Let `g'(x)gt 0 and f'(x) lt 0 AA x in R`, then

Options

A. `g(f(x+1))gt g(f(x-1))`
B. `f(g(x-1)) lt f(g(x+1))`
C. `g(f(x+1)) lt g(f(x-1))`
D. `g(g'(x+1)) lt g(g(x+1))`

Answer

To solve the problem, we need to analyze the implications of the given conditions: \( g'(x) > 0 \) (which means \( g(x) \) is an increasing function) and \( f'(x) < 0 \) (which means \( f(x) \) is a decreasing function) for all \( x \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Understanding the Functions**:
- Since \( g'(x) > 0 \), \( g(x) \) is an increasing function. This means that as \( x \) increases, \( g(x) \) also increases.
- Since \( f'(x) < 0 \), \( f(x) \) is a decreasing function. This means that as \( x \) increases, \( f(x) \) decreases.

2. **Comparing \( g(f(x + 1)) \) and \( g(f(x - 1)) \)**:
- Since \( f(x) \) is decreasing, we have \( f(x + 1) < f(x) < f(x - 1) \).
- Therefore, \( f(x + 1) < f(x - 1) \).
- Since \( g(x) \) is increasing, applying \( g \) to both sides gives us:
\[ g(f(x + 1)) < g(f(x - 1)) \]
- This means that \( g(f(x + 1)) < g(f(x - 1)) \).

3. **Comparing \( f(g(x + 1)) \) and \( f(g(x - 1)) \)**:
- Since \( g(x) \) is increasing, we have \( g(x - 1) < g(x) < g(x + 1) \).
- Therefore, \( g(x - 1) < g(x + 1) \).
- Since \( f(x) \) is decreasing, applying \( f \) to both sides gives us:
\[ f(g(x - 1)) > f(g(x + 1)) \]
- This means that \( f(g(x - 1)) > f(g(x + 1)) \).

4. **Comparing \( g(g'(x + 1)) \) and \( g(g(x + 1)) \)**:
- Since \( g'(x) > 0 \), \( g'(x) \) is increasing.
- Therefore, \( g'(x + 1) > g'(x) \).
- Since \( g(x) \) is increasing, we can conclude that:
\[ g(g'(x + 1)) > g(g'(x)) \]

5. **Conclusion**:
- From the analysis above, we can summarize the relationships:
- \( g(f(x + 1)) < g(f(x - 1)) \)
- \( f(g(x - 1)) > f(g(x + 1)) \)
- \( g(g'(x + 1)) > g(g'(x)) \)

### Final Answer:
Based on the comparisons made, we can conclude that the correct option is **C**.