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

We are given the functional equation

$$f(x)+2f\!\left(\frac1x\right)=3x,\qquad x\neq0.$$

Our aim is to determine the set

$$S=\{x\in\mathbb R\;:\;f(x)=f(-x)\}.$$

First we must obtain an explicit expression for $$f(x).$$ To do this we write the given relation once for $$x$$ and once for $$\dfrac1x$$.

For $$x$$ itself we already have

$$f(x)+2f\!\left(\frac1x\right)=3x\qquad\text{(1)}.$$

Now replace $$x$$ by $$\dfrac1x$$ everywhere in (1). Because $$x\neq0$$, this substitution is legitimate and yields

$$f\!\left(\frac1x\right)+2f(x)=\frac3x\qquad\text{(2)}.$$

Thus we have a pair of simultaneous linear equations in the two unknowns

$$A=f(x),\qquad B=f\!\left(\frac1x\right).$$

Written explicitly, the system is

$$\begin{cases} A+2B=3x,\\[4pt] 2A+B=\dfrac3x. \end{cases}$$

We now solve this system. From elementary algebra, when we have

$$\begin{cases} A+2B=C_1,\\ 2A+B=C_2, \end{cases}$$

the solution is obtained by elimination. Multiplying the first equation by $$2$$ gives

$$2A+4B=6x\qquad\text{(3)}.$$

Subtract (2) from (3):

$$\bigl(2A+4B\bigr)-\bigl(2A+B\bigr)=6x-\frac3x.$$

The left‐hand side simplifies to $$3B$$, so

$$3B=6x-\frac3x.$$

Dividing by $$3$$ gives

$$B=2x-\frac1x.$$

Thus

$$f\!\left(\frac1x\right)=2x-\frac1x.$$

We substitute this value of $$B$$ back into equation (1):

$$A+2\left(2x-\frac1x\right)=3x.$$

Expanding the brackets we get

$$A+4x-\frac2x=3x.$$

Now isolate $$A$$ (which is $$f(x)$$):

$$A=3x-4x+\frac2x=-x+\frac2x.$$

So we have obtained an explicit formula valid for every non-zero real number:

$$f(x)=-x+\frac2x,\qquad x\neq0.$$

With the function in hand, we can now find $$S.$$ By definition,

$$f(x)=f(-x).$$

Using our formula for $$f$$ on both sides, we write

$$-x+\frac2x \;=\; -(-x)+\frac2{-x}.$$

Carefully simplifying the right‐hand side:

$$-(-x)=x,\qquad\frac2{-x}=-\frac2x,$$

so the equality becomes

$$-x+\frac2x = x-\frac2x.$$

To clear the denominators we multiply both sides by $$x$$ (remember, $$x\neq0$$):

$$x\!\left(-x+\frac2x\right)=x\!\left(x-\frac2x\right).$$

Performing the multiplication term by term, we obtain

$$-x^2+2 = x^2-2.$$

Now we move all terms to one side:

$$-x^2+2-(x^2-2)=0\quad\Longrightarrow\quad -x^2+2-x^2+2=0.$$

This simplifies to

$$-2x^2+4=0.$$

Dividing by $$-2$$ gives

$$x^2-2=0.$$

Finally, solving for $$x$$ we get

$$x^2=2\quad\Longrightarrow\quad x=\pm\sqrt2.$$

Both solutions are non-zero real numbers, so they are admissible. No other real numbers satisfy the equality $$f(x)=f(-x).$$

Therefore

$$S=\{\sqrt2,\,-\sqrt2\},$$

which clearly contains exactly two elements.

Hence, the correct answer is Option A.