Quiz interactif : Maths Baccalauréat — Algèbre, Analyse & Géométrie

Question 1 sur 10 10:00

[{"id":55211,"question":"Quelle est la solution de l'équation 2x² - 5x + 2 = 0 ?","option_a":"x = 1 ou x = 2","option_b":"x = 0.5 ou x = 2","option_c":"x = -1 ou x = 2","option_d":"x = 1 ou x = -2","option_e":"","option_f":"","bonne_reponse":"B","explication":"Résolvons l'équation : 2x² - 5x + 2 = 0. Le discriminant Δ = (-5)² - 4*2*2 = 25 - 16 = 9. Les solutions sont x = (5 ± √9)\/4, soit x = 2 ou x = 0.5.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":55212,"question":"La fonction f(x) = x³ - 3x est décroissante sur l'intervalle [-1, 1].","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"La dérivée f'(x) = 3x² - 3. Sur [-1, 1], f'(x) ≤ 0, donc la fonction est décroissante sur cet intervalle. L'affirmation est donc fausse.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":55213,"question":"Quelle est la valeur de sin(π\/6) + cos(π\/3) ?","option_a":"1","option_b":"0.5","option_c":"1.5","option_d":"√2\/2","option_e":"","option_f":"","bonne_reponse":"A","explication":"sin(π\/6) = 0.5 et cos(π\/3) = 0.5. Donc sin(π\/6) + cos(π\/3) = 0.5 + 0.5 = 1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":55214,"question":"L'équation x² + y² - 4x + 6y - 3 = 0 représente un cercle de centre (2, -3).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"En complétant le carré : (x-2)² + (y+3)² = 16. Le centre est donc (2, -3) et le rayon est 4. L'affirmation est vraie.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":55215,"question":"Quelle est la dérivée de la fonction f(x) = e^(2x) ?","option_a":"2e^(2x)","option_b":"e^(2x)","option_c":"2xe^(2x)","option_d":"e^x","option_e":"","option_f":"","bonne_reponse":"A","explication":"La dérivée de e^(u(x)) est u'(x)*e^(u(x)). Ici, u(x) = 2x, donc u'(x) = 2. Ainsi, f'(x) = 2e^(2x).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":55216,"question":"La fonction f(x) = ln(x) est définie pour tout x réel.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"La fonction ln(x) est définie uniquement pour x \u003E 0. L'affirmation est donc fausse.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":55217,"question":"Résoudre l'inéquation 3x - 7 \u003E 2x + 1.","option_a":"x \u003E 8","option_b":"x \u003C 8","option_c":"x \u003E -8","option_d":"x \u003C -8","option_e":"","option_f":"","bonne_reponse":"A","explication":"3x - 7 \u003E 2x + 1 ⇒ x \u003E 8. La solution est donc x \u003E 8.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":55218,"question":"Le produit scalaire de deux vecteurs orthogonaux est toujours égal à 0.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Par définition, deux vecteurs orthogonaux ont un produit scalaire nul. L'affirmation est vraie.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":55219,"question":"Quelle est la limite de (x² - 1)\/(x - 1) quand x tend vers 1 ?","option_a":"0","option_b":"1","option_c":"2","option_d":"∞","option_e":"","option_f":"","bonne_reponse":"B","explication":"Factorisons : (x² - 1)\/(x - 1) = (x-1)(x+1)\/(x-1) = x+1 pour x ≠ 1. Donc la limite quand x→1 est 2. La bonne réponse est 2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":55220,"question":"L'intégrale de 0 à π de sin(x) dx est égale à 2.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"L'intégrale de sin(x) est -cos(x). Donc ∫₀^π sin(x) dx = -cos(π) - (-cos(0)) = 1 - (-1) = 2. L'affirmation est vraie.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]