1. Key Concepts and Formulas

   • Critical Points: For a differentiable function $f(x)$, critical points are the values of $x$ where $f'(x) = 0$.
   • Polynomial Derivatives: If $f(x)$ is a polynomial of degree $n$, then $f'(x)$ is a polynomial of degree $n-1$. If $x_1, x_2, \ldots, x_{n-1}$ are the roots of $f'(x)$, then $f'(x)$ can be written as $a(x-x_1)(x-x_2)\ldots(x-x_{n-1})$ for some constant $a$.
   • Integration: The indefinite integral of $x^n$ is $\frac{x^{n+1}}{n+1} + C$, where $C$ is the constant of integration.

2. Step-by-Step Solution

Step 1: Determine the form of the derivative $f'(x)$

• Reasoning: We are given that $f(x)$ is a polynomial of degree four. Its critical points are at $-1, 0, 1$. For a polynomial, critical points occur where the derivative is zero. Therefore, $f'(x)$ must have roots at $-1, 0, 1$. Since $f(x)$ is of degree four, $f'(x)$ must be of degree three.
• Formulation: We can write $f'(x)$ in factored form using its roots and a leading coefficient, say $a$: $f'(x) = a(x - (-1))(x - 0)(x - 1)$ $f'(x) = ax(x+1)(x-1)$ $f'(x) = ax(x^2 - 1)$ $f'(x) = a(x^3 - x)$ Here, $a$ must be a non-zero constant, otherwise $f(x)$ would be a constant and not degree four.

Step 2: Find the polynomial $f(x)$ by integrating $f'(x)$

• Reasoning: To obtain $f(x)$ from its derivative $f'(x)$, we perform indefinite integration. We must also include a constant of integration, $C$.
• Integration: $f(x) = \int f'(x) \, dx$ $f(x) = \int a(x^3 - x) \, dx$ $f(x) = a \int (x^3 - x) \, dx$ $f(x) = a \left( \frac{x^4}{4} - \frac{x^2}{2} \right) + C$

Step 3: Determine the value of $f(0)$

• Reasoning: The set $T$ is defined by the condition $f(x) = f(0)$. We need to find the value of $f(0)$ by substituting $x=0$ into our expression for $f(x)$.
• Calculation: $f(0) = a \left( \frac{0^4}{4} - \frac{0^2}{2} \right) + C$ $f(0) = a(0 - 0) + C$ $f(0) = C$

Step 4: Set up and solve the equation $f(x) = f(0)$

• Reasoning: We are given that $T = \{x \in \mathbb{R} \mid f(x) = f(0)\}$. By substituting the expressions for $f(x)$ and $f(0)$, we can find the elements of $T$.
• Equation: $a \left( \frac{x^4}{4} - \frac{x^2}{2} \right) + C = C$
• Simplification: Subtracting $C$ from both sides, we get: $a \left( \frac{x^4}{4} - \frac{x^2}{2} \right) = 0$ Since $a \neq 0$, we can divide by $a$: $\frac{x^4}{4} - \frac{x^2}{2} = 0$ To solve for $x$, multiply the entire equation by 4 to clear the denominators: $x^4 - 2x^2 = 0$ Factor out the common term $x^2$: $x^2(x^2 - 2) = 0$ This equation holds if $x^2 = 0$ or $x^2 - 2 = 0$. Case 1: $x^2 = 0 \implies x = 0$. Case 2: $x^2 - 2 = 0 \implies x^2 = 2 \implies x = \sqrt{2}$ or $x = -\sqrt{2}$.

Step 5: Identify the elements of set $T$ and calculate the sum of their squares

• Reasoning: The solutions to the equation $f(x) = f(0)$ are the elements of the set $T$. We need to find the sum of the squares of these elements.
• Set T: The elements of $T$ are $0, \sqrt{2}, -\sqrt{2}$. $T = \{0, \sqrt{2}, -\sqrt{2}\}$
• Sum of Squares: $\sum_{x \in T} x^2 = (0)^2 + (\sqrt{2})^2 + (-\sqrt{2})^2$ $\sum_{x \in T} x^2 = 0 + 2 + 2$ $\sum_{x \in T} x^2 = 4$

1. Common Mistakes & Tips

   • Forgetting the constant of integration (C): When integrating to find $f(x)$, always include the constant of integration. It cancels out in this specific problem, but it's a fundamental part of indefinite integration.
   • Assuming $a=1$: The leading coefficient $a$ in $f'(x)$ is not necessarily 1. While it cancels out in the equation $f(x) = f(0)$, it's important to include it in the general form of $f'(x)$.
   • Incomplete factoring: Ensure all roots are found. For $x^2(x^2 - 2) = 0$, remember that $x^2=0$ gives $x=0$, and $x^2-2=0$ gives two distinct roots.

2. Summary The problem involves determining a polynomial of degree four given its critical points. By using the critical points to construct the derivative $f'(x)$, we integrated it to find the general form of $f(x)$, including the constant of integration $C$. The condition $f(x) = f(0)$ simplifies to an equation in $x$ after evaluating $f(0)$. Solving this equation by factoring revealed the elements of the set $T$. Finally, we computed the sum of the squares of these elements.

The final answer is \boxed{4}, which corresponds to option (D).