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 injectivityLet $$f(x_1)=f(x_2)\Rightarrow \dfrac {x_1 -2}{x_2 -3}=\dfrac {x_2-2}{x_2-3}$$$$\Rightarrow \ (x_1-2)(x_2-3)=(x_2-2)(x_1-3)$$$$\Rightarrow \ x_1x_2-3x_1-2x_2+6\ x_1x_2=3x_2-2x_1+6$$$$\Rightarrow \ -3x_1-2x_2=-3x_2 -2x_1$$$$\Rightarrow \ -xx_1=-x_2\Rightarrow x_1=x_2$$For surjectivityLet $$y=\dfrac {x-2}{x-3}\Rightarrow x-2=xy -3y$$$$\Rightarrow \ x(1-y)=2-3y \Rightarrow \ x=\dfrac {2-3y}{1-y}$$$$\Rightarrow \ x=\dfrac {3y-2}{y-1}\in A, \forall y\in B$$So, $$f(x)$$ is surjective functionHence, $$f(x)$$ is a bijective function.Mathematics

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