QJEEM20Mabd73: JEE Main 2020 Mathematics - Class 11 JEEM20Mabd73 Solution: The inverse function of f(x) = , x (-1, 1), is :

Question

Verified Solution • 2025–26 Session moderate

The inverse function of

f(x) =

{{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}

, x

\in

(-1, 1), is :

A ${1 \over 4}{\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$
B ${1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$
C ${1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 + x} \over {1 - x}}} \right)$
D ${1 \over 4}{\log _e}\left( {{{1 + x} \over {1 - x}}} \right)$

Accepted Answer

f(x) =

{{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}

= y

\therefore

{{y + 1} \over {y - 1}} = {{{{2.8}^{2x}}} \over { - {{2.8}^{ - 2x}}}}

\Rightarrow

{{1 + y} \over {1 - y}}

= 8^4x

\Rightarrow

{\log _e}\left( {{{1 + y} \over {1 - y}}} \right)

= 4x

{\log _e}8

\Rightarrow

x =

{1 \over {4{{\log }_e}8}}{\log _e}\left( {{{1 + y} \over {1 - y}}} \right)

\therefore

f^-1(x) =

{1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 + x} \over {1 - x}}} \right)

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