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ENGLISH AS MEDIUM OF INSTRUCTION ASSIGNMENT
Exercise and Solution about Smart Solution
By:
Shofia Hidayah (110210101036)
Dina Rizki A. (110210101081)
DEPARTMENT OF MATHEMATICS EDUCATION
UNIVERSITY OF JEMBER
1
NUMBER THEORY
• Exercise
1. Calculate the value of 20062
−20052
+20042
−20032
+...+42
−32
+22
−12
!
2. Calculate the value of 13
− 23
+ 33
− 43
+ 53
− 63
+ ... + 20053
!
3. Determine the value of (1 − 1
2
)(1 − 1
3
)(1 − 1
4
)...(1 − 1
2004
)!
4. Determine the value of 1
1.2
+ 1
2.3
+ 1
3.4
+ ... + 1
2004.2005
!
5. Determine the value of 1√
1+
√
2
+ 1√
2+
√
3
+ 1√
3+
√
4
... + 1√
9999+
√
10000
!
• Solution
1. 20062
− 20052
+ 20042
− 20032
+ ... + 42
− 32
+ 22
− 12
=(20062
− 20052
) + (20042
− 20032
) + ... + (42
− 32
) + (22
− 12
)
=(2006+2005)(2006−2005)+(2004+2003)(2004−2003)+...+(4+3)(4−
3) + (2 + 1)(2 − 1)
=(2006 + 2005)1 + (2004 + 2003)1 + ... + (4 + 3)1 + (2 + 1)1
=(2006 + 2005 + 2004 + 2003 + ... + 4 + 3 + 2 + 1)
=2006
2
.(2006 + 1)
=2013021
2. 13
− 23
+ 33
− 43
+ 53
− 63
+ ... + 20053
=(13
+ 23
+ 33
+ ... + 20053
) − 2(23
+ 33
+ ... + 20043
)
=(13
+ 23
+ 33
+ ... + 20053
) − 2.23
(13
+ 23
+ 33
+ ... + 10023
)
=(1
2
.2005.2006)2
− 16(1
2
.1002.1003)2
=10032
(20052
− (4.501)2
)
2
=10032
(20052
− 20042
)
=10032
(2005 + 2004)(2005 − 2004)
=10032
.4009
=2019060063
3. (1 − 1
2
)(1 − 1
3
)(1 − 1
4
)...(1 − 1
2004
)
=1
2
.2
3
.3
4
...2002
2003
.2003
2004
= 1
2004
4. 1
1.2
+ 1
2.3
+ 1
3.4
+ ... + 1
2004.2005
1
k(k+1)
= 1
k
− 1
k+1
1
1.2
+ 1
2.3
+ 1
3.4
+ ... + 1
2004.2005
=(1
1
− 1
2
) + (1
2
− 1
3
) + (1
3
− 1
4
) + ... + ( 1
2004
− 1
2005
)
=1 − 1
2005
=2004
2005
5. 1√
1+
√
2
+ 1√
2+
√
3
+ 1√
3+
√
4
... + 1√
9999+
√
10000
!
=(−
√
1 +
√
2) + (−
√
2 +
√
3) + (−
√
3 +
√
4) + ... + (−
√
9999 +
√
10000)
=(−
√
1 +
√
10000)
=(−1) + 100
=99
3
ALGEBRA
• Exercise
1. For each of Real Number a,b prove that a2
+ b2
≥ 2ab !
2. For each of Real Number a,b if a ≥ 0 and b ≥ 0 prove that a+b
2
≥
√
ab !
3. If a,b,c and d in Positive Number prove that (a + c)(b + d) ≥
√
ab+
√
cd!
4. If a and b in Real Positive Number prove that a3+b3
2
≥ (a+b
2
)3
!
5. If a = 12
1
+ 22
3
+ 32
5
+ ... + 10022
2003
and b = 12
3
+ 22
5
+ 32
7
+ ... + 10022
2005
, determine
the integers that closest from a-b !
• Solution
1. (a − b)2
≥ 0
a2
− 2ab + b2
≥ 0
a2
+ b2
≥ 2ab
2. (
√
a −
√
b)2
≥ 0
a − 2
√
ab + b ≥ 0
a + b ≥ 2
√
ab
a+b
2
≥ ab
3. (
√
ad −
√
bc)2
≥ 0
ad − 2
√
adbc + bc ≥ 0
ad + bc ≥ 2
√
adbc
ad + bc + ab + cd ≥ 2
√
adbc + ab + cd
(a + c)(b + d) ≥ (
√
ab +
√
cd)2
(a + c)(b + d) ≥
√
ab +
√
cd
4
4. (a − b)2
≥ 0
(a + b)(a − b)2
≥ 0
a3
+ b3
− a2
b − ab2
≥ 0
a3
+ b3
≥ a2
b + ab2
3a3
+ 3b3
≥ 3a2
b + 3ab2
(3a3
+ 3b3
) + (a3
+ b3
) ≥ (3a2
b + 3ab2
) + (a3
+ b3
)
4a3
+ 4b3
≥ a3
+ b3
+ 3a2
b + 3ab2
4a3
+ 4b3
≥ (a + b)3
4(a3
+ b3
) ≥ (a + b)3
a3
+ b3
≥ (a+b)3
4
a3
+ b3
≥ (a+b)3
22
a3+b3
2
≥ (a+b)3
22.2
a3+b3
2
≥ ((a+b)
2
)3
5. a − b = (12
1
+ 22
3
+ 32
5
+ ... + 10022
2003
) − (12
3
+ 22
5
+ 32
7
+ ... + 10022
2005
)
a − b = 12
1
+ (22
3
− 12
3
) + (32
5
− 22
5
) + ... + (10022
2003
− 10012
2003
) − 10022
2005
a − b = 1 − 10022
2005
+ (1 + 1 + 1 + ... + 1)
a − b = 1002 − 10022
2005
a − b = 1002(2005−1002)
2005
a − b = 1002.1003
2005
a − b = 501
5
GEOMETRY
• Exercise
1. A beam has a size comparison long : width : high = 6 : 3 : 2. If the length
of the space diagonal is 28cm, determine the volume of the beam !
2. A cone is made of a pie-shaped piece of cardboard circle with central angle
288 degrees and the radius of 10 cm. Calculate the volume of a cone is
formed! (use π = 3.14) !
3. A tube with a diameter of 20 cm of water half full. If a ball diameter of
6 cm is inserted into the tube, what is the water level rising? Look at the
picture !
6
4. A pendulum consists of a tube and a half sphere with radius 6 cm as shown
below. If the total structural height of 15 cm and π = 22
7
. Calculate the
volume of the pendulum !
5. Picture below is a tube with a top and bottom half of the ball. If the tube
diameter and 8.4 cm high and 20 cm tube pi = frac 22 7, determine the
surface area of the shaded tube!
7
• Solution
1. Let l = 6x, w = 3x, h = 2x so,
282
= (6x)2
+ (3x)2
+ (2x2
)
784 = 49.x2
⇐⇒ x = 4
So the volume is V = l . w . h = 24.12.8 = 2304
2. For a cone made of pie, then the area will be equal to the broad segment
covers the cone, and the radius of pie will be a line of cones painter.
Area of cardboard pie is:
A = sudut
360o × πr2
A = 288o
360o × π102
A = 80π cm2
Broad blanket cone = Area of pie carton=80π cm2
Painter line = cone radius = 10 cm
Broad blanket cone = πrs
80π = πr10
80 = 10r
r = 8 cm
Next find the cone height using Pythagorean formula
t2
= s2
− r2
= 102
− 82
= 100 − 64 = 36
t =
√
36 = 6 cm
So the volume of the cone is:
V = 1
3
× πr2
× t
8
V = 1
3
× π82
× 6
V = 1
3
× π64 × 6
V = 128π cm
3. First we find volume of a sphere.
Vball = 4
3
πr3
Vball = 4
3
× π × 33
Vball = 4 × π × 32
Vball = 36π cm3
The volume of water that rise is equal to the volume of a sphere. Search
the rising water level by using the volume of water that rose on the tube.
Vwater = π × r2
× t
36π = π × 102
× t
36 = 100.t
t = 36
100
t = 0.36 cm
So the water level rise was 0.36 cm.
4. high of the cone = high cone entirely - fingers bulb
t = 15 cm − 6 cm = 9 cm
Conical of pendulum Volume = Volume + Volume hemispherical
V = 1
3
πr2
t + 1
2
× 4
3
πr3
V = 1
3
× π × 62
× 9 + 2
3
× π × 63
V = π × 36 × 3 + 2 × π × 72
V = 108π + 144π
9
V = 252π
V = 252 × 22
7
V = 36 × 22
V = 792 cm3
5. Tube broad shaded = Area of cylinder - 2 Broad half-ball (without lid)
L = π d t − 2 × 1
2
× 4πr2
L = π × 8, 4 × 20 − 4 × π × (4, 2)2
L = 168π − 4 × π × 17.64
L = 168π − 70, 56π
L = 97, 44π
L = 97, 44 × 22
7
L = 13, 92 × 22
L = 306, 24 cm2
10
PROBABILITY
• Exercise
1. Number of lines that can be made from 8 points are available, with no 3
point line is ...
2. A box contains 5 red marbles and 3 white marbles. Two marbles are drawn
one by one in which marbles are taken first returned again in the box. The
first chance of picking first and second marbles are red is ....
3. Two dice thrown one-eyed six times simultaneously. Probability the emer-
gence of the dice number 5 or number on the dice 10 is ....
4. By how many ways can 4 people sit in chairs around the circular table?
5. In a meeting consisting of 6 people in a circular position. If the chairman
and representatives should always sit next to each other, how many sit
formations that can be formed?
• Solution
1. 8 C3 = 8!
(8−3)!.3!
8 C3 = 8.7.6.5!
5!.3!
8 C3 = 8.7.6
3.2.1
8 C3 = 56
2. P(A ∩ B) = P(A).P(B) = 5
8
.5
8
= 25
64
11
3. Probability the emergence of the dice number 5 is 4
36
Probability the emergence of the dice number 10 is 3
36
So probability the emergence of the dice number 5 or number on the dice
10 is P(A) + P(B) = 4
36
+ 3
36
= 7
36
4. 4 people sit in chairs around the circular table with (4 − 1)! = 3! = 3.2.1 =
6 ways
5. Because of the chairman and deputy should always sit next to each other,
then we think of as a person, so (5 − 1)! = 4! = 4.3.2.1 = 24
For the position of chairman and vice= 2! = 2.
Thus, the formation of which can be formed = 24 x 2 = 48.
12

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Shofiadinasoal

  • 1. ENGLISH AS MEDIUM OF INSTRUCTION ASSIGNMENT Exercise and Solution about Smart Solution By: Shofia Hidayah (110210101036) Dina Rizki A. (110210101081) DEPARTMENT OF MATHEMATICS EDUCATION UNIVERSITY OF JEMBER 1
  • 2. NUMBER THEORY • Exercise 1. Calculate the value of 20062 −20052 +20042 −20032 +...+42 −32 +22 −12 ! 2. Calculate the value of 13 − 23 + 33 − 43 + 53 − 63 + ... + 20053 ! 3. Determine the value of (1 − 1 2 )(1 − 1 3 )(1 − 1 4 )...(1 − 1 2004 )! 4. Determine the value of 1 1.2 + 1 2.3 + 1 3.4 + ... + 1 2004.2005 ! 5. Determine the value of 1√ 1+ √ 2 + 1√ 2+ √ 3 + 1√ 3+ √ 4 ... + 1√ 9999+ √ 10000 ! • Solution 1. 20062 − 20052 + 20042 − 20032 + ... + 42 − 32 + 22 − 12 =(20062 − 20052 ) + (20042 − 20032 ) + ... + (42 − 32 ) + (22 − 12 ) =(2006+2005)(2006−2005)+(2004+2003)(2004−2003)+...+(4+3)(4− 3) + (2 + 1)(2 − 1) =(2006 + 2005)1 + (2004 + 2003)1 + ... + (4 + 3)1 + (2 + 1)1 =(2006 + 2005 + 2004 + 2003 + ... + 4 + 3 + 2 + 1) =2006 2 .(2006 + 1) =2013021 2. 13 − 23 + 33 − 43 + 53 − 63 + ... + 20053 =(13 + 23 + 33 + ... + 20053 ) − 2(23 + 33 + ... + 20043 ) =(13 + 23 + 33 + ... + 20053 ) − 2.23 (13 + 23 + 33 + ... + 10023 ) =(1 2 .2005.2006)2 − 16(1 2 .1002.1003)2 =10032 (20052 − (4.501)2 ) 2
  • 3. =10032 (20052 − 20042 ) =10032 (2005 + 2004)(2005 − 2004) =10032 .4009 =2019060063 3. (1 − 1 2 )(1 − 1 3 )(1 − 1 4 )...(1 − 1 2004 ) =1 2 .2 3 .3 4 ...2002 2003 .2003 2004 = 1 2004 4. 1 1.2 + 1 2.3 + 1 3.4 + ... + 1 2004.2005 1 k(k+1) = 1 k − 1 k+1 1 1.2 + 1 2.3 + 1 3.4 + ... + 1 2004.2005 =(1 1 − 1 2 ) + (1 2 − 1 3 ) + (1 3 − 1 4 ) + ... + ( 1 2004 − 1 2005 ) =1 − 1 2005 =2004 2005 5. 1√ 1+ √ 2 + 1√ 2+ √ 3 + 1√ 3+ √ 4 ... + 1√ 9999+ √ 10000 ! =(− √ 1 + √ 2) + (− √ 2 + √ 3) + (− √ 3 + √ 4) + ... + (− √ 9999 + √ 10000) =(− √ 1 + √ 10000) =(−1) + 100 =99 3
  • 4. ALGEBRA • Exercise 1. For each of Real Number a,b prove that a2 + b2 ≥ 2ab ! 2. For each of Real Number a,b if a ≥ 0 and b ≥ 0 prove that a+b 2 ≥ √ ab ! 3. If a,b,c and d in Positive Number prove that (a + c)(b + d) ≥ √ ab+ √ cd! 4. If a and b in Real Positive Number prove that a3+b3 2 ≥ (a+b 2 )3 ! 5. If a = 12 1 + 22 3 + 32 5 + ... + 10022 2003 and b = 12 3 + 22 5 + 32 7 + ... + 10022 2005 , determine the integers that closest from a-b ! • Solution 1. (a − b)2 ≥ 0 a2 − 2ab + b2 ≥ 0 a2 + b2 ≥ 2ab 2. ( √ a − √ b)2 ≥ 0 a − 2 √ ab + b ≥ 0 a + b ≥ 2 √ ab a+b 2 ≥ ab 3. ( √ ad − √ bc)2 ≥ 0 ad − 2 √ adbc + bc ≥ 0 ad + bc ≥ 2 √ adbc ad + bc + ab + cd ≥ 2 √ adbc + ab + cd (a + c)(b + d) ≥ ( √ ab + √ cd)2 (a + c)(b + d) ≥ √ ab + √ cd 4
  • 5. 4. (a − b)2 ≥ 0 (a + b)(a − b)2 ≥ 0 a3 + b3 − a2 b − ab2 ≥ 0 a3 + b3 ≥ a2 b + ab2 3a3 + 3b3 ≥ 3a2 b + 3ab2 (3a3 + 3b3 ) + (a3 + b3 ) ≥ (3a2 b + 3ab2 ) + (a3 + b3 ) 4a3 + 4b3 ≥ a3 + b3 + 3a2 b + 3ab2 4a3 + 4b3 ≥ (a + b)3 4(a3 + b3 ) ≥ (a + b)3 a3 + b3 ≥ (a+b)3 4 a3 + b3 ≥ (a+b)3 22 a3+b3 2 ≥ (a+b)3 22.2 a3+b3 2 ≥ ((a+b) 2 )3 5. a − b = (12 1 + 22 3 + 32 5 + ... + 10022 2003 ) − (12 3 + 22 5 + 32 7 + ... + 10022 2005 ) a − b = 12 1 + (22 3 − 12 3 ) + (32 5 − 22 5 ) + ... + (10022 2003 − 10012 2003 ) − 10022 2005 a − b = 1 − 10022 2005 + (1 + 1 + 1 + ... + 1) a − b = 1002 − 10022 2005 a − b = 1002(2005−1002) 2005 a − b = 1002.1003 2005 a − b = 501 5
  • 6. GEOMETRY • Exercise 1. A beam has a size comparison long : width : high = 6 : 3 : 2. If the length of the space diagonal is 28cm, determine the volume of the beam ! 2. A cone is made of a pie-shaped piece of cardboard circle with central angle 288 degrees and the radius of 10 cm. Calculate the volume of a cone is formed! (use π = 3.14) ! 3. A tube with a diameter of 20 cm of water half full. If a ball diameter of 6 cm is inserted into the tube, what is the water level rising? Look at the picture ! 6
  • 7. 4. A pendulum consists of a tube and a half sphere with radius 6 cm as shown below. If the total structural height of 15 cm and π = 22 7 . Calculate the volume of the pendulum ! 5. Picture below is a tube with a top and bottom half of the ball. If the tube diameter and 8.4 cm high and 20 cm tube pi = frac 22 7, determine the surface area of the shaded tube! 7
  • 8. • Solution 1. Let l = 6x, w = 3x, h = 2x so, 282 = (6x)2 + (3x)2 + (2x2 ) 784 = 49.x2 ⇐⇒ x = 4 So the volume is V = l . w . h = 24.12.8 = 2304 2. For a cone made of pie, then the area will be equal to the broad segment covers the cone, and the radius of pie will be a line of cones painter. Area of cardboard pie is: A = sudut 360o × πr2 A = 288o 360o × π102 A = 80π cm2 Broad blanket cone = Area of pie carton=80π cm2 Painter line = cone radius = 10 cm Broad blanket cone = πrs 80π = πr10 80 = 10r r = 8 cm Next find the cone height using Pythagorean formula t2 = s2 − r2 = 102 − 82 = 100 − 64 = 36 t = √ 36 = 6 cm So the volume of the cone is: V = 1 3 × πr2 × t 8
  • 9. V = 1 3 × π82 × 6 V = 1 3 × π64 × 6 V = 128π cm 3. First we find volume of a sphere. Vball = 4 3 πr3 Vball = 4 3 × π × 33 Vball = 4 × π × 32 Vball = 36π cm3 The volume of water that rise is equal to the volume of a sphere. Search the rising water level by using the volume of water that rose on the tube. Vwater = π × r2 × t 36π = π × 102 × t 36 = 100.t t = 36 100 t = 0.36 cm So the water level rise was 0.36 cm. 4. high of the cone = high cone entirely - fingers bulb t = 15 cm − 6 cm = 9 cm Conical of pendulum Volume = Volume + Volume hemispherical V = 1 3 πr2 t + 1 2 × 4 3 πr3 V = 1 3 × π × 62 × 9 + 2 3 × π × 63 V = π × 36 × 3 + 2 × π × 72 V = 108π + 144π 9
  • 10. V = 252π V = 252 × 22 7 V = 36 × 22 V = 792 cm3 5. Tube broad shaded = Area of cylinder - 2 Broad half-ball (without lid) L = π d t − 2 × 1 2 × 4πr2 L = π × 8, 4 × 20 − 4 × π × (4, 2)2 L = 168π − 4 × π × 17.64 L = 168π − 70, 56π L = 97, 44π L = 97, 44 × 22 7 L = 13, 92 × 22 L = 306, 24 cm2 10
  • 11. PROBABILITY • Exercise 1. Number of lines that can be made from 8 points are available, with no 3 point line is ... 2. A box contains 5 red marbles and 3 white marbles. Two marbles are drawn one by one in which marbles are taken first returned again in the box. The first chance of picking first and second marbles are red is .... 3. Two dice thrown one-eyed six times simultaneously. Probability the emer- gence of the dice number 5 or number on the dice 10 is .... 4. By how many ways can 4 people sit in chairs around the circular table? 5. In a meeting consisting of 6 people in a circular position. If the chairman and representatives should always sit next to each other, how many sit formations that can be formed? • Solution 1. 8 C3 = 8! (8−3)!.3! 8 C3 = 8.7.6.5! 5!.3! 8 C3 = 8.7.6 3.2.1 8 C3 = 56 2. P(A ∩ B) = P(A).P(B) = 5 8 .5 8 = 25 64 11
  • 12. 3. Probability the emergence of the dice number 5 is 4 36 Probability the emergence of the dice number 10 is 3 36 So probability the emergence of the dice number 5 or number on the dice 10 is P(A) + P(B) = 4 36 + 3 36 = 7 36 4. 4 people sit in chairs around the circular table with (4 − 1)! = 3! = 3.2.1 = 6 ways 5. Because of the chairman and deputy should always sit next to each other, then we think of as a person, so (5 − 1)! = 4! = 4.3.2.1 = 24 For the position of chairman and vice= 2! = 2. Thus, the formation of which can be formed = 24 x 2 = 48. 12