If the functions $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^3}{3}+2\mathrm{bx}+\frac{{\mathrm{ax}}^2}{2}\text{ and }\mathrm{g}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^3}{3}+\mathrm{ax}+{\mathrm{bx}}^2,\mathrm{a}\ne 2\mathrm{b}$ have a common extreme point, then a + 2b + 7 is equal to:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^3}{3}+2\mathrm{bx}+\frac{{\mathrm{ax}}^2}{2}\mathrm{and} \mathrm{g}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^3}{3}+\mathrm{ax}+{\mathrm{bx}}^2,\mathrm{a}\ne 2\mathrm{b}, \\ {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)={\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=0\Rightarrow {\mathrm{x}}^2+2\mathrm{b}+\mathrm{ax}=0,{\mathrm{x}}^2+\mathrm{a}+2\mathrm{bx}=0 \end{gathered}$$
$\Rightarrow$satisfies both equations
 $\therefore$a + 2b = -1
$\Rightarrow$ a + 2b + 7 = 6