Question

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

Answer 1

Answer: (A)

$f:(0,\infty )⟶R$
$f^{\prime }(x)=2-\frac{f(x)}{x}$
$\forall x\in (0,\infty )$
$\frac{dy}{dx}=2-\frac{y}{x}$
$\Rightarrow \frac{dy}{dx}+\frac{y}{x}=2$
I. F $.=e^{\int \frac{1}{x}dx}=e^{\ell nx}=x$
$\Rightarrow xy=\int 2xdx$
$\Rightarrow xy=x^2+c$
$\Rightarrow y=x+\frac{c}{x}$
$\therefore f(x)=x+\frac{c}{x}$
$f(1)=1+c\ne 1$
$\therefore c\ne 0$
$f^{\prime }(x)=1-\frac{c}{x^2}$
$f^{\prime }\left(\frac{1}{x}\right)=1-c{x}^2$
$\underset{x\to 0^{+}}{\lim }{f}^{\prime }\left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{\lim }\left(1-c{x}^2\right)=1$
$\underset{x\to 0^{+}}{\lim }xf\left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{\lim }x\left(\frac{1}{x}+cx\right)$
$=\underset{x\to 0^{+}}{\lim }\left(1+c{x}^2\right)=1$
$\underset{x\to 0^{+}}{\lim }{x}^2{f}^{\prime }(x)=\underset{x\to 0^{+}}{\lim }\left(x^2-c\right)=-c\ne 0$
$f(x)=x+\frac{c}{x}$
$\underset{x\to 0^{+}}{\lim }f(x)\to \pm \infty$
So $f(x)$ is not bounded
So $|f(x)|\le 2$ not possible