Juin 2007 Solution
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ﺗﺼﻴﺢ اﻣﺘﺤﺎن ﻳﻮﻧﻴﻮ 2007
اﻟﺜﺎﻧﻴﺔ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
اﻟﺘﻤﺮﻳﻦ اﻷول /1ﻧﺒﻴﻦ أن ﻣﺮآﺰ اﻟﻔﻠﻜﺔ ) ( Sهﻲ اﻟﻨﻘﻄﺔ ) Ω (1; 2; 3و أن اﻟﺸﻌﺎع ﻳﺴﺎوي 6
M ( x; y ) ∈ ( S ) ⇔ x 2 + y 2 + z 2 − 2 x − 4 y − 6 z + 8 = 0 ⇔ ( x − 1) − 1 + ( y − 2 ) − 4 + ( z − 3) − 9 + 8 = 0 2
2
2
2
⇔ ( x − 1) + ( y − 2 ) + ( z − 3) = 6 = 6 2
2
2
إذن ) ( Sﻓﻠﻜﺔ ﻣﺮآﺰهﺎ اﻟﻨﻘﻄﺔ ) Ω (1; 2;3ﺷﻌﺎﻋﻬﺎ 6 /2ﻧﺘﺤﻘﻖ أن اﻟﻤﺴﺘﻮى ) ( Pﻣﻤﺎس ﻟﻠﻔﻜﺔ
6
=
1− 2 + 2×3 +1
) (S
= ) ) d ( Ω; ( Ρ
= 6 1+1+ 4 6 إذن اﻟﻤﺴﺘﻮى ) ( Pﻣﻤﺎس ﻟﻠﻔﻜﺔ ) ( S
/3أ -ﻧﺤﺪد ﺗﻤﺜﻴﻼ ﺑﺎراﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ ) ∆ ( اﻟﻤﺎر ﻣﻦ Ωو اﻟﻌﻤﻮدي ﻋﻠﻰ
)( P
ﺑﻤﺎ أن ) ( ∆ ) ⊥ ( Pو ( P ) : x − y + 2 z + 1 = 0ﻓﺎن اﻟﻤﺘﺠﻬﺔ ) n (1; −1; 2اﻟﻤﻨﻈﻤﻴﺔ ﻋﻠﻰ ) ( Pﻣﻮﺟﻬﺔ ﻟﻠﻤﺴﺘﻘﻴﻢ و ﺣﻴﺚ أن ) ∆ ( ﻳﻤﺮ ﻣﻦ ) Ω (1; 2;3ﻓﺎن
∈/t
)∆(
x = 1 + t y = 2 − tﺛﻤﺜﻴﻞ ﺑﺎراﻣﺘﺮي ﻟﻠﻤﺴﺘﻘﻴﻢ ) ∆ ( z = 3 + 2t
ب /ﻧﺤﺪد ﻣﺜﻠﻮث إﺣﺪاﺛﻴﺎت ωﻧﻘﻄﺔ ﺗﻤﺎس ) ( Sو ) ( P ωﻧﻘﻄﺔ ﺗﻤﺎس ) ( Sو ) ( Pوﻣﻨﻪ ωﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ) ( S
و
)( P
x = 1 + t y = 2 − t ω ( x; y; z ) ∈ ( P ) ∩ ( S ) ⇔ z = 3 + 2t x − y + 2 z + 1 = 0 x = 1 + t y = 2 − t ⇔ z = 3 + 2t 1 + t − 2 + t + 6 + 4t + 1 = 0 t = −1 x = 0 ⇔ y = 3 z = 1 إذن ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ) ( Sو ) ( Pهﻲ )ω ( 0;3;1 اﻟﺘﻤﺮﻳﻦ اﻟﺜﺎﻧﻲ /1أ -ﻧﻜﺘﺐ ﻋﻞ اﻟﺸﻜﻞ اﻟﺠﺒﺮي اﻟﻌﺪد اﻟﻌﻘﺪي
2
) ( 3 − 2i
= 9 − 4 − 12 i = 5 − 12 i ب -ﻧﺤﻞ ﻓﻲ
( 3 − 2 i )2
اﻟﻤﻌﺎدﻟﺔ z 2 − 2 ( 4 + i ) z + 10 + 20i = 0 :
ﻟﻴﻜﻦ ' ∆ ﻣﻤﻴﺰ اﻟﻤﻌﺎدﻟﺔ :
2
) ∆ ' = ( 4 + i ) − 10 − 20i = 15 + 8i − 10 − 20i = 5 − 12i = ( 3 − 2i 2
وﻣﻨﻪ ﺟﺪرا اﻟﻤﻌﺎدﻟﺔ هﻤﺎ z2 = 4 + i − 3 + 2i = 1 + 3i ; z1 = 4 + i + 3 − 2i = 7 − i
http://arabmaths.ift.fr
إذن } S = {1 + 3 i ; 7 − i c−a /2أ -ﻧﺒﻴﻦ أن = i b−a ) c − a 5 + 9i − 1 − 3i 4 + 6i i ( 6 − 4i = = = =i b − a 7 − i − 1 − 3i 6 − 4i 6 − 4i ب -ﻧﺴﺘﻨﺘﺞ أن اﻟﻤﺜﻠﺚ ABCﻣﺘﺴﺎوي اﻟﺴﺎﻗﻴﻦ و ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ c−a c−a c−a ﺑﻤﺎ أن = i arg ﻓﺎن = i = 1 و = arg i b−a b−a b−a c−a π arg وﻣﻨﻪ c − a = b − aو ] [ 2π ≡ b−a 2
π
)
و ﺑﺎﻟﺘﺎﻟﻲ AC = ABو ] [ 2π 2 إذن ABCﻣﺘﺴﺎوي اﻟﺴﺎﻗﻴﻦ و ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ ﻓﻲ A اﻟﺘﻤﺮﻳﻦ اﻟﺜﺎﻟﺚ 2 x 1 /1ﻧﺘﺤﻘﻖ أن }∀x ∈ − {−1 = x −1+ x +1 x +1 ﻟﺘﻜﻦ }x ∈ − {−1 1 x2 − 1 + 1 x2 = = x +1 x +1 x +1 x2 /2ﻧﺒﻴﻦ أن dx = ln 3 x +1
(
≡ arg AB; AC
x −1+
2
∫0 2
x2 2 x2 1 dx = ∫ x − 1 + dx = − x + ln x + 1 = 2 − 2 + ln 3 = ln 3 0 x +1 x + 1 2 0 3
2
∫0
2
/3ﺑﺎﺳﺘﻌﻤﺎل ﻣﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء ،ﻧﺒﻴﻦ أن ∫0 x ln ( x + 1)dx = 2 ln 3 1 u ' ( x ) = x + 1 )u ( x ) = ln ( x + 1 وﻣﻨﻪ ﻧﻀﻊ 2 v ' ( x ) = x v ' ( x ) = x 2 2
x2 1 2 x2 1 3 ∫ x ln ( x + 1)dx = ln ( x + 1) − dx = 2 ln 3 − ln 3 = ln 3 0 2 2 2 0 2 x + 1
2
∫0
اﻟﺘﻤﺮﻳﻦ اﻟﺮاﺑﻊ 2 ﻧﺤﺴﺐ ) p ( Aو ) P ( Bو ﻧﺒﻴﻦ أن = ) p ( C 7 " Aﻻ ﺗﻮﺟﺪ أﻳﺔ ﺑﻴﺪﻗﺔ ﺗﺤﻤﻞ اﻟﻌﺪد 0ﻣﻦ ﺑﻴﻦ اﻟﺒﻴﺪﻗﺎت اﻟﺜﻼﺛﺔ اﻟﻤﺴﺤﻮﺑﺔ " C3 4 = p ( A ) = 43 35 C7
" Bﺳﺤﺐ ﺛﻼﺛﺔ ﺑﻴﺪﻗﺎت ﺗﺤﻤﻞ أﻋﺪادا ﻣﺨﺘﻠﻔﺔ ﻣﺜﻨﻰ ﻣﺜﻨﻰ " C 1 × C 11 × C 31 9 p (B ) = 3 = 3 35 C7 " Cﻣﺠﻤﻮع اﻻﻋﺪاد اﻟﻤﺴﺠﻠﺔ ﻋﻠﻰ اﻟﺒﻴﺪﻗﺎت اﻟﺜﻼﺛﺔ اﻟﻤﺴﺤﻮﺑﺔ ﻣﻨﻌﺪم " C 33 + C 31 × C 11 × C 31 1+ 9 10 2 = ) p (C = = = 3 35 35 7 C7 ﻣﺴﺄﻟﺔ -(Iاﻟﺪاﻟﺔ gاﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ
ﺑﻤﺎ ﻳﻠﻲ+ x − 1 :
/1ﻧﺤﺴﺐ ) g ' ( xﻟﻜﻞ xﻣﻦ
−x
g ( x) = e
ﺛﻢ ﻧﺴﺘﻨﺘﺞ أن gﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ [∞ [ 0; +و ﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ
]]−∞; 0 http://arabmaths.ift.fr
∈x
ﻟﺘﻜﻦ
g ( x ) = e− x + x − 1
:
g ' ( x ) = −e − x + 1
ﻟﺪﻳﻨﺎ x ≥ 0 ⇔ − x ≤ 0 ⇔ e− x ≤ 1 ⇔ −e− x + 1 ≥ 0 اذن ∀x ∈ [0; +∞[ g ' ( x ) ≥ 0وﻣﻨﻪ gﻋﻠﻰ ﺗﺰاﻳﺪﻳﺔ [∞[0; + ﻟﺪﻳﻨﺎ x ≤ 0 ⇔ − x ≥ 0 ⇔ e− x ≥ 1 ⇔ −e− x + 1 ≤ 0 اذن ∀x ∈ ]−∞; 0] g ' ( x ) ≤ 0و ﻣﻨﻪ gﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ ]]−∞; 0 /2ﻧﺒﻴﻦ أن g ( x ) ≥ 0
∈ ∀xﺛﻢ ﻧﺴﺘﻨﺘﺞ أن e− x + x ≥ 1
∈ ∀x
ﺑﻤﺎ أن gﻋﻠﻰ ﺗﺰاﻳﺪﻳﺔ [∞ [ 0; +و ﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ ] ]−∞; 0و g ( 0 ) = 0 وﻣﻨﻪ g ( x ) ≥ 0
∈ ∀x
أي e− x + x − 1 ≥ 0
إذن e− x + x ≥ 1
∈ ∀x
∈ ∀x x
-(IIاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fﻟﻠﻤﺘﻐﻴﺮ اﻟﺤﻘﻴﻘﻲ اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ: /1ﻧﺒﻴﻦ أن
x + e− x
= )f ( x
= Df
ﺣﺴﺐ /2 (Iﻟﻴﻨﺎ e − x + x ≥ 1 1
/2أ -ﻧﺒﻴﻦ أن
1
1+
xe x *
ﻓﺎن 0ﻗﻴﻤﺔ دﻧﻴﺎ ﻟﺪاﻟﺔ fﻋﻠﻰ
= )f ( x 1
∈: x
∈ ∀xوﻣﻨﻪ e− x + x ≠ 0 *
∈ ∀x x
=
∈ ∀xاذن
= Df
=
x
= )f ( x
ﻟﻴﻜﻦ 1 x 1 + x 1 + x xe xe ب -ﻧﺒﻴﻦ أن lim f ( x ) = 0و lim f ( x ) = 1و ﻧﺆول اﻟﻨﺘﻴﺠﺘﻲ هﻨﺪﺳﻴﺎ
1
∞x →−
ﻟﺪﻳﻨﺎ lim xe x = 0− ﻟﻠﻤﻨﺤﻨﻰ
) (C
ﻟﺪﻳﻨﺎ ∞lim xe x = +
وﻣﻨﻪ = 1
∞x →+
ﻟﻠﻤﻨﺤﻨﻰ
) (C
/3أ -ﻧﺒﻴﻦ أن
ﻟﻴﻜﻦ
( x + 1) e− x
) (x + e
−x 2
∈: x
1 x
xe
1 xe x
x + e− x
lim 1 +
∞x →−
اذن lim f ( x ) = 0و ﻣﻨﻪ ﻣﺤﻮر اﻻﻓﺎﺻﻴﻞ ﻣﺴﺘﻘﻴﻢ ﻣﻘﺎرب ∞x →−
اذن lim f ( x ) = 1وﻣﻨﻪ اﻟﻤﺴﺘﻘﻴﻢ ذا اﻟﻤﻌﺎدﻟﺔ y = 1ﻣﻘﺎرب
lim 1 +
∞x →+
∞x →−
= )f '( x
x
x+e
∞x →+
وﻣﻨﻪ ∞= −
∞x →−
−x
∈ ∀x
= )f ( x
) = e + xe = ( x + 1) e ) (x + e ) (x + e
−x
−x
−x 2
−x 2
−x
(
x + e− x − x 1 − e− x
) (x + e
−x 2
= )f '( x
ب -ﻧﺪرس إﺷﺎرة ) f ' ( xﺛﻢ ﻧﻀﻊ ﺟﺪول اﻟﺘﻐﻴﺮات ﻟﻴﻨﺎ
( x + 1) e− x
) (x + e
−x 2
وﻣﻨﻪ f ' ( x ) ≥ 0
∈ ∀xو ﻣﻨﻪ إﺷﺎرة ) f ' ( xهﻲ إﺷﺎرة x + 1
= )f '( x
[∞ ∀x ∈ [ −1; +و f ' ( x ) ≤ 0
]∀x ∈ ]−∞; −1
ﺟﺪول اﻟﺘﻐﻴﺮات
∞+
∞−
-1 +
0
-
x )f '( x
http://arabmaths.ift.fr
0
1
f
1 1− e /4أ -ﻧﻜﺘﺐ ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ﻟﻠﻤﻨﺤﻨﻰ ) ( Cﻓﻲ اﻟﻨﻘﻄﺔ O
ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ﻟﻠﻤﻨﺤﻨﻰ ) ( Cﻓﻲ اﻟﻨﻘﻄﺔ Oهﻲ y = f ' ( 0 ) xأي y = xﻷن f ' ( 0 ) = 1 ب -ﻧﺘﺤﻘﻖ أن
ﻟﻴﻜﻦ
) xg ( x
g ( x) + 1
∈: x
ﺑﻤﺎ أن g ( x ) ≥ 0 اذن x − f ( x ) ≥ 0
∈ ∀xﺛﻢ ﻧﺪرس اﺷﺎرة ) x − f ( xﻋﻠﻰ
= )x − f ( x
) xg ( x
g ( x) + 1
=)
(
x x + e− x − 1
−1+1
−x
x+e
=
x −x
x+e
x − f ( x) = x −
∈ ∀xﻓﺎن اﺷﺎرة ) x − f ( xهﻲ اﺷﺎرة x
[∞ ∀x ∈ [ 0; +و ∀x ∈ ]−∞; 0] x − f ( x ) ≤ 0
ج -ﻧﺴﺘﻨﺘﺞ اﻟﻮﺿﻊ اﻟﻨﺴﺒﻲ ﻟﻠﻤﻨﺤﻨﻰ ) ( Cو اﻟﻤﺴﺘﻘﻴﻢ ) ∆ ( اﻟﺬي ﻣﻌﺎدﻟﺘﻪ y = x ﻟﺪﻳﻨﺎ x − f ( x ) ≥ 0 ﻟﺪﻳﻨﺎ x − f ( x ) ≤ 0
[∞ ∀x ∈ [0; +و ﻣﻨﻪ ) ∆ ( ﻓﻮق أو ﻳﻘﻄﻊ ) ( Cﻋﻠﻰ [∞[0; + ] ∀x ∈ ]−∞; 0و ﻣﻨﻪ ) ∆ ( ﺗﺤﺖ أو ﻳﻘﻄﻊ ) ( Cﻋﻠﻰ ]]−∞; 0
/5اﻟﺸﻜﻞ
-(IIIاﻟﻤﺘﺘﺎﻟﻴﺔ اﻟﻌﺪدﻳﺔ ) ( unﻣﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ u0 = 1و ) un +1 = f ( unﻟﻜﻞ nﻣﻦ /1ﻧﺒﻴﻦ ﺑﺎﻟﺘﺮﺟﻊ 0 ≤ un ≤ 1ﻟﻜﻞ nﻣﻦ 0 ≤ u0 ≤ 1 ﻣﻦ أﺟﻞ n = 0ﻟﺪﻳﻨﺎ u0 = 1 ﻟﻨﻔﺘﺮض أن 0 ≤ un ≤ 1ﺻﺤﻴﺤﺔ ﺣﺘﻰ اﻟﺮﺗﺒﺔ nﻟﻨﺒﻴﻦ أن 0 ≤ un +1 ≤ 1
ﻟﺪﻳﻨﺎ 0 ≤ un ≤ 1و fﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ [∞ [ −1; +وﻣﻨﻪ )f ( 0 ) ≤ f ( un ) ≤ f (1
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1
≤ 0 ≤ un +1
أي أن ≺ 1 1 + e−1 وﻣﻨﻪ 0 ≤ un ≤ 1ﻟﻜﻞ nﻣﻦ
اذن 0 ≤ un +1 ≤ 1
/2ﻧﺒﻴﻦ أن ) ( unﺗﻨﺎﻗﺼﻴﺔ ﻟﻴﻜﻦ nﻣﻦ
) un g ( u n
g ( un ) + 1
= ) un − un +1 = un − f ( un
و ﺣﻴﺚ أن 0 ≤ un ≤ 1و g ( un ) ≥ 0ﻓﺎن un − un +1 ≥ 0أي un ≥ un +1
إذن ) ( unﺗﻨﺎﻗﺼﻴﺔ
/3ﻧﺴﺘﻨﺘﺞ أن ) ( unﻣﺘﻘﺎرﺑﺔ ﺛﻢ ﻧﺤﺪد ﻧﻬﺎﻳﺘﻬﺎ
ﻟﺪﻳﻨﺎ ) ( unﺗﻨﺎﻗﺼﻴﺔ و ﻣﺼﻐﻮرة وﻣﻨﻪ ) ( unﻣﺘﻘﺎرﺑﺔ
وﺑﻤﺎ أن fﻣﺘﺼﻠﺔ ﻋﻠﻰ ] [ 0;1و ]⊂ [ 0;1 ﻓﻲ
][0;1 g ( x) ⇔ x = 0
ou
=0⇔ x=0
−1
1
;([0;1]) = 0
1+ e
) xg ( x
g ( x) + 1
fﻓﺎن ﻧﻬﺎﻳﺔ ) ( unهﻲ ﺣﻞ اﻟﻤﻌﺎدﻟﺔ f ( x ) = x
⇔ f ( x) = x ⇔ x − f ( x) = 0
إذن lim un = 0
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اﻟﺜﺎﻧﻴﺔ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
اﻟﺘﻤﺮﻳﻦ اﻷول /1ﻧﺒﻴﻦ أن ﻣﺮآﺰ اﻟﻔﻠﻜﺔ ) ( Sهﻲ اﻟﻨﻘﻄﺔ ) Ω (1; 2; 3و أن اﻟﺸﻌﺎع ﻳﺴﺎوي 6
M ( x; y ) ∈ ( S ) ⇔ x 2 + y 2 + z 2 − 2 x − 4 y − 6 z + 8 = 0 ⇔ ( x − 1) − 1 + ( y − 2 ) − 4 + ( z − 3) − 9 + 8 = 0 2
2
2
2
⇔ ( x − 1) + ( y − 2 ) + ( z − 3) = 6 = 6 2
2
2
إذن ) ( Sﻓﻠﻜﺔ ﻣﺮآﺰهﺎ اﻟﻨﻘﻄﺔ ) Ω (1; 2;3ﺷﻌﺎﻋﻬﺎ 6 /2ﻧﺘﺤﻘﻖ أن اﻟﻤﺴﺘﻮى ) ( Pﻣﻤﺎس ﻟﻠﻔﻜﺔ
6
=
1− 2 + 2×3 +1
) (S
= ) ) d ( Ω; ( Ρ
= 6 1+1+ 4 6 إذن اﻟﻤﺴﺘﻮى ) ( Pﻣﻤﺎس ﻟﻠﻔﻜﺔ ) ( S
/3أ -ﻧﺤﺪد ﺗﻤﺜﻴﻼ ﺑﺎراﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ ) ∆ ( اﻟﻤﺎر ﻣﻦ Ωو اﻟﻌﻤﻮدي ﻋﻠﻰ
)( P
ﺑﻤﺎ أن ) ( ∆ ) ⊥ ( Pو ( P ) : x − y + 2 z + 1 = 0ﻓﺎن اﻟﻤﺘﺠﻬﺔ ) n (1; −1; 2اﻟﻤﻨﻈﻤﻴﺔ ﻋﻠﻰ ) ( Pﻣﻮﺟﻬﺔ ﻟﻠﻤﺴﺘﻘﻴﻢ و ﺣﻴﺚ أن ) ∆ ( ﻳﻤﺮ ﻣﻦ ) Ω (1; 2;3ﻓﺎن
∈/t
)∆(
x = 1 + t y = 2 − tﺛﻤﺜﻴﻞ ﺑﺎراﻣﺘﺮي ﻟﻠﻤﺴﺘﻘﻴﻢ ) ∆ ( z = 3 + 2t
ب /ﻧﺤﺪد ﻣﺜﻠﻮث إﺣﺪاﺛﻴﺎت ωﻧﻘﻄﺔ ﺗﻤﺎس ) ( Sو ) ( P ωﻧﻘﻄﺔ ﺗﻤﺎس ) ( Sو ) ( Pوﻣﻨﻪ ωﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ) ( S
و
)( P
x = 1 + t y = 2 − t ω ( x; y; z ) ∈ ( P ) ∩ ( S ) ⇔ z = 3 + 2t x − y + 2 z + 1 = 0 x = 1 + t y = 2 − t ⇔ z = 3 + 2t 1 + t − 2 + t + 6 + 4t + 1 = 0 t = −1 x = 0 ⇔ y = 3 z = 1 إذن ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ) ( Sو ) ( Pهﻲ )ω ( 0;3;1 اﻟﺘﻤﺮﻳﻦ اﻟﺜﺎﻧﻲ /1أ -ﻧﻜﺘﺐ ﻋﻞ اﻟﺸﻜﻞ اﻟﺠﺒﺮي اﻟﻌﺪد اﻟﻌﻘﺪي
2
) ( 3 − 2i
= 9 − 4 − 12 i = 5 − 12 i ب -ﻧﺤﻞ ﻓﻲ
( 3 − 2 i )2
اﻟﻤﻌﺎدﻟﺔ z 2 − 2 ( 4 + i ) z + 10 + 20i = 0 :
ﻟﻴﻜﻦ ' ∆ ﻣﻤﻴﺰ اﻟﻤﻌﺎدﻟﺔ :
2
) ∆ ' = ( 4 + i ) − 10 − 20i = 15 + 8i − 10 − 20i = 5 − 12i = ( 3 − 2i 2
وﻣﻨﻪ ﺟﺪرا اﻟﻤﻌﺎدﻟﺔ هﻤﺎ z2 = 4 + i − 3 + 2i = 1 + 3i ; z1 = 4 + i + 3 − 2i = 7 − i
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إذن } S = {1 + 3 i ; 7 − i c−a /2أ -ﻧﺒﻴﻦ أن = i b−a ) c − a 5 + 9i − 1 − 3i 4 + 6i i ( 6 − 4i = = = =i b − a 7 − i − 1 − 3i 6 − 4i 6 − 4i ب -ﻧﺴﺘﻨﺘﺞ أن اﻟﻤﺜﻠﺚ ABCﻣﺘﺴﺎوي اﻟﺴﺎﻗﻴﻦ و ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ c−a c−a c−a ﺑﻤﺎ أن = i arg ﻓﺎن = i = 1 و = arg i b−a b−a b−a c−a π arg وﻣﻨﻪ c − a = b − aو ] [ 2π ≡ b−a 2
π
)
و ﺑﺎﻟﺘﺎﻟﻲ AC = ABو ] [ 2π 2 إذن ABCﻣﺘﺴﺎوي اﻟﺴﺎﻗﻴﻦ و ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ ﻓﻲ A اﻟﺘﻤﺮﻳﻦ اﻟﺜﺎﻟﺚ 2 x 1 /1ﻧﺘﺤﻘﻖ أن }∀x ∈ − {−1 = x −1+ x +1 x +1 ﻟﺘﻜﻦ }x ∈ − {−1 1 x2 − 1 + 1 x2 = = x +1 x +1 x +1 x2 /2ﻧﺒﻴﻦ أن dx = ln 3 x +1
(
≡ arg AB; AC
x −1+
2
∫0 2
x2 2 x2 1 dx = ∫ x − 1 + dx = − x + ln x + 1 = 2 − 2 + ln 3 = ln 3 0 x +1 x + 1 2 0 3
2
∫0
2
/3ﺑﺎﺳﺘﻌﻤﺎل ﻣﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء ،ﻧﺒﻴﻦ أن ∫0 x ln ( x + 1)dx = 2 ln 3 1 u ' ( x ) = x + 1 )u ( x ) = ln ( x + 1 وﻣﻨﻪ ﻧﻀﻊ 2 v ' ( x ) = x v ' ( x ) = x 2 2
x2 1 2 x2 1 3 ∫ x ln ( x + 1)dx = ln ( x + 1) − dx = 2 ln 3 − ln 3 = ln 3 0 2 2 2 0 2 x + 1
2
∫0
اﻟﺘﻤﺮﻳﻦ اﻟﺮاﺑﻊ 2 ﻧﺤﺴﺐ ) p ( Aو ) P ( Bو ﻧﺒﻴﻦ أن = ) p ( C 7 " Aﻻ ﺗﻮﺟﺪ أﻳﺔ ﺑﻴﺪﻗﺔ ﺗﺤﻤﻞ اﻟﻌﺪد 0ﻣﻦ ﺑﻴﻦ اﻟﺒﻴﺪﻗﺎت اﻟﺜﻼﺛﺔ اﻟﻤﺴﺤﻮﺑﺔ " C3 4 = p ( A ) = 43 35 C7
" Bﺳﺤﺐ ﺛﻼﺛﺔ ﺑﻴﺪﻗﺎت ﺗﺤﻤﻞ أﻋﺪادا ﻣﺨﺘﻠﻔﺔ ﻣﺜﻨﻰ ﻣﺜﻨﻰ " C 1 × C 11 × C 31 9 p (B ) = 3 = 3 35 C7 " Cﻣﺠﻤﻮع اﻻﻋﺪاد اﻟﻤﺴﺠﻠﺔ ﻋﻠﻰ اﻟﺒﻴﺪﻗﺎت اﻟﺜﻼﺛﺔ اﻟﻤﺴﺤﻮﺑﺔ ﻣﻨﻌﺪم " C 33 + C 31 × C 11 × C 31 1+ 9 10 2 = ) p (C = = = 3 35 35 7 C7 ﻣﺴﺄﻟﺔ -(Iاﻟﺪاﻟﺔ gاﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ
ﺑﻤﺎ ﻳﻠﻲ+ x − 1 :
/1ﻧﺤﺴﺐ ) g ' ( xﻟﻜﻞ xﻣﻦ
−x
g ( x) = e
ﺛﻢ ﻧﺴﺘﻨﺘﺞ أن gﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ [∞ [ 0; +و ﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ
]]−∞; 0 http://arabmaths.ift.fr
∈x
ﻟﺘﻜﻦ
g ( x ) = e− x + x − 1
:
g ' ( x ) = −e − x + 1
ﻟﺪﻳﻨﺎ x ≥ 0 ⇔ − x ≤ 0 ⇔ e− x ≤ 1 ⇔ −e− x + 1 ≥ 0 اذن ∀x ∈ [0; +∞[ g ' ( x ) ≥ 0وﻣﻨﻪ gﻋﻠﻰ ﺗﺰاﻳﺪﻳﺔ [∞[0; + ﻟﺪﻳﻨﺎ x ≤ 0 ⇔ − x ≥ 0 ⇔ e− x ≥ 1 ⇔ −e− x + 1 ≤ 0 اذن ∀x ∈ ]−∞; 0] g ' ( x ) ≤ 0و ﻣﻨﻪ gﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ ]]−∞; 0 /2ﻧﺒﻴﻦ أن g ( x ) ≥ 0
∈ ∀xﺛﻢ ﻧﺴﺘﻨﺘﺞ أن e− x + x ≥ 1
∈ ∀x
ﺑﻤﺎ أن gﻋﻠﻰ ﺗﺰاﻳﺪﻳﺔ [∞ [ 0; +و ﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ ] ]−∞; 0و g ( 0 ) = 0 وﻣﻨﻪ g ( x ) ≥ 0
∈ ∀x
أي e− x + x − 1 ≥ 0
إذن e− x + x ≥ 1
∈ ∀x
∈ ∀x x
-(IIاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fﻟﻠﻤﺘﻐﻴﺮ اﻟﺤﻘﻴﻘﻲ اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ: /1ﻧﺒﻴﻦ أن
x + e− x
= )f ( x
= Df
ﺣﺴﺐ /2 (Iﻟﻴﻨﺎ e − x + x ≥ 1 1
/2أ -ﻧﺒﻴﻦ أن
1
1+
xe x *
ﻓﺎن 0ﻗﻴﻤﺔ دﻧﻴﺎ ﻟﺪاﻟﺔ fﻋﻠﻰ
= )f ( x 1
∈: x
∈ ∀xوﻣﻨﻪ e− x + x ≠ 0 *
∈ ∀x x
=
∈ ∀xاذن
= Df
=
x
= )f ( x
ﻟﻴﻜﻦ 1 x 1 + x 1 + x xe xe ب -ﻧﺒﻴﻦ أن lim f ( x ) = 0و lim f ( x ) = 1و ﻧﺆول اﻟﻨﺘﻴﺠﺘﻲ هﻨﺪﺳﻴﺎ
1
∞x →−
ﻟﺪﻳﻨﺎ lim xe x = 0− ﻟﻠﻤﻨﺤﻨﻰ
) (C
ﻟﺪﻳﻨﺎ ∞lim xe x = +
وﻣﻨﻪ = 1
∞x →+
ﻟﻠﻤﻨﺤﻨﻰ
) (C
/3أ -ﻧﺒﻴﻦ أن
ﻟﻴﻜﻦ
( x + 1) e− x
) (x + e
−x 2
∈: x
1 x
xe
1 xe x
x + e− x
lim 1 +
∞x →−
اذن lim f ( x ) = 0و ﻣﻨﻪ ﻣﺤﻮر اﻻﻓﺎﺻﻴﻞ ﻣﺴﺘﻘﻴﻢ ﻣﻘﺎرب ∞x →−
اذن lim f ( x ) = 1وﻣﻨﻪ اﻟﻤﺴﺘﻘﻴﻢ ذا اﻟﻤﻌﺎدﻟﺔ y = 1ﻣﻘﺎرب
lim 1 +
∞x →+
∞x →−
= )f '( x
x
x+e
∞x →+
وﻣﻨﻪ ∞= −
∞x →−
−x
∈ ∀x
= )f ( x
) = e + xe = ( x + 1) e ) (x + e ) (x + e
−x
−x
−x 2
−x 2
−x
(
x + e− x − x 1 − e− x
) (x + e
−x 2
= )f '( x
ب -ﻧﺪرس إﺷﺎرة ) f ' ( xﺛﻢ ﻧﻀﻊ ﺟﺪول اﻟﺘﻐﻴﺮات ﻟﻴﻨﺎ
( x + 1) e− x
) (x + e
−x 2
وﻣﻨﻪ f ' ( x ) ≥ 0
∈ ∀xو ﻣﻨﻪ إﺷﺎرة ) f ' ( xهﻲ إﺷﺎرة x + 1
= )f '( x
[∞ ∀x ∈ [ −1; +و f ' ( x ) ≤ 0
]∀x ∈ ]−∞; −1
ﺟﺪول اﻟﺘﻐﻴﺮات
∞+
∞−
-1 +
0
-
x )f '( x
http://arabmaths.ift.fr
0
1
f
1 1− e /4أ -ﻧﻜﺘﺐ ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ﻟﻠﻤﻨﺤﻨﻰ ) ( Cﻓﻲ اﻟﻨﻘﻄﺔ O
ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ﻟﻠﻤﻨﺤﻨﻰ ) ( Cﻓﻲ اﻟﻨﻘﻄﺔ Oهﻲ y = f ' ( 0 ) xأي y = xﻷن f ' ( 0 ) = 1 ب -ﻧﺘﺤﻘﻖ أن
ﻟﻴﻜﻦ
) xg ( x
g ( x) + 1
∈: x
ﺑﻤﺎ أن g ( x ) ≥ 0 اذن x − f ( x ) ≥ 0
∈ ∀xﺛﻢ ﻧﺪرس اﺷﺎرة ) x − f ( xﻋﻠﻰ
= )x − f ( x
) xg ( x
g ( x) + 1
=)
(
x x + e− x − 1
−1+1
−x
x+e
=
x −x
x+e
x − f ( x) = x −
∈ ∀xﻓﺎن اﺷﺎرة ) x − f ( xهﻲ اﺷﺎرة x
[∞ ∀x ∈ [ 0; +و ∀x ∈ ]−∞; 0] x − f ( x ) ≤ 0
ج -ﻧﺴﺘﻨﺘﺞ اﻟﻮﺿﻊ اﻟﻨﺴﺒﻲ ﻟﻠﻤﻨﺤﻨﻰ ) ( Cو اﻟﻤﺴﺘﻘﻴﻢ ) ∆ ( اﻟﺬي ﻣﻌﺎدﻟﺘﻪ y = x ﻟﺪﻳﻨﺎ x − f ( x ) ≥ 0 ﻟﺪﻳﻨﺎ x − f ( x ) ≤ 0
[∞ ∀x ∈ [0; +و ﻣﻨﻪ ) ∆ ( ﻓﻮق أو ﻳﻘﻄﻊ ) ( Cﻋﻠﻰ [∞[0; + ] ∀x ∈ ]−∞; 0و ﻣﻨﻪ ) ∆ ( ﺗﺤﺖ أو ﻳﻘﻄﻊ ) ( Cﻋﻠﻰ ]]−∞; 0
/5اﻟﺸﻜﻞ
-(IIIاﻟﻤﺘﺘﺎﻟﻴﺔ اﻟﻌﺪدﻳﺔ ) ( unﻣﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ u0 = 1و ) un +1 = f ( unﻟﻜﻞ nﻣﻦ /1ﻧﺒﻴﻦ ﺑﺎﻟﺘﺮﺟﻊ 0 ≤ un ≤ 1ﻟﻜﻞ nﻣﻦ 0 ≤ u0 ≤ 1 ﻣﻦ أﺟﻞ n = 0ﻟﺪﻳﻨﺎ u0 = 1 ﻟﻨﻔﺘﺮض أن 0 ≤ un ≤ 1ﺻﺤﻴﺤﺔ ﺣﺘﻰ اﻟﺮﺗﺒﺔ nﻟﻨﺒﻴﻦ أن 0 ≤ un +1 ≤ 1
ﻟﺪﻳﻨﺎ 0 ≤ un ≤ 1و fﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ [∞ [ −1; +وﻣﻨﻪ )f ( 0 ) ≤ f ( un ) ≤ f (1
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1
≤ 0 ≤ un +1
أي أن ≺ 1 1 + e−1 وﻣﻨﻪ 0 ≤ un ≤ 1ﻟﻜﻞ nﻣﻦ
اذن 0 ≤ un +1 ≤ 1
/2ﻧﺒﻴﻦ أن ) ( unﺗﻨﺎﻗﺼﻴﺔ ﻟﻴﻜﻦ nﻣﻦ
) un g ( u n
g ( un ) + 1
= ) un − un +1 = un − f ( un
و ﺣﻴﺚ أن 0 ≤ un ≤ 1و g ( un ) ≥ 0ﻓﺎن un − un +1 ≥ 0أي un ≥ un +1
إذن ) ( unﺗﻨﺎﻗﺼﻴﺔ
/3ﻧﺴﺘﻨﺘﺞ أن ) ( unﻣﺘﻘﺎرﺑﺔ ﺛﻢ ﻧﺤﺪد ﻧﻬﺎﻳﺘﻬﺎ
ﻟﺪﻳﻨﺎ ) ( unﺗﻨﺎﻗﺼﻴﺔ و ﻣﺼﻐﻮرة وﻣﻨﻪ ) ( unﻣﺘﻘﺎرﺑﺔ
وﺑﻤﺎ أن fﻣﺘﺼﻠﺔ ﻋﻠﻰ ] [ 0;1و ]⊂ [ 0;1 ﻓﻲ
][0;1 g ( x) ⇔ x = 0
ou
=0⇔ x=0
−1
1
;([0;1]) = 0
1+ e
) xg ( x
g ( x) + 1
fﻓﺎن ﻧﻬﺎﻳﺔ ) ( unهﻲ ﺣﻞ اﻟﻤﻌﺎدﻟﺔ f ( x ) = x
⇔ f ( x) = x ⇔ x − f ( x) = 0
إذن lim un = 0
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