CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Let A = R - {3}, B = R - {1}. Let $$f : A \rightarrow B$$ be defined by $$f(x) = (x -2) / (x - 3)$$. Is $$f$$ bijective? Give reasons. 


Solution

$$A=R-\{ 3\} \\ B=R-\{ 1\} \\ f:A\longrightarrow B\quad is\quad defined\quad as\quad f(x)=\left( \cfrac { x-2 }{ x-3 }  \right) \\ Let\quad x,y\epsilon A\quad such\quad that\quad f(x)=f(y)\\ \Longrightarrow \cfrac { x-2 }{ x-3 } =\cfrac { y-2 }{ y-3 } \\ \Longrightarrow (x-2)(y-3)=(x-3)(y-2)\\ \Longrightarrow xy-3x-2y+6=xy-3y-2x+6\\ \Longrightarrow -3x-2y=-3y-2x\\ \Longrightarrow 3x-2x=3y-2y\\ \Longrightarrow x=y$$
$$\therefore f$$ is one-one.
$$Let\hspace{1mm} y\hspace{1mm} \epsilon \hspace{1mm} B=R-\{ 1\} .\hspace{1mm} Then\hspace{1mm} y\neq 1$$.
The function $$f$$ is onto if there exists $$x$$ $$\epsilon$$ $$A$$ such that $$f(x)=y$$.
$$Now,\quad f(x)=y\\ \Longrightarrow \cfrac { x-2 }{ x-3 } =y\\ \Longrightarrow x-2=xy-3y\\ \Longrightarrow x(1-y)=2-3y\\ \Longrightarrow x=\cfrac { 2-3y }{ 1-y } \epsilon A\quad [y\neq 1]$$ 
Thus, for any $$y$$ $$\epsilon$$ $$B$$, there exists $$\cfrac{2-3y}{1-y}$$ $$\epsilon$$ $$A$$ such that
$$f\left( \cfrac { 2-3y }{ 1-y }  \right) =\cfrac { \left( \cfrac { 2-3y }{ 1-y }  \right) -2 }{ \left( \cfrac { 2-3y }{ 1-y }  \right) -3 } =\cfrac { 2-3y-2+2y }{ 2-3y-3+3y } =\cfrac { -y }{ -1 } =y$$
$$\therefore f$$ is onto.
Hence, the function $$f$$ is one-one and onto.

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image