Question

Let $a,b,c$ be non-zero real numbers such that ${\int }_0^1\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx={\int }_0^2\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx$ then the quadratic equation $a{x}^2+bx+c=0$ has-

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

Answer 1

The correct option is B atleast one root $(0,2)$

$1+{\cos }^{8}x$ is always positive

so Area of function $(1+{\cos }^{8}x)(a{x}^2+6x+c)$ from $0$ to $1$ is same as from $0$ to $2$

Two situations cxan happen

Atleast one react in $(0,2)$

$B$ is correct