Let the function `f(x)=x^(2)+x+sinx- cosx+log(1+|x|)` be defined on the interval [0,1]. The odd extension of f(x) to the interval [-1,1] is

Options

A. `x^(2)+x+sinx+cosx-log(1+|x|)`
B. `-x^(2)+x+sinx+cosx-log(1+|x|)`
C. `-x^(2)+x+sinx-cosx+log(1+|x|)`
D. none of these

Answer

To find the odd extension of the function \( f(x) = x^2 + x + \sin x - \cos x + \log(1 + |x|) \) defined on the interval \([0, 1]\), we will follow these steps:

### Step 1: Find \( f(-x) \)
We need to substitute \(-x\) into the function \( f(x) \):

\[ f(-x) = (-x)^2 + (-x) + \sin(-x) - \cos(-x) + \log(1 + |-x|) \]

Calculating each term:
- \((-x)^2 = x^2\)
- \(-x = -x\)
- \(\sin(-x) = -\sin x\)
- \(-\cos(-x) = -\cos x\)
- \(|-x| = |x|\)

Putting it all together:

\[ f(-x) = x^2 - x - \sin x - \cos x + \log(1 + |x|) \]

### Step 2: Find \(-f(-x)\)
Now we need to find \(-f(-x)\):

\[ -f(-x) = -\left(x^2 - x - \sin x - \cos x + \log(1 + |x|)\right) \]

Distributing the negative sign:

\[ -f(-x) = -x^2 + x + \sin x + \cos x - \log(1 + |x|) \]

### Step 3: Define the odd extension
The odd extension \( g(x) \) of the function \( f(x) \) is defined as:

\[ g(x) = \begin{cases} f(x) & \text{for } x \in [0, 1] \\ -f(-x) & \text{for } x \in [-1, 0] \end{cases} \]

Thus, for \( x \in [-1, 0] \):

\[ g(x) = -x^2 + x + \sin x + \cos x - \log(1 + |x|) \]

### Final Result
The odd extension of \( f(x) \) to the interval \([-1, 1]\) is:

\[ g(x) = \begin{cases} x^2 + x + \sin x - \cos x + \log(1 + |x|) & \text{for } x \in [0, 1] \\ -x^2 + x + \sin x + \cos x - \log(1 + |x|) & \text{for } x \in [-1, 0] \end{cases} \]