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

Answer: (A)

$\frac{1}{3x}+\frac{2x^2}{3}$

$\underset{t\to x}{lim}\frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1$
$\Rightarrow x^{2}f\left(x\right)-2xf\left(x\right)+1=0$
$\Rightarrow f\left(x\right)=c{x}^2+\frac{1}{3x}$ also $f\left(1\right)=1$
$\Rightarrow c=\frac{2}{3}$
Hence $f\left(x\right)=\frac{2}{3}{x}^2+\frac{1}{3x}$