A) Invertible and \[{{f}^{-1}}(y)=\frac{2y+1}{y-1}\]
B) Invertible and \[{{f}^{-1}}(y)=\frac{3y+1}{y-1}\]
C) No invertible
D) Invertible and \[{{f}^{-1}}(y)=\frac{2y-1}{y-1}\]
Correct Answer: D
Solution :
Let \[y=f(x)\] \[\Rightarrow y=\frac{x-1}{x-2}\] \[\Rightarrow yx-2y=x-1\] \[\Rightarrow (y-1)x=2y-1\] \[\Rightarrow x={{f}^{-1}}(y)=\frac{2y-1}{y-1}\] So on the given domain the function is invertible and its inverse can be computed as shown above. So, option D is the correct answer.
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