Question

Let $A = R - \left\{ 3 \right\},{\text{ }}B = R - \left\{ 1 \right\}$. Let $f: A \to B$ be defined by $f\left( x \right) = \dfrac{{x - 2}}{{x - 3}}$, is $ f $ bijective? Give reasons.

Answer 1

Hint- If function is both one-one and onto then the function is bijective. So, We proceed our solution first by proving the given function as onto and one-one function.

Complete step-by-step solution -
$f:A \to B$, be defined by $f\left( x \right) = \dfrac{{x - 2}}{{x - 3}}$
Let $x,y \in A$ such that $f\left( x \right) = f\left( y \right)$
$ \Rightarrow \dfrac{{x - 2}}{{x - 3}} = \dfrac{{y - 2}}{{y - 3}}$
Apply cross multiplication
$
  \left( {x - 2} \right)\left( {y - 3} \right) = \left( {y - 2} \right)\left( {x - 3} \right) \\
   \Rightarrow xy - 3x - 2y + 6 = xy - 3y - 2x + 6 \\
   \Rightarrow 3x - 2x = 3y - 2y \\
   \Rightarrow x = y \\
$
$\therefore f $ is one-one.
Let$y \in B = R - \left\{ 1 \right\}$, then$y \ne 1$.
The function $ f $ is onto if there exists $x \in A$ such that $f\left( x \right) = y$.
Now $f\left( x \right) = y$ ……… (1)
$
   \Rightarrow \dfrac{{x - 2}}{{x - 3}} = y \\
   \Rightarrow x - 2 = xy - 3y \\
   \Rightarrow x - xy = 2 - 3y \\
   \Rightarrow x\left( {1 - y} \right) = 2 - 3y \\
   \Rightarrow x = \dfrac{{2 - 3y}}{{1 - y}} \in A,{\text{ }}\left( {y \ne 1} \right) \\
$
Thus, for any $y \in B$, there exists $\dfrac{{2 - 3y}}{{1 - y}} \in A$ such that
$f\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) = \dfrac{{\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) - 2}}{{\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) - 3}} = \dfrac{{2 - 3y - 2 + 2y}}{{2 - 3y - 3 + 3y}} = \dfrac{{ - y}}{{ - 1}} = y..........\left( 2 \right)$
Now, from equation (1) and (2) we can say that $ f $ is onto.
Hence, the function $ f $ is both one-one and onto.
Therefore function is bijective.

Note- In bijective function, there is one-to-one correspondence, in a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set such that there is no unpaired elements than the function is called as bijective function.