The inverse of the function `f:R to {x in R: x lt 1}"given by "f(x)=(e^(x)-e^(-x))/(e^(x)+e^(-x)),` isA. `(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

The inverse of the function `f:R to {x in R: x lt 1}"given by "f(x)=(e

1.	The inverse of the function `f:R to {x in R: x lt 1}"given by "f(x)=(e^(x)-e^(-x))/(e^(x)+e^(-x)),` isA. `(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» Correct Answer - A 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))`