229 lines
10 KiB
TeX
229 lines
10 KiB
TeX
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\documentclass[a4paper,12pt]{article}
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\usepackage{myXsim}
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\usepackage{pgfplots}
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\usetikzlibrary{decorations.markings}
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\pgfplotsset{compat=1.18}
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\title{ DM1 \hfill \Var{ subject.Nom }}
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\tribe{2nd}
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\date{A rendre pour le 2 décembre 2022}
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\duree{}
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\xsimsetup{
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solution/print = false
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}
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\pagestyle{empty}
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\begin{document}
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\maketitle
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Le barème est donné à titre indicatif, il pourra être modifié.
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\begin{exercise}[subtitle={Calculs avec des fractions}, points=5]
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Détailler les calculs suivants et donner le résultat sous la forme d'une fraction irréductible.
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\Block{
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set fractions = {
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"A": random_expression("{a} / {b} + {c} / {d}", ["a!=b", "c!=d", "b > 1", "d > 1"], global_config={"min_max": (0, 10)}),
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"B": random_expression("{a} / {b} + {c}", ["a!=b", "b > 1"], global_config={"min_max": (1, 10)}),
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"C": random_expression("{a} / {b} * {c}", ["a!=b", "b > 1"], global_config={"min_max": (1, 10)}),
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"D": random_expression("{a} / {b} * {c} / {b}", ["a!=b", "c!=b", "b > 1"], global_config={"min_max": (1, 10)}),
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"E": random_expression("{a} / {b} / ({c} / {b})", ["a!=b", "c!=b", "b > 1"], global_config={"min_max": (1, 10)}),
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}
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}
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\begin{multicols}{3}
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\begin{enumerate}
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%- for (l, e) in fractions.items()
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\item $\Var{l} = \Var{e}$
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%- endfor
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\end{enumerate}
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\end{multicols}
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\end{exercise}
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\begin{solution}
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\begin{multicols}{3}
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\begin{enumerate}
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%- for (l, e) in fractions.items()
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\item
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\begin{align*}
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\Var{l} & = \Var{e.simplify().explain() | join(' \\\\ & = ')} \\
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& = \Var{e.simplify().simplified}
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\end{align*}
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%- endfor
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\end{enumerate}
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\end{multicols}
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\end{solution}
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\begin{exercise}[subtitle={Inéquation et tableaux}, points=3]
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Tracer le tableau de signe de la fonction suivante en le démontrant à l'aide de la résolution d'une inéquation.
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%- set f = random_expression("{a}x + {b}", global_config={"min_max":(-20, 20), "rejected":[0, 1]})
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$$f(x) = \Var{f}$$
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\end{exercise}
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\begin{solution}
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Pour déterminer les valeurs de $x$ pour lesquelles $f(x)$ est positive, il faut résoudre l'inéquation
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%- set cst = -f[0]
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%- set coef = f[1]
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%- set racine = cst / coef
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\begin{align*}
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f(x) & \geq 0 \\
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\Var{f} & \geq 0 \\
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\Var{f + cst} &\geq \Var{0 + cst} \\
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%- if coef > 0
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\frac{\Var{f + cst}}{\Var{coef}} &\geq \frac{\Var{cst}}{\Var{coef}} \\
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x &\geq \Var{racine.simplify()} \\
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\end{align*}
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Donc $f(x)$ est positif quand $x$ est supérieur à $\Var{racine}$. On en déduit le tableau de signe
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%- else
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\frac{\Var{f + cst}}{\Var{coef}} &\leq \frac{\Var{cst}}{\Var{coef}} \\
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x &\leq \Var{racine.simplify()} \\
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\end{align*}
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Donc $f(x)$ est positif quand $x$ est inférieur à $\Var{racine}$. On en déduit le tableau de signe
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%- endif
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\begin{center}
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\begin{tikzpicture}
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\tkzTabInit[lgt=2,espcl=1]{$ t $/1,$ f(t) $/1}{, $\Var{racine}$ ,}
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%- if coef > 0
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\tkzTabLine{, -, z, +, }
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%- else
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\tkzTabLine{, +, z, -, }
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%- endif
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\end{tikzpicture}
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\end{center}
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\end{solution}
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\begin{exercise}[subtitle={Vecteurs}, points=3]
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\begin{multicols}{2}
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\begin{enumerate}
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\item Tracer les vecteurs $\vect{z} = \vect{u} + \vect{v}$ et $\vect{y} = 2\vect{u} - \vect{v}$ (le vecteur peut sortir du cadre)
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%- set xa1, ya1 = random_list(["x", "y"], global_config={"min_max": (-5, 5), "rejected":[-2, -1, 0, 1, 2]})
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%- set xa2, ya2 = -xa1, ya1
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%- set aminx = min(0, xa1, xa2, xa1+xa2, 2*xa1-xa2)
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%- set amaxx = max(0, xa1, xa2, xa1+xa2, 2*xa1-xa2)
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%- set aminy = min(0, ya1, ya2, ya1+ya2, 2*ya1-ya2)
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%- set amaxy = max(0, ya1, ya2, ya1+ya2, 2*ya1-ya2)
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\begin{center}
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\begin{tikzpicture}[scale=0.4]
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\draw (\Var{aminx-1}, \Var{aminy-1}) rectangle (\Var{amaxx+1}, \Var{amaxy+1});
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\draw[very thick, ->] (0, 0) -- node [midway, sloped, above] {$\vect{u}$} (\Var{xa1}, \Var{ya1});
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\draw[very thick, -> ] (0, 0) -- node [midway, sloped, above] {$\vect{v}$} (\Var{xa2}, \Var{ya2});
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\end{tikzpicture}
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\end{center}
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\item Tracer la force résultat de la somme des 3 forces exercées sur le point $0$ représenté ci-dessous.
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%- set x1, y1 = random_list(["x", "y"], global_config={"min_max": (-5, 5), "rejected":[-2, -1, 0, 1, 2]})
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%- set x2, y2 = -x1, y1
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%- set x3, y3 = x2, 0
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%- set minx = min(0, x1, x2, x3, x1+x2, x2+x3, x1+x3, x1+x2+x3 )
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%- set maxx = max(0, x1, x2, x3, x1+x2, x2+x3, x1+x3, x1+x2+x3 )
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%- set miny = min(0, y1, y2, y3, y1+y2, y2+y3, y1+y3, y1+y2+y3 )
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%- set maxy = max(0, y1, y2, y3, y1+y2, y2+y3, y1+y3, y1+y2+y3 )
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\begin{center}
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\begin{tikzpicture}[scale=0.4]
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\draw (\Var{minx-1}, \Var{miny-1}) rectangle (\Var{maxx+1}, \Var{maxy+1});
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\draw[very thick, ->] (0, 0) -- node [midway, sloped, above] {$\vect{F_1}$} (\Var{x1}, \Var{y1});
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\draw[very thick, -> ] (0, 0) -- node [midway, sloped, above] {$\vect{F_2}$} (\Var{x2}, \Var{y2});
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\draw[very thick, -> ] (0, 0) -- node [midway, sloped, above] {$\vect{F_3}$} (\Var{x3}, \Var{y3});
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\end{tikzpicture}
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\end{center}
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\end{enumerate}
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\end{multicols}
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\end{exercise}
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\begin{solution}
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\begin{multicols}{2}
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\begin{enumerate}
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\item
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\begin{center}
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\begin{tikzpicture}[scale=0.4]
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\draw (\Var{aminx-1}, \Var{aminy-1}) rectangle (\Var{amaxx+1}, \Var{amaxy+1});
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\draw[very thick, ->] (0, 0) -- node [midway, sloped, above] {$\vect{u}$} (\Var{xa1}, \Var{ya1});
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\draw[very thick, -> ] (0, 0) -- node [midway, sloped, above] {$\vect{v}$} (\Var{xa2}, \Var{ya2});
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\draw[very thick, ->, color=blue] (0, 0)
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-- ++ (\Var{xa1}, \Var{ya1}) node [midway, sloped, above] {$\vect{u}$}
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-- ++ (\Var{xa2}, \Var{ya2}) node [midway, sloped, above] {$\vect{v}$} ;
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\draw[very thick, ->, color=blue] (0, 0) -- node [midway, sloped, left] {$\vect{z}$} (\Var{xa2+xa1}, \Var{ya2+ya1});
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\draw[very thick, ->, color=green] (0, 0)
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-- ++ (\Var{xa1}, \Var{ya1}) node [midway, sloped, above] {$\vect{u}$}
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-- ++ (\Var{xa1}, \Var{ya1}) node [midway, sloped, above] {$\vect{u}$}
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-- ++ (-\Var{xa2}, -\Var{ya2}) node [midway, sloped, above] {$-\vect{v}$} ;
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\draw[very thick, ->, color=green] (0, 0) -- node [midway, sloped, above] {$\vect{y}$} (\Var{2*xa1 - xa2}, \Var{2*ya1 - ya2});
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\end{tikzpicture}
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\end{center}
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\item
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\begin{center}
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\begin{tikzpicture}[scale=0.4]
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\draw (\Var{minx-1}, \Var{miny-1}) rectangle (\Var{maxx+1}, \Var{maxy+1});
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\draw[very thick, ->] (0, 0) -- node [midway, sloped, above] {$\vect{F_1}$} (\Var{x1}, \Var{y1});
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\draw[very thick, -> ] (0, 0) -- node [midway, sloped, above] {$\vect{F_2}$} (\Var{x2}, \Var{y2});
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\draw[very thick, -> ] (0, 0) -- node [midway, sloped, above] {$\vect{F_3}$} (\Var{x3}, \Var{y3});
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\draw[very thick, ->, color=blue] (0, 0)
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--++ (\Var{x1}, \Var{y1}) node [midway, sloped, above] {$\vect{F_1}$}
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--++ (\Var{x2}, \Var{y2}) node [midway, sloped, above] {$\vect{F_2}$}
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--++ (\Var{x3}, \Var{y3}) node [midway, sloped, above] {$\vect{F_3}$}
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;
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\draw[very thick, ->, color=blue] (0, 0) -- (\Var{x1+x2+x3}, \Var{y1+y2+y3}) node [midway, sloped, above] {$\vect{F_1}+\vect{F_2}+\vect{F_3}$};
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\end{tikzpicture}
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\end{center}
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\end{enumerate}
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\end{multicols}
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\end{solution}
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\begin{exercise}[subtitle={Statistiques}, points=5]
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%- set center = random.randint(30, 50)
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%- set qty = random.randint(20, 40)
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%- set dataset = stat.Dataset.random(qty, rd_args=(center, 1.5), nbr_format=int)
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Ci-dessous la taille des poissons péchés lors du dernier challenge PêcheParty.
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\begin{center}
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\Var{dataset.tabular_latex(ceil(qty/15))}
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\end{center}
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\begin{enumerate}
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\item Décrire la série statistique et donner l'effectif total.
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\item Calculer la moyenne, les quartiles, l'écart interquartile et la médiane de cette série statistique.
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\item Quelle est la valeur de l'écart-type de cette série statistique?
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\end{enumerate}
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\end{exercise}
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\begin{solution}
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Dans cette correction les étapes de construction des indicateurs ne sont pas détaillés.
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Tableau des effectifs
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%- set wdataset = stat.WeightedDataset(dataset)
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\begin{center}
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\Var{wdataset.tabular_latex()}
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\end{center}
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La population sont les poissons péchés lors du dernier challenge PêcheParty. Les individus sont les poissons. Le caractère est la taille des poissons.
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\begin{multicols}{2}
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\begin{itemize}
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\item Effectif total : $\Var{dataset.effectif_total()}$
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\item Premier quartile $ Q_1 = \Var{dataset.quartile(1)}$ (position $\Var{dataset.posi_quartile(1)}$)
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\item Médiane $ Me = \Var{dataset.quartile(2)}$ (position $\Var{dataset.posi_quartile(2)}$)
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\item Troisième quartile $ Q_3 = \Var{dataset.quartile(3)}$ (position $\Var{dataset.posi_quartile(3)}$)
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\item interquartile: $Q_3 - Q_1 = \Var{dataset.quartile(3)} - \Var{dataset.quartile(1)} = \Var{dataset.quartile(3) - dataset.quartile(1) }$
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\item Moyenne: $\overline{x} = \Var{dataset.mean()}$
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\item Écart-type: $\sigma = \Var{dataset.sd()}$
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\end{itemize}
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\end{multicols}
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\end{solution}
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\end{document}
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%%% Local Variables:
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%%% mode: latex
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%%% TeX-master: "master"
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%%% End:
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