Question

Let $f\left(x\right)$ be differentiable on the interval $(0,\infty )$ such that $f\left(1\right)=1,$ and $\underset{t\to x}{lim}\frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1$ for each $x>0.$ Then $f\left(x\right)$ 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 $\frac{1}{3x}+\frac{2x^2}{3}$

$\underset{t\to x}{lim}\frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1\Rightarrow x^2{f}^{\prime }\left(x\right)-2xf\left(x\right)+1=0$

$\Rightarrow \frac{x^2{f}^{\prime }\left(x\right)-2xf\left(x\right)}{{\left(x^2\right)}^2}+\frac{1}{x^4}=0$

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

On integrating

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

$f\left(1\right)=1\Rightarrow c=\frac{2}{3}$

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