Question

Let a, b, c be non-zero real numbers such the : ${\int }_0^1(1+{\cos }^{8}x)(a{x}^2+bx+c)dx={\int }_0^2(1+{\cos }^{8}x)(a{x}^2+bx+c)dx,$ 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

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

${\int }_0^1(1+{\cos }^{8}x)(a{x}^2+dx+c)dx={\int }_0^1(1+{\cos }^{8}x)(a{x}^2+bx+c)dx+$

${\int }_1^2(1+{\cos }^8)(a{x}^2+bx+c)dx\therefore {\int }_1^2(1+{\cos }^{8}x)(a{x}^2+bx+cx+c)dx=0$

since 1 + ${\cos }^{8}x$ x is always positive
$={\int }_a^{b}f\left(x\right)dx=0(b>a)$
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)