Question

Let $f :(0, \infty) \rightarrow R$ be a differentiable function such that $f '( x )=2-\frac{ f ( x )}{ x }$ for all $x \in(0, \infty)$ and $f (1) \neq 1$ Then

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

Answer 1

Answer: (A)

$f :(0, \infty) \longrightarrow { R }$
$f ^{\prime}( x )=2-\frac{ f ( x )}{ x }$
$\forall x \in(0, \infty)$
$\frac{ dy }{ dx }=2-\frac{ y }{ x }$
$\Rightarrow \frac{ dy }{ dx }+\frac{ y }{ x }=2$
I. F $.= e ^{\int \frac{1}{ x } dx }= e ^{\ell n x}= x$
$\Rightarrow xy =\int 2 x d x$
$\Rightarrow xy = x ^{2}+ c$
$\Rightarrow y = x +\frac{ c }{ x }$
$\therefore f ( x )= x +\frac{ c }{ x }$
$f (1)=1+ c \neq 1$
$\therefore c \neq 0$
$f '( x )=1-\frac{ c }{ x ^{2}}$
$f '\left(\frac{1}{ x }\right)=1- cx ^{2}$
$\displaystyle\lim _{ x \rightarrow 0^{+}} f'\left(\frac{1}{ x }\right)=\displaystyle\lim _{ x \rightarrow 0^{+}}\left(1- cx ^{2}\right)=1$
$\displaystyle\lim _{ x \rightarrow 0^{+}} xf \left(\frac{1}{ x }\right)=\displaystyle\lim _{ x \rightarrow 0^{+}} x \left(\frac{1}{ x }+ cx \right)$
$=\displaystyle\lim _{ x \rightarrow 0^{+}}\left(1+ cx ^{2}\right)=1$
$\displaystyle\lim _{ x \rightarrow 0^{+}} x ^{2} f '( x )=\displaystyle\lim _{ x \rightarrow 0^{+}}\left( x ^{2}- c \right)=- c \neq 0$
$f(x)=x+\frac{c}{x}$
$\displaystyle\lim _{ x \rightarrow 0^{+}} f ( x ) \rightarrow \pm \infty$
So $f ( x )$ is not bounded
So $| f ( x )| \leq 2$ not possible