S-exp-ii Math Cour De Log
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א
אאא
-Iاﻟﺪوال اﻷﺻﻠﻴﺔ : -1ﻣﺜﺎل :ﻟﺘﻜﻦ Fاﻟﺪاﻟﺔ اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ \ ﺑﻤﺎ ﻳﻠﻲ :
)
أ -أﺣﺴﺐ
(
. F (x ) = Arc tan x + 1 2
2x x + 2x 2 + 2
وﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ \ ﺑﻤﺎ ﻳﻠﻲ :
)
אWא
4
F ′ ( xﻟﻜﻞ xﻣﻦ \ .
ب -ﻟﺘﻜﻦ Gداﻟﺔ ﻋﺪدﻳﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ \ ﺑﻤﺎ ﻳﻠﻲ :
)
= ) f (x
(
G (x ) = Arc tan x 2 + 1 + αﺣﻴﺚ \∈ . α
أﺣﺴﺐ ) . G ′ ( xﻣﺎذا ﺗﺴﺘﻨﺘﺞ؟
ﺟـ -ﺣﺪد اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ اﻟﺘﻲ ﺗﺤﻘﻖ . G ( 0 ) = π :
اﻟﺤﻞ :
′ )x + 1 ( 2x = ) ). F ′(x ) = ( Arc tan ( x + 1 = أ -ﻟﻴﻜﻦ \ ∈ ، xﻟﺪﻳﻨﺎ = f ( x ) : + + x 2 x 2 )1 + ( x + 1 ′ ′ ب -ﻟﻴﻜﻦ \ ∈ ، xﻟﺪﻳﻨﺎ . G ′(x ) = ( Arc tan ( x + 1) + α ) = ( F ( x ) + α ) = F ′ ( x ) = f ( x ) : 2
4
2
2
′
2
2
2
ﺟـ -ﻟﺪﻳﻨﺎ
)
. ∀x ∈ \ : F ′(x ) = f ( xﻧﻘﻮل إن Fداﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ .
ﻟﺘﻜﻦ Gداﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ .إذن :
∀x ∈ \ : (G − F )′ ( x ) =G ′ ( x ) − F ′ ( x ) = f ( x ) − f ( x ) = 0
وﻣﻨﻩ ﻓﺈن :
. ∃α ∈ \ / ∀x ∈ \ : G ( x ) − F ( x ) = α
أي :
. ∃α ∈ \ / ∀x ∈ \ : G ( x ) = F ( x ) + α
وﺑﻤﺎ أن ، G ( 0 ) = π :ﻓﺈن ، F ( 0 ) + α = π :وﺑﻨﺎءا ﻋﻠﻴﻩ ﻧﺠﺪ : 3π 4
=
π
. Arc tan (1) + α = π ⇒ α = π −
4 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ اﻟﺘﻲ ﺗﺤﻘﻖ G ( 0 ) = π :هﻲ اﻟﺪاﻟﺔ اﻟﻤﻌﺮﻓﺔ آﻤﺎ ﻳﻠﻲ :
-2
→ \ :
\ 3π 4
أ -ﺗﻌﺮﻳﻒ :
)
(
6 Arc tan x 2 + 1 +
G
x
ﻟﺘﻜﻦ fداﻟﺔ ﻋﺪدﻳﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ﻣﺠﺎل . Iﻧﻘﻮل إن Fداﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ I؛ إذا آﺎﻧﺖ Fﻗﺎﺑﻠﺔ ﻟﻺﺷﺘﻘﺎق ﻋﻠﻰ Iوآﺎن ﻟﻜﻞ xﻣﻦ F ′ ( x ) = x : I ﻣﻻﺣﻈﺔ :إذا آﺎﻧﺖ Fداﻟﺔ أﺻﻠﻴﺔ ﻟﺪاﻟﺔ fﻋﻠﻰ ﻣﺠﺎل ، Iﻓﺈن fداﻟﺔ داﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ اﻟﻤﺠﺎل . I
ب -ﻣﺜﺎل :
-1ﺣﺪد داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ :
(a
f (x ) = x
; (b
f (x ) = x + 1 2
f (x ) = −5x 3 + 8x + 2 (d ; f (x ) = 7x 4 (c 1 . f (x ) = 2 -2ﺣﺪد داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +ﻣﻊ : x ﻣﺤﻤﺪ اﻟﺤﻴﺎن
-1-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
-3أ -ﺧﺎﺻﻴﺔ ﻣﻘﺒﻮﻟﺔ : آﻞ داﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل I؛ ﺗﻘﺒﻞ داﻟﺔ أﺻﻠﻴﺔ ﻋﻠﻰ اﻟﻤﺠﺎل . I
ب -ﻣﻼﺣﻈﺔ :
ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل Iوﻟﺘﻜﻦ Fداﻟﺔ أﺻﻠﻴﺔ ﻟﻬﺎ ﻋﻠﻰ . Iاﻟﺪوال اﻷﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ Iهﻲ اﻟﺪوال Gاﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ Iﺑﻤﺎ ﻳﻠﻲ : ∀x ∈ I : G (x ) = F (x ) + αﺣﻴﺚ . α ∈ \ : ﺟـ -ﺧﺎﺻﻴﺔ :
ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل Iوﻟﺘﻜﻦ Fداﻟﺔ أﺻﻠﻴﺔ ﻟﻬﺎ ﻋﻠﻰ I؛ وﻟﻴﻜﻦ x 0 ∈ Iو \∈ . y 0
) (
ﺗﻮﺟﺪ داﻟﺔ أﺻﻠﻴﺔ وﺣﻴﺪة Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل Iﺑﺤﻴﺚ . G x 0 = y 0 : د -ﻣﺜﺎل : 1ﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ \ ﺑﻤﺎ ﻳﻠﻲ . f (x ) = 3x 2 − 4x − 5 : -1ﺣﺪد اﻟﺪوال اﻷﺻﻠﻴﺔ Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ . -2ﺣﺪد اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ اﻟﺘﻲ ﺗﺤﻘﻖ . G (1) = 9 : ﻣﺜﺎل : 2ﺣﺪد داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ
2x
2
) (1+ x 2
= ) . f : x 6 f (x
ﻣﺜﺎل : 3ﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎦⎤ I = ⎡⎣ −1,8ﺑﻤﺎ ﻳﻠﻲ : 2x 3 + 8x 2 + 8x − 3 = ) . f (x x 2 + 4x + 4 -1ﺑﻴﻦ أن :
3 2 )( x + 2
: f ( x ) = 2x −
. ∀x ∈ I
-2أ -اﺱﺘﻨﺘﺞ داﻟﺔ أﺻﻠﻴﺔ Fﻟﻠﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل . I ب -ﺣﺪد اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل Iاﻟﺘﻲ ﺗﺤﻘﻖ . G ( 0 ) = 2
.4ﺟﺪول ﻟﻠﺪوال اﻷﺻﻠﻴﺔ ﻟﻠﺪوال اﻻﻋﺘﻴﺎدﻳﺔ :
) أﻧﻈﺮ اﻟﻤﻄﺒﻮع اﻟﻤﺮﻓﻖ (
Logarithme Népérien :
-IIداﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻨﺒﻴﺮي : -1ﺗﻌﺮﻳﻒ :
1
اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ x 6ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +واﻟﺘﻲ ﺗﻨﻌﺪم ﻓﻲ 1ﺗﺴﻤﻰ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻨﺒﻴﺮي x وﻳﺮﻣﺰ ﻟﻬﺎ ﺑﺎﻟﺮﻣﺰ lnأو . Log ﻣﻼﺣﻈﺘﻴﻦ :
أ -ﻳﻜﻮن ln xﻣﻌﺮﻓﺎ إذا وﻓﻘﻁ إذا آﺎن . x > 0
ب. ln1 = 0 -
-2ﺧﺎﺻﻴﺎت : أ -ﺧﺎﺻﻴﺔ أﺳﺎﺳﻴﺔ :
) ∀a ∈ ⎤⎦ 0, +∞ ⎡⎣ , ∀b ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln (ab ) = ln (a ) + ln (b
ﺑﺮهﺎن :ﻟﻴﻜﻦ a > 0و . b > 0ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ Fاﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ : ) ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : F ( x ) = ln (ax ﻟﺪﻳﻨﺎ x 6 ax :داﻟﺔ ﺣﺪودﻳﺔ ،ﻓﻬﻲ ﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق ﻋﻠﻰ \ وﺑﺎﻷﺧﺺ ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞. ⎤⎦ 0, + وﻟﺪﻳﻨﺎ ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : ax > 0 :و lnﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞. ⎤⎦ 0, + إذن Fﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +وﻟﺪﻳﻨﺎ : ﻣﺤﻤﺪ اﻟﺤﻴﺎن
-2-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : F ′ ( x ) = ( ln (ax ) )′ = (ax )′ × ln ′ (ax ) = a × 1 = 1 x
ax
∈ ⎤⎥0, +∞ ⎡⎢ : ( F − ln )′ ( x ) = F ′ ( x ) − ln ′ ( x ) = 1 − 1 = 0
وﻣﻨﻩ ﻓﺈن :
x
α
ﺑﺤﻴﺚ :
) F (1) = ln (a
،ﻓﺈن :
إذن ﻳﻮﺟﺪ ﻋﺪد ﺣﻘﻴﻘﻲ وﺑﻤﺎ أن :
⎣
x
∀x ∈ ⎤⎥0, +∞ ⎡⎢ : F ( x ) − ln ( x ) = α ⎣
) α = ln (a ) − ln (1) = ln (a
) F ( x ) − ln ( x ) = ln (a
وﺑﻬﺬا ﻧﺠﺪ :
⎦
⎦
:
.
أي : ln (ax ) − ln ( x ) = ln (a ) : وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن . ∀x ∈ ⎤ 0, +∞ ⎡ : ln (ax ) = ln (a ) + ln ( x ) :ﻧﻀﻊ ⎦ ⎣ ) ln (ab ) = ln (a ) + ln (b .
ﻟﻴﻜﻦ aو bﻣﻦ ⎣⎡ ∞ . ⎤⎦ 0, +ﺑﻴﻦ أن : ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ : ⎞⎛1 ⎞ ⎛a ln ⎜ ⎟ = ln a − ln b (ii ; ln ⎜ ⎟ = − ln a (i ⎠ ⎝b ⎠ ⎝a ب -ﻧﺘﺎﺋﺞ :
.1ﻟﻜﻞ aو bﻣﻦ ⎣⎡ ∞ ⎤⎦ 0, +وﻟﻜﻞ ` ∈ n؛ ﻟﺪﻳﻨﺎ :
.
.
⎣⎡ ∞∀x ∈ ⎤⎦ 0, +
⎣⎡ ∞∀x ∈ ⎤⎦ 0, +
، x = bﻓﻨﺠﺪ :
) (
ln a 2 = 2 ln a (iii
;
ln(ab ) = ln a + ln b
⎞⎛1 ln ⎜ ⎟ = − ln a ⎠ ⎝a
⎞ ⎛a ln ⎜ ⎟ = ln a − ln b ⎠ ⎝b
ln (a n ) = n ln a
) (
.2ﻟﻜﻞ _∈ rوﻟﻜﻞ ⎣⎡ ∞ a ∈ ⎤⎦ 0, +؛ ﻟﺪﻳﻨﺎ : ﻣﺜﺎل :
. ∀x
. ln a r = r ln a
ﻧﻀﻊ a = ln 2 :و . b = ln 5 .1ﺣﺪد ﺑﺪﻻﻟﺔ aو bﻣﺎ ﻳﻠﻲ :
(i (iii .2ﺣﺪد ﺑﺪﻻﻟﺔ a
-3دراﺱﺔ اﻟﺪاﻟﺔ ln
;
ln10
⎠⎟⎞ ( 5 ) − ln ⎛⎜⎝ 12 و bﻣﺎ ﻳﻠﻲ ln ( 2 ) :
⎞ ⎛ 4 ⎟ ⎠ ⎝ 125
⎜ ln
(ii
ln ⎛⎜ 2 + 2 ⎞⎟ + ln ⎛⎜ 2 − 2 ⎞⎟ (iv ; ln 3
⎠
و
⎝
⎟⎞ ln ⎛⎜ 3 100 ⎠
⎝
⎠
⎝
.
:
ﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ lnهﻲ . D ln = ⎤⎦ 0, +∞ ⎡⎣ : أ -اﻟﻨﻬﺎﻳﺎت : ﻟﻨﺒﻴﻦ أن اﻟﺪاﻟﺔ lnﻻ ﺗﻘﺒﻞ ﻧﻬﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ﻋﻨﺪ اﻻﻧﻬﺎﻳﺔ :ﺑﺎﻟﺨﻠﻒ ،ﻧﻔﺘﺮض أن lim ln ( x ) = l :
∞x →+
ﺣﻴﺚ \ ∈ . lﻧﻌﻠﻢ أن ، ln ( 2x ) = ln ( 2 ) + ln ( x ) :ﻋﻨﺪﻣﺎ ﻳﺆول ، xﻧﺠﺪ l = ln ( 2 ) + l :
إذن ln ( 2 ) = 0 :وهﺬا ﺗﻨﺎﻗﺾ .إذن lim ln ( x ) :ﻏﻴﺮ ﻣﻨﺘﻬﻴﺔ ... ∞x →+
ﺧﺎﺻﻴﺔ ﻣﻘﺒﻮﻟﺔ :
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
∞lim ln x = +
∞x →+
-3-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
∞lim ln x = −
اﺳﺘﻨﺘﺎج :
x →0+
1 ﻧﻀﻊ x
ب -رﺗﺎﺑﺔ اﻟﺪاﻟﺔ ln
⎞⎛1 ∞− ln t = − lim+ ln x = t lim ln ⎜ ⎟ = t lim ∞→+ x →0 ∞⎝ t ⎠ →+
∞ t → +وﻣﻨﻩ ﻓﺈن : = . tإذن + x →0
:
1 ﻧﻌﻠﻢ أن اﻟﺪاﻟﺔ lnﻗﺎﺑﻠﺔ ﻟﻺﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +وﻟﺪﻳﻨﺎ . ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln ′(x ) = > 0 : x إذن lnﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +؛ وﺑﻤﺎ أن lnﻣﺘﺼﻠﺔ ﻋﻠﻰ [∞ ]0, +ﻓﺈﻧﻬﺎ ﺗﻘﺎﺑﻞ ﻣﻦ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +ﻧﺤﻮ اﻟﻤﺠﺎل . \ :
وﻣﻨﻩ ﻓﺈن :ﻣﻼﺣﻈﺔ : 1
ﻣﻼﺣﻈﺔ : 2
⎞\ = ⎡ ⎟ ⎣⎢ ⎠
(
)
. ⎜⎛ ln ⎤⎦ 0, +∞ ⎡⎣ = ⎤ lim+ ln x , lim ln x ⎥⎦ x →0 ∞x →+ ⎝
∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ , ∀y ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln x < ln y ⇔ x < y
ln x = ln y ⇔ x = y
;
∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ , ∀y ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln x > 0 ⇔ x > 1 ; ln x < 0 ⇔ 0 < x < 1 ; ln x = 0 ⇔ x = 1
ﻣﻼﺣﻈﺔ : 3 اﻟﻤﻌﺎدﻟﺔ ln x = 1ﺗﻘﺒﻞ ﺣﻻ وﺣﻴﺪا ﻓﻲ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +ﻳﺮﻣﺰ ﻟﻩ
ﺑﺎﻟﺮﻣﺰ e . ( ln e = 1) . e
ﻋﺪد ﺣﻘﻴﻘﻲ ﻻ ﺟﺬري ﺑﺤﻴﺚ :
⎞ ⎛π
ﻣﺜﺎل .1 : 1ﺣﺪد إﺷﺎرة اﻷﻋﺪاد اﻟﺘﺎﻟﻴﺔ ; ln ⎜ ⎟ ; ln(0,99) : ⎠⎝5 .2ﻧﻌﺘﺒﺮ اﻟﻤﻌﺎدﻟﺔ . ( E ) : ln(x − 2) + ln(x − 3) = 0
)( 2
e 2,718 . ln
أ -ﺣﺪد Dﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﻳﻒ اﻟﻤﻌﺎدﻟﺔ ) . ( E
ب -ﺣﻞ ﻓﻲ \ اﻟﻤﻌﺎدﻟﺔ ) . ( E
ﺟـ -اﺱﺘﻨﺘﺞ ،ﻓﻲ \ ،ﻣﺠﻤﻮﻋﺔ ﺣﻠﻮل اﻟﻤﺘﺮاﺟﺤﺔ. ln(x − 2) + ln(x − 3) > 0 :
ﻣﺜﺎل : 2
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
ﺣﺪد إﺷﺎرة اﻟﻌﺪد اﻟﺤﻘﻴﻘﻲ ) ( x − 2 ) ln ( x − 1ﺣﻴﺚ ⎣⎡ ∞. x ∈ ⎤⎦1, +
-4-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
ﺟـ -اﻟﻔﺮوع اﻟﻼﻧﻬﺎﺋﻴﺔ : (iﻧﻌﻠﻢ أن ∞ . lim+ ln x = −إذن x →0
) ( Clnﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ ﻋﻤﻮدﻳﺎ ﻣﻌﺎدﻟﺘﻩ . x = 0
ln x (iiﺧﺎﺻﻴﺔ= 0 : x ﺑﺮهﺎن :ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fاﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ f : x 6 x − ln x : 1 x −1 = . ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : f ′ ( x ) = 1 − ﻟﺪﻳﻨﺎ Df = ⎤⎦ 0, +∞ ⎡⎣ :و x x إﺷﺎرة ) f ′ ( xﻋﻠﻰ اﻟﻤﺠﺎل [∞ ]0, +هﻲ إﺷﺎرة اﻟﺒﺴﻁ . x − 1وﻣﻨﻩ ﻧﺠﺪ : lim ∞x →+
إذن 1ﻗﻴﻤﺔ دﻧﻴﺎ ﻣﻄﻠﻘﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞. ⎤⎦ 0, + وﻣﻨﻩ ﻓﺈن . ∀t ∈ ⎤⎦ 0, +∞ ⎡⎣ : f (t ) ≥ 1 :أي . ∀t ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln (t ) − t ≥ 1 > 0 : وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن . ∀t ∈ ⎤ 0, +∞ ⎡ : ln t ≥ t :و ﻣﻦ أﺟﻞ ، t = xﻧﺠﺪ : ⎦ ⎣
)(
ln x 2 ≥ x x
⇒ ≥ x
وﻣﻨﻩ ﻧﺴﺘﻨﺘﺞ أن :
⇒ 1 ln x 2
2 x
≤
x
≥) ( x
∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln
.
ln x
≤ . ∀ x ∈ ⎤⎦1, +∞ ⎡⎣ : 0 x
2 ln x = limو ، lim 0 = 0ﻓﺈن : وﺑﻤﺎ أن = 0 : 0 x →+∞ x x →+∞ x ∞x →+
. lim
وﻣﻨﻩ ﻧﺴﺘﻨﺘﺞ أن اﻟﻤﻨﺤﻨﻰ ) ( Clnﻳﻘﺒﻞ ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ ﺑﺠﻮار ∞ +؛ اﺗﺠﺎهﻩ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ . د -ﺗﻘﻌﺮ اﻟﻤﻨﺤﻨﻰ ) : ( Cln
⎛ 1 ⎞′ 1 ⎤ ⎡ ﻟﺪﻳﻨﺎ . ∀x ∈ ⎦ 0, +∞ ⎣ : ln n ′′ ( x ) = ⎜ ⎟ = − 2 < 0 : x ⎠ ⎝x إذن ﺗﻘﻌﺮ ) ( Clnﻣﻮﺟﻩ ﻧﺤﻮ اﻷراﺗﻴﺐ اﻟﺴﺎﻟﺒﺔ .
هـ -إﻧﺸﺎء اﻟﻤﻨﺤﻨﻰ ) : ( Cln ﻟﺪﻳﻨﺎ ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ) (Tﻟﻠﻤﻨﺤﻨﻰ ) ( Clnﻓﻲ اﻟﻨﻘﻄﺔ اﻟﺘﻲ أﻓﺼﻮﻟﻬﺎ 1هﻲ:
y = ln ′(1) ( x − 1) + ln1
أي :
. y = x − 1ﻷن ln ′ (1) = 1 :
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
و
. ln (1) = 0
-5-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
ﻣﻼﺣﻈﺔ :ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +؛ ﻟﺪﻳﻨﺎ ) ( Clnﻳﻮﺟﺪ ﺗﺤﺖ اﻟﻤﻤﺎس ) . (Tإذن :
∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ ; ln x ≤ x −1 -4ﻧﻬﺎﻳﺎت هﺎﻣﺔ :
) ln(1 + x ; =1 x
lim x ln x = 0
x → 0+
lim x →0
.
ln x ; =1 x −1
lim x →1
ﺑﺮهﺎن :
1 ﻧﻀﻊ : x
∞t → + = . tإذن : + x →0
.
⎞1 ⎛1 ln t − و ﻣﻨﻩ ﻓﺈن = 0 : ln ⎜ ⎟ = t lim ∞→+ t ⎠ ⎝t ln x ln x − ln1 . lim = lim ﻟﺪﻳﻨﺎ = ln′ (1) = 1 : x →1 x − 1 x →1 x −1 ﻧﻀﻊ . t = x −1 :إذن t → 0 :
. lim+ x ln x = lim
t →+∞ t
x →0
x →1
) ln (1 + x ln t = و ﻣﻨﻩ ﻓﺈن = 1 : lim x →0 t →1 t − 1 x
. lim
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
-6-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
أﺣﺴﺐ اﻟﻨﻬﺎﻳﺎت اﻟﺘﺎﻟﻴﺔ :
ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ :
lim 2x − 3ln x
(a
; (b
∞x →+
(e
1 + ln x x
(i
) lim+ x ( ln x
2
;
lim+
x →0
2
(m
x
) ln ( ln x (q ln x
(f
; (j
x →0
) ( ln x
lim ( ln x ) + ln x 2
x →0+
)
(
lim x − ln 1 + x 2 ∞x →+ ) lim+ x ( ln x
3
; (n
lim
∞x →+
;
lim ∞x →+
(r
ln x x
; (g
lim+ x 2 ln x
; (h
(k
x →0
ln x x2
;
lim ∞x →+
lim+ x 3 ln x
x →0
ln x x3
(l
lim ∞x →+
) ln (1 + ln x ) ln (1 − sin x p ( ; lim ; (o lim x →1 x →0 ln 1 + tan x x 2 −1 ( )
3
x
; (d
;
x →0
) ( ln x
; (c
1 − ln x lim+ x →0 ln x
ln x lim x →+∞ 1 + ln x
lim
∞x →+
; (s
lim+
x →0
) (
. lim x + ln x 2 ∞x →−
-IIIاﻟﻤﺸﺘﻘﺔ اﻟﻠﻮﻏﺎرﻳﺘﻤﻴﺔ : -1ﺧﺎﺻﻴﺔ : ﻟﺘﻜﻦ uداﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق وﻻ ﺗﻨﻌﺪم ﻋﻠﻰ ﻣﺠﺎل . I ) ′ u ′(x = ) ∀x ∈ I : ln u (x ﻟﺪﻳﻨﺎ : ) u (x
)
(
-2ﺗﻌﺮﻳﻒ : ﻟﺘﻜﻦ uداﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق وﻻ ﺗﻨﻌﺪم ﻋﻠﻰ ﻣﺠﺎل . I
u′ اﻟﺪاﻟﺔ u
ﺗﺴﻤﻰ اﻟﻤﺸﺘﻘﺔ اﻟﻠﻮﻏﺎرﻳﺘﻤﻴﺔ ﻟﻠﺪاﻟﺔ uﻋﻠﻰ اﻟﻤﺠﺎل I
x −1 ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ : 1ﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ : x -1ﺣﺪد Dfﺣﻴﺰ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ . f -2أﺣﺴﺐ ﻧﻬﺎﻳﺎت fﻋﻨﺪ ﻣﺤﺪات . Df -3أﺣﺴﺐ ) f ′(xﻟﻜﻞ xﻣﻦ Df؛ ﺛﻢ ﺿﻊ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ fﻋﻠﻰ . Df
. f (x ) = ln
JG JJG -4أدرس اﻟﻔﺮوع اﻟﻻ ﻧﻬﺎﺋﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ ) ( Cfاﻟﻤﻤﺜﻞ ﻟﻠﺪاﻟﺔ fﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ . O , i , j
)
(
-5أدرس ﺗﻘﻌﺮ اﻟﻤﻨﺤﻨﻰ ) ( Cfﺛﻢ ﺣﺪد ﻧﻘﻄﺔ اﻧﻌﻄﺎف اﻟﻤﻨﺤﻨﻰ ) . ( Cf -6أﻧﺸﺊ ) . ( Cf
اﻟﺠﻮاب :
x −1 .1ﻟﻴﻜﻦ \ ∈ . xﻟﺪﻳﻨﺎ : x ≠0و >0 x x −1 ⇔ x ≠0و ≠0 x x ≠ 0و ⇔ x ≠1 ⎣⎡ ∞Df = \ − {0,1} = ⎤⎦ −∞,0 ⎡⎣ ∪ ⎤⎦ 0,1⎡⎣ ∪ ⎤⎦1, + إذن :
⇔ x ∈Df
.2ﻟﺪﻳﻨﺎ :
x −1 1 = lim ln 1 − = ln1 = 0 ∞x → ± x x
x −1 و ﺑﻮﺿﻊ x ﻣﺤﻤﺪ اﻟﺤﻴﺎن
= ، tﻧﺠﺪ t → 0+و x →1
. lim f (x ) = lim ln ∞x → ±
∞x → ±
x −1 ∞= lim+ ln (t ) = − t →0 x
-7-
. lim f (x ) = lim ln x →1
x →1
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
x −1 ∞= lim ln (t ) = + ∞t →+ x
∞ t → +و وﻟﺪﻳﻨﺎ : + x →0
. lim f (x ) = lim ln x →0
x →0
1 −1 ′ ⎞ ⎛ x −1 1 0 ⎜ ⎟ 2 ⎠ x 1 1 1 1 ⎝ f ′( x ) = − + =− + x =− + 2 x −1 2 x −1 )2 x ( x −1 x x ) −x 2 + x + 2 − ( x + 1)( x − 2 = ) f ′(x = )x ( x − 1 )x ( x −1
.3ﻟﻴﻜﻦ x ∈Df؛ ﻟﺪﻳﻨﺎ :
.4ﺗﺤﺪﻳﺪ اﻟﻔﺮوع اﻟﻻﻧﻬﺎﺋﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ ) : ( Cf 9ﻟﺪﻳﻨﺎ . lim f (x ) = 0 :إذن ) ( Cfﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ أﻓﻘﻴﺎ ﺑﺠﻮار ∞ ±ﻣﻌﺎدﻟﺘﻩ . y = 0 ∞x → ±
9ﻟﺪﻳﻨﺎ . lim f (x ) = +∞ :إذن ) ( Cfﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ ﻋﻤﻮدﻳﺎ ﻣﻌﺎدﻟﺘﻩ . x = 0 x →0
9ﻟﺪﻳﻨﺎ . lim f (x ) = −∞ :إذن ) ( Cfﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ ﻋﻤﻮدﻳﺎ ﻣﻌﺎدﻟﺘﻩ . x = 1 x →1
.5ﻟﻴﻜﻦ . x ∈Dfﻟﺪﻳﻨﺎ :
′ ′ ′ 2 2 ⎛ ⎞2 +x +2 x x 2 x x 1 x x 2 x x 1 − + + − − − + + − ( ) ( ) − x ⎟ ⎜⎜ = 2 = ⎟ − 1 x x ⎜ ⎟ x x 1 − ( ) ⎝ ⎠
)
إﺷﺎرة
)
()
)
( )
(
()
(
)
(
) f ′′ ( x
) ( −2x + 1) ( x ( x −1) ) − ( −x 2 + x + 2 ) ( x −1+ x = ) f ′′ ( x 2 ))( x ( x −1 )− ( 2x − 1) ( x ( x − 1) ) − ( −x 2 + x + 2 ) ( 2x − 1 = ) f ′′ ( x 2 ))( x ( x −1 )( 2x −1) ( −x 2 + x + x 2 − x − 2) −2 ( 2x −1 = ) f ′′ ( x = 2 2 ))( x ( x −1 ))( x ( x −1
f ′′ ( xﻋﻠﻰ Dfهﻲ إﺷﺎرة ، 1 − 2xوﻣﻨﻩ ﻧﺴﺘﻨﺘﺞ أن :
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
-8-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
⎞ ⎛1 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﻨﻘﻄﺔ ⎟ A ⎜ ,0ﻧﻘﻄﺔ اﻧﻌﻄﺎف ﻟﻠﻤﻨﺤﻨﻰ ) . ( Cf ⎠ ⎝2 .6إﻧﺸﺎء اﻟﻤﻨﺤﻨﻰ ) : ( Cf
ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ : 2
ﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ :
(
)
⎧ f ( x ) = ln 1 − x 3 ; x <0 ⎪ ⎨ ⎪f ( x ) = 4x x − 3x 2 ; x ≥ 0 ⎩ .1أ -ﺑﻴﻦ أن fﻣﺘﺼﻠﺔ ﻓﻲ . 0 ب -ﺑﻴﻦ أن fﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق ﻓﻲ . 0 أدرس ﺗﻐﻴﺮات اﻟﺪاﻟﺔ fﻋﻠﻰ . R .2 .3أ -أﺣﺴﺐ اﻟﻨﻬﺎﻳﺘﻴﻦ lim f ( x ) :و ) . lim f ( x ∞x →−
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
∞x →+
-9-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
ب -أدرس اﻟﻔﺮﻋﻴﻦ اﻟﻻﻧﻬﺎﺋﻴﻴﻦ ﻟﻠﻤﻨﺤﻨﻰ ) . ( Cf .4أﻧﺸﺊ اﻟﻤﻨﺤﻨﻰ ) . ( Cf .5ﻟﻴﻜﻦ gﻗﺼﻮر اﻟﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ . I = ⎤⎦ −∞,0 أ -ﺑﻴﻦ أن gﺗﻘﺎﺑﻞ ﻣﻦ اﻟﻤﺠﺎل Iﻧﺤﻮ ﻣﺠﺎل Jﻳﻨﺒﻐﻲ ﺗﺤﺪﻳﺪﻩ. ب -ﺣﺪد
) g −1 ( x
ﻟﻜﻞ ) . x ∈ Jﻳﻨﺠﺰ هﺬا اﻟﺴﺆال ﺑﻌﺪ درس اﻟﺪوال اﻷﺱﻴﺔ ( ؟
.6ﻧﻌﺘﺒﺮ `∈ (u n )nاﻟﻤﺘﺘﺎﻟﻴﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ :
`∈ ; n
⎧ 4 = u0 ⎪ 9 ⎨ 2 ⎪u ⎩ n +1 = 4u n u n − 3u n
4 أ -ﺑﻴﻦ أن ≤ u ≤ 1 : 9 n ب -ﺑﻴﻦ أن `∈ (u n )nﺗﺰاﻳﺪﻳﺔ .
. ∀n ∈ ` :
ﺟـ -ﺑﻴﻦ أن `∈ (u n )nﻣﺘﻘﺎرﺑﺔ وﺣﺪد ﻧﻬﺎﻳﺘﻬﺎ.
-3اﺱﺘﻨﺘﺎج : ﻟﺘﻜﻦ uداﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻺﺷﺘﻘﺎق وﻻ ﺗﻨﻌﺪم ﻋﻠﻰ ﻣﺠﺎل . I
u′ اﻟﺪوال اﻷﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ﻋﻠﻰ Iهﻲ اﻟﺪوال x 6 ln u ( x ) + α :ﺣﻴﺚ \∈ α u
-IVداﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ ﻟﻸﺱﺎس -1ﺗﻌﺮﻳﻒ : ﻟﻴﻜﻦ
a >0
ln x اﻟﺪاﻟﺔ ln a
و
a ≠1
a
.
:
.
x 6اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞، ⎤⎦ 0, +
ﺗﺴﻤﻰ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ ﻟﻸﺱﺎس a \
. logaوﻟﺪﻳﻨﺎ :
→
وﻧﺮﻣﺰ ﻟﻬﺎ ﺑﺎﻟﺮﻣﺰ
⎡ ∞⎤ 0, + ⎦ ⎣
ln x = ) 6 loga ( x ln a
loga :
x
ln x
= ) . ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : loge ( x ﻣﻼﺣﻈﺔ :ﻟﺪﻳﻨﺎ = ln x : ln e داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻨﺒﻴﺮي هﻲ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ ﻟﻸﺱﺎس . e
-2ﺧﺎﺻﻴﺎت :
ﻟﻴﻜﻦ a > 0
و
a ≠1
.
.1ﻟﻜﻞ xو yﻣﻦ ⎣⎡ ∞ ⎤⎦ 0, +وﻟﻜﻞ `∈ n؛ ﻟﺪﻳﻨﺎ : ; loga ( xy ) = loga ( x ) + loga ( y ) (i
(ii
;
(iv
(iii
⎞⎛1 ) loga ⎜ ⎟ = − loga ( x ⎠ ⎝x
.2ﻟﻜﻞ _∈ rوﻟﻜﻞ ⎣⎡ ∞ x ∈ ⎤⎦ 0, +؛ ﻟﺪﻳﻨﺎ :
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
- 10 -
) loga ( x n ) = n loga ( x
)
⎞ ⎛x ⎟⎟ = loga ( x ) − loga ( y ⎠ ⎝y
⎜⎜ loga
) (
) . loga x r = r loga ( x
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
-3رﺗﺎﺑﺔ اﻟﺪاﻟﺔ loga
:
أ -ﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ
loga
هﻲ :
⎣⎡ ∞. Dloga = ⎤⎦ 0, +
ln x
= ) . ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : loga (xإذن : ب -اﻟﻨﻬﺎﻳﺎت :ﻧﻌﻠﻢ أن : ln a ∞ lim+ loga (x ) = −و ∞lim loga (x ) = + 9إذا آﺎن a > 1؛ ﻓﺈن :
∞x →+
x →0
9إذاآﺎن 0 < a < 1؛ ﻓﺈن :
1 1 ln a x
ﺟـ -ﺟﺪول اﻟﺘﻐﻴﺮات :ﻟﺪﻳﻨﺎ :
= ) ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : loga′ ( x
أ -ﺣﺎﻟﺔ : a > 1 ﻟﺪﻳﻨﺎ logaﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ
ﻟﺪﻳﻨﺎ logaﺗﻨﺎﻗﺼﻴﺔ
ﻋﻠﻰ اﻟﻤﺠﺎل ⎡ ∞⎤ 0, + ⎦ ⎣
ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞⎤⎦ 0, +
a
a >1
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
.
ب -ﺣﺎﻟﺔ : 0 < a < 1
د -إﻧﺸﺎء ) ( Clog ﺣﺎﻟﺔ
∞lim loga (x ) = +
x →0+
و
∞lim loga (x ) = −
∞x →+
.
: ،ﻣﺜﻼ :
) (a = 2
⎛ ⎞1 ﺣﺎﻟﺔ ، 0 < a < 1ﻣﺜﻼ ⎜ a = ⎟ : ⎠4 ⎝
- 11 -
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
Logarithme décimal :
-Vداﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻌﺸﺮي : -1ﺗﻌﺮﻳﻒ :
اﻟﺪاﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻤﻴﺔ اﻟﺘﻲ أﺱﺎﺱﻬﺎ 10ﺗﺴﻤﻰ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻌﺸﺮي وﻧﺮﻣﺰ ﻟﻬﺎ ﺑﺎﻟﺮﻣﺰ ) logﻋﻮض . ( log10وﻟﺪﻳﻨﺎ :
→
\
ln x ln10
log (10r ) = r log10
(b
-3ﻣﺜﺎل :
)
.1أﺣﺴﺐ 100 :
;
_ ∈ ∀r
(
2
)
(
⎣⎡ ∞∀ ( x , y ) ∈ ⎤⎦ 0, +
A = logو ) B = log ( 0, 0001و ) . C = log (10000
.2ﺣﻞ ﻓﻲ \ اﻟﻤﻌﺎدﻟﺔ :
ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ :
;
log1 = 0
; log x = log y ⇔ x = y
(c
3
= ) 6 log(x
x
log10 = 1
(a
-2ﻣﻻﺣﻈﺔ :
⎡ ∞⎤ 0, + ⎦ ⎣
log :
10−x = 3
ﺛﻢ أﻋﻁ ﻗﻴﻤﺔ ﻣﻘﺮﺑﺔ ﻟﻠﺤﻞ ﻣﺴﺘﻌﻤﻻ اﻟﻤﺤﺴﺒﺔ .
ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fﻟﻤﺘﻐﻴﺮ ﺣﻘﻴﻘﻲ xﺣﻴﺚ : .1أ -ﺣﺪد Dfﺣﻴﺰ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ . f
)
3 + 2ln ( x x
f (x ) = x +
ب -أﺣﺴﺐ ﻧﻬﺎﻳﺎت fﻋﻨﺪ ﻣﺤﺪات .Df .2أ -أﺣﺴﺐ ) f ′( xﻣﻦ أﺟﻞ آﻞ ﻋﻨﺼﺮ xﻣﻦ .Df ب -أﻋﻁ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ . f
) (
.3ﻟﻴﻜﻦ Cfاﻟﻤﻨﺤﻨﻰ اﻟﻤﻤﺜﻞ ﻟﻠﺪاﻟﺔ
fﻓﻲ اﻟﻤﺴﺘﻮى اﻟﻤﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ )
أ -أدرس اﻟﻔﺮوع اﻟﻻﻧﻬﺎﺋﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ ) ( Cf ب -أدرس ﺗﻘﻌﺮ اﻟﻤﻨﺤﻨﻰ ) . ( Cf ﺟـ -أﻋﻁ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻤﻤﺎس اﻟﻤﻨﺤﻨﻰ ) ( Cfﻓﻲ اﻟﻨﻘﻄﺔ Iذات اﻷﻓﺼﻮل . 3 د -أرﺱﻢ اﻟﻤﻨﺤﻨﻰ ) ( Cfﻣﺒﺮزا ﻧﻘﻄﻩ اﻟﺘﻲ أﻓﺎﺻﻴﻠﻬﺎ هﻲ 3و -2و . -4
JG JJG
(
. O ,i , j
.
ﺑﺎﻟﺘﻮﻓﻴﻖ إﻧﺸﺎء اﷲ
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
- 12 -
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
אאא
-Iاﻟﺪوال اﻷﺻﻠﻴﺔ : -1ﻣﺜﺎل :ﻟﺘﻜﻦ Fاﻟﺪاﻟﺔ اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ \ ﺑﻤﺎ ﻳﻠﻲ :
)
أ -أﺣﺴﺐ
(
. F (x ) = Arc tan x + 1 2
2x x + 2x 2 + 2
وﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ \ ﺑﻤﺎ ﻳﻠﻲ :
)
אWא
4
F ′ ( xﻟﻜﻞ xﻣﻦ \ .
ب -ﻟﺘﻜﻦ Gداﻟﺔ ﻋﺪدﻳﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ \ ﺑﻤﺎ ﻳﻠﻲ :
)
= ) f (x
(
G (x ) = Arc tan x 2 + 1 + αﺣﻴﺚ \∈ . α
أﺣﺴﺐ ) . G ′ ( xﻣﺎذا ﺗﺴﺘﻨﺘﺞ؟
ﺟـ -ﺣﺪد اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ اﻟﺘﻲ ﺗﺤﻘﻖ . G ( 0 ) = π :
اﻟﺤﻞ :
′ )x + 1 ( 2x = ) ). F ′(x ) = ( Arc tan ( x + 1 = أ -ﻟﻴﻜﻦ \ ∈ ، xﻟﺪﻳﻨﺎ = f ( x ) : + + x 2 x 2 )1 + ( x + 1 ′ ′ ب -ﻟﻴﻜﻦ \ ∈ ، xﻟﺪﻳﻨﺎ . G ′(x ) = ( Arc tan ( x + 1) + α ) = ( F ( x ) + α ) = F ′ ( x ) = f ( x ) : 2
4
2
2
′
2
2
2
ﺟـ -ﻟﺪﻳﻨﺎ
)
. ∀x ∈ \ : F ′(x ) = f ( xﻧﻘﻮل إن Fداﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ .
ﻟﺘﻜﻦ Gداﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ .إذن :
∀x ∈ \ : (G − F )′ ( x ) =G ′ ( x ) − F ′ ( x ) = f ( x ) − f ( x ) = 0
وﻣﻨﻩ ﻓﺈن :
. ∃α ∈ \ / ∀x ∈ \ : G ( x ) − F ( x ) = α
أي :
. ∃α ∈ \ / ∀x ∈ \ : G ( x ) = F ( x ) + α
وﺑﻤﺎ أن ، G ( 0 ) = π :ﻓﺈن ، F ( 0 ) + α = π :وﺑﻨﺎءا ﻋﻠﻴﻩ ﻧﺠﺪ : 3π 4
=
π
. Arc tan (1) + α = π ⇒ α = π −
4 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ اﻟﺘﻲ ﺗﺤﻘﻖ G ( 0 ) = π :هﻲ اﻟﺪاﻟﺔ اﻟﻤﻌﺮﻓﺔ آﻤﺎ ﻳﻠﻲ :
-2
→ \ :
\ 3π 4
أ -ﺗﻌﺮﻳﻒ :
)
(
6 Arc tan x 2 + 1 +
G
x
ﻟﺘﻜﻦ fداﻟﺔ ﻋﺪدﻳﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ﻣﺠﺎل . Iﻧﻘﻮل إن Fداﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ I؛ إذا آﺎﻧﺖ Fﻗﺎﺑﻠﺔ ﻟﻺﺷﺘﻘﺎق ﻋﻠﻰ Iوآﺎن ﻟﻜﻞ xﻣﻦ F ′ ( x ) = x : I ﻣﻻﺣﻈﺔ :إذا آﺎﻧﺖ Fداﻟﺔ أﺻﻠﻴﺔ ﻟﺪاﻟﺔ fﻋﻠﻰ ﻣﺠﺎل ، Iﻓﺈن fداﻟﺔ داﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ اﻟﻤﺠﺎل . I
ب -ﻣﺜﺎل :
-1ﺣﺪد داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ :
(a
f (x ) = x
; (b
f (x ) = x + 1 2
f (x ) = −5x 3 + 8x + 2 (d ; f (x ) = 7x 4 (c 1 . f (x ) = 2 -2ﺣﺪد داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +ﻣﻊ : x ﻣﺤﻤﺪ اﻟﺤﻴﺎن
-1-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
-3أ -ﺧﺎﺻﻴﺔ ﻣﻘﺒﻮﻟﺔ : آﻞ داﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل I؛ ﺗﻘﺒﻞ داﻟﺔ أﺻﻠﻴﺔ ﻋﻠﻰ اﻟﻤﺠﺎل . I
ب -ﻣﻼﺣﻈﺔ :
ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل Iوﻟﺘﻜﻦ Fداﻟﺔ أﺻﻠﻴﺔ ﻟﻬﺎ ﻋﻠﻰ . Iاﻟﺪوال اﻷﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ Iهﻲ اﻟﺪوال Gاﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ Iﺑﻤﺎ ﻳﻠﻲ : ∀x ∈ I : G (x ) = F (x ) + αﺣﻴﺚ . α ∈ \ : ﺟـ -ﺧﺎﺻﻴﺔ :
ﻟﺘﻜﻦ fداﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل Iوﻟﺘﻜﻦ Fداﻟﺔ أﺻﻠﻴﺔ ﻟﻬﺎ ﻋﻠﻰ I؛ وﻟﻴﻜﻦ x 0 ∈ Iو \∈ . y 0
) (
ﺗﻮﺟﺪ داﻟﺔ أﺻﻠﻴﺔ وﺣﻴﺪة Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل Iﺑﺤﻴﺚ . G x 0 = y 0 : د -ﻣﺜﺎل : 1ﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ \ ﺑﻤﺎ ﻳﻠﻲ . f (x ) = 3x 2 − 4x − 5 : -1ﺣﺪد اﻟﺪوال اﻷﺻﻠﻴﺔ Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ . -2ﺣﺪد اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ \ اﻟﺘﻲ ﺗﺤﻘﻖ . G (1) = 9 : ﻣﺜﺎل : 2ﺣﺪد داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ
2x
2
) (1+ x 2
= ) . f : x 6 f (x
ﻣﺜﺎل : 3ﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎦⎤ I = ⎡⎣ −1,8ﺑﻤﺎ ﻳﻠﻲ : 2x 3 + 8x 2 + 8x − 3 = ) . f (x x 2 + 4x + 4 -1ﺑﻴﻦ أن :
3 2 )( x + 2
: f ( x ) = 2x −
. ∀x ∈ I
-2أ -اﺱﺘﻨﺘﺞ داﻟﺔ أﺻﻠﻴﺔ Fﻟﻠﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل . I ب -ﺣﺪد اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ Gﻟﻠﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل Iاﻟﺘﻲ ﺗﺤﻘﻖ . G ( 0 ) = 2
.4ﺟﺪول ﻟﻠﺪوال اﻷﺻﻠﻴﺔ ﻟﻠﺪوال اﻻﻋﺘﻴﺎدﻳﺔ :
) أﻧﻈﺮ اﻟﻤﻄﺒﻮع اﻟﻤﺮﻓﻖ (
Logarithme Népérien :
-IIداﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻨﺒﻴﺮي : -1ﺗﻌﺮﻳﻒ :
1
اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ x 6ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +واﻟﺘﻲ ﺗﻨﻌﺪم ﻓﻲ 1ﺗﺴﻤﻰ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻨﺒﻴﺮي x وﻳﺮﻣﺰ ﻟﻬﺎ ﺑﺎﻟﺮﻣﺰ lnأو . Log ﻣﻼﺣﻈﺘﻴﻦ :
أ -ﻳﻜﻮن ln xﻣﻌﺮﻓﺎ إذا وﻓﻘﻁ إذا آﺎن . x > 0
ب. ln1 = 0 -
-2ﺧﺎﺻﻴﺎت : أ -ﺧﺎﺻﻴﺔ أﺳﺎﺳﻴﺔ :
) ∀a ∈ ⎤⎦ 0, +∞ ⎡⎣ , ∀b ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln (ab ) = ln (a ) + ln (b
ﺑﺮهﺎن :ﻟﻴﻜﻦ a > 0و . b > 0ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ Fاﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ : ) ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : F ( x ) = ln (ax ﻟﺪﻳﻨﺎ x 6 ax :داﻟﺔ ﺣﺪودﻳﺔ ،ﻓﻬﻲ ﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق ﻋﻠﻰ \ وﺑﺎﻷﺧﺺ ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞. ⎤⎦ 0, + وﻟﺪﻳﻨﺎ ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : ax > 0 :و lnﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞. ⎤⎦ 0, + إذن Fﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +وﻟﺪﻳﻨﺎ : ﻣﺤﻤﺪ اﻟﺤﻴﺎن
-2-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : F ′ ( x ) = ( ln (ax ) )′ = (ax )′ × ln ′ (ax ) = a × 1 = 1 x
ax
∈ ⎤⎥0, +∞ ⎡⎢ : ( F − ln )′ ( x ) = F ′ ( x ) − ln ′ ( x ) = 1 − 1 = 0
وﻣﻨﻩ ﻓﺈن :
x
α
ﺑﺤﻴﺚ :
) F (1) = ln (a
،ﻓﺈن :
إذن ﻳﻮﺟﺪ ﻋﺪد ﺣﻘﻴﻘﻲ وﺑﻤﺎ أن :
⎣
x
∀x ∈ ⎤⎥0, +∞ ⎡⎢ : F ( x ) − ln ( x ) = α ⎣
) α = ln (a ) − ln (1) = ln (a
) F ( x ) − ln ( x ) = ln (a
وﺑﻬﺬا ﻧﺠﺪ :
⎦
⎦
:
.
أي : ln (ax ) − ln ( x ) = ln (a ) : وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن . ∀x ∈ ⎤ 0, +∞ ⎡ : ln (ax ) = ln (a ) + ln ( x ) :ﻧﻀﻊ ⎦ ⎣ ) ln (ab ) = ln (a ) + ln (b .
ﻟﻴﻜﻦ aو bﻣﻦ ⎣⎡ ∞ . ⎤⎦ 0, +ﺑﻴﻦ أن : ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ : ⎞⎛1 ⎞ ⎛a ln ⎜ ⎟ = ln a − ln b (ii ; ln ⎜ ⎟ = − ln a (i ⎠ ⎝b ⎠ ⎝a ب -ﻧﺘﺎﺋﺞ :
.1ﻟﻜﻞ aو bﻣﻦ ⎣⎡ ∞ ⎤⎦ 0, +وﻟﻜﻞ ` ∈ n؛ ﻟﺪﻳﻨﺎ :
.
.
⎣⎡ ∞∀x ∈ ⎤⎦ 0, +
⎣⎡ ∞∀x ∈ ⎤⎦ 0, +
، x = bﻓﻨﺠﺪ :
) (
ln a 2 = 2 ln a (iii
;
ln(ab ) = ln a + ln b
⎞⎛1 ln ⎜ ⎟ = − ln a ⎠ ⎝a
⎞ ⎛a ln ⎜ ⎟ = ln a − ln b ⎠ ⎝b
ln (a n ) = n ln a
) (
.2ﻟﻜﻞ _∈ rوﻟﻜﻞ ⎣⎡ ∞ a ∈ ⎤⎦ 0, +؛ ﻟﺪﻳﻨﺎ : ﻣﺜﺎل :
. ∀x
. ln a r = r ln a
ﻧﻀﻊ a = ln 2 :و . b = ln 5 .1ﺣﺪد ﺑﺪﻻﻟﺔ aو bﻣﺎ ﻳﻠﻲ :
(i (iii .2ﺣﺪد ﺑﺪﻻﻟﺔ a
-3دراﺱﺔ اﻟﺪاﻟﺔ ln
;
ln10
⎠⎟⎞ ( 5 ) − ln ⎛⎜⎝ 12 و bﻣﺎ ﻳﻠﻲ ln ( 2 ) :
⎞ ⎛ 4 ⎟ ⎠ ⎝ 125
⎜ ln
(ii
ln ⎛⎜ 2 + 2 ⎞⎟ + ln ⎛⎜ 2 − 2 ⎞⎟ (iv ; ln 3
⎠
و
⎝
⎟⎞ ln ⎛⎜ 3 100 ⎠
⎝
⎠
⎝
.
:
ﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ lnهﻲ . D ln = ⎤⎦ 0, +∞ ⎡⎣ : أ -اﻟﻨﻬﺎﻳﺎت : ﻟﻨﺒﻴﻦ أن اﻟﺪاﻟﺔ lnﻻ ﺗﻘﺒﻞ ﻧﻬﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ﻋﻨﺪ اﻻﻧﻬﺎﻳﺔ :ﺑﺎﻟﺨﻠﻒ ،ﻧﻔﺘﺮض أن lim ln ( x ) = l :
∞x →+
ﺣﻴﺚ \ ∈ . lﻧﻌﻠﻢ أن ، ln ( 2x ) = ln ( 2 ) + ln ( x ) :ﻋﻨﺪﻣﺎ ﻳﺆول ، xﻧﺠﺪ l = ln ( 2 ) + l :
إذن ln ( 2 ) = 0 :وهﺬا ﺗﻨﺎﻗﺾ .إذن lim ln ( x ) :ﻏﻴﺮ ﻣﻨﺘﻬﻴﺔ ... ∞x →+
ﺧﺎﺻﻴﺔ ﻣﻘﺒﻮﻟﺔ :
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
∞lim ln x = +
∞x →+
-3-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
∞lim ln x = −
اﺳﺘﻨﺘﺎج :
x →0+
1 ﻧﻀﻊ x
ب -رﺗﺎﺑﺔ اﻟﺪاﻟﺔ ln
⎞⎛1 ∞− ln t = − lim+ ln x = t lim ln ⎜ ⎟ = t lim ∞→+ x →0 ∞⎝ t ⎠ →+
∞ t → +وﻣﻨﻩ ﻓﺈن : = . tإذن + x →0
:
1 ﻧﻌﻠﻢ أن اﻟﺪاﻟﺔ lnﻗﺎﺑﻠﺔ ﻟﻺﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +وﻟﺪﻳﻨﺎ . ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln ′(x ) = > 0 : x إذن lnﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +؛ وﺑﻤﺎ أن lnﻣﺘﺼﻠﺔ ﻋﻠﻰ [∞ ]0, +ﻓﺈﻧﻬﺎ ﺗﻘﺎﺑﻞ ﻣﻦ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +ﻧﺤﻮ اﻟﻤﺠﺎل . \ :
وﻣﻨﻩ ﻓﺈن :ﻣﻼﺣﻈﺔ : 1
ﻣﻼﺣﻈﺔ : 2
⎞\ = ⎡ ⎟ ⎣⎢ ⎠
(
)
. ⎜⎛ ln ⎤⎦ 0, +∞ ⎡⎣ = ⎤ lim+ ln x , lim ln x ⎥⎦ x →0 ∞x →+ ⎝
∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ , ∀y ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln x < ln y ⇔ x < y
ln x = ln y ⇔ x = y
;
∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ , ∀y ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln x > 0 ⇔ x > 1 ; ln x < 0 ⇔ 0 < x < 1 ; ln x = 0 ⇔ x = 1
ﻣﻼﺣﻈﺔ : 3 اﻟﻤﻌﺎدﻟﺔ ln x = 1ﺗﻘﺒﻞ ﺣﻻ وﺣﻴﺪا ﻓﻲ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +ﻳﺮﻣﺰ ﻟﻩ
ﺑﺎﻟﺮﻣﺰ e . ( ln e = 1) . e
ﻋﺪد ﺣﻘﻴﻘﻲ ﻻ ﺟﺬري ﺑﺤﻴﺚ :
⎞ ⎛π
ﻣﺜﺎل .1 : 1ﺣﺪد إﺷﺎرة اﻷﻋﺪاد اﻟﺘﺎﻟﻴﺔ ; ln ⎜ ⎟ ; ln(0,99) : ⎠⎝5 .2ﻧﻌﺘﺒﺮ اﻟﻤﻌﺎدﻟﺔ . ( E ) : ln(x − 2) + ln(x − 3) = 0
)( 2
e 2,718 . ln
أ -ﺣﺪد Dﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﻳﻒ اﻟﻤﻌﺎدﻟﺔ ) . ( E
ب -ﺣﻞ ﻓﻲ \ اﻟﻤﻌﺎدﻟﺔ ) . ( E
ﺟـ -اﺱﺘﻨﺘﺞ ،ﻓﻲ \ ،ﻣﺠﻤﻮﻋﺔ ﺣﻠﻮل اﻟﻤﺘﺮاﺟﺤﺔ. ln(x − 2) + ln(x − 3) > 0 :
ﻣﺜﺎل : 2
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
ﺣﺪد إﺷﺎرة اﻟﻌﺪد اﻟﺤﻘﻴﻘﻲ ) ( x − 2 ) ln ( x − 1ﺣﻴﺚ ⎣⎡ ∞. x ∈ ⎤⎦1, +
-4-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
ﺟـ -اﻟﻔﺮوع اﻟﻼﻧﻬﺎﺋﻴﺔ : (iﻧﻌﻠﻢ أن ∞ . lim+ ln x = −إذن x →0
) ( Clnﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ ﻋﻤﻮدﻳﺎ ﻣﻌﺎدﻟﺘﻩ . x = 0
ln x (iiﺧﺎﺻﻴﺔ= 0 : x ﺑﺮهﺎن :ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fاﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ f : x 6 x − ln x : 1 x −1 = . ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : f ′ ( x ) = 1 − ﻟﺪﻳﻨﺎ Df = ⎤⎦ 0, +∞ ⎡⎣ :و x x إﺷﺎرة ) f ′ ( xﻋﻠﻰ اﻟﻤﺠﺎل [∞ ]0, +هﻲ إﺷﺎرة اﻟﺒﺴﻁ . x − 1وﻣﻨﻩ ﻧﺠﺪ : lim ∞x →+
إذن 1ﻗﻴﻤﺔ دﻧﻴﺎ ﻣﻄﻠﻘﺔ ﻟﻠﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞. ⎤⎦ 0, + وﻣﻨﻩ ﻓﺈن . ∀t ∈ ⎤⎦ 0, +∞ ⎡⎣ : f (t ) ≥ 1 :أي . ∀t ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln (t ) − t ≥ 1 > 0 : وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن . ∀t ∈ ⎤ 0, +∞ ⎡ : ln t ≥ t :و ﻣﻦ أﺟﻞ ، t = xﻧﺠﺪ : ⎦ ⎣
)(
ln x 2 ≥ x x
⇒ ≥ x
وﻣﻨﻩ ﻧﺴﺘﻨﺘﺞ أن :
⇒ 1 ln x 2
2 x
≤
x
≥) ( x
∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : ln
.
ln x
≤ . ∀ x ∈ ⎤⎦1, +∞ ⎡⎣ : 0 x
2 ln x = limو ، lim 0 = 0ﻓﺈن : وﺑﻤﺎ أن = 0 : 0 x →+∞ x x →+∞ x ∞x →+
. lim
وﻣﻨﻩ ﻧﺴﺘﻨﺘﺞ أن اﻟﻤﻨﺤﻨﻰ ) ( Clnﻳﻘﺒﻞ ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ ﺑﺠﻮار ∞ +؛ اﺗﺠﺎهﻩ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ . د -ﺗﻘﻌﺮ اﻟﻤﻨﺤﻨﻰ ) : ( Cln
⎛ 1 ⎞′ 1 ⎤ ⎡ ﻟﺪﻳﻨﺎ . ∀x ∈ ⎦ 0, +∞ ⎣ : ln n ′′ ( x ) = ⎜ ⎟ = − 2 < 0 : x ⎠ ⎝x إذن ﺗﻘﻌﺮ ) ( Clnﻣﻮﺟﻩ ﻧﺤﻮ اﻷراﺗﻴﺐ اﻟﺴﺎﻟﺒﺔ .
هـ -إﻧﺸﺎء اﻟﻤﻨﺤﻨﻰ ) : ( Cln ﻟﺪﻳﻨﺎ ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ) (Tﻟﻠﻤﻨﺤﻨﻰ ) ( Clnﻓﻲ اﻟﻨﻘﻄﺔ اﻟﺘﻲ أﻓﺼﻮﻟﻬﺎ 1هﻲ:
y = ln ′(1) ( x − 1) + ln1
أي :
. y = x − 1ﻷن ln ′ (1) = 1 :
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
و
. ln (1) = 0
-5-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
ﻣﻼﺣﻈﺔ :ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞ ⎤⎦ 0, +؛ ﻟﺪﻳﻨﺎ ) ( Clnﻳﻮﺟﺪ ﺗﺤﺖ اﻟﻤﻤﺎس ) . (Tإذن :
∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ ; ln x ≤ x −1 -4ﻧﻬﺎﻳﺎت هﺎﻣﺔ :
) ln(1 + x ; =1 x
lim x ln x = 0
x → 0+
lim x →0
.
ln x ; =1 x −1
lim x →1
ﺑﺮهﺎن :
1 ﻧﻀﻊ : x
∞t → + = . tإذن : + x →0
.
⎞1 ⎛1 ln t − و ﻣﻨﻩ ﻓﺈن = 0 : ln ⎜ ⎟ = t lim ∞→+ t ⎠ ⎝t ln x ln x − ln1 . lim = lim ﻟﺪﻳﻨﺎ = ln′ (1) = 1 : x →1 x − 1 x →1 x −1 ﻧﻀﻊ . t = x −1 :إذن t → 0 :
. lim+ x ln x = lim
t →+∞ t
x →0
x →1
) ln (1 + x ln t = و ﻣﻨﻩ ﻓﺈن = 1 : lim x →0 t →1 t − 1 x
. lim
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
-6-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
أﺣﺴﺐ اﻟﻨﻬﺎﻳﺎت اﻟﺘﺎﻟﻴﺔ :
ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ :
lim 2x − 3ln x
(a
; (b
∞x →+
(e
1 + ln x x
(i
) lim+ x ( ln x
2
;
lim+
x →0
2
(m
x
) ln ( ln x (q ln x
(f
; (j
x →0
) ( ln x
lim ( ln x ) + ln x 2
x →0+
)
(
lim x − ln 1 + x 2 ∞x →+ ) lim+ x ( ln x
3
; (n
lim
∞x →+
;
lim ∞x →+
(r
ln x x
; (g
lim+ x 2 ln x
; (h
(k
x →0
ln x x2
;
lim ∞x →+
lim+ x 3 ln x
x →0
ln x x3
(l
lim ∞x →+
) ln (1 + ln x ) ln (1 − sin x p ( ; lim ; (o lim x →1 x →0 ln 1 + tan x x 2 −1 ( )
3
x
; (d
;
x →0
) ( ln x
; (c
1 − ln x lim+ x →0 ln x
ln x lim x →+∞ 1 + ln x
lim
∞x →+
; (s
lim+
x →0
) (
. lim x + ln x 2 ∞x →−
-IIIاﻟﻤﺸﺘﻘﺔ اﻟﻠﻮﻏﺎرﻳﺘﻤﻴﺔ : -1ﺧﺎﺻﻴﺔ : ﻟﺘﻜﻦ uداﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق وﻻ ﺗﻨﻌﺪم ﻋﻠﻰ ﻣﺠﺎل . I ) ′ u ′(x = ) ∀x ∈ I : ln u (x ﻟﺪﻳﻨﺎ : ) u (x
)
(
-2ﺗﻌﺮﻳﻒ : ﻟﺘﻜﻦ uداﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق وﻻ ﺗﻨﻌﺪم ﻋﻠﻰ ﻣﺠﺎل . I
u′ اﻟﺪاﻟﺔ u
ﺗﺴﻤﻰ اﻟﻤﺸﺘﻘﺔ اﻟﻠﻮﻏﺎرﻳﺘﻤﻴﺔ ﻟﻠﺪاﻟﺔ uﻋﻠﻰ اﻟﻤﺠﺎل I
x −1 ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ : 1ﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ : x -1ﺣﺪد Dfﺣﻴﺰ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ . f -2أﺣﺴﺐ ﻧﻬﺎﻳﺎت fﻋﻨﺪ ﻣﺤﺪات . Df -3أﺣﺴﺐ ) f ′(xﻟﻜﻞ xﻣﻦ Df؛ ﺛﻢ ﺿﻊ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ fﻋﻠﻰ . Df
. f (x ) = ln
JG JJG -4أدرس اﻟﻔﺮوع اﻟﻻ ﻧﻬﺎﺋﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ ) ( Cfاﻟﻤﻤﺜﻞ ﻟﻠﺪاﻟﺔ fﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ . O , i , j
)
(
-5أدرس ﺗﻘﻌﺮ اﻟﻤﻨﺤﻨﻰ ) ( Cfﺛﻢ ﺣﺪد ﻧﻘﻄﺔ اﻧﻌﻄﺎف اﻟﻤﻨﺤﻨﻰ ) . ( Cf -6أﻧﺸﺊ ) . ( Cf
اﻟﺠﻮاب :
x −1 .1ﻟﻴﻜﻦ \ ∈ . xﻟﺪﻳﻨﺎ : x ≠0و >0 x x −1 ⇔ x ≠0و ≠0 x x ≠ 0و ⇔ x ≠1 ⎣⎡ ∞Df = \ − {0,1} = ⎤⎦ −∞,0 ⎡⎣ ∪ ⎤⎦ 0,1⎡⎣ ∪ ⎤⎦1, + إذن :
⇔ x ∈Df
.2ﻟﺪﻳﻨﺎ :
x −1 1 = lim ln 1 − = ln1 = 0 ∞x → ± x x
x −1 و ﺑﻮﺿﻊ x ﻣﺤﻤﺪ اﻟﺤﻴﺎن
= ، tﻧﺠﺪ t → 0+و x →1
. lim f (x ) = lim ln ∞x → ±
∞x → ±
x −1 ∞= lim+ ln (t ) = − t →0 x
-7-
. lim f (x ) = lim ln x →1
x →1
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
x −1 ∞= lim ln (t ) = + ∞t →+ x
∞ t → +و وﻟﺪﻳﻨﺎ : + x →0
. lim f (x ) = lim ln x →0
x →0
1 −1 ′ ⎞ ⎛ x −1 1 0 ⎜ ⎟ 2 ⎠ x 1 1 1 1 ⎝ f ′( x ) = − + =− + x =− + 2 x −1 2 x −1 )2 x ( x −1 x x ) −x 2 + x + 2 − ( x + 1)( x − 2 = ) f ′(x = )x ( x − 1 )x ( x −1
.3ﻟﻴﻜﻦ x ∈Df؛ ﻟﺪﻳﻨﺎ :
.4ﺗﺤﺪﻳﺪ اﻟﻔﺮوع اﻟﻻﻧﻬﺎﺋﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ ) : ( Cf 9ﻟﺪﻳﻨﺎ . lim f (x ) = 0 :إذن ) ( Cfﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ أﻓﻘﻴﺎ ﺑﺠﻮار ∞ ±ﻣﻌﺎدﻟﺘﻩ . y = 0 ∞x → ±
9ﻟﺪﻳﻨﺎ . lim f (x ) = +∞ :إذن ) ( Cfﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ ﻋﻤﻮدﻳﺎ ﻣﻌﺎدﻟﺘﻩ . x = 0 x →0
9ﻟﺪﻳﻨﺎ . lim f (x ) = −∞ :إذن ) ( Cfﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ ﻋﻤﻮدﻳﺎ ﻣﻌﺎدﻟﺘﻩ . x = 1 x →1
.5ﻟﻴﻜﻦ . x ∈Dfﻟﺪﻳﻨﺎ :
′ ′ ′ 2 2 ⎛ ⎞2 +x +2 x x 2 x x 1 x x 2 x x 1 − + + − − − + + − ( ) ( ) − x ⎟ ⎜⎜ = 2 = ⎟ − 1 x x ⎜ ⎟ x x 1 − ( ) ⎝ ⎠
)
إﺷﺎرة
)
()
)
( )
(
()
(
)
(
) f ′′ ( x
) ( −2x + 1) ( x ( x −1) ) − ( −x 2 + x + 2 ) ( x −1+ x = ) f ′′ ( x 2 ))( x ( x −1 )− ( 2x − 1) ( x ( x − 1) ) − ( −x 2 + x + 2 ) ( 2x − 1 = ) f ′′ ( x 2 ))( x ( x −1 )( 2x −1) ( −x 2 + x + x 2 − x − 2) −2 ( 2x −1 = ) f ′′ ( x = 2 2 ))( x ( x −1 ))( x ( x −1
f ′′ ( xﻋﻠﻰ Dfهﻲ إﺷﺎرة ، 1 − 2xوﻣﻨﻩ ﻧﺴﺘﻨﺘﺞ أن :
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
-8-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
⎞ ⎛1 وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن اﻟﻨﻘﻄﺔ ⎟ A ⎜ ,0ﻧﻘﻄﺔ اﻧﻌﻄﺎف ﻟﻠﻤﻨﺤﻨﻰ ) . ( Cf ⎠ ⎝2 .6إﻧﺸﺎء اﻟﻤﻨﺤﻨﻰ ) : ( Cf
ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ : 2
ﻟﺘﻜﻦ fاﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ :
(
)
⎧ f ( x ) = ln 1 − x 3 ; x <0 ⎪ ⎨ ⎪f ( x ) = 4x x − 3x 2 ; x ≥ 0 ⎩ .1أ -ﺑﻴﻦ أن fﻣﺘﺼﻠﺔ ﻓﻲ . 0 ب -ﺑﻴﻦ أن fﻗﺎﺑﻠﺔ ﻟﻻﺷﺘﻘﺎق ﻓﻲ . 0 أدرس ﺗﻐﻴﺮات اﻟﺪاﻟﺔ fﻋﻠﻰ . R .2 .3أ -أﺣﺴﺐ اﻟﻨﻬﺎﻳﺘﻴﻦ lim f ( x ) :و ) . lim f ( x ∞x →−
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
∞x →+
-9-
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
ب -أدرس اﻟﻔﺮﻋﻴﻦ اﻟﻻﻧﻬﺎﺋﻴﻴﻦ ﻟﻠﻤﻨﺤﻨﻰ ) . ( Cf .4أﻧﺸﺊ اﻟﻤﻨﺤﻨﻰ ) . ( Cf .5ﻟﻴﻜﻦ gﻗﺼﻮر اﻟﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ . I = ⎤⎦ −∞,0 أ -ﺑﻴﻦ أن gﺗﻘﺎﺑﻞ ﻣﻦ اﻟﻤﺠﺎل Iﻧﺤﻮ ﻣﺠﺎل Jﻳﻨﺒﻐﻲ ﺗﺤﺪﻳﺪﻩ. ب -ﺣﺪد
) g −1 ( x
ﻟﻜﻞ ) . x ∈ Jﻳﻨﺠﺰ هﺬا اﻟﺴﺆال ﺑﻌﺪ درس اﻟﺪوال اﻷﺱﻴﺔ ( ؟
.6ﻧﻌﺘﺒﺮ `∈ (u n )nاﻟﻤﺘﺘﺎﻟﻴﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ :
`∈ ; n
⎧ 4 = u0 ⎪ 9 ⎨ 2 ⎪u ⎩ n +1 = 4u n u n − 3u n
4 أ -ﺑﻴﻦ أن ≤ u ≤ 1 : 9 n ب -ﺑﻴﻦ أن `∈ (u n )nﺗﺰاﻳﺪﻳﺔ .
. ∀n ∈ ` :
ﺟـ -ﺑﻴﻦ أن `∈ (u n )nﻣﺘﻘﺎرﺑﺔ وﺣﺪد ﻧﻬﺎﻳﺘﻬﺎ.
-3اﺱﺘﻨﺘﺎج : ﻟﺘﻜﻦ uداﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻺﺷﺘﻘﺎق وﻻ ﺗﻨﻌﺪم ﻋﻠﻰ ﻣﺠﺎل . I
u′ اﻟﺪوال اﻷﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ﻋﻠﻰ Iهﻲ اﻟﺪوال x 6 ln u ( x ) + α :ﺣﻴﺚ \∈ α u
-IVداﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ ﻟﻸﺱﺎس -1ﺗﻌﺮﻳﻒ : ﻟﻴﻜﻦ
a >0
ln x اﻟﺪاﻟﺔ ln a
و
a ≠1
a
.
:
.
x 6اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞، ⎤⎦ 0, +
ﺗﺴﻤﻰ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ ﻟﻸﺱﺎس a \
. logaوﻟﺪﻳﻨﺎ :
→
وﻧﺮﻣﺰ ﻟﻬﺎ ﺑﺎﻟﺮﻣﺰ
⎡ ∞⎤ 0, + ⎦ ⎣
ln x = ) 6 loga ( x ln a
loga :
x
ln x
= ) . ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : loge ( x ﻣﻼﺣﻈﺔ :ﻟﺪﻳﻨﺎ = ln x : ln e داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻨﺒﻴﺮي هﻲ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ ﻟﻸﺱﺎس . e
-2ﺧﺎﺻﻴﺎت :
ﻟﻴﻜﻦ a > 0
و
a ≠1
.
.1ﻟﻜﻞ xو yﻣﻦ ⎣⎡ ∞ ⎤⎦ 0, +وﻟﻜﻞ `∈ n؛ ﻟﺪﻳﻨﺎ : ; loga ( xy ) = loga ( x ) + loga ( y ) (i
(ii
;
(iv
(iii
⎞⎛1 ) loga ⎜ ⎟ = − loga ( x ⎠ ⎝x
.2ﻟﻜﻞ _∈ rوﻟﻜﻞ ⎣⎡ ∞ x ∈ ⎤⎦ 0, +؛ ﻟﺪﻳﻨﺎ :
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
- 10 -
) loga ( x n ) = n loga ( x
)
⎞ ⎛x ⎟⎟ = loga ( x ) − loga ( y ⎠ ⎝y
⎜⎜ loga
) (
) . loga x r = r loga ( x
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
-3رﺗﺎﺑﺔ اﻟﺪاﻟﺔ loga
:
أ -ﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ
loga
هﻲ :
⎣⎡ ∞. Dloga = ⎤⎦ 0, +
ln x
= ) . ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : loga (xإذن : ب -اﻟﻨﻬﺎﻳﺎت :ﻧﻌﻠﻢ أن : ln a ∞ lim+ loga (x ) = −و ∞lim loga (x ) = + 9إذا آﺎن a > 1؛ ﻓﺈن :
∞x →+
x →0
9إذاآﺎن 0 < a < 1؛ ﻓﺈن :
1 1 ln a x
ﺟـ -ﺟﺪول اﻟﺘﻐﻴﺮات :ﻟﺪﻳﻨﺎ :
= ) ∀x ∈ ⎤⎦ 0, +∞ ⎡⎣ : loga′ ( x
أ -ﺣﺎﻟﺔ : a > 1 ﻟﺪﻳﻨﺎ logaﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ
ﻟﺪﻳﻨﺎ logaﺗﻨﺎﻗﺼﻴﺔ
ﻋﻠﻰ اﻟﻤﺠﺎل ⎡ ∞⎤ 0, + ⎦ ⎣
ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞⎤⎦ 0, +
a
a >1
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
.
ب -ﺣﺎﻟﺔ : 0 < a < 1
د -إﻧﺸﺎء ) ( Clog ﺣﺎﻟﺔ
∞lim loga (x ) = +
x →0+
و
∞lim loga (x ) = −
∞x →+
.
: ،ﻣﺜﻼ :
) (a = 2
⎛ ⎞1 ﺣﺎﻟﺔ ، 0 < a < 1ﻣﺜﻼ ⎜ a = ⎟ : ⎠4 ⎝
- 11 -
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
Logarithme décimal :
-Vداﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻌﺸﺮي : -1ﺗﻌﺮﻳﻒ :
اﻟﺪاﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻤﻴﺔ اﻟﺘﻲ أﺱﺎﺱﻬﺎ 10ﺗﺴﻤﻰ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻌﺸﺮي وﻧﺮﻣﺰ ﻟﻬﺎ ﺑﺎﻟﺮﻣﺰ ) logﻋﻮض . ( log10وﻟﺪﻳﻨﺎ :
→
\
ln x ln10
log (10r ) = r log10
(b
-3ﻣﺜﺎل :
)
.1أﺣﺴﺐ 100 :
;
_ ∈ ∀r
(
2
)
(
⎣⎡ ∞∀ ( x , y ) ∈ ⎤⎦ 0, +
A = logو ) B = log ( 0, 0001و ) . C = log (10000
.2ﺣﻞ ﻓﻲ \ اﻟﻤﻌﺎدﻟﺔ :
ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ :
;
log1 = 0
; log x = log y ⇔ x = y
(c
3
= ) 6 log(x
x
log10 = 1
(a
-2ﻣﻻﺣﻈﺔ :
⎡ ∞⎤ 0, + ⎦ ⎣
log :
10−x = 3
ﺛﻢ أﻋﻁ ﻗﻴﻤﺔ ﻣﻘﺮﺑﺔ ﻟﻠﺤﻞ ﻣﺴﺘﻌﻤﻻ اﻟﻤﺤﺴﺒﺔ .
ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ fﻟﻤﺘﻐﻴﺮ ﺣﻘﻴﻘﻲ xﺣﻴﺚ : .1أ -ﺣﺪد Dfﺣﻴﺰ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ . f
)
3 + 2ln ( x x
f (x ) = x +
ب -أﺣﺴﺐ ﻧﻬﺎﻳﺎت fﻋﻨﺪ ﻣﺤﺪات .Df .2أ -أﺣﺴﺐ ) f ′( xﻣﻦ أﺟﻞ آﻞ ﻋﻨﺼﺮ xﻣﻦ .Df ب -أﻋﻁ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ . f
) (
.3ﻟﻴﻜﻦ Cfاﻟﻤﻨﺤﻨﻰ اﻟﻤﻤﺜﻞ ﻟﻠﺪاﻟﺔ
fﻓﻲ اﻟﻤﺴﺘﻮى اﻟﻤﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ )
أ -أدرس اﻟﻔﺮوع اﻟﻻﻧﻬﺎﺋﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ ) ( Cf ب -أدرس ﺗﻘﻌﺮ اﻟﻤﻨﺤﻨﻰ ) . ( Cf ﺟـ -أﻋﻁ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻤﻤﺎس اﻟﻤﻨﺤﻨﻰ ) ( Cfﻓﻲ اﻟﻨﻘﻄﺔ Iذات اﻷﻓﺼﻮل . 3 د -أرﺱﻢ اﻟﻤﻨﺤﻨﻰ ) ( Cfﻣﺒﺮزا ﻧﻘﻄﻩ اﻟﺘﻲ أﻓﺎﺻﻴﻠﻬﺎ هﻲ 3و -2و . -4
JG JJG
(
. O ,i , j
.
ﺑﺎﻟﺘﻮﻓﻴﻖ إﻧﺸﺎء اﷲ
ﻣﺤﻤﺪ اﻟﺤﻴﺎن
- 12 -
اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ
12
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