if `f:RtoR` is defined as `f(x)=e^(|x|)-e^(-x)` then the correct statements(s) is/are

Options

A. `f` is one- one onto function
B. `f` is many one into function
C. range of `f` is `[0,infty]`
D. range of `f` is `(-infty,infty)`

Explanation

To solve the problem, we need to analyze the function \( f(x) = e^{|x|} - e^{-x} \) and determine the properties of this function, specifically whether it is injective (one-to-one) or not.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function is defined as:
\[ f(x) = e^{|x|} - e^{-x} \]
We need to consider two cases for \( x \): when \( x \geq 0 \) and when \( x < 0 \) because of the absolute value.

2. **Case 1: \( x \geq 0 \)**:
In this case, \( |x| = x \), so:
\[ f(x) = e^{x} - e^{-x} \]
This simplifies to:
\[ f(x) = e^{x} - \frac{1}{e^{x}} = e^{x} - e^{-x} \]

3. **Case 2: \( x < 0 \)**:
Here, \( |x| = -x \), so:
\[ f(x) = e^{-x} - e^{-x} = e^{-x} - e^{-x} \]
This simplifies to:
\[ f(x) = e^{-x} - e^{-x} = e^{-x} - e^{-x} = 0 \]

4. **Finding the Values of \( f(x) \)**:
- For \( x \geq 0 \): \( f(x) = e^{x} - e^{-x} \) is always positive and increases as \( x \) increases.
- For \( x < 0 \): \( f(x) = 0 \).

5. **Injectivity Check**:
To check if \( f(x) \) is injective, we need to see if \( f(a) = f(b) \) implies \( a = b \).

- For \( x \geq 0 \), since \( f(x) \) is strictly increasing, if \( f(a) = f(b) \) for \( a, b \geq 0 \), then \( a = b \).
- For \( x < 0 \), \( f(x) = 0 \) for all \( x < 0 \). Therefore, any two negative values will yield the same output \( f(a) = f(b) = 0 \) for \( a, b < 0 \), which means \( f(x) \) is not injective.

6. **Conclusion**:
The function \( f(x) \) is not injective since multiple inputs (all negative numbers) yield the same output (0).

### Final Answer:
The correct statement is that \( f(x) \) is not injective.