Quiz — تصحيح المضوع8.pdf

Question 1 sur 5 10:00

[{"id":698,"question":"Quelle est la dérivée de la fonction f(x) = sin(2x) ?","option_a":"f'(x) = 2cos(2x)","option_b":"f'(x) = cos(2x)","option_c":"f'(x) = 2sin(2x)","option_d":"f'(x) = -2cos(2x)","option_e":"","option_f":"","bonne_reponse":"A","explication":"La dérivée de sin(u) est u'cos(u). Ici u = 2x donc u' = 2. Ainsi f'(x) = 2cos(2x).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":699,"question":"Résoudre l'équation sin(x) = √2\/2 dans l'intervalle [0, 2π].","option_a":"x = π\/4 ou x = 3π\/4","option_b":"x = π\/6 ou x = 5π\/6","option_c":"x = π\/3 ou x = 2π\/3","option_d":"x = 0 ou x = π","option_e":"","option_f":"","bonne_reponse":"B","explication":"Les solutions de sin(x) = √2\/2 dans [0, 2π] sont x = π\/4 + 2kπ ou x = 3π\/4 + 2kπ, mais dans l'intervalle donné, seules π\/4 et 3π\/4 sont valides. Cependant, √2\/2 correspond à sin(π\/4) et sin(3π\/4), donc la réponse correcte est x = π\/4 ou x = 3π\/4.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":700,"question":"Quelle est la limite de (sin(x)\/x) lorsque x tend vers 0 ?","option_a":"0","option_b":"1","option_c":"∞","option_d":"-1","option_e":"","option_f":"","bonne_reponse":"B","explication":"C'est une limite fondamentale en analyse : lim(x→0) sin(x)\/x = 1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":701,"question":"Soit f(x) = cos²(x). Quelle est sa dérivée ?","option_a":"f'(x) = -2sin(x)cos(x)","option_b":"f'(x) = 2cos(x)","option_c":"f'(x) = -sin(2x)","option_d":"f'(x) = sin(2x)","option_e":"","option_f":"","bonne_reponse":"C","explication":"En utilisant la formule de dérivation de cos²(x), on obtient f'(x) = 2cos(x)(-sin(x)) = -sin(2x) (car 2sin(x)cos(x) = sin(2x)).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":702,"question":"Quelle est la solution générale de l'équation cos(x) = -1\/2 ?","option_a":"x = 2π\/3 + 2kπ ou x = 4π\/3 + 2kπ","option_b":"x = π\/3 + 2kπ ou x = 5π\/3 + 2kπ","option_c":"x = π\/6 + 2kπ ou x = 11π\/6 + 2kπ","option_d":"x = π\/2 + 2kπ ou x = 3π\/2 + 2kπ","option_e":"","option_f":"","bonne_reponse":"A","explication":"Les solutions de cos(x) = -1\/2 sont x = 2π\/3 + 2kπ ou x = 4π\/3 + 2kπ pour tout entier k.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]