JEE MAIN - Mathematics Hindi (2020 - 8th January Morning Slot - No. 7)
का उलटा फ़ंक्शन
f(x) = $${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$$, x $$ \in $$ (-1, 1), है :
f(x) = $${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$$, x $$ \in $$ (-1, 1), है :
$${1 \over 4}{\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$
$${1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$
$${1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 + x} \over {1 - x}}} \right)$$
$${1 \over 4}{\log _e}\left( {{{1 + x} \over {1 - x}}} \right)$$
Comments (0)
