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
Solution
<p>$$\begin{aligned}
& F(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R} \\
& \sin 3 x+\cos 3 x \in[-\sqrt{2}, \sqrt{2}] \\
& 2+\sin 3 x+\cos 3 x \in[2-\sqrt{2}, 2+\sqrt{2}] \\
& \Rightarrow \frac{1}{2+\sin 3 x+\cos 3 x} \in\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right] \\
& \Rightarrow a=\frac{1}{2+\sqrt{2}}, b=\frac{1}{2-\sqrt{2}} \\
& \alpha=\frac{a+b}{2}=\frac{\frac{1}{2+\sqrt{2}}+\frac{1}{2-\sqrt{2}}}{2} \\
& =\frac{4}{2 \times 2}=1 \\
& \beta=\sqrt{a b}=\sqrt{\left(\frac{1}{2+\sqrt{2}}\right) \times\left(\frac{1}{2-\sqrt{2}}\right)} \\
& =\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \\
& \Rightarrow \frac{\alpha}{\beta}=\sqrt{2} \\
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
This question is part of PrepWiser's free JEE Main question bank. 195 more solved questions on Sets, Relations and Functions are available — start with the harder ones if your accuracy is >70%.