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JEE Main 202116 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Morning Shift
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2021 · 16 March · Morning Shift · Mathematics · Trigonometric Equations

The number of roots of the equation, (81) ^ ^ 2 x + (81) ^ ^ 2 x =...

The number of roots of the equation, (81)sin⁡2x+(81)cos⁡2x=30{\left(81\right)}^{\sin^{2}x} + {\left(81\right)}^{\cos^{2}x} = 30 in the interval \left[0, \pi\right] is equal to :The number of roots of the equation, (81) ^ ^ 2 x + (81) ^ ^ 2 x = 30 in the interval [0, π] is equal to :

Options

  1. A33
  2. B44
  3. C88
  4. D22

Correct answer

B. 44

Step-by-step solution

{\left(81\right)}^{\sin^{2}x} + {\left(81\right)}^{\cos^{2}x} = 30</p><p></p><p>{\left(81\right)}^{\sin^{2}x} + \dfrac{{\left(81\right)}^{1}}{{\left(81\right)}^{\sin^{2}x}} = 30</p><p></p><p>{\left(81\right)}^{\sin^{2}x} = t</p><p></p><p>t + \dfrac{81}{t} = 30</p><p></p><p>{t}^{2} - 30t + 81 = 0</p><p>(t−3)(t−27)=0</p><p>\left(t - 3\right)\left(t - 27\right) = 0

(81)sin⁡2x=31{\left(81\right)}^{\sin^{2}x} = {3}^{1} or (81)sin⁡2x=33{\left(81\right)}^{\sin^{2}x} = {3}^{3}

34sin⁡2x=31{3}^{4\sin^{2}x} = {3}^{1} or 34sin⁡2x=33{3}^{4\sin^{2}x} = {3}^{3}

sin⁡2x=14\sin^{2}x = \dfrac{1}{4} or sin⁡2x=34\sin^{2}x = \dfrac{3}{4}

The number of roots of the equation, (81) ^ ^ 2 x + (81) ^ ^ 2 x = 30 in the interval [0, π] is equal to :

Hence, total 44 solution.

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