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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
     Let the function\[f\]be defined by\[f(x)=\frac{2x+1}{1-3x}\]Then\[{{f}^{-1}}(x)\]is:

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


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