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\begin{document}



\title{Asie - 10 juin 2024}
\author{Spécialité mathématiques - Baccalauréat}
\date{}



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


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

\pause
\begin{enumerate}
 \item Le tableau de variations de $f$ semble être le suivant~: \pause

 \begin{center}
\begin{tikzpicture}[scale=1]
\tkzTabInit[ lgt=2, deltacl=0.8, espcl=2.5]%
{$x$ /0.7,
$f(x)$ /2}
{ $0$  , $\pgfmathprintnumber{1.5}$ , $5$}
\tkzTabVar {-/ , +/ , -/}
\end{tikzpicture}
\end{center}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item Au point $A$, la courbe $\mathscr{C}$ semble traverser sa tangente. \pause On peut donc conjecturer que~: \pause
 $$\boxed{\text{La courbe $\mathscr{C}$ semble présenter un point d'inflexion en $A$}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item La fonction $f$ est croissante sur $[0\,;\,1,5]$ et décroissante sur $[1,5\,;\,+\infty[$, \pause sa fonction dérivée est donc positive sur $[0\,;\,1,5]$ \pause et négative sur $[1,5\,;\,+\infty[$. \pause On en déduit que~:
 $$\boxed{\text{$\mathscr{C}_2$ est la courbe représentative de $f'$}}$$
\pause
 La fonction $f$ est concave sur $[0\,;\,2,5]$ \pause et convexe sur $[2,5\,;\,+\infty[$, \pause sa fonction dérivée seconde est donc négative sur $[0\,;\,2,5]$ \pause et positive sur $[2,5\,;\,+\infty[$. \pause On en déduit que~:
 $$\boxed{\text{$\mathscr{C}_1$ est la courbe représentative de $f''$}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item La fonction $f$ est négative sur $[0\,;\,0,5]$ et positive sur $[0,5\,;\,+\infty[$ \pause donc si $F$ est une primitive de $f$ \pause alors $F$ est décroissante sur $[0\,;\,0,5]$ et croissante sur $[0,5\,;\,+\infty[$. \pause On en déduit que~:
 $$\boxed{\text{La courbe $\mathscr{C}_3$ ne peut pas être la courbe d'une primitive de $f$}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{0}
 \item
 \begin{enumerate}[(a)]
  \item La fonction $f$ est dérivable, comme produit de fonctions dérivables, sur $\left[0\,;\,+\infty\right[$ et pour tout $x\in\left[0\,;\,+\infty\right[$~: \pause
  \begin{eqnarray*}
   f'(x) & = & \pause 4\times \mathrm{e}^{-x+1}+(4x-2)\times (-\mathrm{e}^{-x+1})\\ \pause
         & = & 4\mathrm{e}^{-x+1}-(4x-2)\mathrm{e}^{-x+1}\\ \pause
         & = & (4-4x+2)\mathrm{e}^{-x+1}\\ \pause
         & = & (-4x+6)\mathrm{e}^{-x+1} \pause
  \end{eqnarray*}
Soit~:
$$\boxed{f'(x)=(-4x+6)\mathrm{e}^{-x+1}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{0}
 \item
 \begin{enumerate}[(b)]
  \item \label{ch18_ex1_qB1b} Pour tout $x\in\left[0\,;\,+\infty\right[$, \pause $\mathrm{e}^{-x+1}>0$ \pause donc $f'(x)$ est du signe de $-4x+6$. \pause On a alors le tableau~: \pause

 \begin{center}
\begin{tikzpicture}[scale=1]
\tkzTabInit[ lgt=2, deltacl=0.8, espcl=2.5]%
{$x$ /0.7,
$f'(x)$ /0.7,
$f(x)$ /2}
{ $0$  , $\pgfmathprintnumber{1.5}$ , $5$}
\tkzTabLine {  , + , z , - ,  }
\tkzTabVar {-/$-2\mathrm{e}$ , +/$\dfrac{4}{\sqrt{\mathrm{e}}}$ , -/$0$}
\end{tikzpicture}
\end{center}
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{0}
 \item
 \begin{enumerate}[(c)]
  \item La fonction $f'$ est dérivable sur $\left[0\,;\,+\infty\right[$ et, pour tout $x\in\left[0\,;\,+\infty\right[$~: \pause
  \begin{eqnarray*}
   f''(x) & = & \pause -4\times \mathrm{e}^{-x+1}+(-4x+6)\times (-\mathrm{e}^{-x+1})\\ \pause
          & = & (-4+4x-6)\mathrm{e}^{-x+1}\\ \pause
          & = & (4x-10)\mathrm{e}^{-x+1} \pause
  \end{eqnarray*}
Sur l'intervalle $\left[0\,;\,+\infty\right[$, \pause $\mathrm{e}^{-x+1}>0$ \pause donc $f''(x)$ est du signe de $(4x-10)$. \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,
$f$ /0.7}
{ $0$  , $\pgfmathprintnumber{2.5}$ , $+\infty$}
\tkzTabLine {  , - , z , +  }
\tkzTabLine {  , \text{concave} , t , \text{convexe}  }
\end{tikzpicture}
\end{center}
 \pause
On en déduit que~:
$$\boxed{\text{Le point d'abscisse $2,5$ est un point d'inflexion}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item La fonction $F$ est dérivable sur $\left[0\,;\,+\infty\right[$ et, pour tout $x\in\left[0\,;\,+\infty\right[$~: \pause
  \begin{eqnarray*}
   F'(x) & = & \pause a\mathrm{e}^{-x+1}+(ax+b)\times(-\mathrm{e}^{-x+1})\\ \pause
         & = & (a-ax-b)\mathrm{e}^{-x+1}\\ \pause
         & = & (-ax+a-b)\mathrm{e}^{-x+1} \pause
  \end{eqnarray*}
  On veut que $F'(x)=f(x)$, \pause on a alors, par identification des coefficients~: \pause
   $$\begin{cases}-a=4\\a-b=-2\end{cases} \Longleftrightarrow \pause \begin{cases}a=-4\\-4-b=-2\end{cases}$$ \pause
   Soit~:
   $$\boxed{\begin{cases}a=-4\\b=-2\end{cases}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item On a~:
  \begin{eqnarray*}
   I & = & \pause \int_{\frac{3}{2}}^8 f(x)\,\mathrm{d}x\\ \pause
     & = & \Big[ F(x) \Big]_{\frac{3}{2}}^8\\ \pause
     & = & \Big[ (-4x-2)\mathrm{e}^{-x+1} \Big]_{\frac{3}{2}}^8\\ \pause
     & = & (-4\times 8-2)\mathrm{e}^{-8+1}-\left(-4\times \dfrac{3}{2}-2\right)\mathrm{e}^{-\frac{3}{2}+1}\\ \pause
     & = & -34\mathrm{e}^{-7}+8\mathrm{e}^{-\frac{1}{2}} \pause
  \end{eqnarray*}
Soit~:
  $$\boxed{I=-34\mathrm{e}^{-7}+8\mathrm{e}^{-\frac{1}{2}}\approx 4,82}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(a)]
  \item On a $f\left(\dfrac{3}{2}\right)= \pause \dfrac{4}{\sqrt{\mathrm{e}}} \pause \approx 2,43$. \pause La hauteur du point de départ $D$ est donc de~: \pause
  $$\boxed{2,43~\text{mètres}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(b)]
  \item La surface, en $\text{m}^2$, du mur est donnée par l'intégrale calculée dans la question précédente, \pause elle est donc d'environ $4,82~\text{m}^2$. \pause Elle souhaite en couvrir 75~\%, soit $4,82\times 0,75\approx \pause 3,62~\text{m}^2$. \pause Une bombe aérosol couvre $0,8~\text{m}^2$ \pause et $\dfrac{3,62}{0,8}\approx 4,5$. \pause Il faudra donc~: \pause
  $$\boxed{\text{5 bombes}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}




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

\pause
\begin{enumerate}
 \item On a $\overrightarrow{AB}\begin{pmatrix}1\\0\\-1\end{pmatrix}$ \pause et $\overrightarrow{AC}\begin{pmatrix}-3\\4\\1 \end{pmatrix}$. \pause Les vecteurs $\overrightarrow{AB}$ et $\overrightarrow{AC}$ ne sont pas colinéaires donc~: \pause
 $$\boxed{\text{Les points $A$, $B$ et $C$ ne sont pas alignés}}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item \label{ch18_ex2_q2a} On a, de plus, $\overrightarrow{AD}\begin{pmatrix}1\\4\\-3\end{pmatrix}$. \pause On peut alors remarquer que~: \pause
  $$\overrightarrow{AD}=4\overrightarrow{AB}+\overrightarrow{AC}$$ \pause
  On en déduit que les vecteurs $\overrightarrow{AB}$, $\overrightarrow{AC}$ et $\overrightarrow{AD}$ sont coplanaires et donc que~: \pause
$$\boxed{\text{Les points $A$, $B$, $C$ et $D$ sont coplanaires}}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item On a $\overrightarrow{AB}\begin{pmatrix}1\\0\\-1\end{pmatrix}$ \pause et $\overrightarrow{CD}\begin{pmatrix}4\\0\\-4\end{pmatrix}$. \pause On remarque que $\overrightarrow{CD}=4\overrightarrow{AB}$ \pause donc les vecteurs $\overrightarrow{AB}$ et $\overrightarrow{CD}$ sont colinéaires. \pause On en déduit que les droites $(AB)$ et $(CD)$ sont parallèles et donc que~: \pause
  $$\boxed{\text{Le quadrilatère $ABDC$ est un trapèze de bases $[AB]$ et $[CD]$}}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(a)]
  \item On a~:
  \begin{itemize}
   \item $\overrightarrow{n}.\overrightarrow{AB}= \pause 2\times 1+1\times 0+2\times (-1) \pause =2-2 \pause =0$ \pause
   \item $\overrightarrow{n}.\overrightarrow{AC}= \pause 2\times (-3)+1\times 4+2\times 1 \pause =-6+4+2 \pause =0$ \pause
  \end{itemize}
Le vecteur $\overrightarrow{n}$ est donc orthogonal à deux vecteurs non colinéaires du plan $(ABC)$. \pause On en déduit que~:
$$\boxed{\text{Le vecteur $\overrightarrow{n}$ est un vecteur normal au plan $(ABC)$}}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(b)]
  \item Le plan $(ABC)$ admet donc une équation cartésienne de la forme~: \pause
  $$2x+y+2z+d=0 \quad \text{avec $d\in\mathbb{R}$}$$ \pause
  De plus, le point $A$ appartient à ce plan donc ses coordonnées vérifient l'équation, \pause soit $2\times 3-1+2\times 1+d=0$. \pause On a donc $7+d=0$ \pause d'où $d=-7$. \pause Le plan $(ABC)$ admet donc pour équation cartésienne~: \pause
  $$\boxed{2x+y+2z-7=0}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(c)]
  \item La droite $\Delta$ passe par le point $S(2\,;\,1\,;\,4)$ \pause et est dirigée par le vecteur $\overrightarrow{n}\begin{pmatrix}2\\1\\2 \end{pmatrix}$, \pause elle admet donc pour représentation paramétrique~: \pause
  $$\boxed{\begin{cases}x=2+2t\\y=1+t\\z=4+2t\end{cases}}\quad t\in\mathbb{R}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(d)]
  \item \label{ch18_ex2_q2f} On injecte les expression de la représentation paramétrique de $\Delta$ dans l'équation cartésienne du plan $(ABC)$~: \pause
  \begin{eqnarray*}
   \hspace{-0.5cm}2(2+2t)+1+t+2(4+2t)-7=0 & \Longleftrightarrow & \pause 4+4t+1+t+8+4t-7=0\\ \pause
                             & \Longleftrightarrow & 9t=-6\\ \pause
                             & \Longleftrightarrow & t=-\dfrac{6}{9}\\ \pause
                             & \Longleftrightarrow & t=-\dfrac{2}{3} \pause
  \end{eqnarray*}
Le point $I$ est donc le point de paramètre $t=-\dfrac{2}{3}$ \pause dans la représentation paramétrique de $\Delta$, \pause c'est-à-dire le point de coordonnées~: \pause
$$\left(2+2\times \left(-\dfrac{2}{3}\right)\,;\,1+\left(-\dfrac{2}{3}\right)\,;\,4+2\times \left(-\dfrac{2}{3}\right)\right)$$ \pause
\leftskip=1cm
Soit~:
$$\boxed{I\left(\dfrac{2}{3}\,;\,\dfrac{1}{3}\,;\,\dfrac{8}{3}\right)}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

On a alors~:
\begin{eqnarray*}
 SI & = & \pause \sqrt{\left(\dfrac{2}{3}-2\right)^2+\left(\dfrac{1}{3}-1\right)^2+\left(\dfrac{8}{3}-4\right)^2}\\ \pause
    & = & \sqrt{\left(-\dfrac{4}{3}\right)^2+\left(-\dfrac{2}{3}\right)^2+\left(-\dfrac{4}{3}\right)^2}\\ \pause
    & = & \sqrt{\dfrac{16}{9}+\dfrac{4}{9}+\dfrac{16}{9}}\\ \pause
    & = & \sqrt{\dfrac{36}{9}}\\ \pause
    & = & \sqrt{4}\\ \pause
    & = & 2 \pause
\end{eqnarray*}
Soit~:
$$\boxed{SI=2}$$

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(a)]
  \item La droite $(CD)$ passe par le point $C(0\,;\,3\,;\,2)$ et admet le vecteur $\overrightarrow{CD}\begin{pmatrix}4\\0\\-4\end{pmatrix}$ pour vecteur directeur. \pause Elle admet donc pour représentation paramétrique~: \pause
  $$\boxed{\begin{cases}x=4t'\\y=3\\z=2-4t'\end{cases}}\quad t\in\mathbb{R}$$ \pause
  On peut alors remarquer que le point $H$ appartient à la droite $(CD)$ \pause car c'est le point de paramètre $t'=\dfrac{3}{4}$. \pause De plus, on a $\overrightarrow{HB}\begin{pmatrix}1\\-4\\1 \end{pmatrix}$ \pause donc $\overrightarrow{HB}.\overrightarrow{CD}= \pause 1\times 4+(-4)\times 0+1\times(-4) \pause =0$. \pause La droite $(HB)$ est donc orthogonale à la droite $(CD)$. \pause On a donc montré que~: \pause
  $$\boxed{\text{$H(3\,;\,3\,;\,-1)$ est le projeté orthogonal de $B$ sur $(CD)$}}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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


On a alors~:
\begin{eqnarray*}
 HB & = & \pause \sqrt{\left(4-3\right)^2+\left(-1-3\right)^2+\left(0-(-1)\right)^2}\\ \pause
    & = & \sqrt{1^1+(-4)^2+1^2}\\ \pause
    & = & \sqrt{1+16+1}\\ \pause
    & = & \sqrt{18}\\ \pause
    & = & \sqrt{9\times 2}\\ \pause
    & = & 3\sqrt{2} \pause
\end{eqnarray*}
Soit~:
$$\boxed{HB=3\sqrt{2}}$$

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(b)]
  \item On a $AB= \pause \sqrt{1^2+0^2+(-1)^2} \pause =\sqrt{2}$ \pause et $CD= \pause \sqrt{4^2+0^2+(-4)^2}=\sqrt{32} \pause =4\sqrt{2}$. \pause L'aire du trapèze $ABDC$ est alors~:
  \begin{eqnarray*}
  \mathscr{A}_{ABDC} & = & \pause \dfrac{AB + CD}{2}\times HB\\ \pause
                     & = & \dfrac{\sqrt{2}+ 4\sqrt{2}}{2}\times 3\sqrt{2}\\ \pause
                     & = & \dfrac{5\sqrt{2}}{2}\times 3\sqrt{2}\\ \pause
                     & = & \dfrac{5\times 2\times 3}{2}\\ \pause
                     & = & 15 \pause
  \end{eqnarray*}
  Soit~:
  $$\boxed{\mathscr{A}_{ABDC}=15}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item Pour calculer le volume $\mathscr{V}_{SABDC}$ de la pyramide $SABDC$, on utilise la base $ABCD$ et la hauteur correspondante $SI$. \pause On a alors~:
 \begin{eqnarray*}
  \mathscr{V}_{SABDC} & = & \pause \dfrac{1}{3}\times \mathscr{A}_{ABDC}\times SI\\ \pause
      & = \pause \dfrac{1}{3}\times 15\times 2\\ \pause
      & = \pause 10 \pause
  \end{eqnarray*}
 Soit~:
 $$\boxed{\mathscr{V}_{SABDC}=10}$$
\end{enumerate}


\end{frame}







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

\pause
\begin{enumerate}
 \item D'après l'énoncé, la probabilité que l'individu prélevé ait déjà été infecté par la COVID 19 est~: \pause
 $$\boxed{P(I)=0,057}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item On répète 100 fois, de façon identique et indépendante, une épreuve de Bernoulli dont la probabilité de succès est $0,057$. \pause La variable aléatoire $X$ est égale au nombre de succès donc~: \pause
  $$\boxed{\text{$X$ suit une loi binomiale de paramètres $n=100$ et $p=0,057$}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item $X$ suivant une loi binomiale, son espérance est $E(X)= \pause n\times p \pause =100\times 0,057$, \pause soit~:
  $$\boxed{E(X)=5,7}$$ \pause
  Cela signifie que le nombre moyen de personnes ayant déjà été infectées sur un échantillon de 100 personnes et de $5,7$.
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(c)]
  \item \label{ch18_ex3_qA2c} Il s'agit de calculer $P(X=0)$. \pause Or $P(X=0)= \pause (1-0,057)^{100} \pause =0,943^{100} \pause \approx 0,0028$. \pause La probabilité qu'il n'y ait aucune personne infectée dans l'échantillon est donc~: \pause
  $$\boxed{P(X=0)\approx 0,0028}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(d)]
  \item \label{ch18_ex3_qA2d} Il s'agit de calculer $P(X\geqslant 2)$. \pause On obtient, à l'aide de la calculatrice, que la probabilité qu'il y ait au moins 2 personnes infectées dans l'échantillon est~: \pause
  $$\boxed{P(X\geqslant 2)\approx 0,9801}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(e)]
  \item On obtient, à l'aide de la calculatrice~: \pause
  \begin{itemize}
   \item $P(X\leqslant 8)\approx 0,88 \pause <0,9$ \pause
   \item $P(X\leqslant 9)\approx 0,94 \pause >0,9$ \pause
  \end{itemize}
Le plus petit entier $n$ tel que $P(X\leqslant n)>0,9$ est donc~: \pause
$$\boxed{n=9}$$ \pause
  Cela signifie que l'on est sûr à plus de 90~\% qu'il y a moins de 9 personnes infectées dans l'échantillon de 100 personnes.
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\pause
\begin{enumerate}
 \item On complète l'arbre de la façon suivante~: \pause


  \begin{center}
\psset{nodesep=0mm,levelsep=25mm,treesep=12mm}
\pstree[treemode=R]{\Tdot}
{
\pstree
{\Tdot~[tnpos=a]{$I$}\taput{\footnotesize$0,057$ }}
{
\Tdot~[tnpos=r]{$T$}\taput{\footnotesize$0,8$ }
\Tdot~[tnpos=r]{$\overline{T}$}\tbput{\footnotesize$0,2$ }
}
\pstree
{\Tdot~[tnpos=a]{$\overline{I}$}\tbput{\footnotesize$0,943$ }}
{
\Tdot~[tnpos=r]{$T$}\taput{\footnotesize$0,01$}
\Tdot~[tnpos=r]{$\overline{T}$}\tbput{\footnotesize$0,99$ }
}
}
\end{center}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item Les événements $I$ et $\overline{I}$ forment une partition de l'univers. \pause On a alors, d'après la formule des probabilités totales~: \pause
\begin{eqnarray*}
 P(T) & = & \pause P(I)\times P_I(T)+P(\overline{I})\times P_{\overline{I}}(T)\\ \pause
      & = & 0,057\times 0,8+0,943\times 0,01\\ \pause
      & = & 0,05503 \pause
\end{eqnarray*}
On a donc bien~:
$$\boxed{P(T)=0,05503}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Il s'agit de calculer $P_T(I)$~: \pause
\begin{eqnarray*}
 P_T(I) & = & \pause \dfrac{P(I\cap T)}{P(T)}\\ \pause
        & = & \dfrac{0,057\times 0,8}{0,05503}\\ \pause
        & \approx & 0,8286 \pause
\end{eqnarray*}
La probabilité qu'un individu ait été infecté sachant que son test est positif est donc~: \pause
$$\boxed{P_T(I)\approx 0,8286}$$
\end{enumerate}


\end{frame}



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

\pause
En prenant les même notation que dans la partie B, on a maintenant $P(T)=0,2944$. \pause Soit $x=P(I)$ la probabilité cherchée. \pause 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]{$I$}\taput{\footnotesize$x$ }}
{
\Tdot~[tnpos=r]{$T$}\taput{\footnotesize$0,8$ }
\Tdot~[tnpos=r]{$\overline{T}$}\tbput{\footnotesize$0,2$ }
}
\pstree
{\Tdot~[tnpos=a]{$\overline{I}$}\tbput{\footnotesize$1-x$ }}
{
\Tdot~[tnpos=r]{$T$}\taput{\footnotesize$0,01$}
\Tdot~[tnpos=r]{$\overline{T}$}\tbput{\footnotesize$0,99$ }
}
}
\end{center}



\end{frame}





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

D'après la formule des probabilités totales, on a~: \pause
\begin{eqnarray*}
 P(T) & = & \pause x\times 0,8+(1-x)\times 0,01\\ \pause
      & = & 0,8x+0,01-0,01x\\ \pause
      & = & 0,79x+0,01 \pause
\end{eqnarray*}
Or $P(T)=0,2944$, \pause il s'agit alors de résoudre une équation~: \pause
\begin{eqnarray*}
 0,79x+0,01=0,2944 & \Longleftrightarrow & \pause 0,79x=0,2844\\ \pause
                   & \Longleftrightarrow & x=\dfrac{0,2844}{0,79}\\ \pause
                   & \Longleftrightarrow & x=0,36 \pause
\end{eqnarray*}

La probabilité que la personne ait déjà été infectée est donc~: \pause
$$\boxed{P(I)=0,36}$$

\end{frame}





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

\begin{enumerate}
 \item \textbf{Affirmation 1~: Faux}\\ \pause
 Toute suite décroissante et minorée par $0$ est convergente mais la limite n'est pas nécessairement $0$. \pause En effet, considérons la suite $(u_n)$ définie, pour tout $n\in\mathbb{N}^*$ par~:
 $$u_n=2+\dfrac{1}{n}$$ \pause
 Cette suite est décroissante et minorée par $2$, \pause donc a fortiori minorée par $0$. \pause Et elle ne converge pas vers $0$ mais vers $2$.
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item \textbf{Affirmation 2~: Vrai}\\ \pause
 Pour tout $n\in\mathbb{N}$, on a $\dfrac{-9^n+3^n}{7^n}= \pause -\dfrac{9^n}{7^n}+\dfrac{3^n}{7^n} \pause =-\left(\dfrac{9}{7}\right)^n+\left(\dfrac{3}{7}\right)^n$. \pause Or $\dfrac{9}{7}>1$ \pause donc $\lim\limits_{n\to+\infty}\left(\dfrac{9}{7}\right)^n=+\infty$ \pause et $-1<\dfrac{3}{7}<1$ \pause donc $\lim\limits_{n\to+\infty}\left(\dfrac{3}{7}\right)^n=0$. \pause On en déduit que~:
 $$\lim\limits_{n\to+\infty}\dfrac{-9^n+3^n}{7^n}=-\infty$$ \pause
 Et par le théorème de majoration $\lim\limits_{n\to+\infty}u_n=-\infty$.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item \textbf{Affirmation 3~: Vrai}\\ \pause
 La fonction \pyv{terme(4)} renvoie la valeur de $1+0+1+2+3 \pause =7$.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item \label{ch18_ex4_q4} \textbf{Affirmation 4~: Faux}\\ \pause
\begin{itemize}
 \item La valeur du prix A est égale à $\np{1000}\times 15$, \pause soit \np{15000}~euros. \pause
 \item La valeur $S$ du prix B est égale à la somme des 15 premiers termes d'une suite géométrique de premier terme $1$ et de raison $2$. \pause On a donc~:
 $$S= \pause 1\times \dfrac{1-2^{15}}{1-2} \pause=\np{32767}$$
\end{itemize}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item \label{ch18_ex4_q5} \textbf{Affirmation 5~: Vrai}\\ \pause
Sur l'intervalle $\left[1\,;\,+\infty\right[$, la fonction $\ln$ est positive \pause donc plus $n$ est grand, \pause plus l'intégrale est grande. \pause La suite $(v_n)$ est donc croissante.
\end{enumerate}

\end{frame}




\end{document}









