Key Concepts and Formulas

• Rolle's Theorem: If a function $f(x)$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.
• Integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $C$ is the constant of integration.
• Polynomial Properties: Polynomial functions are continuous and differentiable everywhere.

Step-by-Step Solution

1. Understand the Problem and Identify the Goal

We are given the quadratic equation $ax^2 + bx + c = 0$ and the condition $2a + 3b + 6c = 0$. Our goal is to determine the interval in which the quadratic equation is guaranteed to have at least one root. We will use Rolle's Theorem to achieve this.

2. Construct an Auxiliary Function

To apply Rolle's Theorem, we need a function $f(x)$ such that $f'(x) = ax^2 + bx + c$. We find $f(x)$ by integrating $ax^2 + bx + c$. $f(x) = \int (ax^2 + bx + c) dx = \frac{ax^3}{3} + \frac{bx^2}{2} + cx + K$ We can choose the constant of integration $K = 0$ for simplicity. So, our auxiliary function is: $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$ Why this step? Working backward from the derivative to a function whose values at two points are equal lets us use Rolle's Theorem.

3. Evaluate the Auxiliary Function at Specific Points to Satisfy $f(a)=f(b)$ Condition

We need to find two points, $x_1$ and $x_2$, such that $f(x_1) = f(x_2)$. We will try simple points like $x=0$ and $x=1$ due to the coefficients in the condition $2a+3b+6c=0$.

Let's evaluate $f(x)$ at $x=0$: $f(0) = \frac{a(0)^3}{3} + \frac{b(0)^2}{2} + c(0) = 0$

Now, let's evaluate $f(x)$ at $x=1$: $f(1) = \frac{a(1)^3}{3} + \frac{b(1)^2}{2} + c(1) = \frac{a}{3} + \frac{b}{2} + c$ To relate this to the given condition $2a+3b+6c=0$, we can find a common denominator: $f(1) = \frac{2a}{6} + \frac{3b}{6} + \frac{6c}{6} = \frac{2a+3b+6c}{6}$ Now, substitute the given condition $2a+3b+6c=0$: $f(1) = \frac{0}{6} = 0$ Therefore, $f(0) = f(1) = 0$. Why this step? This step is crucial to satisfy the third condition of Rolle's Theorem, $f(a)=f(b)$. By finding that $f(0)=0$ and $f(1)=0$, we have established $f(0)=f(1)$, which means we can apply Rolle's Theorem on the interval $[0,1]$.

4. Check for Continuity and Differentiability

Our auxiliary function $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$ is a polynomial function.

• Polynomials are continuous everywhere on $R$. Therefore, $f(x)$ is continuous on the closed interval $[0, 1]$.
• Polynomials are differentiable everywhere on $R$. Therefore, $f(x)$ is differentiable on the open interval $(0, 1)$. Why this step? These are the first two essential conditions for Rolle's Theorem to be applicable. Without them, the theorem cannot guarantee the existence of a root.

5. Apply Rolle's Theorem

Since $f(x)$ satisfies all three conditions of Rolle's Theorem on the interval $[0,1]$:

1. $f(x)$ is continuous on $[0,1]$.
2. $f(x)$ is differentiable on $(0,1)$.
3. $f(0) = f(1) = 0$.

Therefore, by Rolle's Theorem, there exists at least one value $x_0 \in (0,1)$ such that $f'(x_0) = 0$.

6. Relate Back to the Original Quadratic Equation

We defined $f'(x)$ as the quadratic expression $ax^2 + bx + c$. So, $f'(x) = ax^2 + bx + c$. Since $f'(x_0) = 0$ for some $x_0 \in (0,1)$, it means: $ax_0^2 + bx_0 + c = 0$ This implies that the quadratic equation $ax^2 + bx + c = 0$ has at least one root $x_0$ in the open interval $(0,1)$. Therefore, it also has at least one root in the closed interval $[0,1]$.

7. Conclusion and Option Selection

The quadratic equation $ax^2 + bx + c = 0$ has at least one root in $[0,1]$. Comparing this with the given options, option (A) matches our conclusion.

Common Mistakes & Tips

• Auxiliary Function: Choose $f(x)$ such that $f'(x)$ is the expression whose roots you are trying to find. If the problem asks for roots of $P(x)$, then $f(x) = \int P(x) dx$.
• Matching the Condition: The given condition (e.g., $2a+3b+6c=0$) is key to finding the interval $[a,b]$ where $f(a)=f(b)$. Often, you'll need to manipulate $f(a)$ and $f(b)$ to make them match the given condition.
• Interval Type: Rolle's Theorem guarantees a root in the open interval $(a,b)$. However, if a root is in $(a,b)$, it is also considered to be in the closed interval $[a,b]$. The options usually provide closed intervals.

Summary

This problem demonstrates the application of Rolle's Theorem to prove the existence of roots. By constructing an auxiliary function $f(x)$ whose derivative is the given quadratic expression and strategically choosing points where $f(x)$ takes the same value (using the given condition $2a+3b+6c=0$), we can conclude that the quadratic equation must have at least one root in the interval $(0,1)$ and, consequently, in $[0,1]$.

Final Answer

The final answer is \boxed{[0,1]}, which corresponds to option (A).