Question

If $a$, $b$, $c$ be non-zero real numbers such that ${\int }_0^1(1+{\cos }^{8}x)(a{x}^2+bx+c)dx={\int }_0^2(1+{\cos }^{8}x)(a{x}^2+bx+c)dx=0$. Then the equation $a{x}^2+bx+c=0$ will have

• A. one root between $0$ and $1$ and another between $1$ and $2$
• B. both the roots between $0$ and $1$
• C. both the roots between $1$ and $2$
• D. None of the above

Answer 1

The correct option is A one root between $0$ and $1$ and another between $1$ and $2$

Let

$\varphi \left(x\right)={\int }_0^x(1+{\cos }^{8}t)(a{t}^2+bt+c)dt$
${\varphi }^{\prime }\left(x\right)=(1+{\cos }^{8}x)(a{x}^2+bx+c)$

Here,

$\varphi \left(x\right)=0$

Also,

$\varphi \left(1\right)={\int }_0^1(1+{\cos }^{8}t)(a{t}^2+bt+c)dt$
$\Rightarrow \varphi \left(1\right)=0$
$\Rightarrow \varphi \left(0\right)=\varphi \left(1\right)$

So, by Rolle's theorem, there exits at least one root between $0$ and $1$ such that ${\varphi }^{\prime }\left(x\right)=0$
$\Rightarrow a{x}^2+bx+c$ has at least one root between $0$ and $1$.

Also,

$\varphi \left(2\right)={\int }_0^2(1+{\cos }^{8}t)(a{t}^2+bt+c)dt$
$\Rightarrow \varphi \left(2\right)=0$
$\Rightarrow \varphi \left(1\right)=\varphi \left(2\right)$

So, by Rolle's theorem, there exits at least one root between $1$ and $2$ such that ${\varphi }^{\prime }\left(x\right)=0$
$\Rightarrow a{x}^2+bx+c$ has at least one root between $1$ and $2$.