JEE Main 2016 — Limits Question with Solution

Let $L= \underset{t\to x}{\lim }\frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1$

Applying L'Hospital Rule

$L= \underset{t\to x}{\lim }\frac{2t f\left(x\right)-x^2{f}^{\prime }\left(t\right)}{1}=1$

Or $2x f\left(x\right)-x^2{f}^{\prime }\left(x\right)=1$ $\Rightarrow \frac{dy}{dx}+\left(-\frac{2}{x}\right)y=-\frac{1}{x^2}$..........$\left(1\right)$

Solving the above linear differential equation we get  

Integrating factor $=e^{\int -\frac{2}{x}dx}=e^{-2\ln x}=\frac{1}{x^2}$

After multiplying the equation $\left(1\right)$ by $\frac{1}{x^2}$, and simplifying the equation $\left(1\right)$ becomes,

$\frac{{\displaystyle d}}{dx}\left(\frac{y}{x^2}\right)=-\frac{1}{x^4}$

$\Rightarrow \frac{\mathrm{y}}{{\mathrm{x}}^2}=\int -\frac{1}{{\mathrm{x}}^4}dx=\frac{1}{3{\mathrm{x}}^3}+c$

$\Rightarrow y=\frac{1}{3x}+c{x}^2$

Putting $x=1, y=1$, we get

$c=\frac{2}{3}$.

$\therefore f\left(x\right)=\frac{2}{3}{x}^2+\frac{1}{3x}$

Put $x=\frac{3}{2}$

$f\left(\frac{3}{2}\right)=\frac{2}{3} \times {\left(\frac{3}{2}\right)}^2+\frac{2}{3\times 3}$

$=\frac{3}{2}+\frac{2}{9}=\frac{27+4}{18}=\frac{31}{18}$.