‫مر‬ ‫ا‬

‫ل‬ ‫ت وا‬ ‫ا‬

‫اﻟﻨﻬﺎﻳﺎت واﻻﺗﺼﺎل‬
(‫ ﲤﺮﻦ‬16) 1 ‫اﻟﺴﻠﺴ‬
:1
f ( x ) = x 2 − 2x cos ( x ) + 1 : ‫ا‬f ‫ا ا ا د‬
cos ( x )
lim = 0 ‫أن‬ .1
x →+∞ x
: ! " # $‫ ا‬.2
lim f ( x ) = +∞ .‫أ‬
x →+∞

f (x )
lim = 1 .‫ب‬
x →+∞ x

:2
: ‫ا‬ '# ‫ أدرس ا‬، * *+ !‫ را‬m &
lim 4x + 2x − 5 + mx − 1 .1
2
x →+∞

lim 4x 2 + 2x − 5 + mx − 1 .2
x →−∞

:3
x + x + ......... + x − n
( n ∈ ℕ∗ )
2 n
f (x ) = : ‫ا‬x * *, ‫ ا‬- f ‫ا ا‬
x −1
x k −1
( k ∈ ℕ∗ ) lim
x →1 x − 1
: '# ‫ د ا‬+ .1
lim f ( x ) : '# ‫ " ا‬# $‫ ا‬.2
x →1

:4
: ‫' ت ا‬# ‫أدرس ا‬
tan ( x ) − 1
limπ .1
x → 2cos ( x ) − 2
4

cos ( 2x ) − cos ( 4x )
lim .2
x →0 x2

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sin ( x ) − sin ( a )
( * *+ ‫د‬ a) lim .3
x →a cos ( x ) − cos ( a )

:5
: ‫ ا‬x * *, ‫ ا‬- ‫ا ا ا د‬f &
 2
f ( x ) = x sin   ; x ≠ 0
 x 

 f ( 0) = 0
0 2*# ‫ ا‬f ‫أن ا ا‬ .1
.ℝ 3 4 ! f ‫أن ا ا‬ .2

:6
: ‫ ل ا وال ا‬47‫أدرس ا‬
2x − 1
f (x ) = x 2 − 2 x (2 f ( x ) = 2 (1
x − 6x + 5
x −1
f (x ) = E (x ) (4 f (x ) = (3
x −1

:7
: ‫ ا‬x * *, ‫ ا‬- f ‫ا ا‬

( x ) = ( x 2 − 1) sin 
1 
f  ;x ≠ 1
  x −1 

 f (1) = 1
f ‫ا ا‬8 7 9! D f ‫ د‬+ (1
( ∀x ∈ ]0,2[ ) f ( x ) − f (1) ≤ 3 x − 1 : ‫أن‬ (2
x 0 = 1 2*# ‫ا‬ 4 ! f ‫ " أن ا ا‬# $‫( ا‬3

:8
: 7:‫ ل ا وال ا‬47‫أدرس ا‬
 1 
f ( x ) = x 3 − 2x 2 + 1 (2 f ( x ) = cos  3  (1
x 
f ( x ) = 1 + x 2 (4 f ( x ) = sin ( x 2 + x ) (3

2/18 Math.ma – 9/2017

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:9
x 0 2*# ‫ا‬ 7:‫أدرس ' ا وال ا‬
f ( x ) = 3x 3 + x − 1 x 0 = +∞ (1
x 2 − 6x + 5
f (x ) = x 0 = 1 (2
( x − 1)( x + 3)
3x 2 + x − 1
f (x ) = x 0 = +∞ (3
x +1
x 2 +1
f (x ) = − x x 0 = +∞ (4
x +1
x +1 x
f (x ) = 2 x 0 = −1 (5
x −1
1 1
f (x ) = + 2 x 0 = 0 (6
x x
x −1
3
f (x ) = x 0 =1 (7
( x − 1) ( 3x − 1)
2

x + 1 −1
f (x ) = x 0 = 0 (8
x

: 10
x0 =0 7:‫أدرس ' ا وال ا‬
E (x )
f (x ) = (1
x
x − sin x
f (x ) = (2
x + sin x
tan ( 3x )
f (x ) = (3
sin ( 2x )
tan x − sin x
f (x ) = (4
x3

: 11
x 0 2*# ‫ا‬ 7:‫أدرس ' ا وال ا‬
f ( x ) = x − sin x x 0 = +∞ (1
f ( x ) = E ( x ) x 0 = +∞ (2

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: 12
ℝ @AB‫ ا‬3 C+ @ *7 f ( x ) = 0 ‫أن ا د‬
f ( x ) = 2x 3 + 5x − 4 (1
f ( x ) = 1 + sin x − x (2
1 1
f (x ) = − cos x (3
( x + 1) 2
2

: 13
4
f (x ) = x 4 −:‫ب‬ ‫ا ا ا د ا‬f &
x
[1, 2] @AB‫ ا‬3 C+ @ *7 f ( x ) = x ‫أن ا د‬

: 14
I 3 4 ! ‫ و‬ℝ ! I ‫ ل‬9! 3 ! ‫د‬ ‫ دا‬f &
( ∀x ∈ I ) f ( x ) < 0  ‫ أو‬( ∀x ∈ I ) f ( x ) > 0  : # ‫ن‬J I ‫ م‬#7 I f D E ‫ إذا‬H ‫أ‬

: 15
. [ 0,1] , [ 0,1] ! ! ‫ [ و‬0,1] ‫ ل‬9 ‫ ا‬3 4! ‫د‬ ‫ دا‬f &
f (α n ) = α nn : L , [ 0,1] ! α n K n ! n @& H ‫أ‬

: 16
. [a;b ] , [a;b ] ‫ ل‬9! ! ! ‫ و‬4 ! ‫ دا‬f &
[a;b ] ‫ ل‬9 ‫@ ا‬AB‫ ا‬3 C+ @ *7 f ( x ) = x ‫أن ا د‬

4/18 Math.ma – 9/2017

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1 ‫"! ا‬#

cos ( x ) 1
∀x ∈ ℝ ∗ ≤ ‫إذن‬ ∀x ∈ ℝ cos ( x ) ≤ 1 : # .1
x x
cos ( x ) 1
lim = 0 ‫ن‬J lim = 0 ‫أن‬ ‫و‬
x →+∞ x x →+∞ x
-‫ أ‬.2
lim f ( x ) = lim x 2 − 2x cos ( x ) + 1
x →+∞ x →+∞

 cos x 1 
= lim x 2 1 − 2 + 2  = +∞
x →+∞
 x x 

 xlim x 2 = +∞
→+∞

 cos x
xlim = 0 : ‫ن‬B
→+∞ x

 1
 x →+∞ x 2 = 0
lim

-‫ب‬
f (x ) x 2 − 2x cos ( x ) + 1
lim = lim
x →+∞ x x →+∞ x
 cos x 1  =1
x 2 1 − 2 + 2 cos x 1
x 1− 2 + 2
 x x  x x
= lim = lim
x →+∞ x x →+∞ x

2 ‫"! ا‬#

 2 5 1
lim 4x 2 + 2x − 5 + mx − 1 = lim x  4 + − 2 + m −  : # .1
x →+∞ x →+∞ x x x 

: m > −2 ‫ ن‬E‫إذا‬
lim 4x 2 + 2x − 5 + mx − 1 = +∞
x →+∞

5/18 Math.ma – 9/2017

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  2 5 1
 xlim  4+ − 2 +m −  = 2+m >0
→+∞
  x x x  : ‫ن‬B
 lim x = +∞
 x →+∞

: m < −2 ‫ ن‬E ‫إذا‬

lim 4x 2 + 2x − 5 + mx − 1 = −∞
x →+∞

  2 5 1
 xlim  4+ − 2 +m −  = 2+m <0
→+∞
  x x x  : ‫ن‬B
 lim x = +∞
 x →+∞

: m = −2 ‫ ن‬E ‫إذا‬
 5
x 2− 
4x + 2x − 5 −4x
2 2
 x  −1
lim 4x 2 + 2x − 5 − 2x − 1 = lim − 1 = lim −1 =
x →+∞ x →+∞
4x 2 + 2x − 5 + 2x x →+∞  2 5  2
x  4 + − 2 + 2
 x x 
 2 5 1
lim 4x 2 + 2x − 5 + mx − 1 = lim x  − 4 + − 2 + m −  : # .2
x →−∞ x →−∞ x x x 

: m > 2 ‫ ن‬E ‫إذا‬
lim 4x 2 + 2x − 5 + mx − 1 = −∞
x →−∞

  2 5 1
 xlim  − 4+ − 2 +m −  = m −2>0
→−∞
  x x x  : ‫ن‬B
 lim x = −∞
 x →−∞

: m < 2 ‫ ن‬E ‫إذا‬

lim 4x 2 + 2x − 5 + mx − 1 = +∞
x →−∞

  2 5 1
 xlim  − 4+ − 2 +m −  = m −2< 0
→−∞
  x x x  : ‫ن‬B
 lim x = −∞
 x →−∞

: m = 2 ‫ ن‬E ‫إذا‬

6/18 Math.ma – 9/2017

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 5
x 2− 
4 x + 2 x − 5 −4 x
2 2
 x  −3
lim 4x 2 + 2x − 5 + 2x − 1 = lim − 1 = lim −1 =
x →−∞ x →−∞
4 x 2 + 2x − 5 − 2 x x →−∞  2 5  2
x  − 4 + − 2 − 2
 x x 

3 ‫"! ا‬#

x k −1
= 1 + x + x 2 + ........ + x k −1 : # ℝ \ {1} ! x @& .1
x −1
x k −1
lim = k ‫ن‬J H#! ‫و‬
x →1 x − 1

: ℝ \ {1} ! ‫ ا‬4# x & .2

x + x 2 + ......... + x n − n x − 1 + x 2 − 1 + ........ + x n − 1 n x k − 1
f (x ) = = =∑
x −1 x −1 k =1 x − 1
n
n ( n + 1) x k −1
lim f ( x ) = ∑ k = ‫ن‬J lim = k ‫أن‬
x →1 2 x →1 x − 1
k =1

4 ‫"! ا‬#

: "' ‫ ) ( ل '"" ا‬:1 % ‫ ط‬.1

π π
( h → 0 ) ⇔  x →  : ‫ إذن‬# x = + h MN
 4 4

7/18 Math.ma – 9/2017

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π 
tan  + h  − 1
tan x − 1 4 
limπ = lim
x → 2cos x − 2 h →0 π 
4 2cos  + h  − 2
4 
π 
tan   + tan ( h )
4 −1
π 
1 − tan   tan ( h )
= lim 4
h →0  π  π  
2  cos   cos ( h ) − sin   sin ( h )  − 2
 4 4 
2 tan ( h )
= lim
h →0
(1 − tan ( h ) ) ( 2 ( cos ( h ) − 1) − 2 sin ( h ) )
tan ( h )
2
= lim h
(1 − tan ( h ) )  2  h 2 .h − 2 h( ) 
h →0   cos ( h ) − 1  sin h 
   
2
=−
2

: * + ‫د ا‬-( ‫( ل ا‬ ) :2 % ‫ط‬
π
‫ * ق‬PC A "cos" ‫" و‬tan" ‫ا ا‬
4

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π 
tan ( x ) − tan  
tan ( x ) − 1 4
limπ = limπ
x→ 2cos ( x ) − 2 x→ 
4 2 cos ( x ) − cos
 π 
4
  
  4 
π 
tan ( x ) − tan  
4
π
x−
= limπ 4
 π 
 cos ( x ) − cos  4  
x→
4

2  
 π 
 x− 
 4 
π 
tan′  
= 4
π 
2cos′  
4
π 
1 + tan 2  
= 4
π 
−2sin  
4
2
=
− 2

5 ‫"! ا‬#

2
f ( x ) = x . sin   : ℝ∗ ! x @& # .1
x 
2
f ( x ) ≤ x ‫ن‬J sin   ≤ 1 ‫و أن‬
x 
lim f ( x ) = 0 = f ( 0 ) ‫ن‬J lim x = 0 ‫ أن‬L + ‫و‬
x →0 x →0

. 0 2*# ‫ا‬ 4! f ‫ن ا ا‬J H#! ‫و‬

x 0 ∈ ℝ∗ & .2
x0 4 ! f1:x ֏ x ‫ا ا‬

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2
x0 4 ! f 2 :x ֏ ‫ا ا‬: #
x
f 2 (x 0 ) 4 ! f 3 : x ֏ sin x ‫و ا ا‬
x0 4 ! f 3 f 2 ‫إذن ا ا‬
x0 4 ! f = f 1 × ( f 3 f 2 ) ‫ ا ا‬H#! ‫و‬
ℝ∗ 3 4! f ‫و‬
.ℝ 3 4 ! f ‫ن‬J 0 4 ! f ‫ أن ا ا‬L + ‫و‬

6 ‫"! ا‬#

2x − 1
f (x ) = (1
x − 6x + 5
2

D f = ]−∞,1[ ∪ ]1,5[ ∪ ]5, +∞[ #

‫ر‬SK ‫ ' دا‬B 'Q 7R+3 4! f ‫ا ا‬
x −1
f (x ) = (2
x −1
D f = [ 0,1[ ∪ ]1, +∞[ #
Df 3 ‫ ص‬4, ‫ و‬ℝ 3 4 ! g : x ֏ x −1 ‫ا ا‬
Df 3 ‫ ص‬4U ‫ و‬ℝ + 3 4 ! h : x ֏ x −1 ‫ا ا‬
Df 3 h (x ) ≠ 0 ‫و‬
Df 3 4! ‫ رج دا‬UE D f 3 4! f ‫إذن ا ا‬
f ( x ) = x 2 + 2 x (3
Df = ℝ #
ℝ 3 4 ! g :x ֏ x 2 ‫ا ا‬
ℝ 3 4 ! h :x ֏ 2 x ‫ا ا‬
ℝ 3 4 ! ‫ دا‬f ‫ن‬J f = g + h ‫أن‬ ‫إذن‬
f ( x ) = E ( x ) (4
Df = ℝ #
( k ∈ ℤ ) [k , k + 1[ ‫ ل‬9 ‫ ا‬3
f (x ) = k : #

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W ‫ ر دا‬4A ' B [ k , k + 1[ ‫ ل‬9 ‫ ا‬3 4! f ‫إذن ا ا‬

k ∈ ℤ M! k 2*# ‫ ا‬f ‫ ل‬47‫ا‬
f ( x ) = k − 1 : # [ k − 1, k [ ‫ ل‬9 ‫ ا‬3
lim f ( x ) = lim− k − 1 = k − 1
x →k − x →k

k ‫ر‬X ‫ا‬3 4 ! Y f ‫ن‬J f ( k ) = k ‫أن‬ ‫و‬

‫ رھ‬X 3 4 ! Y ‫' و‬# 3 4! f # ℤ ! 2* @E ‫ و‬ℝ − ℤ 3 4! f ‫إذن ا ا‬

7 ‫"! ا‬#

x ∈ D f ⇔ x ≠ 1 ‫أو‬ x =1 : # * *+ ‫دا‬ x & (1

D f = ℝ ‫ن‬J H#! ‫و‬
  1
 ∀x ∈ ]0;2[  sin ≤ 1 ‫ن‬J ( ∀a ∈ ℝ ) sin a ≤ 1 ‫أن‬ (2
 x ≠1  x −1

( ∀x ∈ ]0;2[ \ {1}) f ( x ) = x 2 − 1 sin

1
x −1
≤ x 2 − 1 ‫ن‬J H#! ‫و‬

x 2 −1 = x −1 x +1 : # ‫و‬
x ∈ ]0, 2[ ⇒ x + 1 ≤ 3
x ∈ ]0, 2[ ⇒ x 2 − 1 ≤ 3 x − 1

( ∀x ∈ ]0,2[ ) f ( x ) − f (1) ≤ 3 x − 1 H#! ‫و‬

lim x − 1 = 0 ‫∀ ( و‬x ∈ ]0, 2[ ) f ( x ) − f (1) ≤ 3 x − 1 ‫أن‬ (3
x →1

x 0 = 1 2*# ‫ا‬ 4! f ‫ن ا ا‬J H#! ‫ و‬lim f ( x ) = f (1) ‫ن‬J

x →1

8 ‫"! ا‬#

 1 
f ( x ) = cos  3  (1
x 
D f = ℝ∗ #
1
h :x ֏ ‫ و‬g : x ֏ cos ( x ) L + f = g h #
x3

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x 0 ∈ ℝ∗ &
( x 0 ∈ D h ‫ر و‬SK ‫ ' دا‬B ) x 0 4 ! ‫ دا‬h
(ℝ 3 4 ! g ‫ن‬B ) h ( x 0 ) 4 ! ‫ دا‬g
x0 4 ! f H#! ‫ و‬x 0 4 ! g h ‫إذن‬
Df 3 4! f ‫و‬
f ( x ) = x 3 − 2x 2 + 1 (2
Df = ℝ #
h : x ֏ x 3 − 2x 2 + 1 ‫ و‬g : x ֏ x L + f = g h #
x0 ∈ℝ &
( ‫ ود‬+ ‫ ' دا‬B ) x 0 4 ! ‫ دا‬h
(ℝ 3 4! *2 ‫ا‬ * ‫ن دا ا‬B ) h ( x 0 ) 4 ! ‫ دا‬g
ℝ 3 4 ! f ‫ن‬J H#! ‫ و‬x 0 4 ! f ‫إذن‬
f ( x ) = sin ( x 2 + x ) (3
Df = ℝ #
h : x ֏ x 2 + x ‫ و‬g : x ֏ sin ( x ) L + f = g h #
x0 ∈ℝ &
( ‫ ود‬+ ‫ ' دا‬B ) x 0 4 ! ‫ دا‬h
(ℝ 3 4 ! "sin" ‫ن‬B ) h ( x 0 ) 4 ! ‫ دا‬g
ℝ 3 4 ! f ‫ن‬J H#! ‫ و‬x 0 4 ! f ‫إذن‬
f ( x ) = 1 + x 2 (4
Df = ℝ #
h : x ֏1+ x 2 ‫ و‬g : x ֏ x L + f = g h #
x0 ∈ℝ &
( ‫ ود‬+ ‫ ' دا‬B ) x 0 4 ! ‫ دا‬h
( h ( x 0 ) > 0 ‫ و‬ℝ+ 3 4 ! ‫ دا‬g ‫ن‬B ) h ( x 0 ) 4 ! ‫ دا‬g
ℝ 3 4 ! f ‫ن‬J H#! ‫ و‬x 0 4 ! f ‫إذن‬

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9 ‫"! ا‬#

(1
lim f ( x ) = lim 3x + x − 1 = lim 3x = +∞
3 3
x →+∞ x →+∞ x →+∞

lim f ( x ) = lim 3x + x − 1 = lim 3x 3 = −∞

3
x →−∞ x →−∞ x →−∞

lim f ( x ) = lim
( x − 1)( x − 5 ) = lim x − 5 = −1 (2
x →1 x →1 ( x − 1)( x + 3 ) x →1 x + 3

(3
3x 2
lim f ( x ) = lim = lim 3x = +∞
x →+∞ x →+∞ x x →+∞

3x 2
lim f ( x ) = lim = lim 3x = −∞
x →−∞ x →−∞ x x →−∞

(4
x 2 +1 1− x
f (x ) = −x =
x +1 x +1
−x
lim f ( x ) = lim = −1
x →+∞ x →+∞ x

(5

lim f ( x ) = lim
x +1 x
= lim
( x + 1) x = lim x = 1
x − 1 xx→− >−1 ( x + 1)( x − 1)
x →−1 x →−1 2 1 x →−1 x − 1 2
x >−1 x >−1 x >−1

x +1 x − ( x + 1) x −x −1
lim f ( x ) = lim = lim = lim =
x − 1 xx→− <−1 ( x + 1)( x − 1)
x →−1 x →−1 2 1 x →−1 x − 1 2
x <−1 x <−1 x <−1

1 1 x +1 1
f (x ) = + 2 = 2 = ( x + 1) . 2 (6
x x x x
1
lim f ( x ) = lim ( x + 1) . 2 = +∞
x →0 x →0 x
x 3 −1 ( x − 1) ( x 2 + x + 1) x 2 + x +1
f ( )
x = = = (7
( x − 1) ( 3x − 1) ( x − 1) ( 3x − 1) ( x − 1)( 3x − 1)
2 2

x 2 + x +1 1 3
lim+ f ( x ) = lim+ × = × +∞ = +∞
x →1 x →1 3x − 1 x −1 2
x + x +1
2
1 3
lim− f ( x ) = lim− × = × −∞ = −∞
x →1 x →1 3x − 1 x −1 2

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1 ‫ ار‬9 ' @ *7 I f ‫ن‬J lim+ f ( x ) ≠ lim− f ( x ) ‫أن‬

x →1 x →1

lim f ( x ) = lim
x +1 −1
= lim
( x +1 −1 )( x +1 +1 ) = lim 1
=
1
(8
x →0 x →0 x x →0
x ( x +1 +1 ) x →0 x +1 +1 2

10 ‫"! ا‬#

(1
−1
f (x ) = ‫ إذن‬E ( x ) = −1 : [ −1,0[ ‫ ل‬9 ‫ ا‬3
x
−1
lim f ( x ) = lim = +∞
x →0 x →0 x
x <0 x <0

0
f (x ) = = 0 ‫ إذن‬E ( x ) = 0 : [ 0,1[ ‫ ل‬9 ‫ ا‬3
x
lim f ( x ) = lim 0 = 0
x →0 x →0
x >0 x >0

0 ‫ ار‬9 ' @ *7 I f ‫إذن‬

(2
y

-4 -3 -2 -1 0 1 2 3 4 x

-1

-2

-3

sin x = −x ⇔ x = 0 : #
D f = ℝ ∗ ‫إذن‬

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 sin x  sin x
x 1 −  1−
f (x ) =  =
x x
 sin x  1 + sin x
x 1 + 
 x  x
sin x
lim f ( x ) = 0 ‫ إذن‬lim = 1 ‫[ أن‬
x →0 x →0 x

tan ( 3x )
f (x ) = (3
sin ( 2x )
k π  π k π 
Df = ℝ −  / k ∈ ℤ ∪  + / k ∈ ℤ
 2  6 3 
3 tan 3x 2x  π   π
f (x ) = × × :  − ,0  ∪  0,  ‫ ل‬9 ‫ ا‬3
2 3x sin 2x  6   6 
2x X tan 3x tan X
lim = lim = 1 ‫ و‬lim = lim =1 : #
x →0 sin 2x X →0 sin X x →0 3x X →0 X
3
lim f ( x ) = : ‫إذن‬
x →0 2
tan x − sin x
f (x ) = (4
x3
π 
D f = ℝ∗ −  + k π / k ∈ ℤ 
2 
tan x − cos x tan x tan x  1 − cos x  1 1
f (x ) = 3
= × 2  = 1× =
x x  x  2 2

11 ‫"! ا‬#

−1 ≤ − sin x ≤ 1 ‫ إذن‬−1 ≤ sin x ≤ 1 ‫( [ أن‬1

x − 1 ≤ x − sin x ≤ x + 1 ‫إذن‬
x − 1 ≤ f ( x ) ‫إذن‬
lim f ( x ) = +∞ ‫ إذن‬lim x − 1 = +∞ : #
x →+∞ x →+∞

x ∈ ℝ & (2
E (x ) ≤ x < E (x ) +1 #
x − 1 < E ( x ) ‫إذن‬

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x − 1 < f ( x ) ‫إذن‬
lim f ( x ) = +∞ ‫ن‬J lim x − 1 = +∞ ‫أن‬
x →+∞ x →+∞

f ( x ) ≤ x ‫ أي‬E ( x ) ≤ x #
lim f ( x ) = −∞ ‫ن‬J lim x = −∞ ‫أن‬
x →−∞ x →−∞

12 ‫"! ا‬#

f ( x ) = 2x 3 + 5x − 4 : # (1
( ‫ ود‬+ ‫ ' دا‬B ) ℝ 3 4 ! f ‫ و‬Df = ℝ : #
[ −1,1] ‫ ل‬9 ‫ا‬
f ( −1) ≤ 0 ≤ f (1) ‫ إذن‬f (1) = 3 ‫ و‬f ( −1) = −11 : #
L , c ∈ [ −1,1] @AB‫ ا‬3 K 2 $ ‫ ا * [ ا‬#‫] ! ھ‬X+ H J [ −1,1] ‫ل‬9 ‫ ا‬3 4 ! f ‫أن‬ ‫و‬
f (c ) = 0
ℝ C+ @AB‫ ا‬3 @ *7 f ( x ) = 0 ‫د‬ ‫ " أن ا‬# X
f ( x ) = 1 + sin x − x # (2
ℝ 3 4 ! f ‫ و‬Df = ℝ : #
f (π ) < 0 < f ( 0 ) ‫ إذن‬f (π ) = 1 − π ‫ و‬f ( 0 ) = 1 : #
L , c ∈ [ 0, π ] @AB‫ ا‬3 K 2 $ ‫ ا * [ ا‬#‫] ! ھ‬X+ H J [0,π ] ‫ل‬9 ‫ ا‬3 4 ! f ‫أن‬ ‫و‬
f (c ) = 0
ℝ C+ @AB‫ ا‬3 @ *7 f ( x ) = 0 ‫د‬ ‫ " أن ا‬# X ‫إذن‬
1 1
f (x ) = − cos x # (3
( x + 1) 2
2

Df 3 4 ! f ‫ و‬D f = ℝ − {−1} : #
[0,2π ] ‫ ل‬9 ‫ا‬
1 1 1
f ( 2π ) < 0 < f ( 0 ) ‫ إذن‬f ( 2π ) = − ‫ و‬f (0) = : #
( 2π + 1)
2
2 2
L , c ∈ [ 0,2π ] @AB‫ ا‬3 K 2 $ ‫ ا * [ ا‬#‫] ! ھ‬X+ H J [0,2π ] ‫ ل‬9 ‫ا‬3 4 ! f ‫أن‬ ‫و‬
f (c ) = 0
ℝ C+ @AB‫ ا‬3 @ *7 f ( x ) = 0 ‫د‬ ‫ " أن ا‬# X ‫إذن‬

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13 ‫"! ا‬#

4
f (x ) = x 4 − : #
x
ℝ∗ 3 4 ! f ‫ و‬D f = ℝ∗ : #
4
f (x ) = x ⇔ x4− =x
x
4
⇔ x4− −x =0
x
⇔ x 5 −x 2 −4=0
g :x ֏ x 5 −x 2 −4
g (1) ≤ 0 ≤ g ( 2 ) ‫ إذن‬g ( 2 ) = 24 ‫ و‬g (1) = −4 : #
L , c ∈ [1,2] @AB‫ ا‬3 K 2 $ ‫ ا * [ ا‬#‫] ! ھ‬X+ H J [1,2] ‫ل‬9 ‫ ا‬3 4 ! g ‫أن‬ ‫و‬
g (c ) = 0
g (c ) = 0 ⇔ c 5 − c 2 − 4 = 0
4
⇔ c4 − =c : #
c
⇔ f (c ) = c
f (c ) = c L , c ∈ [1,2] @AB‫ ا‬3 K ‫إذن‬
[1, 2] ‫ ل‬9 ‫ا‬ @AB‫ ا‬3 C+ @ *7 f ( x ) = x ‫د‬ ‫إذن ا‬

14 ‫"! ا‬#

I ‫ ل‬9! 3 4 ! f ‫دا‬
( ∀x ∈ I ) f ( x ) ≠ 0  ‫ و‬( ∃x ∈ I ) f ( x ) ≥ 0  ‫ و‬( ∃x ∈ I ) f ( x ) ≤ 0  ‫ ض أن‬Q
( ∃x ∈ I ) f ( x ) > 0  ‫ و‬( ∃x ∈ I ) f ( x ) < 0  ‫ن‬J I 3 ‫ م‬#7 I f ‫أن‬
f (x 1) > 0 ‫ و‬f (x 2 ) < 0 : L , I ! x 2 ‫ و‬x 1 K ‫إذن‬
f ( x 2 ) < 0 < f ( x 1 ) ‫ا‬S&‫ و ھ‬x 1 ≠ x 2 ‫! ا ا _ أن‬
I x 2 ‫ و‬x 1 ‫ي ط ه‬S ‫ ل ا‬9 ‫ ا‬: ‫ن‬J I ! x 2 ‫ و‬x 1 ‫أن‬
x 2 ‫ و‬x 1 ‫ي ط ه‬S ‫ ل ا‬9 ‫ ا‬3 4 ! f ‫ إذن‬I 3 4! f #
f (c ) = 0 L , ‫ ل‬9 ‫ا ا‬S‫ ھ‬3 ‫إ‬ # c K H#! ‫و‬

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x 2 ‫ و‬x 1 ‫ي ط ه‬S ‫ ل ا‬9 ‫ا‬ @AB‫ ا‬3 C+ @ *7 f ( x ) = 0 ‫د‬ ‫إذن ا‬

‫ ت‬2 8 U! ‫ا‬S‫ و ھ‬I 3 ‫ م‬#7 f ‫إذن‬
( ∀x ∈ I ) f ( x ) < 0  ‫ أو‬( ∀x ∈ I ) f ( x ) > 0  : # ‫ن‬J I ‫ م‬#7 I f D E ‫إذن إذا‬

15 ‫"! ا‬#

ℕ∗ ! ‫ ا‬4# n &
g (x ) = f (x ) − x n : ‫ا‬x * *, ‫ ا‬- g ‫ا ا ا د‬
[0,1] 3 4! ‫ ع دا‬9 E [ 0,1] ‫ ل‬9 ‫ ا‬3 4! g ‫ا ا‬
g (1) = f (1) − 1 ≤ 0 ‫ و‬g ( 0 ) = f ( 0 ) ≥ 0 # ‫و‬
‫أي‬ g (α n ) = 0 : L , [0,1] ! αn K 2 $ ‫ ا * [ ا‬#‫] ! ھ‬X+ H#! ‫ و‬g ( 0 ) × g (1) ≤ 0 ‫إذن‬
f (α n ) = α nn

16 ‫"! ا‬#

g (x ) = f (x ) − x : [a , b ] 3 ‫ ا‬g ‫ا ا‬
( [a , b ] 3 4 ! ‫ ع دا‬9 E ) [a,b ] 3 4! g ‫ا ا‬
f (b ) ∈ [a;b ] ‫ و‬f ( a ) ∈ [a;b ] ‫ن‬J [a;b ] , [a;b ] ! ! ‫ دا‬f ‫أن‬
g (b ) ≤ 0 ‫ و‬g (a ) ≥ 0 ‫ إذن‬f (b ) − b ≤ 0 ‫ و‬f ( a ) − a ≥ 0 ‫ أي‬f (b ) ≤ b ‫ و‬a ≤ f ( a ) H#! ‫و‬
g ( a ) × g (b ) ≤ 0 : ‫و‬
[a;b ] ‫ ل‬9 ‫ا‬ @AB‫ ا‬3 C+ @ *7 g ( x ) = 0 ‫د‬ ‫ا‬: 2 $ ‫ ا * [ ا‬#‫] ! ھ‬X+ ‫إذن‬
. [a;b ] ‫ ل‬9 ‫ا‬ @AB‫ ا‬3 C+ @ *7 f ( x ) = x ‫د‬ ‫ا‬: ‫و‬

つづく

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