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 $f\left(1\right)\ne 1$. Then

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

Answer 1

The correct option is C $|f\left(x\right)|\le 2$ for all $x\in (0,2)$

$f^{\prime }\left(x\right)+\frac{f\left(x\right)}{x}=2$
$\text{I}\text{.F}=e^{\int \frac{1}{x}dx}=e^{\log x}=x$
$f\left(x\right)\cdot x=\int 2xdx+c$
$f\left(x\right).x=x^2+c$
$f\left(x\right)=x+\frac{c}{x}$
$\because f\left(1\right)\ne 1$
$1+\frac{c}{1}\ne 1c\ne 0$
$f^{\prime }\left(x\right)=1-\frac{c}{x^2}$
$\Rightarrow f^{\prime }\left(\frac{1}{x}\right)=1-c{x}^2$
(1) ${lim}_{x\to 0^{+}}{f}^{\prime }\left(\frac{1}{x}\right)={lim}_{x\to 0^{+}}(1-c{x}^2)=1$
(2) ${lim}_{x\to 0^{+}}x\cdot f\left(\frac{1}{x}\right)={lim}_{x\to 0^{+}}x(\frac{1}{x}+cx)$
= ${lim}_{x\to 0^{+}}(1+c{x}^2)=1$
(3) ${lim}_{x\to 0^{+}}{x}^2{f}^{\prime }\left(x\right)={lim}_{x\to 0^{+}}{x}^2(1-\frac{c}{x^2})$
= ${lim}_{x\to 0^{+}}(x^2-c)=-c$
(4) $f\left(x\right)=x+\frac{c}{x}$