Question

Let $ f(x) $ be differentiable on the interval $ (0, \infty) $ such that $ f(1) = 1 $ and $ lim_{t \rightarrow x} \frac {t^2f(x)-x^2f(t)}{t-x} =1 $ for each $ x>0 $ . Then, $ f(x) $ is

• A. $ \frac{1}{3x} + \frac{2x^2}{3} $
• B. $ -\frac{1}{3x} + \frac{4x^2}{3} $
• C. $ -\frac{1}{x} + \frac{2}{x^2} $
• D. $ \frac{1}{x} $

Answer 1

The Correct Option is A

Given, $\displaystyle\lim_{t\to x} \frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1 $
Apply L' Hospital rule $\displaystyle\lim_{t\to x} \frac{2t\,f \left(x\right)-x^{2}f'\left(t\right)}{1}=1 $
$\Rightarrow 2x f \left(x\right)-x^{2}f'\left(x\right)=1 $
$\Rightarrow x^{2}f' \left(x\right)-2x f\left(x\right)=-1$
$\Rightarrow f'\left(x\right)-\frac{2}{x} f \left(x\right)=-\frac{1}{x^{2}}$
It is a linear differential equation
$\therefore IF =e^{\int \frac{2}{x}dx}$
$=e^{-2\,log\,x}$
$=e^{log \left(\frac{1}{x^{2}}\right)}$
$=\frac{1}{x^{2}}$
$\therefore$ Solution is
$f \left(x\right)\times\frac{1}{x^{2}}=\int-\frac{1}{x^{2}}\times\frac{1}{x^{2}} dx+C $
$f \left(x\right)\times\frac{1}{x^{2}}=\frac{1}{3x^{3}}+C $
$\Rightarrow f\left(x\right)=\frac{1}{3x}+Cx^{2}$
At $f\left(1\right)=1$
$\Rightarrow 1=\frac{1}{3}+C $
$\Rightarrow C=\frac{2}{3}$
$\therefore f\left(x\right)=\frac{1}{3x}+\frac{2}{3}x^{2}$