Question

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

• 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. $|f\left(x\right)|\le 2$ for all $x\in (0,2)$

Answer 1

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

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

$y\left(x\right)=\int 2xdx+c$
$\Rightarrow f\left(x\right)=x+\frac{c}{x}$
Given that $f\left(1\right)\ne 1\Rightarrow c\ne 0$

$\Rightarrow f\left(x\right)=x+\frac{c}{x},\forall c\in R-\left\{0\right\}$

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

Option $2$:
$\underset{x\to 0^{+}}{lim}xf\left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{lim}x(\frac{1}{x}+cx)=\underset{x\to 0^{+}}{lim}(1+c{x}^2)=1$

Option $3$:
$\underset{x\to 0^{+}}{lim}{x}^2{f}^{\prime }\left(x\right)=\underset{x\to 0^{+}}{lim}{x}^2(1-\frac{c}{x^2})=\underset{x\to 0^{+}}{lim}(x^2-c)\ne 0$

Option $4$:
$|f\left(x\right)|\ge 2\sqrt{c}$
For $c=4,|f\left(x\right)|\ge 4$, which contradicts the given option.