A) \[\frac{x-1}{3x+2}\]
B) \[\frac{3x+2}{x-1}\]
C) \[\frac{x+1}{3x-2}\]
D) \[\frac{2x+1}{1-3x}\]
Correct Answer: A
Solution :
Let \[y=\frac{2x+1}{1-3x}\] \[\Rightarrow \] \[(1-3x)y=(2x+1)\] \[\Rightarrow \] \[x(2+3y)=y-1\] \[\Rightarrow \] \[x=\frac{y-1}{2+3y}\] \[\therefore \] \[{{f}^{-1}}(x)=\frac{x-1}{2+3x}\]You need to login to perform this action.
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