`f:RrarrR" defined by "f(x)=(1)/(2)x|x|+cos+1` is

Options

A. one-one and onto
B. one-one and into
C. many-one and onto
D. many-one and into

Explanation

To determine the nature of the function \( f(x) = \frac{1}{2} x |x| + \cos x + 1 \), we will analyze its properties step by step.

### Step 1: Understanding the function
The function is defined as:
\[ f(x) = \frac{1}{2} x |x| + \cos x + 1 \]
Here, \( \frac{1}{2} x |x| \) is a piecewise function because of the absolute value. We can rewrite it as:
- For \( x \geq 0 \): \( f(x) = \frac{1}{2} x^2 + \cos x + 1 \)
- For \( x < 0 \): \( f(x) = -\frac{1}{2} x^2 + \cos x + 1 \)

### Step 2: Finding the range of the function
To find the range, we will analyze the behavior of \( f(x) \) as \( x \) approaches positive and negative infinity.

- As \( x \to \infty \):
\[ f(x) \to \frac{1}{2} x^2 + \cos x + 1 \to \infty \]
- As \( x \to -\infty \):
\[ f(x) \to -\frac{1}{2} x^2 + \cos x + 1 \to -\infty \]

Since \( f(x) \) approaches both \( -\infty \) and \( \infty \), the range of \( f(x) \) is \( (-\infty, \infty) \).

### Step 3: Checking if the function is onto
Since the range of \( f(x) \) is \( (-\infty, \infty) \), we can conclude that the function is onto.

### Step 4: Checking if the function is one-to-one
To determine if the function is one-to-one, we need to find the derivative \( f'(x) \) and check its sign.

#### Step 4.1: Finding the derivative
For \( x \geq 0 \):
\[ f'(x) = \frac{d}{dx} \left(\frac{1}{2} x^2 + \cos x + 1\right) = x - \sin x \]
For \( x < 0 \):
\[ f'(x) = \frac{d}{dx} \left(-\frac{1}{2} x^2 + \cos x + 1\right) = -x - \sin x \]

#### Step 4.2: Analyzing the derivative
- For \( x \geq 0 \):
Since \( x \geq 0 \) and \( \sin x \) is always less than or equal to \( x \) for \( x \geq 0 \), we have:
\[ f'(x) = x - \sin x \geq 0 \]
This implies \( f'(x) > 0 \) for all \( x \geq 0 \), meaning the function is increasing in this interval.

- For \( x < 0 \):
We have \( -x \) which is positive and \( -\sin x \) which is also positive. Thus, \( f'(x) = -x - \sin x > 0 \) for all \( x < 0 \).

Since \( f'(x) > 0 \) for all \( x \), the function is strictly increasing.

### Conclusion
Since \( f(x) \) is both onto and one-to-one, we conclude that the function is a **one-to-one and onto function**.

### Final Answer
The function \( f(x) = \frac{1}{2} x |x| + \cos x + 1 \) is a **one-to-one and onto function**.

---