Q

# 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

101 Views
N

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

Exams
Articles
Questions