# Prove that

Question:

Let $A=R-\{2\}$ and $B=R-\{1\} .$ If $f: A \rightarrow B: f(x)=\frac{x-1}{x-2}$, show that $f$ is one-one and onto. Hence, find $f^{-}$

1.

Solution:

To Show: that $\mathrm{f}$ is one-one and onto

To Find: Inverse of $f$

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function $f: A \rightarrow B$ is said to be a one-one function or injective mapping if different

elements of A have different images in B. Thus for $x_{1}, x_{2} \in A \& f\left(x_{1}\right), f\left(x_{2}\right) \in B, f\left(x_{1}\right)=f\left(x_{2}\right) \leftrightarrow x_{1}=x_{2}$ or $x_{1} \neq$ $x_{2} \leftrightarrow f\left(x_{1}\right) \neq f\left(x_{2}\right)$

onto function: If range $=$ co-domain then $f(x)$ is onto functions.

So, We need to prove that the given function is one-one and onto.

Let $\mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{Q}$ and $\mathrm{f}(\mathrm{x})=\frac{x-1}{x-2} .$ So $\mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right) \rightarrow \frac{x_{1}-1}{x_{1}-2}=\frac{\left(x_{2}-1\right)}{x_{2}-2}$, on solving we get $\rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}$

So $f\left(x_{1}\right)=f\left(x_{2}\right) \leftrightarrow x_{1}=x_{2}, f(x)$ is one-one

Given co-domain of $f(x)$ is $R-\{1\}$

Let $\mathrm{y}=\mathrm{f}(\mathrm{x})=\frac{x-1}{x-2}$, So $\mathrm{x}=\frac{2 y-1}{y-1}[$ Range of $\mathrm{f}(\mathrm{x})=$ Domain of $\mathrm{y}$ ]

So Domain of $y=$ Range of $f(x)=R-\{1\}$

Hence, Range of $f(x)=$ co-domain of $f(x)=R-\{1\}$.

So, $f(x)$ is onto function

As it is a bijective function. So it is invertible

Invers of $f(x)$ is $f^{-1}(y)=\frac{2 y-1}{y-1}$