kem akademik sept 16

Mathematics SPM1

BENGKEL KECEMERLANGAN

AKADEMIK

MATEMATIK (1449) SPM 2016

SMK MUKAH20 SEPTEMBER 2016

0900 – 1030 Disediakan oleh: Panitia Matematik

Mathematics SPM2

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

RENUNG-RENUNGKAN…

Then,KNOWLEDGE = 11 + 14 + 15 + 23 + 12 + 5 + 4 + 7 + 5 = 96%HARDWORK = 8 + 1 + 18 + 4 + 23 + 15 + 18 + 11 = 98% Both are important, but fall just short of 100% But,ATTITUDE = 1 + 20 + 20 + 9 + 20 + 21 + 4 + 5 = 100%

ANALISIS JAWAPAN SPM Tahun A B C D

2005 8 11 11 10

2006 10 10 10 10

2007 6 15 12 11

2008 10 11 9 10

2009 9 10 11 10

2010 10 9 10 11

ANALISIS JAWAPAN SPM Tahun A B C D

2011 11 10 10 9

2012 11 8 10 11

2013 9 10 11 10

2014 10 10 11 9

2015 9 9 11 11

2016 8 – 11

Mathematics SPM5

Question 1KETAKSAMAAN LINEAR

[ 3 marks]

Mathematics SPM6

Symbol Definition Line

> Lebih besar daripada Garis putus-putus------------------

< Kurang daripada

Lebih besar dan sama dengan

Garis solid

Kurang daripada dan sama dengan

Mathematics SPM7

Example 5 : (SPM Nov 2005)On the graph in the answer space, shade the region which satisfies the three inequalities y 2x + 10 , x < 5 and y 10 .

Answer :

x < 5

√ K1

√ K2

Mathematics SPM8

Example 6 : State the three inequalities that satisfied the shaded region in the graph below.

Answer : (i) 3y x + 12 ; (ii) y 2x + 4 ; (iii) x < 2√

N1√ N1

√ N1

Mathematics SPM9

LATIHAN: (SPM Jun 2016)Pada graf tersebut, lorek rantau yang memuaskan ketiga-tiga ketaksamaan y x + 8 , y x and y 6 . (3 markah)

Answer :

Mathematics SPM10

JAWAPAN: (SPM Nov 2016)

Mathematics SPM11

Question 2Persamaan Linear

[ 4 marks]

Mathematics SPM12

Example 7 : (SPM Nov 2015)

Calculate the value of x and of y that satisfy the following simultaneous linear equations:

82445

yxyx

Mathematics SPM13

6366

44104044285)1(int)3(

)3(28:)2()2(82)1(445

yyyyyyosubstitute

yxfromyxyx

Method 1 : Substitution method

√ K1√ N1

4

628:)3(int6

x

oysubstitute

√ N1√ K1

Mathematics SPM14

Method 2 : Elimination method

4123:)1()3(

)1(445)3(1642

2)2(:)2()2(82)1(445

xx

yxyx

fromyxyx

√ K1√ N1

6122824

)2(int4

yyy

oxsubstitute

√ K1

√ N1

Mathematics SPM15

6,4,64

3624

61

85418442

61

84

5142

14251

84

2145

yxso

yx

yx

Method 3 : Kaedah Matriks

√ K2

√ N1 √ N1

Mathematics SPM16

Check answer : Calculator Scientific

MODE 5: EQN 1 : anX+bnY=cn

Mathematics SPM17

Example 8 :

Diberi jumlah 50 tiket telah dijual dengan harga RM 2080 dalam satu konsert. Jika harga satu tiket ialah RM35 ataupun RM50, cari bilangan tiket bagi harga RM35 dan bilangan tiket bagi harga RM50 yang telah dijual.

Penyelesaian :

Katakan x = bilangan tiket bagi harga RM35 y = bilangan tiket bagi harga RM50

2080503550

yx

yx

22,28,2228

330420

151

208050

135150

1355011

208050

503511

yxmaka

yx

yx

Method 3 : Kaedah Matriks

√ K2

√ N1 √ N1

18 Mathematics SPM

Mathematics SPM19

Question 4Lines and Planes in 3-Dimensions

[ 3 marks]

Mathematics SPM20

WON TechniqueExample 12 :Given that V and W are midpoints of UT and PS. Name the angle between line VQ and the base

PQRS.

Mathematics SPM21

WON TechniqueStep 1 : Arrange the line and the plane in two rows. Then draw 3 boxes.

VQPQRSW

Step 2 : Find out the same alphabet. Slash the same alphabet.

VQPQRSW

Mathematics SPM22

WON TechniqueStep 3 : Write V in the first box.

VQPQRSW

Step 4 : Look at V in the diagram. Choose the slash alphabet

nearest to V. Write in the centre box. VQPQRSW

V

V Q

Mathematics SPM23

WON TechniqueStep 5 : Look at V in the diagram. Choose the non-slash alphabet nearest to V. Write in the last box.

VQPQRSW

VQW

V Q W

Mathematics SPM24

LATIHAN : (SPM Jun 2016)Diagram 4 shows a right pyramid with height 7cm. M is the midpoint of AB. Given AB = 4cm and BC = 6cm.

(a)On the diagram, draw the orthogonal projection of the line ME and the base ABCD.

(b) Calculate the angle between the line ME and the plane ABCD. (3 marks)

Mathematics SPM25

JAWAPAN : (SPM Jun 2016)

Mathematics SPM26

Question 7Gradient and Area under a Graph

[ 5 ~ 6 marks]

Mathematics SPM27

(I) Distance-time Graph(a) Find distance or period of time, when object

stationary

(b) Gradient = Speed = Rate of change of distance

(c) Average speed =

Mathematics SPM28

(II) Speed-time Graph(a) Find speed or period of time, when object at ‘uniform

speed’

(b) Gradient = Acceleration = Rate of change of speed

(c) Average speed =

* (d) Total distance travelled = Area under the graph

Mathematics SPM29

LATIHAN : (SPM Jun 2016)Diagram 9 shows a speed time-graph for the movement of two particles, A and B, for a period of 30 seconds. The graph PR represents the movement of particle A and the graph PQR represents the movement of particle B. Both particles start at the same point and move along the same route.

Diagram 9

Mathematics SPM30

Example 15 : (SPM Jun 2016)(a) State the uniform speed, in ms-1, of particle B.

(b) Calculate the rate of change of speed, in ms-2, of particle B.

12 ms-1

2

2124

12306012

ms

Particle A

Particle B

(6, 12)

(30, 0)

√ N1

√ K1

√ N1

Mathematics SPM31

Example 15 : (SPM Jun 2016)(c) Find the difference between the distance, in m, travelled by particle A and particle B for the period of 30 seconds.Distance travelled by particle A = Area under the

graph PR =

12

30

m180

123021

√ K1

Lakarkan bentuk geometri

Mathematics SPM32

Example 15 : (SPM Jun 2016)(c) Find the difference between the distance, in m, travelled by particle A and particle B for the period of 30 seconds.Distance travelled by particle B = Area under the

graph PQR =

12

30

m216

12)306(21

6

The difference = 216 – 180 = 36 m

√ K1√ N1

Lakarkan bentuk geometri

Mathematics SPM33

Question 8Solid Geometry ( Volume )

[ 4 marks]

Mathematics SPM34

Example 16 : (SPM Nov 2007)Diagram 6 shows a solid, formed by joining a cylinder to a right prism. Trapezium AFGB is the uniform cross-section of the prism. AB = BC = 9 cm. The height of the cylinder is 6 cm and its diameter is 7 cm.

Diagram 6

Calculate the volume, in cm3, of the solid.[Use ]7

22

9 cm

9 cm

6 cm 7

cm

Mathematics SPM35

Example 16 : (SPM Nov 2007)

(i) Volume of cylinder

231

627

722 2

2

hr(ii) Volume of prism

756

9812921

Ah

(iii) Volume of the solid = 231 + 756

= 987 cm3

√ K1

√ K1

√ N1

√ K1

Mathematics SPM36

LATIHAN : (SPM Jun 2016)A container contains 1386 cm3 of water. Then of the water is poured into the right prism container as shown in Diagram 3. Trapezium ABCD is the uniform cross section of the prism.

Find the depth, in cm, of the water. (3 marks)[Use ]

722

43

Mathematics SPM37

JAWAPAN : (SPM Jun 2016)

Mathematics SPM38

Question 10Matrices

[ 6 marks]

Mathematics SPM39

Inverse matrix

Let , then .

dcba

A

No Inverse matrix If a matrix has no inverse, then ad – bc = 0 .

ad – bc = 0

acbd

bcadA 11

Mathematics SPM40

LATIHAN: (SPM Jun 2016)

(a) It is given that is the inverse matrix of .

Find the value of p and of q.

;

q

p5

3

3547

p = 4 q = 7√

N1√ N1

Mathematics SPM41

Example 18 : (SPM Jun 2016)(b) Write the following simultaneous linear equations as a matrix form :

7x + 4y = 55x + 3y = 3

Hence, using matrix method, calculate the value of x and of y.

Answer :

43

37553453

35

7543

54371

35

3547

yx

yx

yx

yx

x = 3 , y = 4

√ K1

√ K1

√ N1

√ N1

Mathematics SPM42

Question 11Probability II[ 5 ~ 6 marks]

Mathematics SPM43

LATIHAN: (SPM Jun 2016)Diagram 11 shows two boxes, P and Q. Box P contains five tokens labelled with letter and box Q contains three tokens labelled with number. Two tokens are picked at random. The first token is picked from box P and the second token is picked from box Q.

Diagram 11

Mathematics SPM44

JAWAPAN: (SPM Jun 2016)(a) List all the elements in the sample space.

6 8 9K (K , 6) (K , 8) (K , 9)E (E , 6) (E , 8) (E , 9)L (L , 6) (L , 8) (L , 9)A (A , 6) (A , 8) (A , 9)H (H , 6) (H , 8) (H , 9)

n(S) = 15 √

P2

Mathematics SPM45

JAWAPAN: (SPM Jun 2016)(b) By listing down all the possible outcomes of the event, find the probability that (i) a token labelled with letter A and a token labelled with an even number are picked,

X = { (A, 6), (A, 8) }P(X) =

152

√ K1√ N1

6 8 9K (K , 6) (K , 8) (K , 9)E (E , 6) (E , 8) (E , 9)L (L , 6) (L , 8) (L , 9)A (A , 6) (A , 8) (A , 9)H (H , 6) (H , 8) (H , 9)

√

√

Mathematics SPM46

JAWAPAN: (SPM Jun 2016)(b) By listing down all the possible outcomes of the event, find the probability that (ii) a token labelled with a vowel or a token labelled with number 9 is picked.

6 8 9K (K , 6) (K , 8) (K , 9)E (E , 6) (E , 8) (E , 9)L (L , 6) (L , 8) (L , 9)A (A , 6) (A , 8) (A , 9)H (H , 6) (H , 8) (H , 9)

X = { (K, 9), (E, 6), (E, 8), (E, 9), (L, 9), (A, 6), (A, 8), (A, 9), (H, 9)}

P(X) = 159

√ K1√ N1

√ √ √ √ √

√ √

√ √

SEMOGA BERJAYA

& SEKIAN