Function f : `R rarr R`, f(x) `= x + |x|`, is

Options

A. One-one
B. Onto
C. One-one onto
D. Many one

Answer

To determine the nature of the function \( f(x) = x + |x| \), we will analyze it step by step.

### Step 1: Understand the function definition
The function is defined as:
\[ f(x) = x + |x| \]
The absolute value function \( |x| \) behaves differently depending on whether \( x \) is positive or negative.

### Step 2: Break down the function based on the sign of \( x \)
1. **Case 1: \( x \geq 0 \)**
- Here, \( |x| = x \)
- Thus, \( f(x) = x + x = 2x \)

2. **Case 2: \( x < 0 \)**
- Here, \( |x| = -x \)
- Thus, \( f(x) = x - x = 0 \)

### Step 3: Write the piecewise function
From the above cases, we can express \( f(x) \) as a piecewise function:
\[ f(x) = \begin{cases} 2x & \text{if } x \geq 0 \\ 0 & \text{if } x < 0 \end{cases} \]

### Step 4: Analyze the function for injectivity (one-one)
To check if the function is one-one, we can look at the outputs for different inputs:
- For \( x \geq 0 \): The function \( f(x) = 2x \) is clearly one-one since it is a linear function with a positive slope.
- For \( x < 0 \): The function outputs \( f(x) = 0 \) for all negative \( x \).

Since multiple values of \( x < 0 \) yield the same output \( f(x) = 0 \), the function is not one-one.

### Step 5: Analyze the function for surjectivity (onto)
To check if the function is onto, we need to compare the range of \( f(x) \) with its codomain (which is \( \mathbb{R} \)):
- The range of \( f(x) \) is:
- For \( x \geq 0 \), \( f(x) \) takes values from \( 0 \) to \( +\infty \) (since \( 2x \) can take any non-negative value).
- For \( x < 0 \), \( f(x) = 0 \).

Thus, the range of \( f(x) \) is \( [0, +\infty) \).

Since the codomain is \( \mathbb{R} \) (which includes negative values), and the range does not cover all real numbers, the function is not onto.

### Conclusion
Based on the analysis:
- The function is **not one-one** (many values of \( x < 0 \) map to the same output).
- The function is **not onto** (the range does not cover all real numbers).

Therefore, we conclude that the function \( f(x) = x + |x| \) is a **many-one function**.