JEE MAIN - Mathematics (2024 - 9th April Evening Shift - No. 3)
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
$$\pi$$
$$\sqrt{\pi}$$
$$\sqrt{2}$$
2
Explanation
$$\begin{aligned}
& F(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R} \\
& \sin 3 x+\cos 3 x \in[-\sqrt{2}, \sqrt{2}] \\
& 2+\sin 3 x+\cos 3 x \in[2-\sqrt{2}, 2+\sqrt{2}] \\
& \Rightarrow \frac{1}{2+\sin 3 x+\cos 3 x} \in\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right] \\
& \Rightarrow a=\frac{1}{2+\sqrt{2}}, b=\frac{1}{2-\sqrt{2}} \\
& \alpha=\frac{a+b}{2}=\frac{\frac{1}{2+\sqrt{2}}+\frac{1}{2-\sqrt{2}}}{2} \\
& =\frac{4}{2 \times 2}=1 \\
& \beta=\sqrt{a b}=\sqrt{\left(\frac{1}{2+\sqrt{2}}\right) \times\left(\frac{1}{2-\sqrt{2}}\right)} \\
& =\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \\
& \Rightarrow \frac{\alpha}{\beta}=\sqrt{2} \\
\end{aligned}$$
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