corrigé DM, diag 2nde, chapitre 2nde, calcul mental 1ere
authorDenise sur Titasmo <denise.maurice@normalesup.org>
Wed, 23 Mar 2016 17:43:53 +0000 (18:43 +0100)
committerDenise sur Titasmo <denise.maurice@normalesup.org>
Wed, 23 Mar 2016 17:43:53 +0000 (18:43 +0100)
15 files changed:
1ere/DM_6/dm6.tex
1ere/DM_6/dm6_corrige.pdf
1ere/DM_6/dm6_enonce.pdf
1ere/calcul_mental/derivee.pdf [new file with mode: 0644]
1ere/calcul_mental/derivee.tex [new file with mode: 0644]
2nde/AP/systemes_diag2.pdf [new file with mode: 0644]
2nde/AP/systemes_diag2.tex [new file with mode: 0644]
2nde/AP/systemes_diag2_2pages.pdf [new file with mode: 0644]
2nde/chapitre8_droites/65p190.pdf [new file with mode: 0644]
2nde/chapitre8_droites/65p190.tex [new file with mode: 0644]
2nde/chapitre9_inequations/chap9.pdf [new file with mode: 0644]
2nde/chapitre9_inequations/chap9.tex [new file with mode: 0644]
2nde/chapitre9_inequations/graphe.pdf [new file with mode: 0644]
2nde/chapitre9_inequations/graphe.svg [new file with mode: 0644]
2nde/projet_prog/prog_DM5/corrige

index 21f9938..d972f06 100644 (file)
@@ -24,7 +24,8 @@ Deux clients entrent dans son restaurant, et on considère que leurs choix de pl
 \item Quelle est la probabilité que les deux clients choisissent la raclette ?
 \[ P(RR) = \frac{3}{10} \times \frac{3}{10} = 0,09 \]
 \item Quelle est la probabilité que les deux clients choisissent le même plat ? \reponse{
-\[ P( \text{\og deux fois le même plat \fg}) = P(\{RR; CC; TT \} ) = 0,09 + \frac{1}{5} \times \frac{1}{5} + \frac{1}{2} \times \frac{1}{2} = 0,09 + 0,04 + 0,25 = 0,38\]}
+\[ P( \text{\og deux fois le même plat \fg}) = P(\{RR; CC; TT \} ) \]
+\[= 0,09 + \frac{1}{5} \times \frac{1}{5} + \frac{1}{2} \times \frac{1}{2} = 0,09 + 0,04 + 0,25 = 0,38\]}
 \item Quelle est la probabilité qu'aucun client ne choisisse de choucroute ? \reponse{
 \[P(\text{\og aucune choucroute \fg}) = P(\{ RR; RT; TR; TT \}) \]
 \[ = 0,09 + \frac{3}{10} \times \frac{1}{2} + \frac{1}{2} \times \frac{3}{10} + 0,25 = 0,09 + 0,15 + 0,15 + 0,25 = 0,64 \]
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diff --git a/1ere/calcul_mental/derivee.pdf b/1ere/calcul_mental/derivee.pdf
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diff --git a/1ere/calcul_mental/derivee.tex b/1ere/calcul_mental/derivee.tex
new file mode 100644 (file)
index 0000000..d0f72c2
--- /dev/null
@@ -0,0 +1,42 @@
+\input{../../header_beamer.tex}
+
+\begin{document}
+\slide{Dérivées de fonctions usuelles}
+
+\slide{1) \[
+f(x) = 3x-2
+\]}
+\slide{2) 
+\[
+g(x) = x^2
+\]}
+
+\slide{3)
+\[
+h(x) = 42
+\]
+}
+
+\slide{4)
+\[
+i(x) = \sqrt x
+\]
+}
+
+\slide{5)
+\[
+j(x) = x^{512}
+\]
+}
+
+\slide{
+\begin{enumerate}
+\item $f(x) = 3x-2$ \reponse{2}{$f'(x) = 3$}
+\item $g(x) = x^2$ \reponse{3}{$g'(x) = 2x$}
+\item $h(x) = 42$ \reponse{4}{$h'(x) = 0$}
+\item $i(x) = \sqrt x$ \reponse{5}{$i'(x) = \frac{1}{2\sqrt x}$}
+\item $j(x) = x^{512}$ \reponse{6}{$j'(x) = 512x^{511}$}
+\end{enumerate}
+}
+
+\end{document}
diff --git a/2nde/AP/systemes_diag2.pdf b/2nde/AP/systemes_diag2.pdf
new file mode 100644 (file)
index 0000000..2df29c8
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diff --git a/2nde/AP/systemes_diag2.tex b/2nde/AP/systemes_diag2.tex
new file mode 100644 (file)
index 0000000..df198d3
--- /dev/null
@@ -0,0 +1,47 @@
+\input{../../header_a5.tex}
+
+\begin{document}
+Résoudre les systèmes d'équation suivants (utiliser l'espace de la feuille, recto et verso, pour effectuer les calculs, puis reporter les réponses dans le tableau).
+\[ 
+\begin{array}{l|l|l|l}
+~A & ~B & ~C & ~D\\
+\hline
+\systeme{
+2x+5y = 3 \\
+x = -1}
+&
+\systeme{y=x+9 \\
+y = -2x
+}
+&
+\systeme{x + y = 5 \\
+x - y = 4
+}
+&
+\systeme{
+-x + 6y = 4\\
+7x-2y = -8
+} \\
+\hline
+x = & x= & x= & x= \\
+y= & y=& y=& y=\\
+\hline
+\end{array}
+\]
+
+%\systeme{
+%x + y + z = -4 \\
+%x + 2y + 3z = 8 \\
+%5y-6z = -212
+%}
+
+
+% x = -1, y = 1
+% x = -3, y=6
+
+% x = 9/2, y=1/2
+
+% x = -1, y=1/2
+
+\end{document}
diff --git a/2nde/AP/systemes_diag2_2pages.pdf b/2nde/AP/systemes_diag2_2pages.pdf
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index 0000000..79b7afb
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diff --git a/2nde/chapitre8_droites/65p190.pdf b/2nde/chapitre8_droites/65p190.pdf
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index 0000000..9f51bf4
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diff --git a/2nde/chapitre8_droites/65p190.tex b/2nde/chapitre8_droites/65p190.tex
new file mode 100644 (file)
index 0000000..19af279
--- /dev/null
@@ -0,0 +1,28 @@
+\input{../../header_a5.tex}
+
+\begin{document}
+\exo{65p190}
+\begin{enumerate}
+\setcounter{enumi}{1}
+\item Déterminer les coordonnées de leur point d'intersection :
+
+On doit résoudre le système :
+
+\systeme{y=x\\y=-2x+4}
+
+On obtient : \[x = -2x+4 \equi 3x = 4 \equi x = \frac{4}{3} \]
+En remplaçant dans la première équation, on obtient :
+\[ y= \frac{4}{3}\]
+Donc $K\left(\frac{4}{3}; \frac{4}{3} \right) $
+
+\item On calcule les coefficients directeurs de $(AK)$ et $(AQ)$ :
+
+Pour $(AK)$, $a = \frac{ 0 - \frac{4}{3}}{8 - \frac{4}{3}} = \frac{ \frac{-4}{3}}{ \frac{20}{3}} = \frac{-4}{20} = -\frac{1}{5}$
+
+Pour $(AQ)$, $a = \frac{0-2}{8-(-2)} = \frac{-2}{10} = -\frac{1}{5}$
+
+Les points $A$, $K$ et $Q$ sont donc alignés.
+
+On sait que le point $K$ est sur les droites $(OP)$ et $(BR)$ par les questions précédentes, et il appartient aussi à la droite $(AQ)$ puisque $A$, $K$ et $Q$ sont alignés : les droites droites se coupent donc en un même point, le point $K$.
+\end{enumerate}
+\end{document}
diff --git a/2nde/chapitre9_inequations/chap9.pdf b/2nde/chapitre9_inequations/chap9.pdf
new file mode 100644 (file)
index 0000000..89d2be7
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diff --git a/2nde/chapitre9_inequations/chap9.tex b/2nde/chapitre9_inequations/chap9.tex
new file mode 100644 (file)
index 0000000..4f6cbc2
--- /dev/null
@@ -0,0 +1,106 @@
+\input{../../header.tex}
+
+\title{\vspace{-1cm}Chapitre 9 -- Inéquations \vspace{-1.5cm}}
+\date{}
+
+\begin{document}
+\maketitle
+\begin{defn}
+Résoudre une inéquation, c'est \dotfill
+\smallskip
+\lignepoint
+\end{defn}
+
+\section{Résolution graphique (rappels)}
+Pour résoudre graphiquement une inéquation $f(x) >k$, on cherche l'ensemble des points de la courbe $\C$ qui sont au dessus (strictement) de la droite d'équation $y=k$. La solution est l'ensemble des abscisses de ces points.
+
+\begin{minipage}{7cm}
+\includegraphics[width=7cm]{graphe.pdf}
+\end{minipage}
+\begin{minipage}{10cm}
+Résoudre l'équation $f(x) >0$ : 
+\vspace{1.5cm}
+
+Étudier le signe de $f$ : 
+
+\variations
+x & \hspace{8cm} \\
+f(x) & & \\
+\fin
+\end{minipage}
+
+\section{Signe d'un produit}
+\begin{prop}
+Le produit de deux nombres non nuls est strictement positif si, et seulement si \dotfill
+\smallskip
+\lignepoint
+
+Il est strictement négatif si, et seulement si \dotfill
+\end{prop}
+
+Étude du signe d'une fonction à l'aide d'un \souligne{tableau de signes} : on souhaite étudier le signe de $f$ définie par $f(x) = (2x-1)(-x+1)$. Résoudre ensuite l'inéquation $f(x) \leq 0$
+
+\begin{itemize2}
+\item Pour $2x-1$ :
+\vspace{1.7cm}
+\item Pour $-x+1$ :
+\vspace{1.7cm}
+\end{itemize2}
+
+\begin{tabular}{|c|p{0.5cm}p{7cm}p{1cm}|}
+\hline
+$x$ & $-\infty$ & \vspace{0.5cm} & $+\infty$ \\
+\hline
+$2x-1$ & \vspace{0.5cm}& & \\
+\hline
+$-x + 1$ &\vspace{0.5cm} & & \\
+\hline
+$f(x)$ & &\vspace{0.5cm} & \\
+\hline
+\end{tabular}
+
+\bigskip
+On lit dans le tableau la solution de l'inéquation $f(x) \leq 0$ : 
+
+
+\newpage
+\section{Signe d'un quotient}
+
+\begin{prop}
+Le quotient de deux nombres non nuls est strictement positif si, et seulement si \dotfill
+\smallskip
+\lignepoint
+
+Il est strictement négatif si, et seulement si \dotfill
+\end{prop}
+
+Concrètement, on procède exactement comme pour le produit, avec les valeurs interdites (division par zéro) à surveiller. Dans un tableau de signe, la valeur interdite se note par \dotfill
+
+\bigskip
+
+On souhaite étudier le signe de $g$ définie par $g(x) = \frac{2x-1}{-x+1}$. Résoudre ensuite l'inéquation $g(x) \leq 0$
+
+\begin{itemize2}
+\item Pour $2x-1$ :
+\vspace{1.7cm}
+\item Pour $-x+1$ :
+\vspace{1.7cm}
+\end{itemize2}
+
+\begin{tabular}{|c|p{0.5cm}p{7cm}p{1cm}|}
+\hline
+$x$ & $-\infty$ & \vspace{0.5cm} & $+\infty$ \\
+\hline
+$2x-1$ & \vspace{0.5cm}& & \\
+\hline
+$-x + 1$ &\vspace{0.5cm} & & \\
+\hline
+$g(x)$ & &\vspace{0.5cm} & \\
+\hline
+\end{tabular}
+
+\bigskip
+On lit dans le tableau la solution de l'inéquation $g(x) \leq 0$ : 
+
+
+\end{document}
diff --git a/2nde/chapitre9_inequations/graphe.pdf b/2nde/chapitre9_inequations/graphe.pdf
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index 1da3e27..57df600 100644 (file)
@@ -1,9 +1,9 @@
 Disp "Bienvenue"
 Pause 
 ClrHome
-Disp "1:Nouvelle partie"
-Disp "2:Charger partie"
-Disp "3:Quitter"
+Disp "1 : Nouvelle partie"
+Disp "2 : Charger partie"
+Disp "3 : Quitter"
 Input A
 If A=3
 Then
@@ -61,9 +61,10 @@ randInt(5,10)->W
 Disp "Force ennemi",G
 Disp "Agilité ennemi",B
 Disp "PV ennemi",W
-Pause 
-rand->J
-If J<0.5+0.05*A
+Pause
+
+rand->R
+If R<0.5+0.05*A
 Then
 W-F->W
 Disp "Ennemi touché"
@@ -71,25 +72,14 @@ Disp "PV restants",W
 Else
 Disp "Raté !"
 End
-Pause 
-If W÷0
-Then
-Goto D
-Else
-randNor0,1)->R
+
+rand->R
 If R<0.5+0.05*B
 Then
 V-G->V
-Disp "IL ATTAQUE"
-Disp "TA VIE RESTANTE",V
+Disp "L'ennemi a touché"
+Disp "Vie restante",V
 Else
-Disp "OUF!",V
+Disp "L'ennemi a raté !"
 End
-Pause 
-End
-Lbl D
-If W>V
-Then
-Disp "PERDU"
-Else
-Disp "GAGNE"
+