Question

If $f\left(x\right)+2f\left(\frac{1}{x}\right)=3x,x\ne 0andS=xϵR:f\left(x\right)=f(-x)$; 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

The correct option is C contains exactly two elements

We have, $f\left(x\right)+2f\left(\frac{1}{x}\right)=3x,x\ne 0$ . . . (i)
On replacing $x$ by $\frac{1}{x}$ in the above equation, we get
$f\left(\frac{1}{x}\right)+2f\left(x\right)=\frac{3}{x}$
$\Rightarrow 2f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{3}{x}$ . . . (ii)
On multiplying Eq. (ii) by 2 and subtracting Eq. (i) from Eq. (ii), we get
$3f\left(x\right)=\frac{6}{x}-3x$
$\Rightarrow f\left(x\right)=\frac{2}{x}-x$
Now, consider $f\left(x\right)=f(-x)$
$\Rightarrow \frac{2}{x}-x=-\frac{2}{x}+x\Rightarrow \frac{4}{x}=2x\Rightarrow 2x^2=4\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}$
Hence, S contains exactly two elements.