Juin 2003 Solution
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اﻣﺘﺤﺎن ﻳﻮﻧﻴﻮ2003
اﻟﺤـــﻞ اﻟﺘﻤﺮﻳﻦ1 2
-1ﻧﺤﺴﺐ ﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء I = ∫ ln ( x ) dx 1
2 1 2 2 2 dx = [ x ln x ]1 − ∫ dx = [ x ln x ]1 − [ x ]1 = 2ln 2 − 1 1 x
-2أﺡﺴﺐ اﻟﺘﻜﺎﻣﻞ x e x dx
ln 4
∫0
2
2
× I = ∫ ln ( x ) dx = [ x ln x ]1 − ∫ x 2
1
1
) ﻳﻤﻜﻦ وﺿﻊ ( t = e x
ﻟﻴﻜﻦ ]x ∈ [ 0;ln 4 ﻧﻀﻊ
2 dx 2 أي dx = dt t = e xوﻣﻨﻪ x = 2ln tو ﺑﺎﻟﺘﺎﻟﻲ = t dt t
t = 1 x =0 ⇔ t = 2 x = ln 4 2 2 2 إذن x e x dx = ∫ ( 2ln t ) × t × dt = 4∫ ln (t ) dt = 4I = 8ln 2 − 4 1 1 t
ln 4
∫0
اﻟﺘﻤﺮﻳﻦ2 ﻟﻴﻜﻦ Ωآﻮن اﻹﻣﻜﺎﻧﻴﺎت
-1ﻧﺤﺴﺐ ) p ( Aو ) p ( B ﻟﺪﻳﻨﺎ " : Aﻟﻠﻜﺮﺕﻴﻦ اﻟﻤﺴﺤﻮﺑﺘﻴﻦ ﻧﻔﺲ اﻟﻠﻮن"
cardA C 62 + C 22 15 + 1 16 4 = ) p (A = = = = card Ω 28 28 7 C 82 ﻟﺪﻳﻨﺎ " : Bﺟﺪاء اﻟﻌﺪدﻳﻦ اﻟﻤﺴﺠﻠﻴﻦ ﻋﻠﻰ اﻟﻜﺮﺕﻴﻦ اﻟﻤﺴﺤﻮﺑﺘﻴﻦ ﻣﻨﻌﺪم "
cardB C 42 + C 41 × C 41 6 + 16 22 11 = ) p (B = = = = card Ω 28 28 14 C 82 -2ﻧﺤﺪد ﻗﺎﻧﻮن اﺡﺘﻤﺎل اﻟﻤﺘﻐﻴﺮ اﻟﻌﺸﻮاﺋﻲ .X
}X ( Ω ) = {0;1; 2;3 6 3 = 28 14
12 3 = 28 7 4+3 1 = 28 4
=
=
C 42
=
C 82
C 41 × C 31 C 82
C 41 × C 11 + C 32 C 82 3 28
=
=
=
C 31 × C 11 C 82
) card ( X = 0
=
card Ω
)card ( X = 1 card Ω
) card ( X = 2 card Ω
=
= )p ( X = 0
= )p ( X = 2
)card ( X = 3 card Ω
= )p ( X = 1
= )p ( X = 3
ﻗﺎﻧﻮن اﺡﺘﻤﺎل X 3
2
1
0
3 28
1 4
3 7
3 14
xi
)
p (X = x i
اﻟﺘﻤﺮﻳﻦ3
ﻧﻌﺘﺒﺮ اﻟﻤﻌﺎدﻟﺔ mz 2 − 2z + m = 0
1+ i -1ﻧﺒﻴﻦ أن ﺡﻠﻲ اﻟﻤﻌﺎدﻟﺔ ) ( Eهﻤﺎ m
∈z =' z
(E ) : 1− i و m
اﻟﻤﻤﻴﺰ اﻟﻤﺨﺘﺼﺮ ﻟﻠﻤﻌﺎدﻟﺔ ) ( Eهﻮ = 1 − 2 = −1 = i 2 وﻣﻨﻪ
1+ i m
=' z
1− i و m
'z -2ﻧﻜﺘﺐ آﻞ ﻣﻦ ' zو " zو "z
2
=" z
d = 1 − m ⋅ m = 1 − mﻷن m = 2
=" z
ﻋﻞ اﻟﺸﻜﻞ اﻟﻤﺘﻠﺜﻲ.
π ;2 1 + i 4 π =' zو = = 1; − α m 2;α 4
π ;2 − 1 − i 4 π =" z = = 1; − − α m 4 2;α
π 1; − α 'z 4 = 1; π = z" π 2 α ;1 − − 4 -3ﻧﺒﻴﻦ أن اﻟﺮﺑﺎﻋﻲ OABCﻣﺮﺑﻊ . ' zو " zو " z '+ zأﻟﺤﺎق اﻟﻨﻘﻂ Aو Bو Cﻋﻠﻰ اﻟﺘﻮاﻟﻲ
] [ 2π
−2i −π ' z "− z ≡ ) ≡ arg m ≡ arg(−i arg OC ; AB ≡ arg 2 ' z "+ z 2 m
)
(
إذن ) (OC ) ⊥ ( AB −2i 2 2 2 = = " OC = z '+ zو = 2 = ﻟﺪﻳﻨﺎ = 2 m m 2 2
= ' AB = z "− z
وﻣﻨﻪ AB = OC إذن اﻟﺮﺑﺎﻋﻲ OABCﻣﺮﺑﻊ . اﻟﺘﻤﺮﻳﻦ4 -1ﺡﺪد ﺕﻤﺜﻴﻼ ﺑﺎراﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ ) . ( D اﻟﻤﺴﺘﻘﻴﻢ ) ( Dﻣﺎر ﻣﻦ ) A ( 2;0;2و ﻋﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻮى ) ( Pذا اﻟﻤﻌﺎدﻟﺔ . x + y − z + 3 = 0
وﻣﻨﻪ اﻟﻤﺘﺠﻬﺔ ) u (1;1; −1اﻟﻤﻨﻈﻤﻴﺔ ﻋﻠﻰ ) ( Pﻣﻮﺟﻬﺔ ﻟـ ) ( D
إذن
x = 2 + t y = tﺕﻤﺜﻴﻞ ﺑﺎراﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ ) ( D z = 2 − t
∈t
-2ﻧﺤﺪد إﺡﺪاﺙﻴﺎت . B
= 2 +t x = 3 =t y =1 ⇔ = 2 −t z =1 −3= 0 t = 1
= 2 +t =t = 2 −t
x y ⇔ z + y −z −3= 0 3t
x y B (x ; y ;z ) ∈(D ) ∩ (P ) ⇔ z x
إذن )B ( 3;1;1
) (S
-3أ -ﻧﺤﺪد ﺵﻌﺎع اﻟﻔﻠﻜﺔ ﻟﺪﻳﻨﺎ = 3
2+0−2−3 12 + ( −1) + 12 2
= ) ) AB = d ( A ; ( P 2
3 + 22 = 7
ﻟﻴﻜﻦ rﺵﻌﺎع اﻟﻔﻠﻜﺔ ب -ﻧﺤﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺕﻴﺔ ﻟﻠﻔﻠﻜﺔ
=r
) (S
( S ) : ( x − 2 )2 + y 2 + ( z − 2 ) = 7 ( S ) : x 2 + y 2 + z 2 − 4x − 4z + 1 = 0 اﻟﻤﺴﺄﻟﺔ
(
)
f ( x ) = ln 1 − x 3 ; x ≺ 0 f ( x ) = 4x x − 3x 2 x ≥ 0 - 1أ -ﻧﺒﻴﻦ أن اﻟﺪاﻟﺔ fﻣﺘﺼﻠﺔ ﻓﻲ .0
; f ( 0) = 0
lim+ f ( x ) = lim+ 4x x − 3x 2 = 0
x →0
x →0
(
)
; lim− f ( x ) = lim− ln 1 − x 3 = 0 x →0
) lim f ( x ) = lim f ( x ) = f ( 0اذن fﻣﺘﺼﻠﺔ ﻓﻲ .0 x →0 −
x →0 +
ب -ﻧﺒﻴﻦ أن اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺵﺘﻘﺎق ﻓﻲ 0
x x − 3x 2 = lim+ = lim+ x − 3x = 0 x →0 x →0 x
) (t = − x 3
إذن
3 2
t =0
× ) ) = lim ln (1 + t t
+
(
ln 1 − x 3
t →0
x
= lim− x →0
)f ( x ) − f ( 0 x −0 )f ( x ) − f ( 0 x −0
lim
x →0 +
lim−
x →0
fﻗﺎﺑﻠﺔ ﻟﻼﺵﺘﻘﺎق ﻓﻲ 0و f ' ( 0 ) = 0
-2ﻧﺒﻴﻦ أن fﺕﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ اﻟﻤﺠﺎﻟﻴﻦ [ ]−∞;0و [∞ [1; +و ﺕﺰاﻳﺪﻳﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ]. [ 0;1 *
)
(
− 6x = 4 x + 2 x − 6 x = 6 x 1 − x
2x x
∀x ∈ ]0; +∞[ f ' ( x ) = 4 x +
x →0
(
إﺵﺎرة ) f ' ( xﻋﻠﻰ [∞ ]0;+هﻲ إﺵﺎرة )
. 1− x
ﻟﺪﻳﻨﺎ ∀x ∈ [ 0;1] 1 − x ≥ 0وﻣﻨﻪ fﺕﺰاﻳﺪﻳﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ]. [ 0;1 ∀x ∈ [1; +∞[ 1 − x ≤ 0و ﻣﻨﻪ *
−3x 2 = ) f '( x 1− x 3 ﻟﺪﻳﻨﺎ 0 إذن
fﺕﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ [∞[1;+
[∀x ∈ ]−∞;0
∀x ∈ ]−∞;0[ −3x 2 ≺ 0 ; 1 − x 3وﻣﻨﻪ 0
) ∀x ∈ ]−∞;0[ f ' ( x
fﺕﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ []−∞;0
-3أ -ﻧﺤﺴﺐ ) lim f ( x
∞x →−
;
)
. lim f ( x ∞x →+
4 ∞− 3 = − lim f ( x ) = lim x 2 ∞x →+ ∞x →+ x
(
)
∞lim f ( x ) = lim ln 1 − x 3 = +
∞x →−
∞x →−
)
ب -ﻧﺘﺤﻘﻖ ﻣﻦ أﻧﻪ ﻟﻜﻞ ; x ≺ 0
(
ln 1 − x −3 x
+
) ln ( − x x
=3
) f (x x
ﻟﻴﻜﻦ [x ∈ ]−∞;0
) ) = ln (1 − x ) = f ( x 3
x
) (1 − x −3
x
3
x
) = ln ( −x ) + ln (1 − x ) = ln ( −x −3
3
x
ج -أدرس اﻟﻔﺮﻋﻴﻦ اﻟﻼﻧﻬﺎﺋﻴﻴﻦ ﻟﻠﻤﻨﺤﻨﻰ ) . (C
∞lim f ( x ) = −
∞x →+
4 = lim x ; ∞− 3 = − x →+∞ x
) f (x x
lim
∞x →+
إذن ) (Cﻳﻘﺒﻞ ﻓﺮﻋﺎ ﺵﻠﺠﻤﻴﺎ ﻓﻲ اﺕﺠﺎﻩ ﻣﺤﻮر اﻷراﺕﻴﺐ
∞lim f ( x ) = +
∞x →−
(
)
1 × ln 1 − x −3 = 0 x
+
) ln ( − x x
= lim 3 ∞x →−
) f (x x
lim
∞x →−
إذن ) (Cﻳﻘﺒﻞ ﻓﺮﻋﺎ ﺵﻠﺠﻤﻴﺎ ﻓﻲ اﺕﺠﺎﻩ ﻣﺤﻮر اﻷﻓﺎﺹﻴﻞ -4ﻧﻨﺸﺊ اﻟﻤﻨﺤﻨﻰ ) . (C
16 9
= x
(
)
4x x − 3x 2 = 0 ⇔ x x 4 − 3 x = 0 ⇔ x = 0 ou
)
(
−3x 2 + x 3
) (1 − x
3 2
= ) ∀x ∈ ]−∞;0[ f "( x
ﻋﻠﻰ [ ]−∞;0اﻟﻤﻨﺤﻨﻰ ) (Cﻳﻘﺒﻞ ﻧﻘﻄﺔ اﻧﻌﻄﺎف ﻓﻲ اﻟﻨﻘﻄﺔ ذات اﻷﻓﺼﻮل − 3 2
(
ln 1 − x −3 x
+
) ln ( − x x
3
-5أ -ﻧﺒﻴﻦ أن hﺕﻘﺎﺑﻞ ﻣﻦ [ ]−∞;0ﻧﺤﻮ ﻣﺠﺎل Jﻳﺠﻴﺐ ﺕﺤﺪﻳﺪﻩ . ﻟﺪﻳﻨﺎ hﻗﺼﻮر اﻟﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل [ ]−∞;0و ﻣﻨﻪ
hﻣﺘﺼﻠﺔ و ﺕﻨﺎﻗﺼﻴﺔ ﻗﻄﻌﺎ ﻋﻠﻰ []−∞;0
و [∞ h (]−∞;0[) = ]0; +وﻣﻨﻪ hﺕﻘﺎﺑﻞ ﻣﻦ [ ]−∞;0ﻧﺤﻮ اﻟﻤﺠﺎل [∞J = ]0; +
ب -ﺡﺪد )
h −1 ( xﻟﻜﻞ xﻣﻦ . J
ﻟﻴﻜﻦ [∞ x ∈ ]0; +و [y ∈ ]−∞;0 h −1 ( x ) = y ⇔ h ( y ) = x
)
(
⇔ ln 1 − y 3 = x ⇔ y 3 = 1−e x ) [∞x ∈ ]0; + إذن
[( y ∈ ]−∞;0
h −1 ( x ) = y ⇔ y = − 3 e x − 1 [∞∀x ∈ ]0; +
h −1 ( x ) = − 3 e x − 1
-6ﻧﻌﺘﺒﺮ اﻟﻤﺘﺘﺎﻟﻴﺔ ) (u nاﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ:
4 9 أ-
4 ﻧﺒﻴﻦ ﺑﺎﻟﺘﺮﺟﻊ أن ≤ u n ≤ 1 9
= u 0و u n +1 = 4u n u n − 3u n2
∈ ∀n
∈ ∀n
4 4 ﻣﻦ أﺟﻞ n = 0ﻟﺪﻳﻨﺎ = u 0إذن ≤ u 0 ≤ 1 9 9 4 4 ﻟﻨﻔﺘﺮض أن ≤ u n ≤ 1ﻟﻨﺒﻴﻦ أن ≤ u n +1 ≤ 1 9 9 4 ﻟﺪﻳﻨﺎ fﺕﺰاﻳﺪﻳﺔ ﻋﻠﻰ ] [ 0;1وﻣﻨﻪ )f ( ) ≤ f (u n ) ≤ f (1 9 4 16 4 =) ( f و ﺑﺎﻟﺘﺎﻟﻲ ≤ u n +1 ≤ 1ﻷن f (1) = 1و 9 27 9 4 إذن ≤ u n ≤ 1 9
4 16 ; ≺ 9 27
∈ ∀n
ﻧﺒﻴﻦ أن اﻟﻤﺘﺘﺎﻟﻴﺔ ) (u nﺕﺰاﻳﺪﻳﺔ.
ب-
( )− 1
)
u n +1 − u n = 4u n u n − 3u n2 − u n = u n −3u n + 4 u n − 1
4 ﻟﺪﻳﻨﺎ ≤ u n ≤ 1 9
un
()
∈ ∀n
(
= u n −3 u n + 1
1 2 ∈ ∀nوﻣﻨﻪ ≺ ≤ u n ≤ 1 3 3
وﺑﺎﻟﺘﺎﻟﻲ u n − 1 ≤ 0 ; −3 u n + 1 ≺ 0 إذن u n +1 − u n ≥ 0
∈ ∀n وﻣﻨﻪ ) (u nﺕﺰاﻳﺪﻳﺔ.
∈ ∀n
ج -ﻧﺴﺘﻨﺘﺞ أن اﻟﻤﺘﺘﺎﻟﻴﺔ ) (u nﻣﺘﻘﺎرﺑﺔ ﺙﻢ أﺡﺴﺐ ﻧﻬﺎﻳﺘﻬﺎ. ﺑﻤﺎ أن ) (u nﺕﺰاﻳﺪﻳﺔ و ﻣﻜﺒﻮرة ﻓﺎن ) (u nﻣﺘﻘﺎرﺑﺔ.
ﻟﺪﻳﻨﺎ ) u n +1 = f (u n
9 9 4 ;1 ﻋﻠﻰ ﻣﺘﺼﻠﺔ f و f ∈ ∀nو ;1 ⊂ ;1 9 4 4
4 lim u nهﻮ ﺡﻞ ﻟﻠﻤﻌﺎدﻟﺔ f ( l ) = lﻓﻲ 9 ;1
)
f ( l ) = l ⇔ 4l l − 3l 2 − l = 0
l −l = 0 1 9
= l
()
l = 1 ou
4 وﺡﻴﺚ l ∈ ;1ﻓﺎن l = 1 9
(
⇔ l −3 l + 1 ⇔ l = 0 ou
وﻣﻨﻪ lim u n = 1
اﻟﺤـــﻞ اﻟﺘﻤﺮﻳﻦ1 2
-1ﻧﺤﺴﺐ ﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء I = ∫ ln ( x ) dx 1
2 1 2 2 2 dx = [ x ln x ]1 − ∫ dx = [ x ln x ]1 − [ x ]1 = 2ln 2 − 1 1 x
-2أﺡﺴﺐ اﻟﺘﻜﺎﻣﻞ x e x dx
ln 4
∫0
2
2
× I = ∫ ln ( x ) dx = [ x ln x ]1 − ∫ x 2
1
1
) ﻳﻤﻜﻦ وﺿﻊ ( t = e x
ﻟﻴﻜﻦ ]x ∈ [ 0;ln 4 ﻧﻀﻊ
2 dx 2 أي dx = dt t = e xوﻣﻨﻪ x = 2ln tو ﺑﺎﻟﺘﺎﻟﻲ = t dt t
t = 1 x =0 ⇔ t = 2 x = ln 4 2 2 2 إذن x e x dx = ∫ ( 2ln t ) × t × dt = 4∫ ln (t ) dt = 4I = 8ln 2 − 4 1 1 t
ln 4
∫0
اﻟﺘﻤﺮﻳﻦ2 ﻟﻴﻜﻦ Ωآﻮن اﻹﻣﻜﺎﻧﻴﺎت
-1ﻧﺤﺴﺐ ) p ( Aو ) p ( B ﻟﺪﻳﻨﺎ " : Aﻟﻠﻜﺮﺕﻴﻦ اﻟﻤﺴﺤﻮﺑﺘﻴﻦ ﻧﻔﺲ اﻟﻠﻮن"
cardA C 62 + C 22 15 + 1 16 4 = ) p (A = = = = card Ω 28 28 7 C 82 ﻟﺪﻳﻨﺎ " : Bﺟﺪاء اﻟﻌﺪدﻳﻦ اﻟﻤﺴﺠﻠﻴﻦ ﻋﻠﻰ اﻟﻜﺮﺕﻴﻦ اﻟﻤﺴﺤﻮﺑﺘﻴﻦ ﻣﻨﻌﺪم "
cardB C 42 + C 41 × C 41 6 + 16 22 11 = ) p (B = = = = card Ω 28 28 14 C 82 -2ﻧﺤﺪد ﻗﺎﻧﻮن اﺡﺘﻤﺎل اﻟﻤﺘﻐﻴﺮ اﻟﻌﺸﻮاﺋﻲ .X
}X ( Ω ) = {0;1; 2;3 6 3 = 28 14
12 3 = 28 7 4+3 1 = 28 4
=
=
C 42
=
C 82
C 41 × C 31 C 82
C 41 × C 11 + C 32 C 82 3 28
=
=
=
C 31 × C 11 C 82
) card ( X = 0
=
card Ω
)card ( X = 1 card Ω
) card ( X = 2 card Ω
=
= )p ( X = 0
= )p ( X = 2
)card ( X = 3 card Ω
= )p ( X = 1
= )p ( X = 3
ﻗﺎﻧﻮن اﺡﺘﻤﺎل X 3
2
1
0
3 28
1 4
3 7
3 14
xi
)
p (X = x i
اﻟﺘﻤﺮﻳﻦ3
ﻧﻌﺘﺒﺮ اﻟﻤﻌﺎدﻟﺔ mz 2 − 2z + m = 0
1+ i -1ﻧﺒﻴﻦ أن ﺡﻠﻲ اﻟﻤﻌﺎدﻟﺔ ) ( Eهﻤﺎ m
∈z =' z
(E ) : 1− i و m
اﻟﻤﻤﻴﺰ اﻟﻤﺨﺘﺼﺮ ﻟﻠﻤﻌﺎدﻟﺔ ) ( Eهﻮ = 1 − 2 = −1 = i 2 وﻣﻨﻪ
1+ i m
=' z
1− i و m
'z -2ﻧﻜﺘﺐ آﻞ ﻣﻦ ' zو " zو "z
2
=" z
d = 1 − m ⋅ m = 1 − mﻷن m = 2
=" z
ﻋﻞ اﻟﺸﻜﻞ اﻟﻤﺘﻠﺜﻲ.
π ;2 1 + i 4 π =' zو = = 1; − α m 2;α 4
π ;2 − 1 − i 4 π =" z = = 1; − − α m 4 2;α
π 1; − α 'z 4 = 1; π = z" π 2 α ;1 − − 4 -3ﻧﺒﻴﻦ أن اﻟﺮﺑﺎﻋﻲ OABCﻣﺮﺑﻊ . ' zو " zو " z '+ zأﻟﺤﺎق اﻟﻨﻘﻂ Aو Bو Cﻋﻠﻰ اﻟﺘﻮاﻟﻲ
] [ 2π
−2i −π ' z "− z ≡ ) ≡ arg m ≡ arg(−i arg OC ; AB ≡ arg 2 ' z "+ z 2 m
)
(
إذن ) (OC ) ⊥ ( AB −2i 2 2 2 = = " OC = z '+ zو = 2 = ﻟﺪﻳﻨﺎ = 2 m m 2 2
= ' AB = z "− z
وﻣﻨﻪ AB = OC إذن اﻟﺮﺑﺎﻋﻲ OABCﻣﺮﺑﻊ . اﻟﺘﻤﺮﻳﻦ4 -1ﺡﺪد ﺕﻤﺜﻴﻼ ﺑﺎراﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ ) . ( D اﻟﻤﺴﺘﻘﻴﻢ ) ( Dﻣﺎر ﻣﻦ ) A ( 2;0;2و ﻋﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻮى ) ( Pذا اﻟﻤﻌﺎدﻟﺔ . x + y − z + 3 = 0
وﻣﻨﻪ اﻟﻤﺘﺠﻬﺔ ) u (1;1; −1اﻟﻤﻨﻈﻤﻴﺔ ﻋﻠﻰ ) ( Pﻣﻮﺟﻬﺔ ﻟـ ) ( D
إذن
x = 2 + t y = tﺕﻤﺜﻴﻞ ﺑﺎراﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ ) ( D z = 2 − t
∈t
-2ﻧﺤﺪد إﺡﺪاﺙﻴﺎت . B
= 2 +t x = 3 =t y =1 ⇔ = 2 −t z =1 −3= 0 t = 1
= 2 +t =t = 2 −t
x y ⇔ z + y −z −3= 0 3t
x y B (x ; y ;z ) ∈(D ) ∩ (P ) ⇔ z x
إذن )B ( 3;1;1
) (S
-3أ -ﻧﺤﺪد ﺵﻌﺎع اﻟﻔﻠﻜﺔ ﻟﺪﻳﻨﺎ = 3
2+0−2−3 12 + ( −1) + 12 2
= ) ) AB = d ( A ; ( P 2
3 + 22 = 7
ﻟﻴﻜﻦ rﺵﻌﺎع اﻟﻔﻠﻜﺔ ب -ﻧﺤﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺕﻴﺔ ﻟﻠﻔﻠﻜﺔ
=r
) (S
( S ) : ( x − 2 )2 + y 2 + ( z − 2 ) = 7 ( S ) : x 2 + y 2 + z 2 − 4x − 4z + 1 = 0 اﻟﻤﺴﺄﻟﺔ
(
)
f ( x ) = ln 1 − x 3 ; x ≺ 0 f ( x ) = 4x x − 3x 2 x ≥ 0 - 1أ -ﻧﺒﻴﻦ أن اﻟﺪاﻟﺔ fﻣﺘﺼﻠﺔ ﻓﻲ .0
; f ( 0) = 0
lim+ f ( x ) = lim+ 4x x − 3x 2 = 0
x →0
x →0
(
)
; lim− f ( x ) = lim− ln 1 − x 3 = 0 x →0
) lim f ( x ) = lim f ( x ) = f ( 0اذن fﻣﺘﺼﻠﺔ ﻓﻲ .0 x →0 −
x →0 +
ب -ﻧﺒﻴﻦ أن اﻟﺪاﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻼﺵﺘﻘﺎق ﻓﻲ 0
x x − 3x 2 = lim+ = lim+ x − 3x = 0 x →0 x →0 x
) (t = − x 3
إذن
3 2
t =0
× ) ) = lim ln (1 + t t
+
(
ln 1 − x 3
t →0
x
= lim− x →0
)f ( x ) − f ( 0 x −0 )f ( x ) − f ( 0 x −0
lim
x →0 +
lim−
x →0
fﻗﺎﺑﻠﺔ ﻟﻼﺵﺘﻘﺎق ﻓﻲ 0و f ' ( 0 ) = 0
-2ﻧﺒﻴﻦ أن fﺕﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ اﻟﻤﺠﺎﻟﻴﻦ [ ]−∞;0و [∞ [1; +و ﺕﺰاﻳﺪﻳﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ]. [ 0;1 *
)
(
− 6x = 4 x + 2 x − 6 x = 6 x 1 − x
2x x
∀x ∈ ]0; +∞[ f ' ( x ) = 4 x +
x →0
(
إﺵﺎرة ) f ' ( xﻋﻠﻰ [∞ ]0;+هﻲ إﺵﺎرة )
. 1− x
ﻟﺪﻳﻨﺎ ∀x ∈ [ 0;1] 1 − x ≥ 0وﻣﻨﻪ fﺕﺰاﻳﺪﻳﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ]. [ 0;1 ∀x ∈ [1; +∞[ 1 − x ≤ 0و ﻣﻨﻪ *
−3x 2 = ) f '( x 1− x 3 ﻟﺪﻳﻨﺎ 0 إذن
fﺕﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ [∞[1;+
[∀x ∈ ]−∞;0
∀x ∈ ]−∞;0[ −3x 2 ≺ 0 ; 1 − x 3وﻣﻨﻪ 0
) ∀x ∈ ]−∞;0[ f ' ( x
fﺕﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ []−∞;0
-3أ -ﻧﺤﺴﺐ ) lim f ( x
∞x →−
;
)
. lim f ( x ∞x →+
4 ∞− 3 = − lim f ( x ) = lim x 2 ∞x →+ ∞x →+ x
(
)
∞lim f ( x ) = lim ln 1 − x 3 = +
∞x →−
∞x →−
)
ب -ﻧﺘﺤﻘﻖ ﻣﻦ أﻧﻪ ﻟﻜﻞ ; x ≺ 0
(
ln 1 − x −3 x
+
) ln ( − x x
=3
) f (x x
ﻟﻴﻜﻦ [x ∈ ]−∞;0
) ) = ln (1 − x ) = f ( x 3
x
) (1 − x −3
x
3
x
) = ln ( −x ) + ln (1 − x ) = ln ( −x −3
3
x
ج -أدرس اﻟﻔﺮﻋﻴﻦ اﻟﻼﻧﻬﺎﺋﻴﻴﻦ ﻟﻠﻤﻨﺤﻨﻰ ) . (C
∞lim f ( x ) = −
∞x →+
4 = lim x ; ∞− 3 = − x →+∞ x
) f (x x
lim
∞x →+
إذن ) (Cﻳﻘﺒﻞ ﻓﺮﻋﺎ ﺵﻠﺠﻤﻴﺎ ﻓﻲ اﺕﺠﺎﻩ ﻣﺤﻮر اﻷراﺕﻴﺐ
∞lim f ( x ) = +
∞x →−
(
)
1 × ln 1 − x −3 = 0 x
+
) ln ( − x x
= lim 3 ∞x →−
) f (x x
lim
∞x →−
إذن ) (Cﻳﻘﺒﻞ ﻓﺮﻋﺎ ﺵﻠﺠﻤﻴﺎ ﻓﻲ اﺕﺠﺎﻩ ﻣﺤﻮر اﻷﻓﺎﺹﻴﻞ -4ﻧﻨﺸﺊ اﻟﻤﻨﺤﻨﻰ ) . (C
16 9
= x
(
)
4x x − 3x 2 = 0 ⇔ x x 4 − 3 x = 0 ⇔ x = 0 ou
)
(
−3x 2 + x 3
) (1 − x
3 2
= ) ∀x ∈ ]−∞;0[ f "( x
ﻋﻠﻰ [ ]−∞;0اﻟﻤﻨﺤﻨﻰ ) (Cﻳﻘﺒﻞ ﻧﻘﻄﺔ اﻧﻌﻄﺎف ﻓﻲ اﻟﻨﻘﻄﺔ ذات اﻷﻓﺼﻮل − 3 2
(
ln 1 − x −3 x
+
) ln ( − x x
3
-5أ -ﻧﺒﻴﻦ أن hﺕﻘﺎﺑﻞ ﻣﻦ [ ]−∞;0ﻧﺤﻮ ﻣﺠﺎل Jﻳﺠﻴﺐ ﺕﺤﺪﻳﺪﻩ . ﻟﺪﻳﻨﺎ hﻗﺼﻮر اﻟﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل [ ]−∞;0و ﻣﻨﻪ
hﻣﺘﺼﻠﺔ و ﺕﻨﺎﻗﺼﻴﺔ ﻗﻄﻌﺎ ﻋﻠﻰ []−∞;0
و [∞ h (]−∞;0[) = ]0; +وﻣﻨﻪ hﺕﻘﺎﺑﻞ ﻣﻦ [ ]−∞;0ﻧﺤﻮ اﻟﻤﺠﺎل [∞J = ]0; +
ب -ﺡﺪد )
h −1 ( xﻟﻜﻞ xﻣﻦ . J
ﻟﻴﻜﻦ [∞ x ∈ ]0; +و [y ∈ ]−∞;0 h −1 ( x ) = y ⇔ h ( y ) = x
)
(
⇔ ln 1 − y 3 = x ⇔ y 3 = 1−e x ) [∞x ∈ ]0; + إذن
[( y ∈ ]−∞;0
h −1 ( x ) = y ⇔ y = − 3 e x − 1 [∞∀x ∈ ]0; +
h −1 ( x ) = − 3 e x − 1
-6ﻧﻌﺘﺒﺮ اﻟﻤﺘﺘﺎﻟﻴﺔ ) (u nاﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ:
4 9 أ-
4 ﻧﺒﻴﻦ ﺑﺎﻟﺘﺮﺟﻊ أن ≤ u n ≤ 1 9
= u 0و u n +1 = 4u n u n − 3u n2
∈ ∀n
∈ ∀n
4 4 ﻣﻦ أﺟﻞ n = 0ﻟﺪﻳﻨﺎ = u 0إذن ≤ u 0 ≤ 1 9 9 4 4 ﻟﻨﻔﺘﺮض أن ≤ u n ≤ 1ﻟﻨﺒﻴﻦ أن ≤ u n +1 ≤ 1 9 9 4 ﻟﺪﻳﻨﺎ fﺕﺰاﻳﺪﻳﺔ ﻋﻠﻰ ] [ 0;1وﻣﻨﻪ )f ( ) ≤ f (u n ) ≤ f (1 9 4 16 4 =) ( f و ﺑﺎﻟﺘﺎﻟﻲ ≤ u n +1 ≤ 1ﻷن f (1) = 1و 9 27 9 4 إذن ≤ u n ≤ 1 9
4 16 ; ≺ 9 27
∈ ∀n
ﻧﺒﻴﻦ أن اﻟﻤﺘﺘﺎﻟﻴﺔ ) (u nﺕﺰاﻳﺪﻳﺔ.
ب-
( )− 1
)
u n +1 − u n = 4u n u n − 3u n2 − u n = u n −3u n + 4 u n − 1
4 ﻟﺪﻳﻨﺎ ≤ u n ≤ 1 9
un
()
∈ ∀n
(
= u n −3 u n + 1
1 2 ∈ ∀nوﻣﻨﻪ ≺ ≤ u n ≤ 1 3 3
وﺑﺎﻟﺘﺎﻟﻲ u n − 1 ≤ 0 ; −3 u n + 1 ≺ 0 إذن u n +1 − u n ≥ 0
∈ ∀n وﻣﻨﻪ ) (u nﺕﺰاﻳﺪﻳﺔ.
∈ ∀n
ج -ﻧﺴﺘﻨﺘﺞ أن اﻟﻤﺘﺘﺎﻟﻴﺔ ) (u nﻣﺘﻘﺎرﺑﺔ ﺙﻢ أﺡﺴﺐ ﻧﻬﺎﻳﺘﻬﺎ. ﺑﻤﺎ أن ) (u nﺕﺰاﻳﺪﻳﺔ و ﻣﻜﺒﻮرة ﻓﺎن ) (u nﻣﺘﻘﺎرﺑﺔ.
ﻟﺪﻳﻨﺎ ) u n +1 = f (u n
9 9 4 ;1 ﻋﻠﻰ ﻣﺘﺼﻠﺔ f و f ∈ ∀nو ;1 ⊂ ;1 9 4 4
4 lim u nهﻮ ﺡﻞ ﻟﻠﻤﻌﺎدﻟﺔ f ( l ) = lﻓﻲ 9 ;1
)
f ( l ) = l ⇔ 4l l − 3l 2 − l = 0
l −l = 0 1 9
= l
()
l = 1 ou
4 وﺡﻴﺚ l ∈ ;1ﻓﺎن l = 1 9
(
⇔ l −3 l + 1 ⇔ l = 0 ou
وﻣﻨﻪ lim u n = 1
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