Question

Let \[ A=\mathbb{R}-\{3\} \] and \[ B=\mathbb{R}-\{1\}. \] A function \(f:A\to B\) is defined by \[ f(x)=\frac{x-2}{x-3}. \] Find whether \(f\) is one-one and onto. 

Hint:

To test for 'onto', express \(x\) in terms of \(y\). If the resulting expression is defined for all \(y\) in the codomain and lies within the domain, the function is surjective.
For rational functions, the one-one check always involves a cross-multiplication step.

Answer 1

Concept:
• A function is one-one (injective) if \(f(x_1) = f(x_2) \implies x_1 = x_2\) for all \(x_1, x_2\) in the domain.
• A function is onto (surjective) if for every element \(y\) in the codomain, there exists an \(x\) in the domain such that \(f(x) = y\).

Step 1: Check for one-one property
Let \(x_1, x_2 \in A\) such that \(f(x_1) = f(x_2)\):
\[ \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \] Cross-multiplying:
\[ (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \] \[ x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6 \] Subtracting common terms from both sides:
\[ -3x_1 - 2x_2 = -3x_2 - 2x_1 \] \[ 3x_2 - 2x_2 = 3x_1 - 2x_1 \] \[ x_2 = x_1 \] Since \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function is one-one.

Step 2: Check for onto property
Let \(y \in B\). We check if there exists \(x \in A\) such that \(f(x) = y\):
\[ y = \frac{x - 2}{x - 3} \] \[ y(x - 3) = x - 2 \] \[ xy - 3y = x - 2 \] \[ xy - x = 3y - 2 \] \[ x(y - 1) = 3y - 2 \] \[ x = \frac{3y - 2}{y - 1} \] Since \(y \in B = \mathbb{R} - \{1\}\), \(y \neq 1\), so \(x\) is always defined.
Now check if \(x = 3\) is possible:
If \(\frac{3y - 2}{y - 1} = 3 \implies 3y - 2 = 3y - 3 \implies -2 = -3\) (Impossible).
Thus, for every \(y \in B\), there is a corresponding \(x \in A\). The function is onto.