Quiz — exercice 1 (résolution fonctions exponentieles).pdf

Question 1 sur 10 20:00

[{"id":32170,"question":"Quelle est la solution de l'équation \u003Cem\u003Ee^{x} = 5\u003C\/em\u003E ?","option_a":"A. \u003Cem\u003Ex = ln(5)\u003C\/em\u003E","option_b":"B. \u003Cem\u003Ex = 5\u003C\/em\u003E","option_c":"C. \u003Cem\u003Ex = e^5\u003C\/em\u003E","option_d":"D. \u003Cem\u003Ex = 1\/5\u003C\/em\u003E","option_e":"","option_f":"","bonne_reponse":"a","explication":"La solution de \u003Cem\u003Ee^x = k\u003C\/em\u003E est \u003Cem\u003Ex = ln(k)\u003C\/em\u003E pour \u003Cem\u003Ek \u003E 0\u003C\/em\u003E.","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. \u003Cem\u003Ex = ln(5)\u003C\/em\u003E\", \"b\": \"B. \u003Cem\u003Ex = 5\u003C\/em\u003E\", \"c\": \"C. \u003Cem\u003Ex ","_debug_options_count":4},{"id":32171,"question":"La dérivée de \u003Cem\u003Ef(x) = e^{-x}\u003C\/em\u003E est \u003Cem\u003Ef'(x) = -e^{-x}\u003C\/em\u003E.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La dérivée de \u003Cem\u003Ee^{u(x)}\u003C\/em\u003E est \u003Cem\u003Eu'(x) e^{u(x)}\u003C\/em\u003E. Ici, \u003Cem\u003Eu(x) = -x\u003C\/em\u003E, donc \u003Cem\u003Eu'(x) = -1\u003C\/em\u003E et \u003Cem\u003Ef'(x) = -e^{-x}\u003C\/em\u003E.","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":32172,"question":"Quelle est la limite de \u003Cem\u003Ef(x) = e^{x} - x\u003C\/em\u003E quand \u003Cem\u003Ex → +∞\u003C\/em\u003E ?","option_a":"A. \u003Cem\u003E+∞\u003C\/em\u003E","option_b":"B. \u003Cem\u003E0\u003C\/em\u003E","option_c":"C. \u003Cem\u003E-∞\u003C\/em\u003E","option_d":"D. \u003Cem\u003E1\u003C\/em\u003E","option_e":"","option_f":"","bonne_reponse":"a","explication":"Quand \u003Cem\u003Ex → +∞\u003C\/em\u003E, \u003Cem\u003Ee^x\u003C\/em\u003E domine \u003Cem\u003Ex\u003C\/em\u003E, donc la limite est \u003Cem\u003E+∞\u003C\/em\u003E.","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. \u003Cem\u003E+∞\u003C\/em\u003E\", \"b\": \"B. \u003Cem\u003E0\u003C\/em\u003E\", \"c\": \"C. \u003Cem\u003E-∞\u003C\/em\u003E\",","_debug_options_count":4},{"id":32173,"question":"La fonction \u003Cem\u003Ef(x) = e^{2x+1}\u003C\/em\u003E est-elle strictement croissante sur \u003Cem\u003Eℝ\u003C\/em\u003E ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La dérivée \u003Cem\u003Ef'(x) = 2e^{2x+1} \u003E 0\u003C\/em\u003E pour tout \u003Cem\u003Ex\u003C\/em\u003E, donc \u003Cem\u003Ef\u003C\/em\u003E est strictement croissante.","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":32174,"question":"Quelle est la solution de l'inéquation \u003Cem\u003Ee^{x} ≤ 1\/e\u003C\/em\u003E ?","option_a":"A. \u003Cem\u003Ex ≤ -1\u003C\/em\u003E","option_b":"B. \u003Cem\u003Ex ≤ 0\u003C\/em\u003E","option_c":"C. \u003Cem\u003Ex ≥ 1\u003C\/em\u003E","option_d":"D. \u003Cem\u003Ex ≤ 1\u003C\/em\u003E","option_e":"","option_f":"","bonne_reponse":"a","explication":"\u003Cem\u003Ee^{x} ≤ 1\/e\u003C\/em\u003E équivaut à \u003Cem\u003Ee^{x} ≤ e^{-1}\u003C\/em\u003E, donc \u003Cem\u003Ex ≤ -1\u003C\/em\u003E car la fonction exponentielle est croissante.","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. \u003Cem\u003Ex ≤ -1\u003C\/em\u003E\", \"b\": \"B. \u003Cem\u003Ex ≤ 0\u003C\/em\u003E\", \"c\": \"C. \u003Cem\u003Ex","_debug_options_count":4},{"id":32175,"question":"La limite de \u003Cem\u003Ef(x) = (e^{x} - 1)\/x\u003C\/em\u003E quand \u003Cem\u003Ex → 0\u003C\/em\u003E est égale à 1.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"C'est une forme indéterminée \u003Cem\u003E0\/0\u003C\/em\u003E. En utilisant le taux d'accroissement de l'exponentielle, on trouve \u003Cem\u003Elim_{x→0} (e^x - 1)\/x = 1\u003C\/em\u003E.","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":32176,"question":"Quelle est la dérivée de \u003Cem\u003Ef(x) = e^{x^2}\u003C\/em\u003E ?","option_a":"A. \u003Cem\u003E2x e^{x^2}\u003C\/em\u003E","option_b":"B. \u003Cem\u003Ee^{x^2}\u003C\/em\u003E","option_c":"C. \u003Cem\u003Ex e^{x^2}\u003C\/em\u003E","option_d":"D. \u003Cem\u003E2 e^{x^2}\u003C\/em\u003E","option_e":"","option_f":"","bonne_reponse":"a","explication":"La dérivée de \u003Cem\u003Ee^{u(x)}\u003C\/em\u003E est \u003Cem\u003Eu'(x) e^{u(x)}\u003C\/em\u003E. Ici, \u003Cem\u003Eu(x) = x^2\u003C\/em\u003E, donc \u003Cem\u003Eu'(x) = 2x\u003C\/em\u003E et \u003Cem\u003Ef'(x) = 2x e^{x^2}\u003C\/em\u003E.","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. \u003Cem\u003E2x e^{x^2}\u003C\/em\u003E\", \"b\": \"B. \u003Cem\u003Ee^{x^2}\u003C\/em\u003E\", \"c\": \"C. \u003Cem","_debug_options_count":4},{"id":32177,"question":"La fonction \u003Cem\u003Ef(x) = e^{-x^2}\u003C\/em\u003E admet-elle un maximum en \u003Cem\u003Ex = 0\u003C\/em\u003E ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La dérivée \u003Cem\u003Ef'(x) = -2x e^{-x^2}\u003C\/em\u003E s'annule en \u003Cem\u003Ex = 0\u003C\/em\u003E. Le signe de \u003Cem\u003Ef'\u003C\/em\u003E change de positif à négatif en \u003Cem\u003Ex = 0\u003C\/em\u003E, donc c'est un maximum.","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":32178,"question":"Quelle est la valeur de \u003Cem\u003Ee^{ln(3)} + e^{ln(2)}\u003C\/em\u003E ?","option_a":"A. \u003Cem\u003E5\u003C\/em\u003E","option_b":"B. \u003Cem\u003E6\u003C\/em\u003E","option_c":"C. \u003Cem\u003E3 + 2\u003C\/em\u003E","option_d":"D. \u003Cem\u003Eln(6)\u003C\/em\u003E","option_e":"","option_f":"","bonne_reponse":"b","explication":"\u003Cem\u003Ee^{ln(a)} = a\u003C\/em\u003E pour tout \u003Cem\u003Ea \u003E 0\u003C\/em\u003E, donc \u003Cem\u003Ee^{ln(3)} + e^{ln(2)} = 3 + 2 = 5\u003C\/em\u003E.","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. \u003Cem\u003E5\u003C\/em\u003E\", \"b\": \"B. \u003Cem\u003E6\u003C\/em\u003E\", \"c\": \"C. \u003Cem\u003E3 + 2\u003C\/em\u003E\", \"","_debug_options_count":4},{"id":32179,"question":"La fonction \u003Cem\u003Ef(x) = e^{x}\/x\u003C\/em\u003E admet-elle une asymptote verticale en \u003Cem\u003Ex = 0\u003C\/em\u003E ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Quand \u003Cem\u003Ex → 0^+\u003C\/em\u003E, \u003Cem\u003Ee^x → 1\u003C\/em\u003E et \u003Cem\u003E1\/x → +∞\u003C\/em\u003E, donc \u003Cem\u003Ef(x) → +∞\u003C\/em\u003E. La fonction n'est pas définie en \u003Cem\u003Ex = 0\u003C\/em\u003E, mais admet une asymptote verticale.","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}]