A) Is given by \[\frac{1}{3x-5}\]
B) Is given by \[\frac{x+5}{3}\]
C) Does not exist because f is not one-one
D) Does not exist because f is not onto
Correct Answer: B
Solution :
Let \[f(x)=y\,\,\Rightarrow \,\,x={{f}^{-1}}(y).\] Hence\[f(x)=y=3x-5\,\,\Rightarrow \,\,x=\frac{y+5}{3}\,\Rightarrow \,{{f}^{-1}}(y)=x=\frac{y+5}{3}\] \[\therefore \,\,{{f}^{-1}}(x)=\frac{x+5}{3}\] Also f is one-one and onto, so \[{{f}^{-1}}\] exists and is given by \[{{f}^{-1}}(x)=\frac{x+5}{3}\].
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