Q29 / 48 · JEE Advanced 2025 (Paper 1) · Mathematics
For all $x>0$, let $y_1(x), y_2(x)$, and $y_3(x)$ be the functions satisfying
$\begin{aligned} & \frac{d y_1}{d x}-(\sin x)^2 y_1=0, y_1(1)=5 \\ & \frac{d y_2}{d x}-(\cos x)^2 y_2=0, y_2(1)=\frac{1}{3} \\ & \frac{d y_3}{d x}-\left(\frac{2-x^3}{x^3}\right) y_3=0, y_3(1)=\frac{3}{5 e}\end{aligned}$
respectively. Then $\lim _{x \rightarrow 0^{+}} \frac{y_1(x) y_2(x) y_3(x)+2 x}{e^{3 x} \sin x}$ is equal to _____
Correct answer: 2
Solution
$\begin{aligned} & \frac{d y_1}{y_1}+\frac{d y_2}{y_2}+\frac{d y_3}{y_3}=\left(\sin ^2 x+\cos ^2 x+\frac{2-x^3}{x^3}\right) d x \\ & \ln \left(y_1 y_2 y_3\right)=\frac{-1}{x^2}+C \\ & \ln \left(y_1(x) y_2(x) y_3(x)\right)=\frac{-1}{x^2}+C \\ & \ln \left(5 \cdot \frac{1}{3} \cdot \frac{3}{5 e}\right)=\frac{-1}{x^2}+C \\ & \therefore C=0\end{aligned}$
$\begin{aligned} & y_1(x) y_2(x) y_3(x)=e^{\frac{-1}{x^2}} \\ & \lim _{x \rightarrow 0} \frac{e^{\frac{-1}{x^2}}+2 x}{e^{3 x} \sin x} \\ & \lim _{x \rightarrow 0} \frac{1}{e^{3 x+\frac{1}{x^2}} \sin x}+\lim _{x \rightarrow 0} \frac{2 x}{e^{3 x} \sin x}\end{aligned}$
$\lim _{x \rightarrow 0} \frac{1}{e^{3 x} \cdot \frac{\sin x}{x} \cdot \frac{e^{\frac{1}{x^2}}}{\frac{1}{x}}}+2$
$=0+2 \quad\left\{\lim _{\mathrm{y} \rightarrow \infty} \frac{\mathrm{e}^{\mathrm{y}^2}}{\mathrm{y}}=\frac{\mathrm{e}^{\mathrm{y}^2} \cdot 2 \mathrm{y}}{1}=\infty\right\}$
$=2$
Hence, the correct answer is 2.