From 17b4db0db5d17f8b0425cdb85985a24580fa4db0 Mon Sep 17 00:00:00 2001 From: Lafrite Date: Sun, 13 Sep 2015 06:59:26 +0200 Subject: [PATCH] Examples --- example/1_2ndDeg.tex | 117 ---------------- example/1_corr_DM_0302.tex | 256 +++++++++++++++++------------------ example/1_example.tex | 114 ---------------- example/1_play.tex | 0 example/all_corr_DM_0302.pdf | Bin 67929 -> 85694 bytes example/tpl_corr_DM_0302.tex | 96 ++++++------- 6 files changed, 176 insertions(+), 407 deletions(-) delete mode 100644 example/1_2ndDeg.tex delete mode 100644 example/1_example.tex delete mode 100644 example/1_play.tex diff --git a/example/1_2ndDeg.tex b/example/1_2ndDeg.tex deleted file mode 100644 index aca36ee..0000000 --- a/example/1_2ndDeg.tex +++ /dev/null @@ -1,117 +0,0 @@ -\documentclass[a4paper,10pt]{article} -\RequirePackage[utf8x]{inputenc} -\RequirePackage[francais]{babel} -\RequirePackage{amssymb} -\RequirePackage{amsmath} -\RequirePackage{amsfonts} -\RequirePackage{subfig} -\RequirePackage{graphicx} -\RequirePackage{color} - - - -% Title Page -\title{Calcul littéral et statistiques} -\date{\today} - -\begin{document} -\maketitle - - -\section{Polynômes} - - - - - Résoudre l'équation suivante - \begin{eqnarray*} - 3 x^{ 2 } + 6 x + 3 & = & 0 - \end{eqnarray*} - - Solution: - - - - On commence par calculer le discriminant de $P(x) = 3 x^{ 2 } + 6 x + 3$. - \begin{eqnarray*} - \Delta & = & b^2-4ac \\ - \Delta & = & 6^{ 2 } - 4 \times 3 \times 3 \\ -\Delta & = & 36 - 4 \times 9 \\ -\Delta & = & 36 - 36 \\ -\Delta & = & 0 - \end{eqnarray*} - - - Comme $\Delta = 0$ donc $P$ a une racine - - \begin{eqnarray*} - x_1 = \frac{-b}{2a} = \frac{-6}{2\times 3} = -1 \\ - \end{eqnarray*} - - La solution de $3 x^{ 2 } + 6 x + 3 = 0$ est donc $\mathcal{S} = \left\{ -1\right\}$ - - - - - - \bigskip - ~\dotfill - \bigskip - - - - - Résoudre l'équation suivante - \begin{eqnarray*} - x^{ 2 } + 4 x + 2 & = & - 9 x^{ 2 } + 9 x + 5 - \end{eqnarray*} - - Solution: - - On commence par se ramener à une équation de la forme $ax^2+bx+c = 0$. - - - - \begin{align*} - & & x^{ 2 } + 4 x + 2 = - 9 x^{ 2 } + 9 x + 5 \\ - & \Leftrightarrow & x^{ 2 } + 4 x + 2 - ( - 9 x^{ 2 } + 9 x + 5 )= 0 \\ - & \Leftrightarrow & x^{ 2 } + 4 x + 2 + 9 x^{ 2 } - 9 x - 5= 0 \\ - & \Leftrightarrow & ( 1 + 9 ) x^{ 2 } + ( 4 - 9 ) x + 2 - 5= 0 \\ - & \Leftrightarrow & 10 x^{ 2 } - 5 x - 3= 0 - \end{align*} - - On cherche maintenant à résoudre l'équation $10 x^{ 2 } - 5 x - 3 = 0$. - - - - On commence par calculer le discriminant de $P(x) = 10 x^{ 2 } - 5 x - 3$. - \begin{eqnarray*} - \Delta & = & b^2-4ac \\ - \Delta & = & -5^{ 2 } - 4 \times 10 \times ( -3 ) \\ -\Delta & = & 25 - 4 \times ( -30 ) \\ -\Delta & = & 25 - ( -120 ) \\ -\Delta & = & 145 - \end{eqnarray*} - - - comme $\Delta = 145 > 0$ donc $P$ a deux racines - - \begin{eqnarray*} - x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{-5 - \sqrt{145}}{2 \times 10} = - \frac{\sqrt{145}}{20} + \frac{1}{4} \\ - x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{-5 + \sqrt{145}}{2 \times 10} = \frac{1}{4} + \frac{\sqrt{145}}{20} - \end{eqnarray*} - - Les solutions de l'équation $10 x^{ 2 } - 5 x - 3 = 0$ sont donc $\mathcal{S} = \left\{ - \frac{\sqrt{145}}{20} + \frac{1}{4}; \frac{1}{4} + \frac{\sqrt{145}}{20} \right\}$ - - - - - -\end{document} - -%%% Local Variables: -%%% mode: latex -%%% TeX-master: "master" -%%% End: - - \ No newline at end of file diff --git a/example/1_corr_DM_0302.tex b/example/1_corr_DM_0302.tex index 34d355d..be24f8c 100644 --- a/example/1_corr_DM_0302.tex +++ b/example/1_corr_DM_0302.tex @@ -26,78 +26,72 @@ Le barème est donné à titre indicatif, il pourra être modifié. Vous rendrez \begin{eqnarray*} - 8 x^{ 2 } + 5 x - 2 & > &0 \\ + 6 x^{ 2 } + 7 x + 7 & > &0 \\ \end{eqnarray*} \begin{solution} - On commence par calculer le discriminant de $P(x) = 8 x^{ 2 } + 5 x - 2$. + On commence par calculer le discriminant de $P(x) = 6 x^{ 2 } + 7 x + 7$. \begin{eqnarray*} \Delta & = & b^2-4ac \\ - \Delta & = & 5^{ 2 } - 4 \times 8 ( -2 ) \\ -\Delta & = & 25 - 4 ( -16 ) \\ -\Delta & = & 25 - ( -64 ) \\ -\Delta & = & 89 + \Delta & = & 7^{ 2 } - 4 \times 6 \times 7 \\ +\Delta & = & 49 - 4 \times 42 \\ +\Delta & = & 49 - 168 \\ +\Delta & = & -119 \end{eqnarray*} - comme $\Delta = 89 > 0$ donc $P$ a deux racines - - \begin{eqnarray*} - x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{5 - \sqrt{89}}{2 \times 8} = -0.9 \\ - x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{5 + \sqrt{89}}{2 \times 8} = 0.28 - \end{eqnarray*} - + Alors $\Delta = -119 < 0$ donc $P$ n'a pas de racine. - Comme $a = 8$, on en déduit le tableau de signe de $P$ - \begin{center} - \begin{tikzpicture} - \tkzTabInit[espcl=2]% - {$x$/1, $P$/2}% - {$-\infty$, -0.9 , 0.28 , $+\infty$} - \tkzTabLine{, +, z, -, z , +,} - \end{tikzpicture} - \end{center} + Comme $a = 6$, on en déduit le tableau de signe de $P$ + %\begin{center} + % \begin{tikzpicture} + % \tkzTabInit[espcl=2]% + % {$x$/1, $P$/2}% + % {$-\infty$, $+\infty$} + % \tkzTabLine{, +,} + % \end{tikzpicture} + %\end{center} On regarde maintenant où sont les $+$ dans le tableau de signe pour résoudre l'inéquation. \end{solution} \begin{eqnarray*} - - 3 x^{ 2 } + 2 x + 4 & \leq &0 \\ + - 6 x^{ 2 } + 10 x + 1 & \leq &0 \\ \end{eqnarray*} \begin{solution} - On commence par calculer le discriminant de $Q(x) = - 3 x^{ 2 } + 2 x + 4$. + On commence par calculer le discriminant de $Q(x) = - 6 x^{ 2 } + 10 x + 1$. \begin{eqnarray*} \Delta & = & b^2-4ac \\ - \Delta & = & 2^{ 2 } - 4 ( -3 ) \times 4 \\ -\Delta & = & 4 - 4 ( -12 ) \\ -\Delta & = & 4 - ( -48 ) \\ -\Delta & = & 52 + \Delta & = & 10^{ 2 } - 4 -6 \times 1 \\ +\Delta & = & 100 - 4 \times ( -6 ) \\ +\Delta & = & 100 - ( -24 ) \\ +\Delta & = & 124 \end{eqnarray*} - comme $\Delta = 52 > 0$ donc $Q$ a deux racines + comme $\Delta = 124 > 0$ donc $Q$ a deux racines \begin{eqnarray*} - x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{2 - \sqrt{52}}{2 \times -3} = 1.54 \\ - x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{2 + \sqrt{52}}{2 \times -3} = -0.87 + x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{10 - \sqrt{124}}{2 \times -6} = \frac{5}{6} + \frac{\sqrt{31}}{6} \\ + x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{10 + \sqrt{124}}{2 \times -6} = - \frac{\sqrt{31}}{6} + \frac{5}{6} \end{eqnarray*} - Comme $a = -3$, on en déduit le tableau de signe de $Q$ - \begin{center} - \begin{tikzpicture} - \tkzTabInit[espcl=2]% - {$x$/1, $Q$/2}% - {$-\infty$, -0.87 , 1.54 , $+\infty$} - \tkzTabLine{, -, z, +, z , -,} - \end{tikzpicture} - \end{center} + Comme $a = -6$, on en déduit le tableau de signe de $Q$ + %\begin{center} + % \begin{tikzpicture} + % \tkzTabInit[espcl=2]% + % {$x$/1, $Q$/2}% + % {$-\infty$, - \frac{\sqrt{31}}{6} + \frac{5}{6} , \frac{5}{6} + \frac{\sqrt{31}}{6} , $+\infty$} + % \tkzTabLine{, -, z, +, z , -,} + % \end{tikzpicture} + %\end{center} On regarde maintenant où sont les $-$ dans le tableau de signe pour résoudre l'inéquation. \end{solution} \begin{eqnarray*} - 8 x^{ 2 } + 5 x - 2 & \geq & - 3 x^{ 2 } + 2 x + 4 + 6 x^{ 2 } + 7 x + 7 & \geq & - 6 x^{ 2 } + 10 x + 1 \end{eqnarray*} @@ -105,43 +99,37 @@ Le barème est donné à titre indicatif, il pourra être modifié. Vous rendrez \begin{solution} On commence par se ramener à une équation de la forme $ax^2 + bx + c \geq 0$. \begin{eqnarray*} - 8 x^{ 2 } + 5 x - 2 \geq - 3 x^{ 2 } + 2 x + 4 & \Leftrightarrow & 8 x^{ 2 } + 5 x - 2 - (- 3 x^{ 2 } + 2 x + 4) \geq 0 \\ - & \Leftrightarrow & 8 x^{ 2 } + 5 x - 2 - ( - 3 x^{ 2 } + 2 x + 4 )\geq 0 \\ - & \Leftrightarrow & 8 x^{ 2 } + 5 x - 2 + 3 x^{ 2 } - 2 x - 4\geq 0 \\ - & \Leftrightarrow & ( 8 + 3 ) x^{ 2 } + ( 5 + ( -2 ) ) x + ( -2 ) + ( -4 )\geq 0 \\ - & \Leftrightarrow & 11 x^{ 2 } + 3 x - 6\geq 0 + 6 x^{ 2 } + 7 x + 7 \geq - 6 x^{ 2 } + 10 x + 1 & \Leftrightarrow & 6 x^{ 2 } + 7 x + 7 - (- 6 x^{ 2 } + 10 x + 1) \geq 0 \\ + & \Leftrightarrow & 6 x^{ 2 } + 7 x + 7 - ( - 6 x^{ 2 } + 10 x + 1 )\geq 0 \\ + & \Leftrightarrow & 6 x^{ 2 } + 7 x + 7 + 6 x^{ 2 } - 10 x - 1\geq 0 \\ + & \Leftrightarrow & ( 6 + 6 ) x^{ 2 } + ( 7 - 10 ) x + 7 - 1\geq 0 \\ + & \Leftrightarrow & 12 x^{ 2 } - 3 x + 6\geq 0 \end{eqnarray*} - Ensuite on étudie le signe de $R(X) = 11 x^{ 2 } + 3 x - 6$. + Ensuite on étudie le signe de $R(X) = 12 x^{ 2 } - 3 x + 6$. \begin{eqnarray*} \Delta & = & b^2-4ac \\ - \Delta & = & 3^{ 2 } - 4 \times 11 ( -6 ) \\ -\Delta & = & 9 - 4 ( -66 ) \\ -\Delta & = & 9 - ( -264 ) \\ -\Delta & = & 273 + \Delta & = & -3^{ 2 } - 4 \times 12 \times 6 \\ +\Delta & = & 9 - 4 \times 72 \\ +\Delta & = & 9 - 288 \\ +\Delta & = & -279 \end{eqnarray*} - comme $\Delta = 273 > 0$ donc $R$ a deux racines - - \begin{eqnarray*} - x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{3 - \sqrt{273}}{2 \times 11} = -0.89 \\ - x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{3 + \sqrt{273}}{2 \times 11} = 0.61 - \end{eqnarray*} - + Alors $\Delta = -279 < 0$ donc $R$ n'a pas de racine. - Comme $a = 11$, on en déduit le tableau de signe de $R$ - \begin{center} - \begin{tikzpicture} - \tkzTabInit[espcl=2]% - {$x$/1, $R$/2}% - {$-\infty$, -0.89 , 0.61 , $+\infty$} - \tkzTabLine{, +, z, -, z , +,} - \end{tikzpicture} - \end{center} + Comme $a = 12$, on en déduit le tableau de signe de $R$ + %\begin{center} + % \begin{tikzpicture} + % \tkzTabInit[espcl=2]% + % {$x$/1, $R$/2}% + % {$-\infty$, $+\infty$} + % \tkzTabLine{, +,} + % \end{tikzpicture} + %\end{center} On regarde maintenant où sont les $+$ dans le tableau de signe pour résoudre l'inéquation. @@ -153,83 +141,95 @@ Le barème est donné à titre indicatif, il pourra être modifié. Vous rendrez \begin{parts} - \part $f:x\mapsto - 10 x^{ 3 } + x^{ 2 } - 7 x + 5$ + \part $f:x\mapsto - 2 x^{ 3 } - 4 x^{ 2 } + x + 8$ \begin{solution} Pour avoir les variations de $f$, il faut connaître le signe de sa dérivé. On dérive $P$ \begin{eqnarray*} - f'(x) & = & 3 ( -10 ) x^{ 2 } + 2 \times 1 x + 1 ( -7 ) \\ -f'(x) & = & - 30 x^{ 2 } + 2 x - 7 + f'(x) & = & 3 \times ( -2 ) x^{ 2 } + 2 \times ( -4 ) x + 1 \times 1 \\ +f'(x) & = & - 6 x^{ 2 } - 8 x + 1 \end{eqnarray*} On étudie le signe de $P'$ - Ensuite on étudie le signe de $f'(x) = - 30 x^{ 2 } + 2 x - 7$. + Ensuite on étudie le signe de $f'(x) = - 6 x^{ 2 } - 8 x + 1$. \begin{eqnarray*} \Delta & = & b^2-4ac \\ - \Delta & = & 2^{ 2 } - 4 ( -30 ) ( -7 ) \\ -\Delta & = & 4 - 4 \times 210 \\ -\Delta & = & 4 - 840 \\ -\Delta & = & -836 + \Delta & = & -8^{ 2 } - 4 -6 \times 1 \\ +\Delta & = & 64 - 4 \times ( -6 ) \\ +\Delta & = & 64 - ( -24 ) \\ +\Delta & = & 88 \end{eqnarray*} - Alors $\Delta = -836 < 0$ donc $f'$ n'a pas de racine. + comme $\Delta = 88 > 0$ donc $f'$ a deux racines + + \begin{eqnarray*} + x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{-8 - \sqrt{88}}{2 \times -6} = - \frac{2}{3} + \frac{\sqrt{22}}{6} \\ + x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{-8 + \sqrt{88}}{2 \times -6} = - \frac{\sqrt{22}}{6} - \frac{2}{3} + \end{eqnarray*} + - Comme $a = -30$, on en déduit le tableau de signe de $f'$ - \begin{center} - \begin{tikzpicture} - \tkzTabInit[espcl=2]% - {$x$/1, Signe de $f' $/2}% - {$-\infty$, $+\infty$} - \tkzTabLine{, -,} - \end{tikzpicture} - \end{center} + Comme $a = -6$, on en déduit le tableau de signe de $f'$ + %\begin{center} + % \begin{tikzpicture} + % \tkzTabInit[espcl=2]% + % {$x$/1, Signe de $f' $/2}% + % {$-\infty$, - \frac{2}{3} + \frac{\sqrt{22}}{6} , - \frac{\sqrt{22}}{6} - \frac{2}{3} , $+\infty$} + % \tkzTabLine{, -, z, +, z , -,} + % \end{tikzpicture} + %\end{center} \end{solution} - \part $g:x\mapsto - 9 x^{ 3 } - 8 x^{ 2 } - 5 x - 2$ + \part $g:x\mapsto - 10 x^{ 3 } - 6 x^{ 2 } + 8 x + 7$ \begin{solution} Pour avoir les variations de $g$, il faut connaître le signe de sa dérivé. On dérive $P$ \begin{eqnarray*} - g'(x) & = & 3 ( -9 ) x^{ 2 } + 2 ( -8 ) x + 1 ( -5 ) \\ -g'(x) & = & - 27 x^{ 2 } - 16 x - 5 + g'(x) & = & 3 \times ( -10 ) x^{ 2 } + 2 \times ( -6 ) x + 1 \times 8 \\ +g'(x) & = & - 30 x^{ 2 } - 12 x + 8 \end{eqnarray*} On étudie le signe de $P'$ - Ensuite on étudie le signe de $g'(x) = - 27 x^{ 2 } - 16 x - 5$. + Ensuite on étudie le signe de $g'(x) = - 30 x^{ 2 } - 12 x + 8$. \begin{eqnarray*} \Delta & = & b^2-4ac \\ - \Delta & = & ( -16 )^{ 2 } - 4 ( -27 ) ( -5 ) \\ -\Delta & = & 256 - 4 \times 135 \\ -\Delta & = & 256 - 540 \\ -\Delta & = & -284 + \Delta & = & -12^{ 2 } - 4 -30 \times 8 \\ +\Delta & = & 144 - 4 \times ( -240 ) \\ +\Delta & = & 144 - ( -960 ) \\ +\Delta & = & 1104 \end{eqnarray*} - Alors $\Delta = -284 < 0$ donc $g'$ n'a pas de racine. + comme $\Delta = 1104 > 0$ donc $g'$ a deux racines + + \begin{eqnarray*} + x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{-12 - \sqrt{1104}}{2 \times -30} = - \frac{1}{5} + \frac{\sqrt{69}}{15} \\ + x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{-12 + \sqrt{1104}}{2 \times -30} = - \frac{\sqrt{69}}{15} - \frac{1}{5} + \end{eqnarray*} + - Comme $a = -27$, on en déduit le tableau de signe de $g'$ - \begin{center} - \begin{tikzpicture} - \tkzTabInit[espcl=2]% - {$x$/1, Signe de $g' $/2}% - {$-\infty$, $+\infty$} - \tkzTabLine{, -,} - \end{tikzpicture} - \end{center} + Comme $a = -30$, on en déduit le tableau de signe de $g'$ + %\begin{center} + % \begin{tikzpicture} + % \tkzTabInit[espcl=2]% + % {$x$/1, Signe de $g' $/2}% + % {$-\infty$, - \frac{1}{5} + \frac{\sqrt{69}}{15} , - \frac{\sqrt{69}}{15} - \frac{1}{5} , $+\infty$} + % \tkzTabLine{, -, z, +, z , -,} + % \end{tikzpicture} + %\end{center} \end{solution} - \part $h:x\mapsto - 7 x^{ 2 } - 9 x + 3 - f(x)$ + \part $h:x\mapsto - 7 x^{ 2 } - 5 x - 5 - f(x)$ @@ -237,51 +237,51 @@ g'(x) & = & - 27 x^{ 2 } - 16 x - 5 \begin{solution} On commence par simplifier l'expression de $h$ \begin{eqnarray*} - h(x) & = & - 7 x^{ 2 } - 9 x + 3 - f(x) \\ - h(x) & = & - 7 x^{ 2 } - 9 x + 3 - ( - 10 x^{ 3 } + x^{ 2 } - 7 x + 5 ) \\ -h(x) & = & - 7 x^{ 2 } - 9 x + 3 + 10 x^{ 3 } - x^{ 2 } + 7 x - 5 \\ -h(x) & = & 10 x^{ 3 } + ( ( -7 ) + ( -1 ) ) x^{ 2 } + ( ( -9 ) + 7 ) x + 3 + ( -5 ) \\ -h(x) & = & 10 x^{ 3 } - 8 x^{ 2 } - 2 x - 2 + h(x) & = & - 7 x^{ 2 } - 5 x - 5 - f(x) \\ + h(x) & = & - 7 x^{ 2 } - 5 x - 5 - ( - 2 x^{ 3 } - 4 x^{ 2 } + x + 8 ) \\ +h(x) & = & - 7 x^{ 2 } - 5 x - 5 + 2 x^{ 3 } + 4 x^{ 2 } - x - 8 \\ +h(x) & = & 2 x^{ 3 } + ( -7 + 4 ) x^{ 2 } + ( -5 - 1 ) x - 5 - 8 \\ +h(x) & = & 2 x^{ 3 } - 3 x^{ 2 } - 6 x - 13 \end{eqnarray*} Pour avoir les variations de $h$, il faut connaître le signe de sa dérivé. On dérive $P$ \begin{eqnarray*} - h'(x) & = & 3 \times 10 x^{ 2 } + 2 ( -8 ) x + 1 ( -2 ) \\ -h'(x) & = & 30 x^{ 2 } - 16 x - 2 + h'(x) & = & 3 \times 2 x^{ 2 } + 2 \times ( -3 ) x + 1 \times ( -6 ) \\ +h'(x) & = & 6 x^{ 2 } - 6 x - 6 \end{eqnarray*} On étudie le signe de $P'$ - Ensuite on étudie le signe de $h'(x) = 30 x^{ 2 } - 16 x - 2$. + Ensuite on étudie le signe de $h'(x) = 6 x^{ 2 } - 6 x - 6$. \begin{eqnarray*} \Delta & = & b^2-4ac \\ - \Delta & = & ( -16 )^{ 2 } - 4 \times 30 ( -2 ) \\ -\Delta & = & 256 - 4 ( -60 ) \\ -\Delta & = & 256 - ( -240 ) \\ -\Delta & = & 496 + \Delta & = & -6^{ 2 } - 4 \times 6 \times ( -6 ) \\ +\Delta & = & 36 - 4 \times ( -36 ) \\ +\Delta & = & 36 - ( -144 ) \\ +\Delta & = & 180 \end{eqnarray*} - comme $\Delta = 496 > 0$ donc $h'$ a deux racines + comme $\Delta = 180 > 0$ donc $h'$ a deux racines \begin{eqnarray*} - x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{-16 - \sqrt{496}}{2 \times 30} = -0.1 \\ - x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{-16 + \sqrt{496}}{2 \times 30} = 0.64 + x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{-6 - \sqrt{180}}{2 \times 6} = - \frac{\sqrt{5}}{2} + \frac{1}{2} \\ + x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{-6 + \sqrt{180}}{2 \times 6} = \frac{1}{2} + \frac{\sqrt{5}}{2} \end{eqnarray*} - Comme $a = 30$, on en déduit le tableau de signe de $h'$ - \begin{center} - \begin{tikzpicture} - \tkzTabInit[espcl=2]% - {$x$/1, Signe de $h' $/2}% - {$-\infty$, -0.1 , 0.64 , $+\infty$} - \tkzTabLine{, +, z, -, z , +,} - \end{tikzpicture} - \end{center} + Comme $a = 6$, on en déduit le tableau de signe de $h'$ + %\begin{center} + % \begin{tikzpicture} + % \tkzTabInit[espcl=2]% + % {$x$/1, Signe de $h' $/2}% + % {$-\infty$, - \frac{\sqrt{5}}{2} + \frac{1}{2} , \frac{1}{2} + \frac{\sqrt{5}}{2} , $+\infty$} + % \tkzTabLine{, +, z, -, z , +,} + % \end{tikzpicture} + %\end{center} \end{solution} \end{parts} diff --git a/example/1_example.tex b/example/1_example.tex deleted file mode 100644 index 43c7b28..0000000 --- a/example/1_example.tex +++ /dev/null @@ -1,114 +0,0 @@ -\documentclass[a4paper,10pt]{article} -\RequirePackage[utf8x]{inputenc} -\RequirePackage[francais]{babel} -\RequirePackage{amssymb} -\RequirePackage{amsmath} -\RequirePackage{amsfonts} -\RequirePackage{subfig} -\RequirePackage{graphicx} -\RequirePackage{color} - -% Title Page -\title{Calcul littéral et statistiques} -\date{\today} - -\begin{document} -\maketitle - - -\section{Polynômes} - - - - Résoudre l'équation suivante - \begin{eqnarray*} - - 3 x^{ 2 } + 6 x - 3 & = & 0 - \end{eqnarray*} - - Solution: - - On commence par calculer le discriminant - - \begin{eqnarray*} - \Delta & = & b^2-4ac \\ - \Delta & = & 6^{ 2 } - 4 \times ( -3 ) \times ( -3 ) \\ -\Delta & = & 36 - ( -12 ) \times ( -3 ) \\ -\Delta & = & 36 - 36 \\ -\Delta & = & 0 - \end{eqnarray*} - - - - Alors $\Delta = 0 = 0$ donc il y a une solution - - - - \begin{eqnarray*} - x_1 = \frac{-b}{2a} = \frac{ -6 }{ 2 \times ( -3 ) } = \frac{ -6 }{ -6 } = \frac{ 6 }{ 6 } = 1 = \frac{ -6 }{ -6 } - \end{eqnarray*} - - Les solutions sont donc $\mathcal{S} = \left\{ \frac{ -6 }{ -6 }\right\}$ - - - - \bigskip - ~\dotfill - \bigskip - - - - - Résoudre l'équation suivante - \begin{eqnarray*} - - 7 x^{ 2 } - 7 x + 9 & = & - 2 x^{ 2 } + x - 9 - \end{eqnarray*} - - Solution: - - On commence par se ramener à une équation de la forme $ax^2+bx+c = 0$. - - \begin{eqnarray*} - - 7 x^{ 2 } - 7 x + 9 = - 2 x^{ 2 } + x - 9 & \Leftrightarrow & - 7 x^{ 2 } - 7 x + 9 - (- 2 x^{ 2 } + x - 9) = 0 \\ - & \Leftrightarrow & - 7 x^{ 2 } + 2 x^{ 2 } - 7 x - x + 9 + 9= 0 \\ - & \Leftrightarrow & ( ( -7 ) + 2 ) x^{ 2 } + ( ( -7 ) + ( -1 ) ) x + 9 + 9= 0 \\ - & \Leftrightarrow & - 5 x^{ 2 } - 8 x + 18= 0 - \end{eqnarray*} - - - On cherche maintenant à résoudre l'équation $- 5 x^{ 2 } - 8 x + 18 = 0$. - - On commence par calculer le discriminant - - \begin{eqnarray*} - \Delta & = & b^2-4ac \\ - \Delta & = & ( -8 )^{ 2 } - 4 \times ( -5 ) \times 18 \\ -\Delta & = & 64 - ( -20 ) \times 18 \\ -\Delta & = & 64 - ( -360 ) \\ -\Delta & = & 424 - \end{eqnarray*} - - - - Alors $\Delta = 424 > 0$ donc il y a deux solutions - - - - - \begin{eqnarray*} - x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{-8 - \sqrt{424}}{2 \times -5} = 1.26 \\ - x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{-8 + \sqrt{424}}{2 \times -5} = -2.86 - \end{eqnarray*} - - Les solutions sont donc $\mathcal{S} = \left\{ 1.26; -2.86 \right\}$ - - - - -\end{document} - -%%% Local Variables: -%%% mode: latex -%%% TeX-master: "master" -%%% End: - - \ No newline at end of file diff --git a/example/1_play.tex b/example/1_play.tex deleted file mode 100644 index e69de29..0000000 diff --git a/example/all_corr_DM_0302.pdf b/example/all_corr_DM_0302.pdf index b9d634dc5b7ba569aac5c2b06ef590a09f073435..01cb9897c48f3995c138a0879af69bc97d0420ee 100644 GIT binary patch delta 51616 zcmZ^pV{~2Zy0BxcI1L-ycB96&)7Z8(8?$j^+g4-Sw#^0&8hw4=v-cVAch1@Wp0UPU 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