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 \[y=f(x)\] Þ \[y=\frac{2x+1}{1-3x}\] Þ \[y-3xy=2x+1\] Þ \[x=\frac{y-1}{3y+2}\] Þ \[{{f}^{-1}}(y)=\frac{y-1}{3y+2}\Rightarrow {{f}^{-1}}(x)=\frac{x-1}{3x+2}\].
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