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QCM Fonctions numériques QCM Fonctions n u m e r i q u e s ( c o n t i n u i t é , d e r i v i b a b i l i t é , . . ) Q u e s t i o n 1 : x d é s i g n e la partie e n t i è r e de x . Soit φ une fonction d é f i n i e sur 0 , φ x = 1 x x alors on a : φ est b o r n é e φ est constante sur 0 φ est croissante sur 0 φ est continue sur 0 Q u e s t i o n 2 : On pose f ( x ) = cos x 1 x 2 pour tout x non n u l . Alors : lim x 0 f ( x ) = e lim x 0 f ( x ) = e 1 lim x 0 f ( x ) = e 1 2 lim x 0 f ( x ) = e 1 2 Q u e s t i o n 3 : Soit f une fonction d é rivable et croissante sur 0 , 1 . On note f 1 sa fonction r é c i p r o q u e et f 0 = 0 . on pose F x = 0 x f t dt et G x = 0 f x f 1 t dt alors on a : F x = G x F x = G x F x = xf x + G x F x = xf x G x Q u e s t i o n 4 : On pose g x = 1 cosx x pour tout x different de 0 et g 0 = 0 et G sa primitive qui s ' a n n u l e en 0 . G est croissante sur * G x x 2 4 G x x 2 4 G est impaire Q u e s t i o n 5 : On pose φ x = x 2 x exp t t dt pour tout x de * . lim x 0 φ x = l n 2 lim x 0 φ x = 2 l n 2 lim x 0 φ x = 2 lim x 0 φ x = 0 lim x 0 φ x = 2 l n 2 Q u e s t i o n 6 : Soit la fonction f definie sur son domaine de definition D f par f x = 1 3 x x avec x d é s i g n e la partie e n t i è r e de x . On a : D f = * et lim x 0 f x = 0 D f = , 0 3 , + et lim x 0 f x = 0 D f = , 0 0 , 3 / 2 3 , + et lim x 0 f x = 0 D f = , 0 3 , + et lim x 0 f x = 1 Q u e s t i o n 7 : la limite de ϕ x = 1 x 2 1 sin 2 x quand x tend vers 0 vaut : 0 1 / 6 1 / 3 1 / 6 + Question 8 On consid é re la fonction φ t = arctan t + arctan 1 t ; t 0 on a : la fonction φ est cons tan te la fonction φ ne prend que deux valeurs distincts sur son domaine de d é finition la fonction φ prend plus de deux valeurs distincts sur son domaine de d é finition la fonction φ est paire Question 9 lim x π 4 1 tan x 2 1 + cos 4 x vaut : 1 2 2 1 2 2 2 1 2 Q u e s t i o n 10 Soit g une fonction definie d e ] 1 , + [ à admettant l ' a l l u r e r e p r e s e n t é comme s u i t : A l o r s : g x = xln 1 + 1 x g x = x + ln x + 1 g x = x ln x + 1 g x = xln x + 1 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