If `F(x) and G(x)` are even and odd extensions of the functions `f(x) = x|x|+ sin|x|+ xe^x`, where `x in (0, 1), g(x) = cos|x| + x^2-x`, is where `x in (0, 1)` respectively to the ars interval `(-1, 0)` then `F(x)+G(x) `in `(-1,0)` is

Options

A. `sinx +cosx +xe^(-x)`
B. `-(sin x +cos x +xe^(-x))`
C. `-(sin x +cos x +x + xe^(-x))`
D. `-(sin x +cos x +x^(2)+xe^(-x))`

Explanation

To solve the problem, we need to find \( F(x) + G(x) \) for \( x \in (-1, 0) \), where \( F(x) \) is the even extension of \( f(x) \) and \( G(x) \) is the odd extension of \( g(x) \).

### Step-by-Step Solution:

1. **Identify the functions**:
- Given \( f(x) = x|x| + \sin|x| + xe^x \) for \( x \in (0, 1) \).
- Given \( g(x) = \cos|x| + x^2 - x \) for \( x \in (0, 1) \).

2. **Find the even extension \( F(x) \)**:
- The even extension of a function \( f(x) \) is defined as:
\[ F(x) = \begin{cases} f(x) & \text{if } x \geq 0 \\ f(-x) & \text{if } x < 0 \end{cases} \]
- For \( x < 0 \):
\[ F(x) = f(-x) = (-x)|-x| + \sin|-x| + (-x)e^{-x} \]
Since \( |-x| = -x \) for \( x < 0 \):
\[ F(x) = -x^2 + \sin(-x) - xe^{-x} = -x^2 - \sin x - xe^{-x} \]

3. **Find the odd extension \( G(x) \)**:
- The odd extension of a function \( g(x) \) is defined as:
\[ G(x) = \begin{cases} g(x) & \text{if } x \geq 0 \\ -g(-x) & \text{if } x < 0 \end{cases} \]
- For \( x < 0 \):
\[ G(x) = -g(-x) = -\left(\cos|-x| + (-x)^2 - (-x)\right) \]
Since \( |-x| = -x \):
\[ G(x) = -\left(\cos x + x^2 + x\right) = -\cos x - x^2 - x \]

4. **Combine \( F(x) \) and \( G(x) \)**:
- Now we can find \( F(x) + G(x) \):
\[ F(x) + G(x) = (-x^2 - \sin x - xe^{-x}) + (-\cos x - x^2 - x) \]
- Combine like terms:
\[ F(x) + G(x) = -2x^2 - \sin x - \cos x - x - xe^{-x} \]

5. **Final Expression**:
- Therefore, the expression for \( F(x) + G(x) \) in the interval \( (-1, 0) \) is:
\[ F(x) + G(x) = -2x^2 - \sin x - \cos x - x - xe^{-x} \]