Question

Let $a,b,c$ be non-zero real numbers such that $\underset{0}{\overset{1}{\int }}(1+{\cos }^{8}x)(a{x}^2+bx+c)$ Then the quadratic equation $a{x}^2+bx+c=0$ has

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

Answer 1

Answer: (B)

Let $f^{\prime }\left(x\right)=\left(1+co{s}^{8}x\right)\left(a{x}^2+bx+c\right)$
we are given
$\underset{0}{\overset{1}{\int }}{f}^{\prime }\left(x\right)dx=\underset{0}{\overset{2}{\int }}{f}^{\prime }\left(x\right)dx$
$=\underset{0}{\overset{1}{\int }}{f}^{\prime }\left(x\right)dx+\underset{0}{\overset{2}{\int }}{f}^{\prime }\left(x\right)dx$
$\Rightarrow \underset{1}{\overset{2}{\int }}{f}^{\prime }\left(x\right)dx=0$
$\Rightarrow {\left|f\left(x\right)\right|}_1^2=0$
$\Rightarrow f\left(2\right)-f\left(1\right)=0$
$\Rightarrow f\left(1\right)=f\left(2\right)$
Since $f^{\prime }\left(x\right)$ exists,
$\therefore f\left(x\right)$ satisfies all the conditions of Rolle’s Thm. in $\left[1,2\right]$.
Hence $\exists s$ at least one real value of $x$ say $\alpha$, such that $f^{\prime }\left(\alpha \right)=0$
i.e., $\left(1+co{s}^8\alpha \right)\left(a{\alpha }^2+ba+c\right)=0$
But $1+co{s}^8\alpha \ne 0$.
$\therefore a{\alpha }^2+ba+c=0$
Thus $a{x}^2+bx+c=0$ has at least one root in $(1,2)$.