Question

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

• A. is an empty set.
• B. contains exactly one element.
• C. contains exactly two elements.
• D. contains more than two elements.

Answer 1

Answer: (C)

$f\left(x\right) + 2f \left(\frac{1}{x}\right) = 3x$
Replace x by $ \frac{1}{x} , f\left(\frac{1}{x}\right) + 2f \left(x\right) = \frac{3}{x}$
$ \Rightarrow \frac{3x-f\left(x\right)}{2} = \frac{\frac{3}{x} - 2f\left(x\right)}{1} $
$ \Rightarrow 3x -f\left(x\right) = \frac{6}{x} - 4 f\left(x\right) $
$ \Rightarrow f\left(x\right) = \frac{2}{x} - x$
$ f\left(x\right) = f\left(-x\right) $
$\Rightarrow \frac{2}{x} - x = - \frac{2}{x} + x $
$ \Rightarrow \frac{4}{x} = 2x $
$\Rightarrow x = \pm \sqrt{2} $