Question

Let A =R -{3}, B=R -{1}. If $f:A\to B$ be defined by $f\left(x\right)=\frac{x-2}{x-3},\forall x\in A$. Then, show that f is bijective.

Answer 1

Given that, A=R -{3}, B=R -{1}
$f:A\to B$ is defined by $f\left(x\right)=\frac{x-2}{x-3},\forall x\in A$
For injectivity
Let $f\left(x_1\right)=f\left(x_2\right)\Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}$
$\Rightarrow (x_1-2)(x_2-3)=(x_2-2)(x_1-3)\Rightarrow x_1{x}_2-3x_1-2x_2+6=x_1{x}_2-3x_2-2x_1+6\Rightarrow -3x_1-2x_2=-3x_2-2x_1\Rightarrow -x_1=-x_2\Rightarrow x_1=x_2$
So, f(x)is an injective function.

For surjectivity
Let $y=\frac{x-2}{x-3}\Rightarrow x-2=xy-3y$
$\Rightarrow x(1-y)=2-3y\Rightarrow x=\frac{2-3y}{1-y}$
$\Rightarrow x=\frac{3y-2}{y-1}\in A,\forall y\in B$ [codomain]
So, f(x)is surjective function.

Hence, f(x) is a bijective function