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. $\underset{x\to 0^{+}}{lim}{f}^{\prime }\left(\frac{1}{x}\right)=1$
• 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. $\left|f\left(x\right)\right|\le 2$ for all $X\in (0,2)$

Answer 1

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

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, $IF=e^{\int \frac{1}{x}dx}=e^{\log x}=x$
$\therefore$ Required solution is $y.\left(IF\right)=\int Q\left(IF\right)dx+C$
$\Rightarrow y\left(x\right)=\int 2\left(x\right)dx+C$
$\Rightarrow yx=x^2+C$
$\therefore y=x+\frac{C}{x}$ $[\because C\ne 0$, as $f\left(1\right)\ne 1]$
(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 correct.
(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 (c) is correct.
(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.