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SULIT3472/1 2012 Maths Catch Network www.maths-catch.comMODUL PERSEDIAAN 2 ADD MATHS [From 60 to A+] Hak Cipta Terpelihara
USAHA +DOA+TAWAKAL
FOKUS A+SPM
2012
1MATHS CATCHNETWORK
MATHS Catch
MOHD RAJAEI BIN MOHAMAD ALI
MODUL PERSEDIAAN TERAKHIR
SPM 2012
ADDITIONAL MATHEMATICS
CONTENT/ISI KANDUNGANBAHAGIAN 1A. Preview Modul Persediaan Matematik Tambahan SPM (MPT SPM) 2B. Preview Keseluruhan tentang Matematik Tambahan SPM Kertas 1 dan Kertas 2, 2012 3BAHAGIAN 2
A. Tips Peperiksaan Terakhir Kertas 1 / Last exam Tips PAPER 1 5B. Cadangan Soalan Ramalan Pilihan Kertas 1 / Suggestion Question PAPER 1 23C. Tips Peperiksaan Terakhir Kertas 2/ Last Exam Tips PAPER 2 40D. Cadangan Soalan Ramalan Pilihan Kertas 2 / Suggestion Question PAPER 2 51
IMPROVE from 60 TO SCORE A+
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MATHS Catch
LAST EXAM TIPS PAPER 12012
ADDITIONAL MATHEMATICS
[Improve From 60 to A+]
MOHD RAJAEI BIN MOHAMAD ALI
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USAHA +DOA+TAWAKAL
FOKUS A+SPM
2012
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MATHS Catch
QUESTION SPM08 SPM09 SPM10 SPM11
Q1 Function Function Function Function
Q2 Function Function Function Function
Q3 Function Function Function Function
Q4 Quadratic Equation Quadratic Equation Quadratic Function Quadratic Function
Q5 Quadratic Function Quadratic Function Quadratic Equation Quadratic EquationQ6 Quadratic Function Quadratic Function Quadratic Function Quadratic Function
Q7 Indices & Logarithms Indices & Logarithms Indices & Logarithms Indices & Logarithms
Q8 Indices & Logarithms Indices & Logarithms Indices & Logarithms Indices & Logarithms
Q9 Progression Progression Progression Progression
Q10 Progression Progression Progression Progression
Q11 Progression Progression Progression Progression
Q12 Linear Law Circular Measure Linear Law Linear Law
Q13 Coordinate Geometry Vectors Coordinate Geometry Coordinate Geometry
Q14 Coordinate Geometry Vectors Coordinate Geometry Coordinate Geometry
Q15 Vectors Coordinate Geometry Vectors Vectors
Q16 Vectors Trigonometric Function Vectors VectorsQ17 Trigonometric Function Trigonometric Function Circular Measure Circular Measure
Q18 Circular Measure Integration Trigonometric Function Trigonometric Function
Q19 Differentiation Differentiation Differentiation Differentiation
Q20 Differentiation Differentiation Differentiation Differentiation
Q21 integration Integration Differentiation Differentiation
Q22 Statistics Permuatation&Combination Statistics Statistics
Q23 Permuatation&Combination Probability Permuatation&Combination Permuatation&Combination
Q24 Probability Statistics Probability Probability
Q25 Probability Distribution Probability Distribution Probability Distribution Probability Distribution
FORMAT KERTAS 1 ADD
MATHS SPMJUMLAH SOALAN : 25
JUMLAH MARKAH : 80
Analisis Soalan Peperiksaan
Sebenar SPM 2008-2011 mengikutsusunan soalan dan topik
PERHATIAN:Sila fahamkan format Soalan kertas 1 ini sebaik mungkin.Inilah Kunci Kejayaan Sebenar untuk menjawab dengan baik kertas Matematik Tambahan Ini
TAHUKAH ANDA!Kertas 1 Matematik Tambahan mengandungi 25 soalan dan membawa markah sebanyak 80%. Kertas 1 ini juga dikenali sebagai lubuk emas untuk mengorek markah semaksimum yang mungkindalam matematik tambahan.Ini kerana Soalan-soalan yang ditanya adalah asas-asas sahaja jika nak dibandingkan dengan Kertas 2.Jika anda selalu gagal dalam peperiksaan apa kata tumpukan pada
kertas 1 ini sebagai persediaan terakhir sebelum memasuki dewan peperiksaan sebenar nanti..Disinilah perbezaan gred akan berlaku sama ada anda akan dapat A,B,C atau D..anda masih bolehmeningkat dengan mendadak dari D ke B atau dari C ke A,dan boleh juga meningkat dari D ke A jika kena dengan caranya.
Satu masalah besar pelajar adalah TIDAK TAHU NAK BERMULA ,apabila ingin menjawab sesuatu soalan matematik tambahan.Punca utama adalah kerana pelajar-pelajar ini gagal
mengenalpasti NAMA TAJUK dan gagal MENANGKAP KATAKUNCI yang terdapat didalam soalan menyebabkan pelajar tidak tahu apakah formula yang sesuai digunakan dan bagaimananak bermula. Oleh itu modul ini disusun bagi membantu pelajar mengatasi masalah utama ini
*Tajuk diwarnakan hitam adalah
tajuk-tajuk dari Tingkatan 4.*
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USAHA +DOA+TAWAKAL
FOKUS A+SPM
2012
4MATHS CATCHNETWORK
MATHS Catch
Jika anda lihat stat istik bagi kertas 1 diatas,boleh dikatakan tajuk-tajuk yang keluar adalah sama sahaja setiap tahun.Walaubagaimanapun ini tidak bermakna Soalan yang ditanya
adalah sama.Soalan yang ditanya Kadang-kadang bagi sesuatu tajuk kecil soalanya mempunyai sampai 2-3 bentuk Soalan..Disebabkan modul ini dirangka khas untuk membantupelajar yang lemah maka MC buat andaian,pelajar terbabit sudah pasti tidak tahu apakah soalan yang wajar didahulukan dan yang mana wajar dikemudiankan.
Dicadangkan anda membaca segala tips yang diberikan sekurang-kurangnya dua kali sebelum peperiksaan bermula.Mudah-mudahan segala tips yang disampaikan dapat
,menyuntik semangat untuk anda terus mengulangkaji sehinggalah disaat yang terakhir.Tidak ada yang mustahil j ika anda berjaya mengulangkaji secara tersusun apa yang telah
dibekalkan ,kami yakin anda mampu mengubah gred pencapaian Matematik tambahan dari E ke C dan dari C ke A+.
Secara Umumnya Bab yang akan keluar didalam Kertas 1 Matematik Tambahan mengandungi 16 bab kesemuanya daripada 21 bab bagi tingkatan 4 dan 5. Berikut merupakan
tajuk-tajuk yang akan Keluar didalam kertas 1 mengikut FORMAT peperiksaan sebenar SPM
Berdasarkan kajian dan pemerhatian Kebiasaanya Format Kertas 1 untuk SPM tajuknya adalah seperti berikut:
SUSUNAN SOALAN DAN BAB KEBIASAAN DALAM FORMAT KERTAS 1 SPM
QUESTION CHAPTER MARKS
Q1 Function 3
Q2 Function 4
Q3 Function 3
Q4 Quadratic Equation 2Q5 Quadratic Function 3
Q6 Quadratic Function 3
Q7 Indices & Logarithms 3
Q8 Indices & Logarithms 3
Q9 Progression 2
Q10 Progression 3
Q11 Progression 2
Q12 Linear Law 4
Q13 Coordinate Geometry 4
Q14 Coordinate Geometry 3
Q15 Vectors 2
Q16 Vectors 4
Q17 Trigonometric Function 3Q18 Circular Measure 4
Q19 Differentiation 3
Q20 Differentiation 4
Q21 integration 3
Q22 Statistics 3
Q23 Permuatation&Combination 4
Q24 Probability 4
Q25 Probability Distribution 4
TOTAL 80
Untuk membantu pelajar memahami dengan lebih baik tajuk,bentukdan format soalan,maka MC sudah sediakan modul dinamakan Last
Exam Tips [PAPER 1]TOLONG JANGAN ABAIKANSEMUA
SOALAN DAN KONSEP yang diterangkan didalam modulini..Semoga modul 2012 ini membantu ANDA SEMUA.selamat
Berjaya.
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2012
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MATHS Catch
Kebiasaanya Q1-Q3 adalah dari tajuk function.Untuk menjawab soalan ini anda WAJIB tahu 4 perkara ini
a) Function Relationb) Absolute Functionc) Composite Functiond) Inverse Function
CASE 1:FUNCTION RELATION/FUNCTION NOTAION
Domain = {4, 9, 16}
Codomain = {2, 3, 4, 5}Object = 4,9,16
Image = 2, 3, 4
Range = {2, 3, 4} *Image yang mempunyai objek sahaja*
Relation between set A and B? = xxf )(
EXAM TIPSUntuk (b) Sangat penting.jika anda mahu
hilangkan modulus l l maka Jawapanyamestilah dipecah kepada dua iaitu (+) dan (-).
CASE 2:ABSOLUTE FUNCTION
CASE 3: INVERSE FUNCTION
Remarks:Soalansubtopic Inverse function inibelum pernah tak keluar.Oleh itu sila berikan fokus
yang lebih bagi subtopic ini
Exam Tips 3Langkah 1:tambahkan sendiri y [penting]
Langkah 2 :terbalikkan kedudukan y dan x.ini
bertujuan untuk menghilangkan1
p kepadapsahaja
Langkah 3:Cari nilai y.maka nila y yanganda perolehi itulah inverse function
EXAM TIPS: QUESTION 1-3 FUNCTIONS
CASE 4: COMPOSITE FUNCTION
The functions offand g are defined asf:xx 4andfg :x 4x + 3. Find the function g.
[4 marks]
Answer:
Givenf(x) =x 4 andfg(x) = 4x + 3.fg(x) =f[g(x)]
= g(x) 4g(x) 4 = 4x + 3g(x) = 4x + 7
g :x 4x + 7
EXAM TIPS
Soalan ini sangat penting. Ideanya untukmendapatkan nilai g pelajar perlulah
membuat pemecahan nilai yang
besar.iaitu fg(x) dan bukanya f(x)
FOKUS
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EXAM TIPS: QUESTION 4 QUADRATIC EQUATIONS
CASE 1: Given one roots
Jawapan
CASE 2: Given two roots
Given -3 and are the roots of an equations 4x2 + bx + c = 0 .Find the
value of b and c
Jawapan
03114
03124
04
33
4
1
4
13
2
2
2
xx
xxx
xxx
xx
Compare with 4x2 + bx + c = 0
Thus b = 11 , c = -3
4
1
CASE 3: Given two EQUALS roots
Jawapan
EXAM TIPSKatakuncinya adalah
one of the roots.
Roots bermaksud nilaix tersebut.
Diberix = -2 .gantikan dalam
persamaan diberi.maka anda akan
dapat nilai p
EXAM TIPSKatakuncinya adalah
Apabila diberikan 2 roots
Roots bermaksud nilaix tersebut.
Dalam soalan ini ada 2 nilai roots.
4
1
3
x
x
Langkah 1Langkah yang perlu anda lakukan adalah
pindahkan -3 ke sebelah kiri begitu juga nilai
pindahkan juga ke sebelah kiri
Langkah 2
Bandingkan persamaan yang anda perolehi
dengan yang diberi didalam soalan.
EXAM TIPSKatakuncinya adalah
Two equal roots
LANGKAH 1
-Tukarkan quadratic equation diberidalam bentuk (general form)
-Jadikan diakhir nilai = 0
LANGKAH 2
Kenal pasti formula yang perlu
digunakan.ada 3 sila buatpilihan.contoh disebelah katakuncinya
adalah two equal roots.lain-lain sila
Lihat dibawah
Quadratic EquationsTwo real roots b2-4ac > 0
Two equal roots b
2
-4ac = 0No real roots b2-4ac < 0
**Buat pilihan menggunakan 3
persamaan ini**
CASE 4: STRAIGHT LINE+ CURVE
(SIMULTANEOUS EQUATION)
The straight liney = 7x + 6 does notintersect with the curvey= x2 + 9x + n.Find the range of values ofn.
[3 marks]
Answer:y = 7x + 6
y= x2 + 9x + nx2 + 9x + n = 7x + 6x2 + 2x + n 6 = 0The equation does not intersectb2 4ac < 0(2)2 4(1)(n 6) < 04 + 4n 24 < 01 + n 6 < 0n < 5
EXAM TIPSKatakuncinya adalah
-Does not intersect bermaksud b2 4ac < 0-diberi Straight line dan curve equation
bermaksud penyelsaianya mestilah
menggunakan simultaneous equation
LANGKAH 1Gunakan kaedah simultaneous equatiuon dengan
cara samakan kedua-dua nilai y tersebut
LANGKAH 2Pindahkan semua nilai supaya disebelah
kanann a bersamaan = 0LANGKAH 3Kenal pasti formula yang perlu
digunakan.ada 3 sila buat pil ihan.contoh
disebelah katakuncinya adalah does notintersect = no real roots.lain-lain sila
Lihat dibawah
Quadratic Equations
Two real roots b2-4ac > 0
Two equal roots b2-4ac = 0
No real roots b2-4ac < 0
FOKUS
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EXAM TIPS: QUESTION 5-6 QUADRATIC FUNCTIONS
CASE 1: QF and Graphs
Jawapan
2).(
3).(
2).(
3)(2)( 2
xc
nb
ma
mxxf
y min
x min
CASE 2: inequalities
Solve inequality 452 xx
EXAM TIPS
LANGKAH 1Sila hafal general form quadratic function ini.
nmxaxf 2
)()(
--kemudian bandingkan dengan equation yang diberi
LANGKAH 2SANGAT PENTING sila bandingkan
3)(2)( 2 mxxf
nmxaxf 2)()(
Jawapan (a)
Oleh itu 3n .(ingat n adalah y min)
Jawapan (b)
Untuk nilai m .tandanya anda perlulah mengambil
berlawanan dari yang diberi.jika .dalam kes diatas x
min -2.maka Jawapan m = 2 sahaja.
Jawapan (c)
untuk Soalan c axis symmetry adalah garisan yang
merentasi paksi x.dan pastikan anda meletakx=-2 dan JANGAN letak =-2 sahaja.perkataan x itu
merujuk kepada persamaan (equation)
2
n = adalah nilai
y max atau min
1
m = adalah nilai
x max atau min
EXAM TIPS -Sangat Penting Tajuk Ini!
LANGKAH 1-Jadikan equation yang diberi dalam bentuk = 0-dapatkan 2 nilai x tersebut
LANGKAH 2Lorekkan graphs bahagian dalam kerana dalam soalanmenunjukkan < (less)
Quadratic besar dari 0 [ > 0 ]
Quadratic kurang dari 0 [
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LOGARITHMS
CASE 3 : BASE DIBERI ADALAH SAMA (Guna Teknik Factorise)
Solve the equation log5 (7x + 2) = log5 (3x + 4) + 1. [3 marks]
log5 (7x+ 2) log5 (3x + 4) = 1
log57x + 2
3x + 4= 1
7x + 2
3x + 4= 5
7x + 2 = 5(3x + 4)
7x + 2 = 15x + 20
x=
9
4
CASE 4 : BASE DIBERI TIDAK SAMA (Guna Teknik Samakan base + Factorise)
Given that log25s log5t= 0, express s in terms oft. [3 marks]
log25s log5t= 0log25s = log5t
log5s
log5 25= log5t
log5s
2 log5 5= log5t
log5s = 2 log5t
log5s = log5t2
s = t2
EXAM TIPS
Dalam Soalan case ini base nya adalah sama iaitu
5..Pelajar hanya perlu fikir bagaimana cara untukfactorise kanya sahaja.
Langkah 1
Pindahkan log5 (3x + 4) ke sebelah kiri
Langkah 2
Factorisekan kedua-dua log tersebut menggunakan
Laws of log iaitu
y
xLogyLogxLog aaa
Langkah 3Pindahkan log5 ke sebelah kanan bersamaan 5
1
Laws of indices
c
a
ay
x
cy
xLog
EXAM TIPSSoalan case ini base nya TIDAK SAMA.Langkah
pertama mestilah pelajar perlu fikir untuk samakan
base nya dahulu kemudian barulah guna Teknik
factorise untuk menyelesaikanya.
LOGARITHMSCASE 5 : INDICES KEPADA LOG
Solve the equation 85x + 3 = 74x.
Answer:
85x + 3 = 74x
log10 85x + 3 = log10 7
4x
(5x + 3) log10 8 = 4x log10 75x + 3
4x=
log10 7
log10 8
5x + 34x
= 0.9358
5x + 3 = 3.7432x(5 3.7432)x= 3
x = 38.7432
= 0.3431
EXAM TIPS
Teknik diguna apabila nilai disebelah kiri dankanan tiada factor sepunya atau tiada sifir yang
sesuai digunakan.
**Kiri (8) dan kanan (7)
Cara penyelesainya.pelajar perlulah tambahkan
log disebelah kiri dan kanan
FOKUS
LOGARITHMS (Mirip Trial Selangor 2012)
CASE 6: INDICES KEPADA LOG
Solve the equation6
13.2 xx
Answer:
1
7782.0
7782.0
7782.07782.0
7782.0)47712.0()30103.0(
7782.03log2log
61log3log2log
6
1log)3.2log(
x
x
xx
xx
xx
xx
FOKUS
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EXAM TIPS: QUESTION 9-11 PROGRESSION
CASE 1: A.P & G.P (Value x)
A.P
G.PEXAM TIPSCaranya adalah sama seperti diatasyANGBERBEZA adalah yang ini menggunakan
formula GP sahaja.
G.P (Jawapan a)Langkah 1
Jika anda susun nilai yang diberi ianya akan
menjadi begini.
20x , 4x , 20x
1T 2T 3T
Langkah 2
2
3
1
2
T
T
T
T
*Dari cara ini andaHANYA akan perolehi
nilai x sahaja.*
Langkah 3
Gunakan formula ini untuk dapatkan ,r
1
2
T
Tr
Jawapan (b) : Gunakan formula sahaja
EXAM TIPSA.P
Langkah 1 :Jika anda susun nilai yang diberi
ianya akan menjadi begini.
x7 , x2 , x
13
1T 2T 3T
Langkah 2
2312 TTTTd
*Dari cara ini andaHANYA akan perolehi
nilai x sahaja.*
Langkah 3
Gunakan formula ini untuk dapatkan ,d
12 TTd
CASE 2: A.P & G.P (Sum to infinity)
EXAM TIPSSelain soalan dari case 1. Case 2 ini juga
sangat popular teruatamanya dalam
percubaan negeri SPM 2011.
***Per11,SBP11,Ter11,Phg11,Negeri 9,Sabah 11***
Jawapan (a)
Formula G.P1 nn arT
Jawapan (b)
Gunakan formula sum to infinity.
RINGKASAN
PELAJAR WAJIB TAHU MENGGUNAKAN FORMULA DIBAWAH
A.P
(1) x)nilaimencariuntuk(gunaatau 231212 TTTTdTTd
(2) dnaTn )1( *digunakan apabila sesuatu soalan menyebut tentang
-sesuatunth term (cthnya Given nth term,How many number between..,Find number of term.)
(3) ])1(2[2
dnan
Sn *digunakan apabila sesuatu soalan menyebut tentang Sum of term,
G.P
(1)x)nilaimencariuntuk(gunaatau
2
3
1
2
1
2
T
T
T
T
T
Tr
(2) 1 nn arT ***digunakan sama seperti A.P
(3) 1rnilaisyaratnya1
)1(
r
raS
n
n
(4) 1rnilaisyaratnya1
)1(
r
raS
n
n
(5)
r
a
1
S **digunakan apabila sesuatu soalan menyebut tentang
- Sum to infinity
-angka dalam bentuk decimal [cthnya 0.575757] penyelesaianya 0.57 , 0.0057 , 0.000057- contoh 2 : [ 1.12121212 ] penyelesaianya 1 + , 0.12 , 0.0012 , 0.000012
r
a
1
1S **Kedah Trial 11**
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EXAM TIPS: QUESTION 12 LINEAR LAW
CASE 1: xyLog 1010 logagaints EXAM TIPSPelajar perlulah tahu persamaan umum
linear law iaitu cmxy Kebiasaanya Soalan yang akan disoal adalahexpress equationdan find coordinate
sahaja.
Trick penyelesaian untuk SEMUA jenis soalan
linear Law adalah SAMA sahaja.
Langkah 1Fikir bagaimana cara nak tukar equation
y dari maklumatdiberi KEPADA
y seperti didalam graphs
Caranya tambahkan10log pada equation
yang diberi disebelah kiri dan kanan
Langkah 2Gunakan konsep log iaitu
nmmn aaa logloglog
Langkah 3Jadikan dalam bentuk y = mx + c
Untuk memudahkan senaraikan sepertidibawah
y =
x =
m =c =
dari cth disebelah
erceptyc
gradientm
xx
yy
int
x-x
y-y
GraphspadaLihatlog
GraphspadaLihatlog
12
12
10
10
***Trial Kedah11Kelantan11, Selangor11,
Negeri sembilan11*****
CASE 2: xx
yagaints
EXAM TIPS
Trick penyelesaianADALAH SAMAsahaja seperti dalam case 1
Langkah 1
Fikir bagaimana cara nak tukar equationy dari maklumatdiberi KEPADA-
y seperti didalam graphs
Equation dalam maklumat tiada x dibawahy.oleh itu anda perlulah tambahkan x
dibawahnya supaya sama dengan graphs
Langkah 2Jadikan dalam bentuk y = mx + c
Untuk memudahkan senaraikan seperti
dibawah
y =
x =m =
c =
dari contoh disebelah
erceptyc
gradientm
xx
x
yy
int
x-x
y-y
GraphspadaLihat
GraphspadaLihat
12
12
***Trial Terengganu11 Pahang11,
Sarawak11, Melaka11*****
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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 TIPSKatakunci adalah perpendicular (900)
Langkah 1Tuliskan formula yang ingin diguna
121 mm
***m = gradient
*Kebiasanya 1 gradient diberi (1m ) dan
satu lagi perlu dicari ( 2m )*
Langkah 2
Cari gradient1
m melalui equation
yang diberikan
cmxy
xy
denganBandingkan
2
3
1
Maka3
11 m
Langkah 3
Dapatkan 2m melalui formula
3
13
1
1
2
2
21
m
m
mm
Langkah 4Cari Equation PQ
63
)0(36
)( 11
xy
xy
xxmyy
CASE 2: PARALLEL
EXAM TIPSKatakunci adalah parallel
Bermaksud 21 mm Langkah 1Cari gradient bagi kedua-dua persamaan
diberi.
1
1
pxy
ypx
Gradientnya,m1 = p
Untuk equation ini gradientnya boleh
diperolehi dari formula
a
bm
b
y
a
x 2,1
CASE 3: AREAEXAM TIPSKatakunci adalah Area
Jangan lupa ulang semula coordinatepertama dan letakkan di kedudukan terakhir
a = x - interceptb = y - intercept
**bermakna dari soalan4
3 2 m
Langkah 2
Samakan keduanya
4
3
21
p
mm
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CASE 4: COLLINEAR
CASE 5: RATIO
EXAM TIPSKatakunci adalah point Q divide PR with
ratio 2:1
nm
myny
nm
mxnxyx 2121 ,),(
EXAM TIPSKatakunci Collinear
Bermaksud segaris.maknanya tiada luas.
(Area = 0)
CASE 6: DISTANCE/LOCUS
EXAM TIPS
Katakunci disini adalah 2PQ=QR
*Ianya bermaksud jarak QR bersamaan 2 PQ.*
**Penyelesaianya gunakan formula
distance/locus
2
12
2
12
2
12
2
12 )()()()(2
2
yyxxyyxx
QRPQ
0
0
a
a
a
a
aa
a
1
21
21
21210
21212
10
)]2(17)7(3)(4[)]4(717)3)(2[(2
10
4
2
17
7
34
2
2
10
EXAM TIPS: QUESTION 15-16 - VECTOR
Dari analisis soalan tahun-tahun lepas menunjukkan ada 2 BENTUK soalan dari bab vector ini.Bentuk 1 : soalan dari ayat. Bentuk 2 : soalan dari gambarajah
2 BENTUK DIATAS AKAN SECARA SELANG SELI AKAN MENYOAL 4 PERKARA DIBAWAH
a) Express in terms of-Lihat contoh. 2012b) Find Unit vector
22.
yx
yjxivectorunit
2012
c) Magnitude Vector ada modulus 22 yxr 2012d) Vector ada parallel Atau collinear
CASE 1: SOALAN DARI AYAT (Unit Vector)
Given that a~
= ( )72 and b~ = ( )1
13 , find
(a) the vector a~
b~
.
(b) the unit vector in the direction ofa~
b~
.
(a) a~
b~
= ( )72 ( )1
13
= ( )815 (b) Unit vector in the direction ofa
~ b
~
17
15j+8i
15+8
15j+8i
22
EXAM TIPS
Unit Vector
22yx
yjxi
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CASE 2: SOALAN DARI AYAT (MagnitudVector)Given that a
~= 3i + 9j and b
~= 2i + kj, find
(a) a~
b~
in the formxi +yj.
(b) the value ofkif |a~
b~
| = 13.
Jawapan
(a) a b~
= 3i + 9j (2i + kj) = 3i + 9j + 2i kj= 5i + (9 k)j
(b) |a~
b~
| = 13
13)9(5 22 k = 13
25 + (9 k)2 = 169(9 k)2 = 144
9 k = 12k = 21
or
9 k = 12k = 3CASE 3: SOALAN DARI AYAT (parallel)Given that a~
= (2k-1)i + 3j and b~
= 4i + 5j, find k if
2a+3b is parallel to y-axis
Answer
Langkah 1
: cari hasil 2a +3b dahulu.
21
104
15
12
6
24
5
43
3
122
k
k
k
Langkah 2
Parallel to yaxis bermaksud = (0,y)
yq
k
qba
0
21
104
Langkah 3
Dapatkan nilai k
(4k+10 = q(0)4k =-10
k =-5/2
CASE 4: SOALAN DARI GAMBARAJAH (unit Vector)
Diagram 9 shows vector OR
drawn on a Cartesian plane.
Diagram 9
(a)Express OR
in the form ( )xy .
(b)
Find the unit vector in the direction ofOR
.
Answer
(a)OR
= ( )125 (b)
Unit vector in the direction ofOR
= (12i~
+5j~
) / 122 + 52
= (12i~
+5j~
) / 13
EXAM TIPS
Magnitude Vector
22yxr
EXAM TIPS
Parallel Vector
qba ,q is scalar...
EXAM TIPS
Unit Vector
22yx
yjxi
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EXAM TIPS: QUESTION 17 TRIGONOMETRY FUNCTION
CASE 1: Six Trigonometry Functions of Any Angles
1 Given cosx= 0.4761 and 90 x 270, find the value of(a) x
(b) secx
[2 marks]Answer:
(a) cosx= 0.4761cos = 0.4761 = 61 34'x = 180 + 61 34'
= 241 34'
(b)secx =
1
cosx
=1
0.4761
= 2.1
2 Solve the equation 20 sin2x= sinx+ 24 sin 30 for 0 x 360
[4 marks]
Answer:
20 sin2x= sinx + 24 sin 3020 sin2x= sinx + 1220 sin2x + sinx 12 = 0(4 sin 3)(5 sin + 4) = 0
sinx =3
4
x = 48 35', 131 25'or
sinx= 4
5
x = 233 8', 306 52'
Pelajar hasruslah menghafal dan tahu cara menggunakan formula dibawah
Pelajar hasruslah tahu cara menggunakan formula dibawah
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EXAM TIPS: QUESTION 18 CIRCULAR MEASURE
Exam Tips
1. 0180. rad 2. Arc, rs 3. Area, 2
2
1r
**Soalan dalam paper 1 biasanya tajuk ini
simple,basic dan tidak berbelit-belit.Persedian
perlu dibuat Hafal FORMULA diatas dan tahu
cara menggunakanya,sudah mengcukupi.
*Jika Soalan bertanya tentan angle pastikan
Jawapan anda berikan dalam bentuk radian.*
3 Solve the equation 5 cos 2x = 13 sinx 9 for 0 x 360[2 marks]
Answer:
5 cos 2x = 13 sinx 95(1 2 sin2x) = 13 sinx 95 + 10 sin2x = 13 sinx 910 sin2x 13 sinx + 4 = 0(5 sin 4)(2 sin 1) = 0
sinx =4
5
x = 53 8', 126 52'
or
sinx =1
2
x = 30, 150
Pelajar hasruslah tahu cara menggunakan formula dibawah
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EXAM TIPS: QUESTION 19 INTEGRATION
7
4
62
4
2
25
62
24182
182
)8(3
18)([3
18])(3[
5
2
2
5
2
2
5
2
5
2
2
5
2
5
2
5
2
k
k
xk
xk
dxx
k
xdxkdxxg
dxkxxg
Exam Tips**Tajuk ni sangat mudah dalam paper 1Kebiasaanya ada 2 format soalan.
Format yang pertamaLANGKAH 1:Anda perlu tahu buat PEMECAHAN dr formula
asal .
LANGKAH 2:Kebiasaan dari 2 pecahan tadi.satu Jawapan anda
boleh dapat secara terus dari soalan.Satu pecahan lagi Jawapan
anda boleh perolehi dengan cara integrate
FORMAT KEDUAContoh Soalan
Given that ,Find
(a)
Penyelesaianya.
Jika diperhatikan perbezaanya,nilai limit pada maklumat yangdiberi adalah 6,2 manakala pada soalan adalah terbalik dari nilai
tersebut iaitu 2,6.
Apa yang perlu anda buat adalah tambahkan nilaivesahaja.maka jawapanya adalah (-7)
*Soalan ini diramalkan akan keluar*L
6
2
7)( dxxf
2
6
)( dxxf
EXAM TIPS: QUESTION 20 STATISTICS
Exam Tips**Dalam bab ini terlalu banyak formula perlu difahami.bukan
Lupakan yang lain.WAJIB HAFAL!3 formula Dibawah sudah mengcukupi.
FORMULA
a) Standard Deviation, 22 )(xN
x
b) Mean,N
xx
c) Variance, 222 )(xN
x
ISTILAH (Sangat Penting)
-Sum of Set Square Number= 2x -Sum of Set Number=x
CASE 1: MEAN & STANDART DEVIATION
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CASE 2: NEW MEAN,NEW MEDIAN,NEW MODE
Diagram 1 shows the scores obtained by a group of
players in a game.10, 15, 6, 15, 5, 17, 14
Diagram 1
(a) Find the mean, median and mode for the scores.
(b) If the scores in Diagram 1 is multiplied by 9, and
then added 7, find the mean, median and mode for
the scores.
Answer
(a)Mean, x =
xN
=
10 + 15 + 6 + 15 + 5 + 17 + 14
7
=82
7
= 11.71
Median = 14
Mode = 15
(b) New mean = 9(11.71) + 7 = 112.39
New median = 9(14) + 7 = 133New mode = 9(15) + 7 = 142
Exam Tips
**Case ini sangat penting untuk SPM 2011
Jawapan (a)Gunakan formula sahaja
**Mean
Mean, x =xN
**Median
Pelajar perlu susun mengikut susunanmenaik terlebih dahulu
5, 6, 10, 14, 15, 15, 17
Median = nilai ditengah-tengah = 14
Mode
Kekerapan tertingi = 15
Jawapan (b)Percayalah soalan ini sangat senang!!
Katakunci adalahMultiplied by 9, and then added 7
New mean
New median
New mode
EXAM TIPS: QUESTION 21 PROBABILITY
EXAM TIPS
Soalan Probability ini sebenarnya Sangatmudah.
Jika contoh disebelah diambil pelajar perlutumpukan pada bilangan kad dipilih (TWO
CARD ARE DRAWN) dan satu lagi silafokuskan pada katakunci warna yang sama(same color)
Kita buat kesimpulan 2 kad dipilih mestilahwarnaya adalah sama.oleh itu
kebarangkalianya (Probability) adalah seperti
berikut
(G, G) + (Y, Y) + (R, R)
,
105
34
14
4
15
5
14
6
15
7
14
2
15
3
Tolak 1 kerana tadianda sudah pilih satu
cardMaka 5-1=4
PERHATIAN:
Soalan diatas meminta andadapatkan probability bagi warna
kad yang sama (same color)
Bagaimana jika soalan memintapada anda warna yang tidak sama( NOT same color)
Jangan panic mudah sahaja.notsame color bermakna kedua-
duanya BUKAN warna yang sama.
Apa yang perlu dilakukan adalah(1- same color)
105
71
105
341
Total Card (R+Y+G) = 15
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EXAM TIPS: QUESTION 22 DIFFERENTIATION
CASE 1: Equation Tangent
Find the equation of the tangent to the curve
y = (2x 5) (x 4) at the point (4, 0).
Answer:
y = (2x 5)(x 4)= 2x2 13x + 20
dy
dx= 4x 13
At point (4, 0),dy
dx= 4(4) 13
= 3
The gradient of the tangent at point (4, 0) = 3
Equation:
y = 3(x 4)y = 3x12
CASE 2: Minimum Point
The curvey = 3x2 + 6x 7 has aminimum point atx = k, where kis a
constant.
Find the value k.
Answer:
y = 3x2 + 6x 7dy
dx= 6x + 6
Wheny is minimum,dy
dx= 0
6x + 6 = 0
x= 1k= 1
CASE 3: Rate of Changes
The volume of a sphere increases at the rate of
44.8 cm3 s1. Find the radius of the sphere atthe instant when its radius is increasing at a rate
of 0.7 cm s1
.
[Volume of sphere, V=4
3r3]
Answer:
V=4
3r3
dV
dr= 4r2
dV
dt= 44.8
dr
dt= 0.7
CASE 4: Small of Changes
Find the small change in volume of a sphere
when the radius increases from 6 cm to 5.93 cm.
[Volume of sphere, V=4
3r3]
Answer:
V=4
3r3
dV
dr= 4r2
r= 5.93 6
Vr
dx
dy
VdV
dr r
= 4r2 0.07= 4(6)2 0.07= 10.08 cm3
dV
dt=
dV
dr
dr
dt
44.8 = 4r2 0.7
4r2 =44.80.7
4r2 = 64
r2 = 16
r= 4 cm
Exam Tips** Katakunci adalah EQUATION tangent curve.
**Formulanya adalah )( 11 xxmyy
** gradientdxdym
Langkah 1
Didalam soalan coordinate ),( 11 yx sudah
diberi iaitu (4, 0)...Oleh itu pelajar hanya perlu
dapatkan nilai gradient sahaja
Langkah 2
Gunaklan formula )( 11 xxmyy untukmendapatkan equation
Exam Tips** Katakunci adalah MINIMUM POINT
**Formulanya adalah 0dx
dy
Langkah 1Pelajar perlu gunakan kaedah differentiation
untuk dapatkan nilai
dx
dy
Langkah 2
Minimum point akan menjadikann nilai
dx
dy
adalah 0
Langkah 3
Dapatkan nilai x.
Exam Tips
** Katakunci adalah rate dan unit cm3 s1.
**Formulanya adalahdV
dt=
dV
dr
dr
dt
dV
dr= pelajar perlu buat differentiation
dV
dt= Volume changes of t = 44.8
dr
dt=Radius changes of t =0.7
Exam Tips** Katakunci adalah Small Change inVolume
**Formulanya adalahVr
dV
dr
TIPS SANGAT PENTING!!**Khas untuk pelajar yang tak faham langsung
soalan ini mahukan apa dan tak tahu langsung nak
pakai formula apa.
**Apa yang perlu anda lakukan adalah buat
differentiation sahaja..Soalan ini adalah tajuk
differentiation.apa jua soalan yang ditanya pasti
akan meminta pelajar membuat
dx
dy
*Dalam case 3 dan 4 pelajar perlu membuat
dV
dr
dV
drsama sahaja dengan
dx
dy
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EXAM TIPS: QUESTION 23 PERMUTATION & COMBINATION
CASE 1: PERMUTATION
*Trial Sabah 2011*
RAMALAN 1How many 3-digit numbers that are greater than 400 can be
formed using the digits 1, 2, 3, 4, and 5 without repetition?
Answer:
= 12p x 2
4p = 24
RAMALAN 2
How many 4-digit even numbers can be formed using the
digits 1, 2, 3, 4, and 5 without repetition?
Answer:
= 12p x 3
4p
= 48
RAMALAN 3How many 4-letter codes can be formed using the letters in
the word 'GRACIOUS' without repetition such that the firstletter is a vowel?
Answer:
= 14p x 3
7p
= 840
CASE 2: PERMUTATION
** Trial Kelantan dan Terengganu 2011**
RAMALAN 1
Diagram shows seven letter cards.
A five-letter code is to be formed using five of these cards.
Find
a) The number of different five-letter codes that can be
formed,b) The number of different five-letter codes which end with a
consonant.
Answer:
a) 57p = 2520
b) 46p x 1
4p
= 1440
RAMALAN 2
Diagram shows six numbered cards.
A four-digit number is to be formed by using four of these
cards.How manya) different numbers can be formed?b) different odd numbers can be formed?Answer:
a) 6P4 = 360
b) 5P3 x4P1
= 240
ROFINU M
987541
Exam Tips** Katakunci adalah 3-digit numbers dan greater than 400
Langkah 1:
Buat 2 garis seperti dibawah
d
c
b
a PP .
b = adalah digit pertama [gerenti = 1]
a = Total digit mengikut syarat diberikan.SyaratHow many number
greater than 400.Jawapan 4 dan 5 } 2 nombor 1
2
p
d = adalah baki digit yang tinggal [dalam soalan sebelah = 2 digit.
1 digit sudah diambil maka baki yang tinggal adalah 2]
c = total digit keseluruhan yang tinggal = 4
Asalnya ada 5 nombor.tetapi 1 nombor sudah diambil maka total bakinya
adalah 4 saha a.
etT
TotalP arg
1
2p
24p
2
4p
Exam TipsEven Number = 2 dan 4 } Total Digit pertama = 2
Baki digit [4-1 = 3 digit]
Digit pertama.Gerenti 1
Exam TipsJawapan (a)
Ingat kembali formulanya etTTotalP
arg
Target adalah four-digit number
Total adalah = 6 digit
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CASE 3: COMBINATION
** Trial Johor,Melaka,Perak,SBP dan Selangor 2011**
RAMALAN 1A committee of 3 boys and 3 girls are to be formed from 10 boys and 11 girls.
In how many ways can the committee be formed?
Answer:
3
10C x 3
11C
= 19800
RAMALAN 2A badminton team consists of 8 students. The team will be chosen from a group of 8 boys
and 5 girls. Find the number of teams that can be formed such that each team consists ofa) 5 boys,b) not more than 2 girls.
Answer:
a)5
8C x
3
5C = 560
b) If the team consists of 8 boys and 0 girl 88C x 0
5C = 1
If the team consists of 7 boys and 1 girl 78C x 1
5C = 40
If the team consists of 6 boys and 2 girl 68C x 2
5C = 280
The number of teams that can be formed = 1 + 40 + 280= 321
RAMALAN 3 (SPM2011)
A debating team consists of 6 students. These 6 students are chosen from 2 monitors, 3
assistant monitors and 5 prefects.
a) there is no restriction,
b) the team contains only 1 monitor and exactly 3 prefects.
Answer:
a) 610C = 210
b) 12C x 3
5C x 2
3C = 60
Exam Tips
BOY GIRL
Step 1: C C
Ingat formulaetT
TotalC arg !
Step 2 : 310C
311C
Total Boy = 10 Total Boy = 11
Dipilih = 3 (Target Dipilih = 3 (Target)
Exam TipsJawapan (a)
No restriction = tiada syarat automatic gunakan formula asaletT
TotalC arg
Exam TipsJawapan (b)
TARGET 6 students ..Ingat formulaetT
Total
C arg
Ada Syarat
Langkah 1 :Isikan target mengikut syarat diberikan .iaitu M = 1,P = 3 automatik AM = 2
MONITOR PREFERCTS ASSISTANT MONITOR
1C 3C
2C
LAngkah 2 : Isikan total dari yang diberikan iaitu total M = 2,P = 5 ,AM =3
MONITOR PREFERCTS ASSISTANT MONITOR
1
2C
x 3
5C x 2
3C = 60
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CASE 5:
** Trial Pahang,Perlis 2011**
RAMALAN 1Diagram shows five cards of different letters.
a) Find the number of possible arrangements, in a row, of all the cards.b) Find the number of these arrangements in which the letters A and N are
side by side.
Answer:
a) 5! = 120
b) 4! x 2!= 48
R A J I N
Diagram below shows a standard normal
distribution graph.
The probability represented by the area of the
shaded region is 0.803
a) Find the value of )( kZP b) X is a continuous random variable
which is normally distributed with a
mean of and a standard deviation of
2.If the value of X is 85 when the
scoreZ is k,find the value of
.[3 marks]
EXAM TIPS: QUESTION 25 PROBABILITY DISTRIBUTION
Exam Tips
Apa yang pelajar perlu faham jumlah bagi
kedua-dua luas dibawah graph adalah 0.5 +0.5 =1
Answer
(a)
0985.02
803.01)(
kZP
(b)
Hafal Formula ini
XZ
X Random variable Mean Standard deviation
scoreZ
42.82
2
8529.1
XZ
Area = 0.5 Area = 0.5
Perlu dibahagi 2 kerana
nilai k didalamgambarajah ada 2Sebab total = 1
Exam Tips
Nilai z mestilah diperolehi dari dalam graph.
Lihat muka surat disebelah
EXAM TIPS: QUESTION 25BINOMIAL DISTRIBUTION
FOKUS 2012
FOKUS 2012
1 The discrete random variableXhas a binomial probability distributionwith n = 4, where n is the number of trials. Table 1 shows the probability
distribution ofX.
x P(X=x)
01
16
11
4
2 k
31
4
41
16
:
Find
(a) the value ofk.
(b) P(X 3).[3 marks]
[3 markah]
Answer:
(a)k +
1
16
1
4
1
4
1
16= 1
k =3
8
(b)P(X 3) =
1
4+
1
16
=5
16
Exam Tips
TOTAL Probability = 1
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SUGGESTION QUESTIONPAPER 1
[Improve From 60 to A+]
Bagaimana Z =1.29?dan bukan =0.0985?
Ini kerana nilai z tersebut anda perlu dapatkan dari dalam table.Didalam peperiksaan table ini akn
diberikan kepada anda.cara untuk dapatkan adalah seperti diatas. Z=1.29
****Begitu ramai pelajar melakukan kesilapan mudah ini.pastikan anda bukan termasuk didalam
golongan ini.****
42.82
2
8529.1
XZ
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SUGGESTION QUESTION 1-3: FUNCTION
1 Diagram 1 shows the relation between set
Mand setNin the arrow diagram.
State the
(a) images of 12
(b) objects of 2
(c) domain of this relation
(d) range of this relation
[2 marks]
Answer:
(a) 2, 3
(b) 6, 12, 16
(c) {6, 12, 16}
(d) {2, 3}
2 Given {(16, 2), (20, 2), (20, 5)}.
State the
(a) images of 20
(b) objects of 2(c) domain of this relation
(d) range of this relation
[2 marks]Answer:
(a) 2, 5
(b) 16, 20
(c) {16, 20}
(d) {2, 5}
4 Given the functionf:x |x2 5|.(a) Find the images of 6, 6 and 7.
(b) Find objects which have the image of4.
[4 marks]
Answer:
f(6) = |(6)2 5|= |31|
= 31f(6) = |(6)2 5|
= |31|
= 31
f(7) = |(7)2 5|= |44|
= 44
f(x) = 4
|x2 5| = 4So,
x2 5 = 4x2 = 9
x= 3, 3and
(x2 5) = 4x2 = 1
x= 1, 1
3 Given functionf:x 8x 2, find the(a) image of 7(b) object which has the image 34
[3 marks]
Answer:
(a) f(7) = 8(7) 2
= 54(b) f(x) = 34
8x 2 = 348x = 32x = 4
5 The fuctionfis defined asf:xx + 4.Find
(a)1
(2)
(b) 1(x)
[3 marks]
Answer:
(a) Letf1
(2) = k
Sof(k) = 2k+ 4 = 2
k= 2Therefore,f
1(2) = 2(b) Letf
1(x) =y
Sof(y) =x
y + 4 =x
y =x 4Therefore,f
1(x) =x 4
6 The fuctionsfand g are defined asf:x 1 8x and g :x 2x 5.Find gf
1(x).
[3 marks]
Answer:
Letf1
(x) =ySof(y) =x
1 8y =x
y =1 x
8
Thereforef1(x) =1 x
8
gf1(x) = g(f1(x))
= g(1 x
8)
= 2(1 x
8) 5
=
x 21
4
7 The functions offand g are defined asf
:xx + 3 and g :x 4 x. Find thecomposite function ofgfandfg.
[3 marks]Answer:
Givenf(x) =x + 3 and g(x) = 4 x.gf(x) = g(f(x))
= g(x + 3)
= 4 (x + 3)= x + 1
fg(x) =f(g(x))
=f(4 x)= (4 x) + 3= x + 7
8 The functions offand g are defined asf
:xx 9 and g :x 9x 3. Find thecomposite function ofgfandfg.
[3 marks]
Answer:
Givenf(x) =x 9 and g(x) = 9x 3.gf(x) = g(f(x))
= g(x 9)= 9(x 9) 3= 9x 84
fg(x) =f(g(x))
=f(9x 3)= (9x 3) 9= 9x 12
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SUGGESTION QUESTION 4: QUADRATIC EQUATION
1It is given that
5
2is one of the roots of the
quadratic equation 4x2 8x + q = 0. Findthe value ofq.
[2 marks]
Answer:
4(5
2)2 8(
5
2) + q = 0
25 20 + q = 0q= 5
2 Form the quadratic equation which has the
roots 3 and3
4.
[4 marks]
Answer:
x2 (3 +3
4) + (3)(
3
4) = 0
x2 (9
4)x+ (
9
4) = 0
4x2 + 9x 9 = 0
3 The quadratic equation 4x2 + mx + n = 0has roots 8 and 9. Find the values ofmand n.
[3 marks]
Answer:
(x a)(x b) = 0(x 8)(x + 9) = 0x2 +x 72 = 04x2 4x + 288 = 0Therefore,
m= 4 and n = 288
4 The quadratic equation 4x2 +px + q = 0 has
roots 1 and 5. Find the values ofp and q.
[4 marks]
Answer:
(x a)(x b) = 0(x + 1)(x + 5) = 0
x2 + 6x + 5 = 0
4x2 + 24x + 20 = 0
Therefore,
p = 24 and q = 20
5 The straight liney = 5x + 6 does not
intersect with the curvey = 6x2 + 5x + q.
Find the range of values ofq.
[3 marks]
Answer:
y = 5x + 6
y = 6x2 + 5x + q
6x2 + 5x + q = 5x + 6
6x2 + q 6 = 0The equation does not have real roots
b2 4ac < 0(0)2 4(6)(q 6) < 0 24q + 144 < 024q< 144q > 6
7 The quadratic equation
(4q 3)x2 2x 4 = 0 has twodifferent roots, q is a constant.
Find the range of values ofq.
[3 marks]
Answer:
(4q 3)x2 2x 4 = 0The equation has two different roots
b2 4ac > 0(2)2 4(4q 3)(4) > 04 64q 48 > 064q > 44
q> 11
16
6 The straight liney= 6 4x does notintersect with the curvey= x2 +x + m.Find the range of values ofm.
[3 marks]
Answer:
y= 6 4xy= x2 +x + mx2 +x + m= 6 4xx2 + 5x + m 6 = 0The equation does not have real roots
b2 4ac < 0(5)2 4(1)(m 6) < 025 + 4m 24 < 04m< 1
m< 1
4
8 The quadratic equation 8x2px + 2 =x
has two equal roots,p is a constant.
Find the values ofp.
[3 marks]Answer:
8x2px + 2 =x8x2+ (p 1)x + 2 = 0The equation has two equal rootsb2 4ac = 0(p 1)2 4(8)(2) = 0p2 + 2p+ 1 64 = 0p2 + 2p 63 = 0(p 7)(p + 9) = 0p = 7 orp= 9
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SUGGESTION QUESTION 5-6: QUADRATIC FUNCTION
1 Diagram 3 shows the graph of a quadraticfunctionf(x) = (x + s)2 7, where s is aconstant.
The curvey =f(x) has a minimum point
(7, t), where tis a constant. State
(a) the value ofs
(b) the value oft
(c) the equation of the axis ofsymmetry
[3 marks]
Answer:
(a) s= 7(b) t= 7( ) x = 7
2 Diagram 4 shows the graph of a quadraticfunctionf(x) = 2(x + m)2 7, where m is aconstant.
The curvey =f(x) has a minimum point (4,
n), where n is a constant. State
(a) the value ofm
(b) the value ofn
(c) the equation of the axis of symmetry
[3 marks]
Answer:
(a) = 4(b) = 7(c) x = 4
3Diagram 1 shows the graph of a quadraticfunctionf(x) = 5(x + h)2 + 6, where h is a
constant.
Diagram 1
The curvey =f(x) has a minimum point (8, k),
is a constant. State
(a) the value ofh
(b) the value ofk
(c) he equation of the axis of symmetry
[3
Answer:
(a) h= 8(b) k= 6
(c) x = 8
4 Diagram 2 shows the graph of a quadraticfunctionf(x) = (x + s)2 6, where s is aconstant.
Diagram 2
The curvey =f(x) has a minimum point
(6, t), where tis a constant. State
(a) the value ofs
(b) the value oft
(c) th equation of the axis of symmetry
Answer:
(a) s = 6(b) t = 6(c) x = 6
7 Find the range of the values ofx suchthat (4x 2)(x+ 4) 2x + 8.
[3 marks]
Answer:
(4x 2)(x+ 4) 2x + 84x2 18x 8 2x + 84x2 16x 16 04(x2 + 4x+ 4) 04(x + 2)(x+ 2) 0
The range of values ofx isx 2 orx 2
5 Find the range of the values ofx such that
x(x 4) 4.
[3 marks]
Answer:
x(x 4) 4x2 4x+ 4 0(x + 2)(x+ 2) 0
The range of values ofx isx 2 orx 2
6 Find the range of the values ofx such that
x(x 7) 12. [3 marks]
Answer:x(x 7) 12x2 7x+ 12 0(x + 4)(x+ 3) 0
The range of values ofx isx 3 orx 4
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SUGGESTION QUESTION 7-8: INDICES & LOG
1 Solve the equation 2x + 3 2x + 2 = 8.
Answer:
2x + 3 2x + 2 = 82x23 2x22 = 238(2x) 4(2x) = 234(2x) = 23
2x + 2 = 23
x + 2 = 3
x = 1
2 Solve the equation 2x + 7 2x + 6 = 64.
[3 marks]
Answer:
2x + 7 2x + 6 = 642x27 2x26 = 26128(2x) 64(2x) = 2664(2
x) = 26
2x + 6 = 26
x + 6 = 6
x = 0
3 Solve the equation 32x 2 = 8
5x.
[3 marks]
Answer:
32x 2 = 85x
25(x 2)
= 23(5x)
5x 10 = 15x(5 + 15)x = 10
x =1
2
4 Solve the equation 254x + 5 = 125
5 5x.
[3 marks]
Answer:
254x + 5 = 1255 5x
52(4x + 5) = 53(5 5x)
8x+ 10 = 15 15x(8 + 15)x= 15 10
x =5
23
SUGGESTION QUESTION 9-11: PROGRESSION
1 Three consecutive terms of an
arithmetic progression are 13 x, 8, 13+ 3x. Find the common difference of the
progression. [4 marks]
Answer
13 + 3x 8 = 8 (13 x)2x = 16
x = 8
d= 8 (13 (8))= 3
2 The first three terms of an arithmetic
progressino are 7h + 2, k, 5h + 12.
(a) Express kin terms ofh.
(b) Find the 6th term of the
progression in terms ofh.
[4 marks]
Answer:
(a) k (7h+ 2) = (5h+ 12) k2k= 12h + 14k= 6h + 7
(b) a= 7h + 2d= k+ 7h 2
= (6h + 7) + 7h 2= h + 5
T6 = a + 5d
= 7h + 2 + 5(h + 5)= 7h + 2 + 5h + 25= 2h + 27
3 The first three terms of a geometric
progression arex, 24, 96.
Find
(a) the value ofx.
(b) the sum from the 4th term to the
6th term.
[3 marks]
Answer:
(a) r = 9624
= 4
x
24=
24
96
x =24
96 24
= 6
(b)S6 =
6(46 1)4 1
= 8190
S3 = 6 + 24 + 96= 126
Sum = 8190 126= 8064
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SUGGESTION QUESTION 12 : LINEAR LAW
1 The variablesx andy are related by the
equationy =px5
, wherep is a constant.
(a) Convert the equationy =px5
to
linear form.(b) Diagram 1 shows the straight lineobtained by plotting
log10y against log10x.
Diagram 1
Find the value of
(i) log10p.
(ii) q.
Answer:
(a) log10
= 5 (b) (i) log10p = Yintercept
= 7
(ii) log10y= 5(3) + 7= 8
q = log10y= 8
2 The variablesx andy are related by the
equationy =px4
, wherep is a constant.
(a) Convert the equationy =px4
to
linear form.(b) Diagram 2 shows the straight lineobtained by plotting
log10y against log10x.
Diagram 2
Find the value of
(i) log10p.
(ii) q.
Answer:
(a) l
(b) (i) log10p = Yintercept= 7
(ii) log10y= 4(4) + 7= 9
q = log10y= 9
3 The variablesx andy are related by the
equationy2 = 6x(x 6).Diagram 3 shows the straight line graph
obtained by plottingy2
xagainstx.
Diagram 3
Find the value ofp and q.
[2 marks]
Answer:
y2 = 6x(x 6)y2
x= 6(x 6)
y2
x= 6x 36
Wheny2
x= 0,
0 = 6x 36x = 6p =x = 6
Whenx = 3,y2
x = 6(3) 36= 18
q =y2
x= 18
4 The variablesx andy are related by the
equationy2 = 2x(x 6).Diagram 4 shows the straight line graph
obtained by plottingy2
xagainstx.
Diagram 4
Find the value ofp and q. [2 marks]
Answer:
y2 = 2x(x 6)y2
x= 2(x 6)
y2
x= 2x 12
Wheny2
x= 0,
0 = 2x 12
x = 6p =x = 6
Whenx = 2,y2
x= 2(2) 12
= 8
q =y2
x= 8
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SUGGESTION QUESTION 13-14: COORDINATE GEOMETRY
1 Find the equation of the straight line
which passes through P(5, 5) and isparallel to the straight line joining Q(7,
3) danR(6, 7).[3 marks]
Answer:
Gradient ofQR
=7 + 3
6 7
= 10
13
Equation:
y 5 = 10
13(x 5)
13y + 65 = 10x 5010x + 13y 115 = 0
2 Find the equation of the straight line which
passes throughA(4, 1) and isperpendicular to the straight line joining
B(9, 7) dan C(1, 4).[3 marks]
Answer:
Gradient ofBC
=4 + 71 + 9
=3
8
(3
8)m2= 1
m2= 8
3
Equation:
y 1 = 8
3(x + 4)
3y 3 = 8x 328x + 3y + 29 = 0
4 The pointJ(1, 11) divides a straight line joining P(7, 5) and Q(5, 14) in the ratio e :f.Find the values ofe andf.
[3 marks]
Answer:
(1, 11) = (7f 5ee +f
,5f+ 14e
e +f)
7f 5ee +f
= 1
7f 5e= ef8f= 4ee
f= 2
e = 2 andf= 1
5 Find the equation of the locus of moving pointA such that its distances from P(9, 7) and Q(
Answer:
LetA = (x,y)
GivenAP =AQ
(x + 9)2 + (y + 7)2 = (x + 7)2 + (y + 9)2
Squaring both sides,
(x + 9)2 + (y + 7)2 = (x + 7)2 + (y + 9)2
x2 + 18x + 81 +y2 + 14y + 49 =x2 + 14x + 49 +y2 + 18y + 81
4x 4y = 0
3 Given the area of a triangle with verticesA(3, 10),B(9, 4) and C(4,j) is 81 unit2.Find the possible values ofj.
[3 marks]
Answer:Area ofABC
=1
2|310
94
4
j
310
|= 81
1
2
((12) 9j+ 40 (90) 16 + 3j) =
811
2(102 6j) = 81
6j= 162 102j= 10 orj = 44
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SUGGESTION QUESTION 15-16: VECTORS
1Given that a
~= 10( )56 and b~ = ( )
45
48, find
(a) the vector 10a~
b~
.
(b) the unit vector in the direction of 10a~
b~
.
[4 marks]
Answer:
(a) 10a~
b~
= 10( )56 ( )45
48
= ( )512 (b) Unit vector in the direction of 10a
~ b
~
= (5i~ +12j~) / 52 + 122
= (5i~
+12j~
) / 13
3 Given that a~
= 4i + 3j and b~
= 4i + kj, find
(a) a~
b~
in the formxi +yj.
(b) the value ofkif |a~
b~
| = 17.
[4 marks]
Answer:
(a) a b~
= 4i + 3j (4i + kj) = 4i + 3j + 4i kj = 8i + (3 k)j
(b) |a~
b~
| = 17
82 + (3 k)2 = 1764 + (3 k)2 = 289(3 k)2 = 225
3 k = 15 k = 18
or
3 k = 15= 12
5Diagram 8 shows vector OP
drawn on a
Cartesian plane.
Express in the formxi +yj.
(a)OP
(b)Find the unit vector in the direction of
[Answer:
(a)OP
= 4i + 3j
(b) Unit vector in the direction of OP= (4i
~+3j
~) / 42 + 32
= (4i~
+3j~
) / 5
=4
5i +
3
5j
6Diagram 9 shows vector OA
drawn on a
Cartesian plane.
Diagram 9
(a)Express OA
in the form ( )xy .
(b) Find the unit vector in the
direction ofOA
.
[3 marks]
Answer:
(a)OA
= ( )56
(b) Unit vector in the direction of
= (5i~
+6j~
) / 52 + 62
= (5i~
+6j~
) / 7.81
2 The vectors a~
and b~
are non-zero and non-
parallel.
It is given that (h 7)a~
= (k 9)b~
where h
and kare constants.
Find the value of
(a) h
(b) k
[2 marks]Answer:
a~
and b~
are non-parallel.
Therefore (h 7)a~
= (k 9)b~
is true only when (h
7)a~
and (k 9)b~
are zero vector.
Since a~
and b~
are not zero vector, therefore (h 7)
and (k 9) must be zero, (h 7) = (k 9) = 0.(a) h 7 = 0
= 7
(b) k 9 = 0= 9
4It is given that OA
= 12a
~+ 7b
~, OB
= 32a
~+
21b~
and OC
= 78a~
14b~
.
(a)FindAB
andAC
.
(b) Hence, show that pointsA,B and C
are collinear.[4 marks]
Answer:
(a)AB
=AO
+ OB
= (12a~
+ 7b~
) 32a~
+ 21b~
= 44a~
+ 14b~
AC
=AO
+ OC
= (12a
~+ 7b
~) + 78a
~ 14b
~
= 66a~
21b~
(b)AC
= 3
2(44a
~+ 14b
~)
= 3
2AB
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SUGGESTION QUESTION 17: TRIGONOMETRIC FUNCTION
1 Given tanx= 0.6486 and 0 x180, find the value of
(a) x
(b) cotx
[2 marks]Answer:
(a) tanx= 0.6486tan = 0.6486 = 32 58'x= 180 32 58'
= 147 2'
(b)cotx =
1
tanx
=1
0.6486
= 1.542
2 Given sinx= 0.8026 and 0 x 180, findthe value of
(a) x
(b) cosecx
[2 marks]Answer:
(a) sinx= 0.8026sin = 0.8026 = 53 23'x= 180 53 23'
= 126 37'
(b)cosecx =
1
sinx
=1
0.8026
= 1.246
3 Solve the equation 16 sin2x = 0 sinx 18 sin 30 for 0 x 360[4 marks]
Answer:
16 sin2x = 0 sinx 18 sin 3016 sin2x = 0 sinx 916 sin2x sinx + 9 = 0(4 sin 3)(4sin 3) = 0
sinx =3
4
x = 48 35', 131 25'or
sinx= 3
4
x = 228 35', 311 25'
4 Solve the equation tanx + sinx = 0 for0 x 360
a[4 marks]
Answer:
tanx + sinx = 0sinx
cosx+ sinx = 0
sinx + cosx sinx = 0sinx(1 + cosx) = 0
sinx = 0
x = 0, 180, 360or
1 + cosx = 0
cosx= 1x = 180
5 Solve the equation 8 cos2x = 6 cosx
2 cos 60 for 0 x 360[4 marks]
Answer:
8 cos2x = 6 cosx 2 cos 608 cos2x = 6 cosx 18 cos2x 6 cosx + 1 = 0(2 cos 1)(4 cos 1) = 0
cosx =1
2
x = 60, 300
or
cosx =1
4
x = 75 31', 284 29'
6 Solve the equation 21 cos 2x = 46 sinx 33 for 0 x 360[2 marks]
Answer:
21 cos 2x = 46 sinx 3321(1 2 sin2x) = 46 sinx 3321 + 42 sin2x = 46 sinx 3342 sin2x 46 sinx + 12 = 021 sin2x 23 sinx + 6 = 0(3 sin 2)(7 sin 3) = 0
sinx =2
3
x = 41 49', 138 11'
or
sinx =3
7
x = 25 23', 154 37'
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SUGGESTION QUESTION 18: CIRCULAR MEASURE
1 Diagram 2 shows two arcs,AD andBC, of
two concentric circles with centre O and
having radius OA and OB.
Diagram 2
Find
(a) the angle , in radian.(b) the perimeter of the shaded region,ABCD.
[4 marks]
Answer:
(a) Length of arcBC= r8 = (2 + 16)
=8
18
= 0.44 rad.
(b) Length of arcAD = r
= 2 0.44= 0.88 cm
Perimeter of the shaded region
= 8 + 0.88 + 16 + 16
= 40.88 cm
2 Diagram 10 shows the sector OAB of a
circle with centre O.
Diagram 10
Find
(a) the angle in radians,
(b) the area of the shaded region[4 marks]
Answer:
(a)cosBOC=
12
15
= 0.8
BOC= 36.87
= 36.87
180
= 0.6436 rad.
(b)
Area of sector OAB =
1
2r
2
=1
2(15)2(0.6436)
= 72.4 cm2
Area of triangle OCB
=1
2 9 12
= 54 cm2
Area of the shaded region = 72.4 54
= 18.4 cm2
1Given that
1
2f(x) dx = 7, find
(a)the value of
2
1f(x) dx.
(b)the value ofkif
1
2[kx + 3f(x)] dx
=315
16.
[4 marks]
Answer:
(a)2
1f(x) dx=
1
2f(x) dx
= 7(b)
1
2[kx + 3f(x)] dx =
315
16
1
2kxdx+ 3
1
2f(x) dx =
315
16
k[x2
2]1
2+ 3(7) =
315
16
k(1
2 2) + 21 =
315
16
k=7
8
2Given that
9
4f(x) dx = 1, find
(a)the value of
4
9f(x) dx.
(b)the value ofkif
9
4[kx 3f(x)] dx
= 62.
[4 marks]Answer:
(a)4
9f(x) dx =
9
4f(x) dx
= 1(b)
9
4[kx 3f(x)] dx = 62
9
4kx dx 3
9
4 f(x) dx = 62
k[x2
2]9
4 3(1) = 62
k(81
2 8) 3 = 62
k = 2
SUGGESTION QUESTION 19: INTEGRATION
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SUGGESTION QUESTION 20: STATISTICS
1 A set of data consists of 9 numbers. The sum of the numbers is 27 and the sum of the squares
of the numbers is 2916. Find for the 9 numbers,
(a) the mean,
(b) the standard deviation.
[3 marks]
Answer:
(a)Mean =
27
9
= 3
(b)Standard deviation =
2916
9- 32
= 315
= 17.748
2 A set of data consists of 3 numbers. The sum of the numbers is 27 and the sum of the squares
of the numbers is 2916. Find for the 3 numbers,
(a) the mean,
(b) the standard deviation.
[3 marks]
Answer:
(a)Mean =
27
3
= 9
(b)Standard deviation =
2916
3- 92
= 891
= 29.85
3 Diagram 6 shows the mass of a group of students.
32, 30, 49, 25, 8, 35, 36, 5
Diagram 6
Find the interquartile range for the data.
Answer:
Rearrange the data,
25 28 30 32 35 36 49 53
First quartile =28 + 30
2= 29
Third quartile =36 + 49
2= 42.5
Interquartile range
= 42.5 29= 13.5
4 Diagram 7 shows the mass of a group of students.
44, 53, 26, 49, 43, 33, 38, 30
Diagram 7
Find the interquartile range for the data.
Answer:
Rearrange the data,26 30 33 38 43 44 49 53
First quartile = 30 + 332 = 31.5
Third quartile =44 + 49
2= 46.5
Interquartile range
= 46.5 31.5= 15
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5 Diagram 1 shows the scores obtained by a group of players in a game.
1, 10, 1, 9, 3, 4, 2
(a) Find the mean, median and mode for the scores.
(b) If the scores in Diagram 1 is multiplied by 5, and then added 4, find the mean,
median and mode for the scores.
[3 marks]Answer:
(a)Mean,x =
xN
=1 + 10 + 1 + 9 + 3 + 4 + 2
7
=30
7
= 4.29
Median = 3Mode = 1
(b) New mean = 5(4.29) + 4 = 25.45New median = 5(3) + 4 = 19
New mode = 5(1) + 4 = 9
6 Diagram 2 shows the scores obtained by a group of players in a game.
9, 18, 16, 20, 15, 9, 4
(a) Find the mean, median and mode for the scores.
(b) If the scores in Diagram 2 is multiplied by 3, and then added 6, find the mean,
median and mode for the scores.
[3 marks]
Answer:
(a)Mean,x =
xN
=9 + 18 + 16 + 20 + 15 + 9 + 4
7
=91
7
= 13Median = 15
Mode = 9
(b) New mean = 3(13) + 6 = 45
New median = 3(15) + 6 = 51
New mode = 3(9) + 6 = 33
SUGGESTION QUESTION 21: PROBABILITY
1 A bag contains 7 brown marbles, 2 yellow marbles and 6 purple marbles. Two marbles
are drawn from the bag at random, one after another. Find the probability that
(a) both marbles are brown.
(b) one marble is yellow and the other is purple.
[4 marks]Answer:
(a) Probability
=7
15
6
14
=1
5
(b) Probability
=2
15
6
14+
6
15
2
14
=4
35
2 A bag contains 6 blue marbles, 5 red marbles and 6 white marbles. Two marbles aredrawn from the bag at random, one after another. Find the probability that
(a) both marbles are blue.
(b) one marble is red and the other is white.
[4 marks]
Answer:
(a) Probability
=6
17
5
16
= 15136
(b) Probability
=5
17
6
16+
6
17
5
16
=15
68
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3In a shooting game, the probability for a player to shoot the target for each try is
2
3. If Hock
Seng tried the game for three times, find the probability that she shot at least one target in the
game.
[4 marks]
Answer:
LetA = Shot the target andB = Miss the target
Probability
= 1 P(BBB)
= 1 (1
3
1
3
1
3)
= 1 1
27
=26
27
4In a shooting game, the probability for a player to shoot the target for each try is
1
10. If Jabah
tried the game for three times, find the probability that she shot at least one target in the game.
[4 marks]
Answer:
LetA = Shot the target andB = Miss the target
Probability
= 1 P(BBB)
= 1 (9
10 9
10 9
10 )
= 1 729
1000
=271
1000
SUGGESTION QUESTION 22: PPERMUTATION & COMBINATION
1 How many 4-digit numbers greater than 2000 can be formed from the digits 1, 2, 3, 4,
5 and 6 if no repitition of digits is allowed?
[3 marks]
Answer:
5 5 4 3
Number of ways = 5 5 4 3
= 300
2 How many 3-digit numbers can be formed from the digits 3, 4, 5, 6 and 7 if the
numbers are
(a) less than 400?
(b) odd numbers?
[4 marks]
Answer:
(a) 1 4 3
Number of ways = 1 4 3
= 12
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C
3 There are 2 different t-shirts and 2 trousers on a cupboard. Calcualte the number of differentways to arrange all the clothes in a row if
(a) no condition is imposed.
(b) all the trousers are next to each other.
[4 marks]Answer:
(a) Number of ways = 4!
= 24
(b) Number of ways = 3! 2!
= 6 2
= 12
4 A florist want to choose 5 roses from a box of 5 red roses and 5 white roses to decorate a hamper.Calculate the number of ways the roses can be chosen if
(a) there is no restriction.
(b) there are more red roses than white roses chosen.
[4 marks]
Answer:
(a) Number of ways = 10C5
= 252
(b) The possible ways:5 red roses and 0 white rose
4 red roses and 1 white rose3 red roses and 2 white rose
Number of ways
= 5C5 5C5 +
5C4 5C4 +
5C3 5C3
= 126
5 A school wants to choose 4 students from a group of 6 boys and 5 girls to participatein a national mathematics contest. Calculate the number of ways the students can be
chosen if
(a) there is no restriction.
(b) the students chosen consists of 2 boys and 2 girls.
[4 marks]
Answer:
(a) Number of ways = 11C4
= 330
(b) Number of ways = 6C2 5C2
= 150
6 A florist want to choose 3 roses from a box of 6 red roses and 4 white roses to decorate
a hamper. Calculate the number of ways the roses can be chosen if
(a) there is no restriction.
(b) there are more red roses than white roses chosen.
[4 marks]
Answer:
(a) Number of ways = 10C3
= 120
(b) The possible ways:
3 red roses and 0 white rose2 red roses and 1 white rose
Number of ways
= 6C3 4C3 +
6C2 4C2
= 80
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C
SUGGESTION QUESTION 23: DIFFERENTIATION
1 Find the equation of the tangent to the curvey= x6 + 2x5 at the point (2, 0).
[3 marks]
Answer:
y= x6 + 2x5dy
dx= 6x5 + 10x4
At point (2, 0),dy
dx= 6(2)5 + 10(2)4
= 32The gradient of the tangent at point (2, 0) = 32
Equation:
y= 32(x 2)y= 32x + 64
2 Find the equation of the tangent to the curvey= (2x + 3)(x+ 6) at the point (3, 27).
[3 marks]
Answer:
y= (2x + 3)(x + 6)= 2x2 9x + 18
dy
dx= 4x 9
At point (3, 27),dy
dx= 4(3) 9
= 3The gradient of the tangent at point (3, 27) = 3
Equation:y 27 = 3(x + 3)y = 3x + 36
3Given that the curvey =f(x) and
dy
dx= 4px + 10, wherep is a constant.
The gradient of the curve atx= 2 is 18.Find the value ofp.
[3 marks]
Answer:
dy
dx= 4px + 10
Whenx= 2,dy
dx= 18
4p(2) + 10 = 18p = 1
4 Giveny= (4x + 6)(x 5). find
(a) dydx
(b) the value ofx wheny is maximum.
(c) the maximum value ofy.
[4 marks]
Answer:
(a) y= (4x + 6)(x 5)= 4x2 + 26x 30
dy
dx= 8x + 26
(b)
Wheny is maximum,
dy
dx = 08x + 26 = 0
x =13
4
(c)y= (4(
13
4) + 6)(1(
13
4) 5)
=49
4
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5 The radius of a balloon in the shape of a sphere increases at the rate of 6 cm s1. Find the rate
of change of the volume of the balloon when the radius is 10 cm.
[Volume of sphere, V=4
3r3]
[3 marks]
Answer:
V=4
3r3
dV
dr= 4r2
dV
dt=
dV
dr
dr
dt
= 4r2 6= 24r2
When r= 10,dV
dt= 24(10)2
= 2400 cm3 s1
6 The volume of a sphere increases at the rate of 19.2 cm3 s1. Find the radius of the sphere atthe instant when its radius is increasing at a rate of 0.3 cm s
1.
[Volume of sphere, V=4
3r3]
[3 marks]
Answer:
V=4
3r3
dV
dr= 4r2
dV
dt= 19.2dr
dt= 0.3
dV
dt=
dV
dr
dr
dt
19.2 = 4r2 0.3
4r2 =19.20.3
4r2 = 64r2 = 16
r= 4 cm
7Two variables,x andy, are related by the equationy =
7x2
.
Express the approximate change iny, in terms ofp, whenxchanges from 8 to 8 + p,wherep is a small value.
[3 marks]
Answer:
y =7x2
dydx
= 14x3
Whenx= 8dy
dx=
14
(8)3
= 7
256
x= (8 +p) + 8=p
y =dy
dx x
= 7
256p
8Two variables,x andy, are related by the equationy =
14
x2.
Express the approximate change iny, in terms ofp, whenx changes from 3 to 3 +p,
wherep is a small value.
[3 marks]
Answer:
y =14
x2
dy
dx=
28x3
Whenx = 3dy
dx=
28(3)3
= 28
27
x = (3 +p) 3=p
y =dy
dx x
= 28
27p
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C
SUGGESTION QUESTION 25: PROBABILITY DISTRIBUTION
1 Diagram 1 shows a standard normal dist ribution graph.
Diagram 1
The probability represented by the area of the shaded region is 0.7852.
(a) Find the value ofk.
(b) Xis a continuous random variable which is normally distributed with a mean of 79
and a standard deviation of 16. Find the value ofXwhen thez-score is k.[3 marks]
Answer:
(a) P(Z > k) = 1 0.7852= 0.2148
P(Z > 0.79) = 0.2148
k= 0.79(b) = 79, = 16
X 7916
= 0.79
X = 79 + 16 0.79
= 91.64
12 Diagram 2 shows a standard normal dist ribution graph.
Diagram 2
The probability represented by the area of the shaded region is 0.2486.
(a) Find the value ofk.
(b) Xis a continuous random variable which is normally distributed with a
mean of 51 and a standard deviation of 5. Find the value ofXwhen thez-
score is k.
[3 marks]
Answer:
(a) P(Z> k) = 0.5 0.2486= 0.2514
P(Z> 0.67) = 0.2514
k= 0.67
(b) = 51, = 5X 51
5= 0.67
X= 51 + 5 0.67
= 54.35
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LAST EXAM TIPS PAPER 2
2012
MINI BENGKELADDITIONAL MATHEMATICS
[Improve from 60 to A+]
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KUMPULAN SASARAN:Pelajar Yang mendapat Markah Kurang dari 40%
FORMAT KERTAS 2 ADD MATHS SPM 2012
BAHAGIAN A (6 Soalan Wajib Jawab semua) Total Markah =40
BAHAGIAN B (5 soalan Wajib Jawab 4 sahaja) Total Markah = 40
BAHIAGAN C (4 soalan Wajib Jawab 2 sahaja) Total Markah = 20
Berikut merupakan statistic soalan peperiksaan sebenar yang pernah keluar didalam
Peperiksaan dari 2006- 2010..Sila LIHAT apakah PERSAMAAN dan PEMBEZAAN dari
Segi tajuk.
Section A (Answer all question) Total Marks= 40 marks
Bahagian A mengandungi 6 SOALAN WAJIB yang mesti dijawab.
SPM07 SPM08 SPM09 SPM10 SPM11
1 SimultaneousEquation
SimultaneousEquation
SimultaneousEquation
SimultaneousEquation
SimultaneousEquation
2 Coordinate
Geometry
Quadratic
Functions
Quadratic
Equation
Trigonometric
Function
Trigonometric
Function
3 TrigonometricFunction
Progressions Differentiation+Integration
Progressions Progressions
4 Differentiation+Integration
TrigonometricFunction
TrigonometricFunction
Integration Integration
5 Statistics Statistics Vector GeometryCoordinate
GeometryCoordinate
6 Progressions Vector Progressions Statistics Statistics
Section B (Answer 4 Question only from 5 question) Setiap Soalan=10 markah X 4
Total Marks=40 marks
SPM07 SPM08 SPM09 SPM10 SPM11
7 Linear Law Integration Integration Linear Law Linear Law
8 Vector Linear Law Linear Law Differentiation Differentiation
9 Circular
Measure
Circular
Measure
Geometry
Coordinate
Vector Vector
10 Integration Geometry
Coordinate
Circular
Measure
Probablility
Distribution
Probablility
Distribution
11 Probablility
Distribution
Probablility
Distribution
Probablility
Distribution
Circular
Measure
Circular
Measure
Section C (Answer 2 Question only from 4 question) Total Marks=20 marks
SPM07 SPM08 SPM09 SPM10 SPM11
12 Solution ofTriangle
Solution of
Triangle
Solution of
Triangle
Solution of
Triangle
Solution of
Triangle
13 Index Number Index Number Index Number Index Number Index Number
14 LinearProgramming
Linear
Programming
Linear
Programming
Linear
Programming
Linear
Programming
15 Motion Alongstraight Line
Motion Along
straight Line
Motion Along
straight Line
Motion Along
straight Line
Motion Along
straight Line
Remark:Mengikut statistic 2005-2011 bab yang tidak pernah keluar didalam kertas dua adalah,
indices & logarihms, permuatation & combination serta probability.
Adakah Markah Anda selalu dibawah 40% untuk KERTAS 2?.
Untuk sekadar mencapai 40% dalam Matematik Tambahan adalah tidak sesukar mana.Pelajartidak perlu pun untuk mencapai markah penuh bagi sesuatu Soalan,memadailah dapat curi
separuh dari markah penuh bagi sesuatu Soalan.Oleh itu kepada pelajar yang selalu mendapat
markah kurang dari 40% apa kata anda ulangkaji modul dibawah ini berulang kali,sekurangnya 1
kali tanpa melihat jawapan.Jika anda sudah pun mengulangkaji mengikut saranan ,selepas inijangan terkejut dengan keputusan peperiksaan anda sendiri.Walaupun target anda hanya lulus
(40%) kami jamin anda boleh mencapai markah yang lebih baik lagi.Semoga Berjaya.
Secara Umumnya Bab yang akan keluar didalam Kertas 2 Matematik Tambahan mengandungi 18
bab kesemuanya daripada 21 bab bagi tingkatan 4 dan 5.Banyak bukan.? Masih Sempat? Ya!walaupun masa hanya 3 hari percayalah anda masih berupaya dengan mengikuti tip -tips mudah
yang akan MC berikan nanti.
Remarks:Disebabkan MODUL From 60 to A+ ini khas untuk pelajar kategori sederhana dan cemerlang
maka dibawah ini merupakan Tajuk penuh yang perlu anda ulangkaji
Lihat Jadual Dibawah
BAHAGIAN A (6 Soalan Wajib Jawab semua) TOTAL MARKS= 40
NO TAJUK FULL MARKS
1 Simultaneous Equation 5 markah
2 Trigonometry Function 6 markah
3 Progression 6 markah
4 Vector 8 markah
5 Statistic 7 markah
6 Differemtiation + Integration 8 markah
6 Backup (Quadratic Eq+Quadratic Func)
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EXAM TIPS: QUESTION 3-PROGRESSION Form 5
EXAM TIPSPercayalah soalan sebegini tidak sesukar mana
Soalan (a)Langkah 1:bina progression.sekurangnya 3
sebutan
_________ , _______ , ________
*untuk membina 3 sebutan ini anda perlu
semak dahulu soalan mahu AREA,volumeatau perimeter. *
*Dalam soalan ini,ianya meminta pelajar
dapatkan area rectangles.*
Langkah 2 : Cari common ratio.Tentukan GP
or AP.supaya pelajar tahu apakah formula
yang perlu digunakan
2
3
1
2 .T
Tatau
T
Tr
*Ikuti 2 langkah ini sudah mencukupi.bukan
sahaja lulus tapi anda akan dapat lebih dari
itu.*
Soalan (b) iApabila x = 60 y = 40.progressionnya adalah
2400 , 600 , 1501 nn arT
Soalan (b) ii
r
aS
1
EXAM TIPS: QUESTION 4-VECTOR Form 5
EXAM TIPS
Untuk Kertas 2,tajuk Vector kebiasanyaa maklumat
yang diberi akan mengandungi bentukRATIO atauFRACTION.
Soalan ditanya kebiasaan ADA 3 Bahagian.
TOLONG FAHAM FORMAT SOALAN INI
SANGAT PENTING.KEBIASAAN SAMA
SAJA SOALAN DAN CARANYA
Bentuk soalanya.berkait-kait.ini bermakna jika a (i)
anda salah maka a(ii) juga akan salah.begitulah
berikutnya.Ini kerana untuk mendapatkan Jawapan
a(ii) anda perlukan Jawapan daripada a (i)
Ini bermakna
Jawapan a (i) dapatkan maklumat dari soalan
Jawapan a (ii)-sebahagianya dari soalan dan satu
lagi PASTI dari Jawapan a (ii)
Jawapan b (i) dan b (ii)juga samasahaja.jawapanya dibentuk dari Jawapan dari a
**Soalan (a) biasanya 3 markah.dan penyelesainya
tidaklah sesukar mana.pelajar seharusnya skor. **
Jawapan (a) (i)PQ = PO+OQ
= - 4y +3OA
= -4y + 3 (8x)= 24 x4 y
Jawapan (a) (ii)
Jawapan (b) (i)
PB = mPA= m (PO+OA)
= m (-4y +8x)
= 8mx -4my
Jawapan (b) (ii)BC = nOC
= n (6x+3y)
= 6nx + 3ny
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EXAM TIPS: QUESTION 6 QUADRATIC FUNCTION
Jawapan (a)Coordinate A?Jika diperhatikan coordinate A adalah (0,y)
Maka y yang ingin di cari adalah y intercept..Dari persamaan diberi y intercept adalah 11
Maka A= (0,11)
Jawapan (b)
]114
)2
[()(
]1122
[)(
]1122
[)(
11)(
22
22
2
22
2
2
kkxxf
kkkxxxf
kxxxf
kxxxf
Anda perlu tukar dari cbxaxxf 2)( kepada CTS
qpxa 2)(
Langkah 1: faktorkan nilai a (letak diluar)[jika ada]
Langkah 2:tambahkan formula ini22
22
Langkah 3: gantikan ruangan kosong dengan nilai
disebelah x. dalam soalan ini adalah nilai k
Langkah 4: jadikan qpxa 2)(
*Nilai p = x min* = 6,32
kk
*Nilai q = y min*
2
114