Let a,b,c be non-zero real numbers such that  $\begin{array}{l}{\int }_0^1 \left(e^{-x}+e^x\right)\left(a{x}^2+bx+c\right)dx ={\int }_0^2 \left(e^{-x}+e^x\right)\left(a{x}^2+bx+c\right)dx\\ \end{array}$ 

Then the quadratic equation $a{x}^2+bx+c=0$ has

Let  $f\left(x\right)=\left({\mathrm{e}}^{-\mathrm{x}}+{\mathrm{e}}^x\right)\left(a{x}^2+bx+c\right)$

We have ${\int }_0^2 f\left(x\right)dx={\int }_0^1 f\left(x\right)dx+{\int }_1^2 f\left(x\right)dx$

$\Rightarrow {\int }_1^2 f\left(x\right)dx=0$

If $f\left(x\right)>0(<0)\mathrm{\forall }x\in [1,2]$ then ${\int }_1^2 f\left(x\right)dx>0(<0)$

Thus, f (x) = (e^–x + e^x) (ax² + bx + c) must be positive for some value of x in [1, 2] and must be negative for some
value of x in [1, 2]. As e^–x + e^x ≥ 2, it follows that if g(x) =ax² + bx + c, then there exists some $\alpha ,\beta \in [1,2]$ such that
g(a) > 0 and g($\beta$) < 0. Since g is continuous on R, there exists some $\gamma$ between $\alpha$ and $\beta$ such that g($\gamma$) = 0. Thus, ax² + bx + c = 0 has at least one root in (1, 2).