If f(x) is a twice differentiable function such that `f'' (x) =-f(x),f'(x)=g(x),h(x)=f^2(x)+g^2(x) and h(10)=10` , then h (5) is equal to

Options

A. 5
B. 15
C. 10
D. 17

Answer

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

1. **Given Information**:
- \( f''(x) = -f(x) \)
- \( f'(x) = g(x) \)
- \( h(x) = f^2(x) + g^2(x) \)
- \( h(10) = 10 \)

2. **Differentiate \( h(x) \)**:
We need to find \( h(5) \). First, we differentiate \( h(x) \):
\[ h'(x) = \frac{d}{dx}(f^2(x) + g^2(x)) \]
Using the chain rule:
\[ h'(x) = 2f(x)f'(x) + 2g(x)g'(x) \]

3. **Substituting \( g(x) \) and \( g'(x) \)**:
Since \( g(x) = f'(x) \), we can find \( g'(x) \):
\[ g'(x) = f''(x) \]
From the problem, we know \( f''(x) = -f(x) \). Thus:
\[ g'(x) = -f(x) \]

4. **Substituting back into \( h'(x) \)**:
Now, substitute \( g(x) \) and \( g'(x) \) back into \( h'(x) \):
\[ h'(x) = 2f(x)f'(x) + 2f'(x)(-f(x)) \]
This simplifies to:
\[ h'(x) = 2f(x)f'(x) - 2f(x)f'(x) = 0 \]

5. **Conclusion about \( h(x) \)**:
Since \( h'(x) = 0 \), this indicates that \( h(x) \) is a constant function. Therefore, \( h(x) \) does not change with \( x \).

6. **Using the given value of \( h(10) \)**:
We know that \( h(10) = 10 \). Since \( h(x) \) is constant:
\[ h(5) = h(10) = 10 \]

Thus, the final answer is:
\[ h(5) = 10 \]