# Addmaths Spm(Mpt 60 to a+)

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SULIT3472/1 2012 Maths Catch Network www.maths-catch.comMODUL PERSEDIAAN 2 ADD MATHS [From 60 to A+] Hak Cipta Terpelihara

USAHA +DOA+TAWAKAL

FOKUS A+SPM

2012

1MATHS CATCHNETWORK

MATHS Catch

MODUL PERSEDIAAN TERAKHIR

SPM 2012

CONTENT/ISI KANDUNGANBAHAGIAN 1A. Preview Modul Persediaan Matematik Tambahan SPM (MPT SPM) 2B. Preview Keseluruhan tentang Matematik Tambahan SPM Kertas 1 dan Kertas 2, 2012 3BAHAGIAN 2

A. Tips Peperiksaan Terakhir Kertas 1 / Last exam Tips PAPER 1 5B. Cadangan Soalan Ramalan Pilihan Kertas 1 / Suggestion Question PAPER 1 23C. Tips Peperiksaan Terakhir Kertas 2/ Last Exam Tips PAPER 2 40D. Cadangan Soalan Ramalan Pilihan Kertas 2 / Suggestion Question PAPER 2 51

IMPROVE from 60 TO SCORE A+

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SULIT3472/1 2012 Maths Catch Network www.maths-catch.comMODUL PERSEDIAAN 2 ADD MATHS [From 60 to A+] Hak Cipta Terpelihara

USAHA +DOA+TAWAKAL

FOKUS A+SPM

2012

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MATHS Catch

LAST EXAM TIPS PAPER 12012

[Improve From 60 to A+]

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SULIT3472/1 2012 Maths Catch Network www.maths-catch.comMODUL PERSEDIAAN 2 ADD MATHS [From 60 to A+] Hak Cipta Terpelihara

USAHA +DOA+TAWAKAL

FOKUS A+SPM

2012

3MATHS CATCHNETWORK

MATHS Catch

QUESTION SPM08 SPM09 SPM10 SPM11

Q1 Function Function Function Function

Q2 Function Function Function Function

Q3 Function Function Function Function

Q7 Indices & Logarithms Indices & Logarithms Indices & Logarithms Indices & Logarithms

Q8 Indices & Logarithms Indices & Logarithms Indices & Logarithms Indices & Logarithms

Q9 Progression Progression Progression Progression

Q10 Progression Progression Progression Progression

Q11 Progression Progression Progression Progression

Q12 Linear Law Circular Measure Linear Law Linear Law

Q13 Coordinate Geometry Vectors Coordinate Geometry Coordinate Geometry

Q14 Coordinate Geometry Vectors Coordinate Geometry Coordinate Geometry

Q15 Vectors Coordinate Geometry Vectors Vectors

Q16 Vectors Trigonometric Function Vectors VectorsQ17 Trigonometric Function Trigonometric Function Circular Measure Circular Measure

Q18 Circular Measure Integration Trigonometric Function Trigonometric Function

Q19 Differentiation Differentiation Differentiation Differentiation

Q20 Differentiation Differentiation Differentiation Differentiation

Q21 integration Integration Differentiation Differentiation

Q22 Statistics Permuatation&Combination Statistics Statistics

Q23 Permuatation&Combination Probability Permuatation&Combination Permuatation&Combination

Q24 Probability Statistics Probability Probability

Q25 Probability Distribution Probability Distribution Probability Distribution Probability Distribution

MATHS SPMJUMLAH SOALAN : 25

JUMLAH MARKAH : 80

Analisis Soalan Peperiksaan

Sebenar SPM 2008-2011 mengikutsusunan soalan dan topik

PERHATIAN:Sila fahamkan format Soalan kertas 1 ini sebaik mungkin.Inilah Kunci Kejayaan Sebenar untuk menjawab dengan baik kertas Matematik Tambahan Ini

TAHUKAH ANDA!Kertas 1 Matematik Tambahan mengandungi 25 soalan dan membawa markah sebanyak 80%. Kertas 1 ini juga dikenali sebagai lubuk emas untuk mengorek markah semaksimum yang mungkindalam matematik tambahan.Ini kerana Soalan-soalan yang ditanya adalah asas-asas sahaja jika nak dibandingkan dengan Kertas 2.Jika anda selalu gagal dalam peperiksaan apa kata tumpukan pada

kertas 1 ini sebagai persediaan terakhir sebelum memasuki dewan peperiksaan sebenar nanti..Disinilah perbezaan gred akan berlaku sama ada anda akan dapat A,B,C atau D..anda masih bolehmeningkat dengan mendadak dari D ke B atau dari C ke A,dan boleh juga meningkat dari D ke A jika kena dengan caranya.

Satu masalah besar pelajar adalah TIDAK TAHU NAK BERMULA ,apabila ingin menjawab sesuatu soalan matematik tambahan.Punca utama adalah kerana pelajar-pelajar ini gagal

mengenalpasti NAMA TAJUK dan gagal MENANGKAP KATAKUNCI yang terdapat didalam soalan menyebabkan pelajar tidak tahu apakah formula yang sesuai digunakan dan bagaimananak bermula. Oleh itu modul ini disusun bagi membantu pelajar mengatasi masalah utama ini

tajuk-tajuk dari Tingkatan 4.*

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USAHA +DOA+TAWAKAL

FOKUS A+SPM

2012

4MATHS CATCHNETWORK

MATHS Catch

Jika anda lihat stat istik bagi kertas 1 diatas,boleh dikatakan tajuk-tajuk yang keluar adalah sama sahaja setiap tahun.Walaubagaimanapun ini tidak bermakna Soalan yang ditanya

adalah sama.Soalan yang ditanya Kadang-kadang bagi sesuatu tajuk kecil soalanya mempunyai sampai 2-3 bentuk Soalan..Disebabkan modul ini dirangka khas untuk membantupelajar yang lemah maka MC buat andaian,pelajar terbabit sudah pasti tidak tahu apakah soalan yang wajar didahulukan dan yang mana wajar dikemudiankan.

Dicadangkan anda membaca segala tips yang diberikan sekurang-kurangnya dua kali sebelum peperiksaan bermula.Mudah-mudahan segala tips yang disampaikan dapat

,menyuntik semangat untuk anda terus mengulangkaji sehinggalah disaat yang terakhir.Tidak ada yang mustahil j ika anda berjaya mengulangkaji secara tersusun apa yang telah

dibekalkan ,kami yakin anda mampu mengubah gred pencapaian Matematik tambahan dari E ke C dan dari C ke A+.

Secara Umumnya Bab yang akan keluar didalam Kertas 1 Matematik Tambahan mengandungi 16 bab kesemuanya daripada 21 bab bagi tingkatan 4 dan 5. Berikut merupakan

tajuk-tajuk yang akan Keluar didalam kertas 1 mengikut FORMAT peperiksaan sebenar SPM

Berdasarkan kajian dan pemerhatian Kebiasaanya Format Kertas 1 untuk SPM tajuknya adalah seperti berikut:

SUSUNAN SOALAN DAN BAB KEBIASAAN DALAM FORMAT KERTAS 1 SPM

QUESTION CHAPTER MARKS

Q1 Function 3

Q2 Function 4

Q3 Function 3

Q7 Indices & Logarithms 3

Q8 Indices & Logarithms 3

Q9 Progression 2

Q10 Progression 3

Q11 Progression 2

Q12 Linear Law 4

Q13 Coordinate Geometry 4

Q14 Coordinate Geometry 3

Q15 Vectors 2

Q16 Vectors 4

Q17 Trigonometric Function 3Q18 Circular Measure 4

Q19 Differentiation 3

Q20 Differentiation 4

Q21 integration 3

Q22 Statistics 3

Q23 Permuatation&Combination 4

Q24 Probability 4

Q25 Probability Distribution 4

TOTAL 80

Untuk membantu pelajar memahami dengan lebih baik tajuk,bentukdan format soalan,maka MC sudah sediakan modul dinamakan Last

Exam Tips [PAPER 1]TOLONG JANGAN ABAIKANSEMUA

SOALAN DAN KONSEP yang diterangkan didalam modulini..Semoga modul 2012 ini membantu ANDA SEMUA.selamat

Berjaya.

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USAHA +DOA+TAWAKAL

FOKUS A+SPM

2012

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MATHS Catch

Kebiasaanya Q1-Q3 adalah dari tajuk function.Untuk menjawab soalan ini anda WAJIB tahu 4 perkara ini

a) Function Relationb) Absolute Functionc) Composite Functiond) Inverse Function

CASE 1:FUNCTION RELATION/FUNCTION NOTAION

Domain = {4, 9, 16}

Codomain = {2, 3, 4, 5}Object = 4,9,16

Image = 2, 3, 4

Range = {2, 3, 4} *Image yang mempunyai objek sahaja*

Relation between set A and B? = xxf )(

EXAM TIPSUntuk (b) Sangat penting.jika anda mahu

hilangkan modulus l l maka Jawapanyamestilah dipecah kepada dua iaitu (+) dan (-).

CASE 2:ABSOLUTE FUNCTION

CASE 3: INVERSE FUNCTION

Remarks:Soalansubtopic Inverse function inibelum pernah tak keluar.Oleh itu sila berikan fokus

yang lebih bagi subtopic ini

Exam Tips 3Langkah 1:tambahkan sendiri y [penting]

Langkah 2 :terbalikkan kedudukan y dan x.ini

bertujuan untuk menghilangkan1

Langkah 3:Cari nilai y.maka nila y yanganda perolehi itulah inverse function

EXAM TIPS: QUESTION 1-3 FUNCTIONS

CASE 4: COMPOSITE FUNCTION

The functions offand g are defined asf:xx 4andfg :x 4x + 3. Find the function g.

[4 marks]

Givenf(x) =x 4 andfg(x) = 4x + 3.fg(x) =f[g(x)]

= g(x) 4g(x) 4 = 4x + 3g(x) = 4x + 7

g :x 4x + 7

EXAM TIPS

Soalan ini sangat penting. Ideanya untukmendapatkan nilai g pelajar perlulah

membuat pemecahan nilai yang

besar.iaitu fg(x) dan bukanya f(x)

FOKUS

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MATHS Catch

EXAM TIPS: QUESTION 4 QUADRATIC EQUATIONS

CASE 1: Given one roots

Jawapan

CASE 2: Given two roots

Given -3 and are the roots of an equations 4x2 + bx + c = 0 .Find the

value of b and c

Jawapan

03114

03124

04

33

4

1

4

13

2

2

2

xx

xxx

xxx

xx

Compare with 4x2 + bx + c = 0

Thus b = 11 , c = -3

4

1

CASE 3: Given two EQUALS roots

Jawapan

one of the roots.

Roots bermaksud nilaix tersebut.

Diberix = -2 .gantikan dalam

persamaan diberi.maka anda akan

dapat nilai p

Apabila diberikan 2 roots

Roots bermaksud nilaix tersebut.

Dalam soalan ini ada 2 nilai roots.

4

1

3

x

x

Langkah 1Langkah yang perlu anda lakukan adalah

pindahkan -3 ke sebelah kiri begitu juga nilai

pindahkan juga ke sebelah kiri

Langkah 2

Bandingkan persamaan yang anda perolehi

dengan yang diberi didalam soalan.

Two equal roots

LANGKAH 1

-Tukarkan quadratic equation diberidalam bentuk (general form)

LANGKAH 2

Kenal pasti formula yang perlu

digunakan.ada 3 sila buatpilihan.contoh disebelah katakuncinya

Lihat dibawah

Quadratic EquationsTwo real roots b2-4ac > 0

Two equal roots b

2

-4ac = 0No real roots b2-4ac < 0

**Buat pilihan menggunakan 3

persamaan ini**

CASE 4: STRAIGHT LINE+ CURVE

(SIMULTANEOUS EQUATION)

The straight liney = 7x + 6 does notintersect with the curvey= x2 + 9x + n.Find the range of values ofn.

[3 marks]

y= x2 + 9x + nx2 + 9x + n = 7x + 6x2 + 2x + n 6 = 0The equation does not intersectb2 4ac < 0(2)2 4(1)(n 6) < 04 + 4n 24 < 01 + n 6 < 0n < 5

-Does not intersect bermaksud b2 4ac < 0-diberi Straight line dan curve equation

bermaksud penyelsaianya mestilah

menggunakan simultaneous equation

LANGKAH 1Gunakan kaedah simultaneous equatiuon dengan

cara samakan kedua-dua nilai y tersebut

LANGKAH 2Pindahkan semua nilai supaya disebelah

kanann a bersamaan = 0LANGKAH 3Kenal pasti formula yang perlu

digunakan.ada 3 sila buat pil ihan.contoh

disebelah katakuncinya adalah does notintersect = no real roots.lain-lain sila

Lihat dibawah

Two real roots b2-4ac > 0

Two equal roots b2-4ac = 0

No real roots b2-4ac < 0

FOKUS

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2012

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EXAM TIPS: QUESTION 5-6 QUADRATIC FUNCTIONS

CASE 1: QF and Graphs

Jawapan

2).(

3).(

2).(

3)(2)( 2

xc

nb

ma

mxxf

y min

x min

CASE 2: inequalities

Solve inequality 452 xx

EXAM TIPS

LANGKAH 1Sila hafal general form quadratic function ini.

nmxaxf 2

)()(

--kemudian bandingkan dengan equation yang diberi

LANGKAH 2SANGAT PENTING sila bandingkan

3)(2)( 2 mxxf

nmxaxf 2)()(

Jawapan (a)

Oleh itu 3n .(ingat n adalah y min)

Jawapan (b)

Untuk nilai m .tandanya anda perlulah mengambil

berlawanan dari yang diberi.jika .dalam kes diatas x

min -2.maka Jawapan m = 2 sahaja.

Jawapan (c)

untuk Soalan c axis symmetry adalah garisan yang

merentasi paksi x.dan pastikan anda meletakx=-2 dan JANGAN letak =-2 sahaja.perkataan x itu

2

y max atau min

1

x max atau min

EXAM TIPS -Sangat Penting Tajuk Ini!

LANGKAH 1-Jadikan equation yang diberi dalam bentuk = 0-dapatkan 2 nilai x tersebut

LANGKAH 2Lorekkan graphs bahagian dalam kerana dalam soalanmenunjukkan < (less)

Quadratic besar dari 0 [ > 0 ]

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LOGARITHMS

CASE 3 : BASE DIBERI ADALAH SAMA (Guna Teknik Factorise)

Solve the equation log5 (7x + 2) = log5 (3x + 4) + 1. [3 marks]

log5 (7x+ 2) log5 (3x + 4) = 1

log57x + 2

3x + 4= 1

7x + 2

3x + 4= 5

7x + 2 = 5(3x + 4)

7x + 2 = 15x + 20

x=

9

4

CASE 4 : BASE DIBERI TIDAK SAMA (Guna Teknik Samakan base + Factorise)

Given that log25s log5t= 0, express s in terms oft. [3 marks]

log25s log5t= 0log25s = log5t

log5s

log5 25= log5t

log5s

2 log5 5= log5t

log5s = 2 log5t

log5s = log5t2

s = t2

EXAM TIPS

Dalam Soalan case ini base nya adalah sama iaitu

5..Pelajar hanya perlu fikir bagaimana cara untukfactorise kanya sahaja.

Langkah 1

Pindahkan log5 (3x + 4) ke sebelah kiri

Langkah 2

Factorisekan kedua-dua log tersebut menggunakan

Laws of log iaitu

y

xLogyLogxLog aaa

Langkah 3Pindahkan log5 ke sebelah kanan bersamaan 5

1

Laws of indices

c

a

ay

x

cy

xLog

EXAM TIPSSoalan case ini base nya TIDAK SAMA.Langkah

pertama mestilah pelajar perlu fikir untuk samakan

base nya dahulu kemudian barulah guna Teknik

factorise untuk menyelesaikanya.

LOGARITHMSCASE 5 : INDICES KEPADA LOG

Solve the equation 85x + 3 = 74x.

85x + 3 = 74x

log10 85x + 3 = log10 7

4x

(5x + 3) log10 8 = 4x log10 75x + 3

4x=

log10 7

log10 8

5x + 34x

= 0.9358

5x + 3 = 3.7432x(5 3.7432)x= 3

x = 38.7432

= 0.3431

EXAM TIPS

Teknik diguna apabila nilai disebelah kiri dankanan tiada factor sepunya atau tiada sifir yang

sesuai digunakan.

**Kiri (8) dan kanan (7)

Cara penyelesainya.pelajar perlulah tambahkan

log disebelah kiri dan kanan

FOKUS

LOGARITHMS (Mirip Trial Selangor 2012)

Solve the equation6

13.2 xx

1

7782.0

7782.0

7782.07782.0

7782.0)47712.0()30103.0(

7782.03log2log

61log3log2log

6

1log)3.2log(

x

x

xx

xx

xx

xx

FOKUS

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USAHA +DOA+TAWAKAL

FOKUS A+SPM

2012

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MATHS Catch

EXAM TIPS: QUESTION 9-11 PROGRESSION

CASE 1: A.P & G.P (Value x)

A.P

formula GP sahaja.

G.P (Jawapan a)Langkah 1

Jika anda susun nilai yang diberi ianya akan

20x , 4x , 20x

1T 2T 3T

Langkah 2

2

3

1

2

T

T

T

T

*Dari cara ini andaHANYA akan perolehi

nilai x sahaja.*

Langkah 3

Gunakan formula ini untuk dapatkan ,r

1

2

T

Tr

Jawapan (b) : Gunakan formula sahaja

EXAM TIPSA.P

Langkah 1 :Jika anda susun nilai yang diberi

x7 , x2 , x

13

1T 2T 3T

Langkah 2

2312 TTTTd

*Dari cara ini andaHANYA akan perolehi

nilai x sahaja.*

Langkah 3

Gunakan formula ini untuk dapatkan ,d

12 TTd

CASE 2: A.P & G.P (Sum to infinity)

EXAM TIPSSelain soalan dari case 1. Case 2 ini juga

sangat popular teruatamanya dalam

percubaan negeri SPM 2011.

***Per11,SBP11,Ter11,Phg11,Negeri 9,Sabah 11***

Jawapan (a)

Formula G.P1 nn arT

Jawapan (b)

Gunakan formula sum to infinity.

RINGKASAN

PELAJAR WAJIB TAHU MENGGUNAKAN FORMULA DIBAWAH

A.P

(1) x)nilaimencariuntuk(gunaatau 231212 TTTTdTTd

(2) dnaTn )1( *digunakan apabila sesuatu soalan menyebut tentang

-sesuatunth term (cthnya Given nth term,How many number between..,Find number of term.)

(3) ])1(2[2

dnan

Sn *digunakan apabila sesuatu soalan menyebut tentang Sum of term,

G.P

(1)x)nilaimencariuntuk(gunaatau

2

3

1

2

1

2

T

T

T

T

T

Tr

(2) 1 nn arT ***digunakan sama seperti A.P

(3) 1rnilaisyaratnya1

)1(

r

raS

n

n

(4) 1rnilaisyaratnya1

)1(

r

raS

n

n

(5)

r

a

1

S **digunakan apabila sesuatu soalan menyebut tentang

- Sum to infinity

-angka dalam bentuk decimal [cthnya 0.575757] penyelesaianya 0.57 , 0.0057 , 0.000057- contoh 2 : [ 1.12121212 ] penyelesaianya 1 + , 0.12 , 0.0012 , 0.000012

r

a

1

1S **Kedah Trial 11**

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EXAM TIPS: QUESTION 12 LINEAR LAW

CASE 1: xyLog 1010 logagaints EXAM TIPSPelajar perlulah tahu persamaan umum

linear law iaitu cmxy Kebiasaanya Soalan yang akan disoal adalahexpress equationdan find coordinate

sahaja.

Trick penyelesaian untuk SEMUA jenis soalan

Langkah 1Fikir bagaimana cara nak tukar equation

y seperti didalam graphs

yang diberi disebelah kiri dan kanan

Langkah 2Gunakan konsep log iaitu

nmmn aaa logloglog

Langkah 3Jadikan dalam bentuk y = mx + c

Untuk memudahkan senaraikan sepertidibawah

y =

x =

m =c =

dari cth disebelah

erceptyc

xx

yy

int

x-x

y-y

12

12

10

10

***Trial Kedah11Kelantan11, Selangor11,

Negeri sembilan11*****

CASE 2: xx

yagaints

EXAM TIPS

Trick penyelesaianADALAH SAMAsahaja seperti dalam case 1

Langkah 1

Fikir bagaimana cara nak tukar equationy dari maklumatdiberi KEPADA-

y seperti didalam graphs

Equation dalam maklumat tiada x dibawahy.oleh itu anda perlulah tambahkan x

dibawahnya supaya sama dengan graphs

Langkah 2Jadikan dalam bentuk y = mx + c

Untuk memudahkan senaraikan seperti

dibawah

y =

x =m =

c =

dari contoh disebelah

erceptyc

xx

x

yy

int

x-x

y-y

12

12

***Trial Terengganu11 Pahang11,

Sarawak11, Melaka11*****

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USAHA +DOA+TAWAKAL

FOKUS A+SPM

2012

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EXAM TIPS: QUESTION 13-14 COORDINATE GEOMETRY

Berdasarkan ANALISIS 13 soalan Percubaan Negeri SPM 2012 menunjukkan tajuk ini

BENTUK 1

-perpendicular

-parallel-Area

BENTUK 2

-collinear-ratio

-Distance/locus

CASE 1: PERPENDICULAR

Langkah 1Tuliskan formula yang ingin diguna

121 mm

*Kebiasanya 1 gradient diberi (1m ) dan

satu lagi perlu dicari ( 2m )*

Langkah 2

m melalui equation

yang diberikan

cmxy

xy

denganBandingkan

2

3

1

Maka3

11 m

Langkah 3

Dapatkan 2m melalui formula

3

13

1

1

2

2

21

m

m

mm

Langkah 4Cari Equation PQ

63

)0(36

)( 11

xy

xy

xxmyy

CASE 2: PARALLEL

Bermaksud 21 mm Langkah 1Cari gradient bagi kedua-dua persamaan

diberi.

1

1

pxy

ypx

diperolehi dari formula

a

bm

b

y

a

x 2,1

CASE 3: AREAEXAM TIPSKatakunci adalah Area

Jangan lupa ulang semula coordinatepertama dan letakkan di kedudukan terakhir

a = x - interceptb = y - intercept

**bermakna dari soalan4

3 2 m

Langkah 2

Samakan keduanya

4

3

21

p

mm

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FOKUS A+SPM

2012

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CASE 4: COLLINEAR

CASE 5: RATIO

EXAM TIPSKatakunci adalah point Q divide PR with

ratio 2:1

nm

myny

nm

mxnxyx 2121 ,),(

EXAM TIPSKatakunci Collinear

(Area = 0)

CASE 6: DISTANCE/LOCUS

EXAM TIPS

*Ianya bermaksud jarak QR bersamaan 2 PQ.*

**Penyelesaianya gunakan formula

distance/locus

2

12

2

12

2

12

2

12 )()()()(2

2

yyxxyyxx

QRPQ

0

0

a

a

a

a

aa

a

1

21

21

21210

21212

10

)]2(17)7(3)(4[)]4(717)3)(2[(2

10

4

2

17

7

34

2

2

10

EXAM TIPS: QUESTION 15-16 - VECTOR

Dari analisis soalan tahun-tahun lepas menunjukkan ada 2 BENTUK soalan dari bab vector ini.Bentuk 1 : soalan dari ayat. Bentuk 2 : soalan dari gambarajah

2 BENTUK DIATAS AKAN SECARA SELANG SELI AKAN MENYOAL 4 PERKARA DIBAWAH

a) Express in terms of-Lihat contoh. 2012b) Find Unit vector

22.

yx

yjxivectorunit

2012

c) Magnitude Vector ada modulus 22 yxr 2012d) Vector ada parallel Atau collinear

CASE 1: SOALAN DARI AYAT (Unit Vector)

Given that a~

= ( )72 and b~ = ( )1

13 , find

(a) the vector a~

b~

.

(b) the unit vector in the direction ofa~

b~

.

(a) a~

b~

= ( )72 ( )1

13

= ( )815 (b) Unit vector in the direction ofa

~ b

~

17

15j+8i

15+8

15j+8i

22

EXAM TIPS

Unit Vector

22yx

yjxi

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CASE 2: SOALAN DARI AYAT (MagnitudVector)Given that a

~= 3i + 9j and b

~= 2i + kj, find

(a) a~

b~

in the formxi +yj.

(b) the value ofkif |a~

b~

| = 13.

Jawapan

(a) a b~

= 3i + 9j (2i + kj) = 3i + 9j + 2i kj= 5i + (9 k)j

(b) |a~

b~

| = 13

13)9(5 22 k = 13

25 + (9 k)2 = 169(9 k)2 = 144

9 k = 12k = 21

or

9 k = 12k = 3CASE 3: SOALAN DARI AYAT (parallel)Given that a~

= (2k-1)i + 3j and b~

= 4i + 5j, find k if

2a+3b is parallel to y-axis

Langkah 1

: cari hasil 2a +3b dahulu.

21

104

15

12

6

24

5

43

3

122

k

k

k

Langkah 2

Parallel to yaxis bermaksud = (0,y)

yq

k

qba

0

21

104

Langkah 3

Dapatkan nilai k

(4k+10 = q(0)4k =-10

k =-5/2

CASE 4: SOALAN DARI GAMBARAJAH (unit Vector)

Diagram 9 shows vector OR

drawn on a Cartesian plane.

Diagram 9

(a)Express OR

in the form ( )xy .

(b)

Find the unit vector in the direction ofOR

.

(a)OR

= ( )125 (b)

Unit vector in the direction ofOR

= (12i~

+5j~

) / 122 + 52

= (12i~

+5j~

) / 13

EXAM TIPS

Magnitude Vector

22yxr

EXAM TIPS

Parallel Vector

qba ,q is scalar...

EXAM TIPS

Unit Vector

22yx

yjxi

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EXAM TIPS: QUESTION 17 TRIGONOMETRY FUNCTION

CASE 1: Six Trigonometry Functions of Any Angles

1 Given cosx= 0.4761 and 90 x 270, find the value of(a) x

(b) secx

(a) cosx= 0.4761cos = 0.4761 = 61 34'x = 180 + 61 34'

= 241 34'

(b)secx =

1

cosx

=1

0.4761

= 2.1

2 Solve the equation 20 sin2x= sinx+ 24 sin 30 for 0 x 360

[4 marks]

20 sin2x= sinx + 24 sin 3020 sin2x= sinx + 1220 sin2x + sinx 12 = 0(4 sin 3)(5 sin + 4) = 0

sinx =3

4

x = 48 35', 131 25'or

sinx= 4

5

x = 233 8', 306 52'

Pelajar hasruslah menghafal dan tahu cara menggunakan formula dibawah

Pelajar hasruslah tahu cara menggunakan formula dibawah

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EXAM TIPS: QUESTION 18 CIRCULAR MEASURE

Exam Tips

1. 0180. rad 2. Arc, rs 3. Area, 2

2

1r

**Soalan dalam paper 1 biasanya tajuk ini

simple,basic dan tidak berbelit-belit.Persedian

perlu dibuat Hafal FORMULA diatas dan tahu

cara menggunakanya,sudah mengcukupi.

*Jika Soalan bertanya tentan angle pastikan

Jawapan anda berikan dalam bentuk radian.*

3 Solve the equation 5 cos 2x = 13 sinx 9 for 0 x 360[2 marks]

5 cos 2x = 13 sinx 95(1 2 sin2x) = 13 sinx 95 + 10 sin2x = 13 sinx 910 sin2x 13 sinx + 4 = 0(5 sin 4)(2 sin 1) = 0

sinx =4

5

x = 53 8', 126 52'

or

sinx =1

2

x = 30, 150

Pelajar hasruslah tahu cara menggunakan formula dibawah

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EXAM TIPS: QUESTION 19 INTEGRATION

7

4

62

4

2

25

62

24182

182

)8(3

18)([3

18])(3[

5

2

2

5

2

2

5

2

5

2

2

5

2

5

2

5

2

k

k

xk

xk

dxx

k

xdxkdxxg

dxkxxg

Exam Tips**Tajuk ni sangat mudah dalam paper 1Kebiasaanya ada 2 format soalan.

Format yang pertamaLANGKAH 1:Anda perlu tahu buat PEMECAHAN dr formula

asal .

LANGKAH 2:Kebiasaan dari 2 pecahan tadi.satu Jawapan anda

boleh dapat secara terus dari soalan.Satu pecahan lagi Jawapan

anda boleh perolehi dengan cara integrate

FORMAT KEDUAContoh Soalan

Given that ,Find

(a)

Penyelesaianya.

tersebut iaitu 2,6.

*Soalan ini diramalkan akan keluar*L

6

2

7)( dxxf

2

6

)( dxxf

EXAM TIPS: QUESTION 20 STATISTICS

Exam Tips**Dalam bab ini terlalu banyak formula perlu difahami.bukan

Lupakan yang lain.WAJIB HAFAL!3 formula Dibawah sudah mengcukupi.

FORMULA

a) Standard Deviation, 22 )(xN

x

b) Mean,N

xx

c) Variance, 222 )(xN

x

ISTILAH (Sangat Penting)

-Sum of Set Square Number= 2x -Sum of Set Number=x

CASE 1: MEAN & STANDART DEVIATION

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CASE 2: NEW MEAN,NEW MEDIAN,NEW MODE

Diagram 1 shows the scores obtained by a group of

players in a game.10, 15, 6, 15, 5, 17, 14

Diagram 1

(a) Find the mean, median and mode for the scores.

(b) If the scores in Diagram 1 is multiplied by 9, and

then added 7, find the mean, median and mode for

the scores.

(a)Mean, x =

xN

=

10 + 15 + 6 + 15 + 5 + 17 + 14

7

=82

7

= 11.71

Median = 14

Mode = 15

(b) New mean = 9(11.71) + 7 = 112.39

New median = 9(14) + 7 = 133New mode = 9(15) + 7 = 142

Exam Tips

**Case ini sangat penting untuk SPM 2011

Jawapan (a)Gunakan formula sahaja

**Mean

Mean, x =xN

**Median

Pelajar perlu susun mengikut susunanmenaik terlebih dahulu

5, 6, 10, 14, 15, 15, 17

Median = nilai ditengah-tengah = 14

Mode

Kekerapan tertingi = 15

Jawapan (b)Percayalah soalan ini sangat senang!!

New mean

New median

New mode

EXAM TIPS: QUESTION 21 PROBABILITY

EXAM TIPS

Soalan Probability ini sebenarnya Sangatmudah.

CARD ARE DRAWN) dan satu lagi silafokuskan pada katakunci warna yang sama(same color)

berikut

(G, G) + (Y, Y) + (R, R)

,

105

34

14

4

15

5

14

6

15

7

14

2

15

3

Tolak 1 kerana tadianda sudah pilih satu

cardMaka 5-1=4

PERHATIAN:

Soalan diatas meminta andadapatkan probability bagi warna

Bagaimana jika soalan memintapada anda warna yang tidak sama( NOT same color)

Jangan panic mudah sahaja.notsame color bermakna kedua-

duanya BUKAN warna yang sama.

Apa yang perlu dilakukan adalah(1- same color)

105

71

105

341

Total Card (R+Y+G) = 15

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EXAM TIPS: QUESTION 22 DIFFERENTIATION

CASE 1: Equation Tangent

Find the equation of the tangent to the curve

y = (2x 5) (x 4) at the point (4, 0).

y = (2x 5)(x 4)= 2x2 13x + 20

dy

dx= 4x 13

At point (4, 0),dy

dx= 4(4) 13

= 3

The gradient of the tangent at point (4, 0) = 3

Equation:

y = 3(x 4)y = 3x12

CASE 2: Minimum Point

The curvey = 3x2 + 6x 7 has aminimum point atx = k, where kis a

constant.

Find the value k.

y = 3x2 + 6x 7dy

dx= 6x + 6

Wheny is minimum,dy

dx= 0

6x + 6 = 0

x= 1k= 1

CASE 3: Rate of Changes

The volume of a sphere increases at the rate of

44.8 cm3 s1. Find the radius of the sphere atthe instant when its radius is increasing at a rate

of 0.7 cm s1

.

[Volume of sphere, V=4

3r3]

V=4

3r3

dV

dr= 4r2

dV

dt= 44.8

dr

dt= 0.7

CASE 4: Small of Changes

Find the small change in volume of a sphere

when the radius increases from 6 cm to 5.93 cm.

[Volume of sphere, V=4

3r3]

V=4

3r3

dV

dr= 4r2

r= 5.93 6

Vr

dx

dy

VdV

dr r

= 4r2 0.07= 4(6)2 0.07= 10.08 cm3

dV

dt=

dV

dr

dr

dt

44.8 = 4r2 0.7

4r2 =44.80.7

4r2 = 64

r2 = 16

r= 4 cm

Exam Tips** Katakunci adalah EQUATION tangent curve.

Langkah 1

Didalam soalan coordinate ),( 11 yx sudah

diberi iaitu (4, 0)...Oleh itu pelajar hanya perlu

Langkah 2

Gunaklan formula )( 11 xxmyy untukmendapatkan equation

Exam Tips** Katakunci adalah MINIMUM POINT

dy

Langkah 1Pelajar perlu gunakan kaedah differentiation

untuk dapatkan nilai

dx

dy

Langkah 2

dx

dy

Langkah 3

Dapatkan nilai x.

Exam Tips

** Katakunci adalah rate dan unit cm3 s1.

dt=

dV

dr

dr

dt

dV

dr= pelajar perlu buat differentiation

dV

dt= Volume changes of t = 44.8

dr

Exam Tips** Katakunci adalah Small Change inVolume

dV

dr

TIPS SANGAT PENTING!!**Khas untuk pelajar yang tak faham langsung

soalan ini mahukan apa dan tak tahu langsung nak

pakai formula apa.

**Apa yang perlu anda lakukan adalah buat

differentiation.apa jua soalan yang ditanya pasti

akan meminta pelajar membuat

dx

dy

*Dalam case 3 dan 4 pelajar perlu membuat

dV

dr

dV

drsama sahaja dengan

dx

dy

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EXAM TIPS: QUESTION 23 PERMUTATION & COMBINATION

CASE 1: PERMUTATION

*Trial Sabah 2011*

RAMALAN 1How many 3-digit numbers that are greater than 400 can be

formed using the digits 1, 2, 3, 4, and 5 without repetition?

= 12p x 2

4p = 24

RAMALAN 2

How many 4-digit even numbers can be formed using the

digits 1, 2, 3, 4, and 5 without repetition?

= 12p x 3

4p

= 48

RAMALAN 3How many 4-letter codes can be formed using the letters in

the word 'GRACIOUS' without repetition such that the firstletter is a vowel?

= 14p x 3

7p

= 840

CASE 2: PERMUTATION

** Trial Kelantan dan Terengganu 2011**

RAMALAN 1

Diagram shows seven letter cards.

A five-letter code is to be formed using five of these cards.

Find

a) The number of different five-letter codes that can be

formed,b) The number of different five-letter codes which end with a

consonant.

a) 57p = 2520

b) 46p x 1

4p

= 1440

RAMALAN 2

Diagram shows six numbered cards.

A four-digit number is to be formed by using four of these

cards.How manya) different numbers can be formed?b) different odd numbers can be formed?Answer:

a) 6P4 = 360

b) 5P3 x4P1

= 240

ROFINU M

987541

Exam Tips** Katakunci adalah 3-digit numbers dan greater than 400

Langkah 1:

Buat 2 garis seperti dibawah

d

c

b

a PP .

b = adalah digit pertama [gerenti = 1]

a = Total digit mengikut syarat diberikan.SyaratHow many number

greater than 400.Jawapan 4 dan 5 } 2 nombor 1

2

p

d = adalah baki digit yang tinggal [dalam soalan sebelah = 2 digit.

1 digit sudah diambil maka baki yang tinggal adalah 2]

c = total digit keseluruhan yang tinggal = 4

Asalnya ada 5 nombor.tetapi 1 nombor sudah diambil maka total bakinya

etT

TotalP arg

1

2p

24p

2

4p

Exam TipsEven Number = 2 dan 4 } Total Digit pertama = 2

Baki digit [4-1 = 3 digit]

Digit pertama.Gerenti 1

Exam TipsJawapan (a)

Ingat kembali formulanya etTTotalP

arg

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CASE 3: COMBINATION

** Trial Johor,Melaka,Perak,SBP dan Selangor 2011**

RAMALAN 1A committee of 3 boys and 3 girls are to be formed from 10 boys and 11 girls.

In how many ways can the committee be formed?

3

10C x 3

11C

= 19800

RAMALAN 2A badminton team consists of 8 students. The team will be chosen from a group of 8 boys

and 5 girls. Find the number of teams that can be formed such that each team consists ofa) 5 boys,b) not more than 2 girls.

a)5

8C x

3

5C = 560

b) If the team consists of 8 boys and 0 girl 88C x 0

5C = 1

If the team consists of 7 boys and 1 girl 78C x 1

5C = 40

If the team consists of 6 boys and 2 girl 68C x 2

5C = 280

The number of teams that can be formed = 1 + 40 + 280= 321

RAMALAN 3 (SPM2011)

A debating team consists of 6 students. These 6 students are chosen from 2 monitors, 3

assistant monitors and 5 prefects.

a) there is no restriction,

b) the team contains only 1 monitor and exactly 3 prefects.

a) 610C = 210

b) 12C x 3

5C x 2

3C = 60

Exam Tips

BOY GIRL

Step 1: C C

Ingat formulaetT

TotalC arg !

Step 2 : 310C

311C

Total Boy = 10 Total Boy = 11

Dipilih = 3 (Target Dipilih = 3 (Target)

Exam TipsJawapan (a)

No restriction = tiada syarat automatic gunakan formula asaletT

TotalC arg

Exam TipsJawapan (b)

TARGET 6 students ..Ingat formulaetT

Total

C arg

Langkah 1 :Isikan target mengikut syarat diberikan .iaitu M = 1,P = 3 automatik AM = 2

MONITOR PREFERCTS ASSISTANT MONITOR

1C 3C

2C

LAngkah 2 : Isikan total dari yang diberikan iaitu total M = 2,P = 5 ,AM =3

MONITOR PREFERCTS ASSISTANT MONITOR

1

2C

x 3

5C x 2

3C = 60

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CASE 5:

** Trial Pahang,Perlis 2011**

RAMALAN 1Diagram shows five cards of different letters.

a) Find the number of possible arrangements, in a row, of all the cards.b) Find the number of these arrangements in which the letters A and N are

side by side.

a) 5! = 120

b) 4! x 2!= 48

R A J I N

Diagram below shows a standard normal

distribution graph.

The probability represented by the area of the

a) Find the value of )( kZP b) X is a continuous random variable

which is normally distributed with a

mean of and a standard deviation of

2.If the value of X is 85 when the

scoreZ is k,find the value of

.[3 marks]

EXAM TIPS: QUESTION 25 PROBABILITY DISTRIBUTION

Exam Tips

Apa yang pelajar perlu faham jumlah bagi

kedua-dua luas dibawah graph adalah 0.5 +0.5 =1

(a)

0985.02

803.01)(

kZP

(b)

Hafal Formula ini

XZ

X Random variable Mean Standard deviation

scoreZ

42.82

2

8529.1

XZ

Area = 0.5 Area = 0.5

Perlu dibahagi 2 kerana

nilai k didalamgambarajah ada 2Sebab total = 1

Exam Tips

Nilai z mestilah diperolehi dari dalam graph.

Lihat muka surat disebelah

EXAM TIPS: QUESTION 25BINOMIAL DISTRIBUTION

FOKUS 2012

FOKUS 2012

1 The discrete random variableXhas a binomial probability distributionwith n = 4, where n is the number of trials. Table 1 shows the probability

distribution ofX.

x P(X=x)

01

16

11

4

2 k

31

4

41

16

:

Find

(a) the value ofk.

(b) P(X 3).[3 marks]

[3 markah]

(a)k +

1

16

1

4

1

4

1

16= 1

k =3

8

(b)P(X 3) =

1

4+

1

16

=5

16

Exam Tips

TOTAL Probability = 1

FOKUS 2012

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SUGGESTION QUESTIONPAPER 1

[Improve From 60 to A+]

Bagaimana Z =1.29?dan bukan =0.0985?

Ini kerana nilai z tersebut anda perlu dapatkan dari dalam table.Didalam peperiksaan table ini akn

****Begitu ramai pelajar melakukan kesilapan mudah ini.pastikan anda bukan termasuk didalam

golongan ini.****

42.82

2

8529.1

XZ

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SUGGESTION QUESTION 1-3: FUNCTION

1 Diagram 1 shows the relation between set

Mand setNin the arrow diagram.

State the

(a) images of 12

(b) objects of 2

(c) domain of this relation

(d) range of this relation

[2 marks]

(a) 2, 3

(b) 6, 12, 16

(c) {6, 12, 16}

(d) {2, 3}

2 Given {(16, 2), (20, 2), (20, 5)}.

State the

(a) images of 20

(b) objects of 2(c) domain of this relation

(d) range of this relation

(a) 2, 5

(b) 16, 20

(c) {16, 20}

(d) {2, 5}

4 Given the functionf:x |x2 5|.(a) Find the images of 6, 6 and 7.

(b) Find objects which have the image of4.

[4 marks]

f(6) = |(6)2 5|= |31|

= 31f(6) = |(6)2 5|

= |31|

= 31

f(7) = |(7)2 5|= |44|

= 44

f(x) = 4

|x2 5| = 4So,

x2 5 = 4x2 = 9

x= 3, 3and

(x2 5) = 4x2 = 1

x= 1, 1

3 Given functionf:x 8x 2, find the(a) image of 7(b) object which has the image 34

[3 marks]

(a) f(7) = 8(7) 2

= 54(b) f(x) = 34

8x 2 = 348x = 32x = 4

5 The fuctionfis defined asf:xx + 4.Find

(a)1

(2)

(b) 1(x)

[3 marks]

(a) Letf1

(2) = k

Sof(k) = 2k+ 4 = 2

k= 2Therefore,f

1(2) = 2(b) Letf

1(x) =y

Sof(y) =x

y + 4 =x

y =x 4Therefore,f

1(x) =x 4

6 The fuctionsfand g are defined asf:x 1 8x and g :x 2x 5.Find gf

1(x).

[3 marks]

Letf1

(x) =ySof(y) =x

1 8y =x

y =1 x

8

Thereforef1(x) =1 x

8

gf1(x) = g(f1(x))

= g(1 x

8)

= 2(1 x

8) 5

=

x 21

4

7 The functions offand g are defined asf

:xx + 3 and g :x 4 x. Find thecomposite function ofgfandfg.

Givenf(x) =x + 3 and g(x) = 4 x.gf(x) = g(f(x))

= g(x + 3)

= 4 (x + 3)= x + 1

fg(x) =f(g(x))

=f(4 x)= (4 x) + 3= x + 7

8 The functions offand g are defined asf

:xx 9 and g :x 9x 3. Find thecomposite function ofgfandfg.

[3 marks]

Givenf(x) =x 9 and g(x) = 9x 3.gf(x) = g(f(x))

= g(x 9)= 9(x 9) 3= 9x 84

fg(x) =f(g(x))

=f(9x 3)= (9x 3) 9= 9x 12

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1It is given that

5

2is one of the roots of the

quadratic equation 4x2 8x + q = 0. Findthe value ofq.

[2 marks]

4(5

2)2 8(

5

2) + q = 0

25 20 + q = 0q= 5

2 Form the quadratic equation which has the

roots 3 and3

4.

[4 marks]

x2 (3 +3

4) + (3)(

3

4) = 0

x2 (9

4)x+ (

9

4) = 0

4x2 + 9x 9 = 0

3 The quadratic equation 4x2 + mx + n = 0has roots 8 and 9. Find the values ofmand n.

[3 marks]

(x a)(x b) = 0(x 8)(x + 9) = 0x2 +x 72 = 04x2 4x + 288 = 0Therefore,

m= 4 and n = 288

4 The quadratic equation 4x2 +px + q = 0 has

roots 1 and 5. Find the values ofp and q.

[4 marks]

(x a)(x b) = 0(x + 1)(x + 5) = 0

x2 + 6x + 5 = 0

4x2 + 24x + 20 = 0

Therefore,

p = 24 and q = 20

5 The straight liney = 5x + 6 does not

intersect with the curvey = 6x2 + 5x + q.

Find the range of values ofq.

[3 marks]

y = 5x + 6

y = 6x2 + 5x + q

6x2 + 5x + q = 5x + 6

6x2 + q 6 = 0The equation does not have real roots

b2 4ac < 0(0)2 4(6)(q 6) < 0 24q + 144 < 024q< 144q > 6

(4q 3)x2 2x 4 = 0 has twodifferent roots, q is a constant.

Find the range of values ofq.

[3 marks]

(4q 3)x2 2x 4 = 0The equation has two different roots

b2 4ac > 0(2)2 4(4q 3)(4) > 04 64q 48 > 064q > 44

q> 11

16

6 The straight liney= 6 4x does notintersect with the curvey= x2 +x + m.Find the range of values ofm.

[3 marks]

y= 6 4xy= x2 +x + mx2 +x + m= 6 4xx2 + 5x + m 6 = 0The equation does not have real roots

b2 4ac < 0(5)2 4(1)(m 6) < 025 + 4m 24 < 04m< 1

m< 1

4

8 The quadratic equation 8x2px + 2 =x

has two equal roots,p is a constant.

Find the values ofp.

8x2px + 2 =x8x2+ (p 1)x + 2 = 0The equation has two equal rootsb2 4ac = 0(p 1)2 4(8)(2) = 0p2 + 2p+ 1 64 = 0p2 + 2p 63 = 0(p 7)(p + 9) = 0p = 7 orp= 9

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1 Diagram 3 shows the graph of a quadraticfunctionf(x) = (x + s)2 7, where s is aconstant.

The curvey =f(x) has a minimum point

(7, t), where tis a constant. State

(a) the value ofs

(b) the value oft

(c) the equation of the axis ofsymmetry

[3 marks]

(a) s= 7(b) t= 7( ) x = 7

2 Diagram 4 shows the graph of a quadraticfunctionf(x) = 2(x + m)2 7, where m is aconstant.

The curvey =f(x) has a minimum point (4,

n), where n is a constant. State

(a) the value ofm

(b) the value ofn

(c) the equation of the axis of symmetry

[3 marks]

(a) = 4(b) = 7(c) x = 4

3Diagram 1 shows the graph of a quadraticfunctionf(x) = 5(x + h)2 + 6, where h is a

constant.

Diagram 1

The curvey =f(x) has a minimum point (8, k),

is a constant. State

(a) the value ofh

(b) the value ofk

(c) he equation of the axis of symmetry

[3

(a) h= 8(b) k= 6

(c) x = 8

4 Diagram 2 shows the graph of a quadraticfunctionf(x) = (x + s)2 6, where s is aconstant.

Diagram 2

The curvey =f(x) has a minimum point

(6, t), where tis a constant. State

(a) the value ofs

(b) the value oft

(c) th equation of the axis of symmetry

(a) s = 6(b) t = 6(c) x = 6

7 Find the range of the values ofx suchthat (4x 2)(x+ 4) 2x + 8.

[3 marks]

(4x 2)(x+ 4) 2x + 84x2 18x 8 2x + 84x2 16x 16 04(x2 + 4x+ 4) 04(x + 2)(x+ 2) 0

The range of values ofx isx 2 orx 2

5 Find the range of the values ofx such that

x(x 4) 4.

[3 marks]

x(x 4) 4x2 4x+ 4 0(x + 2)(x+ 2) 0

The range of values ofx isx 2 orx 2

6 Find the range of the values ofx such that

x(x 7) 12. [3 marks]

Answer:x(x 7) 12x2 7x+ 12 0(x + 4)(x+ 3) 0

The range of values ofx isx 3 orx 4

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SUGGESTION QUESTION 7-8: INDICES & LOG

1 Solve the equation 2x + 3 2x + 2 = 8.

2x + 3 2x + 2 = 82x23 2x22 = 238(2x) 4(2x) = 234(2x) = 23

2x + 2 = 23

x + 2 = 3

x = 1

2 Solve the equation 2x + 7 2x + 6 = 64.

[3 marks]

2x + 7 2x + 6 = 642x27 2x26 = 26128(2x) 64(2x) = 2664(2

x) = 26

2x + 6 = 26

x + 6 = 6

x = 0

3 Solve the equation 32x 2 = 8

5x.

[3 marks]

32x 2 = 85x

25(x 2)

= 23(5x)

5x 10 = 15x(5 + 15)x = 10

x =1

2

4 Solve the equation 254x + 5 = 125

5 5x.

[3 marks]

254x + 5 = 1255 5x

52(4x + 5) = 53(5 5x)

8x+ 10 = 15 15x(8 + 15)x= 15 10

x =5

23

SUGGESTION QUESTION 9-11: PROGRESSION

1 Three consecutive terms of an

arithmetic progression are 13 x, 8, 13+ 3x. Find the common difference of the

progression. [4 marks]

13 + 3x 8 = 8 (13 x)2x = 16

x = 8

d= 8 (13 (8))= 3

2 The first three terms of an arithmetic

progressino are 7h + 2, k, 5h + 12.

(a) Express kin terms ofh.

(b) Find the 6th term of the

progression in terms ofh.

[4 marks]

(a) k (7h+ 2) = (5h+ 12) k2k= 12h + 14k= 6h + 7

(b) a= 7h + 2d= k+ 7h 2

= (6h + 7) + 7h 2= h + 5

T6 = a + 5d

= 7h + 2 + 5(h + 5)= 7h + 2 + 5h + 25= 2h + 27

3 The first three terms of a geometric

progression arex, 24, 96.

Find

(a) the value ofx.

(b) the sum from the 4th term to the

6th term.

[3 marks]

(a) r = 9624

= 4

x

24=

24

96

x =24

96 24

= 6

(b)S6 =

6(46 1)4 1

= 8190

S3 = 6 + 24 + 96= 126

Sum = 8190 126= 8064

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SUGGESTION QUESTION 12 : LINEAR LAW

1 The variablesx andy are related by the

equationy =px5

, wherep is a constant.

(a) Convert the equationy =px5

to

linear form.(b) Diagram 1 shows the straight lineobtained by plotting

log10y against log10x.

Diagram 1

Find the value of

(i) log10p.

(ii) q.

(a) log10

= 5 (b) (i) log10p = Yintercept

= 7

(ii) log10y= 5(3) + 7= 8

q = log10y= 8

2 The variablesx andy are related by the

equationy =px4

, wherep is a constant.

(a) Convert the equationy =px4

to

linear form.(b) Diagram 2 shows the straight lineobtained by plotting

log10y against log10x.

Diagram 2

Find the value of

(i) log10p.

(ii) q.

(a) l

(b) (i) log10p = Yintercept= 7

(ii) log10y= 4(4) + 7= 9

q = log10y= 9

3 The variablesx andy are related by the

equationy2 = 6x(x 6).Diagram 3 shows the straight line graph

obtained by plottingy2

xagainstx.

Diagram 3

Find the value ofp and q.

[2 marks]

y2 = 6x(x 6)y2

x= 6(x 6)

y2

x= 6x 36

Wheny2

x= 0,

0 = 6x 36x = 6p =x = 6

Whenx = 3,y2

x = 6(3) 36= 18

q =y2

x= 18

4 The variablesx andy are related by the

equationy2 = 2x(x 6).Diagram 4 shows the straight line graph

obtained by plottingy2

xagainstx.

Diagram 4

Find the value ofp and q. [2 marks]

y2 = 2x(x 6)y2

x= 2(x 6)

y2

x= 2x 12

Wheny2

x= 0,

0 = 2x 12

x = 6p =x = 6

Whenx = 2,y2

x= 2(2) 12

= 8

q =y2

x= 8

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SUGGESTION QUESTION 13-14: COORDINATE GEOMETRY

1 Find the equation of the straight line

which passes through P(5, 5) and isparallel to the straight line joining Q(7,

3) danR(6, 7).[3 marks]

=7 + 3

6 7

= 10

13

Equation:

y 5 = 10

13(x 5)

13y + 65 = 10x 5010x + 13y 115 = 0

2 Find the equation of the straight line which

passes throughA(4, 1) and isperpendicular to the straight line joining

B(9, 7) dan C(1, 4).[3 marks]

=4 + 71 + 9

=3

8

(3

8)m2= 1

m2= 8

3

Equation:

y 1 = 8

3(x + 4)

3y 3 = 8x 328x + 3y + 29 = 0

4 The pointJ(1, 11) divides a straight line joining P(7, 5) and Q(5, 14) in the ratio e :f.Find the values ofe andf.

[3 marks]

(1, 11) = (7f 5ee +f

,5f+ 14e

e +f)

7f 5ee +f

= 1

7f 5e= ef8f= 4ee

f= 2

e = 2 andf= 1

5 Find the equation of the locus of moving pointA such that its distances from P(9, 7) and Q(

LetA = (x,y)

GivenAP =AQ

(x + 9)2 + (y + 7)2 = (x + 7)2 + (y + 9)2

Squaring both sides,

(x + 9)2 + (y + 7)2 = (x + 7)2 + (y + 9)2

x2 + 18x + 81 +y2 + 14y + 49 =x2 + 14x + 49 +y2 + 18y + 81

4x 4y = 0

3 Given the area of a triangle with verticesA(3, 10),B(9, 4) and C(4,j) is 81 unit2.Find the possible values ofj.

[3 marks]

=1

2|310

94

4

j

310

|= 81

1

2

((12) 9j+ 40 (90) 16 + 3j) =

811

2(102 6j) = 81

6j= 162 102j= 10 orj = 44

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SUGGESTION QUESTION 15-16: VECTORS

1Given that a

~= 10( )56 and b~ = ( )

45

48, find

(a) the vector 10a~

b~

.

(b) the unit vector in the direction of 10a~

b~

.

[4 marks]

(a) 10a~

b~

= 10( )56 ( )45

48

= ( )512 (b) Unit vector in the direction of 10a

~ b

~

= (5i~ +12j~) / 52 + 122

= (5i~

+12j~

) / 13

3 Given that a~

= 4i + 3j and b~

= 4i + kj, find

(a) a~

b~

in the formxi +yj.

(b) the value ofkif |a~

b~

| = 17.

[4 marks]

(a) a b~

= 4i + 3j (4i + kj) = 4i + 3j + 4i kj = 8i + (3 k)j

(b) |a~

b~

| = 17

82 + (3 k)2 = 1764 + (3 k)2 = 289(3 k)2 = 225

3 k = 15 k = 18

or

3 k = 15= 12

5Diagram 8 shows vector OP

drawn on a

Cartesian plane.

Express in the formxi +yj.

(a)OP

(b)Find the unit vector in the direction of

(a)OP

= 4i + 3j

(b) Unit vector in the direction of OP= (4i

~+3j

~) / 42 + 32

= (4i~

+3j~

) / 5

=4

5i +

3

5j

6Diagram 9 shows vector OA

drawn on a

Cartesian plane.

Diagram 9

(a)Express OA

in the form ( )xy .

(b) Find the unit vector in the

direction ofOA

.

[3 marks]

(a)OA

= ( )56

(b) Unit vector in the direction of

= (5i~

+6j~

) / 52 + 62

= (5i~

+6j~

) / 7.81

2 The vectors a~

and b~

are non-zero and non-

parallel.

It is given that (h 7)a~

= (k 9)b~

where h

and kare constants.

Find the value of

(a) h

(b) k

a~

and b~

are non-parallel.

Therefore (h 7)a~

= (k 9)b~

is true only when (h

7)a~

and (k 9)b~

are zero vector.

Since a~

and b~

are not zero vector, therefore (h 7)

and (k 9) must be zero, (h 7) = (k 9) = 0.(a) h 7 = 0

= 7

(b) k 9 = 0= 9

4It is given that OA

= 12a

~+ 7b

~, OB

= 32a

~+

21b~

and OC

= 78a~

14b~

.

(a)FindAB

andAC

.

(b) Hence, show that pointsA,B and C

are collinear.[4 marks]

(a)AB

=AO

+ OB

= (12a~

+ 7b~

) 32a~

+ 21b~

= 44a~

+ 14b~

AC

=AO

+ OC

= (12a

~+ 7b

~) + 78a

~ 14b

~

= 66a~

21b~

(b)AC

= 3

2(44a

~+ 14b

~)

= 3

2AB

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SUGGESTION QUESTION 17: TRIGONOMETRIC FUNCTION

1 Given tanx= 0.6486 and 0 x180, find the value of

(a) x

(b) cotx

(a) tanx= 0.6486tan = 0.6486 = 32 58'x= 180 32 58'

= 147 2'

(b)cotx =

1

tanx

=1

0.6486

= 1.542

2 Given sinx= 0.8026 and 0 x 180, findthe value of

(a) x

(b) cosecx

(a) sinx= 0.8026sin = 0.8026 = 53 23'x= 180 53 23'

= 126 37'

(b)cosecx =

1

sinx

=1

0.8026

= 1.246

3 Solve the equation 16 sin2x = 0 sinx 18 sin 30 for 0 x 360[4 marks]

16 sin2x = 0 sinx 18 sin 3016 sin2x = 0 sinx 916 sin2x sinx + 9 = 0(4 sin 3)(4sin 3) = 0

sinx =3

4

x = 48 35', 131 25'or

sinx= 3

4

x = 228 35', 311 25'

4 Solve the equation tanx + sinx = 0 for0 x 360

a[4 marks]

tanx + sinx = 0sinx

cosx+ sinx = 0

sinx + cosx sinx = 0sinx(1 + cosx) = 0

sinx = 0

x = 0, 180, 360or

1 + cosx = 0

cosx= 1x = 180

5 Solve the equation 8 cos2x = 6 cosx

2 cos 60 for 0 x 360[4 marks]

8 cos2x = 6 cosx 2 cos 608 cos2x = 6 cosx 18 cos2x 6 cosx + 1 = 0(2 cos 1)(4 cos 1) = 0

cosx =1

2

x = 60, 300

or

cosx =1

4

x = 75 31', 284 29'

6 Solve the equation 21 cos 2x = 46 sinx 33 for 0 x 360[2 marks]

21 cos 2x = 46 sinx 3321(1 2 sin2x) = 46 sinx 3321 + 42 sin2x = 46 sinx 3342 sin2x 46 sinx + 12 = 021 sin2x 23 sinx + 6 = 0(3 sin 2)(7 sin 3) = 0

sinx =2

3

x = 41 49', 138 11'

or

sinx =3

7

x = 25 23', 154 37'

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SUGGESTION QUESTION 18: CIRCULAR MEASURE

1 Diagram 2 shows two arcs,AD andBC, of

two concentric circles with centre O and

Diagram 2

Find

(a) the angle , in radian.(b) the perimeter of the shaded region,ABCD.

[4 marks]

(a) Length of arcBC= r8 = (2 + 16)

=8

18

(b) Length of arcAD = r

= 2 0.44= 0.88 cm

= 8 + 0.88 + 16 + 16

= 40.88 cm

2 Diagram 10 shows the sector OAB of a

circle with centre O.

Diagram 10

Find

(b) the area of the shaded region[4 marks]

(a)cosBOC=

12

15

= 0.8

BOC= 36.87

= 36.87

180

(b)

Area of sector OAB =

1

2r

2

=1

2(15)2(0.6436)

= 72.4 cm2

Area of triangle OCB

=1

2 9 12

= 54 cm2

Area of the shaded region = 72.4 54

= 18.4 cm2

1Given that

1

2f(x) dx = 7, find

(a)the value of

2

1f(x) dx.

(b)the value ofkif

1

2[kx + 3f(x)] dx

=315

16.

[4 marks]

(a)2

1f(x) dx=

1

2f(x) dx

= 7(b)

1

2[kx + 3f(x)] dx =

315

16

1

2kxdx+ 3

1

2f(x) dx =

315

16

k[x2

2]1

2+ 3(7) =

315

16

k(1

2 2) + 21 =

315

16

k=7

8

2Given that

9

4f(x) dx = 1, find

(a)the value of

4

9f(x) dx.

(b)the value ofkif

9

4[kx 3f(x)] dx

= 62.

(a)4

9f(x) dx =

9

4f(x) dx

= 1(b)

9

4[kx 3f(x)] dx = 62

9

4kx dx 3

9

4 f(x) dx = 62

k[x2

2]9

4 3(1) = 62

k(81

2 8) 3 = 62

k = 2

SUGGESTION QUESTION 19: INTEGRATION

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SUGGESTION QUESTION 20: STATISTICS

1 A set of data consists of 9 numbers. The sum of the numbers is 27 and the sum of the squares

of the numbers is 2916. Find for the 9 numbers,

(a) the mean,

(b) the standard deviation.

[3 marks]

(a)Mean =

27

9

= 3

(b)Standard deviation =

2916

9- 32

= 315

= 17.748

2 A set of data consists of 3 numbers. The sum of the numbers is 27 and the sum of the squares

of the numbers is 2916. Find for the 3 numbers,

(a) the mean,

(b) the standard deviation.

[3 marks]

(a)Mean =

27

3

= 9

(b)Standard deviation =

2916

3- 92

= 891

= 29.85

3 Diagram 6 shows the mass of a group of students.

32, 30, 49, 25, 8, 35, 36, 5

Diagram 6

Find the interquartile range for the data.

Rearrange the data,

25 28 30 32 35 36 49 53

First quartile =28 + 30

2= 29

Third quartile =36 + 49

2= 42.5

Interquartile range

= 42.5 29= 13.5

4 Diagram 7 shows the mass of a group of students.

44, 53, 26, 49, 43, 33, 38, 30

Diagram 7

Find the interquartile range for the data.

Rearrange the data,26 30 33 38 43 44 49 53

First quartile = 30 + 332 = 31.5

Third quartile =44 + 49

2= 46.5

Interquartile range

= 46.5 31.5= 15

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5 Diagram 1 shows the scores obtained by a group of players in a game.

1, 10, 1, 9, 3, 4, 2

(a) Find the mean, median and mode for the scores.

(b) If the scores in Diagram 1 is multiplied by 5, and then added 4, find the mean,

median and mode for the scores.

(a)Mean,x =

xN

=1 + 10 + 1 + 9 + 3 + 4 + 2

7

=30

7

= 4.29

Median = 3Mode = 1

(b) New mean = 5(4.29) + 4 = 25.45New median = 5(3) + 4 = 19

New mode = 5(1) + 4 = 9

6 Diagram 2 shows the scores obtained by a group of players in a game.

9, 18, 16, 20, 15, 9, 4

(a) Find the mean, median and mode for the scores.

(b) If the scores in Diagram 2 is multiplied by 3, and then added 6, find the mean,

median and mode for the scores.

[3 marks]

(a)Mean,x =

xN

=9 + 18 + 16 + 20 + 15 + 9 + 4

7

=91

7

= 13Median = 15

Mode = 9

(b) New mean = 3(13) + 6 = 45

New median = 3(15) + 6 = 51

New mode = 3(9) + 6 = 33

SUGGESTION QUESTION 21: PROBABILITY

1 A bag contains 7 brown marbles, 2 yellow marbles and 6 purple marbles. Two marbles

are drawn from the bag at random, one after another. Find the probability that

(a) both marbles are brown.

(b) one marble is yellow and the other is purple.

(a) Probability

=7

15

6

14

=1

5

(b) Probability

=2

15

6

14+

6

15

2

14

=4

35

2 A bag contains 6 blue marbles, 5 red marbles and 6 white marbles. Two marbles aredrawn from the bag at random, one after another. Find the probability that

(a) both marbles are blue.

(b) one marble is red and the other is white.

[4 marks]

(a) Probability

=6

17

5

16

= 15136

(b) Probability

=5

17

6

16+

6

17

5

16

=15

68

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3In a shooting game, the probability for a player to shoot the target for each try is

2

3. If Hock

Seng tried the game for three times, find the probability that she shot at least one target in the

game.

[4 marks]

LetA = Shot the target andB = Miss the target

Probability

= 1 P(BBB)

= 1 (1

3

1

3

1

3)

= 1 1

27

=26

27

4In a shooting game, the probability for a player to shoot the target for each try is

1

10. If Jabah

tried the game for three times, find the probability that she shot at least one target in the game.

[4 marks]

LetA = Shot the target andB = Miss the target

Probability

= 1 P(BBB)

= 1 (9

10 9

10 9

10 )

= 1 729

1000

=271

1000

SUGGESTION QUESTION 22: PPERMUTATION & COMBINATION

1 How many 4-digit numbers greater than 2000 can be formed from the digits 1, 2, 3, 4,

5 and 6 if no repitition of digits is allowed?

[3 marks]

5 5 4 3

Number of ways = 5 5 4 3

= 300

2 How many 3-digit numbers can be formed from the digits 3, 4, 5, 6 and 7 if the

numbers are

(a) less than 400?

(b) odd numbers?

[4 marks]

(a) 1 4 3

Number of ways = 1 4 3

= 12

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C

3 There are 2 different t-shirts and 2 trousers on a cupboard. Calcualte the number of differentways to arrange all the clothes in a row if

(a) no condition is imposed.

(b) all the trousers are next to each other.

(a) Number of ways = 4!

= 24

(b) Number of ways = 3! 2!

= 6 2

= 12

4 A florist want to choose 5 roses from a box of 5 red roses and 5 white roses to decorate a hamper.Calculate the number of ways the roses can be chosen if

(a) there is no restriction.

(b) there are more red roses than white roses chosen.

[4 marks]

(a) Number of ways = 10C5

= 252

(b) The possible ways:5 red roses and 0 white rose

4 red roses and 1 white rose3 red roses and 2 white rose

Number of ways

= 5C5 5C5 +

5C4 5C4 +

5C3 5C3

= 126

5 A school wants to choose 4 students from a group of 6 boys and 5 girls to participatein a national mathematics contest. Calculate the number of ways the students can be

chosen if

(a) there is no restriction.

(b) the students chosen consists of 2 boys and 2 girls.

[4 marks]

(a) Number of ways = 11C4

= 330

(b) Number of ways = 6C2 5C2

= 150

6 A florist want to choose 3 roses from a box of 6 red roses and 4 white roses to decorate

a hamper. Calculate the number of ways the roses can be chosen if

(a) there is no restriction.

(b) there are more red roses than white roses chosen.

[4 marks]

(a) Number of ways = 10C3

= 120

(b) The possible ways:

3 red roses and 0 white rose2 red roses and 1 white rose

Number of ways

= 6C3 4C3 +

6C2 4C2

= 80

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SUGGESTION QUESTION 23: DIFFERENTIATION

1 Find the equation of the tangent to the curvey= x6 + 2x5 at the point (2, 0).

[3 marks]

y= x6 + 2x5dy

dx= 6x5 + 10x4

At point (2, 0),dy

dx= 6(2)5 + 10(2)4

= 32The gradient of the tangent at point (2, 0) = 32

Equation:

y= 32(x 2)y= 32x + 64

2 Find the equation of the tangent to the curvey= (2x + 3)(x+ 6) at the point (3, 27).

[3 marks]

y= (2x + 3)(x + 6)= 2x2 9x + 18

dy

dx= 4x 9

At point (3, 27),dy

dx= 4(3) 9

= 3The gradient of the tangent at point (3, 27) = 3

Equation:y 27 = 3(x + 3)y = 3x + 36

3Given that the curvey =f(x) and

dy

dx= 4px + 10, wherep is a constant.

The gradient of the curve atx= 2 is 18.Find the value ofp.

[3 marks]

dy

dx= 4px + 10

Whenx= 2,dy

dx= 18

4p(2) + 10 = 18p = 1

4 Giveny= (4x + 6)(x 5). find

(a) dydx

(b) the value ofx wheny is maximum.

(c) the maximum value ofy.

[4 marks]

(a) y= (4x + 6)(x 5)= 4x2 + 26x 30

dy

dx= 8x + 26

(b)

Wheny is maximum,

dy

dx = 08x + 26 = 0

x =13

4

(c)y= (4(

13

4) + 6)(1(

13

4) 5)

=49

4

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5 The radius of a balloon in the shape of a sphere increases at the rate of 6 cm s1. Find the rate

of change of the volume of the balloon when the radius is 10 cm.

[Volume of sphere, V=4

3r3]

[3 marks]

V=4

3r3

dV

dr= 4r2

dV

dt=

dV

dr

dr

dt

= 4r2 6= 24r2

When r= 10,dV

dt= 24(10)2

= 2400 cm3 s1

6 The volume of a sphere increases at the rate of 19.2 cm3 s1. Find the radius of the sphere atthe instant when its radius is increasing at a rate of 0.3 cm s

1.

[Volume of sphere, V=4

3r3]

[3 marks]

V=4

3r3

dV

dr= 4r2

dV

dt= 19.2dr

dt= 0.3

dV

dt=

dV

dr

dr

dt

19.2 = 4r2 0.3

4r2 =19.20.3

4r2 = 64r2 = 16

r= 4 cm

7Two variables,x andy, are related by the equationy =

7x2

.

Express the approximate change iny, in terms ofp, whenxchanges from 8 to 8 + p,wherep is a small value.

[3 marks]

y =7x2

dydx

= 14x3

Whenx= 8dy

dx=

14

(8)3

= 7

256

x= (8 +p) + 8=p

y =dy

dx x

= 7

256p

8Two variables,x andy, are related by the equationy =

14

x2.

Express the approximate change iny, in terms ofp, whenx changes from 3 to 3 +p,

wherep is a small value.

[3 marks]

y =14

x2

dy

dx=

28x3

Whenx = 3dy

dx=

28(3)3

= 28

27

x = (3 +p) 3=p

y =dy

dx x

= 28

27p

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SUGGESTION QUESTION 25: PROBABILITY DISTRIBUTION

1 Diagram 1 shows a standard normal dist ribution graph.

Diagram 1

The probability represented by the area of the shaded region is 0.7852.

(a) Find the value ofk.

(b) Xis a continuous random variable which is normally distributed with a mean of 79

and a standard deviation of 16. Find the value ofXwhen thez-score is k.[3 marks]

(a) P(Z > k) = 1 0.7852= 0.2148

P(Z > 0.79) = 0.2148

k= 0.79(b) = 79, = 16

X 7916

= 0.79

X = 79 + 16 0.79

= 91.64

12 Diagram 2 shows a standard normal dist ribution graph.

Diagram 2

The probability represented by the area of the shaded region is 0.2486.

(a) Find the value ofk.

(b) Xis a continuous random variable which is normally distributed with a

mean of 51 and a standard deviation of 5. Find the value ofXwhen thez-

score is k.

[3 marks]

(a) P(Z> k) = 0.5 0.2486= 0.2514

P(Z> 0.67) = 0.2514

k= 0.67

(b) = 51, = 5X 51

5= 0.67

X= 51 + 5 0.67

= 54.35

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LAST EXAM TIPS PAPER 2

2012

[Improve from 60 to A+]

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KUMPULAN SASARAN:Pelajar Yang mendapat Markah Kurang dari 40%

FORMAT KERTAS 2 ADD MATHS SPM 2012

BAHAGIAN A (6 Soalan Wajib Jawab semua) Total Markah =40

BAHAGIAN B (5 soalan Wajib Jawab 4 sahaja) Total Markah = 40

BAHIAGAN C (4 soalan Wajib Jawab 2 sahaja) Total Markah = 20

Berikut merupakan statistic soalan peperiksaan sebenar yang pernah keluar didalam

Peperiksaan dari 2006- 2010..Sila LIHAT apakah PERSAMAAN dan PEMBEZAAN dari

Segi tajuk.

Section A (Answer all question) Total Marks= 40 marks

Bahagian A mengandungi 6 SOALAN WAJIB yang mesti dijawab.

SPM07 SPM08 SPM09 SPM10 SPM11

1 SimultaneousEquation

SimultaneousEquation

SimultaneousEquation

SimultaneousEquation

SimultaneousEquation

2 Coordinate

Geometry

Functions

Equation

Trigonometric

Function

Trigonometric

Function

3 TrigonometricFunction

Progressions Differentiation+Integration

Progressions Progressions

4 Differentiation+Integration

TrigonometricFunction

TrigonometricFunction

Integration Integration

5 Statistics Statistics Vector GeometryCoordinate

GeometryCoordinate

6 Progressions Vector Progressions Statistics Statistics

Section B (Answer 4 Question only from 5 question) Setiap Soalan=10 markah X 4

Total Marks=40 marks

SPM07 SPM08 SPM09 SPM10 SPM11

7 Linear Law Integration Integration Linear Law Linear Law

8 Vector Linear Law Linear Law Differentiation Differentiation

9 Circular

Measure

Circular

Measure

Geometry

Coordinate

Vector Vector

10 Integration Geometry

Coordinate

Circular

Measure

Probablility

Distribution

Probablility

Distribution

11 Probablility

Distribution

Probablility

Distribution

Probablility

Distribution

Circular

Measure

Circular

Measure

Section C (Answer 2 Question only from 4 question) Total Marks=20 marks

SPM07 SPM08 SPM09 SPM10 SPM11

12 Solution ofTriangle

Solution of

Triangle

Solution of

Triangle

Solution of

Triangle

Solution of

Triangle

13 Index Number Index Number Index Number Index Number Index Number

14 LinearProgramming

Linear

Programming

Linear

Programming

Linear

Programming

Linear

Programming

15 Motion Alongstraight Line

Motion Along

straight Line

Motion Along

straight Line

Motion Along

straight Line

Motion Along

straight Line

Remark:Mengikut statistic 2005-2011 bab yang tidak pernah keluar didalam kertas dua adalah,

indices & logarihms, permuatation & combination serta probability.

Adakah Markah Anda selalu dibawah 40% untuk KERTAS 2?.

Untuk sekadar mencapai 40% dalam Matematik Tambahan adalah tidak sesukar mana.Pelajartidak perlu pun untuk mencapai markah penuh bagi sesuatu Soalan,memadailah dapat curi

separuh dari markah penuh bagi sesuatu Soalan.Oleh itu kepada pelajar yang selalu mendapat

markah kurang dari 40% apa kata anda ulangkaji modul dibawah ini berulang kali,sekurangnya 1

kali tanpa melihat jawapan.Jika anda sudah pun mengulangkaji mengikut saranan ,selepas inijangan terkejut dengan keputusan peperiksaan anda sendiri.Walaupun target anda hanya lulus

(40%) kami jamin anda boleh mencapai markah yang lebih baik lagi.Semoga Berjaya.

Secara Umumnya Bab yang akan keluar didalam Kertas 2 Matematik Tambahan mengandungi 18

bab kesemuanya daripada 21 bab bagi tingkatan 4 dan 5.Banyak bukan.? Masih Sempat? Ya!walaupun masa hanya 3 hari percayalah anda masih berupaya dengan mengikuti tip -tips mudah

yang akan MC berikan nanti.

Remarks:Disebabkan MODUL From 60 to A+ ini khas untuk pelajar kategori sederhana dan cemerlang

maka dibawah ini merupakan Tajuk penuh yang perlu anda ulangkaji

BAHAGIAN A (6 Soalan Wajib Jawab semua) TOTAL MARKS= 40

NO TAJUK FULL MARKS

1 Simultaneous Equation 5 markah

2 Trigonometry Function 6 markah

3 Progression 6 markah

4 Vector 8 markah

5 Statistic 7 markah

6 Differemtiation + Integration 8 markah

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EXAM TIPS: QUESTION 3-PROGRESSION Form 5

EXAM TIPSPercayalah soalan sebegini tidak sesukar mana

Soalan (a)Langkah 1:bina progression.sekurangnya 3

sebutan

_________ , _______ , ________

*untuk membina 3 sebutan ini anda perlu

semak dahulu soalan mahu AREA,volumeatau perimeter. *

*Dalam soalan ini,ianya meminta pelajar

dapatkan area rectangles.*

Langkah 2 : Cari common ratio.Tentukan GP

or AP.supaya pelajar tahu apakah formula

yang perlu digunakan

2

3

1

2 .T

Tatau

T

Tr

*Ikuti 2 langkah ini sudah mencukupi.bukan

sahaja lulus tapi anda akan dapat lebih dari

itu.*

Soalan (b) iApabila x = 60 y = 40.progressionnya adalah

2400 , 600 , 1501 nn arT

Soalan (b) ii

r

aS

1

EXAM TIPS: QUESTION 4-VECTOR Form 5

EXAM TIPS

Untuk Kertas 2,tajuk Vector kebiasanyaa maklumat

yang diberi akan mengandungi bentukRATIO atauFRACTION.

Soalan ditanya kebiasaan ADA 3 Bahagian.

TOLONG FAHAM FORMAT SOALAN INI

SANGAT PENTING.KEBIASAAN SAMA

SAJA SOALAN DAN CARANYA

Bentuk soalanya.berkait-kait.ini bermakna jika a (i)

anda salah maka a(ii) juga akan salah.begitulah

berikutnya.Ini kerana untuk mendapatkan Jawapan

a(ii) anda perlukan Jawapan daripada a (i)

Ini bermakna

Jawapan a (i) dapatkan maklumat dari soalan

Jawapan a (ii)-sebahagianya dari soalan dan satu

lagi PASTI dari Jawapan a (ii)

Jawapan b (i) dan b (ii)juga samasahaja.jawapanya dibentuk dari Jawapan dari a

**Soalan (a) biasanya 3 markah.dan penyelesainya

tidaklah sesukar mana.pelajar seharusnya skor. **

Jawapan (a) (i)PQ = PO+OQ

= - 4y +3OA

= -4y + 3 (8x)= 24 x4 y

Jawapan (a) (ii)

Jawapan (b) (i)

PB = mPA= m (PO+OA)

= m (-4y +8x)

= 8mx -4my

Jawapan (b) (ii)BC = nOC

= n (6x+3y)

= 6nx + 3ny

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EXAM TIPS: QUESTION 6 QUADRATIC FUNCTION

Jawapan (a)Coordinate A?Jika diperhatikan coordinate A adalah (0,y)

Maka y yang ingin di cari adalah y intercept..Dari persamaan diberi y intercept adalah 11

Maka A= (0,11)

Jawapan (b)

]114

)2

[()(

]1122

[)(

]1122

[)(

11)(

22

22

2

22

2

2

kkxxf

kkkxxxf

kxxxf

kxxxf

Anda perlu tukar dari cbxaxxf 2)( kepada CTS

qpxa 2)(

Langkah 1: faktorkan nilai a (letak diluar)[jika ada]

Langkah 2:tambahkan formula ini22

22

Langkah 3: gantikan ruangan kosong dengan nilai

disebelah x. dalam soalan ini adalah nilai k