Quiz : Maîtrisez les fonctions dérivées en 3ème année secondaire

Question 1 sur 10 10:00

[{"id":43156,"question":"Quelle est la dérivée de la fonction f(x) = 3x² + 2x - 5 ?","option_a":"6x + 2","option_b":"3x² + 2","option_c":"6x + 2x - 5","option_d":"3x + 2","option_e":"","option_f":"","bonne_reponse":"A","explication":"La dérivée d’une fonction polynomiale s’obtient en appliquant la règle : (ax^n)' = n*ax^(n-1). Ici, f'(x) = 6x + 2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":43157,"question":"La dérivée de la fonction f(x) = e^(2x) est f'(x) = 2e^(2x).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"C’est vrai ! La dérivée de e^(u(x)) est u'(x)*e^(u(x)). Ici, u(x) = 2x, donc u'(x) = 2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":43158,"question":"Quelle est la dérivée de la fonction f(x) = (x² + 1)\/(x - 1) ?","option_a":"(2x(x-1) - (x²+1))\/(x-1)²","option_b":"(2x)\/(x-1)","option_c":"(x²+1)\/(x-1)²","option_d":"2x - 1","option_e":"","option_f":"","bonne_reponse":"A","explication":"On utilise la règle du quotient : (u\/v)' = (u'v - uv')\/v². Ici, u = x²+1 et v = x-1, donc u' = 2x et v' = 1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":43159,"question":"Si f'(x) = 0 pour tout x, alors la fonction f est constante.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"C’est vrai ! Si la dérivée est nulle partout, la fonction ne varie pas, donc elle est constante.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":43160,"question":"Quelle est la dérivée de la fonction f(x) = ln(3x + 2) ?","option_a":"3\/(3x + 2)","option_b":"1\/(3x + 2)","option_c":"3x\/(3x + 2)","option_d":"1\/x","option_e":"","option_f":"","bonne_reponse":"B","explication":"La dérivée de ln(u(x)) est u'(x)\/u(x). Ici, u(x) = 3x+2, donc u'(x) = 3. Ainsi, f'(x) = 3\/(3x+2).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":43161,"question":"La fonction f(x) = x³ - 3x² + 2x est croissante sur l’intervalle [-1, 1].","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"Faux ! On calcule f'(x) = 3x² - 6x + 2. En testant x = 0, f'(0) = 2 \u003E 0, mais en x = 1, f'(1) = -1 \u003C 0. La fonction n’est pas croissante sur tout l’intervalle.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":43162,"question":"Quelle est la dérivée de la fonction f(x) = sin(2x) ?","option_a":"2cos(2x)","option_b":"cos(2x)","option_c":"2sin(2x)","option_d":"-2cos(2x)","option_e":"","option_f":"","bonne_reponse":"A","explication":"La dérivée de sin(u(x)) est u'(x)*cos(u(x)). Ici, u(x) = 2x, donc u'(x) = 2. Ainsi, f'(x) = 2cos(2x).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":43163,"question":"Si f'(a) \u003E 0, alors la fonction f est croissante en x = a.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"C’est vrai ! Le signe de la dérivée en un point indique le sens de variation de la fonction : positif = croissante, négatif = décroissante.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":43164,"question":"Quelle est la dérivée de la fonction f(x) = √(x² + 1) ?","option_a":"x\/√(x² + 1)","option_b":"1\/√(x² + 1)","option_c":"(2x)\/√(x² + 1)","option_d":"x²\/√(x² + 1)","option_e":"","option_f":"","bonne_reponse":"A","explication":"On utilise la règle de la racine : (√u)' = u'\/(2√u). Ici, u = x²+1, donc u' = 2x. Ainsi, f'(x) = 2x\/(2√(x²+1)) = x\/√(x²+1).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":43165,"question":"La fonction f(x) = x⁴ - 4x³ + 6x² a un point critique en x = 1.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Vrai ! Un point critique est où f'(x) = 0. f'(x) = 4x³ - 12x² + 12x = 4x(x² - 3x + 3). En x = 1, f'(1) = 4(1 - 3 + 3) = 4 ≠ 0. Correction : la question est fausse. f'(1) = 4 ≠ 0, donc x=1 n'est pas un point critique.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]