Question

Let $a, b, c$ be non-zero real numbers such that $\int\limits _0^1 (1+\cos^8x) (ax^2+bx+c)$ Then the quadratic equation $ax^2 + bx + c = 0$ 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 $f'\left(x\right)=\left(1+cos^{8}\,x\right)\left(ax^{2}+bx+c\right)$
we are given
$\int\limits_{0}^{1} f'\left(x\right)dx=\int\limits_{0}^{2} f' \left(x\right)dx$
$=\int\limits_{0}^{1} f'\left(x\right)dx+\int\limits_{0}^{2} f' \left(x\right)dx$
$\Rightarrow \int\limits_{1}^{2} f' \left(x\right)dx=0$
$\Rightarrow \left|f\left(x\right)\right|_{1}^{2}=0$
$\Rightarrow f \left(2\right)-f\left(1\right)=0$
$\Rightarrow f\left(1\right)=f\left(2\right)$
Since $f'\left(x\right)$ 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 $f' \left(\alpha\right)=0$
i.e., $\left(1+cos^{8}\, \alpha\right) \left(a \alpha^{2}+ba+c\right)=0$
But $1 +cos^{8} \alpha \ne0$.
$\therefore \, a \alpha^{2}+ba+c=0$
Thus $ax^{2} + bx+c=0$ has at least one root in $(1, 2)$.