Mathematics · 2020
JEE Main · 8 January 2020, Shift 1 · Q51
The inverse function of f(x)=(8^2 x-8^-2 x)/(8^2 x+8^-2 x), x ∈(-1,1), is \ \ \ \.
The inverse function of
$$f(x)=\frac{8^{2 x}-8^{-2 x}}{8^{2 x}+8^{-2 x}}, x \in(-1,1),
$$
is \_\_\_\_.
Official answer
(4)
$\displaystyle \frac{1}{4}\left(\log _8 \mathrm{e}\right) \log _{\mathrm{e}}\left(\frac{1+x}{1-x}\right)$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.