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

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(B) Let $f(x) = (ax^2 + bx + c)(1 + \cos^8 x)$.Given $\int_0^1 f(x) \, dx = \int_0^2 f(x) \, dx$.This implies $\int_0^2 f(x) \, dx - \int_0^1 f(x) \,

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

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(B) Let $f(x) = (ax^2 + bx + c)(1 + \cos^8 x)$.Given $\int_0^1 f(x) \, dx = \int_0^2 f(x) \, dx$.This implies $\int_0^2 f(x) \, dx - \int_0^1 f(x) \, dx = 0$,which simplifies to $\int_1^2 f(x) \, dx = 0$.Since $f(x)$ is a continuous function and the integral over the interval $(1, 2)$ is zero,$f(x)$ must change sign in the interval $(1, 2)$ unless $f(x) = 0$ for all $x \in (1, 2)$.Since $a, b, c$ are non-zero,$ax^2 + bx + c$ is not identically zero. Also,$(1 + \cos^8 x) \ge 1 > 0$ for all $x$.Thus,$f(x)$ must take both positive and negative values in $(1, 2)$.By the Intermediate Value Theorem,there exists at least one $x_0 \in (1, 2)$ such that $f(x_0) = 0$.Since $(1 + \cos^8 x) 
eq 0$,we must have $ax_0^2 + bx_0 + c = 0$.Therefore,the quadratic equation $ax^2 + bx + c = 0$ has at least one root in $(1, 2)$,which is contained 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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