Let the range of the function $$f(x) = \frac{1}{2 + \sin 3x + \cos 3x}, x \in \mathbb{R}$$ be $$[a, b]$$. If $$\alpha$$ and $$\beta$$ are respectively the A.M. and the G.M. of $$a$$ and $$b$$, then $$\frac{\alpha}{\beta}$$ is equal to

$$f(x) = \frac{1}{2+\sin 3x+\cos 3x}$$. Range of $$\sin 3x+\cos 3x = \sqrt{2}\sin(3x+\pi/4)$$ is $$[-\sqrt{2},\sqrt{2}]$$.

Denominator range: $$[2-\sqrt{2}, 2+\sqrt{2}]$$. So $$f$$ range: $$[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}]$$.

$$a = \frac{1}{2+\sqrt{2}} = \frac{2-\sqrt{2}}{2}$$, $$b = \frac{1}{2-\sqrt{2}} = \frac{2+\sqrt{2}}{2}$$.

AM: $$\alpha = \frac{a+b}{2} = \frac{(2-\sqrt{2}+2+\sqrt{2})/2}{2} = \frac{2}{2} = 1$$.

GM: $$\beta = \sqrt{ab} = \sqrt{\frac{(2-\sqrt{2})(2+\sqrt{2})}{4}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$$.

$$\alpha/\beta = 1/(1/\sqrt{2}) = \sqrt{2}$$.

The correct answer is Option 4: $$\sqrt{2}$$.