Let $$a,\,b,\,c$$   be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$     has-

Solution :
$$\eqalign{ & {\text{Given,}} \cr & \int_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx} \cr & = \int_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx} \cr & = \int_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx} + \int_1^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx} \cr & \Rightarrow \int_1^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx} = 0 \cr} $$
Now we know that if $$\int_\alpha ^\beta {f\left( x \right)dx} = 0$$    then it means that $$f\left( x \right)$$  is $$+ve$$  on some part of $$\left( {\alpha ,\,\beta } \right)$$   and $$-ve$$  on other part of $$\left( {\alpha ,\,\beta } \right).$$
But here $${1 + {{\cos }^8}x}$$    is always $$+ve,$$
$$\therefore a{x^2} + bx + c$$    is $$+ve$$  on some part of $$\left[ {1,\,2} \right]$$  and $$-ve$$  on other Part $$\left[ {1,\,2} \right]$$
$$\therefore a{x^2} + bx + c = 0$$     has at least one root in $$\left( {1,\,2} \right)$$.
$$ \Rightarrow a{x^2} + bx + c = 0$$     has at least one root in $$\left( {0,\,2} \right)$$