Let g(x)=f(sin x)+ f(cosx), then g(x) is decreasing on:

Options

A. `(pi/4,pi/2)`
B. `(0,pi/4)`
C. `(0,pi/2)`
D. `(pi/2,pi/2)`

Explanation

To determine where the function \( g(x) = f(\sin x) + f(\cos x) \) is decreasing, we need to analyze the derivative \( g'(x) \).

### Step 1: Find the derivative \( g'(x) \)
Using the chain rule, we differentiate \( g(x) \):
\[ g'(x) = f'(\sin x) \cdot \cos x + f'(\cos x) \cdot (-\sin x) \]
This simplifies to:
\[ g'(x) = f'(\sin x) \cos x - f'(\cos x) \sin x \]

### Step 2: Determine when \( g'(x) < 0 \)
For \( g(x) \) to be decreasing, we need \( g'(x) < 0 \):
\[ f'(\sin x) \cos x < f'(\cos x) \sin x \]

### Step 3: Analyze the intervals
We need to analyze the behavior of \( g'(x) \) in the interval \( [0, \frac{\pi}{2}] \):
- At \( x = 0 \):
\[ g'(0) = f'(\sin 0) \cos 0 - f'(\cos 0) \sin 0 = f'(0) \cdot 1 - f'(1) \cdot 0 = f'(0) \]
- At \( x = \frac{\pi}{4} \):
\[ g'\left(\frac{\pi}{4}\right) = f'\left(\frac{1}{\sqrt{2}}\right) \cdot \frac{1}{\sqrt{2}} - f'\left(\frac{1}{\sqrt{2}}\right) \cdot \frac{1}{\sqrt{2}} = 0 \]
- At \( x = \frac{\pi}{2} \):
\[ g'\left(\frac{\pi}{2}\right) = f'(\sin \frac{\pi}{2}) \cdot 0 - f'(\cos \frac{\pi}{2}) \cdot 1 = -f'(0) \]

### Step 4: Determine the intervals of decrease
From the analysis:
- \( g'(x) > 0 \) in the interval \( (0, \frac{\pi}{4}) \)
- \( g'(x) < 0 \) in the interval \( (\frac{\pi}{4}, \frac{\pi}{2}) \)

Thus, \( g(x) \) is decreasing on the interval \( \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \).

### Conclusion
The function \( g(x) \) is decreasing on the interval \( \left( 0, \frac{\pi}{4} \right) \).