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The inverse function of f(x) = $${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$$, x $$ \in $$ (-1, 1), is :

JEE · Math · previous-year question

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

Answer

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

Explanation

f(x) = $${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$$ = y $$ \therefore $$ $${{y + 1} \over {y - 1}} = {{{{2.8}^{2x}}} \over { - {{2.8}^{ - 2x}}}}$$ $$ \Rightarrow $$ $${{1 + y} \over {1 - y}}$$ = 84x $$ \Rightarrow $$ $${\log _e}\left( {{{1 + y} \over {1 - y}}} \right)$$ = 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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