Final Exam of first Semester 1434-1435
Department of Mathematics
Level: 6
Article: Real Analysis-I (Math 3460)
Number of pages: 2
Time: 3 hours
Kingdom of Saudi Arabia
Ministry of Higher Education
University of Salman bin Abdul Aziz
Faculty of Science and Humanities Aflaj
Girls Section
Solve the following questions
Question 1: (20 Degree)
1- Choose true or false and correct the false.
True / False
(4 degree)
The set S is said to be bounded above if there exists a number u  R such that
T
F
u  s  sS
The set S is said to be denumerable (or Countably finite) if there exists a bijection
T
F
of N onto N .
T
F
T
F
lim (
n
 3 n 1
)
is divergent.
4
If 0  x  y , then 0 
1 1

y x
Solution:
The set S is said to be bounded above if there exists a number u  R such that
s  u  sS .
The set S is said to be denumerable (or Countably infinite) if there exists a bijection of N onto
2- Determine the element of set C  {x  R :
S.
2x  1
 1} (4 degee).
x2
Solution:
2x  1
2x  1 x  2
2x  1  x  2
x 1
1  0 


0

0

0
x2
x2 x2
x2
x2
Therefore we have either x  1  0 and
Or x  1  0 and
x20
 x  1
x20
 x  1
and
and
x  2
x  2
We conclude that C  {x  R : 2  x  1} .
3- Let S be a nonempty subset of R that is bounded above, and let a be any number in
R . Define the set a  S  {a  s : s  S} . Prove that Sup(a  S )  a  SupS. (6 degrees)
Solution:
1
If we let u  sup S , then x  u
 x  S . So that a  x  a  u therefor a  u is
upper bound of a  S 
 sup( a  S )  a  u .
If t is any upper bound of the set a  S . Then
a  x  t  xS 
 x  t  a  x  S
Then t  a is upper bound of S .
u  sup S  t  a 
 a  u  t . But t is upper bound of a  S .
a  u  sup( a  S ) ‫ نحصل على‬sup( a  S ) ‫ بــ‬t ‫نستبدل‬
sup( a  S )  a  u  a  sup S
4- Show that the function f ( x) 
(6 degrees)
Solution:
f ( x )  f (c ) 
1
is not uniformly continuous on the interval (0, ) .
x
1 1
xc

 

x c
xc
xc
‫حيث أن‬
x  c   ‫ وان‬  0 ‫نفرض أن‬
  x  c   
 c    x  c   ‫ فان‬x  c   ‫وبما أن‬
1
1
1
1
c   x 
 

 
x c 
xc c(c   )

  ‫ومن هذا ينتج أن‬
‫ ومنها فان‬f ( x)  f (c) 
c (c   )

 
    (c 2  c ) 
   c 2  c
c (c   )
c 2
. c ‫ تعتمد على‬ ,  ‫( نالحظ أن العالقة بين‬0, ) ‫ في‬c ‫ متصلة عند‬f ‫ وهنا نستنتج أن‬ 
1  c
c ‫نحاول نتخلص من‬
c 2
, c  0}  0
‫ ليست‬f ‫ تحقق الشرط المطلوب فان‬  0 ‫ وهذا يفيد أنه ال يوجد‬  Inf {
1  c
. (0, ) ‫متصلة اتصاال منتظما عل الفترة‬
Question 2: (18 degree)
1- State the definition (i)   neighborhood- (ii) uniformly continuous). (4 degrees)
Solution:
2
(i) Let a  R and
  0 then the   neighborhood of the set
V (a)  {x  R : x  a   }
  x  a   
 a    x  a  
‫ دالةة مصلة ة االة ن ممص مة‬f ‫ دالةة اة ن‬f : D 
 R ‫ وأن‬R ‫ مجموعة جزئيةة مة‬D ‫نفرض أن‬
‫ بحيث اذا ك ن اي ا ن‬  0 ‫ يوجد‬  0 ‫ اذا احقق الشرط الص لي لكل‬D ‫ع ى‬
)ii(
y x  
 f ( y )  f ( x)  
x0 ‫نح ي الفرق بي انال ل وانال ل الممص م انال ل ك ن الشرط يصحقق عمد موضع م مسبق ب لعدد‬
y x  
 x, y ‫انال ل الممص م الشرط ن يصوقف عمد عدد دون غيره اهو لكل‬
2- Let a, b  R. Show that a  b  a  b .(4 degrees)
Solution:
 a  a  a and  b  b  b then
 ( a  b )  a  b  a  b 
 a  b  a  b
3- Show that if the function f ( x)  sin
1
, for x  0 does not limit at x  0 (by
x
using sequence)(4 degrees).
Solution:
1
1

 lim xn  lim
0
n
n   n
n
1
1
f ( x)  sin 
 f ( xn )  sin
x
xn
xn 
lim f ( xn )  lim sin
n0
n0
1
 lim sin n  0
1
n0
n
lim f ( x n )  lim sin n  0 ‫النهاية موجودة‬
n0
3
n0
yn 
1
2 n 

2
lim f ( y n )  lim sin( 2n 
n0
n0

2
) 1
lim f ( xn )  lim f ( yn )
n0
n0
.‫اذن النهاية غير موجودة‬
 x

4- Study the continuous the function f ( x)   x
 1
Solution:
 xx

f ( x)   xx
1

lim
x 0 
, x  0
at point 0.(6 degrees)
, x0
, x0
, x0
, x0
x
 1,
x
lim
x 0 
x
 1
x
lim f ( x)  lim f ( x)
x 0 
x 0 
.‫اذن الدالة غير مصل ة عمد اللفر‬
Question 3: (22 degree)
1- State the definition (i)Riemann sum - (ii)Limit Sequence (4 degrees)
Solution:
‫ [ نعةةرم مجمةةوم نيم ة ن ل دالةةة‬a, b] ‫ اجزئةةة لهةةله الفصةةر‬P
‫ دالةةة معراةةة ع ةةى الفصةةر ولةةصك‬f ‫) لةةصك‬i(
S n ‫ ونرمز له ب لرمز‬f
n
[ xi 1, xi ], (i  1,2,, n) ‫ عدد اخصي ني يمصمي ل فصر‬i ‫ حيث‬S n   f (i )xi
i 1
c ‫ نه يةةة عمةةد‬f ‫ نقةةول ان ل دالةةة‬D ‫ نقطةةة اةةراكم ل مجموعةةة‬c ‫ وان‬f : D 
 R ‫) نفةةرض أن‬ii(
0 xc  
 f ( x)  L   ‫ يحقق‬  0 ‫ يوجد‬  0 ‫وقيمصه اذا ك ن لكل‬
4
3
2- Find the value x0 which satisfy Mean-value theorem of calculus
 (x
2
 2 x  1) dx .
1
(4 degrees)
Solution:
3
2
1 ( x  2 x  1)dx  (3  1) f ( x0 )  2 f ( x0 )
f ( x)  x 2  2 x  1
f ( x0 )  2 x02  2 x0  1
3
2
2
 ( x  2 x  1)dx  2( x0  2 x0  1)
1
x3 2 x 2

x
3
2
3
1
 2( x02  2 x0  1)
1
(9  9  3)  (  1  1)  2( x02  2 x0  1)
3
1
3   2( x02  2 x0  1)
3
8
 2( x02  2 x0  1)
3
4
4
x02  2 x0  1  
 x02  2 x0  1   0
3
3
6  36  12
3 x02  6 x0  1  0 
 x0 
6
4- Using the definition of Derivative and Limiting Theorem Prove that f ( x)  x 2 is
Differentiable on interval [0,1] . (4 degrees)
Solution:
‫ يكون‬x0  (0,1) ‫نفرض أن‬
f ( x)  f ( x0 )
x 2  x02
( x  x0 )( x  x0 )
f ( x0 )  lim
 lim
 lim
 2 x0
x  x0
x  x0 x  x
x  x0
x  x0
x  x0
0
[0 ,1 ] ‫عمد نقطة المه ية يكون‬
5
f ( x)  f (0)
x2
f (0)  lim
 lim

 lim x  0
x 0  0
x0
x 0  x
x 0 
f ( x)  f (1)
x2  1
( x  1)( x  1)
f (1)  lim
 lim

 lim
2
x 1 0
x 1 x  1
x 1
x 1
x 1
[0,1] ‫اذن اكون الدالة ق ب ة ل صف ضل ع ى الفصر المغ قة‬
1
5- By the Archimedean Property if a  0 then lim (
)  0.
1  na
Solution:
1
 0 ‫ ا ن‬  0, a  0 ‫لكل‬
a 
1
1
1
1
N
 a    
  



a 
N
aN
a.N
1
1

1  na na
1
1
1


  ‫ اذن‬n  N ‫ وك نت‬n  N ‫واذا ك نت‬
1  na na Na
1
1
0  
 lim
0
n   1  na
1  na
‫ب سصخدام ن رية انخميدس‬
6- Let X  (an ) be defined by
1
an  2  an  an 1  an , Show that X is Cauchy
2
sequence in R .
Solution:
k  a2  a1 ‫نفرض أن‬
1
1
a2  a1  k
2
2
1
1 1
n2
 a4  a3  a3  a2  ( )( )k
2
2 2
1
a n  2  a n 1  ( ) n k
2
1
a n  3  a n  2  ( ) n 1 k
2
n 1
 a3  a2 
‫ بحيث أن‬N ‫ اي‬n, m ‫وني عددي‬
6
am  an  am  am 1  am 1  am  2  am  2  am  3  am  3    an
 am  am 1  am 1  am  2  am  2  am  3    an 1  an
m2
m 3
m4
n
1
1
1
1
1
am  an   
k   k  
k    k   
2
2
2
2
2
1
1
1
1
k
1 1
 m  2 k  m  3 k    n k  n 1 k  n 1 (1   2  )
2 2
2
2
2
2
2

k
1
4k
(
)


lim
0
n  2n
2 n 1 1  1
2
End Questions - Good luck
Shadia El-Naggar
Signature
Head of the department
Dr. Ahmed Abdulrahman
7
n 1
k