JEE 2022 — Mathematics PYQ

Solution

1. Integrating Factor:

   • Denominator $D=2{\cos }^{4}x-\cos 2x=\frac{1+{\cos }^{2}2x}{2}$.

   • $P(x)=\frac{2\sqrt{2}}{1+{\cos }^{2}2x}$.

   • Integration shows $IF=e^{-{\tan }^{-1}(\sqrt{2}\cot 2x)}$.

2. Solve the Equation:

   • Multiplying by $IF$: $\frac{d}{dx}(y\cdot e^{-{\tan }^{-1}(\sqrt{2}\cot 2x)})=x$.

   • $y\cdot e^{-{\tan }^{-1}(\sqrt{2}\cot 2x)}=\frac{x^2}{2}+C$.

   • Using $y(\frac{\pi }{4})=\frac{{\pi }^2}{32}$ where $\cot (\pi /2)=0$, we find $C=0$.

3. Find $\alpha$:

   • At $x=\frac{\pi }{3}$, $\cot (2\pi /3)=-\frac{1}{\sqrt{3}}$.

   • $y(\frac{\pi }{3})=\frac{(\pi /3{)}^2}{2}{e}^{{\tan }^{-1}(\sqrt{2}\cdot -\frac{1}{\sqrt{3}})}=\frac{{\pi }^2}{18}{e}^{-{\tan }^{-1}(\sqrt{2/3})}$.

   • Thus, $\alpha =\sqrt{\frac{2}{3}}$.

   • $3{\alpha }^2=3\times \frac{2}{3}=2$.

Correct Answer: 2