ISI2019-MMA-25

Question

Let $a,b,c$ be non-zero real numbers such that 

$\int_{0}^{1} (1 + \cos^8x)(ax^2 + bx +c)dx = \int_{0}^{2}(1+ \cos^8x)(ax^2 + bx + c) dx =0$

Then the quadratic equation $ax^2 + bx +c=0$ has 

• A. no roots in $(0,2)$
• B. one root in $(0,2)$ and one root outside this interval
• C. one repeated root in $(0,2)$
• D. two distinct real roots in $(0,2)$

Answer 1

Answer is D. The equations given in the question imply two things:

$\int_{0}^{1} f(x) dx = 0 \\ \int_{1}^{2} f(x) dx = 0$

by using properties of definite integrals

where $f(x) = (1+ \cos ^8 x)(ax^2+bx+c)$

Let us consider the first equation. We know that definite integration is the area under the curve with area above x axis as positive and below x axis as negative. When you are getting the integration as $0$, that means curve cuts the x - axis atleast once in the given interval . but
$1 + \cos ^8 x > 0$ as $1>0$ and $\cos ^8 x \geq 0$ (even power is always nonegative)
That means quadratic equation has atleast one root in each of the intervals $(0,1)$ and $(1,2)$ .

But it is a quadratic equation which cannot have more than two distinct real roots .

So $D$ is correct.