The S-I model is used to study the spread of infectious diseases across different scenarios. In this question, it is assumed that we are studying a homogeneous population in a closed community of constant size $N$ throughout the period of consideration. The population is divided into two groups - one group of infected individuals who have the diseases and another group of healthy and susceptible individuals who becomes infected when they come into some form of contact with the other group of infected individuals. Using $x$ and $y$ to represent the number of infected individuals and number of healthy and susceptible individuals at time $t$ respectively, we can model the situation using the following differential equation
$\frac{\text{d}x}{\text{d}t} = kxy, \quad\quad\quad\quad \text{(I)}$
where $k$ is a positive constant.

1. Express $y$ in terms of $N$ and $x$, and rewrite the differential equation.

Since the population size is constant at $N$, the total number of individuals is the sum of infected and healthy individuals:
$x + y = N$
$y = N - x$
Substituting this into equation (I):
$\frac{\text{d}x}{\text{d}t} = kx(N - x)$

2. Separate the variables to prepare for integration.

$\frac{1}{x(N - x)} \frac{\text{d}x}{\text{d}t} = k$
$\int \frac{1}{x(N - x)} \,\text{d}x = \int k \,\text{d}t$

3. Use partial fractions to split the integrand.

Let $\frac{1}{x(N - x)} = \frac{P}{x} + \frac{Q}{N - x}$
$1 = P(N - x) + Qx$
Comparing coefficients or substituting values for $x$ gives $P = \frac{1}{N}$ and $Q = \frac{1}{N}$.
$\frac{1}{N} \int \left( \frac{1}{x} + \frac{1}{N - x} \right) \,\text{d}x = \int k \,\text{d}t$

4. Integrate both sides and combine logarithmic terms.

$\frac{1}{N} \left( \ln x - \ln(N - x) \right) = kt + C$
Note: Modulus signs are not strictly necessary as $0 < x < N$ (the number of infected is positive and cannot exceed the total population).
$\frac{1}{N} \ln \left( \frac{x}{N - x} \right) = kt + C$
$\ln \left( \frac{x}{N - x} \right) = Nkt + NC$

5. Convert to exponential form to isolate the fraction.

$\frac{x}{N - x} = \text{e}^{Nkt + NC}$
$\frac{x}{N - x} = D\text{e}^{Nkt}$
where $D = \text{e}^{NC}$ is an arbitrary constant.

6. Substitute the initial condition to find the constant $D$.

Given that $x = \alpha$ when $t = 0$:
$\frac{\alpha}{N - \alpha} = D\text{e}^{0}$
$D = \frac{\alpha}{N - \alpha}$
Substituting $D$ back into the equation:
$\frac{x}{N - x} = \left( \frac{\alpha}{N - \alpha} \right) \text{e}^{Nkt}$

7. Rearrange the equation to make $x$ the subject in the required format.

Taking the reciprocal on both sides:
$\frac{N - x}{x} = \left( \frac{N - \alpha}{\alpha} \right) \text{e}^{-Nkt}$
$\frac{N}{x} - 1 = \left( \frac{N}{\alpha} - 1 \right) \text{e}^{-Nkt}$
$\frac{N}{x} = 1 + \left( \frac{N}{\alpha} - 1 \right) \text{e}^{-Nkt}$
$x = \frac{N}{1 + \left( \frac{N}{\alpha} - 1 \right)\text{e}^{-Nkt}}$

8. State the values of constants $A$ and $B$.

Comparing with $x = \frac{A}{1 + (B - 1)\text{e}^{-Nkt}}$:
$A = N$
$B - 1 = \frac{N}{\alpha} - 1 \implies B = \frac{N}{\alpha}$

1. Determine the values of constants A, B, and the exponent coefficient from the given data.

From part (i), $A = N$ and $B = \frac{N}{\alpha}$.
Given $\alpha = 1$ and $N = 206$:
$A = 206$
$B = \frac{206}{1} = 206$
The coefficient for $t$ is $Nk = 206 \times (8.1835 \times 10^{-4}) = 0.1685801 \approx 0.169$ (to 3 s.f.).
The equation of the solution curve is:
$x = \frac{206}{1 + 205\text{e}^{-0.169t}}$

2. Identify key features to sketch the relevant part of the solution curve.

The context considers time $t \ge 0$.
At $t = 0$, $x = 1$. The curve starts at $(0, 1)$.
As $t \to \infty$, $\text{e}^{-0.169t} \to 0$, so $x \to 206$. There is a horizontal asymptote at $x = 206$.
Since $x$ models cumulative infections, it is strictly increasing.
The sketch will have a horizontal axis labeled $t$ and a vertical axis labeled $x$. It will show a logistic 'S' shape in the first quadrant, starting at $(0, 1)$ and flattening out as it approaches the dotted line $x = 206$.

3. State the long-term behavior of $x$ for large $t$.

As $t \to \infty$, the term $\text{e}^{-0.169t}$ tends to $0$.
$x = \frac{206}{1 + 0} = 206$
For large values of $t$, $x$ approaches $206$. (The entire population eventually becomes infected.)