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\begin{document}



\title{Amérique du nord - 22 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

 \bigskip
 \begin{center}
\psset{nodesep=0mm,levelsep=25mm,treesep=12mm}
\pstree[treemode=R]{\Tdot}
{
\pstree
{\Tdot~[tnpos=a]{$N$}\taput{\footnotesize $0,2286$}}
{
\Tdot~[tnpos=r]{$R$}\taput{\footnotesize $0,0808$}
\Tdot~[tnpos=r]{$\overline{R}$}\tbput{\footnotesize $0,9192$}
}
\pstree
{\Tdot~[tnpos=a]{$\overline{N}$}\tbput{\footnotesize $0,7714$}}
{
\Tdot~[tnpos=r]{$R$}\taput{\footnotesize $0,0127$}
\Tdot~[tnpos=r]{$\overline{R}$}\tbput{\footnotesize $0,9873$}
}
}
\end{center}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item Il s'agit de calculer $P(N\cap R)$~: \pause
 \begin{eqnarray*}
  P(N\cap R) & = & \pause P(N)\times P_N(R)\\ \pause
             & = & 0,2286\times 0,0808\\ \pause
             & \approx & 0,0185 \pause
 \end{eqnarray*}
La probabilité que le véhicule soit neuf et hybride rechargeable est donc~: \pause
$$\boxed{P(N\cap R)\approx 0,0185}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Il s'agit de calculer $P(R)$. \pause Les événements $N$ et $\overline{N}$ forment une partition de l'univers donc, \pause d'après la formule des probabilités totales~: \pause
 \begin{eqnarray*}
  P(R) & = & \pause P(N)\times P_N(R)+P(\overline{N})\times P_{\overline{N}}(R)\\ \pause
       & = & 0,2286\times 0,0808+0,7714\times 0,0127\\ \pause
       & \approx & 0,0283 \pause
 \end{eqnarray*}
La probabilité que le véhicule soit hybride rechargeable est donc~: \pause
$$\boxed{P(R)\approx 0,0283}$$
\end{enumerate}


\end{frame}


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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item \label{ch15_ex1_qA4} Il s'agit de calculer $P_R(N)$~: \pause
 \begin{eqnarray*}
  P_R(N) & = & \pause \dfrac{P(N\cap R)}{P(R)}\\ \pause
         & = & \dfrac{0,2286\times 0,0808}{0,2286\times 0,0808+0,7714\times 0,0127}\\ \pause
         & \approx & 0,6534 \pause
 \end{eqnarray*}
La probabilité que le véhicule soit neuf sachant qu'il est hybride rechargeable est donc~: \pause
$$\boxed{P_R(N)\approx 0,6534}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \item On répète 500 fois, de façon identique et indépendante, une épreuve de Bernoulli dont la probabilité de succès est égale à $0,65$. \pause La variable aléatoire $X$ est égale au nombre de succès donc~: \pause
 $$\boxed{\text{$X$ suit une loi binomiale de paramètres $n=500$ et $p=0,65$}}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item \label{ch15_ex1_qB2} Il s'agit de calculer $P(X=325)$. \pause On obtient, à l'aide de la calculatrice, \pause que la probabilité qu'exactement 325 de ces véhicules soient neufs est~: \pause
 $$\boxed{P(X=325)\approx 0,0374}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item On obtient, à l'aide de la calculatrice~: \pause
 $$\boxed{P(X\geqslant 325)\approx 0,5206}$$ \pause
 Cela signifie qu'il y a plus d'une chance sur deux qu'au moins 325 véhicules soient neufs.
\end{enumerate}


\end{frame}



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

\pause
\begin{enumerate}
 \item Soit $Y$ la variable aléatoire égale au nombre de véhicule neufs parmi les $n$ véhicules. \pause La variable aléatoire $Y$ suit alors une loi binomiale de paramètres $n$ et $p=0,65$. \pause L'événement \og tous les véhicules sont d'occasion \fg{} correspond à l'événement $Y=0$. \pause Or~:
 $$P(Y=0)= \pause \binom{n}{0}\times 0,65^0\times (1-0,65)^{n-0}$$ \pause
 Soit~:
 $$\boxed{p_n=0,35^n}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item L'événement \og au moins un des véhicule est neuf \fg{} est lévénement contraire de l'événement \og tous les véhicules sont d'occasion \fg{}, \pause on a donc $q_n=1-p_n$, \pause soit~:
 $$q_n=1-0,35^n$$ \pause
 On a alors~:
 \begin{eqnarray*}
  q_n\geqslant 0,9999 & \Longleftrightarrow & \pause 1-0,35^n\geqslant 0,9999 \\ \pause
            & \Longleftrightarrow & -0,35^n\geqslant -0,0001 \\ \pause
            & \Longleftrightarrow & 0,35^n\leqslant 0,0001 \\ \pause
            & \Longleftrightarrow & \ln\left(0,35^n\right)\leqslant \ln(0,0001) \\ \pause
            & \Longleftrightarrow & n\ln(0,35)\leqslant \ln(0,0001) \\ \pause
            & \Longleftrightarrow & n\geqslant \dfrac{\ln(0,0001)}{\ln(0,35)} \quad  \textsl{(car $\ln(0,35)<0$)} \pause
 \end{eqnarray*}

\leftskip=1.5cm

Or $\dfrac{\ln(0,0001)}{\ln(0,35)}\approx 8,77$ \pause donc la plus petite valeur de $n$ telle que $q_n\geqslant 0,9999$ est~: \pause
$$\boxed{n=9}$$
\end{enumerate}


\end{frame}




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

\pause
\begin{enumerate}
 \item On a~:
 $$\boxed{F(3\,;\,0\,;\,1)} \pause \qquad \boxed{H(0\,;\,1\,;\,1)} \pause \qquad \boxed{M\left(\dfrac{3}{2}\,;\,1\,;\,0\right)}$$
\end{enumerate}


\end{frame}



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

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


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item On en déduit que le plan $(HMF)$ admet une équation cartésienne de la forme~: \pause
  $$2x+6y+3z+d=0 \quad \text{avec $d\in\mathbb{R}$}$$ \pause
  De plus, le point $H(0\,;\,1\,;\,1)$ appartient au plan $(HMF)$ \pause donc ses coordonnées vérifient l'équation, \pause soit $2\times 0+6\times 1+3\times 1+d=0$ \pause et donc $d=-9$. \pause Une équation cartésienne du plan $(HMF)$ est donc~: \pause
  $$\boxed{2x+6y+3z-9=0}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(c)]
  \item D'après son équation cartésienne, \pause le plan $\mathscr{P}$ admet pour vecteur normal \pause le vecteur $\overrightarrow{n'}\begin{pmatrix}5\\15\\-3 \end{pmatrix}$. \pause Les vecteurs $\overrightarrow{n}$ et $\overrightarrow{n'}$ ne sont pas colinéaires donc~: \pause
  $$\boxed{\text{Les plans $\mathscr{P}$ et $(HMF)$ ne sont pas parallèles}}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item La droite $(DG)$ admet pour représentation paramétrique~: \pause
 $$\begin{cases}x=3t\\y=1\\z=t \end{cases} \quad t\in\mathbb{R}$$ \pause
 On injecte les expressions de la représentation paramétrique dans l'équation cartésienne du plan $(HMF)$~: \pause
 \begin{eqnarray*}
  2\times 3t+6\times 1+3\times t-9=0 & \Longleftrightarrow & \pause  6t+6+3t-9=0\\ \pause
                    & \Longleftrightarrow & 9t=3\\ \pause
                    & \Longleftrightarrow & t=\dfrac{1}{3} \pause
 \end{eqnarray*}

\leftskip=1.5cm

Le point $N$ est donc le point de paramètre $t=\dfrac{1}{3}$ dans la représentation paramétrique de la droite $(DG)$, \pause soit~:
$$\boxed{N\left(1\,;\,1\,;\,\dfrac{1}{3}\right)}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item \label{ch15_ex2_q4} On a $\overrightarrow{RG}\begin{pmatrix}0\\3/4\\1/2\end{pmatrix}$ \pause donc le vecteur $\overrightarrow{RG}$ n'est pas colinéaire au vecteur $\overrightarrow{n}$. \pause On en déduit que $\overrightarrow{RG}$ n'est pas orthogonal au plan $(HMF)$ \pause et donc que~: \pause
 $$\boxed{\text{$R$ n'est pas le projeté orthogonal de $G$ sur le plan $(HMF)$}}$$
\end{enumerate}


\end{frame}







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

\pause
\begin{enumerate}
 \setcounter{enumi}{0}
 \item La fonction $g$ est dérivable sur $\left[0\,;\,1\right]$ et pour tout $x\in\left[0\,;\,1\right]$~: \pause
 $$g'(x)= \pause 2-2x \pause =2(1-x)$$ \pause
 Pour tout $x\in\left[0\,;\,1\right[$, on a $g'(x)>0$ \pause donc~:
 $$\boxed{\text{$g$ est strictement croissante sur $\left[0\,;\,1\right]$}}$$ \pause
 Et on a~: \pause
 $$\boxed{g(0)= \pause 0} \pause \quad \text{et} \quad \boxed{g(1)= 1}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item On a~:
 \begin{itemize}
  \item $u_1= \pause g\left(u_0\right) \pause =g\left(\dfrac{1}{2}\right) \pause=2\times \dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2 \pause =1-\dfrac{1}{4} \pause =\dfrac{3}{4}$ \pause
  \item $u_2= \pause g\left(u_1\right) \pause =g\left(\dfrac{3}{4}\right) \pause =2\times \dfrac{3}{4}-\left(\dfrac{3}{4}\right)^2 \pause =\dfrac{3}{2}-\dfrac{9}{16} \pause =\dfrac{15}{16}$ \pause
 \end{itemize}
Soit~:
$$\boxed{u_1=\dfrac{3}{4}} \pause \quad \text{et} \quad \boxed{u_2=\dfrac{15}{16}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Montrons par récurrence que, pour tout entier naturel $n$, on a~: $0<u_n<u_{n+1}<1$. \pause
 \begin{itemize}
  \item \textbf{Initialisation~:}\\ \pause
  Pour $n=0$, \pause on a $u_0=0,5$ \pause et $u_1=0,75$. \pause On a donc bien~:
  $$0<0,5<0,75<1$$ \pause
  La propriété est donc 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 $0<u_n<u_{n+1}<1$. \pause On a alors, en appliquant la fonction $g$ qui est strictement croissante sur $\left[0\,;\,1\right]$~: \pause
  $$g(0)<g(u_n)<g(u_{n+1})<g(1)$$ \pause
  Soit~:
  $$0<u_{n+1}<u_{n+2}<1$$ \pause

\leftskip=1.5cm
\vspace{-0.3cm}
  La propriété est donc vraie au rang $n+1$. \pause
  \item \textbf{Conclusion~:}\\ \pause
  La propriété est vraie pour $n=0$ et elle est héréditaire, \pause elle est donc vraie pour tout $n\in\mathbb{N}$.
 \end{itemize}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item \label{ch15_ex3_q3} D'après la question précédente, la suite $(u_n)$ est~: \pause
 \begin{itemize}
  \item croissante (car $u_n<u_{n+1}$ pour tout $n\in\mathbb{N}$) \pause
  \item majorée par $1$ (car $u_n<1$ pour tout $n\in\mathbb{N}$) \pause
 \end{itemize}
On en déduit que~: \pause
$$\boxed{\text{La suite $(u_n)$ est convergente}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item La fonction $g$ étant continue sur $\left[0\,;\,1\right]$, \pause on sait que la limite $l$ est un point fixe de $g$, \pause c'est-à-dire une solution de l'équation $g(x)=x$. \pause Résolvons cette équation, soit $x\in\left[0\,;\,1\right]$~: \pause
 \begin{eqnarray*}
  g(x)=x & \Longleftrightarrow & \pause 2x-x^2=x\\ \pause
         & \Longleftrightarrow & x-x^2=0\\ \pause
         & \Longleftrightarrow & x(1-x)=0\\ \pause
         & \Longleftrightarrow & x=0 \quad \text{ou} \quad x=1
 \end{eqnarray*} \pause
Or $u_0=\dfrac{1}{2}$ \pause et la suite $(u_n)$ est croissante, \pause elle ne peut donc pas converger vers $0$. \pause On en déduit que~: \pause
$$\boxed{l=1}$$
\end{enumerate}

\end{frame}



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

\vspace{-0.5cm}
\begin{enumerate}
 \setcounter{enumi}{5}
 \item Pour tout $n\in\mathbb{N}$, on a~: \pause
 \vspace{-0.2cm}
 \begin{eqnarray*}
  v_{n+1} & = & \pause \ln(1-u_{n+1})\\ \pause
          & = & \ln(1-g(u_n))\\ \pause
          & = & \ln(1-2u_n+u_n^2)\\ \pause
          & = & \ln\left((1-u_n)^2\right)\\ \pause
          & = & 2\ln(1-u_n)\\ \pause
          & = & 2v_n \pause
 \end{eqnarray*}
\vspace{-0.2cm}
Et comme $v_0= \pause \ln\left(1-\dfrac{1}{2}\right) \pause =\ln\left(\dfrac{1}{2}\right) \pause =-\ln(2)$ \pause, on en déduit que~: \pause
$$\boxed{\text{La suite $(v_n)$ est géométrique de raison $2$ et de $1^{\text{er}}$ terme $v_0=-\ln(2)$}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{6}
 \item On en déduit que, pour tout $n\in\mathbb{N}$, \pause $v_n=v_0\times 2^n$ \pause, soit~:
 $$\boxed{v_n=-\ln(2)\times 2^n}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{7}
 \item \label{ch15_ex3_q8} Pour tout $n\in\mathbb{N}$, on a $v_n=\ln(1-u_n)$ \pause donc, en appliquant la fonction exponentielle~: \pause
 $$\mathrm{e}^{v_n}=1-u_n$$ \pause
 Et donc~:
 $$u_n=1-\mathrm{e}^{v_n}$$ \pause
 Soit~:
 $$\boxed{u_n=1-\mathrm{e}^{-\ln(2)\times 2^n}}$$ \pause

 Et comme $\lim\limits_{n\to +\infty}\left(-\ln(2)\times 2^n\right)=-\infty$ \pause donc par composition, \pause $\lim\limits_{n\to +\infty}\left(\mathrm{e}^{-\ln(2)\times 2^n}\right)=0$ \pause d'où~:
 $$\boxed{\lim\limits_{n\to +\infty}u_n=1}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{8}
 \item On peut compléter le script de la façon suivante~: \pause

\begin{pyverbatim}[][frame=single]
def seuil() :
    n = 0
    u = 0.5
    while u < 0.95 :
        n = n+1
        u = 2*u - u**2
    return n
\end{pyverbatim}
\end{enumerate}

\end{frame}





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

\begin{enumerate}
 \item Il s'agit de résoudre l'équation $f(x)=0$. \pause Soit $x\in\left]0\,;\,+\infty\right[$, on a~: \pause
 \begin{eqnarray*}
  f(x)=0 & \Longleftrightarrow & \pause a\ln(x)=0\\ \pause
         & \Longleftrightarrow & \ln(x)=0 \quad \text{(car $a\neq 0$)}\\ \pause
         & \Longleftrightarrow & x=1 \pause
 \end{eqnarray*}
L'abscisse du point d'intersection de la courbe $\mathscr{C}$ et de l'axe des abscisses est donc~: \pause
$$\boxed{x=1}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \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\left(1\times \ln(x)+x\times \dfrac{1}{x}-1\right)\\ \pause
        & = & a\left(\ln(x)+1-1\right)\\ \pause
        & = & a\ln(x)\\ \pause
        & = & f(x) \pause
 \end{eqnarray*}
 On en déduit que~: \pause
 $$\boxed{\text{$F$ est une primitive de $f$ sur $\left]0\,;\,+\infty\right[$}}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Il s'agit de calculer l'intégrale entre $1$ et $x_0$ de la fonction $f$~: \pause
 \begin{eqnarray*}
  \int_1^{x_0}f(x)\,\mathrm{d}x & = & \pause \int_1^{x_0}a\ln(x)\,\mathrm{d}x\\ \pause
                  & = & \Big[a(x\ln(x)-x)\Big]_1^{x_0}\\ \pause
                  & = & a(x_0\ln(x_0)-x_0)-a(1\times \ln(1)-1)\\ \pause
                  & = & a(x_0\ln(x_0)-x_0)+a\\ \pause
                  & = & a(x_0\ln(x_0)-x_0+1) \pause
 \end{eqnarray*}
 L'aire du domaine grisé est donc~: \pause
 $$\boxed{\mathscr{A}=a(x_0\ln(x_0)-x_0+1)}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item Déterminons l'équation réduite de la tangente $T$. \pause La fonction $f$ est dérivable sur $\left]0\,;\,+\infty\right[$ et $f'(x)= \pause \dfrac{a}{x}$. \pause
 \begin{itemize}
  \item $f(x_0)= \pause a\ln(x_0)$ \pause
  \item $f'(x_0)= \pause \dfrac{a}{x_0}$ \pause
 \end{itemize}
La tangente $T$ admet donc pour équation~: \pause
$$y=\dfrac{a}{x_0}\left(x-x_0\right)+a\ln(x_0)$$ \pause
Soit~:
$$y=\dfrac{a}{x_0}x-a+a\ln(x_0)$$
\end{enumerate}


\end{frame}



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


L'ordonnée à l'origine est donc $-a+a\ln(x_0)$, \pause il s'agit de l'ordonnée du point $A$. \pause Et comme l'ordonnée du point $B$ est $a\ln(x_0)$, on a~: \pause
\begin{eqnarray*}
 AB & = & \pause y_B-y_A \\ \pause
    & = & a\ln(x_0)-(-a+a\ln(x_0))\\ \pause
    & = & a\ln(x_0)+a-a\ln(x_0)\\ \pause
    & = & a \pause
\end{eqnarray*}
La longueur $AB$ est donc égale à une constante. \pause Plus précisément~:
$$\boxed{AB=a}$$



\end{frame}




\end{document}









