CERAMAH SPM Kelas Tam Cuti Mei 2014

MATHEMATICS

SPM 2014EN TAN CHUN NAM

Format for SPM MathematicsMARKS: 40 marksNo. of Questions :40 questionsDuration: 75 minutes (1.8 minutes every question)INSTRUCTIONSAnswer all questionsScientific Calculator is allowed

Format for SPM MathematicsMARKS: 100 marksNo. of Questions :Section A : 11 questionsSection B : 5 questionsINSTRUCTIONSAnswer all questionsScientific Calculator is allowed

Format for SPM MathematicsINSTRUCTIONSSection A[52 marks - 78 minutes]11 questions, answer all questionSection B[48 marks - 72 minutes]Choose 4 questions out of 5.(18 minutes every question)

TOTAL =140 MARKS

No Topic No. of questions

1 Polygon I & II 22 Transformations 1 & II 23 Algebraic Expressions I, II, III 14 Algebraic Formulae 15 Algebraic Fractions 16 Linear Equations I & II 1


7 Trigonometry I 18 Indices 29 Linear Inequalities 110 Statistic I & II 3


1 Standard Form 3 @ 42 Set 33 The Straight Line 24 Circles III 15 Trigonometry II 2 @ 36 Lines and Planes in 3-Dimensions 17 Angle of Elevation and Depressions 1 @ 2


1 Number Bases 22 Matrices 1 @ 23 Variations 2 @ 34 Probability I & II 2 @ 35 Bearing 16 Earth as a Sphere 1 @ 27 Graph of Functions 1

No Topic MarksSPM 2013

1 Set 32 Simultaneous Linear

Equations4

3 Quadratic Equations 34 Lines and Planes in 3

Dimensions3

SPM kertas 2


5 Straight Lines 56 Circles 67 Mathematical

Reosening6

8 Probability 5

SPM kertas 2


9 Matrices 610 Gradient and Area

under a Graph6

11 Solid Geometry 5

SPM

No Topic Marks2011 2012 SPM

201312 Graph Functions kubic kuad kub1213 Planes and Elevasion 1214 Earth as a sphere 1215 Transformations 1216 Statistic. HISTOG

12P.KE

12PKE12

SPM

QUADRATIC EQUATION

REARRANGE TO GENERAL FORM OF QUADRATIC EQUATION

02 cbxax

Factorise 0

State the values of x

7,31

0713

n

nn

07203

014763)21(7632

22

nn

nnnnnn

Solve the quadratic equation nnn 21763 2

Factors must be given by using the

whole number

Solve the quadratic equation 22 5 33

k k

K1

K1

N1 N1

Solve the quadratic equation 22 5 33

k k

K1

K0

N0 N0

101

10

K1

K1

N1 N1

Elimination method

SIMULTANEOUS LINEAR EQUATIONS

Substitution method

Matrix method

4 MARKS

Substitution method

qpqp 627276

965 qp1

2

49630135

96)627(5

qqqqq

3)4(627

pp

1 2substitute into

Correct operation

276 qp965 qp

K1

K1

N1 N1

101

10

K1

N1

5 8 38 3

5

x yyx

8 3 1 55 2

y y

16 6 5 5011 66

66 611

y yy

y

8 3(6) 10 25 5

x

K1 N1

Substitution method

ExampleExample

Shade the setShade the setI

VIVIII IVII

C

B

A

CBA )(

1. Label each part.

2. Identify the shaded region2. Identify the shaded region

(A B) U C(A B) U C

I

VIVIII IVII

C

B

A

III, IV U IV, V

U

III, IV, V

3. Shade the region mark with3. Shade the region mark with III, IV, VIII, IV, V

(A B) U C

U

101

10

1

2 3 4

P’ = 1,3,4Q = 1,2,3P’∩ Q = 1,3

1

2 3 4

P = 2 Q’ = 4P ∪Q’= 2,4R = 3,4(P∪Q’)∩R= 4

1

2 3P1

P2

C

A D

CB

6

G

H

F

E 6

6

LINES AND PLANES IN 3 DIMENSION

Name the angle between the line CE and the plane EFGH

C E

E F G H

E GTan θ= 6 √72

θ= 35.26°

K 2

N1

P1

Find the angle between the plane JFE and the plane DEF.L

E

DF

J

M5

513

5

Answer : JMD

22 513

5

Tan JMD

JMD'3722

62.22o

o

atau

L

E

DF

J

M5

513

5

THE STRAIGHT LINE – 6 MARKSREMEMBER :

1. Gradient

2. Equation of a line21

21

xxyym

cmxy

REMEMBER :

3. Parallel lines , same gradient 21 mm

THE STRAIGHT LINE

REMEMBER :

5. x-intercept , substitute

6. y-intercept , substitute

0y

0x

THE STRAIGHT LINE

21

m

21

m

P1

Solid Geometry & Volume

Combination of two solid geometry

1. Determine the two solids involved

2.Choose the operations + or -

3.Write the correct formulae

4.Substitute the values of r, h, d

Use 722

Example 1 A hemisphere PQR has taken out from the cylinder.

Find the volume of the remaining solid.

32

32 rhr

555722

32855

722

P

Q

R

10 cm

8 cm

Cylinder - hemisphere

32366

The diagram shows a solid cone with radius 9 cmand height 14 cm. A cylinder with radius 3 cm and height 7 cm is taken out of the solid.

Calculate the volume,in cm3 , of theremaining solid.

Use 722

Example 2Example 2

Diagram 3 shows a solid cone with radius 9 cm andheight 14 cm. A cylinder with radius 3 cm and height 7cm is taken out of the solid.

14 cm

9 cm

7 cm

3 cmhrhrV 22

31

73722149

722

31 22

9901981188 V

CIRCLES : Perimeter and Area

722

1. Use the correct formulae

2. Substitute with the correct values.

rP 2 rP 2360

2rA 2

360rA

V

S

P

R6 cm

T

6 cm

9 cm

30O

7 cm

6 cmU

Q

W

a) Find the perimeter of the shaded region

b) Calculate the area of the shaded region

(a).

Perimeter = PU + UV + VQ + ( PQ – ST ) + SWT

7218

7131@ 14.31@

27

7222

3601807966

7222

360309

7218

V

SP

R6 cm

T

6 cm

9 cm

30O

7 cm6 cm

U

Q

W

(b).

2

2

6722

36030

27

722

360180912

21

28709

28925@ 32.25@

V

SP

R6 cm

T

6 cm

9 cm

30O

7 cm6 cm

U

Q

W

Area = area of triangle – area of semicircle – area of the sector

120 22 180 22 60 227 7 3.5 3.5 3.5 3.5360 7 360 7 360 7

A

K1 K1

N1

MATHEMATICAL REASONING

9229

Is the following sentence a statement ? Give your reason.

Statement Not acceptedyes

A statement.It is a false statement.

answer

Make a conclusion for the number sequence below

712

312

112

012

3

2

1

0

...,3,2,1,0,12 nn

3 dots

2 MARKS

-1 marks

2n ─ 1

2n ─ 1 , n = 0 , 1 , 2 , 3 , ..

2n ─ 1 , n = 0 , 1 , 2 , 3 ,

2n ─ 1 , n = 0 , 1 , 2 ,

12553 pjikahanyadanjikap

Write two implication

Jika , maka p = 125

Jika p = 125 ,maka

53 p

53 p2 marks

, maka p = 125

jika p = 125 ,

53 p

53 pno mark

Premise1 : All A are BPremise 2 : C is AConclusion : C is B

PQRST has a hexagon

Complete the following argument

164 x

Premise1 : If 4x = 16 , then x = 4Premise 2 : Conclusion :

4x

Premise1 : If A , then BPremise 2 : A trueConclusion : B true

Complete the following argument

Premise1 : If 4x = 16 , then x = 4Premise 2 : Conclusion :

Premise1 : If A , then BPremise 2 : A FalseConclusion : B False

4x164 x

A Statement

R P then P R P IfP R P then R P If

P1

P1

P1

KEBARANGKALIAN

)()()(

SnAnAP

10. Diagram 9 shows two boxes , P and Q . Box P contains four cards labeled with letters and box Q

contains three cards labeled with numbers.

TSEB 764

Two cards are picked at random, a card from box P and another card from box Q .a)List the sample space and the outcomes of the events . b) Hence , find the probability that(i)a card labeled with letter E and a card labelled with an even number are picked (ii)(ii) a card lebelled with letter E or a card labelled with an even number are picked

P Q

a) {(B, 4), (B, 6), (B, 7), (E, 4), (E, 6), (E, 7), (S, 4), (S, 6), (S, 7), (T, 4), (T, 6), (T, 7)}

Notes : 1. Accept 8 correct listings for 1 mark

b) i) {(E, 4), (E, 6)}

ii) {(E, 4), (E, 6), (E, 7), (B, 4), (B, 6), (S, 4), (S, 6), (T, 4), T,6)}

61@

122

43@

129

11

1

11

1

1

1

1

1

35

241012

1

52

13215

1

171

m 1n

621193

yx

yx

69

21131

yx

69

21131

yx

69

11132

)3(11)2(11

yx

69

11132

)3(11)2(11

yx

30

930

311

6991818

311

yx

30

930

311

6991818

311

yx

3,0 yx

36

4231

nm

23,

23

36

1234

101

nm

nm

P1

K1

N1 N1

36

4231

nm

23,

23

1234

36

101

nm

nm

P0

K0

N0 N0

36

4231

nm

23,

23

1234

36

101

nm

nm

P1

K0

N0 N0

object of Area Image of Area 2 k

K1

K1

N1

2014 ???2013 POLIGON KEKERAPAN2012 POLIGON KEKERAPAN2011 HISTORAM2010 OGIF2009 HISTOGRAM2008 POLIGON KEKERAPAN 2007 OGIF2006 POLIGON KEKERAPAN2005 HISTOGRAM2004 HISTOGRAM2003 OGIF

Marks Midpoint Frequency20 – 24 22 525 - 29 27 730 - 34 32 835 - 39 37 1040 - 44 42 645 - 49 47 450 - 54 52 2

962524476421037832727522

84.10696

10257

Marks Midpoint Frequency20 – 24 22 525 - 29 27 730 - 34 32 835 - 39 37 1040 - 44 42 645 - 49 47 450 - 54 52 2

OGIF

020406080

100120

SEMPADAN ATAS

KEK

ERA

PAN

LO

NG

GO

KA

N

Upper boundary

Cumulative frequency Full marks

OGIF

020406080

100120

SEMPADAN ATAS

KEK

ERA

PAN

LO

NG

GO

KA

N

Upper boundary

Cumulative frequency

less marks

OGIF

0102030405060708090

100110

MARKAH

KEK

ERA

PAN

LO

NG

GO

KA

N

limit

Cumulative frequency

No mark

Upper boundary

TABLE FOR HISTOGRAM

CLASSfrequency

fmidpoint x

Upper boundary

36 – 40 41 – 4546 – 5051 – 55 56 – 60 61 – 65 66 – 70 71 – 75

JUMLAH 50

0461210558

3843485358636873

40.545.550.555.560.565.570.575.5

Full mark

40.5 45.5 50.5 55.5 60.5 65.5 70.5 75.5

Full mark

41 -

45

46 -

50

51 -

55

56 -

60

61 -

65

61 -

70

71 -

75

Full mark

-1 mark

Should have gap

41 - 45 46 - 50 51 - 5556 - 60

61 - 6561 - 70 71 - 75

no mark

30 – 3435 – 3940 – 4445 – 4950 - 54

273237424752

578

10642

22 5 25 7 32 8 37 10 42 6 47 4 52 242

1469 34.9842

P1 P1 P1 P1

K1

K1

K2 N1

K1

K1

N1

6

5

3 K1 K1 K1 N1

7

7

8

8

7

8

52

5

5

K1 K1 N2

8

6

8

6

8

6

3 3

5

33

3

3 K1 K1 N2

63125W

0 180 E180 W 125 W180

?

55 E63

30

63125W 55 E63

30

25 E

P2P1

63125W 55 E63

30

25 E

6354

N1K1 K1

63125W 55 E63

30

25 E

N1K1K1

63125W 55 E63

30

25 E

63

63125W 55 E63

30

25 E

63 SpeedceDisTime

TimeceDisSpeed

tan

tan

43.16510

18.8377

N1K1

K1

CERAMAH SPM Kelas Tam Cuti Mei 2014

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