Step 1: Understanding the Concept:
This problem uses the Mean Value Theorem for Integrals or the fundamental properties of continuous functions. If the integral of a function over two overlapping intervals is equal, the integral over the difference of those intervals must be zero.
Step 2: Key Formula or Approach:
1. Property: \( \int_0^2 f(x) dx = \int_0^1 f(x) dx + \int_1^2 f(x) dx \).
2. If \( \int_1^2 w(x)P(x) dx = 0 \) where \( w(x)>0 \), then \( P(x) \) must change sign in \( (1, 2) \).
Step 3: Detailed Explanation:
Let \( w(x) = 1 + \cos^8 x \) and \( P(x) = ax^2 + bx + c \).
The given condition is:
\[ \int_{0}^{1} w(x)P(x) dx = \int_{0}^{2} w(x)P(x) dx \]
Subtracting the left side from the right:
\[ \int_{0}^{2} w(x)P(x) dx - \int_{0}^{1} w(x)P(x) dx = 0 \]
\[ \int_{1}^{2} w(x)P(x) dx = 0 \]
We know that \( w(x) = 1 + \cos^8 x \) is always positive because \( \cos^8 x \ge 0 \).
If the integral of the product \( w(x)P(x) \) over \( [1, 2] \) is zero, and \( w(x) \) never changes sign, then \( P(x) \) must change its sign at least once within the interval \( (1, 2) \).
If a continuous function \( P(x) \) changes its sign in an interval, by the Intermediate Value Theorem, there must be at least one value \( x_0 \in (1, 2) \) such that \( P(x_0) = 0 \).
Thus, the quadratic equation \( ax^2 + bx + c = 0 \) has at least one real root in the interval \( (1, 2) \).
Step 4: Final Answer:
The quadratic equation has at least one root in (1, 2), corresponding to option (B).