Devoir commun de seconde Mathématiques

Exercice 2 :(5 points)1° f(x) = x 2 – 6 x + 9 – 2 (3 x – x 2 + 12 – 4 x) = x 2 – 6 x + 9 – 6 x + 2 x 2 – 24 + 8 x = 3 x 2 – 4 x – 15 .2° f(x) = (x – 3) 2 + 2 (x + 4) (x – 3) = (x – 3) ((x – 3) + 2 (x + 4))= (x – 3) (x – 3 + 2 x + 8)= (x – 3) (3 x + 5)3° a) f(x) = – 15 ⇔ 3 x 2 – 4 x – 15 = – 15⇔ 3 x 2 – 4 x = 0⇔ x (3 x – 4) = 0⇔ x = 0 ou x = 4 3S = ⎩⎪⎨ ⎪ ⎧0 , 4 3 ⎭⎪⎬ ⎪ ⎫b) f(x) > 0 ⇔ (x – 3) ( 3 x + 5) > 0.S = ] – ∞ , – 5/3 [ ∪ ] 3 , + ∞ [4° a) g(x) = (x – 3) 2b) f(x) < g(x) ⇔ (x – 3) (3 x + 5) < (x – 3) 2⇔ (x – 3) (3 x + 5) – (x – 3) 2 < 0⇔ (x – 3) [ (3 x + 5) – (x – 3)] < 0⇔ (x – 3) (2 x + 8) < 0S = ] – 4 , 3 [x – ∞ –5/3 3 + ∞x– 3 – 0 + + +3 x + 5 – – 0 +f(x) + 0 – 0 +x – ∞ – 4 3 + ∞x – 3 – – 0 +2 x + 8 – 0 + ++ 0 – 0 +c) f(x) (x – 3) (3 x + 5)< 3 ⇔g(x) (x – 3) 2 < 3⇔ 3 x + 5x – 3 – 3 < 0⇔3 x + 5 – 3 (x – 3)x – 3⇔ 1 4x – 3 < 0S = ] – ∞ , 3 [< 0x – ∞ 3 + ∞x – 3 – 0 ++ –NExercice 3 : (3,5 points)2° a) MN ⎯⎯→= MA ⎯⎯→+ AN ⎯⎯→= – 3 ⎯⎯→AB + 3 AD⎯⎯→2⎯⎯→BI = BA ⎯⎯→+ AD ⎯⎯→+ ⎯⎯→DI = – AB ⎯⎯→+ AD ⎯⎯→+ 1 2⎯⎯→DC = – AB ⎯⎯→+ AD ⎯⎯→+ 1 2⎯⎯→AB = – 1 2⎯⎯→AB + AD⎯⎯→DICb)⎯⎯→MN = 3 ⎯⎯→BI donc les vecteurs MN ⎯⎯→et ⎯⎯→BI sont colinéairesLes droites (MN) et (BI) sont donc parallèles.3° a) CM ⎯⎯→= CA ⎯⎯→+ AM ⎯⎯→= CB ⎯⎯→+ BA ⎯⎯→+ 3 2⎯⎯→AB= – AD ⎯⎯→– AB ⎯⎯→+ 3 2⎯⎯→AB = 1 2⎯⎯→AB – AD⎯⎯→ABM⎯⎯→CN = CA ⎯⎯→+ AN ⎯⎯→= – AC ⎯⎯→+ 3 AD ⎯⎯→= – AB ⎯⎯→– AD ⎯⎯→+ 3 AD ⎯⎯→= – AB ⎯⎯→+ 2 AD⎯⎯→b)⎯⎯→CN = – 2 CN. ⎯⎯→les vecteurs CM ⎯⎯→et CN ⎯⎯→sont colinéaires.Les points C, M et N sont donc alignés.