Question

Let \(A= \mathbb{R} -{3} - B = \mathbb{R} -{1}\)
Let \(f:A\rightarrow B\) be defined by
\(f(x) = \dfrac{x-2}{x-3} \forall x \in A\) . Then show that \(f\) is bijective.

Answer 1

Testing whether \(f\) is injective

Given : \(A=R-{3}, B=R-{1}\)

\(f:A\rightarrow B\) be defined by \(f(x) = \dfrac{x-2}{x-3} \forall x \in A\)

Let \(f(x_1) = f(x_2)\)

\(\Rightarrow \dfrac{x_{1}-2}{x_{1}-3} = \dfrac{x_{2}-2}{x_{2}-3}\)
\(\Rightarrow 1+ \dfrac{1}{x_{1}-3}=1+ \dfrac{1}{x_{1}-3}\)

\(\Rightarrow \dfrac{1}{x_{1}-3}=\dfrac{1}{x_{2}-3}\)

\(\Rightarrow x_{1} -3 = x_{2}-3\)

\(\Rightarrow x_{1} = x_2\)

\(\therefore f(x) \text { is an injective function }\)

Testing whether \(f\) surjective

Let \(y= \dfrac{x-2}{x-3}\)

\(\Rightarrow yx-3y=x-2\)
\(\Rightarrow yx-x=3y-2\)

\(\Rightarrow x= \dfrac{(3y-2)}{(y-1)}\)

Thus, \(y \in R -{1}\)

\(\therefore\) Range \((f) = R-{1} =\) co - domian (f)

\(\therefore f (x) \text { is surjective function }\)


\(\therefore f (x) \text { is both injective and surjective function }\)

therefore f(x) is bijective function .

Hence proved.