Let`f'(sin x)lt0 and f''(sin x) gt0 forall x in (0,(pi)/(2))` and g(x) =f(sinx)+f(cosx)
If x = 3 is the only point of minima in its neighborhood and x=4 is neither a point of maxima nor a point minima, then which of the following can be true?

Options

A. `a gt 0, b lt 0`
B. `a lt , b lt 0`
C. `a gt 0, b in R`
D. none of these

Explanation

To solve the problem, we need to analyze the given conditions about the function \( f \) and its derivatives, as well as the function \( g(x) = f(\sin x) + f(\cos x) \).

### Step-by-Step Solution:

1. **Understanding the Conditions**:
- We are given that \( f'(\sin x) < 0 \) for all \( x \) in \( (0, \frac{\pi}{2}) \). This implies that \( f \) is a decreasing function on the interval \( (0, 1) \) since \( \sin x \) varies from 0 to 1 in this interval.
- We are also given that \( f''(\sin x) > 0 \) for all \( x \) in \( (0, \frac{\pi}{2}) \). This means that \( f' \) is increasing on the interval \( (0, 1) \).

2. **Analyzing the Function \( g(x) \)**:
- The function \( g(x) = f(\sin x) + f(\cos x) \) combines the values of \( f \) at \( \sin x \) and \( \cos x \).
- Since \( \sin x \) and \( \cos x \) are both decreasing in the interval \( (0, \frac{\pi}{2}) \), and \( f \) is decreasing, \( g(x) \) will also have properties influenced by the behavior of \( f \).

3. **Identifying Minima and Maxima**:
- We know that \( x = 3 \) is a point of minima in its neighborhood, meaning \( g'(3) = 0 \) and \( g''(3) > 0 \).
- We also know that \( x = 4 \) is neither a point of maxima nor minima, indicating that \( g'(4) \neq 0 \).

4. **Finding Values of \( f \)**:
- Since \( x = 3 \) is a local minimum, we can express the continuity condition:
\[ f(3^-) = f(3^+) \implies f(3) = 2a + b \]
- For \( x = 4 \), since it is neither a maximum nor a minimum, we can express:
\[ f(4^+) = f(4^-) \implies f(4) = 4a + b \]

5. **Setting Up Equations**:
- From the conditions, we can set up equations based on the values of \( f \):
- For \( x = 3 \): \( 2a + b = 3 \)
- For \( x = 4 \): \( 4a + b = -b \) or \( 2a + b = 3 \)

6. **Solving the Equations**:
- From the equations, we can deduce:
- \( 2a + b = 3 \)
- \( 4a + b = -b \) leads to \( 4a + 2b = 0 \) or \( 2a + b = 0 \).
- Solving these equations simultaneously gives us the values of \( a \) and \( b \).

7. **Conclusion**:
- After solving the equations, we find that the function \( f(x) \) is continuous but not differentiable at the points we analyzed. This leads us to conclude that \( a > b \) and \( b < 0 \).