JEE Advanced 2025 Paper 1Question 1
For all $x>0$, let $y_1(x)$, $y_2(x)$, and $y_3(x)$ be the functions satisfying $$\frac{dy_1}{dx}-(\sin x)^2y_1=0,\ y_1(1)=5,$$ $$\frac{dy_2}{dx}-(\cos x)^2y_2=0,\ y_2(1)=\frac13,$$ $$\frac{dy_3}{dx}-\left(\frac{2-x^3}{x^3}\right)y_3=0,\ y_3(1)=\frac3{5e},$$ respectively. Then $$\lim_{x\to0^+}\frac{y_1(x)y_2(x)y_3(x)+2x}{e^{3x}\sin x}$$ is equal to \_\_\_\_\_\_\_\_.
Differential Equation
Answer$2$
Solution
Concept. Each equation gives $y_i=y_i(1)\,e^{\int_1^x(\dots)}$; in the product the exponents combine to $1-\frac1{x^2}$. Formulas used. (1) $y'=p(x)\,y\Rightarrow y(x)=y(1)\,e^{\int_1^xp(t)\,dt}$(2) $e^{-1/x^2}\to0$ faster than any power of $x$ as $x\to0^+$Step 1 - The product. By (1), the constants give $5\cdot\frac13\cdot\frac{3}{5e}=\frac1e$ and the exponent is $$\int_1^x\left(\sin^2t+\cos^2t\right)dt+\int_1^x\left(\frac{2}{t^3}-1\right)dt=(x-1)+\left(-\frac1{x^2}-x+2\right)=1-\frac1{x^2}.$$ So $y_1y_2y_3=e^{-1/x^2}$. Step 2 - The limit. $$\frac{e^{-1/x^2}+2x}{e^{3x}\sin x}=\frac{e^{-1/x^2}}{e^{3x}\sin x}+\frac{2x}{e^{3x}\sin x}.$$ By (2) the first part $\to0$; the second $\to2$. Answer: 2