The solutions of the equation | cases 1 + ^ 2 x & ^ 2 x & ^ 2 x ^ 2 x & 1 + ^ 2 x & ^ 2 x 4 2x & 4 2x & 1 + 4 2x cases | = 0, (0 < x < π), are

Correct answer: D. (7π)/(12), (11π)/(12) | cases 1 + ^ 2 x & ^ 2 x & ^ 2 x ^ 2 x & 1 + ^ 2 x & ^ 2 x 4 2x & 4 2x & 1 + 4 2x cases | = 0 use R _ 1 → R _ 1 + R _ 2 + R _ 3 ⇒ (2 + 4 2 x)| cases 1 & 1 & 1 ^ 2 x & 1 + ^ 2 x & ^ 2 x 4 2x & 4 2x & 1 + 4 2x cases | = 0 use C _ 1 → C _ 1 - C _ 3 and C _ 2 → C _ 2 - C _ 3 ⇒ (2 + 4 2 x)| cases 0 & 0 & 1 0 & 1 & ^ 2 x - 1 & - 1 & 1 + 4 2x cases | = 0 ⇒ (2 + 4 2 x) = 0 ⇒ 2x = - (1)/(2) ⇒ 2x = π + (π)/(6), 2π - (π)/(6) x = (π)/(2) + (π)/(12), π - (π)/(12)

JEE Main 202118 MarchMorning Shift

Exam: JEE Main
Year: 2021
Date: 18 March
Shift: Morning Shift
Subject: Mathematics
Chapter: Trigonometric Equations

JEE Main 2021 · 18 March · Morning Shift · Mathematics · Trigonometric Equations

The solutions of the equation | cases 1 + ^ 2 x & ^ 2 x & ^ 2 x ^ 2...

The solutions of the equation

\left|

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

\right| = 0,

\left(0 < x < \pi\right),$ are

Options

A $\dfrac{\pi}{12}, \dfrac{\pi}{6}$
B $\dfrac{\pi}{6}, \dfrac{5\pi}{6}$
C $\dfrac{5\pi}{12}, \dfrac{7\pi}{12}$
D $\dfrac{7\pi}{12}, \dfrac{11\pi}{12}$

Correct answer

D. $\dfrac{7\pi}{12}, \dfrac{11\pi}{12}$

Step-by-step solution

$\left|$ \begin{cases}1 + \sin^{2}x & \sin^{2}x & \sin^{2}x \\ \cos^{2}x & 1 + \cos^{2}x & \cos^{2}x \\ 4\sin 2x & 4\sin 2x & 1 + 4\sin 2x\end{cases} $\right| = 0$

use ${R}_{1} \to {R}_{1} + {R}_{2} + {R}_{3}$

$⇒ \left(2 + 4\sin 2 x\right)\left|$ \begin{cases}1 & 1 & 1 \\ \cos^{2}x & 1 + \cos^{2}x & \cos^{2}x \\ 4\sin 2x & 4\sin 2x & 1 + 4\sin 2x\end{cases} $\right| = 0$

use ${C}_{1} \to {C}_{1} - {C}_{3} \text{\text{ and } } {C}_{2} \to {C}_{2} - {C}_{3}$

$⇒ \left(2 + 4\sin 2 x\right)\left|$ \begin{cases}0 & 0 & 1 \\ 0 & 1 & \cos^{2}x \\ - 1 & - 1 & 1 + 4\sin 2x\end{cases} $\right| = 0$

$⇒ \left(2 + 4\sin 2 x\right) = 0$

$⇒ \sin 2x = - \dfrac{1}{2}$

$⇒ 2x = \pi + \dfrac{\pi}{6}, 2\pi - \dfrac{\pi}{6}$

$x = \dfrac{\pi}{2} + \dfrac{\pi}{12}, \pi - \dfrac{\pi}{12}$

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