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Question

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


Solution

Given that $$A=R-\left\{3\right\}, B=R-\left\{1\right\}$$. 
$$f:A\to B$$ be defined by $$f(x)=\dfrac {x-2}{x-3}\forall x\in A$$.
For injectivity
Let $$f(x_1)=f(x_2)\Rightarrow \dfrac {x_1 -2}{x_2 -3}=\dfrac {x_2-2}{x_2-3}$$
$$\Rightarrow \ (x_1-2)(x_2-3)=(x_2-2)(x_1-3)$$
$$\Rightarrow \ x_1x_2-3x_1-2x_2+6\ x_1x_2=3x_2-2x_1+6$$
$$\Rightarrow \ -3x_1-2x_2=-3x_2 -2x_1$$
$$\Rightarrow \ -xx_1=-x_2\Rightarrow x_1=x_2$$
For surjectivity
Let $$y=\dfrac {x-2}{x-3}\Rightarrow x-2=xy -3y$$
$$\Rightarrow \ x(1-y)=2-3y \Rightarrow \ x=\dfrac {2-3y}{1-y}$$
$$\Rightarrow \ x=\dfrac {3y-2}{y-1}\in A, \forall y\in B$$
So, $$f(x)$$ is surjective function
Hence, $$f(x)$$ is a bijective function.

Mathematics

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