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 $f'(c) = 0$.
• Power Rule of Differentiation: $\frac{d}{dx}(x^n) = nx^{n-1}$.
• Polynomial Properties: Polynomials are continuous and differentiable everywhere.

Step-by-Step Solution

Step 1: Define the given function and identify its roots.

Let $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x$. We are given that $f(x) = 0$ has a positive root $x = \alpha$, meaning $f(\alpha) = 0$. We also need to find another root to apply Rolle's Theorem.

Notice that $f(0) = a_n (0)^n + a_{n-1} (0)^{n-1} + \dots + a_1 (0) = 0$. Thus, $x=0$ is also a root of $f(x)$.

Why: We are setting up the conditions to use Rolle's Theorem. We need two points where the function has the same value.

Step 2: Verify the conditions of Rolle's Theorem on the interval $[0, \alpha]$.

Since $f(x)$ is a polynomial, it is continuous on $[0, \alpha]$ and differentiable on $(0, \alpha)$. Also, $f(0) = 0$ and $f(\alpha) = 0$, so $f(0) = f(\alpha)$. Thus, all conditions of Rolle's Theorem are satisfied.

Why: We are verifying that Rolle's Theorem is applicable.

Step 3: Apply Rolle's Theorem.

By Rolle's Theorem, there exists a $c \in (0, \alpha)$ such that $f'(c) = 0$.

Why: This is the direct application of Rolle's Theorem, which guarantees the existence of a root for the derivative within the specified interval.

Step 4: Find the derivative of $f(x)$.

We have $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x$. Taking the derivative with respect to $x$, we get:

$f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1.$

Why: We need to find the derivative to relate it to the second given equation.

Step 5: Relate the derivative to the second equation and interpret the root.

The second equation given in the problem is $n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1 = 0$, which is precisely $f'(x) = 0$. Since we know there exists a $c \in (0, \alpha)$ such that $f'(c) = 0$, the second equation has a root $x = c$ where $0 < c < \alpha$. This means that the second equation has a positive root that is smaller than $\alpha$.

Why: We are connecting the result from Rolle's Theorem to the problem's question. The root $c$ is positive because it lies in the interval $(0, \alpha)$, and it is smaller than $\alpha$.

Common Mistakes & Tips

• Forgetting to check Rolle's Theorem conditions: Always ensure the function is continuous, differentiable, and has the same value at the endpoints before applying Rolle's Theorem.
• Misunderstanding the conclusion of Rolle's Theorem: Rolle's Theorem guarantees the existence of at least one root of the derivative within the interval, not necessarily the only root.
• Not recognizing $f(0) = 0$: This is a key step to apply Rolle's Theorem

Summary

By defining $f(x)$ as the given polynomial expression, we found that $f(0) = f(\alpha) = 0$. Applying Rolle's Theorem on the interval $[0, \alpha]$, we showed that there exists a $c \in (0, \alpha)$ such that $f'(c) = 0$. Since $f'(x)$ corresponds to the second given equation, this implies that the second equation has a positive root smaller than $\alpha$.

Final Answer

The final answer is \boxed{smaller than \alpha}, which corresponds to option (B).