JEE Advanced 2025 — Paper 1 — Question 27

$\quad \frac{d y_{1}}{y_{1}}+\frac{d y_{2}}{y_{2}}+\frac{d y_{3}}{y_{3}}=\left(\sin ^{2} x+\cos ^{2} x+\frac{2-x^{3}}{x^{3}}\right) d x$

$\ln \left(y_{1} y_{2} y_{3}\right)=\frac{-1}{x^{2}}+C$

$\ln \left(\mathrm{y}_{1}(\mathrm{x}) \mathrm{y}_{2}(\mathrm{x}) \mathrm{y}_{3}(\mathrm{x})\right)=\frac{-1}{\mathrm{x}^{2}}+\mathrm{C}$

$\ln \left(5 \cdot \frac{1}{3} \cdot \frac{3}{5 \mathrm{e}}\right)=\frac{-1}{\mathrm{x}^{2}}+\mathrm{C}$

$\therefore \mathrm{C}=0$

$y_{1}(x) y_{2}(x) y_{3}(x)=e^{\frac{-1}{x^{2}}}$

$\lim _{x \rightarrow 0} \frac{e^{\frac{-1}{x^{2}}}+2 x}{e^{3 x} \sin x}$

$\lim _{x \rightarrow 0} \frac{1}{e^{3 x+\frac{1}{x^{2}}} \sin x}+\lim _{x \rightarrow 0} \frac{2 x}{e^{3 x} \sin x}$

$\lim _{x \rightarrow 0} \frac{1}{e^{3 x} \cdot \frac{\sin x}{x} \cdot \frac{e^{\frac{1}{x^{2}}}}{\frac{1}{x}}}+2$

$=0+2 \quad\left\{\lim _{y \rightarrow \infty} \frac{e^{y^{2}}}{y}=\frac{e^{y^{2}} \cdot 2 y}{1}=\infty\right\}$

$=2$