Let the range of the function

f(x)=frac{1} | vedclass

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Question

$ f(x) \frac{1}{2+\sin 3 x+\cos 3 x} $

$ {\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right]} $

$ \frac{\alpha}{\beta}=\frac{a+b}{2 \sqrt{a

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

$ f(x) \frac{1}{2+\sin 3 x+\cos 3 x} $

$ {\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}ight]} $

$ \frac{\alpha}{\beta}=\frac{a+b}{2 \sqrt{a b}}=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}ight) $

$ =\frac{1}{2}\left(\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}+\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}ight) $

$ =\frac{(2-\sqrt{2})+(2+\sqrt{2})}{2 	imes \sqrt{2}}=\sqrt{2}$

Let the range of the function

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