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JCECE Engineering JCECE 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) \[\frac{3x+2}{x-1}\]

    C) \[\frac{x+1}{3x-2}\]                       

    D) \[\frac{2x+1}{1-3x}\]

    Correct Answer: A

    Solution :

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


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