Question

Let $A=R−\\{3\\}$ and $B=R−\\{1\\}$. Consider the function $f: A→B$ defined by $f(x)=\bigg(\frac{x-2}{x-3}\bigg)$. Is f one-one and onto? Justify your answer.

Answer 1

$A=R−\\{3\\}$, $B=R−\\{1\\}$
$f: A→B$ is defined as $f(x)=\bigg(\frac{x-2}{x-3}\bigg)$.

Let $x, y ∈ A$ such that $f(x)=f(y)$

$⇒\frac{x-2}{x-3}=\frac{y-2}{y-3}$
$⇒(x-2)(y-3)=(y-2)(x-3)$
$⇒xy-3x-2y+6=xy-3y-2x+6$
$⇒-3x-2y=-3y-2x$
$⇒3x-2x=3y-2y$
$⇒x=y$
∴ f is one-one.

Let $y ∈B = R − \\{1\\}$. Then, $y ≠ 1$.
The function f is onto if there exists $x ∈A$ such that $f(x) = y$.

Now,
$f(x)=y$
$⇒\frac{x-2}{x-3}=y$
$⇒x-2=xy-3y$
$⇒x(1-y)=-3y+2$
$⇒x=\frac{2-3y}{1-y} ∈ A$ $[y≠1]$

Thus, for any $y ∈ B$, there exists $\frac{2-3y}{1-y} ∈A$ such that
$f\big(\frac{2-3y}{1-y}\big)$
$=\frac{\big(\frac{2-3y}{1-y}\big)-2}{\big(\frac{2-3y}{1-y}\big)-3}$
$=\frac{2-3y-2+2y}{2-3y-3+3y}$
$=\frac{-y}{-1}$
$=y$.
$\therefore$ f is onto.

Hence, function f is one-one and onto.