Check Function \(f(x)=\dfrac{2x+3}{x-3}\) on Rational Numbers
📺 Video Explanation
📝 Question
Check whether the function
\[ f:\mathbb{Q}\setminus\{3\}\to\mathbb{Q},\quad f(x)=\frac{2x+3}{x-3} \]
is:
- injection (one-one)
- surjection (onto)
- bijection
✅ Solution
🔹 Step 1: Check Injection (One-One)
Assume:
\[ f(x_1)=f(x_2) \]
Then:
\[ \frac{2x_1+3}{x_1-3}=\frac{2x_2+3}{x_2-3} \]
Cross multiply:
\[ (2x_1+3)(x_2-3)=(2x_2+3)(x_1-3) \]
Expand:
\[ 2x_1x_2-6x_1+3x_2-9=2x_1x_2-6x_2+3x_1-9 \]
Simplify:
\[ -6x_1+3x_2=-6x_2+3x_1 \]
\[ -9x_1+9x_2=0 \]
\[ x_1=x_2 \]
✔ Hence, function is one-one.
🔹 Step 2: Check Surjection (Onto)
Let:
\[ y\in\mathbb{Q} \]
Need:
\[ \frac{2x+3}{x-3}=y \]
Cross multiply:
\[ 2x+3=yx-3y \]
Rearrange:
\[ x(2-y)=-(3+3y) \]
So:
\[ x=\frac{-3(1+y)}{2-y} \]
This is rational whenever:
\[ y\neq2 \]
But if:
\[ y=2 \]
Then equation becomes:
\[ 2x+3=2x-6 \]
Impossible.
So:
\[ 2 \] is not in range.
❌ Not onto.
🎯 Final Answer
\[ \boxed{\text{f is one-one but not onto}} \]
So:
✔ Injection
❌ Surjection
❌ Bijection
🚀 Exam Shortcut
- Rational functions often need cross multiplication
- Check special value making denominator zero
- One missing codomain value breaks onto