Let a, b, c be non-zero real numbers such tha | vedclass

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(B) Let $f(x) = ax^2 + bx + c$ and $g(x) = 1 + \cos^8 x$. The given equation is $\int_{0}^{1} g(x)f(x) dx = \int_{0}^{2} g(x)f(x) dx$. This can be rew

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at least one root in $(0, 2)$

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(B) Let $f(x) = ax^2 + bx + c$ and $g(x) = 1 + \cos^8 x$. The given equation is $\int_{0}^{1} g(x)f(x) dx = \int_{0}^{2} g(x)f(x) dx$. This can be rewritten as $\int_{0}^{1} g(x)f(x) dx = \int_{0}^{1} g(x)f(x) dx + \int_{1}^{2} g(x)f(x) dx$. Subtracting $\int_{0}^{1} g(x)f(x) dx$ from both sides,we get $\int_{1}^{2} g(x)f(x) dx = 0$. Since $g(x) = 1 + \cos^8 x$ is always positive for all real $x$,the integral of $g(x)f(x)$ over the interval $(1, 2)$ can only be zero if $f(x)$ changes sign in the interval $(1, 2)$. By the Intermediate Value Theorem,if a continuous function $f(x)$ changes sign in an interval $(1, 2)$,there must exist at least one point $c \in (1, 2)$ such that $f(c) = 0$. Since $(1, 2) \subset (0, 2)$,the quadratic equation $ax^2 + bx + c = 0$ must have at least one root in $(0, 2)$.

Let $a, b, c$ be non-zero real numbers such that $\int_{0}^{1} (1 + \cos^8 x)(ax^2 + bx + c) dx = \int_{0}^{2} (1 + \cos^8 x)(ax^2 + bx + c) dx$. Then the quadratic equation $ax^2 + bx + c = 0$ has:

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