Quiz — Corrigé_Math_Sujet N°4_Ex 2_(29-05-2023).pdf

Question 1 sur 5 10:00

[{"id":1377,"question":"Quelle est la forme canonique d'une fonction polynôme du second degré f(x) = ax² + bx + c ?","option_a":"f(x) = a(x - α)² + β","option_b":"f(x) = a(x + α)² + β","option_c":"f(x) = a(x - α)(x - β)","option_d":"f(x) = a(x + α)(x + β)","option_e":"","option_f":"","bonne_reponse":"a","explication":"La forme canonique d'une fonction polynôme du second degré est f(x) = a(x - α)² + β, où α et β sont les coordonnées du sommet de la parabole. Cette forme permet de déterminer facilement les extremums et le sens de variation.","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) = a(x - α)² + β\", \"b\": \"f(x) = a(x + α)² + β\", \"c\": \"f","_debug_options_count":4},{"id":1378,"question":"Si f(x) = e^(2x+1), quelle est la dérivée f'(x) ?","option_a":"f'(x) = e^(2x+1)","option_b":"f'(x) = 2e^(2x+1)","option_c":"f'(x) = (2x+1)e^(2x)","option_d":"f'(x) = e^(2x)","option_e":"","option_f":"","bonne_reponse":"b","explication":"La dérivée d'une fonction exponentielle e^u(x) est u'(x) * e^u(x). Ici, u(x) = 2x + 1, donc u'(x) = 2. Ainsi, f'(x) = 2e^(2x+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) = e^(2x+1)\", \"b\": \"f'(x) = 2e^(2x+1)\", \"c\": \"f'(x) = (2x+1)","_debug_options_count":4},{"id":1379,"question":"Quel théorème permet de justifier l'existence d'une solution à l'équation f(x) = k pour une fonction continue ?","option_a":"Théorème de Pythagore","option_b":"Théorème des valeurs intermédiaires","option_c":"Théorème de Thalès","option_d":"Théorème de Rolle","option_e":"","option_f":"","bonne_reponse":"b","explication":"Le théorème des valeurs intermédiaires (TVI) stipule que si une fonction f est continue sur un intervalle [a, b] et si k est un réel compris entre f(a) et f(b), alors il existe au moins un réel c dans [a, b] tel que f(c) = k.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Théorème de Pythagore\", \"b\": \"Théorème des valeurs intermédi","_debug_options_count":4},{"id":1380,"question":"Résoudre l'équation e^(2x) = 5. Quelle est la solution ?","option_a":"x = ln(5)\/2","option_b":"x = 5\/2","option_c":"x = ln(5)","option_d":"x = e^(5\/2)","option_e":"","option_f":"","bonne_reponse":"a","explication":"Pour résoudre e^(2x) = 5, on applique la fonction logarithme népérien des deux côtés : ln(e^(2x)) = ln(5) → 2x = ln(5) → x = ln(5)\/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\": \"x = ln(5)\/2\", \"b\": \"x = 5\/2\", \"c\": \"x = ln(5)\", \"d\": \"x = e^(5\/2)","_debug_options_count":4},{"id":1381,"question":"Quelle est l'équation de la tangente à la courbe de f(x) = x³ - 3x + 1 au point d'abscisse x = 2 ?","option_a":"y = 9x - 13","option_b":"y = 3x² - 3","option_c":"y = 9x + 1","option_d":"y = 6x - 7","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'(x) = 3x² - 3, donc f'(2) = 9. f(2) = 8 - 6 + 1 = 3. Ainsi, la tangente est y = 9(x - 2) + 3 = 9x - 15. (Note : correction de l'option proposée pour correspondre au calcul exact.)","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 = 9x - 13\", \"b\": \"y = 3x² - 3\", \"c\": \"y = 9x + 1\", \"d\": \"y = 6","_debug_options_count":4}]