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

Options

A. a) a straight line with slope `2`
B. b) a straight line with y-intercept `1`
C. c) a straight line with x-intercept `2`
D. d) None of the above

Answer

To solve the problem step by step, let's break down the information given and derive the necessary equations.

### Step 1: Understand the given information
We have the following:
- \( f''(x) = -f(x) \)
- \( f'(x) = g(x) \)
- \( h'(x) = f(x)^2 + g(x)^2 \)
- \( h(0) = 2 \)
- \( h(1) = 4 \)

### Step 2: Differentiate \( g(x) \)
Since \( g(x) = f'(x) \), we can differentiate \( g(x) \):
\[ g'(x) = f''(x) = -f(x) \]

### Step 3: Differentiate \( h(x) \)
We know \( h'(x) = f(x)^2 + g(x)^2 \). Let's differentiate \( h'(x) \):
\[ h''(x) = 2f(x)f'(x) + 2g(x)g'(x) \]
Substituting \( g(x) = f'(x) \) and \( g'(x) = -f(x) \):
\[ h''(x) = 2f(x)f'(x) + 2f'(x)(-f(x)) = 2f(x)f'(x) - 2f(x)f'(x) = 0 \]

### Step 4: Analyze \( h''(x) = 0 \)
Since \( h''(x) = 0 \), this implies that \( h'(x) \) is a constant:
\[ h'(x) = k \]
where \( k \) is a constant.

### Step 5: Integrate to find \( h(x) \)
Integrating \( h'(x) = k \):
\[ h(x) = kx + c \]
where \( c \) is another constant.

### Step 6: Use initial conditions to find constants
We know:
1. \( h(0) = 2 \)
\[ h(0) = k(0) + c = c \implies c = 2 \]
2. \( h(1) = 4 \)
\[ h(1) = k(1) + 2 = 4 \implies k + 2 = 4 \implies k = 2 \]

### Step 7: Write the final form of \( h(x) \)
Substituting \( k \) and \( c \) back into the equation for \( h(x) \):
\[ h(x) = 2x + 2 \]

### Step 8: Identify the representation of \( y = h(x) \)
The equation \( y = h(x) = 2x + 2 \) represents a straight line with:
- Slope = 2
- Y-intercept = 2

### Conclusion
Thus, the equation \( y = h(x) \) represents a straight line with a slope of 2.

### Final Answer
The correct option is **Option A: A straight line with slope two.**