Correction Devoir de synthèse n°2 4ème Mathématiques Mr Kouch 06 03 12

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Lycée Dar Chaabene Correction devoir de synthèse n°2 4ème_Math Exercice 1 :

1)

z'= 2iz 1 i

a) k = |2i| = 2 Vrai b) Ω 2

ab + b 2i(-1+ i) -1-i -3i -3

z = = = =1+ i

-3 -3

1-|a|

Faux Remarque b =-1-i (-1-i)(1+ 2i) 1 3= = - i

1-a 1-2i 5 5 5

c) D : y = x W Î D

soit A(-1, -1)ÎD, f(A)= A’Û

z

_A'= 2(-1 + i) – 1 – i = -zi – 3 ÛA’(-3, -3) ÎD Vrai Equation : Wuuuur uuuurM' 2 M= W Û z – 1 – i = 2(z – 1 – i) Û

2iz -1-i -1-i = 2z - 2i - 2iÛ2z - 2- 2i

Û2iz - 2- 2i = 2z - 2 - 2i Ûz = -izÛ x – iy = -ix + y Û y = x

2) a) 2012 = 190 ´ 10 + 112 Þ -2012 = 190 ´ (-11) + 78 Þ -2012 = (-190) ´ 11 + 78 Þ q = 11 Faux b) p Ù (2p + 1) = 1 car 2p + 1 – 2p = 1 Faux c) 31premier _31/530 _{1 5}30 _{1 31}

[ ]

₅30 _{1 31}

[ ]

31 5=1 n ü_Û _{- Þ} _º _Þ _º ý Ù _þ Vrai Exercice 2 : 1) S : similitude directe / S(J) = B / S(D) = K a)

(

JD,BK

) (

AD,BA 2π

)

( )

(

BC,BA 2π

)

( )

π

( )

2π 3 º º º

uur uuur_$ uuur uuur_$ uuur uuur_$ 1 2 BK 2BA π k = = = 4×cos = 2 JD BC 3 b) CD = AKuuur uuurÞJ= C*K CBK équilatéral et (CA) ^ (BK)

(

) (

)

( )

1 2 π CB OB CJ,CB CK,CB 2π 2π et = = 2 S(C) = C 3 OJ CB

Þ uur uuur$ º uuur uuur$ º Û

2) f antidéplacement / f(0) = A / f(A) = A’

a) fof(O) = A’ ¹ D Þ f est une symétrique glissante u = DA'= DC = AB1

2

r uuuur uuur uuur D’=(JI).

b) f(K) = S(JI) ot (K)=S (A) =CABuuur (JI)

3) g = fos

a) g = sim ind de rapport k = 2 ´ 1 = 2

b) gog(D) = ?

g(D) = fos(D) = f(K) = C

g(C) = fos(C) = f(C) = B Þ gog(D) = B

c) h(W, 4)(D) = B ÞΩB - 4ΩD = 0uuur uuur r ÛΩ by(B,1) et (D, -4)

BΩ = BD4 3 uuur uuur d) D = biss int de

(

W Wuuur uuurD, C$

)

Exercice 3 : (E) : 7x – 4y = 13 1) a) 7 ´ 3 – 4 ´ 2 = 21 – 8 = 13 b) SZ2 = {(3 + 4k ; 2 + 7k) ; k Î Z} 2) a = 4n + 3 ; b = 7n + 2 a) Si d/a et d/bÞd / 7a - 4bÞd/21-8Þd/13 b) a Ù b = 13 Þ13 et 13a bÞ13/2a - b 13 /Þ n+ Þ + º4 n 4 0 13

[ ]

Þ ºn 9 13

[ ]

Inversement Si n º 9[13] Þ 4n + 3 º 0 [13] et 7n + 2 º 0 [13] 13/a et 13/b 13/a b or a b/13 a b =13 Þ Þ Ù Ù Þ Ù 3) a) 101_{º 10[13] ; 10}2_{º 9[13] ; 10}3_{º 12[13] ; 10}4_{º [13] ; 10}5_{º 4[13] ; 10}6_{º 1[13]} b) n = 20122012_{; 2012 º 10[13] Þ 2012}2012_{º 10}2_{[13] º 9[13] Þ a Ù b = 13} 4) a = 4n +3 n = 9 +13k 7a -4b =13 b = 7n + 2 b = 63+91k a b =13 a b =13 1893 91k 1949 1956 b 2012 1956 b 2012 20,... k 21,... ì ì ì _ï _ï ï _Ù _Ûï _Ûï í í _Ù í _£ _£ ï _{£ £} ï ï î _ï _{< £} _ï _{£ £} î î (a, b) = (1131, 1976) k = 21 Þ n = 282 Exercice 5 : f(x) = x2_{– 2lnx – 1 ; x > 0}

1) f est dérivable sur ]0, +¥[ et on a f’(x) = 2x – 2=2(x -1)2

x x lim f(x) x x®+¥ =+ ¥ 2) a) l Î ]0, 1] A(l) = 1 3 3 1 ₂ _λ λ λ x 4 λ x - 2ln x -1dx = - 2x ln + 2x - x = - + 2λ ln λ - λ 3 3 3 é ù ê ú ë û

ò

b) 0 4 lim A(λ) = 3 l_®+ 3) n ³ 2 x 0 1 +¥ f’(x) - + f(x) +¥ +¥ 0

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a) 1 £ k £n – 1 0 <k t k +1 1 n£ £ n £ et f décroissante sur ]0, 1] Þ k +1n k_n k +1 k 1 k +1 1 k f f(t) f f f(t)dt f n n n n n n æ ö_£ _£ æ ö_Þ æ ö_£ _£ æ ö ç ÷ ç ÷ ç ÷ ç ÷ è ø è ø è ø

ò

è ø b) 2_n 1_n 3 n 2_n 1-1 n -1 k=1 k =1 1 1-1 n 1 2 1 1 k =1 f f(t)dt f( ) n 1 n n 1 3 1 2 1 k +1 1 1 k k = 2 f f(t)dt f f et f n n n n n n n n n 1 1 1-1 k = n -1 f(1) f(t)dt f n n n ü æ ö £_{ç ÷} £ _ï è ø _ï ï æ ö æ ö æ ö æ ö æ ö £_{è ø}ç ÷ £ ç ÷_{è ø} ýÞ ç_è _ø÷ ç ÷_{è ø}£ ç ÷_{è ø} ï ï æ ö £ £ _ç _÷ï è ø þ

ò

å

ò

4) n ( )k 1 ( )1 ( )1 ( )1 n n n n n n n n n n k=1 1 1 1 1 4

S = f S ( ) d'autre part S f ( ) A S A + f limS =

n

å

³ -n n £ Þ £ n Þ+¥ 3 Exercice 5: 1) x ₂ e dt F(x) = ; n e (1+ ln t) ³

ò

J = p e3

( )

3 e 1 dx = π×F e x(1+lnx)

ò

2) g(x) = tan x ; xÎ 0, 2 p é é ê ê ë ë a) b) g’(x) ¹ 0 -1 -1 2 2 1 1 1 (g )'(x) = = = g'(g )(x) 1+ tan t 1+ x 3) H(x) = ln ₂ 1 dt x e 1+ t x " ³

ò

a)

( )

3 3 -1

( )

-1 2 1 dt π π π H(e) = 0 ; H e = = g 3 -g (1) = - = 3 4 12 1+ t

ò

b)

[

(

)

[