\documentclass[10pt,mathserif]{beamer}
\usepackage{eulervm}
\usepackage{concrete}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[francais]{babel}
\usepackage{amsmath,amssymb,amsfonts}
\usepackage{mathrsfs}
\usepackage{pstricks-add}
\usepackage{pst-tree}
\usepackage{multicol}
\usepackage{tikz,tkz-tab}
\usepackage{pythontex}
\usepackage{tikz}
\usepackage{nccmath}
\usepackage{mathtools}
%\usepackage{pifont}
%\usepackage{bm}
\usepackage[np]{numprint}
\usetheme{Warsaw}

\setbeamertemplate{enumerate items}[default]
\beamertemplatenavigationsymbolsempty

\addtobeamertemplate{footline}{
  \setlength\unitlength{1ex}
  \begin{picture}(0,0)
    \put(0,3){\makebox(0,0)[bl]{
    \includegraphics[scale=0.13]{fleur_terminale.eps}}}
%     \put(130,3){\makebox(0,0)[bl]{
%     \includegraphics[scale=0.3]{nom_vinsu.eps}}
%     }
  \end{picture}
}{}




\begin{document}



\title{Amérique du nord - 21 mai 2024}
\author{Spécialité mathématiques - Baccalauréat}
\date{}



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


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

\pause
\begin{enumerate}
 \item On peut représenter la situation par l'arbre suivant~: \pause

 \begin{center}
\psset{nodesep=0mm,levelsep=25mm,treesep=12mm}
\pstree[treemode=R]{\Tdot}
{
\pstree
{\Tdot~[tnpos=a]{$R$}\taput{\footnotesize $0,07$}}
{
\Tdot~[tnpos=r]{$E$}\taput{\footnotesize $0,8$}
\Tdot~[tnpos=r]{$\overline{E}$}\tbput{\footnotesize $0,2$}
}
\pstree
{\Tdot~[tnpos=a]{$\overline{R}$}\tbput{\footnotesize $0,93$}}
{
\Tdot~[tnpos=r]{$E$}\taput{\footnotesize $0,4$}
\Tdot~[tnpos=r]{$\overline{E}$}\tbput{\footnotesize $0,6$}
}
}
\end{center}

\pause
Et on a~:
\begin{eqnarray*}
 P(R\cap E) & = & \pause P(R)\times P_R(E)\\ \pause
            & = & 0,07\times 0,8\\ \pause
            & = & 0,056 \pause
\end{eqnarray*}
\vspace{-0.3cm}
Soit~:
$$\boxed{P(R\cap E)=0,056}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item Il s'agit de calculer $P(E)$. \pause Les événements $R$ et $\overline{R}$ forment une partition de l'univers. \pause On a alors, d'après la formule des probabilités totales~: \pause
 \begin{eqnarray*}
  P(E) & = & \pause P(R)\times P_R(E)+P(\overline{R})\times P_{\overline{R}}(E)\\ \pause
       & = & 0,07\times 0,8+0,93\times 0,4\\ \pause
       & = & 0,428 \pause
 \end{eqnarray*}
La probabilité de tirer une épée est donc~:
$$\boxed{P(E)=0,428}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Il s'agit de calculer $P_E(R)$~: \pause
 \begin{eqnarray*}
  P_E(R) & = & \pause \dfrac{P(R\cap E)}{P(E)}\\ \pause
         & = & \dfrac{0,056}{0,428}\\ \pause
         & \approx & 0,131 \pause
 \end{eqnarray*}
La probabilité de tirer un objet rare sachant que c'est une épée est donc~:
$$\boxed{P_E(R)\approx 0,131}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \item On répète 30 fois, de façon identique et indépendante, une épreuve de Bernoulli dont la probabilité de succès est $0,07$. \pause La variable aléatoire $X$ compte le nombre de succès donc~: \pause
 $$\boxed{\text{$X$ suit une loi binomiale de paramètres $n=30$ et $p=0,07$}}$$ \pause
 On sait alors que son espérance est $E(X)= \pause n\times p \pause =30\times 0,07 \pause =2,1$ \pause, soit~:
 $$\boxed{E(X)=2,1}$$
\end{enumerate}



\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item \label{ch14_ex1_qB2} On a $P(X<6)=P(X\leqslant 5)$ \pause et on obtient, à l'aide de la calculatrice, $P(X\leqslant 5)\approx 0,984$ \pause , soit~:
 $$\boxed{P(X<6)\approx 0,984}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item On a~:
 \begin{itemize}
  \item $P(X\geqslant 2)\approx 0,63\geqslant 0,5$ \pause
  \item $P(X\geqslant 3)\approx 0,35< 0,5$ \pause
 \end{itemize}
La plus grande valeur de $k$ telle que $P(X\geqslant k)\geqslant 0,5$ est donc~: \pause
$$\boxed{k=2}$$ \pause
Cela signifie que sur les 30 objet tirés, il y a plus de 50~\% de chances d'avoir au moins 2 objets rares.
\end{enumerate}


\end{frame}


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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item \label{ch14_ex1_qB4} Soit $Y$ la variable aléatoire égale au nombre d'objets rares parmi les $N$ objets tirés. \pause $Y$ suit alors une loi binomiale de paramètres $N$ et $p=0,07$. \pause La probabilité qu'un joueur obtienne au moins un objet rare est alors~: \pause
 $$P(Y\geqslant 1)= \pause 1-P(Y=0) \pause =1-0,93^n$$ \pause
 Il s'agit alors de résoudre une inéquation~: \pause

 \begin{eqnarray*}
  P(Y\geqslant 1)\geqslant 0,95 & \Longleftrightarrow & \pause 1-0,93^N\geqslant 0,95\\ \pause
                                & \Longleftrightarrow & 0,93^N\leqslant 0,05\\ \pause
                                & \Longleftrightarrow & \ln\left(0,93^N\right)\leqslant \ln(0,05)\\ \pause
                                & \Longleftrightarrow & N\ln\left(0,93\right)\leqslant \ln(0,05)\\ \pause
                                & \Longleftrightarrow & N\geqslant \dfrac{\ln(0,05)}{\ln(0,93)} \quad \textsl{(car $\ln(0,93)<0$)} \pause
 \end{eqnarray*}

\leftskip=1.5cm

 Or $\dfrac{\ln(0,05)}{\ln(0,93)}\approx 41,3$ \pause donc le nombre minimum d'objets à tirer pour atteindre l'objectif est~:
 $$\boxed{N=42}$$
\end{enumerate}


\end{frame}




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

\begin{enumerate}
 \item \textbf{Réponse (c)}\\ \pause
 La droite $(AB)$ passe par le point $A(1\,;\,0\,;\,3)$ et admet le vecteur $\overrightarrow{AB}\begin{pmatrix}3\\1\\-3\end{pmatrix}$ pour vecteur directeur. \pause Elle admet donc pour représentation paramétrique~: \pause
 $$\begin{cases}x=1+3t\\y=t\\z=3-3t\end{cases} \quad \text{avec $t\in\mathbb{R}$}$$
\end{enumerate}



\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item \textbf{Réponse (d)}\\ \pause
 Le point $R(-3\,;\,-9\,;\,7)$ appartient à la droite $(d)$, \pause il s'agit du point de paramètre $t=-\dfrac{3}{2}$.
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item \textbf{Réponse (b)}\\ \pause
 La droite $(d)$ est dirigée par le vecteur $\overrightarrow{u}\begin{pmatrix}4\\6\\-2 \end{pmatrix}$ \pause et la droite $(d')$ est dirigée par le vecteur $\overrightarrow{v}\begin{pmatrix}3\\-2\\1 \end{pmatrix}$. \pause Les vecteurs $\overrightarrow{u}$ et $\overrightarrow{v}$ ne sont pas colinéaires \pause donc les droites ne sont pas parallèles. \pause Déterminons alors leur intersection~:
\end{enumerate}


\end{frame}


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

\begin{small}

\vspace{-0.6cm}

 \begin{eqnarray*}
  \begin{cases}3+4t=-2+3k\\6t=-1-2k\\4-2t=1+k \end{cases} & \Longleftrightarrow & \pause \begin{cases}3+4t=-2+3(3-2t)\\6t=-1-2(3-2t)\\k=3-2t \end{cases}\\ \pause
    & \Longleftrightarrow & \begin{cases}3+4t=-2+9-6t\\6t=-1-6+4t\\k=3-2t \end{cases}\\ \pause
    & \Longleftrightarrow & \begin{cases}10t=4\\2t=-7\\k=3-2t \end{cases}\\ \pause
    & \Longleftrightarrow & \begin{dcases}t=\dfrac{2}{5}\\t=-\dfrac{7}{2}\\k=3-2t \end{dcases}
 \end{eqnarray*} \pause

\vspace{-0.3cm}

\leftskip=2cm

Ce système n'admet aucune solution donc les droites n'ont aucun point d'intersection. \pause Finalement les droites $(d)$ et $(d')$ ne sont ni parallèles ni sécantes, \pause elles ne sont donc pas coplanaires.

\end{small}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item \label{ch14_ex2_q4} \textbf{Réponse (a)}\\ \pause
 Le plan $(P)$ est perpendiculaire à la droite $(d)$, \pause il admet donc pour vecteur normal un vecteur directeur de la droite $(d)$, \pause soit par exemple le vecteur de coordonnées $\begin{pmatrix}2\\3\\-1 \end{pmatrix}$. \pause Il admet donc une équation cartésienne de la forme~:
 $$2x+3y-z+d=0 \quad \text{avec $d\in\mathbb{R}$}$$ \pause
 Et comme le point $I(2\,;\,1\,;\,0)$ appartient au plan, ses coordonnées vérifient l'équation. \pause On a alors $2\times 2+3\times 1-0+d=0$ \pause soit $d=-7$. \pause Le plan $(P)$ admet donc pour équation cartésienne~:
 $$2x+3y-z-7=0$$
\end{enumerate}


\end{frame}




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

\begin{enumerate}
 \item \label{ch14_ex3_qA1} On lit graphiquement que le coefficient directeur de la tangente $(T)$ est égal à $3$ \pause, on a donc~:
 $$\boxed{f'(1)=3}$$ \pause
 Et comme $(T)$ coupe l'axe des ordonnées en $-4$ \pause , son ordonnée à l'origine est égale à $-4$. \pause L'équation réduite de $(T)$ est donc~:
 $$\boxed{y=3x-4}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item La fonction $f$ semble~: \pause
 \begin{itemize}
  \item concave sur $\left]0\,;\,1\right]$ \pause
  \item convexe sur $\left[1\,;\,+\infty\right[$ \pause
 \end{itemize}
La fonction semble changer de convexité en $1$, \pause le point $A$ semble donc être un point d'inflexion (la courbe traverse la tangente).
\end{enumerate}


\end{frame}




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

\vspace{-1cm}

\pause
\begin{enumerate}
 \item On a~:
 $$\boxed{\lim\limits_{x\to +\infty}f(x)=+\infty} \pause \quad \text{car}\quad \begin{dcases}\lim\limits_{x\to +\infty}\ln(x^2)=+\infty\\\lim\limits_{x\to +\infty}\dfrac{1}{x}=0\end{dcases}$$ \pause
 Et pour tout $x\in \left]0\,;\,+\infty\right[$, on a $\ln(x^2)=2\ln(x)$ \pause donc~:
 $$f(x)=2x\ln(x)-\dfrac{1}{x}$$ \pause

\vspace{-1cm}

 On en déduit que~: \pause
 $$\boxed{\lim\limits_{x\to 0}f(x)=-\infty} \pause \quad \text{car}\quad \begin{dcases}\lim\limits_{x\to 0}x\ln(x)=0\quad \textsl{(par croissances comparées)}\\\lim\limits_{x\to 0}\dfrac{1}{x}=+\infty\end{dcases}$$
\end{enumerate}


\end{frame}




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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item \label{ch14_ex3_qB2}
 \begin{enumerate}[(a)]
  \item La fonction $f$ est dérivable sur $\left]0\,;\,+\infty\right[$ et, \pause pour tout $x\in\left]0\,;\,+\infty\right[$~: \pause
  \begin{eqnarray*}
   f'(x) & = & \pause  1\times \ln(x^2)+x\times \dfrac{2x}{x^2}-\left(-\dfrac{1}{x^2}\right)\\ \pause
         & = & \ln(x^2)+2+\dfrac{1}{x^2} \pause
  \end{eqnarray*}
Soit~:
$$\boxed{f'(x)=\ln(x^2)+2+\dfrac{1}{x^2}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}




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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item La fonction $f'$ est dérivable sur $\left]0\,;\,+\infty\right[$ \pause et, pour tout $x\in\left]0\,;\,+\infty\right[$~: \pause
  \begin{eqnarray*}
   f''(x) & = & \pause \dfrac{2}{x}-\dfrac{2}{x^3}\\ \pause
          & = & \dfrac{2x^2-2}{x^3}\\ \pause
          & = & \dfrac{2(x^2-1)}{x^3}\\ \pause
          & = & \dfrac{2(x+1)(x-1)}{x^3} \pause
  \end{eqnarray*}
Soit~:
  $$\boxed{f''(x)=\dfrac{2(x+1)(x-1)}{x^3}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(a)]
  \item Sur $\left]0\,;\,+\infty\right[$, \pause $2(x+1)>0$ \pause et $x^3>0$ \pause donc $f''(x)$ est du signe de $x-1$, \pause d'où 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}
{ $0$  , $1$ , $+\infty$}
\tkzTabLine { d , - , z , + , }
\tkzTabLine { d , \text{concave} , t , \text{convexe}  }
\end{tikzpicture}
\end{center}
 \end{enumerate}

\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(b)]
  \item La fonction $f''$ étant la dérivée de la fonction $f'$, \pause le signe de $f''$ permet de déterminer les variations de $f'$. \pause 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,
$f'(x)$ /2}
{ $0$ , $1$ , $+\infty$}
\tkzTabLine { d , - , z , + , }
\tkzTabVar {D+/ , -/$3$ , +/}
\end{tikzpicture}
\end{center} \pause

La fonction $f'$ atteint donc son minimum en $1$ et ce minimum vaut $3$. \pause On en déduit que $f'(x)\geqslant 3$ \pause et, a fortiori, $f(x)>0$ \pause pour tout $x\in\left]0\,;\,+\infty\right[$ \pause et donc que~:
$$\boxed{\text{La fonction $f$ est strictement croissante sur $\left]0\,;\,+\infty\right[$}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(a)]
  \item Sur l'intervalle $\left]0\,;\,+\infty\right[$, \pause la fonction $f$ est continue \pause et strictement croissante. \pause De plus $\lim\limits_{x\to 0}f(x)=-\infty$ \pause 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 l'équation $f(x)=0$ admet une unique solution $\alpha$ sur $\left]0\,;\,+\infty\right[$.
 \end{enumerate}

\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(b)]
  \item On obtient, à l'aide de la calculatrice~: \pause
  \begin{itemize}
   \item $f(1,32)\approx -0,02<0$ \pause
   \item $f(1,33)\approx 0,007>0$ \pause
  \end{itemize}
Donc~:
$$\boxed{1,32<\alpha<1,33}$$ \pause

On sait que $f(\alpha)=0$, c'est-à-dire~: \pause
$$\alpha \ln(\alpha^2)-\dfrac{1}{\alpha}=0$$ \pause
Puis~:
$$\alpha \ln(\alpha^2)=\dfrac{1}{\alpha}$$ \pause
Donc~:
$$\ln(\alpha^2)=\dfrac{1}{\alpha^2}$$ \pause
Et enfin, en appliquant la fonction exponentielle~: \pause
$$\boxed{\alpha^2=\text{exp}\left(\dfrac{1}{\alpha^2}\right)}$$
 \end{enumerate}

\end{enumerate}


\end{frame}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Exercice 4}

\begin{enumerate}
 \item On a~:
 \begin{eqnarray*}
  I_0 & = & \pause \int_0^{\pi}\sin(x)\,\mathrm{d}x \\ \pause
      & = & \Big[-\cos(x)\Big]_0^{\pi}\\ \pause
      & = & -\cos(\pi)-(-\cos(0))\\ \pause
      & = & -(-1)+1\\ \pause
      & = & 2 \pause
 \end{eqnarray*}
Soit~:
$$\boxed{I_0=2}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item Soit $n\in\mathbb{N}$, pour tout $x\in\left[0\,;\,\pi\right]$, \pause on a $\mathrm{e}^{-nx}>0$ \pause et $\sin(x)\geqslant 0$ \pause donc~:
  $$\mathrm{e}^{-nx}\sin(x)\geqslant 0$$ \pause
  Donc~:
  $$\int_0^{\pi}\mathrm{e}^{-nx}\sin(x)\,\mathrm{d}x\geqslant 0$$ \pause
  Soit~:
  $$\boxed{I_n\geqslant 0}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item Pour tout entier naturel $n$, on a~: \pause
  \begin{eqnarray*}
   I_{n+1}-I_n & = & \pause \int_0^{\pi}\mathrm{e}^{-(n+1)x}\sin(x)\,\mathrm{d}x-\int_0^{\pi}\mathrm{e}^{-nx}\sin(x)\,\mathrm{d}x\\ \pause
        & = & \int_0^{\pi}\mathrm{e}^{-(n+1)x}\sin(x)-\mathrm{e}^{-nx}\sin(x)\,\mathrm{d}x\\ \pause
        & = & \int_0^{\pi}\mathrm{e}^{-nx}\sin(x)\left(\mathrm{e}^{-x}-1\right)\,\mathrm{d}x \pause
  \end{eqnarray*}
Or, pour tout $x\in\left[0\,;\,\pi\right]$, $\mathrm{e}^{-nx}\sin(x)\geqslant 0$ \pause et $\mathrm{e}^{-x}\leqslant 1$ \pause car $x\leqslant 0$. \pause On en déduit que $\mathrm{e}^{-nx}\sin(x)\left(\mathrm{e}^{-x}-1\right)\leqslant 0$ \pause puis~:
$$\int_0^{\pi}\mathrm{e}^{-nx}\sin(x)\left(\mathrm{e}^{-x}-1\right)\,\mathrm{d}x\leqslant 0$$ \pause
Soit~:
$$\boxed{I_{n+1}-I_n\leqslant 0}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(c)]
  \item D'après les questions précédentes, la suite $(I_n)$ est~: \pause
  \begin{itemize}
   \item décroissante car $I_{n+1}-I_n\leqslant 0$ pour tout $n\in\mathbb{N}$ \pause
   \item minorée par $0$ car $I_n\geqslant 0$ pour tout $n\in\mathbb{N}$ \pause
  \end{itemize}
On en déduit que~:
$$\boxed{\text{La suite $(I_n)$ converge}}$$
 \end{enumerate}
\end{enumerate}


\end{frame}




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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(a)]
  \item Soit $n\in\mathbb{N}$, pour tout $x\in\left[0\,;\,\pi\right]$, on a~: \pause
  $$\mathrm{e}^{-nx}\sin(x)\leqslant \mathrm{e}^{-nx} \pause \quad \textsl{(car $\sin(x)\leqslant 1$)}$$ \pause
  Donc par croissance de l'intégrale~: \pause
  $$\int_0^{\pi}\mathrm{e}^{-nx}\sin(x)\,\mathrm{d}x\leqslant \int_0^{\pi}\mathrm{e}^{-nx}\,\mathrm{d}x$$ \pause
  Soit~:
  $$\boxed{I_n\leqslant \int_0^{\pi}\mathrm{e}^{-nx}\,\mathrm{d}x}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(b)]
  \item Pour tout entier naturel $n\geqslant 1$, on a~: \pause
  \begin{eqnarray*}
   \int_0^{\pi}\mathrm{e}^{-nx}\,\mathrm{d}x & = & \pause \left[-\dfrac{1}{n}\mathrm{e}^{-nx}\right]_0^{\pi}\\ \pause
        & = & -\dfrac{1}{n}\mathrm{e}^{-n\pi}-\left(-\dfrac{1}{n}\mathrm{e}^{0}\right)\\ \pause
        & = & \dfrac{\mathrm{e}^{-n\pi}+1}{n} \pause
  \end{eqnarray*}
Soit~:
  $$\boxed{\int_0^{\pi}\mathrm{e}^{-nx}\,\mathrm{d}x=\dfrac{1-\mathrm{e}^{-n\pi}}{n}}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(c)]
  \item D'après les questions précédentes, pour tout $n\in\mathbb{N}$, on a~: \pause
  $$0\leqslant I_n\leqslant \dfrac{1-\mathrm{e}^{-n\pi}}{n}$$ \pause
  Or $\lim\limits_{n\to+\infty}\dfrac{1-\mathrm{e}^{-n\pi}}{n}=0$ \pause donc, d'après le théorème des gendarmes~: \pause
  $$\boxed{\lim\limits_{n\to+\infty}I_n=0}$$
 \end{enumerate}
\end{enumerate}


\end{frame}




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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(a)]
  \item \begin{itemize}
  \item \textbf{Première intégration par parties~:}\\ \pause
  On pose~:
  $$\begin{cases}u(x)=\mathrm{e}^{-nx}\\ v'(x)=\sin(x)\end{cases} \pause \quad \text{et} \quad \begin{cases}u'(x)=-n\mathrm{e}^{-nx}\\ v(x)=-\cos(x)\end{cases}$$ \pause
  On a alors, par intégration par parties~: \pause
  \begin{eqnarray*}
   I_n & = & \pause \Big[-\cos(x)\mathrm{e}^{-nx}\Big]_0^{\pi}-\int_0^{\pi}n\mathrm{e}^{-nx}\cos(x)\,\mathrm{d}x\\ \pause
       & = & -\cos(\pi)\mathrm{e}^{-n\pi}+\cos(0)\mathrm{e}^{0}-n\int_0^{\pi}\mathrm{e}^{-nx}\cos(x)\,\mathrm{d}x\\ \pause
       & = & \mathrm{e}^{-n\pi}+1-nJ_n \pause
  \end{eqnarray*}
Soit~:
$$\boxed{I_n=1+\mathrm{e}^{-n\pi}-nJ_n}$$

\end{itemize}
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{small}
\begin{itemize}
\item \textbf{Deuxième intégration par parties~:}\\ \pause
  On pose~:
  $$\begin{cases}u(x)=\sin(x)\\ v'(x)=\mathrm{e}^{-nx}\end{cases} \pause \quad \text{et} \quad \begin{cases}u'(x)=\cos(x)\\ v(x)=-\dfrac{1}{n}\mathrm{e}^{-nx}\end{cases}$$ \pause
  On a alors, par intégration par parties~: \pause
  \begin{eqnarray*}
   I_n & = & \pause \Big[-\dfrac{1}{n}\mathrm{e}^{-nx}\sin(x)\Big]_0^{\pi}+\int_0^{\pi}\dfrac{1}{n}\mathrm{e}^{-nx}\cos(x)\,\mathrm{d}x\\ \pause
       & = & -\dfrac{1}{n}\mathrm{e}^{-n\pi}\sin(\pi)+-\dfrac{1}{n}\mathrm{e}^{0}\sin(0)+\dfrac{1}{n}\int_0^{\pi}\mathrm{e}^{-nx}\cos(x)\,\mathrm{d}x\\ \pause
       & = & \dfrac{1}{n}J_n \pause
  \end{eqnarray*}
Soit~:
$$\boxed{I_n=\dfrac{1}{n}J_n}$$
\end{itemize}
\end{small}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(b)]
  \item On en déduit que~: \pause
$$1+\mathrm{e}^{-n\pi}-nJ_n=\dfrac{1}{n}J_n$$ \pause
Puis~:
$$\left(1+\dfrac{1}{n}\right)J_n=1+\mathrm{e}^{-n\pi}$$ \pause
Soit~:
$$\dfrac{n^2+1}{n}J_n=1+\mathrm{e}^{-n\pi}$$ \pause
Et donc~:
$$J_n=\dfrac{n(1+\mathrm{e}^{-n\pi})}{n^2+1}$$ \pause
Et comme $I_n=\dfrac{1}{n}J_n$, on a~: \pause
   $$\boxed{I_n=\dfrac{1+\mathrm{e}^{-n\pi}}{n^2+1}}$$
 \end{enumerate}
\end{enumerate}


\end{frame}




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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item On peut compléter le script de la façon suivante~: \pause

\begin{pyverbatim}[][frame=single]
from math import *
def seuil() :
    n = 0
    I = 2
    while I > 0.1 :
        n=n+1
        I=(1+exp(-n*pi))/(n*n+1)
    return n
\end{pyverbatim}

\end{enumerate}


\end{frame}





\end{document}









