Let `f : R to R` be defined by `f(x) =(|x| -1)/(|x|+1)` is

Options

A. onto but not one-one
B. both one-one and onto
C. one-one but not onto
D. neither one-one nor onto

Answer

To determine the properties of the function \( f(x) = \frac{|x| - 1}{|x| + 1} \), we will analyze whether it is one-to-one (injective) and onto (surjective).

### Step 1: Understand the Function
The function is defined as:
\[ f(x) = \frac{|x| - 1}{|x| + 1} \]
This function is defined for all real numbers \( x \).

### Step 2: Analyze the Function for Different Cases
Since the function involves the absolute value, we will consider two cases: when \( x \) is non-negative and when \( x \) is negative.

1. **Case 1: \( x \geq 0 \)**
- Here, \( |x| = x \).
- Thus, the function simplifies to:
\[ f(x) = \frac{x - 1}{x + 1} \]

2. **Case 2: \( x < 0 \)**
- Here, \( |x| = -x \).
- Thus, the function simplifies to:
\[ f(x) = \frac{-x - 1}{-x + 1} = \frac{-1 - x}{1 - x} \]

### Step 3: Determine Injectivity (One-to-One)
To check if \( f(x) \) is one-to-one, we need to see if it is either always increasing or always decreasing.

1. **For \( x \geq 0 \)**:
- Consider the derivative of \( f(x) = \frac{x - 1}{x + 1} \):
\[ f'(x) = \frac{(1)(x + 1) - (x - 1)(1)}{(x + 1)^2} = \frac{(x + 1) - (x - 1)}{(x + 1)^2} = \frac{2}{(x + 1)^2} \]
- Since \( f'(x) > 0 \) for all \( x \geq 0 \), the function is increasing in this interval.

2. **For \( x < 0 \)**:
- Consider the derivative of \( f(x) = \frac{-x - 1}{-x + 1} \):
\[ f'(x) = \frac{(-1)(-x + 1) - (-x - 1)(-1)}{(-x + 1)^2} = \frac{(x - 1) - (x + 1)}{(-x + 1)^2} = \frac{-2}{(-x + 1)^2} \]
- Since \( f'(x) < 0 \) for all \( x < 0 \), the function is decreasing in this interval.

Since \( f(x) \) is increasing for \( x \geq 0 \) and decreasing for \( x < 0 \), it is not one-to-one (it fails the horizontal line test).

### Step 4: Determine Surjectivity (Onto)
Next, we check if the function is onto. The range of \( f(x) \) can be determined by analyzing the limits:

1. As \( x \to 0 \):
\[ f(0) = \frac{0 - 1}{0 + 1} = -1 \]

2. As \( x \to \infty \):
\[ f(x) \to \frac{x - 1}{x + 1} \to 1 \]

3. As \( x \to -\infty \):
\[ f(x) \to \frac{-x - 1}{-x + 1} \to 1 \]

Thus, the function approaches \( -1 \) as \( x \to 0 \) and approaches \( 1 \) as \( x \to \infty \) or \( -\infty \). Therefore, the range of \( f(x) \) is \( (-1, 1) \).

### Step 5: Conclusion
The codomain of \( f(x) \) is \( \mathbb{R} \) (all real numbers), while the range is \( (-1, 1) \). Since the range is not equal to the codomain, the function is not onto.

### Final Result
The function \( f(x) = \frac{|x| - 1}{|x| + 1} \) is neither one-to-one nor onto.

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