A function f,R `rarr` R given by f(x)=x+`sqrt(x^(2))` is

Options

A. injective
B. surjective
C. bijective
D. none of these

Answer

To determine the nature of the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = x + \sqrt{x^2} \), we will analyze whether it is injective (one-to-one), surjective (onto), or bijective (both).

### Step 1: Simplify the function
The function can be simplified as follows:
\[ f(x) = x + \sqrt{x^2} \]
The term \( \sqrt{x^2} \) can be expressed as \( |x| \) (the absolute value of \( x \)). Therefore, we can rewrite the function:
\[ f(x) = x + |x| \]

### Step 2: Analyze the function for different cases of \( x \)
We will consider two cases based on the value of \( x \):

**Case 1: \( x \geq 0 \)**
- If \( x \) is non-negative, then \( |x| = x \).
- Thus, \( f(x) = x + x = 2x \).

**Case 2: \( x < 0 \)**
- If \( x \) is negative, then \( |x| = -x \).
- Thus, \( f(x) = x - x = 0 \).

### Step 3: Determine the output of the function
From the above analysis, we can summarize the function as:
\[ f(x) = \begin{cases} 0 & \text{if } x < 0 \\ 2x & \text{if } x \geq 0 \end{cases} \]

### Step 4: Check if the function is injective (one-to-one)
A function is injective if different inputs produce different outputs.

- For \( x < 0 \), \( f(x) = 0 \) for all \( x < 0 \). Thus, it is not injective because multiple values of \( x \) yield the same output (0).
- For \( x \geq 0 \), \( f(x) = 2x \) is injective since for every distinct \( x_1, x_2 \geq 0 \), \( f(x_1) \neq f(x_2) \).

Since the function is not injective overall, we conclude that it is not a one-to-one function.

### Step 5: Check if the function is surjective (onto)
A function is surjective if every element in the codomain has a pre-image in the domain.

- The codomain of \( f \) is all real numbers \( \mathbb{R} \).
- The range of \( f \) is \( [0, \infty) \) because:
- For \( x < 0 \), \( f(x) = 0 \).
- For \( x \geq 0 \), \( f(x) \) can take any value from \( 0 \) to \( \infty \).

Since the range \( [0, \infty) \) does not cover all real numbers (it does not include negative numbers), the function is not surjective.

### Conclusion
Since the function \( f(x) \) is neither injective nor surjective, we conclude that it is neither bijective.

### Final Answer
The function \( f(x) = x + \sqrt{x^2} \) is neither injective nor surjective.

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