If f is twice differentiable function such that
`f''(x) =-f(x)`, and `f'(x) = g(x), h(x) =[f(x)]^(2) + [g(x)]^(2)` amd h(5)=11, then h(10)=

Options

A. 22
B. 11
C. 15
D. none

Answer

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

1. \( f''(x) = -f(x) \)
2. \( f'(x) = g(x) \)
3. \( h(x) = [f(x)]^2 + [g(x)]^2 \)
4. \( h(5) = 11 \)

We need to find \( h(10) \).

### Step 1: Analyze the differential equation
The equation \( f''(x) = -f(x) \) is a second-order linear differential equation. The general solution for this type of equation is:

\[ f(x) = A \cos(x) + B \sin(x) \]

where \( A \) and \( B \) are constants determined by initial conditions.

### Step 2: Find \( g(x) \)
Since \( g(x) = f'(x) \), we differentiate \( f(x) \):

\[ f'(x) = -A \sin(x) + B \cos(x) = g(x) \]

### Step 3: Find \( h(x) \)
Now, we can express \( h(x) \):

\[ h(x) = [f(x)]^2 + [g(x)]^2 \]

Substituting \( f(x) \) and \( g(x) \):

\[ h(x) = [A \cos(x) + B \sin(x)]^2 + [-A \sin(x) + B \cos(x)]^2 \]

### Step 4: Expand \( h(x) \)
Expanding both squares:

\[ h(x) = (A^2 \cos^2(x) + 2AB \cos(x) \sin(x) + B^2 \sin^2(x)) + (A^2 \sin^2(x) - 2AB \sin(x) \cos(x) + B^2 \cos^2(x)) \]

Combining like terms:

\[ h(x) = A^2 (\cos^2(x) + \sin^2(x)) + B^2 (\sin^2(x) + \cos^2(x)) = A^2 + B^2 \]

### Step 5: Determine constants using \( h(5) = 11 \)
Since \( h(x) = A^2 + B^2 \) is constant for all \( x \), we can use the given condition:

\[ h(5) = A^2 + B^2 = 11 \]

### Step 6: Find \( h(10) \)
Since \( h(x) \) is constant for all \( x \):

\[ h(10) = A^2 + B^2 = 11 \]

### Final Answer
Thus, \( h(10) = 11 \).