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: Define the function based on the absolute value
The absolute value function \( |x| \) can be defined as:
- \( |x| = x \) when \( x \geq 0 \)
- \( |x| = -x \) when \( x < 0 \)

Using this definition, we can rewrite \( f(x) \) in two cases:

1. **Case 1**: When \( x \geq 0 \)
\[ f(x) = x + |x| = x + x = 2x \]

2. **Case 2**: When \( x < 0 \)
\[ f(x) = x + |x| = x - x = 0 \]

### Step 2: 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 3: Analyze the function for injectivity (1-1)
A function is considered one-to-one (injective) if different inputs produce different outputs.

- For \( x \geq 0 \):
- The function \( f(x) = 2x \) is a linear function with a positive slope, which means it is strictly increasing. Therefore, for any two different values \( x_1 \) and \( x_2 \) in this range, \( f(x_1) \neq f(x_2) \).

- For \( x < 0 \):
- The function \( f(x) = 0 \) is constant. This means that for any \( x < 0 \), \( f(x) = 0 \). Therefore, multiple values of \( x \) (e.g., \( -1, -2, -3 \)) all map to the same output \( 0 \).

Since the function is constant for \( x < 0 \), it is not one-to-one overall.

### Step 4: Analyze the function for surjectivity (onto)
A function is onto (surjective) if every possible output in the codomain is covered by some input in the domain.

- The output for \( x \geq 0 \) is \( f(x) = 2x \), which covers all non-negative real numbers \( [0, \infty) \).
- The output for \( x < 0 \) is always \( 0 \).

Thus, the function does not cover negative outputs, meaning it is not onto.

### Conclusion
The function \( f(x) = x + |x| \) is:
- Not one-to-one (many-to-one) because multiple inputs (all negative \( x \)) yield the same output (0).
- Not onto because it does not cover all real numbers (specifically, it does not cover negative numbers).

Thus, the final answer is that the function is **many-one**.