Let `F(x) =(f(x/2))^(2) +(g(x/2))^(2). F(5)=5` and `f''(x) =-f(x), g(x) = f'(x)`, then `F(10)` is equal to:

Options

A. 5
B. 10
C. 0
D. 3

Answer

To solve the problem step by step, we start with the given information:

1. **Function Definition**:
\[ F(x) = (f(x/2))^2 + (g(x/2))^2 \]
where \( g(x) = f'(x) \) and \( f''(x) = -f(x) \).

2. **Given Condition**:
\[ F(5) = 5 \]

3. **Objective**:
We need to find \( F(10) \).

### Step 1: Analyze the Function \( F(x) \)

We start by differentiating \( F(x) \):
\[ F'(x) = 2f(x/2) \cdot f'(x/2) \cdot \frac{1}{2} + 2g(x/2) \cdot g'(x/2) \cdot \frac{1}{2} \]
This simplifies to:
\[ F'(x) = f(x/2) \cdot f'(x/2) + g(x/2) \cdot g'(x/2) \]

### Step 2: Substitute \( g(x) \)

Since \( g(x) = f'(x) \), we can express \( g'(x) \) as:
\[ g'(x) = f''(x) \]
Thus, we can rewrite \( F'(x) \):
\[ F'(x) = f(x/2) \cdot f'(x/2) + f'(x/2) \cdot f''(x/2) \]

### Step 3: Substitute \( f''(x) \)

Using the condition \( f''(x) = -f(x) \):
\[ F'(x) = f(x/2) \cdot f'(x/2) + f'(x/2) \cdot (-f(x/2)) \]
This simplifies to:
\[ F'(x) = f'(x/2) \cdot (f(x/2) - f(x/2)) = 0 \]

### Step 4: Conclusion from the Derivative

Since \( F'(x) = 0 \), it indicates that \( F(x) \) is a constant function.

### Step 5: Use Given Condition

From the problem, we know:
\[ F(5) = 5 \]
Since \( F(x) \) is constant, we have:
\[ F(x) = 5 \quad \text{for all } x \]

### Step 6: Find \( F(10) \)

Thus, we conclude:
\[ F(10) = 5 \]

### Final Answer

\[ \boxed{5} \]

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