Understanding the Problem

We are given the functional equation: $f\left(x\right)+2f\left(\frac{1}{x}\right)=3x$, for all $x\ne 0$.

We are also told that $S=\{x\in ℝ:f(x)=f(-x\left)\right\}$ and we need to determine which of the given options correctly describes the set S.

Step 1: Find the Function f(x)

The key to solving this is to find an expression for f(x). Notice that the equation involves both f(x) and f(1/x). We can create a system of equations by substituting x with 1/x.

Step 1a: Write the original equation.

Equation (1): $f\left(x\right)+2f\left(\frac{1}{x}\right)=3x$

Step 1b: Substitute x with 1/x in Equation (1).

This gives us a new equation: $f\left(\frac{1}{x}\right)+2f\left(x\right)=3\cdot \frac{1}{x}$
Which simplifies to:
Equation (2): $2f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{3}{x}$

Step 1c: Solve the system of equations (1) and (2) for f(x).

Let $a=f\left(x\right)$ and $b=f\left(\frac{1}{x}\right)$.
Our system is:
(1) $a+2b=3x$
(2) $2a+b=\frac{3}{x}$

Multiply equation (1) by 2: $2a+4b=6x$
Now subtract equation (2) from this result:
$(2a+4b)-(2a+b)=6x-\frac{3}{x}$
$3b=6x-\frac{3}{x}$
$b=2x-\frac{1}{x}$

Substitute the value of b back into equation (1):
$a+2(2x-\frac{1}{x})=3x$
$a+4x-\frac{2}{x}=3x$
$a=3x-4x+\frac{2}{x}$
$a=\frac{2}{x}-x$

Therefore, the function is:
$f\left(x\right)=\frac{2}{x}-x$

Step 2: Analyze the Set S = {x ∈ ℝ : f(x) = f(–x)}

We need to find all real numbers x (where x ≠ 0) such that f(x) is equal to f(-x).

Step 2a: Write the equation f(x) = f(-x).

Substitute our found function into this equation:
$\frac{2}{x}-x=\frac{2}{(-x)}-(-x)$
Simplify the right-hand side:
$\frac{2}{x}-x=-\frac{2}{x}+x$

Step 2b: Solve for x.

Bring all terms to one side of the equation:
$\frac{2}{x}-x+\frac{2}{x}-x=0$
$\frac{4}{x}-2x=0$

Multiply both sides by x to eliminate the denominator (remember x ≠ 0):
$4-2x^2=0$
$2x^2=4$
$x^2=2$
$x=\pm \sqrt{2}$

Step 2c: Interpret the solution.

We found two distinct real solutions: $x=\sqrt{2}$ and $x=-\sqrt{2}$. Both are valid as they are not zero. Therefore, the set S contains exactly these two elements.

Final Answer

The set S contains exactly two elements.

Related Topics & Formulae

Functional Equations: These are equations where the variables are functions. Solving them often involves clever substitutions (like replacing x with 1/x or -x) to create a system of equations that can be solved for the unknown function.

Even and Odd Functions: The set S is asking for the points where the function is even, i.e., where f(x) = f(-x). A function that is even for all x in its domain is called an even function (e.g., f(x)=x²). Our function f(x)=2/x - x is neither purely even nor purely odd, but it coincides with its reflection at