Let the range of the function f(x)=(1)/(2+ 3 x+ 3 x), x ∈ ℝ be [a, b] . If α and β are respectively the A.M. and the G.M. of a and b , then (α)/(β) is equal to

Correct answer: D. 2 aligned & f(x) (1)/(2+ 3 x+ 3 x) & [ 1 2+ 2 , 1 2- 2 ] & (α)/(β)= a+b 2 a b =(1)/(2)( (a)/(b) + (b)/(a) ) & =(1)/(2)( 2- 2 2+ 2 + 2+ 2 2- 2 ) & = (2- 2 )+(2+ 2 ) 2 × 2 = 2 aligned

JEE Main 20249 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Functions

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

Let the range of the function

f(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R}

[a, b]

. If

\alpha

and

\beta

are respectively the A.M. and the G.M. of

a

and

b

, then

\frac{\alpha}{\beta}

is equal to

Options

A $\pi$
B $\sqrt{\pi}$
C2
D $\sqrt{2}$

Correct answer

D. $\sqrt{2}$

Step-by-step solution

\begin{aligned} & f(x) \frac{1}{2+\sin 3 x+\cos 3 x} \\ & {\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right]} \\ & \frac{\alpha}{\beta}=\frac{a+b}{2 \sqrt{a b}}=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right) \\ & =\frac{1}{2}\left(\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}+\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}\right) \\ & =\frac{(2-\sqrt{2})+(2+\sqrt{2})}{2 \times \sqrt{2}}=\sqrt{2}\end{aligned}

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