JEE MAIN - Mathematics (2020 - 8th January Morning Slot - No. 7)
The inverse function of
f(x) = $${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$$, x $$ \in $$ (-1, 1), is :
f(x) = $${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$$, x $$ \in $$ (-1, 1), is :
$${1 \over 4}{\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$
$${1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$
$${1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 + x} \over {1 - x}}} \right)$$
$${1 \over 4}{\log _e}\left( {{{1 + x} \over {1 - x}}} \right)$$
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) = $${1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 + x} \over {1 - x}}} \right)$$
$$ \therefore $$ $${{y + 1} \over {y - 1}} = {{{{2.8}^{2x}}} \over { - {{2.8}^{ - 2x}}}}$$
$$ \Rightarrow $$ $${{1 + y} \over {1 - y}}$$ = 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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