Question

Let A=R-3 and B=R-1.Consider the function f: A \[\to\] B defined by \[\text{f(x)=}\dfrac{(x-2)}{(x-3)}.\] Is f one-one and onto?

Answer 1

Hint: We consider \[{{x}_{1}}\] and \[{{x}_{2}}\] from the range of A. Put the value of \[{{x}_{1}}\] and \[{{x}_{2}}\] in \[\text{f(x)}\] and make
\[\text{f(}{{\text{x}}_{1}}\text{)=f(}{{\text{x}}_{2}}\text{)}\] . If \[{{x}_{1}}\] and \[{{x}_{2}}\] becomes equal then, the given function is a one-one function. Assume, \[\text{y=}\dfrac{(x-2)}{(x-3)}\] and then find the value of x in terms of y. Then, put the value of x in the expression\[\text{f(x)=}\dfrac{(x-2)}{(x-3)}\]. If we get, \[\text{f(x)=y}\] then our function is onto.

Complete step by step answer:
We have, A=R-3 and B=R-1 and f: A \[\to\] B
such that, \[\text{f(x)=}\dfrac{(x-2)}{(x-3)}.\]
For a one-one function, we need to prove \[{{x}_{1}}={{x}_{2}}\] .
\[\text{f(}{{\text{x}}_{1}}\text{)=f(}{{\text{x}}_{2}}\text{)}\]
\[\begin{align}
  & \Rightarrow \dfrac{({{x}_{1}}-2)}{({{x}_{1}}-3)}=\dfrac{({{x}_{2}}-2)}{({{x}_{2}}-3)} \\
 & \Rightarrow ({{x}_{1}}-2)({{x}_{2}}-3)=({{x}_{1}}-3)({{x}_{2}}-2) \\
 & \Rightarrow -3{{x}_{1}}-2{{x}_{2}}=-3{{x}_{2}}-2{{x}_{1}} \\
 & \Rightarrow -{{x}_{1}}=-{{x}_{2}} \\
 & \Rightarrow {{x}_{1}}={{x}_{2}} \\
\end{align}\]
So, \[\text{f(x)}\] is a one-one function.
\[\begin{align}
  & \text{f(x)=}\dfrac{(x-2)}{(x-3)} \\
 & \text{y=}\dfrac{(x-2)}{(x-3)} \\
 & \Rightarrow y(x-3)=(x-2) \\
 & \Rightarrow xy-3y=x-2 \\
 & \Rightarrow x(y-1)=3y-2 \\
 & \Rightarrow x=\dfrac{(3y-2)}{(y-1)} \\
\end{align}\]
We have \[\text{f(x)=}\dfrac{(x-2)}{(x-3)}.\]
Putting the value of x in \[\text{f(x)=}\dfrac{(x-2)}{(x-3)}\], we get
\[\text{f(x)=}\dfrac{(\dfrac{(3y-2)}{(y-1)}-2)}{(\dfrac{(3y-2)}{(y-1)}-3)}\]
\[=\dfrac{\dfrac{(3y-2)}{(y-1)}-2}{\dfrac{(3y-2)}{(y-1)}-3}\]
\[\begin{align}
  & =\dfrac{\dfrac{(3y-2)-2(y-1)}{(y-1)}}{\dfrac{(3y-2)-3(y-1)}{(y-1)}} \\
 & =y \\
 & f(x)=y \\
\end{align}\]
f(x) is onto.

Therefore, we can say that f(x) is one-one and onto.

Note: In this question, one can make mistakes in proving onto function. Just follow the basic steps for it. For onto function we should have preimage. For that assume, \[\text{y=}\dfrac{(x-2)}{(x-3)}\] and then find the value of x in terms of y. Then, put the value of x in the expression\[\text{f(x)=}\dfrac{(x-2)}{(x-3)}\]. If we get, \[\text{f(x)=y}\] then our function is onto.