Let f:(0, ) R be a differentiable function such that f^ (x)=2-f(x… - Vidyadip

MCQ

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

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

✓

Answer

Correct option: A.

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

a
$f:(0, \infty) \rightarrow R , f^{\prime}(x)=2-\frac{f(x)}{x}$

Now, the linear differential equation is

$f^{\prime}(X)+f(x) / x=2$

The integrating factor is

$e^{\int \frac{1}{x} d x}=e^{\ln x}=x$

Therefore,

$\int d(x f(x))=\int 2 x d x+c$

$x f(x)=x^2+c \Rightarrow f(x)=\left(x+\frac{c}{x}\right) \forall x \in(0, \infty)$

Now,

$f(1) \neq 1 \Rightarrow 1 \neq 1+c \Rightarrow c \neq 0$

and

$f^{\prime}(x)=1-\frac{c}{x^2} \Rightarrow \lim _{x \rightarrow 0^{+}} f^{\prime}(x)=\lim _{x \rightarrow 0^{+}}\left(1-\frac{c}{x^2}\right)=1$

$\Rightarrow \lim _{x \rightarrow 0^{+}} f^{\prime}\left(\frac{1}{x}\right)=\lim _{x \rightarrow 0^{+}}\left(1-c x^2\right)=1$

Now,

$\lim _{x \rightarrow 0^{+}} x f\left(\frac{1}{x}\right)=\lim _{x \rightarrow 0^{+}} x\left(\frac{1}{x}+c x\right)=\lim _{x \rightarrow 0^{+}}\left(1+\left(c x^2\right)\right)=1$

Now,

$\lim _{x \rightarrow 0^{+}} x^2 f^{\prime}(x)=\lim _{x \rightarrow 0^{+}} x^2\left(1-\frac{c}{x^2}\right)$

$=\lim _{x \rightarrow 0^{+}}\left(x^2-c\right)=-c \neq 0$

We cannot say anything about $f x() \leq 2 \forall x \in(0,2)$ because we do not know the value of $c$.

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