Quiz : Maîtrisez les fonctions de deux variables en Terminale Scientifique

Question 1 sur 10 10:00

[{"id":61807,"question":"Soit la fonction f(x,y) = x² + y². Quelle est la valeur de la dérivée partielle ∂f\/∂x en (1,2) ?","option_a":"0","option_b":"1","option_c":"2","option_d":"4","option_e":"","option_f":"","bonne_reponse":"C","explication":"La dérivée partielle ∂f\/∂x = 2x. En (1,2), ∂f\/∂x = 2*1 = 2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":61808,"question":"L'équation ∂f\/∂x = 0 et ∂f\/∂y = 0 définit toujours un extremum local.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"Ces équations définissent un point critique, mais pas nécessairement un extremum (ce peut être un point selle).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":61809,"question":"Pour la fonction f(x,y) = x² - y², le point (0,0) est un :","option_a":"Minimum local","option_b":"Maximum local","option_c":"Point selle","option_d":"Aucun des trois","option_e":"","option_f":"","bonne_reponse":"C","explication":"Le hessien H = (2)(-2) - (0)² = -4 \u003C 0, donc (0,0) est un point selle.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":61810,"question":"La dérivée partielle ∂f\/∂y mesure :","option_a":"La variation de f par rapport à x en gardant y constant","option_b":"La variation de f par rapport à y en gardant x constant","option_c":"La variation totale de f","option_d":"La pente de la tangente à la courbe de niveau","option_e":"","option_f":"","bonne_reponse":"B","explication":"La dérivée partielle ∂f\/∂y mesure comment f change lorsque y varie, x restant fixe.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":61811,"question":"Si f(x,y) = xy, alors le point (0,0) est :","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"Le point (0,0) est un point selle car le hessien H = (0)(0) - (1)² = -1 \u003C 0.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":61812,"question":"La fonction f(x,y) = x² + 2y² - 4x + 6y admet un minimum en :","option_a":"(2, -1.5)","option_b":"(1, -3)","option_c":"(0,0)","option_d":"(-2, 1.5)","option_e":"","option_f":"","bonne_reponse":"A","explication":"Les dérivées partielles sont ∂f\/∂x = 2x - 4 et ∂f\/∂y = 4y + 6. Le point critique est (2, -1.5), et le hessien confirme qu'il s'agit d'un minimum.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":61813,"question":"Le gradient d'une fonction f(x,y) est un vecteur :","option_a":"Perpendiculaire aux courbes de niveau de f","option_b":"Parallèle aux courbes de niveau de f","option_c":"Tangent aux courbes de niveau de f","option_d":"Nul en tous les points","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le gradient ∇f = (∂f\/∂x, ∂f\/∂y) est perpendiculaire aux courbes de niveau de f.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":61814,"question":"Pour f(x,y) = x³ + y³ - 3xy, le point (1,1) est :","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"Le point (1,1) est un point selle car le hessien H = (6x)(6y) - (-3)² = 36 - 9 = 27 \u003E 0 en (1,1), mais la dérivée seconde mixte est négative (f_xy = -3), donc c'est un point selle.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":61815,"question":"La dérivée directionnelle de f en un point dans la direction d'un vecteur unitaire u est donnée par :","option_a":"∇f · u","option_b":"∇f × u","option_c":"|∇f|","option_d":"∇f \/ |∇f|","option_e":"","option_f":"","bonne_reponse":"A","explication":"La dérivée directionnelle est le produit scalaire du gradient ∇f et du vecteur unitaire u.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":61816,"question":"Si f(x,y) = e^(x+y), alors ∂²f\/∂x∂y est égal à :","option_a":"e^(x+y)","option_b":"0","option_c":"e^x","option_d":"e^y","option_e":"","option_f":"","bonne_reponse":"A","explication":"∂f\/∂x = e^(x+y), donc ∂²f\/∂x∂y = e^(x+y).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]