Juillet 2007 Solution
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ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ2007
ﺍﻝﺸﻌﺏ :ﺍﻝﻌﻠﻭﻡ ﺍﻝﺘﺠﺭﻴﺒﻴﺔ ﺍﻷﺼﻴﻠﺔ ﺍﻝﻌﻠﻭﻡ ﺍﻝﺘﺠﺭﻴﺒﻴﺔ ﺍﻝﻌﻠﻭﻡ ﺍﻝﺯﺭﺍﻋﻴﺔ
ا اول : ﻓﻲ ﺍﻝﻔﻀﺎﺀ ﺍﻝﻤﻨﺴﻭﺏ ﺇﻝﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﻤﻨﻅﻡ ) (o, i , j , kﻝﺩﻴﻨﺎ ﺍﻝﻨﻘﻁ ) A(2, 0, −1ﻭ )B (2, 4, 2 ﻭ ) C (3,3, 3ﻭ ﺍﻝﻔﻠﻜﺔ ) ( Sﺍﻝﺘﻲ ﻤﻌﺎﺩﻝﺘﻬﺎ ﺍﻝﺩﻴﻜﺎﺭﺘﻴﺔ ﻫﻲ x 2 + y 2 + z 2 − 4 x − 4 y − 8 z + 20 = 0 : 1
( 1ﻨﺒﻴﻥ ﺍﻥ ﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ ) (Sﻫﻲ ﺍﻝﻨﻘﻁﺔ ) Ω ( 2, 2, 4ﺃﻥ ﺸﻌﺎﻋﻬﺎ ﻴﺴﺎﻭﻱ 2
∀M ( x, y, z ) ∈ ( S ) ⇔ x 2 + y 2 + z 2 − 4 x − 4 y − 8 z + 20 = 0 ⇔ ( x 2 − 4 x) + ( y 2 − 4 y ) + ( z 2 − 8 z ) + 20 = 0 ⇔ ( x 2 − 4 x + 4) − 4 + ( y 2 − 4 y + 4) − 4 + ( z 2 − 8 z + 16) − 16 + 20 = 0 ⇔ ( x − 2) 2 + ( y − 2) 2 + ( z − 4) 2 = 2 2 ﺇﺫﻥ ﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ ) (Sﻫﻲ ﺍﻝﻨﻘﻁﺔ ) Ω ( 2, 2, 4ﻭ ﺸﻌﺎﻋﻬﺎ .R= 2
0,75
1
( 2ﻨﺒﻴﻥ ﺃﻥ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ ) ( Pﻫﻲ x − y + z − 1 = 0 : ﻤﻌﺎﺩﻝﺔ ﺍﻝﻤﺴﺘﻭﻯ ) ( Pﺘﻜﺘﺏ ﻋﻠﻰ ﺍﻝﺸﻜل ax + by + cz + d = 0ﺤﻴﺙ ) n (a, b, cﻤﺘﺠﻬﺔ ﻤﻨﻅﻤﻴﺔ ﻋﻠﻴﻪ. ﻝﺩﻴﻨﺎ ) B(2, 4, 2ﻭ ) C (3,3, 3ﺇﺫﻥ )BC (1, −1,1 ﻝﺩﻴﻨﺎ ﺍﻝﻤﺴﺘﻭﻯ ) ( Pﻋﻤﻭﺩﻱ ﻋﻠﻰ ﺍﻝﻤﺴﺘﻘﻴﻡ ) (BCﺇﺫﻥ ﺍﻝﻤﺘﺠﻬﺔ ) BC (1, −1,1ﻤﻨﻅﻤﻴﺔ ﻋﻠﻰ )( P ﻭﻤﻨﻪ ﻓﺎﻥ ﻤﻌﺎﺩﻝﺔ ) ( Pﻫﻲ x − y + z + d = 0 ﻝﺩﻴﻨﺎ ﺍﻝﻤﺴﺘﻭﻯ ) ( Pﻴﻤﺭ ﻤﻥ ﺍﻝﻨﻘﻁﺔ ) A(2, 0, −1ﺇﺫﻥ 2 − 0 + ( −1) + d = 0ﺃﻱ d = −1 ﺇﺫﻥ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ ) ( Pﻫﻲ . x − y + z − 1 = 0 : ( 3ﺃ – ﻨﺒﻴﻥ ﺃﻥ ﺍﻝﻤﺴﺘﻭﻯ ) ( Pﻴﻘﻁﻊ ﺍﻝﻔﻠﻜﺔ ) ( Sﻭﻓﻕ ﺩﺍﺌﺭﺓ ) (Γﺸﻌﺎﻋﻬﺎ ﻴﺴﺎﻭﻱ .1
ﻝﺩﻴﻨﺎ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ ) ( Pﻫﻲ x − y + z − 1 = 0 :ﻭﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ ) (Sﻫﻲ ﺍﻝﻨﻘﻁﺔ ) Ω ( 2, 2, 4 3 ﻝﺩﻴﻨﺎ = 3 3
=
2 − 2 + 4 −1
= )) d (Ω, ( P
ﻭﻝﺩﻴﻨﺎ
R=2
1 + (−1) + 1 ﺒﻤﺎ ﺃﻥ d (Ω, ( P )) ≺ Rﺇﺫﻥ ﺍﻝﻤﺴﺘﻭﻯ ) ( Pﻴﻘﻁﻊ ﺍﻝﻔﻠﻜﺔ ) ( Sﻭﻓﻕ ﺩﺍﺌﺭﺓ ) (Γﺸﻌﺎﻋﻬﺎ rﺤﻴﺙ :
0,25
2
2
2
r = R 2 − d 2 = 22 − 32 = 4 − 3 = 1 ﺏ -ﻨﺤﺩﺩ ﺘﻤﺜﻴﻼ ﺒﺎﺭﺍ ﻤﺘﺭﻴﺎ ﻝﻠﻤﺴﺘﻘﻴﻡ ) ∆( ﺍﻝﻤﺎﺭ ﻤﻥ Ωﻭ ﺍﻝﻌﻤﻭﺩﻱ ﻋﻠﻰ ) . ( P ﻝﺩﻴﻨﺎ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ ) ( Pﻫﻲ x − y + z − 1 = 0 :ﺇﺫﻥ ) n (1, −1,1ﻤﺘﺠﻬﺔ ﻤﻨﻅﻤﻴﺔ ﻋﻠﻴﻪ. ﻝﺩﻴﻨﺎ ﺍﻝﻤﺴﺘﻘﻴﻡ ) ∆( ﻋﻤﻭﺩﻱ ﻋﻠﻰ ) ( Pﺇﺫﻥ ) n (1, −1,1ﻤﻭﺠﻬﺔ ﻝﻠﻤﺴﺘﻘﻴﻡ ) ∆( . ﺇﺫﻥ ﺍﻝﺘﻤﺜﻴل ﺍﻝﺒﺎراﻤﺘﺭﻱ ﻝﻠﻤﺴﺘﻘﻴﻡ ) ∆( ﺍﻝﻤﺎﺭ ﻤﻥ ﺍﻝﻨﻘﻁﺔ ) Ω ( 2, 2, 4ﻭ ﺍﻝﻤﻭﺠﻪ ﺒﺎﻝﻤﺘﺠﻬﺔ ) n (1, −1,1ﻫﻭ: x = 2 + t y = 2 −t z = 4 + t
0,5
ﺝ -ﻨﺤﺩﺩ ﻤﺜﻠﻭﺙ ﺇﺤﺩﺍﺜﻴﺎﺕ ﺍﻝﻨﻘﻁﺔ ωﻤﺭﻜﺯ ﺍﻝﺩﺍﺌﺭﺓ ). (Γ ωﻤﺭﻜﺯ ﺍﻝﺩﺍﺌﺭﺓ ) (Γﻫﻲ ﺘﻘﺎﻁﻊ ) ∆( ﻭ ) . ( P
ض ) (1) (2
)∆( ∈ ωو ){ω} = (∆) ∩ ( P) ⇔ ω ∈ ( P
x = 2 + t (2) : y=2-tو ⇔ (1) : x − y + z − 1 = 0 z = 4 + t (2 + t ) − (2 − t )(2 + t ) − (2 − t ) + (4 + t ) − 1 = 0 : t = −1
اذ :م ري 1
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ﻨﻌﻭﺽ ﻗﻴﻤﺔ t = -1ﻓﻲ ) (2ﻨﺤﺼل ﻋﻠﻰ x = 2 + (−1) = 1 y = 2 − (−1) = 3 z = 4 + (−1) = 3
ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ2007
ﺇﺫﻥ ). ω (1, 3,3
ا ا : ﻴﺤﺘﻭﻱ ﻜﻴﺱ ﻋﻠﻰ ﺜﻼﺙ ﺒﻴﺩ ﻗﺎﺕ ﺒﻴﻀﺎﺀ ﻭ ﺃﺭﺒﻊ ﺒﻴﺩ ﻗﺎﺕ ﺴﻭﺩﺍﺀ ) ﻻ ﻴﻤﻜﻥ ﺍﻝﺘﻤﻴﻴﺯ ﺒﻴﻥ ﺍﻝﺒﻴﺩ ﻗﺎﺕ ﺒﺎﻝﻠﻤﺱ(. ﻨﺴﺤﺏ ﻋﺸﻭﺍﺌﻴﺎ ﻭﻓﻲ ﺁﻥ ﻭﺍﺤﺩ ﺜﻼﺙ ﺒﻴﺩ ﻗﺎﺕ ﻤﻥ ﺍﻝﻜﻴﺱ . ﻝﺩﻴﻨﺎ card (Ω) = C73 = 35 0,75
(1ﺍﻝﺤﺩﺙ " Aﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺒﻴﺩﻗﺘﻴﻥ ﺒﺎﻝﻀﺒﻁ ﻝﻭﻨﻬﻤﺎ ﺃﺒﻴﺽ "ﺃﻱ
) ( B, B, N
card ( A) 12 = ﺇﺫﻥ ﻝﺩﻴﻨﺎ card ( A) = C32 ⋅ C41 = 12 card (Ω) 35 (2ﺍﻝﺤﺩﺙ " Bﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺜﻼﺙ ﺒﻴﺩ ﻗﺎﺕ ﻤﻥ ﻨﻔﺱ ﺍﻝﻠﻭﻥ ".ﺍﻱ ) ( N , N , Nأو = )p ( A
0,75
1
) ( B, B, B
card ( B ) 5 1 = card ( B ) = C33 + C43 = 1 + 4 = 5ﺇﺫﻥ = card (Ω) 35 7 (3ﺍﻝﺤﺩﺙ " Cﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺒﻴﺩﻗﺔ ﺒﻴﻀﺎﺀ ﻋﻠﻰ ﺍﻷﻗل " ﺍﻝﺤﺩﺙ ﺍﻝﻤﻀﺎﺩ " Cﻋﺩﻡ ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺃﻴﺔ ﺒﻴﺩﻗﺔ ﺒﻴﻀﺎﺀ " ﻴﻌﻨﻲ )ﺍﻝﺒﻴﺩ ﻗﺎﺕ ﺍﻝﺜﻼﺙ ﺍﻝﻤﺴﺤﻭﺒﺔ ﺴﻭﺩﺍﺀ( card (C ) 4 = ) P(Cﺇﺫﻥ : ﻝﺩﻴﻨﺎ card (C ) = C43 = 4ﺇﺫﻥ : = card (Ω) 35 = ) P( B
) p (C ) = 1 − p (C 4 = 1− 35 31 = 35 ا ا :
1
1 ﻝﺘﻜﻥ ) (unﺍﻝﻤﺘﺘﺎﻝﻴﺔ ﺍﻝﻤﻌﺭﻓﺔ ﺒﻤﺎ ﻴﻠﻲ u0 = 2 :ﻭ ) un +1 = (un − 4n − 1ﻝﻜل nﻤﻥ . ℕ 5 ﻨﻀﻊ vn = un + n − 1ﻝﻜل nﻤﻥ . ℕ 1 ( 1ﻨﺒﻴﻥ ﺃﻥ ) (vnﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ . 5 ∀n ∈ ℕ: vn +1 = un +1 + (n + 1) − 1 1 = (un − 4n − 1) + n 5 1 )= (un − 4n − 1 + 5n 5 1 )= (un + n − 1 5 1 = vn 5
0,5 1 ﺍﺩﻥ ) (vnﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ 5 ( 2ﺃ – ﻨﺤﺴﺏ vnﺒﺩﻻﻝﺔ . n
0,5
=q
n
1 1 ﻝﺩﻴﻨﺎ ) (vnﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ = qﻭﺤﺩﻫﺎ ﺍﻷﻭل v0 = u0 + 0 − 1 = 2 − 1 = 1ﺇﺫﻥ vn = v0 ⋅ q nﺃﻱ vn = 5 5
اذ :م ري 2
2007ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ . lim un ﺜﻡ ﺤﺴﺎﺏn ﺒﺩﻻﻝﺔun ﺍﺴﺘﻨﺘﺎﺝ-ﺏ x →+∞
n
1 un = − n + 1 ﻭﻤﻨﻪ ﻓﺎﻥ. un = vn − n + 1 ﺇﺫﻥvn = un + n − 1 ﻝﺩﻴﻨﺎ 5
0,5
n
1 1 lim un = −∞ ﺇﺫﻥlim ( − n + 1) = −∞ ﻭ ﻝﺩﻴﻨﺎlim = 0 ﺇﺫﻥ−1 ≺ ≺ 1 ﻝﺩﻴﻨﺎ x →+∞ x →+∞ 5 x →+∞ 5 S n = u0 + u1 + ............. + un ﻭTn = v0 + v1 + .............. + vn (3 1 1 Tn = 5 − n : ﻨﺒﻴﻥ ﺃﻥ 4 5 Tn = v0 + v1 + .............. + vn = v0 ⋅
1
1 − q n +1 1− q n +1
1 1− 5 = 1⋅ 1 1− 5 n +1 5 1 = 1 − 4 5 n +1 1 1 = 5 − 5 4 5 1 1 = 5 − 5 ⋅ n +1 4 5 1 1 = 5 − n 4 5 (n + 1)(n − 2) ﻨﺒﻥ ﺃﻥ 2 : ﺍﺫﻥun = vn − n + 1 ﻝﺩﻴﻨﺎ
S n = Tn − S n = u0 + u1 + ............. + un
= ( v0 − (−1) ) + ( v1 − 0 ) + ( v2 − 1) + ................ + ( vn − (n − 1) )
= ( v0 + v1 + .............. + vn ) − ( (−1) + 0 + 1 + 2 + ............. + (n − 1) ) = Tn − = Tn −
( (−1) + (n − 1) )( n + 1)
2 ( n + 1)( n − 2 ) 2
:ا ا ا ( 2 + 2i ) = −2 + 4 2i : ( ﻨﺘﺤﻘﻕ ﻤﻥ ﺃﻥ1 2
( 2 + 2i )2 = 2 + 2 2 ⋅ 2i + ( 2i ) 2
0,25
2
= 2 + 4 2i − 4 = −2 + 4 2i
م ري: اذ 3
ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ2007
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ 0,75
(2ﻨﺤل ﻓﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻝﻌﻘﺩﻴﺔ ℂﺍﻝﻤﻌﺎﺩﻝﺔ z 2 − ( 2 + 2) z + 2 + 2 − 2i = 0 : ﻝﺩﻴﻨﺎ ﻤﻤﻴﺯ ﺍﻝﻤﻌﺎﺩﻝﺔ ﻫﻭ
)
(
2
)
2 + 2 − 4 2 + 2 − 2i
(
∆ = −
= 2 + 4 2 + 4 − 8 − 4 2 + 4 2i = −2 + 4 2i = ( 2 + 2i ) 2
إذن ا د ه) = 1 − i :
2 + 2i
(
2 +2−
=)
= z1و 2 + 1 + 2i
2 (3ﻝﺩﻴﻨﺎ ﺍﻝﻌﺩﺩﻴﻥ ﺍﻝﻌﻘﺩﻴﻴﻥ z1 = 1 − iﻭ . z2 = 1 + 2 + i
2 + 2i
(
2 +2+
2
= z2
ﺃ – ﻨﺤﺩﺩ ﺍﻝﺸﻜل ﺍﻝﻤﺜﻠﺜﻲ ﻝﻠﻌﺩﺩ ﺍﻝﻌﻘﺩﻱ . z1 z1 = 1 − i = 2 #$%إذن : 0,5
2 2 π π π π z1 = 1 − i = 2 − i = 2 cos + i sin = 2 cos − + i sin − 2 4 4 4 4 2 ﺏ – ﻨﺒﻴﻥ ﺃﻥ z2 ) z1.z2 = 2 z2 :ﻫﻭ ﻤﺭﺍﻓﻕ ﺍﻝﻌﺩﺩ . ( z2 ﻝﺩﻴﻨﺎ :
)
1
(
z1 ⋅ z2 = (1 − i ) 1 + 2 + i
= 1+ 2 + i − i − i 2 +1
) )
z1.z2 = 2 z2 #$%ﺍﺫﻥ 2 z2 [ 2π ] :
و* أن ] arg ( z2 ) ≡ − arg ( z2 ) [ 2πو
arg ( z1 .z2 ) ≡ argﺃﻱ
] ( 2 ) ≡ 0 [2π
ﺝ – ﻨﺤﺩﺩ ﻋﻤﺩﺓ ﻝﻠﻌﺩﺩ z2 0,5
ﻝﺩﻴﻨﺎ ] arg( z1 ) + 2 arg( z2 ) ≡ 0 [ 2π
2 +1− i
(
= 2
= 2 z2
ا(''#ج arg( z1 ) + 2 arg( z2 ) ≡ 0 [ 2π ] :
(
= 2+ 2 −i 2
ﻭ
] [ 2π
π 4
arg
] ( 2 ) + arg ( z ) [2π 2
arg( z1 ) + arg( z2 ) ≡ arg
ن ] arg( z1 ) + 2 arg( z2 ) ≡ 0 [ 2π
arg ( z1 ) ≡ −ﺇﺫﻥ
] [ 2π
π 8
arg ( z2 ) ≡ −
: (Iﻝﺩﻴﻨﺎ gﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ ﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰ [∞ ]0, +ﺒﻤﺎ ﻴﻠﻲ : ( x − 1) 2 (1ﻨﺒﻴﻥ ﺃﻥ x2
1 − 2 ln x x
. g ( x) = x −
= ) g '( xﻝﻜل xﻤﻥ [∞ ]0, +ﺜﻡ ﻨﺴﺘﻨﺘﺞ ﻤﻨﺤﻰ ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ gﻋﻠﻰ [∞. ]0, +
1
اذ :م ري
4
ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ2007
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ 1 g '( x) = 1 + 2 − 2 ln x x 1 1 ⋅= 1+ 2 − 2 x x 2 x + 1 − 2x = x2 2
)( x − 1 x2
∀x ∈ ]0, +∞[ :
=
#$%ﻝﻜل xﻤﻥ [∞]0, + إذن ∀x ∈ ]0, +∞[ : g ' ( x ) ≥ 0و*' ن ا%ا -. gا $%$ا,ل [∞]0, + #$% (2ا%ا gﺘﺯﺍﻴﺩﻴﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل [∞ ]0, +و / 0ا,ل ] ]0,1إذن )∀x ∈ ]0,1] ⇒ 0 ≺ x ≤ 1 ⇒ g ( x) ≤ g (1 * أن g (1) = 0ن 23 x 1 g ( x) ≤ 0ا,ل ]]0,1 #$%ا%ا gﺘﺯﺍﻴﺩﻴﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل [∞ [1, +ﺇﺫﻥ )∀x ∈ [1, +∞[ ⇒ 1 ≤ x ⇒ g (1) ≤ g ( x ﺒﻤﺎ ﺃﻥ g (1) = 0ن 23 x 1 g ( x) ≥ 0ا,ل [∞[1, + ( x − 1) ≥ 0ﻭ . x 2 ≻ 0 2
0,5
1 (IIﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ fﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰ [∞ ]0, +ﺒﻤﺎ ﻴﻠﻲ − (ln x) 2 − 2 : x (ln x) 2 ) limﻴﻤﻜﻥ ﻭﻀﻊ ( t = xﺜﻡ ﻨﺤﺴﺏ )lim f ( x (1ﺃ – ﻨﺒﻴﻥ ﺃﻥ ∞x →+ ∞x →+ x ﻨﻀﻊ t = xﺇﺫﻥ x = t 2ﻋﻨﺩﻤﺎ ∞ x → +ﻓﺎﻥ ∞t → + f ( x) = x +
0,75
2
ﻝﺩﻴﻨﺎ 2
0,25
ln t = 4 t 2
2
) ( ln t ) = ( 2ln t =
) ( ln x
2 2
t2
t2
2
) ( ln x x
ln t ln t = lim 4 = 0 #$% tlimإذن = 0 ∞x →+ t ∞→+ ∞→+ x t t 2 1 1 ( ln x ) 2 2 f ( x ) = x + − ( ln x ) − 2 = x 1 + 2 − − x x x x 2 ) ( ln x 1 2 limو lim 2 = 0ﺇﺫﻥ ∞lim f ( x ) = + ﻝﺩﻴﻨﺎ lim = 0و = 0 ∞x →+ x →+∞ x ∞x →+ x ∞→+ x x 1 ب – ﻨﺘﺤﻘﻕ ﻤﻥ ﺃﻥ f ( ) = f ( x) :ﻝﻜل xﻤﻥ [∞. ]0, + x ﻝﻜل xﻤﻥ [∞ ]0, +ﻝﺩﻴﻨﺎ
lim
2
1 1 1 1 f = + − ln − 2 x x 1 x x 1 2 = + x − ( − ln x ) − 2 x 1 2 = + x − ( ln x ) − 2 x )= f ( x
اذ :م ري
5
0,5
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ج – ﻨﺤﺴﺏ )lim f ( x
ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ2007
x →0 x≻0
1 1 ﻨﻀﻊ = tﺇﺫﻥ ﻋﻨﺩﻤﺎ x → 0+ﻓﺎﻥ ∞ t → +ﻭ ﻤﻨﻪ ﻓﺎﻥ ∞lim f ( x) = lim f = lim f (t ) = + x →0 x →0 t ∞→+ x x x ≻0 x≻0 ﺇﺫﻥ ﺍﻝﻤﻨﺤﻨﻰ ) (Cﻴﻘﺒل ﻤﻘﺎﺭﺒﺎ ﺭﺃﺴﻲ ﻤﻌﺎﺩﻝﺘﻪ x = 0 ﺩ – -ﻨﺒﻴﻥ ﺃﻥ ) (Cﻴﻘﺒل ﻓﺭﻋﺎ ﺸﻠﺠﻤﻴﺎ ﺍﺘﺠﺎﻫﻪ ﺍﻝﻤﻘﺎﺭﺏ ﻫﻭ ﺍﻝﻤﺴﺘﻘﻴﻡ ﺍﻝﺫﻱ ﻤﻌﺎﺩﻝﺘﻪ ﻫﻲ y = x :
0,5
ﻝﺩﻴﻨﺎ ∞lim f ( x ) = +
∞x →+
2 1 ( ln x ) 2 lim f ( x ) = lim 1 + 2 − ﻭ− =1 ∞x →+ x →+∞ x x x
1 ∞− (ln x) 2 − 2 = − ∞x →+ x →+∞ x ﻫﻲ y = x : )g ( x (2ﺒﻴﻥ ﺃﻥ : = ) f '( xﻝﻜل xﻤﻥ [∞]0, + x
lim f ( x ) − x = limﺇﺫﻥ ) (Cﻴﻘﺒل ﻓﺭﻋﺎ ﺸﻠﺠﻤﻴﺎ ﺍﺘﺠﺎﻫﻪ ﺍﻝﻤﻘﺎﺭﺏ ﻫﻭ ﺍﻝﻤﺴﺘﻘﻴﻡ ﺍﻝﺫﻱ ﻤﻌﺎﺩﻝﺘﻪ
1 ' ) − 2 ( ln x )( ln x x2 1 1 ⋅ ) = 1 − 2 − 2 ( ln x x x 1 1 = x − − 2 ln x x x )g ( x = x
1,5
f '( x) = 1−
ﺇﺸﺎﺭﺓ ) f ' ( xﻫﻲ ﺇﺸﺎﺭﺓ ) g ( x ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ f
∞+
0
1
φ
+ ∞+
0 (3ا## 1
اذ :م ري 6
x )f '( x
∞+
)f ( x
0,5
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ2007 ( 4ﺃ -ﻨﺒﻴﻥ ﺃﻥ ﺍﻝﺩﺍﻝﺔ G : x ln x − xﺩﺍﻝﺔ ﺃﺼﻠﻴﺔ ﻝﻠﺩﺍﻝﺔ g : x → ln xﻋﻠﻰ [∞]0, + ﻝﺩﻴﻨﺎ ∀x ∈ ]0, +∞[ : G ' ( x ) = x 'ln x + x ( ln x ) '− 1 1 = ln x + x ⋅ − 1 x = ln x + 1 − 1 = ln x
ﺇﺫﻥ ﺍﻝﺩﺍﻝﺔ Gﺩﺍﻝﺔ ﺃﺼﻠﻴﺔ ﻝﻠﺩﺍﻝﺔ .g
ﺏ -ﺒﺎﺴﺘﻌﻤﺎل ﻤﻜﺎﻤﻠﺔ ﺒﺎﻷﺠﺯﺍﺀ ،ﻨﺒﻴﻥ ﺃﻥ (ln x) 2 dx = e − 2 : 0,75
u ( x ) = ( ln x )2 ﻨﻀﻊ v ' ( x ) = 1
ﺇﺫﻥ
e
∫
1
ln x u ' ( x ) = 2 x v ( x ) = x
ﺇﺫﻥ ln x 2 ⋅ x dx x
e
e
− 1 ∫1
2
2
) ∫ ( ln x ) dx = x ( ln x e
1
e
2 = x ( ln x ) − 2 ∫ ln x dx 1 1 e
e 2 = x ( ln x ) − 2 [ x ln x − x ]1 1 e
) )) − 2 (( e ln e − e ) − (1ln1 − 1 0,75
2
(
)= e ( ln e ) − 1( ln1 2
=e−2 ﺝ – ﻤﺴﺎﺤﺔ ﺤﻴﺯ ﺍﻝﻤﺴﺘﻭﻯ ﺍﻝﻤﺤﺼﻭﺭ ) (Cﻭ ﻤﺤﻭﺭ ﺍﻷﻓﺎﺼﻴل ﻭ ﺍﻝﻤﺴﺘﻘﻴﻤﻴﻥ ﺍﻝﻠﺫﻴﻥ ﻤﻌﺎﺩﻝﺘﺎﻫﻤﺎ x = 1 :ﻭ x = e
ﻝﺩﻴﻨﺎ
fﺩﺍﻝﺔ ﻤﻭﺠﺒﺔ ﻭ ﻤﺘﺼﻠﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل ] [1, eﺇﺫﻥ ﺍﻝﻤﺴﺎﺤﺔ ﺍﻝﻤﻁﻠﻭﺒﺔ ﻫﻲ
e e 1 A = ∫ f ( x ) dx = ∫ x + − (ln x) 2 − 2 dx 1 1 x e e 1 2 = ∫ x + − 2 dx − ∫ ( ln x ) dx 1 1 x e
x2 ) = + ln x − 2 x − ( e − 2 2 1 e2 1 = + ln e − 2e − + ln1 − 2 − e + 2 2 2 e2 1 = + 1 − 2e − + 2 − e + 2 2 2 2 e 9 = − 3e + 2 2 e2 9 = Aﺒﻭﺤﺩﺓ ﻗﻴﺎﺱ ﺍﻝﻤﺴﺎﺤﺔ ﺇﺫﻥ − 3 e + 2 2
اذ :م ري
7
ﺍﻝﺸﻌﺏ :ﺍﻝﻌﻠﻭﻡ ﺍﻝﺘﺠﺭﻴﺒﻴﺔ ﺍﻷﺼﻴﻠﺔ ﺍﻝﻌﻠﻭﻡ ﺍﻝﺘﺠﺭﻴﺒﻴﺔ ﺍﻝﻌﻠﻭﻡ ﺍﻝﺯﺭﺍﻋﻴﺔ
ا اول : ﻓﻲ ﺍﻝﻔﻀﺎﺀ ﺍﻝﻤﻨﺴﻭﺏ ﺇﻝﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﻤﻨﻅﻡ ) (o, i , j , kﻝﺩﻴﻨﺎ ﺍﻝﻨﻘﻁ ) A(2, 0, −1ﻭ )B (2, 4, 2 ﻭ ) C (3,3, 3ﻭ ﺍﻝﻔﻠﻜﺔ ) ( Sﺍﻝﺘﻲ ﻤﻌﺎﺩﻝﺘﻬﺎ ﺍﻝﺩﻴﻜﺎﺭﺘﻴﺔ ﻫﻲ x 2 + y 2 + z 2 − 4 x − 4 y − 8 z + 20 = 0 : 1
( 1ﻨﺒﻴﻥ ﺍﻥ ﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ ) (Sﻫﻲ ﺍﻝﻨﻘﻁﺔ ) Ω ( 2, 2, 4ﺃﻥ ﺸﻌﺎﻋﻬﺎ ﻴﺴﺎﻭﻱ 2
∀M ( x, y, z ) ∈ ( S ) ⇔ x 2 + y 2 + z 2 − 4 x − 4 y − 8 z + 20 = 0 ⇔ ( x 2 − 4 x) + ( y 2 − 4 y ) + ( z 2 − 8 z ) + 20 = 0 ⇔ ( x 2 − 4 x + 4) − 4 + ( y 2 − 4 y + 4) − 4 + ( z 2 − 8 z + 16) − 16 + 20 = 0 ⇔ ( x − 2) 2 + ( y − 2) 2 + ( z − 4) 2 = 2 2 ﺇﺫﻥ ﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ ) (Sﻫﻲ ﺍﻝﻨﻘﻁﺔ ) Ω ( 2, 2, 4ﻭ ﺸﻌﺎﻋﻬﺎ .R= 2
0,75
1
( 2ﻨﺒﻴﻥ ﺃﻥ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ ) ( Pﻫﻲ x − y + z − 1 = 0 : ﻤﻌﺎﺩﻝﺔ ﺍﻝﻤﺴﺘﻭﻯ ) ( Pﺘﻜﺘﺏ ﻋﻠﻰ ﺍﻝﺸﻜل ax + by + cz + d = 0ﺤﻴﺙ ) n (a, b, cﻤﺘﺠﻬﺔ ﻤﻨﻅﻤﻴﺔ ﻋﻠﻴﻪ. ﻝﺩﻴﻨﺎ ) B(2, 4, 2ﻭ ) C (3,3, 3ﺇﺫﻥ )BC (1, −1,1 ﻝﺩﻴﻨﺎ ﺍﻝﻤﺴﺘﻭﻯ ) ( Pﻋﻤﻭﺩﻱ ﻋﻠﻰ ﺍﻝﻤﺴﺘﻘﻴﻡ ) (BCﺇﺫﻥ ﺍﻝﻤﺘﺠﻬﺔ ) BC (1, −1,1ﻤﻨﻅﻤﻴﺔ ﻋﻠﻰ )( P ﻭﻤﻨﻪ ﻓﺎﻥ ﻤﻌﺎﺩﻝﺔ ) ( Pﻫﻲ x − y + z + d = 0 ﻝﺩﻴﻨﺎ ﺍﻝﻤﺴﺘﻭﻯ ) ( Pﻴﻤﺭ ﻤﻥ ﺍﻝﻨﻘﻁﺔ ) A(2, 0, −1ﺇﺫﻥ 2 − 0 + ( −1) + d = 0ﺃﻱ d = −1 ﺇﺫﻥ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ ) ( Pﻫﻲ . x − y + z − 1 = 0 : ( 3ﺃ – ﻨﺒﻴﻥ ﺃﻥ ﺍﻝﻤﺴﺘﻭﻯ ) ( Pﻴﻘﻁﻊ ﺍﻝﻔﻠﻜﺔ ) ( Sﻭﻓﻕ ﺩﺍﺌﺭﺓ ) (Γﺸﻌﺎﻋﻬﺎ ﻴﺴﺎﻭﻱ .1
ﻝﺩﻴﻨﺎ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ ) ( Pﻫﻲ x − y + z − 1 = 0 :ﻭﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ ) (Sﻫﻲ ﺍﻝﻨﻘﻁﺔ ) Ω ( 2, 2, 4 3 ﻝﺩﻴﻨﺎ = 3 3
=
2 − 2 + 4 −1
= )) d (Ω, ( P
ﻭﻝﺩﻴﻨﺎ
R=2
1 + (−1) + 1 ﺒﻤﺎ ﺃﻥ d (Ω, ( P )) ≺ Rﺇﺫﻥ ﺍﻝﻤﺴﺘﻭﻯ ) ( Pﻴﻘﻁﻊ ﺍﻝﻔﻠﻜﺔ ) ( Sﻭﻓﻕ ﺩﺍﺌﺭﺓ ) (Γﺸﻌﺎﻋﻬﺎ rﺤﻴﺙ :
0,25
2
2
2
r = R 2 − d 2 = 22 − 32 = 4 − 3 = 1 ﺏ -ﻨﺤﺩﺩ ﺘﻤﺜﻴﻼ ﺒﺎﺭﺍ ﻤﺘﺭﻴﺎ ﻝﻠﻤﺴﺘﻘﻴﻡ ) ∆( ﺍﻝﻤﺎﺭ ﻤﻥ Ωﻭ ﺍﻝﻌﻤﻭﺩﻱ ﻋﻠﻰ ) . ( P ﻝﺩﻴﻨﺎ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ ) ( Pﻫﻲ x − y + z − 1 = 0 :ﺇﺫﻥ ) n (1, −1,1ﻤﺘﺠﻬﺔ ﻤﻨﻅﻤﻴﺔ ﻋﻠﻴﻪ. ﻝﺩﻴﻨﺎ ﺍﻝﻤﺴﺘﻘﻴﻡ ) ∆( ﻋﻤﻭﺩﻱ ﻋﻠﻰ ) ( Pﺇﺫﻥ ) n (1, −1,1ﻤﻭﺠﻬﺔ ﻝﻠﻤﺴﺘﻘﻴﻡ ) ∆( . ﺇﺫﻥ ﺍﻝﺘﻤﺜﻴل ﺍﻝﺒﺎراﻤﺘﺭﻱ ﻝﻠﻤﺴﺘﻘﻴﻡ ) ∆( ﺍﻝﻤﺎﺭ ﻤﻥ ﺍﻝﻨﻘﻁﺔ ) Ω ( 2, 2, 4ﻭ ﺍﻝﻤﻭﺠﻪ ﺒﺎﻝﻤﺘﺠﻬﺔ ) n (1, −1,1ﻫﻭ: x = 2 + t y = 2 −t z = 4 + t
0,5
ﺝ -ﻨﺤﺩﺩ ﻤﺜﻠﻭﺙ ﺇﺤﺩﺍﺜﻴﺎﺕ ﺍﻝﻨﻘﻁﺔ ωﻤﺭﻜﺯ ﺍﻝﺩﺍﺌﺭﺓ ). (Γ ωﻤﺭﻜﺯ ﺍﻝﺩﺍﺌﺭﺓ ) (Γﻫﻲ ﺘﻘﺎﻁﻊ ) ∆( ﻭ ) . ( P
ض ) (1) (2
)∆( ∈ ωو ){ω} = (∆) ∩ ( P) ⇔ ω ∈ ( P
x = 2 + t (2) : y=2-tو ⇔ (1) : x − y + z − 1 = 0 z = 4 + t (2 + t ) − (2 − t )(2 + t ) − (2 − t ) + (4 + t ) − 1 = 0 : t = −1
اذ :م ري 1
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ﻨﻌﻭﺽ ﻗﻴﻤﺔ t = -1ﻓﻲ ) (2ﻨﺤﺼل ﻋﻠﻰ x = 2 + (−1) = 1 y = 2 − (−1) = 3 z = 4 + (−1) = 3
ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ2007
ﺇﺫﻥ ). ω (1, 3,3
ا ا : ﻴﺤﺘﻭﻱ ﻜﻴﺱ ﻋﻠﻰ ﺜﻼﺙ ﺒﻴﺩ ﻗﺎﺕ ﺒﻴﻀﺎﺀ ﻭ ﺃﺭﺒﻊ ﺒﻴﺩ ﻗﺎﺕ ﺴﻭﺩﺍﺀ ) ﻻ ﻴﻤﻜﻥ ﺍﻝﺘﻤﻴﻴﺯ ﺒﻴﻥ ﺍﻝﺒﻴﺩ ﻗﺎﺕ ﺒﺎﻝﻠﻤﺱ(. ﻨﺴﺤﺏ ﻋﺸﻭﺍﺌﻴﺎ ﻭﻓﻲ ﺁﻥ ﻭﺍﺤﺩ ﺜﻼﺙ ﺒﻴﺩ ﻗﺎﺕ ﻤﻥ ﺍﻝﻜﻴﺱ . ﻝﺩﻴﻨﺎ card (Ω) = C73 = 35 0,75
(1ﺍﻝﺤﺩﺙ " Aﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺒﻴﺩﻗﺘﻴﻥ ﺒﺎﻝﻀﺒﻁ ﻝﻭﻨﻬﻤﺎ ﺃﺒﻴﺽ "ﺃﻱ
) ( B, B, N
card ( A) 12 = ﺇﺫﻥ ﻝﺩﻴﻨﺎ card ( A) = C32 ⋅ C41 = 12 card (Ω) 35 (2ﺍﻝﺤﺩﺙ " Bﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺜﻼﺙ ﺒﻴﺩ ﻗﺎﺕ ﻤﻥ ﻨﻔﺱ ﺍﻝﻠﻭﻥ ".ﺍﻱ ) ( N , N , Nأو = )p ( A
0,75
1
) ( B, B, B
card ( B ) 5 1 = card ( B ) = C33 + C43 = 1 + 4 = 5ﺇﺫﻥ = card (Ω) 35 7 (3ﺍﻝﺤﺩﺙ " Cﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺒﻴﺩﻗﺔ ﺒﻴﻀﺎﺀ ﻋﻠﻰ ﺍﻷﻗل " ﺍﻝﺤﺩﺙ ﺍﻝﻤﻀﺎﺩ " Cﻋﺩﻡ ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺃﻴﺔ ﺒﻴﺩﻗﺔ ﺒﻴﻀﺎﺀ " ﻴﻌﻨﻲ )ﺍﻝﺒﻴﺩ ﻗﺎﺕ ﺍﻝﺜﻼﺙ ﺍﻝﻤﺴﺤﻭﺒﺔ ﺴﻭﺩﺍﺀ( card (C ) 4 = ) P(Cﺇﺫﻥ : ﻝﺩﻴﻨﺎ card (C ) = C43 = 4ﺇﺫﻥ : = card (Ω) 35 = ) P( B
) p (C ) = 1 − p (C 4 = 1− 35 31 = 35 ا ا :
1
1 ﻝﺘﻜﻥ ) (unﺍﻝﻤﺘﺘﺎﻝﻴﺔ ﺍﻝﻤﻌﺭﻓﺔ ﺒﻤﺎ ﻴﻠﻲ u0 = 2 :ﻭ ) un +1 = (un − 4n − 1ﻝﻜل nﻤﻥ . ℕ 5 ﻨﻀﻊ vn = un + n − 1ﻝﻜل nﻤﻥ . ℕ 1 ( 1ﻨﺒﻴﻥ ﺃﻥ ) (vnﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ . 5 ∀n ∈ ℕ: vn +1 = un +1 + (n + 1) − 1 1 = (un − 4n − 1) + n 5 1 )= (un − 4n − 1 + 5n 5 1 )= (un + n − 1 5 1 = vn 5
0,5 1 ﺍﺩﻥ ) (vnﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ 5 ( 2ﺃ – ﻨﺤﺴﺏ vnﺒﺩﻻﻝﺔ . n
0,5
=q
n
1 1 ﻝﺩﻴﻨﺎ ) (vnﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ = qﻭﺤﺩﻫﺎ ﺍﻷﻭل v0 = u0 + 0 − 1 = 2 − 1 = 1ﺇﺫﻥ vn = v0 ⋅ q nﺃﻱ vn = 5 5
اذ :م ري 2
2007ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ . lim un ﺜﻡ ﺤﺴﺎﺏn ﺒﺩﻻﻝﺔun ﺍﺴﺘﻨﺘﺎﺝ-ﺏ x →+∞
n
1 un = − n + 1 ﻭﻤﻨﻪ ﻓﺎﻥ. un = vn − n + 1 ﺇﺫﻥvn = un + n − 1 ﻝﺩﻴﻨﺎ 5
0,5
n
1 1 lim un = −∞ ﺇﺫﻥlim ( − n + 1) = −∞ ﻭ ﻝﺩﻴﻨﺎlim = 0 ﺇﺫﻥ−1 ≺ ≺ 1 ﻝﺩﻴﻨﺎ x →+∞ x →+∞ 5 x →+∞ 5 S n = u0 + u1 + ............. + un ﻭTn = v0 + v1 + .............. + vn (3 1 1 Tn = 5 − n : ﻨﺒﻴﻥ ﺃﻥ 4 5 Tn = v0 + v1 + .............. + vn = v0 ⋅
1
1 − q n +1 1− q n +1
1 1− 5 = 1⋅ 1 1− 5 n +1 5 1 = 1 − 4 5 n +1 1 1 = 5 − 5 4 5 1 1 = 5 − 5 ⋅ n +1 4 5 1 1 = 5 − n 4 5 (n + 1)(n − 2) ﻨﺒﻥ ﺃﻥ 2 : ﺍﺫﻥun = vn − n + 1 ﻝﺩﻴﻨﺎ
S n = Tn − S n = u0 + u1 + ............. + un
= ( v0 − (−1) ) + ( v1 − 0 ) + ( v2 − 1) + ................ + ( vn − (n − 1) )
= ( v0 + v1 + .............. + vn ) − ( (−1) + 0 + 1 + 2 + ............. + (n − 1) ) = Tn − = Tn −
( (−1) + (n − 1) )( n + 1)
2 ( n + 1)( n − 2 ) 2
:ا ا ا ( 2 + 2i ) = −2 + 4 2i : ( ﻨﺘﺤﻘﻕ ﻤﻥ ﺃﻥ1 2
( 2 + 2i )2 = 2 + 2 2 ⋅ 2i + ( 2i ) 2
0,25
2
= 2 + 4 2i − 4 = −2 + 4 2i
م ري: اذ 3
ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ2007
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ 0,75
(2ﻨﺤل ﻓﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻝﻌﻘﺩﻴﺔ ℂﺍﻝﻤﻌﺎﺩﻝﺔ z 2 − ( 2 + 2) z + 2 + 2 − 2i = 0 : ﻝﺩﻴﻨﺎ ﻤﻤﻴﺯ ﺍﻝﻤﻌﺎﺩﻝﺔ ﻫﻭ
)
(
2
)
2 + 2 − 4 2 + 2 − 2i
(
∆ = −
= 2 + 4 2 + 4 − 8 − 4 2 + 4 2i = −2 + 4 2i = ( 2 + 2i ) 2
إذن ا د ه) = 1 − i :
2 + 2i
(
2 +2−
=)
= z1و 2 + 1 + 2i
2 (3ﻝﺩﻴﻨﺎ ﺍﻝﻌﺩﺩﻴﻥ ﺍﻝﻌﻘﺩﻴﻴﻥ z1 = 1 − iﻭ . z2 = 1 + 2 + i
2 + 2i
(
2 +2+
2
= z2
ﺃ – ﻨﺤﺩﺩ ﺍﻝﺸﻜل ﺍﻝﻤﺜﻠﺜﻲ ﻝﻠﻌﺩﺩ ﺍﻝﻌﻘﺩﻱ . z1 z1 = 1 − i = 2 #$%إذن : 0,5
2 2 π π π π z1 = 1 − i = 2 − i = 2 cos + i sin = 2 cos − + i sin − 2 4 4 4 4 2 ﺏ – ﻨﺒﻴﻥ ﺃﻥ z2 ) z1.z2 = 2 z2 :ﻫﻭ ﻤﺭﺍﻓﻕ ﺍﻝﻌﺩﺩ . ( z2 ﻝﺩﻴﻨﺎ :
)
1
(
z1 ⋅ z2 = (1 − i ) 1 + 2 + i
= 1+ 2 + i − i − i 2 +1
) )
z1.z2 = 2 z2 #$%ﺍﺫﻥ 2 z2 [ 2π ] :
و* أن ] arg ( z2 ) ≡ − arg ( z2 ) [ 2πو
arg ( z1 .z2 ) ≡ argﺃﻱ
] ( 2 ) ≡ 0 [2π
ﺝ – ﻨﺤﺩﺩ ﻋﻤﺩﺓ ﻝﻠﻌﺩﺩ z2 0,5
ﻝﺩﻴﻨﺎ ] arg( z1 ) + 2 arg( z2 ) ≡ 0 [ 2π
2 +1− i
(
= 2
= 2 z2
ا(''#ج arg( z1 ) + 2 arg( z2 ) ≡ 0 [ 2π ] :
(
= 2+ 2 −i 2
ﻭ
] [ 2π
π 4
arg
] ( 2 ) + arg ( z ) [2π 2
arg( z1 ) + arg( z2 ) ≡ arg
ن ] arg( z1 ) + 2 arg( z2 ) ≡ 0 [ 2π
arg ( z1 ) ≡ −ﺇﺫﻥ
] [ 2π
π 8
arg ( z2 ) ≡ −
: (Iﻝﺩﻴﻨﺎ gﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ ﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰ [∞ ]0, +ﺒﻤﺎ ﻴﻠﻲ : ( x − 1) 2 (1ﻨﺒﻴﻥ ﺃﻥ x2
1 − 2 ln x x
. g ( x) = x −
= ) g '( xﻝﻜل xﻤﻥ [∞ ]0, +ﺜﻡ ﻨﺴﺘﻨﺘﺞ ﻤﻨﺤﻰ ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ gﻋﻠﻰ [∞. ]0, +
1
اذ :م ري
4
ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ2007
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ 1 g '( x) = 1 + 2 − 2 ln x x 1 1 ⋅= 1+ 2 − 2 x x 2 x + 1 − 2x = x2 2
)( x − 1 x2
∀x ∈ ]0, +∞[ :
=
#$%ﻝﻜل xﻤﻥ [∞]0, + إذن ∀x ∈ ]0, +∞[ : g ' ( x ) ≥ 0و*' ن ا%ا -. gا $%$ا,ل [∞]0, + #$% (2ا%ا gﺘﺯﺍﻴﺩﻴﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل [∞ ]0, +و / 0ا,ل ] ]0,1إذن )∀x ∈ ]0,1] ⇒ 0 ≺ x ≤ 1 ⇒ g ( x) ≤ g (1 * أن g (1) = 0ن 23 x 1 g ( x) ≤ 0ا,ل ]]0,1 #$%ا%ا gﺘﺯﺍﻴﺩﻴﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل [∞ [1, +ﺇﺫﻥ )∀x ∈ [1, +∞[ ⇒ 1 ≤ x ⇒ g (1) ≤ g ( x ﺒﻤﺎ ﺃﻥ g (1) = 0ن 23 x 1 g ( x) ≥ 0ا,ل [∞[1, + ( x − 1) ≥ 0ﻭ . x 2 ≻ 0 2
0,5
1 (IIﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ fﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰ [∞ ]0, +ﺒﻤﺎ ﻴﻠﻲ − (ln x) 2 − 2 : x (ln x) 2 ) limﻴﻤﻜﻥ ﻭﻀﻊ ( t = xﺜﻡ ﻨﺤﺴﺏ )lim f ( x (1ﺃ – ﻨﺒﻴﻥ ﺃﻥ ∞x →+ ∞x →+ x ﻨﻀﻊ t = xﺇﺫﻥ x = t 2ﻋﻨﺩﻤﺎ ∞ x → +ﻓﺎﻥ ∞t → + f ( x) = x +
0,75
2
ﻝﺩﻴﻨﺎ 2
0,25
ln t = 4 t 2
2
) ( ln t ) = ( 2ln t =
) ( ln x
2 2
t2
t2
2
) ( ln x x
ln t ln t = lim 4 = 0 #$% tlimإذن = 0 ∞x →+ t ∞→+ ∞→+ x t t 2 1 1 ( ln x ) 2 2 f ( x ) = x + − ( ln x ) − 2 = x 1 + 2 − − x x x x 2 ) ( ln x 1 2 limو lim 2 = 0ﺇﺫﻥ ∞lim f ( x ) = + ﻝﺩﻴﻨﺎ lim = 0و = 0 ∞x →+ x →+∞ x ∞x →+ x ∞→+ x x 1 ب – ﻨﺘﺤﻘﻕ ﻤﻥ ﺃﻥ f ( ) = f ( x) :ﻝﻜل xﻤﻥ [∞. ]0, + x ﻝﻜل xﻤﻥ [∞ ]0, +ﻝﺩﻴﻨﺎ
lim
2
1 1 1 1 f = + − ln − 2 x x 1 x x 1 2 = + x − ( − ln x ) − 2 x 1 2 = + x − ( ln x ) − 2 x )= f ( x
اذ :م ري
5
0,5
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ج – ﻨﺤﺴﺏ )lim f ( x
ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ2007
x →0 x≻0
1 1 ﻨﻀﻊ = tﺇﺫﻥ ﻋﻨﺩﻤﺎ x → 0+ﻓﺎﻥ ∞ t → +ﻭ ﻤﻨﻪ ﻓﺎﻥ ∞lim f ( x) = lim f = lim f (t ) = + x →0 x →0 t ∞→+ x x x ≻0 x≻0 ﺇﺫﻥ ﺍﻝﻤﻨﺤﻨﻰ ) (Cﻴﻘﺒل ﻤﻘﺎﺭﺒﺎ ﺭﺃﺴﻲ ﻤﻌﺎﺩﻝﺘﻪ x = 0 ﺩ – -ﻨﺒﻴﻥ ﺃﻥ ) (Cﻴﻘﺒل ﻓﺭﻋﺎ ﺸﻠﺠﻤﻴﺎ ﺍﺘﺠﺎﻫﻪ ﺍﻝﻤﻘﺎﺭﺏ ﻫﻭ ﺍﻝﻤﺴﺘﻘﻴﻡ ﺍﻝﺫﻱ ﻤﻌﺎﺩﻝﺘﻪ ﻫﻲ y = x :
0,5
ﻝﺩﻴﻨﺎ ∞lim f ( x ) = +
∞x →+
2 1 ( ln x ) 2 lim f ( x ) = lim 1 + 2 − ﻭ− =1 ∞x →+ x →+∞ x x x
1 ∞− (ln x) 2 − 2 = − ∞x →+ x →+∞ x ﻫﻲ y = x : )g ( x (2ﺒﻴﻥ ﺃﻥ : = ) f '( xﻝﻜل xﻤﻥ [∞]0, + x
lim f ( x ) − x = limﺇﺫﻥ ) (Cﻴﻘﺒل ﻓﺭﻋﺎ ﺸﻠﺠﻤﻴﺎ ﺍﺘﺠﺎﻫﻪ ﺍﻝﻤﻘﺎﺭﺏ ﻫﻭ ﺍﻝﻤﺴﺘﻘﻴﻡ ﺍﻝﺫﻱ ﻤﻌﺎﺩﻝﺘﻪ
1 ' ) − 2 ( ln x )( ln x x2 1 1 ⋅ ) = 1 − 2 − 2 ( ln x x x 1 1 = x − − 2 ln x x x )g ( x = x
1,5
f '( x) = 1−
ﺇﺸﺎﺭﺓ ) f ' ( xﻫﻲ ﺇﺸﺎﺭﺓ ) g ( x ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ f
∞+
0
1
φ
+ ∞+
0 (3ا## 1
اذ :م ري 6
x )f '( x
∞+
)f ( x
0,5
ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ2007 ( 4ﺃ -ﻨﺒﻴﻥ ﺃﻥ ﺍﻝﺩﺍﻝﺔ G : x ln x − xﺩﺍﻝﺔ ﺃﺼﻠﻴﺔ ﻝﻠﺩﺍﻝﺔ g : x → ln xﻋﻠﻰ [∞]0, + ﻝﺩﻴﻨﺎ ∀x ∈ ]0, +∞[ : G ' ( x ) = x 'ln x + x ( ln x ) '− 1 1 = ln x + x ⋅ − 1 x = ln x + 1 − 1 = ln x
ﺇﺫﻥ ﺍﻝﺩﺍﻝﺔ Gﺩﺍﻝﺔ ﺃﺼﻠﻴﺔ ﻝﻠﺩﺍﻝﺔ .g
ﺏ -ﺒﺎﺴﺘﻌﻤﺎل ﻤﻜﺎﻤﻠﺔ ﺒﺎﻷﺠﺯﺍﺀ ،ﻨﺒﻴﻥ ﺃﻥ (ln x) 2 dx = e − 2 : 0,75
u ( x ) = ( ln x )2 ﻨﻀﻊ v ' ( x ) = 1
ﺇﺫﻥ
e
∫
1
ln x u ' ( x ) = 2 x v ( x ) = x
ﺇﺫﻥ ln x 2 ⋅ x dx x
e
e
− 1 ∫1
2
2
) ∫ ( ln x ) dx = x ( ln x e
1
e
2 = x ( ln x ) − 2 ∫ ln x dx 1 1 e
e 2 = x ( ln x ) − 2 [ x ln x − x ]1 1 e
) )) − 2 (( e ln e − e ) − (1ln1 − 1 0,75
2
(
)= e ( ln e ) − 1( ln1 2
=e−2 ﺝ – ﻤﺴﺎﺤﺔ ﺤﻴﺯ ﺍﻝﻤﺴﺘﻭﻯ ﺍﻝﻤﺤﺼﻭﺭ ) (Cﻭ ﻤﺤﻭﺭ ﺍﻷﻓﺎﺼﻴل ﻭ ﺍﻝﻤﺴﺘﻘﻴﻤﻴﻥ ﺍﻝﻠﺫﻴﻥ ﻤﻌﺎﺩﻝﺘﺎﻫﻤﺎ x = 1 :ﻭ x = e
ﻝﺩﻴﻨﺎ
fﺩﺍﻝﺔ ﻤﻭﺠﺒﺔ ﻭ ﻤﺘﺼﻠﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل ] [1, eﺇﺫﻥ ﺍﻝﻤﺴﺎﺤﺔ ﺍﻝﻤﻁﻠﻭﺒﺔ ﻫﻲ
e e 1 A = ∫ f ( x ) dx = ∫ x + − (ln x) 2 − 2 dx 1 1 x e e 1 2 = ∫ x + − 2 dx − ∫ ( ln x ) dx 1 1 x e
x2 ) = + ln x − 2 x − ( e − 2 2 1 e2 1 = + ln e − 2e − + ln1 − 2 − e + 2 2 2 e2 1 = + 1 − 2e − + 2 − e + 2 2 2 2 e 9 = − 3e + 2 2 e2 9 = Aﺒﻭﺤﺩﺓ ﻗﻴﺎﺱ ﺍﻝﻤﺴﺎﺤﺔ ﺇﺫﻥ − 3 e + 2 2
اذ :م ري
7
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