Given,
$2 f(x)+f\left(\frac{1}{x}\right)=4 x$
Replace $x$ by $1 / x$, we get,
$2 f(1 / x)+f(x)=\frac{4}{x}$
On multiply by 2 in Eq. (i) and then subtracting Eq. (ii), we get
$\begin{aligned}
4 f(x)+2 f\left(\frac{1}{x}\right) & =8 x \\
f(x)+2 f(1 / x) & =\frac{4}{x} \\
-\quad-\quad- & -\frac{8 x^2-4}{x} \\
3 f(x)=8 x-\frac{4}{x} & =\frac{8 x^2-4}{3 x}=\frac{4\left(2 x^2-1\right)}{3 x} \\
f(x) & =\frac{4\left(2 x^2-1\right)}{-3 x} \\
f(-x) &
\end{aligned}$
Again given,
$\begin{array}{rlrl}
& \frac{4\left(2 x^2-1\right)}{3 x} =-\left[\frac{4\left(2 x^2-1\right)}{3 x}\right] \\
\Rightarrow & 2\left[\frac{4\left(2 x^2-1\right)}{3 x}\right] =0 \\
\Rightarrow & 2 x^2-1 =0 \text { but } x \neq 0 \\
\Rightarrow & x = \pm \frac{1}{\sqrt{2}} \text { but } x \neq 0 \\
& \Rightarrow S =(-1 / \sqrt{2}, 1 / \sqrt{2}\}
\end{array}$
$f(x)=f(-x)$