Question

Let $f:A\to B$ 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 :

• A. invertible and $f^{-1}\left(y\right)=\frac{3y-1}{y-1}$
• B. invertible and $f^{-1}\left(y\right)=\frac{2y-1}{y-1}$
• C. invertible and $f^{-1}\left(y\right)=\frac{2y+1}{y-1}$
• D. Not invertible

Answer 1

Answer: (B)

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

As you can see that it is satisfying the domain conditions perfectly that is not containing $\left\{2\right\}$ and we have to check whether the range which becomes a domain for the inverse also satisfies or not.

$y=f\left(x\right)$

$x=f^{-1}\left(y\right)$

$y=\frac{x-1}{x-2}=1+\frac{1}{x-2}$

$x=2+\frac{1}{y-1}=\frac{2y-1}{y-1}$

you can see that range also satisfies the condition that is not containing $\left\{1\right\}$.So we can say that is invertible.