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

Question

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

Accepted Answer

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

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

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

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

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

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

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

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

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

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

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

Hence, $S$ contains exactly two elements.

Solutions · 1