Question

Let $a,b,c$ be non – zero real numbers such that
${\int }_0^1(1+co{s}^{8}x)(a{x}^2+bx+c)dx={\int }_0^2(1+co{s}^{8}x)(a{x}^2+bx+c)dx.$
Then the quadratic equation $a{x}^2+bx+c=0$ has

• A. No root in (0, 2)
• B. At least one root in (0,2)
• C. A double root in (0, 2)
• D. Two imaginary roots

Answer 1

The correct option is B At least one root in (0,2)

Given ${\int }_0^1(1+co{s}^{8}x)(a{x}^2+bx+c)dx={\int }_0^2(1+co{s}^{8}x)(a{x}^2+bx+c)dx\Rightarrow {\int }_0^2(1+co{s}^{8}x)(a{x}^2+bx+c)dx={\int }_0^1(1+co{s}^{8}x)(a{x}^2+bx+c)dx+{\int }_1^2(1+co{s}^{8}x)(a{x}^2+bx+c)dx\Rightarrow {\int }_1^2(1+co{s}^{8}x)(a{x}^2+bx+c)dx=0$
Now we know that if ${\int }_{\alpha }^{\beta }f\left(x\right)dx=0$ then it means that f(x)is +ve on some part of $(\alpha ,\beta )$ and -ve on other part of $(\alpha ,\beta ).$
But here $1+co{s}^{8}x$ is always +ve.
$\therefore a{x}^2+bx+c$ is positive on some part of [1,2] and negative on other part of [1,2]
$\therefore a{x}^2+bx=0$ has at least one root in (1,2)
$\Rightarrow a{x}^2+bx+c=0$ has at least one root in (0,2)