A value of θ ∈ (0, (π)/(3)), for which | cases 1 + ^ 2 θ & ^ 2 θ & 4 6θ ^ 2 θ & 1 + ^ 2 θ & 4 6θ ^ 2 θ & ^ 2 θ & 1 + 4 6θ cases | = 0, is

Correct answer: A. (π)/(9) R _ 1 → R _ 1 - R _ 2 | cases 1 & - 1 & 0 ^ 2 θ & 1 + ^ 2 θ & 4 6θ ^ 2 θ & ^ 2 θ & 1 + 4 6θ cases | = 0 R _ 2 → R _ 2 - R _ 3 | cases 1 & - 1 & 0 0 & 1 & - 1 ^ 2 θ & ^ 2 θ & 1 + 4 6θ cases | = 0 ⇒ (1 + 4 6θ) + ^ 2 θ + 1( ^ 2 θ) = 0 1 + 2 6θ = 0 ⇒ 6θ = - (1)/(2) 6θ = (2π)/(3) ⇒ θ = (π)/(9)

JEE Main 201912 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2019 · 12 April · Evening Shift · Mathematics · Determinants

A value of θ ∈ (0, (π)/(3)), for which | cases 1 + ^ 2 θ & ^ 2 θ &...

A value of

\theta \in

\left(0, \dfrac{\pi}{3}\right),

for which \left|

\begin{cases}1 + \cos^{2}\theta & \sin^{2}\theta & 4\cos 6\theta \\ \cos^{2}\theta & 1 + \sin^{2}\theta & 4\cos 6\theta \\ \cos^{2}\theta & \sin^{2}\theta & 1 + 4\cos 6\theta\end{cases}

\right| = 0,

Options

A $\dfrac{\pi}{9}$
B $\dfrac{7\pi}{24}$
C $\dfrac{7\pi}{36}$
D $\dfrac{\pi}{18}$

Correct answer

A. $\dfrac{\pi}{9}$

Step-by-step solution

{R}_{1} \to {R}_{1} - {R}_{2}

\left|

\begin{cases}1 & - 1 & 0 \\ \cos^{2}\theta & 1 + \sin^{2}\theta & 4\cos 6\theta \\ \cos^{2}\theta & \sin^{2}\theta & 1 + 4\cos 6\theta\end{cases}

\right| = 0

<br/>

{R}_{2} \to {R}_{2} - {R}_{3}

<br/>\left|

\begin{cases}1 & - 1 & 0 \\ 0 & 1 & - 1 \\ \cos^{2}\theta & \sin^{2}\theta & 1 + 4\cos 6\theta\end{cases}

\right| = 0

⇒ \left(1 + 4\cos 6\theta\right) + \sin^{2}\theta + 1\left(\cos^{2}\theta\right) = 0

1 + 2\cos 6\theta = 0 ⇒ \cos 6\theta = - \dfrac{1}{2}

6\theta = \dfrac{2\pi}{3} ⇒ \theta = \dfrac{\pi}{9}

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