ADDMATHS-SPM(MPT-60%-TO-A+)

MATHS Catch

USAHA +DOA+TAWAKAL

MODUL PERSEDIAAN TERAKHIR

SPM

FOKUS A+

2012

SPM 2012

ADDITIONAL MATHEMATICS CONTENT/ISI KANDUNGAN BAHAGIAN 1

A. Preview Modul Persediaan Matematik Tambahan SPM (MPT SPM) B. Preview Keseluruhan Keseluruhan tentang Matematik Tambahan SPM Kertas 1 dan Kertas 2, 2012

2 3

BAHAGIAN 2

A. B. C. D.

Tips Peperiksaan Terakhir Terakhir Kertas 1 / Last exam Tips PAPER 1 Cadangan Soalan Ramalan Pilihan Kertas 1 / Suggestion Question PAPER 1 Tips Peperiksaan Terakhir Kertas 2/ Last Exam Tips PAPER 2 Cadangan Soalan Ramalan Pilihan Kertas 2 / Suggestion Question PAPER 2

5 23 40 51

IMPROVE from ‘60’ TO SCORE A+ [Lihat halaman sebelah] SULIT 3472/1 2012 Maths Catch Network © www.maths-catch.com MODUL PERSEDIAAN PERSEDIAAN 2 ADD MATHS [From [From 60 to A+]

MOHD RAJAEI BIN MOHAMAD ALI

Hak Cipta Terpelihara

MATHS CATCH

NETWORK

1

MATHS Catch

USAHA +DOA+TAWAKAL

MODUL PERSEDIAAN TERAKHIR

SPM

FOKUS A+

2012

SPM 2012

ADDITIONAL MATHEMATICS CONTENT/ISI KANDUNGAN BAHAGIAN 1

A. Preview Modul Persediaan Matematik Tambahan SPM (MPT SPM) B. Preview Keseluruhan Keseluruhan tentang Matematik Tambahan SPM Kertas 1 dan Kertas 2, 2012

2 3

BAHAGIAN 2

A. B. C. D.

Tips Peperiksaan Terakhir Terakhir Kertas 1 / Last exam Tips PAPER 1 Cadangan Soalan Ramalan Pilihan Kertas 1 / Suggestion Question PAPER 1 Tips Peperiksaan Terakhir Kertas 2/ Last Exam Tips PAPER 2 Cadangan Soalan Ramalan Pilihan Kertas 2 / Suggestion Question PAPER 2

5 23 40 51

IMPROVE from ‘60’ TO SCORE A+ [Lihat halaman sebelah] SULIT 3472/1 2012 Maths Catch Network © www.maths-catch.com MODUL PERSEDIAAN PERSEDIAAN 2 ADD MATHS [From [From 60 to A+]

MOHD RAJAEI BIN MOHAMAD ALI


MATHS CATCH

NETWORK

1

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

ADDITIONAL MATHEMATICS [Improve From 60 to A+]

LAST EXAM TIPS PAPER 1 2012 MOHD RAJAEI BIN MOHAMAD ALI

[Lihat halaman sebelah] SULIT 3472/1 2012 Maths Catch Network © www.maths-catch.com MODUL PERSEDIAAN PERSEDIAAN 2 ADD MATHS [From [From 60 to A+]


MATHS CATCH

NETWORK

2

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

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 mungkin dalam 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 boleh meningkat 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 d an gagal MENANGKAP KATAKUNCI yang terdapat didalam soalan menyebabkan pelajar tidak tahu apakah formula yang sesuai digunakan dan bagaimana nak bermula. Oleh itu modul ini disusun bagi membantu pelajar mengatasi masalah utama ini SPM’08 QUESTION Function Q1 Function Q2 Q3 Function Q4 Quadratic Equation Q5 Quadratic Function Quadratic Function Q6 Indices & Logarithms Q7 Indices & Logarithms Q8 Progression Q9 Q10 Progression Q11 Progression Q12 Linear Law Coordinate Geometry Q13 Coordinate Geometry Q14 Vectors Q15 Vectors Q16 Q17 Trigonometric Function Q18 Circular Measure Differentiation Q19 Differentiation Q20 integration Q21 Statistics Q22 Q23 Permuatation&Combination Q24 Probability Q25 Probability Distribution

SPM’09

SPM’10

Function Function Function Quadratic Equation Quadratic Function Quadratic Function Indices & Logarithms Indices & Logarithms Progression Progression Progression Circular Measure Vectors Vectors Coordinate Geometry Trigonometric Function Trigonometric Function Integration Differentiation Differentiation Integration Permuatation&Combination Probability Statistics Probability Distribution

Function Function Function Quadratic Function Quadratic Equation Quadratic Function Indices & Logarithms Indices & Logarithms Progression Progression Progression Linear Law Coordinate Geometry Coordinate Geometry Vectors Vectors Circular Measure Trigonometric Function Differentiation Differentiation Differentiation Statistics Permuatation&Combination Probability Probability Distribution

SPM’11 Function Function Function Quadratic Function Quadratic Equation Quadratic Function Indices & Logarithms Indices & Logarithms Progression Progression Progression Linear Law Coordinate Geometry Coordinate Geometry Vectors Vectors Circular Measure Trigonometric Function Differentiation Differentiation Differentiation Statistics Permuatation&Combination Probability Probability Distribution

FORMAT KERTAS 1 ADD MATHS SPM JUMLAH SOALAN : 25 JUMLAH MARKAH : 80 Analisis Soalan Peperiksaan Sebenar SPM 2008-2011 mengikut susunan soalan dan topik

*Tajuk diwarnakan hitam adalah tajuk-tajuk dari Tingkatan 4.*

[Lihat halaman sebelah] SULIT 3472/1 2012 Maths Catch Network © www.maths-catch.com MODUL PERSEDIAAN 2 ADD MATHS [From 60 to A+]


MATHS CATCH

NETWORK

3

MATHS Catch SPM 2012

USAHA +DOA+TAWAKAL

FOKUS A+ Jika anda lihat stat istik bagi kertas 1 diatas,boleh dikatakan tajuk-tajuk yang keluar adalah sama sahaja setiap t ahun.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 k has untuk membantu pelajar 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 jika 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 d aripada 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 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25

CHAPTER Function Function Function Quadratic Equation Quadratic Function Quadratic Function Indices & Logarithms Indices & Logarithms Progression Progression Progression Linear Law Coordinate Geometry Coordinate Geometry Vectors Vectors Trigonometric Function Circular Measure Differentiation Differentiation integration Statistics Permuatation&Combination Probability Probability Distribution

TOTAL

MARKS 3 4 3 2 3 3 3 3 2 3 2 4 4 3 2 4 3 4 3 4 3 3

Untuk membantu pelajar memahami dengan lebih baik tajuk,bentuk dan format soalan,maka MC sudah sediakan modul dinamakan “Last Exam Tips [PAPER 1]”TOLONG JANGAN ABAIKAN SEMUA SOALAN DAN KONSEP yang diterangkan didalam modul ini..Semoga modul 2012 ini membantu ANDA SEMUA.selamat Berjaya.

4 4 4 80



MATHS CATCH

NETWORK

4


USAHA +DOA+TAWAKAL

FOKUS A+ EXAM TIPS: QUESTION 1-3 –FUNCTIONS Kebiasaanya Q1-Q3 adalah dari tajuk function.Untuk menjawab soalan ini anda WAJIB tahu 4 perkara ini a) Function Relation CASE 3: INVERSE FUNCTION b) Absolute Function c) Composite Function d) Inverse Function

Exam Tips 3 Langkah 1: tambahkan sendiri „y‟ [penting] Langkah 2 : terbalikkan kedudukan y dan x.ini bertujuan untuk menghilangkan sahaja

CASE 1:FUNCTION RELATION/FUNCTION NOTAION Domain Codomain Object Image Range

= {4, 9, 16} = {2, 3, 4, 5} = 4,9,16 = 2, 3, 4 = {2, 3, 4} *Image yang mempunyai objek sahaja*

p 

1

k epada p

Langkah 3: Cari nilai „y‟.maka nila y yang anda perolehi itulah inverse function

Relation between set A and B? = f ( x)  x Remarks: Soalan subtopic “Inverse function ini belum pernah tak keluar.Oleh itu sila berikan fokus yang lebih bagi subtopic ini”

CASE 2:ABSOLUTE FUNCTION EXAM TIPS Untuk (b) Sangat penting.jika anda mahu hilangkan modulus “ l l “ maka Jawapanya mestilah dipecah kepada dua iaitu (+) dan (-).

CASE 4: COMPOSITE FUNCTION

FOKUS

The functions of f a nd g a re defined as f : x → x − 4 and fg : x → 4 x+ 3. Find th e function g . [4 marks] Answer: Given f ( x) = x − 4 and fg ( x) = 4 x+ 3. fg ( x) = f [ g ( x)] = g ( x) − 4 g ( x) − 4 = 4 x + 3 g ( x) = 4 x + 7 ∴ g : x → 4 x + 7

EXAM TIPS Soalan ini sangat penting. Ideanya untuk mendapatkan nilai g pelajar perlulah membuat pemecahan nilai yang besar.iaitu fg(x) dan bukanya f(x)



MATHS CATCH

NETWORK

5

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

EXAM TIPS: QUESTION 4 –QUADRATIC EQUATIONS

CASE 3: Given two EQUALS roots EXAM TIPS Katakuncinya adalah Two equal roots

CASE 1: Given one roots EXAM TIPS Katakuncinya adalah one of the roots. Roots bermaksud nilai x tersebut. Diberi x = -2 .gantikan dalam persamaan diberi.maka anda akan dapat nilai p

Jawapan

LANGKAH 1 -Tukarkan quadratic equation diberi dalam bentuk (general form) -Jadikan diakhir nilai = 0 Jawapan

LANGKAH 2 Kenal pasti formula yang perlu digunakan.ada 3 sila buat pilihan.contoh disebelah katakuncinya adalah two equal roots .lain-lain sila Lihat dibawah Quadratic Equations 2 b -4ac > 0 Two real roots 2 Two equal roots b -4ac = 0 2 No real roots b -4ac < 0

CASE 2: Given two roots Given -3 and

1 4

are the roots of an equations 4x2 + bx + c = 0 . Find the

value of b and c

EXAM TIPS Katakuncinya adalah Apabila diberikan 2 roots

Jawapan

1  x  3  x  



Roots bermaksud nilai x tersebut. Dalam soalan ini ada 2 nilai roots. x  3

4 

1 3 x 2  x  3 x   0 4 4 4 x 2  x  12 x  3  0 4 x 2  11 x  3  0

Compare with 4x2 + bx + c = 0 Thus b = 11 , c = -3

x 

1 4

Langkah 1 Langkah yang perlu anda lakukan adalah pindahkan -3 ke sebelah kiri begitu juga nilai ¼ pindahkan juga ke sebelah kiri

Langkah 2 Bandingkan persamaan yang anda perolehi dengan yang diberi didalam soalan.

3472/1 2012 Maths Catch Network © www.maths-catch.com MODUL PERSEDIAAN 2 ADD MATHS [From 60 to A+]

CASE 4: STRAIGHT LINE+ CURVE (SIMULTANEOUS EQUATION)

FOKUS

The straight line y = 7 x+ 6 does not intersect with the curve y = − x2 + 9 x + n. Find the range of values of n. [3 marks] Answer: y = 7 x + 6 y = − x2 + 9 x + n − x2 + 9 x + n = 7 x + 6 − x2 + 2 x + n − 6 = 0 The equation does not intersect b2 − 4ac < 0 (2)2 − 4(−1)( n − 6) < 0 4 + 4n − 24 < 0 1+n−6<0 n < 5

LANGKAH 3 Kenal pasti formula yang perlu digunakan.ada 3 sila buat pil ihan.contoh disebelah katakuncinya adalah does not intersect = no real roots.lain-lain sila Lihat dibawah Quadratic Equations 2 Two real roots b -4ac > 0 2 Two equal roots b -4ac = 0 2 Hak Cipta Terpelihara b -4ac 0 No real roots

**Buat pilihan menggunakan 3 persamaan ini**

EXAM TIPS Katakuncinya adalah -Does not intersect bermaksud b2 − 4ac < 0 -diberi Strai ght li ne dan curve equation bermak sud penyelsaianya mestil ah menggunakan simul taneous equation

LANGKAH 1 Gunakan kaedah simultaneous equatiuon dengan cara samakan kedua-dua nilai y tersebut LANGKAH 2 Pindahkan semua nilai supaya disebelah kanann a bersamaan = 0

[Lihat halaman sebelah] SULIT MATHS CATCH

NETWORK

6

MATHS Catch

USAHA +DOA+TAWAKAL

FOKUS A+

SPM 2012

EXAM TIPS: QUESTION 5-6 –QUADRATIC FUNCTIONS

CASE 1: QF and Graphs

EXAM TIPS

FOKUS

LANGKAH 1 Sila hafal general form quadratic function ini.

f ( x)  a( x  m)2  n 1 m = adalah nilai

2 n = adalah nilai

x max atau min

y max atau min

--kemudian bandingkan dengan equation yang diberi

LANGKAH 2 SANGAT PENTING sila bandingkan

x min y min

f ( x)  2( x  m)2  3 f ( x)  a( x  m)2  n

EXAM TIPS: QUESTION 7-8 –Indices & Log CASE 1 : Pelajar Perlu Faktorise x  2

Solve the equation 2

2

x 1

8

( 2 x  2 2 )  ( 2 x  2 1 )  8

factorise nilai “x” dan bukan lagi nombor s eperti di

2 x ( 2 2  21 )  8

Lihat contoh disebelah anda perlu factorisekan nilai 2 x Soalan ini juga kebiasaanya bermain dengan sifir.contoh disebelah adalah sifir 2 dan dihujungnya adalah nilai 8.Ini kerana 8 juga merupakan factor bagi 2

tingkatan 3 dahulu.

2 x ( 4  2)  8 2 x ( 2)  8 2 x 

**Tips Penting!! ada kemungkinan Soalan boleh bertanya menggunakan sifir 3 dan dihujungkan adalah nilai 9,27,81 atau 243. Jika sifir 4 dihujungnya adalah nilai 16 atau 64 Jika sifir 5 dihujungnya adalah nilai 25 atau 125 Sifir 6 dan ketas jarang ditanya k erana nilainya akn mejadi terlalu besar.

8 2

2 4 x

Jawapan (a) Oleh itu n  3 .(ingat n adalah y min)

Jawapan f ( x)  2( x  m)  3

Jawapan (b)

(a).m  2

Untuk nilai m .tandanya anda perlulah mengambil berlawanan dari yang diberi.jika .dalam kes diatas x min -2.maka Jawapan m = 2 sahaja.

2

(b).n  3 (c). x  2

Jawapan (c) untuk Soalan c axis symmetry adalah garisan yang merentasi paksi x.dan pastikan anda meletak x=-2 dan JANGAN letak =-2 sahaja.perkataan x itu merujuk kepada persamaan (equation)

CASE 2: inequalities 2 Solve inequality 5 x  x  4

EXAM TIPS -Sangat Penting Tajuk Ini! LANGKAH 1 -Jadikan equation yang diberi dalam bentuk = 0 -dapatkan 2 nilai x tersebut LANGKAH 2 Lorekkan graphs bahagian dalam kerana dalam soalan menunjukkan < (less)

EXAM TIPS Pelajar perlulah menguasai konsep asas matematik semasa ditingkatan 3 dahulu iaitu “Factorise”.Cuma kalini ianya agak tinggi sedikit.pelajar perlulah

bandingkan 2 x  2 2 x  2 CASE 2 : Guna Konsep Sifir. Solve the equation 27

2 x  3



1 x  2

9 

3 3( 2 x 3)  9 x  2 

EXAM TIPS Sifir yang digunakan adalah sifir 3 Ini kerana nilai 9 dan 27 mempunyai nilai sepunyanya adalah 3

1 2

 2( x  2)  12   3     ( x  2 ) 3 Bandingkan

3(2 x  3)  ( x  2) x  1

Quadratic besar dari “0” [ > 0 ]

[Lihat halaman sebelah] SULIT

Quadratic kurang dari “0” [ <0] 3472/1 2012 Maths Catch Network © www.maths-catch.com MODUL PERSEDIAAN 2 ADD MATHS [From 60 to A+]


MATHS CATCH

NETWORK

7

MATHS Catch

USAHA +DOA+TAWAKAL

SPM

FOKUS A+

2012

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

FOKUS LOGARITHMS CASE 5 : INDICES KEPADA LOG

Solve the equation log 5 (7 x+ 2) = log 5 (3 x+ 4) + 1. [3 marks]

log5 (7 x + 2) − log 5 (3 x+ 4) = 1 7 x + 2 log5 = 1 3 x + 4 7 x + 2 = 5 3 x + 4

7 x+ 2 = 5(3 x+ 4) 7 x+ 2 = 15 x+ 20 9 x = − 4

EXAM TIPS

Solve the equation 8 −5 x + 3 = 7 4 x. Answer:

Dalam Soalan case ini base nya adalah sama iaitu 5..Pelajar hanya perlu fikir bagaimana cara untuk factorise kanya sahaja.

8−5 x + 3 = 7 4 x log10 8 −5 x + 3 = log 10 7 4 x

Langkah 1 Pindahkan log5 (3 x+ 4) ke sebelah kiri Langkah 2 Factorisekan kedua-dua log tersebut menggunakan  x  Laws of log iaitu Log a x  Log a y  Log a    y  Langkah 3 Pindahkan log5 ke sebelah kanan bersamaan 5 1  x  Log a    c Laws of indices  y   x  c  y   a  

Given that log 25 s − log5 t = 0, express si n terms of t . [3 marks]

EXAM TIPS Soalan case ini base nya TIDAK SAMA.Langkah pertama mestilah pelajar perlu fikir untuk samakan base nya dahulu kemudian barulah guna Teknik factorise untuk menyelesaikanya.

**Kiri (8) dan kanan (7) Cara penyelesainya.pelajar perlulah t ambahkan log disebelah kiri dan kanan

= 4 x log10 8 −5 x + 3 = 0.9358 4 x −5 x+ 3 = 3.7432 x

(−5 − 3.7432) x = −3 −3 x = −8.7432 = 0.3431

LOGARITHMS (Mirip Trial Selangor 2012) CASE 6: INDICES KEPADA LOG



FOKUS

1 6

Answer: log( 2 x .3 x )  log

1 6

log 2  log 3  log x

log25 s − log5 t = 0 log25 s = log 5 t log5 s = log 5 t log 5 25 log5 s = log 5 t 2 log5 5 log5 s= 2 log 5 t log5 s = log 5 t 2 2 ∴ s = t

Teknik diguna apabila nilai disebelah kiri dan kanan tiada factor sepunya atau tiada sifir yang sesuai digunakan.

(−5 x+ 3) log 10 8 = 4 x log 10 7 −5 x + 3 log10 7

Solve the equation 2 x.3 x

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

EXAM TIPS

x

1

6 x log 2  x log 3  0.7782 x (0.30103)  x(0.47712)  0.7782 0.7782 x  0.7782 x 

 0.7782 0.7782

 1



MATHS CATCH

NETWORK

8

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

EXAM TIPS: QUESTION 9-11 –PROGRESSION

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

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

EXAM TIPS A.P Langkah 1 :Jika anda susun nilai yang diberi ianya akan menjadi begini. 7  x , 2 x , 13  x

T 2

T 1

EXAM TIPS Selain soalan dari case 1. Case 2 ini juga sangat popular teruatamanya dalam percubaan negeri SPM 2011. ***Per‟11,SBP‟11,Ter‟11,Phg‟11, Negeri 9,Sabah „11***

T 3

Jawapan (a)

Formula G.P n 1 T n  ar 

Langkah 2 d  T 2  T 1  T 3  T 2 *Dari cara ini anda HANYA akan perolehi nilai x sahaja.*

Jawapan (b) Gunakan formula sum to infinity.

Langkah 3 Gunakan formula ini untuk dapatkan ,d d  T 2  T 1

G.P

EXAM TIPS Caranya adalah sama seperti diatasyANG BERBEZA adalah yang ini menggunakan formula GP sahaja.

G.P (Jawapan a) Langkah 1 Jika anda susun nilai yang diberi ianya akan menjadi begini. x  20 , x  4 , x  20

T 1

T 2

T 3

RINGKASAN PELAJAR WAJIB TAHU MENGGUNAKAN FORMULA DIBAWAH A.P (1) d  T 2  T 1

atau d  T 2  T 1  T 3  T 2 (guna untuk mencari nilai x) (2) T n  a  (n  1)d *digunakan apabila sesuatu soalan menyebut tentang -sesuatu n th term (cthnya Given nth term,How many number between..,Find number of term.)

(3) S  n [2a  (n  1)d ] * digunakan apabila sesuatu soalan menyebut tentang Sum of term, n 2

Langkah 2 T 2 T 1



T 3 T 2

*Dari cara ini anda HANYA akan perolehi nilai x sahaja.*

Langkah 3 Gunakan formula ini untuk dapatkan ,r r 

T 2 T 1

Jawapan (b) : Gunakan formula sahaja


G.P (1)

r 

T 2

atau

T 1

T 2 T 1



T 3 T 2

(guna untuk mencari nilai x)

(2) T n  ar n 1 * **digunakan sama seperti A.P n (3) S  a(r  1) syaratnyanilai r  1 n r  1

n (4) S  a(1  r ) syaratnyanilai r  1 n

1  r

(5) S  a **digunakan apabila sesuatu soalan menyebut tentang  1  r [Lihat halaman sebelah] - Sum to infinity SULIT -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 Hak Cipta Terpelihara S 1

a MATHS

CATCH

9

**Kedah Trial „11** NETWORK


USAHA +DOA+TAWAKAL

FOKUS A+

CASE 2: EXAM TIPS: QUESTION 12 –LINEAR LAW

CASE 1: Log 10 y againts log10 x

y x

againts x EXAM TI PS Trick penyelesaian ADALAH SAMA sahaja seperti dalam case 1

EXAM TI PS Pelajar perlulah tahu persamaan umum linear law iaitu y  mx  c

Langkah 1 Fikir bagaimana cara nak tukar equation y dari maklumat d iberi KEPADAy seperti didalam graphs

Kebiasaanya Soalan yang akan disoal adalah “express equation” dan “find coordinate” . sahaja Trick penyelesaian untuk SEMUA jenis soalan linear Law adalah SAMA sahaja.

Equation dalam maklumat tiada x dibawah y.oleh itu anda perlulah tambahkan x dibawahnya supaya sama dengan graphs

Langkah 1 Fikir bagaimana cara nak tukar equation y dari maklumat d iberi KEPADA y seperti didalam graphs

Langkah 2 Jadikan dalam bentuk y = mx + c Untuk memudahkan senaraikan seperti dibawah y = x = m= c=

Caranya tambahkan log10 p ada equation yang diberi disebelah kiri dan kanan

Langkah 2 Gunakan konsep log iaitu

log a mn  log a m  log a n

dari contoh disebelah

Langkah 3

y 

y

Lihat p ada Graphs

Jadikan dalam bentuk y = mx + c Untuk memudahkan senaraikan seperti dibawah y = x = m= c=

x x  x

dari cth disebelah

* * * Trial Terengganu‟ 11 Pahang ‟11 , Sarawak‟11, Melaka‟11 * * * * *

y  log10 y

Lihat pada Graphs

x  log10 x

Lihat pada Graphs

m  gradient



Lihat pada Graphs

m  gradient



y 2 - y1 x 2 - x1

c  y  int ercept

y 2 - y1 x 2 - x1

c  y  int ercept * * * Trial Kedah‟ 11 Kelantan‟11, Selangor‟11, Negeri sembilan‟11* * * * *




MATHS CATCH

NETWORK

10


USAHA +DOA+TAWAKAL

FOKUS A+

CASE 2: PARALLEL

EXAM TIPS: QUESTION 13-14 –COORDINATE GEOMETRY Berdasarkan ANALISIS 13 soalan Percubaan Negeri SPM 2012 menunjukkan tajuk ini terbahagi kepada 2 bentuk.

BENTUK 1 -perpendicular -parallel -Area

BENTUK 2 -collinear -ratio -Distance/locus

CASE 1: PERPENDICULAR

EXAM TI PS Katakunci adalah perpendicular (90 0 ) Langkah 1 Tuliskan formula yang ingin diguna

m1  m2  1 ***m = gradient *Kebiasanya 1 gradient diberi ( m1 ) dan satu lagi perlu dicari ( m2 )*

EXAM TI PS Katakunci adalah parallel Bermaksud m1  m2 Langkah 1 Cari gradient bagi kedua-dua persamaan diberi. px  y  1

y  px  1 Gradientnya,m1 = p

a = x - intercept b = y - intercept

**bermakna dari soalan m  3 2 4

Langkah 2 Samakan keduanya m1  m2

p 

3 4

Langkah 2

Cari gradient

elalui equation m1 m

yang diberikan 1 y  x  2 3 Bandingkan dengan y  mx  c

Untuk equation ini gradientnya boleh diperolehi dari formula x y b   1,  m2   a b a

CASE 3: AREA

EXAM TI PS Katakunci adalah Area

Maka m  1 1 3

Jangan lupa ulang semula coordinate pertama dan letakkan di kedudukan terakhir

Langkah 3 Dapatkan

elalui formula m2 m

m1  m2  1 1

 m2  1 3 m2  3 Langkah 4 Cari Equation PQ

y  y1  m( x  x1 ) y  6  3( x  0) y

 3 x  6



MATHS CATCH

NETWORK

11

MATHS Catch SPM

FOKUS A+

2012

CASE 4: COLLINEAR

USAHA +DOA+TAWAKAL EXAM TI PS Katakunci Collinear Bermaksud segaris.maknanya tiada luas. (Area = 0)

CASE 6: DISTANCE/LOCUS

EXAM TI PS Katakunci disini adalah 2PQ=QR

0 0 0 0

1 2

a

7

2 4

3

17

1 2 1

2 4

0

*Ianya bermaksud jarak QR bersamaan 2 PQ.* **Penyelesaianya gunakan formula distance/locus

[(2)(3)  17 a  7( 4)]  [ 4(a)  3(7)  17(2)]

2 PQ  QR

21a  21

2 ( x2  x1 ) 2  ( y2  y1 ) 2 

2 0  21a  21 21

( x2  x1 ) 2  ( y2  y1 ) 2

EXAM TIPS: QUESTION 15-16 - VECTOR

a

21 1 a

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

CASE 5: RATIO

2 BENTUK DIATAS AKAN SECARA SELANG SELI AKAN MENYOAL 4 PERKARA DIBAWAH EXAM TI PS Katakunci adalah point Q divide PR with ratio 2:1  nx  mx2 ny1  my2  ( x, y)   1 ,  m  n   m  n

a) b)

“Express in terms of”-Lihat contoh. Find Unit vector

unit .vector 

c)

Magnitude Vector ada modulus

d)

Vector ada parallel Atau collinear

xi  yj

012 2 2012

x 2  y 2

r  x 2  y 2

2012

CASE 1: SOALAN DARI AYAT (Unit Vector) 7 −1 Given that a = and b = 2 −13 , find ~ ~ (a) the vector a − b. ~ ~ (b) the unit vector in the direction of a − b. ~ ~

()

( )

(a) a − b ~ 7 −1 = − −13 2 8 = 15 (b) Unit vector in the direction of a − b ~ ~

() ( ) ( )



8 + 15 8i + 15j

2


17


Unit Vector xi  yj  x 2  y 2

8i + 15j 2



EXAM TIPS


MATHS CATCH

NETWORK

12

MATHS Catch SPM

FOKUS A+

2012

EXAM TIPS

USAHA +DOA+TAWAKAL CASE 2: SOALAN DARI AYAT (MagnitudVector) j , find Given that a = 3 i + 9 ja nd b = −2i + k ~ ~ (a) a − b in the form xi + y j. ~ (b) the value of k if | a − b| = 13. ~ ~

CASE 4: SOALAN DARI GAMBARAJAH (unit Vector)

→

Diagram 9 shows vector ORd rawn on a Cartesian plane.

Magnitude Vector

r  x 2  y 2

Jawapan (a)

(b)

a − b ~

= 3i + 9j − (−2i + kj) = 3i + 9j + 2i − kj = 5i + (9 − k)j |a − b| = 13 ~

~

13 52  (9  k ) 2  13 =

25 + (9 − k)2 = 169 (9 − k)2 = 144

CASE 3: SOALAN DARI AYAT (parallel) Given that a= (2k-1) i + 3 j and b = 4 i + 5 j, find k if ~ ~ 2a+3b is parallel to y-axis

9 − k = −12

(a)

k = 21 or

(b)

9 − k = 12 k = −3

(b)

 2 k  1   4   2  3    3  5       4k  2   12          6   15   4k  10       21 

a  qb

 4k  10   0    21    q  y      

→

Express ORi n the form

( x y). →

Find the unit vector in the direction of OR.

Answer (a) → OR =

Answer Langkah 1 : cari hasil 2a +3b dahulu.

Langkah 2 Parallel to y – axis bermaksud = (0,y)

Diagram 9

EXAM TIPS 12 5

( )

Unit vector in the direction of OR = (12 i +5 j) / 122 + 5 2 ~ ~ = (12 i +5 j) / 13 ~ ~

EXAM TIPS Parallel Vector a  qb , q i s scalar...

Langkah 3 Dapatkan nilai k (4k+10 = q(0) 4k =-10 k =-5/2



Unit Vector xi  yj  x 2  y 2

→


MATHS CATCH

NETWORK

13

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

EXAM TIPS: QUESTION 17 –TRIGONOMETRY FUNCTION CASE 1: Six Trigonometry Functions of Any Angles Given cos x = −0.4761 and 90° ≤ x ≤ 270°, find the value of 1 (a) x (b) sec x

2

[4 marks] Answer: [2 marks]

Answer: (a) cos x = −0.4761

cos α = 0.4761 α = 61° 34'

x= 180° + 61° 34'

(b)

Solve the equation 20 sin x = −sin x + 24 sin 30° for 0° ≤ x ≤ 360°

= 241° 34' 1 sec x = cos x 1 =

−0.4761 = −2.1

20 sin2 x = −sin x+ 24 sin 30° 20 sin2 x = −sin x + 12 20 sin2 x+ sin x − 12 = 0

(4 sin − 3)(5 sin + 4) = 0 3 4 x= 48° 35', 131° 25' or 4 sin x = − 5 x= 233° 8', 306° 52' sin x =

Pelajar hasruslah tahu cara menggunakan formula dibawah

Pelajar hasruslah menghafal dan tahu cara menggunakan formula dibawah



MATHS CATCH

NETWORK

14

MATHS Catch

USAHA +DOA+TAWAKAL

FOKUS A+

SPM

EXAM TIPS: QUESTION 18 – CIRCULAR MEASURE

2012

3

Solve the equation −5 cos 2 x= 13 sin x − 9 for 0° ≤ x ≤ 360°

Exam Tips

[2 marks]

−5 cos 2 x= 13 sin x − 9 −5(1 − 2 sin 2 x) = 13 sin x − 9 −5 + 10 sin 2 x= 13 sin x − 9 10 sin2 x − 13 sin x+ 4=0 (5 sin − 4)(2 sin − 1) = 0 4 5 x= 53° 8', 126° 52' or 1 sin x = 2 x= 30°, 150° sin x =

2.

 180 0 Arc, s  r 

3.

Area,

1.

Answer:

 .rad

1 2

2

r 

**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.*

Pelajar hasruslah tahu cara menggunakan formula dibawah



MATHS CATCH

NETWORK

15

MATHS Catch

USAHA +DOA+TAWAKAL

FOKUS A+

SPM 2012

EXAM TIPS: QUESTION 19 – INTEGRATION

EXAM TIPS: QUESTION 20 – STATISTICS CASE 1: MEAN & STANDART DEVIATION

Exam Tips 5



[3 g ( x ) kx]dx  18

2

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

Format yang pertama LANGKAH 1: Anda perlu tahu buat PEMECAHAN dr formula 5

5





3 [ g ( x )dx k xdx  18 2

2

 x  5 3(8)  k   dx  18 2 2 2  5



2

 x 2   k   5  18  24  2 2  x 2   k   5  6  2 2  25 4  k   6  2 2  k 

4 7

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 KEDUA Contoh Soalan 6 Given that f ( x)dx  7



Exam Tips ,Find

2

**Dalam bab ini terlalu banyak formula perlu dif ahami.bukan Lupakan yang lain.WAJIB HAFAL!3 formula Dibawah sudah mengcukupi.

FORMULA

ISTILAH (Sangat Penting)

2

(a)

 f ( x)dx 6

Penyelesaianya. Jika diperhatikan perbezaanya,nilai limit pada maklumat yang diberi adalah 6,2 manakala pada soalan adalah terbalik dari nilai tersebut iaitu 2,6. Apa yang perlu anda buat adalah tambahkan nilai – ve sahaja.maka jawapanya adalah (-7) *Soalan ini diramalkan akan keluar*L

a)

Standard Deviation,  

b)



c)

Mean, x 

Variance,

 x N

2

 ( x) 2

-Sum of Set Square Number= -Sum of Set Number=

x

2

 x

N



2



 x N

2

 ( x) 2


 x


MATHS CATCH

NETWORK

16

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+ EXAM TIPS: QUESTION 21 –PROBABILITY

2012

EXAM TIPS Soalan Probability ini sebenarnya Sangat mudah. Jika contoh disebelah diambil pelajar perlu tumpukan pada bilangan kad dipilih (TWO CARD ARE D RAWN) dan satu lagi sila fokuskan pada katakunci warna yang sama (same color) Kita buat kesimpulan 2 kad dipilih mestilah warnaya adalah sama.oleh itu kebarangkalianya (Probability) adalah seperti berikut

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. Answer (a)

Mean, x¯ =

∑x

N 10 + 15 + 6 + 15 + 5 + 17 + 14 = 7 82 = 7 = 11.71 Median = 1 4 Mode = 15 (b) New mean = 9(11.71) + 7 = 112.39 New median = 9(14)+ 7 = 133 New mode = 9(15) + 7 = 142





Exam Tips **Case ini sangat penting untuk SPM 2011

Jawapan (a) Gunakan formula sahaja **Mean

Mean, x¯ =



∑x N

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

**Median Pelajar perlu susun mengikut s usunan menaik terlebih dahulu 5, 6, 10, 14, 15, 15, 17

Tolak 1 kerana tadi anda sudah pilih satu card Maka 5-1=4

Median = nilai ditengah-tengah = 14 Mode Kekerapan tertingi = 15

Jawapan (b) Percayalah soalan ini sangat senang!!

Katakunci adalah “Multiplied by 9, and then added 7 ” New mean New median

PERHATIAN: Soalan diatas meminta anda dapatkan probability bagi warna kad yang sama (same color) Bagaimana jika soalan meminta pada anda warna yang tidak sama ( NOT same color) Jangan panic mudah sahaja.not same color bermakna keduaduanya BUKANw arna yang sama. Apa yang perlu dilakukan adalah (1- same color) 

,

 3 2   7 6   5 4           15 14   15 14   15 14 





34 105





New mode

 1 

Total Card (R+Y+G) = 15

34

105 71

105



MATHS CATCH

NETWORK

17

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

CASE 3: Rate of Changes

EXAM TIPS: QUESTION 22 –DIFFERENTIATION

CASE 1: Equation Tangent Find the equation of the tangent to the curve

y = (2 x − 5) ( x

− 4) at the point (4, 0).

Exam Tips atakunci adalah EQUATIONt angent curve. **K **Formulanya adalah y  y1  m( x  x1 ) ** m 

Answer:

dy dx

 gradient Answer:

Langkah 1 Didalam soalan coordinate

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

( x1 , y1 ) sudah

diberi iaitu (4, 0)...Oleh itu pelajar hanya perlu dapatkan nilai gradient sahaja

Langkah 2 Gunaklan formula y  y1

At point (4, 0), dy = 4(4) − 13 dx

The volume of a sphere increases at the rate of 44.8π cm3 s −1. Find the radius of the sphere at the instant when its radius is increasing at a rate of 0.7 cm s −1. 4 [Volume of sphere, V = πr 3] 3

 m( x  x1 ) u ntuk

mendapatkan equation

V =

4 3 πr 3

dV = 4πr 2 dr dV = 44.8π dt dr = 0.7 dt

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

Equation:

y = 3( x − 4) = 3x – 12 y

CASE 2: Minimum Point 2

The curve y = 3 x + 6 x − 7 has a minimum point at x = k , where k is a constant. Find the value k .

dx

2

y = 3 x + 6 x − 7 dy = 6 x + 6 dx

6 x+ 6=0 x = −1 ∴ k = −1

Langkah 1

Minimum point akan menjadikann nilai dy dy = 0 dx

dx

adalah 0

Langkah 3

dV = pelajar perlu buat differentiation dr dV = Volume changes of t = 44.8π dt dr = Radius changes of t =0.7 dt

CASE 4: Small of C hanges

Exam Tips **K atakunci adalah “Small Change in

Find the small change in volume of a sphere when the radius increases from 6 cm to 5.93 cm. 4 [Volume of sphere, V = πr 3] 3

Volume

dV = 4πr 2 dr δr = 5.93 − 6

δV δr ≈

Langkah 2

dV dV dr = × dt dr dt

4r 2 = 64 r 2 = 16 r = 4 cm

4 V = πr 3 3

dx

**Formulanya adalah

44.8π 4πr 2 = 0.7

Answer:

Pelajar perlu gunakan kaedah differentiation untuk dapatkan nilai dy

Answer:

When yi s minimum,

Exam Tips **K atakunci adalah MINIMUM POINT **Formulanya adalah dy  0

dV dV dr = × dt dr dt 44.8π = 4π r 2× 0.7

Exam Tips **K atakunci adalah “rate ” dan unit cm3 s−1.

dy

**Formulanya adalah

δV dV δr ≈ 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 sahaja..Soalan ini adalah tajuk differentiation.apa jua soalan yang ditanya pasti akan meminta pelajar membuat dy dx

dx

dV δV ≈ dr × δr = 4πr 2 × −0.07

= 4π(6)2 × −0.07 = −10.08π cm3

*Dalam case 3 dan 4 pelajar p erlu membuat dV sama sahaja dengan dr

dV dr

dy dx

Dapatkan nilai x.




MATHS CATCH

NETWORK

18

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

EXAM TIPS: QUESTION 23 –PERMUTATION & COMBINATION

** Trial Kelantan dan Terengganu 2011**

Total

CASE 1: PERMUTATION

P T arg et

RAMALAN 1 Diagram shows seven letter cards.

*Trial Sabah 2011*

U

RAMALAN 1 How many 3-digit numbers that are greater than 400 can be formed using the digits 1, 2, 3, 4, and 5 without repetition? Answer:

2

= p1 x = 24

4

p 2

Exam Tips **K atakunci adalah 3-digit numbers dan greater than 400 Langkah 1: Buat 2 garis seperti dibawah

P b  c. P d

CASE 2: PERMUTATION

2

p1

b = adalah digit pertama [gerenti = 1] a = Total digit mengikut syarat diberikan.Syarat “How many number ” .Jawapan 4 dan 5 } 2 nombor greater than 400

p1

F

I

O

M

R

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

2

N

4

p2

Answer: a) b)

d = adalah baki digit yang tinggal [dalam soalan sebelah = 2 digit. 1 digit sudah diambil maka baki yang tinggal adalah 2]

7

p5 = 2520

6

p 4 x 4 p1

= 1440

4

p2 c = total digit keseluruhan yang tinggal = 4 Asalnya ada 5 nombor.tetapi 1 nombor sudah diambil maka total bakinya adalah 4 s aha a.

RAMALAN 2

RAMALAN 2 Diagram shows six numbered cards. How many 4-digit even numbers can be f ormed using the digits 1, 2, 3, 4, and 5 without repetition?

Exam Tips

Answer: =

2

p1 x

Even Number = 2 dan 4 } Total Digit pertama = 2 4

p3

= 48

Baki digit [4-1 = 3 digit] Digit pertama.Gerenti 1

RAMALAN 3 How many 4-letter codes can be formed using the letters in the word 'GRACIOUS' without repetition such that the first letter is a vowel?

1

4

7

5

8

9

A four-digit number is to be formed by using four of these cards.How many Exam Tips a) different numbers can be formed? Jawapan (a) b) different odd numbers can be formed? Ingat kembali formulanya

Answer: a) b)

6

P4 = 360 P3 x 4P1 = 240 5

Total

P T arg et

Target adalah four-digit number Total adalah = 6 digit

Answer: =

4

p1 x

7

p3


= 840



MATHS CATCH

NETWORK

19


USAHA +DOA+TAWAKAL

FOKUS A+

RAMALAN 2 A 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 of a) 5 boys, b) not more than 2 girls.

CASE 3: COMBINATION ** Trial Johor,Melaka,Perak,SBP dan Selangor 2011**

RAMALAN 1 A 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? Answer: 10

C 3 x

Answer: a)

8

C 5 x 5C 3

b)

If the team consists of 8 boys and 0 girl

= 560

11

C 3

If the team consists of 7 boys and 1 girl



= 19800 If the team consists of 6 boys and 2 girl

8





8 8

C 8 x 5C 0 = 1 5

C 7 x C 1

= 40

5

C 6 x C 2

= 280

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

Exam Tips BOY Step 1:

Ingat formula Step 2 :

Total Boy = 10 Dipilih = 3 (Target

GIRL

C

C

RAMALAN 3 (SPM2011) A debating team consists of 6 students. These 6 students are chosen from 2 monitors, 3 assistant monitors and 5 prefects.

Total

C T arg et !

10

C 3

a) there is no restriction, b) the team contains only 1 monitor and exactly 3 prefects.



11

C 3

Total Boy = 11 Dipilih = 3 (Target)

Answer: a) b)

Exam Tips Jawapan (a) No restriction = tiada syarat automatic gunakan formula asal

Total

C T arg et

10

2

C 6

= 210 3

5

C 1 x C 3 x

C 2 =

60

Exam Tips Jawapan (b) TARGET 6 students ..Ingat formula

Total

C T arg et

Ada Syarat Langkah 1 :Isikan target mengikut syarat diberikan .iaitu M = 1,P = 3 automatik AM = 2 MONITOR PREFERCTS ASSISTANT MONITOR

C 1

C 3

C 2

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

C 1

x

5

C 3

x

3

C 2 =

60



MATHS CATCH

NETWORK

20


USAHA +DOA+TAWAKAL

FOKUS A+

FOKUS 2012

EXAM TIPS: QUESTION 25 –PROBABILITY DISTRIBUTION

CASE 5: ** Trial Pahang,Perlis 2011**

RAMALAN 1 Diagram shows five cards of different letters.

R

J

A a) b)

I

Exam Tips Apa yang pelajar perlu faham jumlah bagi kedua-dua luas dibawah graph adalah 0.5 + 0.5 =1

Diagram below shows a standard normal distribution graph.

Area = 0.5

Area = 0.5

N

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

Answer: a) b)

(a) 5! 4! x 2!

= 120 = 48

The probability represented by the area of t he shaded region is 0.803 a) Find the value of P ( Z  k )

EXAM TIPS: QUESTION 25 –BINOMIAL DISTRIBUTION

b)

FOKUS 2012 1

The discrete random variable X h as a binomial probability distribution with n= 4, where ni s the number of trials. Table 1 shows the probability distribution of X . P(X = x ) x Find 1 (a) the value of k . 0 16 (b) P( X ≥ 3). [3 marks] 1 1 [3 markah] 4 Answer: k 2 3 4

1 4 1 16

:

1 1 1 1 − − − =1 16 4 4 16 3 k= 8 1 1 (b) P( X ≥ 3) = + 4 16 5 = 16 (a)

P ( Z  k ) 

1  0.803 2

 0.0985

Perlu dibahagi 2 kerana nilai k didalam gambarajah ada 2

X is a continuous random variable which is normally distributed with a mean of  a nd a standard deviation of 2.If the value of X is 85 when the Z  score is k ,find the value of  .[3 marks]

Sebab total = 1

(b) Hafal Formula ini

Z 

X   

X  Random variable   Mean

Exam Tips



Nilai z mestilah diperolehi dari dalam graph. Lihat muka surat disebelah



Z 

k+

Standard deviation

Z  score X  

1.29 

 85  

2   82.42

Exam Tips TOTAL Probability = 1

FOKUS 2012 [Lihat halaman sebelah] SULIT



MATHS CATCH

NETWORK

21

MATHS Catch SPM

FOKUS A+

2012 Bagaimana Z =1.29?dan bukan =0.0985?

USAHA +DOA+TAWAKAL

Z 

X  

 85   2   82.42

1.29 

SUGGESTION QUESTION PAPER 1 [Improve From 60 to A+] Ini kerana nilai z tersebut anda p erlu dapatkan dari dalam table.Didalam peperiksaan table ini akn diberikan kepada anda.cara untuk dapatkan adalah s eperti diatas. Z=1.29

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



MATHS CATCH

NETWORK

22

MATHS Catch SPM

FOKUS A+

2012

1

USAHA +DOA+TAWAKAL

5

SUGGESTION QUESTION 1-3: FUNCTION Diagram 1 shows the relation bet ween set

2

M a nd set N i n the arrow diagram.

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) Let f −1(2) = k So f (k ) = 2 k + 4=2 k = −2

Therefore, f −1(2) = −2 (b) Let f −1( x) = y So f ( y) = x + 4 = x = x − 4 Therefore, f −1( x) = x − 4

[2 marks] State the (a) images of 12 (b) objects of 2 (c) domain of this relation (d) range of this relation

(a) (b) (c) (d) [2 marks]

Answer: (a) 2, 3 (b) 6, 12, 16 (c) {6, 12, 16} (d) {2, 3}

3

4

Given function f : x → −8 x − 2, find the (a) image of −7 (b) object which has the image −34 [3 marks] Answer:

(a)

f(−7) = −8(−7) − 2

(b)

f(x) = −34 −8x − 2 = −34 −8x = −32

= 54

x=4

The fuction f i s defined as f : x → x+ 4. Find (a) −1(2) (b) −1( x) [3 marks] Answer:

Answer: 2, 5 16, 20 {16, 20} {2, 5}

= |31| = 31 (6) = |(6)2 − 5| = |31| = 31 (7) = |(7)2 − 5| = |44| = 44 ( x) = 4

|x2 − 5| = 4

So,

x2 − 5 = 4 x2 = 9 x = −3, 3

y =

1 − x 8

Therefore f −1( x) =

1 − x 8

gf −1( x) = g ( f −1( x)) 1 − x = g ( )

8

1 − x

= −2( 8 ) − 5 x − 21

Given the function f : x → |x2 − 5|. (a) Find the images of −6, 6 and 7. (b) Find objects which have the image of 4. [4 marks] Answer:

(−6) = |(−6) − 5|

6 The fuctions f a nd g a re defined as f : x → 1 − 8 xa nd g : x → −2 x − 5. Find gf −1( x). [3 marks] Answer: −1 Let f ( x) = y So f ( y) = x 1 − 8 y = x

=

8

7

The functions of f a nd g a re defined as f : x → x+ 3 and g : x → 4 − x. Find the composite function of gf a nd fg . [3 marks] Answer: Given f ( x) = x+ 3 and g ( x) = 4 − x. gf ( x) = g ( f ( x)) = g ( x+ 3) = 4 − ( x + 3) = − x + 1 g ( x) = f ( g ( x)) = f (4 − x) = (4 − x) + 3 = − x + 7

4 The functions of f a nd g a re defined as f : x → x − 9 and g : x → 9 x − 3. Find the composite function of gf a nd fg . [3 marks] Answer: Given f ( x) = x − 9 and g ( x) = 9 x − 3. gf ( x) = g ( f ( x)) = g ( x − 9) = 9( x − 9) − 3 = 9 x − 84 g ( x) = f ( g ( x)) = f (9 x − 3) = (9 x − 3) − 9 = 9 x − 12

and

−( x2 − 5) = 4 x2 = 1 x = −1, 1



MATHS CATCH

NETWORK

23

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

5

SUGGESTION QUESTION 4: QUADRATIC EQUATION

The straight line y = 5 x+ 6 does not intersect with the curve y = 6 x2 + 5 x + q. Find the range of values of q.

7 1

5 It is given that i s one of the roots of the 2 quadratic equation 4 x2 − 8 x + q= 0. Find the value of q. [2 marks] Answer:

2

[3 marks]

Form the quadratic equation which has the 3 roots −3 and 4 .

[3 marks]

Answer: 3 3 ) + (−3)(4 ) = 0 4 9 9 x2 − (− ) x + (− ) = 0 4 4 4 x2 + 9 x − 9 = 0 x2 − (−3 +

q = −5

(−4q − 3) x2 − 2 x − 4 = 0 has two different roots, qi s a constant. Find the range of values of q.

Answer:

[4 marks]

5 5 4( ) 2 − 8( ) + q = 0 2 2 25 − 20 + q = 0

The quadratic equation

y = 5 x + 6 y = 6 x2 + 5 x + q 6 x2 + 5 x + q = 5 x + 6 6 x2 + q − 6 = 0

Answer:

The equation does not have real roots b2 − 4ac < 0 (0)2 − 4(6)(q − 6) < 0 − 24q+ 144 < 0

(−2)2 − 4(−4 q − 3)(−4) > 0 4 − 64 q − 48 > 0 −64q > 44

(−4q − 3) x2 − 2 x − 4 = 0 The equation has two different roots b2 − 4ac > 0

q>−

−24q < −144

11 16

q > 6

8 3

The quadratic equation −4 x2 + mx + n = 0 has roots 8 and −9. Find t he values of m

4

The quadratic equation 4 x2 + px + q= 0 has roots −1 and −5. Find the values of pa nd q.

and n. [4 marks]

6

The straight line y = 6 − 4 xd oes not intersect with the curve y = − x2 + x + m. Find the range of values of m.

[3 marks]

[3 marks]

Answer: Answer: ( x − a)( x − b) = 0 ( x − 8)( x+ 9) = 0 x2 + x − 72 = 0 −4 x2 − 4 x+ 288 = 0 Therefore, m = −4 and n= 288

( x − a)( x − b) = 0 ( x+ 1)( x+ 5) = 0 x2 + 6 x+ 5=0 4 x2 + 24 x+ 20 = 0 Therefore, p= 24 and q= 20

Answer: y = 6 − 4 x y = − x2 + x + m − x2 + x + m = 6 − 4 x − x2 + 5 x + m − 6 = 0

The equation does not have real roots b2 − 4ac < 0 (5)2 − 4(−1)( m − 6) < 0 25 + 4 m − 24 < 0 4m < −1 1 m<− 4



The quadratic equation 8 x − px+ 2 = x has two equal roots, pi s a constant. Find the values of p. [3 marks] Answer: 8 x2 − px+ 2 = x 8 x2 + (− p − 1) x+ 2=0 The equation has two equal roots b2 − 4ac = 0

(− p − 1)2 − 4(8)(2) = 0 p2 + 2 p + 1 − 64 = 0 p2 + 2 p − 63 = 0 ( p − 7)( p+ 9) = 0 p= 7 or p = −9


NETWORK

24

MATHS Catch FOKUS A+

SPM 2012

1

USAHA +DOA+TAWAKAL

SUGGESTION QUESTION 5-6: QUADRATIC FUNCTION Diagram 3 shows the graph of a quadratic function f ( x) = ( x + s)2 − 7, where si s a constant.

2

3 Diagram 1 shows the graph of a quadratic function f ( x) = 5( x + h)2+ 6, where h is a constant. 4 Diagram 2 shows the graph of a quadratic function f ( x) = ( x + s)2 − 6, where s is a constant.

Diagram 4 shows the graph of a quadratic unction f ( x) = 2( x + m)2 − 7, where m is a constant.

Diagram 1

The curve y = f ( x) has a minimum point (7, t ), where t i s a constant. State (a) the value of s (b) the value of t (c) the equation of the axis of symmetry [3 marks] Answer: (a) s = −7 (b) t = −7 ( ) x = 7

5

Find the range of the values of xs uch that

The curve y = f ( x) has a minimum point (4, n), where ni s a constant. State (a) the value of m (b) the value of n (c) the equation of the axis of symmetry [3 marks] Answer:

(a) (b) (c)

= −4 = −7

The curve y = f ( x) has a minimum point (8, k ), is a constant. State (a) he value of h (b) he value of k (c) he equation of the axis of symmetry [3 Answer: (a) (b) (c)

x( x − 4) ≥ −4.

Find the range of the values of x such that x( x − 7) ≥ −12. [3 marks]

[3 marks]

Answer: x( x − 7) ≥ −12

Answer: x( x − 4) ≥ −4

x2 − 7 x + 12 ≥ 0 ( x+ 4)( x + 3) ≥ 0

The curve y = f ( x) has a minimum point (6, t ), where t i s a constant. State (a) the value of s (b) the value of t (c) th equation of the axis of symmetry Answer:

x=4

7 6

h = −8 k = 6 x = 8

Diagram 2

Find the range of the values of xs uch that (−4 x − 2)( x + 4) ≤ −2 x+ 8. [3 marks] Answer:

(a) s = −6 (b) t = −6 (c) x = 6

(−4 x − 2)( x + 4) ≤ −2 x + 8 −4 x2 − 18 x − 8 ≤ −2 x + 8 −4 x2 − 16 x − 16 ≤ 0 −4( x2 + 4 x + 4) ≤ 0 −4( x+ 2)( x + 2) ≤ 0

x2 − 4 x + 4 ≥ 0 ( x+ 2)( x + 2) ≥ 0

The range of values of x is x ≤ 2 or x ≥ 2

The range of values of x is x ≤ 3 or x ≥ 4

The range of values of x is

x ≤ −2 or x ≥ −2




NETWORK

25


SPM 2012

1

USAHA +DOA+TAWAKAL

SUGGESTION QUESTION 7-8: INDICES & LOG Solve the equation 2 x + − 2 x + = 8.

2

SUGGESTION QUESTION 9-11: PROGRESSION

Solve the equation 2 x + − 2 x + = 64.

1

arithmetic progression are 13 − x, 8, −13

[3 marks] Answer:

x +

Three consecutive terms of an

3

+ 3 x. Find the common difference of t he progression. [4 marks]

Find

Answer

(a)

the value of x.

−13 + 3 x − 8 = 8 − (13 − x)

(b)

the sum from the 4th term to the 6th term.

Answer: 2 x + 7 − 2 x + 6 = 64 2 x27 − 2 x26 = 2 6 128(2 x) − 64(2 x) = 26 64(2 x) = 26 2 x + 6 = 2 6 x+ 6=6 x = 0

x +

2 −2 =8 2 x23 − 2 x22 = 2 3 8(2 x) − 4(2 x) = 23 4(2 x) = 23 2 x + 2 = 2 3 x+ 2=3 x = 1

2 x = 16 x = 8 d = 8 − (13 − (8))

[3 marks]

=3 Answer:

2

The first three terms of an arithmetic progressino are −7h + 2, k , −5h+ 12.

(a)

3

Solve the equation 32 x − 2 = 8 −5 x.

4

Answer: 32 x − 2 = 8 −5 x 25( x − 2) = 2 3(−5 x) 5 x − 10 = −15 x (5 + 15) x = 10 1 x = 2

(a)

Express k i n terms of h.

(b)

Find the 6th term of the progression in terms of h.

Solve the equation 25 4 x + 5 = 125 5 − 5 x.

[3 marks]

[3 marks] Answer: 254 x + 5 = 125 5 − 5 x 2(4 x + 5) 5 = 5 3(5 − 5 x) 8 x + 10 = 15 − 15 x (8 + 15) x = 15 − 10 5 x = 23

[4 marks] Answer:

(b)

(a)

(b)

k − (−7h + 2) = (−5 h + 12) − k 2k = −12h+ 14 k = −6h + 7 a = −7h + 2 d = k + 7 h − 2 = (−6h+ 7) + 7 h − 2 = h + 5 T 6 = a + 5 d = −7h+ 2 + 5( h+ 5) = −7h+ 2 + 5 h+ 25 = −2h+ 27


The first three terms of a geometric progression are x, 24, 96.


96 24 =4 x 24 = 24 96 24 x= × 24 96 =6 r=

S6 =

6(46 − 1)

4−1

= 8190 S3 = 6 + 24 + 96 = 126

Sum = 8190 − 126 = 8064


NETWORK

26

MATHS Catch

USAHA +DOA+TAWAKAL

FOKUS A+

SPM

3

The variables xa nd ya re related by the equation y2 = 6 x( x − 6). Diagram 3 shows the straight line graph

2012

SUGGESTION QUESTION 12 : LINEAR LAW

obtained by plotting

1

2

The variables xa nd ya re related by the equation y = px−5, where pi s a constant. (a) Convert the equation y = px−5 to linear form. (b) Diagram 1 shows the straight line obtained by plotting log10 ya gainst log 10 x.

y2 a gainst x. x

4

The variables xa nd ya re related by the equation y2 = 2 x( x − 6). Diagram 4 shows the straight l ine graph

The variables xa nd ya re related by the equation y = px−4, where pi s a constant. (a) Convert the equation y = px−4 to linear form. (b) Diagram 2 shows the straight line obtained by plotting log10 ya gainst log 10 x.

obtained by plotting

y2 a gainst x. x

Diagram 3 Find the value of pa nd q. [2 marks]

Diagram 4

Answer:

Find the value of pa nd q. [2 marks]

2

Diagram 1

Diagram 2

Find the value of (i) log p.

Find the value of (i) log p.

(ii)

(ii)

10

10

q.

Answer:

(a)

q.

Answer:

log10

(a)

l

= −5

(b)

(i) (ii)

log10 p = Y −intercept =7 log10 y = −5(3) + 7

Answer: y2 = 2 x( x − 6) y2 = 2( x − 6) x 2 y = 2 x − 12 x y2 When = 0, x 0 = 2 x − 12 x = 6 p = x = 6 When x = 2, y2 = 2(2) − 12 x

= −18 (b)

(i)

log10 p = Y −intercept =7

= −8

q= log 10 y = −8

y = 6 x( x − 6) y2 = 6( x − 6) x 2 y = 6 x − 36 x y2 When = 0, x 0 = 6 x − 36 x = 6 p = x = 6 When x= 3, y2 = 6(3) − 36 x q =

y2 = −18 x

= −8 (ii)

log10 y = −4(4) + 7

q =

= −9 q= log 10 y

y2 = −8 x

= −9



MATHS CATCH

NETWORK

27

MATHS Catch FOKUS A+ SUGGESTION QUESTION 13-14: COORDINATE GEOMETRY

SPM 2012

1

USAHA +DOA+TAWAKAL

Find the equation of the straight line hich passes through P (5, 5) and is arallel to the straight line joining Q(7,

2

−3) dan R(−6, 7).

Find the equation of the straight line which asses through A(−4, 1) and is erpendicular to the straight line joining

B(−9, −7) dan C (−1, −4).

[3 marks] Answer:

4

[3 marks]

7 f − 5e 5 f + 14 e (−1, 11) = ( e + f , e + f ) 7 f − 5e = −1 e + f

Answer: Gradient of BC

Gradient of QR 7+3 =

−6 − 7 10

= −13 Equation: 10 y − 5 = − ( x − 5) 13 −13 y+ 65 = 10 x − 50 10 x + 13 y − 115 = 0

3

Given the area of a triangle with vertices

A(−3, 10), B(−9, 4) and C (4, j) is 81 unit 2. Find the possible values of j.

=

−4 + 7 −1 + 9

7 f − 5e = −e − f 8 f = 4 e

3 = 8 3 ( ) m2 = −1 8 8 m2 = − 3 Equation: 8 − 1 = − 3 ( x+ 4) 3 y − 3 = −8 x − 32 8 x + 3 y+ 29 = 0

[3 marks] Answer: Area of ABC 1 −3 −9 4 −3 = | | = ±81 2 10 4 j 10 1 ((−12) − 9 j + 40 − (−90) − 16 + 3 j) = 2 ±81 1 (102 − 6 j) = ±81 2

The point J (−1, 11) divides a straight line joining P (7, 5) and Q(−5, 14) in the ratio e : f . Find the values of ea nd f . [3 marks] Answer:

e = 2 f 2 and f = 1 ∴ e=

5

Find the equation of the locus of moving point As uch that its distances from P (−9, −7) and Q(− Answer:

Let A = ( x, y) Given AP = 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 + 18 x+ 81 + y2 + 14 y+ 49 = x2 + 14 x+ 49 + y2 + 18 y+ 81 4 x − 4 y = 0

−6 j = ±162 − 102 = −10 or j= 44



MATHS CATCH

NETWORK

28


USAHA +DOA+TAWAKAL

FOKUS A+

5

SUGGESTION QUESTION 15-16: VECTORS

→

Diagram 8 shows vector OP d rawn on a Cartesian plane.

6 1

(a)

5 45 Given that a = 10 and b = , find 6 48 ~ ~ (a) he vector 10 a − b. ~ ~ (b) he unit vector in the direction of 10 a ~ − b. ~ [4 marks] Answer: 10a − b ~ ~ 5 45 = 10 − 48 6 5 = 12 Unit vector in the direction of 10a − b ~ ~ = (5i+ 12j ) / 52 + 122 ~ ~ = (5i+ 12j ) / 13 ~ ~

()

( )

() ( ) ( )

(b)

2

The vectors aa nd ba re non-zero and non~ ~ arallel. It is given that ( h − 7)a = ( k − 9)b where h ~ ~ and k a re constants.

3

(a)

(b)

j, find Given that a = 4 i + 3 j and b = −4i + k ~ ~ (a) a − b in the form xi + y j. ~ ~ (b) the value of k if | a − b| = 17. ~ ~ [4 marks] Answer: a − b ~

= 4i + 3j − (−4i + kj) = 4i + 3j + 4i − kj = 8i + (3 − k)j |a − b| = 17

~

Express in the form xi + y j. (a) → OP

(b)

~ 82 + ( 3 − k )2 = 17

3 − k = −15 = 18 or

3 − k = 15 = −12

Find the value of (a) h (b) k [2 marks] Answer: a and ba re non-parallel. ~ ~ Therefore (h − 7)a = ( k − 9)b is true only when ( h ~ ~ − 7)a and ( k − 9)b are zero vector. ~ ~ Since aa nd ba re 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

Find the unit vector in the direction of

→

→

Answer: (a) → OP= 4i + 3j (b) → Unit vector in the direction of OP = (4i +3j ) / 42 + 32 ~ ~ = (4i +3j ) / 5 ~ ~ 4 3 = i + j 5 5

It is given that OA = 12 a + 7 b , OB = −32 a + ~ ~ ~

(a) (b)

→

Express OAi n the form

( x y).

Find the unit vector in the

→

direction of OA. [3 marks] Answer:

(a)

→

OA =

(56)

→

21b and OC = 78 a − 14b . ~ ~ ~ (a) → → Find AB and AC . (b) Hence, show that points A, Ba nd C are collinear. [4 marks] Answer: (a)

→

→

(b)

Unit vector in the direction of = (5i +6 j) / 52 + 6 2 ~ ~ = (5i +6 j) / 7.81 ~ ~

→

AB = AO + OB = −(12a + 7 b) − 32 a + 21 b

~

~

~

~

= −44a + 14 b ~

→ →

→

~

AC = AO + OC = −(12a + 7 b) + 78 a − 14b

~ ~ = 66a − 21b ~ ~

~

~

(b)

→

3 (−44a + 14 b) 2 ~ ~

AC = −

3→ = − 2 AB



Diagram 9 [

64 + (3 − k)2 = 289 (3 − k)2 = 225

4

→

Diagram 9 shows vector OAd rawn on a Cartesian plane.


MATHS CATCH

NETWORK

29


SPM 2012

1

USAHA +DOA+TAWAKAL

SUGGESTION QUESTION 17: TRIGONOMETRIC FUNCTION Given tan x = −0.6486 and 0° ≤ x ≤ 180°, find the value of (a) x (b) cot x [2 marks] Answer:

2

(a) tan x = −0.6486

tan α = 0.6486 α = 32° 58' x = 180° − 32° 58'

(b)

= 147° 2' 1 cot x = tan x 1 =

−0.6486 = −1.542

Given sin x = −0.8026 and 0° ≤ x ≤ 180°, find the value of (a) x (b) cosec x [2 marks] Answer:

4

0° ≤ x ≤ 360°

sin α = 0.8026 α = 53° 23' x = 180° − 53° 23'

Solve the equation 8 cos 2 x = 6 cos x

− 2 cos 60° for 0° ≤ x ≤ 360°

= 126° 37' 1 cosec x = sin x 1 =

Answer:

tan x+ sin x = 0 sin x + sin x = 0 cos x sin x+ cos x sin x = 0 sin x(1 + cos x) = 0

8 cos2 x = 6 cos x − 2 cos 60° 8 cos2 x = 6 cos x − 1 8 cos2 x − 6 cos x+ 1=0

or 1 + cos x = 0 cos x = −1 x= 180°

6

[4 marks]

Answer:

sin x = 0 x= 0°, 180°, 360°

−0.8026 = −1.246

5

a[4 marks]

(a) sin x = −0.8026

(b)

Solve the equation tan x+ sin x= 0 for

(2 cos − 1)(4 cos − 1) = 0 1 2 x= 60°, 300° or 1 cos x = 4 x= 75° 31', 284° 29' cos x =

Solve the equation −21 cos 2 x= 46 sin x − 33 for 0° ≤ x ≤ 360° [2 marks]

3

Answer:

Solve the equation −16 sin x= 0 sin x − 18 sin 30° for 0° ≤ x ≤ 360° [4 marks] Answer:

−16 sin2 x= 0 sin x − 18 sin 30° −16 sin2 x= 0 sin x − 9 −16 sin2 xs in x+ 9 = 0 (4 sin − 3)(−4 sin − 3) = 0 3 4 x= 48° 35', 131° 25' or 3 sin x = − 4 x= 228° 35', 311° 25' sin x =

−21 cos 2 x= 46 sin x − 33 −21(1 − 2 sin 2 x) = 46 sin x − 33 −21 + 42 sin 2 x= 46 sin x − 33 42 sin2 x − 46 sin x+ 12 = 0 21 sin2 x − 23 sin x+ 6=0 (3 sin − 2)(7 sin − 3) = 0 2 3 x= 41° 49', 138° 11' or 3 sin x = 7 x= 25° 23', 154° 37' sin x =



MATHS CATCH

NETWORK

30

MATHS Catch

USAHA +DOA+TAWAKAL

FOKUS A+

SPM 2012

SUGGESTION QUESTION 18: CIRCULAR MEASURE 1

Diagram 2 shows two arcs, ADa nd BC , of wo concentric circles with centre Oa nd aving radius OAa nd OB.

2

SUGGESTION QUESTION 19: INTEGRATION

Diagram 10 shows the sector OABo fa circle with centre O.

1

1

Given that ∫2 f ( x) dx= 7, find (a)

2

2

(a)

the value of ∫ 1 f ( x) dx.

Diagram 10

2 1 ∫1 f ( x) dx = −∫2 f ( x) dx

(b)

1 315 ∫2 [ kx + 3 f ( x)] dx = 16 1 1 315 ∫2 kx dx + 3∫2 f ( x) dx = 16 x2 1 315 k [ ] + 3(7) = 2 2 16 1 315 k ( − 2) + 21 = 2 16 7 k = 8

Diagram 2 Find (a) the angle θ, in radian. (b) the perimeter of the shaded region, ABCD. [4 marks] Answer:

Find (a) the angle θ in radians, (b) the area of the shaded region [4 marks] Answer:

(a) (a)

Length of arc BC = r θ

BOC = 36.87°

∠

8

(b)

Perimeter of the shaded region = 8 + 0.88 + 16 + 16 = 40.88 cm

4

9

(a)

∫9 f(x) dx = −∫ 4f (x) dx = −1

(b)

∫4 [kx − 3f(x)] dx = 62

9 9

9

∫4 kx dx − 3∫ 4 f (x) dx = 62 x2 9 k[ ] − 3(1) = 62 2 4 81 k( − 8) − 3 = 62 2 k=2

π θ = 36.87 × 180

θ = 18 = 0.44 rad. Length of arc AD = r θ = 2 × 0.44 = 0.88 cm

12 15

= 0.8

8 = (2 + 16)θ

(b)

cos∠ BOC =

= −7

4

the value of ∫ 9 f ( x) dx.

9 the value of k if ∫ [ kx − 3 f ( x)] dx 4 = 62. [4 marks] Answer:

1 the value of k if ∫ [ kx + 3 f ( x)] dx 2 315 = . 16 [4 marks] Answer: (b)

(a)

9

Given that ∫4 f ( x) dx= 1, find

= 0.6436 rad. (b)

Area of sector OAB =

1 2 r θ 2

1 = (15) 2(0.6436) 2 = 72.4 cm2 Area of triangle OCB 1 = × 9 × 12 2 = 54 cm2 Area of the shaded region = 72.4 − 54 = 18.4 cm2



MATHS CATCH

NETWORK

31


SPM 2012

1

(a)

USAHA +DOA+TAWAKAL

SUGGESTION QUESTION 20: STATISTICS 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] Answer: Mean =

27 9

Standard deviation =

Diagram 6 Find the interquartile range for the data. Answer:

Rearrange the data, 25 28 30 32 35 36 49 53 28 + 30 First quartile = = 29 2 36 + 49 Third quartile = = 42.5 2 Interquartile range

=3

(b)

3 Diagram 6 shows the mass of a group of students. 32, 30, 49, 25, 8, 35, 36, 5

2916 - 32 9

= 315 = 17.748

= 42.5 − 29 = 13.5

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, the mean,

(b)

the standard deviation.

Answer: Mean =

Diagram 7 Find the interquartile range for the data.

(a)

Answer: [3 marks]

(a)

4 Diagram 7 shows the mass of a group of students. 44, 53, 26, 49, 43, 33, 38, 30

27 3

=9

Rearrange the data, 26 30 33 38 43 44 49 53 30 + 33 First quartile = = 31.5 2 44 + 49 Third quartile = = 46.5 2 Interquartile range

= 46.5 − 31.5 (b)

Standard deviation =

2916 - 92 3

= 15

= 891 = 29.85



MATHS CATCH

NETWORK

32

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+ SUGGESTION QUESTION 21: PROBABILITY

2012

5

Diagram 1 shows the scores obtained by a group of players in a game. 1, 10, 1, 9, 3, 4, 2

1 (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. [3 marks] Answer:

(a)

Mean, x ¯ =

6

N

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] Answer: (a)

Mean, x ¯ =

(a) Probability 7 6 = × 15 14 1 = 5 (b) Probability 2 6 6 2 = × + × 15 14 15 14 4 = 35

∑ x

1 + 10 + 1 + 9 + 3 + 4 + 2 = 7 30 = 7 = 4.29 Median = 3 Mode = 1 (b) New mean = 5(4.29) + 4 = 25.45 New median = 5(3) + 4 = 19 New mode = 5(1) + 4 = 9

∑ x N

9 + 18 + 16 + 20 + 15 + 9 + 4 = 7 91 = 7 = 13 Median = 15 Mode = 9 (b) New mean = 3(13) + 6 = 45 New median = 3(15) + 6 = 51 New mode = 3(9) + 6 = 33

A bag contains 7 brown marbles, 2 yellow marbles and 6 purple marbles. T wo marbles are drawn from the bag at random, one after another. Find the probability t hat (a) both marbles are brown. (b) one marble is yellow and the other is purple. [4 marks] Answer:

2

A bag contains 6 blue marbles, 5 red marbles and 6 white marbles. Two marbles are drawn from the bag at random, one aft er another. Find the probability that (a) both marbles are blue. (b) one marble is red and the other is white. [4 marks] Answer: (a) Probability 6 5 = × 17 16 15 = 136 (b) Probability 5 6 6 5 = × + × 17 16 17 16 15 = 68




MATHS CATCH

NETWORK

33

MATHS Catch SPM

FOKUS A+

2012

3

2 . If Hock 3 Seng tried the game for three times, find the probability that she shot at least one target in t he game. [4 marks] Answer:

USAHA +DOA+TAWAKAL

SUGGESTION QUESTION 22: PPERMUTATION & COMBINATION 1

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

[3 marks] Answer:

Let A= Shot the target and B= Miss the target Probability = 1 − P( BBB) 1 1 1 = 1 − ( 3 × 3 × 3 ) 1 = 1 − 27 26 = 27

5

1 In a shooting game, the probability for a player to shoot the target for each try is . If Jabah 10 ried the game for three times, find the probability that she shot at least one target in the game. [4 marks] Answer: Let A= Shot the target and B= Miss the target Probability = 1 − P( BBB) 9 9 9 = 1 − ( 10 × 10 × 10 ) 729 = 1 − 1000 271 = 1000

5

4

3

Number of ways = 5 × 5 × 4 × 3 = 300

2 4

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?

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] Answer:

(a)

1

4

3

Number of ways = 1 × 4 × 3 = 12



MATHS CATCH

NETWORK

34

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

3

There are 2 different t-shirts and 2 trousers on a cupboard. Calcualte the number of different ways to arrange all the clothes in a row if (a) o condition is imposed. (b) all the trousers are next to each other. [4 marks] Answer:

5

(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 r oses can be chosen if (a) there is no restriction. (b) there are more red roses than white roses chosen. [4 marks] Answer: (a) Number of ways = 1 C 5 = 252 (b) The possible ways: 5 red roses and 0 white rose 4 red roses and 1 white rose 3 red roses and 2 white rose Number of ways = 5C 5 × 5C 5 + 5C 4 × 5C 4 + 5C 3 × 5C 3 = 126

A school wants to choose 4 students from a group of 6 boys and 5 girls to participate in a national mathematics contest. Calculate the number of ways the students can be chosen if (a) here is no restriction. (b) he students chosen consists of 2 boys and 2 girls. [4 marks] Answer: (a) Number of ways = 11C 4 = 330 (b) Number of ways = 6C 2 × 5C 2 = 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) here is no restriction. (b) here are more red roses than white roses chosen. [4 marks] Answer: (a) Number of ways = 10C 3 = 120 (b) The possible ways: 3 red roses and 0 white rose 2 red roses and 1 white rose Number of ways = 6C 3 × 4C 3 + 6C 2 × 4C 2 = 80



MATHS CATCH

NETWORK

35

MATHS Catch SPM

FOKUS A+

2012

1

USAHA +DOA+TAWAKAL

SUGGESTION QUESTION 23: DIFFERENTIATION

3

Given that the curve y = f ( x) and

dy = −4 px+ 10, where pi s a constant. dx

The gradient of the curve at x = −2 is 18. Find the value of p.

Find the equation of the tangent to the curve y = − x6 + 2 x5a t the point (2, 0). [3 marks]

[3 marks]

Answer: Answer:

= − x6 + 2 x5 dy = −6 x5 + 10 x4 dx

dy = −4 px+ 10 dx dy When x = −2, = 18 dx

At point (2, 0), dy = −6(2)5+ 10(2) dx

4

−4 p(−2) + 10 = 18 p = 1

= −32 The gradient of the tangent at point (2, 0) = −32 4

Equation:

= −32( x − 2) = −32 x+ 64 2

dx

Find the equation of the tangent to the curve y = (−2 x + 3)( x + 6) at the point (−3, 27). [3 marks] Answer:

= (−2 x + 3)( x+ 6) = −2 x2 − 9 x+ 18 dy = −4 x − 9 dx

Given y = (−4 x + 6)( x − 5). find (a) dy (b) the value of xw hen yi s maximum. (c) the maximum value of y. [4 marks] Answer: (a) y = (−4 x+ 6)( x − 5) = −4 x2 + 26 x − 30 dy = −8 x+ 26 dx

(b)

When yi s maximum,

At point (−3, 27),

−8 x+ 26 = 0

dy = −4(−3) − 9 dx

x =

=3

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

− 27 = 3( x+ 3)

(c)

13 4

y = (−4(

=

dy = 0 dx

13 13 ) + 6)(1( ) − 5) 4 4

49 4

= 3 x+ 36



MATHS CATCH

NETWORK

36

MATHS Catch

USAHA +DOA+TAWAKAL

FOKUS A+

SPM

7

2012

5

=

7

δ x = (−8 + p) + 8 = p dy

δ y = dx × δ x 7

= −256 p

the instant when its radius is increasing at a rate of 0.3 cm s −1. 4 [Volume of sphere, V = πr 3] 3

[3 marks]

2

19.2π

4πr = 0.3 4r 2= 64 r 2 = 16 r = 4 cm

−7

x2

= −256

The volume of a sphere increases at the rate of 19.2π cm s −1. Find the radius of the s phere at

dV = 4πr 2 dr dV = 19.2π dt dr = 0.3 dt dV dV dr = × dt dr dt 19.2π = 4π r 2× 0.3

8

Two variables, xa nd y, are related by t he equation y =

14 .

x2

Express the approximate change in y, in terms of p, when xc hanges from 3 to 3 + p, here pi s a small value. [3 marks] Answer: =

14 x2

dy −28 = dx x3 When x = 3 dy −28 = dx (3)3

28

= −27 δ x= (3 + p) − 3 = p dy

δ y = dx × δ x 28

= −27 p



.

dy 14 = dx x3 When x = −8 dy 14 = dx (−8)3

= 2400π cm3 s −1

Answer: 4 V = πr 3 3

−7 x2

Express the approximate change in y, in terms of p, when x changes from −8 to −8 + p, here pi s a small value. [3 marks] Answer:

The radius of a balloon in the shape of a sphere increases at the rate of 6 cm s −1. Find the rate of change of the volume of the balloon when the radius is 10 cm. 4 [Volume of sphere, V = πr 3] 3 [3 marks] Answer: 4 V = πr 3 3 dV = 4πr 2 dr dV dV dr = × dt dr dt = 4πr 2 × 6 = 24πr 2 When r = 10, dV = 24π(10)2 dt

6

Two variables, xa nd y, are related by t he equation y =


MATHS CATCH

NETWORK

37


USAHA +DOA+TAWAKAL

FOKUS A+ SUGGESTION QUESTION 25: PROBABILITY DISTRIBUTION 12

1

Diagram 2 shows a standard normal dist ribution graph.

Diagram 1 shows a standard normal dist ribution graph.

Diagram 2

Diagram 1 The probability represented by the area of t he shaded region is 0.7852. (a) Find the value of k . X i s a continuous random variable which is normally distributed with a mean of 79 (b) and a standard deviation of 16. Find t he value of X w hen the z -score is k . [3 marks] Answer:

(a)

(b)

The probability represented by the area of t he shaded region is 0.2486. (a)

Find the value of k .

(b)

X i s a continuous random variable which is normally distributed with a mean of 51 and a st andard deviation of 5. Find the value of X w hen the z score is k .

[3 marks]

P(Z > k) = 1 − 0.7852 = 0.2148 P(Z > 0.79) = 0.2148 ∴ k = −0.79

Answer:

μ = 79, σ = 16 X − 79

(a)

= 0.79

16 X = 79 + 16 × 0.79 = 91.64 (b)

P( Z > k) = 0.5 − 0.2486 = 0.2514 P( Z > 0.67) = 0.2514 0.67 ∴ k =

μ = 51, σ = 5 X − 51

= 0.67 5 X = 51 + 5 × 0.67 = 54.35



MATHS CATCH

NETWORK

38

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

MINI BENGKEL ADDITIONAL MATHEMATICS [Improve from 60 to A+]

LAST EXAM TIPS PAPER 2 2012 [Lihat halaman sebelah] SULIT 3472/1 2012 Maths Catch Network © www.maths-catch.com MODUL PERSEDIAAN 2 ADD MATHS [From 60 to A+]


MATHS CATCH

NETWORK

39

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

Section C (Answer 2 Question only from 4 q uestion) Total Marks=20 marks KUMPULAN SASARAN:Pelajar Yang mendapat Markah Kurang dari 40% FORMAT KERTAS 2 ADD MATHS SPM 2012 BAHAGIAN A (6 Soalan Wajib Jawab semua) BAHAGIAN B (5 soalan Wajib Jawab 4 sahaja) BAHIAGAN C (4 soalan Wajib Jawab 2 sahaja)

13

Total Markah =40 Total Markah = 40 Total Markah = 20

Section A (Answer all question) Total Marks= 40 marks Bahagian A mengandungi 6 SOALAN WAJIB yang mesti dijawab. SPM’07 SPM’ 08 SPM’09 SPM’10 Simultaneous Equation Quadratic Functions Progressions

5

Simultaneous Equation Coordinate Geometry Trigonometric Function Differentiation +Integration Statistics

Trigonometric Function Statistics

Simultaneous Equation Quadratic Equation Differentiation +Integration Trigonometric Function Vector

6

Progressions

Vector

Progressions

2 3 4

15

SPM’ 08

SPM’09

SPM’11

Simultaneous Equation Trigonometric Function Progressions

Simultaneous Equation Trigonometric Function Progressions

Integration

Integration

Geometry Coordinate Statistics

Geometry Coordinate Statistics

SPM’10

SPM’11

7

Linear Law

Integration

Integration

Linear Law

Linear Law

8

Vector

Linear Law

Linear Law

Differentiation

Differentiation

9

Circular Measure Integration

Circular Measure Geometry Coordinate Probablility Distribution

Geometry Coordinate Circular Measure Probablility Distribution

10 11

Probablility Distribution

Vector Probablility Distribution Circular Measure

Programming Motion Along straight Line

SPM’08

SPM’09

SPM’10

SPM’11

Solution of Triangle Index Number




Linear Programming 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?.

Section B (Answer 4 Question only from 5 question) Setiap Soalan=10 markah X 4 Total Marks=40 marks SPM’07


14 Linear

Berikut merupakan statistic soalan peperiksaan sebenar yang pernah keluar didalam Peperiksaan dari 2006- 2010..Sila LIHAT apakah PERSAMAANd an PEMBEZAAN dari Segi tajuk.

1

SPM’07

12

Vector Probablility Distribution Circular Measure

Untuk sekadar mencapai 40% dalam Matematik Tambahan adalah tidak sesukar mana.Pelajar tidak 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 i ni berulang kali,sekurangnya 1 kali tanpa melihat jawapan.Jika anda sudah pun mengulangkaji mengikut saranan ,selepas ini jangan 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 Lihat Jadual Dibawah BAHAGIAN A (6 Soalan Wajib Jawab semua) TOTAL MARKS= 40 NO TAJUK FULL MARKS Simultaneous Equation 5 markah 1 Trigonometry Function 6 markah 2

3 4 5 6

Progression Vector

6

Backup (Quadratic Eq+Quadratic Func)

6 markah 8 markah

Statistic

7 markah

Differemtiation + Integration

8 markah



MATHS CATCH

NETWORK

40


USAHA +DOA+TAWAKAL

FOKUS A+ BAHAGIAN B (5 soalan- Wajib Jawab 4 sahaja) TOTAL MARKAH =40 NO TAJUK MARKAH Linear Law

10 markah

Geometry Coordinate Circular Measure Integration (backup) Probability Distribution

10 markah 10 markah 10 markah 10 markah

BAHIAGAN C (4 soalan- Wa jib Jawab 2 sahaja) NO TAJUK

10 markah 10 markah

Linear Programming

10 markah

EXAM TIPS: Simultaneous Equation Form 4 Solve the simultaneous equations x + 2 y = −5 , x − y = −3 1

x = −5 − 2(−2)

= −1 14 Substitute y = − i nto (3) 3 14 x = −5 − 2(− ) 3 13 = 3 13 14 , y = − ∴ x = −1, y = −2 or x = 3 3

for

0  x  2

GRAPHS COS Persamaan umum graphs COS adalah y=cos x. sudut pertamanya

[5 marks] 2

Sketch the3graph of y 

EXAM TIPS 

x = −5 − 2 y −−−−( 3)

Substitute (3) into (2). (−5 − 2 y)2 − y2 = −3 4 y2 + 20 y + 25 − y2 = −3 3 y2 + 20 y+ 28 = 0 ( y+ 2)(3 y+ 14) = 0 2 y = −2 or y = −4 3 Substitute y = −2 into (3)

 2 sin 2 x

1 Sudut pertama   2

LAST EXAM TIPS PAPER 2 SECTION A

From (1),

Sketch the graph of y

MARKAH

Solution of Triangle Index Number (Backup)

x + 2 y = −5 −−−− (1) x2 − y2 = −3 −−−− (2)

EXAM TIPS: QUESTION 2- Trigonometric Function Form 5





Soalan ini Sangat mudah.walaubagaimanapun ramai pelajar ditolak markah jika tidak berhatihati. Kunci UTAMA adalah anda perlu membentuk EQUATION 3 Equation 3 datangnya dari equation 1.Equation 1 dipilih kerana Ianya lebih mudah dari equation 2

TIPS PENTING! *Didalam Add Maths Jawapan diakhir nanti anda akan mendapat 4 JAWAPAN..Ianya berbeza didalam matematik biasa iaitu 2 jawapan sahaja. x  ................., x  ................

y  ................., y  ................ Ramai pelajar hilang markah disini. INGAT! ada 4 jawapan.2 nilai x dan 2 nilai y.

Ini Tajuk Paling Mudah Dalam Add Maths.Anda Wajib Score

TARGET

1 4





2 3

3 4

cos 2 x for

0  x  2

PENTING

Case 1 Jika soalanya cos 2x ..sudut pertamanya adalah



5 4

2 x  7



4

2



1 2

1  , maka, x   4

Case 2 Jika soalanya cos

3

3

2 pertamanya adalah 3

1 Sudut pertama   4

2

x 

1 2

x ..sudut

1

 , maka, x   3

2

Sudut kedua   4

Sudut ketiga 

*Dapatkan nilai x tersebut* 3 4




4/5


adalah 1  2


MATHS CATCH

NETWORK

41


USAHA +DOA+TAWAKAL

FOKUS A+

EXAM TIPS: QUESTION 4-VECTOR Form 5

EXAM TIPS: QUESTION 3-PROGRESSION Form 5 EXAM TIPS

EXAM TIPS Percayalah soalan sebegini tidak sesukar mana Soalan (a) Langkah 1: bina progression.sekurangnya 3 sebutan

Untuk Kertas 2,tajuk Vector kebiasanyaa maklumat yang diberi akan mengandungi bentuk RATIOa tau FRACTION. Soalan ditanya kebiasaan ADA 3B ahagian.

_________ , _______ , ________

TOLONG FAHAM FORMAT SOALAN INI SANGAT PENTING.KEBIASAAN SAMA SAJA SOALAN DAN CARANYA

*untuk membina 3 sebutan ini anda perlu semak dahulu soalan mahu AREA,volume atau perimeter. * *Dalam soalan ini,ianya meminta pelajar dapatkan area rectangles .*

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)

Langkah 2 : C ari common ratio.Tentukan GP or AP.supaya pelajar tahu apakah formula yang perlu digunakan

r 

T 2 T 1

.atau 

T 3 T 2

*Ikuti 2 langkah ini sudah mencukupi.bukan sahaja lulus tapi anda akan dapat l ebih dari itu.* Soalan (b) i

Jawapan (a) (i) PQ = PO+OQ = - 4y +3OA = -4y + 3 (8x) = 24 x – 4 y

Jawapan (a) (ii)

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 sama sahaja.jawapanya dibentuk dari Jawapan dari a

Apabila x = 60 y = 40.progressionnya adalah

2400 , 600 , 150 T n  ar n 1

**Soalan (a)b iasanya 3 markah.dan penyelesainya tidaklah sesukar mana.pelajar seharusnya skor. **

Jawapan (b) (i) PB = mPA = m (PO+OA) = m (-4y +8 x) = 8mx -4my Soalan (b) ii

S  

a 1  r

Jawapan (b) (ii) BC = nOC = n (6x+3y) = 6nx + 3ny



MATHS CATCH

NETWORK

42


USAHA +DOA+TAWAKAL

FOKUS A+ CASE 2 : Mean,Interquatile and median EXAM TIPS: QUESTION 5-STATISTICS Form 4

CASE 1: Median

EXAM TIPS Soalan ini diramalkan akan keluar nanti.Apa yang perlu anda tahu adalah MENGGUNAKAN formula ini  N    F  2  C median  L   f m      WAJIB HAFAL!! 1. L=lower boundary of the MEDIAN CLASS 2. N=Total Frequency 3. F=cumulative frequency BEFOREm edian Class 4.

f m =frequency of the median class

5.

C=size class interval

Table1 show the distribution of the scores of 60 student in a test Score Frequency

1 -10 11 -20

10 16

21 -30 31 -40 41 -50

11 12 11

1. 2.

L=lower boundary of the MEDIAN CLASS=20.5 [Dari class median (21-30) ambil 21-0.5] N=Total Frequency=5+8+12+k+6+5 =

k  36

3. F=cumulative frequency BEFOREm edian Class=13 Anda perlu tambahankan 5+8=13. Nilai 5 dan 8 dipilih kerana BEFOREm edian class hanya ada nilai 5 dan 8 sahaja. 4. f m =frequency of the median class=12 5.C = size class =10 [c=Upper boundary-lower boundary =30.5-20.5=10] Gunakan formu la diberi .itu sahaja. Soalan (b) Minta anda lukis graphs.oleh itu nilai mode mesti diambil dari GRAPH juga



x 

(a) Find the mean of scores [3 marks] (b) Without drawing an ogive, find the Median and interquatile range the score [5 marks] 41  50  45.5 Jawapan (a) 2 

x  Soalan (a) LANGKAH 1~SANGAT PENTING!! Anda perlu kenalpasti median class terlebih dahulu.didalam soalan sudah diberitahu * median adalah 28.maka class median adalah 21-30. Inilah kunci soalan berbentuk begini.

EXAM TIPS Soalan (a) Soalan ini diramalkan juga akan keluar nanti.Apa yang perlu anda tahu adalah MENGGUNAKAN formula ini Mean:

 fx  f

WAJIB HAFAL!  fx  hasil .tambah... f  titik .tengah

 f  hasil .tambah... f

5.5(10)  15.5(16)  25.5(11)  35.5(12)  45.5(11)



60 1510

EXAM TIPS Soalan (b) Dapatkan class median terlebih dahulu. Class median =

60

 25.17

Jawapan (b) Median?

Class median 

 N    F  2  C median  L   f m        60     2   26    10  20.5   11        24.14 Interquatile = Third Quatile class = 31-40  3   (60)  37  4  10 Q3  30.5    12

 

Total Frequency 2



2

 30

Oleh itu class median adalah 21-30 ___________________________________

Interquatile = Q3-Q1  3   N  F  4 C Q3  L    f m    ,

 1   N  F  4 C Q1  L    f m   

First Quatile class = 11-20  1   (60)  10  4  10   Q1 10.5 16  

 

 37.17

 

 

 13.63

Interquatile = Q3-Q1 = 37.17 – 13.63 = 23.54 [Lihat halaman sebelah]

SULIT


60


MATHS CATCH

NETWORK

43


USAHA +DOA+TAWAKAL

FOKUS A+ EXAM TIPS: QUESTION 6 –QUADRATIC FUNCTION EXAM TIPS~Penting!

BAHAGIAN B

General Form quadratic equation ialah

f ( x)  a( x  p)  q 2

*Nilai q = y maks atau min* *Nilai p = x maks atau min*

PENTING

r

r

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) Anda perlu tukar dari f ( x)

Jawapan (b)

Dalam kertas 2 ,janganlah anda abaikan completing the square.ianya Sangat penting.Jika soalan tajuk ini keluar didalam SPM boleh dikatakan 99% Ianya pasti akan menyoal “completing the square ini”.

Linear Law

10 markah

Geometry Coordinate Circular Measure Integration Probability

10 markah 10 markah 10 markah 10 markah

Ked‟12 Per‟12 ,Sel‟11,Mel‟11, Jhr‟12Sarawak‟12

 ax2  bx  c kepada CTS

2 a( x  p)  q

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

f ( x)  x 2  kx  11

Langkah 2 :tambahkan formula ini

 

2

 

2

f ( x)  [ x 2  kx        11]  2   2  2

2

 k   k  f ( x)  [ x 2  kx        11]  2   2 

k k 2 f ( x)  [( x  ) 2   11] 2 4

x min*

BAHAGIAN B (5 soalan- Wajib Jawab 4 sahaja) TOTAL MARKAH =40 NO TAJUK MARKAH

y min*

2

         2   2 

2

Langkah 3: gantikan ruangan kosong dengan nilai disebelah x. dalam soalan ini adalah nilai k Langkah 4: jadikan a( x  p)2  q

*Nilai p = x min* =  k  3, k  6 2  k 2  11 r  4 *Nilai q = y min*  (6)2 11  4 2



MATHS CATCH

NETWORK

44

MATHS Catch SPM

FOKUS A+

USAHA +DOA+TAWAKAL EXAM TIPS: QUESTION 7 –LINEAR LAW

CASE 2:

y x

2012

againts . x

CASE 1: Log 10 y againts . x 2

EXAM TIPS Linear law mempunyai pelbagai bentuk.walau bagaimanapun caranya adalah sama sahaja..Apa yang sama?

EXAM TIPS Soalan (a) 



Soalan (a)  

Mesti kena bina table baru. Nak bina table ini Lihat pada maklumat didalam soalan.Soalan mahukan anda bina table

Log 10 y againts . x  

2

Buatlah table baru Plotkan graph mengikut scale yang diberi.pastikan sekurangnya 5 point anda adalah betul

*6 markah sudah ditangan ketika ini*

Soalan (b) Ramai yang melakukan kesilapan disini Sebenarnya apa yang perlu anda lakukan adalah MENUKAR equation yang diberi menjadi y  mx  c

Sama seperti disebelah Mesti kena bina table baru. Nak bina table ini Lihat pada maklumat didalam soalan.Soalan mahukan anda bina table

y x  

againts . x

Buatlah table baru Plotkan graph mengikut scale yang diberi.pastikan sekurangnya 5 point anda adalah betul

*6 markah sudah ditangan ketika ini*

Soalan (b) Soalan berikan persamaanya adalah

k y  2hx 2  x h apa yang perlu anda lakukan adalah MENUKAR equation yang diberi menjadi y = mx + c iaitu seperti disebelah anda kemudian seharusnya tahu menterjemahkan ini

anda seharusnya tahu menterjemahkan ini y  log 10 y

x  x 2 c  y  intercept  log 10 h m

gradient  log 10 k

*Jawapan ini mestilah diperolehi daripada graphs anda dan bukanya melalui kaedah formula.* SBP‟11,Terangganu‟11,SPM‟03

y 

m

gradient  2h

c  y  intercept 

k h

*Jawa an ini mestilah di erolehi dari ada ra hs



y

x x  x


MATHS CATCH

NETWORK

45


FOKUS A+

USAHA +DOA+TAWAKAL

EXAM TIPS: QUESTION 8 –GEOMETRY COORDINATE EXAM TIPS Tajuk ini adalah dalam kategori yang sukar.Dari 10 markah untuk lulus seharusnya pelajar mencuri sekurang-kurang 3 markah dari soalan ini.bagaimana caranya? Melalui kajian soalan peperiksaan Sebenar SPM 2005-2011 serta 13 soalan percubaan negeri 2011 menunjukkan soalan geometri koordinat ini mengandungi satu kata kunci yang sangat popular iaitu perpendicul ar bermaksud 90 0 Anda wajib tahu katakunci ini kerana darinya akan terbit formula ini

m1  m2  1 Soalan (a)-i Bagi soalan disebelah m1 merujuk kepada gradient QS m2 merujuk kepada gradient OS anda akan perolehi h disitu. **cara untuk mendapatkan coordinate ada 3 cara.** Cara 1 :Coordinate jenis Intercept (y atau x) Cara 2: Coordinate Jenis Float – Pakai simulatnaeous equation Cara 3.Coordinate jenis ratio – pakai formula point dividing segment

Soalan (a) - ii Katakuncinya adalah coordinate S adalah jenis “FLOAT” [Cara 2] Untuk jenis ini penyelesainya adalah melalui kaedah simultaneous equation

Soalan (b) - i Katakuncinya adalah coordinate Q berada dalam ratio SR: R Q Penyelesaianya adalah melalui formula dibawah..[diberi dalam peperiksaan]  nx  mx2 ny1  my 2  ( x, y )   1 ,  m  n   mn Soalan (b) - ii Katakuncinya equation straight line maka pelajar wajib tahu formulanya adalah

y  y1  m( x  x1 ) Soalan (c) Katakuncinya LOCUS maka pelajar wajib tahu f ormulanya adalah QT = 2 TS

( x2  x1)2  ( y2  y1)2  2 ( x2  x1 )2  ( y2  y1)2



MATHS CATCH

NETWORK

46

MATHS Catch SPM

FOKUS A+

USAHA +DOA+TAWAKAL EXAM TIPS: QUESTION 9 –CIRCULAR MEASURE

2012

EXAM TIPS Dalam bab ini pelajar perlu tahu :kebiasanyaa soalan (a) dan (b) soalan yang akan disoal adalah sama ada sudut  dalam radian atau length.. Untuk menghadapi soalan sebegini pelajar mestilah membuat persediaan dengan mengetahui konsep segitiga seperti dibawah.Segitiga ada 3 jenis semuanya. TRIANGLE

JENIS A

JENIS B

Sisi sama

  60

JENIS C

0

Semua sudut adalah sama

Perpendicular triange Theorem Phytagoras – Untuk mendapatkan length sahaja

Sudut Tidak Sama -Belajar detail dalam Solution of triangle

Trigonometetry [SOH CAH TOA] – Untuk mendapatkan - Length Sudut  -

Soalan (a) Soalan ini mahukan sudut..melalui gambar rajah kita tahu ia nya merupakan segitiga jenis B (perpendicular triangle) Gunakan SOH CAH TOA untuk mendapatkan sudut tersebut

Soalan (b) Soalan ini barulah pelajar perlu menggunakan formula asas bagi t ajuk circular measure ini iaitu s  r  Soalan (c) Soalan ini barulah pelajar perlu menggunakan formula asas bagi t ajuk circular measure ini iaitu A 

1 2

r 2



MATHS CATCH

NETWORK

47

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

BAHAGIAN C (4 soalan- Wajib Jawab 2 sahaja) NO TAJUK MARKAH Solution of Triangle 10 markah Index Number 10 markah TARGET MARKAH BOLEH DICAPAI

EXAM TIPS Soalan ini terbahagi kepada dua bahAgian TARGET MARKAH 2/10 2/10 4/20

Bahagian 1 Soalan (a)(i) dan soalan pada (a) (ii) Kebiasanya Soalan ini akan menyoal tentang LENGTH dan ANGLE Formula SIN

a sin A

EXAM TIPS: QUESTION 11 – SOLUTION OF TRIANGLE



b sin B



c sin C

Formula Cos

a 2  b 2  c 2  2bcCosA Tips Penting!!! Jika Soalan di a (i) menyoal untuk dapatkan angle,  Utamakan formula sin dahulu DANp astikan Ianya mengandungi sekurangnya 3 maklumat.dari 3 maklumat ini pastikan sekurangnya satu,angle diberi. Soalan di a (i) seperti disebelah menyoal untuk dapatkan length.Utamakan

formula cos…

Sekiranya di a (i) tadi a nda sudah menggunakan formula cos (seperti contoh disebelah) maka Jawapan untuk a (ii) pasti perlu menggunakan formula sin pula.begitulah sebaliknya.



MATHS CATCH

NETWORK

48

MATHS Catch SPM

FOKUS A+

USAHA +DOA+TAWAKAL EXAM TIPS: QUESTION 12 – INDEX NUMBER

2012

EXAM TIPS Secara umumnya pelajar mestilah tahu index harga adalah satu kajian harga pada 3 keadaan berbeza. Keadaan 1 : Base Year Year [tahun sebelum cthnya 2005] Keadaan 2 : Tahun sekarang [contohnya 2007] Keadaan 3 : Tahun akan datang [contohnya 2015] Soalan (a) Gunakan Formula ini Q Pr ice. Index, I  1 Q0 Bagi kes disebelah Q1  2007, Q0  2005 .Gunakan formula ini utk dapatkan nilai x dan y. Soalan (b)

Composite Index, I 

 I W  W i

i

i

*W = weightage . *Composite Index ini adalah pada masa sekarang ,iaitu pada tahun 2007 dan based year adalah 2005*

Soalan (c) Soalan ini minta anda harga pada tahun 2009 Anda perlu faham begini Tahun 2005= tahun asas sebagai perbandingan Tahun 2007 = tahun sekarang ini diukur, Tahun 2009 = adalah tahun akan datang.(Future) Gunakan formula ini Composite. Index. Future Future  Composite . Index Index. Now  %(Increase)

Perhatian % Increase Penigkatan

= 20% + 100% = 120%

Jika soalan adalah penurunan penurunan (decrease) % Decrease =100 % -20 % =80 %



MATHS CATCH

NETWORK

49

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+ SUGGESTION QUESTION 1: SIMULTANEOUS EQUATION

2012

1

Solve the simultaneous equations x + 2 y+ 2=0 x2 + xy − 84 = 0

[5 marks] Answer: x + 2 y + 2 = 0 −−−− (1) x2 + xy − 84 = 0 −−−− (2)

From (1),

x = −2 y − 2 −−− (3)

Substitute (3) into (2),

SUGGESTION QUESTION PAPER 2 [Improve from 60 to A+]

y − 84 = 0 (−2 y − 2)2 + (−2 y − 2) y 4 y2 + 8 y + 4 − 2 y2 − 2 y − 84 = 0 2 y2 + 6 y − 80 = 0 2 + 3 y − 40 = 0 ( y 8) = 0 y − 5)( y y+ ∴ y= 5 or y = −8

When y = 5, x = −2(5) − 2

= −12

When y = −8, x = −2(−8) − 2 = 14 x = −12, y= 5 or x = 14, y = −8

∴



MATHS CATCH

NETWORK

50

MATHS Catch 1 SPM (a) 2012

USAHA +DOA+TAWAKAL

(i) Sketch FOKUS A+ the graph of y= sin 2 x for 0 ≤ x ≤ 2π. (ii)

Hence, using the same axes, draw a suitable straight line to fint fint the number number of solutions to the equation 3 (2 sin x − 4 sin x+ 4 sin x cos

2 x

) = − 1 for 0 ≤ x ≤ 2π. 2 π

State the number of solutions. [6 marks] Answer: (a) (i)

SUGGESTION QUESTION 2: TRIGONOMETRIC FUNCTION

x

2

(a)

Sketch the graph of y = 3 cos 3 x for 0 ≤ x ≤ π.

(b)

[3 marks] Hence, using using the same axes, draw a suitable straight line to fint the number of of solutions to the equation

−6 x π + 4 − 3 cos 3 x = 0 for 0 ≤ x ≤ π.

State the number of solutions. [3 marks] Answer:

(a)

(ii)

x

x

3 (2 sin x − 4 sin x+ 4 sin x cos 2 ) = − 1 2 π 3 sin 2 x = sin 2 x = =

x

x

x

π−1 1

3π − 3 1

3π − 3

When x = 0, y = −

(b) 1 3

When x = π, y = 0 Number of solutions = 4

−6 x π + 4 − 3 cos 3 x = 0 −6 x 3 cos 3 x = π + 4 When x = 0, y = 4 When x = π, y = −2 Number of solutions = 3



MATHS CATCH

NETWORK

51

MATHS Catch SPM

FOKUS A+

2012

1

USAHA +DOA+TAWAKAL

SUGGESTION QUESTION 3: PROGRESSION

Diagram 1 shows the side elevation of part of stairs built of wood blocks.

Diagram 1 The length of the first block is 789 cm. The length of each subsequent block is 30 cm less than the prec The height of each block is 1 8 cm. (a) If the height of the stairs is 1.62 m, calculate (i) the length of the topmost block. (ii) the total length of the blocks. (b)

Calculate the maximum height of the stairs.

Answer:

(a)

(i)

a= 789, d = − 30

162 18 =9

n =

T 9 = 789 + 8(−30)

(ii)

(b)

= 549 cm 9 S 9 = ( 789 + 549) 2 = 6021 cm

T n > 0

789 + ( n − 1)(−30) > 0 789 − 30 n+ 30 > 0 30 n < 819 n< 27.3 n= 27 Maximum height = 27 × 18 = 486 cm



MATHS CATCH

NETWORK

52

MATHS Catch

USAHA +DOA+TAWAKAL

FOKUS A+

SPM

SUGGESTION QUESTION 4: VECTOR

2012

4

SUGGESTION QUESTION 5: STATISTICS

Diagram 4 shows a triangle ABC . The point Dl ies on AC a nd the point E l ies on AB. The straight line BDi ntersects the straight line CE at the point F .

→

→

→ →

1

Table 1 shows the marks obtained by 28 students in a test. Marks Number

50 − 59 60 − 69 70 − 79 80 − 89 90 − 99

→ →

It is given that ∠ ACB= 90°, AC = 16 x, CB = 14 y, AC : DC = 5 : 1 and AB : AE = 2 : 1. ~ ~ (a) Express in terms of xa nd y. ~ ~ (i) → BD

(ii)

→

→

→

→ → BD = CD− CB 1→ → = CA− CB 5 1 (−16 x) − 14 y 5 ~ ~ 16 = − 5 x − 14 y ~ ~ =

→ → →

(ii)

CE = CB + BE

→ 1→

= CB +

BA

2 → 1 → → = CB + ( CA− CB) 2 1 = 14 y + (−16 x − 14 y) ~ 2 ~ ~ = −8 x + 7 y ~ ~ (b)

→

→

→

→

CF = hCE BF = kBD

→ →

→

BF = CF − CB

→

3

→

Given that |x| = 5 units and | y| = 4 units, find | AB|. ~ ~

→

(i)

→

Using CF = hCE a nd BF = kBD, where ha nd k a re constants, find the value of ha nd k . [5 marks]

(c)

(a)

→

4 9

Given that the median mark is 79.5, find the values of x and y. Hence, state the modal class. [6 marks] Answer:

CE

[3 marks] (b)

x

→ →

kBD = hCE − CB

16 k (− x − 14 y) = h(−8 x + 7 y) − 14 y 5~ ~ ~ ~ ~ 16 − 5 kx − 14ky = −8hx + (7 h − 14) y ~ ~ ~ ~ 16 − 5 k = −8h 5 k = h −−−− (1) 2 −14k = 7 h − 14 −−−− (2) Substitute (1) into (2), 5 −14(2 h) = 7h − 14

16 x − 14 y) = h(−8 x + 7 y) − 14 y 5~ ~ ~ ~ ~

x

x

4 9

x + 4 x+ 13

y

x + y + 13

3

x + y + 16

79.5 = 79.5 + [

14 − ( x+ 13) y

] 10

1 − x 12 − x )10 0 = 10 − 10 x 10 x= 10 x = 1 y = 12 − 1 = 11

(c)

→

→

→

|AB|2 = | AC |2 + | CB|2 2 = (16| x|) + (14| y|)2

Modal class = 80 − 89

~ ~ = [16(5)] 2+ [14(4)] 2 = 9536

→


|AB| = 97.65 unit


Cumulative frequency

0=(

h =

k (−

Number

x + y+ 16 = 28 x + y= 12 y = 12 − x

−42h = −14 1 3 5 k = h 2 5 1 = ( ) 2 3 5 = 6

Marks

50 − 59 60 − 69 70 − 79 80 − 89 90 − 99


MATHS CATCH

NETWORK

53

MATHS Catch

USAHA +DOA+TAWAKAL

FOKUS A+

SPM 2012

2

Table 2 shows the marks obtained by 74 students in a test. Marks

3 Number

30 − 39 40 − 49 50 − 59 60 − 69 70 − 79

x

18 y

19 3

Table 2 Given that the median mark is 54.5, find the values of x and y. Hence, state the modal class. [6 marks] Answer:

Marks

30 − 39 40 − 49 50 − 59 60 − 69 70 − 79

Number

Cumulative frequency

x

x

18

x+ 18

y

x + y + 18

19 3

x + y + 37 x + y + 40

x + y+ 40 = 74 x + y= 34 = 34 − x

54.5 = 49.5 + [

37 − ( x+ 18) y

] 10

19 − x 34 − x) 10 170 − 5 x = 190 − 10 x 5=(

Table 3 shows the frequency distribution of the ages of residents in a n apartment. Age (years) Number 0−4 2 22 5−9 10 − 14 32 15 − 19 10 20 − 24 4 Table 3 Calculate the variance and the standard deviation of this distribution. [7 marks] Answer:

Age

0−4 5−9 10 − 14 15 − 19 20 − 24

f

Midpoint, x 2 7 12 17 22

2 22 32 10 4 ∑ f = 70 800 Mean, x ¯ = 70 = 11.43 1377.15 Variance, σ2 = 70 = 19.67

fx

4 154 384 170 88 ∑ fx= 800

2

f (x − ¯ x )

177.85 431.75 10.4 310.25 446.9 ∑ f ( x − x ¯ )2= 1377.15

Standard deviation, σ = 19.67 = 4.44

5 x = 20 x = 4

= 34 − 4 = 30

Modal class = 50 − 59



MATHS CATCH

NETWORK

54


SPM 2012

1

USAHA +DOA+TAWAKAL

SUGGESTION QUESTION 6: QUADRATIC FUNCTION Given y = − x + 2 mx − 5mh as a maximum value of 36. (a) By using completing the square method, express f(x) in the form a(x+p) 2+ q and finds the two possible values of m. [4 marks] (b) With the values of m, sketch the two graphs for y = − x2 + 2 mx − 5mo n the same axes. [2 marks] (c) State the coordinates of the maximum point of the graph y = − x2 + 2 mx − 5m. [1 mark] Answer:

(a)

(c)

Given y = x − 2mx + 7 mh as a minimum value of 10. (a) By using completing the square method, express f(x) in the form a(x+p) 2+ q and inds the two possible values of m. [4 marks] With the values of m, sketch the two graphs for y = x2 − 2mx + 7 mo n the same axes. [2 marks] (c) State the coordinates of the minimum point of the graph y = x2 − 2mx + 7 m. [1 mark] Answer: (b)

y = − x + 2 mx − 5m = −( x2 − 2mx + 5 m) −2m 2 −2m 2 = −( x2 − 2mx + ( ) − ( ) + 5 m)

(a)

m2 − 5m − 36 = 0 (m − 9)(m+ 4) = 0 m= 9 or m = −4 y = −( x − 9)2+ 36 y = −( x + 4) 2+ 36

(b)

2 = −(( x − m)2 − m2 + 5 m) = −( x − m)2 + m2 − 5m Maximum value = m2 − 5m m2 − 5m= 36

(b)

2

= x2 − 2mx + (

2

(9, 36) and (−4, 36)

= x2 − 2mx + 7 m

(c)

−2m

−2m

) 2 − ( ) 2 + 7 m 2 2 2 2 = ( x − m) − m + 7 m Minimum value = − m2 + 7 m −m2 + 7 m= 10 m2 − 7m+ 10 = 0 (m − 2)(m − 5) = 0 m= 2 or m = 5 = ( x − 2)2+ 10 = ( x − 5)2+ 10

(2, 10) and (5, 10)



MATHS CATCH

NETWORK

55

MATHS Catch SPM

USAHA +DOA+TAWAKAL

FOKUS A+

2012

SUGGESTION QUESTION 6: DIFFERENTIATION (BACKUP)

*Trial Kelantan 2011*



MATHS CATCH

NETWORK

56


SPM

SUGGESTION QUESTION 7: LINEAR EQUATION

2012

1

USAHA +DOA+TAWAKAL

Table 1 shows the values of t wo variables, xa nd y, obtained from an experiment. The variables xa nd ya re related by th e equation y = 3 px2 +

3

5

7

9

11

13

y

2.7

5.6

9

13.5

18.8

24.7

4

5

6

7

8

11.8

19.7

30

42

56.5

72.5

y x

Plot a gainst x, by using a scale of 2 cm to 1 units on the x-axis and 2 cm to 1 units on th

(b) Use your graph in (a) to find the value of (i) p (ii) q (iii) y when x = 6 Answer:

x

3

5

7

9

11

13

y x

0.9

1.12

1.29

1.5

1.71

1.9

p x q y p = 3 px + x q y = 3 px2 +

(i)

3

y

line of best fit.

(a)

Answer:

(b)

p x, where pa nd qa re constants. q

x

(a)

y Plot a gainst x, by using a scale of 2 cm to 2 units on the x-axis and 2 cm to 0.2 units on x y the a xis. Hence, draw the line of best fit. x

(b) Use your graph in (a) to find the value of (i) (ii) q (iii) when x = 9

(a)

Table 3 shows the values of t wo variables, xa nd y, obtained from an experiment. The variable s the equation y = 4 px2 +

p x, where p and qa re constants. q

x

(a)

2

3 p= Gradient of the graph = 0.1 p= 0.0333

(ii) p = Y −intercept q

0.0333 = 0.6

q q= 0.0555 (iii) x= 4.4 y = 1 x y= 4.4

(b)

x

3

4

5

6

7

8

y x

3.93

4.92

6

7

8.07

9.06

p x q y p = 4 px + x q y = 4 px2 +

(i)

4 p= Gradient of the graph =1 p= 0.25

(ii) p q

= Y −intercept

0.25 = 1

q q= 0.25 (iii) x= 6.6 y = 7.6 x 50.16 y=



MATHS CATCH

NETWORK

57

MATHS Catch

USAHA +DOA+TAWAKAL

FOKUS A+

SPM 2012

3

4

Use graph paper to answer this question.

Use graph paper to answer this question.

Table 4 shows the values of t wo variables, xa nd y, obtained from an experiment. The variabl es xa nd

Table 7 shows the values of two variables, xa nd y, obtained from an experiment. The variables

the equation y =

the equation y =

p9 x , where pa nd qa re constants. q

1 2.2

x y

2.5 4.4

4 8.5 Table 4

5.5 15.5

7 29.5

x

1

2.5

4

y

3.8

7.2

11.7

5.5

7

8.5

21.9 Table 7

38

74.1

(a) Plot log10 ya gainst x, by using a scale of 2 cm to 1 units on the x-axis and 2 cm to 0.2 units on the Hence, draw the line of best fit.

(a) Plot log10 ya gainst x, by using a scale of 2 cm to 1 units on the x-axis and 2 cm to 0.2 uni Hence, draw the line of best fit.

(b) Use your graph in (a) to find the value of

(b) Use your graph in (a) to find the value of

(i) (ii) q (iii) when x= 4.4.

(i) p (ii) q (iii) yw hen x= 4.7.

Answer:

Answer: (a)

(a)

(b) (c)

p9 x , where pa nd qa re constants. q

x

1

2.5

4

5.5

7

8.5

log10 y

0.34

0.64

0.93

1.19

1.47

1.78

p x q log10 y = 9 x(log10 p) + (−log10q) = 9(log10 p) x + (−log10 q) (i) 9 log10 p= Gradient of the graph

y =

= 0.19 log10 p= 0.0211 p= 1.05 (ii) −log10 q = Y −intercept = 0.16 log10 q = −0.16 p= 0.692 (iii) x= 4.4 log10 y = 1 y= 10

(b) (c)

x

1

2.5

4

5.5

7

8.5

log10 y

0.58

0.86

1.07

1.34

1.58

1.87

p x q log10 y = 9 x(log10 p) + (−log10q) = 9(log10 p) x + (−log10 q) (i) 9 log10 p= Gradient of the graph

y =

= 0.17 log10 p= 0.0189 1.044 p= (ii) −log10 q = Y −intercept = 0.41 log10 q = −0.41 p= 0.389 (iii) x= 4.7 log10 y= 1.2 y= 15.849



MATHS CATCH

NETWORK

58

ADDMATHS-SPM(MPT-60%-TO-A+)

Recommend Documents