Question

\(f:\mathbb{R}\rightarrow\mathbb{R}\) is a function defined by \[ f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \] Then \(f\) is

• A. One–one and into
• B. One–one not into
• C. Onto but not one–one
• D. Neither one–one nor onto

Hint:

For functions involving exponentials:
Check monotonicity using derivatives to test one–one property
Always compare range with codomain to test onto/into
\(\tanh x\) has range \((-1,1)\)

Answer 1

The Correct Option is A

Step 1: Rewrite the function: \[ f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=\tanh x \]
Step 2: Check one–one property The derivative of \(f(x)=\tanh x\) is: \[ f'(x)=\operatorname{sech}^2 x>0 \quad \forall x\in\mathbb{R} \] Since \(f'(x)\) is always positive, \(f(x)\) is strictly increasing. Hence, \(f\) is one–one. 
Step 3: Check onto property For all real \(x\), [ -1]
Step 4: Since the codomain is \(\mathbb{R}\) but the range is only \((-1,1)\), the function does not cover all real numbers. Hence, \(f\) is into but not onto.