MATHS Catch

Transcription

MATHS Catch

MATHS Catch
SPM
2012
USAHA +DOA+TAWAKAL
FOKUS A+
LAST EXAM TIPS PAPER 2
MATHS Catch
MATHEMATICS SPM
Improve to A+
MOHD RAJAEI BIN MOHAMAD ALI
©Hak Cipta Terpelihara
1
1449/1 2012 Maths Catch Network © www.maths-catch.com
[Lihat halaman sebelah]
SULIT
MATHS Catch
SPM
2012
USAHA +DOA+TAWAKAL
FOKUS A+
FORMAT KERTAS 2 MATEMATIK SPM 2012
JUMLAH SOALAN : 15 YANG WAJIB DIJAWAB
JUMLAH MARKAH : 100
Last Exam Tips for Mathematics SPM 2012
Exam Year: November 2012
Reference: The analysis is base on last 5 year National SPM exam paper 2005-2011 and State
trial Exam 2011
Disclaimer/Penafian:
The exam tips provided are base on pure forecast and assumptions. Maths Catch Network and
www.maths-catch.com will not be liable for any inaccuracy of the information. Students are not
encouraged to rely 100% on the tips to score in SPM exams. Students are advised to study hard
for their exam. Students can use the tips as a guide. All the materials have not gone for been
proof reading or editing process.
Format SPM Mathematics Exam 2011
PAPER
TIME
QUESTION
1
1 Hour 15 min
Objective
2
2 hour 30 min
Subjective
NO
1
2
3
4
5
6
7
8
9
ANSWER
Option A,B,C,D
Working Steps
Senarai Topik Matematik 1449 (Kertas 1 dan 2)
FORM 1-3
FORM 4
Circle I and II
Standard Form
Polygons I and II
Quadratic Expression and
Equation
Solid and Volumes
Sets
Transformations I and II
Mathematical Reasoning
Trigonometry I
The Straight Line
Algebraic Expressions I and II
Statistic III
10
Algebraic Fractions
Algebraic Formulae
Linear Equations (1 and 2
unknowns)
Indices
11
Linear Inequalities
12
13
Statistics I and II
Arc Length & Area of Sector
Probability I
Circles III
Trigonometry II
Angles of Elevation and
Depression
Lines and Plane in 3
Dimensions
TOTAL Q
40
20
MARKS
40%
100%
FORM5
Number Bases
Graphs of Function II
Transformation III
Matrices
Variations
Gradient and Area Under
a Graph
Probability II
Bearing
Earth as a Sphere
Plans and Elevations
2
SULIT
MATHS Catch
SPM
2012
USAHA +DOA+TAWAKAL
FOKUS A+
Analisis Kertas 2 Soalan Matematik SPM 2005-2011
Topic
SECTION A (Question 1-11)
1
Linear Equation
2
Quadratic Equation
Sets (Shade Venn Diagrams)
3
Region for Inequalities
4
Mathematical Reasoning
5
The straight Line
6
Probability II
7
Arc Length & Area of Sector
8
Volume of Solids
 Pyramids + Half Cylinder
 Cuboids + Half Cylinders
 Cones + Cuboids
 Pyramid + Prism
 Cones + Hemisphere
9
Matrices
10
Gradient and Area Under a Graphs
 Speed-Time Graphs
 Distance-Time Graphs
11
Lines & Planes in 3-D
SECTION B (Question 1-11)
12
Graphs of Function II
 Quadratic
 Cubic
 Reciprocal
13
Transformation III
14
Statistics III
 Ogive
 Histogram
 Frequency Polygon
15
Plans and Elevations
 Prism + Cuboids
 Cuboids +Half Cylinder, Prism
 Prism + Prism
16
Earth as a Sphere
SPM’06
SPM’07
SPM’08
SPM’09
SPM’10
SPM’11
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Answer Four Only
3
SULIT
MATHS Catch
SPM
2012
USAHA +DOA+TAWAKAL
FOKUS A+
TABURAN SOALAN PERCUBAAN NEGERI 2012 DAN SOALAN RAMALAN 2012
MATEMATIK SPM KERTAS 2
TAHUKAH ANDA!
Kertas 2 Matematik SPM ini mengandungi 15 soalan yang Wajib dijawab dan membawa markah
sebanyak 100%. Kertas 2 ini juga dikenali sebagai LUBUK EMAS untuk mengorek markah
semaksimum yang mungkin dalam Matematik SPM.Ini kerana Soalan-soalan yang ditanya adalah
lebih mudah sedikit daripada kertas 1.Walaupun Kertas 2 merupakan soalan berbentuk subjektif
tetapi percayalah Ianya adalah lebih mudah dari kertas 1.Jika anda selalu gagal dalam peperiksaan
apa kata tumpukan pada kertas 2 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 dari C ke A jika kena dengan
caranya.
Daripada statistik Soalan Peperiksaan Sebenar 2005-2011 dan statistic soalan Percubaan negeri
2012,maka berikut merupakan tajuk-tajuk yang perlu CALON SPM 2012 beri tumpuan bagi tahun
2012 dalam KERTAS 2.
SOALAN 1
Inequalities
Sets
MRSM’12, Kedah’12, Negeri Sembilan’12, Perak’12
SBP 2012, SPM 2011
RAMALAN 2012
SOALAN 2
QUADRATIC EQUATION
Fraction
Simple
 Perak’12, Terengganu’12, Melaka’12, Penang’12 RAMALAN 2012
 SBP 2012, Kedah’12, MRSM’12, Negeri Sembilan’12, SPM2011
SOALAN 3
LINEAR EQUATION
Fraction
Simple
 Perak’12, Terengganu’12, Melaka’12, Penang’12, SPM2011
 SBP 2012, Kedah’12, MRSM’12, Negeri Sembilan’12
RAMALAN 2012
SOALAN 4
LINES & PLANES IN 3D
Lines + Plane
Plane + Plane
 Kedah’12, Perak’12, Penang’12
MRSM’12, Negeri Sembilan’12, Terengganu’12, Melaka’12, SBP’12, SPM 2011
RAMALAN 2012
SOALAN 5
MATHEMATICAL REASONING
Quantifier ‘All’ and ‘Some’  MRSM’12, Melaka’12
Operation on Statement
SBP’12, Kdh’12, Prk’12, N.Sembilan’12, Trg’12, Mlk’12, Png’12, SPM2011
Implication
 SBP’12, Kdh’12, MRSM, Prk’12, Trg’12, Png’12, SPM2011
Arguments
 SBP’12, Kdh’12, N.Sembilan’12, Trg’12, SPM2011
Deduction and Induction  Prk’12, N.Sembilan’12, Mlk’12, Png’12
RAMALAN 2012
RAMALAN 2012
RAMALAN 2012
SOALAN 6
SOLID GEOMETRY
Prism + Cylinder
Prism + cone
Cylinder + Cone
Hemispeher + Cone
Half Cylinder + Triangle
Half Cylinder + Cuboid
Kedah’12, Penang’12
Perak’12, negeri Sembilan’12
MRSM’12
Melaka’12
Terengganu’12
SBP’12, SPM2011
RAMALAN 2012
4
SULIT
MATHS Catch
SPM
2012
USAHA +DOA+TAWAKAL
FOKUS A+
SOALAN 7
THE STRAIGHT LINE
Find Equation + x-intercept
Find Equation + y-intercept
RAMALAN 2012
 Kedah’12, MRSM’12, Perak’12, Terengganu’12, SBP’12
Negeri Sembilan’12, Melaka’12
SOALAN 8
MATRICES
Soalan (a)
-Matrices Indentities
-No inverse matrices or
Simple matrice
SOALAN 9
CIRCLES
Sector + Sector
Sector + Semicircles
RAMALAN 2012
 Kedah’12, MRSM’12, N.Sembilan’12, Tereng’12, Melaka, SPM2011
 Perak’12 , Penang’12, SBP’12
RAMALAN 2012
 Kedah’12, Penang, SBP’12
MRSM’12, Negeri Sembilan’12, Perak’12, Terengganu’12, Melaka’12, SPM2011
SOALAN 10
GRADIENT AND AREA UNDER A GRAPHS
 Kedah’12, MRSM’12, N.Sembilan’12, Penang’12, SBP’12, SPM2011
Perak’12, Terengganu’12, Melaka’12
RAMALAN 2012
Speed Vs Time
Disntace Vs Time
SOALAN 11
PROBABILITY II
Diagram/graphs
Sentences
RAMALAN 2012
 Kedah’12, MRSM’12, Perak’12, Terengganu’12, Melaka’12, SBP’12, SPM2011
Negeri Sembilan’12, Penang’12
SOALAN 12
GRAPHS OF FUNCTION II
Graphs Quadratic x2
Graphs cubix x 3
Graphs reciprocal
1
x
 Kedah’12, MRSM’12, Terengganu’12
 SPM2011
 Negeri Sembilan’12, Perak’12, Melak’12, Penang’12, SBP’12
RAMALAN 2012
SOALAN 13
TRANSFORMATION III
Translation + Reflection
Translation + Rotation
Reflection + Rotation
 Kedah’12, Negeri Sembilan’12, Melaka’12, Terengganu’12, SPM2011
MRSM’12, Perak’12, SBP’12
RAMALAN 2012
Penang’12
SOALAN 14
STATISTICS III
Graphs Ogive
Graphs Histogram
Graphs Frequency polygon
 Kedah’12
Negeri Sembilan’12, Terengganu’12, SPM2011
 MRSM’12, Perak’12, Melaka’12, Penang’12, SBP’12
RAMALAN 2012
5
SULIT
MATHS Catch
FOKUS A+
SPM
2012
USAHA +DOA+TAWAKAL
EXAM TIPS: QUESTION 1 – INEQUALITIES
EXAM TIPS
Soalan bagi tajuk ini meminta kepada pelajar membuat lorekkan sahaja.Pelajar perlu membuat lorekkan yang memenuhi ketiga-tiga
persamaan yang diberikan.
Perkara penting yang perlu pelajar FAHAMI antaranya adalah maksud symbol-simbol ini
< Atau > [pelajar perlu buat garis putus-putus] contoh
----------------------------------------___________________________
 Atau  [pelajar perlu buat garis terus] contoh
SPM’05
1 soalan keluar
SPM’06
SPM’07
1 soalan keluar
SPM’08
SPM’09
1 soalan keluar
SPM’10
SPM’11
SOALAN 1
Inequalities
1
MRSM’12, Kedah’12, Negeri Sembilan’12, Perak’12
Diagram 1 shows the graph of y = −4x + 4 and x = 0.
RAMALAN 2012
Shade the region that satisfies y ≥
3
x − 3, y > −x and y
5
≤ 3.
3
4
Diagram 3 shows the graph of y = − x − 4 and x = 3.
3
Shade the region that satisfies y ≤ −4x + 4, y ≤ −x and x
> 0.
2
Diagram 2 shows the graph of y =
3
x − 3 and y = −x.
5
Diagram 3
4
Shade the region that satisfies y ≥ − x − 4, y < x and x
3
< 3.
Diagram 2
6
SULIT
MATHS Catch
USAHA +DOA+TAWAKAL
FOKUS A+
SPM
2012
EXAM TIPS: QUESTION 2 – QUADRATIC EQUATIONS
SPM’07
1 soalan
keluar 4 x 2  15  17 x
SPM’08
1 soalan Keluar
x 1 
6  3x
2x
SPM’09
SPM’10
SPM’11
1 soalan keluar
1 soalan keluar
1 soalan keluar
x 2  4 x  9  2( x  3)
5x 2  4 x  3(2  x)
4 x( x  4)  9  16 x
SOALAN 2
QUADRATIC EQUATION
Fraction
Simple
FRACTION QUESTION
1
2
3
4
5
RAMALAN 2012
SIMPLE (DIRECT QUESTION)
3  9x
Solve the equation x  2 =
.
4x
12  9x
.
2x
20  7x
.
3x
Solve the quadratic equation
4(4x2  1)
= 7x.
9
1
Solve the equation 4x2 = 7(10x + 9) + 11.
2
Solve the equation 2x2 = 3(7x + 2)  16.
3
Solve the equation 5x2 = 6(2x + 7)  10.
4
Using factorisation, solve the quadratic
equation 3x2 + 16 = 26x.
5
Solve the quadratic equation x(x + 1) = 6 +
2x.
4x2  4x
Solve the equation
= 5.
7x  7
7
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MATHS Catch
FOKUS A+
SPM
2012
USAHA +DOA+TAWAKAL
EXAM TIPS: QUESTION 3 – LINEAR EQUATION
SPM’07
Solve
g  2h  1
4 g  3h  18
SPM’08
Solve
3
y  3
2
4 x  y  16
x
SPM’09
Solve
2
x  4 x  9  2( x  3)
SPM’10
Solve
5m  6n  13
m  2n  1
SPM’11
Solve
m  3n  12
2
mn2
3
SOALAN 3
LINEAR EQUATION
Fraction
Simple
SIMPLE (Direct Question)
1
FRACTION QUESTION
Calculate the value of s and of t that satisfy the
following simultaneous linear equations:
1
Calculate the value of m and of n that satisfy the
m  4n = 4
5
m − 3n = 1
4
2
Calculate the value of x and of y that satisfy the
x + 2y = 4
3
x − 2y = 1
2
3
m  4n = 4
3
m − n = 1
4
4
s + 2t = 2
5
s + 3t = 1
2
5
Calculate the value of p and of q that satisfy the
p + 3q = 3
5
p+q=1
3
st=4
7s + 2t = 10
2
mn=3
2m + n = 3
3
Calculate the value of x and of y that satisfy the
xy=6
x+y=2
4
st=7
3s + 4t = 7
5
RAMALAN 2012
mn=2
5m + 4n = 1
EXAM TIPS
Kebiasaanya soalan ini markahnya adalah 4
markah..KUNCI soalan ini pelajar perlu
membuat persamaan 3
8
SULIT
MATHS Catch
USAHA +DOA+TAWAKAL
FOKUS A+
SPM
2012
EXAM TIPS: QUESTION 4 – LINES & PLANE IN 3D
SOALAN 4
Lines + Plane
Plane + Plane
1
 Kedah’12, Perak’12, Penang’12
MRSM’12, Negeri Sembilan’12, Terengganu’12, Melaka’12, SBP’12, SPM 2011
Diagram 1 shows a cuboid.
RAMALAN 2012
(c) Calculate the angle between the line BH and the
base ABCD.
4
Diagram 4 shows a right prism with the right-angled
triangle PQT as its uniform cross-section. V is the
midpoint of QR.
Diagram 1
(a) Name the orthogonal projection of the line QW on
the plane QRVU.
(b) Name the angle between the plane PQW and the
plane PQUT.
(c) Calculate the angle between the line QW and the
base PQRS.
2
Diagram 4
(a) Calculate the length of PV.
(b) Calculate the angle between the line TV and the
plane PQRS.
(c) Calculate the angle between the plane PVT and the
plane PSUT.
5
Diagram 5 shows a cuboid. X is the midpoint of QR.
Diagram 2
(a) Name the orthogonal projection of the line QW on
the plane QRVU.
(b) Name the angle between the plane PQW and the
plane PQUT.
(c) Calculate the angle between the line QW and the
base PQRS.
3
Diagram 5
(a) Calculate the length of SX.
(b) Calculate the angle between the line WX and the
base PQRS.
(c) Name the angle between the plane QXWT and the
plane PSWT.
Diagram 3
(a) Name the orthogonal projection of the line BH on
the plane BCGF.
(b) Name the angle between the plane ABH and the
plane ABFE.
9
SULIT
MATHS Catch
SPM
2012
FOKUS A+
USAHA +DOA+TAWAKAL
EXAM TIPS: QUESTION 5 – MATHEMATICAL REASONING
Quantifier ‘All’ and ‘Some’  MRSM’12, Melaka’12
Operation on Statement
SBP’12, Kdh’12, Prk’12, N.Sembilan’12, Trg’12, Mlk’12, Png’12, SPM2011
Implication
 SBP’12, Kdh’12, MRSM, Prk’12, Trg’12, Png’12, SPM2011
Arguments
 SBP’12, Kdh’12, N.Sembilan’12, Trg’12, SPM2011
Deduction and Induction  Prk’12, N.Sembilan’12, Mlk’12, Png’12
RAMALAN 2012
RAMALAN 2012
RAMALAN 2012
1
(a) Determine whether the following statements is true or false.
(i) 6 ÷ 5 = 1 and 82 = 64
(ii) The elements of set X = {10, 15, 20} are divisible by 3 or the elements of set Y = {32, 36, 40} are multiples of 4.
(b) Write down Premise 2 to complete the following argument:
Premise 1:
If m is greater than zero, then m is a positive number.
Premise 2:
Conclusion: 10 is a positive number.
(c) Write two implications from the sentence below:
4x > 24 if and only if x > 6
2
(i) 7 ÷ 3 = 4 and 102 = 100
(ii) The elements of set S = {12, 15, 18} are divisible by 5 or the elements of set T = {16, 20, 24} are multiples of 4.
Premise 1:
Premise 2:
2n > 18 if and only if n > 9
3
(i) 15 ÷ 3 = 5 and 72 = 14
(ii) The elements of set M = {18, 20, 22} are divisible by 2 or the elements of set N = {12, 16, 20} are multiples of 3.
Premise 1:
If p is greater than zero, then p is a positive number.
Premise 2:
5n > 25 if and only if n > 5
4
(i) 5 ÷ 2 = 3 and 52 = 25
(ii) The elements of set X = {24, 27, 30} are divisible by 3 or the elements of set Y = {20, 22, 24} are multiples of 5.
Premise 1:
Premise 2:
9p > 63 if and only if p > 7
5
(i) 14 ÷ 2 = 7 and 42 = 8
(ii) The elements of set S = {20, 22, 24} are divisible by 3 or the elements of set T = {16, 20, 24} are multiples of 4.
Premise 1:
If p is greater than zero, then p is a positive number.
Premise 2:
9q > 54 if and only if q > 6
10
SULIT
MATHS Catch
SPM
2012
USAHA +DOA+TAWAKAL
FOKUS A+
EXAM TIPS: QUESTION 6 – SOLID GEOMETRY
(VOLUME)
SOLID
GEOMETRY
Prism + Cylinder
Prism + cone
Cylinder + Cone
Hemispeher + Cone
Half Cylinder + Triangle
Half Cylinder + Cuboid
1
Kedah’12, Penang’12
Perak’12, negeri Sembilan’12
MRSM’12
Melaka’12
Terengganu’12
SBP’12, SPM2011
RAMALAN 2012
Diagram 1 shows a combined solid consists of a right
prism and a cylinder. Trapezium ABCD is the uniform
cross section of the prism.
2
Diagram 2 shows a combined solid consists of a right
prism and a pyramid which are joined at the plane
PQUT. Trapezium PQRS is the uniform cross section of
the prism. XUV is a straight line.
Diagram 1
Diagram 2
Given AB = 18 cm, DC = 22 cm, BC = 15 cm and FB =
14 cm. The diameter and height of the cylinder is 7 cm
and 10 cm respectively.
Calculate the volume, in cm3, of the solid.
22
[Use π =
]
7
Volume Cone
=
1 2
r h
3
Volume Cylinder
=
Volume Sphere
=
r 2 h
4 3
r
3
Volume Prism
=
Volume Pyramid
=
Given PQ = 22 cm, SR = 19 cm, TP = 6 cm and XU = 9
cm.
(a) Calculate the volume, in cm3, of the pyramid.
(b) Calculate the length, in cm, of QR if the volume
of the combined solid is 1380 cm3.
Area.surface  length
1
 Base. Area  height
3
NOTE
**Sila gantikan Bongkah Silinder dan juga pyramid yang diberikan dengan cone**
11
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MATHS Catch
USAHA +DOA+TAWAKAL
FOKUS A+
SPM
2012
EXAM TIPS: QUESTION 7 – THE STRAIGHT LINE
THE STRAIGHT LINE
Find Equation + x-intercept
Find Equation + y-intercept
Equation + x-intercept
1
RAMALAN 2012
 Kedah’12, MRSM’12, Perak’12, Terengganu’12, SBP’12
Negeri Sembilan’12, Melaka’12
Given that the equation of the straight line CD is 11x +
5y = −95, Find
(a) the equation of the straight line AB,
(b) the x-intercept of the straight line AB.
In Diagram 1, the straight line WX is parallel to the
straight line YZ.
4
straight line YZ.
Diagram 1
Diagram 4
Given that the equation of the straight line YZ is −14x −
6y = −8, Find
(a) the equation of the straight line WX,
(b) the x-intercept of the straight line WX.
2
Given that the equation of the straight line YZ is 13x +
12y = −93, Find
straight line YZ.
5
Given that the equation of the straight line YZ is 15x +
10y = 85, Find
Diagram 2
Given that the equation of the straight line YZ is −11x −
8y = −69, Find
3
In Diagram 3, the straight line AB is parallel to the
straight line CD.
straight line YZ.
EXAM TIPS
Didalam peperiksaan 3 soalan yang paling popular ditanya dan
dijangkakan keluar pada 2012 adalah 3 soalan iaitu:
Find
-Equation
- x intercept
- y intercept
Berikut merupakan tips mudah perlu pelajar fahami untuk menguasai bab
ini
Persedian perlu dibuat.Tahu 3 perkara ini

Gradient= m  y 2  y1 atau m   y  int ercept
x  int ercept
x 2  x1

Equation y  mx  c
INGAT !!!!!!
y  intercept bermakna coordinatennya (0,

x  intercept bermakna coordinatenya

y)
(x,0)
12
SULIT
MATHS Catch
FOKUS A+
SPM
2012
USAHA +DOA+TAWAKAL
EXAM TIPS: QUESTION 8 – MATRICES
PREVIEW Analisis Soalan Peperiksaan SEBENAR 2009 – 2011
SPM’09
SPM’10
Question No 11
(c) Given inverse matrix of
Question No 8
 2  3

 4  5
(a) Given matrix A  

SPM’11
Question No 8
 3 2 1 0


5   0 1 
(e) Given M 
6

Find M
 4  1
 5 1

is..k 

2
5


  2 m
Find inverse matrix of A
Find m and k
(b) Find x and y using matrix
(f)
2x  3y  7
Find x and y using matrix
(d) Find x and y using matrix
4x  y  7
4 x  5 y  13
3x  2 y  3
6x  5 y  9
2 x  5 y  2
SOALAN 8
MATRICES
Soalan (a)
-Matrices Indentities
-No inverse matrices or
Simple matrice
1
It is given that S =
that ST =
 Kedah’12, MRSM’12, N.Sembilan’12, Tereng’12, Melaka, SPM2011
 Perak’12 , Penang’12, SBP’12
(63 37) and T = x(76 3y) such
x and y.
4
(10 10).
(a) Find the values of x and y.
(b) Write the following simultaneous linear equations
as matrix equation:
It is given that P =
that PQ =
(10 01).
9x  7y = 9
7x + 5y = 4
Hence, using matrix method, calculate the value of
x and y.
It is given that S =
that ST =
(
2 3
7 8
)
x and y.
8 3
and T = x
y 2
( ) such
5
(10 10).
(a) Find the values of x and y.
as matrix equation:
It is given that X =
that XY =
(
)
n 7
and Y = m
4 5
(
(16 27) and N = s(6t 12 ) such
(10 10).
(a) Find the values of s and t.
as matrix equation:
x  2y = 4
6x + 7y = 8
x and y.
5 7
4 5
It is given that M =
that MN =
2x  3y = 8
7x + 8y = 8
3
(97 75) and Q = s(75 9t ) such
(a) Find the values of s and t.
as matrix equation:
3x  3y = 7
6x + 7y = 10
2
RAMALAN 2012
x and y.
) such
(10 10).
(a) Find the values of m and n.
as matrix equation:
5x  7y = 8
4x + 5y = 1
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EXAM TIPS: QUESTION 9– CIRCLE (AREA)
CIRCLES
Sector + Sector
RAMALAN 2012
 Kedah’12, Penang, SBP’12
MRSM’12, Negeri Sembilan’12, Perak’12, Terengganu’12, Melaka’12, SPM2011
Sector + Sector
1
Diagram 1 shows two sectors OPQ and ORS with the
same centre O. OSQ is a straight line.
4
It is given that ∠POQ = 60° and ∠ROS = 45°.
22
Using π =
, calculate
7
(a) the perimeter, in cm, of the sector OPQ,
(b) the area, in cm2, of the shaded region.
2
In Diagram 4, OPQ is a sector of a circle with centre O
and RSTU is a semicircle with centre R. PSRUO is a
straight line.
Diagram 4
It is given that PO = 35 cm, RS = 7 cm and ∠POQ =
48°.
22
Use π =
, and give the answer correct to two decimal
7
places.
Calculate
(a) the area, in cm2, of the shaded region.
(b) the perimeter, in cm, of the shaded region.
Diagram 2 shows two sectors OAB and OCD with the
same centre O. ODB is a straight line.
5
Diagram 2
Diagram 5 shows two sectors OPQ and ORST with the
same centre O. OUT is a semicircle with diameter OT
and OT = 2PO. POT and OQR are straight lines.
It is given that ∠AOB = 63° and ∠COD = 42°.
22
Using π =
, calculate
7
(a) the perimeter, in cm, of the sector OAB,
3
Diagram 3 shows two sectors OAB and OCD with the
same centre O. ODB is a straight line.
Diagram 5
PO = 21 cm and ∠POQ = 60°.
22
Using π =
, calculate
7
(a) the perimeter, in cm, of the whole diagram,
Diagram 3
It is given that ∠AOB = 63° and ∠COD = 45°.
22
Using π =
, calculate
7
(a) the perimeter, in cm, of the sector OAB,
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EXAM TIPS
2 Soalan akan ditanya disini.iaitu perimeter dan area.
Adalah dicadangkan anda membentu kjawapan anda
mengikut cara yang ditunjukkan Dibawah.contohnya
Soalan (a) -perimeter
Buat perkara ini > OE + arc EF + arc FG + GH +HO
(Sontoh sahaja)
Tujuannya untuk mengurangkan kesilapan dan
mengelakkan ada maklumat tertinggal.
Soalan (b) -Area
Sebaiknya untuk soalan area anda melukis gambara rajah
dahulu.
FORMULA PENTING
Soalan ini agak sukar bagi
sesetengah pelajar.Soalan ini adalah
soalan yang wajib akan keluar
didalam peperiksaan.
Jika anda tak tahu langsung MC
cadangkan anda lakukan perkara
dibawah
Jawapan (a)
Formula lengkung (arc curve)


 2  r
360
Jawapan (b)
+
-
= shade region
Luas (Area) = 

360
  r2
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EXAM TIPS: QUESTION 10– GRADIENT & AREA UNDER GRAPHS
SOALAN 10
GRADIENT AND AREA UNDER A GRAPHS
Speed Vs Time
Distance Vs Time
 Kedah’12, MRSM’12, N.Sembilan’12, Penang’12, SBP’12, SPM2011
RAMALAN 2012
Perak’12, Terengganu’12, Melaka’12
Distance Vs Time
1
Diagram 1 shows the distance-time graph of Shukor's
journey from a recreation park to his house by bicycle.
The graph PQRS represents the journey of the bus from
town W to town X. The graph AB represents the journey
of the taxi from town X to town W. The bus leaves town
W and the taxi leaves town X at the same time and they
travel along the same road.
(a) State the length of time, in minutes, during which
the bus is stationery.
(b) (i) If the journey starts at 12.00 p.m., at what time
do the vehicles meet?
(ii) Find the distance, in km, from town X when
the vehicles meet.
(c) Calculate the average speed, in km h−1, of the bus
for the whole journey.
Diagram 1
4
(a) State the time interval, in minutes, when Shukor
was at rest.
(b) Calculate the average speed, in m/minute, of
Shukor's journey in the first 50 minutes.
(c) Given that the average speed for the whole journey
is 60 m/minute, find the value of y.
2
Diagram 4 shows the distance-time graph of the
journey of a car and a bus.
Diagram 2 shows the distance-time graph of Jason's
journey from a recreation park to his house by bicycle.
Diagram 4
The graph ABCD represents the journey of the car from
town P to town Q. The graph WX represents the journey
of the bus from town Q to town P. The car leaves town
P and the bus leaves town Q at the same time and they
travel along the same road.
(a) State the length of time, in minutes, during which
the car is stationery.
(b) (i) If the journey starts at 8.00 a.m., at what time
do the vehicles meet?
(ii) Find the distance, in km, from town Q when
the vehicles meet.
(c) Calculate the average speed, in km h−1, of the car
for the whole journey.
Diagram 2
(a) State the time interval, in minutes, when Jason was
at rest.
(b) Calculate the average speed, in m/minute, of
Jason's journey in the first 40 minutes.
(c) Given that the average speed for the whole journey
is 45 m/minute, find the value of y.
3
Diagram 3 shows the distance-time graph of the
journey of a bus and a taxi.
Diagram 3
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Speed Vs Time
5
Diagram 5 shows a speed-time graph for the movement
of a particle for a period of 36 seconds.
EXAM TIPS
Tajuk ini Sangat penting.ianya
terbahagi kepada dua iaitu:
 Graphs Speed Vs. Time
 Graphs Distance vs. Time
Kebiasanyaa soalan Speed vs time akan
disoal.2012 besar kemungkinanan
graphs Distance vs. Timepula akan
ditanya
Diagram 5
(a) State the uniform speed, in m s−1, of the particle.
(b) The distance travelled by the particle with uniform
speed is 1 152 m. Calculate
(i) the value of t,
(ii) the average speed, in m s−1, of the particle for
the period of 36 seconds.
Jika anda tak tahu langsung MC
cadangkan lihat dibawah
Anda WAJIB TAHU!!
Jawapan (a)
Katkunci
-Uniform speed
-rest
-stationery
Berlaku ketika graph nya mendatar.
Jawapan (b)
Katakunci:
-Rate Decrease of speed
-Rate Increase of Speed
-Rate of change
Salah satu dari 3 katakunci ini
(kadar perubahan halaju,pecutan)ianya ada dua shj MENAIK atau
MENURUN
rate.of .change 
Final .velocity  Initial .velocity
time
Jawapan (c)
Distance=Area under a graphs
*Untuk mencari area kebiasanyaa akan
menggunakan formula area triangle
dan juga area trapezium.*
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EXAM TIPS: QUESTION 11 – PROBABILITY II
PROBABILITY II
Diagram/graphs
Sentences
1
 Kedah’12, MRSM’12, Perak’12, Terengganu’12, Melaka’12, SBP’12, SPM2011
Negeri Sembilan’12, Penang’12
Diagram 1 shows eight labelled cards in two boxes, M
and N.
RAMALAN 2012
letter Z are picked,
(b) a card with a number which is a multiple of 10 or a
card labelled with letter D are picked.
4
Diagram 4 shows seven letter cards.
Diagram 4
All these cards are put into a box. Two cards are picked
at random, one after another, without replacement.
Calculate the probability that
(a) the first card drawn is not a letter card H,
(b) the first card drawn is a letter card I and the second
card is a letter card L,
(c) the two cards drawn are of the same letter.
Diagram 1
A card is picked at random from each of the boxes. By
listing the outcomes, find the probability that
(a) both cards are labelled with a letter,
(b) one card is labelled with a letter and the other card
is labelled with a number.
5
2
Diagram 2 shows six labelled cards in two boxes, S and
T.
Diagram 5 shows eight labelled cards in two boxes, P
and Q.
Diagram 2
Diagram 5
(a) both cards are labelled with a number,
3
(a) both cards are labelled with a number,
Diagram 3 shows two cards labelled with numbers in
box M and three cards labelled with letters in box N.
Diagram 3
Yahya picks a card at random from box M and then
picks a card at random from box N. By listing the
sample of all the possible outcomes of the event, find
the probability that
(a) a card with an odd number and a card labelled with
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SECTION B
EXAM TIPS: QUESTION 12– GRAPHS OF FUNCTION II
SPM’07
Solve
g  2h  1
4 g  3h  18
SPM’08
Solve
3
y  3
2
4 x  y  16
x
SPM’09
Solve
2
x  4 x  9  2( x  3)
SPM’10
Solve
5m  6n  13
m  2n  1
SPM’11
Solve
m  3n  12
2
mn2
3
Graphs Quadratic x2
Graphs cubix x 3
Graphs reciprocal
1
1
x
 Kedah’12, MRSM’12, Terengganu’12
 SPM2011
 Negeri Sembilan’12, Perak’12, Melak’12, Penang’12, SBP’12
(a) Complete Table 1 in the answer space for the
18
equation y = − .
x
2
(b) For this part of the question, use the graph paper
provided. You may use a flexible curve rule.
RAMALAN 2012
11
equation y = − .
x
By using a scale of 2 cm to 1 unit of the x-axis and
2 cm to 2 units on the y-axis, draw the graph of y =
18
− bagi 2 ≤ x ≤ 7.
x
11
− bagi 1 ≤ x ≤ 6.
x
(c) From your graph, find
(i) the value of y when x = 4,
(ii) the value of x when y = −8.2,
(i) the value of y when x = 3.8,
(ii) the value of x when y = −4,
(d) Draw a suitable straight line on your graph to find
all the values of x which satisfy the equation 2x2 −
13x = −18 for 2 ≤ x ≤ 7. State the values of x.
10x + 11 = 0 for 1 ≤ x ≤ 6. State the values of x.
(a)
x
y
2
(a)
2.5 3 4 4.5 5 6 6.5 7
−7.2
−4.5 −4 −3.6 −3 −2.8 −2.6
Table 1
x 1 1.5 2 2.5 3 4 5 5.5 6
y −11 −7.3 −5.5 −4.4
−2.8 −2.2
−1.8
Table 2
(b)
(b)
(c)
(c)
(d)
(d)
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3
4
19
equation y = − .
x
17
equation y = − .
x
19
− bagi −4 ≤ x ≤ 4.
x
17
− bagi 1 ≤ x ≤ 6.
x
all the values of x which satisfy the equation −3x2
− 3x + 19 = 0 for −4 ≤ x ≤ 4. State the values of x.
17x = −17 for 1 ≤ x ≤ 6. State the values of x.
(a)
(a)
x
y
−4 −3 −2 −1 1 1.5 2 3 4
4.8 6.3
19
−12.7 −9.5 −6.3 −4.8
Table 3
x
y
(b)
(b)
(c)
(c)
(d)
(d)
1
1.5 2 3 4 4.5 5 5.5 6
−11.3 −8.5 −5.7 −4.3 −3.8 −3.4
−2.8
Table 4
Note
**Pelajar digalak ulangkaji sendiri tajuk cubic dan quadratic**
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EXAM TIPS: QUESTION 13– TRANSFORMATION III
SOALAN 13
TRANSFORMATION III
Translation + Reflection
Translation + Rotation
Reflection + Rotation
1
(a)
 Kedah’12, Negeri Sembilan’12, Melaka’12, Terengganu’12, SPM2011
MRSM’12, Perak’12, SBP’12
RAMALAN 2012
Penang’12
Transformation R is a translation
(−1−2).
Transformation S is an anticlockwise rotation of 90°
about the centre (0, 0).
State the coordinates of the image of point (−1, 1)
under each of the following transformations:
(i) S
(ii) R
(iii) R2
(b) Diagram 1 shows three quadrilaterals EFGH, JKMN
and PQRS on a Cartesian plane.
Diagram 2
(i) PQRS is the image of EFGH under the combined
transformation BA. Describe in full the
transformations:
(a) A
(b) B
(ii) Given that quadrilateral EFGH represents a
region of area 18.7 cm2, calculate the area, in
cm2, of the region represented by the shaded
region.
Diagram 1
3
transformations:
(a) A
(b) B
region.
2
(a)
Transformation A is a translation
(a)
Transformation P is a translation
(35).
Transformation Q is an anticlockwise rotation of 90°
about the centre (−2, −2).
State the coordinates of the image of point (−5, −5)
(i) Q
(ii) P
(iii) P2
(−1−1).
Transformation B is an anticlockwise rotation of 90°
State the coordinates of the image of point (3, 2)
(i) B
(ii) A
(iii) A2
Diagram 3
transformations:
(a) A
(b) B
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region.
4
(a)
(b) Diagram 5 shows three quadrilaterals ABCD, EFGH
and JKMN on a Cartesian plane.
Transformation M is a translation
(−31).
Transformation N is an clockwise rotation of 90°
State the coordinates of the image of point (4, 2)
(i) N
(ii) M
(iii) M2
(b) Diagram 4 shows three quadrilaterals ABCD, EFGH
Diagram 5
(i) JKMN is the image of ABCD under the combined
transformation QP. Describe in full the
transformations:
(a) P
(b) Q
(ii) Given that quadrilateral ABCD represents a
region.
Diagram 4
(i) PQRS is the image of ABCD under the combined
transformation KJ. Describe in full the
transformations:
(a) J
(b) K
(ii) Given that quadrilateral ABCD represents a
region.
5
(a)
Transformation R is a translation
(−31).
Transformation S is an anticlockwise rotation of 90°
about the centre (−1, 1).
State the coordinates of the image of point (1, −1)
(i) S
(ii) R
(iii) R2
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EXAM TIPS: QUESTION 14– STATISTICS III
SOALAN 14
STATISTICS III
 Kedah’12
Negeri Sembilan’12, Terengganu’12, SPM2011
 MRSM’12, Perak’12, Melaka’12, Penang’12, SBP’12
Graphs Ogive
Graphs Histogram
Graphs Frequency polygon
1
The data in Diagram 1 shows the amount of money, in
RM, spent by 40 students in a month.
41 82 73 92 55 58 77 56
81 62 94 73 61 56 74 35
93 52 75 77 81 93 61 63
97 51 75 62 54 76 87 41
96 54 76 88 56 61 64 88
Diagram 1
2
(a) Based on the data in Diagram 1 and by using a
class interval of 9, complete Table 1 in the answer
space.
(b) Based on Table 2, calculate the estimated mean of
money spent by a student.
(c) By using the scales of 2 cm to RM9 on the
horizontal axis and 2 cm to 1 student on the
vertical axis, draw a frequency polygon for the
data.
(c) By using the scales of 2 cm to RM8 on the
horizontal axis and 2 cm to 1 student on the
vertical axis, draw a frequency polygon for the
data.
(d) Based on the frequency polygon in (c), state one
piece of information about the expenditure.
Frequency
Midpoint
3
39
The data in Diagram 2 shows the amount of money, in
RM, spent by 36 students in a month.
57 46 37 67 52 80 50 67
66 39 64 85 50 59 55 71
66 47 48 65 46 57 69 78
31 56 62 74 46 70 32 44
49 78 77 84
Diagram 2
(a) Based on the data in Diagram 2 and by using a
class interval of 8, complete Table 2 in the answer
space.
(b) Based on Table 1, calculate the estimated mean of
money spent by a student.
Class
interval
35 − 43
44 − 52
RAMALAN 2012
(d) Based on the frequency polygon in (c), state one
piece of information about the expenditure.
Class
interval
30 − 37
38 − 45
Frequency
Midpoint
3
33.5
Table 1
Table 2
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EXAM TIPS: QUESTION 14– PLAN AND ELAVATION
1
You are not allowed to use graph paper to answer this
question.
(a) Diagram 1.1 shows a solid right prism with
rectangular base EFGH on a horizontal plane. The
surface FGPNKJ is the uniform cross-section o the
prism. JF, NK and PG are vertical edges. Rectangle
IJKL is a horizontal plane and rectangle MNPQ is an
inclined plane.
Diagram 2.1
Draw to full scale, the elevation of the solid on a
vertical plane parallel to CD as viewed from X.
(b) A solid cuboid is cut and removed from the solid in
Diagram 2.1. The remaining solid is shown in
Diagram 2.2. Rectangle LMNP is a horizontal plane.
MC = 3 cm, QL = 3 cm and LM = 6 cm.
Diagram 1.1
Draw to full scale, the plan of the solid.
(b) A half-cylinder solid of diameter 4 cm is joined to the
prism in Diagram 1.1 at the plane IRSL. The length of
IR is 2 cm. The combined solid is shown in Diagram
1.2.
Diagram 2.2
Draw to full scale,
(i) the plan of the remaining solid.
(ii) the elevation of the remaining solid on a vertical
plane parallel to BC as viewed from Y.
Diagram 1.2
Draw to full scale,
(i) the elevation of the combined solid on a vertical
plane parallel to EF as viewed from X.
(ii) the elevation of the combined solid on a vertical
plane parallel to FG as viewed from Y.
2
EXAM TIPS
Untuk pengetahuan anda inilah tajuk paling mudah sekali dan
tiada sebarang formula matematik yang perlu anda gunakan.apa
yang perlu anda lakukan hanyalah sekadar melukis apa yang
anda Nampak mengikut ARAH PANDANGAN (view) yang
diberikan.
You are not allowed to use graph paper to answer this
question.
(a) Diagram 2.1 shows a solid right prism with
rectangular base BCDE on a horizontal plane. The
surface CDJHG is the uniform cross-section of the
prism. CG and DJ are vertical eeges. Rectangle HJKI
is an inclined plane and rectangle FGHI is a horizontal
plane.
BAGAIMANA jika anda tiada idea?
MC cadangkan anda pusing-pusing soalan tersebut dan cuba
fikirkan arah pandangan yang diminta.tiada tips istimewa yang
dapat diberikan melain yang tadi sahaja.
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JAWAPAN
SOALAN 1
INEQUALITIES
RAMALAN 2012
SOALAN 2
QUADRATIC EQUATION
1
RAMALAN 2012
FRACTION QUESTION
1
3  9x
x2=
4x
(x  2)4x = 3  9x
4x2  8x = 3  9x
4x2 + x  3 = 0
(4x  3)(x + 1) = 0
(4x  3) = 0 or (x + 1) = 0
3
x = or x = 1
4
12  9x
2x
(x  2)2x = 12  9x
2x2  4x = 12  9x
2x2 + 5x  12 = 0
(2x  3)(x + 4) = 0
(2x  3) = 0 or (x + 4) = 0
3
x = or x = 4
2
2
x2=
3
x1=
4
4(4x2  1)
= 7x
9
4(4x2  1) = 63x
16x2  4 = 63x
16x2  63x  4 = 0
(x  4)(16x + 1) = 0
(x  4) = 0 or (16x + 1) = 0
1
x = 4 or x = 
16
5
4x2  4x
=5
7x  7
4x2  4x = 35x  35
4x2  39x + 35 = 0
(4x  35)(x  1) = 0
(4x  35) = 0 or (x  1) = 0
35
x=
or x = 1
4
2
3
20  7x
3x
(x  1)3x = 20  7x
3x2  3x = 20  7x
3x2 + 4x  20 = 0
(x  2)(3x + 10) = 0
(x  2) = 0 or (3x + 10) = 0
10
x = 2 or x = 
3
SIMPLE QUESTION
4x2 = 7(10x + 9) + 11
1
4x2 = 70x + 63 + 11
4x2  70x  74 = 0
2x2  35x  37 = 0
(2x  37)(x + 1) = 0
(2x  37) = 0 or (x + 1) = 0
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x=
2
3
37
or x = 1
2
x = 6 + (2)
x=4
∴ x = 4, y = 2
2x2 = 3(7x + 2)  16
2x2 = 21x + 6  16
2x2  21x + 10 = 0
(x  10)(2x  1) = 0
(x  10) = 0 or (2x  1) = 0
1
x = 10 or x =
2
5x2 = 6(2x + 7)  10
5x2 = 12x + 42  10
5x2  12x  32 = 0
(x  4)(5x + 8) = 0
(x  4) = 0 or (5x + 8) = 0
8
x = 4 or x = 
5
4
3x2 + 16 = 26x
3x2 + 26x + 16 = 0
(3x + 2)(x + 8) = 0
(3x + 2) = 0 or (x + 8) = 0
2
x =  or x = 8
3
5
x(x + 1) = 6 + 2x
x2 + x = 6 + 2x
x2  x  6 = 0
(x  3)(x + 2) = 0
(x  3) = 0 or (x + 2) = 0
x = 3 or x = 2
SOALAN 3
LINEAR EQUATION
3
st=7
s = 7 + t -------------- (1)
3s + 4t = 7 ------------- (2)
3(7 + t) + 4t = 7
21 + 3t + 4t = 7
7t = 14
t = 2
s = 7 + (2)
s=5
∴ s = 5, t = 2
5
mn=2
m = 2 + n -------------- (1)
5m + 4n = 1 ------------- (2)
5(2 + n) + 4n = 1
10 + 5n + 4n = 1
9n = 9
n = 1
m = 2 + (1)
m=1
∴ m = 1, n = 1
FRACTION QUESTION
1
m  4n = 4 --------------- (1)
5
m − 3n = 1 --------------- (2)
4
From (1): m = 4 + 4n --------------- (3)
Substitute (3) into (2):
5
(4 + 4n) − 3n = 1
4
5 + 5n  3n = 1
2n = 4
n=2
RAMALAN 2012
SIMPLE QUESTION
1
st=4
s = 4 + t -------------- (1)
7s + 2t = 10 ------------- (2)
7(4 + t) + 2t = 10
28 + 7t + 2t = 10
9t = 18
t = 2
s = 4 + (2)
s=2
∴ s = 2, t = 2
2
4
m = 4 + 4(2)
m = 4 + 8
m=4
∴ n = 2, m = 4
2
mn=3
m = 3 + n -------------- (1)
2m + n = 3 ------------- (2)
2(3 + n) + n = 3
6 + 2n + n = 3
3n = 3
n = 1
m = 3 + (1)
m=2
∴ m = 2, n = 1
x + 2y = 4 --------------- (1)
3
x − 2y = 1 --------------- (2)
2
From (1): x = 4  2y --------------- (3)
3
(4  2y) − 2y = 1
2
6  3y  2y = 1
5y = 5
y = −1
x = 4 − 2(−1)
x = 4 − (−2)
x = −2
xy=6
x = 6 + y -------------- (1)
x + y = 2 ------------- (2)
1(6 + y) + y = 2
6+y+y=2
2y = 4
y = 2
∴ y = −1, x = −2
3
m  4n = 4 --------------- (1)
3
m − n = 1 --------------- (2)
4
From (1): m = 4 + 4n --------------- (3)
26
SULIT
MATHS Catch
USAHA +DOA+TAWAKAL
FOKUS A+
SPM
2012
= 0.509
∠SQW = 26° 59'
The angle between the line QW and the base PQRS
is 26° 59'.
3
(4 + 4n) − n = 1
4
3 + 3n  n = 1
2n = 2
n=1
2
(a) Orthogonal projection of the line QW on the plane
QRVU is VW.
(b) The angle between the plane PQW and the plane
PQUT is ∠TPW.
(c) The angle between the line QW and the base PQRS
is ∠SQW.
QS2 = PQ2 + QR2
= 81 + 16
= 97
QS = 9.85 cm
SW
tan ∠SQW =
QS
6
=
9.85
= 0.609
∠SQW = 31° 20'
The angle between the line QW and the base PQRS
is 31° 20'.
3
(a) Orthogonal projection of the line BH on the plane
BCGF is GH.
(b) The angle between the plane ABH and the plane
ABFE is ∠EAH.
(c) The angle between the line BH and the base ABCD
is ∠DBH.
BD2 = AB2 + BC2
= 9 + 169
= 178
BD = 13.34 cm
DH
tan ∠DBH =
BD
12
=
13.34
= 0.9
∠DBH = 41° 59'
The angle between the line BH and the base ABCD
is 41° 59'.
4
(a) PV2 = PQ2 + QV2
= 64 + 36
= 100
PV = 10 cm
The length of PV is 10 cm.
(b) The angle between the line TV and the plane PQRS
is ∠PVT.
PT
tan ∠PVT =
PV
11
=
10
= 1.1
∠PVT = 47° 44'
The angle between the line TV and the plane PQRS
is 47° 44'.
(c) Let W = midpoint of PS. The angle between the
plane PVT and the plane PSUT is ∠VPW.
WV
tan ∠VPW =
PW
8
=
6
= 1.333
∠VPW = 53° 7'
The angle between the plane PVT and the plane
m = 4 + 4(1)
m = 4 + 4
m=0
∴ n = 1, m = 0
4
s + 2t = 2 --------------- (1)
5
s + 3t = 1 --------------- (2)
2
From (1): s = 2  2t --------------- (3)
5
(2  2t) + 3t = 1
2
5  5t + 3t = 1
2t = 4
t = −2
s = 2 − 2(−2)
s = 2 − (−4)
s=2
∴ t = −2, s = 2
5
p + 3q = 3 --------------- (1)
5
p + q = 1 --------------- (2)
3
From (1): p = 3  3q --------------- (3)
5
(3  3q) + q = 1
3
5  5q + q = 1
4q = 4
q=1
p = 3 − 3(1)
p=3−3
p=0
∴ q = 1, p = 0
SOALAN 4
1
RAMALAN 2012
(a) Orthogonal projection of the line QW on the plane
QRVU is VW.
(b) The angle between the plane PQW and the plane
PQUT is ∠TPW.
(c) The angle between the line QW and the base PQRS
is ∠SQW.
QS2 = PQ2 + QR2
= 169 + 144
= 313
QS = 17.69 cm
SW
tan ∠SQW =
QS
9
=
17.69
27
SULIT
MATHS Catch
USAHA +DOA+TAWAKAL
FOKUS A+
SPM
2012
SOALAN 6
SOLID GEOMETRY
PSUT is 53° 7'.
5
(a) SX2 = XR2 + RS2
= 36 + 64
= 100
SX = 10 cm
The length of SX is 10 cm.
(b) The angle between the line WX and the base PQRS
is ∠SXW.
SW
tan ∠SXW =
SX
10
=
10
=1
∠SXW = 45°
The angle between the line WX and the base PQRS
is 45°.
(c) The angle between the plane QXWT and the plane
PSWT is ∠PTQ.
PQ
tan ∠PTQ =
PT
8
=
10
= 0.8
∠PTQ = 38° 40'
The angle between the plane QXWT and the plane
PSWT is 38° 40'.
Volume of the solid
22 7
1
= [ × ( )2 × 10] + [ × (18 + 22) × 14 × 15
7
2
2
= 385 + 4200
= 4585 cm3
2
(a)
(a) (i) False
(ii) True
(b) Premise 2:
9 is greater than zero.
(c) Implication 1: If 2n > 18, then n > 9
Implication 2: If n > 9, then 2n > 18
3
(a) (i) False
(ii) True
(b) Premise 2:
(c) Implication 1: If 5n > 25, then n > 5
Implication 2: If n > 5, then 5n > 25
4
(a) (i) False
(ii) True
(b) Premise 2:
(c) Implication 1: If 9p > 63, then p > 7
Implication 2: If p > 7, then 9p > 63
5
(a) (i) False
(ii) True
(b) Premise 2:
(c) Implication 1: If 9q > 54, then q > 6
Implication 2: If q > 6, then 9q > 54
1
× 22 × 6 × 9
3
SOALAN 7
THE STRAIGHT LINE
1
(a) (i) False
(ii) True
(b) Premise 2:
(c) Implication 1: If 4x > 24, then x > 6
Implication 2: If x > 6, then 4x > 24
2
Volume of pyramid =
= 396 cm3
(b) Volume of pyramid
= 1380 cm3
1
× (22 + 19) × QR × 6 + 396 = 1380
2
123 × QR = 984
QR = 8 cm
RAMALAN 2012
SOALAN 5
1
RAMALAN 2012
1
RAMALAN 2012
(a) −14x − 6y = −8
−6y = 14x − 8
7
4
y=− x+
3
3
7
m=−
3
7
c = y − (− )(x)
3
7
= −5 − (− )(−2)
3
2
= −9
3
7
The equation of the straight line WX is y = − x −
3
2
9
3
7
2
(b)
y=− x−9
3
3
7
2
0=− x−9
3
3
7
2
− x = −(−9 )
3
3
29
3
=
×−
3
7
1
= −4
7
1
The x-intercept of the straight line WX is −4
7
2
(a) −11x − 8y = −69
−8y = 11x − 69
11
69
y=− x+
8
8
11
m=−
8
11
c = y − (− )(x)
8
11
= −3 − (− )(2)
8
1
=−
4
The equation of the straight line WX is y = −
11
x−
8
28
SULIT
MATHS Catch
USAHA +DOA+TAWAKAL
FOKUS A+
SPM
2012
1
4
(b)
11
1
x−
8
4
11
1
0=− x−
8
4
11
1
− x = −(− )
8
4
1
8
= ×−
4
11
2
=−
11
y=−
The x-intercept of the straight line WX is −
3
7
12
=− ×−
12
13
7
=
13
The x-intercept of the straight line WX is
5
2
11
(a) 11x + 5y = −95
5y = −11x − 95
11
y = − x − 19
5
11
m=−
5
11
c = y − (− )(x)
5
11
= 7 − (− )(−7)
5
2
= −8
5
The equation of the straight line AB is y = −
8
(b)
(a) 15x + 10y = 85
10y = −15x + 85
3
17
y=− x+
2
2
3
m=−
2
3
c = y − (− )(x)
2
3
= 9 − (− )(−7)
2
1
= −1
2
3
The equation of the straight line WX is y = − x −
2
1
1
2
3
1
(b)
y=− x−1
2
2
3
1
0=− x−1
2
2
3
1
− x = −(−1 )
2
2
3
2
= ×−
2
3
= −1
The x-intercept of the straight line WX is −1
11
x−
5
2
5
11
2
x−8
5
5
11
2
0=− x−8
5
5
11
2
− x = −(−8 )
5
5
42
5
=
×−
5
11
9
= −3
11
y=−
SOALAN 8
MATRICES
1
(a) S
1
7 3
3(7) − (3)(6) 6 3
1 7 3
=
3 6 3
T = S−1
7 y
1 7 3
x
=
6 3
3 6 3
1
∴x= ,y=3
3
( ) ( )
(b)
13
x+
12
7
12
(b)
( )
( )
(a) 13x + 12y = −93
12y = −13x − 93
13
31
y=− x−
12
4
13
m=−
12
13
c = y − (− )(x)
12
13
= 6 − (− )(−5)
12
7
=
12
The equation of the straight line WX is y = −
RAMALAN 2012
−1
=
9
The x-intercept of the straight line AB is −3
11
4
7
13
(63 37)(xy) = (107 )
(xy) = 31 (76 33)(107 )
+ 3(10)
(xy) = 31 (7(7)
6(7) + 3(10))
(xy) = 31 (7972)
x
y
()
13
7
y=− x+
12
12
13
7
0=− x+
12
12
13
7
− x = −( )
12
12
(xy)
∴x =
13 (79)

=
1 (72)
3 
79 
=3 
 24 
79
, y = 24
3
29
SULIT
MATHS Catch
SPM
2012
USAHA +DOA+TAWAKAL
FOKUS A+
2
(a) S
Q = P−1
5 t
1
s
=
4
7 9
1
∴s= ,t=7
4
−1
(
(
(b)
)
)
( ) (
(b)
)
(27 38)(xy) = (88)
(xy) = 51 (78 23 )(88)
8(8) + 3(8)
(xy) = 51 (7(8)
+ (2)(8))
(xy) = 51 (40
40 )
()
(xy)
()
5
7 2
1
1(7) − (2)(6) 6 1
1 7 2
=
5 6 1
N = M−1
t 2
1 7 2
s
=
5 6 1
6 1
1
∴s= ,t=7
5
(
5 7
1
5(5) − (7)(4) 4 5
1 5 7
=
3 4 5
Y = X−1
n 7
1 5 7
m
=
3 4 5
4 5
1
∴m= ,n=5
3
(
(b)
(
)
(b)
(54 75)(xy) = (81)
(xy) = 31 (45 57 )(18)
5(8) + 7(1)
(xy) = 31 (4(8)
+ (5)(1))
(xy) = 31 (47
37 )
x
y
()
(xy)
)
) (
)
)
(16 27)(xy) = (84 )
(xy) = 51 (67 12 )(84 )
7(4) + 2(8)
(xy) = 51 (6(4)
+ (1)(8))
12
(xy) = 51 (16
)
x
y
()
13 (47)

=
 1 (37) 
3

47
− 3 

=
 37 
3 
∴x = −
4
)
)
) (
(
=
(a) X
(
73
99
,y=
4
4
(a) M−1
−1
(
14 (73)

=
 1 (99) 
4

73
− 4 

=
 99 
4 
∴x = −
() ( )
=
(97 75)(xy) = (94)
(xy) = 41 (75 97 )(49)
5(9) + 7(4)
(xy) = 41 (7(9)
+ (9)(4))
(xy) = 41 (73
99 )
x
y
15 (40)
x

=
y
 1 (40) 
5

x
−8
=
y
8
∴x = −8, y = 8
3
) (75 97 )
(
8 3
1
=
2(8) − (3)(7) 7 2
1 8 3
=
5 7 2
T = S−1
8 3
1 8 3
x
=
y 2
5 7 2
1
∴ x = , y = 7
5
(xy)
∴x =
 15 (12) 

=
1 (16)
5

12
5 

=
−16 
 5
12
16
,y=−
5
5
47
37
,y=
3
3
(a) P−1
5 7
1
9(5) − (7)(7) 7 9
1 5 7
=
4 7 9
(
=
(
)
)
30
SULIT
MATHS Catch
SPM
2012
USAHA +DOA+TAWAKAL
FOKUS A+
SOALAN 9
CIRCLES
1
2
3
63° 22
× × 212
360° 7
4851 2
=
cm
20
Area of sector OCD
45° 22
=
× × 72
360° 7
77 2
=
cm
4
Area of the shaded region
4851 77
=
−
20
4
3
= 223 cm2
10
=
RAMALAN 2012
(a) Length of arc PQ
60°
22
=
×2×
× 28
360°
7
88
=
cm
3
Perimeter
88
=
+ 28 × 2
3
1
= 85 cm
3
(b) Area of sector OPQ
60° 22
=
× × 282
360° 7
1232 2
=
cm
3
Area of sector ORS
45° 22
=
× × 142
360° 7
= 77 cm2
1232
=
− 77
3
2
= 333 cm2
3
(a) Length of arc AB
63°
22
=
×2×
× 28
360°
7
154
=
cm
5
Perimeter
154
=
+ 28 × 2
5
4
= 86 cm
5
(b) Area of sector OAB
63° 22
=
× × 282
360° 7
2156 2
=
cm
5
Area of sector OCD
42° 22
=
× × 212
360° 7
1617 2
=
cm
10
2156 1617
=
−
5
10
1 2
= 269 cm
2
(a) Length of arc AB
63°
22
=
×2×
× 21
360°
7
231
=
cm
10
Perimeter
231
=
+ 21 × 2
10
1
= 65 cm
10
(b) Area of sector OAB
4
(a) Area of sector OPQ
48° 22
=
× × 352
360° 7
= 513.33 cm2
Area of semicircle RSTU
1 22
= × × 72
2 7
= 77 cm2
= 513.33 − 77
= 436.33 cm2
(b) Length of arc PQ
48°
22
=
×2×
× 35
360°
7
= 29.33 cm
Length of arc 29.33
1
22
= ×2×
× 72
2
7
= 22 cm
Perimeter
= 29.33 + 22 + 35 + (35 − (7 × 2))
= 107.33 cm
5
(a) Length of arc PQ
60°
22
=
×2×
× 21
360°
7
= 22 cm
Length of arc RST
120°
22
=
×2×
× 42
360°
7
= 88 cm
Perimeter
= 22 + 88 + 21 + 42
= 173 cm
(b) Area of sector POQ
60° 22
=
× × 212
360° 7
= 231 cm2
Area of sector ORST
120° 22
=
× × 422
360° 7
= 1 848 cm2
Area of semicircle OUT
1 22
= × × 212
2 7
= 693 cm2
= 231 + 1 848 − 693
= 1 386 cm2
31
SULIT
MATHS Catch
SPM
2012
USAHA +DOA+TAWAKAL
FOKUS A+
SOALAN 11
PROBABILITY II
SOALAN 10
GRADIENT AND AREA UNDER A
RAMALAN 2012
GRAPHS
1
(a) 100 − 50 = 50 minutes
4750
(b)
Laju =
50
= 95 m/minute
y
(c)
60 =
100
y = 60 × 100
= 9000 m
2
(a) 90 − 40 = 50 minutes
2400
(b)
Laju =
40
= 60 m/minute
y
(c)
45 =
90
y = 45 × 90
= 6525 m
3
(a) 70 − 30 = 40 minutes
(b) (i) 12.00 p.m. + 55 minutes = 12.55 p.m.
(ii) 91 − 19 = 72 km
(c) Average Speed
91
=
130
7
=
km minute−1
10
7
=
× 60
10
= 42 km h-1
4
5
1
RAMALAN 2012
Sample space, S
= {(E, J), (E, V), (E, Z), (E, 4), (1, J), (1, V), (1, Z), (1,
4), (2, J), (2, V), (2, Z), (2, 4), (3, J), (3, V), (3, Z), (3,
4)}
(a) A = Event of both cards are labelled with a letter
= {(E, J), (E, V), (E, Z)}
P(A)
n(A)
=
n(S)
3
=
16
(b) B = Event of one card is labelled with a letter and
the other card is labelled with a number
= {(E, 4), (1, J), (1, V), (1, Z), (2, J), (2, V), (2, Z),
(3, J), (3, V), (3, Z)}
P(B)
n(B)
=
n(S)
10
=
16
5
=
8
2
Sample space, S
= {(C, N), (C, 2), (C, 3), (F, N), (F, 2), (F, 3), (1, N), (1,
2), (1, 3)}
(a) A = Event of both cards are labelled with a number
= {(1, 2), (1, 3)}
P(A)
n(A)
=
n(S)
2
=
9
= {(C, 2), (C, 3), (F, 2), (F, 3), (1, N)}
P(B)
n(B)
=
n(S)
5
=
9
(a) 90 − 40 = 50 minutes
(b) (i) 8.00 a.m. + 61 minutes = 9.01 a.m.
(ii) 70 − 22 = 48 km
(c) Average Speed
70
=
150
7
=
km minute−1
15
7
=
× 60
15
= 28 km h-1
3
(a) Uniform speed
= 72 m s−1
(b) (i) 1 152 = 72 × (t − 12)
1 152
t=
+ 12
72
= 28
(ii) Distance
1
1
= × (132 +72) × 12 + 72 × 16 + × 72 × 8
2
2
= 2664 m
Average speed
2664
=
36
= 74 m s−1
Sample space, S
= {(59, D), (59, J), (59, Z), (60, D), (60, J), (60, Z)}
(a) A = Event of picking a card with an odd number
and a card labelled with letter Z
= {(59, Z)}
P(A)
n(A)
=
n(S)
1
=
6
(b) B = Event of picking a card with a number which
is a multiple of 10 or a card labelled with letter D
= {(59, D), (60, D), (60, J), (60, Z)}
P(B)
n(B)
=
n(S)
4
=
6
32
SULIT
MATHS Catch
USAHA +DOA+TAWAKAL
FOKUS A+
SPM
2012
=
4
2
3
6
12
1
=
2
=
Sample space, S
= {(H, Q), (H, L), (H, Y), (H, H), (H, R), (H, I), (Q, H),
(Q, L), (Q, Y), (Q, H), (Q, R), (Q, I), (L, H), (L, Q), (L,
Y), (L, H), (L, R), (L, I), (Y, H), (Y, Q), (Y, L), (Y, H),
(Y, R), (Y, I), (H, H), (H, Q), (H, L), (H, Y), (H, R), (H,
I), (R, H), (R, Q), (R, L), (R, Y), (R, H), (R, I), (I, H),
(I, Q), (I, L), (I, Y), (I, H), (I, R)}
SOALAN 12
1
(a) A = Event that the first card drawn is not a letter
card H.
= {(Q, H), (Q, L), (Q, Y), (Q, H), (Q, R), (Q, I),
(L, H), (L, Q), (L, Y), (L, H), (L, R), (L, I), (Y, H),
(Y, Q), (Y, L), (Y, H), (Y, R), (Y, I), (R, H), (R,
Q), (R, L), (R, Y), (R, H), (R, I), (I, H), (I, Q), (I,
L), (I, Y), (I, H), (I, R)}
n(A) = 30
∴ P(A)
n(A)
=
n(S)
30
=
42
5
=
7
(b) B = Event that the first card drawn is a letter card I
and the second card is a letter card L.
= {(I, L)}
n(B) = 1
∴ P(B)
n(B)
=
n(S)
1
=
42
(c) C = Event that the two cards drawn are of the same
letter.
= {(H, H), (H, H)}
n(C) = 2
∴ P(C)
n(C)
=
n(S)
2
=
42
1
=
21
5
(a)
x
y
RAMALAN 2012
2 2.5 3 4 4.5 5 6 6.5 7
−9 −7.2 −6 −4.5 −4 −3.6 −3 −2.8 −2.6
(b)
(c)(i) y = −4.4
(ii) x = 2.2
(d) 2x2 − 13x = −18
18
2x − 13 = −
x
y = 2x − 13
x = 2, 4.5
2
Sample space, S
= {(B, T), (B, X), (B, Y), (B, 2), (B, 3), (B, 4), (1, T),
(1, X), (1, Y), (1, 2), (1, 3), (1, 4)}
(a)
x 1 1.5 2 2.5 3 4 5 5.5 6
y −11 −7.3 −5.5 −4.4 −3.7 −2.8 −2.2 −2 −1.8
(b)
(a) A = Event of both cards are labelled with a number
= {(1, 2), (1, 3), (1, 4)}
P(A)
n(A)
=
n(S)
3
=
12
1
=
4
= {(B, 2), (B, 3), (B, 4), (1, T), (1, X), (1, Y)}
P(B)
n(B)
=
n(S)
33
SULIT
MATHS Catch
USAHA +DOA+TAWAKAL
FOKUS A+
SPM
2012
x = −3.1, 2.1
4
(a)
x 1 1.5 2
3 4 4.5 5 5.5 6
y −17 −11.3 −8.5 −5.7 −4.3 −3.8 −3.4 −3.1 −2.8
(b)
(c)(i) y = −2.5
(ii) x = 2.7
(d) 2x2 − 10x = −11
11
2x − 10 = −
x
y = 2x − 10
x = 1.6, 3.4
3
(a)
x −4 −3 −2 −1 1 1.5 2
3 4
y 4.8 6.3 9.5 19 −19 −12.7 −9.5 −6.3 −4.8
(c)(i) y = −4
(ii) x = 1.1
(d) 3x2 − 17x = −17
17
3x − 17 = −
x
y = 3x − 17
x = 1.3, 4.4
(b)
SOALAN 13
TRANSFORMATION III
(c)(i) y = −4.5
(ii) x = 2.9
(d) −3x2 − 3x = −19
19
−3x − 3 = −
x
y = −3x − 3
RAMALAN 2012
1
(a) (i) (−1, −1)
(ii) (−2, −1)
(iii) (−3, −3)
(b) (i) (a) A : Reflection in the line x = 11
(b) B : Enlargement with scale factor of 3
about centre (7, 1)
(ii) Area of image
= k2 × Area of object
= 32 × 27.3
= 245.7 cm2
= 245.7 − 27.3
= 218.4 cm2
2
(a) (i) (1, 4)
(ii) (2, 1)
(iii) (1, 0)
(b) (i) (a) A : Reflection in the line x = −8
about centre (−3, 5)
(ii) Area of image
= 32 × 18.7
34
SULIT
MATHS Catch
USAHA +DOA+TAWAKAL
FOKUS A+
SPM
2012
= 168.3 cm2
= 168.3 − 18.7
= 149.6 cm2
3
4
5
Total frequency
= 3 + 2 + 10 + 4 + 9 + 6 + 6
= 40
(a) (i) (1, −5)
(ii) (−2, 0)
(iii) (1, 5)
(b) (i) (a) A : Reflection in the line x = −5
(ii) Area of image
= 32 × 28.2
= 253.8 cm2
= 253.8 − 28.2
= 225.6 cm2
Estimated mean of money
2 784
=
40
= RM69.60
(c)
(a) (i) (1, −1)
(ii) (1, 3)
(iii) (−2, 4)
(b) (i) (a) J : Reflection in the line x = 3
(b) K : Enlargement with scale factor of 2
about centre (1, 2)
(ii) Area of image
= 22 × 14.3
= 57.2 cm2
= 57.2 − 14.3
= 42.9 cm2
(d) Modal class of the expenditure is 53 − 61.
(a) (i) (1, 3)
(ii) (−2, 0)
(iii) (−5, 1)
(b) (i) (a) P : Reflection in the line x = 1
(b) Q : Enlargement with scale factor of 2
(ii) Area of image
= 22 × 14.5
= 58 cm2
= 58 − 14.5
= 43.5 cm2
SOALAN 14
STATISTICS III
1
= 2 784
2
(a)
Class
Frequency
Midpoint
interval
30 − 37
3
33.5
38 − 45
2
41.5
46 − 53
9
49.5
54 − 61
5
57.5
62 − 69
8
65.5
70 − 77
4
73.5
78 − 85
5
81.5
(b) Total (midpoint of class × frequency)
= 33.5(3) + 41.5(2) + 49.5(9) + 57.5(5) + 65.5(8) +
73.5(4) + 81.5(5)
= 2 142
Total frequency
=3+2+9+5+8+4+5
= 36
RAMALAN 2012
Estimated mean of money
2 142
=
36
= RM59.50
(a)
Class
Frequency
Midpoint
interval
35 − 43
3
39
44 − 52
2
48
53 − 61
10
57
62 − 70
4
66
71 − 79
9
75
80 − 88
6
84
89 − 97
6
93
(b) Total (midpoint of class × frequency)
= 39(3) + 48(2) + 57(10) + 66(4) + 75(9) + 84(6) +
93(6)
(c)
35
SULIT
MATHS Catch
SPM
2012
USAHA +DOA+TAWAKAL
FOKUS A+
(ii
)
2 (a)
(d) Modal class of the expenditure is 46 − 53.
SOALAN 15
PLAN & ELEVATION
RAMALAN 2012
1 (a)
(b (i)
)
(b (i)
)
(ii
)
Prepared By:
CIKGU MOHD RAJAEI BIN MOHAMAD ALI
©Hak Cipta Terpelihara
Reference: The analysis is base on last 6 year National SPM exam paper 2005-2011 and State trial Exam 2011
Disclaimer/Penafian:
The exam tips provided are base on pure forecast and assumptions. Maths Catch Network and
www.maths-catch.com will not be liable for any inaccuracy of the information. Students are not encouraged to rely
100% on the tips to score in SPM exams. Students are advised to study hard for their exam. Students can use the
tips as a guide. All the materials have not gone for been proof reading or editing process.
36
SULIT

MATHS Catch

Transcription

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