Question

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 \(ax^{2}+bx+c=0\) has:

• A. no solution in $(0, 2)$
• B. at least one root in $(1, 2)$
• C. two imaginary roots
• D. two roots in $(0, 2)$

Hint:

Whenever you see a definite integral problem where the area from $0 \rightarrow 1$ equals the area from $0 \rightarrow 2$, it tells you that the net area gained between 1 and 2 is exactly zero. For a continuous function to trap zero net area, it *must* cross the x-axis somewhere inside that interval!

Answer 1

The Correct Option is B

Concept: According to Rolle's Theorem for integrals, if a continuous function $g(x)$ satisfies $\int_{a}^{b} g(x)\,dx = 0$, there must exist at least one real point $k \in (a, b)$ where the integrand itself vanishes: $g(k) = 0$. Step 1: Rearrange the definite integral limits into a single equation.
We are given the following identity equation involving definite integrals: \[ \int_{0}^{1}(1+\cos^{8}x)(ax^{2}+bx+c)\,dx = \int_{0}^{2}(1+\cos^{8}x)(ax^{2}+bx+c)\,dx \] Using standard calculus integration properties, split the right-hand integral across the intermediate boundary point $x = 1$: \[ \int_{0}^{2} g(x)\,dx = \int_{0}^{1} g(x)\,dx + \int_{1}^{2} g(x)\,dx \] Substitute this split integral representation back into the primary balance equation: \[ \int_{0}^{1} g(x)\,dx = \int_{0}^{1} g(x)\,dx + \int_{1}^{2} g(x)\,dx \]

Step 2: Isolate the remaining non-zero integral segment.
Subtract the shared integral term $\int_{0}^{1} g(x)\,dx$ from both sides of the equation: \[ \int_{1}^{2}(1+\cos^{8}x)(ax^{2+bx+c})\,dx = 0 \]

Step 3: Analyze the signs of the individual factor components.
Let us look at the mathematical properties of the first factor component, $(1 + \cos^8 x)$, inside the interval $x \in [1, 2]$:
• For any real number $x$, the even power cosine expression is always non-negative: $\cos^8 x \ge 0$.
• Adding 1 ensures that the entire first factor is strictly positive: \[ 1 + \cos^8 x \ge 1 > 0 \quad \forall x \in [1, 2] \]

Step 4: Apply Rolle's Theorem to locate the quadratic roots.
For the total integrated area to collapse exactly to zero, the remaining quadratic factor component $(ax^2 + bx + c)$ cannot stay entirely positive or entirely negative throughout the interval—it must cross the zero axis to balance out the areas. By Rolle's Theorem for integrals, there must exist at least one real coordinate point $x_0 \in (1, 2)$ where the integrand vanishes: \[ (1+\cos^{8}x_0)(ax_0^{2}+bx_0+c) = 0 \] Since $(1 + \cos^8 x_0) \neq 0$, the quadratic equation must satisfy: \[ ax_0^2 + bx_0 + c = 0 \quad \text{for some } x_0 \in (1, 2) \] This proves that the equation $ax^2 + bx + c = 0$ is guaranteed to have at least one root in the open interval $(1, 2)$, matching option (B).