Question

Let $f:(0,\infty )\to R$ be a differentiable function such that $f^{\prime }\left(x\right)=2-\frac{f\left(x\right)}{x}$ for all $x\in (0,\infty )$ and $f\left(1\right)\ne 1$

• A. $\underset{x\to 0}{lim}{f}^{\prime }\left(\frac{1}{x}\right)=1$
• B. $\underset{x\to 0}{lim}xf\left(\frac{1}{x}\right)=2$
• C. $\underset{x\to 0}{lim}{x}^2{f}^{\prime }\left(x\right)=0$
• D. $\left|f\left(x\right)\right|\le 2forallx\in (0,2)$

Answer 1

The correct option is A $\underset{x\to 0}{lim}{f}^{\prime }\left(\frac{1}{x}\right)=1$

$f^{\prime }\left(x\right)=2-\frac{f\left(x\right)}{x}⟹x{f}^{\prime }\left(x\right)+f\left(x\right)=2x$

$d\left(xf\right(x\left)\right)=2xdx$

Integrating on both sides

$xf\left(x\right)=x^2+c⟹f\left(x\right)=x+\frac{c}{x}$

$f^{\prime }\left(x\right)=1-\frac{c}{x^2}$

$\underset{x\to 0}{lim}{f}^{\prime }(\frac{1}{x})=\underset{x\to 0}{lim}(1-c{x}^2)=1$

$\underset{x\to 0}{lim}xf(\frac{1}{x})=\underset{x\to 0}{lim}(1+x^2)=1$

$\underset{x\to 0}{lim}{x}^2{f}^{\prime }\left(x\right)=\underset{x\to 0}{lim}(x^2-c)=-c$

$AM\ge GM$

$⟹|f\left(x\right)|\ge 2\sqrt{c}$