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\begin{document}



\title{Centres étrangers - 6 juin 2024}
\author{Spécialité mathématiques - Baccalauréat}
\date{}



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


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

\pause
\begin{enumerate}
 \item Il s'agit du nombre de triplets dans un ensemble à $8$ éléments. \pause Le nombre de tirages possibles est donc~: \pause
 $$\boxed{8^3=512}$$
\end{enumerate}


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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item Il s'agit du nombre de triplets d'éléments distincts dans un ensemble à $8$ éléments. \pause Le nombre de tirages sans répétition de numéro est donc~: \pause
  $$\boxed{\dfrac{8!}{(8-3)!}=\dfrac{8!}{5!}=8\times 7\times 6=336}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item Le nombre de tirages contenant au moins une répétition de numéro est donc~: \pause
  $$\boxed{512-336=176}$$
 \end{enumerate}
\end{enumerate}

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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item La variable aléatoire $X_1$ suit une loi uniforme sur l'ensemble des entiers compris entre $1$ et $8$. \pause Autrement dit sa loi est donnée par le tableau~: \pause
 \begin{center}
 \renewcommand{\arraystretch}{1.8}
 \begin{tabular}{|c|c|c|c|c|c|c|c|c|}
  \hline
  $k$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ \\ \hline
  $P(X=k)$ & $\mfrac{1}{8}$ & $\mfrac{1}{8}$ & $\mfrac{1}{8}$ & $\mfrac{1}{8}$ & $\mfrac{1}{8}$ & $\mfrac{1}{8}$ & $\mfrac{1}{8}$ & $\mfrac{1}{8}$ \\ \hline
 \end{tabular}
 \end{center}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item Calculons l'espérance $E(X_1)$ de la variable aléatoire $X_1$~: \pause
\begin{eqnarray*}
\hspace{-1.5cm}  E(X_1) & = & \pause 1\times \dfrac{1}{8}+2\times \dfrac{1}{8}+3\times \dfrac{1}{8}+4\times \dfrac{1}{8}+5\times \dfrac{1}{8}+6\times \dfrac{1}{8}+7\times \dfrac{1}{8}+8\times \dfrac{1}{8}\\ \pause
        & = & \dfrac{1}{8}+\dfrac{2}{8}+\dfrac{3}{8}+\dfrac{4}{8}+\dfrac{5}{8}+\dfrac{6}{8}+\dfrac{7}{8}+\dfrac{8}{8}\\ \pause
        & = & \dfrac{36}{8}\\ \pause
        & = & \dfrac{9}{2} \pause
 \end{eqnarray*}
Soit~:
$$\boxed{E(X_1)=\dfrac{9}{2}=4,5}$$
\end{enumerate}

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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item On a $S=X_1+X_2+X_3$ \pause donc, par linéarité de l'espérance, l'espérance de $S$ est~: \pause
 \begin{eqnarray*}
  E(S) & = & \pause E(X_1+X_2+X_3)\\ \pause
       & = & E(X_1)+E(X_2)+E(X_3)\\ \pause
       & = & 4,5+4,5+4,5\\ \pause
       & = & 13,5 \pause
 \end{eqnarray*}
Soit~:
$$\boxed{E(S)=13,5}$$
\end{enumerate}

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

\begin{enumerate}
 \setcounter{enumi}{5}
 \item \label{ch17_ex1_q6} La seule façon d'obtenir une somme égale à $24$ \pause est d'obtenir $8$ pour chacun des trois jetons. \pause On a donc~:
 \begin{eqnarray*}
  P(S=24) & = & \pause P\left((X_1=8)\cap(X_2=8)\cap(X_3=8)\right) \\ \pause
          & = & P(X_1=8)\times P(X_2=8)\times P(X_3=8) \quad \textsl{(par indépendance)}\\ \pause
          & = & \dfrac{1}{8}\times \dfrac{1}{8}\times \dfrac{1}{8}\\ \pause
          & = & \dfrac{1}{512} \pause
 \end{eqnarray*}
Soit~:
$$\boxed{P(S=24)=\dfrac{1}{512}}$$
\end{enumerate}

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

\begin{enumerate}
 \setcounter{enumi}{6}
 \item
 \begin{enumerate}[(a)]
  \item Les tirages permettant d'obtenir une somme supérieure ou égale à 22 sont~: \pause $(8\,;\,8\,;\,8)$, $(7\,;\,8\,;\,8)$, $(8\,;\,7\,;\,8)$, $(8\,;\,8\,;\,7)$, $(7\,;\,7\,;\,8)$, $(7\,;\,8\,;\,7)$, $(8\,;\,7\,;\,7)$, $(6\,;\,8\,;\,8)$, $(8\,;\,6\,;\,8)$ et $(8\,;\,8\,;\,6)$. \pause Soit~:
  $$\boxed{\text{10 tirages}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{6}
 \item
 \begin{enumerate}[(b)]
  \item Tous les tirages étant équiprobables, \pause la probabilité de gagner un lot est donc~: \pause
  $$\boxed{P(S\geqslant 22)=\dfrac{10}{512}=\dfrac{5}{256}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}





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

\begin{enumerate}
 \item
 \begin{enumerate}[(a)]
  \item On a~: \pause
  $$\boxed{\lim\limits_{\substack{x\to 1\\x<1}}f(x)=-\infty} \pause \quad \text{car} \quad \begin{cases}\lim\limits_{\substack{x\to 1\\x<1}}\mathrm{e}^x=\mathrm{e}>0\\\lim\limits_{\substack{x\to 1\\x<1}}(x-1)=0^-\end{cases}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \item
 \begin{enumerate}[(b)]
  \item On en déduit que~: \pause
  $$\boxed{\text{La courbe $\mathscr{C}$ admet une asymptote verticale d'équation $x=1$}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item On a~:
  $$\boxed{\lim\limits_{x\to -\infty}f(x)=0} \pause \quad \text{car} \quad \begin{cases}\lim\limits_{x\to -\infty}\mathrm{e}^x=0\\\lim\limits_{x\to -\infty}(x-1)=-\infty \end{cases}$$
\end{enumerate}

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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(a)]
  \item La fonction $f$ est dérivable sur $\left]-\infty\,;\,1\right[$, comme quotient de fonctions dérivables \pause et, pour tout $x\in\left]-\infty\,;\,1\right[$, on a~: \pause
  \begin{eqnarray*}
   f'(x) & = & \pause \dfrac{\mathrm{e}^x\times (x-1)-\mathrm{e}^x\times 1}{\left(x-1\right)^2}\\ \pause
         & = & \dfrac{(x-1-1)\mathrm{e}^x}{\left(x-1\right)^2} \pause
  \end{eqnarray*}
Soit~:
$$\boxed{f'(x)=\dfrac{(x-2)\mathrm{e}^x}{\left(x-1\right)^2}}$$
 \end{enumerate}

\end{enumerate}

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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(b)]
  \item Pour tout $x\in\left]-\infty\,;\,1\right[$, \pause on a $\mathrm{e}^x>0$, \pause $x-2<0$ \pause et $\left( x-1\right)^2>0$ \pause donc $f'(x)>0$. \pause On en déduit le tableau~: \pause

  \begin{center}
\begin{tikzpicture}[scale=1]
\tkzTabInit[ lgt=2, deltacl=0.5, espcl=4]%
{$x$ /0.7,
$f'(x)$ /0.7,
$f(x)$ /2.1}
{ $-\infty$  , $1$ }
\tkzTabLine {  , - , d  }
\tkzTabVar {+/$0$ , -D/$-\infty$ }
\end{tikzpicture}
\end{center}
 \end{enumerate}

\end{enumerate}

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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(a)]
  \item \label{ch17_ex2_q4b} Pour tout $x\in\left]-\infty\,;\,1\right[$, \pause $\mathrm{e}^x>0$, \pause $(x-1)^3<0$ (car $x-1<0$) \pause et $x^2-4x+5>0$ (car c'est un polynôme du second degré sans racine réelle). \pause On en déduit que, pour tout $x\in\left]-\infty\,;\,1\right[$, $f''(x)<0$ \pause et donc que~: \pause
  $$\boxed{\text{La fonction $f$ est concave sur $\left]-\infty\,;\,1\right[$}}$$
 \end{enumerate}

\end{enumerate}

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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(b)]
  \item On a~:
  \begin{itemize}
   \item $f(0)= \pause \dfrac{\mathrm{e}^0}{0-1} \pause =-1$ \pause
   \item $f'(0)= \pause \dfrac{-2\mathrm{e}^0}{(-1)^2} \pause =-2$ \pause
  \end{itemize}
  La tangente $T$ à la courbe $\mathscr{C}$ au point d'abscisse $0$ admet donc pour équation $y=-2(x-0)-1$ \pause soit~:
  $$\boxed{y=-2x-1}$$
 \end{enumerate}

\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(c)]
  \item La fonction $f$ étant concave sur $\left]-\infty\,;\,1\right[$, \pause sa courbe représentative est en-dessous de ses tangentes. \pause En particulier $\mathscr{C}$ est en-dessous de $T$. \pause On a donc, pour tout $x\in\left]-\infty\,;\,1\right[$, \pause $f(x)\leqslant -2x-1$ \pause soit~:
  $$\dfrac{\mathrm{e}^x}{x-1}\leqslant -2x-1$$ \pause
  Et donc, en multipliant par $x-1$, qui est négatif~: \pause
  $$\boxed{\mathrm{e}^x\geqslant (-2x-1)(x-1)}$$
 \end{enumerate}

\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item
 \begin{enumerate}[(a)]
  \item Sur l'intervalle $\left]-\infty\,;\,1\right[$, \pause la fonction $f$ est continue et strictement décroissante. \pause De plus $\lim\limits_{x\to -\infty}f(x)=0$ et $\lim\limits_{x\to 1}f(x)=-\infty$. \pause Or $-2\in\left]-\infty\,;\,0\right[$ \pause donc, d'après le théorème des valeurs intermédiaires, \pause l'équation $f(x)=-2$ admet une unique solution $\alpha$ sur $\left]-\infty\,;\,1\right[$.
 \end{enumerate}

\end{enumerate}

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\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{4}
 \item
 \begin{enumerate}[(b)]
  \item On a~:
  \begin{itemize}
   \item $f(0,31)\approx -1,98 \pause >-2$ \pause
   \item $f(0,32)\approx -2,03 \pause <-2$ \pause
  \end{itemize}
On en déduit~:
$$\boxed{0,31<\alpha <0,32}$$
 \end{enumerate}

\end{enumerate}

\end{frame}





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

\begin{enumerate}
 \item On a~:
 $$\boxed{I\left(\dfrac{1}{2}\,;\,0\,;\,0\right)} \pause \quad \text{et} \quad \boxed{J\left(1\,;\,1\,;\,\dfrac{1}{2}\right)}$$
\end{enumerate}


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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item On a $\overrightarrow{EJ}\begin{pmatrix}1\\1\\-\mfrac{1}{2}\end{pmatrix}$, \pause $\overrightarrow{FH}\begin{pmatrix}-1\\1\\0 \end{pmatrix}$ \pause et $\overrightarrow{FI}\begin{pmatrix}-\mfrac{1}{2}\\0\\-1 \end{pmatrix}$. \pause On a alors~:
 \begin{itemize}
  \item $\overrightarrow{EJ}.\overrightarrow{FH}= \pause 1\times (-1)+1\times 1+\left(-\dfrac{1}{2}\right)\times 0 \pause =-1+1 \pause =0$ \pause
  \item $\overrightarrow{EJ}.\overrightarrow{FI}= \pause 1\times \left(-\dfrac{1}{2}\right)+1\times 0+\left(-\dfrac{1}{2}\right)\times (-1) \pause =-\dfrac{1}{2}+\dfrac{1}{2} \pause =0$ \pause
 \end{itemize}
 Le vecteur $\overrightarrow{EJ}$ est donc orthogonal à deux vecteurs non colinéaires du plan $(FHI)$. \pause On en déduit que~:
 $$\boxed{\text{Le vecteur $\overrightarrow{EJ}$ est normal au plan $(FHI)$}}$$
\end{enumerate}

\end{frame}



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\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{2}
 \item D'après la question précédente, le plan $(FHI)$ admet une équation cartésienne de la forme~: \pause
 $$-2x-2y+z+d=0 \quad \text{avec $d\in\mathbb{R}$}$$ \pause
 De plus, le point $F(1\,;\,0\,;\,1)$ appartient au plan donc ses coordonnées vérifient l'équation. \pause On a donc $-2-0+1+d=0$ \pause soit $d=1$. \pause Le plan $(FHI)$ admet donc pour équation cartésienne~: \pause
 $$\boxed{-2x-2y+z+1=0}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item La droite $(EJ)$ passe par le point $E(0\,;\,0\,;\,1)$ \pause et admet le vecteur $\overrightarrow{EJ}\begin{pmatrix}1\\1\\-\mfrac{1}{2}\end{pmatrix}$ pour vecteur directeur. \pause Elle admet donc pour représentation paramétrique~: \pause
 $$\boxed{\begin{dcases}x=t\\y=t\\z=1-\dfrac{1}{2}t\end{dcases}} \quad t\in\mathbb{R}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item
 \begin{enumerate}[(a)]
  \item Il s'agit de déterminer les coordonnées du point d'intersection de la droite $(EJ)$ et du plan $(FHI)$. \pause On injecte les coordonnées de la représentation paramétrique de $(EJ)$ dans l'équation cartésienne du plan $(FHI)$~: \pause
  \begin{eqnarray*}
   -2t-2t+1-\dfrac{1}{2}t+1=0 & \Longleftrightarrow & \pause -\dfrac{9}{2}t=-2\\ \pause
                              & \Longleftrightarrow & t=2\times \dfrac{2}{9}\\ \pause
                              & \Longleftrightarrow & t=\dfrac{4}{9} \pause
  \end{eqnarray*}
  Le point $K$ est donc le point de paramètre $t=\dfrac{4}{9}$ dans la représentation paramétrique de la droite $(EJ)$, \pause soit~:
  $$\boxed{K\left(\dfrac{4}{9}\,;\,\dfrac{4}{9}\,;\,\dfrac{7}{9}\right)}$$
 \end{enumerate}

\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item
 \begin{enumerate}[(b)]
  \item Soit $L\left(\dfrac{1}{2}\,;\,0\,;\,1\right)$ le milieu du segment $[EF]$, \pause projeté orthogonal du point $I$ sur le plan $(EFH)$. \pause Soit $\mathscr{A}_{EFH}$ l'aire du triangle rectangle $EFH$, on a~: \pause
  $$\mathscr{A}_{EFH}= \pause \dfrac{EF\times EH}{2} \pause =\dfrac{1\times 1}{2} \pause =\dfrac{1}{2}$$ \pause
  En choisissant le triangle $EFH$ comme base, \pause la hauteur correspondante est la longueur $IL$, \pause le volume $\mathscr{V}_{EFHI}$ de la pyramide $EFHI$ est alors égal à~: \pause
  $$\mathscr{V}_{EFHI}= \pause \dfrac{1}{3}\times \mathscr{A}_{EFH}\times IL \pause =\dfrac{1}{3}\times \dfrac{1}{2}\times 1 \pause =\dfrac{1}{6}$$ \pause
  Le volume de la pyramide $EFHI$ est donc, en $\text{cm}^3$~: \pause
  $$\boxed{\mathscr{V}_{EFHI}=\dfrac{1}{6}}$$
 \end{enumerate}

\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item
 \begin{enumerate}[(c)]
  \item D'autre part, on peut exprimer le volume de la pyramide $EFHI$ en choisissant le triangle $FHI$ pour base. \pause La hauteur correspondante est alors la longueur $EK$. \pause Or~:
  \begin{eqnarray*}
   EK & = & \pause \sqrt{\left(\dfrac{4}{9}-0\right)^2+\left(\dfrac{4}{9}-0\right)^2+\left(\dfrac{7}{9}-1\right)^2}\\ \pause
      & = & \sqrt{\dfrac{16}{81}+\dfrac{16}{81}+\dfrac{4}{81}}\\ \pause
      & = & \sqrt{\dfrac{36}{81}}\\ \pause
      & = & \dfrac{6}{9}\\ \pause
      & = & \dfrac{2}{3}
  \end{eqnarray*}
 \end{enumerate}

\end{enumerate}


\end{frame}



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

On a alors, en notant $\mathscr{A}_{FHI}$ l'aire du triangle $FHI$~: \pause
  $$\mathscr{V}_{EFHI}= \pause \dfrac{1}{3}\times \mathscr{A}_{FHI}\times EK \pause =\dfrac{1}{3}\times \mathscr{A}_{FHI}\times \dfrac{2}{3} \pause =\dfrac{2}{9}\mathscr{A}_{FHI}$$ \pause
  On en déduit l'égalité~: \pause
  $$\dfrac{2}{9}\mathscr{A}_{FHI}=\dfrac{1}{6}$$ \pause
  Soit $\mathscr{A}_{FHI}= \pause \dfrac{1}{6}\times \dfrac{9}{2}$ \pause et donc l'aire du triangle $FHI$, en $\text{cm}^3$, est~: \pause
  $$\boxed{\mathscr{A}_{FHI}=\dfrac{3}{4}}$$

\end{frame}





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

\begin{enumerate}
 \item \label{ch17_ex4_q1} La fonction $f$ est dérivable sur l'intervalle $\left[0\,;\,+\infty \right[$ et pour tout $x\in\left[0\,;\,+\infty \right[$, on a~: \pause
 $$f'(x)=\dfrac{1}{2\sqrt{x+1}}$$ \pause
 Pour tout $x\in\left[0\,;\,+\infty \right[$, $f'(x)\geqslant 0$ \pause donc~:
 $$\boxed{\text{La fonction $f$ est croissante sur l'intervalle $\left[0\,;\,+\infty \right[$}}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item Pour tout $x\in\left[0\,;\,+\infty \right[$, on a~: \pause
 \begin{eqnarray*}
  f(x)-x & = & \pause \sqrt{x+1}-x\\ \pause
         & = & \dfrac{(\sqrt{x+1}-x)(\sqrt{x+1}+x)}{\sqrt{x+1}+x}\\ \pause
         & = & \dfrac{\left(\sqrt{x+1}\right)^2-x^2}{\sqrt{x+1}+x}\\ \pause
         & = & \dfrac{x+1-x^2}{\sqrt{x+1}+x}\\ \pause
 \end{eqnarray*}
Soit~:
 $$\boxed{f(x)-x=\dfrac{-x^2+x+1}{\sqrt{x+1}+x}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item \label{ch17_ex4_qA3} Pour tout $x\in\left[0\,;\,+\infty \right[$, on a~: \pause
 \begin{eqnarray*}
  f(x)=x & \Longleftrightarrow & \pause f(x)-x=0\\ \pause
         & \Longleftrightarrow & \dfrac{-x^2+x+1}{\sqrt{x+1}+x}=0\\ \pause
         & \Longleftrightarrow & -x^2+x+1=0 \pause
 \end{eqnarray*}
 Il s'agit d'un polynôme du second degré qui admet pour discriminant $\Delta=5$ \pause et qui a donc deux racines~: \pause
 $$x_1=\dfrac{-1-\sqrt{5}}{-2} \pause \quad \text{et} \quad x_2=\dfrac{-1+\sqrt{5}}{-2}$$ \pause
 Soit~:
$$x_1=\dfrac{1+\sqrt{5}}{2} \pause \quad \text{et} \quad x_2=\dfrac{1-\sqrt{5}}{2}$$ \pause

\leftskip=1.5cm

Or $x_2$ n'appartient pas à l'intervalle $\left[0\,;\,+\infty \right[$ \pause donc l'équation $f(x)=x$ admet pour unique solution sur l'intervalle $\left[0\,;\,+\infty \right[$~: \pause
 $$\boxed{\ell=\dfrac{1+\sqrt{5}}{2}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{0}
 \item Montrons par récurrence que, pour tout entier naturel $n$, on a $1\leqslant u_{n+1}\leqslant u_n$. \pause
 \begin{itemize}
  \item \textbf{Initialisation~:}\\ \pause
  Pour $n=0$, on a $u_0=5$ \pause et $u_1= \pause f(5) \pause =\sqrt{6} \pause \approx 2,45$. \pause On a donc bien $1\leqslant u_1\leqslant u_0$. \pause La propriété est vraie au rang $n=0$. \pause
  \item \textbf{Hérédité~:}\\ \pause
  Supposons la propriété vraie pour un certain rang $n\in\mathbb{N}$, \pause c'est-à-dire~:
  $$1\leqslant u_{n+1}\leqslant u_n$$ \pause
  On a, en appliquant la fonction $f$, qui est croissante sur $\left[0\,;\,+\infty \right[$~: \pause
  $$f(1)\leqslant f(u_{n+1})\leqslant f(u_n)$$ \pause
  Soit~:
  $$\sqrt{2}\leqslant u_{n+2}\leqslant u_{n+1}$$ \pause

\leftskip=0.7cm

  Et donc, a fortiori~:
  $$1\leqslant u_{n+2}\leqslant u_{n+1}$$ \pause
  La propriété est donc vraie au rang $n+1$.
 \end{itemize}
\end{enumerate}

\end{frame}



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

\begin{itemize}
 \item \textbf{Conclusion~:}\\ \pause
  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 donc montré que, pour tout $n\in\mathbb{N}$~: \pause
  $$\boxed{1\leqslant u_{n+1}\leqslant u_n}$$
\end{itemize}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item D'après la question précedente, la suite $(u_n)$ est~: \pause
 \begin{itemize}
  \item décroissante \pause (car $u_{n+1}\leqslant u_n$ pour tout $n\in\mathbb{N}$). \pause
  \item minorée par $1$ \pause (car $u_n\geqslant 1$ pour tout $n\in\mathbb{N}$). \pause
 \end{itemize}
 On en déduit que~: \pause
$$\boxed{\text{La suite $(u_n)$ est convergente}}$$
\end{enumerate}

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\begin{enumerate}
 \setcounter{enumi}{2}
 \item La suite $(u_n)$ est définie par la relation de récurrence~: \pause
 $$u_{n+1}=f(u_n)$$ \pause
 Or on a vu, dans la question précédente, que cette suite converge vers une limite $l$. \pause Et comme la fonction $f$ est continue sur $\left[0\,;\,+\infty \right[$, \pause par passage à la limite dans la relation précédente, on a~: \pause
 $$l=f(l)$$ \pause
 Autrement dit, $l$ est solution de l'équation $f(x)=x$. \pause Et d'après la question \ref{ch17_ex4_qA3} de la partie A, cette équation admet pour unique solution $\ell=\dfrac{1+\sqrt{5}}{2}$. \pause On en déduit donc que~: \pause
 $$\boxed{\text{La suite $(u_n)$ converge vers $\ell=\dfrac{1+\sqrt{5}}{2}$}}$$
\end{enumerate}

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\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(a)]
  \item \label{ch17_ex4_qB4a} La commande seuil(2) renvoie \pause le premier indice à partir duquel l'écart entre $u_n$ et $\ell$ est strictement inférieur à $10^{-2}$. \pause On obtient, à l'aide de la calculatrice~: \pause
 \begin{itemize}
  \item $u_4-\ell\approx 0,022 \pause >0,01$ \pause
  \item $u_5-\ell\approx 0,007  \pause<0,01$ \pause
 \end{itemize}
C'est donc à partir de $n=5$ que $|u_n-\ell|<0,02$. \pause La valeur renvoyée par \pyv{seuil(2)} est donc~: \pause
$$\boxed{\text{\pyv{5}}}$$
 \end{enumerate}
\end{enumerate}

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\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(b)]
  \item Cela signifie que c'est à partir de $u_9$ \pause que l'écart entre les termes de la suite et sa limite devient strictement plus petit que $10^{-4}$. \pause Autrement dit~:
 $$\boxed{\text{Pour tout $n\geqslant 9$, $|u_n-\ell|<0,0001$}}$$
 \end{enumerate}
\end{enumerate}

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