Quiz — serie passe bande 1.pdf

Question 1 sur 10 20:00

[{"id":18369,"question":"Quelle est la limite de la fonction f(x) = (3x² + 2x - 1)\/(x² - 4) quand x tend vers l'infini ?","option_a":"A. 3","option_b":"B. 0","option_c":"C. -1\/4","option_d":"D. +∞","option_e":"","option_f":"","bonne_reponse":"a","explication":"En divisant numérateur et dénominateur par x², on obtient f(x) = (3 + 2\/x - 1\/x²)\/(1 - 4\/x²). Quand x tend vers l'infini, les termes en 1\/x et 1\/x² tendent vers 0, donc la limite est 3\/1 = 3.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"A. 3\", \"b\": \"B. 0\", \"c\": \"C. -1\/4\", \"d\": \"D. +∞\"}}","_debug_options_count":4},{"id":18370,"question":"La fonction f(x) = x³ - 3x + 2 est strictement croissante sur ℝ.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"La dérivée f'(x) = 3x² - 3 s'annule en x = ±1. La fonction n'est pas strictement croissante sur tout ℝ car elle admet des points critiques (minimum local en x=1).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":18371,"question":"Résoudre l'équation ln(x² - 5x + 6) = 0 dans ℝ.","option_a":"A. x = 1 ou x = 4","option_b":"B. x = 2 ou x = 3","option_c":"C. x = -1 ou x = 6","option_d":"D. Aucune solution","option_e":"","option_f":"","bonne_reponse":"d","explication":"ln(y) = 0 ⇒ y = 1. Donc x² - 5x + 6 = 1 ⇒ x² - 5x + 5 = 0. Le discriminant est 25 - 20 = 5 \u003E 0, mais les solutions x = (5 ± √5)\/2 ne vérifient pas x² - 5x + 6 \u003E 0 (car 1 \u003E 0). Aucune solution réelle n'est valide.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"d\", \"options\": {\"a\": \"A. x = 1 ou x = 4\", \"b\": \"B. x = 2 ou x = 3\", \"c\": \"C. x = -1 ou ","_debug_options_count":4},{"id":18372,"question":"Soit la suite définie par u₀ = 2 et uₙ₊₁ = (uₙ + 3)\/(uₙ + 1). Cette suite est convergente.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La suite est bornée et monotone (décroissante à partir de n=1), donc elle converge vers l'unique point fixe de la fonction, solution de l'équation x = (x + 3)\/(x + 1), soit x² - 2x - 3 = 0 ⇒ x = 3 (car x \u003E 0).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":18373,"question":"Quel est le produit scalaire des vecteurs u(2; -1; 3) et v(-1; 4; 0) dans l'espace ?","option_a":"A. -6","option_b":"B. 2","option_c":"C. 6","option_d":"D. -2","option_e":"","option_f":"","bonne_reponse":"b","explication":"Le produit scalaire est calculé par u·v = (2)(-1) + (-1)(4) + (3)(0) = -2 -4 + 0 = -6. Aucune des options ne correspond, donc la bonne réponse est A (-6).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"A. -6\", \"b\": \"B. 2\", \"c\": \"C. 6\", \"d\": \"D. -2\"}}","_debug_options_count":4},{"id":18374,"question":"La fonction f(x) = e^(2x) - 3e^x + 2 admet un minimum en x = ln(1.5).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La dérivée f'(x) = 2e^(2x) - 3e^x s'annule en e^x = 3\/2 ⇒ x = ln(1.5). La dérivée seconde f''(x) = 4e^(2x) - 3e^x \u003E 0 pour x = ln(1.5), donc c'est un minimum.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":18375,"question":"Résoudre l'inéquation (x² - 4x + 3)\/(x - 2) \u003E 0.","option_a":"A. x ∈ ]1; 2[ ∪ ]3; +∞[","option_b":"B. x ∈ ]-∞; 1[ ∪ ]2; 3[","option_c":"C. x ∈ ]1; 3[","option_d":"D. x ∈ ]-∞; 1[ ∪ ]3; +∞[","option_e":"","option_f":"","bonne_reponse":"a","explication":"Le numérateur x² - 4x + 3 = (x-1)(x-3) s'annule en x=1 et x=3. Le dénominateur x-2 s'annule en x=2. Le tableau de signes montre que l'inéquation est vérifiée pour x ∈ ]1; 2[ ∪ ]3; +∞[.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"A. x ∈ ]1; 2[ ∪ ]3; +∞[\", \"b\": \"B. x ∈ ]-∞; 1[ ∪ ]2; ","_debug_options_count":4},{"id":18376,"question":"Soit f une fonction dérivable sur ℝ telle que f(0) = 1 et f'(0) = -2. L'équation de la tangente à la courbe de f en x=0 est y = -2x + 1.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"L'équation de la tangente en x=0 est y = f'(0)(x - 0) + f(0) = -2x + 1. L'affirmation est donc vraie.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":18377,"question":"Quelle est la valeur de la somme S = Σ (de k=1 à n) k² ?","option_a":"A. n(n+1)\/2","option_b":"B. n(n+1)(2n+1)\/6","option_c":"C. n²(n+1)²\/4","option_d":"D. n(n+1)","option_e":"","option_f":"","bonne_reponse":"b","explication":"La formule de la somme des carrés est S = n(n+1)(2n+1)\/6. Les autres options correspondent à la somme des entiers (A), la somme des cubes (C) ou la somme des entiers (D).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"A. n(n+1)\/2\", \"b\": \"B. n(n+1)(2n+1)\/6\", \"c\": \"C. n²(n+1)²\/4\", \"","_debug_options_count":4},{"id":18378,"question":"La fonction f(x) = |x² - 4| est dérivable en x = 2.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"En x=2, f(x) = x² - 4 pour x ≥ 2 et f(x) = -(x² - 4) pour x \u003C 2. La dérivée à gauche est -4x = -8 et à droite est 4x = 8. Les dérivées à gauche et à droite ne sont pas égales, donc f n'est pas dérivable en x=2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4}]