Juin 2004 Solution

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SAID BOUZAWIT - lycée Abdelali Benchakroune

: ‫اﻟﺘﻤﺮﯾﻦ اﻷول‬

x  y  z  4 y  2 z  2  x  y  2   z  1  3 2

2

2

2

2

‫( ﻟﺪﯾﻨﺎ‬1

2

 0,2,1‫ ﻓﻠﻜﺔ ﻣﺮﻛﺰھﺎ‬S ‫ ﻣﻌﺎدﻟﺔ دﯾﻜﺎرﺗﯿﺔ ل‬x 2  y  2   z  1  3 2

‫إذن‬

2

. r  3 ‫و ﺷﻌﺎﻋﮭﺎ‬ . A  3  A  S -‫أ‬

(2

. M x, y, z  P   A. AM  0  x  y  z  0 -‫ب‬  . d  2  B  Q ‫ و‬Q  : x  y  z  d  0  Q  ‫ ﻣﻨﻈﻤﯿﺔ ﻋﻠﻰ‬n 1,1,1 -‫( أ‬3 0  2 1 2 3 ‫ وﻓﻖ داﺋﺮة ﺷﻌﺎﻋﮭﺎ‬S  ‫ ﯾﻘﻄﻊ‬Q   d  r  d , Q   -‫ب‬  3 3 . Q ‫ ﻋﻠﻰ‬ ‫ اﻟﻤﺴﻘﻂ اﻟﻌﻤﻮدي ل‬H a, b, c ‫ و ﻣﺮﻛﺰھﺎ‬R  r 2  d 2 

2 2 3

a  t b  2  t  . H 1 , 7 ,  1  t  1  t  IR /  : ‫إذن‬ 3 3 3 3 c   1  t  a  b  c  2  0









: ‫اﻟﺘﻤﺮﯾﻦ اﻟﺜﺎﻧﻲ‬

.   2 2 1  i   2i  1  i  ‫ و‬  16  161  i   16i 2



2







(1

z"  2 i  2 2 1  i   2 2  2  2 2 i ‫ و‬z '  2 i  2 2 1  i   2 2  2  2 2 i . z '  z 2 ‫ و‬z"  z1  Rez"  0     . b  2,  ‫ و‬a  2,  (2  4  2 -‫( أ‬3

OC  OA  OB  aff C   aff A aff B  OA a  1 . OA  OB  OB b ‫ ﻣﺘﻮازي اﻷﺿﻼع‬OBCA  OC  OA  OB -‫ب‬  . argz1   e1 , OC 2  . ‫ ﻣﻌﯿﻦ‬OBCA  OA  OB





17

SAID BOUZAWIT - lycée Abdelali Benchakroune

















. e1 , OC  arg b   1 OB , OA  3 2   e1 , OC  e1 , OB  OB , OC 2  2

8

pB  

C C 1  3 6 C9 3 5

3 4

: ‫اﻟﺘﻤﺮﯾﻦ اﻟﺜﺎﻟﺚ‬ ، pA 

1 2

1 3 3 9

C C C 41 2  7 C

(1

("‫ "ﻻ ﺗﻮﺟﺪ أي ﺑﯿﺪﻗﺔ ﺣﻤﺮاء ﻣﻦ ﺑﯿﻦ اﻟﺒﯿﺪﻗﺎت اﻟﻤﺴﺤﻮﺑﺔ‬: C ) pC   1  p C  pC   . pA  B  

C3 16 5  p C  63  21 C 9 21

C 21C11C 21 1  " B1 R1 N 1 ": A  B (2 3 21 C9

: ‫اﻟﺘﻤﺮﯾﻦ اﻟﺮاﺑﻊ‬

1 e e 11 1    1 x -‫( أ‬1 x x e 1 1 e 1 e e 1 ‫ ﻣﺘﻤﺎﺛﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺼﻔﺮ‬IR -‫ب‬ 1 2 1 1   x  IR, f  x   1  x   x  1  x  21  x    f x  ‫و‬ 2 2 e 1  e 1 . lim f x    (2 .

x

x  IR,

x

x

x  

2 1 2e x 1  e 2 x  2e x  1  1  ex 1  -‫( أ‬3 . x  IR, f ' x         x 2 ex 1 2 2  e x  1 2  2  e  1 







: ‫ إذن‬، IR ‫ ﺗﻨﺎﻗﺼﯿﺔ ﻗﻄﻌﺎ ﻋﻠﻰ‬f

1





 f ' x   0 ‫ ﻟﺪﯾﻨﺎ‬IR  ‫ ﻣﻦ‬x ‫ ﻟﻜﻞ‬-‫ب‬

1 2 x x  0 ‫ ﯾﻌﻨﻲ‬f x   f 0   x  0 : ‫ إذن‬، IR  ‫ ﺗﻨﺎﻗﺼﯿﺔ ﻗﻄﻌﺎ ﻋﻠﻰ‬f -‫ج‬ 2 e 1 2 1 . x  IR  : 1  x  x ‫و ﻣﻨﮫ‬ e 1 2   1  . lim e x    lim  f x  1  x   0 (4 x   x    2   1 y  1  x ‫ ﻣﻌﺎدﻟﺘﮫ‬  ‫ ﯾﻘﺒﻞ ﻣﻘﺎرﺑﺎ ﻣﺎﺋﻼ ﺑﺠﻮار‬C  2

18

SAID BOUZAWIT - lycée Abdelali Benchakroune

‫( اﻟﻤﻨﺤﻨﻰ‬5

1 t 1 1 1  dt   e 1 dx       dt  ln   ‫إذن‬ x 1 1  e e 1 t e 1 t  t   2 

.

0

dx  

dt  t  e  x -‫( أ‬6 t

0 1    e  1 5  e 1   2 ln .    f x  dx   x  x 2   2 ln   um  -‫ب‬ 1 4  1   2  4  2  0

-II . U 0  1  0 : n  0 ‫( ﻣﻦ أﺟﻞ‬1

. U n 1  1 

1

2 e

U n 1

 0 ‫أي‬

2 e

U n 1

 1 ‫ وﻣﻨﮫ‬eU n  1  2 ‫ إذن‬U n  0 ‫ﻧﻔﺘﺮض أن‬

. n  IN

2 1  U n ‫ ( ﻓﻨﺠﺪ‬U n  0 ) x  U n ‫ ﻧﻀﻊ‬. x  IR  e 1 2 Un

: ‫إذن‬

Un  0

2 1  x ‫ ﻟﺪﯾﻨﺎ‬-‫( أ‬2 e 1 2 1 . IN ‫ ﻣﻦ‬n ‫ ﻟﻜﻞ‬U n 1  U n ‫أي‬ 2 1 1 .‫ ﻣﺘﺘﺎﻟﯿﺔ ﺗﻨﺎﻗﺼﯿﺔ‬U n   U n 1  U n   U n  0  U n 1  U n ‫ ﻟﺪﯾﻨﺎ‬-‫ب‬ 2 2 : 1

x

0

1 U 0  1     1 : n  0 ‫( ﻣﻦ أﺟﻞ‬3 2 1 .( U n 1  U n ‫)ﻷن‬ 2

1 U n 1    2

n 1

‫وﻣﻨﮫ‬

19

1 1 Un    2 2

n 1

1 ‫ إذن‬U n    2

n

‫ﻧﻔﺘﺮض أن‬

SAID BOUZAWIT - lycée Abdelali Benchakroune 1 . IN ‫ ﻣﻦ‬n ‫ ﻟﻜﻞ‬U n    2 n

. lim U n  0

20

n

‫إذن‬ n

1 1  lim   0 ‫ و‬0  U n    ‫ﻟﺪﯾﻨﺎ‬ 2 2

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