Check Injective / Surjective
🎥 Video Explanation
📝 Question
Let \( f:\mathbb{R} \to \mathbb{R} \) be defined by
\[ f(x)=\frac{e^{|x|}-e^{-x}}{e^x+e^{-x}} \]
- A. bijection
- B. injection only
- C. surjection only
- D. neither
✅ Solution
🔹 Step 1: Case-wise Simplification
Case 1: \(x \ge 0\)
\[ |x|=x \]
\[ f(x)=\frac{e^x – e^{-x}}{e^x + e^{-x}} \]
\[ f(x)=\tanh x \]
—Case 2: \(x < 0\)
\[ |x|=-x \]
\[ f(x)=\frac{e^{-x} – e^{-x}}{e^x + e^{-x}}=0 \]
—🔹 Step 2: Final Form
\[ f(x)= \begin{cases} \tanh x, & x \ge 0 \\ 0, & x < 0 \end{cases} \]
—🔹 Step 3: Check Injective
For all \(x<0\), \(f(x)=0\).
Multiple inputs → same output ⇒ ❌ Not injective
—🔹 Step 4: Check Surjective
Range:
\[ [0,1) \]
Codomain is \(\mathbb{R}\), not fully covered.
❌ Not surjective
—🔹 Final Answer
\[ \boxed{\text{Option D: neither injection nor surjection}} \]