Question

Let \(f : \mathbb{R} \to \mathbb{R}\) be a function defined by \(f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}}\), then:

• A. \(f\) is both one-to-one and onto
• B. \(f\) is one-to-one but not onto
• C. \(f\) is onto but not one-to-one
• D. \(f\) is neither one-to-one nor onto

Hint:

For functions, check injectivity (one-to-one) and surjectivity (onto) by analyzing their behavior and range. If either is violated, the function is neither one-to-one nor onto.

Answer 1

The Correct Option is D

Step 1: The provided function is:

\( f(x) = \frac{e^{|x|} - e^{-x}}{e^{|x|} + e^{-x}} \). Let's start by understanding how this function behaves by looking at its components.

Step 2: We must determine if the function is one-to-one (injective) and onto (surjective).

• A function is one-to-one (injective) if each input yields a unique output; that is, if \( f(a) = f(b) \) implies \( a = b \).
• A function is onto (surjective) if for every possible output in the codomain \( \mathbb{R} \), there exists an input \( x \) such that \( f(x) = y \).

Step 3: Analyze the function for injectivity:

• For \( f(x) \) to be injective, distinct inputs must produce distinct outputs. However, the term \( e^{|x|} \) causes the function to give the same results for positive and negative \( x \) values, therefore violating injectivity. For instance, \( f(-x) = f(x) \), indicating the function isn't one-to-one.

Step 4: Analyze the function for surjectivity:

• The function \( f(x) \) has a restricted range. The output is bounded between \(-1\) and \( 1 \) because both \( e^{|x|} \) and \( e^{-x} \) increase exponentially, keeping the output within this range. As a result, the function cannot cover all of \( \mathbb{R} \), and thus it's not onto.

Step 5: Consequently, the function is neither one-to-one nor onto.