File: /home/mmtprep/public_html/mathzen.mmtprep.com/assets/2F21-1-tW6MHAW7.js.map
{"version":3,"file":"2F21-1-tW6MHAW7.js","sources":["../../src/exercices/2e/2F21-1.js"],"sourcesContent":["import { antecedentParDichotomie, courbe, courbeInterpolee } from '../../lib/2d/courbes.js'\nimport { droiteParPointEtPente } from '../../lib/2d/droites.js'\nimport { point } from '../../lib/2d/points.js'\nimport { repere } from '../../lib/2d/reperes.js'\nimport { segment } from '../../lib/2d/segmentsVecteurs.js'\nimport { texteParPosition } from '../../lib/2d/textes.js'\nimport { choice, combinaisonListes } from '../../lib/outils/arrayOutils'\nimport { numAlpha, sp } from '../../lib/outils/outilString.js'\nimport { prenom, prenomM } from '../../lib/outils/Personne'\nimport { texPrix, texteGras } from '../../lib/format/style'\nimport { stringNombre, texNombre } from '../../lib/outils/texNombre'\nimport Exercice from '../deprecatedExercice.js'\nimport Decimal from 'decimal.js'\n\nimport { mathalea2d } from '../../modules/2dGeneralites.js'\n\nimport { listeQuestionsToContenu, randint } from '../../modules/outils.js'\nimport { exp } from 'mathjs'\nexport const titre = 'Model a situation using a function'\nexport const dateDePublication = '14/02/2023'\nexport const dateDeModifImportante = '24/01/2024'\n/**\n * Description didactique de l'exercice\n * @author Gilles Mora\n * Référence\n*/\nexport const uuid = '5621b'\nexport const ref = '2F21-1'\nexport default function ModeliserParUneFonction () {\n Exercice.call(this) // Héritage de la classe Exercice()\n this.consigne = ''\n this.nbQuestions = 1\n this.nbQuestionsModifiable = false\n this.nbCols = 1 // Uniquement pour la sortie LaTeX\n this.nbColsCorr = 1 // Uniquement pour la sortie LaTeX\n this.sup = 1\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 // 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', 'typeE4', 'typeE5', 'typeE10']//\n } else if (this.sup === 2) {\n typeDeQuestionsDisponibles = ['typeE3', 'typeE6', 'typeE7']//\n } else if (this.sup === 3) {\n typeDeQuestionsDisponibles = ['typeE8', 'typeE9']//\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, question1, question2, question3, question4, question5, question6, correction1,\n correction2, correction3, correction4, correction5, correction6, cpt = 0; i < this.nbQuestions && cpt < 50;) {\n // Main loop where i+1 corresponds to the question number\n const nomF = [\n ['f'], ['g'], ['h'], ['u'],\n ['v'], ['w']\n ]\n switch (listeTypeQuestions[i]) { // Suivant le type de question, le contenu sera différent\n case 'typeE1':// salle de sport deux formules\n {\n const a = randint(10, 12)\n const dec1 = choice([0, 0.25, 0.5, 0.75, 1])\n const b = (new Decimal(randint(5, 6))).add(dec1)\n const b1 = Math.round(b * 100) / 100\n const c = randint(25, 30)\n const dec2 = choice([0, 0.25, 0.5, 0.75, 1])\n const d = (new Decimal(randint(2, 3))).add(dec2)\n const d1 = Math.round(d * 100) / 100\n const P = prenomM()\n const T = randint(30, 70)\n const e = randint(25, 30)\n const o = texteParPosition('$O$', -0.3, -0.3, 'medium', 'black', 1)\n const A = point(0.5 * (c - a) / (b - d), 0.1 * (a + b * (c - a) / (b - d)))\n const Ax = point(A.x, 0)\n const sAAx = segment(A, Ax)\n sAAx.epaisseur = 2\n sAAx.pointilles = 5\n const TexteX = texteParPosition('Number of sessions', 13, 0.5, 'medium', 'black', 1.5)\n const TexteY = texteParPosition('Price (€)', 1.3, 11.5, 'medium', 'black', 1.5)\n const r1 = repere({\n xMin: 0,\n yMin: 0,\n yMax: 120,\n xMax: 30,\n xUnite: 0.5,\n yUnite: 0.1,\n thickHauteur: 0.1,\n xLabelMin: 1,\n yLabelMin: 10,\n xLabelMax: 29,\n yLabelMax: 110,\n yThickDistance: 10,\n xThickDistance: 2,\n axeXStyle: '->',\n axeYStyle: '->',\n grilleYDistance: 10,\n grilleSecondaire: true,\n grilleSecondaireYDistance: 10,\n grilleSecondaireXDistance: 10,\n grilleSecondaireYMin: 0,\n grilleSecondaireYMax: 120,\n grilleSecondaireXMin: 0,\n grilleSecondaireXMax: 30\n\n })\n const f = x => a + b1 * x\n const g = x => c + d1 * x\n const graphique = mathalea2d({\n xmin: -1,\n xmax: 30,\n ymin: -1,\n ymax: 12,\n pixelsParCm: 30,\n scale: 0.7,\n style: 'margin: self'\n }, courbe(f, {\n repere: r1,\n color: 'blue',\n epaisseur: 2\n })\n , courbe(g, {\n repere: r1,\n color: 'red',\n epaisseur: 2\n }), TexteX, TexteY, r1, o, sAAx)\n this.introduction = ` In a gym, two options are offered:<br>${texteGras('Formula A:')} monthly subscription of $${a}$ € then $${texPrix(b)}$ € per session;<br>${texteGras('Formula B:')} monthly subscription of $${c}$ € then $${texPrix(d)}$ € per session.<br> br>The number of monthly sessions cannot exceed $${e}$. <br>We note $x$ the number of monthly sessions of a subscriber, $f(x)$ the price paid with formula A and $g(x)$ the price paid with formula B.<br>`\n question1 = ' Give the definition set of the functions $f$ and $g$.'\n question2 = ' Express in terms of $x$, $f(x)$, then $g(x)$.'\n question3 = `${P} chooses a plan but does not want to spend more than $${T}$ € for a month. What formula should I recommend if he wants to do as many sports sessions as possible during the month?`\n question4 = ' From how many monthly sessions is formula B more advantageous?'\n\n correction1 = `\n Le nombre minimal de séances dans le mois est $0$ et le nombre maximal est $${e}$, donc l'definition set of functions $f$ and $g$ is the set of integers in the interval $[0\\\\,,\\\\,${e}]$. `\n correction2 = `The packages include a fixed subscription and a specific price for a session. <br>Thus, the monthly amount for a formula is: Subscription + Cost of a session $\\\\times$ Number of sessions. <br>The function $f$ is defined by $f(x)=${a}+${texPrix(b)}x$ and the function $g$ is defined by $g(x)=${c}+${texPrix(d)}x$. `\n correction3 = ` We are looking for the maximum number of sessions that we can do with $${T}$ € with formulas A and B.<br>For formula A, we are looking for $x$ such that $f(x)\\\\leqslant${T}$. <br>$\\\\begin{aligned}${a}+${texPrix(b)}x&\\\\leqslant${T}\\\\\\\\${texPrix(b)}x&\\\\leqslant ${T}-${a}${sp(8)} \\\\text{(We subtract ${a} from each member)} \\\\\\\\${texPrix(b)}x&\\\\leqslant ${T - a}\\\\\\\\x&\\\\leqslant \\\\dfrac{${T - a}}{${texPrix(b)}}${sp(8)}\\\\text{(We divide each member by ${texPrix(b)})}\\\\end{aligned}$<br>The largest integer less than or equal to $\\\\dfrac{${T - a}}{${texPrix(b)}}$ is $${Math.floor((T - a) / b)}$.<br>With formula A, ${P} will be able to do at most $${Math.floor((T - a) / b)}$ sessions.<br><br>For formula B, we look for $x$ such that $g(x)\\\\leqslant${T}$.<br>$\\\\begin{aligned}${c}+${texPrix(d)}x&\\\\leqslant${T}\\\\\\\\${texPrix(d)}x&\\\\leqslant ${T}-${c}${sp(8)} \\\\text{(We subtract ${c} from each member)} \\\\\\\\${texPrix(d)}x&\\\\leqslant ${T - c}\\\\\\\\x&\\\\leqslant \\\\dfrac{${T - c}}{${texPrix(d)}}${sp(8)} \\\\text{(We divide each member by ${texPrix(d)})}\\\\end{aligned}$<br> The largest integer less than or equal to $\\\\dfrac{${T - c}}{${texPrix(d)}}$ is $${Math.floor((T - c) / d)}$.<br>With formula B, ${P} will be able to do at most $${Math.floor((T - c) / d)}$ sessions.<br><br>`\n\n if (Math.floor((T - c) / d) === Math.floor((T - a) / b)) {\n correction3 += `The two formulas allow as many sessions with a budget of $${T}$ €.<br><br>`\n } else {\n correction3 += `${texteGras('Conclusion :')} ${Math.floor((T - c) / d) > Math.floor((T - a) / b) ? 'Formula B' : 'Formula A'} allows you to do more sessions, it is more advantageous for ${P}. `\n }\n\n correction4 = ` Formula B is more advantageous than formula A when $g(x)$ is strictly less than $f(x)$.<br>${sp(8)} $\\\\begin{aligned}${c}+${texPrix(d)}x&<${a}+${texPrix(b)}x\\\\\\\\ ${texPrix(d)}x&< ${a}+${texPrix(b)}x-${c}${sp(8)}\\\\text{(We subtract ${c} from each member)} \\\\\\\\${texPrix(d)}x-${texPrix(b)}x&< ${a - c}${sp(8)}\\\\text{(We subtract ${texPrix(b)} }x\\\\text{ from each member)} \\\\\\\\${texPrix(d - b)}x&<${a - c}\\\\\\\\x&> \\\\dfrac{${a - c}}{${texPrix(d - b)}}${sp(8)}\\\\text{(We divide by } ${texNombre(d - b, 2)} < 0 \\\\text{ each member)}\\\\\\\\x&> \\\\dfrac {${c - a}}{${texPrix(b - d)}} \\\\end{aligned}$<br>The largest integer greater than $\\\\dfrac{${c - a}}{${texPrix(b - d)}}$ is $${Math.floor((a - c) / (d - b)) + 1}$.<br>Formula B is more interesting than formula A from $${Math.floor((a - c) / (d - b)) + 1}$ sessions.<br><br>We find this result with a graph: below, we have drawn the graphic representation of $f$ in blue and that of $g$ in red.<br>The red line goes well below the blue from the integer $x=${Math.floor((a - c) / (d - b)) + 1}$.${graphique}`\n this.listeQuestions = [question1, question2, question3, question4]\n this.listeCorrections = [correction1, correction2, correction3, correction4]\n }\n break\n case 'typeE2':// location de voitures\n {\n const a = randint(80, 120) // forfait\n const c = new Decimal(randint(41, 65, [50, 60])).div(100)// prix /km\n const c1 = Math.round(c * 100) / 100\n const km = randint(7, 10) * 100// km max\n const d = randint(50, 400)// nbre km\n const prix = new Decimal(c).mul(d).add(a)// prix payé\n const prix2 = Math.round(prix, 0)\n const o = texteParPosition('$O$', -0.3, -0.3, 'medium', 'black', 1)\n const A = point(0.01 * (prix - a) / c, 0.01 * prix)\n const Ax = point(A.x, 0)\n const Ay = point(0, A.y)\n const sAAx = segment(A, Ax)\n const sAAy = segment(A, Ay)\n sAAx.epaisseur = 2\n sAAx.pointilles = 5\n sAAy.epaisseur = 2\n sAAy.pointilles = 5\n const TexteX = texteParPosition('km', 9, 0.5, 'medium', 'black', 1.5)\n const TexteY = texteParPosition('Price (€)', 1.2, 5.5, 'medium', 'black', 1.5)\n const TexteVal1 = texteParPosition(`${texNombre(d, 0)}`, 0.01 * (prix - a) / c, -1, 'medium', 'black', 1.5)\n const TexteVal2 = texteParPosition(`${stringNombre(prix, 2)}`, -1.8, prix * 0.01, 'medium', 'black', 1.5)\n const r1 = repere({\n xMin: 0,\n yMin: 0,\n yMax: 600,\n xMax: 1000,\n xUnite: 0.01,\n yUnite: 0.01,\n thickHauteur: 0.1,\n xLabelMin: 100,\n yLabelMin: 100,\n xLabelMax: 900,\n yLabelMax: 590,\n yThickDistance: 100,\n xThickDistance: 100,\n axeXStyle: '->',\n axeYStyle: '->',\n grilleYDistance: 100,\n grilleXDistance: 100,\n grilleSecondaire: true,\n grilleSecondaireYDistance: 100,\n grilleSecondaireXDistance: 100,\n grilleSecondaireYMin: 0,\n grilleSecondaireYMax: 600,\n grilleSecondaireXMin: 0,\n grilleSecondaireXMax: 1000\n })\n\n const f = x => a + c1 * x\n const Cg = droiteParPointEtPente(point(0, prix2), 0, '', 'red')\n Cg.epaisseur = 2\n const graphique = mathalea2d({\n xmin: -3,\n xmax: 11,\n ymin: -1.5,\n ymax: 6,\n pixelsParCm: 30,\n scale: 1,\n style: 'margin: self'\n }, courbe(f, {\n repere: r1,\n color: 'blue',\n epaisseur: 2\n }), Cg, TexteX, TexteY\n , r1, o, sAAx, sAAy, TexteVal1, TexteVal2)\n this.introduction = ` A private vehicle rental company offers the following rate for a rental weekend:<br>${texteGras('WEEKEND RATE:')} flat rate of $${a}$ € then $${texNombre(c, 2)}$ € per km traveled (within the limit of $${texNombre(km, 0)}$ km).<br>We note $x$ as the number of km traveled by a customer during a weekend and we consider the function $T$ which associates the price paid by the customer with each value of $x$.<br>`\n question1 = 'Give the definition set of the function $T$.'\n question2 = ' Express $T(x)$ in terms of $x$.'\n question3 = ` Solve the equation $T(x)=${texNombre(prix, 2)}$. Interpret this result in the context of the exercise. `\n correction1 = ` We cannot do more than $${texNombre(km)}$ km during the weekend, so the definition set of the function $T$ is $[0\\\\,,\\\\,${km}]$. `\n correction2 = ` The price includes a fixed rate and a rate per km traveled. <br>Thus, the rental amount is: Package + Cost of one km $\\\\times$ Number of km traveled, i.e. $T(x)=${a}+${texNombre(c, 2)}x$. `\n correction3 = ` We solve the equation $T(x)=${texNombre(prix, 2)}$.<br>$\\\\begin{aligned}${a}+${texNombre(c, 2)}x&=${texNombre(prix, 2)}\\\\\\\\${texNombre(c, 2)}x&= ${texNombre(prix, 2)}-${a}${sp(8)} \\\\text{(We subtract ${a} from each member) } \\\\\\\\x&=\\\\dfrac{${texNombre(prix - a, 2)}}{${texNombre(c, 2)}}${sp(8)}\\\\text{(We divide each member by ${texNombre(c, 2)})}\\\\\\\\x&=${texNombre(d, 0)}\\\\end{aligned}$<br>The equation has the solution $${texNombre(d, 2)}$.<br>We can say that when the price paid for the rental is $${texNombre(prix, 2)}$ €, the customer traveled $${texNombre(d, 0)}$ km during the weekend.<br>We find this result graphically. Below, the blue line represents the function $f$. <br>`\n correction3 += `${graphique}`\n this.listeQuestions = [question1, question2, question3]\n this.listeCorrections = [correction1, correction2, correction3]\n }\n break\n case 'typeE3':// distance de freinage\n {\n const a = new Decimal(randint(2011, 2035)).div(10) //\n const a1 = Math.round(a * 100) / 100 //\n const b = randint(30, 80)\n const v = randint(70, 100) //\n const o = texteParPosition('$O$', -0.3, -0.3, 'medium', 'black', 1)\n const TexteX = texteParPosition('v (in km/h)', 12, 0.5, 'medium', 'black', 1.5)\n const TexteY = texteParPosition('d (in m)', 1.8, 9.5, 'medium', 'black', 1.5)\n const A = point(0.1 * Math.sqrt(a * b), 0.1 * b)\n const Ax = point(A.x, 0)\n const sAAx = segment(A, Ax)\n sAAx.epaisseur = 2\n sAAx.pointilles = 5\n const TexteVal1 = texteParPosition(`${texNombre(Math.round(Math.sqrt(b * a), 0))}`, A.x, -1, 'medium', 'black', 1.5)\n const TexteVal2 = texteParPosition(`${texNombre(b, 0)}`, -1.5, A.y, 'medium', 'black', 1.5)\n const r1 = repere({\n xMin: 0,\n yMin: 0,\n yMax: 100,\n xMax: 130,\n xUnite: 0.1,\n yUnite: 0.1,\n thickHauteur: 0.1,\n xLabelMin: 10,\n yLabelMin: 10,\n xLabelMax: 120,\n yLabelMax: 90,\n yThickDistance: 10,\n xThickDistance: 10,\n axeXStyle: '->',\n axeYStyle: '->',\n grilleYDistance: 10,\n grilleXDistance: 10,\n grilleSecondaire: true,\n grilleSecondaireYDistance: 10,\n grilleSecondaireXDistance: 10,\n grilleSecondaireYMin: 0,\n grilleSecondaireYMax: 100,\n grilleSecondaireXMin: 0,\n grilleSecondaireXMax: 130\n })\n const f = x => x ** 2 / a1\n const Cg = droiteParPointEtPente(point(0, b), 0, '', 'red')\n Cg.epaisseur = 2\n const graphique = mathalea2d({\n xmin: -2.5,\n xmax: 14.5,\n ymin: -2,\n ymax: 10,\n pixelsParCm: 30,\n scale: 0.75,\n style: 'margin: self'\n }, courbe(f, {\n repere: r1,\n color: 'blue',\n epaisseur: 2\n }), Cg, TexteX, TexteY\n , r1, o, sAAx, TexteVal1, TexteVal2)\n this.introduction = ` On any dry road, the braking distance in meters of a car is modeled as follows: <br>By noting $v$ the speed of the vehicle (in km/h), its braking distance $d(v) $ (in m) is given by the square of its speed divided by $${texNombre(a, 1)}$. `\n question1 = ' Give the expression for $d(v)$ as a function of $v$.'\n question2 = ` Calculate to the nearest meter the braking distance of the car if it is traveling at $${v}$ km/h. `\n question3 = ' Is braking distance proportional to speed?'\n question4 = ` The braking distance of this car was $${b}$ m. What was his speed in km/h rounded to the nearest unit?`\n correction1 = ` The square of the speed is $v^2$, so the function $d$ is defined by: $d(v)=\\\\dfrac{v^2}{${texNombre(a, 1)}}$. `\n correction2 = ` $d(${v})=\\\\dfrac{${v}^2}{${texNombre(a, 1)}}\\\\simeq ${Math.round(v ** 2 / a, 0)}$. The braking distance is approximately $${Math.round(v ** 2 / a, 0)}$. `\n correction3 = ' The braking distance is not proportional to the speed because the function $d$ is not a linear function. It does not reflect a situation of proportionality.'\n correction4 = ` We look for $v$ such that $d(v)=${b}$.<br>$\\\\begin{aligned}\\\\dfrac{v^2}{${texNombre(a, 1)}}&=${b}\\\\\\\\v^2&=${b} \\\\times ${texNombre(a, 2)} ${sp(8)} \\\\text{(We multiply each member by ${texNombre(a, 1)})} \\\\\\\\v^2&= ${texNombre(b * a, 2)}\\\\\\\\v&= -\\\\sqrt{${texNombre(b * a, 2)}} ${sp(8)} \\\\text{or} ${sp(8)} v= \\\\sqrt{${texNombre(b * a, 2)}}${sp(8)}\\\\text{(two numbers have the square } ${texNombre(b * a, 2)} \\\\text{)}\\\\end{aligned}$<br>Since $v$ is a positive number, we deduce $v= \\\\sqrt{${texNombre(b * a, 2)}}\\\\simeq ${Math.round(Math.sqrt(b * a), 0)}$.<br>When the braking distance of the car is $${b}$ m, its speed is then approximately $${Math.round(Math.sqrt(b * a), 0)}$ km/h.<br>Here is the curve representative of the function $d$ with the solution of the previous question read graphically. <br>`\n correction4 += `${graphique}`\n this.listeQuestions = [question1, question2, question3, question4]\n this.listeCorrections = [correction1, correction2, correction3, correction4]\n }\n break\n\n case 'typeE4':// abonnement à une revue\n {\n const nom = choice(nomF)\n const a = randint(6, 10) * 1000 //\n const b = choice([40, 50, 80, 100])\n const c = randint(31, 49) * 100 //\n const d = randint(30, 39) * 10\n const o = texteParPosition('$O$', -0.3, -0.3, 'medium', 'black', 1)\n const TexteX = texteParPosition('Subscription price (in €)', 11, -2, 'medium', 'black', 1.5)\n const TexteY = texteParPosition('Number of subscribers', 1.5, 10.5, 'medium', 'black', 1.5)\n const A = point(0.05 * (a - c) / b, 0.001 * c)\n const Ax = point(A.x, 0)\n const sAAx = segment(A, Ax)\n sAAx.epaisseur = 2\n sAAx.pointilles = 5\n const TexteVal1 = texteParPosition(`${stringNombre((a - c) / b, 1)}`, A.x, -1.5, 'medium', 'black', 1.5)\n const TexteVal2 = texteParPosition(`${c}`, -2.5, A.y, 'medium', 'black', 1.5)\n const r1 = repere({\n xMin: 0,\n yMin: 0,\n yMax: 10000,\n xMax: 300,\n xUnite: 0.05,\n yUnite: 0.001,\n thickHauteur: 0.1,\n xLabelMin: 50,\n yLabelMin: 1000,\n xLabelMax: 300,\n yLabelMax: 9000,\n yThickDistance: 1000,\n yLabelEcart: 1,\n xThickDistance: 50,\n axeXStyle: '->',\n axeYStyle: '->',\n grilleYDistance: 1000,\n grilleXDistance: 50,\n grilleSecondaire: true,\n grilleSecondaireYDistance: 500,\n grilleSecondaireXDistance: 50,\n grilleSecondaireYMin: 0,\n grilleSecondaireYMax: 10000,\n grilleSecondaireXMin: 0,\n grilleSecondaireXMax: 300\n })\n const f = x => a - b * x\n const Cg = droiteParPointEtPente(point(0, c), 0, '', 'red')\n Cg.epaisseur = 2\n const graphique = mathalea2d({\n xmin: -4,\n xmax: 17,\n ymin: -3,\n ymax: 11.5,\n pixelsParCm: 25,\n scale: 0.7,\n style: 'margin: self'\n }, courbe(f, {\n repere: r1,\n color: 'blue',\n epaisseur: 2\n }), Cg, TexteX, TexteY\n , r1, o, sAAx, TexteVal1, TexteVal2)\n this.introduction = ` The number of subscribers to a magazine depends on the price of the subscription to this magazine, price expressed in euros.<br>We consider that we have the relationship: <br>number of subscribers $= ${texNombre(a)} - ${b} \\\\times$ subscription price in euros.<br>Let $${nom}$ be the function which gives the number of subscribers according to the price of the annual subscription to this magazine. `\n question1 = `Determine the algebraic expression for $${nom}$. Specify the variable. `\n question2 = ' What can we say about the number of subscribers when the subscription price increases?'\n question3 = ` Explain why the subscription price should not be $${d}$ €. Determine the definition set of the function $${nom}$. `\n question4 = ` The subscription director wants $${texNombre(c)}$ to subscribe to the magazine. How much should the subscription cost?`\n question5 = ' We obtain the revenue from the sale of $x$ subscriptions by multiplying the number of subscribers by the price of a subscription. <br>Express the revenue as a function of the subscription price in expanded form. '\n\n correction1 = ` Noting $x$ for the variable, the algebraic expression for $${nom}$ is: $${nom}(x)=${texNombre(a)}-${b}x$. `\n correction2 = ` The relationship $${nom}(x)=${texNombre(a)}-${b}x$ shows that when the price of subscription $x$ increases, the number of subscribers $${nom}(x)$ decreases. <br>More precisely, for each increase of $1$ €, the number of subscribers decreases by $${b}$ (coefficient in front of $x$). `\n correction3 = ` For an amount of $${d}$ € of the subscription, we obtain $${nom}(${d})=${texNombre(a, 0)}-${b}\\\\times ${d}=${texNombre(a - b * d, 0)}$.<br>We would then obtain a negative number of subscribers which is impossible. We cannot therefore set the subscription amount at $${d}$ €.<br>We seek the value of $x$ giving a zero number of subscribers by solving the equation $${nom}(x)=0$: <br>$\\\\begin{aligned}${texNombre(a)}-${b}x&=0\\\\\\\\- ${b}x&= -${texNombre(a)}${sp(8)} \\\\text{(We subtract ${texNombre(a)} from each member)} \\\\\\\\x&=\\\\dfrac{${texNombre(-a)}}{ ${-b}}${sp(8)}\\\\text{(We divide by } ${-b} \\\\text{ each member)}\\\\\\\\x&=\\\\dfrac{${texNombre(a)}}{${b}}\\\\\\\\x&=${texNombre(a / b, 2)}\\\\end{aligned}$ <br>We deduce that the amount of the subscription must be between $0$ € and $${texNombre(a / b, 2)}$ €. <br>Consequently the definition set of the function $${nom}$ is: $[0\\\\,,\\\\,${texNombre(a / b, 2)}]$. `\n\n correction4 = ` We look for the value of $x$ so that $${nom}(x)=${texNombre(c)}$.<br>$\\\\begin{aligned}${texNombre(a)}-${b}x&=${texNombre(c)}\\\\\\\\- ${b}x&= ${texNombre(c)}-${a}${sp(8)} \\\\text{(On subtracts ${texNombre(a)} from each member)} \\\\\\\\x&=\\\\dfrac{${texNombre(-a + c)}}{${-b}}${sp(8)}\\\\text{(We divide by } ${-b} \\\\text{ each member)}\\\\\\\\x&=\\\\dfrac {${texNombre(a - c)}}{${b}}\\\\\\\\x&=${texNombre((a - c) / b, 2)}\\\\end{aligned}$<br>To have $${texNombre(c)}$ subscribers, the subscription director must set the subscription price at $${texPrix((a - c) / b)}$ €. `\n\n correction5 = ` As $x$ denotes the amount of the subscription and $${nom}(x)$ the number of subscribers, the product of the number of subscribers by the price of a subscription is $${nom}(x)\\\\times x$, i.e. $(${texNombre(a)}-${b}x)\\\\times x$.<br>Its expanded expression is: $${texNombre(a)}x-${b}x^2$.<br><br>Here is the graphical representation of the function $${nom}$ (in blue) with the graphic answer to the ${numAlpha(3)} question: <br>`\n correction5 += `${graphique}`\n this.listeQuestions = [question1, question2, question3, question4, question5]\n this.listeCorrections = [correction1, correction2, correction3, correction4, correction5]\n }\n break\n\n case 'typeE5':// station service\n {\n const a = new Decimal(randint(150, 200)).div(100) //\n const a1 = Math.round(a * 100) / 100 //\n const b = randint(3, 6)\n const c = choice([40, 45, 50, 55, 60, 65, 70])\n const d = randint(b, c)\n const prix = new Decimal(a).mul(d)\n const prix1 = Math.round(prix * 100) / 100\n const P = prenom()\n const nom = choice(nomF)\n const o = texteParPosition('$O$', -0.3, -0.3, 'medium', 'black', 1)\n const TexteX = texteParPosition('Number of liters', 11, -1.7, 'medium', 'black', 1.5)\n const TexteY = texteParPosition('Price paid (in €)', 1.2, 11.5, 'medium', 'black', 1.5)\n const A = point(0.2 * d, 0.08 * prix)\n const Ax = point(A.x, 0)\n const sAAx = segment(A, Ax)\n sAAx.epaisseur = 2\n sAAx.pointilles = 5\n const TexteVal1 = texteParPosition(`${texNombre(d, 2)}`, A.x, -1, 'medium', 'black', 1.5)\n const TexteVal2 = texteParPosition(`${stringNombre(prix, 2)}`, -2, A.y, 'medium', 'black', 1.5)\n const r1 = repere({\n xMin: 0,\n yMin: 0,\n yMax: 140,\n xMax: 70,\n xUnite: 0.2,\n yUnite: 0.08,\n thickHauteur: 0.1,\n xLabelMin: 10,\n yLabelMin: 10,\n xLabelMax: 70,\n yLabelMax: 130,\n yLabelEcart: 0.8,\n yThickDistance: 10,\n xThickDistance: 10,\n axeXStyle: '->',\n axeYStyle: '->',\n grilleYDistance: 10,\n grilleXDistance: 10,\n grilleSecondaire: true,\n grilleSecondaireYDistance: 10,\n grilleSecondaireXDistance: 10,\n grilleSecondaireYMin: 0,\n grilleSecondaireYMax: 140,\n grilleSecondaireXMin: 0,\n grilleSecondaireXMax: 70\n })\n const f = x => a1 * x\n const Cg = droiteParPointEtPente(point(0, prix1), 0, '', 'red')\n Cg.epaisseur = 2\n const graphique = mathalea2d({\n xmin: -3,\n xmax: 16,\n ymin: -2.5,\n ymax: 12.5,\n pixelsParCm: 30,\n scale: 0.7,\n style: 'margin: self'\n }, courbe(f, {\n repere: r1,\n color: 'blue',\n epaisseur: 2\n }), Cg, TexteX, TexteY\n , r1, o, sAAx, TexteVal1, TexteVal2)\n this.introduction = ` In a gas station, the price of 95 unleaded gasoline is $${texNombre(a)}$ € per liter.<br>In this station, it is not possible to take less than $${b}$ liters of gasoline.<br>${P} refuels his car at this gas station. The tank of his car is empty and can contain a maximum of $${c}$ liters.<br>We note $x$ as the number of liters that ${P} takes to fill up his car's tank. <br>We consider the function $${nom}$ which associates with each value of $x$, the price paid in euros by ${P}. `\n\n question1 = `Give the definition set of the function $${nom}$. `\n question2 = ` Determine the algebraic expression of the function $${nom}$ (i.e. the expression of $${nom}(x)$ as a function of $x$). `\n question3 = ' Is the price paid proportional to the number of liters put in the tank? Justify.'\n question4 = ` Solve the equation $${nom}(x)=${texNombre(prix, 2)}$. Interpret this result in the context of the exercise. `\n correction1 = ` The minimum liters that ${P} can put is $${b}$ and the maximum is $${c}$. <br>The definition set of $${nom}$ is therefore $[${b}\\\\,,\\\\,${c}]$. `\n correction2 = ` To obtain the price paid, multiply the number of liters by the price of one liter. <br>Thus, the algebraic expression of $${nom}$ is: $${nom}(x)=${texNombre(a, 2)}\\\\times x$, or $${nom}(x)=${texNombre(a, 2)}x$. `\n correction3 = ` The price paid is proportional to the number of liters. The $${nom}$ function is a linear function reflecting a situation of proportionality. `\n correction4 = ` We are looking for $x$ such that $${nom}(x)=${texNombre(prix, 2)}$.<br>$\\\\begin{aligned}${texNombre(a, 2)}x&=${texNombre(prix, 2)}\\\\\\\\x&=\\\\dfrac{${texNombre(prix, 2)}}{${texNombre(a, 2)}} ${sp(8)} \\\\text{( We divide each member by ${texNombre(a, 2)})} \\\\\\\\x&= ${d}\\\\end{aligned}$<br>For $${d}$ liters put in the tank, the cost is $${texNombre(prix, 2)}$ €.<br>Here is the curve representative of the function $${nom}$ with the solution of the previous question read graphically.<br>The function $${nom}$ is represented by a straight line passing through the origin of the benchmark (characteristic of a situation of proportionality).<br>`\n correction4 += `${graphique}`\n this.listeQuestions = [question1, question2, question3, question4]\n this.listeCorrections = [correction1, correction2, correction3, correction4]\n }\n break\n\n case 'typeE6':// restaurateur recette après une hausse\n {\n const a = randint(18, 22) //\n const b = randint(28, 31) * 10\n const c = randint(7, 11)\n const d = randint(20, 25)\n const h = randint(2, 6)\n const nom = choice(nomF)\n\n this.introduction = ` The owner of a restaurant knows perfectly well that, in his establishment, the number of seats during the midday meal depends on the price of his menu.<br>The market study he had carried out made it possible to model the link between the price of the menu and the number of seats in the following way:<br>$\\\\bullet$ By selling $${a}$ € his menu (initially proposed price), he serves $${b}$ seats.<br>$\\\\bullet$ Each increase of $1$ € in the price of the menu reduces the number of place settings by $${c}$.<br>We note $x$ the amount of the proposed increase in the price of the menu (in €) compared to the initial price which was $${a}$ €. We admit that $0 \\\\leqslant x \\\\leqslant ${d}$. `\n question1 = `Give the definition set of the function $${nom}$. `\n question2 = ` For an increase in $${h}$ €, give the price of the menu, the number of seats served and the restaurant owner's recipe (in €) (obtained by the product of the price of a menu and the number of seats served). `\n question3 = ' Express the price of the menu as a function of $x$ after an increase of $x$ €.'\n question4 = ' Express as a function of $x$ the number of place settings served after an increase of $x$ €.'\n question5 = ` Deduce the revenue $${nom}(x)$ achieved after an increase in the menu price of $x$ € and show that it can be expressed in the form: $${nom}(x) = ${-c}x^2 + ${b - a * c}x + ${a * b}$. `\n correction1 = ` The definition set is given by the statement ($0 \\\\leqslant x \\\\leqslant ${d}$). <br>The definition set of $${nom}$ is therefore $[0\\\\,,\\\\,${d}]$. `\n\n correction2 = ` After an increase of $${h}$ €:<br>$\\\\bullet$ the price of the menu is $${a}+${h}=${a + h}$ €;<br>$\\\\bullet$ the number of place settings is $${b}-10\\\\times ${h}=${b - 10 * h}$ ;<br>$\\\\bullet$ the recipe is $${a + h}\\\\times ${b - 10 * h}=${(a + h) * (b - 10 * h)}$. `\n\n correction3 = `The price of the menu after an increase of $x$ € is $${a}+x$. `\n\n correction4 = ` Since with each increase of $1$ €, the number of covers decreases by $${c}$, we deduce that the number of covers after an increase of $x$ € is $${b}-${c}\\\\times x$ i.e. $${b}- ${c}x$. `\n\n correction5 = `The recipe is given by the product of the price of a menu and the number of menus, i.e. $(${a}+x)\\\\times (${b}-${c}x)=${a * b}-${a * c}x+${b}x-${c}x^2=-${c}x^2+ ${b - a * c}x+${a * b}$. `\n this.listeQuestions = [question1, question2, question3, question4, question5]\n this.listeCorrections = [correction1, correction2, correction3, correction4, correction5]\n }\n break\n\n case 'typeE7':// la moto\n {\n const a = new Decimal(randint(-5, -2)) //\n const b = new Decimal(randint(-15, -10)).div(10)\n const c = new Decimal(randint(-39, -25)).div(10)\n const P = prenomM()\n const nom = choice(nomF)\n const o = texteParPosition('$O$', -0.3, -0.3, 'medium', 'black', 1)\n const TexteX = texteParPosition('Time (in s)', 6, -0.7, 'medium', 'black', 1.2)\n const TexteY = texteParPosition('Height (in m)', 1.5, 7, 'medium', 'black', 1.2)\n const r1 = repere({\n xMin: 0,\n yMin: 0,\n yMax: 3,\n xMax: 5,\n xUnite: 2,\n yUnite: 2,\n axeXStyle: '->',\n axeYStyle: '->',\n grilleX: false,\n grilleY: false,\n xThickMax: 0,\n yThickMax: 0\n\n })\n const f = x => -0.5 * (x + 1) * (x - 4)\n\n const graphique = mathalea2d({\n xmin: -1,\n xmax: 13,\n ymin: -1.2,\n ymax: 7.5,\n pixelsParCm: 20,\n scale: 0.7,\n style: 'margin: self'\n }, courbe(f, {\n repere: r1,\n xMin: 0,\n xMax: 4,\n color: 'blue',\n epaisseur: 2\n }), TexteX, TexteY\n , r1, o)\n this.introduction = ` During a motocross race, after crossing a ramp, ${P} performed a motorcycle jump.<br>We note $t$ the duration (in seconds) of this jump. The jump begins as soon as ${P} leaves the ramp, that is to say when $t=0$.<br>The height (in meters) is determined as a function of the duration $t$ by the following $${nom}$ function: $${nom}(t)=(${texNombre(a, 3)}t${texNombre(b, 2)})(t${texNombre(c, 2)})$<br>Here is the representative curve of this $${nom}$ function:<br>`\n this.introduction += `${graphique}`\n question1 = ` Calculate $${nom}(4)$. What can we deduce from this?`\n question2 = ` How high is ${P} when it leaves the ramp?`\n question3 = ` How long does the ${P} jump last?`\n question4 = ` Expand and collapse the $${nom}$ expression. `\n\n correction1 = `${numAlpha(0)} $${nom}(4)=(${texNombre(a, 3)}\\\\times 4${texNombre(b, 2)})(4 ${texNombre(c, 2)})=${texNombre(a.mul(4).plus(b), 2)}\\\\times ${texNombre(c.plus(4), 2)}=${texNombre((a.mul(4).plus(b)) * (c.plus(4)), 2)}$<br>As the result is negative, we deduce that the jump lasts less than $4$ seconds. `\n correction2 = ` The height of the start of the jump is given by: $${nom}(0)=(${texNombre(a, 3)}\\\\times 0${texNombre(b, 2)})(0 ${texNombre(c, 2)})=${texNombre(b.mul(c), 2)}$.<br>${P} is $${texNombre(b.mul(c), 2)}$ meters at the start of the jump. `\n correction3 = ` The jump starts at $t=0$ and ends when ${P} ends up on the ground, i.e. when the height is zero. <br>Thus, the jump time is given by the positive solution of the equation $(${texNombre(a, 3)}t${texNombre(b, 2)})(t${texNombre(c, 2)})=0$<br>This is a zero product equation which has two solutions: $t_1= ${texNombre(-b.div(a), 2)}$ and $t_2= ${texNombre(-c, 2)}$. <br>The jump lasts $${texNombre(-c, 2)}$ seconds. `\n correction4 = `We develop using the double distributivity:<br>$${nom}(t)=(${texNombre(a, 3)}t${texNombre(b, 2)})(t${texNombre(c, 2)})=${texNombre(a, 3)}t^2+${texNombre(a.mul(c), 3)}t${texNombre(b, 2)}t+${texNombre(b.mul(c), 2)}=${texNombre(a, 3)}t^2+${texNombre(a.mul(c).plus(b), 2)}t+${texNombre(b.mul(c), 2)}$. `\n this.listeQuestions = [question1, question2, question3, question4]\n this.listeCorrections = [correction1, correction2, correction3, correction4]\n }\n break\n\n case 'typeE8':// pression artérielle\n {\n const a = randint(100, 120)\n const b0 = randint(90, 98)\n const a1 = randint(10, 14) * 10\n const b1 = randint(60, 75) * 2\n const a2 = randint(25, 27) * 10\n const b2 = randint(48, 53) * 2\n const a3 = randint(38, 42) * 10\n const b3 = randint(55, 60) * 2\n const a4 = randint(50, 52) * 10\n const b4 = randint(43, 47) * 2\n const a5 = randint(56, 65) * 10\n const b5 = b4 + 5\n const nom = choice(nomF)\n const o = texteParPosition('$O$', 0, 15.5, 'medium', 'black', 1)\n const TexteX = texteParPosition('Blood pressure in mmHg', 150 * 0.03, 155 * 0.2, 'medium', 'black', 1.2)\n const TexteY = texteParPosition('Time (in ms)', 670 * 0.03, 72 * 0.2, 'medium', 'black', 1.2)\n\n const r1 = repere({\n xMin: 0,\n yMin: 80,\n yMax: 150,\n xMax: 800,\n xUnite: 0.03,\n yUnite: 0.2,\n xThickDistance: 50,\n yThickDistance: 10,\n xLabelMin: 0,\n yLabelMin: 80,\n yLabelEcart: 1,\n grilleXDistance: 50,\n grilleYDistance: 10,\n grilleXMin: 0,\n grilleYMin: 80,\n grilleXMax: 800,\n grilleYMax: 150,\n grilleSecondaireX: true,\n grilleSecondaireXDistance: 50,\n grilleSecondaireXMin: 0,\n grilleSecondaireXMax: 800,\n grilleSecondaireY: true,\n grilleSecondaireYDistance: 2,\n grilleSecondaireYMin: 80,\n grilleSecondaireYMax: 150\n })\n\n const gr = courbeInterpolee(\n [\n [0, b0], [a1, b1], [a2, b2], [a3, b3], [a4, b4], [a5, b5]\n ],\n {\n color: 'blue',\n epaisseur: 2,\n repere: r1,\n xMin: 0,\n xMax: 650\n })\n\n const graphique = mathalea2d({\n xmin: -2,\n xmax: 24,\n ymin: 15,\n ymax: 32,\n pixelsParCm: 20,\n scale: 0.5,\n style: 'margin: self'\n }, TexteX, TexteY\n , r1, o, gr)\n this.introduction = ` Arterial tonometry provides continuous measurement of blood pressure. The examination provides information on the condition of the patient's arteries in the context of the development of high blood pressure. <br>A recording of the measurements makes it possible to assess the arterial pressure curve.<br>We note $${nom}$ the function which at time $t$ in milliseconds (ms) associates the radial arterial pressure $${nom}(t)$ in millimeters of mercury (mmHg), measured at rest in a patient suspected of heart failure. We give the representative curve of $${nom}$ below.<br>`\n this.introduction += ` <br>${graphique}`\n question1 = `What is the definition set of $${nom}$. `\n question2 = ' Which inequality has as its solution set the time interval during which blood pressure is greater than or equal to $130$ mmHg?'\n question3 = ' Determine the measured systolic value, that is to say the maximum value of blood pressure.'\n question4 = ' Determine the measured diastolic value, that is to say the minimum value of blood pressure.'\n question5 = ' A patient is in hypertension when the systolic pressure is greater than or equal to $140 mmHg or the diastolic pressure is greater than or equal to $90 mmHg.<br>Is this patient in hypertension? Justify. '\n question6 = ` The $${nom}$ function was represented over a time interval corresponding to that of a patient's heartbeat.<br>We speak of tachycardia when, at rest, the number of heartbeats is greater than $100$ per minute. <br>Based on this examination, can we estimate that the patient suffers from tachycardia?`\n\n correction1 = `The definition set of $${nom}$ is $[0\\\\,,\\\\, ${a5}]$. `\n correction2 = ` The inequality whose solution set is the time interval during which blood pressure is greater than or equal to $${a}$ mmHg is $${nom}(t)\\\\geqslant ${a}$. `\n correction3 = ` The measured systolic value is given by the ordinate of the highest point of the curve: $${b1}$ mmHg. `\n correction4 = ` The diastolic value measured is given by the ordinate of the lowest point of the curve: $${b4}$ mmHg. `\n correction5 = ` The systolic value is $${b1}$ mmHg, the diastolic value is $${b4}$ mmHg. <br>`\n if (b1 >= 140 || b4 >= 90) {\n if (b1 >= 140 || b4 < 90) {\n correction5 += `As $${b1} \\\\geqslant 140$, the patient has high blood pressure.<br>`\n } else {\n correction5 += `As $${b4} \\\\geqslant 90$, the patient has high blood pressure.<br>`\n }\n } else {\n correction5 += `As $${b4} < 90$ and $${b1} < 140$, the patient is not in high blood pressure.<br>`\n }\n correction6 = `\n L'time interval is $[0\\\\,,\\\\, ${a5}]$, the time d'un battement de cœur est donc $${a5}$ ms.<br>\n Comme $${a5}$ ms $=${texNombre(a5 / 1000, 3)}$ s, en notant $n$ le nombre de battements en $1$ minute, on obtient le tableau de proportionnalité suivant :<br>\n $\\\\begin{array}{|c|c|c|}\\n`\n correction6 += '\\\\hline\\n'\n correction6 += '\\n\\\\text{Number of beats} &1 &n \\\\\\\\\\n'\n correction6 += '\\\\hline\\n'\n correction6 += `\\n \\\\text{Time (in s)}&${texNombre(a5 / 1000, 3)}&60 \\\\\\\\\\n`\n correction6 += '\\\\hline\\n'\n correction6 += '\\\\end{array}\\n$'\n\n correction6 += `<br>$n=\\\\dfrac{60\\\\times 1}{${texNombre(a5 / 1000, 3)}}\\\\simeq ${texNombre(60 * 1000 / a5, 0)}$.<br>`\n if (60 * 1000 / a5 > 100) {\n correction6 += `As $${texNombre(60 * 1000 / a5, 0)}>100$, this patient suffers from tachycardia. `\n } else {\n correction6 += `As $${texNombre(60 * 1000 / a5, 0)}\\\\leqslant 100$, this patient does not suffer from tachycardia. `\n }\n this.listeQuestions = [question1, question2, question3, question4, question5, question6]\n this.listeCorrections = [correction1, correction2, correction3, correction4, correction5, correction6]\n }\n break\n\n case 'typeE9':// alcool dans le sang\n {\n const a = randint(17, 21) / 10 //\n const b = randint(-10, -5) / 10 //\n const h = choice([11, 12, 13, 17, 18]) //\n const nom = choice(nomF)\n const o = texteParPosition('$O$', -0.3, -0.3, 'medium', 'black', 1)\n const f = x => a * x * exp(b * x)\n const fprime = x => (a + a * b * x) * exp(b * x)\n const Cg = droiteParPointEtPente(point(0, 5), 0, '', 'red')\n Cg.epaisseur = 2\n const s0 = antecedentParDichotomie(0, 7, fprime, 0, 0.01)\n const s1 = antecedentParDichotomie(0, s0 * 1.5, f, 0.5, 0.01)\n const s2 = antecedentParDichotomie(s0 * 1.5, 6 * 1.5, f, 0.5, 0.01)\n const r1 = repere({\n xMin: 0,\n yMin: 0,\n yMax: f(-1 / b) + 0.2,\n xMax: 10,\n xUnite: 1.5,\n yUnite: 10,\n axeXStyle: '->',\n axeYStyle: '->',\n xThickDistance: 1,\n yThickDistance: 0.1,\n xLabelMin: 0,\n yLabelMin: 0,\n yLabelEcart: 1,\n xLabelEcart: 0.6,\n grilleXDistance: 1,\n grilleYDistance: 1,\n grilleXMin: 0,\n grilleYMin: 0,\n grilleXMax: 10,\n grilleYMax: f(s0) + 0.2\n })\n\n const A = point(s0 * 1.5, f(s0) * 10)\n const Ax = point(A.x, 0)\n const sAAx = segment(A, Ax)\n sAAx.epaisseur = 2\n sAAx.pointilles = 5\n const Ay = point(0, A.y)\n const sAAy = segment(A, Ay)\n sAAy.epaisseur = 2\n sAAy.pointilles = 5\n const B = point(s1 * 1.5, f(s1) * 10)\n const Bx = point(B.x, 0)\n const sBBx = segment(B, Bx)\n sBBx.epaisseur = 2\n sBBx.pointilles = 5\n\n const C = point(s2 * 1.5, f(s2) * 10)\n const Cx = point(C.x, 0)\n const sCCx = segment(C, Cx)\n sCCx.epaisseur = 2\n sCCx.pointilles = 5\n const sBxCx = segment(Bx, Cx, 'red')\n sBxCx.epaisseur = 5\n const Texte1 = texteParPosition(`Max = ${stringNombre(Math.round(f(s0) * 100) / 100)}`, -3, A.y, 'medium', 'red', 1.2)\n const Texte2 = texteParPosition(`${stringNombre(Math.round(s0 * 10) / 10)}`, A.x, -1.3, 'medium', 'red', 1.2)\n const Texte3 = texteParPosition(`${stringNombre(Math.round(s1 * 10) / 10)}`, B.x, -1.3, 'medium', 'red', 1.2)\n const Texte4 = texteParPosition(`${stringNombre(Math.round(s2 * 10) / 10)}`, C.x, -1.3, 'medium', 'red', 1.2)\n const graphique = mathalea2d({\n xmin: -2,\n xmax: 16,\n ymin: -1,\n ymax: (f(s0) + 0.2) * 10,\n pixelsParCm: 30,\n scale: 0.7,\n style: 'margin: self'\n }, [courbe(f, {\n repere: r1,\n xMin: 0,\n xMax: 10,\n color: 'blue',\n epaisseur: 2\n }),\n r1, o])\n const graphiqueCorr = mathalea2d({\n xmin: -5,\n xmax: 16,\n ymin: -2.5,\n ymax: (f(s0) + 0.2) * 10,\n pixelsParCm: 30,\n scale: 0.7,\n style: 'margin: self'\n }, [courbe(f, {\n repere: r1,\n xMin: 0,\n xMax: 9,\n color: 'blue',\n epaisseur: 2\n }),\n r1, o, sAAx, sAAy, sCCx, sBBx, sBxCx, Texte1, Texte2, Texte3, Texte4, Cg,\n r1, o])\n this.introduction = `The Highway Code prohibits any driving of a vehicle when the blood alcohol level is greater than or equal to $0.5$ g/L.<br>The blood alcohol level of a person during the $10$ hours following the consumption of a certain quantity of alcohol is modeled by the function $${nom}$.<br>$\\\\bullet$ $t$ represents the time (expressed in hours) elapsed since alcohol consumption;<br>$\\\\bullet$ $${nom} (t)$ represents the blood alcohol level (expressed in g/L) of this person.<br>We give the graphical representation of the $${nom}$ function in a reference frame. <br>`\n this.introduction += `${graphique}`\n question1 = `\n À quel instant le taux d’alcoolémie de cette personne est-il maximal ? Quelle est alors sa valeur ? Arrondir\n au centième. `\n question2 = `Graphically solve the inequality $${nom}(t)>0.5$. `\n question3 = ` At time $t=0$, it was $${h}$ h. At what time, to the minute, can the motorist get back behind the wheel without being in violation?`\n correction1 = ` The maximum blood alcohol level is reached when $t=${texNombre(Math.round(s0 * 10) / 10, 1)}$. Its value is approximately $${texNombre(Math.round(f(s0) * 100) / 100, 2)}$. `\n correction2 = ` The solutions of the inequality $${nom}(t)>0.5$ are the abscissa of the points of the curve which are located strictly below the line of equation $y=0.5$. <br>This inequality has the solution set $]${texNombre(Math.round(s1 * 10) / 10, 1)}\\\\,,\\\\,${texNombre(Math.round(s2 * 10) / 10, 1)}[$. <br>`\n\n if (Math.round(s2 * 10) / 10 === 2 || Math.round(s2 * 10) / 10 === 3 || Math.round(s2 * 10) / 10 === 4 || Math.round(s2 * 10) / 10 === 5 || Math.round(s2 * 10) / 10 === 6) {\n correction3 = ` ${numAlpha(2)} 11111The motorist can return to the road (without being in violation) $${Math.round(s2 * 10) / 10} \\\\text{ h }$ after consuming alcohol, i.e. at $${Math.round(s2 * 10) / 10 + h} \\\\text{ h}$.<br><br>`\n } else {\n correction3 = ` ${numAlpha(2)} $${texNombre(Math.round(s2 * 10) / 10, 1)}\\\\text{ h } =${Math.floor(s2)} \\\\text{ h } +${texNombre(Math.round(s2 * 10) / 10 - Math.floor(s2))}\\\\text{ h }$.<br>Or, $${texNombre(Math.round(s2 * 10) / 10 - Math.floor(s2))}\\\\text{ h }=${texNombre(Math.round(s2 * 10) / 10 - Math.floor(s2))}\\\\times 60 \\\\text{ min }=${texNombre((Math.round(s2 * 10) / 10 - Math.floor(s2)) * 60)} \\\\text{ min }$.<br>The motorist can return to the road (without being in violation) $${Math.floor(s2)} \\\\text{ h }$ and $${texNombre((Math.round(s2 * 10) / 10 - Math.floor(s2)) * 60)} \\\\text{ min }$ after consumption alcohol, i.e. $${Math.floor(s2 + h)} \\\\text{ h }$ and $${texNombre((Math.round(s2 * 10) / 10 - Math.floor(s2)) * 60)} \\\\text{ min}$.<br><br>`\n }\n\n correction3 += `${graphiqueCorr}`\n this.listeQuestions = [question1, question2, question3]\n this.listeCorrections = [correction1, correction2, correction3]\n }\n break\n\n case 'typeE10':// silo à grain\n {\n const m = new Decimal(randint(27, 38, 30)).add(choice([0.2, 0.4, 0.6, 0.8])) // kg de grains mangés par jour\n const p = 5 * m * randint(6, 11)// capacité du silo\n const j = randint(15, 25, 20)// nbre de jours pour l'image\n const ant = p - m * randint(12, 17)\n const rest = randint(251, 299)\n\n this.introduction = ` A chicken farmer decides to fill his grain silo.<br>By noting $t$ the number of days elapsed after filling his grain silo and $f(t)$ the mass (in kg) remaining after $t $ days, we have: $f(t)=${texNombre(p, 0)}-${texNombre(m, 1)}t$<br>`\n question1 = ` ${numAlpha(0)} Calculate the image of $${j}$ by $f$. Interpret the result in the context of the exercise.<br>${numAlpha(1)} Calculate the antecedent of $${texNombre(ant, 0)}$ by $f$.<br>`\n question2 = 'Knowing that the breeder had filled his silo to its maximum capacity, what is the capacity (in kg) of the silo?<br>'\n question3 = ' After how many days will the farmer run out of grain? Justify.<br>'\n question4 = ' How much grain in kg do chickens consume in a day?<br>'\n question5 = ` One day, foxes killed half of the chickens, halving the amount of grain consumed per day.<br>He had $${rest}$ kg of grain left. <br>Give the function which models the quantity of grains remaining as a function of the number of days. <br>We will denote $g$ for this function. `\n correction1 = ` ${numAlpha(0)} $f(${j})=${texNombre(p, 0)}-${texNombre(m, 1)}\\\\times ${j}=${texNombre(p - m * j, 2)}$.<br>After $${j}$ days, $${texNombre(p - m * j, 2)}$ kg of grain remains in the silo.<br><br>${numAlpha(1)} antecedent of $${texNombre(ant, 0)}$ is the solution to the equation $f(x)=${texNombre(ant, 0)}$. <br><br>$\\\\begin{aligned}${texNombre(p, 0)}-${texNombre(m, 1)}t&=${texNombre(ant, 0)}\\\\\\\\-${texNombre(m, 1)}t&=${texNombre(ant, 0)}-${texNombre(p, 0)}\\\\\\\\t&=\\\\dfrac{${texNombre(ant - p, 1)}}{-${texNombre(m, 1)}}\\\\\\\\t&= ${texNombre((p - ant) / m, 0)}\\\\end{aligned}$<br>The antecedent of $${texNombre(ant, 0)}$ is $${texNombre((p - ant) / m, 0)}$. `\n correction2 = ` The capacity of the silo is given by $f(0)$. <br>As $f(0)=${texNombre(p, 0)}-${texNombre(m, 1)}\\\\times 0=${texNombre(p, 0)}$, the capacity of the silo is $${texNombre(p, 0)}$ kg. `\n correction3 = `We are looking for $t$ such that $f(t)=0$.<br><br>$\\\\begin{aligned}${texNombre(p, 0)}-${texNombre(m, 1)}t&=0\\\\\\\\-${texNombre(m, 1)}x&=-${texNombre(p, 0)}\\\\\\\\t&=\\\\dfrac{${texNombre(-p, 0)}}{-${texNombre(m, 1)}}\\\\\\\\t&=${texNombre(-p / (-m), 0)}\\\\end{aligned}$<br>After $${texNombre(p / m, 0)}$ days, the breeder will run out of grain. `\n correction4 = ` Every day chickens consume $${texNombre(m, 1)}$ kg of grains. <br>For example, the mass of grains eaten on the first day is given by $f(0)-f(1)$.<br><br>$\\\\begin{aligned}f(0)-f(1) )&=(${texNombre(p, 0)}-${texNombre(m, 1)}\\\\times 0)-(${texNombre(p, 0)}-${texNombre(m, 1)}\\\\times 1)\\\\\\\\&=${texNombre(p, 0)}-${texNombre(p - m, 1)}\\\\\\\\&=${texNombre(m, 1)}\\\\end{aligned}$`\n correction5 = ` The function $g$ is given by: <br>$g(t)=(\\\\text{mass in the silo})-(\\\\text{mass consumed by the chickens in one day})\\\\times t$ .<br>Thus, $g(t)=${rest}-${texNombre(m / 2, 1)}t$. `\n this.listeQuestions = [question1, question2, question3, question4, question5]\n this.listeCorrections = [correction1, correction2, correction3, correction4, correction5]\n }\n break\n }\n\n cpt++\n }\n listeQuestionsToContenu(this)\n }\n this.besoinFormulaireNumerique = ['Choice of questions', 3, '1: With affine functions\\n2: With second degree polynomial functions\\n3: With a 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