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%Tapuscrit : Denis Vergès
%Corrigé : Isabelle Leyraud 
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pdfauthor = {LEYRAUD},
pdfsubject = {BTS Métropole},
pdftitle = {SIO épreuve facultative 12 mai 2015},
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\begin{document}
\setlength\parindent{0mm}
\rhead{\textbf{A. P{}. M. E. P{}.}}
\lhead{\small Brevet de technicien supérieur SIO}
\lfoot{\small{Corrigé de l'épreuve facultative - Métropole}}
\rfoot{\small{13 mai  2015}}
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\marginpar{\rotatebox{90}{\textbf{A. P{}. M. E. P{}.}}}

\begin{center} {\Large \textbf{Corrigé du  BTS SIO - Métropole - mai 2015}} \\ \textbf{Épreuve facultative}
\end{center}

\vspace{0,25cm}

\textbf{Exercice 1 \hfill 10 points}


\textbf{Partie A}

\medskip

\begin{enumerate}
\item 
	\begin{enumerate}
		\item Les défauts $a$ et $b$ pouvant se présenter de façon indépendante, les évènements $A$ et $B$ sont indépendants donc on a bien  l'égalité  $P(A \cap B) = P(A) \times P(B)$.
		\item $P(A\cup B)= P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)\times P(B)=0,02+0,01-0,02\times 0,01= \np{0,0298}$.\\
		Ainsi la probabilité qu'une batterie produite soit défectueuse est de $\np{0,0298}$.
	\end{enumerate}
\item 
	\begin{enumerate}
		\item On a ici une expérience aléatoire répétée 100 fois de manière indépendante (assimilé à un tirage avec remise). Chaque expérience a deux issues, le succès \og être défectueuse \fg{}  ayant une probabilité de $\np{0,0298}$ et l'échec \og ne pas être défectueuse \fg{}. $X$ compte de nombre de succès donc $X$ sut la loi binomiale de paramètre 100 et $0,0298$ : $X\hookrightarrow \mathcal{B}(100\:;\:0,0298)$
		\item $P(X \geqslant 3) = 1 - P(X < 3) = 1 - P(X \leq 2)\approx \np{0,5757}$ (en utilisant la calculatrice). 
		
La probabilité qu'il y ait au moins 3 pièces avec un défaut est de $\np{0,5757}$ à $10^{-4}$ près
	\end{enumerate}
\end{enumerate}

\bigskip

\textbf{Partie B}

\medskip

\begin{enumerate}
\item $Y\hookrightarrow \mathcal{N}(80,10))$, avec la calculatrice, on obtient  $P( 60 \leqslant Y \leqslant 100)\approx 0,9545$ à $10^{-4}$ près
\item \emph{\textbf{Remarque : question hors programme !}}

$P(Y \geqslant  h) = 0,95$ équivaut à $1-P(Y <h) = 0,95$ et par continuité de $Y$ on a $1-P(Y \leqslant h)=0,95$ soit $P(Y\leqslant h)=0,05$.
 
Avec la calculatrice on obtient $h\approx 63,55$.

Cela signifie que la probabilité pour le temps de charge d'une batterie soit   supérieur à $63,55$~min est de $0,95$.
\end{enumerate}

\bigskip

\textbf{Partie C}

\medskip

\begin{enumerate}
\item %En arrondissant à la quatrième décimale,justifier que $\lambda$ s'exprime en jour$^{-1}$ par: $\lambda = \np{0,0005}$.
Comme $T \hookrightarrow \mathcal{E}(\lambda)$, le temps moyen de bon fonctionnement est l'espérance de $T$ soit : $E(T) = \dfrac{1}{\lambda}$.

Ainsi $\np{1900} =\dfrac{1}{\lambda}$, soit $\lambda=\dfrac{1}{\np{1900}}\approx \np{0,0005}$ à $10^{-4}$ près et $\lambda$ est exprimé en  jour$^{-1}$ (car $E(T)$ s'exprime en jours).
\item $P(T\geqslant \np{4000})= \text{e}^{-\lambda\times 4000}= \text{e}^{- \np{0,0005}\times \np{4000}}\approx \np{0,1353}$ à $10^{-4}$ près.

La probabilité que l'écran fonctionne encore correctement après \np{4000} jours
d 'utilisation est de $\np{0,1353}$.
\item  $P(T \leqslant  t) = 0,7$ équivaut à  $1 - \text{e}^{- \np{0,0005} t} = 0,7$\quad $\Leftrightarrow\quad \text{e}^{- \np{0,0005}t}= 0,3$ \quad$\Leftrightarrow\quad - \np{0,0005}t = \ln(0,3)$ \quad $\Leftrightarrow\quad t = \dfrac{-\ln(0,3)}{\np{0,0005}}$ \\soit $t\approx \np{2408}$ à l'unité près.

Cela signifie que la probabilité qu'une défaillance ait lieu dans les \np{2408} premiers jours est de 0,7.
\end{enumerate}

\vspace{0,5cm}

\newpage
\textbf{Exercice 2 \hfill 10 points}

\bigskip

\textbf{A. Étude d'une fonction}

\medskip

\begin{enumerate}
\item 
	\begin{enumerate}
		\item Pour tout réel $x$ de l'intervalle [1~;~6,5], $f$ est dérivable et on a 
		$f'(x) = - 2\times 2 x + 20 - 0 - 16\dfrac{1}{x}= - 4x + 20-\dfrac{16}{x}=\dfrac{-4x^2 + 20x - 16}{x}$
		Or $\dfrac{-4(x - 1)(x - 4)}{x}=\dfrac{-4(x^2 - 4x - x + 4)}{x}=\dfrac{-4x^2 + 20x - 16}{x} = f'(x)$.
		
		Ainsi on a bien $f'(x)=\dfrac{-4(x-1)(x-4)}{x}$
\begin{multicols}{2}
		\item Étude  du signe de $f'(x)$ sur l'intervalle [1~;~6,5].\\
\begin{variations}
x 		&~~  1	& \hspace*{1cm} & 4 			&\hspace*{0.7cm}&6,5~~ \\\hline
- 4 	& \ga- 	&\l 			&  -\\\hline
x-1 	& \z 	&+ 				&\l 			&  +\\\hline
x - 4	& \ga- 	&\z 			&  +\\\hline
x 		& \ga+ 	&\l 			&  +\\\hline
\text{signe~de}~f'(x) & \ga+ &\z& - &\\\hline
\end{variations}

		\item D'après la question précédente on obtient le tableau de variation de la fonction $f$ suivant :

\begin{variations}
x 		&~~  1	& \hspace*{1cm} & 4 &\hspace*{0.7cm}&6,5~~ \\\hline
\m{f(x)}&0  &\c &\h{7,8} 		&\d	& -2,5\\\hline
\end{variations}
		
\end{multicols}
	\end{enumerate}
\item 
	\begin{enumerate}
		\item 
\begin{tabularx}{\linewidth}{|c|*{7}{>{\centering \arraybackslash}X|}}\hline
$x$		&1 	&2 		&3 		&4 		&5 		&6		&6,5\\ \hline
$f(x)$	&0	&2,9	&6,4	&7,8	&6,2	&1,3	&$- 2,5$\\ \hline
\end{tabularx}

		\item Courbe $\mathcal{C}_f$ dans le repère \Oij.
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	\end{enumerate}	
\item $F(x) = - \dfrac{2}{3} x^3 + 10x^2 - 2x - 16x \ln (x)$ donc on a :\\
$F'(x)= - \dfrac{2}{3} \times 3 x^2 + 10\times 2x - 2 - 16\left( x \times \dfrac{1}{x}+1\times \ln(x) \right)
=-2x^2+20x-2-16-16\ln(x)=-2x^2+20x-18-16\ln(x)=f(x)$\\
Ainsi la fonction $F$ est une primitive de la fonction $f$ sur l'intervalle [1~;~6,5].
\end{enumerate}

\bigskip

\textbf{B. Applications à l'économie}

\medskip

\begin{enumerate}
\item 
	\begin{enumerate}
		\item $f$ est continue sur $[4\:;\: 6,5]$ avec $f(4)>0$ et $f(6,6)<0$ donc l'équation $f(q)=0$ admet une solution sur $[4\:;\: 6,5]$ d'après le théorème des valeurs intermédiaires. De plus $f$ est strictement décroissante sur cet intervalle donc la solution est unique.\\
On trouve  $f(q) = 0$ sur  l'intervalle [4~;~6,5] pour $q\approx 6,19$
		\item Ainsi l'entreprise réalise un bénéfice ($f(q)>0$) si elle fabrique jusqu'à 619 pièces.
	\end{enumerate}
\item Afin d'obtenir le bénéfice maximal, l'entreprise doit fabrique 400 pièces (maximum de $f$ pour $q=4$ d'après la partie A).\\
 Ce bénéfice maximal est alors de $f(4)=7,8$ milliers d'euro (soit 7800 euros à la centaine d'euros près)
\item \[ B_m = \dfrac{1}{5,5} \times \displaystyle\int_1^{6,5}f(x)\:\text{d}x
=\dfrac{1}{5,5}\times \left[ F(x)\right]_1^{6,5}=\dfrac{1}{5,5}\times \left(F(6,5)-F(1)\right)\approx 4,4\]
	
Le bénéfice moyen est de 4,4 milliers d'euros, arrondi à la centaine d'euro.
\end{enumerate}
\end{document}