Let f be a twice differentiable function such that
`f''(x)=-f(x)" and "f'(x)=g(x)."If "h'(x)=[f(x)]^(2)+[g(x)]^(2),`
`h(1)=8" and "h(0)=2," then "h(2)=`

Options

A. 1
B. 2
C. 3
D. none of these

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the given information
We know that:
- \( f''(x) = -f(x) \)
- \( f'(x) = g(x) \)
- \( h'(x) = [f(x)]^2 + [g(x)]^2 \)
- \( h(1) = 8 \)
- \( h(0) = 2 \)

### Step 2: Differentiate \( h'(x) \)
We start by differentiating \( h'(x) \):
\[ h'(x) = [f(x)]^2 + [g(x)]^2 \]
Using the product rule, we differentiate:
\[ h''(x) = 2f(x)f'(x) + 2g(x)g'(x) \]
Substituting \( f'(x) = g(x) \):
\[ h''(x) = 2f(x)g(x) + 2g(x)g'(x) \]

### Step 3: Find \( g'(x) \)
From \( f''(x) = -f(x) \) and knowing that \( f'(x) = g(x) \), we differentiate \( g(x) \):
\[ g'(x) = f''(x) = -f(x) \]
Substituting this into our expression for \( h''(x) \):
\[ h''(x) = 2f(x)g(x) + 2g(x)(-f(x)) = 2f(x)g(x) - 2f(x)g(x) = 0 \]
Thus, we find:
\[ h''(x) = 0 \]

### Step 4: Integrate \( h''(x) \)
Since \( h''(x) = 0 \), we can integrate to find \( h'(x) \):
\[ h'(x) = C \]
where \( C \) is a constant.

### Step 5: Integrate \( h'(x) \)
Integrating again gives:
\[ h(x) = Cx + C_1 \]
where \( C_1 \) is another constant.

### Step 6: Use the initial conditions to find constants
Using \( h(0) = 2 \):
\[ h(0) = C(0) + C_1 = C_1 = 2 \]

Now using \( h(1) = 8 \):
\[ h(1) = C(1) + C_1 = C + 2 = 8 \]
This simplifies to:
\[ C = 8 - 2 = 6 \]

### Step 7: Write the final expression for \( h(x) \)
Now we have:
\[ h(x) = 6x + 2 \]

### Step 8: Calculate \( h(2) \)
To find \( h(2) \):
\[ h(2) = 6(2) + 2 = 12 + 2 = 14 \]

### Final Answer
Thus, the value of \( h(2) \) is:
\[ \boxed{14} \]