( ) ( ) ) ( ) ( ) ( )E
ﺗﺼﺤﻴﺢ ﻓﺮض ﺷﻬﺮ أآﺘﻮﺑﺮ 2004
ﻣﻮﻗﻊ اﻟﺮﻳﺎﺿﻴﺎت ﺑﺎﻟﺜﺎﻧﻮي
ﺗﻤﺮﻳﻦ1
-1ﻧﺒﻴﻦ ﺑﺎﻟﺘﺮﺟﻊ أن 4n + 6n − 1ﺗﻘﺒﻞ اﻟﻘﺴﻤﺔ ﻋﻠﻰ 9
* ﻣﻦ أﺟﻞ n = 0ﻟﺪﻳﻨﺎ 40 + 6 × 0 − 1 = 0وﻣﻨﻪ 40 + 6 × 0 − 1ﺗﻘﺒﻞ اﻟﻘﺴﻤﺔ ﻋﻠﻰ 9
* ﻟﻨﻔﺘﺮض أن 4n + 6n − 1ﺗﻘﺒﻞ اﻟﻘﺴﻤﺔ ﻋﻠﻰ 9ﻋﺒﺎرة ﺻﺤﻴﺤﺔ ﺣﺘﻰ اﻟﺮﺗﺒﺔ n
ﻧﺒﻴﻦ أن 4n+1 + 6 ( n + 1) − 1ﺗﻘﺒﻞ اﻟﻘﺴﻤﺔ ﻋﻠﻰ 9
4n+1 + 6 ( n + 1) − 1 = 4n ⋅ 4 + 6n + 5
= 4 n ⋅ 4 + 6n ⋅ 4 − 3 ⋅ 6 n + 9 − 4
(
)
)= 4 4n + 6n − 1 − 9 ( 2n − 1
ﺑﻤﺎ أن اﻟﻌﺪدﻳﻦ 4n + 6n − 1و ) 9 ( 2n − 1ﻳﻘﺒﻼن ﻣﻌﺎ اﻟﻘﺴﻤﺔ ﻋﻠﻰ 9
ﻓﺎن 4n+1 + 6 ( n + 1) − 1ﺗﻘﺒﻞ اﻟﻘﺴﻤﺔ ﻋﻠﻰ 9
إذن 9 4n + 6n − 1
-2ﻟﻴﻜﻦ
*
∈ ∀n
un = 77777......7
∈ nﻧﻀﻊ
)(n
)
رﻗ ﻢ ﻣﺴ ﺎوﻳﺔ ل 7
(
7
ﻧﺒﻴﻦ ﺑﺎﻟﺘﺮﺟﻊ أن 10n − 1
9
)
*
= un
(
∈ ∀n
7 1
* ﻟﺪﻳﻨﺎ u1 = 7و 10 − 1 = 7
9
7
* ﻧﻔﺘﺮض أن un = 10n − 1ﻋﺒﺎرة ﺻﺤﻴﺤﺔ ﺣﺘﻰ اﻟﺮﺗﺒﺔ nرﻗﻢ ﻣﺴﺎوﻳﺔ ﻟـ7
9
7
ﻧﺒﻴﻦ أن un+1 = 10n+1 − 1
9
= un+1
77777......7
) ( n+1رﻗ ﻢ ﻣﺴ ﺎوﻳﺔ ل 7
)
)
إذن اﻟﻌﺒﺎرة ﺻﺤﻴﺤﺔ ﻣﻦ أﺟﻞ n = 1
(
(
un+1 = 77777......7 + 7 ⋅ 10n
وﻣﻨﻪ
)(n
7
7
7
أي أن = 10n − 1 + 7 ⋅ 10n = 10n − 1 + 9 ⋅ 10n = 10 × 10n − 1
9
9
9
رﻗ ﻢ ﻣﺴ ﺎوﻳﺔ ل 7
)
( )
(
)
7 n+1
وﻣﻨﻪ 10 − 1
9
7
إذن un = 10n − 1
9
)
(
)
(
un+1
= un+1
(
*
∈ ∀n
ﺗﻤﺮﻳﻦ2
ﻟﻴﻜﻦ Aو Bﺟﺰﺋﻴﻦ ﻣﻦ اﻟﻤﺠﻤﻮﻋﺔ Eﺣﻴﺚ A ⊂ B
ﻧﺤﻞ ﻓﻲ ) Ρ ( Eاﻟﻤﻌﺎدﻟﺔ X ∩ B = X ∪ A
ﻟﺘﻜﻦ Sﻣﺠﻤﻮﻋﺔ ﺣﻠﻮل اﻟﻤﻌﺎدﻟﺔ
* ﻟﺪﻳﻨﺎ A ⊂ X ∪ Aو X ⊂ X ∪ A
و ﺣﻴﺚ أن X ∩ B = X ∪ Aﻓﺎن A ⊂ X ∩ Bو X ⊂ X ∩ B
وﺑﺎﻟﺘﺎﻟﻲ A ⊂ Xو X ⊂ Bوﻣﻨﻪ A ⊂ X ⊂ B
1
Moustaouli Mohamed
site http://site.voila.fr/arabmaths
وﻣﻨﻪ }S ⊂ { X ∈ P ( E ) / A ⊂ X ⊂ B
* ﻟﻴﻜﻦ ) X ∈ P ( Eﺣﻴﺚ A ⊂ X ⊂ B
ﻟﺪﻳﻨﺎ X ∪ A = X
et
X ∩ B = Xوﻣﻨﻪ X ∩ B = X ∪ Aأي Xﺣﻞ ﻟﻠﻤﻌﺎدﻟﺔ
و ﺑﺎﻟﺘﺎﻟﻲ { X ∈ P ( E ) / A ⊂ X ⊂ B} ⊂ S
إذن }S = { X ∈ P ( E ) / A ⊂ X ⊂ B
ﺗﻤﺮﻳﻦ3
3
−1
ﻧﻌﺘﺒﺮ اﻟﺘﻄﺒﻴﻖ fاﻟﻤﻌﺮف ﻣﻦ 2 ; +∞ ﻧﺤﻮ 4 ; +∞
ﺑـ f ( x ) = x 2 + x + 1
ﻧﺒﻴﻦ أن fﺗﻘﺎﺑﻠﻲ و ﻧﺤﺪد اﻟﺘﻘﺎﺑﻞ اﻟﻌﻜﺴﻲ
−1
f
−1
3
ﻟﻴﻜﻦ x ∈ ; +∞ و y ∈ ; +∞
2
4
f ( x ) = y ⇔ x2 + x + 1 = y
2
1
3
⇔ x+ = y−
2
4
3
4
y−
3 1
−
4 2
1
=
2
y−
⇔ x+
=⇔x
3 1
1
3
و ﺣﻴﺚ أن y ∈ ; +∞ ﻓﺎن − ≥ −
4 2
2
4
y−
−1
3
;
∞+
∈
ﻓﻲ
وﺣﻴﺪا
ﺣﻼ
ﺗﻘﺒﻞ
f
x
=
y
اﻟﻤﻌﺎدﻟﺔ
y
(
)
2
اذن ﻟﻜﻞ 4 ; +∞
3
−1
fﺗﻘﺎﺑﻞ ﻣﻦ 2 ; +∞ ﻧﺤﻮ 4 ; +∞
3 1
و −
4 2
f −1 ( x ) = x −
3
∀x ∈ ; +∞
4
ﺗﻤﺮﻳﻦ4
→
ﻧﻌﺘﺒﺮ اﻟﺘﻄﺒﻴﻘﻴﻦ fو gاﻟﻤﻌﺮﻓﻴﻦ ﺑـ:
x → x +1
f :
→
g:
و 0 ;x = 0
x→
x −1 ; x ≥ 1
-1ﻧﺪرس ﺗﺒﺎﻳﻨﻴﺔ و ﺷﻤﻮﻟﻴﺔ و ﺗﻘﺎﺑﻠﻴﺔ آﻞ ﻣﻦ fو g
* ﻟﺪﻳﻨﺎ
f ( x) = f ( y) ⇒ x + 1 = y + 1 ⇒ x = y
2
∈ ) ∀ ( x; y
إذن fﺗﺒﺎﻳﻨﻲ
2
Moustaouli Mohamed
site http://site.voila.fr/arabmaths
ﻟﺪﻳﻨﺎ f ( x ) ≠ 0
∈ ∀xوﻣﻨﻪ 0ﻻ ﻳﻘﺒﻞ ﺳﺎﺑﻘﺎ ﻓﻲ
ﺑﻮاﺳﻄﺔ f
وﻣﻨﻪ fﻟﻴﺲ ﺷﻤﻮﻟﻲ ،ﻟﻴﺲ ﺗﻘﺎﺑﻠﻲ
* -ﻟﺪﻳﻨﺎ و g ( 0 ) = g (1) = 0و 0 ≠ 1إذن g
ﻟﻴﺲ ﺗﺒﺎﻳﻨﻲ ،ﻟﻴﺲ ﺗﻘﺎﺑﻠﻲ
∈y
-ﻟﻴﻜﻦ
g ( x) = y ⇔ x −1 = y ⇔ x = y + 1
∈ yﻓﺎن
وﺣﻴﺚ
*
∈ y + 1وﻣﻨﻪ اﻟﻤﻌﺎدﻟﺔ g ( x ) = yﺗﻘﺒﻞ ﻋﻠﻰ اﻷﻗﻞ ﺣﻞ ﻓﻲ
إذن gﺷﻤﻮﻟﻲ
-2ﻧﺤﺪد f
gو g
f
∈x
ﻟﺘﻜﻦ
g f ( x ) = g ( f ( x ) ) = g ( x + 1) = x + 1 − 1 = x
*
* إذا آﺎن
∈ xﻓﺎن g ( x ) = f ( g ( x ) ) = f ( x − 1) = x − 1 + 1
*
f
g ( 0) = f (0) = 1
f
ﺗﻤﺮﻳﻦ5
-1
ﻧﺤﻞ ﻓﻲ ] ]π ; 4π
)
اﻟﻤﺘﺮاﺟﺤﺔ 2 − 3 sin x − 6 ≺ 0
(
4sin 2 x − 2
)
ﻟﻴﻜﻦ ' ∆ اﻟﻤﻤﻴﺰ اﻟﻤﺨﺘﺼﺮ ﻟﺜﻼﺛﻴﺔ اﻟﺤﺪود 2 − 3 X − 6
2
)
2+ 3
(
2
2
= +4 6 = 3 + 2 +2 6
3
2
و
وﻣﻨﻪ ﺟﺪرا ﺛﻼﺛﻴﺔ اﻟﺤﺪود هﻤﺎ
2
2
2
)
2− 3
(
(
4X 2 − 2
= '∆
−
2
3
2 − 3 X − 6 = 4 X −
X+
وﻣﻨﻪ
2
2
)
(
4X 2 − 2
2
3
2 − 3 sin x − 6 ≺ 0 ⇔ 4 sin x −
sin
x
+
و ﺑﺎﻟﺘﺎﻟﻲ ≺ 0
2
2
)
11π
4
11π
3
=x
ou
10π
3
=x
=x
ou
ou
5π
3
=x
2
9π
=⇔x
2
4
ou
= sin x
3
4π
=⇔x
2
3
(
4sin 2 x − 2
] x ∈ ]π ; 4π
sin x = −
] x ∈ ]π ; 4π
ﺟﺪول اﻹﺷﺎرة
3
Moustaouli Mohamed
site http://site.voila.fr/arabmaths
4π
3
π
x
2
sin x −
2
3
sin x +
2
2
3
4 sin x −
sin x +
2
2
-
5π
3
9π
4
-
-
+
0
-
0 +
-
0
+
0
11π
4
0
+
10π
3
0
-
+
-
0
+
0
11π
3
4π
-
-
+
0
-
0 +
-
0
+
0 -
4π 5π 9π 11π 10π 11π
S = π ; ∪ ; ∪
;
; 4π
∪
3 3
3 3 4 4
1
1
≥
[ اﻟﻤﺘﺮاﺟﺤﺔ0; 2π ] ﻧﺤﻞ ﻓﻲ-2
cos x sin x
π π 3π 3π
x ∈ 0; ∪ ; π ∪ π ; ∪ ; 2π ﻟﻴﻜﻦ
2 2 2 2
1
1
1
1
≥
⇔
−
≥0
cos x sin x
cos x sin x
sin x − cos x
⇔
≥0
cos x ⋅ sin x
x ∈ [ 0; 2π ]
x
0
sin x
0
π
π
4
2
+
+
cos x
+
+
sin x − cos x
-
0
sin x − cos x
sin x ⋅ cos x
-
0
sin x = cos x ⇔ x =
π
+
4
3π
2
5π
4
2π
-
-
-
-
+
+
+
0
-
-
+
-
+
0
-
+
0
0
x=
ou
-
π π 5π
S = ; ∪ π ;
4
4 2
site http://site.voila.fr/arabmaths
5π
4
π
0
0
+
3π
∪ 2 ; 2π
Moustaouli Mohamed
4
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