If the equation ${\mathrm{x}}^2+5+4\cos (\mathrm{\alpha x}+\mathrm{\beta })=2\mathrm{x}$ has at least one solution where $\mathrm{\alpha },\mathrm{\beta }\in [2,5]$ then the value of $\left(\mathrm{\alpha }+\mathrm{\beta }\right)$ equal to

$\begin{array}{r}{\mathrm{x}}^2+5+4\cos (\mathrm{\alpha x}+\mathrm{\beta })=2\mathrm{x}\\ \Rightarrow (\mathrm{x}-1{)}^2+4=-4\cos (\mathrm{\alpha x}+\mathrm{\beta })\end{array}$

Here $(\mathrm{x}-1{)}^2+4\ge 4,-4\cos (\mathrm{\alpha x}+\mathrm{\beta })\le 4$

But, the given equation has at least one solution, then

$\begin{array}{l}(\mathrm{x}-1{)}^2+4=4=-4\cos (\mathrm{\alpha x}+\mathrm{\beta })\\ \therefore \mathrm{x}=1\Rightarrow \cos (\mathrm{\alpha }+\mathrm{\beta })=-1\\ \Rightarrow \mathrm{\alpha }+\mathrm{\beta }=3\mathrm{\pi }=9.42(\because \mathrm{\alpha }+\mathrm{\beta }\in [4,10\left]\right)\end{array}$