Explanation

(b)
Consider $f(x) = \int_{0}^{x}{(1 + \cos^{8}{x)\left( ax^{2} + bx + c \right)dx}}$
Obviously, $f(x)$ is continuous and differentiable on [1, 2] and (1, 2) respectively.
Also, $f(1) = f(2)$ (given)
$\therefore\$ By Rolle’s theorem there exist at least one point $k \in (1,\ 2)$ such that $f^{'}(k) = 0$
Now, $f^{'}(x) = (1 + \cos^{8}{x)(ax^{2} + bx + c)}$
$f^{'}(k) = 0$
$\Rightarrow (1 + \cos^{8}{k)\left( ak^{2} + bk + c \right) = 0}$
$\Rightarrow ak^{2} + bk + c = 0\ \ \ \ \lbrack as\ (1 + \cos^{8}{k) \neq 0\rbrack\ }$
$\therefore x = k$ is root of $ax^{2} + bx + c = 0,$
Where $k \in (1,\ 2)$
Hence, at least one root $\in (1,\ 2)$