From b242056b1ebc5b05e990eec12db28b495f8e2bb2 Mon Sep 17 00:00:00 2001 From: Bertrand Benjamin Date: Wed, 1 Oct 2025 17:22:55 +0200 Subject: [PATCH] feat(1G_math): DS2 --- .../Evaluations/DS_2025-10-08/exercises.tex | 359 ++++++++++++++++++ .../Evaluations/DS_2025-10-08/solution.pdf | Bin 0 -> 31142 bytes .../Evaluations/DS_2025-10-08/solution.tex | 33 ++ 1G_math/Evaluations/DS_2025-10-08/sujet.pdf | Bin 0 -> 35148 bytes 1G_math/Evaluations/DS_2025-10-08/sujet.tex | 31 ++ 5 files changed, 423 insertions(+) create mode 100644 1G_math/Evaluations/DS_2025-10-08/exercises.tex create mode 100644 1G_math/Evaluations/DS_2025-10-08/solution.pdf create mode 100644 1G_math/Evaluations/DS_2025-10-08/solution.tex create mode 100644 1G_math/Evaluations/DS_2025-10-08/sujet.pdf create mode 100644 1G_math/Evaluations/DS_2025-10-08/sujet.tex diff --git a/1G_math/Evaluations/DS_2025-10-08/exercises.tex b/1G_math/Evaluations/DS_2025-10-08/exercises.tex new file mode 100644 index 0000000..9892473 --- /dev/null +++ b/1G_math/Evaluations/DS_2025-10-08/exercises.tex @@ -0,0 +1,359 @@ +\begin{exercise}[subtitle={Automatismes},step={1}, origin={Divers}, topics={}, tags={ polynômes, représentation graphique, suite, trigonométrie }, points={4}] + L'exercice suivante est QCM. Une seule des 4 réponses est juste. Il n'est pas demandé de justifier. Une bonne réponse rapporte un point et un mauvaise réponse ne fait ni perdre ni gagner de points. + \begin{enumerate} + % développement + \item Quelle est la forme factorisée de $f(x) = 0.5(x-2)^2 -8$? + \begin{multicols}{2} + \begin{enumerate}[label=Réponse \Alph*: , leftmargin=*] + \item $0.5x^2-2x-6$ + \item $0.5(x-6)(x+2)$ + \item $0.5(x+10)(x-6)$ + \item $0.5(x-10)(x+6)$ + \end{enumerate} + \end{multicols} + % Calcul des termes d'une suite par récurrence avec n + \item Soit $(u_n)$ une suite définie par $u_0 = \frac{3}{2}$ et pour tout $n \in \N$, $u_{n+1} = \frac{2}{3}u_n + n$. Quelle est la valeur de $u_3$? + \begin{multicols}{4} + \begin{enumerate}[label=Réponse \Alph*: , leftmargin=*] + \item $\frac{28}{9}$ + \item $\frac{37}{9}$ + \item $\frac{41}{9}$ + \item $\frac{31}{9}$ + \end{enumerate} + \end{multicols} + % Résolution d'inéquation + \item Résoudre dans $\R$ l'inéquation : $\frac{2x-3}{4} \geq \frac{2x+3}{6}$ + \begin{multicols}{4} + \begin{enumerate}[label=Réponse \Alph*: , leftmargin=*] + \item $x \geq \frac{17}{2}$ + \item $x \leq \frac{17}{2}$ + \item $x \geq \frac{15}{2}$ + \item $x \geq \frac{19}{2}$ + \end{enumerate} + \end{multicols} + % Mesure principale d'un angle en radian + \item Quelle est la mesure principale de l'angle $\frac{17\pi}{4}$ radians? + \begin{multicols}{4} + \begin{enumerate}[label=Réponse \Alph*: , leftmargin=*] + \item $\frac{\pi}{4}$ + \item $\frac{3\pi}{4}$ + \item $-\frac{\pi}{4}$ + \item $\frac{5\pi}{4}$ + \end{enumerate} + \end{multicols} + % Retrouver un point sur le cercle trigo à partir d'un angle + % \item + % \begin{minipage}[t]{0.6\textwidth} + % Sur le cercle trigonométrique ci-contre, quel point correspond à l'angle $\frac{5\pi}{6}$ radians? + % \begin{multicols}{2} + % \begin{enumerate}[label=Réponse \Alph*: , leftmargin=*] + % \item A + % \item B + % \item C + % \item D + % \end{enumerate} + % \end{multicols} + % \end{minipage} + % \begin{minipage}{0.5\textwidth} + % \begin{center} + % \begin{tikzpicture}[scale=2] + % % Cercle trigonométrique + % \draw[thick] (0,0) circle (1); + % % Axes + % \draw[->] (-1.3,0) -- (1.3,0); + % \draw[->] (0,-1.3) -- (0,1.3); + % % Points remarquables + % \node[circle, fill=black, inner sep=2pt, label=right:A] (A) at (1,0) {}; + % \node[circle, fill=black, inner sep=2pt, label=above right:B] (B) at ({cos(30)},{sin(30)}) {}; + % \node[circle, fill=black, inner sep=2pt, label=above left:C] (C) at ({cos(150)},{sin(150)}) {}; + % \node[circle, fill=black, inner sep=2pt, label=below left:D] (D) at ({cos(210)},{sin(210)}) {}; + % \end{tikzpicture} + % \end{center} + % \end{minipage} + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item On développe : $f(x) = 0.5(x-2)^2 - 8 = 0.5(x^2 - 4x + 4) - 8 = 0.5x^2 - 2x + 2 - 8 = 0.5x^2 - 2x - 6$ + + Puis on factorise : $0.5x^2 - 2x - 6 = 0.5(x^2 - 4x - 12) = 0.5(x-6)(x+2)$ + + \textbf{Réponse B} + + \item $u_0 = \frac{3}{2}$ + + $u_1 = \frac{2}{3} \times \frac{3}{2} + 0 = 1 + 0 = 1$ + + $u_2 = \frac{2}{3} \times 1 + 1 = \frac{2}{3} + 1 = \frac{5}{3}$ + + $u_3 = \frac{2}{3} \times \frac{5}{3} + 2 = \frac{10}{9} + 2 = \frac{10}{9} + \frac{18}{9} = \frac{28}{9}$ + + \textbf{Réponse A} + + \item $\frac{2x-3}{4} \geq \frac{2x+3}{6}$ + + On multiplie par 12 : $3(2x-3) \geq 2(2x+3)$ + + $6x - 9 \geq 4x + 6$ + + $2x \geq 15$ + + $x \geq \frac{15}{2}$ + + \textbf{Réponse C} + + \item $\frac{17\pi}{4} = \frac{16\pi + \pi}{4} = 4\pi + \frac{\pi}{4}$ + + Or $4\pi = 2 \times 2\pi$ donc la mesure principale est $\frac{\pi}{4}$ + + \textbf{Réponse A} + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Exemplaires}, step={1}, origin={E3C}, topics={}, tags={ polynômes, représentation graphique, suite, trigonométrie }, points={3}] + Lors du lancement d’un hebdomadaire, 1 200 exemplaires ont été vendus. + + Une étude de marché prévoit une progression des ventes de 2\% chaque semaine. + + On modélise le nombre d'hebdomadaires vendus par une suite $(u_n)$ où $u_n$ représente le nombre d'exemplaires vendus durant la $n$-ième semaine après le début de l'opération. + On a donc $u_0 = 1 200$. + \begin{enumerate} + \item Calculer $u_2$ et interpréter le résultat dans le contexte de l'exercice. + \item Écrire un programme Python qui permet de calculer $u_{10}$ (il n'est pas demandé de calculer cette valeur). + \end{enumerate} +\end{exercise} +\begin{solution} + \begin{enumerate} + \item On a $u_0 = 1200$. Chaque semaine, les ventes augmentent de 2\%, donc $u_{n+1} = u_n \times 1.02$ + + $u_1 = 1200 \times 1.02 = 1224$ + + $u_2 = 1224 \times 1.02 = 1248.48$ + + Interprétation : Lors de la deuxième semaine après le lancement, environ 1248 exemplaires seront vendus. + + \item Programme Python : + \begin{verbatim} +u = 1200 +for i in range(10): + u = u * 1.02 +print(u) + \end{verbatim} + + Ou plus directement : + \begin{verbatim} +u_10 = 1200 * (1.02)**10 +print(u_10) + \end{verbatim} + + Résultat : $u_{10} = 1200 \times 1.02^{10} \approx 1462.85$ exemplaires + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={polynômes}, step={1}, origin={ma tête}, topics={}, tags={ polynômes, représentation graphique, suite, trigonométrie }, points={5}] + On considère la fonction polynôme $f$ définie sur $\R$ par + \[ + f(x) = -2x^2 + 4x + 6 + \] + \begin{enumerate} + \item Déterminer les coefficients du polynôme $f$. + % \item Démontrer que $f(x) = -2(x-1)^2 + 8$ + % \item En déduire le tableau de variation de la fonction $f$ ainsi que les coordonnées de son extremum. + \item Démontrer que 3 est une racine de $f$ + \item Démontrer que $f(x) = -2(x-3)(x+1)$ + \item Déterminer le tableau de signe de $f(x)$. + \item Tracer l'allure de la fonction $f(x)$ et placer les éléments remarquables de la fonction sur le graphique. + \end{enumerate} +\end{exercise} +\begin{solution} + \begin{enumerate} + \item $f(x) = -2x^2 + 4x + 6$ est de la forme $ax^2 + bx + c$ avec : + \begin{itemize} + \item $a = -2$ + \item $b = 4$ + \item $c = 6$ + \end{itemize} + + \item $f(3) = -2 \times 3^2 + 4 \times 3 + 6 = -18 + 12 + 6 = 0$ + + Donc 3 est une racine de $f$. + + \item On développe $-2(x-3)(x+1)$ : + + $-2(x-3)(x+1) = -2(x^2 + x - 3x - 3) = -2(x^2 - 2x - 3) = -2x^2 + 4x + 6 = f(x)$ + + Donc $f(x) = -2(x-3)(x+1)$ + + \item Les racines de $f$ sont 3 et $-1$ (puisque $f(x) = -2(x-3)(x+1)$) + + Le coefficient dominant est $a = -2 < 0$, donc la parabole est tournée vers le bas. + + \begin{center} + \begin{tabular}{|c|ccccccc|} + \hline + $x$ & $-\infty$ & & $-1$ & & $3$ & & $+\infty$ \\ + \hline + $f(x)$ & & $-$ & 0 & $+$ & 0 & $-$ & \\ + \hline + \end{tabular} + \end{center} + + \item Sur le graphique, on place : + \begin{itemize} + \item Les racines : $(-1, 0)$ et $(3, 0)$ + \item L'ordonnée à l'origine : $(0, 6)$ + \item Le sommet : $x_s = \frac{-1+3}{2} = 1$, $f(1) = -2 + 4 + 6 = 8$, donc $(1, 8)$ + \item La parabole est tournée vers le bas + \end{itemize} + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Radians}, step={1}, origin={ma tête}, topics={}, tags={ polynômes, représentation graphique, suite, trigonométrie }, points={3}] + \begin{minipage}{0.6\textwidth} + Sur le cercle trigonométrique ci-contre, placer les points suivants et donner la mesure de l'angle associé en degrés. + \begin{enumerate} + \item Le point $A$ est l'image de $\frac{2\pi}{3}$ radians. + \item Le point $B$ est l'image de $-\frac{3\pi}{4}$ radians. + \item Le point $C$ est l'image de $\frac{7\pi}{6}$ radians. + \item Le point $D$ est l'image de $\frac{7\pi}{2}$ radians. + \end{enumerate} + \end{minipage} + \hfill + \begin{minipage}{0.35\textwidth} + \begin{center} + \begin{tikzpicture}[scale=2] + % Cercle trigonométrique + \draw[thick] (0,0) circle (1); + % Axes + \draw[->] (-1.3,0) -- (1.3,0); + \draw[->] (0,-1.3) -- (0,1.3); + % Rayons multiples de π/4 + \foreach \angle in {0, 45, 90, 135, 180, 225, 270, 315} { + \draw[gray, thin] (0,0) -- (\angle:1); + } + % Rayons multiples de π/6 + \foreach \angle in {30, 60, 120, 150, 210, 240, 300, 330} { + \draw[gray, thin] (0,0) -- (\angle:1); + } + \end{tikzpicture} + \end{center} + \end{minipage} +\end{exercise} +\begin{solution} + \begin{enumerate} + \item $\frac{2\pi}{3}$ radians $= \frac{2\pi}{3} \times \frac{180°}{\pi} = \frac{2 \times 180°}{3} = 120°$ + + Le point $A$ se situe dans le deuxième quadrant, à $120°$ de l'axe des abscisses (sens trigonométrique). + + \item $-\frac{3\pi}{4}$ radians $= -\frac{3\pi}{4} \times \frac{180°}{\pi} = -\frac{3 \times 180°}{4} = -135°$ + + Le point $B$ se situe dans le troisième quadrant, à $-135°$ (ou $225°$ en sens positif). + + \item $\frac{7\pi}{6}$ radians $= \frac{7\pi}{6} \times \frac{180°}{\pi} = \frac{7 \times 180°}{6} = 210°$ + + Le point $C$ se situe dans le troisième quadrant, à $210°$ de l'axe des abscisses. + + \item $\frac{7\pi}{2}$ radians $= \frac{7\pi}{2} \times \frac{180°}{\pi} = \frac{7 \times 180°}{2} = 630°$ + + Or $630° = 360° + 270° = 270°$ (mesure principale) + + Le point $D$ se situe sur l'axe des ordonnées négatif, à $270°$ (ou $-90°$). + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Le virus!}, step={1}, points=5] + On s'intéresse à la propagation d'une maladie dans une ville de \np{150000} habitants. La fonction $f$ définie sur l'intervalle $\intFF{0}{40}$ par + \begin{align*} + f(x) &= -30 x^2 + 1650 x - 13500 + \end{align*} + modélise le nombre de personnes touchées par la maladie au bout de $x$ jours de suivi de la propagation. + \begin{enumerate} + \item \textit{On donne en annexe la courbe représentative de la fonction $f$. Répondre aux questions ci-dessous par lecture graphique. Les résultats seront justifiés en commentant le travail réalisé sur le graphique et en y laissant les traits de construction.} + \begin{enumerate} + \item Déterminer le nombre de personnes touchées par la maladie au bout de 15 jours de suivi de la propagation. + \item Le conseil municipal a décidé de fermer les crèches de la ville lorsque plus de 10\% de la population est touchée par la maladie. Pendant combien de jours les crèches ont-elles été fermées? + \end{enumerate} + + \item \textit{Les questions suivantes ne devront pas être justifiées par lecture graphique.} + \begin{enumerate} + \item Démontrer que $f(x) = -30(x-27.5)^2 + 9187.5$ + \item En déduire le tableau de variations de la fonction $f$. + \item Au bout de combien de jours, l'épidémie a-t-elle été à son maximum? Combien de personnes étaient alors touchées? + \end{enumerate} + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Par lecture graphique : + \begin{enumerate} + \item À $x = 15$ jours, on lit $f(15) \approx 7500$ personnes touchées. + + \item 10\% de 150 000 habitants = 15 000 personnes. + + On cherche quand $f(x) > 15000$, mais sur le graphique on voit que $f(x)$ ne dépasse jamais 12 500, donc le maximum est inférieur à 10\%. + + Les crèches n'ont jamais été fermées. + \end{enumerate} + + \item + \begin{enumerate} + \item On développe $-30(x-27.5)^2 + 9187.5$ : + + $-30(x-27.5)^2 + 9187.5 = -30(x^2 - 55x + 756.25) + 9187.5$ + + $= -30x^2 + 1650x - 22687.5 + 9187.5$ + + $= -30x^2 + 1650x - 13500 = f(x)$ + + \item La fonction $f$ est sous forme canonique $a(x-\alpha)^2 + \beta$ avec $a = -30 < 0$, $\alpha = 27.5$ et $\beta = 9187.5$. + + La parabole est tournée vers le bas, donc $f$ est croissante sur $[0; 27.5]$ et décroissante sur $[27.5; 40]$. + + \begin{center} + \begin{tabular}{|c|ccccc|} + \hline + $x$ & 0 & & 27.5 & & 40 \\ + \hline + & & & $9187.5$ & & \\ + $f(x)$ & & $\nearrow$ & & $\searrow$ & \\ + & $-13500$ & & & & $-10500$ \\ + \hline + \end{tabular} + \end{center} + + \item L'épidémie atteint son maximum au sommet de la parabole, soit au bout de $27.5$ jours. + + À ce moment-là, $f(27.5) = 9187.5 \approx 9188$ personnes sont touchées. + \end{enumerate} + \end{enumerate} +\end{solution} + +\begin{annexe} + \begin{center} + \begin{tikzpicture} + \begin{axis}[ + width=14cm, + height=7cm, + xmin=0, xmax=50, + ymin=0, ymax=12500, + xtick distance=5, + ytick distance=2500, + minor tick num=4, + grid=both, + minor grid style={gray!30}, + xlabel={\textit{Nombre de jours}}, + ylabel={\textit{Nombre de personnes touchées}}, + xlabel style={below}, + ylabel style={above}, + scaled y ticks=false, + yticklabel style={/pgf/number format/fixed, /pgf/number format/1000 sep={\,}}, + thick + ] + \addplot[blue, very thick, domain=0:50, samples=100] {-30*x^2 + 1650*x - 13500}; + \end{axis} + \end{tikzpicture} + + \end{center} +\end{annexe} diff --git a/1G_math/Evaluations/DS_2025-10-08/solution.pdf b/1G_math/Evaluations/DS_2025-10-08/solution.pdf new file mode 100644 index 0000000000000000000000000000000000000000..c044f5faac86741e93ead8a8fb86ff4c800e6d89 GIT binary patch literal 31142 zcmce81F$X2w&k{Md!LQ7ZQHhO+qU=Fwr$(CZQC|Z-}`_6ez)V_`2C_I-dho?){3mE z%$zYtWzLynWRc1ViBQwiutJeeF7+)Atpx857W|$r@M~ z8#)otDmoiD{YQz2t%EfI?Vrd0xPqR5RzSekjX?8{Tj>bs>HhPpO+YKH@8n?qXF2_U zp81dCALR0KZI@GrnyC}SI=|9$3|$A}hEpv|4* zd$RQJB2C*ckj=27NfUgI@4=sMwh~V^?kI7g3mwvv({3i=k@6-HZhySJ;fLqtK7Cij zjKBh~qEa8g2b|r%1>dpWdHaxspV7tR>G9+Rpuq2J9!S~2cLVs%2pk_^296(iA2yIu zGe8@o5x9gUZW$=s#i=u7ebHSI@MHpqIbiodjte6|sz!1D%y>ClAUXZU#){ z=JpO!_=eqe=j(R=Sz>H0muZ{&njV5N@O!~|Dv&8I%BTM9=RFw@X8hA81MErJ{TURf zua9fp7(F{E58Sp!@Mr71p@%n+W+gQWWKj8X3!P-$wzHS10ljt|W690J?+v19Y1)AB z*OoKKpV6C`rZWXXpl#cS@j0uDHA96*8O+GY%hFLB@?pVo-t)GM56HM)a6|yQZaInX 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