SULIT 1449/2 - · PDF fileLuas trapezium = 2. 1 ... tagon sekata. [5 marks / markah] Answ)...

SULIT 1449/2

[Lihat sebelah 1449/22009 SULIT

NAMA :………………………………………………………………………………

ANGKA GILIRAN :…………………………………………… TINGKATAN : 5……….

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA

SEKOLAH MENENGAH NEGERI KEDAH DARUL AMAN.

PEPERIKSAAN PERCUBAAN SPM 2009 1449/2 MATHEMATICS Kertas 2 10 September 2009

22

1jam Dua jam tiga puluh minit

Kertas soalan ini mengandungi 32 halaman bercetak.

Untuk Kegunaan Pemeriksa

Bahagian Soalan Markah Penuh

Markah Diperoleh

1 3

2 4

3 4

4 4

5 5

6 5

7 6

8 6

9 5

10 6

A

11 4

12 12

13 12

14 12

15 12

B

16 12

Jumlah

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. Tulis nama, tingkatan dan angka giliran anda pada ruang yang disediakan.

2. Kertas soalan ini adalah dalam

dwibahasa. 3. Soalan dalam bahasa Inggeris

mendahului soalan yang sepadan dalam bahasa Melayu.

4. Calon dibenarkan menjawab

keseluruhan atau sebahagian soalan sama ada dalam bahasa Inggeris atau bahasa Melayu.

5. Calon dikehendaki membaca arahan

di halaman belakang kertas soalan ini.

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2

2

MATHEMATICAL FORMULAE RUMUS MATEMATIK

The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used.

Rumus-rumus berikut boleh membantu anda untuk menjawab soalan. Simbol-simbol yang diberi adalah yang biasa digunakan.

RELATIONS PERKAITAN

1 nmnm aaa

10

Pythagoras Theorem Teorem Pithagoras

2 2c a b

2 nmnm aaa

11

)(

)()(

Sn

AnAP

3 mnnm aa )(

12 ( ') 1 ( )P A P A

4

ac

bd

bcadA

11 13 12

12

xx

yym

5

Distance / Jarak = 212

212 )()( yyxx

14 -intercept

-intercept

ym

x

pintasan-ym

pintasan-x

6 Midpoint / Titik tengah, 1 2 1 2( , ) ,2 2

x x y yx y

[[

7 Average speed = distance travelled

time taken

Purata laju =

jarak yang dilalui

masa yang diambil

8

Mean = sum of data

number of data

Min =

hasil tambah nilai data

bilangan data

9 Mean =sum of (class mark frequency)

sum of frequencies

Min = hasil tambah (nilai titik tengah kelas × kekerapan)

hasil tambah kekerapan

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3

SHAPES AND SPACE

1 Area of trapezium

BENTUK DAN RUANG

= 1

× sum of parallel sides × height2

Luas trapezium

= 2

1 hasil tambah dua sisi selari tinggi

Scale factor,

14 'PA

kPA

Faktor skala, 'PA

kPA

2 Circumference of circle = d = 2r

15

age = k2 area of object

3 r

4 rh 2jt

5 Surface area of sphere = 4 r

6 Volume of right prism = cross sectional area length

Volume of cylinder = r h

8 Volume of cone =

Lilitan bulatan = d = 2j Area of imLuas imej = k2 luas objek

Area of circle = 2 Luas bulatan = j 2

Curved surface area of cylinder = 2Luas permukaan melengkung silinder =

2

Luas permukaan sfera = 4j 2

Isipadu prisma tegak = Luas keratan rentas panjang

7 2

Isipadu silinder = j 2 t

21π

3r h

Isipadu kon = tj 2π3

1

9 Volume of sphere = 34π

3r

Isipadu sfera = 3π3

4j

10 Volume of right pyramid = 1

× base area × height3

Isipadu piramid tegak = 1

× × 3

luas tapak tinggi

11 Sum of interior angles of a polygon = (n – 2) 180o 0o

12

Hasil tambah sudut pedalaman poligon = (n – 2) 18

arc length angle subtended at centre

circumference of circle 360

360

panjang lengkok sudut pusat

lilitan bulatan

13 area of sector angle subtended at centre

area of circle 360

360

luas sektor sudut pusat

luas bulatan

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For

Examiner’s Use

Untuk

Kegunaan

[ 52 marks / markah ]

Answer all is section. Jawab semua soalan dalam bahagian ini.

Pemeriksa

Section A Bahagian A

questions in th

1 hich satisfy the three

rantau yang memuaskan ketiga-tiga

[3 marks / markah]

hich satisfy the three

rantau yang memuaskan ketiga-tiga

[3 marks / markah]

On the graph in tinequalities 2y

Pada graf di ruang jawapan, lorekkan

in tinequalities y

Pada graf di ruang jawapan, lorekkan

he answer space, shade the region whe answer space, shade the region w4, 3 and 1.x y x x 2 4, 3 and 1.x y x x

ketaksamaan 2 4, 3 1.y x y x dan x ketaksamaan 2 4, 3 1.y x y x dan x

Answer / Jawapan: Answer / Jawapan:

x O

y

– 4

– 2 2

– 4

6

2

– 2

y = 2x + 4

y = – x – 3

4

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5

2 te the value of p and of q that satisfy the following simultaneous r equations:

Hitungkan nilai p dan nilai q yang memuaskan persamaan linear serentak

4 8p q

Calculalinea

berikut:

3 2 17p q

[4 marks / markah] nswer / Jawapan:

A

3 olve the quadratic equation:

elesaikan persamaan kuadratik:

S

S

12 113

2

xx

x

[4 marks / markah]

Answer / Jawapan:

For Examiner’s

Use

Untuk Kegunaan

Pemeriksa

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For

Examiner’s Use

Kegunaan Pemeriksa

4 Diagram 4 shows a composite solid, formed by the combination of a cone and a right prism. Trapezium ABFE is the uniform cross-section of the prism. AB = BC = 8 cm and AE = 6 cm. The height of the cone is 10 cm and the diameter of its base is 7 cm.

.

sing

Untuk

Rajah 4 menunjukkan sebuah gabungan pepejal yang dibentuk daripada cantuman sebuah kon dan sebuah prisma tegak. Trapezium ABFE ialah keratan rentas seragam prisma itu. AB = BC = 8 cm dan AE = 6 cm. Tinggi kon ialah 10 cm dan diameter tapaknya ialah 7 cm

[4 marks / markah]

U22

π7

, calculate the volum 3, o e composite solid.

enggunakan

e, in cm f th

M22

π7

, hitung isipadu dalam cm3, cantuman pepejal itu.

Answer / Jawa

pan:

A

B

C

E

F

G

Diagram 4 Rajah 4

13 cm

H

D

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5 (a) State whether the following compound statement is true or false.

Nyatakan sama ada pernyataan berikut adalah benar atau palsu.

b) Write down two implications based on the following compound atement:

uliskan dua implikasi berdasarkan pernyataan berikut:

c) interior angles of a regular polygon of sides is 2 180n .

ake one conclus m of the interior angles f a regular pentagon.

isi ialah 180n .

tagon sekata.

[5 marks / markah]

Answ

) ..................................................................................................................

) Implication 1 / Implikasi 1:

.........................................................................................

c) Conclusion

For Examiner’s

Use


(st

T

( It is given that the sum of the n

M ion by deduction on the suo

Diberi bahawa jumlah sudut pedalaman sebuah poligon sekata dengan n s 2

Buat satu kesimpulan secara deduksi tentang jumlah sudut edalaman sebuah penp

er / Jawapan:

(a

(b

.........................

..................................................................................................................

Implication 2 / Implikasi 2:

..................................................................................................................

..................................................................................................................

(

/ Kesimpulan:

..................................................................................................................

..................................................................................................................

33 27 and 21 1x x x 33 27 dan 21 1x x x

P Q P if and only if P Q .

P Q P jika dan hanya jika P Q .

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For Examiner’s

Use


uation of

.

ind.

ari,

) the equation of the straight l

ersamaan garis lurus RS,

e x-intercept of the straight line RS.

[5 marks / markah]

6 Diagram 6 shows a rhombus PQRS drawn on a Cartesian plane. The eqstraight line PQ is y = 2x + 5.

Rajah 6 menunujukkan rombus PQRS yang dilukis pada suatu satah CartesanPersamaan garis lurus PQ ialah y = 2x + 5.

y

F

C

(a

x

R (8, 1) P

Q

S

O

Diagram 6 Rajah 6

ine RS,

p

(b) th

pintasan-x bagi garis lurus RS.

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Answer / Jawapan:

6 (a)

(b)

For Examiner’s

Use


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For Examiner’s

Use


7 POS and sector PTO with

sektor POS dan sektor PTO

t is g OP = 14 cm, SPO = 45 and OPT = 36.

= 14 cm, SPO dan OPT = 36.

sing

Dc

iagram 7 shows semicircle PQR with centre O, sector entre P.

Rajah 7 menunjukkan semi bulatan PQR berpusat O,berpusat P.

I

iven that

Diberi bahawa OP

P O

Q

T

S

R

Diagram 7 Rajah 7

= 45

U22

π7

, calculate

Menggunakan 22

π7

, hitung

(a) th eter, in cm, of t

e perim he whole diagram,

perimeter, dalam cm, seluruh rajah itu.

(b) the area, in cm2, of the shaded region.

[6 marks / markah]

luas, dalam cm2, kawasan yang berlorek.

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11

Answer / Jawapan:

7 (a)

(b)

For Examiner’s

Use


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Examiner’s


8

(a) Given that For

Use

4 2 3 2 11

0

3 5 4 0 1mk

, find the value of m and of k.

Diberi

4 2 3 2 1 01

3 5 4 0 1mk

, cari nilai m dan nilai k.

( llowing simultaneous linear

Tulis persamaan linear serentak berikut dalam bentuk persamaan matriks:

3x

5x + 4y = – 11

[6 marks / markah]

Answer /Jawapan:

(a

( )

b) Write the fo equation as matrix equation:

+ 2y = – 7

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

Seterusnya, menggunakan kaedah matriks, hitung nilai x dan nilai y.

)

b

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13

9

card is picked at random from box E and then a card is picked at random om box F.

atu kad dipilih rawak daripada kotak E dan k satu kad ula dipilih seca daripada kotak F.

y listing the sample space of all the possible outcomes of the event, find

n, cari kebarangkalian

,

gan huruf P dipilih.

[5 marks / markah] Answer

(b)

For Examiner’s

Use


Diagram 9 shows seven cards labelled with letters in box E and box F.

Rajah 9 menunjukkan tujuh kad huruf di dalam kotak E dan kotak F.

Afr

S secara ra rawak

emudian p

Bthe probability that

Dengan menyenaraikan ruang sampel bagi semua kesudahan peristiwa ang mungkiy

(a) both cards are labelled with same letters

kedua-dua kad dilabel dengan huruf yang sama,

(b) only one of the cards is labelled with the letter P are picked.

hanya sekeping kad dilabel den

/ Jawapan:

(a)

Box F

P M R U P S R

Box E Kotak E Kotak E

agram 9 ajah 9

DiR

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For Examiner’s

Use


10

a) State the uniform speed, in ms – 1, of the particle.

Nya an laju seragam, dala – 1, zarah itu.

b) Calculate the rate of change of speed, in ms – 2, of the particle in the first 14 seconds.

Hitungkan kadar perubahan laju, dalam ms – 2, zarah itu dalam 14 saat

c) travelled for the period

ilai t, jika jumlah jarak yang dilalui dalam tempoh t saat itu

[6 marks / markah]

Diagram 10 shows the speed–time graph of a particle for a period of t seconds.

Rajah 10 menunjukkan graf laju–masa bagi suatu zarah dalam tempoh t saat. (

tak m ms (

yang pertama.

( Calculate the value of t, if the total distanceof t seconds is 443 metres.

Hitungkan nialah 443 meter.

0 18 14

Speed (ms

5

–

32

10

1) – 1

Diagram 10 Rajah 10

Laju (ms )

t Time (s) Masa (s)

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Answer / Jawapan: 10 (a)

(b)

(c)

For Examiner’s

Use


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For Examiner’s

Use


11 VW and QP = 6 cm.

lah titik tengah bagi sisi VW

) Name the angle between the line PM and plane TUVW.

Namakan sudut di antara garis PM dengan satah TUVW.

) Calculate the angle between the line PM and the plane TUVW.

Hitung sudut di antara garis PM dengan satah TUVW. [4 marks / markah]

(a)

Diagram 11 shows a cuboid. M is the midpoint of the side

Rajah 11 menunjukkan sebuah kuboid. M iadan QP = 6 cm.

(a

(b

Answer / Jawapan:

(b)

P S

M

V U

Diagram 11 Rajah 11

Q R

W T

12 cm

4 cm

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17

2

Section B Bahagian B

[48 marks / markah]

Answer any four questions from this section empat soalan daripada bahagian ini.

(a) 2 in the an for the equation 22 5 4y x x1

Jawab mana-mana

Complete Table 1 swer space by writing down the values o 1 and x = 2.

Lengkapkan Jadual wapan bagi persamaan 22 5y x – 1 dan x = 2.

/ markah]

page 19.

Anda boleh meng

scale of 1 cm to 1 unit on the x

f y when x = –

12 di ruang ja4x dengan menulis nilai-nilai y apabila x =

[2 marks

(b) For this part of question, use the graph paper provided on You may use a flexible curve rule.

Untuk ceraian soalan ini, gunakan kertas graf yang disediakan pada halaman 19.

gunakan pembaris fleksibel.

By using a -axis and 1 cm to 2 units on the y-axis, draw the graph of 22 5 4y x x for 2 4x .

engan menggunakan skala 1 cmD kepada 1 unit pada paksi-x dan 1 cm kepada 2 unit pada paksi-y, lukiskan graf 22 5 4y x x untuk

2 4x . [4 marks / markah]

(c)

From your graph, find

Daripada graf anda, carikan

(i) the value of y when x = 2 5 ,

y

nilai apabila x = 2 5 ,

i) the values of x when y = 5.

nilai nilai x apabila[3 marks / markah]

d) x which 2 8 0x x

(i

- y = 5.

( Draw a suitable straight line on the graph in 12 (b) to find value of

satisfy the equation 2 7 for 2 4x .

State

the value of x.

Lukis satu garis lurus yang sesuai pada graf di 12 (b) untuk mencari nilai x yang memuaskan persamaan 22x 7 8 0x untuk

2 4x .

[3 marks / markah]

For Examiner’s

Use


Nyatakan nilai x itu.

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Answer / Jawapan: (a)

22 5 4y x x

32 4 x – 2 – 15 – 1 0 1 2

y – 14 – 8 4 7 – 05 – 8

Table 12 Jadual 12

(b) Ref grap pag

Rujuk graf di halaman 19.

er h on e 19.

(i) y = ………………………………….

(c)

(ii) x = ………………………………… , x = …………………………………

For Examiner’s

Use


(d)

x = ……………………………………………

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For Examiner’s

Use


Graph for Question 12 Graf untuk Soalan 12

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20

For xaminer’s

Use


13

(a) Transformation T is a translation6

@2009 SULIT

E

2

.

Transformation P is a refl e line 3yection in th .

Transformation R is a rotation of 180 about the centre (4, 1).

Penjelmaan T ialah translasi 6

2

.

Penjelmaan P ialah pantulan pada paksi 3y .

pada pusat (4, 1).

Find the coordinates of the image of point (1, 5) under the following

ut:

(ii)

[4 marks / markah]

(b)

13 menunjukkan tiga trapezium, ABCD, PQRS dan PTUV, dilukis a tu satah Cartesan.

Penjelmaan R ialah putaran 180

transformations:

ari koordinat imej bagi titik (1, 5) di bawah penjelmaan berikC

(i) P,

R,

(ii) PT.

Diagram 13 shows three trapeziums, ABCD, PQRS and PTUV, drawn on aian plane. Cartes

Rajah pad sua

T

R

Q

U

V

S

P

O

2

8

2 4

y

x – 2 – 4 – 6

6

4

Diagram 13 Rajah 13

– 2

6

B A

C

D

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21

(i) PTUV is the image of ABCD under a combined transformation MN.

PTUV ialah imej bagi ABCD di bawah gabungan penjelmaan MN.

Describe in full the transformation

Huraikan selengkapnya penjelmaan

N,

(ii)

(a)

(b) M. It is given that the trapezium PTUV represents a region of area 54 4 cm2.

2

akili luas 54 4

Calculate the area, in cm , of the region represented by the shaded region.

Diberi bahawa trapezium PTUV mew cm2. n n luas, dalam cm2, kawasan yang diwakili oleh rantau o

[8 marks / markah] nsw J

(a) (i)

(ii)

)

)

For Examiner’s

Use


Hituberl

gkarek.

A er / awapan:

(iii)

(b)

(i)

(a

(b

(ii)

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For Examiner’s

Use


14 The data in Diagram 14 shows the marks scored by 40 students in C -cu icular Evaluation.

Data dalam Rajah 14 menunjukkan markah bagi 40 orang pelajar dalam Penilaian Kokurikulum.

33 42 72 57 43 67 83 65 79 85

a

o rr

75 58 45 75 68 78 49 76 63 43

82 65 74 56 78 51 35 84 55 69

92 32 63 94 52 85 68 59 88 77

Diagram 14

ah

(a) (i) as n t ata m T 14 the on page 24.

erd rka ata , lengkapkan Jadual 1 ad an wapan yang disediakan halaman 24.

[4 rks arkah] (ii) Based on Table 14 in 1 e the estimated mean of the mark

scored by a student.

[3 marks / markah] (b) For t graph paper provided on page 25.

ntuk ceraian soalan ini, gunakan kertas graf yang disediakan di m

By u cm to 10 marks on the horizontal axis and 2 cm

eng n skala 2 cm kepada 10 markah pada paksi engufuk dan 2 cm kepada 1 pelajar pada paksi men

ah]

(c) marks for the Co-curricular Evaluation.

Raj 14

B ed o he d , co plete able in answer space provided

B asa n d itu 4 p a ru g ja

ma / m

4(a), calculat

Berdasarkan Jadual 14 di 14(a), hitung min anggaran markah bagi seorang murid.

his part of the question, use the

Uhala an 25.

ing the scale of 2

s to1 student on the vertical axis, draw a histogram based on Table 14.

an menggunakaDm cancang, lukis satu histogram berdasarkan Jadual 14.

[4 marks / mark

Based on a histogram in 14(b), state the number of students who scored more than 70

Berdasarkan histogram di 14(b), nyatakan bilangan murid yang mendapat markah lebih daripada 70 di dalam Penilaian Kokurikulum.

[1 mark / markah]

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BLANK PAGE HALAMAN KOSONG

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For Examiner’s

Use


Answer / Jawapan: (a) (i)

Marks Markah

Mid-point Titik tengah

Frequency Kekerapan

31 – 40 355

Table 14 adual 14

(ii)

(b) Refer graph on page 25.

Rujuk graf pada halaman 25.

)

J

(c

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For Examiner’s

Use


Graph for Question 14 Graf untuk Soalan 14

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Examiner’s Use


15 You are not allowed to use graph paper to answer this question.

Anda tidak dibenarkan menggunakan kertas graf untuk menjawab soalan ini.

(a) Diagram 15(i) shows a solid with rectangular base ADEK on horizontal plane. ABJK is trapezium. DC, EF and HL are vertical edges, LF = 2 cm. Rectangle CFLM is a horizontal plane and BMHJ is an inclined plane.

Rajah 15(i) menunjukkan sebuah pepejal dengan tapak segiempat tepat. ADEK terletak di atas tapak mengufuk. Permukaan ABJK ialah trapezium.

M h condong.

raw full scale, the plan of the solid.

ukis dengan skala penuh, pelan pepejal itu. [3 marks / markah]

For

Tepi DC, EF dan HL adalah tegak, LF = 2 cm. Segiempat tepat CFLialah satah mengufuk dan segiempat tepat BMHJ ialah sata

D

L

A

B

C

L

H K

J

3 cm

M

5 cm

6 cm

E

F

D

8 cm

Diagram 15(i) Rajah 15(i)

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Answ 15 (a

r ner’s

Use


er / Jawapan:

)

FoExami

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For Examiner’s

Use


(b) A half-cylinder solid of diameter GN = 6 cm is joined to the solid in diagram 15(ii) at the plane EFLHNR. The combined solid is shown in Diagram 15 (ii).

Sebuah pepejal berbentuk separuh silinder berdiameter GN = 6 cm dicantumkan kepada pepejal pada Rajah 15 (i) pada satah EFLHNR. Gabungan pepejal adalah seperti ditunjukkan pada Rajah 15 (ii).

raw full scale,

kis dengan skala penuh,

) the elevation of the combined cal plane parallel to DE as viewed from P.

dongakan gabungan pepejal itu pada satah mencancang yang selari dengan DE sebagaimana dilihat dari P.

[4 marks / markah]

of the combined solid on a vertical plane parallel to AD as viewed

epejal itu pada satah mencancang yang selari dengan

[5 marks / markah]

A

D

L

B

H

K

E

F

D

C

L

J

3 cm

M

5 cm

8 cm

Diagram 15(ii) Rajah 15(ii)

G

Q

P

N

R

u

(i solid on a verti

(ii) the elevationfrom Q.

dongakan gabungan pAD sebagaimana dilihat dari Q.

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Answe 15

For miner’s Use

Untuk

Kegunaan eriksa

r / Jawapan:

(b) (i), (ii)

Exa

Pem

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For Examiner’s

Use


16 P(75S, 82W), Q(75S, 18E), K and L are four points on the surface of the earth. K lies due north of P and LP is the diameter of the earth.

P(75S, 82B), Q(75S, 18T), K dan L ialah empat titik pada permukaan bumi. K berada di utara P dan LP ialah diameter bumi.

(a) Find the position of L.

Carikan kedudukan bagi L. [3 marks / markah]

(b) Given the distance from P to K measured along the meridian is

n ialah 6 300 batu

itung latitud bagi K.

(c) Calculate the shortest distance, in nautical miles, from K to L measured

itungkan jarak terpendek, dalam batu nautika, dari K ke L diukur

(d) rom Q and flew due west to P and then flew due speed for the whole flight wa

alculate the total time, in hours, taken for the whole flight.

dari Q arah ke barat ke P dan kemudian

itung jumlah masa, dalam jam, yang diambil bagi seluruh penerbangan itu.

6 300 nautical miles.

Calculate the latitude of K. Diberi jarak dari P ke K diukur sepanjang meridianautika.

H[3 marks / markah]

along the surface of the earth.

Hsepanjang permukaan bumi.

[2 marks / markah] An aeroplane took off fnorth to K. The average s 650 knots.

C Sebuah kapal terbang berlepasterbang arah ke utara ke K. Purata laju seluruh penerbangan kapal terbang itu ialah 650 knot.

H

[4 marks / markah]

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SULIT 1449/2


31

Answer / Jawapan: 16 (a)

(b)

(c)

(d)

END OF QUESTION PAPER KERTAS SOALAN TAMAT

For Examiner’s

Use


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SULIT 1449/2

1449/2 @2009 SULIT

32

INFORMATION FOR CANDIDATES MAKLUMAT UNTUK CALON

1. This question paper consists of two sections: Section A and Section B.

Kertas soalan ini mengandungi dua bahagian: Bahagian A dan Bahagian B.

2. Answer all questions in Section A and four questions from Section B. Jawab semua soalan d soalan dalam Bahagian B.

3. Write your answer in th ion paper. Tulis jawapan anda pada ruang yang disediakan dalam kertas soalan ini.

4. Show your working.Tunjukkan langkah-langk nda. Ini boleh membantu anda untuk mendapatkan markah.

5. hen write

ian tulis

. .

7. bagi setiap soalan dan ceraian soalan ditunjukkan dalam

urungan.

8. vided on pages 2 to 3.

9. ure mathematical can be used.

10.

11. ertas soalan ini kepada pengawas peperiksaan pada akhir peperiksaan.

alam Bahagian A dan empat

e spaces provided in the quest

It may help you to get marks. ah penting dalam kerja mengira a

If you wish to change your answer, cross out the answer that you have done. Tdown the new answer. Jika anda hendak menukar jawapan, batalkan jawapan yang telah dibuat. Kemudjawapan yang baru.

6. The diagrams in the questions provided are not drawn to scale unless statedRajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan The marks allocated for each question and sub-part of a question are shown in brackets. Markah yang diperuntukkan k A list of formulae is proSatu senarai rumus disediakan di halaman 2 hingga 3.

booklet of four-figABuku sifir matematik empat angka boleh digunakan.

ou may use a non-programmable scientific calculator. YAnda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram. Hand this question paper to the invigilator at the end of the examination. erahkan kS

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SULIT

1449/1 © 2009 SULIT

1449/1 (PP) Matematik Kertas 1 Peraturan Pemarkahan 10 September 2009

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA SEKOLAH MENENGAH

NEGERI KEDAH DARUL AMAN.

PEPERIKSAAN PERCUBAAN SPM 2009 MATHEMATICS

Kertas 1

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Peraturan pemarkahan ini mengandungi 2 halaman bercetak

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SULIT

1449/1 © 2009 SULIT


NEGERI KEDAH DARUL AMAN

PEPERIKSAAN PERCUBAAN SPM 2009 1449/1 MATHEMATICS ANSWER FOR PAPER 1

1 D 11 B 21 D 31 A

2 D 12 A 22 C 32 B

3 C 13 A 23 B 33 A

4 A 14 C 24 A 34 C

5 C 15 B 25 D 35 C

6 D 16 A 26 C 36 D

7 C 17 D 27 C 37 B

8 B 18 B 28 A 38 D

9 D 19 A 29 B 39 A

10 B 20 D 30 D 40 B

ANALISIS

A 10

B 10

C 9

D 11

NOTA: MARKAH PELAJAR = 100140

)21(

KK

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SULIT 1449/2(PP)

[Lihat sebelah 1449/2(PP) SULIT

1449/2 (PP) Matematik Kertas 2 Peraturan Pemarkahan 10 September 2009


NEGERI KEDAH DARUL AMAN.

PEPERIKSAAN PERCUBAAN SPM 2009 MATHEMATICS

Kertas 2

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Peraturan pemarkahan ini mengandungi 18 halaman bercetak

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SULIT 1449/2(PP)

1449/2(PP) SULIT

2

Section A [ 52 marks]

Question Solution and Mark Scheme Marks

1

The line x = 1 correctly drawn. The region correctly shaded. Note : Award P1 to shaded region bounded by any 2 correct lines. (check one vertex from any two correct lines)

K1

P2

3

O

y

x – 4

– 4

6

– 2 2

2

– 2

y = 2x + 4

y = – x – 3

4

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SULIT 1449/2(PP)


3


2

3 12 2p q 4 or 6 4 3p q 4 or equivalent 14 7q or 7 4p 2 or equivalent OR

8 4p q or 17 3

2

pq

or equivalent (K1)

14 7q or 14 84p or equivalent (K1) OR

2 4 81

3 1 17(1) 2 4 3

p

q

or equivalent (K2)

6p

1

2q or 0 5

Note :

1. 6

1

2

p

q

as final answer, award N1.

2.

2 41

3 1(1) 2 4 3

or equivalent seen, award (K1).

K1

K1

N1

N1

4

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SULIT 1449/2(PP)

1449/2(PP) SULIT

4


3

22 5 12x x 0

2 3 4x x 0 or equivalent

OR

25 5 4(2)( 12)

2(2)

(K1)

4x

31 5

2x or

Note : 1. Accept without ‘= 0’.

2. Accept three terms on the same side, in any order.

3. Accept 34

2x x

or 1 5 4x x with

3

1 5, 42

x or x for Kk2.

K1

K1

N1

N1

4

4

2

1 22 710

3 7 2

113 8 6 8

2

2

1 22 710

3 7 2

+ 113 8 6 8

2

1897

3 or

1632

3 or 632 3

Note :

1. Accept π for K mark. 2. Correct answer from incomplete working, award Kk2.

K1

K1

K1

N1

4

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SULIT 1449/2(PP)


5


False / Palsu P1

If , then P Q P P Q K1

5(a)

(b)

If , then P Q P Q P K1

(c) 5 2 180 or 3 180 K1

540 N1 5

Note : Accept answer without working for K1N1

6(a)

(b)

mPQ = 2

1

8

y

x

= *( )PQm or 1 = *( )PQm (8) + c or equivalent

2 15y x or equivalent.

0 2 15x

157 5

2x or

Note :

1. Attempt to substitute 0y in *( )a or equivalent award K1. 2. Accept answer without working from correct equation from

(a) for K1N1.

3. Do not accept 15

, 02

or

7 5 , 0 .

P1

K1

N1

K1

N1

5

7(a)

45 22

2360 7

14 or 180 22

2 1360 7

4

45 22

2 14360 7

+ 180 22

2360 7

14 + 14 + 14

83

K1

K1

N1

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SULIT 1449/2(PP)

1449/2(PP) SULIT

6


7(b)

180 22

14 14360 7

or 36 22

14 14360 7

or 45 22

14 14360 7

180 22

14 14360 7

36 22

14 14360 7

+ 45 22

14 14360 7

1617

5 or

2323

5 or 323 4

Note :

1. Accept π for K mark. 2. Correct answer from incomplete working, award Kk2.

K1

K1

N1

6

8(a)

(b)

2k

5m

Note :

4 21

5 33 4 5 2

seen, award P1

3 2 7

5 4 11

x

y

4 2 71

5 3 113 4 5 2

x

y

or 7

*11

x Inverse

y Matrix

3x

1y

Note :

1. Do not accept *

inverse 3 2

matrix 5 4

or . *

inverse 1 0

matrix 0 1

2. as final answer, award N1.

3

1

x

y

3. Do not accept any solution solved not using matrix method.

P1

P1

P1

K1

N1

N1

6

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SULIT 1449/2(PP)


7


9

(a)

(b)

{(U, P), (U, M), (U, R), (P, P), (P, M), (P, R), (S, P), (S, M), (S, R), (R, P), (R, M), (R, R)} {(P, P), (R, R)} 2

12 or

1

6

{(U, P), (P, M), (P, R), (S, P), (R, P)} 5

12

Note : 1. Accept any correct method for K1. 2. Accept answer without working for K1N1.

P1

K1

N1

K1

N1

5

10(a) 10

P1

(b) 32 5 5 32

14 0 0 14or

K1

27 12

1 115 15

or or 8 N1

Note : Accept answer without working for K1N1.

(c) 1 15 32 14 32 10 4 18 10 443

2 2t

or equivalent method

K2

28 N1 6

Note : 1

5 32 142 or 1

32 10 42 or 18 10t or

259 or 84 , award K1.

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SULIT 1449/2(PP)

1449/2(PP) SULIT

8


11(a)

(b)

PMT or TMP

2 2

12tanθ

3 4

or equivalent

67 38 or 67 22 '

Note : 2 23 4 or 5 seen, award K1.

P1

K2

N1

4

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SULIT 1449/2(PP)


9


12(a)

(b)

(c)(i)

(ii)

(d)

3

6

Note : K only meant for table value. Graph Axes drawn in correct direction, uniform scales in 2 4x and . 14 8y 7 points and *2 points correctly plotted or curve passes through these points for and 2 4x 14 8y . A smooth and continuous curve without any straight line and passes through all 9 correct points using the given scales for 2 4x and . 14 8y Note : 1. 7 or 8 points correctly plotted, award K1.

2. Ignore curve out of range.

3 8 4 2y

0 9 1 1x

1 4 1 6x

Identify equation 2y x 4 or equivalent. Straight line correctly drawn. 2y x 4

Value of x : 1 0 0 8x

Note : 1. Allow N marks if values of x shown on the graph. 2. Values x obtained by calculation, award N0.

K1

K1

P1

K2 (does not

depend on P)

N1

(depends on P and K)

P1

P1

P1

K1

K1

N1 (dep 2nd K1)

2

4

3

3

12

1. Allow P marks if value of x and y are shown on the graph.

2. Value of x and y obtained by calculation of from wrong graph award P0.

Note:

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SULIT 1449/2(PP) 10

1449/2(PP) SULIT

8

y

6

– 2 – 1 O 1

– 2

– 4

– 6

– 8

– 12

– 10

– 14

2 3 4 x

2

4

y = – 2x – 4

y = – 2 x2 + 5x + 4

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SULIT 1449/2(PP)


11


13(a)(i) (1, 1) P1

(ii) (7, –3) P1

(iii) (–5, –1) P2 4

Note : (–5, 7) seen, award P1.

(b)(i)(a) N: Rotation of 90 clockwise about centre (0, –1). P3

Note :

1. P2 : Rotation 90 clockwise or Rotation about centre (0, –1),

2. P1 : Rotation.

(b) M: Enlargement at centre (2, 1) with scale factor 2 P3

Note :

1. P2 : Enlargement at centre (2, 1) or Enlargement scale factor 2

2. P1 : Enlargement.

OR

N: Enlargement at centre (2, 1) with scale factor 2 (P3)

Note :

1. P2 : Enlargement at centre (2, 1) or Enlargement scale factor 2

2. P1 : Enlargement.

M: Rotation of 90 clockwise about centre (0, –1) (P3)

Note :

1. P2 : Rotation 90 clockwise or Rotation about centre (0, –1),

2. P1 : Rotation.

(ii) 2

54 454 4

*(2)

K1

40 8 N1 8

12

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SULIT 1449/2(PP)

1449/2(PP) SULIT

12


14(a)(i)

(ii) (b)

(c)

Marks : (II to VII)

Mid point : (II to VII)

Frequency : (I to VII)

Note : Allow one mistake in frequency for P1.

(35 5 3) (45 5 5) (55 5 7) (65 5 8) (75 5 9) (85 5 6) (95 5 2)

40

or 2630

40

263

4 or

365

4 or 65 75

Axes drawn in correct direction and uniform scale for 35 and 0 y 9. 5 95 5x *7 points correctly plotted Note : *5 or *6 points correctly plotted or bar passes through using at least 6 correct mid-point, award K1. Correct bar passes all 7 correct points for using given scales 35 . 5 95 5x

*17 from *histogram only.

P1

P1

P2

K2

N1

K1

K2

N1

N1

4

3

4

1

12

Marks Mid-point Frequency Markah Titik tengah Kekerapan

31 40 35 5 3 I

41 50 45 5 5 II

51 60 55 5 7 III

61 70 65 5 8 IV

71 80 75 5 9 V

81 90 85 5 6 VI

91 100 95 5 2 VII

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SULIT 1449/2(PP)


13

Frequency

9

1

8

7

6

5

4

3

2

0 45.5 35.5

55.5 65.5 75.5 85.5

95.5

30.5 40.5

Marks

50.5 60.5 70.5 80.5 90.5 100.5

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SULIT 1449/2(PP)

1449/2(PP) SULIT

14


15 Note :

(1) Accept drawing only (not sketch).

(2) Accept diagrams with wrong labels or without labels. (3) Accept correct rotation of diagrams. (4) Lateral inversions are not accepted. (5) If more than 3 diagrams are drawn, award mark to the correct ones only. (6) For extra lines (dotted or solid) except construction lines, no mark is awarded. (7) If other scales are used with accuracy of 0.2 cm one way, deduct 1 mark from the N mark obtained, for each part attempted. (8) Accept small gaps extensions at the corners. For each part attempted :

(i) If cm, deduct 1 mark from the N mark obtained. 0 4

(ii) If > cm, no N mark is awarded. 0 4 (9) If the construction lines cannot be differentiated from the actual lines: (i) Dotted line : If outside the diagram, award the N

mark. If inside the diagram, award N0. (ii) Solid line : If outside the diagram, award N0.

If inside the diagram, no mark is awarded.

(10) For double lines or non-collinear or bold lines, deduct 1 mark from the N mark obtained, for each part attempted.

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SULIT 1449/2(PP)


15


15(a)

Correct shape with rectangles BJFC and MHFC All solid lines. BC BM BJ > MC Measurement correct to 0 2 cm (one way) and all angles at vertices of rectangles = 90 1

K1

K1 dep K1

N1 dep K1K1

3

B J

M H

C F

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SULIT 1449/2(PP)

1449/2(PP) SULIT

16


15(b)(i)

Correct shape with the triangle GFC , rectangles CDEF and GEYX All solid lines

DY > YX > DE > EY = EF = FG = DC

Measurement correct to 0 2 cm (one way) and all angles at the vertices of rectangles = 90 1

K1

K1 dep K1

N2 dep K1K1

4

G

F C

D E

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SULIT 1449/2(PP)


17


15(b)(ii)

Correct shape with rectangles ADGJ, BMHJ and CGHM All solid lines Note : Ignore NR N and R joined with dashed line to form rectangle NRDG and

N lies between J and H

AD > DG = JH > JB = BA > HG Measurement correct to 0 2 cm (one way) and all angles at the vertices of rectangles = 90 1

K1

K1 dep K1

K1 dep K1K1

N2 dep K1K1K1

5

12

N G H J

M B C

R A D

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SULIT 1449/2(PP)

1449/2(PP) SULIT

18


16 (a)

(b)

(c)

(d)

(75N, 98E) // (75U, 98T) Note: 98 or E or T , award P1

6300

60 or 105

6300

7560

30N // 30U

180 30 75 60 or 75 60

4500

82 18 60 cos 75 or 100 60 cos 75

Note: 82 18 or 100 or cos 75 multiply NOT divide used

correctly, award K1

[ 82 18 60 cos 75]* 6300

650

12 hours 08 or 12 hours 5 minutes

END OF MARK SCHEME

P1P2

K1

K1

N1

K1

N1

K2

K1

N1

3

3

2

4

12

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