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.