Révisons les maths : Algèbre, Analyse et Géométrie pour la 4ème année secondaire

Question 1 sur 10 10:00

[{"id":25398,"question":"Quelle est la solution de l'équation 2x² - 5x + 2 = 0 ?","option_a":"x = 1 ou x = 2","option_b":"x = 2 ou x = 1\/2","option_c":"x = -1 ou x = -2","option_d":"x = 1\/2 ou x = 2","option_e":"","option_f":"","bonne_reponse":"B","explication":"La solution s'obtient en utilisant la formule du discriminant Δ = b² - 4ac. Ici, Δ = 25 - 16 = 9, donc x = (5 ± 3)\/4, soit x = 2 ou x = 1\/2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":25399,"question":"La fonction f(x) = x³ - 3x est croissante sur l'intervalle [-2, 2].","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"La dérivée f'(x) = 3x² - 3 s'annule en x = ±1. La fonction est décroissante sur [-1, 1], donc elle n'est pas croissante sur [-2, 2].","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":25400,"question":"Quel est le produit scalaire de deux vecteurs u(2, -1) et v(-3, 4) ?","option_a":"-10","option_b":"10","option_c":"-2","option_d":"2","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le produit scalaire se calcule par u·v = (2)(-3) + (-1)(4) = -6 - 4 = -10.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":25401,"question":"La limite de la fonction f(x) = (x² - 4)\/(x - 2) lorsque x tend vers 2 est égale à :","option_a":"0","option_b":"4","option_c":"2","option_d":"indéterminée","option_e":"","option_f":"","bonne_reponse":"C","explication":"En factorisant, f(x) = (x - 2)(x + 2)\/(x - 2) = x + 2 pour x ≠ 2. Donc lim(x→2) f(x) = 4.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":25402,"question":"L'équation de la tangente à la courbe de f(x) = x² + 3x au point d'abscisse x = 1 est :","option_a":"y = 5x - 1","option_b":"y = 5x + 1","option_c":"y = x + 5","option_d":"y = 2x + 3","option_e":"","option_f":"","bonne_reponse":"A","explication":"La pente est f'(1) = 2(1) + 3 = 5. L'équation est y = f(1) + f'(1)(x - 1) = 4 + 5(x - 1) = 5x - 1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":25403,"question":"La probabilité d'obtenir un nombre pair en lançant un dé équilibré est de 1\/2.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Un dé a 6 faces : 1, 2, 3, 4, 5, 6. Les nombres pairs sont 2, 4, 6, soit 3 faces. La probabilité est donc 3\/6 = 1\/2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":25404,"question":"Quel est le développement limité de la fonction f(x) = e^x au voisinage de 0 à l'ordre 2 ?","option_a":"1 + x + x²\/2","option_b":"1 + x - x²\/2","option_c":"1 - x + x²\/2","option_d":"x + x²\/2","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le développement limité de e^x est 1 + x + x²\/2 + o(x²).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":25405,"question":"La fonction f(x) = 1\/x est définie sur ℝ.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"La fonction f(x) = 1\/x n'est pas définie en x = 0, donc elle n'est pas définie sur tout ℝ.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":25406,"question":"Quel est le résultat de l'intégrale ∫(3x² + 2x) dx ?","option_a":"x³ + x² + C","option_b":"x³ + x + C","option_c":"3x³ + x² + C","option_d":"x³ + 2x² + C","option_e":"","option_f":"","bonne_reponse":"A","explication":"L'intégrale de 3x² est x³ et celle de 2x est x². Donc ∫(3x² + 2x) dx = x³ + x² + C.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":25407,"question":"La fonction f(x) = ln(x) est définie pour tout x ∈ ℝ.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"La fonction ln(x) est définie uniquement pour x \u003E 0.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]