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I have a doubt, kindly clarify. - Sets, Relations and Functions - JEE Main-4

Let f:A\rightarrow B be a function defined as f\left ( x \right )=\frac{x-1}{x-2} where A=R-\left ( 2 \right ) and B=R-\left ( 1 \right ). Then f is :

  • Option 1)

    invertible and f^{-1}\left ( y \right )=\frac{3y-1}{y-1}

  • Option 2)

    invertible and f^{-1}\left ( y \right )=\frac{2y-1}{y-1}

  • Option 3)

    invertible and f^{-1}\left ( y \right )=\frac{2y+1}{y-1}

  • Option 4)

    not invertible

 
Answers (2)
101 Views
N neha
H Himanshu

As we learned,

 

f\left ( x \right )=\frac{x-1}{x-2}=y

\Rightarrow \: xy-2y=x-1

\Rightarrow \: x\left ( y-1 \right )=2y-1\: \Rightarrow \: x=\frac{2y-1}{y-1}

The function is invertible and f^{-1}\cdot \left ( y \right )=\frac{2y-1}{y-1}


Option 1)

invertible and f^{-1}\left ( y \right )=\frac{3y-1}{y-1}

Option 2)

invertible and f^{-1}\left ( y \right )=\frac{2y-1}{y-1}

Option 3)

invertible and f^{-1}\left ( y \right )=\frac{2y+1}{y-1}

Option 4)

not invertible

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