Maths Terminale : Testez vos connaissances en séries d'exercices
🧠 Quiz 10 questions 10 min
QUIZ INTERACTIFDiff. 5/10
Corrigé complet de la Série 1 de Maths Terminale avec exercices corrigés et explications détaillées pour réviser efficacement le programme officiel.
Question 1 sur 10 10:00
[{"id":32578,"question":"Quelle est la dérivée de la fonction f(x) = 3x² + 2x - 5 ?","option_a":"6x + 2","option_b":"3x² + 2","option_c":"6x² + 2x","option_d":"6x + 2x - 5","option_e":"","option_f":"","bonne_reponse":"A","explication":"La dérivée d'une fonction polynôme se calcule terme par terme : (3x²)' = 6x, (2x)' = 2 et (-5)' = 0. Donc f'(x) = 6x + 2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":32579,"question":"L'équation e^x = 3 admet une solution réelle.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"La fonction exponentielle est strictement croissante et prend toutes les valeurs de ]0, +∞[, donc elle admet une solution unique à e^x = 3.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":32580,"question":"Quelle est la limite de la suite u_n = (2n² + 3n) \/ (n² - 1) quand n tend vers +∞ ?","option_a":"2","option_b":"1","option_c":"0","option_d":"+∞","option_e":"","option_f":"","bonne_reponse":"A","explication":"En divisant numérateur et dénominateur par n², on obtient (2 + 3\/n) \/ (1 - 1\/n²). Quand n → +∞, 3\/n → 0 et 1\/n² → 0, donc la limite est 2\/1 = 2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":32581,"question":"Soit Z = 3 + 4i un nombre complexe. Quel est son module ?","option_a":"5","option_b":"7","option_c":"√7","option_d":"√(3² + 4²)","option_e":"","option_f":"","bonne_reponse":"D","explication":"Le module d'un nombre complexe a + bi est √(a² + b²). Ici, |Z| = √(3² + 4²) = √(9 + 16) = √25 = 5.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":32582,"question":"La fonction f(x) = x³ + x est impaire.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Une fonction est impaire si f(-x) = -f(x). Ici, f(-x) = (-x)³ + (-x) = -x³ - x = -(x³ + x) = -f(x). Donc f est impaire.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":32583,"question":"Résolvez l'inéquation ln(x) \u003E 1.","option_a":"x \u003E e","option_b":"x \u003E 1","option_c":"x \u003E 0","option_d":"x \u003E 1\/e","option_e":"","option_f":"","bonne_reponse":"A","explication":"La fonction ln est croissante et ln(e) = 1. Donc ln(x) \u003E 1 ⇔ x \u003E e.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":32584,"question":"Quelle est la solution de l'équation 2cos(x) = √3 ?","option_a":"x = π\/6 + 2kπ","option_b":"x = π\/3 + 2kπ","option_c":"x = π\/4 + 2kπ","option_d":"x = π\/2 + 2kπ","option_e":"","option_f":"","bonne_reponse":"B","explication":"2cos(x) = √3 ⇔ cos(x) = √3\/2 ⇔ x = π\/6 + 2kπ ou x = -π\/6 + 2kπ (k ∈ ℤ). La solution principale est x = π\/6 + 2kπ.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":32585,"question":"La suite v_n = (-1)^n est convergente.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"La suite v_n alterne entre -1 et 1 et n'a pas de limite finie. Elle est donc divergente.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":32586,"question":"Quelle est la primitive de f(x) = 5x^4 ?","option_a":"x^5 + C","option_b":"5x^5 + C","option_c":"x^5\/5 + C","option_d":"20x^3 + C","option_e":"","option_f":"","bonne_reponse":"A","explication":"La primitive de x^n est x^(n+1)\/(n+1) + C. Donc ∫5x^4 dx = 5 * (x^5\/5) + C = x^5 + C.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":32587,"question":"Soit A et B deux événements d'un univers Ω. Si P(A ∩ B) = 0, alors A et B sont incompatibles.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Deux événements sont incompatibles si leur intersection est vide, c'est-à-dire P(A ∩ B) = 0. Donc l'affirmation est vraie.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]
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