int_{0}^{1} (1 + |sin x|)(ax^2 + bx + c) dx = | vedclass

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(A) Given the equation: $\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx = \int_{0}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$. We can rewrite the right side

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At least one real root between $(1, 2)$

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(A) Given the equation: $\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx = \int_{0}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$. We can rewrite the right side as: $\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx + \int_{1}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$. Subtracting the integral from $0$ to $1$ from both sides,we get: $\int_{1}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx = 0$. Let $f(x) = ax^2 + bx + c$ and $g(x) = 1 + |\sin x|$. Since $g(x) > 0$ for all $x \in [1, 2]$,for the integral of the product $f(x)g(x)$ to be zero over the interval $[1, 2]$,the function $f(x)$ must change its sign at least once in the interval $(1, 2)$. By the Intermediate Value Theorem,if a continuous function changes sign in an interval,it must have at least one root in that interval. Therefore,$ax^2 + bx + c = 0$ has at least one real root between $(1, 2)$.

$\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx = \int_{0}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$. Then,the location of the roots of $ax^2 + bx + c = 0$ is:

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