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\begin{document}



\title{Métropole - 20 juin 2024}
\author{Spécialité mathématiques - Baccalauréat}
\date{}



\begin{frame}
\titlepage
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Exercice 1}

\pause
\begin{enumerate}
 \item On sait que~: \pause
 \begin{itemize}
  \item 91,7~\% des étudiants ont répondu \og oui \fg{} donc~: \pause
  $$\boxed{P(Q)=0,917}$$ \pause
  \item 65~\% des étudiants ayant échoué ont répondu \og non \fg{} donc~: \pause
  $$\boxed{P_{\overline{R}}(\overline{Q})=0,65}$$
 \end{itemize}
\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item On complète l'arbre de la façon suivante~: \pause

  \bigskip

  \begin{center}
\psset{nodesep=0mm,levelsep=20mm,treesep=10mm}
\pstree[treemode=R]{\Tdot}
{
\pstree
{\Tdot~[tnpos=a]{$R$}\taput{\footnotesize $x$}}
{
\Tdot~[tnpos=r]{$Q$}\taput{\footnotesize $0,98$}
\Tdot~[tnpos=r]{$\overline{Q}$}\tbput{\footnotesize $0,02$}
}
\pstree
{\Tdot~[tnpos=a]{$\overline{R}$}\tbput{\footnotesize $1-x$}}
{
\Tdot~[tnpos=r]{$Q$}\taput{\footnotesize $0,35$}
\Tdot~[tnpos=r]{$\overline{Q}$}\tbput{\footnotesize $0,65$}
}
}
\end{center}
 \end{enumerate}
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item Les événements $R$ et $\overline{R}$ forment une partition de l'univers donc, \pause d'après la formule des probabilités totales~: \pause
  \begin{eqnarray*}
   P(Q) & = & \pause P(R)\times P_R(Q)+P(\overline{R})\times P_{\overline{R}}(Q)\\ \pause
        & = & x\times 0,98+(1-x)\times 0,35\\ \pause
        & = & 0,98x+0,35-0,35x\\ \pause
        & = & 0,63x+0,35 \pause
  \end{eqnarray*}
Or, d'après l'énoncé, on sait que $P(Q)=0,917$, \pause il s'agit alors de résoudre une équation~: \pause
\begin{eqnarray*}
 0,63x+0,35=0,917 & \Longleftrightarrow & \pause 0,63x=0,917-0,35\\ \pause
                  & \Longleftrightarrow & 0,63x=0,567\\ \pause
                  & \Longleftrightarrow & x=\dfrac{0,567}{0,63}\\ \pause
                  & \Longleftrightarrow & x=0,9 \pause
\end{eqnarray*}
\hspace{1cm} On a donc bien~:
$$\boxed{x=0,9}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Il s'agit de calculer $P_Q(R)$~: \pause
 \begin{eqnarray*}
  P_Q(R) & = & \pause \dfrac{P(R\cap Q)}{P(Q)}\\ \pause
         & = & \dfrac{P(R)\times P_R(Q)}{P(Q)}\\ \pause
         & = & \dfrac{0,9\times 0,98}{0,917}\\ \pause
         & \approx & 0,962
 \end{eqnarray*} \pause
Sachant que l'étudiant a répondu \og oui \fg{} à la question, la probabilité qu'il ait réussi l'examen est donc~: \pause
$$\boxed{P_Q(R)\approx 0,962}$$
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{3}
 \item \label{ch13_ex1_q4} Il s'agit de déterminer un entier $k$ tel que $P(N\geqslant k)\approx 0,65$. \pause On obtient, à l'aide de la calculatrice~: \pause
 $$\boxed{P(N\geqslant 12)\approx 0,65}$$ \pause
 Afin qu'environ 65~\% des étudiants soient récompensés, \pause elle doit attribuer les récompenses à partir de la note de 12.
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\vspace{-0.3cm}
\begin{enumerate}
 \setcounter{enumi}{4}
 \item La variable aléatoire $N$ suit une loi binomiale de paramètres $n=20$ et $p=0,615$. \pause Son espérance est donc $E(N)=v n\times p \pause =20\times 0,615 \pause =12,3$ \pause et sa variance $V(N)= \pause n\times p\times (1-p) \pause =20\times 0,615\times 0,385 \pause =4,7355$.\\ \pause
 Les variables aléatoires $N_1$, $N_2$, \dots, $N_{10}$ suivent toutes la même loi que $N$, on a donc~: \pause
 $$E(S)= \pause 10\times E(N) \pause =10\times 12,3 \pause =123$$ \pause
 Et comme les variables aléatoires $N_1$, $N_2$, \dots, $N_{10}$ sont indépendantes, on a~: \pause
 $$V(S)= \pause 10\times V(N) \pause =4,7355\times 10 \pause =47,355$$ \pause
 Soit~:
 $$\boxed{E(S)=123} \pause \quad \text{et} \quad \boxed{V(S)=47,355}$$
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{5}
 \item
 \begin{enumerate}[(a)]
  \item La variable aléatoire $M$ modélise~: \pause
  $$\boxed{\text{La moyenne du groupe des dix étudiants}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{5}
 \item
 \begin{enumerate}[(b)]
  \item On sait alors, d'après le cours, \pause que $E(M)= \pause E(N)$ \pause et que $V(M)= \pause \dfrac{1}{10}\times V(N)$ \pause , soit~:
  $$\boxed{E(M)=12,3} \pause \quad \text{et} \quad \boxed{V(M)=0,47355}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{5}
 \item
 \begin{enumerate}[(c)]
  \item \label{ch13_ex1_q6c} D'après l'inégalité de Bienaymé-Tchebychev, on a~: \pause
  $$P\left(|M-E(M)|\geqslant 2\right)\leqslant \dfrac{V(M)}{2^2}$$ \pause
  Soit~:
  $$P\left(|M-12,3|\geqslant 2\right)\leqslant \dfrac{0,47355}{4}$$ \pause
  Et par passage à l'événement contraire~:
  $$P\left(|M-12,3|< 2\right)\leqslant 1-\dfrac{0,47355}{4}$$ \pause
  Soit~:
  $$P\left(12,3-2< M< 12,3+2\right)\geqslant 0,8816125$$ \pause
  Et donc~:
 $$\boxed{P\left(10,3< M< 14,3\right)\geqslant 0,8816125}$$ \pause

 \leftskip=0.7cm

La probabilité que la moyenne des notes de dix étudiants pris au hasard soit strictement comprise entre $10,3$ et $14,3$ est donc supérieure à $0,8$.
 \end{enumerate}
\end{enumerate}

\end{frame}





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Exercice 2 - Partie A}

\pause
\begin{enumerate}
 \item Alain ajoute $15$~g, \pause soit $\np{15000}$~mg \pause et la piscine contient $50~\text{m}^3$ \pause soit $\np{50000}$~L. \pause On a alors~:
 $$\dfrac{\np{15000}}{\np{50000}}= \pause 0,3$$ \pause
 L'ajout de $15$~g pour $50~\text{m}^3$ \pause fait donc bien augmenter le taux de $0,3~\text{mg}.\text{L}^{-1}$.
\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item Montrons par récurrence que pour tout entier naturel $n$, $v_n\leqslant v_{n+1}\leqslant 4$. \pause
  \begin{itemize}
   \item \textbf{Initialisation~:}\\ \pause
   Pour $n=0$, on a $v_0=0,7$ \pause et $v_1=0,944$. \pause On a donc $v_0\leqslant v_1\leqslant 4$ \pause et la propriété est vraie au rang $n=0$. \pause
   \item \textbf{Hérédité~:}\\ \pause
   Supposons la propriété vraie pour un certain rang $n\in\mathbb{N}$, c'est-à-dire~: \pause
   $$v_n\leqslant v_{n+1}\leqslant 4$$ \pause
   On a alors, en multipliant par $0,92$ qui est positif~: \pause
   $$0,92v_n\leqslant 0,92v_{n+1}\leqslant 3,68$$ \pause
   Puis, en ajoutant $0,3$~: \pause
   $$0,92v_n+0,3\leqslant 0,92v_{n+1}+0,3\leqslant 3,98$$ \pause
   Soit~:
   $$v_{n+1}\leqslant v_{n+2}\leqslant 3,98$$ \pause
   Et donc a fortiori~:
   $$v_{n+1}\leqslant v_{n+2}\leqslant 4$$ \pause
   La propriété est donc vraie au rang $n+1$. \pause
   \item \textbf{Conclusion~:}\\
   La propriété est vraie au rang $n=0$ et elle est héréditaire, \pause elle est donc vraie pour tout $n\in\mathbb{N}$. \pause On a montré que, pour tout $n\in\mathbb{N}$~: \pause
   $$\boxed{v_n\leqslant v_{n+1}\leqslant 4}$$
  \end{itemize}
 \end{enumerate}
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item \label{ch13_ex2_qA3b} D'après la question précédente, pour tout $n\in\mathbb{N}$~: \pause
  \begin{itemize}
   \item $v_n\leqslant v_{n+1}$ \pause donc la suite $(v_n)$ est croissante. \pause
   \item $v_n\leqslant 4$ \pause donc la suite $(v_n)$ est majorée par $4$. \pause
  \end{itemize}
 La suite $(v_n)$ est croissante et majorée \pause donc elle converge vers une limite $l$ inférieure ou égale à $4$.\\ \pause
 De plus, on sait qu'alors $(v_n)$ converge vers un point fixe, \pause c'est-à-dire un réel $l$ qui vérifie $0,92l+0,3=l$. \pause Résolvons cette équation~: \pause
 \begin{eqnarray*}
  0,92l+0,3=l & \Longleftrightarrow & \pause 0,08l=0,3\\ \pause
              & \Longleftrightarrow & l=\dfrac{0,3}{0,08}\\ \pause
              & \Longleftrightarrow & l=3,75 \pause
 \end{eqnarray*}
La suite $(v_n)$ converge donc vers~: \pause
$$\boxed{l=3,75}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{2}
 \item À long terme, le taux de chlore ne sera pas conforme à la préconisation des piscinistes car il va se rapprocher de $3,75~\text{mg}.\text{L}^{-1}$ \pause et dépassera donc $3~\text{mg}.\text{L}^{-1}$.
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]

\begin{enumerate}
 \setcounter{enumi}{3}
 \item On complète l'algorithme de la façon suivante~: \pause
 \begin{pyverbatim}[][frame=single]
def alerte_chlore(s) :
  n = 0
  v = 0.7
  while v <= s :
    n = n+1
    v = 0,92*v + 0.3
  return n
\end{pyverbatim}
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{4}
 \item \label{ch13_ex2_qA5} En exécutant l'instruction \pyv{alerte_chlore(3)} \pause on obtient la valeur~: \pause
 $$\boxed{17}$$ \pause
 Cela signifie que la taux de chlore dépassera $3~\text{mg}.\text{L}^{-1}$ après 17 jours de traitement.
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Partie B}

\pause
\begin{enumerate}
 \item Résolvons l'équation différentielle $y'=-0,08y+\dfrac{q}{50}$. \pause
 \begin{itemize}
  \item L'équation homogène associée $y'=-0,08y$ admet pour solutions les fonctions de la forme~: \pause
  $$x\mapsto C\mathrm{e}^{-0,08x} \quad \text{avec $C\in\mathbb{R}$}$$ \pause
  \item On a $\dfrac{\mfrac{q}{50}}{0,08}= \pause \dfrac{q}{50\times 0,08} \pause =\dfrac{q}{4}$ \pause donc l'équation $y'=-0,08y+\dfrac{q}{50}$ \pause admet pour solution particulière constante la fonction~: \pause
  $$x\mapsto \dfrac{q}{4}$$ \pause
  \item Les solutions de l'équation différentielle $y'=-0,08y+\dfrac{q}{50}$ \pause sont donc les fonctions de la forme~: \pause
  $$\boxed{x\mapsto C\mathrm{e}^{-0,08x}+\dfrac{q}{4}} \quad \text{avec $C\in\mathbb{R}$}$$
 \end{itemize}
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item On a~:
  $$\boxed{\lim\limits_{x\to +\infty}f(x)=\dfrac{q}{4}} \pause \quad \text{car} \quad \lim\limits_{x\to +\infty}\mathrm{e}^{-0,08x}=0$$
 \end{enumerate}
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item Il s'agit de déterminer les valeurs de $C$ et $q$ telles que $f(0)=0,7$ et $\lim\limits_{x\to +\infty}f(x)=2$~: \pause
  \begin{eqnarray*}
   \begin{dcases}f(0)=0,7\\\lim\limits_{x\to +\infty}f(x)=2\end{dcases} & \Longleftrightarrow & \pause \begin{dcases}C+\dfrac{q}{4}=0,7\\\dfrac{q}{4}=2\end{dcases}\\ \pause
   & \Longleftrightarrow & \begin{dcases}C=0,7-\dfrac{q}{4}\\q=8\end{dcases}\\ \pause
   & \Longleftrightarrow & \begin{cases}C=-1,3\\q=8\end{cases} \pause
  \end{eqnarray*}
Soit~:
$$\boxed{\begin{cases}C=-1,3\\q=8\end{cases}}$$ \pause

\leftskip=0.7cm

On a alors~:
$$\boxed{f(x)=-1,3\mathrm{e}^{-0,08x}+2}$$
 \end{enumerate}
\end{enumerate}

\end{frame}








%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Exercice 3 - Partie A}

\pause
\begin{enumerate}
 \item La courbe $\mathscr{C}_f$ passe par le point $B(-1\,;\,-2)$ donc~: \pause
 $$\boxed{f(-1)=-2}$$ \pause
 La tangente au point d'abscisse $-1$ passe par les points $B(-1\,;\,-2)$ et $A(0\,;\,-1)$. \pause Son coefficient directeur est donc égal à~: \pause
 $$\dfrac{-1-(-2)}{0-(-1)}=\dfrac{1}{1}=1$$ \pause
 Soit~:
 $$\boxed{f'(-1)=1}$$
\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{1}
 \item La courbe $\mathscr{C}_f$ semble être \pause concave puis convexe.
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{2}
 \item A priori, la courbe $\mathscr{C}_f$ ne coupe l'axe des abscisse qu'en un point. \pause On peut donc conjecturer que l'équation $f(x)=0$ admet une unique solution~: \pause
 $$\boxed{x\approx 0,1}$$
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Partie B}

\pause
\begin{enumerate}
 \item On a~:
 $$\boxed{\lim\limits_{x\to -2}f(x)=-\infty} \pause \quad \text{car} \quad \begin{cases}\lim\limits_{x\to -2}(x^2-2x-1)=-1\\\lim\limits_{x\to -2}\ln(x+2)=-\infty \end{cases}$$ \pause
 La courbe $\mathscr{C}_f$ admet donc~: \pause
 $$\boxed{\text{Une asymptote verticale d'équation $x=-2$}}$$
\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{1}
 \item La fonction $f$ est dérivable sur $\left]-2\,;\,+\infty\right[$ et, pour tout $x>-2$~: \pause
 \begin{eqnarray*}
  f'(x) & = & \pause 2x+2+\dfrac{1}{x+2}\\ \pause
        & = & \dfrac{(2x+2)(x+2)+1}{x+2}\\ \pause
        & = & \dfrac{2x^2+4x+2x+4+1}{x+2}\\ \pause
        & = & \dfrac{2x^2+6x+5}{x+2} \pause
 \end{eqnarray*}
Soit~:
 $$\boxed{f'(x)=\dfrac{2x^2+6x+5}{x+2}}$$
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Le polynôme $2x^2+6x+5$ \pause admet pour discriminant $\Delta=-4$, \pause il n'a donc aucune racine réelle et est strictement positif sur $\mathbb{R}$. \pause De plus, $x+2$ est strictement positif sur $\left]-2\,;\,+\infty\right[$ \pause donc $f'(x)>0$ sur $\left]-2\,;\,+\infty\right[$. \pause On en déduit que~: \pause
 $$\boxed{\text{$f$ est strictement croissante sur $\left]-2\,;\,+\infty\right[$}}$$

\pause
 On a donc le tableau~: \pause

 \begin{center}
\begin{tikzpicture}[scale=1]
\tkzTabInit[ lgt=1.8, deltacl=0.5, espcl=4.8]%
{$x$ /0.7,
$f'(x)$ /0.7,
$f(x)$ /2}
{ $-2$ , $+\infty$}
\tkzTabLine { d , + ,   }
\tkzTabVar {D-/$-\infty$, +/$+\infty$}
\end{tikzpicture}
\end{center}
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{3}
 \item Sur l'intervalle $\left]-2\,;\,+\infty\right[$, \pause la fonction $f$ est continue et strictement croissante. \pause De plus $\lim\limits_{x\to -\infty}f(x)=-\infty$ \pause et $\lim\limits_{x\to +\infty}f(x)=+\infty$. \pause Or $0\in\left]-\infty\,;\,+\infty\right[$ donc, d'après le théorème des valeurs intermédiaires, \pause l'équation $f(x)=0$ admet une unique solution $\alpha$ sur l'intervalle $\left]-2\,;\,+\infty\right[$.\\ \pause
 On a~:
 \begin{itemize}
  \item $f(0,11)\approx -0,02 \pause <0$ \pause
  \item $f(0,12)\approx 0,0058 \pause >0$ \pause
 \end{itemize}
Donc~:
$$\boxed{0,11<\alpha<0,12}$$
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{4}
 \item $f$ est strictement croissante sur $\left]-2\,;\,+\infty\right[$ \pause et s'annule en $\alpha$. \pause On a donc le tableau~: \pause

 \begin{center}
\begin{tikzpicture}[scale=1]
\tkzTabInit[ lgt=2, deltacl=0.8, espcl=2.5]%
{$x$ /0.7,
$f(x)$ /1}
{ $-2$ , $\alpha$, $+\infty$}
\tkzTabLine { d , - , z , + , }
\end{tikzpicture}
\end{center}
\end{enumerate}

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{5}
 \item \label{ch13_ex3_qB6} La fonction $f'$ est dérivable sur $\left]-2\,;\,+\infty\right[$ et, pour tout $x\in \left]-2\,;\,+\infty\right[$~: \pause
 \begin{eqnarray*}
  f''(x) & = & \pause \dfrac{(4x+6)(x+2)-(2x^2+6x+5)\times 1}{(x+2)^2}\\ \pause
         & = & \dfrac{4x^2+8x+6x+12-2x^2-6x-5}{(x+2)^2}\\ \pause
         & = & \dfrac{2x^2+8x+7}{(x+2)^2} \pause
 \end{eqnarray*}
 Le polynôme $2x^2+8x+7$ admet deux racines réelles~: \pause
 $$x_1=\dfrac{-4-\sqrt{2}}{2} \pause \approx -2,71 \pause \quad \text{et} \quad x_2=\dfrac{-4+\sqrt{2}}{2} \pause \approx -1,29$$
\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

Et il est positif à l'extérieur de l'intervalle des racines. \pause Et comme $(x+2)^2$ est positif, on a le tableau~: \pause
 \begin{center}
 \begin{tikzpicture}[scale=1]
\tkzTabInit[ lgt=2, deltacl=0.6, espcl=2.5]%
{$x$ /0.7,
$f''(x)$ /0.7,
$f$ /0.7}
{ $-2$  , $x_2$ , $+\infty$}
\tkzTabLine { d , - , z , +  }
\tkzTabLine { d , \text{concave} , t , \text{convexe}  }
\end{tikzpicture}
\end{center} \pause
La dérivée seconde s'annule en changeant de signe en $x_2$. \pause La courbe $\mathscr{C}_f$ admet donc un unique point d'inflexion au point d'abscisse~: \pause
$$\boxed{\dfrac{-4+\sqrt{2}}{2}}$$

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Partie C}

\pause
\begin{enumerate}
 \item On a $J(0\,;\,1)$ \pause et $M(x\,;\,\ln(x+2))$ \pause donc, pour tout $x>-2$~: \pause
 \begin{eqnarray*}
  h(x) & = & \pause JM^2\\ \pause
       & = & (x-0)^2+(\ln(x+2)-1)^2 \pause
 \end{eqnarray*}
Soit~:
$$\boxed{h(x)=x^2+\left[\ln(x+2)-1\right]^2}$$
\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item Pour tout $x\in\left]-2\,;\,+\infty\right[$, $x+2>0$ \pause donc $h'(x)$ est du signe de $f(x)$. \pause On a donc le tableau~: \pause

 \begin{center}
\begin{tikzpicture}[scale=1]
\tkzTabInit[ lgt=2, deltacl=0.8, espcl=2.7]%
{$x$ /0.7,
$h'(x)$ /0.7,
$h(x)$ /1.8}
{ $-2$ , $\alpha$ , $+\infty$}
\tkzTabLine { d , - , z , + ,   }
\tkzTabVar {D+/ , -/$h(\alpha)$ , +/}
\end{tikzpicture}
\end{center}
 \end{enumerate}
\end{enumerate}

\end{frame}



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\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item La distance $JM$ étant positive, \pause elle est minimale lorsque son carré est minimal, \pause c'est-à-dire lorsque la valeur de $h(x)$ est minimale. \pause D'après le tableau de variations, $h$ atteint son minimum en $\alpha$. \pause On en déduit que~: \pause
  $$\boxed{\text{$JM$ est minimale pour $x=\alpha$}}$$
 \end{enumerate}
\end{enumerate}

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\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(a)]
  \item On sait que $f(\alpha)=0$ \pause , c'est-à-dire que~:
  $$\alpha^2+2\alpha-1+\ln(\alpha+2)=0$$ \pause
  Et donc que~: \pause
  $$\boxed{\ln(\alpha+2)=1-2\alpha-\alpha^2}$$
 \end{enumerate}
\end{enumerate}

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\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(b)]
  \item \begin{itemize}
   \item La fonction $g$ est dérivable sur $\left]-2\,;\,+\infty\right[$ et pour tout $x\in\left]-2\,;\,+\infty\right[$, \pause $g'(x)=\dfrac{1}{x+2}$. \pause Le coefficient directeur de la tangente à $\mathscr{C}_g$ au point $M_{\alpha}$ est donc~: \pause
   $$g'(\alpha)= \pause \dfrac{1}{\alpha+2}$$
   \item On a $J(0\,;\,1)$ \pause et $M_{\alpha}(\alpha\,;\,g(\alpha))$, \pause soit $M_{\alpha}(\alpha\,;\,\ln(\alpha+2))$ \pause et donc, d'après la question précédente $M_{\alpha}(\alpha\,;\,1-2\alpha-\alpha^2)$. \pause Calculons le coefficient directeur $m_{\alpha}$ de la droite $(JM_{\alpha})$~: \pause
   \begin{eqnarray*}
    m_{\alpha} & = & \pause \dfrac{y_{M_{\alpha}}-y_J}{x_{M_{\alpha}}-x_J}\\ \pause
      & = & \dfrac{1-2\alpha-\alpha^2-1}{\alpha-0}\\ \pause
      & = & \dfrac{-2\alpha-\alpha^2}{\alpha}\\ \pause
      & = & \dfrac{\alpha(-2-\alpha)}{\alpha}\\ \pause
      & = & -2-\alpha\\ \pause
      & = & -(\alpha+2)
   \end{eqnarray*}
  \end{itemize}
 \end{enumerate}
\end{enumerate}


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\begin{itemize}
 \item Calculons alors le produit des coefficients directeurs~: \pause
 $$g'(\alpha)\times m_{\alpha}= \pause \dfrac{1}{\alpha+2}\times \left(-(\alpha+2)\right) \pause =-1$$ \pause
 On en déduit que~: \pause
 $$\hspace{-1cm}\boxed{\text{La tangente à $\mathscr{C}_g$ au point $M_{\alpha}$ et la droite $(JM_{\alpha})$ sont perpendiculaires}}$$
\end{itemize}


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\begin{frame}{Exercice 4}

\pause
\begin{itemize}
 \item \textbf{Affirmation 1~: Vrai}\\ \pause
On a $\overrightarrow{AC}\begin{pmatrix}4-2\\4-0\\1-0\end{pmatrix}$ \pause soit $\overrightarrow{AC}\begin{pmatrix}2\\4\\1 \end{pmatrix}$ \pause et $\overrightarrow{AD}\begin{pmatrix}0-2\\0-0\\4-0\end{pmatrix}$ \pause soit $\overrightarrow{AD}\begin{pmatrix}-2\\0\\4 \end{pmatrix}$. \pause Les vecteurs $\overrightarrow{AC}$ et $\overrightarrow{AD}$ ne sont pas colinéaires \pause donc les points $A$, $C$ et $D$ définissent un plan.
\end{itemize}


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Soit alors $\overrightarrow{n}\begin{pmatrix}8\\-5\\4 \end{pmatrix}$ \pause un vecteur normal au plan d'équation $8x-5y+4z-16=0$. \pause On a~:
\begin{itemize}
 \item $\overrightarrow{n}.\overrightarrow{AC}= \pause 8\times 2+(-5)\times 4+4\times 1 \pause =16-20+4 \pause =0$ \pause
 \item $\overrightarrow{n}.\overrightarrow{AD}= \pause 8\times (-2)+(-5)\times 0+4\times 4 \pause =-16+0+16 \pause =0$ \pause
\end{itemize}
Le plan $(ACD)$ admet donc une équation de la forme $8x-5y+4z+d=0$. \pause Et comme le point $A(2\,;\,0\,;\,0)$, par exemple, appartient à ce plan, \pause ses coordonnées vérifient l'équation \pause donc $8\times 2-5\times 0+4\times 0+d=0$ \pause soit $d=-16$. \pause Le plan $(ACD)$ admet donc bien pour équation cartésienne \pause $8x-5y+4z-16=0$.

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\begin{itemize}
 \item \textbf{Affirmation 2~: Faux}\\ \pause
Le plan $(ABC)$ admet pour équation $8x-5y+4z-16=0$. \pause Or les coordonnées du point $B$ ne vérifient pas cette équation, \pause en effet $8\times 0-5\times 4+4\times 3-16= \pause -24 \pause \neq 0$. \pause Le point $B$ n'appartient donc pas au plan $(ACD)$. \pause On en déduit que les points $A$, $B$, $C$ et $D$ ne sont pas coplanaires.
\end{itemize}

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\begin{itemize}
 \item \textbf{Affirmation 3~: Vrai}\\ \pause
La droite $(AC)$ passe par $A(2\,;\,0\,;\,0)$ \pause et admet le vecteur $\overrightarrow{AC}\begin{pmatrix}2\\4\\1 \end{pmatrix}$ pour vecteur directeur, \pause elle admet donc pour représentation paramétrique~: \pause
$$\begin{cases}x=2+2t\\y=4t\\z=t\end{cases}\quad \text{avec $t\in\mathbb{R}$}$$ \pause
La droite $(BH)$ passe par $B(0\,;\,4\,;\,3)$ \pause et admet le vecteur $\overrightarrow{BH}\begin{pmatrix}-1\\-3\\-1 \end{pmatrix}$ \pause pour vecteur directeur, elle admet donc pour représentation paramétrique~: \pause
$$\begin{cases}x=-t'\\y=4-3t'\\z=3-t'\end{cases}\quad \text{avec $t\in\mathbb{R}$}$$
\end{itemize}


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Déterminons l'intersection de $(AC)$ et $(BH)$~: \pause
\begin{eqnarray*}
 \begin{cases}2+2t=-t'\\4t=4-3t'\\t=3-t'\end{cases} & \Longleftrightarrow & \pause \begin{cases}2+2(3-t')=-t'\\4(3-t')=4-3t'\\t=3-t'\end{cases}\\ \pause
  & \Longleftrightarrow & \begin{cases}2+6-2t'=-t'\\12-4t'=4-3t'\\t=3-t'\end{cases}\\ \pause
  & \Longleftrightarrow & \begin{cases}t'=8\\t'=8\\t=-5\end{cases} \pause
\end{eqnarray*}

\leftskip=1.8cm

Les droites $(AC)$ et $(BH)$ sont donc sécantes. \pause Le point d'intersection est le point de paramètre $t=-5$ dans la représentation paramétrique de $(AC)$ \pause et de paramètre $t'=8$ dans celle de $(BH)$, \pause c'est-à-dire le point de coordonnées~: \pause
$$\boxed{(-8\,;\,-20\,;\,-5)}$$

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\begin{itemize}
 \item \textbf{Affirmation 4~: Vrai}\\ \pause
 Le point $H$ appartient au plan $(ABC)$, \pause en effet $-1-1+2\times 2-2= \pause 0$. \pause On a $\overrightarrow{HD}\begin{pmatrix}0-(-1)\\0-1\\4-2\end{pmatrix}$ \pause soit $\overrightarrow{HD}\begin{pmatrix}1\\-1\\2 \end{pmatrix}$. \pause D'après son équation cartésienne, le plan $(ABC)$ admet le vecteur $\begin{pmatrix}1\\-1\\2 \end{pmatrix}$ pour vecteur normal, \pause soit le vecteur $\overrightarrow{HD}$. \pause Finalement, $(HD)$ est perpendiculaire au plan $(ABC)$ \pause et $H$ appartient au plan $(ABC)$ \pause donc $H$ est le projeté orthogonal de $D$ sur le plan $(ABC)$.
\end{itemize}

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