Quiz — 683202f59e754_correction sujet 3 partie II ex2+ex3.pdf

Question 1 sur 10 20:00

[{"id":80804,"question":"Quelle est la dérivée de la fonction f(x) = 3x² + 5x - 2 ?","option_a":"f'(x) = 6x + 5","option_b":"f'(x) = 3x + 5","option_c":"f'(x) = 6x² + 5","option_d":"f'(x) = 3x² + 5","option_e":"","option_f":"","bonne_reponse":"a","explication":"La dérivée d'une fonction polynomiale se calcule en appliquant la règle : (ax^n)' = n*ax^(n-1). Pour f(x) = 3x² + 5x - 2, on obtient f'(x) = 6x + 5.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"f'(x) = 6x + 5\", \"b\": \"f'(x) = 3x + 5\", \"c\": \"f'(x) = 6x² + 5\", ","_debug_options_count":4},{"id":80805,"question":"La fonction f(x) = x³ - 3x a-t-elle un extremum en x = 1 ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Pour vérifier un extremum, on calcule f'(x) = 3x² - 3. En x = 1, f'(1) = 0 et f''(1) = 6 \u003E 0, donc la fonction admet un minimum en x = 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":80806,"question":"Quelle est la dérivée de f(x) = (2x + 1)\/(x - 3) ?","option_a":"f'(x) = 2\/(x - 3)","option_b":"f'(x) = -7\/(x - 3)²","option_c":"f'(x) = (2x - 7)\/(x - 3)²","option_d":"f'(x) = 2\/(x - 3)²","option_e":"","option_f":"","bonne_reponse":"c","explication":"En utilisant la formule de dérivation d'un quotient : (u\/v)' = (u'v - uv')\/v². Ici, u = 2x + 1, v = x - 3, u' = 2, v' = 1. On obtient f'(x) = (2(x - 3) - (2x + 1)(1))\/(x - 3)² = -7\/(x - 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\": \"f'(x) = 2\/(x - 3)\", \"b\": \"f'(x) = -7\/(x - 3)²\", \"c\": \"f'(x) = (2","_debug_options_count":4},{"id":80807,"question":"La fonction f(x) = x⁴ - 4x³ est-elle croissante sur l'intervalle [0 ; 3] ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"On calcule f'(x) = 4x³ - 12x² = 4x²(x - 3). Sur [0 ; 3], f'(x) ≤ 0, donc la fonction est décroissante sur cet intervalle.","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":80808,"question":"Quelle est l'équation de la tangente à la courbe de f(x) = x² + 2x en x = 1 ?","option_a":"y = 4x + 1","option_b":"y = 4x - 1","option_c":"y = 2x + 3","option_d":"y = 2x - 1","option_e":"","option_f":"","bonne_reponse":"a","explication":"L'équation de la tangente en x = a est y = f'(a)(x - a) + f(a). Ici, f(1) = 3, f'(x) = 2x + 2, donc f'(1) = 4. L'équation est y = 4(x - 1) + 3 = 4x - 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\": \"y = 4x + 1\", \"b\": \"y = 4x - 1\", \"c\": \"y = 2x + 3\", \"d\": \"y = 2x -","_debug_options_count":4},{"id":80809,"question":"La dérivée de f(x) = e^(2x) est-elle f'(x) = 2e^(2x) ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La dérivée de e^(u(x)) est u'(x)e^(u(x)). Ici, u(x) = 2x, donc u'(x) = 2. Ainsi, f'(x) = 2e^(2x).","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":80810,"question":"Quelle est la dérivée de f(x) = ln(3x + 1) ?","option_a":"f'(x) = 1\/(3x + 1)","option_b":"f'(x) = 3\/(3x + 1)","option_c":"f'(x) = 3x\/(3x + 1)","option_d":"f'(x) = 1\/(x + 1)","option_e":"","option_f":"","bonne_reponse":"b","explication":"La dérivée de ln(u(x)) est u'(x)\/u(x). Ici, u(x) = 3x + 1, donc u'(x) = 3. Ainsi, f'(x) = 3\/(3x + 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\": \"f'(x) = 1\/(3x + 1)\", \"b\": \"f'(x) = 3\/(3x + 1)\", \"c\": \"f'(x) = 3x\/","_debug_options_count":4},{"id":80811,"question":"La fonction f(x) = sin(x) a-t-elle une dérivée nulle en x = π\/2 ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La dérivée de sin(x) est cos(x). En x = π\/2, cos(π\/2) = 0, donc f'(π\/2) = 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":80812,"question":"Quelle est la dérivée de f(x) = √(x² + 1) ?","option_a":"f'(x) = x\/√(x² + 1)","option_b":"f'(x) = 2x\/√(x² + 1)","option_c":"f'(x) = x\/(2√(x² + 1))","option_d":"f'(x) = 1\/√(x² + 1)","option_e":"","option_f":"","bonne_reponse":"a","explication":"En utilisant la formule de dérivation d'une racine : (√u)' = u'\/(2√u). Ici, u = x² + 1, donc u' = 2x. Ainsi, f'(x) = 2x\/(2√(x² + 1)) = x\/√(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\": \"f'(x) = x\/√(x² + 1)\", \"b\": \"f'(x) = 2x\/√(x² + 1)\", \"c\": \"f'","_debug_options_count":4},{"id":80813,"question":"La fonction f(x) = x³ - 3x² + 2x a-t-elle un point d'inflexion en x = 1 ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"Un point d'inflexion correspond à un changement de concavité, donc à f''(x) = 0. On calcule f''(x) = 6x - 6. En x = 1, f''(1) = 0, mais il faut vérifier le changement de signe de f''(x) autour de x = 1. Comme f''(x) change de signe en x = 1, il s'agit bien d'un point d'inflexion.","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}]