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b/TST/DM/2102_DM2/TST1/01_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill AIOUAZ Ahmed} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.8$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.2} + } + child {node {$1$} + edge from parent + node[above] {0.8} + } + edge from parent + node[above] {0.2} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.2} + } + child {node {$1$} + edge from parent + node[above] {0.8} + } + edge from parent + node[above] {0.2} + } + edge from parent + node[above] {0.2} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.2} + } + child {node {$1$} + edge from parent + node[above] {0.8} + } + edge from parent + node[above] {0.2} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.2} + } + child {node {$1$} + edge from parent + node[above] {0.8} + } + edge from parent + node[above] {0.2} + } + edge from parent + node[above] {0.8} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.8^1 \times 0.2^2 \approx 0.096 + \] + \item + \[ + P(X = 0) = 0.2^3 \approx 0.008 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.8^2 \times 0.2^1 + 0.8^3 \approx 0.896 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.008$ & $0.096$ & $0.384$ &$0.512$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.008 + 1 \times 0.096 + 2 \times 0.384 + 3 \times 0.512 = 2.4 + \] + On peut donc estimer qu'il y aura en moyenne $2.4$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 27$ + \item $20^x = 45$ + \item $0.55^x \leq 27$ + \item $3 \times 0.31^x = 37$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(27)$ + \item $x = \frac{\log(45)}{\log(20)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(27)}{\log(0.55)}$ + + \item Il faut penser à faire la division à par $3$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(12.33)}{\log(0.31)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = - 10x^3 + 720x^2 - 11400x - 9$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(38)$ et $f'(10)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = - 30x^2 + 1440x - 11400$ + \item + \begin{align*} + f'(38) &= - 30 \times 38^{2} + 1440 \times 38 - 11400\\&= - 30 \times 1444 + 54720 - 11400\\&= - 43320 + 43320\\&= 0 + \end{align*} + \begin{align*} + f'(10) &= - 30 \times 10^{2} + 1440 \times 10 - 11400\\&= - 30 \times 100 + 14400 - 11400\\&= - 3000 + 3000\\&= 0 + \end{align*} + Donc $x = 38$ et $x=10$ sont des racines de $f'(x) = - 30x^2 + 1440x - 11400$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = -30 (x - 38)(x-10) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/02_2102_DM2.tex b/TST/DM/2102_DM2/TST1/02_2102_DM2.tex new file mode 100644 index 0000000..0b9d502 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/02_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill BAHBAH Zakaria} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.25$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.75} + } + child {node {$1$} + edge from parent + node[above] {0.25} + } + edge from parent + node[above] {0.75} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.75} + } + child {node {$1$} + edge from parent + node[above] {0.25} + } + edge from parent + node[above] {0.75} + } + edge from parent + node[above] {0.75} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.75} + } + child {node {$1$} + edge from parent + node[above] {0.25} + } + edge from parent + node[above] {0.75} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.75} + } + child {node {$1$} + edge from parent + node[above] {0.25} + } + edge from parent + node[above] {0.75} + } + edge from parent + node[above] {0.25} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.25^1 \times 0.75^2 \approx 0.422 + \] + \item + \[ + P(X = 0) = 0.75^3 \approx 0.422 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.25^2 \times 0.75^1 + 0.25^3 \approx 0.157 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.422$ & $0.422$ & $0.141$ &$0.016$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.422 + 1 \times 0.422 + 2 \times 0.141 + 3 \times 0.016 = 0.75 + \] + On peut donc estimer qu'il y aura en moyenne $0.75$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 6$ + \item $4^x = 46$ + \item $0.74^x \leq 39$ + \item $4 \times 0.52^x = 19$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(6)$ + \item $x = \frac{\log(46)}{\log(4)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(39)}{\log(0.74)}$ + + \item Il faut penser à faire la division à par $4$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(4.75)}{\log(0.52)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = 10x^3 - 600x^2 - 15000x + 35$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(50)$ et $f'(-10)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 30x^2 - 1200x - 15000$ + \item + \begin{align*} + f'(50) &= 30 \times 50^{2} - 1200 \times 50 - 15000\\&= 30 \times 2500 - 60000 - 15000\\&= 75000 - 75000\\&= 0 + \end{align*} + \begin{align*} + f'(-10) &= 30 \times - 10^{2} - 1200(- 10) - 15000\\&= 30 \times 100 + 12000 - 15000\\&= 3000 - 3000\\&= 0 + \end{align*} + Donc $x = 50$ et $x=-10$ sont des racines de $f'(x) = 30x^2 - 1200x - 15000$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 30 (x - 50)(x--10) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/03_2102_DM2.tex b/TST/DM/2102_DM2/TST1/03_2102_DM2.tex new file mode 100644 index 0000000..a9ff777 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/03_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill BALLOFFET Kenza} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.83$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.17} + } + child {node {$1$} + edge from parent + node[above] {0.83} + } + edge from parent + node[above] {0.17} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.17} + } + child {node {$1$} + edge from parent + node[above] {0.83} + } + edge from parent + node[above] {0.17} + } + edge from parent + node[above] {0.17} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.17} + } + child {node {$1$} + edge from parent + node[above] {0.83} + } + edge from parent + node[above] {0.17} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.17} + } + child {node {$1$} + edge from parent + node[above] {0.83} + } + edge from parent + node[above] {0.17} + } + edge from parent + node[above] {0.83} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.83^1 \times 0.17^2 \approx 0.072 + \] + \item + \[ + P(X = 0) = 0.17^3 \approx 0.005 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.83^2 \times 0.17^1 + 0.83^3 \approx 0.923 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.005$ & $0.072$ & $0.351$ &$0.572$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.005 + 1 \times 0.072 + 2 \times 0.351 + 3 \times 0.572 = 2.49 + \] + On peut donc estimer qu'il y aura en moyenne $2.49$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 15$ + \item $6^x = 15$ + \item $0.63^x \leq 2$ + \item $8 \times 0.02^x = 23$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(15)$ + \item $x = \frac{\log(15)}{\log(6)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(2)}{\log(0.63)}$ + + \item Il faut penser à faire la division à par $8$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(2.88)}{\log(0.02)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = 3x^3 + 18x^2 - 1485x - 39$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(11)$ et $f'(-15)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 9x^2 + 36x - 1485$ + \item + \begin{align*} + f'(11) &= 9 \times 11^{2} + 36 \times 11 - 1485\\&= 9 \times 121 + 396 - 1485\\&= 1089 - 1089\\&= 0 + \end{align*} + \begin{align*} + f'(-15) &= 9 \times - 15^{2} + 36(- 15) - 1485\\&= 9 \times 225 - 540 - 1485\\&= 2025 - 2025\\&= 0 + \end{align*} + Donc $x = 11$ et $x=-15$ sont des racines de $f'(x) = 9x^2 + 36x - 1485$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 9 (x - 11)(x--15) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/04_2102_DM2.tex b/TST/DM/2102_DM2/TST1/04_2102_DM2.tex new file mode 100644 index 0000000..e789856 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/04_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill BENHATTAL Chakir} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.16$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.84} + } + child {node {$1$} + edge from parent + node[above] {0.16} + } + edge from parent + node[above] {0.84} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.84} + } + child {node {$1$} + edge from parent + node[above] {0.16} + } + edge from parent + node[above] {0.84} + } + edge from parent + node[above] {0.84} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.84} + } + child {node {$1$} + edge from parent + node[above] {0.16} + } + edge from parent + node[above] {0.84} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.84} + } + child {node {$1$} + edge from parent + node[above] {0.16} + } + edge from parent + node[above] {0.84} + } + edge from parent + node[above] {0.16} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.16^1 \times 0.84^2 \approx 0.339 + \] + \item + \[ + P(X = 0) = 0.84^3 \approx 0.593 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.16^2 \times 0.84^1 + 0.16^3 \approx 0.069 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.593$ & $0.339$ & $0.065$ &$0.004$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.593 + 1 \times 0.339 + 2 \times 0.065 + 3 \times 0.004 = 0.48 + \] + On peut donc estimer qu'il y aura en moyenne $0.48$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 31$ + \item $11^x = 39$ + \item $0.65^x \leq 46$ + \item $2 \times 0.75^x = 11$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(31)$ + \item $x = \frac{\log(39)}{\log(11)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(46)}{\log(0.65)}$ + + \item Il faut penser à faire la division à par $2$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(5.5)}{\log(0.75)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = - 4x^3 + 354x^2 - 9120x + 5$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(40)$ et $f'(19)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = - 12x^2 + 708x - 9120$ + \item + \begin{align*} + f'(40) &= - 12 \times 40^{2} + 708 \times 40 - 9120\\&= - 12 \times 1600 + 28320 - 9120\\&= - 19200 + 19200\\&= 0 + \end{align*} + \begin{align*} + f'(19) &= - 12 \times 19^{2} + 708 \times 19 - 9120\\&= - 12 \times 361 + 13452 - 9120\\&= - 4332 + 4332\\&= 0 + \end{align*} + Donc $x = 40$ et $x=19$ sont des racines de $f'(x) = - 12x^2 + 708x - 9120$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = -12 (x - 40)(x-19) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/05_2102_DM2.tex b/TST/DM/2102_DM2/TST1/05_2102_DM2.tex new file mode 100644 index 0000000..a2766ae --- /dev/null +++ b/TST/DM/2102_DM2/TST1/05_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill CLAIN Avinash} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.29$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.71} + } + child {node {$1$} + edge from parent + node[above] {0.29} + } + edge from parent + node[above] {0.71} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.71} + } + child {node {$1$} + edge from parent + node[above] {0.29} + } + edge from parent + node[above] {0.71} + } + edge from parent + node[above] {0.71} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.71} + } + child {node {$1$} + edge from parent + node[above] {0.29} + } + edge from parent + node[above] {0.71} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.71} + } + child {node {$1$} + edge from parent + node[above] {0.29} + } + edge from parent + node[above] {0.71} + } + edge from parent + node[above] {0.29} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.29^1 \times 0.71^2 \approx 0.439 + \] + \item + \[ + P(X = 0) = 0.71^3 \approx 0.358 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.29^2 \times 0.71^1 + 0.29^3 \approx 0.203 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.358$ & $0.439$ & $0.179$ &$0.024$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.358 + 1 \times 0.439 + 2 \times 0.179 + 3 \times 0.024 = 0.87 + \] + On peut donc estimer qu'il y aura en moyenne $0.87$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 8$ + \item $10^x = 44$ + \item $0.23^x \leq 48$ + \item $3 \times 0.81^x = 10$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(8)$ + \item $x = \frac{\log(44)}{\log(10)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(48)}{\log(0.23)}$ + + \item Il faut penser à faire la division à par $3$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(3.33)}{\log(0.81)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = 9x^3 - 837x^2 + 19872x - 40$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(46)$ et $f'(16)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 27x^2 - 1674x + 19872$ + \item + \begin{align*} + f'(46) &= 27 \times 46^{2} - 1674 \times 46 + 19872\\&= 27 \times 2116 - 77004 + 19872\\&= 57132 - 57132\\&= 0 + \end{align*} + \begin{align*} + f'(16) &= 27 \times 16^{2} - 1674 \times 16 + 19872\\&= 27 \times 256 - 26784 + 19872\\&= 6912 - 6912\\&= 0 + \end{align*} + Donc $x = 46$ et $x=16$ sont des racines de $f'(x) = 27x^2 - 1674x + 19872$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 27 (x - 46)(x-16) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/06_2102_DM2.tex b/TST/DM/2102_DM2/TST1/06_2102_DM2.tex new file mode 100644 index 0000000..9e37e72 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/06_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill COLASSI Alexis} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.0$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {1.0} + } + child {node {$1$} + edge from parent + node[above] {0.0} + } + edge from parent + node[above] {1.0} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {1.0} + } + child {node {$1$} + edge from parent + node[above] {0.0} + } + edge from parent + node[above] {1.0} + } + edge from parent + node[above] {1.0} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {1.0} + } + child {node {$1$} + edge from parent + node[above] {0.0} + } + edge from parent + node[above] {1.0} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {1.0} + } + child {node {$1$} + edge from parent + node[above] {0.0} + } + edge from parent + node[above] {1.0} + } + edge from parent + node[above] {0.0} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.0^1 \times 1.0^2 \approx 0.0 + \] + \item + \[ + P(X = 0) = 1.0^3 \approx 1.0 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.0^2 \times 1.0^1 + 0.0^3 \approx 0.0 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $1.0$ & $0.0$ & $0.0$ &$0.0$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 1.0 + 1 \times 0.0 + 2 \times 0.0 + 3 \times 0.0 = 0.0 + \] + On peut donc estimer qu'il y aura en moyenne $0.0$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 18$ + \item $16^x = 40$ + \item $0.19^x \leq 17$ + \item $6 \times 0.07^x = 45$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(18)$ + \item $x = \frac{\log(40)}{\log(16)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(17)}{\log(0.19)}$ + + \item Il faut penser à faire la division à par $6$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(7.5)}{\log(0.07)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = 6x^3 - 27x^2 - 972x + 2$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(9)$ et $f'(-6)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 18x^2 - 54x - 972$ + \item + \begin{align*} + f'(9) &= 18 \times 9^{2} - 54 \times 9 - 972\\&= 18 \times 81 - 486 - 972\\&= 1458 - 1458\\&= 0 + \end{align*} + \begin{align*} + f'(-6) &= 18 \times - 6^{2} - 54(- 6) - 972\\&= 18 \times 36 + 324 - 972\\&= 648 - 648\\&= 0 + \end{align*} + Donc $x = 9$ et $x=-6$ sont des racines de $f'(x) = 18x^2 - 54x - 972$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 18 (x - 9)(x--6) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/07_2102_DM2.tex b/TST/DM/2102_DM2/TST1/07_2102_DM2.tex new file mode 100644 index 0000000..6620a9e --- /dev/null +++ b/TST/DM/2102_DM2/TST1/07_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill COUBAT Alexis} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.1$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + edge from parent + node[above] {0.9} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + edge from parent + node[above] {0.1} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.1^1 \times 0.9^2 \approx 0.243 + \] + \item + \[ + P(X = 0) = 0.9^3 \approx 0.729 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.1^2 \times 0.9^1 + 0.1^3 \approx 0.028 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.729$ & $0.243$ & $0.027$ &$0.001$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.729 + 1 \times 0.243 + 2 \times 0.027 + 3 \times 0.001 = 0.3 + \] + On peut donc estimer qu'il y aura en moyenne $0.3$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 40$ + \item $17^x = 2$ + \item $0.14^x \leq 21$ + \item $3 \times 0.62^x = 26$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(40)$ + \item $x = \frac{\log(2)}{\log(17)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(21)}{\log(0.14)}$ + + \item Il faut penser à faire la division à par $3$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(8.67)}{\log(0.62)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = 6x^3 - 189x^2 - 828x - 3$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(23)$ et $f'(-2)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 18x^2 - 378x - 828$ + \item + \begin{align*} + f'(23) &= 18 \times 23^{2} - 378 \times 23 - 828\\&= 18 \times 529 - 8694 - 828\\&= 9522 - 9522\\&= 0 + \end{align*} + \begin{align*} + f'(-2) &= 18 \times - 2^{2} - 378(- 2) - 828\\&= 18 \times 4 + 756 - 828\\&= 72 - 72\\&= 0 + \end{align*} + Donc $x = 23$ et $x=-2$ sont des racines de $f'(x) = 18x^2 - 378x - 828$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 18 (x - 23)(x--2) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/08_2102_DM2.tex b/TST/DM/2102_DM2/TST1/08_2102_DM2.tex new file mode 100644 index 0000000..117c697 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/08_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill COULLON Anis} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.37$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.63} + } + child {node {$1$} + edge from parent + node[above] {0.37} + } + edge from parent + node[above] {0.63} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.63} + } + child {node {$1$} + edge from parent + node[above] {0.37} + } + edge from parent + node[above] {0.63} + } + edge from parent + node[above] {0.63} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.63} + } + child {node {$1$} + edge from parent + node[above] {0.37} + } + edge from parent + node[above] {0.63} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.63} + } + child {node {$1$} + edge from parent + node[above] {0.37} + } + edge from parent + node[above] {0.63} + } + edge from parent + node[above] {0.37} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.37^1 \times 0.63^2 \approx 0.441 + \] + \item + \[ + P(X = 0) = 0.63^3 \approx 0.25 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.37^2 \times 0.63^1 + 0.37^3 \approx 0.31 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.25$ & $0.441$ & $0.259$ &$0.051$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.25 + 1 \times 0.441 + 2 \times 0.259 + 3 \times 0.051 = 1.11 + \] + On peut donc estimer qu'il y aura en moyenne $1.11$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 10$ + \item $19^x = 35$ + \item $0.59^x \leq 32$ + \item $4 \times 0.92^x = 16$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(10)$ + \item $x = \frac{\log(35)}{\log(19)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(32)}{\log(0.59)}$ + + \item Il faut penser à faire la division à par $4$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(4.0)}{\log(0.92)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = x^3 - 45x^2 + 648x - 33$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(18)$ et $f'(12)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 3x^2 - 90x + 648$ + \item + \begin{align*} + f'(18) &= 3 \times 18^{2} - 90 \times 18 + 648\\&= 3 \times 324 - 1620 + 648\\&= 972 - 972\\&= 0 + \end{align*} + \begin{align*} + f'(12) &= 3 \times 12^{2} - 90 \times 12 + 648\\&= 3 \times 144 - 1080 + 648\\&= 432 - 432\\&= 0 + \end{align*} + Donc $x = 18$ et $x=12$ sont des racines de $f'(x) = 3x^2 - 90x + 648$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 3 (x - 18)(x-12) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/09_2102_DM2.tex b/TST/DM/2102_DM2/TST1/09_2102_DM2.tex new file mode 100644 index 0000000..dba5511 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/09_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill DINGER Sölen} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.53$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.47} + } + child {node {$1$} + edge from parent + node[above] {0.53} + } + edge from parent + node[above] {0.47} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.47} + } + child {node {$1$} + edge from parent + node[above] {0.53} + } + edge from parent + node[above] {0.47} + } + edge from parent + node[above] {0.47} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.47} + } + child {node {$1$} + edge from parent + node[above] {0.53} + } + edge from parent + node[above] {0.47} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.47} + } + child {node {$1$} + edge from parent + node[above] {0.53} + } + edge from parent + node[above] {0.47} + } + edge from parent + node[above] {0.53} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.53^1 \times 0.47^2 \approx 0.351 + \] + \item + \[ + P(X = 0) = 0.47^3 \approx 0.104 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.53^2 \times 0.47^1 + 0.53^3 \approx 0.545 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.104$ & $0.351$ & $0.396$ &$0.149$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.104 + 1 \times 0.351 + 2 \times 0.396 + 3 \times 0.149 = 1.59 + \] + On peut donc estimer qu'il y aura en moyenne $1.59$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 22$ + \item $20^x = 6$ + \item $0.11^x \leq 14$ + \item $8 \times 0.45^x = 46$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(22)$ + \item $x = \frac{\log(6)}{\log(20)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(14)}{\log(0.11)}$ + + \item Il faut penser à faire la division à par $8$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(5.75)}{\log(0.45)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = - 4x^3 - 48x^2 + 2880x + 18$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(12)$ et $f'(-20)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = - 12x^2 - 96x + 2880$ + \item + \begin{align*} + f'(12) &= - 12 \times 12^{2} - 96 \times 12 + 2880\\&= - 12 \times 144 - 1152 + 2880\\&= - 1728 + 1728\\&= 0 + \end{align*} + \begin{align*} + f'(-20) &= - 12 \times - 20^{2} - 96(- 20) + 2880\\&= - 12 \times 400 + 1920 + 2880\\&= - 4800 + 4800\\&= 0 + \end{align*} + Donc $x = 12$ et $x=-20$ sont des racines de $f'(x) = - 12x^2 - 96x + 2880$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = -12 (x - 12)(x--20) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/10_2102_DM2.tex b/TST/DM/2102_DM2/TST1/10_2102_DM2.tex new file mode 100644 index 0000000..a988d3e --- /dev/null +++ b/TST/DM/2102_DM2/TST1/10_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill EYRAUD Cynthia} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.14$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.86} + } + child {node {$1$} + edge from parent + node[above] {0.14} + } + edge from parent + node[above] {0.86} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.86} + } + child {node {$1$} + edge from parent + node[above] {0.14} + } + edge from parent + node[above] {0.86} + } + edge from parent + node[above] {0.86} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.86} + } + child {node {$1$} + edge from parent + node[above] {0.14} + } + edge from parent + node[above] {0.86} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.86} + } + child {node {$1$} + edge from parent + node[above] {0.14} + } + edge from parent + node[above] {0.86} + } + edge from parent + node[above] {0.14} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.14^1 \times 0.86^2 \approx 0.311 + \] + \item + \[ + P(X = 0) = 0.86^3 \approx 0.636 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.14^2 \times 0.86^1 + 0.14^3 \approx 0.054 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.636$ & $0.311$ & $0.051$ &$0.003$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.636 + 1 \times 0.311 + 2 \times 0.051 + 3 \times 0.003 = 0.42 + \] + On peut donc estimer qu'il y aura en moyenne $0.42$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 5$ + \item $20^x = 40$ + \item $0.85^x \leq 38$ + \item $2 \times 0.24^x = 24$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(5)$ + \item $x = \frac{\log(40)}{\log(20)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(38)}{\log(0.85)}$ + + \item Il faut penser à faire la division à par $2$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(12.0)}{\log(0.24)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = - 3x^3 + 49.5x^2 + 5580x - 3$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(31)$ et $f'(-20)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = - 9x^2 + 99x + 5580$ + \item + \begin{align*} + f'(31) &= - 9 \times 31^{2} + 99 \times 31 + 5580\\&= - 9 \times 961 + 3069 + 5580\\&= - 8649 + 8649\\&= 0 + \end{align*} + \begin{align*} + f'(-20) &= - 9 \times - 20^{2} + 99(- 20) + 5580\\&= - 9 \times 400 - 1980 + 5580\\&= - 3600 + 3600\\&= 0 + \end{align*} + Donc $x = 31$ et $x=-20$ sont des racines de $f'(x) = - 9x^2 + 99x + 5580$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = -9 (x - 31)(x--20) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/11_2102_DM2.tex b/TST/DM/2102_DM2/TST1/11_2102_DM2.tex new file mode 100644 index 0000000..0521f45 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/11_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill FERREIRA Léo} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.05$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.95} + } + child {node {$1$} + edge from parent + node[above] {0.05} + } + edge from parent + node[above] {0.95} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.95} + } + child {node {$1$} + edge from parent + node[above] {0.05} + } + edge from parent + node[above] {0.95} + } + edge from parent + node[above] {0.95} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.95} + } + child {node {$1$} + edge from parent + node[above] {0.05} + } + edge from parent + node[above] {0.95} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.95} + } + child {node {$1$} + edge from parent + node[above] {0.05} + } + edge from parent + node[above] {0.95} + } + edge from parent + node[above] {0.05} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.05^1 \times 0.95^2 \approx 0.135 + \] + \item + \[ + P(X = 0) = 0.95^3 \approx 0.857 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.05^2 \times 0.95^1 + 0.05^3 \approx 0.007 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.857$ & $0.135$ & $0.007$ &$0.0$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.857 + 1 \times 0.135 + 2 \times 0.007 + 3 \times 0.0 = 0.15 + \] + On peut donc estimer qu'il y aura en moyenne $0.15$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 39$ + \item $19^x = 43$ + \item $0.24^x \leq 2$ + \item $5 \times 0.52^x = 16$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(39)$ + \item $x = \frac{\log(43)}{\log(19)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(2)}{\log(0.24)}$ + + \item Il faut penser à faire la division à par $5$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(3.2)}{\log(0.52)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = 2x^3 - 171x^2 + 2592x + 12$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(48)$ et $f'(9)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 6x^2 - 342x + 2592$ + \item + \begin{align*} + f'(48) &= 6 \times 48^{2} - 342 \times 48 + 2592\\&= 6 \times 2304 - 16416 + 2592\\&= 13824 - 13824\\&= 0 + \end{align*} + \begin{align*} + f'(9) &= 6 \times 9^{2} - 342 \times 9 + 2592\\&= 6 \times 81 - 3078 + 2592\\&= 486 - 486\\&= 0 + \end{align*} + Donc $x = 48$ et $x=9$ sont des racines de $f'(x) = 6x^2 - 342x + 2592$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 6 (x - 48)(x-9) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/12_2102_DM2.tex b/TST/DM/2102_DM2/TST1/12_2102_DM2.tex new file mode 100644 index 0000000..6248f18 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/12_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill FILALI Zakaria} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.01$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.99} + } + child {node {$1$} + edge from parent + node[above] {0.01} + } + edge from parent + node[above] {0.99} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.99} + } + child {node {$1$} + edge from parent + node[above] {0.01} + } + edge from parent + node[above] {0.99} + } + edge from parent + node[above] {0.99} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.99} + } + child {node {$1$} + edge from parent + node[above] {0.01} + } + edge from parent + node[above] {0.99} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.99} + } + child {node {$1$} + edge from parent + node[above] {0.01} + } + edge from parent + node[above] {0.99} + } + edge from parent + node[above] {0.01} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.01^1 \times 0.99^2 \approx 0.029 + \] + \item + \[ + P(X = 0) = 0.99^3 \approx 0.97 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.01^2 \times 0.99^1 + 0.01^3 \approx 0.0 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.97$ & $0.029$ & $0.0$ &$0.0$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.97 + 1 \times 0.029 + 2 \times 0.0 + 3 \times 0.0 = 0.03 + \] + On peut donc estimer qu'il y aura en moyenne $0.03$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 39$ + \item $19^x = 15$ + \item $0.73^x \leq 39$ + \item $8 \times 0.15^x = 20$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(39)$ + \item $x = \frac{\log(15)}{\log(19)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(39)}{\log(0.73)}$ + + \item Il faut penser à faire la division à par $8$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(2.5)}{\log(0.15)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = 8x^3 - 264x^2 - 20160x - 32$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(42)$ et $f'(-20)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 24x^2 - 528x - 20160$ + \item + \begin{align*} + f'(42) &= 24 \times 42^{2} - 528 \times 42 - 20160\\&= 24 \times 1764 - 22176 - 20160\\&= 42336 - 42336\\&= 0 + \end{align*} + \begin{align*} + f'(-20) &= 24 \times - 20^{2} - 528(- 20) - 20160\\&= 24 \times 400 + 10560 - 20160\\&= 9600 - 9600\\&= 0 + \end{align*} + Donc $x = 42$ et $x=-20$ sont des racines de $f'(x) = 24x^2 - 528x - 20160$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 24 (x - 42)(x--20) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/13_2102_DM2.tex b/TST/DM/2102_DM2/TST1/13_2102_DM2.tex new file mode 100644 index 0000000..d20b88e --- /dev/null +++ b/TST/DM/2102_DM2/TST1/13_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill FOIGNY Romain} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.25$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.75} + } + child {node {$1$} + edge from parent + node[above] {0.25} + } + edge from parent + node[above] {0.75} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.75} + } + child {node {$1$} + edge from parent + node[above] {0.25} + } + edge from parent + node[above] {0.75} + } + edge from parent + node[above] {0.75} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.75} + } + child {node {$1$} + edge from parent + node[above] {0.25} + } + edge from parent + node[above] {0.75} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.75} + } + child {node {$1$} + edge from parent + node[above] {0.25} + } + edge from parent + node[above] {0.75} + } + edge from parent + node[above] {0.25} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.25^1 \times 0.75^2 \approx 0.422 + \] + \item + \[ + P(X = 0) = 0.75^3 \approx 0.422 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.25^2 \times 0.75^1 + 0.25^3 \approx 0.157 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.422$ & $0.422$ & $0.141$ &$0.016$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.422 + 1 \times 0.422 + 2 \times 0.141 + 3 \times 0.016 = 0.75 + \] + On peut donc estimer qu'il y aura en moyenne $0.75$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 13$ + \item $17^x = 26$ + \item $0.75^x \leq 40$ + \item $9 \times 0.86^x = 44$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(13)$ + \item $x = \frac{\log(26)}{\log(17)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(40)}{\log(0.75)}$ + + \item Il faut penser à faire la division à par $9$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(4.89)}{\log(0.86)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = 2x^3 - 84x^2 + 960x + 28$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(8)$ et $f'(20)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 6x^2 - 168x + 960$ + \item + \begin{align*} + f'(8) &= 6 \times 8^{2} - 168 \times 8 + 960\\&= 6 \times 64 - 1344 + 960\\&= 384 - 384\\&= 0 + \end{align*} + \begin{align*} + f'(20) &= 6 \times 20^{2} - 168 \times 20 + 960\\&= 6 \times 400 - 3360 + 960\\&= 2400 - 2400\\&= 0 + \end{align*} + Donc $x = 8$ et $x=20$ sont des racines de $f'(x) = 6x^2 - 168x + 960$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 6 (x - 8)(x-20) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/14_2102_DM2.tex b/TST/DM/2102_DM2/TST1/14_2102_DM2.tex new file mode 100644 index 0000000..c8a03e6 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/14_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill HIPOLITO DA SILVA Andréa} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.1$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + edge from parent + node[above] {0.9} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + edge from parent + node[above] {0.1} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.1^1 \times 0.9^2 \approx 0.243 + \] + \item + \[ + P(X = 0) = 0.9^3 \approx 0.729 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.1^2 \times 0.9^1 + 0.1^3 \approx 0.028 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.729$ & $0.243$ & $0.027$ &$0.001$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.729 + 1 \times 0.243 + 2 \times 0.027 + 3 \times 0.001 = 0.3 + \] + On peut donc estimer qu'il y aura en moyenne $0.3$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 36$ + \item $12^x = 44$ + \item $0.08^x \leq 43$ + \item $4 \times 0.93^x = 46$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(36)$ + \item $x = \frac{\log(44)}{\log(12)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(43)}{\log(0.08)}$ + + \item Il faut penser à faire la division à par $4$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(11.5)}{\log(0.93)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = - 7x^3 + 294x^2 - 4032x - 33$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(12)$ et $f'(16)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = - 21x^2 + 588x - 4032$ + \item + \begin{align*} + f'(12) &= - 21 \times 12^{2} + 588 \times 12 - 4032\\&= - 21 \times 144 + 7056 - 4032\\&= - 3024 + 3024\\&= 0 + \end{align*} + \begin{align*} + f'(16) &= - 21 \times 16^{2} + 588 \times 16 - 4032\\&= - 21 \times 256 + 9408 - 4032\\&= - 5376 + 5376\\&= 0 + \end{align*} + Donc $x = 12$ et $x=16$ sont des racines de $f'(x) = - 21x^2 + 588x - 4032$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = -21 (x - 12)(x-16) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/15_2102_DM2.tex b/TST/DM/2102_DM2/TST1/15_2102_DM2.tex new file mode 100644 index 0000000..a64a555 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/15_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill HUMBERT Rayan} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.68$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.32} + } + child {node {$1$} + edge from parent + node[above] {0.68} + } + edge from parent + node[above] {0.32} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.32} + } + child {node {$1$} + edge from parent + node[above] {0.68} + } + edge from parent + node[above] {0.32} + } + edge from parent + node[above] {0.32} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.32} + } + child {node {$1$} + edge from parent + node[above] {0.68} + } + edge from parent + node[above] {0.32} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.32} + } + child {node {$1$} + edge from parent + node[above] {0.68} + } + edge from parent + node[above] {0.32} + } + edge from parent + node[above] {0.68} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.68^1 \times 0.32^2 \approx 0.209 + \] + \item + \[ + P(X = 0) = 0.32^3 \approx 0.033 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.68^2 \times 0.32^1 + 0.68^3 \approx 0.758 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.033$ & $0.209$ & $0.444$ &$0.314$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.033 + 1 \times 0.209 + 2 \times 0.444 + 3 \times 0.314 = 2.04 + \] + On peut donc estimer qu'il y aura en moyenne $2.04$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 8$ + \item $7^x = 21$ + \item $0.19^x \leq 12$ + \item $9 \times 0.76^x = 6$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(8)$ + \item $x = \frac{\log(21)}{\log(7)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(12)}{\log(0.19)}$ + + \item Il faut penser à faire la division à par $9$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(0.67)}{\log(0.76)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = 7x^3 - 483x^2 + 10584x - 2$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(28)$ et $f'(18)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 21x^2 - 966x + 10584$ + \item + \begin{align*} + f'(28) &= 21 \times 28^{2} - 966 \times 28 + 10584\\&= 21 \times 784 - 27048 + 10584\\&= 16464 - 16464\\&= 0 + \end{align*} + \begin{align*} + f'(18) &= 21 \times 18^{2} - 966 \times 18 + 10584\\&= 21 \times 324 - 17388 + 10584\\&= 6804 - 6804\\&= 0 + \end{align*} + Donc $x = 28$ et $x=18$ sont des racines de $f'(x) = 21x^2 - 966x + 10584$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 21 (x - 28)(x-18) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/16_2102_DM2.tex b/TST/DM/2102_DM2/TST1/16_2102_DM2.tex new file mode 100644 index 0000000..687d1e5 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/16_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill MASSON Grace} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.17$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.83} + } + child {node {$1$} + edge from parent + node[above] {0.17} + } + edge from parent + node[above] {0.83} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.83} + } + child {node {$1$} + edge from parent + node[above] {0.17} + } + edge from parent + node[above] {0.83} + } + edge from parent + node[above] {0.83} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.83} + } + child {node {$1$} + edge from parent + node[above] {0.17} + } + edge from parent + node[above] {0.83} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.83} + } + child {node {$1$} + edge from parent + node[above] {0.17} + } + edge from parent + node[above] {0.83} + } + edge from parent + node[above] {0.17} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.17^1 \times 0.83^2 \approx 0.351 + \] + \item + \[ + P(X = 0) = 0.83^3 \approx 0.572 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.17^2 \times 0.83^1 + 0.17^3 \approx 0.077 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.572$ & $0.351$ & $0.072$ &$0.005$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.572 + 1 \times 0.351 + 2 \times 0.072 + 3 \times 0.005 = 0.51 + \] + On peut donc estimer qu'il y aura en moyenne $0.51$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 37$ + \item $6^x = 32$ + \item $0.89^x \leq 36$ + \item $7 \times 0.02^x = 44$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(37)$ + \item $x = \frac{\log(32)}{\log(6)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(36)}{\log(0.89)}$ + + \item Il faut penser à faire la division à par $7$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(6.29)}{\log(0.02)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = - 8x^3 + 96x - 33$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(2)$ et $f'(-2)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = - 24x^2 + 96$ + \item + \begin{align*} + f'(2) &= - 24 \times 2^{2} + 96\\&= - 24 \times 4 + 96\\&= - 96 + 96\\&= 0 + \end{align*} + \begin{align*} + f'(-2) &= - 24 \times - 2^{2} + 96\\&= - 24 \times 4 + 96\\&= - 96 + 96\\&= 0 + \end{align*} + Donc $x = 2$ et $x=-2$ sont des racines de $f'(x) = - 24x^2 + 96$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = -24 (x - 2)(x--2) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/17_2102_DM2.tex b/TST/DM/2102_DM2/TST1/17_2102_DM2.tex new file mode 100644 index 0000000..cb0e40d --- /dev/null +++ b/TST/DM/2102_DM2/TST1/17_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill MOKHTARI Nissrine} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.1$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + edge from parent + node[above] {0.9} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.9} + } + child {node {$1$} + edge from parent + node[above] {0.1} + } + edge from parent + node[above] {0.9} + } + edge from parent + node[above] {0.1} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.1^1 \times 0.9^2 \approx 0.243 + \] + \item + \[ + P(X = 0) = 0.9^3 \approx 0.729 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.1^2 \times 0.9^1 + 0.1^3 \approx 0.028 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.729$ & $0.243$ & $0.027$ &$0.001$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.729 + 1 \times 0.243 + 2 \times 0.027 + 3 \times 0.001 = 0.3 + \] + On peut donc estimer qu'il y aura en moyenne $0.3$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 45$ + \item $2^x = 25$ + \item $0.83^x \leq 40$ + \item $8 \times 0.07^x = 42$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(45)$ + \item $x = \frac{\log(25)}{\log(2)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(40)}{\log(0.83)}$ + + \item Il faut penser à faire la division à par $8$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(5.25)}{\log(0.07)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = - 4x^3 + 384x^2 - 9216x - 22$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(48)$ et $f'(16)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = - 12x^2 + 768x - 9216$ + \item + \begin{align*} + f'(48) &= - 12 \times 48^{2} + 768 \times 48 - 9216\\&= - 12 \times 2304 + 36864 - 9216\\&= - 27648 + 27648\\&= 0 + \end{align*} + \begin{align*} + f'(16) &= - 12 \times 16^{2} + 768 \times 16 - 9216\\&= - 12 \times 256 + 12288 - 9216\\&= - 3072 + 3072\\&= 0 + \end{align*} + Donc $x = 48$ et $x=16$ sont des racines de $f'(x) = - 12x^2 + 768x - 9216$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = -12 (x - 48)(x-16) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/18_2102_DM2.tex b/TST/DM/2102_DM2/TST1/18_2102_DM2.tex new file mode 100644 index 0000000..e85a2d9 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/18_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill MOUFAQ Amine} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.95$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.05} + } + child {node {$1$} + edge from parent + node[above] {0.95} + } + edge from parent + node[above] {0.05} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.05} + } + child {node {$1$} + edge from parent + node[above] {0.95} + } + edge from parent + node[above] {0.05} + } + edge from parent + node[above] {0.05} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.05} + } + child {node {$1$} + edge from parent + node[above] {0.95} + } + edge from parent + node[above] {0.05} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.05} + } + child {node {$1$} + edge from parent + node[above] {0.95} + } + edge from parent + node[above] {0.05} + } + edge from parent + node[above] {0.95} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.95^1 \times 0.05^2 \approx 0.007 + \] + \item + \[ + P(X = 0) = 0.05^3 \approx 0.0 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.95^2 \times 0.05^1 + 0.95^3 \approx 0.992 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.0$ & $0.007$ & $0.135$ &$0.857$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.0 + 1 \times 0.007 + 2 \times 0.135 + 3 \times 0.857 = 2.85 + \] + On peut donc estimer qu'il y aura en moyenne $2.85$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 26$ + \item $11^x = 31$ + \item $0.01^x \leq 10$ + \item $5 \times 0.24^x = 3$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(26)$ + \item $x = \frac{\log(31)}{\log(11)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(10)}{\log(0.01)}$ + + \item Il faut penser à faire la division à par $5$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(0.6)}{\log(0.24)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = 4x^3 - 192x^2 - 9216x - 28$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(48)$ et $f'(-16)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 12x^2 - 384x - 9216$ + \item + \begin{align*} + f'(48) &= 12 \times 48^{2} - 384 \times 48 - 9216\\&= 12 \times 2304 - 18432 - 9216\\&= 27648 - 27648\\&= 0 + \end{align*} + \begin{align*} + f'(-16) &= 12 \times - 16^{2} - 384(- 16) - 9216\\&= 12 \times 256 + 6144 - 9216\\&= 3072 - 3072\\&= 0 + \end{align*} + Donc $x = 48$ et $x=-16$ sont des racines de $f'(x) = 12x^2 - 384x - 9216$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 12 (x - 48)(x--16) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/19_2102_DM2.tex b/TST/DM/2102_DM2/TST1/19_2102_DM2.tex new file mode 100644 index 0000000..6f92bd5 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/19_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill ONAL Yakub} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.97$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.03} + } + child {node {$1$} + edge from parent + node[above] {0.97} + } + edge from parent + node[above] {0.03} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.03} + } + child {node {$1$} + edge from parent + node[above] {0.97} + } + edge from parent + node[above] {0.03} + } + edge from parent + node[above] {0.03} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.03} + } + child {node {$1$} + edge from parent + node[above] {0.97} + } + edge from parent + node[above] {0.03} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.03} + } + child {node {$1$} + edge from parent + node[above] {0.97} + } + edge from parent + node[above] {0.03} + } + edge from parent + node[above] {0.97} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.97^1 \times 0.03^2 \approx 0.003 + \] + \item + \[ + P(X = 0) = 0.03^3 \approx 0.0 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.97^2 \times 0.03^1 + 0.97^3 \approx 0.998 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.0$ & $0.003$ & $0.085$ &$0.913$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.0 + 1 \times 0.003 + 2 \times 0.085 + 3 \times 0.913 = 2.91 + \] + On peut donc estimer qu'il y aura en moyenne $2.91$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 10$ + \item $19^x = 8$ + \item $0.75^x \leq 2$ + \item $5 \times 0.57^x = 24$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(10)$ + \item $x = \frac{\log(8)}{\log(19)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(2)}{\log(0.75)}$ + + \item Il faut penser à faire la division à par $5$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(4.8)}{\log(0.57)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = - 5x^3 - 7.5x^2 + 1080x - 33$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(8)$ et $f'(-9)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = - 15x^2 - 15x + 1080$ + \item + \begin{align*} + f'(8) &= - 15 \times 8^{2} - 15 \times 8 + 1080\\&= - 15 \times 64 - 120 + 1080\\&= - 960 + 960\\&= 0 + \end{align*} + \begin{align*} + f'(-9) &= - 15 \times - 9^{2} - 15(- 9) + 1080\\&= - 15 \times 81 + 135 + 1080\\&= - 1215 + 1215\\&= 0 + \end{align*} + Donc $x = 8$ et $x=-9$ sont des racines de $f'(x) = - 15x^2 - 15x + 1080$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = -15 (x - 8)(x--9) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/20_2102_DM2.tex b/TST/DM/2102_DM2/TST1/20_2102_DM2.tex new file mode 100644 index 0000000..37ad3f6 --- /dev/null +++ b/TST/DM/2102_DM2/TST1/20_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill SORIANO Laura} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.76$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.24} + } + child {node {$1$} + edge from parent + node[above] {0.76} + } + edge from parent + node[above] {0.24} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.24} + } + child {node {$1$} + edge from parent + node[above] {0.76} + } + edge from parent + node[above] {0.24} + } + edge from parent + node[above] {0.24} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.24} + } + child {node {$1$} + edge from parent + node[above] {0.76} + } + edge from parent + node[above] {0.24} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.24} + } + child {node {$1$} + edge from parent + node[above] {0.76} + } + edge from parent + node[above] {0.24} + } + edge from parent + node[above] {0.76} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.76^1 \times 0.24^2 \approx 0.131 + \] + \item + \[ + P(X = 0) = 0.24^3 \approx 0.014 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.76^2 \times 0.24^1 + 0.76^3 \approx 0.855 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.014$ & $0.131$ & $0.416$ &$0.439$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.014 + 1 \times 0.131 + 2 \times 0.416 + 3 \times 0.439 = 2.28 + \] + On peut donc estimer qu'il y aura en moyenne $2.28$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 14$ + \item $14^x = 36$ + \item $0.74^x \leq 6$ + \item $10 \times 0.11^x = 18$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(14)$ + \item $x = \frac{\log(36)}{\log(14)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(6)}{\log(0.74)}$ + + \item Il faut penser à faire la division à par $10$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(1.8)}{\log(0.11)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = 4x^3 - 228x^2 + 1632x + 16$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(34)$ et $f'(4)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 12x^2 - 456x + 1632$ + \item + \begin{align*} + f'(34) &= 12 \times 34^{2} - 456 \times 34 + 1632\\&= 12 \times 1156 - 15504 + 1632\\&= 13872 - 13872\\&= 0 + \end{align*} + \begin{align*} + f'(4) &= 12 \times 4^{2} - 456 \times 4 + 1632\\&= 12 \times 16 - 1824 + 1632\\&= 192 - 192\\&= 0 + \end{align*} + Donc $x = 34$ et $x=4$ sont des racines de $f'(x) = 12x^2 - 456x + 1632$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 12 (x - 34)(x-4) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/21_2102_DM2.tex b/TST/DM/2102_DM2/TST1/21_2102_DM2.tex new file mode 100644 index 0000000..d6b3cfa --- /dev/null +++ b/TST/DM/2102_DM2/TST1/21_2102_DM2.tex @@ -0,0 +1,203 @@ +\documentclass[a5paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tasks} + +% Title Page +\title{DM2 \hfill VECCHIO Léa} +\tribe{TST} +\date{\hfillÀ render pour le Mercredi 24 février} + +\xsimsetup{ + solution/print = false +} + +\begin{document} +\maketitle + +\begin{exercise}[subtitle={Loi binomiale}] + Trois personnes s'apprêtent à passer le portique de sécurité. On suppose que pour chaque personne la probabilité que le portique sonne est égale à $0.55$. + + Soit $X$ la variable aléatoire donnant le nombre de personnes faisant sonner le portique, parmi les 3 personnes de ce groupe. + \begin{enumerate} + \item Tracer l'arbre représentant le situation. + \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres. + \item Quelle est la probabilité qu'une seule personne fasse sonner le portique? + \item Calculer puis interpréter les probabilités suivantes + \[ + P(X = 0) \qquad \qquad P(X \geq 2) + \] + \item Calculer l'espérance de $X$ et interpréter le résultat. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item + \begin{tikzpicture}[sloped] + \node {.} + child {node {$0$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.45} + } + child {node {$1$} + edge from parent + node[above] {0.55} + } + edge from parent + node[above] {0.45} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.45} + } + child {node {$1$} + edge from parent + node[above] {0.55} + } + edge from parent + node[above] {0.45} + } + edge from parent + node[above] {0.45} + } + child[missing] {} + child[missing] {} + child[missing] {} + child { node {$1$} + child {node {$0$} + child {node {$0$} + edge from parent + node[above] {0.45} + } + child {node {$1$} + edge from parent + node[above] {0.55} + } + edge from parent + node[above] {0.45} + } + child[missing] {} + child {node {$1$} + child {node {$0$} + edge from parent + node[above] {0.45} + } + child {node {$1$} + edge from parent + node[above] {0.55} + } + edge from parent + node[above] {0.45} + } + edge from parent + node[above] {0.55} + } ; + \end{tikzpicture} + \item Chaque personne a 2 possibilités (1: fait sonner ou 2: ne fait pas sonner) et l'on fait passer 3 personnes ce qui correspond à une répétition identique et aléatoire. On peut donc modéliser la situation par une loi binomiale. + \[ + X \sim \mathcal{B}(3; 0.76) + \] + \item Probabilité qu'une seule personne fasse sonner le portique. On voit qu'il y a 3 branches qui correspondent à cette situation dont + \[ + P(X = 1) = 3 \times 0.55^1 \times 0.45^2 \approx 0.334 + \] + \item + \[ + P(X = 0) = 0.45^3 \approx 0.091 + \] + \[ + P(X \geq 2) = P(X = 2) + P(X = 3) = 3 \times 0.55^2 \times 0.45^1 + 0.55^3 \approx 0.574 + \] + + \item Il faut d'abord tracer le tableau résumant la loi de probabilité: + \begin{center} + \begin{tabular}{|c|*{4}{c|}} + \hline + Valeur & 0 & 1 & 2 & 3 \\ + \hline + Probabilité & $0.091$ & $0.334$ & $0.408$ &$0.166$ \\ + \hline + \end{tabular} + \end{center} + On peut alors calculer l'espérance + \[ + E[X] = 0 \times 0.091 + 1 \times 0.334 + 2 \times 0.408 + 3 \times 0.166 = 1.65 + \] + On peut donc estimer qu'il y aura en moyenne $1.65$ personnes qui feront sonner le portique sur les 3 personnes. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Équation puissance}] + Résoudre les équations et inéquations suivantes + \begin{multicols}{2} + \begin{enumerate} + \item $10^x = 14$ + \item $7^x = 47$ + \item $0.18^x \leq 40$ + \item $3 \times 0.21^x = 25$ + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + Les solutions ci-dessous ne sont pas justifiée car l'ordinateur ne sait pas faire. Par contre, vous vous devez savoir justifier vos réponses! + \begin{enumerate} + \item $x = \log(14)$ + \item $x = \frac{\log(47)}{\log(7)}$ + \item Il faut faire attention quand on divise par un log car ce dernier peut être négatif ce qui est le cas ici. Il faut donc pense à changer le sens de l'inégalité. + + $x \geq \frac{\log(40)}{\log(0.18)}$ + + \item Il faut penser à faire la division à par $3$ avant d'utiliser le log car sinon, on ne peut pas utiliser la formule $\log(a^n) = n\times \log(a)$. + + $x = \frac{\log(8.33)}{\log(0.21)}$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Étude de fonctions}] + Soit $f(x) = x^3 - 64.5x^2 + 360x - 44$ une fonction définie sur $\R$. + \begin{enumerate} + \item Calculer $f'(x)$ la dérivée de $f(x)$. + \item Calculer $f'(40)$ et $f'(3)$. + \item En déduire une forme factorisée de $f'(x)$. + \item Étudier le signe de $f'(x)$ et en déduire les variations de $f(x)$. + \item Est-ce que la fonction $f(x)$ admet un maximum ou un minimum? Si oui, calculer sa valeur. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Dérivée de $f(x)$: $f'(x) = 3x^2 - 129x + 360$ + \item + \begin{align*} + f'(40) &= 3 \times 40^{2} - 129 \times 40 + 360\\&= 3 \times 1600 - 5160 + 360\\&= 4800 - 4800\\&= 0 + \end{align*} + \begin{align*} + f'(3) &= 3 \times 3^{2} - 129 \times 3 + 360\\&= 3 \times 9 - 387 + 360\\&= 27 - 27\\&= 0 + \end{align*} + Donc $x = 40$ et $x=3$ sont des racines de $f'(x) = 3x^2 - 129x + 360$. + \item On en déduit la forme factorisée suivante + \[ + f'(x) = 3 (x - 40)(x-3) + \] + \item Pas de correction disponible + \item À causes des branches extérieurs, la fonction $f(x)$ n'a pas de maximum ou de minimum. + \end{enumerate} +\end{solution} + + + +%\printsolutionstype{exercise} + + + +\end{document} + +%%% Local Variables: +%%% mode: latex +%%% TeX-master: "master" +%%% End: diff --git a/TST/DM/2102_DM2/TST1/all_2102_DM2.pdf b/TST/DM/2102_DM2/TST1/all_2102_DM2.pdf new file mode 100644 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