Question

If $\underset{2}{\overset{4}{\int }}(1+{\sin }^{4}x)(a{x}^2+bx+c)dx=\underset{2}{\overset{6}{\int }}(1+{\sin }^{4}x)(a{x}^2+bx+c)dx,$ then which of the following statement is always correct for the equation $a{x}^2+bx+c=0?$
(where $a,b,c$ are non-zero constants)

• A. no root in $(4,6)$
• B. both roots in $(4,6)$
• C. atleast one root in $(4,6)$
• D. imaginary roots

Answer 1

The correct option is C atleast one root in $(4,6)$

Given : $\underset{2}{\overset{4}{\int }}(1+{\sin }^{4}x)(a{x}^2+bx+c)dx=\underset{2}{\overset{6}{\int }}(1+{\sin }^{4}x)(a{x}^2+bx+c)dx$
$\Rightarrow \underset{4}{\overset{6}{\int }}(1+{\sin }^{4}x)(a{x}^2+bx+c)dx=0\cdots \left(i\right)$
Let $\underset{0}{\overset{t}{\int }}(1+{\sin }^{4}x)(a{x}^2+bx+c)dx=g\left(t\right)$
$\Rightarrow g\left(4\right)-g\left(6\right)=0\Rightarrow g\left(4\right)=g\left(6\right)$
By Rolle's theorem, $g^{\prime }\left(t\right)$ has atleast one root in $(4,6)$
$\Rightarrow (1+{\sin }^{4}t)(a{t}^2+bt+c)=0$
So, $a{t}^2+bt+c=0$ for atleast one value of $t\in (4,6)$
[$\because 1+{\sin }^{4}t\ne 0$]