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(C) Given, $2f(x) + f\left(\frac{1}{x}\right) = 4x$  --- $(i)$ Replacing $x$ with $\frac{1}{x}$, we get: $2f\left(\frac{1}{x}\right) + f(x) = \frac{4}

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$2$

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(C) Given, $2f(x) + f\left(\frac{1}{x}\right) = 4x$  --- $(i)$ Replacing $x$ with $\frac{1}{x}$, we get: $2f\left(\frac{1}{x}\right) + f(x) = \frac{4}{x}$  --- (ii) Multiply equation $(i)$ by $2$: $4f(x) + 2f\left(\frac{1}{x}\right) = 8x$  --- (iii) Subtracting equation (ii) from equation (iii): $(4f(x) + 2f\left(\frac{1}{x}\right)) - (f(x) + 2f\left(\frac{1}{x}\right)) = 8x - \frac{4}{x}$ $3f(x) = 8x - \frac{4}{x} = \frac{8x^2 - 4}{x}$ $f(x) = \frac{4(2x^2 - 1)}{3x}$ Now, for $S = \{x \in R : f(x) = f(-x)\}$: $f(-x) = \frac{4(2(-x)^2 - 1)}{3(-x)} = -\frac{4(2x^2 - 1)}{3x} = -f(x)$ Setting $f(x) = f(-x)$ implies $f(x) = -f(x)$, which means $2f(x) = 0$, so $f(x) = 0$. $\frac{4(2x^2 - 1)}{3x} = 0 \implies 2x^2 - 1 = 0 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$. Thus, $S = \left\{-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right\}$. The number of elements in $S$ is $2$.

If $f: R \setminus \{0\} \rightarrow R$ is such that $2 f(x) + f\left(\frac{1}{x}\right) = 4x$ and $S = \{x \in R : f(x) = f(-x)\}$, then the number of elements in $S$ is

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