Which of the following function(s) has/have the same range ?

Options

A. A. `f(x) = (1)/(1 + x)`
B. B. `f(x) = (1)/(1+x^(2))`
C. C. `f(x) = (1)/(1+ sqrt(x))`
D. D. `f(x) = (1)/(sqrt(3-x))`

Answer

To determine which of the given functions have the same range, we will analyze each function step by step.

### Given Functions:
1. \( f(x) = \frac{1}{1+x} \)
2. \( f(x) = \frac{1}{1+x^2} \)
3. \( f(x) = \frac{1}{1+\sqrt{x}} \)
4. \( f(x) = \frac{1}{\sqrt{3-x}} \)

### Step 1: Analyze the first function \( f(x) = \frac{1}{1+x} \)

Let \( y = \frac{1}{1+x} \).

Rearranging gives:
\[ 1 + x = \frac{1}{y} \implies x = \frac{1}{y} - 1 \]
For this function to be defined, \( y \) cannot be zero, hence:
\[ y \neq 0 \]
As \( x \) approaches \( -1 \), \( y \) approaches \( 1 \) (from the left). As \( x \) approaches \( \infty \), \( y \) approaches \( 0 \) (from the right). Therefore, the range is:
\[ y \in (0, 1) \]

### Step 2: Analyze the second function \( f(x) = \frac{1}{1+x^2} \)

Let \( y = \frac{1}{1+x^2} \).

Rearranging gives:
\[ 1 + x^2 = \frac{1}{y} \implies x^2 = \frac{1}{y} - 1 \]
For this function to be defined, \( y \) must be positive and \( \frac{1}{y} - 1 \geq 0 \) which implies:
\[ y < 1 \quad \text{and} \quad y > 0 \]
Thus, the range is:
\[ y \in (0, 1) \]

### Step 3: Analyze the third function \( f(x) = \frac{1}{1+\sqrt{x}} \)

Let \( y = \frac{1}{1+\sqrt{x}} \).

Rearranging gives:
\[ 1 + \sqrt{x} = \frac{1}{y} \implies \sqrt{x} = \frac{1}{y} - 1 \]
For this function to be defined, \( y \) must be positive and \( \frac{1}{y} - 1 \geq 0 \) which gives:
\[ y < 1 \quad \text{and} \quad y > 0 \]
Thus, the range is:
\[ y \in (0, 1) \]

### Step 4: Analyze the fourth function \( f(x) = \frac{1}{\sqrt{3-x}} \)

Let \( y = \frac{1}{\sqrt{3-x}} \).

Rearranging gives:
\[ \sqrt{3-x} = \frac{1}{y} \implies 3 - x = \frac{1}{y^2} \implies x = 3 - \frac{1}{y^2} \]
For this function to be defined, \( 3 - x > 0 \) implies \( x < 3 \). Thus, \( y \) must be positive, and \( \sqrt{3-x} \) cannot be zero:
\[ y > 0 \]
As \( x \) approaches \( 3 \), \( y \) approaches \( \infty \). Therefore, the range is:
\[ y \in (0, \infty) \]

### Conclusion:
- The ranges for the functions are:
- \( f(x) = \frac{1}{1+x} \): \( (0, 1) \)
- \( f(x) = \frac{1}{1+x^2} \): \( (0, 1) \)
- \( f(x) = \frac{1}{1+\sqrt{x}} \): \( (0, 1) \)
- \( f(x) = \frac{1}{\sqrt{3-x}} \): \( (0, \infty) \)

The functions that have the same range are:
- **B** and **C**: \( f(x) = \frac{1}{1+x^2} \) and \( f(x) = \frac{1}{1+\sqrt{x}} \).