solution outline kpmt2010 q1

8/8/2019 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





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


1 2 x y


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





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


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


1 2 x y

0

0.5

(2, 0)(−2, 0) x

y





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


1 2 x y


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





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)


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





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


FIRST METHOD:

428

1 2 x

x y

4.5

(4, 4)

5

0






4

0

2)4

28

1(54 dx

x x

3

8 m2

SECOND METHOD: Using Geometer’s Sketchpad.


FIRST METHOD: x x y2

1

8

1 2

x

0.5

(2, 0)(−2, 0)

y






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






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





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





(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.

solution outline kpmt2010 q1

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