Let `f(x) = (x)/(1+|x|), x in R`, then f is

Options

A. One-one
B. Even
C. Decreasing
D. Many one

Answer

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

### Step 1: Check if the function is one-to-one (injective)

To check if \( f \) is one-to-one, we need to verify if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

Assume \( f(x_1) = f(x_2) \):
\[ \frac{x_1}{1 + |x_1|} = \frac{x_2}{1 + |x_2|} \]

Cross-multiplying gives:
\[ x_1(1 + |x_2|) = x_2(1 + |x_1|) \]

Expanding both sides:
\[ x_1 + x_1 |x_2| = x_2 + x_2 |x_1| \]

Rearranging terms:
\[ x_1 - x_2 = x_2 |x_1| - x_1 |x_2| \]

Now, if we assume \( x_1 \neq x_2 \), we can analyze the implications of the right side. If \( x_1 \) and \( x_2 \) have the same sign, we can factor out terms, leading to a contradiction. Thus, we conclude that \( x_1 = x_2 \).

**Conclusion**: The function \( f \) is one-to-one.

### Step 2: Check if the function is onto (surjective)

To check if \( f \) is onto, we need to see if every \( y \) in the codomain has a corresponding \( x \) in the domain such that \( f(x) = y \).

Let \( y = f(x) \):
\[ y = \frac{x}{1 + |x|} \]

Rearranging gives:
\[ y(1 + |x|) = x \]
\[ y + y|x| = x \]

This can be rearranged to find \( |x| \):
\[ y|x| = x - y \]

This shows that for every \( y \), we can find an \( x \) such that \( f(x) = y \). Thus, the function is onto.

**Conclusion**: The function \( f \) is onto.

### Step 3: Check if the function is increasing or decreasing

To check if the function is increasing or decreasing, we can find the derivative \( f'(x) \).

Using the quotient rule:
\[ f'(x) = \frac{(1 + |x|)(1) - x \cdot \frac{d}{dx}(|x|)}{(1 + |x|)^2} \]

Calculating \( \frac{d}{dx}(|x|) \):
- For \( x \geq 0 \), \( \frac{d}{dx}(|x|) = 1 \)
- For \( x < 0 \), \( \frac{d}{dx}(|x|) = -1 \)

Thus, we can analyze the derivative in two cases:

1. **For \( x \geq 0 \)**:
\[ f'(x) = \frac{(1 + x)(1) - x(1)}{(1 + x)^2} = \frac{1}{(1 + x)^2} > 0 \]

2. **For \( x < 0 \)**:
\[ f'(x) = \frac{(1 - x)(-1) - x(-1)}{(1 - x)^2} = \frac{-1 + x + x}{(1 - x)^2} = \frac{-1 + 2x}{(1 - x)^2} \]
This derivative is negative for \( x < 0 \).

**Conclusion**: The function is increasing for \( x \geq 0 \) and decreasing for \( x < 0 \).

### Final Answer

The function \( f(x) = \frac{x}{1 + |x|} \) is one-to-one and onto, but it is not strictly increasing or decreasing across its entire domain.