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

Question

Let the range of the function $$f(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R}$$ 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

$$\begin{aligned}
& F(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R} \
& \sin 3 x+\cos 3 x \in[-\sqrt{2}, \sqrt{2}] \
& 2+\sin 3 x+\cos 3 x \in[2-\sqrt{2}, 2+\sqrt{2}] \
& \Rightarrow \frac{1}{2+\sin 3 x+\cos 3 x} \in\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right] \
& \Rightarrow a=\frac{1}{2+\sqrt{2}}, b=\frac{1}{2-\sqrt{2}} \
& \alpha=\frac{a+b}{2}=\frac{\frac{1}{2+\sqrt{2}}+\frac{1}{2-\sqrt{2}}}{2} \
& =\frac{4}{2 \times 2}=1 \
& \beta=\sqrt{a b}=\sqrt{\left(\frac{1}{2+\sqrt{2}}\right) \times\left(\frac{1}{2-\sqrt{2}}\right)} \
& =\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \
& \Rightarrow \frac{\alpha}{\beta}=\sqrt{2} \
\end{aligned}$$

Solutions · 1