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,

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

The function $f(x)$ satisfies the differential equation

$\frac{d f(x)}{d x}+\frac{1}{x} f(x)=2$            .....(i)

This is a linear differential equation with I.F. = $e^{\int \frac{1}{x} d x}=e^{\log x}=x$

Multiplying (i) by Integrating factor = x and integrating, we obtain

$x f(x)=x^2+C$

$\Rightarrow f(x)=x+\frac{C}{x}$

$\Rightarrow f'(x)=1-\frac{C}{x^2}$

$\Rightarrow f'\left(\frac{1}{x}\right)=1-C x^2$

It is given that $f(1) \neq 1$. So, $C \neq 0$.

∴   $\lim\limits_{x \rightarrow 0^{+}} f'\left(\frac{1}{x}\right)=\lim\limits_{x \rightarrow 0^{+}}\left(1-C x^2\right)=1,$

$\lim\limits_{x \rightarrow 0^{+}} x^2 f'(x)=\lim\limits_{x \rightarrow 0^{+}}\left(x^2-C\right)=-C, $

and, $\lim\limits_{x \rightarrow 0^{+}} x f\left(\frac{1}{x}\right)=\lim\limits_{x \rightarrow 0^{+}} x\left(\frac{1}{x}+C x\right)=1$

Thus, option (a) is correct.

We observe that for $C=1, f(x)=x+\frac{1}{x} \geq 2$.

So, $|f(x)| \leq 2$ is not true. This can also be observed from the fact that $\lim\limits_{x \rightarrow 0^{+}} f(x) \rightarrow \infty$ for $C>0$.