Quiz — 664f2bb1ed6cd_énoncé_prototypes d'examen 6.pdf

Question 1 sur 5 10:00

[{"id":2225,"question":"Quelle est la limite de la suite définie par uₙ = (2n² + 3n - 1)\/(5n² - 2n + 4) lorsque n tend vers l'infini ?","option_a":"A) 2\/5","option_b":"B) 0","option_c":"C) +∞","option_d":"D) 1","option_e":"","option_f":"","bonne_reponse":"a","explication":"Pour trouver la limite d'une suite rationnelle, on compare les termes de plus haut degré au numérateur et au dénominateur. Ici, les termes dominants sont 2n² et 5n², donc la limite est 2\/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\": \"A) 2\/5\", \"b\": \"B) 0\", \"c\": \"C) +∞\", \"d\": \"D) 1\"}}","_debug_options_count":4},{"id":2226,"question":"Résolvez l'équation différentielle y' + 2y = 0 avec la condition initiale y(0) = 3.","option_a":"A) y = 3e^(-2x)","option_b":"B) y = 3e^(2x)","option_c":"C) y = -2x + 3","option_d":"D) y = 3 - 2x","option_e":"","option_f":"","bonne_reponse":"a","explication":"Cette équation est de la forme y' + ay = 0. Sa solution générale est y = Ce^(-ax). Avec y(0) = 3, on trouve C = 3, donc y = 3e^(-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\": \"A) y = 3e^(-2x)\", \"b\": \"B) y = 3e^(2x)\", \"c\": \"C) y = -2x + 3\", \"","_debug_options_count":4},{"id":2227,"question":"Soit Z = 3 - 4i un nombre complexe. Quel est son module ?","option_a":"A) 5","option_b":"B) 7","option_c":"C) √7","option_d":"D) 25","option_e":"","option_f":"","bonne_reponse":"a","explication":"Le module d'un nombre complexe Z = a + bi est donné par |Z| = √(a² + b²). Ici, |Z| = √(3² + (-4)²) = √(9 + 16) = √25 = 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\": \"A) 5\", \"b\": \"B) 7\", \"c\": \"C) √7\", \"d\": \"D) 25\"}}","_debug_options_count":4},{"id":2228,"question":"Un sac contient 5 boules rouges et 3 boules bleues. On tire deux boules successivement sans remise. Quelle est la probabilité d'obtenir deux boules rouges ?","option_a":"A) 5\/14","option_b":"B) 25\/64","option_c":"C) 10\/28","option_d":"D) 5\/8","option_e":"","option_f":"","bonne_reponse":"c","explication":"La probabilité de tirer une boule rouge au premier tirage est 5\/8. Sans remise, celle du second tirage est 4\/7. La probabilité conjointe est (5\/8) × (4\/7) = 20\/56 = 5\/14.","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) 5\/14\", \"b\": \"B) 25\/64\", \"c\": \"C) 10\/28\", \"d\": \"D) 5\/8\"}}","_debug_options_count":4},{"id":2229,"question":"Quelle est la dérivée de la fonction f(x) = x²e^(3x) ?","option_a":"A) 2xe^(3x) + 3x²e^(3x)","option_b":"B) e^(3x)(2x + 3x²)","option_c":"C) 2xe^(3x)","option_d":"D) 3x²e^(3x)","option_e":"","option_f":"","bonne_reponse":"b","explication":"On utilise la règle de dérivation d'un produit : (uv)' = u'v + uv'. Ici, u = x² et v = e^(3x). u' = 2x et v' = 3e^(3x). Donc f'(x) = 2xe^(3x) + x² × 3e^(3x) = e^(3x)(2x + 3x²).","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) 2xe^(3x) + 3x²e^(3x)\", \"b\": \"B) e^(3x)(2x + 3x²)\", \"c\": \"C) ","_debug_options_count":4}]