Let | θ (60-θ) (60+θ)| ≤ (1)/(8), θ [0,2 π] . Then, the sum of all θ [0,2 π] , where 3 θ attains its maximum value, is :

Correct answer: C. 6 π We know that ( θ)( ( 6 0 ^ - θ ) ( (60^ +θ)=(1)/(4) 3 θ.. So equation reduces to |(1)/(4) 3 θ| ≤ (1)/(8) aligned & | 3 θ| ≤ (1)/(2) & -(1)/(2) ≤ 3 θ ≤ (1)/(2) aligned maximum value of 3 θ=(1)/(2) , here aligned & 3 θ=2 n π ± (π)/(3) & θ= 2 n π 3 ± (π)/(9) aligned As θ ∈[0,2 π] possible values are θ=\ (π)/(9), (5 π)/(9), (7 π)/(9), (11 π)/(9), (13 π)/(9), (17 π)/(9)\ Whose sum is (π)/(9)+(5 π)/(9)+(7 π)/(9)+(11 π)/(9)+(13 π)/(9)+(17 π)/(9)=(54 π)/(9)=6 π

JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Trigonometric Ratios & Identities

Let | θ (60-θ) (60+θ)| ≤ (1)/(8), θ [0,2 π] . Then, the sum of all...

Let

|\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8}, \theta \epsilon[0,2 \pi]

. Then, the sum of all

\theta \epsilon[0,2 \pi]

, where

\cos 3 \theta

attains its maximum value, is :

Options

A $15 \pi$
B $18 \pi$
C $6 \pi$
D $9 \pi$

Correct answer

C. $6 \pi$

Step-by-step solution

We know that

(\cos \theta)\left(\cos( 6 0 ^ { \circ } - \theta ) \left(\cos \left(60^{\circ}+\theta\right)=\frac{1}{4} \cos 3 \theta\right.\right.

So equation reduces to

\left|\frac{1}{4} \cos 3 \theta\right| \leq \frac{1}{8}

\begin{aligned} & \Rightarrow|\cos 3 \theta| \leq \frac{1}{2} \\ & \Rightarrow-\frac{1}{2} \leq \cos 3 \theta \leq \frac{1}{2} \end{aligned}

\Rightarrow

maximum value of

\cos 3 \theta=\frac{1}{2}

, here

\begin{aligned} & \Rightarrow 3 \theta=2 \mathrm{n} \pi \pm \frac{\pi}{3} \\ & \theta=\frac{2 \mathrm{n} \pi}{3} \pm \frac{\pi}{9} \end{aligned}

\theta \in[0,2 \pi]

possible values are

\theta=\left\{\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9}\right\}

Whose sum is

\frac{\pi}{9}+\frac{5 \pi}{9}+\frac{7 \pi}{9}+\frac{11 \pi}{9}+\frac{13 \pi}{9}+\frac{17 \pi}{9}=\frac{54 \pi}{9}=6 \pi

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