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\begin{document}



\title{Asie - 11 juin 2024}
\author{Spécialité mathématiques - Baccalauréat}
\date{}



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


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

\pause
\vspace{-0.6cm}
\begin{enumerate}
 \item \begin{itemize}
  \item \textbf{Limite en $0$~:}\\ \pause
  On a~:
  \vspace{-0.2cm}
 $$\boxed{\lim\limits_{x\to 0}f(x)=0} \pause \quad \text{car} \quad \begin{cases}\lim\limits_{x\to 0}x^2=0\\\lim\limits_{x\to 0}x\ln(x)=0 \quad \textsl{(par croissances comparées)}\end{cases}$$ \pause
  \item \textbf{Limite en $+\infty$~:}\\ \pause
  Pour tout $x\in\left]0\,;\,+\infty\right[$, on a~: \pause
 $$f(x)=x^2\left(1-\dfrac{\ln(x)}{x}\right)$$ \pause
 Et donc~:
 \vspace{-0.3cm}
 $$\hspace{-0.9cm}\boxed{\lim\limits_{x\to +\infty}f(x)=+\infty} \pause \quad \text{car} \quad \begin{dcases}\lim\limits_{x\to +\infty}x^2=+\infty\\\lim\limits_{x\to +\infty}\dfrac{\ln(x)}{x}=0 \quad \textsl{(par croissances comparées)}\end{dcases}$$
 \end{itemize}
\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 2x-\left(1\times \ln(x)+x\times \dfrac{1}{x}\right)\\ \pause
      & = & 2x-\left(\ln(x)+1\right)\\ \pause
      & = & 2x-\ln(x)-1 \pause
 \end{eqnarray*}
Soit~:
$$\boxed{f'(x)=2x-\ln(x)-1}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \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 2-\dfrac{1}{x}\\ \pause
         & = & \dfrac{2x-1}{x} \pause
 \end{eqnarray*}
Soit~:
 $$\boxed{f''(x)=\dfrac{2x-1}{x}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item Pour tout $x\in\left]0\,;\,+\infty\right[$, $x>0$ donc $f'(x)$ est du signe de $2x-1$. \pause On en déduit le tableau~: \pause

 \begin{center}
\begin{tikzpicture}[scale=1]
\tkzTabInit[ lgt=2, deltacl=0.8, espcl=3]%
{$x$ /0.8,
$f''(x)$ /0.8,
$f'(x)$ /2}
{ $0$  , $\mfrac{1}{2}$ , $+\infty$}
\tkzTabLine { d , - , z , + ,  }
\tkzTabVar {D+/ , -/$\ln(2)$ , +/}
\end{tikzpicture}
\end{center}

\pause
Le minimum est atteint en $\dfrac{1}{2}$ et~: \pause
$$\hspace{0.5cm} f'\left(\dfrac{1}{2}\right)= \pause 2\times \dfrac{1}{2}-\ln\left(\dfrac{1}{2}\right)-1 \pause =1+\ln(2)-1 \pause=\ln(2)$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item D'après son tableau de variations, la fonction $f'$ atteint son minimum en $\dfrac{1}{2}$ \pause et ce minimum vaut $\ln(2)$. \pause Or $\ln(2)\approx 0,69 \pause >0$ \pause donc $f'(x)>0$ pour tout $x\in\left]0\,;\,+\infty\right[$. \pause On en déduit que~: \pause

$$\boxed{\text{La fonction $f$ est strictement croissante sur $\left]0\,;\,+\infty\right[$}}$$
\end{enumerate}

\end{frame}



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

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


\end{frame}



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

Or, pour tout $x\in\left]0\,;\,+\infty\right[$, $x>0$ \pause donc $g'(x)$ est du signe de $x-1$. \pause Et comme $g(1)= \pause 1-\ln(1) \pause =1$ \pause, on a le tableau~: \pause

 \begin{center}
\begin{tikzpicture}[scale=1]
\tkzTabInit[ lgt=2, deltacl=0.8, espcl=3]%
{$x$ /0.8,
$g'(x)$ /0.8,
$g(x)$ /2}
{ $0$  , $1$ , $+\infty$}
\tkzTabLine { d , - , z , + ,  }
\tkzTabVar {D+/ , -/$1$ , +/}
\end{tikzpicture}
\end{center}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item \label{ch19_ex1_qB2} Pour tout $x\in\left]0\,;\,+\infty\right[$, on a~: \pause
 \begin{eqnarray*}
  f(x)=x & \Longleftrightarrow & \pause  x^2-x\ln(x)=x\\ \pause
         & \Longleftrightarrow & x^2-x\ln(x)-x=0\\ \pause
         & \Longleftrightarrow & x\left(x-\ln(x)-1)\right)=0\\ \pause
         & \Longleftrightarrow & x-\ln(x)-1=0 \quad \textsl{(car $x\neq 0$)}\\ \pause
         & \Longleftrightarrow & x-\ln(x)=1\\ \pause
         & \Longleftrightarrow & g(x)=1 \pause
 \end{eqnarray*}
Or l'équation $g(x)=1$ admet $1$ pour unique solution \pause donc l'équation $f(x)=x$ admet pour ensemble solution sur l'intervalle $\left]0\,;\,+\infty\right[$~: \pause
$$\boxed{\mathscr{S}=\left\lbrace 1 \right\rbrace}$$
\end{enumerate}

\end{frame}



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

\pause
\begin{enumerate}
 \item Montrons par récurrence que, pour tout entier naturel $n$, $\dfrac{1}{2}\leqslant u_n\leqslant u_{n+1}\leqslant 1$. \pause
 \begin{itemize}
  \item \textbf{Initialisation~:}\\ \pause
  Pour $n=0$, on a $u_0=\dfrac{1}{2}$ \pause et~:
  $$u_1= \pause f\left(\dfrac{1}{2}\right) \pause =\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}\ln\left(\dfrac{1}{2}\right) \pause =\dfrac{1}{4}+\dfrac{1}{2}\ln(2) \pause \approx 0,6$$ \pause
  On a donc bien $\dfrac{1}{2}\leqslant u_0\leqslant u_1\leqslant 1$ \pause et la propriété est vraie au rang $n=0$.
 \end{itemize}
\end{enumerate}


\end{frame}



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


\begin{itemize}
 \item \textbf{Hérédité~:}\\ \pause
  Supposons la propriété vraie pour un certain rang $n\in\mathbb{N}$, \pause c'est-à-dire~: \pause
  $$\dfrac{1}{2}\leqslant u_n\leqslant u_{n+1}\leqslant 1$$ \pause
  On a alors, en appliquant la fonction $f$ qui est croissante sur l'intervalle $\left[\dfrac{1}{2}\,;\,1\right]$~: \pause
  $$f\left(\dfrac{1}{2}\right)\leqslant f(u_n)\leqslant f(u_{n+1})\leqslant f(1)$$ \pause
  Soit~:
  $$f\left(\dfrac{1}{2}\right)\leqslant u_{n+1}\leqslant u_{n+2}\leqslant 1$$ \pause
  Et comme $f\left(\dfrac{1}{2}\right)\approx 0,6\geqslant \dfrac{1}{2}$, \pause on a, a fortiori~: \pause
  $$\dfrac{1}{2}\leqslant u_{n+1}\leqslant u_{n+2}\leqslant 1$$

  \leftskip=1.5cm

  La propriété est donc vraie au rang $n+1$.
\end{itemize}


\end{frame}



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


\begin{itemize}
  \item \textbf{Conclusion~:}\\ \pause
  La propriété est vraie pour $n=0$ et elle est héréditaire, elle est donc vraie pour tout $n\in\mathbb{N}$. \pause Soit, pour tout $n\in\mathbb{N}$~:
  $$\boxed{\dfrac{1}{2}\leqslant u_n\leqslant u_{n+1}\leqslant 1}$$
\end{itemize}


\end{frame}



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

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

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item \label{ch19_ex1_qC3} On a vu, dans la question \ref{ch19_ex1_qB2} de la partie B, que l'équation $f(x)=x$ admettait $1$ pour unique solution sur $\left]0\,;\,+\infty\right[$. \pause Or $l$ est solution de cette équation donc~: \pause
 $$\boxed{l=1}$$
\end{enumerate}

\end{frame}






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

\pause
\begin{enumerate}
 \item Il s'agit de la probabilité que Léa perde la deuxième partie sachant qu'elle a gagné la première. \pause Or lorsqu'elle gagne une partie, elle gagne la suivante dans 70~\% des cas donc elle la perd dans 30~\% des cas. \pause On a donc~:
 $$\boxed{P_{G_1}(D_2)=0,3}$$
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item On complète l'arbre de la façon suivante~: \pause

 \bigskip

 \begin{center}
\psset{nodesep=0mm,levelsep=25mm,treesep=12mm}
\pstree[treemode=R]{\Tdot}
{
\pstree
{\Tdot~[tnpos=a]{$G_1$}\taput{\footnotesize$0,5$ }}
{
\Tdot~[tnpos=r]{$G_2$}\taput{\footnotesize$0,7$ }
\Tdot~[tnpos=r]{$D_2$}\tbput{\footnotesize$0,3$ }
}
\pstree
{\Tdot~[tnpos=a]{$D_1$}\tbput{\footnotesize$0,5$ }}
{
\Tdot~[tnpos=r]{$G_2$}\taput{\footnotesize$0,2$ }
\Tdot~[tnpos=r]{$D_2$}\tbput{\footnotesize$0,8$ }
}
}
\end{center}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item Il s'agit de calculer $P(G_2)$. \pause Les événements $G_1$ et $D_1$ forment une partition de l'univers donc, \pause d'après la formule des probabilités totales~: \pause
\begin{eqnarray*}
 P(G_2) & = & \pause P(G_1)\times P_{G_1}(G_2)+P(D_1)\times P_{D_1}(G_2)\\ \pause
        & = & 0,5\times 0,7+0,5\times 0,2\\ \pause
        & = & 0,45 \pause
\end{eqnarray*}
Soit~:
$$\boxed{g_2=0,45}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(a)]
  \item On complète l'arbre de la façon suivante~: \pause

\bigskip

 \begin{center}
\psset{nodesep=0mm,levelsep=25mm,treesep=12mm}
\pstree[treemode=R]{\Tdot}
{
\pstree
{\Tdot~[tnpos=a]{$G_n$}\taput{\footnotesize$g_n$ }}
{
\Tdot~[tnpos=r]{$G_{n+1}$}\taput{\footnotesize$0,7$ }
\Tdot~[tnpos=r]{$D_{n+1}$}\tbput{\footnotesize$0,3$ }
}
\pstree
{\Tdot~[tnpos=a]{$D_n$}\tbput{\footnotesize$1-g_n$ }}
{
\Tdot~[tnpos=r]{$G_{n+1}$}\taput{\footnotesize$0,2$ }
\Tdot~[tnpos=r]{$D_{n+1}$}\tbput{\footnotesize$0,8$ }
}
}
\end{center}
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(b)]
  \item Les événements $G_n$ et $D_n$ forment une partition de l'univers donc, \pause d'après la formule des probabilités totales~: \pause
 \begin{eqnarray*}
  P(G_{n+1}) & = & \pause P(G_n)\times P_{G_n}(G_{n+1})+P(D_n)\times P_{D_n}(G_{n+1})\\ \pause
             & = & g_n\times 0,7+(1-g_n)\times 0,2\\ \pause
             & = & 0,7g_n+0,2-0,2g_n\\ \pause
             & = & 0,5g_n+0,2 \pause
 \end{eqnarray*}
Soit, pour tout entier naturel $n$ non nul~: \pause
$$\boxed{g_{n+1}=0,5g_n+0,2}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item
 \begin{enumerate}[(a)]
  \item \label{ch19_ex2_q5a} Pour tout entier naturel $n$ non nul, on a~: \pause
  \begin{eqnarray*}
   v_{n+1} & = & \pause g_{n+1}-0,4\\ \pause
           & = & 0,5g_n+0,2-0,4\\ \pause
           & = & 0,5g_n-0,2\\ \pause
           & = & 0,5\left(g_n-0,4\right)\\ \pause
           & = & 0,5v_n \pause
  \end{eqnarray*}
Et comme $v_1=g_1-0,4=0,5-0,4=0,1$, on en déduit que~: \pause
$$\boxed{\text{La suite $(v_n)$ est géométrique de raison $0,5$ et de $1^{\text{er}}$ terme $v_1=0,1$}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{4}
 \item
 \begin{enumerate}[(b)]
  \item D'après la question précédente, on a pour tout $n\in\mathbb{N}^*$, \pause $v_n=v_1\times 0,5^{n-1}$ \pause , soit~:
  $$v_n=0,1\times 0,5^{n-1}$$ \pause
  Et comme $v_n=g_n-0,4$ \pause , on a $g_n=v_n+0,4$ \pause , soit pour tout $n\in\mathbb{N}^*$~: \pause
$$\boxed{g_n=0,1\times 0,5^{n-1}+0,4}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{5}
 \item Pour tout $n\in\mathbb{N}^*$, on a~: \pause
 \begin{eqnarray*}
  g_{n+1}-g_n & = & \pause  0,1\times 0,5^{n}+0,4-(0,1\times 0,5^{n-1}+0,4)\\ \pause
              & = & 0,1\times 0,5^{n}+0,4-0,1\times 0,5^{n-1}-0,4\\ \pause
              & = & 0,1\times 0,5^{n}-0,1\times 0,5^{n-1}\\ \pause
              & = & 0,1\times 0,5^{n-1}(0,5-1)\\ \pause
              & = & 0,1\times 0,5^{n-1}\times (-0,5)\\ \pause
              & = & -0,1\times 0,5^n \pause
 \end{eqnarray*}
On en déduit que, pour tout $n\in\mathbb{N}^*$, $g_{n+1}-g_n<0$ \pause et donc que~: \pause
$$\boxed{\text{La suite $(g_n)$ est strictement décroissante}}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{6}
 \item On a $\lim\limits_{n\to+\infty}\left(0,5^{n-1}\right)=0$ \pause car $-1<0,5<1$ \pause donc~:
 $$\boxed{\lim\limits_{n\to+\infty}g_n=0,4}$$ \pause
 Cela signifie qu'à long terme, la probabilité que Léa gagne une partie sera proche de $0,4$. \pause Ou encore que Léa gagnera environ 40~\% des parties.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{7}
 \item Pour tout $n\in\mathbb{N}^*$, on a~: \pause
 \begin{eqnarray*}
  g_n-0,4\leqslant 0,001 & \Longleftrightarrow & \pause 0,1\times 0,5^{n-1}+0,4-0,4\leqslant 0,001\\ \pause
                         & \Longleftrightarrow & 0,1\times 0,5^{n-1}\leqslant 0,001\\ \pause
                         & \Longleftrightarrow & 0,5^{n-1}\leqslant 0,01\\ \pause
                         & \Longleftrightarrow & \ln\left(0,5^{n-1}\right)\leqslant \ln(0,01)\\ \pause
                         & \Longleftrightarrow & (n-1)\ln(0,5)\leqslant \ln(0,01)\\ \pause
                         & \Longleftrightarrow & n-1\geqslant \dfrac{\ln(0,01)}{\ln(0,5)} \quad \textsl{(car $\ln(0,5)<0$)} \pause
 \end{eqnarray*}
Or $\dfrac{\ln(0,01)}{\ln(0,5)}\approx 6,6$ \pause donc $n-1\geqslant 7$. \pause On en déduit que le plus petit entier $n$ tel que $g_n-0,4\leqslant 0,001$ est~: \pause
$$\boxed{n=8}$$
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{8}
 \item On peut compléter l'algorithme de la façon suivante~: \pause

\begin{pyverbatim}[][frame=single]
def seuil(e):
    g = 0.5
    n = 1
    while g > 0.4 + e :
        g = 0.5*g + 0.2
        n = n+1
    return (n)
\end{pyverbatim}
\end{enumerate}

\end{frame}







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

\begin{enumerate}
 \item \label{ch19_ex3_q1} \textbf{Affirmation 1~: Vrai}\\ \pause
 On a~:
 $$\lim\limits_{n\to +\infty}\dfrac{3n^2+4n+7}{6n^2+1} \pause =\lim\limits_{n\to +\infty}\dfrac{3n^2}{6n^2} \pause =\lim\limits_{n\to +\infty}\dfrac{3}{6} \pause =\dfrac{1}{2}$$ \pause
 Donc, d'après le théorème des gendarmes, \pause $\lim\limits_{n\to +\infty}u_n=\dfrac{1}{2}$.
\end{enumerate}


\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item \label{ch19_ex3_q2} \textbf{Affirmation 2~: Faux}\\ \pause
 On peut remarquer que, sur l'intervalle $[-1\,;\,3]$, la fonction dérivée $h'$ est croissante puis décroissante. \pause On en déduit que la fonction $h$ est convexe puis concave.
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{2}
 \item \label{ch19_ex3_q3} \textbf{Affirmation 3~: Vrai} \pause
 \begin{itemize}
  \item Calculons d'abord le nombre total de codes possibles. \pause Le nombre de possibilités pour le choix des 4 chiffres est égal au nombre de $4$-uplets dans un ensemble à $10$ éléments \pause (les 10 chiffres compris entre $0$ et $9$). \pause Il y en a donc $10^4 \pause =\np{10000}$. \pause Le nombre de possibilités pour le choix des 2 lettres est égal au nombre de $2$-uplets d'éléments distincts dans un ensemble à $3$ éléments \pause (les 3 lettres A, B et C). \pause Il y en a donc $3\times 2 \pause =6$. \pause Le nombre de codes possibles est donc $\np{10000}\times 6 \pause =\np{60000}$. \pause
  \item Déterminons maintenant le nombre de codes ne contenant aucun $0$. \pause Le nombre de possibilités pour le choix des 4 chiffres est égal au nombre de $4$-uplets dans un ensemble à $9$ éléments \pause (les 9 chiffres compris entre $1$ et $9$). \pause Il y en a donc $9^4 \pause =\np{6561}$ \pause Le nombre de possibilités pour le choix des 2 lettres est égal au nombre de $2$-uplets d'éléments distincts dans un ensemble à $3$ éléments \pause (les 3 lettres A, B et C). \pause Il y en a donc $3\times 2 \pause =6$. \pause

  \leftskip=1cm

  Le nombre de codes possibles ne contenant aucun $0$ est donc \pause $\np{6561}\times 6 \pause =\np{39366}$. \pause
 \item Finalement, le nombre de codes contenant au moins un $0$ est égal à \pause $\np{60000}-\np{39366} \pause =\np{20634}$.
 \end{itemize}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item \label{ch19_ex3_q4} \textbf{Affirmation 4~: Vrai}\\ \pause
La fonction $f$ est dérivable sur $\left]0\,;\,+\infty\right[$ et, pour tout $x\in\left]0\,;\,+\infty\right[$, on a~: \pause
$$f'(x)= \pause 1\times \ln(x)+x\times \dfrac{1}{x}=\ln(x)+1$$ \pause
On a alors, pour tout $x\in\left]0\,;\,+\infty\right[$~: \pause
\begin{eqnarray*}
 x\times f'(x)-f(x) & = & \pause x\left(\ln(x)+1\right)-x\ln(x)\\ \pause
                    & = & x\ln(x)+x-x\ln(x)\\ \pause
                    & = & x \pause
\end{eqnarray*}
Donc la fonction $f$ est bien solution sur $\left]0\,;\,+\infty\right[$ de l'équation différentielle $xy'-y=x$.
\end{enumerate}

\end{frame}





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

\pause
\begin{enumerate}
 \item On a~: \pause
 \begin{itemize}
  \item $2\times 1+2\times 0-3\times 1+1= \pause 2-3+1 \pause =0$ \pause donc~:
  $$\boxed{\text{Le point $A$ appartient au plan $(P)$}}$$ \pause
  \item $2\times 2+2\times (-1)-3\times 1+1= \pause 4-2-3+1 \pause =0$ \pause donc~:
  $$\boxed{\text{Le point $B$ appartient au plan $(P)$}}$$
  \item $2\times (-4)+2\times (-6)-3\times 5+1= \pause -8-12-15+1 \pause =-34 \pause \neq 0$ \pause donc~:
  $$\boxed{\text{Le point $C$ n'appartient pas au plan $(P)$}}$$
 \end{itemize}
\end{enumerate}


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\begin{frame}

\vspace{-0.7cm}
\begin{enumerate}
 \setcounter{enumi}{1}
 \item \label{ch19_ex4_qA2} Il s'agit de montrer que le point $C'$ appartient au plan $(P)$ et que le vecteur $\overrightarrow{CC'}$ est orthogonal au plan $(P)$. \pause
 \begin{itemize}
  \item On a $2\times 0+2\times (-2)-3\times (-1)+1= \pause -2+3-1 \pause =0$ \pause donc le point $C'$ appartient au plan $(P)$. \pause
  \item On a $\overrightarrow{CC'}\begin{pmatrix}4\\4\\-6\end{pmatrix}$ \pause et le plan $(P)$ admet pour vecteur normal le vecteur $\overrightarrow{n}\begin{pmatrix}2\\2\\-3\end{pmatrix}$. \pause On remarque que $\overrightarrow{CC'}=2\overrightarrow{n}$ \pause donc les vecteurs $\overrightarrow{CC'}$ et $\overrightarrow{n}$ sont colinéaires. \pause On en déduit que le vecteur $\overrightarrow{CC'}$ est orthogonal au plan $(P)$. \pause
 \end{itemize}
On en déduit que~: \pause
\vspace{-0.1cm}
$$\boxed{\text{Le point $C'$ est le projeté orthogonal du point $C$ sur le plan $(P)$}}$$
\end{enumerate}

\end{frame}



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\begin{frame}

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

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{3}
 \item \label{ch19_ex4_qA4} Le point $H$ appartient à la droite $(AB)$, \pause il existe donc un réel $t$ tel que $H(1+t\,;\,-t\,;\,1)$. \pause On a alors $\overrightarrow{HC}\begin{pmatrix}-5-t\\-6+t\\4 \end{pmatrix}$ \pause et donc~:
 \begin{eqnarray*}
  \hspace{-1.2cm}\text{$\overrightarrow{AB}$ et $\overrightarrow{HC}$ sont orthogonaux} & \Longleftrightarrow & \pause \overrightarrow{AB}.\overrightarrow{HC}=0\\ \pause
    & \Longleftrightarrow & 1\times (-5-t)+(-1)\times (-6+t)+0\times 4=0\\ \pause
    & \Longleftrightarrow & -5-t+6-t=0\\ \pause
    & \Longleftrightarrow & 2t=1\\ \pause
    & \Longleftrightarrow & t=\dfrac{1}{2} \pause
 \end{eqnarray*}
Le point $H$ est donc le point de paramètre $t=\mfrac{1}{2}$ \pause , soit~:
$$\boxed{H\left(\dfrac{3}{2}\,;\,-\dfrac{1}{2}\,;\,1\right)}$$
\end{enumerate}

\end{frame}



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

\pause
\begin{enumerate}
 \setcounter{enumi}{0}
 \item On a~:
 \begin{eqnarray*}
  \|\overrightarrow{HC}\| & = & \pause \sqrt{\left(-\dfrac{11}{2}\right)^2+\left(-\dfrac{11}{2}\right)^2+4^2}\\ \pause
            & = & \sqrt{\dfrac{121}{4}+\dfrac{121}{4}+16}\\ \pause
            & = & \sqrt{\dfrac{153}{2}} \pause
 \end{eqnarray*}
Soit~:
$$\boxed{\|\overrightarrow{HC}\|=\sqrt{\dfrac{153}{2}}}$$
\end{enumerate}

\end{frame}



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


\begin{enumerate}
 \setcounter{enumi}{1}
 \item On a $AB= \pause \sqrt{1^2+(-1)^2} \pause =\sqrt{2}$ \pause et $[CH]$ est la hauteur issue de $C$ dans le triangle $ABC$. \pause On a donc~:
 \begin{eqnarray*}
  S & = & \pause \dfrac{AB\times CH}{2}\\ \pause
    & = & \dfrac{\sqrt{2}\times \sqrt{\mfrac{153}{2}}}{2}\\ \pause
    & = & \dfrac{\mfrac{\sqrt{2}\times \sqrt{153}}{\sqrt{2}}}{2}\\ \pause
    & = & \dfrac{\sqrt{153}}{2} \pause
 \end{eqnarray*}
Soit~:
$$\boxed{S=\dfrac{3\sqrt{17}}{2}}$$
\end{enumerate}

\end{frame}



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


\begin{enumerate}
 \setcounter{enumi}{0}
 \item Dans le triangle $CHC'$, rectangle en $C'$, on a~: \pause
 \begin{eqnarray*}
  \cos\left(\widehat{CHC'}\right) & = & \pause \dfrac{HC'}{HC}\\ \pause
                                  & = & \dfrac{\sqrt{\mfrac{17}{2}}}{\sqrt{\mfrac{153}{2}}}\\ \pause
                                  & = & \dfrac{\sqrt{17}\times \sqrt{2}}{\sqrt{2}\times \sqrt{153}}\\ \pause
                                  & = & \sqrt{\dfrac{17}{153}}\\ \pause
                                  & = & \sqrt{\dfrac{1}{9}}\\ \pause
                                  & = & \dfrac{1}{3} \pause
 \end{eqnarray*}
Soit~:
$$\boxed{\cos(\alpha)=\dfrac{1}{3}}$$
\end{enumerate}

\end{frame}





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\begin{frame}

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item On a $\overrightarrow{C'H}\begin{pmatrix}\mfrac{3}{2}\\[1.5ex]\mfrac{3}{2}\\[1.5ex]2 \end{pmatrix}$ \pause et $\overrightarrow{AB}\begin{pmatrix}1\\-1\\0 \end{pmatrix}$ \pause donc~:
  $$\overrightarrow{C'H}.\overrightarrow{AB}= \pause \dfrac{3}{2}\times 1+\dfrac{3}{2}\times (-1)+(-2)\times 0 \pause =\dfrac{3}{2}-\dfrac{3}{2} \pause =0$$ \pause
  Les vecteurs $\overrightarrow{C'H}$ et $\overrightarrow{AB}$ sont donc orthogonaux. \pause On en déduit que~: \pause
  $$\boxed{\text{Les droites $(C'H)$ et $(AB)$ sont perpendiculaires}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item Dans le triangle $ABC'$, $HC'$ est la hauteur issue de $C'$. \pause On a donc~:
  $$S'= \pause \dfrac{AB\times HC'}{2} \pause =\dfrac{\sqrt{2}\times \sqrt{\dfrac{17}{2}}}{2} \pause =\dfrac{\sqrt{17}}{2}$$ \pause
  Soit~:
  $$\boxed{S'=\dfrac{\sqrt{17}}{2}}$$
 \end{enumerate}
\end{enumerate}

\end{frame}



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

\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(c)]
  \item \label{ch19_ex4_C2c} On peut remarquer que~: \pause
  $$\boxed{S'=S\times \cos(\alpha)}$$
 \end{enumerate}
\end{enumerate}

\end{frame}




\end{document}








