Quiz — Calcul asymptotique.pdf

Question 1 sur 10 20:00

[{"id":64676,"question":"Quelle est la limite de la fonction f(x) = (2x² + 3x - 1)\/(x² - 5) lorsque x tend vers l'infini ?","option_a":"A. 0","option_b":"B. 2","option_c":"C. +∞","option_d":"D. -∞","option_e":"","option_f":"","bonne_reponse":"b","explication":"En divisant numérateur et dénominateur par x², on obtient (2 + 3\/x - 1\/x²)\/(1 - 5\/x²). Lorsque x tend vers l'infini, les termes en 1\/x et 1\/x² tendent vers 0, donc la limite est 2\/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. 2\", \"c\": \"C. +∞\", \"d\": \"D. -∞\"}}","_debug_options_count":4},{"id":64677,"question":"La fonction f(x) = ln(x) admet-elle une asymptote verticale en x = 0 ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La fonction ln(x) n'est pas définie en x = 0, mais sa limite lorsque x tend vers 0+ est -∞. Elle admet donc une asymptote verticale d'équation x = 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":64678,"question":"Quel développement limité d'ordre 2 de la fonction f(x) = e^x au voisinage de 0 est correct ?","option_a":"A. 1 + x + x²\/2","option_b":"B. 1 + x - x²\/2","option_c":"C. x + x²\/2","option_d":"D. 1 - x + x²\/2","option_e":"","option_f":"","bonne_reponse":"a","explication":"Le développement limité d'ordre 2 de e^x au voisinage de 0 est 1 + x + x²\/2 + o(x²).","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. 1 + x + x²\/2\", \"b\": \"B. 1 + x - x²\/2\", \"c\": \"C. x + x²\/2\", ","_debug_options_count":4},{"id":64679,"question":"La fonction f(x) = x³ - 2x + 1 admet-elle une asymptote oblique ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"Une fonction polynôme n'admet jamais d'asymptote oblique, car son comportement à l'infini est dominé par le terme de plus haut degré.","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":64680,"question":"Quelle est la limite de la fonction f(x) = (x² + 1)\/x lorsque x tend vers 0 ?","option_a":"A. 0","option_b":"B. 1","option_c":"C. +∞","option_d":"D. -∞","option_e":"","option_f":"","bonne_reponse":"c","explication":"Lorsque x tend vers 0, le numérateur tend vers 1 et le dénominateur tend vers 0. La fonction tend donc vers +∞ ou -∞ selon le signe de x. Ici, la limite est +∞.","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. +∞\", \"d\": \"D. -∞\"}}","_debug_options_count":4},{"id":64681,"question":"Le développement limité d'ordre 1 de la fonction f(x) = sin(x) au voisinage de 0 est :","option_a":"A. x - x³\/6","option_b":"B. x","option_c":"C. 1 - x²\/2","option_d":"D. x + x²\/2","option_e":"","option_f":"","bonne_reponse":"b","explication":"Le développement limité d'ordre 1 de sin(x) au voisinage de 0 est x + o(x), soit simplement x pour l'ordre 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\": \"A. x - x³\/6\", \"b\": \"B. x\", \"c\": \"C. 1 - x²\/2\", \"d\": \"D. x + x²","_debug_options_count":4},{"id":64682,"question":"La fonction f(x) = 1\/x admet-elle une asymptote horizontale ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La limite de f(x) = 1\/x lorsque x tend vers ±∞ est 0. La droite d'équation y = 0 est donc une asymptote horizontale.","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":64683,"question":"Quelle est la limite de la fonction f(x) = (e^x - 1)\/x lorsque x tend vers 0 ?","option_a":"A. 0","option_b":"B. 1","option_c":"C. e","option_d":"D. +∞","option_e":"","option_f":"","bonne_reponse":"b","explication":"Cette limite est une forme indéterminée 0\/0. En utilisant le développement limité de e^x au voisinage de 0 (1 + x + o(x)), on obtient (1 + x - 1)\/x = 1 + o(1), donc la limite est 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\": \"A. 0\", \"b\": \"B. 1\", \"c\": \"C. e\", \"d\": \"D. +∞\"}}","_debug_options_count":4},{"id":64684,"question":"Le développement limité d'ordre 2 de la fonction f(x) = cos(x) au voisinage de 0 est :","option_a":"A. 1 - x²\/2","option_b":"B. 1 + x²\/2","option_c":"C. x - x³\/6","option_d":"D. 1 - x + x²\/2","option_e":"","option_f":"","bonne_reponse":"a","explication":"Le développement limité d'ordre 2 de cos(x) au voisinage de 0 est 1 - x²\/2 + o(x²).","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. 1 - x²\/2\", \"b\": \"B. 1 + x²\/2\", \"c\": \"C. x - x³\/6\", \"d\": \"D.","_debug_options_count":4},{"id":64685,"question":"La fonction f(x) = x ln(x) admet-elle une asymptote verticale en x = 0 ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"La fonction f(x) = x ln(x) n'est pas définie en x = 0, mais sa limite lorsque x tend vers 0+ est 0 (car x ln(x) tend vers 0). Elle n'admet donc pas d'asymptote verticale en x = 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\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4}]