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

Concept: Bijection Proof — Show injectivity (one-one) by proving $f(x_1)=f(x_2) \Rightarrow x_1=x_2$, and surjectivity (onto) by showing for every $y \in B$ there exists $x \in A$ with $f(x)=y$.

Step 1 (One-one):

Let $f(x_1)=f(x_2)$. Then

Cross-multiplying: $(x_1-2)(x_2-3) = (x_2-2)(x_1-3)$. Expanding and simplifying gives $x_1 x_2 -3x_1 -2x_2 +6 = x_1 x_2 -3x_2 -2x_1 +6$, which reduces to $-3x_1 -2x_2 = -3x_2 -2x_1$, i.e. $x_1 = x_2$. So $f$ is one-one.

Step 2 (Onto):

For any $y \in B$ (so $y \neq 1$), solve $y = \frac{x-2}{x-3}$ for $x$. …

Why a Bijection Proof Works

When a function is defined by a rational expression like $\frac{x-2}{x-3}$, the natural instinct is to check one-one by setting $f(x_1) = f(x_2)$ and solving, and check onto by solving $f(x) = y$ for $x$ in terms of $y$. Both steps are algebraic, but the real insight is this: if you can solve $f(x) = y$ uniquely for every $y$ in the codomain, you have simultaneously proven one-one and onto. That single equation gives you the inverse function, and an invertible function is exactly a bijection.

The domain excludes $x = 3$ (division by zero), and the codomain excludes $y = 1$ because the horizontal asymptote of $f$ is $y = 1$ — the function never actually reaches that value. So the mapping is perfectly tailored: $f$ sends every real number except $3$ to every real number except $1$, exactly once each.

Step-by-Step Solution

1. Check one-one (injectivity).

Assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in A$. Then:

Cross-multiply (valid since denominators are non-zero in $A$):

Expand both sides:

Cancel $x_1 x_2$ and $6$ from both sides:

Bring terms together:

So $x_1 = x_2$. Hence $f$ is one-one.

2. Check onto (surjectivity).

We need to show: for every $y \in B = \mathbb{R} - \{1\}$, there exists some $x \in A = \mathbb{R} - \{3\}$ such that $f(x) = y$.

Set $y = \frac{x-2}{x-3}$ and solve for $x$:

Bring $x$ terms together:

$x(y - 1) = 3y - 2$ …

Method: One-one and onto for a rational function on a restricted domain

For $f(x)=\frac{ax+b}{cx+d}$ with excluded points, prove injectivity by clearing fractions and surjectivity by inverting — while checking the preimage avoids the forbidden value.

Steps

Step 1: One-one — set $f(x_1)=f(x_2)$, cross-multiply, expand, and cancel; the cross terms drop and leave $x_1=x_2$. (Shortcut: rewrite $\frac{x-2}{x-3}=1+\frac{1}{x-3}$, so equal outputs force equal denominators.) …

Common Mistakes

Mistake 1: Forgetting to verify the solved preimage stays in the domain.

Why it's wrong: onto requires $x=\frac{3y-2}{y-1}\in A=\mathbf{R}-\{3\}$; without checking $x\neq 3$ the proof is incomplete. Correct approach: show $\frac{3y-2}{y-1}=3$ leads to $-2=-3$, impossible, so $x\neq 3$ always.

Mistake 2: Ignoring why $y=1$ is excluded from the codomain. …