If f is twice differentiable such that f"(x)=-f(x)
`f'(x)=g(x),h'(x)=[f(x)]^(2)+[g(x)]^(2)andh(0)=2,`
`h(1)=4,` then the equation y=h(x) represents.

Options

A. curve of degree 2
B. curve passing through origin
C. a straight line with slope 2
D. a straight line with slope `-2`

Answer

To solve the problem step by step, we will analyze the given equations and derive the necessary functions.

### Step 1: Understand the given equations
We have:
1. \( f''(x) = -f(x) \)
2. \( f'(x) = g(x) \)
3. \( h'(x) = [f(x)]^2 + [g(x)]^2 \)
4. \( h(0) = 2 \)
5. \( h(1) = 4 \)

### Step 2: Differentiate \( h(x) \)
From the equation \( h'(x) = [f(x)]^2 + [g(x)]^2 \), we can differentiate \( h'(x) \) to find \( h''(x) \):
\[ h''(x) = 2f(x)f'(x) + 2g(x)g'(x) \]
Since \( g(x) = f'(x) \), we have \( g'(x) = f''(x) \). Thus, substituting \( g'(x) \) into the equation gives:
\[ h''(x) = 2f(x)f'(x) + 2g(x)f''(x) \]

### Step 3: Substitute \( f''(x) \)
Using \( f''(x) = -f(x) \), we can rewrite the equation:
\[ h''(x) = 2f(x)f'(x) + 2g(x)(-f(x)) \]
Substituting \( g(x) = f'(x) \):
\[ h''(x) = 2f(x)f'(x) - 2f'(x)f(x) = 0 \]

### Step 4: Conclusion about \( h''(x) \)
Since \( h''(x) = 0 \), this indicates that \( h'(x) \) is a constant. Therefore, we can express \( h(x) \) as:
\[ h(x) = ax + b \]
where \( a \) and \( b \) are constants.

### Step 5: Use initial conditions to find constants
We know:
1. \( h(0) = 2 \)
2. \( h(1) = 4 \)

Using \( h(0) = 2 \):
\[ h(0) = a(0) + b = b = 2 \]

Using \( h(1) = 4 \):
\[ h(1) = a(1) + b = a + 2 = 4 \]
This gives:
\[ a + 2 = 4 \implies a = 2 \]

### Step 6: Write the final form of \( h(x) \)
Thus, we have:
\[ h(x) = 2x + 2 \]

### Step 7: Identify the representation of \( h(x) \)
The equation \( y = h(x) = 2x + 2 \) represents a straight line with a slope of 2.