Which of the functions defined below is one one function ?

Options

A. `f(x) = ((x^(3)+1)/(x^(4)+x^(2)+1)), (x in R)`
B. `g(x) = x + (1)/(x), (x in R^(+))`
C. `h(x) = x^(2) + 4x - 5(x in R)`
D. `f(x) = e^(-x), (x in R, x le 0)`

Answer

To determine which of the given functions is a one-one function, we will analyze each function step by step.

### Step 1: Analyze the First Function
**Function:** \( f(x) = \frac{x^3 + 1}{x^4 + x^2 + 1} \)

1. **Assume:** \( f(x_1) = f(x_2) \)
\[ \frac{x_1^3 + 1}{x_1^4 + x_1^2 + 1} = \frac{x_2^3 + 1}{x_2^4 + x_2^2 + 1} \]
2. **Cross-multiply:**
\[ (x_1^3 + 1)(x_2^4 + x_2^2 + 1) = (x_2^3 + 1)(x_1^4 + x_1^2 + 1) \]
3. **Conclusion:** This equation is complex, and we cannot directly determine if the function is one-one. We will check the derivative to see if it is always increasing or decreasing.

### Step 2: Find the Derivative of the First Function
1. **Differentiate:**
\[ f'(x) = \frac{(x^4 + x^2 + 1)(3x^2) - (x^3 + 1)(4x^3 + 2x)}{(x^4 + x^2 + 1)^2} \]
2. **Analysis:** Since the derivative is complicated, we cannot conclude if it is always positive or negative. Thus, the first function is not one-one.

### Step 3: Analyze the Second Function
**Function:** \( g(x) = x + \frac{1}{x} \) where \( x > 0 \)

1. **Assume:** \( g(x_1) = g(x_2) \)
\[ x_1 + \frac{1}{x_1} = x_2 + \frac{1}{x_2} \]
2. **Rearranging gives:**
\[ x_1 - x_2 = \frac{1}{x_2} - \frac{1}{x_1} \]
3. **Differentiate:**
\[ g'(x) = 1 - \frac{1}{x^2} \]
4. **Critical Point:** Set \( g'(x) = 0 \) to find critical points.
\[ 1 - \frac{1}{x^2} = 0 \Rightarrow x^2 = 1 \Rightarrow x = 1 \]
5. **Test intervals:**
- For \( x < 1 \), \( g'(x) < 0 \) (decreasing).
- For \( x > 1 \), \( g'(x) > 0 \) (increasing).

### Conclusion for Second Function
Since \( g(x) \) is not always increasing or decreasing, it is not a one-one function.

### Step 4: Analyze the Third Function
**Function:** \( h(x) = x^2 + 4x - 5 \)

1. **Differentiate:**
\[ h'(x) = 2x + 4 \]
2. **Analysis:**
- \( h'(x) \) is always positive for all \( x \) (since \( 2x + 4 > 0 \) for all \( x \)).
- Therefore, \( h(x) \) is always increasing.

### Conclusion for Third Function
Since \( h(x) \) is always increasing, it is a one-one function.

### Step 5: Analyze the Fourth Function
**Function:** \( f(x) = e^{-x} \) where \( x \leq 0 \)

1. **Differentiate:**
\[ f'(x) = -e^{-x} \]
2. **Analysis:**
- \( f'(x) < 0 \) for all \( x \), indicating that \( f(x) \) is always decreasing.

### Conclusion for Fourth Function
Since \( f(x) \) is always decreasing, it is a one-one function.

### Final Answer
The one-one functions from the options provided are:
- **Third Function:** \( h(x) = x^2 + 4x - 5 \)
- **Fourth Function:** \( f(x) = e^{-x} \)