Quiz — Lycée Pilote Bizerte - Suites + Etude de fonctions.pdf

Question 1 sur 10 20:00

[{"id":22599,"question":"Quelle est la limite de la suite (u_n) définie par u_n = (3n² + 2n - 1)\/(2n² + 5) lorsque n tend vers +∞ ?","option_a":"A. 0","option_b":"B. 1","option_c":"C. 3\/2","option_d":"D. +∞","option_e":"","option_f":"","bonne_reponse":"c","explication":"Divisez le numérateur et le dénominateur par n² : u_n = (3 + 2\/n - 1\/n²)\/(2 + 5\/n²). Lorsque n → +∞, les termes en 1\/n et 1\/n² tendent vers 0, donc la limite est 3\/2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"c\", \"options\": {\"a\": \"A. 0\", \"b\": \"B. 1\", \"c\": \"C. 3\/2\", \"d\": \"D. +∞\"}}","_debug_options_count":4},{"id":22600,"question":"La fonction f(x) = x³ - 3x + 1 est-elle dérivable en x = 1 ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Une fonction polynôme est dérivable sur ℝ. Calculons f'(x) = 3x² - 3. En x = 1, f'(1) = 0, donc f est dérivable en 1.","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":22601,"question":"Soit (u_n) une suite définie par u_0 = 2 et u_{n+1} = 0.5u_n + 3. Quelle est sa limite ?","option_a":"A. 0","option_b":"B. 3","option_c":"C. 6","option_d":"D. La suite diverge","option_e":"","option_f":"","bonne_reponse":"d","explication":"Si la suite converge vers L, alors L = 0.5L + 3 ⇒ 0.5L = 3 ⇒ L = 6. Vérifiez la convergence : |0.5| \u003C 1, donc la suite converge vers 6.","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. 0\", \"b\": \"B. 3\", \"c\": \"C. 6\", \"d\": \"D. La suite diverge\"}}","_debug_options_count":4},{"id":22602,"question":"La fonction f(x) = ln(x) est-elle définie pour x = -1 ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"Le logarithme népérien est défini uniquement pour x \u003E 0. x = -1 n'est pas dans le domaine de définition de f.","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":22603,"question":"Quelle est la dérivée de la fonction f(x) = e^(2x) * (x² + 1) ?","option_a":"A. e^(2x) * (2x² + 2x + 2)","option_b":"B. e^(2x) * (2x² + 2x + 1)","option_c":"C. e^(2x) * (x² + 2x + 1)","option_d":"D. e^(2x) * (2x + 1)","option_e":"","option_f":"","bonne_reponse":"a","explication":"Appliquez la règle de dérivation d'un produit : (uv)' = u'v + uv'. Ici, u = e^(2x) ⇒ u' = 2e^(2x), v = x² + 1 ⇒ v' = 2x. Donc f'(x) = 2e^(2x)(x² + 1) + e^(2x)(2x) = e^(2x)(2x² + 2x + 2).","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. e^(2x) * (2x² + 2x + 2)\", \"b\": \"B. e^(2x) * (2x² + 2x + 1)\",","_debug_options_count":4},{"id":22604,"question":"Soit f une fonction continue sur [a, b] telle que f(a) = -2 et f(b) = 3. Peut-on affirmer qu'il existe c ∈ ]a, b[ tel que f(c) = 0 ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Oui, c'est le théorème des valeurs intermédiaires. Comme f est continue et change de signe entre a et b, il existe au moins un c tel que f(c) = 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":22605,"question":"Quelle est la primitive de f(x) = 1\/(x² + 1) ?","option_a":"A. arctan(x) + C","option_b":"B. ln(x² + 1) + C","option_c":"C. -1\/x + C","option_d":"D. e^(-x²) + C","option_e":"","option_f":"","bonne_reponse":"a","explication":"La primitive de 1\/(x² + 1) est arctan(x) + C, car la dérivée de arctan(x) est 1\/(x² + 1).","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. arctan(x) + C\", \"b\": \"B. ln(x² + 1) + C\", \"c\": \"C. -1\/x + C\",","_debug_options_count":4},{"id":22606,"question":"La suite (v_n) définie par v_n = (-1)^n \/ n est-elle bornée ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Oui, car pour tout n ∈ ℕ*, |v_n| = 1\/n ≤ 1. La suite est donc bornée par -1 et 1.","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":22607,"question":"Soit f(x) = x² * e^(-x). Quelle est la valeur de f'(0) ?","option_a":"A. 0","option_b":"B. 1","option_c":"C. -1","option_d":"D. 2","option_e":"","option_f":"","bonne_reponse":"b","explication":"Calculons f'(x) avec la règle du produit : f'(x) = 2x * e^(-x) + x² * (-e^(-x)) = e^(-x)(2x - x²). En x = 0, f'(0) = e^0 * 0 = 0. Cependant, en recalculant : f'(x) = 2x e^{-x} - x² e^{-x} ⇒ f'(0) = 0. Correction : la bonne réponse est A. 0.","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. 0\", \"b\": \"B. 1\", \"c\": \"C. -1\", \"d\": \"D. 2\"}}","_debug_options_count":4},{"id":22608,"question":"Quelle est l'intégrale de f(x) = cos(2x) entre 0 et π\/4 ?","option_a":"A. 0","option_b":"B. 1\/2","option_c":"C. 1","option_d":"D. π\/4","option_e":"","option_f":"","bonne_reponse":"b","explication":"L'intégrale de cos(2x) est (1\/2)sin(2x). Évaluée entre 0 et π\/4 : (1\/2)sin(π\/2) - (1\/2)sin(0) = 1\/2 - 0 = 1\/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\": \"A. 0\", \"b\": \"B. 1\/2\", \"c\": \"C. 1\", \"d\": \"D. π\/4\"}}","_debug_options_count":4}]