Feat: devoir pour les sti2d
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TST_sti2d/DS/DS_21_02_25/DOC-sujet.pdf
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TST_sti2d/DS/DS_21_02_25/DOC-sujet.pdf
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TST_sti2d/DS/DS_21_02_25/DS_21_02_25.tex
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TST_sti2d/DS/DS_21_02_25/DS_21_02_25.tex
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\documentclass[a4paper, 12pt]{article}
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\usepackage[francais,bloc,completemulti]{automultiplechoice}
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\usepackage{etex}
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\usepackage{tkz-fct}
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\geometry{left=10mm,right=10mm, top=25mm}
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\begin{document}
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\baremeDefautS{b=1,m=0}
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\element{expComplexe}{
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\begin{question}{Algébrique vers exponentielle}
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La forme exponentielle du nombre $z = \sqrt{3} - i$ est
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\begin{reponseshoriz}
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\bonne{$2e^{-\frac{\pi}{6}i}$}
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\mauvaise{$e^{-\frac{\pi}{3}i}$}
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\mauvaise{$3e^{\frac{\pi}{6}i}$}
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\mauvaise{$2e^{\frac{\pi}{4}i}$}
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\end{reponseshoriz}
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\end{question}
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}
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\element{expComplexe}{
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\begin{question}{Exponentielle vers algébrique}
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La forme algébrique du nombre $z = 2e^{\frac{\pi}{3}i}$ est
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\begin{reponseshoriz}
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\bonne{$1 + \sqrt{3}i$}
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\mauvaise{$\sqrt{3} + i$}
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\mauvaise{$\sqrt{2} + \sqrt{2}i$}
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\mauvaise{$\frac{1}{2} - \frac{\sqrt{3}}{2} i$}
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\end{reponseshoriz}
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\end{question}
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}
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\element{expComplexe}{
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\begin{question}{Multiplication complexes}
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Soit $z_A = 2e^{\frac{\pi}{2}i}$ et $z_B = 4e^{\pi i}$. Alors $z_A \times z_B$ vaut
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\begin{reponseshoriz}
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\bonne{$8e^{i \frac{3\pi}{2}}$}
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\mauvaise{$8e^{i\pi}$}
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\mauvaise{$2e^{-i^2}$}
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\mauvaise{impossible}
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\end{reponseshoriz}
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\end{question}
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}
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\element{expComplexe}{
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\begin{question}{Quotient complexes}
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Soit $z_A = 3e^{\frac{\pi}{6}i}$ et $z_B = e^{\frac{\pi}{2}i}$. Alors $\frac{z_A}{z_B}$ vaut
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\begin{reponseshoriz}
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\bonne{$3e^{-i\frac{\pi}{3}}$}
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\mauvaise{$3e^{i\frac{\pi}{3}}$}
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\mauvaise{$3e^{-i\times0}$}
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\end{reponseshoriz}
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\end{question}
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}
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\setgroupmode{expComplexe}{withreplacement}
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\element{exponentielle}{
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\begin{question}{Dérivation}
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Soit $f(x) = (4x - 2)e^{5x}$ alors sa dérivée est
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\begin{reponseshoriz}
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\bonne{$f'(x) = e^{5x}(20x-6)$}
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\mauvaise{$f'(x) = (4x+2)e^{5x}$}
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\mauvaise{$f'(x) = 20e^{5x}$}
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\mauvaise{$f'(x) = 4 + 5e^{5x}$}
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\end{reponseshoriz}
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\end{question}
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}
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\element{exponentielle}{
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\begin{question}{Primitive}
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Soit $g(x) = 24e^{-6x}$ alors sa primitive est
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\begin{reponseshoriz}
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\bonne{$F(x) = -4e^{-6x}$}
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\mauvaise{$F(x) = -144e^{-6x}$}
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\mauvaise{$F(x) = \frac{1}{-6}e^{-6x}$}
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\mauvaise{$F(x) = 24 - 6e^{-6x}$}
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\end{reponseshoriz}
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\end{question}
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}
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\element{exponentielle}{
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\begin{question}{Calculer intégrale}
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La valeur exacte de $\displaystyle \int_0^5 e^{2x} \; dx$ vaut
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\begin{reponseshoriz}
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\bonne{$0.5(e^{10} - 1)$}
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\mauvaise{$0.5e^{5} - 0.5$}
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\mauvaise{$2(e^{10} - 1)$}
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\mauvaise{$2e^{10}$}
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\end{reponseshoriz}
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\end{question}
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}
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\element{exponentielle}{
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\begin{question}{Vérifier une primitive}
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Soit $f(x) = (3x^2 + 2x + 3)e^{3x}$. Alors une primitive de $f(x)$ est
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\begin{reponseshoriz}
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\bonne{$F(x) = (x^2 + 1)e^{3x} + 100$}
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\mauvaise{$F(x) = (9x + 6)e^{3x}$}
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\mauvaise{$F(x) = (x^2+2)e^{3x}$}
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\end{reponseshoriz}
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\end{question}
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}
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\setgroupmode{exponentielle}{withreplacement}
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\exemplaire{2}{
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\noindent{\bf QCM \hfill DS 6}
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\begin{minipage}{.4\linewidth}
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\centering\Large\bf DS 6 - Tsti2d \\ 25/02/2021
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%\normalsize Durée : 10 minutes.
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\end{minipage}
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\begin{minipage}{.6\linewidth}
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\champnom{%
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\fbox{
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\begin{minipage}{0.8\linewidth}
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Nom, prénom, classe:
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\vspace*{.5cm}\dotfill
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\vspace*{1mm}
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\end{minipage}
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}
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}
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\AMCcodeGridInt[h]{etu}{2}
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\end{minipage}
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\begin{center}\em
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Aucun document n'est autorisé.
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L'usage de la calculatrice est interdit.
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\end{center}
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%%% fin de l'en-tête
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\restituegroupe[4]{expComplexe}
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\restituegroupe[4]{exponentielle}
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%\AMCaddpagesto{2}
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}
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\end{document}
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