ADDITIONAL MATHEMATICS 3472/1 6 5 Find the range of values of x for which x x x (3 1)( 2) 8( 1) Cari...

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PEPERIKSAAN PERCUBAAN BERSAMASIJIL PELAJARAN MALAYSIA 2008

ANJURAN BERSAMA

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUASEKOLAH MENENGAH CAWANGAN NEGERI PERLIS

DAN JABATAN PELAJARAN PERLIS_____________________________________________________________

ADDITIONAL MATHEMATICSPaper 1Kertas 1

Two hoursDua jam

_____________________________________________________________

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. Tulis nama dan tingkatan anda pada ruangan yangdisediakan.

2. Kertas soalan ini adalah dalam dwibahasa.

3. Soalan dalam bahasa Inggeris mendahului soalanyang sepadan dalam bahasa Melayu.

4. Calon dibenarkan menjawab keseluruhan atausebahagian soalan sama ada dalam bahasa Inggerisatau bahasa Melayu.

5. Calon dikehendaki membaca maklumat di halamanbelakang kertas soalan ini.

Kertas soalan ini mengandungi 18 halaman bercetak

3472/1AdditionalMathematicsPaper 1September2008

2 Jam

NAMA : ………………………………………………………….

TINGKATAN : ………………………………………………….

Untuk Kegunaan PemeriksaSoalan Markah

PenuhMarkah

Diperoleh1 22 33 34 45 36 37 38 39 410 311 212 313 314 415 316 217 418 419 320 321 422 323 424 325 4

Jumlah 80

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

the ones commonly used.

Rumus-rumus berikut boleh membantu anda menjawab soalan. Simbol-simbol yang diberi

adalah yang biasa digunakan.

ALGEBRA

1a

acbbx

2

42

2 nmnm aaa

3 nmnm aaa

4 mnnm aa )(

5 nmmn aaa logloglog

6 nmn

maaa logloglog

7 mnm an

a loglog

CALCULUS

1dx

duv

dx

dvu

dx

dyuvy ,

22

,v

dx

dvu

dx

duv

dx

dy

v

uy

3dx

du

du

dy

dx

dy

4 Area under a curve

Luas di bawah lengkung

= b

adxy or (atau)

= b

adyx

5 Volume generated/Isipadu janaan

= dxyb

a2 or (atau)

= b

adyx2

8a

b

c

ca

log

loglog

9 dnaTn )1(

10 ])192[2

dnan

Sn

11 1 nn arT

12

1,1

1

1

)1(

r

r

ra

r

raS

nn

n

13 1,1

rr

aS

GEOMETRY

1 Distance/Jarak = 221

221 )()( yyxx

2 Midpoint/Titik tengah

(x, y) =

2,

22121 yyxx

3 A point dividing a segment of a line/Titik yangMembahagi suatu tembereng garis

(x, y) =

nm

myny

nm

mxnx 2121 ,

4 Area of triangle/ Luas segitiga

= )()(2

1312312133221 yxyxyxyxyxyx

5 22 yxr

622

ˆyx

jyixr


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STATISTICS/STATISTIK

1N

xx

2

f

fxx

3N

xx

2)( =

22

xN

x

4

f

xxf 2)( =

22

xf

fx

5 Cf

FNLm

m

2

1

6 1000

1 Q

QI

7

i

ii

W

IWI

8.)!1(

!

n

nPr

n

9.!)!1(

!

rn

nCr

n

10 )()()()( BAPBPAPBAP

11 1,)( qpqpCrXP rnrr

n

12 Mean/ Min, np

13 npq

14

XZ

TRIGONOMETRY/ TRIGONOMETRI

1 Arc length, rs Panjang lengkuk, js

2 Area of sector, 22

1rA

Luas sector, 22

1jL

3 1cossin 22 AA

4 AA 22 tan1sec

5 AAec 22 cot1cos

6 AAA cossin22sin

7 AAA 22 sincos2cos

= 1cos2 2 A

= A2sin21

8 BABABA sincoscossinsin

9 BABABA sinsincoscoscos

10 BA

BABA

tantan1

tantantan

11A

AA

2tan1

tan22tan

12C

c

B

b

A

a

sinsinsin

13 Abccba cos2222

14 Area of triangle/ Luas segitiga

= Cabsin2

1


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1 Diagram 1 shows the function 2)( xxf .

Rajah 1 menunjukkan fungsi 2)( xxf .

Diagram /Rajah1

(a) State the value of k,Nyatakan nilai k,

(b) By using the value of k in 1(a), rewrite the above relation in orderedpairs.Dengan menggunakan nilai k dalam 1(a), tulis semula hubungan diatas dalam pasangan bertertib. [2 marks /markah]

Answer/Jawapan :(a) k = ____________________

(b) ____________________________________________________________________________________________________

2 Given the function xxf 34: .

Diberi fungsi xxf 34: .

FindCari(a) )2(f ,

(b) the value of x which maps onto itself.

nilai bagi x yang memetakan kepada dirinya sendiri.[3 marks /markah]

Answer/Jawapan :(a) )2(f = _________________

(b) x =____________________

1 •

2 •

k •

6 •

• 1

• 4

• 16

• 36

x f(x)


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3. Given the function 32: xxf and 0,8

: xx

xg .

Diberi fungsi 32: xxf dan 0,8

: xx

xg .

Find )5(1gf .

Cari )5(1gf .

[3 marks/markah]

Answer/Jawapan : ______________________________________________________________________________________________________

4 (a) Solve the following quadratic equation :Selesaikan persamaan kuadratik berikut :

0532 2 xx

(b) The quadratic equation 0122 pxx , where p is a constant , has two

different roots.

Persamaan kuadratik 0122 pxx , di mana p ialah pemalar, mempunyai

dua punca berbeza.

Find the range of values of p.

Cari julat nilai p.[4 marks/4 markah]

Answer / Jawapan : (a) _____________________

(b) ____________________


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5 Find the range of values of x for which )1(8)2)(13( xxx

Cari julat nilai x bagi )1(8)2)(13( xxx

[3 marks/markah]

Answer / Jawapan : ___________________________________________________________________________________________________

6 The quadratic equation khxxf 2

)( , where h and k are constants, has a

minimum point ( 2 , 3 ).

Persamaan kuadratik khxxf 2

)( , di mana h dan k adalah pemalar ,

mempunyai titik minimum ( 2, 3 )

FindCari

(a) the value of h and of k

nilai h dan k

(b) the equation of the axis of symmetrypersamaan bagi paksi simetri

[3 marks/markah]

Answer/ jawapan : (a) h =___________________

k = __________________

(b) ______________________


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7 Solve 42 x =38

1x

Selesaikan 42 x =38

1x

[3 marks/markah]

Answer / Jawapan : ____________________________________________________________________________________________________

8 Solve 3 + log xx 22 log)1(

Selesaikan 3 + log xx 22 log)1(

[3 marks/markah]

Answer / Jawapan : ________________________


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9 Given rm 2log and log tn 2 , express log16

3

8

mnin terms of r and / or t .

Diberi rm 2log dan log tn 2 , ungkapkan log16

3

8

mndalam sebutan r dan/ atau t

[4 marks/markah]

Answer / Jawapan : ____________________________________________________________________________________________________

10 The fourth terms of an arithmetic progression is 5 and the sum of the first eight terms

of the progression is 12 .Sebutan keempat bagi suatu janjang aritmetik ialah 5 dan hasil tambah lapan sebutan

pertama janjang itu ialah 12 .

CalculateHitung

(a) the first term.sebutan pertama

(b) the common difference of the progression.beza sepunya janjang itu

[3 marks/markah]

Answer / Jawapan : (a) ______________________

(b) _____________________


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11. Given a geometric progression 1,2,4,……., 256.Calculate the number of terms of the progression.

Diberi suatu janjang geometri 1,2,4,…….., 256.Hitungkan bilangan sebutan bagi janjang itu.

[2 marks/markah]

Answer / Jawapan : _________________________________________________________________________________________________

11 The first three terms of a geometric progression are ,3m ,m 1m .Find

Tiga sebutan pertama suatu janjang geometri ialah ,3m ,m 1m . Carikan

(a) the value of mnilai m

(b) The common ratio of the progressionnisbah sepunya janjang itu.

[3 marks/markah]

Answer / Jawapan : (a) m = __________________

(b) ______________________


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13 Diagram 2 shows a straight line graph of xy against 2x .

Rajah 2 menunjukkan graf garis lurus xy melawan 2x .

Express y in terms of x .

Ungkapkan y dalam sebutan x.[3 marks/markah]

Answer / Jawapan : ________________________

(2, 4)

(6, 6)

2x

xy

O

Diagram/ Rajah 2


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14 Diagram 3 shows a semicircle ABC with centre O.Rajah 3 menunjukkan satu semibulatan ABC berpusat di O

The length of arc BC is 33 cm and the angle AOB is 0.95 radians.Panjang lengkok B ialah 33 cm dan sudut AOB ialah 0.95 radian.

FindCari(a) the length of OB

panjang OB

(b) the area of sector OAB.luas sektor OAB

[4 marks/markah][use / gunakan = 3.142]

Answer / Jawapan : (a)______________________

(b) ______________________

A

B

C O

0.95rad

Diagram/Rajah 3


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15 In Diagram 4 , the point O is the origin and the straight line AB is perpendicular to thestraight line BC.Rajah 4 menunjukkan titik o ialah titik asalan, dan garis lurus AB berserenjang dengangaris lurus BC

Find the equation of the straight line AB.Cari persamaan garis lurus AB

[3 marks/markah]

Answer / Jawapan : ________________________

y

xO

A

B (2, 5)

C (-1, 3)

Diagram/ Rajah 4


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16 Diagram 5 shows two vectors, PO and QO .

Rajah 5 menunjukkan dua vektor PO dan QO

It is given that jnimQP . Find the values of m and of n.

Diberi bahawa jnimQP . Cari nilai m dan nilai n.

[2 marks/markah]

Answer / Jawapan : m =_____________________

n =____________________________________________________________________________________________________

17. Given that O(0, 0), A( 3,4 ) and B )3,2( .

Diberi O(0, 0), A( 3,4 ) dan B )3,2( .

Find in terms of unit vectors i and j ,

Cari dalam sebutan vektor unit i dan j

(a) AB

(b) the unit vector in the direction of AB

vektor unit dalam arah AB

[4 marks/markah]

Answer / Jawapan : (a)______________________

(b) _____________________

y

P(-2, -5)

Q(6, 4)

x0

Diagram / Rajah 5


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18 Solve the equation 1sincos3 2 xx for 0o x 360o.

Selesaikan persamaan 1sincos3 2 xx bagi 0o x 360o.[4 marks/markah]

Answer / Jawapan : ____________________________________________________________________________________________________

19 Differentiate1

)31(2

2

x

xwith respect to x.

Bezakan1

)31(2

2

x

xterhadap x. [3 marks/markah]

Answer / Jawapan : ____________________________________________________________________________________________________

20 The surface area of a sphere increases at the rate of 15cm 2 s 1 .

Luas permukaan sebuah sfera bertambah dengan kadar 15cm 2 s 1 .

Find the rate of change of the radius, in cm s 1 at the instant when its radius is 10 cm

Cari kadar perubahan jejari sfera itu dalam cm s 1 pada ketika jejarinya ialah 10 cm.

[Surface area of sphere / Luas permukaan sfera = 24 r , use / guna = 3.142][3 marks/markah]

Answer / Jawapan : ________________________


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21 Given that 4

1

16)( dxxg . Find

Diberi 4

1

16)( dxxg . Cari

(a) 1

4

)( dxxg

(b) 4

1

)9)((8

1dxxg

[4 marks/markah]

Answer / Jawapan : (a)______________________

(b) _________________________________________________________________________________________________

22 A set of positive integers consists of 7, 2, 3, m, 5, 8, n and 13. The mean for this set ofintegers is 5.Satu set integer positif terdiri daripada 7, 2, 3, m, 5, 8, n dan 13. Min bagi set integer ituadalah 5.

(a) State a possible value of m and of n.Nyatakan satu nilai yang mungkin bagi m dan n.

(b) Find the mean for m and n.Cari min untuk m dan n.

[3 marks/markah]

Answer / Jawapan : (a) m = __________________

n = _________________

(b) _____________________


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23 A team of 5 students are to be chosen from a group of 8 boys and 6 girls.Satu pasukan yang terdiri daripada 5 orang pelajar dipilih daripada sekumpulan 8orang pelajar lelaki dan 6 orang pelajar perempuan

FindCari(a) the number of ways to form the team ,

bilangan cara pasukan itu dibentuk,

(b) the probability of forming a team of all boys.kebarangkalian pasukan yang dibentuk itu adalah terdiri daripada semua pelajarlelaki

[4 marks/markah]

Answer / Jawapan : (a) ____________________

(b) _________________________________________________________________________________________________

24 Diagram 6 shows 4 letters and 5 digits.Rajah 6 menunjukkan 4 huruf dan 5 digit.

Diagram 6 / Rajah 6

A code is to be formed using the letters and the digits in diagram 6. The code mustconsist of 2 letters followed by 4 digits.Satu kod dibentuk menggunakan huruf dan digit dalam rajah 6. Kod itu mestimempunyai 2 huruf diikuti dengan 4 digit.

Find the number of ways the code can be formed if no letter or digit is repeated in eachcode.Cari bilangan cara kod itu dapat dibentuk jika tiada huruf dan digit boleh diulang dalamsetiap kod.

[3 marks/ markah]

Answer / Jawapan : ________________________

A B C D 2 3 4 5 6


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25 The height of students in a school has a normal distribution with a mean of 148 cm and astandard deviation of 15 cm.

Ketinggian pelajar-pelajar di sebuah sekolah mempunyai taburan normal dengan min148cm dan sisihan piawai 15cm.

FindCari

(a) the Z score for the height of 156 cm,skor Z bagi ketinggian 156cm,

(b) the value of t when P( t < z < t) = 0.383.

nilai bagi t bila P( t < z < t ) = 0.383.[4 marks/ markah]

Answer / Jawapan: (a) ______________________

(b) t =____________________

END OF QUESTION PAPER


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INFORMATION FOR CANDIDATESMAKLUMAT UNTUK CALON

1. This question paper consists of 25 questions.Kertas soalan ini mengandungi 25 soalan.

2. Answer all questions.Jawab semua soalan.

3. Give only one answer for each question.Bagi setiap soalan beri satu jawapan sahaja.

4. Write your answer in the spaces provided in the question paper.

Jawapan anda hendaklah ditulis pada ruang yang disediakan dalam kertas soalan ini.

5. Show your working. It may help you to get marks.

Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini boleh membantuanda untuk mendapatkan markah.

6. If you wish to change your answer, cross out the answer that you have done. Then writedown the new answer.

Jika anda hendak menukar jawapan, batalkan dengan kemas jawapan yang telahdibuat. Kemudian tulis jawapan yang baru.

7. The diagrams in the questions provided are not drawn to scale unless stated.Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.

8. The marks allocated for each question are shown in brackets.Markah yang diperuntukkan bagi setiap soalan ditunjukkan dalam kurungan.

9. A list of formulae is provided on pages 2 and 3.Satu senarai rumus disediakan di halaman 2 dan 3.

10. The normal distribution table is provided.Sifir taburan normal disediakan.

11. You may use a non-programmable scientific calculator.

Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram.

10. Hand in this question paper to the invigilator at the end of the examination.

Serahkan kertas soalan ini kepada pengawas peperiksaan pada akhir peperiksaan.


Mark Scheme For Additional Mathematics Paper 1

Trial Examination (SPM) 2008 Question Distribution Of Marks Sub Marks Total Marks

1.

(a) 4 (b) (1,1) (2,4) (4,16) (6,36)

1 1

2

2.

(a) 2−(b) 1 B1 : xx =− 34

1 2

3

3.

2

B2 : 48

B1 : 2

3)(1 +=− xxf or 4)5(1 =−f

3

4.

(a) 1,25−

(b) > p 2−B1 : > 0 )1)(1(4)2( 2 p−−−−

2

2

4

5.

32

−≤x , 5≥x

B2 : or equivalent

B1 : 0)5)(23( ≥−+ xx

3

6.

(a) ,2−=h 3=k(b) 2=x

1, 1

1

3

7.

2−B2 : 932

4 −−=− xx or equivalent B1 : 2

4−x or 3( )3+x

3

32− 5


8.

78

B2 : or xx 23

2 log)1(2log =− 132 −= x

x

B1 : log or 32 2 12log −x

x

3

9.

3

43 −+ tr

B3 : 2log3 2

r + 2log33

2

t - 2log32log4

2

2

B2 : 8log16log

8loglog

8loglog

2

2

2

32

2

2 −+ nm

B1 : 16logloglog 83

88 −+ nm

4

10.

(a) 26−(b) 7 B1 : or da 35 +=− )72(412 da +=−

1 2

3

11.

9 B1 : or 1,2,4,8,16,32,64,128,256. 18 22 −= n

2

12.

(a) 23

(b) 31

B1 : 32

323

+ or equivalent (substitute m to find r)

1

2

3

13.

x

xy 32+=

B2 : 3221 += xxy

B1 : gradient 21=m or y-intercept = 3

3

14.

(a) 15.06 cm B1 : 95.0−π or π×−

0

00

18043.54180 or 2.192

(b) 248.5 cm2

B1 : 192.206.15* 221 ×× * follow his (a) answer

2

2

4


15. +−= xy or equivalent

B2 :

1632

)2(23 −x or equ5 −=−y ivalent using

gradient = cmxy += B1 : 2

3−

3

16.

1, 1

,8−=m 9−=n 2

2

2

4

17.

)

(b)

(a ji 62 + B1 : jiji 4652 −−−−

103 ji +

or equivalent

B1 : 4062 ji+

18. ) or 90 90 or 990

2sin

(a 000 19.318,81.221,90 '0'00 11318,49221, B3 : 00 441,00 81.41, '

B2 : )(sin3( 0)1− =+x or x ,sin 32−=x 1

B1 :

1sin)sin1(3 2 =+− xx

4

19.

22 )1(4+xx

B2 : 22

22

)1()2)(31()6)(1(

++−+

xxxxx

for differentiating or for

3

B1 : x6 231 x+ x2 differentiating

12 +x

20.

π16

3 or 0.05968 cm s-1

B2 : =dtdr 15)10(8

1 ×π

B1 : 15=dt

dA

3

21. )

(a 16−

(b) 836 or 6.375

B2 : [ ]4189)16( x+ 8

1

1

3

4


B1 : ∫∫ +4

189

4

181 )( dxxg

22. )

1, 1

(a ,1=m 1=n (b) 1 1

3

23.

279 or

(a) 2002 B1 : 5

14C (b) 0.0 7 2002

56

B1 : 58C

2

2

4

24. 440

0 x 12 x 2 or 4 x 3 x 5 x 4 x 3 x 2 or

P or or 12 x …….. or …….x 5!

3

1B2 : 6 4

52

4 PP × B1 : 4

45P2

1

3

4

25. (a) 0.5333

) 0.5 2 : or

(b 3085.0)( => tzP 3085.0)( =−< tzP B

B1 : 0.1915

0.3085

t - t


ANSWER SCHEME TRIAL 08 PAPER 2 ADD MATHS SECTION A

1. 2

525 xyoryx −=−= 1M

12

522

5

1)25(2)25(

22

22

=⎟⎠⎞

⎜⎝⎛ −

−⎟⎠⎞

⎜⎝⎛ −

+

=−−+−

xxxx

oryyyy

1M

( )( )

)3(2)7)(3(4)10()10(

0173

)3(2)8)(3(4)10()10(

0)2)(43(

2

2

−−±−−=

=−−

−−±−−=

=−−

x

orxxor

y

oryy

1M

2,34

== yy 1M

1,37

== xx 1M

2. (a) 1M )50)(16(2006 −+=T = 450 1M

(b) [ ])50)(130()200(22

3030 −+=S

1M = 27 750 1M

(c) [ ] 12350)50)(1()200(22

=−+ nn

1M n = 19 1M Depth = 19 x 3 = 57 1M

3. (a) 22 −+= xpxdxdy 1M

1M 82)2()2( 2 =−+p p = 2 1M

(b) 81

−=m 1M

)2(815 −−=− xy 1M

421

81

+−= xy 1M

or equivalent

4.(a) (i) 55)2(21 2 =−θπr and

22)2( =−θπr 1M Solve simultaneous eqn. 1M r = 5 1M (ii) =θ 1.884 or 1.883 or 1.88 or 107.93o or 107.89o or 107.7o 1M (b) AP = 10 or OP = 15 1M

)884.1)(5(21)884.1()15(

21 22 or 1M

)884.1)(5(21)884.1()15(

21 22 − 1M

= 188.4 or 188.3 or 188 1M


ADDITIONAL MATHEMATICS 3472/1 6 5 Find the range of values of x for which x x x (3 1)( 2) 8( 1) Cari...

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Transcript of ADDITIONAL MATHEMATICS 3472/1 6 5 Find the range of values of x for which x x x (3 1)( 2) 8( 1) Cari...