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 ${{x}_{1}}={{x}_{2}}$ .
$\text{f(}{{\text{x}}_{1}}\text{)=f(}{{\text{x}}_{2}}\text{)}$

& \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{aligned}$$ So, $$\text{f(x)}$$ is a one-one function. $$\begin{aligned} & \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{aligned}$$ We have $$\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{aligned} & =\dfrac{\dfrac{(3y-2)-2(y-1)}{(y-1)}}{\dfrac{(3y-2)-3(y-1)}{(y-1)}} \\\ & =y \\\ & f(x)=y \\\ \end{aligned}$$ f(x) is onto. _Therefore, we can say that f(x) is one-one and onto._ Note: In this question, one can make mistakes in proving onto function. Just follow the basic steps for it. For onto function we should have preimage. For that assume, $$\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.