Question:

Step 1: Simplify the expression in the denominator of the function $f(x)$. Let $g(x) = \sin3x + \cos3x$. We can express this in the form $R\sin(\theta+\phi)$ or $R\cos(\theta-\phi)$. Using the identity $A\sin\theta + B\cos\theta = \sqrt{A^2+B^2}\sin(\theta+\arctan(B/A))$, with $A=1$ and $B=1$, we get: $$ g(x) = \sqrt{1^2+1^2}\left(\frac{1}{\sqrt{2}}\sin3x + \frac{1}{\sqrt{2}}\cos3x\right) $$ $$ g(x) = \sqrt{2}\left(\cos\frac{\pi}{4}\sin3x + \sin\frac{\pi}{4}\cos3x\right) $$ $$ g(x) = \sqrt{2}\sin\left(3x+\frac{\pi}{4}\right) $$ Step 2: Determine the range of the denominator. Since the range of $\sin\left(3x+\frac{\pi}{4}\right)$ is $[-1, 1]$, the range of $g(x) = \sqrt{2}\sin\left(3x+\frac{\pi}{4}\right)$ is $[-\sqrt{2}, \sqrt{2}]$. The denominator of $f(x)$ is $D(x) = 2 + g(x) = 2 + \sin3x + \cos3x$. Therefore, the range of the denominator $D(x)$ is: $$ [2-\sqrt{2}, 2+\sqrt{2}] $$ Step 3: Find the range of the function $f(x)$. The function is $f(x)=\dfrac{1}{2+\sin3x+\cos3x} = \dfrac{1}{D(x)}$. Since $2-\sqrt{2} \approx 2-1.414 = 0.586$, the denominator $D(x)$ is always positive. For a function $h(x) = \frac{1}{y}$ where $y \in [m, M]$ and $m>0$, the range of $h(x)$ is $\left[\frac{1}{M}, \frac{1}{m}\right]$. So, the range $[a,b]$ of $f(x)$ is: $$ \left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right] $$ Step 4: Rationalize the terms to simplify the values of $a$ and $b$. For the lower bound $a$: $$ a = \frac{1}{2+\sqrt{2}} = \frac{1}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2-\sqrt{2}}{2^2 - (\sqrt{2})^2} = \frac{2-\sqrt{2}}{4-2} = \frac{2-\sqrt{2}}{2} = 1-\frac{\sqrt{2}}{2} $$ For the upper bound $b$: $$ b = \frac{1}{2-\sqrt{2}} = \frac{1}{2-\sqrt{2}} \times \frac{2+\sqrt{2}}{2+\sqrt{2}} = \frac{2+\sqrt{2}}{2^2 - (\sqrt{2})^2} = \frac{2+\sqrt{2}}{4-2} = \frac{2+\sqrt{2}}{2} = 1+\frac{\sqrt{2}}{2} $$ So, the range is $[a,b] = \left[1-\frac{\sqrt{2}}{2}, 1+\frac{\sqrt{2}}{2}\right]$. Step 5: Calculate the Arithmetic Mean ($\alpha$) and Geometric Mean ($\beta$) of $a$ and $b$. The Arithmetic Mean $\alpha$ is: $$ \alpha = \frac{a+b}{2} = \frac{\left(1-\frac{\sqrt{2}}{2}\right) + \left(1+\frac{\sqrt{2}}{2}\right)}{2} = \frac{1-\frac{\sqrt{2}}{2}+1+\frac{\sqrt{2}}{2}}{2} = \frac{2}{2} = 1 $$ The Geometric Mean $\beta$ is: $$ \beta = \sqrt{ab} = \sqrt{\left(1-\frac{\sqrt{2}}{2}\right)\left(1+\frac{\sqrt{2}}{2}\right)} $$ Using the difference of squares formula $(x-y)(x+y)=x^2-y^2$: $$ \beta = \sqrt{1^2 - \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{1 - \frac{2}{4}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} $$ Step 6: Calculate the ratio $\dfrac{\alpha}{\beta}$. $$ \frac{\alpha}{\beta} = \frac{1}{1/\sqrt{2}} = \sqrt{2} $$ The final answer is $\boxed{\sqrt{2}}$.