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← JEE Advanced 2025 · Paper 1
MathematicsNumerical+4Question 13 of 48

JEE Advanced 2025 · Paper 1, Mathematics question 13

For all x>0x > 0, let y1(x)y_1(x), y2(x)y_2(x), and y3(x)y_3(x) be the functions satisfying

dy1dx−(sin⁡x)2 y1=0,y1(1)=5,dy2dx−(cos⁡x)2 y2=0,y2(1)=13,dy3dx−(2−x3x3)y3=0,y3(1)=35e,\begin{aligned} &\frac{dy_1}{dx} - (\sin x)^2\,y_1 = 0, & y_1(1) &= 5,\\ &\frac{dy_2}{dx} - (\cos x)^2\,y_2 = 0, & y_2(1) &= \frac13,\\ &\frac{dy_3}{dx} - \left(\frac{2 - x^3}{x^3}\right)y_3 = 0, & y_3(1) &= \frac{3}{5e}, \end{aligned}

respectively. Then

lim⁡x→0+y1(x) y2(x) y3(x)+2xe3xsin⁡x\lim_{x\to0^+}\frac{y_1(x)\,y_2(x)\,y_3(x) + 2x}{e^{3x}\sin x}

is equal to _____.

Answer and worked solution

Official answer: 2

y1y2y_1y_2. Each equation is separable:

  • y1=5e∫1xsin⁡2t dty_1=5e^{\int_1^x\sin^2t\,dt}
  • y2=13e∫1xcos⁡2t dty_2=\frac13e^{\int_1^x\cos^2t\,dt}

Since sin⁡2+cos⁡2=1\sin^2+\cos^2=1, multiplying gives y1y2=53ex−1y_1y_2=\frac53e^{x-1}.

y3y_3. y3′y3=2x3−1\frac{y_3'}{y_3}=\frac{2}{x^3}-1, so ln⁡y3=−1x2−x+c\ln y_3=-\frac1{x^2}-x+c. The condition y3(1)=35ey_3(1)=\frac3{5e} gives y3=35e1−x−1/x2y_3=\frac35e^{1-x-1/x^2}.

Product. y1y2y3=e−1/x2y_1y_2y_3=e^{-1/x^2}, which tends to 0 faster than any power of xx.

lim⁡x→0+e−1/x2+2xe3xsin⁡x=lim⁡2xx=2.\lim_{x\to0^+}\frac{e^{-1/x^2}+2x}{e^{3x}\sin x}=\lim\frac{2x}{x}=2.

Topics:Variable separableLimits
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