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Persediaan

terakhir pelajar terhadap teknik-teknik menjawab, sebelum mereka menduduki setiap kertas peperiksaan. Mengawal pergerakan pelajar. Memberi motivasi.

SUNTIKAN

AKHIR PMR

SUNTIKAN

AKHIR SPM

2009

2009 Matematik Moden

SUNTIKAN TERAKHIR MATEMATIK SPM FINAL REVISION OF MATHEMATICS SPM

MR WAN FADHLURRAHMAN

MDM NORHAYATI

MDM GOH NGOK MEI

1449/1 – 1 hour 15 minutes 40 objective questions 1449/2 – subjective

11 compulsory questions 4 out of 5 questions

2 Hours 30 minutes

SCIENTIFIC CALCULATOR GEOMETRICAL SET

STANDARD FORM

SETS

OPERATION ON SET

CIRCLES III

TRIGONOMETRY II

ANGLE OF ELEVATION AND DEPRESSION

NUMBER BASES

GRAPH OF FUNCTION II

VARIATION

BEARING

QUADRATIC EQUATIONS 4 MARKS

2k − 5 = 3k 3 2

Solve the quadratic equation

2k − 5 = 9k 2

2k − 9k − 5 = 0 2

( 2k + 1) ( k − 5) = 0 1 k =− , k =5 2

5 MARKS

MATHEMATICAL REASONING

MATHEMATICAL REASONING Is the following sentence a statement ? Give your reason.

9 + 2 = 2 −9 Statement

yes

Not accepted

answer

A statement. It is a false statement.

Make a conclusion for the number sequence below

2 −1= 0 0

2 −1=1

Full marks

1

2 −1= 3 2

3 dots

2 −1= 7 3

2 − 1, n = 0 ,1, 2 , 3 , ... n

Write two implication from this compound statement 3

If

p = 5 if and only if p = 125 3

p = 5 , then p = 125

If p = 125 , then 3

3

p =5

p = 5 , then p = 125

If p = 125 ,

3

p =5

Full marks

no mark

3 MARKS

Linear inequlities Shade the region that satisfies the inequalities

Know how to sketch straight line Application of y-intercept Understand the inequality sign < , > for dashed line and ≤ , ≥ for solid line

Shade the region that satisfies the three inequalities y ≤ 2 x + 8 , y ≥ x and y < 8

y=x

0

y = 2x + 8

Know how to sketch straight line y = 8 From , y = 2 x + 8 , y-intercept = 8

y = 2x + 8

y=x 8

y<8 Full marks

y = 2x + 8

y=x 8

less marks

5 MARKS

STRAIGHT LINE EQUATION

THE STRAIGHT LINE – 6 MARKS

REMEMBER :

y1 − y2 1. Gradient m = x1 − x2 2. Equation of a line

y = mx + c x y + = 1 a b

THE STRAIGHT LINE

REMEMBER : 3. Parallel lines , same gradient

m1 = m2

4. Perpendicular lines , the product of their gradients = − 1

m1m2 = −1

THE STRAIGHT LINE

REMEMBER : 5. x-intercept , substitute 6. y-intercept , substitute

y=0 x=0

5 MARKS

PROBABILITY II

n( A ) P( A) = n( S )

Students

Bus

Car

Bicycle

Form 4 Form 5

30 25

6 14

10 15

The table shows the mode of transport taken by 100 students in Forms 4 and 5, to come to school If a student is chosen at random from the whole group, calculate the probability that the student comes to school by car.

20 1 Probability= = 100 5

6 MARKS

MATRICES NOTES 1. When the matrix has no inverse

ad − bc = 0

2. MATRIX FORM 3. Formula of the inverse matrix 4. State the value of x and of y

The inverse of matrix

 5 − 9   is  2 − 3

1  − 3 9   p  q 5

Find the value of p and of q

Use the inverse formula

 − 3 9 1   inverse = 5 × (−3) − (−9) × 2  − 2 5  1  − 3 9 Compare with the  =  3  − 2 5  given inverse matrix

1  − 3 9   p  q 5

∴ p= 3 q= −2

Calculate the value of x and the value of y by using matrix method

2 x − 3 y = −14 − x + 3 y = 13 Form a matrix equation

 2 − 3  x   − 14      =    −1 3   y   13 

Write the inverse formula Infront

Full marks

1  d − b   ad − bc  − c a 

 x  1  3 3   − 14    =      y  3  1 2   13   x   − 1   =    y  4 

∴ x = −1 y=4

5 MARKS

GRADIENT AND AREA UNDER GRAPH

12 MARKS

2006 FREQUENCY POLYGON 2005 HISTOGRAM 2004 HISTOGRAM 2003 OGIVE

STATISTICS III

Marks 20 – 24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54

Midpoint 22 27 32 37 42 47 52

Frequency 5 7 8 10 6 4 2

MEAN TABLE x 92 97 102 107 112 117 122 127

fx 92 x 4 = 368 97 x 10 = 970 102 x 26 = 2652 107 x 24 = 2568 112 x 17 = 1904 117 x 9 = 1053 122 x 4 = 488 127 x 2 = 254 10257

10257 mean = 96

= 106 .84

TABLE FOR HISTOGRAM CLASS

frequency

36 – 40 41 – 45

0 4

38 43

40.5

46 – 50 51 – 55

6 12

48 53

50.5

56 – 60 61 – 65

10 5

58 63

60.5

66 – 70 71 – 75 JUMLAH

5 8

68 73

70.5

f

50

midpoint

x

Upper boundary

45.5 55.5 65.5 75.5

Full mark

40.5

45.5

50.5

55.5

60.5

65.5

70.5

75.5

Full marks

71 - 75

61 - 70

61 - 65

56 - 60

51 - 55

46 - 50

41 - 45

Full mark

hould have gap

Less mark

41 - 45 46 - 50 51 - 55 61 - 65 71 - 75 61 70 56 - 60

no mark

Table for OGIVE CLASS

frequency

f

Cumulative frequency

Upper boundary

36 – 40 41 – 45

0 4

0 4

40.5 45.5

46 – 50 51 – 55

6 12

10 22

50.5 55.5

56 – 60 61 – 65

10 5

32 37

60.5 65.5

66 – 70 71 – 75

5 8

∑

42 50

70.5 75.5

50

Full marks

OGIF

80 60 40 20 99 .5 10 4. 5 10 9. 5 11 4. 5 11 9. 5 12 4. 5 12 9. 5

0 84 .5 89 .5 94 .5

KEKERAPAN LONGGOKAN

Cumulative 120 frequency 100

SEMPADAN ATAS

Upper boundary

100×

100 2

100 4

3 4

Cumulative Frequency

Ogive Of Time Taken For 100 Students To Complete Their Compositions

105 100 95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 0

Finding the third quartile

Finding the median (second quartile)

Finding the first quartile 0

20

40

60

Time ( minutes )

80

100

INTERQUARTILE RANGE

100×

100 2

100 4

3 4

Cumulative Frequency

Ogive Of Time Taken For 100 Students To Complete Their Compositions

Q3 − Q1

105 100 95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 0

Q3 Finding the third quartile

Finding the median (second quartile)

Q1 Finding the first quartile 0

20

40

60

Time ( minutes )

80

100

12 MARKS

15 – 20 MINUTES PER QUESTION

x y

-2 7

-1

-0.5

-2

1 -2

2 3

3 12

4

4.5 33

12 (a) In the table 1 , find the value of m and the value of n for the equation y = 2 x 3 − 12 x + 3 ( 2 marks)

x

-3

-2

-1

0

1

2

y

-15

m

13

3

-7

n

5 42

x

-2

y

7

-1

0

-0.5 1

2

3

-2

3

12

-2

4

4.5 5 33 42

25

12

GRAPHS OF FUNCTIONS 1. Fill in the blanks in the table

y = 2 x − 12 x + 3 3

x y

-2

2

m 11

n −5

12(b). Draw the graphs of functions Scales and the range of x are given

−3 ≤ x ≤ 5 Plot the points accurately

Can use flexible curve

losing marks

Using your own scale 12 marks – 1 mark

12c. Write the answer in the answer spaces provid

x =1.5

12 MARKS

TRANSFORMATIONS III A Combined Transformation RS means transformation S followed by transformation R. - Use the right terminologies - Start the answer with the right transformation - No short form - Describe in full the transformation – with the correct properties.

y 4

E F

y=3

D 2

C A -4

B

H

Rotation

G

180o o

centre ( 0,3 ) -2

2

Describe in full the transformation PQ

4

Enlargement centre( 2,6 ) 6 4

E

H

Scale factor 3

J

2

M

G K 2

4

6

-2 L

8

12 MARKS

PLAN AND ELEVATION

Reminders 

Calculate , find , solve , – all steps are clearly shown

State – only the answer is required





Unit / label – must be correct if written

All steps must be clearly shown. Read the instructions and questions

very carefully . The answer must be in the lowest form, to 4 significant figures and to 2 decimal places. Master the calculator Do not sleep during the exam!