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\begin{document}



\title{Polynésie - 5 septembre 2024 (remplacement)}
\author{Spécialité mathématiques - Baccalauréat}
\date{}



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


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

\pause
\begin{enumerate}
 \setcounter{enumi}{0}
 \item On peut représenter la situation par l'arbre suivant~: \pause

 \medskip
\begin{center}
\psset{nodesep=0mm,levelsep=21mm,treesep=10mm}
\pstree[treemode=R]{\Tdot}
{
\pstree
{\Tdot~[tnpos=a]{$E$}\taput{\footnotesize $0,6$}}
{
\Tdot~[tnpos=r]{$B$}\taput{\footnotesize $0,75$}
\Tdot~[tnpos=r]{$\overline{B}$}\tbput{\footnotesize $0,25$}
}
\pstree
{\Tdot~[tnpos=a]{$\overline{E}$}\tbput{\footnotesize $0,4$}}
{
\Tdot~[tnpos=r]{$B$}\taput{\footnotesize $0,52$}
\Tdot~[tnpos=r]{$\overline{B}$}\tbput{\footnotesize $0,48$}
}
}
 \end{center}
\end{enumerate}


\end{frame}



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

Il s'agit alors de calculer $P(E\cap B)$~: \pause
 \begin{eqnarray*}
  P(E\cap B) & = & \pause P(E)\times P_E(B)\\ \pause
             & = & 0,6\times 0,75\\ \pause
             & = & 0,45 \pause
 \end{eqnarray*}
La probabilité que l'acheteur choisisse un véhicule tout-électrique et qu'il ait la possibilité d'installer une borne de recharge à son domicile est donc~: \pause
$$\boxed{P(E\cap B)=0,45}$$

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item Les événements $E$ et $\overline{E}$ forment une partition de l'univers donc, d'après la formule des probabilités totales~: \pause
 \begin{eqnarray*}
  P(B) & = & \pause P(E)\times P_E(B)+P(\overline{E})\times P_{\overline{E}}(B)\\ \pause
   & = & 0,6\times 0,75+0,4\times 0,52\\ \pause
   & = & 0,45+0,208\\ \pause
   & = & 0,658 \pause
 \end{eqnarray*}
Soit~:
$$\boxed{P(B)=0,658}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Il s'agit de calculer $P_B(E)$~: \pause
 \begin{eqnarray*}
  P_B(E) & = & \pause \dfrac{P(E\cap B)}{P(B)}\\ \pause
         & = & \dfrac{0,45}{0,658}\\ \pause
         & \approx & 0,684 \pause
 \end{eqnarray*}
 La probabilité que l'acheteur choisisse un véhicule tout-électrique sachant qu'il a la possibilité d'installer une borne de recharge à son domicile est donc~: \pause
 $$\boxed{P_B(E)\approx 0,684}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(a)]
  \item On répète 20 fois, de façon identique et indépendante, une épreuve de Bernoulli dont la probabilité de succès (l'acheteur peut installer une borne de recharge à son domicile) est égale à $0,658$. \pause La variable aléatoire $X$ est égale au nombre de succès donc~: \pause
  $$\boxed{\text{$X$ suit une loi binomiale de paramètres $n=20$ et $p=0,658$}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(b)]
  \item \label{ch18_ex1_q4b} On obtient, à l'aide de la calculatrice~: \pause
  $$\boxed{P(X=8)\approx 0,011}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(c)]
  \item Il s'agit de calculer $P(X\geqslant 10)$. \pause On obtient alors, à l'aide de la calculatrice, que la probabilité qu'au moins 10 acheteurs puissent installer une borne de recharge est~: \pause
  $$\boxed{P(X\geqslant 10)\approx 0,955}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(d)]
  \item L'espérance de $X$ est donnée par la formule $E(X)= \pause n\times p \pause =20\times 0,658 \pause =13,16$, \pause soit~:
  $$\boxed{E(X)=13,16}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(e)]
  \item En moyenne, lors de la vente de 20 véhicule, il y a en a 13,16 qui ont la possibilité d'installer une borne de recharge. \pause On a alors $13,16\times \np{1200}= \pause \np{15792}$. \pause Pour la vente de 20 véhicules, la directrice devra donc prévoir d'engager, en moyenne, la somme de~: \pause
  $$\boxed{\np{15792}~\text{\euro{}}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}






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

\pause
\begin{enumerate}
 \setcounter{enumi}{0}
 \item \textbf{Affirmation A~: Vrai}\\ \pause
 La fonction $f$ est dérivable sur $\mathbb{R}$ et, pour tout $x\in\mathbb{R}$~: \pause
 $$f'(x)= \pause \mathrm{e}^x+1$$ \pause
 Pour tout $x\in\mathbb{R}$, $\mathrm{e}^x>0$ \pause donc $f'(x)>0$. \pause La fonction $f$ est donc strictement croissante sur $\mathbb{R}$. \pause De plus~:
 $$\lim\limits_{x\to-\infty}f(x)=-\infty \pause \quad \text{car} \quad \begin{cases}\lim\limits_{x\to-\infty}\mathrm{e}^x=0\\\lim\limits_{x\to-\infty}x=-\infty \end{cases}$$ \pause
 Et~:
 $$\lim\limits_{x\to+\infty}f(x)=+\infty \pause \quad \text{car} \quad \begin{cases}\lim\limits_{x\to+\infty}\mathrm{e}^x=+\infty\\\lim\limits_{x\to+\infty}x=+\infty \end{cases}$$
\end{enumerate}


\end{frame}



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

On a donc bien le tableau~: \pause
 \begin{center}
\begin{tikzpicture}[scale=1]
\tkzTabInit[ lgt=3, espcl=6]%
{$x$ /0.7,
variations de $f$ /2.1}
{ $-\infty$  , $+\infty$}
\tkzTabVar {-/$-\infty$  , +/$+\infty$}
\end{tikzpicture}
\end{center}

\end{frame}



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


\textbf{Affirmation B~: Faux}\\ \pause
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 $-2\in\left]-\infty\,;\,+\infty\right[$ \pause donc, d'après le théorème des valeurs intermédiaires, \pause l'équation $f(x)=-2$ admet une unique solution sur $\mathbb{R}$.

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item \textbf{Affirmation C~: Vrai}\\ \pause
Pour tout $x\in\left]0\,;\,+\infty\right[$, on a~:
\begin{eqnarray*}
 \dfrac{\ln(x)-x^2+2}{3x^2} & = & \pause \dfrac{x^2\left(\mfrac{\ln(x)}{x^2}-1+\mfrac{2}{x^2}\right)}{3x^2}\\ \pause
         & = & \dfrac{\mfrac{\ln(x)}{x^2}-1+\mfrac{2}{x^2}}{3} \pause
\end{eqnarray*}
Or $\lim\limits_{x\to+\infty}\dfrac{2}{x^2}=0$ \pause et, par croissance comparées, $\lim\limits_{x\to+\infty}\dfrac{\ln(x)}{x^2}=0$ \pause donc~:
$$\lim\limits_{x\to+\infty}\dfrac{\ln(x)-x^2+2}{3x^2}=-\dfrac{1}{3}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item \textbf{Affirmation D~: Faux}\\ \pause
Pour tout $x\in\mathbb{R}$, on a $\mathrm{e}^{-x^2+1}>0$ \pause donc $k'(x)>0$. \pause Soit $K$ une primitive de $k$  \pause alors, pour tout $x\in\mathbb{R}$, $K'(x)= \pause k(x) \pause >0$. \pause Donc la fonction $K$ est strictement croissante sur $\mathbb{R}$.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item \label{ch18_ex2_q4} \textbf{Affirmation E~: Vrai}\\ \pause
Soit $g$ la fonction définie sur $\mathbb{R}$ par $g(x)=4\mathrm{e}^{-{\text{\footnotesize$\frac{1}{3}$}}x}+1$. \pause On a alors, pour tout $x\in\mathbb{R}$~: \pause
$$g'(x)= \pause -\dfrac{4}{3}\mathrm{e}^{-{\text{\footnotesize$\frac{1}{3}$}}x}$$ \pause
Et donc~:
\begin{eqnarray*}
3g'(x)+g(x) & = & \pause 3\times \left(-\dfrac{4}{3}\mathrm{e}^{-{\text{\footnotesize$\frac{1}{3}$}}x}\right)+4\mathrm{e}^{-{\text{\footnotesize$\frac{1}{3}$}}x}+1\\ \pause
      & = & -4\mathrm{e}^{-{\text{\footnotesize$\frac{1}{3}$}}x}+4\mathrm{e}^{-{\text{\footnotesize$\frac{1}{3}$}}x}+1\\ \pause
      & = & 1
\end{eqnarray*} \pause
\leftskip=1.3cm
Et comme $g(0)= \pause 4+1 \pause =5$, \pause $g$ est la solution de l'équation différentielle $(E)$ qui vérifie $g(0)=5$.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item \textbf{Affirmation F~: Vrai}\\ \pause
Pour tout $x\in\left[0\,;\,1\right]$, 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~:
\begin{eqnarray*}
 \int_0^1 x\mathrm{e}^{-x}\,\mathrm{d}x & = & \pause \Big[-x\mathrm{e}^{-x}\Big]_0^1-\int_0^1 -\mathrm{e}^{-x}\,\mathrm{d}x\\ \pause
      & = & -\mathrm{e}^{-1}+\Big[-\mathrm{e}^{-x}\Big]_0^1\\ \pause
      & = & -\mathrm{e}^{-1}-\mathrm{e}^{-1}+1\\ \pause
      & = & 1-2\mathrm{e}^{-1} \pause
\end{eqnarray*}
Soit~:
$$\int_0^1 x\mathrm{e}^{-x}\,\mathrm{d}x=1-2\mathrm{e}^{-1}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{0}
 \item Le nombre total de boules d'une pyramide de 4 étages est~: \pause
 $$1^2+2^2+3^2+4^2= \pause 1+4+9+16 \pause =30$$ \pause
 Soit~:
 $$\boxed{\text{30 boules}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item On a $S_5= \pause S_4+5^2 \pause =30+25 \pause =55$. \pause On en déduit que le nombre total de boules d'une pyramide de 5 étages est~: \pause
  $$\boxed{S_5=55}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item On peut compléter le programme de la façon suivante~: \pause
  \begin{center}
\begin{pyverbatim}[][frame=single]
def pyramide(n) :
   S = 0
   for i in range(1,n+1) :
      S = S + i**2
   return S
\end{pyverbatim}
  \end{center}
 \end{enumerate}
\end{enumerate}

\end{frame}




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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(c)]
  \item \label{ch18_ex3_q2c} Pour tout entier naturel $n$, on a d'une part~: \pause
 \begin{eqnarray*}
  \dfrac{n(n+1)(2n+1)}{6}+(n+1)^2 & = & \pause \dfrac{n(n+1)(2n+1)+6(n+1)^2}{6}\\ \pause
                                  & = & \dfrac{(n^2+n)(2n+1)+6(n^2+2n+1)}{6}\\ \pause
                                  & = & \dfrac{2n^3+n^2+2n^2+n+6n^2+12n+6)}{6}\\ \pause
                                  & = & \dfrac{2n^3+9n^2+13n+6)}{6} \pause
 \end{eqnarray*}
Et d'autre part~:
\begin{eqnarray*}
  \dfrac{(n+1)(n+2)\left(2(n+1)+1\right)}{6} & = & \pause \dfrac{(n+1)(n+2)(2n+3)}{6}\\ \pause
                                             & = & \dfrac{(n^2+3n+2)(2n+3)}{6}\\ \pause
                                             & = & \dfrac{2n^3+3n^2+6n^2+9n+4n+6}{6}\\ \pause
                                             & = & \dfrac{2n^3+9n^2+13n+6)}{6} \pause
 \end{eqnarray*}
 \leftskip=0.5cm
 On a donc bien~:
 $$\hspace{0.5cm}\boxed{\dfrac{n(n+1)(2n+1)}{6}+(n+1)^2=\dfrac{(n+1)(n+2)\left(2(n+1)+1\right)}{6}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(d)]
  \item Montrons par récurrence que, pour tout entier $n\geqslant 1$, $S_n=\dfrac{n(n+1)(2n+1)}{6}$. \pause
 \begin{itemize}
  \item \textbf{Initialisation~:}\\ \pause
  Pour $n=1$, on a d'une part $S_1=1$ \pause et d'autre part $\dfrac{1(1+1)(2+1)}{6}= \pause \dfrac{1\times 2\times 3}{6} \pause =1$. \pause La propriété est donc vraie au rang $1$. \pause
  \item \textbf{Hérédité~:}\\ \pause
  Supposons la propriété vraie pour un certain rang $n\geqslant 1$, \pause c'est-à-dire $S_n=\dfrac{n(n+1)(2n+1)}{6}$. \pause On a alors~:
  \begin{eqnarray*}
   S_{n+1} & = & \pause S_n+(n+1)^2\\ \pause
           & = & \dfrac{n(n+1)(2n+1)}{6}+(n+1)^2 \quad \textsl{(hypothèse de récurrence)}\\ \pause
           & = & \dfrac{(n+1)(n+2)\left(2(n+1)+1\right)}{6} \quad \textsl{(d'après 2.(c))} \pause
  \end{eqnarray*}
La propriété est donc vraie au rang $n+1$. \pause
  \item \textbf{Conclusion~:}\\ \pause
  La propriété est vraie pour $n=1$ et elle est héréditaire, \pause elle est donc vraie pour tout $n\in\mathbb{N}^*$. \pause Soit~:
  $$\boxed{S_n=\dfrac{n(n+1)(2n+1)}{6}}$$
 \end{itemize}
 \end{enumerate}
\end{enumerate}

\end{frame}





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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item On obtient, à l'aide de la calculatrice~: \pause
 \begin{itemize}
  \item $S_7=140\leqslant 200$ \pause
  \item $S_8=204> 200$ \pause
 \end{itemize}
Le marchand peut donc construire 7 étages en utilisant~: \pause
$$\boxed{\text{140 oranges}}$$
\end{enumerate}

\end{frame}





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

\begin{enumerate}
 \setcounter{enumi}{0}
 \item On a~:
 $$\boxed{E(0\,;\,0\,;\,1)} \pause \quad \text{et} \quad \boxed{C(1\,;\,1\,;\,0)}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item On a $\overrightarrow{AE}\begin{pmatrix}0\\0\\1 \end{pmatrix}$ \pause et $\overrightarrow{AG}\begin{pmatrix}1\\1\\1 \end{pmatrix}$ \pause donc~:
  \begin{itemize}
   \item $\overrightarrow{DB}.\overrightarrow{AE}= \pause 1\times 0-1\times 0+0\times 1 \pause =0$ \pause
   \item $\overrightarrow{DB}.\overrightarrow{AG}= \pause 1\times 1-1\times 1+0\times 1 \pause =0$ \pause
  \end{itemize}
 Le vecteur $\overrightarrow{DB}$ est orthogonal à deux vecteurs non colinéaires du plan $(GEA)$, \pause on en déduit que~:
 $$\boxed{\text{$\overrightarrow{DB}$ est normal au plan $(GEA)$}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item D'après la question précédente, le plan $(GEA)$ admet une équation cartésienne de la forme~: \pause
  $$x-y+d=0 \quad \text{avec $d\in\mathbb{R}$}$$ \pause
  Et comme le point $A(0\,;\,0\,;\,0)$ appartient à ce plan, ses coordonnées vérifient l'équation, \pause d'où $0-0+d=0$, \pause soit $d=0$. \pause Le plan $(GEA)$ admet donc pour équation~: \pause
  $$\boxed{x-y=0}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(a)]
  \item On a $\overrightarrow{CK}\begin{pmatrix}m-1\\0-1\\0-0 \end{pmatrix}$ \pause soit $\overrightarrow{CK}\begin{pmatrix}m-1\\-1\\0 \end{pmatrix}$ \pause et $\overrightarrow{LE}\begin{pmatrix}0-(1-m)\\0-1\\1-1 \end{pmatrix}$ \pause soit $\overrightarrow{LE}\begin{pmatrix}m-1\\-1\\0 \end{pmatrix}$. \pause On a donc $\overrightarrow{CK}=\overrightarrow{LE}$, \pause on en déduit que~:
  $$\boxed{\text{$CKEL$ est un parallélogramme}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(b)]
  \item On a $\overrightarrow{KC}\begin{pmatrix}1-m\\1\\0 \end{pmatrix}$ \pause et $\overrightarrow{KE}\begin{pmatrix}-m\\0\\1 \end{pmatrix}$ \pause d'où~:
  $$\overrightarrow{KC}.\overrightarrow{KE}= \pause (1-m)\times(-m)+1\times 0+0\times 1$$ \pause
  Soit~:
  $$\boxed{\overrightarrow{KC}.\overrightarrow{KE}=m(m-1)}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(c)]
  \item Le quadrilatère $CKEL$ étant un parallélogramme, c'est un rectangle si et seulement si l'angle $\widehat{CKE}$ est un angle droit, \pause c'est-à-dire si et seulement si les vecteurs $\overrightarrow{KC}$ et $\overrightarrow{KE}$ sont orthogonaux. \pause Or~:
  \begin{eqnarray*}
   \text{$\overrightarrow{KC}$ et $\overrightarrow{KE}$ sont orthogonaux} & \Longleftrightarrow & \pause \overrightarrow{KC}.\overrightarrow{KE}=0\\ \pause
   & \Longleftrightarrow & m(m-1)=0 \quad \textsl{(d'après 3.(b))}\\ \pause
   & \Longleftrightarrow & m=0 \quad \text{ou} \quad m=1 \pause
  \end{eqnarray*}
On a donc montré que~:
$$\boxed{\text{$CKEL$ est un rectangle} \Longleftrightarrow \text{$m=0$ ou $m=1$}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(a)]
  \item On a~:
  \begin{itemize}
   \item $CK= \pause \sqrt{\left(-\dfrac{1}{2}\right)^2+(-1)^2+0^2} \pause =\sqrt{\dfrac{5}{4}} \pause =\dfrac{\sqrt{5}}{2}$ \pause
   \item $KE= \pause \sqrt{\left(-\dfrac{1}{2}\right)^2+0^2+1^2} \pause =\sqrt{\dfrac{5}{4}} \pause =\dfrac{\sqrt{5}}{2}$ \pause
  \end{itemize}
On a $CK=KE$ \pause donc $CKEL$ est un parallélogramme qui a deux côtés adjacents de même longueur. \pause On en déduit que~:
$$\boxed{\text{$CKEL$ est un losange}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(b)]
  \item \label{ch18_ex4_q4b} D'une part, d'après la question 3.(b), en prenant $m=\dfrac{1}{2}$, on a~: \pause
  $$\overrightarrow{KC}.\overrightarrow{KE}= \pause \dfrac{1}{2}\times \left( \dfrac{1}{2}-1\right) \pause =-\dfrac{1}{4}$$ \pause
  D'autre part, on a~:
  \begin{eqnarray*}
  \overrightarrow{KC}.\overrightarrow{KE} & = & \pause KC\times KE\times \cos\left(\widehat{CKE}\right)\\ \pause
   & = & \dfrac{\sqrt{5}}{2}\times \dfrac{\sqrt{5}}{2}\times \cos\left(\widehat{CKE}\right)\\ \pause
   & = & \dfrac{5}{4}\times \cos\left(\widehat{CKE}\right)
   \end{eqnarray*} \pause
On en déduit l'égalité~:
$$\dfrac{5}{4}\times \cos\left(\widehat{CKE}\right)=-\dfrac{1}{4}$$ \pause
Soit~:
$$\cos\left(\widehat{CKE}\right)=-\dfrac{1}{5}$$ \pause
\leftskip=0.7cm
On obtient alors, à l'aide de la calculatrice~: \pause
$$\boxed{\widehat{CKE}\approx 102\degres}$$
 \end{enumerate}
\end{enumerate}

\end{frame}


\end{document}




