Question:

<p><strong>Step 1: Find f(x) explicitly using the functional equation.</strong></p><p>Given: $f(x) + 2f\left(\frac{1}{x}\right) = 3x$ ... (1)</p><p>Substitute $x \to \frac{1}{x}$:</p><p>$f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x}$ ... (2)</p><p><strong>Step 2: Solve the system of equations (1) and (2).</strong></p><p>From equation (1): $f(x) + 2f\left(\frac{1}{x}\right) = 3x$</p><p>From equation (2): $2f(x) + f\left(\frac{1}{x}\right) = \frac{3}{x}$</p><p>Multiply equation (1) by 2: $2f(x) + 4f\left(\frac{1}{x}\right) = 6x$ ... (3)</p><p>Subtract equation (2) from (3):</p><p>$3f\left(\frac{1}{x}\right) = 6x - \frac{3}{x}$</p><p>$f\left(\frac{1}{x}\right) = 2x - \frac{1}{x}$</p><p><strong>Step 3: Find f(x) by substituting back.</strong></p><p>From equation (1): $f(x) = 3x - 2f\left(\frac{1}{x}\right) = 3x - 2\left(2x - \frac{1}{x}\right)$</p><p>$f(x) = 3x - 4x + \frac{2}{x} = -x + \frac{2}{x}$</p><p><strong>Step 4: Find where f(x) = f(-x) (even function property).</strong></p><p>$f(x) = -x + \frac{2}{x}$</p><p>$f(-x) = -(-x) + \frac{2}{-x} = x - \frac{2}{x}$</p><p>Set $f(x) = f(-x)$:</p><p>$-x + \frac{2}{x} = x - \frac{2}{x}$</p><p>$\frac{4}{x} = 2x$</p><p>$4 = 2x^2$</p><p>$x^2 = 2$</p><p>$x = \pm\sqrt{2}$</p><p><strong>Step 5: Verify both solutions are valid.</strong></p><p>Both $x = \sqrt{2}$ and $x = -\sqrt{2}$ are non-zero real numbers, so both satisfy the domain constraint $x \neq 0$.</p><p>Therefore, $S = \{\sqrt{2}, -\sqrt{2}\}$ contains exactly two elements.</p><p><strong>∴ Answer: 4</strong></p>