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\begin{document}



\title{Amérique du nord - 22 mai 2024 (sujet de secours)}
\author{Spécialité mathématiques - Baccalauréat}
\date{}



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


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Exercice 1 - Partie A}

\pause
\begin{enumerate}
 \setcounter{enumi}{0}
 \item On a~:
 $$\boxed{\lim\limits_{x\to-\infty}f(x)=-\infty} \pause \quad \text{car} \quad \begin{cases}\lim\limits_{x\to-\infty}x=-\infty\\ \lim\limits_{x\to-\infty}\mathrm{e}^{-x}=+\infty\\\lim\limits_{x\to-\infty}(2x-1)=-\infty \end{cases}$$ \pause
 Et~:
 $$\hspace{-0.5cm}\boxed{\lim\limits_{x\to+\infty}f(x)=+\infty} \pause \quad \text{car} \quad \begin{cases}\lim\limits_{x\to+\infty}x\mathrm{e}^{-x}=0 \quad \textsl{(par croissances comparées)}\\\lim\limits_{x\to+\infty}(2x-1)=+\infty \end{cases}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item Pour tout $x\in\mathbb{R}$, on a~:
 \begin{eqnarray*}
  f'(x) & = & \pause 1\times \mathrm{e}^{-x}+x\times \left(-\mathrm{e}^{-x}\right)+2\\ \pause
        & = & \mathrm{e}^{-x}-x\mathrm{e}^{-x}+2\\ \pause
        & = & (1-x)\mathrm{e}^{-x}+2 \pause
 \end{eqnarray*}
Soit~:
$$\boxed{f'(x)=(1-x)\mathrm{e}^{-x}+2}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Pour tout $x\in\mathbb{R}$, on a~:
 \begin{eqnarray*}
  f''(x) & = & \pause -1\times \mathrm{e}^{-x}+(1-x)\times \left(-\mathrm{e}^{-x}\right)\\ \pause
        & = & -\mathrm{e}^{-x}+(-1+x)\mathrm{e}^{-x}\\ \pause
        & = & (x-2)\mathrm{e}^{-x} \pause
 \end{eqnarray*}
Soit~:
$$\boxed{f''(x)=(x-2)\mathrm{e}^{-x}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item On a alors 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}
{ $-\infty$  , $2$ , $+\infty$}
\tkzTabLine {  , - , z , +  }
\tkzTabLine {  , \text{concave} , t , \text{convexe}  }
\end{tikzpicture}
\end{center}
\end{enumerate}

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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item 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,
$f'(x)$ /2}
{ $-\infty$ , $2$ , $+\infty$}
\tkzTabLine { , - , z , + , }
\tkzTabVar {+/ , -/$2-\mathrm{e}^{-2}$ , +/}
\end{tikzpicture}
\end{center} \pause
Le minimum de $f'$' est atteint en $2$ et vaut $f'(2)=(1-2)\mathrm{e}^{-2}+2=2-\mathrm{e}^{-2}$.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{5}
 \item On a $2-\mathrm{e}^{-2}\approx 1,86 \pause >0$ \pause et comme il s'agit du minimum de $f'$, \pause on en déduit que, pour tout $x\in\mathbb{R}$~:
 $$\boxed{f'(x)>0}$$ \pause
 Et par conséquent~:
 $$\boxed{\text{La fonction $f$ est strictement croissante sur $\mathbb{R}$}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{6}
 \item Sur $\mathbb{R}$, la fonction $f$ est continue et strictement croissante. \pause De plus $\lim\limits_{x\to-\infty}f(x)=-\infty$ et $\lim\limits_{x\to+\infty}f(x)=+\infty$. \pause Or $0\in\left]-\infty\,;\,+\infty\right[$ \pause donc, d'après le théorème des valeurs intermédiaires, \pause il existe un unique réel $\alpha$ tel que $f(\alpha)=0$.\\ \pause
 On obtient, par balayage à la calculatrice~: \pause
 \begin{itemize}
  \item $f(0,37)\approx -0,004<0$ \pause
  \item $f(0,38)\approx 0,02>0$ \pause
 \end{itemize}
On en déduit l'encadrement~: \pause
$$\boxed{0,37<\alpha<0,38}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{7}
 \item Pour tout $x\in\mathbb{R}$, on a~:
 \begin{eqnarray*}
f(x)-(2x-1) & = & \pause x\mathrm{e}^{-x}+2x-1-(2x-1)\\ \pause
            & = & x\mathrm{e}^{-x}+2x-1-2x+1\\ \pause
            & = & x\mathrm{e}^{-x} \pause
 \end{eqnarray*}
 Pour tout $x\in\mathbb{R}$, $\mathrm{e}^{-x}>0$ \pause donc $x\mathrm{e}^{-x}$ est du signe de $x$. \pause On en déduit que~: \pause
 \begin{itemize}
  \item Sur l'intervalle $\left]-\infty\,;\,0\right[$, $f(x)-(2x-1)<0$ \pause donc $\mathscr{C}_f$ est en-dessous de $\Delta$. \pause
  \item Sur l'intervalle $\left]0\,;\,+\infty\right[$, $f(x)-(2x-1)>0$ \pause donc $\mathscr{C}_f$ est au-dessus de $\Delta$. \pause
  \item En $x=0$, $f(x)-(2x-1)=0$ \pause donc $\mathscr{C}_f$ coupe $\Delta$.
 \end{itemize}
\end{enumerate}

\end{frame}







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

\begin{enumerate}
 \setcounter{enumi}{0}
 \item Pour tout $x\in\mathbb{R}$, on pose~:
 $$\begin{cases}u(x)=x\\v'(x)=\mathrm{e}^{-x}\end{cases} \pause \quad \text{et} \quad \begin{cases}u'(x)=1\\v(x)=-\mathrm{e}^{-x}\end{cases}$$ \pause
 On a alors, par intégration par parties~: \pause
 \begin{eqnarray*}
  I_n & = & \pause \Big[-x\mathrm{e}^{-x}\Big]_1^n+\int_1^n\mathrm{e}^{-x}\,\mathrm{d}x\\ \pause
      & = & -n\mathrm{e}^{-n}+\mathrm{e}^{-1}+\Big[-\mathrm{e}^{-x}\Big]_1^n\\ \pause
      & = & -n\mathrm{e}^{-n}+\mathrm{e}^{-1}-\mathrm{e}^{-n}+\mathrm{e}^{-1}\\ \pause
      & = & 2\mathrm{e}^{-1}-(n+1)\mathrm{e}^{-n} \pause
 \end{eqnarray*}
Soit~:
$$\boxed{I_n=2\mathrm{e}^{-1}-(n+1)\mathrm{e}^{-n}}$$
\end{enumerate}

\end{frame}


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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item Pour tout $x\in\left[1\,;\,n\right]$, $f(x)>2x-1$ \pause et $f(x)-(2x-1)=x\mathrm{e}^{-x}$. \pause On a donc~:
  $$I_n=\int_1^n f(x)-(2x-1)\,\mathrm{d}x$$ \pause
  L'intégrale $I_n$ est donc égale à l'aire, en unités d'aire, du domaine délimité par $\mathscr{C}_f$, \pause $\Delta$ \pause et les droites verticales d'équations $x=1$ et $x=n$. \pause Soit~:
  $$\boxed{\text{L'aire du domaine $D_n$ est $I_n$}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item Par croissances comparées, $\lim\limits_{n\to+\infty}(n+1)\mathrm{e}^{-n}=0$ \pause donc $\lim\limits_{n\to+\infty}I_n=2\mathrm{e}^{-1}$. \pause La limite de l'aire du domaine $D_n$ quand $n$ tend vers $+\infty$ est donc~: \pause
  $$\boxed{\lim\limits_{n\to+\infty}I_n=\dfrac{2}{\mathrm{e}}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Exercice 2}

\begin{enumerate}
 \setcounter{enumi}{0}
 \item \textbf{Affirmation vraie}\\ \pause
 La droite $D$ admet le vecteur $\overrightarrow{u}\begin{pmatrix}-1\\3\\4 \end{pmatrix}$ pour vecteur directeur. \pause La droite $D''$ admet le vecteur $\overrightarrow{u'}\begin{pmatrix}2\\-6\\-8 \end{pmatrix}$ pour vecteur directeur. \pause On remarque que $\overrightarrow{u'}=-2\overrightarrow{u}$ \pause donc les vecteurs $\overrightarrow{u}$ et $\overrightarrow{u'}$ sont colinéaires. \pause On en déduit que les droites $D$ et $D'$ sont parallèles.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item \textbf{Affirmation fausse}\\ \pause
La droite $D$ est dirigée par le vecteur $\overrightarrow{u}\begin{pmatrix}-1\\3\\4 \end{pmatrix}$ \pause et on a $\overrightarrow{AB}\begin{pmatrix}3\\0\\-5 \end{pmatrix}$. \pause On a alors~:
$$\overrightarrow{u}.\overrightarrow{AB}= \pause -1\times 3+3\times 0+4\times (-5) \pause =-23 \pause \neq 0$$ \pause
Les vecteurs $\overrightarrow{u}$ et $\overrightarrow{AB}$ ne sont donc pas orthogonaux. \pause La droite $D$ ne peut donc pas être orthogonale au plan $(ABC)$.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item \textbf{Affirmation fausse}\\ \pause
 Déterminons l'intersection des droites $D$ et $\Delta$. \pause
 \begin{eqnarray*}
 \begin{cases}3-t=-4+2t'\\-2+3t=1-3t'\\1+4t=2+t'\end{cases} & \Longleftrightarrow & \pause \begin{cases}t+2t'=7\\3t+3t'=3\\4t-t'=1\end{cases}\\ \pause
   & \Longleftrightarrow & \begin{cases}t+2t'=7\\-3t'=-18\\-9t'=-27\end{cases}\\ \pause
   & \Longleftrightarrow & \begin{cases}t+2t'=7\\t'=6\\t'=3\end{cases} \pause
 \end{eqnarray*}

\leftskip = 1.5cm

Ce système n'admet aucune solution donc les droites $D$ et $\Delta$ ne sont pas sécantes.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item \textbf{Affirmation vraie}\\ \pause
 On a $\overrightarrow{EF}\begin{pmatrix}2\\-3\\1 \end{pmatrix}$ \pause et, d'après l'équation cartésienne du plan $P$, ce vecteur est un vecteur normal au plan $P$. \pause De plus $2\times (-3)-3\times (-3)+3-6=-6+9+3-6=0$ \pause donc les coordonnées de $F$ vérifient l'équation du plan $P$. \pause Le point $F$ est donc un point du plan $P$ et la droite $(EF)$ est orthogonale à ce plan. \pause On en déduit que $F$ est le projeté orthogonal de $E$ sur le plan $P$.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item \textbf{Affirmation fausse}\\ \pause
 Le plan $P'$ admet le vecteur $\overrightarrow{v}\begin{pmatrix}-3\\1\\-a^2\end{pmatrix}$ pour vecteur normal \pause et la droite $D$ est dirigée par le vecteur $\overrightarrow{u}\begin{pmatrix}-1\\3\\4 \end{pmatrix}$. \pause On a alors~:
 \begin{eqnarray*}
  \text{$D$ et $P''$ sont parallèles} & \Longleftrightarrow & \pause \text{$\overrightarrow{u}$ et $\overrightarrow{v}$ sont orthogonaux}\\ \pause
   & \Longleftrightarrow & \overrightarrow{u}.\overrightarrow{v}=0\\ \pause
   & \Longleftrightarrow & -1\times(-3)+3\times 1+4\times (-a^2)=0\\ \pause
   & \Longleftrightarrow & 6-4a^2=0\\ \pause
   & \Longleftrightarrow & a^2=\dfrac{3}{2}\\ \pause
   & \Longleftrightarrow & a=\sqrt{\dfrac{3}{2}} \quad \text{ou} \quad a=-\sqrt{\dfrac{3}{2}} \pause
 \end{eqnarray*}

\leftskip = 1.5cm

 Il existe donc deux valeurs du paramètre réel $a$ telle que le plan $P'$ d'équation ${-3x+y-a^2z+3=0}$ soit parallèle à la droite $D$.
\end{enumerate}

\end{frame}





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Exercice 3}

\begin{enumerate}
 \setcounter{enumi}{0}
 \item On sait que 2~\% de la population a été contaminée donc~: \pause
 $$\boxed{P(C)=0,02}$$ \pause
 On sait que 90~\% de la population a été vaccinée donc~: \pause
 $$\boxed{P(V)=0,9}$$ \pause
 On sait que parmi les personnes contaminées 62~\% avaient été vaccinées donc~: \pause
 $$\boxed{P_C(V)=0,62}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item On a~:
  \begin{eqnarray*}
   P(C\cap V) & = & \pause P(C)\times P_C(V)\\ \pause
              & = & 0,02\times 0,62\\ \pause
              & = & 0,0124 \pause
  \end{eqnarray*}
Soit~:
$$\boxed{P(C\cap V)=0,0124}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item Les événements $C$ et $\overline{C}$ forment une partition de l'univers donc, d'après la formule des probabilités totales~: \pause
  $$P(V)=P(C\cap V)+P(\overline{C}\cap V)$$ \pause
  D'où~:
  \begin{eqnarray*}
   P(\overline{C}\cap V) & = & \pause  P(V)-P(C\cap V)\\ \pause
                         & = & 0,9-0,0124\\ \pause
                         & = & 0,8876 \pause
  \end{eqnarray*}
  Soit~:
  $$\boxed{P(\overline{C}\cap V)= 0,8876}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Afin de compléter l'arbre, nous avons besoin de la probabilité conditionnelle $P_{\overline{C}}(V)$. \pause On a~:
 $$P_{\overline{C}}(V)= \pause \dfrac{P(\overline{C}\cap V)}{P(\overline{C})} \pause =\dfrac{0,8876}{0,98} \pause \approx 0,9057$$ \pause

 On obtient alors l'arbre suivant~: \pause

 \begin{center}
\psset{nodesep=0mm,levelsep=23mm,treesep=11mm}
\pstree[treemode=R]{\Tdot}
{
\pstree
{\Tdot~[tnpos=a]{$C$}\taput{\footnotesize $0,02$}}
{
\Tdot~[tnpos=r]{$V$}\taput{\footnotesize $0,62$}
\Tdot~[tnpos=r]{$\overline{V}$}\tbput{\footnotesize $0,38$}
}
\pstree
{\Tdot~[tnpos=a]{$\overline{C}$}\tbput{\footnotesize $0,98$}}
{
\Tdot~[tnpos=r]{$V$}\taput{\footnotesize $\approx 0,9057$}
\Tdot~[tnpos=r]{$\overline{V}$}\tbput{\footnotesize $\approx 0,0943$}
}
}
\end{center}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item On a~:
\begin{eqnarray*}
 P_V(C) & = & \pause \dfrac{P(C\cap V)}{P(C)}\\ \pause
        & = & \dfrac{0,0124}{0,9}\\ \pause
        & \approx & 0,0138 \pause
\end{eqnarray*}
La probabilité que la personne ait été contaminée sachant qu'elle avait été vaccinée est donc~: \pause
$$\boxed{P_V(C)\approx 0,0138}$$ \pause

\leftskip = 1.5cm

Autrement dit, parmi les personnes vaccinées, environ $1,38~\%$ avait été vaccinées.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item
 \begin{enumerate}[(a)]
  \item On a~:
 $$P_{\overline{C}}(V)\approx 0,9057 \quad \text{et} \quad P_{\overline{C}}(\overline{V})\approx 0,0943$$ \pause
 L'affirmation est donc \og presque vraie \fg{} \pause car parmi les personnes non contaminées, il y a environ $9,43~\%$ de personnes non vaccinées \pause et environ $90,57~\%$ de personnes vaccinées, \pause soit presque 10 fois plus.
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item
 \begin{enumerate}[(b)]
  \item Il s'agit de calculer $P_V(\overline{C})$~: \pause
 \begin{eqnarray*}
  P_V(\overline{C}) & = & \pause \dfrac{P(\overline{C}\cap V)}{P(V)}\\ \pause
                    & = & \dfrac{0,8876}{0,9}\\ \pause
                    & \approx & 0,9862 \pause
 \end{eqnarray*}
L'affirmation est donc vraie car il y a plus de 98~\% de la population vaccinée qui n'a pas été contaminée.
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{5}
 \item
 \begin{enumerate}[(a)]
  \item On répète 20 fois, de façon identique et indépendante, une épreuve de Bernoulli dont la probabilité de succès est égale à $0,02$. \pause La variable aléatoire $X$ compte le nombre de succès donc~: \pause
  $$\boxed{\text{$X$ suit une loi binomiale de paramètres $n=20$ et $p=0,02$}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{5}
 \item
 \begin{enumerate}[(b)]
  \item Il s'agit de calculer $P(X=4)$~: \pause
  $$P(X=4)= \pause \binom{20}{4}\times 0,02^4\times 0,98^{16} \pause \approx 0,0006$$ \pause
  La probabilité que 4 personnes exactement soient contaminées dans ce groupe de 20 personnes est donc~: \pause
  $$\boxed{P(X=4)\approx 0,0006}$$
 \end{enumerate}
\end{enumerate}

\end{frame}






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Exercice 4 - Partie A}

\pause
\begin{enumerate}
 \setcounter{enumi}{0}
 \item On obtient les valeurs~:
\pause
 \begin{center}
 \setlength{\tabcolsep}{0.5cm}
\renewcommand{\arraystretch}{2}
  \begin{tabular}{|c|c|c|c|c|c|c|}
   \hline
   $n$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ \\ \hline
   $u_n$ & $0$ & $\mfrac{1}{2}$ & $\mfrac{1}{2}$ & $\mfrac{3}{8}$ & $\mfrac{1}{4}$ & $\mfrac{5}{32}$ \\ \hline
  \end{tabular}
 \end{center}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item En étudiant les termes de la suite (obtenus à l'aide de la calculatrice), on peut conjecturer que~: \pause
 $$\boxed{\lim\limits_{n\to+\infty}u_n=0}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{0}
 \item On a $w_0= \pause u_1-\dfrac{1}{2}u_0 \pause =\dfrac{1}{2}-\dfrac{1}{2}\times 0 \pause =\dfrac{1}{2}$, \pause soit~:
 $$\boxed{w_0=\dfrac{1}{2}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \vspace{-1.5cm}
 \item Pour tout entier naturel $n$, on a~: \pause
 \begin{eqnarray*}
  w_{n+1} & = & \pause u_{n+2}-\dfrac{1}{2}u_{n+1}\\ \pause
          & = & u_{n+1}-\dfrac{1}{4}u_n-\dfrac{1}{2}u_{n+1}\\ \pause
          & = & \dfrac{1}{2}u_{n+1}-\dfrac{1}{4}u_n\\ \pause
          & = & \dfrac{1}{2}\left(u_{n+1}-\dfrac{1}{2}u_n\right)\\ \pause
          & = & \dfrac{1}{2}w_n \pause
 \end{eqnarray*}
 \vspace{-0.3cm}
 On en déduit que~:
$$\boxed{\text{La suite $(w_n)$ est géométrique de raison $\dfrac{1}{2}$ et de premier terme $w_0=\dfrac{1}{2}$}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item D'après la question précédente, on a pour tout $n\in\mathbb{N}$, $w_n=w_0\times \left(\dfrac{1}{2}\right)^n \pause=\dfrac{1}{2}\times \left(\dfrac{1}{2}\right)^n$, \pause soit~:
 $$\boxed{w_n=\left(\dfrac{1}{2}\right)^{n+1}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item Pour tout $n\in\mathbb{N}$, on a $w_n=u_{n+1}-\dfrac{1}{2}u_n$ \pause donc $u_{n+1}=w_n+\dfrac{1}{2}u_n$, \pause soit~:
 $$\boxed{u_{n+1}=\left(\dfrac{1}{2}\right)^{n+1}+\dfrac{1}{2}u_n}$$
\end{enumerate}

\end{frame}



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


\begin{enumerate}
 \setcounter{enumi}{4}
 \item Montrons par récurrence que, pour tout $n\in\mathbb{N}$, $u_n=n\left(\dfrac{1}{2}\right)^n$. \pause
 \begin{itemize}
  \item \textbf{Initialisation~:}\\ \pause
  Pour $n=0$, on a d'une part $u_0=0$ \pause et d'autre part $0\times \left(\dfrac{1}{2}\right)^0=0$. \pause On a donc $u_0=0\times \left(\dfrac{1}{2}\right)^0$ \pause et la propriété est vraie au rang $n=0$.
 \end{itemize}
\end{enumerate}


\end{frame}



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

\begin{itemize}
 \item \textbf{Hérédité~:}\\ \pause
  Supposons la propriété vraie pour un certain rang $n\in\mathbb{N}$, \pause c'est-à-dire que $u_n=n\left(\dfrac{1}{2}\right)^n$. On a alors~: \pause
  \begin{eqnarray*}
   u_{n+1} & = & \pause \left(\dfrac{1}{2}\right)^{n+1}+\dfrac{1}{2}u_n\\ \pause
           & = & \left(\dfrac{1}{2}\right)^{n+1}+\dfrac{1}{2}\times n\left(\dfrac{1}{2}\right)^n \quad \textsl{(hypothèse de récurrence)}\\ \pause
           & = & \left(\dfrac{1}{2}\right)^{n+1}+n\left(\dfrac{1}{2}\right)^{n+1}\\ \pause
           & = & \left(\dfrac{1}{2}\right)^{n+1}\left(1+n\right)\\ \pause
           & = & (n+1)\left(\dfrac{1}{2}\right)^{n+1} \pause
  \end{eqnarray*}

\leftskip = 1.5cm

  La propriété est donc vraie au rang $n+1$.

\pause
\item \textbf{Conclusion~:}\\ \pause
La propriété est vraie au rang $n=0$ et elle est héréditaire, elle est donc vraie pour tout $n\in\mathbb{N}$.
\end{itemize}



\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{0}
 \item Pour tout $n\in\mathbb{N}^*$, on a~: \pause
 \begin{eqnarray*}
  u_{n+1}-u_n & = & \pause (n+1)\left(\dfrac{1}{2}\right)^{n+1}-n\left(\dfrac{1}{2}\right)^n\\ \pause
              & = & \left(\dfrac{1}{2}\right)^n\left(\dfrac{1}{2}(n+1)-n\right)\\ \pause
              & = & \left(\dfrac{1}{2}\right)^n\left(\dfrac{1}{2}-\dfrac{1}{2}n\right)\\ \pause
              & = & \left(\dfrac{1}{2}\right)^{n+1}\left(1-n\right) \pause
 \end{eqnarray*}

\leftskip = 1.5cm

Pour tout $n\geqslant 1$, on a $\left(\dfrac{1}{2}\right)^{n+1}>0$ \pause et $1-n\leqslant 0$ \pause d'où $u_{n+1}-u_n\leqslant 0$. \pause On en déduit que~:


$$\hspace{0.2cm}\boxed{\text{La suite $(u_n)$ est décroissante à partir du rang $n=1$}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item D'après la question précédente, la suite $(u_n)$ est décroissante. \pause De plus, pour tout $n\in\mathbb{N}$, $n\geqslant 0$ et $\left(\dfrac{1}{2}\right)^n\geqslant 0$ \pause donc $u_n\geqslant 0$. \pause La suite $(u_n)$ est donc décroissante et minorée par $0$, on en déduit que~: \pause
 $$\boxed{\text{La suite $(u_n)$ est convergente}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Pour tout $l\in\mathbb{R}$, on a~: \pause
 \begin{eqnarray*}
  l=l-\dfrac{1}{4}l & \Longleftrightarrow & \pause \dfrac{1}{4}l=0\\ \pause
                    & \Longleftrightarrow & l=0 \pause
 \end{eqnarray*}
L'équation $l=l-\dfrac{1}{4}l$ admet donc $0$ pour unique solution. \pause On en déduit que~:
$$\boxed{\lim\limits_{n\to+\infty}u_n=0}$$
\end{enumerate}

\end{frame}





\end{document}









