solution outline kpmt2010 q1
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Additional Mathematics Project Work 1 2010
Curriculum Development Division, Ministry of Education Malaysia 1
ADDITIONAL MATHEMATICS PROJECT WORK 1/2010
SOLUTION OUTLINE
Part (a)
QUADRATIC FUNCTION 1
Axis of symmetry, 0 x
FIRST METHOD:
General form:
cbxax y
2
, 5.4c
Passing through (2, 4) 5.42)2(42 ba
Passing through (−2, 4) 5.42)2(42 ba
b = 0 ,8
1a
Quadratic function: 5.48
1 2 x y
4.5
(2, 4)(−2, 4)
0
y
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Additional Mathematics Project Work 1 2010
Curriculum Development Division, Ministry of Education Malaysia 2
SECOND METHOD:
Completing the square: cb xa y 2)( 0,5.4 bc
5.42 ax y
Passing through (2, 4) 5.42 ax y
8
1a
Quadratic function: 5.48
1 2 x y
QUADRATIC FUNCTION 2
Axis of symmetry, 2 x
FIRST METHOD:
General form:
cbxax y 2, 5.4c
Passing through (4, 4), 4)4()4(42 ba
Passing through (2, 4.5), 4)2()2(5.42 ba
8
1a ,
2
1b
Quadratic function: 428
1 2 x
x y
0
2, 4.5
4, 40 , 4
22
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Additional Mathematics Project Work 1 2010
Curriculum Development Division, Ministry of Education Malaysia 3
SECOND METHOD:
Completing the square: cb xa y 2)( , 5.4,2 cb
5.4)2(42 xa
Passing through (0, 4) 5.4)2(4 2 xa
8
1a
Quadratic function: 5.4)2(8
1 2 x y
428
1 2 x
x y
QUADRATIC FUNCTION 3
FIRST METHOD:
General form:
cbxax y 2, 5.0c
5.02 bxax y
Passing through (2, 0) 5.0)2()2(02 ba
Passing through (−2, 0) 5.0)2()2(0 2 ba
0b ,8
1a
Quadratic function: 5.08
1 2 x y
0
0.5
(2, 0)(−2, 0) x
y
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Additional Mathematics Project Work 1 2010
Curriculum Development Division, Ministry of Education Malaysia 4
SECOND METHOD:
By completing the square,
cb xa y 2)( , b = 0, c = 0.5
5.02 ax y
Passing through (2, 0), 5.0)2(02 a
8
1a
Quadratic function: 5.08
1 2 x y
QUADRATIC FUNCTION 4
FIRST METHOD:
General form:
cbxax y 2, 0c
bxax y 2
Passing through (4, 0), )4()4(02
ba
Passing through (2, 0.5), )2()2(5.02
ba
8
1a ,
2
1b
Quadratic function: x x y2
1
8
1 2
(2, 0.5)
(4, 0)0 2
0.5
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Additional Mathematics Project Work 1 2010
Curriculum Development Division, Ministry of Education Malaysia 5
SECOND METHOD:
Completing the square
cb xa y 2)(
5.0)2(2 xa y
Passing through (4, 0), 5.0)24(02 a
8
1a
Quadratic function: x x y2
1
8
1 2
Part (b)
QUADRATIC FUNCTION 1
FIRST METHOD
5.48
1 2 x y
5.4)2(8
1 2 x y
( 0, 4.5)
(2, 4)(−2, 4)
5
y
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Additional Mathematics Project Work 1 2010
Curriculum Development Division, Ministry of Education Malaysia 6
Area of the concrete to be painted = Area of rectangle – Area under the curve
2
2
2)5.4
8
1(20 dx x
3
11720
3
22 m2
SECOND METHOD: Using Geometer’s Sketchpad
QUADRATIC FUNCTION 2
FIRST METHOD:
428
1 2 x
x y
4.5
(4, 4)
5
0
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Additional Mathematics Project Work 1 2010
Curriculum Development Division, Ministry of Education Malaysia 7
Area of the concrete to be painted = Area of rectangle – Area under the curve
4
0
2)4
28
1(54 dx
x x
3
8 m2
SECOND METHOD: Using Geometer’s Sketchpad.
QUADRATIC FUNCTION 3
FIRST METHOD: x x y2
1
8
1 2
x
0.5
(2, 0)(−2, 0)
y
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Additional Mathematics Project Work 1 2010
Curriculum Development Division, Ministry of Education Malaysia 8
Area of the concrete to be painted = Area of rectangle – Area under the curve
2
2
2
2
1
8
14 dx x
38 m2
SECOND METHOD: Using Geometer’s Sketchpad.
QUADRATIC FUNCTION 4FIRST METHOD:
2
1
8
1 2 x y
2
0.5
0
1
2 4
x
y
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Additional Mathematics Project Work 1 2010
Curriculum Development Division, Ministry of Education Malaysia 9
Area of the concrete to be painted = Area of rectangle – Area under the curve
dx x x
4
0
2
2
1
8
14
38 m2
SECOND METHOD: Using Geometer’s Sketchpad
FURTHER EXPLORATION
PART (a)(i)
Shape 1
Cost =3
8 4.0 RM840
= RM896.00
Shape 2
Area = 4(1) – (2
1) (4)
2
1
= 3 m2
Cost = 3 4.0 RM840
= RM1008.00
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Additional Mathematics Project Work 1 2010
Curriculum Development Division, Ministry of Education Malaysia 10
Shape 3
Area = 4(1) – [0.5(1) + 2(2
1)(1.5)(0.5)]
= 2.75 m
2
Cost = 2.75 4.0 RM840
= RM924.00
Shape 4
Area = 4(1) – [0.5(2) + 2(2
1)(1)(0.5)]
= 2.50 m2
Cost = 2.5 4.0 RM840
= RM840.00
Conclusion: Shape 4 can be constructed at the minimum cost of RM840.00.
(ii) Accept any suitable and relevant answers.
PART (b)(i)
k(m)
Surface Area (m2)
(correct to 4 decimal places)0 3.0000
0.25 2.9375
0.50 2.8750
0.75 2.8125
1.00 2.7500
1.25 2.6875
1.50 2.6250
1.75 2.5625
2.00 2.5000
(ii) T 1 = 3.0000, T 2 = 2.9375, T 3 = 2.8750 …,
d = T 2 − T 1 = 2.9375 − 3.0000 = −0.0625
d = T 3 − T 2 = 2.8750 − 2.9375 = −0.0625
…
Arithmetic progression with common difference, d = −0.0625
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Additional Mathematics Project Work 1 2010
Curriculum Development Division, Ministry of Education Malaysia 11
(c) A = 4(0.5) )5.0(2
4
2
12
k
A = 34
k
A =
4
3lim4
k k
= 2 m2
Rectangle
.
REFLECTION
Accept any forms of reflections which should include moral values.