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 type of function \( f(x) = \frac{x}{1 + |x|} \), we will analyze it step by step.

### Step 1: Understand the function definition
The function is defined as:
\[ f(x) = \frac{x}{1 + |x|} \]
where \( x \) is a real number. The absolute value \( |x| \) affects the function differently depending on whether \( x \) is positive or negative.

### Step 2: Analyze the function for \( x \geq 0 \)
When \( x \geq 0 \), the absolute value \( |x| \) is simply \( x \). Thus, the function simplifies to:
\[ f(x) = \frac{x}{1 + x} \]

### Step 3: Analyze the function for \( x < 0 \)
When \( x < 0 \), the absolute value \( |x| \) is \( -x \). Therefore, the function becomes:
\[ f(x) = \frac{x}{1 - x} \]

### Step 4: Check if the function is one-to-one (injective)
To determine if \( f \) is one-to-one, we need to show that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).

#### Case 1: \( x_1, x_2 \geq 0 \)
Assuming \( f(x_1) = f(x_2) \):
\[ \frac{x_1}{1 + x_1} = \frac{x_2}{1 + x_2} \]
Cross-multiplying gives:
\[ x_1(1 + x_2) = x_2(1 + x_1) \]
Expanding both sides:
\[ x_1 + x_1 x_2 = x_2 + x_1 x_2 \]
This simplifies to:
\[ x_1 = x_2 \]

#### Case 2: \( x_1, x_2 < 0 \)
Assuming \( f(x_1) = f(x_2) \):
\[ \frac{x_1}{1 - x_1} = \frac{x_2}{1 - x_2} \]
Cross-multiplying gives:
\[ x_1(1 - x_2) = x_2(1 - x_1) \]
Expanding both sides:
\[ x_1 - x_1 x_2 = x_2 - x_1 x_2 \]
This simplifies to:
\[ x_1 = x_2 \]

#### Case 3: \( x_1 \geq 0 \) and \( x_2 < 0 \)
If \( x_1 \geq 0 \) and \( x_2 < 0 \), then \( f(x_1) \) will be non-negative (since both numerator and denominator are positive), while \( f(x_2) \) will be negative (since numerator is negative and denominator is positive). Thus, \( f(x_1) \neq f(x_2) \).

### Conclusion
Since in all cases \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), we conclude that the function \( f(x) \) is one-to-one (injective).

### Final Answer
Thus, the function \( f(x) = \frac{x}{1 + |x|} \) is **one-to-one** (injective).

---