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JEE Main 202131 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Evening Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2021 · 31 August · Evening Shift · Mathematics · Quadratic Equation

The number of solutions of the equation 32 ^ ^ 2 x + 32 ^ ^ 2 x =...

The number of solutions of the equation 32tan⁡2x+32sec⁡2x=81,0≤x≤π4{32}^{\tan^{2}x} + {32}^{\sec^{2}x} = 81, 0 \le x \le \dfrac{\pi}{4} is :

Options

  1. A00
  2. B22
  3. C11
  4. D33

Correct answer

C. 11

Step-by-step solution

Given, 32tan⁡2x+32sec⁡2x=81,0≤x≤π4{32}^{\tan^{2}x} + {32}^{\sec^{2}x} = 81, 0 \le x \le \dfrac{\pi}{4}

32tan⁡2x+321+tan⁡2x=81{32}^{\tan^{2}x} + {32}^{1 + \tan^{2}x} = 81

32tan⁡2x+32.32tan⁡2x=81{32}^{\tan^{2}x} + 32 . {32}^{\tan^{2}x} = 81

33×32tan⁡2x=8133 \times {32}^{\tan^{2}x} = 81

32tan⁡2x=2711{32}^{\tan^{2}x} = \dfrac{27}{11}

tan⁡2x=ln⁡32(2711)\tan^{2}x = \ln_{32}\left(\dfrac{27}{11}\right)
tan⁡x=ln⁡32(2711)∈(0,1)\tan x = \sqrt{\ln_{32}\left(\dfrac{27}{11}\right)} \in \left(0, 1\right)
One solution in \left[0, \dfrac{\pi}{4}\right]$

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