Let the range of the function f(x) = frac{1}{ | vedclass

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(A) We know that the range of $g(x) = \sin 3x + \cos 3x$ is $[-\sqrt{2}, \sqrt{2}]$.Thus,the denominator $2 + \sin 3x + \cos 3x$ ranges from $[2 - \sq

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$\sqrt{2}$

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(A) We know that the range of $g(x) = \sin 3x + \cos 3x$ is $[-\sqrt{2}, \sqrt{2}]$.Thus,the denominator $2 + \sin 3x + \cos 3x$ ranges from $[2 - \sqrt{2}, 2 + \sqrt{2}]$.Therefore,the range of $f(x) = \frac{1}{2 + \sin 3x + \cos 3x}$ is $[a, b] = \left[\frac{1}{2 + \sqrt{2}}, \frac{1}{2 - \sqrt{2}}\right]$.Rationalizing the endpoints: $a = \frac{2 - \sqrt{2}}{4 - 2} = \frac{2 - \sqrt{2}}{2} = 1 - \frac{1}{\sqrt{2}}$ and $b = \frac{2 + \sqrt{2}}{4 - 2} = \frac{2 + \sqrt{2}}{2} = 1 + \frac{1}{\sqrt{2}}$.Given $\alpha = \frac{a + b}{2}$ and $\beta = \sqrt{ab}$,we need to find $\frac{\alpha}{\beta} = \frac{a + b}{2\sqrt{ab}}$.$a + b = \frac{2 - \sqrt{2} + 2 + \sqrt{2}}{2} = 2$.$ab = \left(\frac{2 - \sqrt{2}}{2}\right) \left(\frac{2 + \sqrt{2}}{2}\right) = \frac{4 - 2}{4} = \frac{2}{4} = \frac{1}{2}$.$\frac{\alpha}{\beta} = \frac{2}{2\sqrt{1/2}} = \frac{1}{1/\sqrt{2}} = \sqrt{2}$.

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:

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