Verified Solution • 2025–26 Session moderate QJEEM24Ma578c: JEE Main 2024 Mathematics - Class 11 JEEM24Ma578c Solution: Let the range of the function be . If and ar respectively the… Let the range of the function f(x)=12+sin3x+cos3x,x∈Rf(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R}f(x)=2+sin3x+cos3x1,x∈R be [a,b][a, b][a,b] . If α\alphaα and β\betaβ ar respectively the A.M. and the G.M. of aaa and bbb , then αβ\frac{\alpha}{\beta}βα is equal to Multiple Choice Options A π\piπ B π\sqrt{\pi}π C 2\sqrt{2}2 D 2 Step-by-Step Verified Solution amp;F(x)=12+sin3x+cos3x,x∈Ramp;sin3x+cos3x∈[−2,2]amp;2+sin3x+cos3x∈[2−2,2+2]amp;⇒12+sin3x+cos3x∈[12+2,12−2]amp;⇒a=12+2,b=12−2amp;α=a+b2=12+2+12−22amp;=42×2=1amp;β=ab=(12+2)×(12−2)amp;=12=12amp;⇒αβ=2\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}amp;F(x)=2+sin3x+cos3x1,x∈Ramp;sin3x+cos3x∈[−2,2]amp;2+sin3x+cos3x∈[2−2,2+2]amp;⇒2+sin3x+cos3x1∈[2+21,2−21]amp;⇒a=2+21,b=2−21amp;α=2a+b=22+21+2−21amp;=2×24=1amp;β=ab=(2+21)×(2−21)amp;=21=21amp;⇒βα=2 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring