Question:

Step 1: Find the first derivatives of the given functions. To find the extreme points of a function, we first need to calculate its derivative. Given functions are $f(x)=\dfrac{x^3}{3}+\dfrac{ax^2}{2}+2bx$ and $g(x)=\dfrac{x^3}{3}+bx^2+ax$. Differentiating $f(x)$ with respect to $x$: $$f'(x) = \frac{d}{dx}\left(\frac{x^3}{3}+\frac{ax^2}{2}+2bx\right) = \frac{3x^2}{3}+\frac{2ax}{2}+2b = x^2+ax+2b$$ Differentiating $g(x)$ with respect to $x$: $$g'(x) = \frac{d}{dx}\left(\frac{x^3}{3}+bx^2+ax\right) = \frac{3x^2}{3}+2bx+a = x^2+2bx+a$$ Step 2: Set the derivatives to zero to find critical points. For a function to have an extreme point at $x=x_0$, its first derivative at that point must be zero. Since $f(x)$ and $g(x)$ have a common extreme point, let this common point be $x_0$. Then: $$f'(x_0) = x_0^2+ax_0+2b = 0 \quad \text{(Equation 1)}$$ $$g'(x_0) = x_0^2+2bx_0+a = 0 \quad \text{(Equation 2)}$$ Step 3: Solve the system of equations for $x_0$. Subtract Equation 2 from Equation 1 to eliminate $x_0^2$: $$(x_0^2+ax_0+2b) - (x_0^2+2bx_0+a) = 0$$ $$ax_0 - 2bx_0 + 2b - a = 0$$ Factor out $x_0$ from the first two terms and $-1$ from the last two terms: $$x_0(a-2b) - (a-2b) = 0$$ Factor out the common term $(a-2b)$: $$(x_0-1)(a-2b) = 0$$ Step 4: Use the given condition to find the value of $x_0$. The problem states that $a \neq 2b$. This implies that $(a-2b) \neq 0$. For the product $(x_0-1)(a-2b)$ to be zero, one of the factors must be zero. Since $(a-2b) \neq 0$, it must be that: $$x_0-1 = 0 \implies x_0 = 1$$ So, the common extreme point occurs at $x=1$. Step 5: Substitute the value of $x_0$ back into one of the derivative equations to find the relation between $a$ and $2b$. Substitute $x_0=1$ into Equation 1: $$1^2 + a(1) + 2b = 0$$ $$1 + a + 2b = 0$$ Rearrange the terms to find the value of $a+2b$: $$a+2b = -1$$ Step 6: Calculate the required expression $a+2b+7$. We have found that $a+2b = -1$. Now substitute this value into the expression $a+2b+7$: $$a+2b+7 = (-1)+7 = 6$$ The final answer is $\boxed{6}$.