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Let $f:A\to B$ be defined by $f(x)=\frac{3x-5}{2x+1}$. Which of the following choices of $A$ and $B$ makes $f$ a bijection?

  1. $A=\mathbb R$ and $B=\mathbb R$
     
  2. $A=\mathbb R-{-\frac12}$ and $B=\mathbb R$
     
  3. $A=\mathbb R-{-\frac12}$ and $B=\mathbb R-{\frac32}$
     
  4. $A=\mathbb R-{\frac32}$ and $B=\mathbb R-{-\frac12}$

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The function is $f(x)=\frac{3x-5}{2x+1}$.

First, $2x+1\neq 0$, so $x\neq -\frac12$. Therefore, the domain cannot be $\mathbb R$. Hence $A=\mathbb R-{-\frac12}$.

Now find the value that $f(x)$ can never take. 

Let $y=\dfrac{3x-5}{2x+1}$.

$\Rightarrow y(2x+1)=3x-5$.

$\Rightarrow 2xy+y=3x-5$.

$\Rightarrow 2xy-3x=-y-5$.

$\Rightarrow x(2y-3)=-(y+5)$.

$\therefore x=\frac{-(y+5)}{2y-3}$.

This is defined only when $2y-3\neq 0$, so $y\neq \frac32$.

Thus, the range of $f$ is $\mathbb R-{\frac32}$.

So if $A=\mathbb R-{-\frac12}$ and $B=\mathbb R-{\frac32}$, then every element of $B$ has exactly one preimage in $A$. Therefore, $f$ is both one-one and onto.

Hence $f:\mathbb R-{-\frac12}\to \mathbb R-{\frac32}$ is a bijection.

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