Function `g(x)=2f((x^(2))/(2))+f(6-x^(2))` for all `xepsilonR` is decreasing and `f^(``)(x)gt0` for all `xepsilonR` then complete set of values of x is

Options

A. `(-2,2)`
B. `(-infty,-2)cup(0,2)`
C. `(2,infty)`
D. `(2,4)`

Explanation

To solve the problem, we need to analyze the function \( g(x) = 2f\left(\frac{x^2}{2}\right) + f(6 - x^2) \) under the conditions that \( g(x) \) is decreasing and \( f'(x) > 0 \) for all \( x \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
- The function \( g(x) \) is decreasing, which means that its derivative \( g'(x) < 0 \) for all \( x \in \mathbb{R} \).
- The function \( f \) is strictly increasing since \( f'(x) > 0 \) for all \( x \).

2. **Finding the Derivative of \( g(x) \)**:
- We compute the derivative \( g'(x) \):
\[ g'(x) = 2 \cdot f'\left(\frac{x^2}{2}\right) \cdot \left(\frac{d}{dx}\left(\frac{x^2}{2}\right)\right) + f'(6 - x^2) \cdot \left(\frac{d}{dx}(6 - x^2)\right) \]
- This simplifies to:
\[ g'(x) = 2 \cdot f'\left(\frac{x^2}{2}\right) \cdot x + f'(6 - x^2) \cdot (-2x) \]
- Thus, we have:
\[ g'(x) = 2x \left( f'\left(\frac{x^2}{2}\right) - f'(6 - x^2) \right) \]

3. **Setting the Derivative Less Than Zero**:
- For \( g(x) \) to be decreasing, we need:
\[ 2x \left( f'\left(\frac{x^2}{2}\right) - f'(6 - x^2) \right) < 0 \]
- This means either \( x > 0 \) and \( f'\left(\frac{x^2}{2}\right) < f'(6 - x^2) \) or \( x < 0 \) and \( f'\left(\frac{x^2}{2}\right) > f'(6 - x^2) \).

4. **Analyzing the Cases**:
- **Case 1**: If \( x > 0 \):
- We have \( f'\left(\frac{x^2}{2}\right) < f'(6 - x^2) \).
- Since \( f' \) is increasing, this implies:
\[ \frac{x^2}{2} < 6 - x^2 \]
- Rearranging gives:
\[ \frac{3x^2}{2} < 6 \implies x^2 < 4 \implies -2 < x < 2 \]
- Since \( x > 0 \), we have \( 0 < x < 2 \).

- **Case 2**: If \( x < 0 \):
- We have \( f'\left(\frac{x^2}{2}\right) > f'(6 - x^2) \).
- Again, since \( f' \) is increasing:
\[ \frac{x^2}{2} > 6 - x^2 \]
- Rearranging gives:
\[ \frac{3x^2}{2} > 6 \implies x^2 > 4 \implies x < -2 \text{ or } x > 2 \]
- Since \( x < 0 \), we have \( x < -2 \).

5. **Combining the Results**:
- From Case 1, we have \( x \in (0, 2) \).
- From Case 2, we have \( x \in (-\infty, -2) \).
- Therefore, the complete set of values of \( x \) for which \( g(x) \) is decreasing is:
\[ x \in (-\infty, -2) \cup (0, 2) \]

### Final Answer:
The complete set of values of \( x \) is \( (-\infty, -2) \cup (0, 2) \).