Prove \(f(x)=\dfrac{x-2}{x-3}\) is a Bijection
📺 Video Explanation
📝 Question
Show that:
\[ f:\mathbb{R}\setminus\{3\}\to\mathbb{R}\setminus\{1\} \]
defined by:
\[ f(x)=\frac{x-2}{x-3} \]
is a bijection.
✅ Solution
🔹 Step 1: Prove One-One (Injective)
Assume:
\[ f(x_1)=f(x_2) \]
Then:
\[ \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} \]
Cross multiply:
\[ (x_1-2)(x_2-3)=(x_2-2)(x_1-3) \]
Expand:
\[ x_1x_2-3x_1-2x_2+6=x_1x_2-3x_2-2x_1+6 \]
Simplify:
\[ -x_1+x_2=0 \]
\[ x_1=x_2 \]
✔ Hence, \(f\) is one-one.
🔹 Step 2: Prove Onto (Surjective)
Let:
\[ y\in\mathbb{R}\setminus\{1\} \]
Need to find:
\[ x\in\mathbb{R}\setminus\{3\} \]
such that:
\[ \frac{x-2}{x-3}=y \]
Solve:
\[ x-2=y(x-3) \]
\[ x-2=yx-3y \]
\[ x-yx=2-3y \]
\[ x(1-y)=2-3y \]
\[ x=\frac{2-3y}{1-y} \]
Since:
\[ y\neq1 \]
denominator is non-zero.
So such \(x\) exists.
Also:
\[ x\neq3 \]
(otherwise denominator of original function undefined).
✔ Hence, onto.
🎯 Final Answer
\[ \boxed{f(x)=\frac{x-2}{x-3}\text{ is bijective}} \]
🚀 Exam Shortcut
- Rational functions: use cross multiplication for injection
- For onto, solve \(f(x)=y\)
- If inverse exists, function is bijection