Key Concepts and Formulas

• Rolle's Theorem: If a function $f$ is continuous on the closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists at least one number $c$ in $(a, b)$ such that $f'(c) = 0$.
• Generalized Rolle's Theorem (or Corollary): If a function $g$ is continuous on $[a, b]$ and $n$ times differentiable on $(a, b)$, and $g$ has $n+1$ distinct roots in $[a, b]$, then there exists at least one number $c$ in $(a, b)$ such that $g^{(n)}(c) = 0$.
• Properties of Continuous and Differentiable Functions: If a function is continuous on $[a, b]$ and differentiable on $(a, b)$, then the sum, difference, product, and quotient (where the denominator is non-zero) of such functions are also continuous and differentiable on the respective intervals.

Step-by-Step Solution

• Step 1: Define an auxiliary function. We are given a continuous function $f$ on $[0, 2]$ such that $f(0) = 0$, $f(1) = 1$, and $f(2) = 2$. We want to analyze the second derivative of $f$. Let's define a new function $h(x) = f(x) - x$.

  • Reasoning: This auxiliary function is constructed to simplify the conditions. If $f(x) = x$, then $h(x) = 0$ for all $x$. The given conditions suggest that $f(x)$ might be "close" to $x$. By subtracting $x$, we aim to find roots for $h(x)$ that can then be used with Rolle's Theorem.

• Step 2: Analyze the properties of the auxiliary function $h(x)$. Since $f(x)$ is continuous on $[0, 2]$ and twice differentiable on $(0, 2)$, and $g(x) = x$ is also continuous and differentiable everywhere, their difference $h(x) = f(x) - x$ will also be continuous on $[0, 2]$ and twice differentiable on $(0, 2)$.

  • Reasoning: The sum/difference of continuous functions is continuous, and the sum/difference of differentiable functions is differentiable.

• Step 3: Evaluate $h(x)$ at the given points. Let's check the values of $h(x)$ at $x=0$, $x=1$, and $x=2$: $h(0) = f(0) - 0 = 0 - 0 = 0$ $h(1) = f(1) - 1 = 1 - 1 = 0$ $h(2) = f(2) - 2 = 2 - 2 = 0$

  • Reasoning: We are checking if the auxiliary function shares the same root properties as implied by the given values of $f(x)$.

• Step 4: Apply Rolle's Theorem to $h(x)$ on the interval $[0, 1]$. Since $h(x)$ is continuous on $[0, 1]$, differentiable on $(0, 1)$, and $h(0) = h(1) = 0$, by Rolle's Theorem, there exists at least one number $c_1 \in (0, 1)$ such that $h'(c_1) = 0$.

  • Reasoning: The conditions for Rolle's Theorem are satisfied, guaranteeing the existence of a point where the derivative is zero.

• Step 5: Apply Rolle's Theorem to $h(x)$ on the interval $[1, 2]$. Similarly, since $h(x)$ is continuous on $[1, 2]$, differentiable on $(1, 2)$, and $h(1) = h(2) = 0$, by Rolle's Theorem, there exists at least one number $c_2 \in (1, 2)$ such that $h'(c_2) = 0$.

  • Reasoning: The conditions for Rolle's Theorem are satisfied again for a different interval, guaranteeing another point where the derivative is zero.

• Step 6: Relate the derivative of $h(x)$ back to the derivative of $f(x)$. We have $h(x) = f(x) - x$. Differentiating $h(x)$ with respect to $x$, we get $h'(x) = f'(x) - 1$.

  • Reasoning: This step connects the derivative of our auxiliary function back to the derivative of the original function $f$.

• Step 7: Use the results from Step 4 and Step 5 with the relation from Step 6. From Step 4, we have $h'(c_1) = 0$, which means $f'(c_1) - 1 = 0$, so $f'(c_1) = 1$. From Step 5, we have $h'(c_2) = 0$, which means $f'(c_2) - 1 = 0$, so $f'(c_2) = 1$. Thus, we have found two distinct points $c_1 \in (0, 1)$ and $c_2 \in (1, 2)$ such that $f'(c_1) = 1$ and $f'(c_2) = 1$.

  • Reasoning: We have established that the derivative of $f$ takes the value 1 at at least two distinct points.

• Step 8: Apply Rolle's Theorem to $f'(x)$ on the interval $[c_1, c_2]$. Consider the function $f'(x)$. Since $f(x)$ is twice differentiable on $(0, 2)$, $f'(x)$ is continuous on $(0, 2)$. Since $f(x)$ is twice differentiable on $(0, 2)$, $f'(x)$ is differentiable on $(0, 2)$. We have found $c_1 \in (0, 1)$ and $c_2 \in (1, 2)$ such that $f'(c_1) = 1$ and $f'(c_2) = 1$. Therefore, $f'(x)$ is continuous on $[c_1, c_2]$ and differentiable on $(c_1, c_2)$, and $f'(c_1) = f'(c_2)$. By Rolle's Theorem applied to $f'(x)$, there exists at least one number $c \in (c_1, c_2)$ such that $(f'(x))'|_{x=c} = 0$.

  • Reasoning: We are now applying Rolle's Theorem to the first derivative of $f$ because we have found two points where $f'(x)$ has the same value. This will allow us to conclude something about the second derivative of $f$.

• Step 9: State the conclusion about $f''(x)$. The derivative of $f'(x)$ is $f''(x)$. So, from Step 8, we have $f''(c) = 0$ for some $c \in (c_1, c_2)$. Since $c_1 \in (0, 1)$ and $c_2 \in (1, 2)$, the interval $(c_1, c_2)$ is a subset of $(0, 2)$. Therefore, there exists some $x \in (0, 2)$ (namely, $x=c$) such that $f''(x) = 0$.

  • Reasoning: This is the direct consequence of applying Rolle's theorem to $f'(x)$.

Common Mistakes & Tips

• Misapplication of Rolle's Theorem: Ensure that all conditions of Rolle's Theorem (continuity on the closed interval, differentiability on the open interval, and equal function values at the endpoints) are met before applying it.
• Confusing Derivatives: Be careful to distinguish between $f(x)$, $f'(x)$, and $f''(x)$. When applying Rolle's Theorem to $f'(x)$, the conclusion is about $(f'(x))'$, which is $f''(x)$.
• Generalization: The problem asks for the existence of a point where $f''(x) = 0$. Proving this for "some" $x$ is sufficient. Options like $f''(x) = 0$ for all $x$ are much stronger claims and would require more specific information about $f(x)$ (e.g., $f(x) = ax+b$).

Summary The problem leverages Rolle's Theorem by constructing an auxiliary function $h(x) = f(x) - x$. This function is shown to have roots at $x=0$, $x=1$, and $x=2$. By applying Rolle's Theorem twice to $h(x)$ on the intervals $[0, 1]$ and $[1, 2]$, we deduce the existence of two points $c_1 \in (0, 1)$ and $c_2 \in (1, 2)$ where $h'(x) = 0$. Since $h'(x) = f'(x) - 1$, this implies $f'(c_1) = 1$ and $f'(c_2) = 1$. Finally, applying Rolle's Theorem to the function $f'(x)$ on the interval $[c_1, c_2]$ leads to the conclusion that there exists a point $c \in (c_1, c_2)$ such that $(f'(x))'|_{x=c} = f''(c) = 0$. This means $f''(x) = 0$ for some $x \in (0, 2)$.

The final answer is \boxed{A}.