx2)+2⋅21​ln∣t+1t−1​∣=c

$$y\mathrm{ln}(\mathrm{tan}{x}^{2})+\mathrm{ln}(\frac{\mathrm{sin}{x}^{2}+\mathrm{cos}{x}^{2}-1}{\mathrm{sin}{x}^{2}+\mathrm{cos}{x}^{2}+1})=c$$yln(tanx2)+ln(sinx2+cosx2+1sinx2+cosx2−1​)=c

Put $y=1$y=1 and $x=\sqrt{\frac{\pi }{6}}$x=6π​​

$$1\mathrm{ln}(\frac{1}{\sqrt{3}})+\mathrm{ln}\frac{(\frac{1}{2}+\frac{\sqrt{3}}{2}-1)}{(\frac{1}{2}+\frac{\sqrt{3}}{2}+1)}=c$$1ln(3​1​)+ln(21​+23​​+1)(21​+23​​−1)​=c

Now at $x=\sqrt{\frac{\pi }{3}}$x=3π​​

$$\Rightarrow y(\mathrm{ln}\sqrt{3})+\mathrm{ln}\frac{(\frac{\sqrt{3}}{2}+\frac{1}{2}-1)}{(\frac{\sqrt{3}}{2}+\frac{1}{2}+1)}=\mathrm{ln}(\frac{1}{\sqrt{3}})+\mathrm{ln}(\frac{\sqrt{3}-1}{\sqrt{3}+3})$$⇒y(ln3​)+ln(23​​+21​+1)(23​​+21​−1)​=ln(3​1​)+ln(3​+33​−1​)

$$\Rightarrow y(\mathrm{ln}\sqrt{3})=\mathrm{ln}(\frac{1}{\sqrt{3}})$$⇒y(ln3​)=ln(3​1​)

$$\Rightarrow y=-1$$⇒y=−1

So, $|y|=1$∣y∣=1