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

The function $f \left(x\right)$satisfies the differential equation $\frac{df\left(x\right)}{dx}+\frac{1}{x}f\left(x\right)=2$                 (i)

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

Multiplying (i) by I.F. $= x$ and integrating, we obtain

$\begin{array}{r}xf\left(x\right)=x^2+C\\ \Rightarrow f\left(x\right)=x+\frac{C}{x}\\ \Rightarrow f\left(x\right)=1-\frac{C}{x^2}\\ \Rightarrow f\left(\frac{1}{x}\right)=1-C{x}^2\end{array}$

It is given that $f\left(1\right)\ne 1,\mathrm{So},\mathrm{C}\ne 0$

$\begin{array}{r}\therefore \underset{x\to 0^{+}}{lim} f^{\mathrm{\prime }}\left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{lim} \left(1-C{x}^2\right)=1,\\ \underset{x\to 0^{+}}{lim} x^2{f}^{\mathrm{\prime }}\left(x\right)=\underset{x\to 0^{+}}{lim} \left(x^2-C\right)=-C,\end{array}$

and , $\underset{x\to 0^{+}}{lim} xf\left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{lim} x\left(\frac{1}{x}+Cx\right)=1$

Thus, option (a) is correct.  

We observe that for $C=1,f\left(x\right)=x+\frac{1}{x}\ge 2$

So,  $\left|f\left(x\right)\right|\le 2$is not true. This can also be observed from the fact 

that $\underset{x\to 0^{+}}{lim} f\left(x\right)\to \mathrm{\infty }$ for $C>0$