If $$f(x)$$ is a non-zero polynomial of degree four, having local extreme points at $$x = -1, 0, 1$$; then the set $$S = \{x \in R : f(x) = f(0)\}$$ contains exactly:

We are told that the polynomial $$f(x)$$ is of degree four and has local extreme points at $$x=-1,\;0,\;1$$. The definition of a local extremum says that the derivative of the function must vanish at each such point. Hence we must have

$$f'(-1)=0,\qquad f'(0)=0,\qquad f'(1)=0.$$

A polynomial of degree four has a derivative of degree three. Writing the derivative in its factored form, and because the three distinct real numbers $$-1,0,1$$ are all its zeros, we can state

$$f'(x)=k\,(x+1)\,x\,(x-1),$$

where $$k\neq0$$ is the leading-coefficient constant needed to match the unknown vertical scaling of the original quartic. Multiplying out the factors inside, we have

$$(x+1)\,x\,(x-1)=x\,(x^2-1)=x^3-x.$$

So the derivative simplifies to

$$f'(x)=k\,(x^3-x).$$

To recover $$f(x)$$ we integrate, remembering to add an integration constant. We use the elementary integration formulas $$\int x^n\,dx=\dfrac{x^{n+1}}{n+1}$$ and $$\int x\,dx=\dfrac{x^2}{2}$$. Thus:

$$

\begin{aligned}

f(x)&=\int f'(x)\,dx =\int k\,(x^3-x)\,dx \\

&=k\left(\int x^3\,dx-\int x\,dx\right) \\

&=k\left(\frac{x^4}{4}-\frac{x^2}{2}\right)+C,

\end{aligned}

$$

where $$C$$ is the constant of integration. Writing this compactly,

$$f(x)=\frac{k}{4}\bigl(x^4-2x^2\bigr)+C.$$

This expression indeed has degree four because $$k\neq0$$. Now we must solve the equation

$$f(x)=f(0)$$

and determine how many rational and irrational solutions it has. First, compute the value at $$x=0$$:

$$f(0)=\frac{k}{4}\bigl(0^4-2\cdot0^2\bigr)+C=C.$$

Setting $$f(x)=f(0)$$ therefore gives

$$\frac{k}{4}\bigl(x^4-2x^2\bigr)+C=C.$$

Subtract $$C$$ from both sides so the constants cancel:

$$\frac{k}{4}\bigl(x^4-2x^2\bigr)=0.$$

Because $$k\neq0$$, we can safely divide by $$\dfrac{k}{4}$$ (a non-zero number), leaving

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

We factor completely:

$$x^2\bigl(x^2-2\bigr)=0.$$

Now use the zero-product property, which states that if a product of factors equals zero then at least one factor must be zero. Thus we obtain two sub-equations:

$$x^2=0\quad\text{or}\quad x^2-2=0.$$

From $$x^2=0$$ we get the single solution

$$x=0.$$

From $$x^2-2=0$$ we get

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

Putting all roots together, the set

$$S=\{x\in\mathbb R:f(x)=f(0)\}$$

is

$$S=\{0,\;\sqrt2,\;-\sqrt2\}.$$

The element $$0$$ is a rational number, while $$\sqrt2$$ and $$-\sqrt2$$ are both irrational numbers. Therefore the set contains exactly two irrational numbers and one rational number.

Hence, the correct answer is Option C.