Check One-One and Onto
🎥 Video Explanation
📝 Question
Let \( f:\mathbb{R} – \{-b\} \to \mathbb{R} – \{1\} \) be defined by
\[ f(x)=\frac{x+a}{x+b}, \quad a \ne b \]
Choose the correct option:
- A. one-one but not onto
- B. onto but not one-one
- C. both one-one and onto
- D. none of these
✅ Solution
🔹 Step 1: Check One-One (Injective)
Assume \(f(x_1)=f(x_2)\):
\[ \frac{x_1+a}{x_1+b}=\frac{x_2+a}{x_2+b} \]
Cross multiply:
\[ (x_1+a)(x_2+b)=(x_2+a)(x_1+b) \]
Expand:
\[ x_1x_2 + bx_1 + ax_2 + ab = x_1x_2 + bx_2 + ax_1 + ab \]
Simplify:
\[ bx_1 + ax_2 = bx_2 + ax_1 \]
\[ (b-a)x_1 = (b-a)x_2 \]
Since \(a \ne b\):
\[ x_1 = x_2 \]
✔️ Function is one-one
—🔹 Step 2: Check Onto (Surjective)
Let \(y = \frac{x+a}{x+b}\)
Solve for \(x\):
\[ y(x+b)=x+a \]
\[ yx + yb = x + a \]
\[ x(y-1)=a – yb \]
\[ x=\frac{a-yb}{y-1} \]
This is defined for all \(y \ne 1\), which matches codomain.
✔️ Function is onto
—🔹 Final Answer
\[ \boxed{\text{Option C: both one-one and onto}} \]