If the function `F: R to R ` is defined by `f(x)=|x|(x-sinx)`, then which o f the following statemnts is true ?

Options

A. f is one-one but NOT onto
B. f is onto but NOT one-one
C. fis BOT one- one and onto
D. f is NEITHER one-one NOR onto

Answer

To solve the problem, we will analyze the function \( f(x) = |x|(x - \sin x) \) step by step.

### Step 1: Understanding the Function
The function is defined as \( f(x) = |x|(x - \sin x) \). The absolute value function \( |x| \) indicates that the function will behave differently for positive and negative values of \( x \).

### Step 2: Break Down the Function
We can break the function into two cases based on the sign of \( x \):
1. **For \( x \geq 0 \)**:
\[ f(x) = x(x - \sin x) = x^2 - x \sin x \]
2. **For \( x < 0 \)**:
\[ f(x) = -x(-x - \sin(-x)) = -x(-x + \sin x) = x^2 - x \sin x \]
Notice that in both cases, the function simplifies to the same expression: \( f(x) = x^2 - x \sin x \).

### Step 3: Check for Odd Function
To check if the function is odd, we compute \( f(-x) \):
\[ f(-x) = | -x | (-x - \sin(-x)) = |x|(-x + \sin x) = x(-x + \sin x) = -x^2 + x \sin x \]
Now, we check if \( f(-x) = -f(x) \):
\[ -f(x) = -(x^2 - x \sin x) = -x^2 + x \sin x \]
Thus, \( f(-x) = -f(x) \), confirming that \( f(x) \) is an odd function.

### Step 4: Determine Increasing/Decreasing Behavior
Next, we find the derivative \( f'(x) \) to analyze the increasing or decreasing nature of the function:
\[ f'(x) = \frac{d}{dx}(x^2 - x \sin x) = 2x - (\sin x + x \cos x) \]
We will analyze \( f'(x) \):
- **For \( x > 0 \)**:
- \( f'(x) = 2x - (\sin x + x \cos x) \)
- Since \( \sin x \) and \( x \cos x \) are both less than \( 2x \) for small \( x \), \( f'(x) > 0 \) for small positive \( x \), indicating \( f(x) \) is increasing.

- **For \( x < 0 \)**:
- The derivative will have a similar form and will be negative, indicating \( f(x) \) is decreasing.

### Step 5: Conclusion on One-One and Onto
Since \( f(x) \) is increasing for \( x > 0 \) and decreasing for \( x < 0 \), it means that the function is one-to-one. Additionally, since \( f(x) \) is an odd function and covers all real values as \( x \) approaches \( \pm \infty \), it is onto as well.

### Final Answer
The function \( f(x) = |x|(x - \sin x) \) is both one-one and onto.