Quiz — 65a781a4dbef1_corrigé_Série N°23 Etudes des fcts.pdf

Question 1 sur 5 10:00

[{"id":1804,"question":"Quelle est la dérivée de la fonction f(x) = x³ - 2x² + 5x - 1 ?","option_a":"f'(x) = 3x² - 4x + 5","option_b":"f'(x) = 3x² - 4x + 1","option_c":"f'(x) = x² - 4x + 5","option_d":"f'(x) = 3x² - 2x + 5","option_e":"","option_f":"","bonne_reponse":"a","explication":"La dérivée d'un polynôme s'obtient en appliquant la formule (x^n)' = n*x^(n-1) à chaque terme. Ainsi, f'(x) = 3x² - 4x + 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) = 3x² - 4x + 5\", \"b\": \"f'(x) = 3x² - 4x + 1\", \"c\": \"f'(x)","_debug_options_count":4},{"id":1805,"question":"Soit f(x) = (2x + 1)\/(x - 3). Quelle est sa dérivée f'(x) ?","option_a":"f'(x) = (2x - 7)\/(x - 3)²","option_b":"f'(x) = (2x + 7)\/(x - 3)²","option_c":"f'(x) = (2x - 7)\/(x + 3)²","option_d":"f'(x) = (2x + 7)\/(x + 3)²","option_e":"","option_f":"","bonne_reponse":"a","explication":"On utilise la formule de dérivation d'un quotient : (u\/v)' = (u'v - uv')\/v². Ici, u = 2x + 1 et v = x - 3, donc f'(x) = (2*(x-3) - (2x+1)*1)\/(x-3)² = (2x - 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\": \"a\", \"options\": {\"a\": \"f'(x) = (2x - 7)\/(x - 3)²\", \"b\": \"f'(x) = (2x + 7)\/(x - 3)²\", \"","_debug_options_count":4},{"id":1806,"question":"Quelle est la condition pour qu'une fonction f admette un extremum local en un point a ?","option_a":"f'(a) = 0 et f''(a) ≠ 0","option_b":"f'(a) = 0","option_c":"f''(a) = 0","option_d":"f(a) = 0","option_e":"","option_f":"","bonne_reponse":"a","explication":"Un extremum local en a nécessite que la dérivée s'annule en ce point (f'(a) = 0) et que la dérivée seconde soit non nulle (f''(a) ≠ 0) pour déterminer la nature de l'extremum.","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'(a) = 0 et f''(a) ≠ 0\", \"b\": \"f'(a) = 0\", \"c\": \"f''(a) = 0\", ","_debug_options_count":4},{"id":1807,"question":"Soit f(x) = x⁴ - 4x³. En quel(s) point(s) cette fonction admet-elle un extremum local ?","option_a":"x = 0 et x = 3","option_b":"x = 1 et x = 2","option_c":"x = 0 et x = 1","option_d":"x = 2 et x = 3","option_e":"","option_f":"","bonne_reponse":"b","explication":"On calcule f'(x) = 4x³ - 12x² = 4x²(x - 3). Les points critiques sont x = 0 et x = 3. En étudiant le signe de f'(x), on trouve que f admet un minimum local en x = 3 et un point d'inflexion 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\": \"x = 0 et x = 3\", \"b\": \"x = 1 et x = 2\", \"c\": \"x = 0 et x = 1\", \"d","_debug_options_count":4},{"id":1808,"question":"Quelle est l'équation de la tangente à la courbe de f(x) = ln(x) au point d'abscisse x = e ?","option_a":"y = (1\/e)x + 1","option_b":"y = (1\/e)x - 1","option_c":"y = x\/e","option_d":"y = (1\/e)x + e","option_e":"","option_f":"","bonne_reponse":"a","explication":"La tangente en x = a a pour équation y = f'(a)(x - a) + f(a). Ici, f'(x) = 1\/x, donc f'(e) = 1\/e et f(e) = 1. Ainsi, l'équation est y = (1\/e)(x - e) + 1 = (1\/e)x - 1 + 1 = (1\/e)x.","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 = (1\/e)x + 1\", \"b\": \"y = (1\/e)x - 1\", \"c\": \"y = x\/e\", \"d\": \"y =","_debug_options_count":4}]