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JEE Main 20208 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2020 · 8 January · Morning Shift · Mathematics · Functions

The inverse function of f(x) = 8 ^ 2x - 8 ^ - 2x 8 ^ 2x + 8 ^ - 2x...

The inverse function of f(x)=82x−8−2x82x+8−2x,x∈(−1,1),f\left(x\right) = \dfrac{{8}^{2x} - {8}^{- 2x}}{{8}^{2x} + {8}^{- 2x}}, x \in \left(- 1, 1\right), is __________.

Options

  1. A14log⁡e(1+x1−x)\dfrac{1}{4}\log_{e}\left(\dfrac{1 + x}{1 - x}\right)
  2. B14log⁡e(1−x1+x)\dfrac{1}{4}\log_{e}\left(\dfrac{1 - x}{1 + x}\right)
  3. C14(log⁡e)log⁡e(1−x1+x)\dfrac{1}{4}\left(\log_{e}\right)\log_{e}\left(\dfrac{1 - x}{1 + x}\right)
  4. D14log⁡8(1+x1−x)\dfrac{1}{4}\log_{8}\left(\dfrac{1 + x}{1 - x}\right)

Correct answer

D. 14log⁡8(1+x1−x)\dfrac{1}{4}\log_{8}\left(\dfrac{1 + x}{1 - x}\right)

Step-by-step solution

y=82x−8−2x82x+8−2xy = \dfrac{{8}^{2x} - {8}^{- 2x}}{{8}^{2x} + {8}^{- 2x}}

⇒1+y1−y=82x8−2x⇒ \dfrac{1 + y}{1 - y} = \dfrac{{8}^{2x}}{{8}^{- 2x}} (Applying componendo-dividendo)

⇒84x=1+y1−y⇒ {8}^{4x} = \dfrac{1 + y}{1 - y}

⇒4x=log8(1+y1−y)⇒ 4x = lo{g}_{8}\left(\dfrac{1 + y}{1 - y}\right)

f−1(x)=14log8(1+x1−x)s{f}^{- 1}\left(x\right) = \dfrac{1}{4}lo{g}_{8}\left(\dfrac{1 + x}{1 - x}\right)s

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