If f(x) is a twice differentiable function such that `f'' (x) =-f,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 with the given information about the functions \( f(x) \), \( g(x) \), and \( h(x) \).

### Step 1: Understand the given functions
We know:
- \( f''(x) = -f(x) \)
- \( g(x) = f'(x) \)
- \( h(x) = f^2(x) + g^2(x) \)

### Step 2: Differentiate \( h(x) \)
We will differentiate \( h(x) \) with respect to \( x \):
\[ h'(x) = \frac{d}{dx}(f^2(x) + g^2(x)) = 2f(x)f'(x) + 2g(x)g'(x) \]
Since \( g(x) = f'(x) \), we have \( g'(x) = f''(x) \).

### Step 3: Substitute \( g(x) \) and \( g'(x) \)
Now substituting \( g(x) \) and \( g'(x) \) into the derivative:
\[ h'(x) = 2f(x)f'(x) + 2f'(x)f''(x) \]
Using \( f''(x) = -f(x) \):
\[ h'(x) = 2f(x)f'(x) + 2f'(x)(-f(x)) = 2f(x)f'(x) - 2f(x)f'(x) \]

### Step 4: Simplify \( h'(x) \)
The terms cancel out:
\[ h'(x) = 0 \]
This indicates that \( h(x) \) is a constant function.

### Step 5: Use the given value of \( h(10) \)
We are given that \( h(10) = 10 \). Since \( h(x) \) is constant, it must be equal to 10 for all \( x \):
\[ h(x) = 10 \quad \text{for all } x \]

### Step 6: Find \( h(5) \)
Since \( h(x) \) is constant:
\[ h(5) = 10 \]

Thus, the final answer is:
\[ \boxed{10} \]