Question

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

• A. no root in $(0,2)$
• B. atleast one root in $(0,2)$
• C. a double root in $(0,2)$
• D. none

Answer 1

Answer: (B)

$\underset{0}{\overset{1}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx$
$=\underset{0}{\overset{1}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx+\underset{1}{\overset{2}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx$
$\therefore \underset{1}{\overset{2}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx=0$
since $1+{\cos }^{8}x$ is always positive
$=\underset{a}{\overset{b}{\int }}f(x)dx=0(b>a)$
means $f(x)$ is positive in some portion and negative in some portion from a to $b$
$\therefore a{x}^2+bx+c$ is positive and negative in $(1,2)$
$\therefore a{x}^2+bx+c$ has a root in $(1,2)$