Question

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

The correct option is C contains exactly two elements.

$f\left(x\right)+2f\left(\frac{1}{x}\right)=3x$

Putting $x$ as $\frac{1}{x}$ then $f\left(\frac{1}{x}\right)+2f\left(x\right)=\frac{3}{x}$

Solving these two equations,

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

Now for $f\left(x\right)=f(-x)$

$\Rightarrow \frac{2}{x}-x=x-\frac{2}{x}$

$\Rightarrow 2\times \frac{2}{x}=2x$

$\Rightarrow x^2=2$

$\therefore x=\sqrt{2},-\sqrt{2}$