Quiz — 6627969f39d8e_énoncé_Série N°4-Etude de fonction-Probabilités.pdf

Question 1 sur 5 10:00

[{"id":3281,"question":"Quelle est la dérivée de la fonction f(x) = e^(2x+1) ?","option_a":"A) 2e^(2x+1)","option_b":"B) e^(2x+1)","option_c":"C) 2xe^(2x+1)","option_d":"D) (2x+1)e^(2x)","option_e":"","option_f":"","bonne_reponse":"a","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) = 2 * e^(2x+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\": \"A) 2e^(2x+1)\", \"b\": \"B) e^(2x+1)\", \"c\": \"C) 2xe^(2x+1)\", \"d\": \"D)","_debug_options_count":4},{"id":3282,"question":"Soit X une variable aléatoire suivant une loi binomiale B(n=5, p=0.3). Quelle est la probabilité P(X=2) ?","option_a":"A) 0.1323","option_b":"B) 0.3087","option_c":"C) 0.6471","option_d":"D) 0.8319","option_e":"","option_f":"","bonne_reponse":"b","explication":"Pour une loi binomiale B(n,p), P(X=k) = C(n,k) * p^k * (1-p)^(n-k). Ici, C(5,2) = 10, p^2 = 0.09, (1-p)^3 = 0.343. Donc P(X=2) = 10 * 0.09 * 0.343 = 0.3087.","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) 0.1323\", \"b\": \"B) 0.3087\", \"c\": \"C) 0.6471\", \"d\": \"D) 0.8319\"}","_debug_options_count":4},{"id":3283,"question":"Quelle est la limite de la fonction f(x) = (x^2 - 1)\/(x - 1) lorsque x tend vers 1 ?","option_a":"A) 0","option_b":"B) 1","option_c":"C) 2","option_d":"D) +∞","option_e":"","option_f":"","bonne_reponse":"c","explication":"La fonction f(x) = (x^2 - 1)\/(x - 1) peut être simplifiée en f(x) = (x+1)(x-1)\/(x-1) = x+1 pour x ≠ 1. Ainsi, la limite lorsque x tend vers 1 est 1+1 = 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\": \"A) 0\", \"b\": \"B) 1\", \"c\": \"C) 2\", \"d\": \"D) +∞\"}}","_debug_options_count":4},{"id":3284,"question":"Soit X une variable aléatoire discrète prenant les valeurs 0, 1 et 2 avec les probabilités respectives 0.2, 0.5 et 0.3. Quel est l'espérance E(X) ?","option_a":"A) 0.8","option_b":"B) 1.1","option_c":"C) 1.3","option_d":"D) 1.5","option_e":"","option_f":"","bonne_reponse":"b","explication":"L'espérance E(X) est calculée par E(X) = Σ x_i * P(X=x_i). Ici, E(X) = 0*0.2 + 1*0.5 + 2*0.3 = 0 + 0.5 + 0.6 = 1.1.","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) 0.8\", \"b\": \"B) 1.1\", \"c\": \"C) 1.3\", \"d\": \"D) 1.5\"}}","_debug_options_count":4},{"id":3285,"question":"Quelle est la dérivée de la fonction f(x) = ln(x^2 + 1) ?","option_a":"A) 2x\/(x^2 + 1)","option_b":"B) 1\/(x^2 + 1)","option_c":"C) 2x","option_d":"D) x\/(x^2 + 1)","option_e":"","option_f":"","bonne_reponse":"a","explication":"La dérivée de ln(u(x)) est u'(x)\/u(x). Ici, u(x) = x^2 + 1, donc u'(x) = 2x. Ainsi, f'(x) = 2x\/(x^2 + 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\": \"A) 2x\/(x^2 + 1)\", \"b\": \"B) 1\/(x^2 + 1)\", \"c\": \"C) 2x\", \"d\": \"D) x","_debug_options_count":4}]