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:

Any linear fractional function of the form $f(x) = \frac{ax+b}{cx+d}$ is always one-one on its natural domain, provided that its determinant condition $ad - bc \neq 0$ is satisfied. Here, $(1)(-3) - (-2)(1) = -1 \neq 0$.

Answer 1

Concept: To evaluate the mathematical nature of the rational function mapping, we perform two independent algebraic verifications:
• One-One Test: Set $f(x_1) = f(x_2)$ and algebraically solve to determine if this requires $x_1 = x_2$.
• Onto Test: Equate the expression to $y$ (where $y \in B$), and solve completely for $x$ in terms of $y$. Check if this value of $x$ is valid and defined within the domain $A$ for all possible values of $y$.

Step 1: Perform the algebraic verification for the one-one (injective) condition.
Let $x_1, x_2 \in A$ such that: \[ f(x_1) = f(x_2) \] Substitute the definition of the function into the expression: \[ \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \] Cross-multiply to clear the denominators: \[ (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \] Expand both polynomial sides: \[ x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6 \] Cancel out the identical terms $x_1 x_2$ and $6$ present on both sides: \[ -3x_1 - 2x_2 = -3x_2 - 2x_1 \] Rearrange the terms to group $x_1$ on the left and $x_2$ on the right: \[ -3x_1 + 2x_1 = -3x_2 + 2x_2 \] \[ -x_1 = -x_2 \quad \implies \quad x_1 = x_2 \] Since $f(x_1) = f(x_2)$ simplifies down to $x_1 = x_2$, the function is one-one.

Step 2: Perform the algebraic verification for the onto (surjective) condition.
Let $y$ be an arbitrary element belonging to the codomain $B = \mathbb{R} - \{1\}$. Set $f(x) = y$: \[ y = \frac{x - 2}{x - 3} \] Cross-multiply to isolate the variable $x$: \[ y(x - 3) = x - 2 \quad \implies \quad xy - 3y = x - 2 \] Move all terms containing $x$ to one side: \[ xy - x = 3y - 2 \] Factor out $x$ on the left-hand side: \[ x(y - 1) = 3y - 2 \quad \implies \quad x = \frac{3y - 2}{y - 1} \] Now evaluate this result: Since the codomain is given as $B = \mathbb{R} - \{1\}$, $y$ can never equal 1. This means the denominator $(y-1)$ is never zero, so $x$ is always a well-defined real number. Could $x$ ever equal 3? Let us check: \[ 3 = \frac{3y - 2}{y - 1} \quad \implies \quad 3(y - 1) = 3y - 2 \quad \implies \quad 3y - 3 = 3y - 2 \quad \implies \quad -3 = -2 \] This is a logical contradiction, which means $x$ can never equal 3. Thus, for every $y \in B$, there exists a valid pre-image $x = \frac{3y - 2}{y - 1}$ that lies within the domain $A$. Therefore, the function is onto.