If a, b, c be non-zero real numbers such that $\int\limits_0^1\left(1+\cos ^8 x\right)\left(a x^2+b x+c\right) d x=\int\limits_0^2\left(1+\cos ^8 x\right)\left(a x^2+b x+c\right) d x=0$, then the equation $a x^2+b x+c=0$ will have
Answer & explanation
Correct answer: option 1
Consider the function $\phi(x)$ given by
$\phi(x)=\int\limits_0^x\left(1+\cos ^8 t\right)\left(a t^2+b t+c\right) d t$
$\Rightarrow \phi^{\prime}(x)=\left(1+\cos ^8 x\right)\left(a x^2+b x+c\right)$ ......(i)
We observe that
$\phi(0)=0$
$\phi(1)=\int\limits_0^1\left(1+\cos ^8 t\right)\left(a t^2+b t+c\right) d t=0$ [Given]
and, $\phi(2)=\int\limits_0^2\left(1+\cos ^8 t\right)\left(a t^2+b t+c\right) d t=0$ [Given]
Therefore, 0, 1 and 2 are the roots of $\phi(x)$.
By Rolle's theorem $\phi^{\prime}(x)=0$ will have at least one real root between 0 and 1 and at least one real root between 1 and 2.