Let $y=y(x)$ be a differentiable function in the interval $(0,\infty )$ such that $y(1)=2,$ and $\lim _{t\to x}\frac{{t}^{2}y\left(x\right)−{x}^{2}y\left(t\right)}{x−t}=3$ for each $x0.$ Then $2y(2)$ is equal to [JEE Main 2026, 24 Jan (Shift 2)]

Question

Let $y=y(x)$ be a differentiable function in the interval $(0,\infty )$ such that $y(1)=2,$ and $\lim _{t\to x}\frac{{t}^{2}y\left(x\right)−{x}^{2}y\left(t\right)}{x−t}=3$ for each $x>0.$ Then $2y(2)$ is equal to [JEE Main 2026, 24 Jan (Shift 2)]

YESPYQ · Accepted Answer

Correct answer: c) $23$. $\lim _{t\to x}\frac{{t}^{2}y\left(x\right)−{x}^{2}y\left(t\right)}{x−t}=3$ Using L'hospital rule : $\lim _{t\to x}\frac{2ty(x)−{x}^{2}{y}^{'}(t)}{−1}=3$ ${x}^{2}{y}^{'}(x)−2xy(x)=3$ $\frac{dy}{dx}−\frac{2y}{x}=\frac{3}{{x}^{2}}$ $\text{I.F. }={e}^{−\int \frac{2}{x}dx}={e}^{−2{\log }_{e}x}=1/{x}^{2}$ $y⋅\frac{1}{{x}^{2}}=\int \frac{3}{{x}^{4}}dx$ $\frac{y}{{x}^{2}}=−\frac{1}{{x}^{3}}+c$ $\Rightarrow y=c{x}^{2}−\frac{1}{x}=y(x)$ $y(1)=2=c−1\Rightarrow c=3$ $y(x)=3{x}^{2}−\frac{1}{x}$ $y(2)=12−\frac{1}{2}\Rightarrow 2f(2)=23$

Let \(y=y(x)\) be a differentiable function in the interval \((0,\infty )\) such that \(y(1)=2,\) and \(\lim _{t\to x}\f…

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