JEE Main 2020 (Online) 8th January Morning ShiftQuestion 1
The inverse function of $f(x)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}}$, $x\in(-1,1)$, is
Inverse Trigonometric Functions
Solution
With $u=8^{2x}$: $f=\frac{u-u^{-1}}{u+u^{-1}}=\tanh(2x\ln8)$. $y=\tanh(2x\ln 8)\Rightarrow2x\ln8=\frac12\ln\frac{1+y}{1-y}\Rightarrow x=\frac{1}{4\ln8}\ln\frac{1+y}{1-y}=\frac14(\log_8e)\ln\frac{1+y}{1-y}$. Option (D).