Jean-Christophe Jameux
4 years ago
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#LyX 2.3 created this file. For more info see http://www.lyx.org/ |
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|
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\begin_layout Title |
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Correction du DST 1 |
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\end_layout |
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|
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\begin_layout Section* |
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\paragraph_spacing onehalf |
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Exercice 1 |
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\end_layout |
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|
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\begin_layout Standard |
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\paragraph_spacing onehalf |
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Certains ont été perturbés, à juste titre, par le fait que la suite commence |
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à |
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\begin_inset Formula $u_{0}$ |
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\end_inset |
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|
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alors que l'énoncé prétend qu'elle commence au rang |
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\begin_inset Formula $1$ |
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\end_inset |
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|
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... |
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Vous trouverez donc ci-dessous la correction pour les deux interprétations |
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de l'énoncé ! |
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\end_layout |
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|
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\paragraph_spacing onehalf |
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Soit la suite |
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\begin_inset Formula $u$ |
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\end_inset |
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|
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telle que |
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\begin_inset Formula $u_{0}=2$ |
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\end_inset |
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|
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et |
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\begin_inset Formula $u_{n+1}=5u_{n}+4$ |
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\end_inset |
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|
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pour tout entier |
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\begin_inset Formula $n\geqslant0$ |
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\end_inset |
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|
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, et soit |
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\begin_inset Formula $v$ |
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\end_inset |
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|
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la suite définie par |
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\begin_inset Formula $v_{n}=u_{n}+1$ |
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\end_inset |
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|
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pour tout entier |
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\begin_inset Formula $n\geqslant0$ |
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\end_inset |
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|
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. |
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\end_layout |
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|
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\begin_deeper |
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\begin_layout Standard |
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\paragraph_spacing onehalf |
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Montrer que la suite |
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\begin_inset Formula $v$ |
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\end_inset |
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|
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est une suite géométrique dont on donnera la raison et le premier terme |
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puis donner une formule pour |
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\begin_inset Formula $v_{n}$ |
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\end_inset |
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|
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puis pour |
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\begin_inset Formula $u_{n}$ |
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\end_inset |
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|
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en fonction de l'entier naturel |
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\begin_inset Formula $n$ |
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\end_inset |
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|
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. |
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\end_layout |
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|
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\begin_layout Itemize |
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\paragraph_spacing onehalf |
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On a, pour tout entier naturel |
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\begin_inset Formula $n$ |
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\end_inset |
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|
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: |
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\begin_inset Formula |
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\[ |
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\begin{array}{rl} |
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v_{n+1} & =u_{n+1}+1\\ |
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& =5u_{n}+4+1\\ |
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& =5u_{n}+5\\ |
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& =5\left(u_{n}+1\right)\\ |
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& 5v_{n} |
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\end{array} |
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\] |
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|
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\end_inset |
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|
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Ce qui montre que la suite |
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\begin_inset Formula $v$ |
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\end_inset |
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|
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est géométrique de raison |
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\begin_inset Formula $5$ |
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\end_inset |
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|
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. |
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Son premier terme étant |
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\begin_inset Formula $v_{0}=u_{0}+1=2+1=3$ |
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\end_inset |
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|
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, on a que, pour tout entier |
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\begin_inset Formula $n\in\mathbb{N}$ |
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\end_inset |
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|
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: |
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\begin_inset Formula |
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\[ |
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v_{n}=3\times5^{n} |
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\] |
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|
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\end_inset |
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|
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On en déduit que, pour tout entier |
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\begin_inset Formula $n\in\mathbb{N}$ |
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\end_inset |
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|
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: |
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\begin_inset Formula |
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\[ |
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u_{n}=v_{n}-1=3\times5^{n}-1 |
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\] |
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|
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\end_inset |
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|
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|
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\end_layout |
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|
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\begin_layout Itemize |
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\paragraph_spacing onehalf |
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Soit la suite |
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\begin_inset Formula $u$ |
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\end_inset |
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|
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telle que |
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\begin_inset Formula $u_{1}=2$ |
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\end_inset |
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|
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et |
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\begin_inset Formula $u_{n+1}=5u_{n}+4$ |
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\end_inset |
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|
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pour tout entier |
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\begin_inset Formula $n\geqslant1$ |
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\end_inset |
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|
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, et soit |
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\begin_inset Formula $v$ |
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\end_inset |
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|
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la suite définie par |
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\begin_inset Formula $v_{n}=u_{n}+1$ |
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\end_inset |
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|
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pour tout entier |
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\begin_inset Formula $n\geqslant1$ |
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\end_inset |
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|
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. |
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\end_layout |
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|
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\begin_deeper |
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\begin_layout Standard |
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\paragraph_spacing onehalf |
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Montrer que la suite |
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\begin_inset Formula $v$ |
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\end_inset |
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|
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est une suite géométrique dont on donnera la raison et le premier terme |
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puis donner une formule pour |
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\begin_inset Formula $v_{n}$ |
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\end_inset |
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|
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puis pour |
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\begin_inset Formula $u_{n}$ |
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\end_inset |
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|
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en fonction de l'entier naturel non nul |
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\begin_inset Formula $n$ |
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\end_inset |
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|
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. |
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\end_layout |
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|
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\begin_layout Itemize |
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\paragraph_spacing onehalf |
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On a, pour tout entier naturel non nul |
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\begin_inset Formula $n$ |
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\end_inset |
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|
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: |
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\begin_inset Formula |
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\[ |
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\begin{array}{rl} |
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v_{n+1} & =u_{n+1}+1\\ |
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& =5u_{n}+4+1\\ |
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& =5u_{n}+5\\ |
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& =5\left(u_{n}+1\right)\\ |
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& 5v_{n} |
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\end{array} |
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\] |
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|
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\end_inset |
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|
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Ce qui montre que la suite |
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\begin_inset Formula $v$ |
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\end_inset |
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|
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est géométrique de raison |
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\begin_inset Formula $5$ |
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\end_inset |
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|
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. |
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Son premier terme étant |
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\begin_inset Formula $v_{1}=u_{1}+1=2+1=3$ |
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\end_inset |
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|
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, on a que, pour tout entier |
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\begin_inset Formula $n\in\mathbb{N}$ |
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\end_inset |
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|
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: |
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\begin_inset Formula |
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\[ |
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v_{n}=3\times5^{n-1} |
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\] |
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|
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\end_inset |
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|
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On en déduit que, pour tout entier |
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\begin_inset Formula $n\in\mathbb{N}$ |
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\end_inset |
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|
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: |
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\begin_inset Formula |
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\[ |
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u_{n}=v_{n}-1=3\times5^{n-1}-1 |
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\] |
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|
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\end_inset |
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|
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|
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\end_layout |
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|
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\end_deeper |
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\end_deeper |
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\begin_layout Standard |
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\paragraph_spacing onehalf |
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|
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\end_layout |
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|
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\begin_layout Section* |
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\paragraph_spacing onehalf |
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Exercice 2 |
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\end_layout |
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|
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\begin_layout Standard |
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\paragraph_spacing onehalf |
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Établir, en faisant apparaitre les calculs, le tableau de signe sur |
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\begin_inset Formula $\mathbb{R}$ |
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\end_inset |
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|
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de la fonction |
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\begin_inset Formula $f:x\mapsto3x^{2}+4x-7$ |
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\end_inset |
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|
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et en déduire les solutions sur |
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\begin_inset Formula $\mathbb{R}$ |
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\end_inset |
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|
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de l'inéquation |
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\begin_inset Formula $3x^{2}+4x+8\leqslant15$ |
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\end_inset |
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|
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. |
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\end_layout |
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|
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\begin_layout Standard |
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\paragraph_spacing onehalf |
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(Si vous ne parvenez pas à le faire par le calcul, une résolution graphique |
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du problème rapportera une partie des points) |
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\end_layout |
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|
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\begin_layout Itemize |
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\paragraph_spacing double |
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\begin_inset Formula $f$ |
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\end_inset |
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|
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est une fonction polynôme du second degré, son déterminant est |
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\begin_inset Formula $\Delta=4^{2}-4\times3\times\left(-7\right)=16+84=100>0$ |
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\end_inset |
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|
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, elle admet donc deux racines : |
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\begin_inset Formula |
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\[ |
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x_{1}=\frac{-4-\sqrt{100}}{2\times3}=-\frac{7}{3}\qquad\text{et}\qquad x_{2}=\frac{-4+\sqrt{100}}{2\times3}=1 |
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\] |
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|
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\end_inset |
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|
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Son coefficient dominant étant |
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\begin_inset Formula $a=3>0$ |
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\end_inset |
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|
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, on en déduit le tableau de signe suivant : |
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\end_layout |
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|
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\begin_deeper |
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\begin_layout Standard |
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\paragraph_spacing onehalf |
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\begin_inset ERT |
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status open |
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|
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\begin_layout Plain Layout |
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|
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|
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\backslash |
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begin{tikzpicture} |
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\end_layout |
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|
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\begin_layout Plain Layout |
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|
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|
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\backslash |
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tkzTabInit [ lgt = 3 ] |
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\end_layout |
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|
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\begin_layout Plain Layout |
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|
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{$x$ / 1 , $3x^{2}+4x-7$ / 1} |
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\end_layout |
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|
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\begin_layout Plain Layout |
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|
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{$- |
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\backslash |
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infty$, $- |
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\backslash |
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frac{7}{3}$, $1$, $+ |
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\backslash |
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infty$} |
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|
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|
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tkzTabLine |
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{, +, z, -, z, +, } |
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|
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\begin_layout Plain Layout |
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|
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\backslash |
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end{tikzpicture} |
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\end_layout |
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|
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\end_inset |
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|
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|
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\end_layout |
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|
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\begin_layout Standard |
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\paragraph_spacing onehalf |
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Ainsi, on a, pour tout réel |
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\begin_inset Formula $x$ |
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\end_inset |
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|
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, les équivalences suivantes : |
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\begin_inset Formula |
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\[ |
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\begin{array}{rl} |
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3x^{2}+4x+8\leqslant15 & \Longleftrightarrow3x^{2}+4x+8-15\leqslant0\\ |
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& \Longleftrightarrow3x^{2}+4x-7\leqslant0\\ |
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& \Longleftrightarrow3x^{2}+4x-7\leqslant0\\ |
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& \Longleftrightarrow-\frac{7}{3}\leqslant x\leqslant1 |
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\end{array} |
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\] |
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|
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\end_inset |
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|
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Les solutions réelles de l'inéquation |
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\begin_inset Formula $3x^{2}+4x+8\leqslant15$ |
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\end_inset |
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|
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sont donc les réels compris, au sens large, entre |
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\begin_inset Formula $-\frac{7}{3}$ |
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\end_inset |
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|
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et |
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\begin_inset Formula $1$ |
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\end_inset |
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|
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, autrement dit, les réels appartenant à l'intervalle |
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\begin_inset Formula $\left[-\frac{7}{3};1\right]$ |
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\end_inset |
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|
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. |
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\end_layout |
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|
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\end_deeper |
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\end_body |
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\end_document |
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Binary file not shown.
@ -1,354 +0,0 @@ |
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#LyX 2.3 created this file. For more info see http://www.lyx.org/ |
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thickness "0.4pt" |
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separation "3pt" |
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shadowsize "4pt" |
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framecolor "black" |
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status open |
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|
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|
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|
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\begin_layout Plain Layout |
||||
Soit la suite |
||||
\begin_inset Formula $u$ |
||||
\end_inset |
||||
|
||||
telle que |
||||
\begin_inset Formula $u_{0}=2$ |
||||
\end_inset |
||||
|
||||
et |
||||
\begin_inset Formula $u_{n+1}=5u_{n}+4$ |
||||
\end_inset |
||||
|
||||
pour tout entier |
||||
\begin_inset Formula $n\geqslant1$ |
||||
\end_inset |
||||
|
||||
, et soit |
||||
\begin_inset Formula $v$ |
||||
\end_inset |
||||
|
||||
la suite définie par |
||||
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\end_inset |
||||
|
||||
pour tout entier |
||||
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||||
\end_inset |
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||||
. |
||||
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||||
|
||||
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||||
Montrer que la suite |
||||
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||||
\end_inset |
||||
|
||||
est une suite géométrique dont on donnera la raison et le premier terme |
||||
puis donner une formule pour |
||||
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||||
\end_inset |
||||
|
||||
puis pour |
||||
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||||
\end_inset |
||||
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||||
en fonction de l'entier naturel |
||||
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||||
\end_inset |
||||
|
||||
. |
||||
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vspace{8mm} |
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\backslash |
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hrule |
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\backslash |
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vspace{8mm} |
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\backslash |
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hrule |
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vspace{8mm} |
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Établir, en faisant apparaitre les calculs, le tableau de signe sur |
||||
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||||
\end_inset |
||||
|
||||
de la fonction |
||||
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||||
\end_inset |
||||
|
||||
et en déduire les solutions sur |
||||
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||||
\end_inset |
||||
|
||||
de l'inéquation |
||||
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||||
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||||
|
||||
. |
||||
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||||
|
||||
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||||
(Si vous ne parvenez pas à le faire par le calcul, une résolution graphique |
||||
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||||
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# TSP - DST 1 - Questions originales |
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|
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a : -2,-3,-4,-5 |
||||
b : 2,4,6,8 |
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c : 1,3,5,7,9 |
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|
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\item L'ensemble des solutions de l'inéquation $#a# x+#b#<#c#$ est : |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse $\mathbb{R}$ |
||||
\item\BonneReponse $]#(c-b)/a#,+\infty[$ |
||||
\item\MauvaiseReponse $]-\infty, #(b-c)/a#]$ |
||||
\end{enumerate} |
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|
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|
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a : -2,-3,-4,-5 |
||||
b : 2,4,6,8 |
||||
c : 1,3,5,7,9 |
||||
|
||||
\item L'ensemble des solutions de l'inéquation $#a#x+#b# \geqslant #c#$ |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse $\mathbb{R}$ |
||||
\item\MauvaiseReponse $[#(c-b)/a#,+\infty[$ |
||||
\item\BonneReponse $]-\infty,#(c-b)/a#]$ |
||||
\end{enumerate} |
||||
|
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|
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a : 2,3,4 |
||||
b : 2,4,6,8 |
||||
c : 1,3,5,7,9 |
||||
|
||||
\item Le nombre $#1/b#$ |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse est solution de l'équation $#b#x+1=0$ |
||||
\item\MauvaiseReponse est solution de l'équation $x+#b#=0$ |
||||
\item\BonneReponse est solution de l'inéquation $#a#x+#c#>0$ |
||||
\end{enumerate} |
||||
|
||||
|
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a : 3,5,6,7 |
||||
b : 4,5,6 |
||||
c : 1,2,3,5 |
||||
|
||||
\item Le nombre $\sqrt{#a#}$ |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse est solution de l'équation $x^{2}+#a#=0$ |
||||
\item\BonneReponse est solution de l'équation $x^{3}-#a#x=0$ |
||||
\item\MauvaiseReponse est solution de l'inéquation $-#b#x+#c#>0$ |
||||
\end{enumerate} |
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|
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|
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|
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|
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a : 8,9,10,11,12 |
||||
b : 2,4,6,8 |
||||
c : 1,3,5,7 |
||||
|
||||
\item Le nombre $#1/a#$ |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse est solution de l'équation $#a-1#x+1=0$ |
||||
\item\BonneReponse est solution de l'équation $#a#x-1=0$ |
||||
\item\MauvaiseReponse est solution de l'inéquation $#b#x+#c#<0$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
a : -2,-3,-4,2,3,5 |
||||
b : 2,4,-2,3,-3 |
||||
c : 3,5 |
||||
d : 2,5 |
||||
e : 6,8 |
||||
|
||||
\item Le couple solution du système $\left\{\begin{array}{c}#c#x+#d#y=#c*a+d*b# \\ x-#e#y=#a-e*b#\end{array}\right.$ est |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse $(#-2*a#;#b#)$ |
||||
\item\MauvaiseReponse $(#b/2# ; #a#)$ |
||||
\item\BonneReponse $(#a# ;#b#)$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
a : 2,3,4,5 |
||||
b : 6,7,8,9 |
||||
|
||||
\item L'ensemble des solutions de l'équation $x^{2}+#b-a# x-#a* b#=0$ est : |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse $\emptyset$ |
||||
\item\BonneReponse $\{#a# ;-#b#\}$ |
||||
\item\MauvaiseReponse $\{#a# ; #b#\}$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
a : 15/2,12/5,6/7 |
||||
b : 3/4,3/8,5/4 |
||||
c : 2,3 |
||||
|
||||
\item $#a# - #b# \times#c#=$ |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse $#(a-b)* c#$ |
||||
\item\MauvaiseReponse $#a-b#$ |
||||
\item\BonneReponse $#a-b* c#$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
a : 13/2,9/5,5/7 |
||||
b : 1/4,1/8,1/3 |
||||
|
||||
\item $\frac{#a#-1}{#b#+1}=$ |
||||
\begin{enumerate} |
||||
\item\BonneReponse $#(a-1)/(b+1)#$ |
||||
\item\MauvaiseReponse $#a/b#$ |
||||
\item\MauvaiseReponse $#(a-1)* (b+1)#$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
a : 2,4,6,8 |
||||
b : 10,12,14,16,18,20 |
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $A$, si $AB=#a#$ et $BC=#b#$, alors |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse $\sin (\hat{B})=#a/b#$ |
||||
\item\BonneReponse $AC=\sqrt{#b^2-a^2#}$ |
||||
\item\MauvaiseReponse $\cos (\hat{C})=#a/b#$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
a : 2,4,6,8 |
||||
b : 10,12,14,16,18,20 |
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $A$, si $AB=#a#$ et $BC=#b#$, alors |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse $\tan (\hat{C})=#a/b#$ |
||||
\item\MauvaiseReponse $AC=\sqrt{#a^2+b^2#}$ |
||||
\item\BonneReponse $\cos (\hat{B})=#a/b#$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
a : 2,4,6,8 |
||||
b : 10,15,20,25,30,35,40 |
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $A$, si $AB=#a#\mathrm{cm}$ et $AC=#b#\mathrm{mm}$, alors $\hat{B}=$ : |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse $#10* a#\times\arctan\left(#b#\right)$ |
||||
\item\BonneReponse $\arctan\left(\frac{#b#}{#10* a#}\right)$ |
||||
\item\MauvaiseReponse $\frac{\arctan\left(#b/10#\right)}{#a#}$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
a : 10,15,20,25,30,35,40 |
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $B$, si $\widehat{A}=#a#^{\circ}$ alors |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse $\widehat{C}=#100-a#^{\circ}$ |
||||
\item\MauvaiseReponse $\widehat{C}=#a+45#^{\circ}$ |
||||
\item\BonneReponse $\widehat{C}=#90-a#^{\circ}$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
a : 50,55,60,65,70,75,80 |
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $B$, si $\widehat{A}=#a#^{\circ}$ alors |
||||
\begin{enumerate} |
||||
\item\MauvaiseReponse $\widehat{C}=#100-a#^{\circ}$ |
||||
\item\MauvaiseReponse $\widehat{C}=#a-45#^{\circ}$ |
||||
\item\BonneReponse $\widehat{C}=#90-a#^{\circ}$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
n : 3,4,5,6,7,8,9,10,11 |
||||
|
||||
\item Si $ABC$ est un triangle rectangle en $B$ tel que $AB=#n^2-1#\mathrm{cm}$ et $BC=#2*n#\mathrm{cm}$, alors le segment $\left[AC\right]$ mesure : |
||||
\begin{enumerate} |
||||
\item\BonneReponse $#n^2+1#\mathrm{cm}$ |
||||
\item\MauvaiseReponse $#n^2+2*n-1#\mathrm{cm}$ |
||||
\item\MauvaiseReponse $#n^2-2* n-1#\mathrm{cm}$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
n : 3,4,5,6,7,8,9,10,11 |
||||
|
||||
\item Si $ABC$ est un triangle rectangle en $B$ tel que $AB=#2*n+1#\mathrm{cm}$ et $AC=#2*n^2+2*n+1#\mathrm{cm}$, alors le segment $\left[BC\right]$ mesure : |
||||
\begin{enumerate} |
||||
\item\BonneReponse $#2*n^2+2*n#\mathrm{cm}$ |
||||
\item\MauvaiseReponse $#n^2+2*n-1#\mathrm{cm}$ |
||||
\item\MauvaiseReponse $#2*n^2+4* n+2#\mathrm{cm}$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
n : 3,4,5,6,7,8,9,10,11 |
||||
|
||||
|
||||
\item Dans quel cas le triangle $ABC$ est-il rectangle ? |
||||
\begin{enumerate} |
||||
\item\BonneReponse $AB=#n^2-1#\mathrm{cm}$, $AC=#n^2+1#\mathrm{cm}$ et $BC=#2* n#\mathrm{cm}$ |
||||
\item\MauvaiseReponse $AB=#n^2-1#\mathrm{cm}$, $AC=#n^2+2* n-1#\mathrm{cm}$ et $BC=#2* n#\mathrm{cm}$ |
||||
\item\MauvaiseReponse $AB=#n^2-1#\mathrm{cm}$, $AC=#n^2-2* n-1#\mathrm{cm}$ et $BC=#2* n#\mathrm{cm}$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
|
||||
|
||||
a : 6,18,30 |
||||
b : 12,24,36 |
||||
k : 2/3,3/2 |
||||
|
||||
\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$, $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. Si on a $AB=#a#\mathrm{cm}$, $AC=#b#\mathrm{cm}$ et $A^{\prime}B^{\prime}=#a*k#\mathrm{mm}$, alors $A^{\prime}C^{\prime}=$ |
||||
\begin{enumerate} |
||||
\item\BonneReponse $#b*k#\mathrm{mm}$ |
||||
\item\MauvaiseReponse $#a/k#\mathrm{cm}$ |
||||
\item\MauvaiseReponse $#b/k#\mathrm{cm}$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
a : 5,15,25 |
||||
b : 10,20,30 |
||||
k : 2/5,5/2 |
||||
|
||||
\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$, $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. Si on a $AB=#a#\mathrm{mm}$, $AC=#b#\mathrm{mm}$ et $A^{\prime}B^{\prime}=#a* k#\mathrm{cm}$, alors $A^{\prime}C^{\prime}=$ |
||||
\begin{enumerate} |
||||
\item\BonneReponse $#10*b*k#\mathrm{mm}$ |
||||
\item\MauvaiseReponse $#b/k#\mathrm{mm}$ |
||||
\item\MauvaiseReponse $#b/k#\mathrm{cm}$ |
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
a : 21,63 |
||||
b : 42,84 |
||||
k : 3/7,7/3 |
||||
|
||||
|
||||
\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. On a $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ si on a : |
||||
\begin{enumerate} |
||||
\item\BonneReponse $AB=#a#\mathrm{m}$, $AC=#b#\mathrm{m}$, $A^{\prime}B^{\prime}=#a*k#\mathrm{cm}$ et $A^{\prime}C^{\prime}=#b*k#\mathrm{cm}$ |
||||
\item\MauvaiseReponse $AB=#a#\mathrm{m}$, $AC=#b#\mathrm{m}$, $A^{\prime}B^{\prime}=#a*k#\mathrm{cm}$ et $A^{\prime}C^{\prime}=#b/k#\mathrm{cm}$ |
||||
\item\MauvaiseReponse $AB=#a#\mathrm{m}$, $AC=#b#\mathrm{m}$, $A^{\prime}B^{\prime}=#a*k#\mathrm{cm}$ et $A^{\prime}C^{\prime}=#a/k#\mathrm{cm}$ |
||||
\end{enumerate} |
||||
|
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\gdef \@abspage@last{3} |
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\documentclass[oneside,twoside]{book} |
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} |
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\begin{document} |
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\chapter*{Contrôle de Mathématiques} |
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%\let\MauvaiseReponse\null |
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%\let\BonneReponse\null |
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\def\MauvaiseReponse#1\par{\textcolor{red}{#1}} |
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\def\BonneReponse#1\par{\textcolor{green}{#1}} |
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\newcounter{question} |
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\setcounter{question}{0} |
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{\bf Merci de répondre sur la grille fournie et de rendre le sujet avec la grille.} |
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\vspace{2em} |
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\begin{enumerate} |
||||
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|
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\item L'ensemble des solutions de l'inéquation $-2 x+2<1$ est : |
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse $]\frac{1}{2},+\infty[$ |
||||
|
||||
\item\MauvaiseReponse $\mathbb{R}$ |
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\item\MauvaiseReponse $]-\infty, -\frac{1}{2}]$ |
||||
|
||||
\end{enumerate} |
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|
||||
|
||||
\item L'ensemble des solutions de l'inéquation $-2x+8 \geqslant 9$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\mathbb{R}$ |
||||
|
||||
|
||||
\item\BonneReponse $]-\infty,-\frac{1}{2}]$ |
||||
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||||
\item\MauvaiseReponse $[-\frac{1}{2},+\infty[$ |
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||||
\end{enumerate} |
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\stepcounter{question} |
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\setcounter{enumi}{\thequestion} |
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||||
\item Le nombre $\frac{1}{2}$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse est solution de l'équation $x+2=0$ |
||||
|
||||
\item\MauvaiseReponse est solution de l'équation $2x+1=0$ |
||||
|
||||
|
||||
\item\BonneReponse est solution de l'inéquation $3x+7>0$ |
||||
|
||||
\end{enumerate} |
||||
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||||
|
||||
|
||||
\item Le nombre $\sqrt{5}$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\BonneReponse est solution de l'équation $x^{3}-5x=0$ |
||||
|
||||
\item\MauvaiseReponse est solution de l'inéquation $-5x+2>0$ |
||||
|
||||
\item\MauvaiseReponse est solution de l'équation $x^{2}+5=0$ |
||||
|
||||
\end{enumerate} |
||||
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||||
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||||
|
||||
\item Le nombre $\frac{1}{8}$ |
||||
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||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\BonneReponse est solution de l'équation $8x-1=0$ |
||||
|
||||
\item\MauvaiseReponse est solution de l'équation $7x+1=0$ |
||||
|
||||
\item\MauvaiseReponse est solution de l'inéquation $2x+7<0$ |
||||
|
||||
\end{enumerate} |
||||
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||||
|
||||
\newpage |
||||
|
||||
\item Le couple solution du système $\left\{\begin{array}{c}5x+5y=10 \\ x-8y=-34\end{array}\right.$ est |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $(4;4)$ |
||||
|
||||
\item\BonneReponse $(-2 ;4)$ |
||||
|
||||
\item\MauvaiseReponse $(2 ; -2)$ |
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item L'ensemble des solutions de l'équation $x^{2}+4 x-32=0$ est : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\emptyset$ |
||||
|
||||
\item\BonneReponse $\{4 ;-8\}$ |
||||
|
||||
\item\MauvaiseReponse $\{4 ; 8\}$ |
||||
|
||||
\end{enumerate} |
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|
||||
\item $\frac{15}{2} - \frac{3}{8} \times3=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\frac{171}{8}$ |
||||
|
||||
\item\MauvaiseReponse $\frac{57}{8}$ |
||||
|
||||
\item\BonneReponse $\frac{51}{8}$ |
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|
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\end{enumerate} |
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|
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|
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|
||||
\item $\displaystyle\frac{\frac{5}{7}-1}{\frac{1}{8}+1}=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse $-\frac{16}{63}$ |
||||
|
||||
\item\MauvaiseReponse $\frac{40}{7}$ |
||||
|
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\item\MauvaiseReponse $-\frac{9}{28}$ |
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|
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\end{enumerate} |
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|
||||
|
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|
||||
\item Dans un triangle $ABC$ rectangle en $A$, si $AB=2$ et $BC=14$, alors |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\sin (\hat{B})=\frac{1}{7}$ |
||||
|
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\item\BonneReponse $AC=\sqrt{192}$ |
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|
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\item\MauvaiseReponse $\cos (\hat{C})=\frac{1}{7}$ |
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|
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\end{enumerate} |
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|
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|
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|
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\item Dans un triangle $ABC$ rectangle en $A$, si $AB=2$ et $BC=20$, alors |
||||
|
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\begin{enumerate} |
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|
||||
\item\MauvaiseReponse $\tan (\hat{C})=\frac{1}{10}$ |
||||
|
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\item\MauvaiseReponse $AC=\sqrt{404}$ |
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|
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\item\BonneReponse $\cos (\hat{B})=\frac{1}{10}$ |
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|
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\end{enumerate} |
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|
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|
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|
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\item Dans un triangle $ABC$ rectangle en $A$, si $AB=2\mathrm{cm}$ et $AC=20\mathrm{mm}$, alors $\hat{B}=$ : |
||||
|
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\begin{enumerate} |
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|
||||
\item\MauvaiseReponse $20\times\arctan\left(20\right)$ |
||||
|
||||
\item\BonneReponse $\arctan\left(\frac{20}{20}\right)$ |
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|
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\item\MauvaiseReponse $\frac{\arctan\left(2\right)}{2}$ |
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|
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\end{enumerate} |
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|
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|
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\item Dans un triangle $ABC$ rectangle en $B$, si $\widehat{A}=10^{\circ}$ alors |
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|
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\begin{enumerate} |
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|
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\item\MauvaiseReponse $\widehat{C}=90^{\circ}$ |
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|
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\item\MauvaiseReponse $\widehat{C}=55^{\circ}$ |
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|
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\item\BonneReponse $\widehat{C}=80^{\circ}$ |
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|
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\end{enumerate} |
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|
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|
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\item Dans un triangle $ABC$ rectangle en $B$, si $\widehat{A}=50^{\circ}$ alors |
||||
|
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\begin{enumerate} |
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|
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\item\MauvaiseReponse $\widehat{C}=50^{\circ}$ |
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|
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\item\MauvaiseReponse $\widehat{C}=5^{\circ}$ |
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|
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\item\BonneReponse $\widehat{C}=40^{\circ}$ |
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|
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\end{enumerate} |
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|
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|
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|
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\item Si $ABC$ est un triangle rectangle en $B$ tel que $AB=15\mathrm{cm}$ et $BC=8\mathrm{cm}$, alors le segment $\left[AC\right]$ mesure : |
||||
|
||||
\begin{enumerate} |
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|
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\item\BonneReponse $17\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $23\mathrm{cm}$ |
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|
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\item\MauvaiseReponse $7\mathrm{cm}$ |
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|
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\end{enumerate} |
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|
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|
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|
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\item Si $ABC$ est un triangle rectangle en $B$ tel que $AB=9\mathrm{cm}$ et $AC=41\mathrm{cm}$, alors le segment $\left[BC\right]$ mesure : |
||||
|
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\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $23\mathrm{cm}$ |
||||
|
||||
\item\BonneReponse $40\mathrm{cm}$ |
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|
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\item\MauvaiseReponse $50\mathrm{cm}$ |
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|
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\end{enumerate} |
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|
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|
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|
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\item Dans quel cas le triangle $ABC$ est-il rectangle ? |
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|
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\begin{enumerate} |
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|
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|
||||
\item\MauvaiseReponse $AB=24\mathrm{cm}$, $AC=34\mathrm{cm}$ et $BC=10\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $AB=24\mathrm{cm}$, $AC=14\mathrm{cm}$ et $BC=10\mathrm{cm}$ |
||||
|
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\item\BonneReponse $AB=24\mathrm{cm}$, $AC=26\mathrm{cm}$ et $BC=10\mathrm{cm}$ |
||||
|
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\end{enumerate} |
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|
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|
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|
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\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$, $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. Si on a $AB=6\mathrm{cm}$, $AC=24\mathrm{cm}$ et $A^{\prime}B^{\prime}=4\mathrm{mm}$, alors $A^{\prime}C^{\prime}=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $9\mathrm{cm}$ |
||||
|
||||
\item\BonneReponse $16\mathrm{mm}$ |
||||
|
||||
\item\MauvaiseReponse $36\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
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|
||||
|
||||
\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$, $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. Si on a $AB=5\mathrm{mm}$, $AC=20\mathrm{mm}$ et $A^{\prime}B^{\prime}=2\mathrm{cm}$, alors $A^{\prime}C^{\prime}=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $50\mathrm{mm}$ |
||||
|
||||
\item\BonneReponse $80\mathrm{mm}$ |
||||
|
||||
\item\MauvaiseReponse $50\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
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|
||||
|
||||
|
||||
|
||||
|
||||
\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. On a $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ si on a : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse $AB=63\mathrm{m}$, $AC=42\mathrm{m}$, $A^{\prime}B^{\prime}=147\mathrm{cm}$ et $A^{\prime}C^{\prime}=98\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $AB=63\mathrm{m}$, $AC=42\mathrm{m}$, $A^{\prime}B^{\prime}=147\mathrm{cm}$ et $A^{\prime}C^{\prime}=18\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $AB=63\mathrm{m}$, $AC=42\mathrm{m}$, $A^{\prime}B^{\prime}=147\mathrm{cm}$ et $A^{\prime}C^{\prime}=27\mathrm{cm}$ |
||||
|
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\end{enumerate} |
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|
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\end{enumerate} |
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\end{document} |
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\documentclass[oneside,twoside]{book} |
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\usepackage[T1]{fontenc} |
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\usepackage[utf8]{inputenc} |
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\usepackage{geometry} |
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\geometry{verbose,tmargin=2cm,bmargin=2cm,lmargin=2cm,rmargin=2cm} |
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\usepackage{amsmath} |
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\usepackage{mhchem} |
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands. |
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\usepackage{amsfonts} |
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\usepackage{bbold} |
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\raggedbottom |
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\AtBeginDocument{ |
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\def\labelitemiv{\(\triangleright\)} |
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} |
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\makeatother |
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|
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\begin{document} |
||||
|
||||
\chapter*{Contrôle de Mathématiques} |
||||
|
||||
%\let\MauvaiseReponse\null |
||||
%\let\BonneReponse\null |
||||
\def\MauvaiseReponse#1\par{\textcolor{red}{#1}} |
||||
\def\BonneReponse#1\par{\textcolor{green}{#1}} |
||||
|
||||
{\bf Merci de répondre sur la grille fournie et de rendre le sujet avec la grille.} |
||||
\vspace{2em} |
||||
\begin{enumerate} |
||||
|
||||
|
||||
|
||||
|
||||
\item L'ensemble des solutions de l'inéquation $-3 x+2<5$ est : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\mathbb{R}$ |
||||
|
||||
\item\BonneReponse $]-1,+\infty[$ |
||||
|
||||
\item\MauvaiseReponse $]-\infty, 1]$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
\item L'ensemble des solutions de l'inéquation $-3x+8 \geqslant 5$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\mathbb{R}$ |
||||
|
||||
\item\MauvaiseReponse $[1,+\infty[$ |
||||
|
||||
\item\BonneReponse $]-\infty,1]$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\item Le nombre $\frac{1}{6}$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse est solution de l'équation $6x+1=0$ |
||||
|
||||
\item\MauvaiseReponse est solution de l'équation $x+6=0$ |
||||
|
||||
\item\BonneReponse est solution de l'inéquation $3x+3>0$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Le nombre $\sqrt{6}$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse est solution de l'équation $x^{2}+6=0$ |
||||
|
||||
\item\BonneReponse est solution de l'équation $x^{3}-6x=0$ |
||||
|
||||
\item\MauvaiseReponse est solution de l'inéquation $-6x+2>0$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
\newpage |
||||
|
||||
\item Le nombre $\frac{1}{8}$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse est solution de l'équation $8x-1=0$ |
||||
|
||||
\item\MauvaiseReponse est solution de l'équation $7x+1=0$ |
||||
|
||||
|
||||
\item\MauvaiseReponse est solution de l'inéquation $8x+3<0$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Le couple solution du système $\left\{\begin{array}{c}5x+2y=-7 \\ x-8y=-35\end{array}\right.$ est |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $(6;4)$ |
||||
|
||||
\item\BonneReponse $(-3 ;4)$ |
||||
|
||||
\item\MauvaiseReponse $(2 ; -3)$ |
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item L'ensemble des solutions de l'équation $x^{2}+2 x-24=0$ est : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse $\{4 ;-6\}$ |
||||
|
||||
\item\MauvaiseReponse $\emptyset$ |
||||
|
||||
|
||||
\item\MauvaiseReponse $\{4 ; 6\}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item $\frac{12}{5} - \frac{5}{4} \times2=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\frac{23}{10}$ |
||||
|
||||
\item\BonneReponse $-\frac{1}{10}$ |
||||
|
||||
\item\MauvaiseReponse $\frac{23}{20}$ |
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item $\displaystyle\frac{\frac{9}{5}-1}{\frac{1}{8}+1}=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $\frac{72}{5}$ |
||||
|
||||
\item\BonneReponse $\frac{32}{45}$ |
||||
|
||||
\item\MauvaiseReponse $\frac{9}{10}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $A$, si $AB=4$ et $BC=10$, alors |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\sin (\hat{B})=\frac{2}{5}$ |
||||
|
||||
|
||||
\item\MauvaiseReponse $\cos (\hat{C})=\frac{2}{5}$ |
||||
|
||||
\item\BonneReponse $AC=\sqrt{84}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $A$, si $AB=4$ et $BC=14$, alors |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\tan (\hat{C})=\frac{2}{7}$ |
||||
|
||||
\item\BonneReponse $\cos (\hat{B})=\frac{2}{7}$ |
||||
|
||||
\item\MauvaiseReponse $AC=\sqrt{212}$ |
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $A$, si $AB=2\mathrm{cm}$ et $AC=35\mathrm{mm}$, alors $\hat{B}=$ : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse $\arctan\left(\frac{35}{20}\right)$ |
||||
|
||||
\item\MauvaiseReponse $20\times\arctan\left(35\right)$ |
||||
|
||||
|
||||
\item\MauvaiseReponse $\frac{\arctan\left(\frac{7}{2}\right)}{2}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $B$, si $\widehat{A}=20^{\circ}$ alors |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\widehat{C}=80^{\circ}$ |
||||
|
||||
\item\MauvaiseReponse $\widehat{C}=65^{\circ}$ |
||||
|
||||
\item\BonneReponse $\widehat{C}=70^{\circ}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $B$, si $\widehat{A}=60^{\circ}$ alors |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\widehat{C}=40^{\circ}$ |
||||
|
||||
\item\BonneReponse $\widehat{C}=30^{\circ}$ |
||||
|
||||
\item\MauvaiseReponse $\widehat{C}=15^{\circ}$ |
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Si $ABC$ est un triangle rectangle en $B$ tel que $AB=35\mathrm{cm}$ et $BC=12\mathrm{cm}$, alors le segment $\left[AC\right]$ mesure : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $47\mathrm{cm}$ |
||||
|
||||
\item\BonneReponse $37\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $23\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Si $ABC$ est un triangle rectangle en $B$ tel que $AB=13\mathrm{cm}$ et $AC=85\mathrm{cm}$, alors le segment $\left[BC\right]$ mesure : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $47\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $98\mathrm{cm}$ |
||||
|
||||
\item\BonneReponse $84\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
\newpage |
||||
|
||||
\item Dans quel cas le triangle $ABC$ est-il rectangle ? |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse $AB=48\mathrm{cm}$, $AC=50\mathrm{cm}$ et $BC=14\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $AB=48\mathrm{cm}$, $AC=62\mathrm{cm}$ et $BC=14\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $AB=48\mathrm{cm}$, $AC=34\mathrm{cm}$ et $BC=14\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$, $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. Si on a $AB=6\mathrm{cm}$, $AC=36\mathrm{cm}$ et $A^{\prime}B^{\prime}=9\mathrm{mm}$, alors $A^{\prime}C^{\prime}=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse $54\mathrm{mm}$ |
||||
|
||||
\item\MauvaiseReponse $4\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $24\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$, $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. Si on a $AB=5\mathrm{mm}$, $AC=30\mathrm{mm}$ et $A^{\prime}B^{\prime}=\frac{25}{2}\mathrm{cm}$, alors $A^{\prime}C^{\prime}=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $12\mathrm{mm}$ |
||||
|
||||
\item\BonneReponse $750\mathrm{mm}$ |
||||
|
||||
\item\MauvaiseReponse $12\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. On a $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ si on a : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $AB=21\mathrm{m}$, $AC=84\mathrm{m}$, $A^{\prime}B^{\prime}=49\mathrm{cm}$ et $A^{\prime}C^{\prime}=36\mathrm{cm}$ |
||||
|
||||
\item\BonneReponse $AB=21\mathrm{m}$, $AC=84\mathrm{m}$, $A^{\prime}B^{\prime}=49\mathrm{cm}$ et $A^{\prime}C^{\prime}=196\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $AB=21\mathrm{m}$, $AC=84\mathrm{m}$, $A^{\prime}B^{\prime}=49\mathrm{cm}$ et $A^{\prime}C^{\prime}=9\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
\end{document} |
||||
@ -1,2 +0,0 @@ |
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\relax |
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\gdef \@abspage@last{3} |
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This is pdfTeX, Version 3.14159265-2.6-1.40.21 (MiKTeX 20.11) (preloaded format=pdflatex 2020.11.23) 5 NOV 2021 16:54 |
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**"./TSP - DST 1 - Sujet C.tex" |
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\documentclass[oneside,twoside]{book} |
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\chapter*{Contrôle de Mathématiques} |
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%\let\MauvaiseReponse\null |
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%\let\BonneReponse\null |
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\def\MauvaiseReponse#1\par{\textcolor{red}{#1}} |
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\def\BonneReponse#1\par{\textcolor{green}{#1}} |
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{\bf Merci de répondre sur la grille fournie et de rendre le sujet avec la grille.} |
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\vspace{2em} |
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\begin{enumerate} |
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\item L'ensemble des solutions de l'inéquation $-4 x+8<9$ est : |
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\begin{enumerate} |
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\item\MauvaiseReponse $\mathbb{R}$ |
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\item\MauvaiseReponse $]-\infty, \frac{1}{4}]$ |
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\item\BonneReponse $]-\frac{1}{4},+\infty[$ |
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\item L'ensemble des solutions de l'inéquation $-5x+4 \geqslant 3$ |
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\begin{enumerate} |
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\item\MauvaiseReponse $\mathbb{R}$ |
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\item\BonneReponse $]-\infty,\frac{1}{5}]$ |
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\item\MauvaiseReponse $[\frac{1}{5},+\infty[$ |
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\item Le nombre $\frac{1}{4}$ |
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\begin{enumerate} |
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\item\MauvaiseReponse est solution de l'équation $4x+1=0$ |
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|
||||
\item\BonneReponse est solution de l'inéquation $4x+5>0$ |
||||
|
||||
\item\MauvaiseReponse est solution de l'équation $x+4=0$ |
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Le nombre $\sqrt{7}$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse est solution de l'équation $x^{3}-7x=0$ |
||||
|
||||
\item\MauvaiseReponse est solution de l'équation $x^{2}+7=0$ |
||||
|
||||
|
||||
\item\MauvaiseReponse est solution de l'inéquation $-5x+3>0$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Le nombre $\frac{1}{10}$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse est solution de l'équation $9x+1=0$ |
||||
|
||||
|
||||
\item\MauvaiseReponse est solution de l'inéquation $2x+3<0$ |
||||
|
||||
\item\BonneReponse est solution de l'équation $10x-1=0$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
\newpage |
||||
|
||||
\item Le couple solution du système $\left\{\begin{array}{c}3x+5y=-24 \\ x-6y=15\end{array}\right.$ est |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $(6;-3)$ |
||||
|
||||
\item\BonneReponse $(-3 ;-3)$ |
||||
|
||||
\item\MauvaiseReponse $(-\frac{3}{2} ; -3)$ |
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
|
||||
\item L'ensemble des solutions de l'équation $x^{2}+3 x-40=0$ est : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\emptyset$ |
||||
|
||||
\item\BonneReponse $\{5 ;-8\}$ |
||||
|
||||
\item\MauvaiseReponse $\{5 ; 8\}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item $\frac{6}{7} - \frac{5}{4} \times2=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse $-\frac{23}{14}$ |
||||
|
||||
\item\MauvaiseReponse $-\frac{11}{14}$ |
||||
|
||||
\item\MauvaiseReponse $-\frac{11}{28}$ |
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item $\displaystyle\frac{\frac{13}{2}-1}{\frac{1}{8}+1}=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $52$ |
||||
|
||||
\item\BonneReponse $\frac{44}{9}$ |
||||
|
||||
\item\MauvaiseReponse $\frac{99}{16}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $A$, si $AB=4$ et $BC=20$, alors |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\sin (\hat{B})=\frac{1}{5}$ |
||||
|
||||
|
||||
\item\MauvaiseReponse $\cos (\hat{C})=\frac{1}{5}$ |
||||
|
||||
\item\BonneReponse $AC=\sqrt{384}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $A$, si $AB=6$ et $BC=10$, alors |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse $\cos (\hat{B})=\frac{3}{5}$ |
||||
|
||||
\item\MauvaiseReponse $\tan (\hat{C})=\frac{3}{5}$ |
||||
|
||||
\item\MauvaiseReponse $AC=\sqrt{136}$ |
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $A$, si $AB=4\mathrm{cm}$ et $AC=35\mathrm{mm}$, alors $\hat{B}=$ : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $40\times\arctan\left(35\right)$ |
||||
|
||||
|
||||
\item\MauvaiseReponse $\frac{\arctan\left(\frac{7}{2}\right)}{4}$ |
||||
|
||||
\item\BonneReponse $\arctan\left(\frac{35}{40}\right)$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $B$, si $\widehat{A}=30^{\circ}$ alors |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\widehat{C}=70^{\circ}$ |
||||
|
||||
\item\BonneReponse $\widehat{C}=60^{\circ}$ |
||||
|
||||
\item\MauvaiseReponse $\widehat{C}=75^{\circ}$ |
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
|
||||
\item Dans un triangle $ABC$ rectangle en $B$, si $\widehat{A}=75^{\circ}$ alors |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\MauvaiseReponse $\widehat{C}=25^{\circ}$ |
||||
|
||||
\item\MauvaiseReponse $\widehat{C}=30^{\circ}$ |
||||
|
||||
\item\BonneReponse $\widehat{C}=15^{\circ}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Si $ABC$ est un triangle rectangle en $B$ tel que $AB=63\mathrm{cm}$ et $BC=16\mathrm{cm}$, alors le segment $\left[AC\right]$ mesure : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
\item\BonneReponse $65\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $79\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $47\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Si $ABC$ est un triangle rectangle en $B$ tel que $AB=19\mathrm{cm}$ et $AC=181\mathrm{cm}$, alors le segment $\left[BC\right]$ mesure : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $98\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $200\mathrm{cm}$ |
||||
|
||||
\item\BonneReponse $180\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item Dans quel cas le triangle $ABC$ est-il rectangle ? |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $AB=99\mathrm{cm}$, $AC=119\mathrm{cm}$ et $BC=20\mathrm{cm}$ |
||||
|
||||
\item\BonneReponse $AB=99\mathrm{cm}$, $AC=101\mathrm{cm}$ et $BC=20\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $AB=99\mathrm{cm}$, $AC=79\mathrm{cm}$ et $BC=20\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$, $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. Si on a $AB=30\mathrm{cm}$, $AC=12\mathrm{cm}$ et $A^{\prime}B^{\prime}=45\mathrm{mm}$, alors $A^{\prime}C^{\prime}=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $20\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $8\mathrm{cm}$ |
||||
|
||||
\item\BonneReponse $18\mathrm{mm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
|
||||
\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$, $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. Si on a $AB=15\mathrm{mm}$, $AC=20\mathrm{mm}$ et $A^{\prime}B^{\prime}=6\mathrm{cm}$, alors $A^{\prime}C^{\prime}=$ |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $50\mathrm{mm}$ |
||||
|
||||
\item\BonneReponse $80\mathrm{mm}$ |
||||
|
||||
\item\MauvaiseReponse $50\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\item On considère deux triangles non plat $ABC$ et $A^\prime B^\prime C^\prime$ tels que $\left(AB\right)//\left(A^{\prime}B^{\prime}\right)$ et $\left(CB\right)//\left(C^{\prime}B^{\prime}\right)$. On a $\left(AC\right)//\left(A^{\prime}C^{\prime}\right)$ si on a : |
||||
|
||||
\begin{enumerate} |
||||
|
||||
|
||||
\item\MauvaiseReponse $AB=21\mathrm{m}$, $AC=42\mathrm{m}$, $A^{\prime}B^{\prime}=49\mathrm{cm}$ et $A^{\prime}C^{\prime}=18\mathrm{cm}$ |
||||
|
||||
\item\MauvaiseReponse $AB=21\mathrm{m}$, $AC=42\mathrm{m}$, $A^{\prime}B^{\prime}=49\mathrm{cm}$ et $A^{\prime}C^{\prime}=9\mathrm{cm}$ |
||||
|
||||
\item\BonneReponse $AB=21\mathrm{m}$, $AC=42\mathrm{m}$, $A^{\prime}B^{\prime}=49\mathrm{cm}$ et $A^{\prime}C^{\prime}=98\mathrm{cm}$ |
||||
|
||||
\end{enumerate} |
||||
|
||||
|
||||
\end{enumerate} |
||||
|
||||
\end{document} |
||||
|
||||
|
||||
|
||||
|
||||
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**.\BTS 2- DST 2 - Sujets.tex |
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\documentclass[oneside,twocolumn,landscape]{book} |
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\usepackage[T1]{fontenc} |
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\usepackage[utf8]{inputenc} |
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\usepackage{geometry} |
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\geometry{verbose,tmargin=1cm,bmargin=1cm,lmargin=2cm,rmargin=2cm} |
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\setcounter{secnumdepth}{3} |
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\setcounter{tocdepth}{3} |
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\usepackage{mathrsfs} |
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\usepackage{amsmath} |
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\usepackage{bbold} |
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\usepackage{xcolor} |
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\newenvironment{Sujet}[1][] |
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{\refstepcounter{NumeroDuSujet}\section*{Numéro du sujet :~\theNumeroDuSujet}\par #1}{\newpage} |
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\newcounter{NumeroDeLaQuestion}[NumeroDuSujet] |
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\newenvironment{VraiFaux}[1][] |
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{\begin{enumerate}\setcounter{enumi}{\theNumeroDeLaQuestion}#1} |
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\newenvironment{QCM}[1][] |
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{\begin{enumerate}\setcounter{enumi}{\theNumeroDeLaQuestion}\item #1} |
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{\end{enumerate}\stepcounter{NumeroDeLaQuestion}} |
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% Pour obtenir les corrigés, ajouter un % devant les deux lignes suivantes et retirer les devant les deux lignes qui suivent. |
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\let\Vrai\item |
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\let\Faux\item |
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|
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%\def\Vrai#1{\item{\color{green}#1}} |
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%\def\Faux#1{\item{\color{red}#1}} |
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\begin{document} |
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% |
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% Copier-coller le code genéré par le générateur de sujets aléatoires ci-dessous. |
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|
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\end{document} |
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@ -0,0 +1,130 @@ |
||||
module Fraction exposing |
||||
( Fraction |
||||
, Resultat |
||||
, Erreur |
||||
, frac |
||||
, num |
||||
, den |
||||
, estEntier |
||||
, opp |
||||
, add |
||||
, neg |
||||
, sub |
||||
, mul |
||||
, inv |
||||
, div |
||||
, exp |
||||
, teX |
||||
, raw |
||||
) |
||||
|
||||
import Arithmetic exposing (gcd) |
||||
|
||||
|
||||
|
||||
type alias Fraction = { num : Int, den : Int } |
||||
|
||||
type alias Resultat = Result Erreur Fraction |
||||
type alias Erreur = String |
||||
|
||||
frac : Int -> Int -> Resultat |
||||
frac a b = |
||||
let |
||||
min = 1 - 2^31 |
||||
max = 2^31 - 1 |
||||
in |
||||
case b of |
||||
0 -> Err "Division par zéro." |
||||
_ -> |
||||
if a > max || b > max || a < min || b < min then |
||||
Err "Certains calculs font intervenir des valeurs trop grandes pour être prises en charge." |
||||
else |
||||
Ok <| simplifier <| Fraction a b |
||||
|
||||
simplifier : Fraction -> Fraction |
||||
simplifier a = |
||||
let |
||||
pgcd = gcd a.num a.den |
||||
sgnDuDen = |
||||
if a.den < 0 then -1 else 1 |
||||
in |
||||
{ a |
||||
| num = sgnDuDen*(a.num // pgcd) |
||||
, den = sgnDuDen*(a.den // pgcd) |
||||
} |
||||
|
||||
num a = a.num |
||||
|
||||
den a = a.den |
||||
|
||||
estEntier a = a.den == 1 |
||||
|
||||
opp : (Fraction -> Fraction -> Resultat) -> Resultat -> Resultat -> Resultat |
||||
opp operation resultat1 resultat2 = |
||||
case (resultat1, resultat2) of |
||||
(Ok fraction1, Ok fraction2) -> operation fraction1 fraction2 |
||||
(Err erreur, _) -> Err erreur |
||||
(_, Err erreur) -> Err erreur |
||||
|
||||
add : Fraction -> Fraction -> Resultat |
||||
add a b = |
||||
let |
||||
pgcd = gcd a.den b.den |
||||
aDenBis = a.den // pgcd |
||||
bDenBis = b.den // pgcd |
||||
in |
||||
frac (a.num*bDenBis + b.num*aDenBis) (a.den*bDenBis) |
||||
|
||||
neg : Fraction -> Fraction |
||||
neg a = Fraction (-a.num) a.den |
||||
|
||||
sub : Fraction -> Fraction -> Resultat |
||||
sub a b = add a (neg b) |
||||
|
||||
mul : Fraction -> Fraction -> Resultat |
||||
mul a b = |
||||
let |
||||
pgcd = gcd a.num b.den |
||||
pgcdBis = gcd b.num a.den |
||||
aNum = a.num // pgcd |
||||
aDen = a.den // pgcdBis |
||||
bNum = b.num // pgcdBis |
||||
bDen = b.den // pgcd |
||||
in |
||||
frac (aNum*bNum) (aDen*bDen) |
||||
|
||||
inv : Fraction -> Resultat |
||||
inv a = |
||||
case a.num of |
||||
0 -> Err "Division par zéro" |
||||
_ -> Ok <| Fraction a.den a.num |
||||
|
||||
div : Fraction -> Fraction -> Resultat |
||||
div a b = Result.andThen (mul a) <| inv b |
||||
|
||||
exp : Fraction -> Fraction -> Resultat |
||||
exp a b = |
||||
let |
||||
sgnDeA = |
||||
if a.num < 0 then -1 else 1 |
||||
sgnDeB = |
||||
if b.num < 0 then -1 else 1 |
||||
in |
||||
if b.den == 1 && b.num < 0 then |
||||
frac ((sgnDeA*a.den)^(sgnDeB*b.num)) ((sgnDeA*a.num)^(sgnDeB*b.num)) |
||||
else if b.den == 1 then |
||||
frac (a.num^b.num) (a.den^b.num) |
||||
else |
||||
Err "Extraction de racine impossible" |
||||
|
||||
|
||||
teX a |
||||
= case a.den of |
||||
1 -> String.fromInt a.num |
||||
_ -> |
||||
if a.num < 0 then |
||||
"-\\frac{" ++ String.fromInt (-a.num) ++ "}{" ++ String.fromInt a.den ++ "}" |
||||
else |
||||
"\\frac{" ++ String.fromInt a.num ++ "}{" ++ String.fromInt a.den ++ "}" |
||||
|
||||
raw a = "(" ++ String.fromInt a.num ++ "/" ++ String.fromInt a.den ++ ")" |
||||
@ -1,58 +0,0 @@ |
||||
module Fractions exposing (..) |
||||
|
||||
import Arithmetic exposing (gcd) |
||||
|
||||
type alias Frac = { num : Int, den : Int } |
||||
|
||||
simplifier a = |
||||
let |
||||
pgcd = gcd a.num a.den |
||||
sgnDuDen = |
||||
if a.den < 0 then -1 else 1 |
||||
in |
||||
Frac (sgnDuDen*a.num // pgcd) (sgnDuDen*a.den // pgcd) |
||||
--{ a | num = a.num // pgcd, den = a.den // pgcd } |
||||
|
||||
estEntier a = .den (simplifier a) == 1 |
||||
|
||||
add a b = |
||||
simplifier <| Frac (a.num*b.den + a.den*b.num) (a.den*b.den) |
||||
|
||||
neg a = simplifier <| Frac (-a.num) a.den |
||||
|
||||
sub a b = simplifier <| add a (neg b) |
||||
|
||||
mul a b = simplifier <| Frac (a.num*b.num) (a.den*b.den) |
||||
|
||||
inv a = simplifier <| Frac a.den a.num |
||||
|
||||
div a b = simplifier <| mul a (inv b) |
||||
|
||||
exp : Frac -> Frac -> Maybe Frac |
||||
exp a b = |
||||
let |
||||
aa = simplifier a |
||||
bb = simplifier b |
||||
sgnDeA = |
||||
if aa.num < 0 then -1 else 1 |
||||
sgnDeB = |
||||
if bb.num < 0 then -1 else 1 |
||||
in |
||||
if bb.den == 1 && bb.num < 0 then |
||||
Just <| Frac ((sgnDeA*aa.den)^(sgnDeB*bb.num)) ((sgnDeA*aa.num)^(sgnDeB*bb.num)) |
||||
else if bb.den == 1 then |
||||
Just <| Frac (aa.num^bb.num) (aa.den^bb.num) |
||||
else |
||||
Nothing |
||||
|
||||
-- À utiliser sur une fraction simplifier ! |
||||
teX a |
||||
= case a.den of |
||||
1 -> String.fromInt a.num |
||||
_ -> |
||||
if a.num < 0 then |
||||
"-\\frac{" ++ String.fromInt (-a.num) ++ "}{" ++ String.fromInt a.den ++ "}" |
||||
else |
||||
"\\frac{" ++ String.fromInt a.num ++ "}{" ++ String.fromInt a.den ++ "}" |
||||
|
||||
raw a = "(" ++ String.fromInt a.num ++ "/" ++ String.fromInt a.den ++ ")" |
||||
@ -1,222 +0,0 @@ |
||||
module ParserTest exposing (..) |
||||
|
||||
import Parser as P exposing (..) |
||||
import Browser |
||||
import Html exposing (Html, button, div, text) |
||||
import Html.Events exposing (onClick) |
||||
import Maybe exposing (Maybe) |
||||
import Result exposing (Result) |
||||
import String as S |
||||
import List as L |
||||
import ParserMaths as PM |
||||
import Fractions as F exposing (Frac) |
||||
|
||||
|
||||
|
||||
-- MAIN |
||||
|
||||
|
||||
main = |
||||
Browser.sandbox { init = init, update = update, view = view } |
||||
|
||||
|
||||
|
||||
-- MODEL |
||||
|
||||
|
||||
type alias Model = Int |
||||
|
||||
|
||||
init : Model |
||||
init = |
||||
0 |
||||
|
||||
|
||||
|
||||
-- UPDATE |
||||
|
||||
|
||||
type Msg |
||||
= Increment |
||||
| Decrement |
||||
|
||||
|
||||
update : Msg -> Model -> Model |
||||
update msg model = |
||||
case msg of |
||||
Increment -> |
||||
model + 1 |
||||
|
||||
Decrement -> |
||||
model - 1 |
||||
|
||||
|
||||
|
||||
-- VIEW |
||||
|
||||
|
||||
view : Model -> Html Msg |
||||
view model = |
||||
div [] [ text unTexte ] |
||||
|
||||
{- |
||||
Sans le withIndent -1, les arbres sous-indentés sautes |
||||
-} |
||||
unTexte = voirArbresParses <| run (withIndent -1 arbres) |
||||
{--} |
||||
""" |
||||
macro |
||||
hh |
||||
|
||||
jj |
||||
""" |
||||
--} |
||||
{-- |
||||
""" |
||||
* |
||||
* |
||||
#1+3# |
||||
* |
||||
* |
||||
* |
||||
* |
||||
* |
||||
* |
||||
* |
||||
* |
||||
* |
||||
* |
||||
* |
||||
* |
||||
* |
||||
* |
||||
* |
||||
""" |
||||
--} |
||||
|
||||
voirArbresParses arbresParsesPotentiels = |
||||
case arbresParsesPotentiels of |
||||
Err erreurs -> deadEndsToStringBis erreurs |
||||
Ok arbresParses -> voirArbres arbresParses |
||||
|
||||
deadEndsToStringBis errs = |
||||
errs |
||||
|> List.map voirErreur |
||||
|> String.concat |
||||
|> (++) "Il y a des problèmes aux endroits suivants :\n" |
||||
|
||||
voirErreur err = |
||||
"Ligne : " ++ String.fromInt err.row |
||||
++ " | Colonne : " ++ String.fromInt err.col |
||||
|
||||
type Arbre = Arbre String (List Arbre) |
||||
|
||||
voirArbre arbr = |
||||
Debug.log "voirArbre " <| |
||||
case arbr of |
||||
Arbre chn arbrs -> chn ++ "\n [" ++ String.concat (List.map voirArbre arbrs) ++ "]" |
||||
|
||||
voirArbres = |
||||
List.map voirArbre >> String.concat |
||||
|
||||
{-| Ce parser change l'indentation courante, cré un arbre puis |
||||
y intègre ses branches grâce à une boucle |
||||
-} |
||||
arbre : Parser Arbre |
||||
arbre = |
||||
let |
||||
suite = |
||||
Debug.log "? :" <| |
||||
flip withIndent |
||||
<| succeed Arbre |
||||
|= chaine |
||||
|= arbres |
||||
in |
||||
getCol |
||||
|> andThen suite |
||||
|> Debug.log "Arbre : " |
||||
|
||||
flip f a b = f b a |
||||
|
||||
arbres = |
||||
let |
||||
sousArbres arbrs = |
||||
let |
||||
boucle = |
||||
succeed ( \arbr -> Loop (arbr :: arbrs) ) |
||||
|= arbre -- lazy (\_ -> arbre) semble inutile malgrè l'appel récursif... |
||||
fin = |
||||
map (\_ -> Done (List.reverse arbrs)) |
||||
suite col_ind = |
||||
oneOf |
||||
[ succeed () |
||||
|. end |
||||
|> fin |
||||
, if Tuple.first col_ind > Tuple.second col_ind then -- if col > ind |
||||
boucle |
||||
else |
||||
succeed () |
||||
|> fin |
||||
] |
||||
in |
||||
succeed Tuple.pair |
||||
|. spaces |
||||
|= getCol |
||||
|= getIndent |
||||
|> andThen suite |
||||
in |
||||
loop [] sousArbres |
||||
|> Debug.log "Début de la boucle arbres " |
||||
|
||||
|
||||
{- |
||||
██████ █████ ██████ ███████ ███████ ██████ ███ ███ █████ ██████ ██████ ██████ |
||||
██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ████ ████ ██ ██ ██ ██ ██ ██ ██ |
||||
██████ ███████ ██████ ███████ █████ ██████ ██ ████ ██ ███████ ██ ██████ ██ ██ |
||||
██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ ██ |
||||
██ ██ ██ ██ ██ ███████ ███████ ██ ██ ██ ██ ██ ██ ██████ ██ ██ ██████ |
||||
-} |
||||
|
||||
chaine : Parser String |
||||
chaine = |
||||
let |
||||
suite txt = |
||||
P.oneOf |
||||
[ succeed (\x -> P.Loop (txt ++ x)) |
||||
|= texte |
||||
{-- |
||||
|= P.oneOf |
||||
[ texte |
||||
, retourAlaLigne |
||||
] |
||||
--} |
||||
, succeed () |
||||
--|. token "\n" |
||||
|> P.map (\_ -> P.Done [txt]) |
||||
] |
||||
tete ls = |
||||
case ls of |
||||
[] -> "" |
||||
l :: lss -> l |
||||
in |
||||
succeed tete |
||||
|= P.loop "" suite |
||||
|> Debug.log "macro " |
||||
|
||||
texte : Parser String |
||||
texte = |
||||
P.getChompedString |
||||
<| succeed () |
||||
|. P.chompIf ( (/=) '\n' ) |
||||
|. P.chompWhile ( (/=) '\n' ) |
||||
|
||||
retourAlaLigne : Parser String |
||||
retourAlaLigne = |
||||
let |
||||
suite ind = |
||||
succeed "\n" |
||||
|. token "\n" |
||||
|. token (S.repeat ind " ") |
||||
in |
||||
getIndent |
||||
|> andThen suite |
||||
@ -0,0 +1,56 @@ |
||||
module Test exposing (..) |
||||
|
||||
import Browser |
||||
import Html exposing (Html, button, div, text) |
||||
import Html.Events exposing (onClick) |
||||
import MiniLatex |
||||
--import MiniLatex.Render exposing(MathJaxRenderOption(..)) |
||||
|
||||
|
||||
-- MAIN |
||||
|
||||
|
||||
main = |
||||
Browser.sandbox { init = init, update = update, view = view } |
||||
|
||||
texteLaTeX = "Pythagoras says: $a^2 + b^2 = c^2$" |
||||
|
||||
|
||||
|
||||
-- MODEL |
||||
|
||||
|
||||
type alias Model = Int |
||||
|
||||
|
||||
init : Model |
||||
init = |
||||
0 |
||||
|
||||
|
||||
|
||||
-- UPDATE |
||||
|
||||
|
||||
type Msg |
||||
= Increment |
||||
| Decrement |
||||
|
||||
|
||||
update : Msg -> Model -> Model |
||||
update msg model = |
||||
case msg of |
||||
Increment -> |
||||
model + 1 |
||||
|
||||
Decrement -> |
||||
model - 1 |
||||
|
||||
|
||||
|
||||
-- VIEW |
||||
|
||||
|
||||
view : Model -> Html Msg |
||||
view model = |
||||
MiniLatex.render "" NoDelay texteLaTeX |
||||
Loading…
Reference in new issue