Question

If $a,b,c$ be non-zero real number 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 $a{x}^2+bx+c$ will have

• A. One root between $0$ and $1$ and other root between $1$ and $2$
• B. both the roots between $0$ and $1$
• C. both the roots between $1$ and $2$
• D. none of these

Answer 1

The correct option is A One root between $0$ and $1$ and other root between $1$ and $2$

Let $f\left(y\right)={\int }_0^y(1+{\cos }^{8}x)({ax}^2+bx+c)dx$

$\Rightarrow f^{\prime }\left(x\right)=(1+{\cos }^{8}y)({ay}^2+by+c)$ ...(1)

Now, $f\left(1\right)={\int }_0^1(1+{\cos }^{8}x)({ax}^2+bx+c)dx=0$

and, $f\left(2\right)={\int }_0^2(1+{\cos }^{8}x)({ax}^2+bx+c)dx=0$

Also, $f\left(x\right)=0$

$\therefore f\left(0\right)=f\left(1\right)=f\left(2\right).$

Now, by Rolle's theorem for $f\left(x\right)$ in $[0,1],$

$f^{\prime }\left(\alpha \right)=0,$ for atleast one $\alpha ,0<\alpha <1$

and by Rolle's theorem for $f\left(x\right)$ in $[1,2],$

$f^{\prime }\left(\beta \right)=0,$ for atleast one $\beta ,1<\beta <2$

From (1), $f^{\prime }\left(\alpha \right)=0\Rightarrow (1+{\cos }^8\alpha )({a\alpha }^2+b\alpha +c)=0$

But $1+{\cos }^8\alpha \ne 0,$

$\therefore {a\alpha }^2+b\alpha +c=0,$

i.e., $\alpha$ is a root of the equation ${ax}^2+bx+c=0.$

Similarly, $f^{\prime }\left(\beta \right)=0\Rightarrow a{\beta }^2+bx+c=0,$

i.e., $\beta$ is a root of the equation ${ax}^2+bx+c=0.$

But the equation ${ax}^2+bx+c=0,$ being a quadratic equation, cannot have more then two roots.

$\therefore$ The equation ${ax}^2+bx+c=0$ has one root $\alpha$

between $0$ and $1$ and other root $\beta$ between one root $1$ and $2.$