We study the evolution of the population of an animal species within a nature reserve. The numbers of this population were recorded in different years. The collected data are presented in the following table:

\begin{center}
\begin{tabular}{|l|c|c|c|c|}
\hline
Year & 2000 & 2005 & 2010 & 2015 \\
\hline
Number of individuals & 50 & 64 & 80 & 100 \\
\hline
\end{tabular}
\end{center}

To anticipate the evolution of this population, the reserve management chose to model the number of individuals as a function of time. For this, it uses a function, defined on the interval $[0; +\infty[$, where the variable $x$ represents the elapsed time, in years, from the year 2000. In its model, the image of 0 by this function equals 50, which corresponds to the number of individuals in the year 2000.

\textbf{Part A. Model 1}

In this part, the reserve management makes the hypothesis that the function sought satisfies the following differential equation:
$$y' = 0.05y - 0.5 \quad (E_1)$$
\begin{enumerate}
  \item Solve the differential equation $(E_1)$ with the initial condition $y(0) = 50$.
  \item Compare the results in the table with those that would be obtained with this model.
\end{enumerate}

\textbf{Part B. Model 2}

In this part, the reserve management makes the hypothesis that the function sought satisfies the following differential equation:
$$y' = 0.05y(1 - 0.00125y)$$

Let $f$ be the function defined on $[0; +\infty[$ by:
$$f(x) = \frac{800}{1 + 15\mathrm{e}^{-0.05x}}$$
and $C$ its representative curve in an orthonormal coordinate system.

Using computer algebra software, the following results were obtained. For the rest of the exercise, these results may be used without proof, except for question 5.

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
 & Instruction & Result \\
\hline
1 & $f(x) := \frac{800}{1 + 15\mathrm{e}^{-0.05x}}$ & $f(x) = \frac{800}{1 + 15\mathrm{e}^{-0.05x}}$ \\
\hline
2 & $f'(x) :=$ Derivative$(f(x))$ & $f'(x) = \frac{600\mathrm{e}^{-0.05x}}{\left(1 + 15\mathrm{e}^{-0.05x}\right)^2}$ \\
\hline
3 & $f''(x) :=$ Derivative$(f'(x))$ & $f''(x) = \frac{30\mathrm{e}^{-0.05x}}{\left(1 + 15\mathrm{e}^{-0.05x}\right)^3}\left(15\mathrm{e}^{-0.05x} - 1\right)$ \\
\hline
4 & Solve$(15\mathrm{e}^{-0.05x} - 1 \geqslant 0)$ & $x \leqslant 20\ln(15)$ \\
\hline
\end{tabular}
\end{center}

\begin{enumerate}
  \item Prove that the function $f$ satisfies $f(0) = 50$ and that for all $x \in \mathbb{R}$:
  $$f'(x) = 0.05f(x)(1 - 0.00125f(x))$$
  We admit that this function $f$ is the unique solution of $(E_2)$ taking the initial value of 50 at 0.
  \item With this new model $f$, estimate the population size in 2050. Round the result to the nearest integer.
  \item Calculate the limit of $f$ as $x \to +\infty$. What can be deduced about the curve $C$? Interpret this limit in the context of this concrete problem.
  \item Justify that the function $f$ is increasing on $[0; +\infty[$.
  \item Prove the result obtained in line 4 of the software.
  \item We admit that the growth rate of the population of this species, expressed in number of individuals per year, is modeled by the function $f'$.\\
  a. Study the convexity of the function $f$ on the interval $[0; +\infty[$ and determine the coordinates of any inflection points of the curve $C$.\\
  b. The reserve management claims: ``According to this model, the growth rate of the population of this species will increase for a little more than fifty years, then will decrease''. Is the management correct? Justify.
\end{enumerate}