This exercise is a multiple choice questionnaire. For each of the six following questions, only one of the four proposed answers is correct.

A wrong answer, multiple answers, or absence of an answer to a question earns neither points nor deducts points.

\begin{enumerate}
  \item Consider the function $g$ defined and differentiable on $]0;+\infty[$ by:
$$g(x) = \ln\left(x^2 + x + 1\right).$$
For every strictly positive real number $x$:\\
a. $g^{\prime}(x) = \frac{1}{2x+1}$\\
b. $g^{\prime}(x) = \frac{1}{x^2+x+1}$\\
c. $g^{\prime}(x) = \ln(2x+1)$\\
d. $g^{\prime}(x) = \frac{2x+1}{x^2+x+1}$

  \item The function $x \longmapsto \ln(x)$ admits as an antiderivative on $]0;+\infty[$ the function:\\
a. $x \longmapsto \ln(x)$\\
b. $x \longmapsto \frac{1}{x}$\\
c. $x \longmapsto x\ln(x) - x$\\
d. $x \longmapsto \frac{\ln(x)}{x}$

  \item Consider the sequence $(a_n)$ defined for all $n$ in $\mathbb{N}$ by:
$$a_n = \frac{1 - 3^n}{1 + 2^n}.$$
The limit of the sequence $(a_n)$ is equal to:\\
a. $-\infty$\\
b. $-1$\\
c. $1$\\
d. $+\infty$

  \item Consider a function $f$ defined and differentiable on $[-2;2]$. The variation table of the function $f^{\prime}$ derivative of the function $f$ on the interval $[-2;2]$ is given by:
\begin{center}
\begin{tabular}{ | c | c c c c | }
\hline
$x$ & $-2$ & $-1$ & $0$ & $2$ \\
\hline
variations of $f^{\prime}$ & $1$ & & & $>_{-2}^{-1}$ \\
\hline
\end{tabular}
\end{center}
The function $f$ is:\\
a. convex on $[-2;-1]$\\
b. concave on $[0;1]$\\
c. convex on $[-1;2]$\\
d. concave on $[-2;0]$

  \item The representative curve of the derivative $f^{\prime}$ of a function $f$ defined on the interval $[-2;4]$ is given above.\\
By graphical reading of the curve of $f^{\prime}$, determine the correct statement for $f$:\\
a. $f$ is decreasing on $[0;2]$\\
b. $f$ is decreasing on $[-1;0]$\\
c. $f$ admits a maximum at $1$ on $[0;2]$\\
d. $f$ admits a maximum at $3$ on $[2;4]$

  \item A stock is quoted at $57\,€$. Its value increases by $3\%$ every month.\\
The Python function \texttt{seuil()} which returns the number of months to wait for its value to exceed $200\,€$ is:\\
a.
\begin{verbatim}
def seuil() :
    m=0
    v=57
    while v < 200 :
        m=m+1
        v = v*1.03
return m
\end{verbatim}
b.
\begin{verbatim}
def seuil() :
    m=0
    v=57
    while v > 200 :
        m=m+1
        v = v*1.03
    return m
\end{verbatim}
c.
\begin{verbatim}
def seuil() :
    v=57
    for i in range (200) :
        v = v*1.03
    return v
\end{verbatim}
d.
\begin{verbatim}
def seuil() :
    m=0
    v=57
    if v<200:
        m=m+1
    else :
        v = v*1.03
    return m
\end{verbatim}
\end{enumerate}