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 $xϵ(0,\infty )$ 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. $\underset{x\to 0^{+}}{lim}{x}^2{f}^{\prime }x=0$
• D. $|f\left(x\right)|\le 2$ for all $xϵ(0,2)$

Answer 1

The correct option is A

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

Here $f^{\prime }\left(x\right)=2-\frac{f\left(x\right)}{x}$
or $\frac{dy}{dx}+\frac{y}{x}=2$
[i.e. linear differential equation in y]
Integrating factor, $IF=e^{\int \frac{1}{x}dx}=e^{lnx}=x$
$\therefore$ Required solution is $y.\left(IF\right)=\int Q\left(IF\right)dx+C$
$\Rightarrow y\left(x\right)=\int 2\left(x\right)dx+C\Rightarrow yx=x^2+C$
$\therefore y=x+\frac{C}{x}$ [$\because C\ne 0,asf\left(1\right)\ne 1$]
a) $\underset{x\to 0^{+}}{lim}{f}^{\prime }\left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{lim}(1-C{x}^2)=1$
$\therefore$ Option (a) is correct

b) $\underset{x\to 0^{+}}{lim}xf\left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{lim}(1+C{x}^2)=1$
$\therefore$ Option (b) is incorrect

c) $\underset{x\to 0^{+}}{lim}{x}^2{f}^{\prime }\left(x\right)=\underset{x\to 0^{+}}{lim}(x^2-C)=-C\ne 0$
$\therefore$ Option (a) is incorrect

d) $f\left(x\right)=x+\frac{c}{x},C\ne 0$
For C > 0, $\underset{x\to 0^{+}}{lim}f\left(x\right)=\infty$
$\therefore$ Function is not bounded in (0,2)
$\therefore$ Option (d) is incorrect