The inverse of the function `f: Rvec{x in R : x<1}` given by `f(x)=(e^x-e^(-x))/(e^x+e^(-x)),` is `1/2log(1+x)/(1-x)` (b) `1/2log(2+x)/(2-x)` `1/2log(1-x)/(1+x)` (d) None of these

Explanation

`f(x)=(e^x-e^(−x))/(e^x+e^(−x​))`


Let `f^(−1)(x)=y ---- ( 1 )`

`⇒ f(y)=x`

`⇒ (e^x-e^(−x))/(e^x+e^(−x​))​=x`


`⇒ (e^(−y)(e^(2y)-1))/(e^(−y)(e^(2y)−1))​=x`


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