Let the range of the function f(x)=(1)/(2+ 3x+ 3x),x R be [a,b].

Question

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

Accepted Answer

$\begin{gathered} F(x)=\frac{1}{2+\sin 3x+\cos 3x},x\in R \ \sin 3x+\cos 3x\in [-\sqrt{2},\sqrt{2}] \ 2+\sin 3x+\cos 3x\in [2-\sqrt{2},2+\sqrt{2}] \ \Rightarrow \frac{1}{2+\sin 3x+\cos 3x}\in [\frac{1}{2+\sqrt{2}},\frac{1}{2-\sqrt{2}}] \ \Rightarrow a=\frac{1}{2+\sqrt{2}},b=\frac{1}{2-\sqrt{2}} \ \alpha =\frac{a+b}{2}=\frac{\frac{1}{2+\sqrt{2}}+\frac{1}{2-\sqrt{2}}}{2} \ =\frac{4}{2	imes 2}=1 \ \beta =\sqrt{ab}=\sqrt{(\frac{1}{2+\sqrt{2}})	imes (\frac{1}{2-\sqrt{2}})} \ =\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \ \Rightarrow \frac{\alpha }{\beta }=\sqrt{2} \end{gathered}$

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