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}:

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

Answer: C. contains exactly two elements. We have, f(x)+2 f\left(\frac{1}{x}ight)=3 x, \quad x 
eq 0 \ldots (i) On replacing x by \frac{1}{x} in the above equation, we get \begin{aligned} & f\left(\frac{1}{x}ight)+2 f(x) =\frac{3}{x} \\ \Rightarrow & \,\, 2 f(x)+f\left(\frac{1}{x}ight) =\frac{3}{x} \,\,\,\,\,...(ii) \end{aligned} On multiplying Eq. (ii) by 2, we get 4 f(x)+2 f\left(\frac{1}{x}ight)=\frac{6}{x}\quad...(iii) and subtracting Eq. (i) from Eq. (iii), we get [4 f(x)+2 f\left(\frac{1}{x}ight)] - [f(x)+2 f\left(\frac{1}{x}ight)]=\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.

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

Answer

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