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 \( f \) is:

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

Hint:

To determine whether a function is one-to-one, check if it is either strictly increasing or strictly decreasing. To check if it is onto, determine if the function can take all possible values in the codomain.

Answer 1

The Correct Option is B

We are given the function: \[ f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}} \] Let's analyze this function to determine whether it is one-to-one (injective) or onto (surjective). ### 
Step 1: Analyze if \( f(x) \) is one-to-one. For \( f(x) \) to be one-to-one, we must check if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). This means the function must not repeat any value for different inputs. - When \( x \geq 0 \), \( |x| = x \), and the function simplifies to \( f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \), which is a continuous and strictly increasing function.
- When \( x < 0 \), \( |x| = -x \), and the function becomes \( f(x) = \frac{e^{-x} - e^x}{e^{-x} + e^x} \), which is continuous and strictly decreasing for \( x < 0 \).
As the function is strictly increasing for \( x \geq 0 \) and strictly decreasing for \( x < 0 \), \( f(x) \) is one-to-one.
### 
Step 2: Analyze if \( f(x) \) is onto. For \( f(x) \) to be onto, it must take every possible value in \( \mathbb{R} \). However, we can observe the following:
- As \( x \to \infty \), \( f(x) \) approaches 1.
- As \( x \to -\infty \), \( f(x) \) approaches -1.
Thus, \( f(x) \) can only take values between -1 and 1, meaning it is not onto because it does not cover all of \( \mathbb{R} \).

 Final Answer: Option (B) One-one but not onto.