Question

Let $a , b , c$ be non-zero real numbers such that ; $\int\limits_0^1\left(1+\cos ^8 x \right)\left( ax ^2+ bx + c \right) dx =\int\limits_0^2\left(1+\cos ^8 x \right)\left( ax { }^2+ bx + c \right) dx$, then the quadratic equation $a x^2+b x+c=0$ 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)

$\int\limits_0^1\left(1+\cos ^8 x\right)\left(a x^2+b x+c\right) d x$
$= \int\limits_0^1\left(1+\cos ^8 x\right)\left(a x^2+b x+c\right) d x+ \int\limits_1^2\left(1+\cos ^8 x\right)\left(a x^2+b x+c\right) d x $
$\therefore \int\limits_1^2\left(1+\cos ^8 x\right)\left(a x^2+b x+c\right) d x=0$
since $1+\cos ^8 x$ is always positive
$=\int\limits_a^b f(x) d x=0 (b >a)$
means $f ( x )$ is positive in some portion and negative in some portion from a to $b$
$\therefore ax ^2+ bx + c$ is positive and negative in $(1,2)$
$\therefore ax ^2+ bx + c$ has a root in $(1,2)$