If f(x)+2f((1)/(x))=3x,x 0, and S=x R:f(x)=f(-x); then S:

Question

If $f(x)+2f(\frac{1}{x})=3x,x
e 0$, and $S={x\in R:f(x)=f(-x)}$; then $S:$

Accepted Answer

We have, $f(x)+2f(\frac{1}{x})=3x,\, x
e 0$
  $\ldots$ (i)

On replacing $x$ by $\frac{1}{x}$ in the above equation, we get

$\begin{gathered} f(\frac{1}{x})+2f(x)=\frac{3}{x} \ \ \Rightarrow \, \, 2f(x)+f(\frac{1}{x})=\frac{3}{x}\, \, \, \, \, ...(ii) \end{gathered}$

On multiplying Eq. (ii) by 2, we get

$4f(x)+2f(\frac{1}{x})=\frac{6}{x}\, ...(iii)$

and subtracting Eq. (i) from Eq. (iii), we get

$[4f(x)+2f(\frac{1}{x})]-[f(x)+2f(\frac{1}{x})]=\frac{6}{x}-3x$

$\Rightarrow 3f(x)=\frac{6}{x}-3x$

$\Rightarrow f(x)=\frac{2}{x}-x$

Now, consider $\, f(x)=f(-x)$

$\begin{gathered} \Rightarrow \frac{2}{x}-x=-\frac{2}{x}+x \ \Rightarrow \frac{4}{x}=2x \ \Rightarrow 2{x}^{2}=4 \ \Rightarrow {x}^{2}=2 \ \Rightarrow x=\pm \sqrt{2} \end{gathered}$

Hence, $S$ contains exactly two elements.

Solutions · 1