File: /home/mmtprep/public_html/mathzen.mmtprep.com/assets/2N60-1-QDcgYVEL.js.map
{"version":3,"file":"2N60-1-QDcgYVEL.js","sources":["../../src/exercices/2e/2N60-1.js"],"sourcesContent":["import { codageAngleDroit } from '../../lib/2d/angles.js'\nimport { milieu, point } from '../../lib/2d/points.js'\nimport { polygone } from '../../lib/2d/polygones.js'\nimport { segment } from '../../lib/2d/segmentsVecteurs.js'\nimport { labelPoint, texteParPosition } from '../../lib/2d/textes.js'\nimport { choice, combinaisonListes } from '../../lib/outils/arrayOutils'\nimport { miseEnEvidence } from '../../lib/outils/embellissements'\nimport { ecritureAlgebrique, ecritureParentheseSiNegatif, reduireAxPlusB, rienSi1 } from '../../lib/outils/ecritures'\nimport { abs } from '../../lib/outils/nombres'\nimport { sp } from '../../lib/outils/outilString.js'\nimport { prenomF } from '../../lib/outils/Personne'\nimport { pgcd } from '../../lib/outils/primalite'\nimport { texPrix, texteGras } from '../../lib/format/style'\nimport { texNombre } from '../../lib/outils/texNombre'\nimport Exercice from '../deprecatedExercice.js'\nimport Decimal from 'decimal.js'\nimport FractionEtendue from '../../modules/FractionEtendue.js'\nimport { mathalea2d, colorToLatexOrHTML } from '../../modules/2dGeneralites.js'\nimport { listeQuestionsToContenu, itemize, randint } from '../../modules/outils.js'\nexport const titre = 'Model a problem using an inequality.'\nexport const dateDePublication = '14/02/2023'\n/**\n * Description didactique de l'exercice\n * @author Gilles Mora\n * Référence\n*/\nexport const uuid = 'd2084'\nexport const ref = '2N60-1'\nexport default function ModeliseInequations () {\n Exercice.call(this) // Héritage de la classe Exercice()\n this.consigne = ''\n this.nbQuestions = 1\n this.nbQuestionsModifiable = true\n this.nbCols = 1 // Uniquement pour la sortie LaTeX\n this.nbColsCorr = 1 // Uniquement pour la sortie LaTeX\n this.sup = 4\n this.tailleDiaporama = 2 // Pour les exercices chronométrés. 50 par défaut pour les exercices avec du texte\n this.spacing = 1.5 // Interligne des questions\n this.spacingCorr = 1.5// Interligne des réponses\n this.nouvelleVersion = function () {\n this.listeQuestions = [] // Liste de questions\n this.listeCorrections = [] // Liste de questions corrigées\n let typeDeQuestionsDisponibles\n if (this.sup === 1) {\n typeDeQuestionsDisponibles = ['typeE1', 'typeE2', 'typeE3']//\n } else if (this.sup === 2) {\n typeDeQuestionsDisponibles = ['typeE4', 'typeE5', 'typeE6']//\n } else if (this.sup === 3) {\n typeDeQuestionsDisponibles = ['typeE7', 'typeE8']\n } else if (this.sup === 4) {\n typeDeQuestionsDisponibles = ['typeE1', 'typeE2', 'typeE3', 'typeE4', 'typeE5', 'typeE6', 'typeE7', 'typeE8']\n }\n //\n const listeTypeQuestions = combinaisonListes(typeDeQuestionsDisponibles, this.nbQuestions) // Tous les types de questions sont posés mais l'ordre diffère à chaque 'cycle'\n for (let i = 0, texte, texteCorr, cpt = 0; i < this.nbQuestions && cpt < 50;) {\n // Main loop where i+1 corresponds to the question number\n switch (listeTypeQuestions[i]) { // Suivant le type de question, le contenu sera différent\n case 'typeE1'://\n { const a = randint(20, 30) //\n const b = randint(a + 5, 50) //\n const c = new Decimal(randint(20, 35)).div(100)\n const d = new Decimal(randint(14, 19)).div(100)\n texte = ` A private vehicle rental company offers two rates:<br>$\\\\bullet$ Rate A: a fixed price of $\\\\$${a}$ and $\\\\$${texNombre(c, 2)}$ per km traveled;<br>$\\\\bullet$ Rate B: one flat rate of $\\\\$${b}$ and $\\\\$${texNombre(d, 2)}$ per km traveled;<br>From how many km (rounded to the nearest unit), is rate B more attractive than rate A?<br>`\n texteCorr = `By noting $x$, the number of km traveled, we have:<br>$\\\\bullet$ With rate A, the price to pay is: $${reduireAxPlusB(c, a)}$;<br>$\\\\bullet$ With rate B , the price to pay is: $${reduireAxPlusB(d, b)}$;<br>Price B is more advantageous than price A when $x$ satisfies: $${reduireAxPlusB(d, b)} < ${reduireAxPlusB(c, a)}$.<br>We solve this inequality: <br>$\\\\begin{aligned}${texNombre(d, 2)}x+${b}&<${texNombre(c, 2)}x+${a}\\\\\\\\${texNombre(d, 2)}x+${b}-${miseEnEvidence(texNombre(c, 2))}${miseEnEvidence('\\\\textit{x}')}&< ${texNombre(c, 2)}x-${miseEnEvidence(texNombre(c, 2))}${miseEnEvidence('\\\\textit{x}')}+${a}\\\\\\\\${texNombre(d - c)}x+${b}&<${a}\\\\\\\\${texNombre(d - c, 2)}x+${b}- ${miseEnEvidence(texNombre(b))}&<${a}-${b}\\\\\\\\${texNombre(d - c, 2)}x&<${a - b}\\\\\\\\\\\\dfrac{${texNombre(d - c, 2)}x}{${miseEnEvidence(texNombre(d - c, 2))}}&>\\\\dfrac{${a - b}}{${miseEnEvidence(texNombre(d - c, 2))}}${sp(7)} \\\\text{We divide by } ${texNombre(d - c, 2)} <0\\\\\\\\x&>\\\\dfrac{${abs(a - b)}}{${texNombre(abs(d - c), 2)}}\\\\end{aligned}$<br>`\n if (Math.round((a - b) / (d - c)) === (a - b) / (d - c)) {\n texteCorr += `As $\\\\dfrac{${abs(a - b)}}{${texNombre(abs(d - c), 2)}}= ${texNombre((a - b) / (d - c), 2)}$, it is therefore for a minimum distance of $${texNombre(Math.ceil((a - b) / (d - c)) + 1, 0)}$ km that rate B is more interesting than rate A. `\n } else {\n texteCorr += ` Like $\\\\dfrac{${abs(a - b)}}{${texNombre(abs(d - c), 2)}}\\\\simeq ${texNombre((a - b) / (d - c), 2)}$, it is therefore for a minimum distance of $${Math.ceil((a - b) / (d - c))}$ km that rate B is more interesting than rate A. `\n }\n }\n break\n\n case 'typeE2':\n {\n const quidam = prenomF()\n const b = randint(90, 120) //\n const a = new Decimal(randint(15, 25)).div(100)\n const budget = randint(20, 35) * 10 //\n\n texte = ` For the monthly rental of a vehicle, a company offers the following rate:<br>Flat rate of $\\\\$${b}$ regardless of the number of km traveled, then a supplement per kilometer traveled of $\\\\$${texNombre(a, 2)}$. <br>${quidam} rents a car from this company. She has a budget of $\\\\$${budget}$ and does not want to exceed it.<br>What is the maximum number of km she can travel without exceeding her budget?`\n texteCorr = `By noting $x$, the number of km traveled, the cost for the monthly rental is given by: $${reduireAxPlusB(a, b)}$.<br>The ${quidam} budget being $\\\\$${budget}$, the number of km $x$ that it will be able to browse must check $${reduireAxPlusB(a, b)}<${budget}$.<br>$\\\\begin{aligned}${reduireAxPlusB(a, b)}&<${budget}\\\\\\\\${texNombre(a, 2)}x+${b}-${miseEnEvidence(b)}&< ${budget}x-${miseEnEvidence(b)}\\\\\\\\${texNombre(a, 2)}x&<${budget - b}\\\\\\\\x&<\\\\dfrac{${budget - b}}{${texNombre(a, 2)}}\\\\end{aligned}$<br>`\n\n texteCorr += `Like $\\\\dfrac{${budget - b}}{${texNombre(a, 2)}}${Math.round((budget - b) / a) === (budget - b) / a ? '=' : '\\\\simeq'} ${texNombre((budget - b) / a, 2)}$, ${quidam} will be able to do a maximum of $${Math.floor((budget - b) / a)}$ km during the month with its budget of $\\\\$${budget}$. `\n }\n break\n case 'typeE3':\n { const PB = new Decimal(randint(7, 25, [10, 20])).div(2)// prix billet\n const EM = randint(70, 150) // nombre entrée matin\n const RT = randint(200, 400) * 10 // recette totale\n texte = `At midday the revenue of a museum amounts to $\\\\$${texNombre(PB * EM, 2)}$ for $${EM}$ entries. The entry price is unique.<br>What must be the minimum number of entries in the second half of the day so that the day's revenue is at least equal to $\\\\$${texNombre(RT)}$ ?<br>Solve this problem by writing and solving an inequality modeling the situation. `\n texteCorr = `The amount of the entry ticket is given by $\\\\$${texNombre(PB * EM, 2)}${sp(1)} \\\\div ${EM}=\\\\$${texPrix(PB, 2)}$ .<br>By noting $x$ the number of entries in the second part of the day, we obtain: $${texNombre(PB * EM, 2)} +${texNombre(PB, 2)}\\\\times x\\\\geqslant ${texNombre(RT)}$.<br>$\\\\begin{aligned}${texNombre(PB * EM, 2)} +${texNombre(PB, 2)} x&\\\\geqslant ${texNombre(RT)}\\\\\\\\${texNombre(PB * EM, 2)} +${texNombre(PB, 2)}x-${miseEnEvidence(texNombre(PB * EM, 2))}&\\\\geqslant ${texNombre(RT)}-${miseEnEvidence(texNombre(PB * EM, 2))}\\\\\\\\${texNombre(PB, 2)}x&\\\\geqslant ${texNombre(RT - PB * EM, 2)} \\\\\\\\x&\\\\geqslant \\\\dfrac{${texNombre(RT - PB * EM, 2)}}{${texNombre(PB, 2)}}\\\\\\\\\\\\end{aligned}$<br>Like $\\\\dfrac{${texNombre(RT - PB * EM, 2)}}{${texNombre(PB, 2)}}${Math.round((RT - PB * EM) / PB) === (RT - PB * EM) / PB ? '=' : '\\\\simeq'} ${texNombre((RT - PB * EM) / PB, 1)}$, it will be necessary at least ${Math.round((RT - PB * EM) / PB) === (RT - PB * EM) / PB ? `$${texNombre((RT - PB * EM) / PB, 0)}$` : `$${texNombre((RT - PB * EM) / PB + 1, 0)}$`} entries so that the day's revenue is at least equal to $\\\\$${texNombre(RT)}$. `\n }\n break\n\n case 'typeE4':\n\n {\n const choix = choice([true, false])\n const l = randint(3, 10)// largeur\n const L = l + randint(3, 10)// longueur\n\n const P = choice([['to the third', 3], ['quarter', 4], ['At the half of it or in the middle', 2], ['to the tenth', 10], ['on the fifth', 5]])\n const f = new FractionEtendue(L * l / 2, P[1] * l / 2 + l / 2).simplifie()\n const f2 = new FractionEtendue(l * L, l * P[1] + l).simplifie()\n const A = point(0, 0, 'A', 'below')\n const B = point(10, 0, 'B', 'below')\n const C = point(10, 6, 'C')\n const M = point(4, 0, 'M', 'below')\n const D = point(0, 6, 'D')\n const objets = []\n const poly1 = polygone([A, M, D], 'black')\n const poly2 = polygone([C, M, B], 'black')\n poly1.couleurDeRemplissage = colorToLatexOrHTML('lightgray')\n poly2.couleurDeRemplissage = colorToLatexOrHTML('lightgray')\n objets.push(segment(A, B), segment(B, C), segment(D, A), segment(C, D), labelPoint(A, B, C, D, M), poly1, poly2)\n objets.push(texteParPosition('x', milieu(A, M).x, milieu(A, M).y - 0.7, 'medium', 'black', 1, 'middle', true),\n texteParPosition(`${texNombre(l)}`, milieu(A, D).x - 0.5, milieu(A, D).y, 'medium', 'black', 1, 'middle', true),\n texteParPosition(`${texNombre(L)}`, milieu(C, D).x, milieu(C, D).y + 0.5, 'medium', 'black', 1, 'middle', true))\n\n texte = ` Let $ABCD$ be a rectangle such that $AD=${l}$ and $DC=${L}$.<br>$M$ is a point of the segment $[AB]$. We note $AM=x$.<br>For what values of $x$ is the area of the triangle $AMD$ ${choix ? 'at most' : 'at least'} equal to ${P[0]} of the area of the triangle $CMB$?<br>`\n texte += mathalea2d({ xmin: -1, ymin: -1, xmax: 12, ymax: 8, pixelsParCm: 20, mainlevee: false, amplitude: 0.5, scale: 0.5, style: 'margin: self' }, objets)\n texteCorr = ` The area of triangle $AMD$ is: ${l % 2 === 0 ? `$\\\\dfrac{x\\\\times ${l}}{2}=${texNombre(l / 2, 0)}x$` : `$\\\\dfrac{x\\\\times ${l}}{2}$`}. <br>As $MB=${L}-x$, the area of triangle $CMB$ is: ${l % 2 === 0 ? `$\\\\dfrac{(${L}-x)\\\\times ${l}}{2}=${texNombre(l / 2, 0)}(${L}-x)$` : `$\\\\dfrac{ (${L}-x)\\\\times ${l}}{2}$`}. <br>The problem therefore comes down to determining the values of $x$ such that: $${l % 2 === 0 ? `${texNombre(l / 2, 0)}x` : `\\\\dfrac{${l}x}{2}`} ${choix ? '\\\\leqslant' : '\\\\geqslant'} ${l % 2 === 0 ? `\\\\dfrac{1}{${P[1]}}\\\\times ${texNombre(l / 2, 0)}(${L}-x)` : `\\\\dfrac{1}{${P[1]}}\\\\times \\\\dfrac{${l}(${L}-x)}{2}`}$. <br>`\n if (l % 2 === 0) {\n texteCorr += `$\\\\begin{aligned}${texNombre(l / 2, 0)}x &${choix ? '\\\\leqslant' : '\\\\geqslant'} \\\\dfrac{${texNombre(l / 2, 0)}(${L}-x)}{${P[1]}}${sp(7)}\\\\\\\\`\n texteCorr += `${texNombre(l / 2, 0)}x \\\\times ${P[1]}&${choix ? '\\\\leqslant' : '\\\\geqslant'} \\\\dfrac{${texNombre(l / 2, 0)}(${L}-x)}{${P[1]}}\\\\times ${P[1]} ${sp(7)}\\\\text{ We multiply by ${P[1]}, the direction of the inequality does not change.}\\\\\\\\`\n texteCorr += ` ${texNombre(P[1] * l / 2, 0)}x &${choix ? '\\\\leqslant' : '\\\\geqslant'} ${texNombre(l / 2, 0)}(${L}-x)\\\\\\\\`\n texteCorr += ` ${texNombre(P[1] * l / 2, 0)}x &${choix ? '\\\\leqslant' : '\\\\geqslant'} ${texNombre(L * l / 2, 0)}-${texNombre(l / 2, 0)}x\\\\\\\\`\n texteCorr += ` ${texNombre(P[1] * l / 2, 0)}x +${texNombre(l / 2, 0)}x&${choix ? '\\\\leqslant' : '\\\\geqslant'} ${texNombre(L * l / 2, 0)}\\\\\\\\`\n texteCorr += ` ${texNombre(P[1] * l / 2 + l / 2, 0)}x &${choix ? '\\\\leqslant' : '\\\\geqslant'} ${texNombre(L * l / 2, 0)}\\\\\\\\`\n texteCorr += ` x &${choix ? '\\\\leqslant' : '\\\\geqslant'} ${f.texFraction}\\\\\\\\`\n\n texteCorr += '\\\\end{aligned}$<br>'\n texteCorr += `The area of triangle $AMD$ is ${choix ? 'at most' : 'at least'} equal to ${P[0]} of the area of triangle $CMB$ for $x\\\\in ${choix ? `\\\\left[0\\\\,,\\\\,${f.texFraction}\\\\right]` : `\\\\left[${f.texFraction}\\\\,,\\\\,${L}\\\\right]`}$`\n } else {\n texteCorr += `$\\\\begin{aligned}\\\\dfrac{${l} x}{2} &${choix ? '\\\\leqslant' : '\\\\geqslant'} \\\\dfrac{${l}(${L}-x)}{${2 * P[1]}}\\\\\\\\`\n texteCorr += `\\\\dfrac{${l} x}{2} \\\\times ${2 * P[1]}&${choix ? '\\\\leqslant' : '\\\\geqslant'} \\\\dfrac{${l}(${L}-x)}{${texNombre(P[1] * 2, 0)}}\\\\times ${2 * P[1]} ${sp(7)}\\\\text{ We multiply by ${2 * P[1]}, the direction of the inequalities does not change. }\\\\\\\\`\n texteCorr += ` ${texNombre(l * P[1], 0)}x &${choix ? '\\\\leqslant' : '\\\\geqslant'} ${l}(${L}-x)\\\\\\\\`\n texteCorr += ` ${texNombre(l * P[1], 0)}x &${choix ? '\\\\leqslant' : '\\\\geqslant'} ${l * L}-${l}x\\\\\\\\`\n texteCorr += ` ${texNombre(l * P[1], 0)}x +${l}x&${choix ? '\\\\leqslant' : '\\\\geqslant'} ${l * L}\\\\\\\\`\n texteCorr += ` ${texNombre(l * P[1] + l, 0)}x &${choix ? '\\\\leqslant' : '\\\\geqslant'} ${l * L}\\\\\\\\`\n texteCorr += ` x &${choix ? '\\\\leqslant' : '\\\\geqslant'} ${f2.texFraction}\\\\\\\\`\n\n texteCorr += '\\\\end{aligned}$<br>'\n texteCorr += `The area of triangle $AMD$ is ${choix ? 'at most' : 'at least'} equal to ${P[0]} of the area of triangle $CMB$ for $x\\\\in ${choix ? `\\\\left[0\\\\,,\\\\,${f.texFraction}\\\\right]` : `\\\\left[${f.texFraction}\\\\,,\\\\,${L}\\\\right]`}$`\n }\n }\n break\n\n case 'typeE5':\n\n {\n const a = randint(8, 15)// longueur en bas\n const b = randint(2, 6)// longueur ajoutée à x\n const P = randint(50, 70)\n const A = point(0, 0, 'A')\n const B = point(10, 0, 'B')\n const C = point(10, 6, 'C')\n const D = point(0, 6, 'D')\n const E = point(0, 2, 'E')\n const F = point(-2, 2, 'F')\n const G = point(-2, 0, 'G')\n const segmentEA = segment(A, E)\n segmentEA.pointilles = 2\n const objets = []\n const poly = polygone([A, B, C, D, E, F, G], 'black')\n\n poly.couleurDeRemplissage = colorToLatexOrHTML('lightgray')\n\n objets.push(poly, segmentEA)\n objets.push(texteParPosition('x', milieu(G, F).x - 0.5, milieu(G, F).y, 'medium', 'black', 1, 'middle', true),\n texteParPosition('x', milieu(G, A).x, milieu(G, A).y - 0.5, 'medium', 'black', 1, 'middle', true),\n texteParPosition(`x+${texNombre(b)}`, milieu(B, C).x + 1, milieu(B, C).y, 'medium', 'black', 1, 'middle', true),\n texteParPosition(`${texNombre(a)}`, milieu(A, B).x, milieu(A, B).y - 0.5, 'medium', 'black', 1, 'middle', true)\n )\n\n texte = ` Consider the figure below (the unit is centimeters). <br>What are the possible values of $x$ so that the perimeter of the figure is greater than $${P}$ cm.<br>`\n texte += mathalea2d({ xmin: -3, ymin: -1, xmax: 12, ymax: 8, pixelsParCm: 20, mainlevee: false, amplitude: 0.5, scale: 0.5, style: 'margin: self' }, objets)\n texteCorr = `The perimeter of the figure is: $x+${a}+(x+${b})+${a}+${b}+x+x=4x+${2 * b + 2 * a}$.<br>The perimeter of the figure must be greater than $${P}$, we are looking for $x $ such as: <br>`\n texteCorr += `$\\\\begin{aligned}4x+${2 * b + 2 * a} &>${P}\\\\\\\\4x &>${P}-${2 * b + 2 * a}\\\\\\\\4x&>${P - 2 * b - 2 * a}\\\\\\\\x&>\\\\dfrac{${P - 2 * b - 2 * a}}{4}`\n texteCorr += '\\\\end{aligned}$<br>'\n\n texteCorr += `Since $\\\\dfrac{${P - 2 * b - 2 * a}}{4}=${texNombre((P - 2 * b - 2 * a) / 4, 2)}$, $x$ must be greater than $${texNombre((P - 2 * b - 2 * a) / 4, 2)}$ cm for the perimeter of the figure to be greater than $${P}$ cm. `\n }\n break\n\n case 'typeE6':\n\n {\n const a = randint(1, 3)// côté carré\n const b = choice([6, 8, 10, 12])// hauteur triangle\n const Aire = randint(50, 70)\n const f = new FractionEtendue(Aire - a ** 2, a + b / 2).simplifie()\n const A = point(0, 0, 'A')\n const B = point(8, 0, 'B')\n const C = point(10, 0, 'C')\n const D = point(10, 2, 'D')\n const E = point(8, 2, 'E')\n const F = point(4, 8, 'F')\n const G = point(0, 2, 'G')\n const H = point(4, 2, 'H')\n const segmentFH = segment(F, H)\n segmentFH.pointilles = 2\n const objets = []\n const poly1 = polygone([A, B, E, G], 'black')\n poly1.couleurDeRemplissage = colorToLatexOrHTML('#7dbdd8')\n const poly2 = polygone([B, C, D, E], 'black')\n poly2.couleurDeRemplissage = colorToLatexOrHTML('#e1ac66')\n const poly3 = polygone([G, E, F], 'black')\n poly3.couleurDeRemplissage = colorToLatexOrHTML('#75ee7e')\n objets.push(poly1, poly2, poly3, segmentFH, codageAngleDroit(F, H, E))\n objets.push(texteParPosition('x', milieu(A, B).x, milieu(A, B).y - 0.7, 'medium', 'black', 1, 'middle', true),\n texteParPosition(`${texNombre(a)}`, milieu(G, A).x - 0.8, milieu(G, A).y, 'medium', 'black', 1, 'middle', true),\n texteParPosition(`${texNombre(a)}`, milieu(B, C).x, milieu(B, C).y - 0.7, 'medium', 'black', 1, 'middle', true),\n texteParPosition(`${texNombre(b)}`, milieu(F, H).x + 0.5, milieu(F, H).y, 'medium', 'black', 1, 'middle', true)\n )\n\n texte = ` Consider the figure below in which the lengths are in cm. <br>What are the possible values of $x$ so that the area of this figure exceeds $${Aire}$ cm$^2$?<br>Solve this problem by modeling it with an inequality.<br>`\n texte += mathalea2d({ xmin: -3, ymin: -1, xmax: 12, ymax: 9, pixelsParCm: 20, mainlevee: false, amplitude: 0.5, scale: 0.5, style: 'margin: self' }, objets)\n texteCorr = `The figure consists of a rectangle, a square and a triangle.<br>$\\\\bullet$ The area of the rectangle is: $${a}\\\\times x=${rienSi1(a)}x$.<br>$\\\\ bullet$ The area of the square is: $${a}\\\\times ${a}=${a ** 2}$.<br>$\\\\bullet$ The area of the triangle is: $\\\\dfrac{${b}\\\\times x}{2}=${texNombre(b / 2, 0)}x $.<br>The problem therefore amounts to finding the values of $x$ verifying: $${rienSi1(a)}x+${a ** 2}+${texNombre(b / 2, 0)}x>${Aire}$, i.e. $${texNombre(a + b / 2, 0)}x+${a ** 2}>${Aire}$.<br>`\n texteCorr += `$\\\\begin{aligned}${texNombre(a + b / 2, 0)}x+${a ** 2} &>${Aire}\\\\\\\\${texNombre(a + b / 2, 0)}x &>${Aire}-${a ** 2}\\\\\\\\${texNombre(a + b / 2, 0)}x &>${Aire - a ** 2}\\\\\\\\x&>\\\\dfrac{${Aire - a ** 2}}{${texNombre(a + b / 2, 0)}}`\n texteCorr += '\\\\end{aligned}$<br>'\n if (pgcd(Aire - a ** 2, a + b / 2) === 1) {\n texteCorr += `$x$ must be greater than $\\\\dfrac{${Aire - a ** 2}}{${texNombre(a + b / 2, 0)}}$ cm for the area of the figure to exceed $${Aire}$ cm$^2$. `\n } else {\n texteCorr += `Since $\\\\dfrac{${Aire - a ** 2}}{${texNombre(a + b / 2, 0)}}=${f.texFraction}$, $x$ must be greater than $${f.texFraction}$ cm for the area of the figure to exceed $${Aire}$ cm$^2$. `\n }\n }\n break\n\n case 'typeE7':\n {\n const a = randint(-10, 10, [-1, 0, 1])\n const b = randint(-10, 10, [-1, 0, 1])\n const c = randint(2, 10)\n const res = randint(-20, 20, 0)\n const f = new FractionEtendue(res - b * c, c * a).simplifie()\n const choix = choice([['strictly superior', '>', '<'], ['strictly inferior', '<', '>'], ['less or equal', '\\\\leqslant', '\\\\geqslant'], ['greater than or equal to', '\\\\geqslant', '\\\\leqslant']])\n texte = ` ${texteGras('Here is a calculation program:')}`\n texte += itemize(['Choose a number', `Multiply this number by $${a}$`, `Add $${b}$`, `Multiply the result by $${c}$`])\n texte += `What numbers should we choose initially to obtain a ${choix[0]} number at $${res}$.<br>`\n\n texteCorr = `By noting $x$ the number chosen at the start, we obtain:<br>$\\\\bullet$ Multiply this number by $${a}$: ${sp(6)}$${a}\\\\times x=${a}x$ ;<br>$\\\\bullet$ Add $${b}$: ${sp(6)}$${a}x${ecritureAlgebrique(b)}$; <br>$\\\\bullet$ Multiply the result by $${c}$:${sp(6)}$${c}\\\\times (${a}x${ecritureAlgebrique(b)})=${texNombre(c * a)}x${ecritureAlgebrique(b * c)}$.<br>We are looking for $x$ such that: <br>`\n texteCorr += `$\\\\begin{aligned}${texNombre(c * a)}x${ecritureAlgebrique(b * c)} &${choix[1]}${res}\\\\\\\\${texNombre(c * a)}x &${choix[1]}${res}${ecritureAlgebrique(-b * c)}\\\\\\\\`\n texteCorr += ` x &${c * a > 0 ? `${choix[1]}` : `${choix[2]}`}\\\\dfrac{${res - b * c}}{${texNombre(c * a)}}\\\\\\\\`\n texteCorr += '\\\\end{aligned}$<br>'\n if (pgcd(res - b * c, c * a) === 1) {\n texteCorr += `We must choose $x${c * a > 0 ? `${choix[1]}` : `${choix[2]}`}${f.texFraction}$ to obtain a ${choix[0]} number at $${res}$. . `\n } else {\n texteCorr += `Since $\\\\dfrac{${res - b * c}}{${texNombre(c * a)}}=${f.texFraction}$, we must choose $x${c * a > 0 ? `${choix[1]}` : `${choix[2]}`}${f.texFraction}$ to obtain a ${choix[0]} number at $${res}$. `\n }\n }\n break\n\n case 'typeE8':\n {\n const a = randint(-10, 10, 0)\n\n const b = randint(-10, 10, 0)\n const f = new FractionEtendue(b * b, a - 2 * b).simplifie()\n const choix = choice([['strictly superior', '>', '<'], ['strictly inferior', '<', '>'], ['less or equal', '\\\\leqslant', '\\\\geqslant'], ['greater than or equal to', '\\\\geqslant', '\\\\leqslant']])\n texte = `We give the following two calculation programs:<br>${texteGras('Program 1:')}<br>`\n texte += itemize(['Choose a number', `Add $${a}$`, 'Multiply the result by the number chosen at the start'])\n texte += `<br>${texteGras('Program 2:')}<br>`\n texte += itemize(['Choose a number', `Add $${b}$`, 'Take the square of the result'])\n texte += `<br>Determine the numbers that must be entered into these two programs so that in the end the result obtained with program 1 is ${choix[0]} to that obtained with program 2.<br><br>`\n texteCorr = `By noting $x$ the number chosen at the start: <br>We obtain with the ${texteGras('program 1:')} <br>$\\\\bullet$ Add $${a}$: ${sp(5)} $x+${ecritureParentheseSiNegatif(a)}$ ;<br>$\\\\bullet$ Multiply the result by the number chosen at the start: ${sp(5)} $x\\\\times(x${ecritureAlgebrique(a)})=x^2${a > 0 ? '+' : '-'}${rienSi1(abs(a))}x$.<br>We obtain with the ${texteGras('program 2:')} <br>$\\\\bullet$ Add $${b}$:${sp(5)} $x+${ecritureParentheseSiNegatif(b)}$ ; <br>$\\\\bullet$ Take the square of the result:${sp(5)} $(x${ecritureAlgebrique(b)})^2=x^2${ecritureAlgebrique(2 * b)}x+${(b * b)}$.<br>The numbers $x$ that must be entered in the two programs so that in the end the result obtained with program 1 is ${choix[0]} to that obtained with program 2 verify: <br>$\\\\begin{aligned}x^2${a > 0 ? '+' : '-'}${rienSi1(abs(a))}x & ${choix[1]} x^2${ecritureAlgebrique(2 * b)}x+${b * b}\\\\\\\\${rienSi1(a)}x & ${choix[1]} ${ecritureAlgebrique(2 * b)}x+${b * b} \\\\\\\\${2 * b > 0 ? `${rienSi1(a)}x- ${2 * b}x` : `${rienSi1(a)}x- (${2 * b}x)`}& ${choix[1]} ${b * b}\\\\\\\\${rienSi1(a - 2 * b)}x& ${choix[1]} ${b * b}\\\\\\\\`\n if (a - 2 * b === 1) { texteCorr += '' } else { texteCorr += ` x &${a - 2 * b > 0 ? `${choix[1]}` : `${choix[2]}`}\\\\dfrac{${b * b}}{${a - 2 * b}}\\\\\\\\` }\n texteCorr += '\\\\end{aligned}$<br>'\n if (pgcd(b * b, a - 2 * b) === 1) {\n texteCorr += `We must choose $x${a - 2 * b > 0 ? `${choix[1]}` : `${choix[2]}`}${f.texFraction}$ so that the result obtained with program 1 is ${choix[0]} to that obtained with program 2. `\n } else {\n texteCorr += `As $\\\\dfrac{${b * b}}{${a - 2 * b}}=${f.texFraction}$, we must choose $x${a - 2 * b > 0 ? `${choix[1]}` : `${choix[2]}`}${f.texFraction}$ so that the result obtained with program 1 is ${choix[0]} to that obtained with program 2. `\n }\n }\n\n break\n }\n if (this.listeQuestions.indexOf(texte) === -1) {\n // If the question has never been asked, we create another one\n this.listeQuestions.push(texte)\n this.listeCorrections.push(texteCorr)\n i++\n }\n cpt++\n }\n listeQuestionsToContenu(this)\n }\n this.besoinFormulaireNumerique = ['Choice of questions', 4, '1: Concrete situation\\n2: Geometric situation\\n3: Calculation program\\n4: Combination of previous 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