Theory of Equations 5 Question 8
8. Let $a, b, c$ be non-zero real numbers such that
$ \begin{aligned} \int _0^{1}\left(1+\cos ^{8} x\right)\left(a x^{2}\right. +b x+c) d x \\ =\int _0^{2}\left(1+\cos ^{8} x\right)\left(a x^{2}+b x+c\right) d x \end{aligned} $
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Answer:
Correct Answer: 8. (b)
Solution:
- Consider,
$ f(x)=\int _0^{1}\left(1+\cos ^{8} x\right)\left(a x^{2}+b x+c\right) d x $
Obviously, $f(x)$ is continuous and differentiable in the interval [1, 2]. Also,
$ f(1)=f(2) $
[given]
$\therefore$ By Rolle’s theorem, there exist atleast one point $k \in(1,2)$, such that $f^{\prime}(k)=0$.
$ \begin{aligned} & \text { Now, } \\ & f^{\prime}(x)=\left(1+\cos ^{8} x\right)\left(a x^{2}+b x+c\right) \\ & f^{\prime}(k)=0 \\ & \Rightarrow \quad\left(1+\cos ^{8} k\right)\left(a k^{2}+b k+c\right)=0 \\ & \Rightarrow \quad a k^{2}+b k+c=0 \quad\left[\operatorname{as}\left(1+\cos ^{8} k\right) \neq 0\right] \\ & \therefore x=k \text { is root of } a x^{2}+b x+c=0, \end{aligned} $
where $k \in(1,2)$
