Question

Let $f$ be defined for all non-zero real numbers $x$ as $f\left(x\right)+2f\left(\frac{1}{x}\right)=3x.$ Then the number of values of $x$ satisfying the equation $f\left(x\right)=f(-x)$ is

Answer 1

$f\left(x\right)+2f\left(\frac{1}{x}\right)=3x\cdots \left(1\right)$
Replacing $x$ by $\frac{1}{x},$ we get
$f\left(\frac{1}{x}\right)+2f\left(x\right)=3\left(\frac{1}{x}\right)$
$\Rightarrow 2f\left(\frac{1}{x}\right)+4f\left(x\right)=6\left(\frac{1}{x}\right)\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we get
$3f\left(x\right)=\frac{6}{x}-3x$
Since $f\left(x\right)=f(-x),$
$\therefore \frac{6}{x}-3x=-\frac{6}{x}+3x$
$\Rightarrow \frac{12}{x}=6x$
$\Rightarrow x^2=2$
$\therefore x=\pm \sqrt{2}$