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

If f(x) + 2f(1/x) = 3x, x ≠ 0 and S = {x ∈ R : f(x) = f(-x)}; then S: $\text{Given: } f(x) + 2f\left(\frac{1}{x}\right) = 3x,\quad x \neq 0$ $S = \{x \in \mathbb{R} : f(x) = f(-x)\}$ $\text{To find: the number of elements in } S$ $(1)\ \text{contains exactly two elements}.\qquad (2)\ \text{contains more than two elements}.$ $(3)\ \text{is an empty set}.\qquad (4)\ \text{contains exactly one element}.$ $1.\ \text{put } x \to \tfrac{1}{x} \text{ to get a second equation}$ $2.\ \text{eliminate } f\!\left(\tfrac{1}{x}\right) \text{ and solve for } f(x)$ $3.\ \text{apply } f(x) = f(-x) \text{ and count elements of } S$ $\text{Replace } x \to \frac{1}{x}:$ $f\!\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x}$ $2f\!\left(\frac{1}{x}\right) + 4f(x) = \frac{6}{x}$ $3f(x) = \frac{6}{x} - 3x$ $f(x) = \frac{2}{x} - x$ $f(-x) = \frac{2}{-x} - (-x) = -\frac{2}{x} + x$ $\frac{2}{x} - x = -\frac{2}{x} + x$ $\frac{4}{x} = 2x$ $4 = 2x^2 \implies x^2 = 2$ $x = \pm\sqrt{2}$ $f(\sqrt{2}) = \frac{2}{\sqrt{2}} - \sqrt{2} = \sqrt{2} - \sqrt{2} = 0$ $f(-\sqrt{2}) = \frac{2}{-\sqrt{2}} + \sqrt{2} = -\sqrt{2} + \sqrt{2} = 0$ $S = \{-\sqrt{2},\ \sqrt{2}\}$ $\therefore\ S \text{ contains exactly two elements} \Rightarrow \boxed{\text{option (1)}}$