Maths Bac : Maîtrisez les fondamentaux du 1er trimestre !

Question 1 sur 10 10:00

[{"id":71942,"question":"Quelle est la solution de l'équation x² - 5x + 6 = 0 ?","option_a":"x = 2 ou x = 3","option_b":"x = -2 ou x = -3","option_c":"x = 1 ou x = 6","option_d":"x = -1 ou x = -6","option_e":"","option_f":"","bonne_reponse":"A","explication":"On factorise l'équation : (x-2)(x-3)=0. Les solutions sont donc x=2 et x=3.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":71943,"question":"La fonction f(x) = x³ - 3x est dérivable sur ℝ.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Toute fonction polynôme est dérivable sur ℝ. La dérivée est f'(x) = 3x² - 3.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":71944,"question":"Quelle est la limite de f(x) = (2x² + 3x - 1)\/(x² - 4) quand x tend vers +∞ ?","option_a":"2","option_b":"0","option_c":"1","option_d":"∞","option_e":"","option_f":"","bonne_reponse":"A","explication":"En divisant numérateur et dénominateur par x², on obtient lim (2 + 3\/x - 1\/x²)\/(1 - 4\/x²) = 2\/1 = 2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":71945,"question":"La dérivée de f(x) = √(x² + 1) est f'(x) = x\/√(x² + 1).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"En appliquant la formule de dérivation des fonctions composées, f'(x) = (1\/2)(x² + 1)^(-1\/2) * 2x = x\/√(x² + 1).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":71946,"question":"Quelle est la solution de l'inéquation x² - 4x + 3 \u003C 0 ?","option_a":"x ∈ ]1 ; 3[","option_b":"x ∈ ]-∞ ; 1[ ∪ ]3 ; +∞[","option_c":"x ∈ ]-∞ ; -1[","option_d":"x ∈ ]-1 ; 1[","option_e":"","option_f":"","bonne_reponse":"A","explication":"On résout l'équation x² - 4x + 3 = 0 → (x-1)(x-3)=0. Le polynôme est négatif entre les racines : x ∈ ]1 ; 3[.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":71947,"question":"La fonction f(x) = 1\/x est continue sur ℝ*.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"La fonction 1\/x est continue sur son domaine de définition ℝ*, mais pas sur ℝ tout entier (elle n'est pas définie en x=0).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":71948,"question":"Quelle est la dérivée de f(x) = e^(2x + 1) ?","option_a":"f'(x) = e^(2x + 1)","option_b":"f'(x) = 2e^(2x)","option_c":"f'(x) = (2x + 1)e^(2x)","option_d":"f'(x) = 2e^(2x + 1)","option_e":"","option_f":"","bonne_reponse":"D","explication":"En appliquant la formule de dérivation des fonctions exponentielles, f'(x) = 2e^(2x + 1).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":71949,"question":"La fonction f(x) = x² - 4x + 5 admet un minimum en x = 2.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"La dérivée f'(x) = 2x - 4 s'annule en x=2. Comme f''(x)=2\u003E0, il s'agit bien d'un minimum.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":71950,"question":"Quelle est la solution de l'équation ln(x) = 2 ?","option_a":"x = e²","option_b":"x = 2","option_c":"x = ln(2)","option_d":"x = 1\/2","option_e":"","option_f":"","bonne_reponse":"A","explication":"En appliquant la fonction exponentielle des deux côtés, on obtient x = e².","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":71951,"question":"La fonction f(x) = |x| est dérivable en x = 0.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"La fonction valeur absolue n'est pas dérivable en x=0 car la dérivée à gauche (-1) ≠ dérivée à droite (1).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]