Key Concepts and Formulas

• Range of $A \sin \theta + B \cos \theta$: For real numbers $A$ and $B$, the expression $A \sin \theta + B \cos \theta$ has a range of $[-\sqrt{A^2+B^2}, \sqrt{A^2+B^2}]$.
• Arithmetic Mean (AM): For two numbers $a$ and $b$, the AM is $\alpha = \frac{a+b}{2}$.
• Geometric Mean (GM): For two non-negative numbers $a$ and $b$, the GM is $\beta = \sqrt{ab}$.
• Reciprocal of an Interval: If $x \in [C, D]$ and $C, D > 0$, then $\frac{1}{x} \in [\frac{1}{D}, \frac{1}{C}]$.

Step-by-Step Solution

Step 1: Determine the range of the trigonometric part of the denominator. We are given the function $f(x)=\frac{1}{2+\sin 3 x+\cos 3 x}$. The denominator contains the expression $\sin 3x + \cos 3x$. This is in the form $A \sin \theta + B \cos \theta$ with $A=1$, $B=1$, and $\theta=3x$. Using the formula for the range of $A \sin \theta + B \cos \theta$, we find $R = \sqrt{A^2+B^2} = \sqrt{1^2+1^2} = \sqrt{2}$. Therefore, the range of $\sin 3x + \cos 3x$ is $[-\sqrt{2}, \sqrt{2}]$.

Step 2: Determine the range of the denominator. The denominator of $f(x)$ is $2+\sin 3x+\cos 3x$. Since $\sin 3x + \cos 3x \in [-\sqrt{2}, \sqrt{2}]$, we can find the range of the denominator by adding 2 to the endpoints of this interval: $2 + (-\sqrt{2}) \le 2+\sin 3x+\cos 3x \le 2 + \sqrt{2}$ So, the range of the denominator is $[2-\sqrt{2}, 2+\sqrt{2}]$.

Step 3: Determine the range of the function $f(x)$. The function is $f(x) = \frac{1}{2+\sin 3x+\cos 3x}$. Let $y = 2+\sin 3x+\cos 3x$. We found that $y \in [2-\sqrt{2}, 2+\sqrt{2}]$. To find the range of $f(x) = \frac{1}{y}$, we need to take the reciprocal of the interval $[2-\sqrt{2}, 2+\sqrt{2}]$. We observe that $2-\sqrt{2} \approx 2-1.414 = 0.586 > 0$ and $2+\sqrt{2} \approx 2+1.414 = 3.414 > 0$. Since both endpoints are positive, we can invert the interval and swap the endpoints: $f(x) \in \left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right]$ This interval is given as $[a, b]$. Thus, $a = \frac{1}{2+\sqrt{2}}$ and $b = \frac{1}{2-\sqrt{2}}$.

Step 4: Rationalize the values of $a$ and $b$. To simplify calculations for the AM and GM, we rationalize the denominators of $a$ and $b$. For $a$: $a = \frac{1}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2-\sqrt{2}}{2^2 - (\sqrt{2})^2} = \frac{2-\sqrt{2}}{4-2} = \frac{2-\sqrt{2}}{2}$ For $b$: $b = \frac{1}{2-\sqrt{2}} \times \frac{2+\sqrt{2}}{2+\sqrt{2}} = \frac{2+\sqrt{2}}{2^2 - (\sqrt{2})^2} = \frac{2+\sqrt{2}}{4-2} = \frac{2+\sqrt{2}}{2}$

Step 5: Calculate the Arithmetic Mean ($\alpha$) of $a$ and $b$. The arithmetic mean is $\alpha = \frac{a+b}{2}$. $\alpha = \frac{\left(\frac{2-\sqrt{2}}{2}\right) + \left(\frac{2+\sqrt{2}}{2}\right)}{2} = \frac{\frac{2-\sqrt{2}+2+\sqrt{2}}{2}}{2} = \frac{\frac{4}{2}}{2} = \frac{2}{2} = 1$ So, $\alpha = 1$.

Step 6: Calculate the Geometric Mean ($\beta$) of $a$ and $b$. The geometric mean is $\beta = \sqrt{ab}$. $\beta = \sqrt{\left(\frac{2-\sqrt{2}}{2}\right) \times \left(\frac{2+\sqrt{2}}{2}\right)} = \sqrt{\frac{(2-\sqrt{2})(2+\sqrt{2})}{4}}$ Using the difference of squares formula $(x-y)(x+y)=x^2-y^2$ in the numerator: $\beta = \sqrt{\frac{2^2 - (\sqrt{2})^2}{4}} = \sqrt{\frac{4-2}{4}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}}$ Rationalizing the denominator: $\beta = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ So, $\beta = \frac{\sqrt{2}}{2}$.

Step 7: Calculate the ratio $\frac{\alpha}{\beta}$. Now we compute the required ratio: $\frac{\alpha}{\beta} = \frac{1}{\frac{\sqrt{2}}{2}} = 1 \times \frac{2}{\sqrt{2}} = \frac{2}{\sqrt{2}}$ To simplify, we multiply the numerator and denominator by $\sqrt{2}$: $\frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

Common Mistakes & Tips

• When finding the range of $\frac{1}{y}$ where $y \in [C, D]$, ensure $C$ and $D$ have the same sign. If $C, D > 0$, then $\frac{1}{y} \in [\frac{1}{D}, \frac{1}{C}]$. If $C, D < 0$, then $\frac{1}{y} \in [\frac{1}{C}, \frac{1}{D}]$.
• Always rationalize the denominators of $a$ and $b$ before calculating their AM and GM. This simplifies the arithmetic considerably.
• The expression $A \sin \theta + B \cos \theta$ can be written as $R \sin(\theta + \phi)$ or $R \cos(\theta - \phi)$, where $R = \sqrt{A^2+B^2}$. This form directly gives the amplitude and hence the range.

Summary

The problem requires finding the range of a reciprocal trigonometric function. We first determined the range of the trigonometric expression in the denominator using the standard formula for $A \sin \theta + B \cos \theta$. Then, we found the range of the denominator by adding the constant term. Taking the reciprocal of this interval gave us the range of the function, $[a, b]$. Finally, we calculated the arithmetic mean ($\alpha$) and geometric mean ($\beta$) of $a$ and $b$, after rationalizing their values, and then computed the ratio $\frac{\alpha}{\beta}$.

The final answer is $\boxed{\sqrt{2}}$ which corresponds to option (C).