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