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\rhead{\textbf{A. P{}. M. E. P.{}}}
\lhead{\small Corrigé du baccalauréat S}
\lfoot{\small{Amérique du Nord}}
\rfoot{\small{juin 2003}}
\pagestyle{fancy}
\thispagestyle{empty}
\begin{center} {\Large\textbf{\decofourleft~Corrigé du baccalauréat  S 
Amérique du Nord~\decofourright\\juin 2003}}\end{center}

\vspace{0,25cm}

\textbf{\textsc{Exercice 1} \hfill 5 points}

\textbf{Commun à tous les candidats}

\medskip

%\emph{Cet exercice est un questionnaire à choix multiples constitué de six questions : chacune comporte trois réponses, une et une seule étant 
%exacte.}
%
%\emph{Les réponses à cet exercice sont à inscrire dans la feuille jointe 
%en annexe, page 5, en cochant pour chaque question la case correspondante à la réponse proposée.}
% 
%\emph{Toute réponse ambigüe sera considérée comme une absence de réponse. Toute réponse exacte entraîne une bonification, toute erreur est pénalisée.}
%
%\medskip
%
%On s'intéresse à la durée de vie, exprimée en années, d'un appareil ménager avant la première panne. On peut  modéliser cette situation par une loi de probabilité $p$ de durée de vie sans vieillissement, définie sur  l'intervalle $[0~;~+ \infty$[. Ainsi, la probabilité d'un intervalle  $[0,~t[$, notée $p([0~;~t[)$, est la probabilité que l'appareil ménager tombe en panne avant l'instant $t$.
%
%Cette loi est telle que $p([0,~t[) = \displaystyle\int_0^t 
%\lambda \text{e}^{- \lambda x}\: \text{d}x$, où $t$ est un nombre réel positif représentant le nombre d'années (loi exponentielle de paramètre $\lambda$, avec $\lambda~>~0$).

\begin{enumerate}
\item %Pour $t \geqslant 0$, la valeur exacte de $p([t,~ +\infty [)$ est :

%\begin{tabular}{l l l l l l}
%    \hspace{2cm} \textbf{a.}& $1 - \text{e}^{-\lambda t}$& \textbf{b.}& $\text{e}^{-\lambda t}$
%    &\textbf{c.}& $1 + 
%    \text{e}^{-\lambda t}$\\
%    \end{tabular}
Réponse \textbf{b.}
\item %La valeur de $t$ pour laquelle on a $p([0,~ t[) =  p ([t,~+\infty [)$ est :

%\begin{tabular}{l l l l l l}
%    \hspace{2cm} 
%    \textbf{a.}& $\dfrac{\ln 2}{\lambda}$&   \textbf{b.}& $\dfrac{\lambda}{\ln 2}$&	
%    \textbf{c.}& $\dfrac{\lambda}{2}$\\
%\end{tabular} 
En égalant les deux intégrales on obtient :

$- \text{e}^{-\lambda t} + 1 = \text{e}^{-\lambda t} \iff 1 = 2\text{e}^{-\lambda t} \iff \text{e}^{-\lambda t} = \dfrac{1}{2}$ et par croissance de la fonction logarithme népérien :

$-\lambda t = \ln \frac{1}{2} \iff -\lambda t = - \ln 2 \iff t = \dfrac{\ln 2}{\lambda}$.

Réponse \textbf{a.}
\item %D’après une étude statistique, la probabilité que l'appareil tombe en panne avant la fin de la première année est 0,18. La valeur exacte de $\lambda$ est alors :

%\begin{tabular}{l l l l l l}
%    \hspace{2cm} 
%    \textbf{a.}&	$\ln \left(\dfrac{50}{41}\right)$&	\textbf{b.}& $\ln 
%\left(\dfrac{41}{50}\right)$& \textbf{c.}&	$\dfrac{\ln (82)}{\ln (100)}$\\
%\end{tabular}
On a donc $p([0,~1[) = \displaystyle\int_0^1 
\lambda \text{e}^{- \lambda x}\: \text{d}x = - \text{e}^{- \lambda} + 1 = 0,18 \iff \text{e}^{- \lambda}  = 0,82 = \dfrac{82}{100} = \dfrac{41}{50} \iff - \lambda = \ln \left(\dfrac{41}{50} \right) \iff \lambda = \ln \left(\dfrac{50}{41} \right)$.

Réponse \textbf{a.}
\item %Sachant que cet appareil n'a connu aucune panne au cours des deux premières années après sa mise en service, la probabilité qu'il ne connaisse aucune panne  l'année suivante est :

%\begin{tabular}{l l l l l l}
%    \hspace{2cm} 
%    \textbf{a.}& $p ([1,~ + \infty[)$& \textbf{b.}& $p ([3, + \infty[)$& 
%    \textbf{c.}& $p ([2~;~3 [$\\
%\end{tabular}
On calcule $\dfrac{p(3~;~+ \infty)}{p(2~;~+ \infty)} = \dfrac{\text{e}^{-3\lambda}}{\text{e}^{-2\lambda}} = \text{e}^{- \lambda} = p(1, + \infty[)$.


 Dans la suite de l'exercice on prendra $\lambda = 0,2$.

\item %La probabilité que l'appareil n'ait pas eu de panne au cours des trois premières années, arrondie à $10^{-4}$ près, est :

%\begin{tabular}{l l l l l l}
%    \hspace{2cm} 
%    \textbf{a.}&	\np{0,5523}&	\textbf{b.}& \np{0,5488}& \textbf{c.}& \np{0,4512}\\
%\end{tabular}
La probabilité cherchée est $p(3, + \infty) =\text{e}^{- 3 \lambda} = \text{e}^{-0,6} \approx \np{0,5488}$.

Réponse \textbf{b.}
\item %Dix appareils neufs de ce type ont été mis en service en même temps. On désigne par $X$ la  variable aléatoire égale au nombre d'appareils qui n'ont pas de panne au cours des trois premières années.

%La valeur la plus proche de la probabilité de l'évènement \og $X = 4$ \fg{} est :

%\begin{tabular}{l l l l l l}
%    \hspace{2cm} 
%    \textbf{a.}& \np{0,5555}& \textbf{b.}& \np{0,8022}& \textbf{c.} & \np{0,1607}\\
%\end{tabular}
On vient de voir que la probabilité pour un appareil de ne pas tomber en panne au cours des trois premières années est $\text{e}^{- 0,6}$

On a un schéma de Bernoulli avec une variable aléatoire $X$ de paramètres $n = 10$ et $p = \text{e}^{- 0,6}$.

D’où $p(X = 4) = \text{C}_4^{10} \left(\text{e}^{- 0,6} \right)^4 \left (1 - \text{e}^{- 0,6}\right)^{6} \approx \np{0,1607}$.

Réponse \textbf{c.}
\end{enumerate}

\vspace{0,5cm}

\textbf{\textsc{Exercice 2} \hfill 5 points}

\textbf{Candidats ayant suivi l'enseignement de spécialité}

\medskip

%Dans le plan complexe rapporté à  un repère orthonormé \Ouv, d'unité graphique 1 cm, on considère  les points A$_0$, A$_1$, A$_2$ d'affixes respectives $z_0 = 5 - 4\text{i},\\ z_1 = - 1- 4\text{i},~ z_2 = - 4 - \text{i}$.
%
%\medskip

\begin{enumerate}
\item 
	\begin{enumerate}
	
		\item %Justifier l'existence d'une unique similitude directe  $S$ telle que $S(\text{A}_0)~=~\text{A}_1$ et $S (\text{A}_1) = \text{A}_2$.
Comme  $S(\text{A}_0)~\ne~\text{A}_1$ et $S (\text{A}_1) \ne \text{A}_2$, on sait qu’il existe une seule similitude directe $S$ transformant A$_0$ en A$_1$ et A$_1$ en A$_2$.
		\item 
		%Établir que l'écriture complexe de $S$ est $z'= \dfrac{1-\text{i}}{2}z + \dfrac{- 3 + \text{i}}{2}$.
On sait que l’écriture complexe de $S$ est $z^{\prime} = az + b$. En l’appliquant à A$_0$ et A$_1$, on obtient 

\[\left\{\begin{array}{l c l}
- 1- 4\text{i}&=&a(5 - 4\text{i}) + b\\
- 4- \text{i}&=&a(- 1- 4\text{i}) + b
\end{array}\right.\]

Par différence on obtient $3 - 3\text{i} = a(6) \iff a = \dfrac{1}{2} - \dfrac{1}{2}\text{i}$.

Puis en remplaçant $a$ dans la première équation :

$- 1- 4\text{i}=\left(\dfrac{1}{2} - \dfrac{1}{2}\text{i} \right)(5 - 4\text{i}) + b \iff b = - 1- 4\text{i}- \left(\dfrac{1}{2} - \dfrac{1}{2}\text{i} \right)(5 - 4\text{i}) =$

$ - 1- 4\text{i} - \dfrac{5}{2} + 2 + 2\text{i} + \dfrac{5}{2}\text{i} = - \dfrac{3}{2} + \dfrac{1}{2}\text{i}$.

L’écriture complexe de $S$ est donc :

$z’ = \left(\dfrac{1}{2} - \dfrac{1}{2}\text{i} \right)z  - \dfrac{3}{2} + \dfrac{1}{2}\text{i}$.
		\item %En déduire le rapport, l'angle et l'affixe $\omega$ du centre $\Omega$ de la similitude $S$.
On calcule $|a| = \dfrac{|1 - \text{i}|}{2}	 = \dfrac{\sqrt{2}}{2}$.

Donc $a = 	\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{2}}{2}  + \text{i}\dfrac{- \sqrt{2}}{2}\right) = \dfrac{\sqrt{2}}{2}\left(\cos (- \frac{\pi}{4}) + \text{i}\sin  (- \frac{\pi}{4}) \right) = \dfrac{\sqrt{2}}{2}\text{e}^{-\text{i}\frac{\pi}{4}}$.

Le rapport de $S$ est $|a|  = \dfrac{\sqrt{2}}{2}$ et son angle est $- \dfrac{\pi}{4}$.

Le centre $\Omega$ est le point invariant de la similitude ; donc 

$\omega = \left(\dfrac{1}{2} - \dfrac{1}{2}\text{i} \right)\omega  - \dfrac{3}{2} + \dfrac{1}{2}\text{i} \iff \omega\left(\dfrac{1}{2} + \dfrac{1}{2}\text{i}\right) = - \dfrac{3}{2} + \dfrac{1}{2}\text{i} \iff \omega = \dfrac{- 3 + \text{i} }{1 + \text{i}} = \dfrac{-3 + 1 + 3\text{i} + \text{i}}{1+1} = \dfrac{- 2 + 4 \text{i}}{2} = - 1 + 2\text{i}.$
		\item %On considère un point $M$, d'affixe $z$ avec $z ­\neq 0$, et son image $M'$, d'affixe $z'$.

%Vérifier la relation : $\omega - z'= \text{i}(z - z')$ ; en déduire la nature du triangle $\Omega MM'$.
$\omega - z^{\prime} =  - 1 + 2\text{i} - \left(\dfrac{1}{2} - \dfrac{1}{2}\text{i} \right)z  - \dfrac{3}{2} - \dfrac{1}{2}\text{i} = \dfrac{1}{2} + \dfrac{3}{2}\text{i} +\dfrac{- 1 + \text{i}}{2}z$.

$\text{i}\left(z - z^{\prime} \right) = \text{i}\left( z - \dfrac{1 - \text{i}}{2}z - \dfrac{- 3 + \text{i}}{2}\right) = \dfrac{1 + 3\text{i}}{2} + \dfrac{- 1 + \text{i}}{2}z $.

Conclusion : $\omega - z^{\prime} = \text{i}\left(z - z’ \right)$.

En terme de modules : $\left|\omega - z’\right| = \left|\text{i}\left(z - z^{\prime} \right)\right| \iff \Omega M’ = MM’$ ; le triangle $\Omega MM'$ est isocèle.

En terme d’arguments : $\left(\vect{M^{\prime}M},~\vect{M^{\prime}\Omega} \right) = \text{arg}\dfrac{\omega - z^{\prime}}{z - z^{\prime}} = \text{arg i} = \dfrac{\pi}{2}$, donc le triangle $\Omega MM^{\prime}$ est rectangle en $M^{\prime}$.
	\end{enumerate}
\item %Pour tout entier naturel $n$, le point A$_{n+1}$, est défini par A$_{n+1} = S(\text{A}_n)$ et on pose $u_n = \text{A}_n\text{A}_{n+1}$.
	\begin{enumerate}
		\item %Placer les points A$_0$,~ A$_1$, A$_2$ et construire géométriquement les points A$_3$,~ A$_4$,~ A$_5$,~ A$_6$.
Pour placer A$_{n+1}$ : ce point est à l’intersection du demi-cercle de diamètre $\left[\Omega \text{A}_n\right]$ et de la médiatrice de $\left[\Omega \text{A}_n\right]$.

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		\item %Démontrer que la suite $(u_n)$ est géométrique.
A$_{n+1}$ est l’image de A$_n$ et $A_{n+2}$ est l’image de A$_{n+1}$ par $S$, donc A$_{n+1}$A$_{n+2} = k$A$_n$A$_{n+1}$ avec $k = \dfrac{\sqrt{2}}{2}$ rapport de la similitude.

On a donc $u_{n+1} = ku_n$ ce qui montre que la suite $\left(u_n\right)$ est g\’eom\’etrique de raison $k = \dfrac{\sqrt{2}}{2}$, de premier terme A$_0$A$_1 = 6$.
	\end{enumerate}
\item %La suite $(v_n)$ est définie sur $\N$ par $v_n = u_0 + 
%u_1 + \cdots + u_n = \displaystyle\sum_{k=0}^{n} u_k$.
	\begin{enumerate}
		\item %Exprimer $v_n$ en fonction de $n$.
On sait que $v_n = u_0 \times \frac{1 - \left(\frac{\sqrt{2}}{2}\right)^{n+1}}{1 - \frac{\sqrt{2}}{2}} = 6 \times \frac{1 - \left(\frac{\sqrt{2}}{2}\right)^{n+1}}{1 - \frac{\sqrt{2}}{2}}$.
		\item %La suite $(v_n)$ est-elle convergente ?
		Comme $- 1 < \frac{\sqrt{2}}{2}< 1$, on a 
		$\displaystyle\lim_{n \to + \infty} \left(\frac{\sqrt{2}}{2}\right)^{n+1} = 0$, $\displaystyle\lim_{n \to + \infty} v_n = \dfrac{6}{1 - \frac{\sqrt{2}}{2}} =\dfrac{12}{2 - \sqrt{2}}$.
		
La suite $\left(v_n\right)$ a pour limite $\dfrac{12}{2 - \sqrt{2}}$. 		
	\end{enumerate}
\item 
	\begin{enumerate}
		\item %Calculer en fonction de $n$ le rayon $r_n$ du cercle circonscrit au triangle $\Omega\text{A}_n\text{A}_{n+1}$.
Le triangle $\Omega\text{A}_n\text{A}_{n+1}$ isocèle rectangle en A$_{n+1}$ est inscrit dans le cercle de centre le milieu de $\left[\text{A}_n\Omega \right]$ ; son rayon $r_n = \dfrac{1}{2}\text{A}_n \Omega = \dfrac{1}{2}\text{A}_n\text{A}_{n+1} \sqrt{2} = \dfrac{\sqrt{2}}{2}u_n$.

Or $u_n = 6 \times \left( \dfrac{\sqrt{2}}{2}\right)^n$, donc $r_n = 6 \times \left( \dfrac{\sqrt{2}}{2}\right)^{n+1}$.
 		\item  %Déterminer le plus petit entier naturel $p$ tel que, pour tout entier naturel $n$ :  si $n > p$ alors $r_n < 10^{-2}$. 
		On a $r_n < 10^{-2} \iff 6 \times \left( \frac{\sqrt{2}}{2}\right)^{n+1}  < 10^{-2} \iff  \left( \frac{\sqrt{2}}{2}\right)^{n+1} < \frac{10^{- 2}}{6}$.
		
Par croissance de la fonction logarithme népérien :

$(n + 1)\ln \left( \dfrac{\sqrt{2}}{2}\right) < \ln \left(\frac{10^{- 2}}{6} \right)  \iff n + 1 > \frac{\ln \dfrac{10^{- 2}}{6}}{\ln \dfrac{\sqrt{2}}{2}}$

Or $\dfrac{\ln \frac{10^{- 2}}{6}}{\ln \frac{\sqrt{2}}{2}} \approx 18,4$, d’où $n > 18$.

Il faut donc choisir $p = 17$.
	\end{enumerate}
\end{enumerate}

\vspace{0,5cm}

\textbf{\textsc{Exercice 2} \hfill 5 points}

\textbf{Candidats n'ayant pas suivi l'enseignement de spécialité}

\medskip

%Le plan est rapporté au repère orthonormé \Ouv{} (unité graphique : 2~cm).
%
%On considère les points A, B et C d'affixes respectives 
%$z_{\text{A}} = - 1 + \text{i}\sqrt{3},$
%
%$z_{\text{B}} =- 1 - \text{i}\sqrt{3}$ et $z_{\text{C}} = 2$.
%
%\medskip

\begin{enumerate}
\item ~%Placer ces points sur un dessin.
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\item 
	\begin{enumerate}
		\item ~%Vérifier que : $\dfrac{z_{\text{B}}  - z_{\text{C}} 
%}{z_{\text{A}} - z_{\text{C}} }	= \text{e}^{\text{i}\frac{\pi}{3}}$.

$\dfrac{z_{\text{B}}  - z_{\text{C}} }{z_{\text{A}} - z_{\text{C}} }	=\dfrac{- 1 - \text{i}\sqrt{3} - 2}{-1 + \text{i}\sqrt{3} - 2} = \dfrac{- 3  - \text{i}\sqrt{3}}{- 3  + \text{i}\sqrt{3}} =\dfrac{3 + \text{i}\sqrt{3}}{3 -\text{i}\sqrt{3}} =  \dfrac{\sqrt{3} + \text{i}}{\sqrt{3} - \text{i}} =  \dfrac{\left(\sqrt{3} + \text{i} \right)^2}{\left(\sqrt{3} + \text{i}\right)\left(\sqrt{3} - \text{i}\right)} =   \dfrac{3 - 1 + 2\text{i}\sqrt{3}}{3 + 1} = \dfrac{1}{2} + \text{i}\dfrac{\sqrt{3}}{2} = \cos \frac{\pi}{3} + \text{i}\sin \frac{\pi}{3}  = \text{e}^{\text{i}\frac{\pi}{3}}$.
		\item %En déduire la nature du triangle ABC.
En prenant les modules des deux membres de l’égalité précédente :

$\left|\dfrac{z_{\text{B}}  - z_{\text{C}} }{z_{\text{A}} - z_{\text{C}} } \right| =\dfrac{\left| z_{\text{B}}  - z_{\text{C}}\right|}{\left| z_{\text{A}} - z_{\text{C}}\right|} = \dfrac{\text{CB}}{\text{CA}} = \left|\text{e}^{\text{i}\frac{\pi}{3} } \right| = 1$, c’est-à-dire CB = CA : le triangle est isocèle en C.

En prenant les arguments des deux membres :

$\left(\vect{\text{CA}},~\vect{\text{CB} }\right)  = \frac{\pi}{3}$. L’angle au sommet a pour mesure $\frac{\pi}{3}$, donc les deux autres aussi, c’est-à-dire que le triangle ABC est équilatéral.
		\item %Déterminer le centre et le rayon du cercle $\Gamma_1$ circonscrit au triangle ABC.
Le triangle étant équilatéral, le centre du cercle circonscrit est aussi le centre de gravité du triangle, soit l’isobarycentre des trois sommets.
		
Donc $z_{\Gamma} = \dfrac{1}{3}(- 1 + \text{i}\sqrt{3} - 1 - \text{i}\sqrt{3} +2) = 0$.

Un rayon est $\Omega\text{A} = 2$. Voir le tracé ci-dessus
%Tracer le cercle $\Gamma_1$.
	\end{enumerate}
\item 
	\begin{enumerate}
		\item %Établir que l'ensemble $\Gamma_2$ des points $M$  d'affixe $z$ qui vérifient 

%$2(z + \overline{z}) + z\overline{z}~=~0$ est un cercle de centre $\Omega$ d'affixe $- 2$. Préciser son rayon. Construire $\Gamma_2$.
Soit $M$ d’affixe $z = x + \text{i}y$.

$2(z + \overline{z})+ z\overline{z}= 0 \iff 2(x + \text{i}y + x - \text{i}y) + (x + \text{i}y)(x - \text{i}y) = 0 \iff 4x + x^2 + y^2 = 0 \iff (x + 2)^{2} - 4 + y^{2} = 0 \iff (x + 2)^2 + y^2 = 4 $.

Ceci montre que les points $M(x~;~y)$ appartiennent au cercle $\Gamma_2$ de centre le point de coordonnées $(- 2~;~0)$ et de rayon 2.
		\item %Vérifier que les points A et B sont éléments de $\Gamma_2$	
A$\left(- 1~;~\sqrt{3}\right) \in \Gamma_{2} \iff  (- 1 + 2)^{2} + 3 = 4 \iff 4 = 4$ vraie.

B$(- 1~;~- \sqrt{3}) \in \Gamma_{2} \iff (- 1 + 2)^{2} + 3 = 4 \iff 4 = 4$ vraie.

les points A et B sont éléments de $\Gamma_2$.
	\end{enumerate}
\item %On appelle $r_1$ la rotation de centre A et d'angle $\dfrac{\pi}{3}$.
	\begin{enumerate}
		\item %Quelles sont les images des points A et B par la rotation $r_1$ ? 
%Construire l'image C$_1$ du point C par la rotation $r_1$ puis calculer son affixe.
L’écriture complexe de la rotation est :

$z_{\text{A}_{1}} - z_{\text{A}} = \text{e}^{\text{i}\frac{\pi}{3}}\left(z -  z_{\text{A}}\right)$.

Donc $z_{\text{A}_{1}} - z_{\text{A}} = \text{e}^{\text{i}\frac{\pi}{3}}\left(z_{\text{A}} - z_{\text{A}}\right) = 0$ donc $z_{\text{A}_{1}} = z_{\text{A}}$ ce qui est normal pour le centre de la rotation !

De même

$z_{\text{B}_{1}} - z_{\text{A}} = \text{e}^{\text{i}\frac{\pi}{3}}\left(z_{\text{B}} - z_{\text{A}}\right) \iff z_{\text{B}_{1}} = - 1 + \text{i}\sqrt{3} + \left(\frac{1}{2} + \text{i}\frac{\sqrt{3}}{2}\right)\left(- 1 -  \text{i}\sqrt{3} - 1 - \text{i}\sqrt{3}\right) = 
- 1 + \text{i}\sqrt{3} - 1 - 3 - \text{i}\sqrt{3}} +  \text{i}\sqrt{3} = 1 - \text{i}\sqrt{3} = z_{\text{C}}$.

On aurait pu dire que ABC étant un triangle équilatéral direct l’image de B dans la rotation de centre A  et d’angle $\frac{\pi}{3}$ est le point C.

Soit C$_{1}$ l’image de C par $r_{1}$. Pour construire ce point on construit le triangle équilatéral direct ACC$_{1}$ ; comme ABC et ACC$_{1}$ sont symétriques autour de la droite (AC), le point C$_{1}$ est le symétrique de B autour de la droite (AC).

$z_{\text{C}_{1}} - z_{\text{A}} = \text{e}^{\text{i}\frac{\pi}{3}}\left(z_{\text{C}} - z_{\text{A}}\right) \iff z_{\text{C}_{1}} = - 1 + \text{i}\sqrt{3}}   + \left(\frac{1}{2} + \text{i}\frac{\sqrt{3}}{2}\right)\left(2\right) =$

$  - 1 + \text{i}\sqrt{3} + 2 + \text{i}\sqrt{3} = 2 + 2\text{i}\sqrt{3}$.
		\item %Déterminer l'image du cercle $\Gamma_2$ par la rotation $r_1$.
Puisque le triangle A$\Omega$O est équilatéral, l’image de $\Gamma $ dans la rotation $r_{1}$ est le point O ; comme il y a conservation des longueurs, l’image du cercle $\Gamma_2$ par la rotation $r_1$ est le cercle $\Gamma_{1}$.
	\end{enumerate}
\item %Soit $r$ une rotation. Pour tout point $M$ d'affixe 
%$z$, on note $M'$ l'image de $M$ par $r$ et $z'$ l'affixe de $M'$.

%On posera : $z'= az + b$, avec $a$ et $b$ des nombres complexes vérifiant $|a| = 1$ et $a ­\neq 1$.

%On suppose que $r$ transforme le cercle $\Gamma_2$ en le cercle $\Gamma_1$.
	\begin{enumerate}
		\item %Quelle est l'image du point $\Omega$ par $r$ ? En déduire une relation  entre $a$ et $b$.
		L’image de $\Omega$ est le point O ; donc $0 = - 2a + b \iff b = 2a$.
		\item %Déterminer en fonction de $a$ l'affixe du point $r(\text{C})$, image du point C par la rotation $r$ ;  en déduire que le point $r(\text{C})$ appartient  un cercle fixe que l'on définira. Vérifier que ce cercle passe par C$_1$.
$z_{r(\text{C})} = 2a + b = 2a + 2a = 4a$.

On sait que $|a| = 1$, donc $|4a| = 4$ : le point $r(\text{C})$ appartient au cercle de centre O et de rayon 4.

On a $\left|z_{1} \right| = \left|2 + 2\text{i}\sqrt{3} \right| = \sqrt{4 + 12} = \sqrt{16} = 4$.

Le point C$_{1}$ appartient bien au cercle de centre O de rayon 4.
	\end{enumerate}
\end{enumerate}

\vspace{0,5cm}

\textbf{\textsc{Problème} \hfill 10 points}

\textbf{Commun à tous les candidats}

\medskip

%\textbf{Partie A : étude d'une fonction  $f$  et construction de sa courbe}
%
%\medskip
% 
%On considère la fonction $f$ définie sur $\R$ par $f(x) = \text{e}^{-x} 
%\ln \left(1 + \text{e}^x\right)$.
%
%On note $\mathcal{C}$ sa courbe représentative dans le plan rapporté 
%au repère orthogonal  \Oij.
%
%L'unité graphique est 1 cm sur l'axe des abscisses et 10 cm sur l'axe des ordonnées.
\begin{enumerate}
\item 
	\begin{enumerate}
		\item %On rappelle que : $\displaystyle\lim_{h \to 0} \dfrac{\ln (1 + h)}{h} = 1$. Déterminer la limite de $f$ en $- \infty$.
On sait que $\displaystyle\lim_{x\to - \infty}\text{e}^{x} = 0$.

Posons $h = \text{e}^x$, donc $f(x) = \dfrac{\ln (1 + h)}{h}$.

D’après le rappel de l’énoncé :

$\displaystyle\lim_{x\to - \infty} f(x) = \displaystyle\lim_{x\to 0} f(h) = 0$.
		\item %Vérifier que pour tout réel $x : f(x) = \dfrac{x}{\text{e}^x} + \text{e}^{-x} \ln \left(1 + \text{e}^{-x}\right)$.
On factorise $\text{e}^x$ dans la parenthèse :

$f(x) = \text{e}^{- x}\ln \left(\text{e}^x\left[\text{e}^{- x} + 1\right]\right) = \text{e}^{- x}\left(\ln \text{e}^x + \ln \left(1 +  \text{e}^{-x} \right) \right) = \text{e}^{- x}\left(x + \ln \left(1 +  \text{e}^{-x} \right) \right) $ et en développant :

$f(x) = x\text{e}^{- x} + \text{e}^{- x} \ln \left(1 +  \text{e}^{-x} \right)  = \dfrac{x}{\text{e}^x } + \text{e}^{- x} \ln \left(1 +  \text{e}^{-x} \right) $.
%Déterminer la limite de $f$ en $+ \infty$.

On sait que $\displaystyle\lim_{x\to + \infty} \dfrac{\text{e}^x}{x} = + \infty$, donc $\displaystyle\lim_{x\to + \infty} \dfrac{x}{\text{e}^x} = 0$.

D’autre part $\displaystyle\lim_{x\to + \infty} \text{e}^{- x} = 0$, d’où $\displaystyle\lim_{x\to + \infty} \ln \left(1 + \text{e}^{- x}\right) = 0$ et comme 

$\displaystyle\lim_{x\to + \infty} \text{e}^{- x} = 0$, par produit $\displaystyle\lim_{x\to + \infty} \text{e}^{- x}\ln \left(1 +  \text{e}^{-x} \right) = 0$.

Conclusion :  $\displaystyle\lim_{x\to + \infty} f(x) = 0$.
		\item %En déduire que la courbe admet deux asymptotes que l'on précisera.
Les résultats précédents montrent que la droite d’équation $y = 1$ est asymptote horizontale à $\mathcal{C}$ au voisinage de moins l’infini et que l’axe des abscisses est asymptote horizontale à $\mathcal{C}$ au voisinage de  l’infini.
	\end{enumerate}
\item %On considère la fonction $g$ définie sur l'intervalle $]- 1~;~+ \infty[$ par :

\[g(t)= \dfrac{t}{1 + t} 
- \ln (1 + t).\]

	\begin{enumerate}
		\item %Démontrer que la fonction $g$ est strictement décroissante sur l'intervalle $[0~;~+ \infty[$.
$g$ somme de fonctions définies et dérivables sur $]- 1~;~+ \infty[$ est dérivable et sur intervalle :
		
$g’(t) = \dfrac{1 + t - t}{(t + 1)^{2}} - \dfrac{1}{1 + t} = \dfrac{1 - (t + 1)}{(t + 1)^{2}} = \dfrac{- t}{(t + 1)^{2}}$.

Pour $t > 0, \:g’(t) < 0$, donc la fonction $g$ est décroissante sur $]0~;~+ \infty[$.
		\item %En déduire le signe de $g(t)$ lorsque $t > 0$.
		On a $g(0)  = 0$, la fonction étant décroissante sur  $]0~;~+ \infty[$, elle est strictement négative sur cet intervalle.
	\end{enumerate}
\item 
	\begin{enumerate}
		\item %Calculer $f'(x)$ et l'exprimer en fonction de 
%$g\left(\text{e}^x\right),\: f'$ désignant la fonction dérivée de $f$.
La fonction $f$ est dérivable comme produit de fonctions dérivables sur $]- 1~;~+ \infty[$ et sur cet intervalle :

$f'(x) = - \text{e}^{-x} \ln \left(1 + \text{e}^x\right) + \text{e}^{-x} \times \dfrac{ \text{e}^{x}}{1 + \text{e}^x}$.

Posons $t = \text{e}^x$, donc $\text{e}^{- x} = \dfrac{1}{\text{e}^x}  = \dfrac{1}{t}$. On peut donc écrire :

$f'(x) = f’(t) = - \dfrac{1}{t}\ln (1 + t) +  \dfrac{1}{1 + t} = \dfrac{1}{t}\left[\dfrac{t}{1 + t} - \ln (1 + t)\right] = \dfrac{1}{t}g(t)$.

Ou encore  $f'(x) = \text{e}^{-x}g\left( \text{e}^x\right)$
		\item %En déduire le sens de variation de la fonction $f$ puis dresser son tableau de variations.
		On sait que pour tout réel $x, \:\text{e}^x > 0$, donc le signe de $f’$ est celui de $g$ sur $]0~;~+ \infty[$, donc $f' < 0$ : la fonction $f$ est décroissante sur $\R$.

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	\end{enumerate}
\item~ %Tracer les asymptotes à la courbe $\mathcal{C}$ et la courbe $\mathcal{C}$.

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\end{enumerate}

\bigskip

\textbf{Partie B : comportements asymptotiques d'une primitive 
\boldmath $F$ \unboldmath  de \boldmath $f$ \unboldmath  sur \boldmath  $\R$ 
\unboldmath}

\medskip

%Soit $F$ la fonction définie sur $\R$ par $F(x) = \displaystyle\int_0^x f(t)\: \text{d}t.$

\medskip

\begin{enumerate}
\item %Étudier le sens de variations de la fonction $F$.
$F$ est la primitive de $f$ qui s’annule en $x = 0$. Donc $F’(x) = f(x)$.

La fonction $f$ est positive sur $\R$, donc la fonction $F$ est strictement croissante sur $\R$.
\item 
	\begin{enumerate}
		\item %Vérifier que, pour tout nombre réel 
%$t,~\dfrac{1}{1 + \text{e}^t} = 1 - \dfrac{\text{e}^t}{1 + \text{e}^t}$	et calculer $\displaystyle\int_0^x \dfrac{1}{1 + \text{e}^t}\: 
%\text{d}t.$

$\dfrac{1}{1 + \text{e}^t} = \dfrac{1+ \text{e}^t -  \text{e}^t}{1 + \text{e}^t}	 = 1 - \dfrac{\text{e}^t}{1 + \text{e}^t}$.

En utilisant le résultat précédent :

$\displaystyle\int_0^x \dfrac{1}{1 + \text{e}^t}\: 
\text{d}t= \displaystyle\int_0^x 1 \:\text{d}t - \displaystyle\int_0^x \dfrac{\text{e}^t}{1 + \text{e}^t} = \left[t - \ln \left(1 + \text{e}^{t}\right)\right]_{0}^{x} = x - \ln \left(1 + \text{e}^{x}\right) + \ln 2$	
		\item %En déduire, à l'aide d'une intégration par parties, le calcul de $F(x)$.
$F(x) = \displaystyle\int_0^x \text{e}^{-t}\ln \left(1 + \text{e}^{t}\right)\: \text{d}t.$
		
Posons : $u(t) = \ln \left(1 + \text{e}^{t}\right)$ ; $\text{d}v = \text{e}^{-t}$
		
Donc $\text{d}u = \dfrac{\text{e}^{t}}{1 + \text{e}^{t}}$ ; $v(t) = - \text{e}^{- t}$.

Toutes ces fonctions sont continues car dérivables ; on peut donc intégrer par parties.

$F(x) = \left[ - \text{e}^{-t}\ln \left(1 + \text{e}^{t}\right)\right]_{0}^{x} + \displaystyle\int_{0}^{x} 	\text{e}^{-t}	\dfrac{\text{e}^{t}}{1 + \text{e}^{t}}\:\text{d}t =$

$- \text{e}^{- x}\ln \left(1 + \text{e}^{x}\right) + \ln 2  + x - \ln \left(1 + \text{e}^{x}\right) + \ln 2  + \ln 2 - \ln 2 =$

$F(x)  = x - \left(1 + \text{e}^{- x} \right)\ln \left(1 + \text{e}^{x}\right) + 2\ln 2$.
		
		\item %Vérifier que $F(x)$ peut s'écrire sous les formes suivantes :

%\[\begin{array}{l l c l}
%(1)& F(x)&=& x - \ln (1+ \text{e}^x) - f(x) + 2\ln 2.\\
%(2)& F(x)& =&\ln \left(\dfrac{\text{e}^x}{1 + \text{e}^x}\right) -  f(x) + 2\ln 2.\\
%\end{array}\]
$\bullet~~$$F(x)  = x - \left(1 + \text{e}^{- x} \right)\ln \left(1 + \text{e}^{x}\right) + 2\ln 2 = x - \ln \left(1 + \text{e}^{x}\right) - \text{e}^{- x} \ln \left(1 + \text{e}^{x}\right) + 2\ln 2 = x - \ln \left(1 + \text{e}^{x}\right)  - f(x) + 2\ln 2$.

$\bullet~~$$F(x) = \ln \text{e}^{x} - \ln \left(1 + \text{e}^{x}\right)  - \text{e}^{- x} \ln \left(1 + \text{e}^{x}\right) - f(x) + 2 \ln 2 = \ln \left(\dfrac{\text{e}^{x}}{1 + \text{e}^{x}} \right)$.
	\end{enumerate}
\item %Déterminer $\displaystyle\lim_{x \to + \infty} F(x)$.
On utilise la deuxième écriture : $\dfrac{\text{e}^{x}}{1 + \text{e}^{x}} = \dfrac{1}{1 + \text{e}^{- x}}$.

Or $\displaystyle\lim_{x \to + \infty}\text{e}^{- x} = 0$ donc $\displaystyle\lim_{x \to + \infty}\dfrac{\text{e}^{x}}{1 + \text{e}^{x}} = 1$, puis $\displaystyle\lim_{x \to + \infty} \ln \left(\dfrac{\text{e}^{x}}{1 + \text{e}^{x}} \right) = \ln 1 = 0$.

On sait d’autre part que $\displaystyle\lim_{x \to + \infty} f(x) = 0$.

D’où finalement : $\displaystyle\lim_{x \to + \infty}F(x) = 2 \ln 2$.
\item %Déterminer $\displaystyle\lim_{x \to - \infty} \left[F(x) - 
%x\right]$. Donner une interprétation graphique de ce résultat.
Calculons $F(x) - x  = - \ln \left(1 + \text{e}^{x}\right) - f(x) + 2 \ln 2$.

On a $\displaystyle\lim_{x \to - \infty}1 + \text{e}^{x} = 1$, puis 
$\displaystyle\lim_{x \to - \infty}\ln \left(1 + \text{e}^{x}\right) = \ln 1 = 0$ et comme $\displaystyle\lim_{x \to - \infty} f(x) = 1$, on a finalement :

$\displaystyle\lim_{x \to - \infty} F(x) - x =  - 1  +2 \ln 2$.

La dernière égalité peut s’écrire $\displaystyle\lim_{x \to - \infty}F(x) - x + 1 - 2\ln 2 = 0$, ce qui signifie que la représentation graphique de $F$ a pour asymptote oblique au voisinage de moins l’infini la droite d’équation $y =  x - 1 + 2\ln 2$.

\end{enumerate}

\bigskip

\textbf{Partie C : étude d'une suite}

%\medskip
%
%Soit $\left(u_n\right)$ la suite définie sur $\N^*$ par :
%
%\[u_n = f(1) + f(2) + \cdots + 
%f(n) = \displaystyle\sum_{k = 1}^n \text{e}^{-k} \ln (1 + \text{e}^ k).\]

\begin{enumerate}
\item %Hachurer sur la représentation graphique un domaine dont l'aire, en unités d'aire, est $u_n$.
Quel que soit $n \in \N$, \: $f_{n}$ est l’aire du rectangle de côtés $1$ et $f(n)$. On peut choisir pour le côté de longueur 1 le segment ayant pour extrémités les points d’abscisses $n- 1$ et $n$.

$u_{n}$ est donc la somme des aires des $n$ premiers rectangles.

Exemple : sur le dessin on a représenté $u_{3}$. 
\item %Déterminer le sens de variation de la suite  $\left(u_n\right)$.
On calcule $u_{n+1} - u_{n} = f(n+1) > 0$ : la suite est croissante (strictement).
\item 
	\begin{enumerate}
		\item %Justifier que, pour tout entier $k$ tel que $1 
%\leqslant k \leqslant n$, on a : 

%\[f(k) \leqslant \displaystyle\int_{k-1}^{k} 
%f(t)\:\text{d}t.\]
La fonction $f$ est décroissante : donc pour tout naturel $k$ non nul et pour $t$ tel que $k- 1 \leqslant t \leqslant k$, par décroissance $f(k) \leqslant f(t)$ et puisque $k - 1 \leqslant k$ :

$\displaystyle\int_{k-1}^{k} f(k)\:\text{d}t \leqslant \displaystyle\int_{k-1}^{k} f(t)\:\text{d}t$, soit $f(k) \leqslant \displaystyle\int_{k-1}^{k} f(t)\:\text{d}t + \cdots + \displaystyle\int_{n-1}^{n} f(t)\:\text{d}t$ soit par linéarité de l’intégration :

$u_{n} \leqslant \displaystyle\int_{0}^{n} f(t)\:\text{d}t$
		\item %Comparer $u_n$ et $F(n)$.
		En sommant les inégalités précédentes  pour $k$ allant de 1 à $n$, on obtient 
		
		$u_{n} = f(1) + f(2) + \cdots +f(n) \leqslant \displaystyle\int_{0}^{1} f(t)\:\text{d}t$ c’est-à-dire $u_{n}\leqslant F(n)$ pour tout $n$ non nul.
	\end{enumerate}
\item %La suite  $\left(u_n\right)$ est-elle convergente ?
La suite $(F(n))$ est croissante et a pour limite $2 \ln 2$ en plus l’infini. On a donc $u_{n} \leqslant F(n) \leqslant 2\ln 2$.

La suite $\left(u_{n}\right)$ est croissante et majorée : elle converge vers une limite $\ell$ telle que 
$\ell \leqslant 2 \ln 2$.
\end{enumerate}
\end{document}
%\vspace{2cm}
%
%\begin{center} 
%
%\textbf{Annexe à rendre avec la copie}
%
%\vspace{0,5cm}
%
%\emph{Réponses à l'exercice} 1 (mettre une croix dans la case 
%correspondant à la réponse choisie)
%
%\vspace{0,5cm}
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%\begin{tabularx}{0.5\linewidth}{|*{4}{>{\centering \arraybackslash}X|}}\hline
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%\end{tabularx}
%\end{center}

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