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

Options

A. `(1)/(2)"log" (1+x)/(1-x)`
B. `(1)/(2)"log" (2+x)/(2-x)`
C. `(1)/(2)"log" (1-x)/(1+x)`
D. None of these

Answer

A

Explanation

Clearly, f is a bijection and hence invertible. Let `f(x)=y "Clearly, "y lt 1`
`"fof"^(-1)(x)=x" for all "xlt 1`
`Rightarrow f(f^(-1)(x))=x`
`(e^(f^(-1)(x))-e^(-f^(-1)(x)))/(e^(f^(-1)(x))+e^(-f^(-1)(x)))=(x)/(1)Rightarrow(2e^(f^(-1))(x))/(-2e^(-f^(-1))(x))=(x+1)/(x-1)`
`e^(2f^(-1)(x))=(1+x)/(1-x) Rightarrow f^(-1)(x)=(1)/(2)"log"((1+x)/(1-x))`
`"Hence", f^(-1):{x in R: x lt 1} to R` is given by
`f^(-1)(x)=(1)/(2)"log"((1+x)/(1-x))`