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Let A= R-\left \{ 2 \right \}and \: B= R-\left \{ 1 \right \}.\; If\;\; f:A\rightarrow B  is a function defined by f\left ( x \right )= \frac{x-1}{x-2},  show that f is one -one and onto. Hence,find f^{-1}.

 

 

 

 
 
 
 
 

Answers (1)
R Ravindra Pindel

f\left ( x \right )= \frac{x-1}{x-2}
Let  x_{1},x_{2}\: \epsilon A   such that  f\left ( x_{1} \right )= f\left ( x_{2} \right )
That is \frac{x_{1}-1}{x_{1}-2}= \frac{x_{2}-1}{x_{2}-2}
\Rightarrow x_{1}x_{2}-2x_{1}-x_{2}+2= x_{1}x_{2}-x_{1}\; \; 2x_{2}+2
\Rightarrow x_{1}= x_{2}
\therefore f\left ( x \right )   is one-one function 
Let\: y= f\left ( x \right ),y\epsilon B
That is y= \frac{x-1}{x-2}\Rightarrow xy-2y= x-1
\Rightarrow xy-x= 2y-1\; \; \Rightarrow x= \frac{2y-1}{y-1}
clearly, y\epsilon B  for all x\epsilon A  That    implies range =  codomain 
so, f is onto.
Also f^{-1}\left ( y \right )= \frac{2y-1}{y-1}    i.e   f^{-1}\left ( x \right )= \frac{2x-1}{x-1}
 

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