If the functions f(x) = x ^ 3 3 + 2bx + a x ^ 2 2 and g(x) = x ^ 3...

Given,

$f\left(x\right) = \dfrac{{x}^{3}}{3} + 2bx + \dfrac{a{x}^{2}}{2}$ and $g\left(x\right) = \dfrac{{x}^{3}}{3} + ax + b{x}^{2}$

Now differentiating the above function we get,

${f}^{'}\left(x\right) = {x}^{2} + 2b + ax$

${g}^{'}\left(x\right) = {x}^{2} + a + 2bx$

Now given, ${f}^{'} ( x ) = 0and{g}^{'} ( x ) = 0$ have a common extreme point,

So, ${x}^{2} + 2b + ax = {x}^{2} + a + 2bx$

$⇒ \left(2b - a\right) - x\left(2b - a\right) = 0$

$∴ x = 1$ is the common extreme point

Put $x = 1$ in ${f}^{'}\left(x\right) = 0$ or ${g}^{'}\left(x\right) = 0$ we get,

$1 + 2b + a = 0$

$⇒ a + 2b + 7 = 6$