seminar math tamb
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Powerpoint SlideTRANSCRIPT
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SEMINAR KECEMERLANGAN SPM
2014
Matematik Tambahan
Mohd Ridwan b. Anang @ Talib
SMK Ulu Sapi, Telupid.
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Penyelesaian Masalah
FahamiMasalah
Rancang Strategi
Semak Jawapan
Pelaksanaan Strategi
•Pilih strategi yang sesuai•Pilih formula yang betul
•Topik/Subtopik ? •Maklumat yang diberi•Apa yang perlu dicari
•Buat pengiraan•Melakar graf
•Membina jadual
•Adakah jawapan boleh diterima akal?
• Adakah jalan/strategi lain yang boleh diguna?
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FORMAT KERTAS 1
Ujian Subjektif: Jumlah soalan: Jumlah Markah : Masa : Tahap Kesukaran : Peralatan Tamb. :
Soalan pendek 25 soalan 80 2 jam R (15) , S(7-8), T(2-3) Kalkulator saintifik, Set
geometri, Jadual taburan kebarangkalian.
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FORMAT KERTAS 2
Soalan Subjektif Bil. Soalan : A (6), B (4/5), C (2/4) Jumlah Markah : 100 Masa : 2 jam 30 minit Tahap Kesukaran : R (6) , S(4-5), T(4-5) Peralatan Tambahan : Kalkulator saintifik, Set geometri, Jadual taburan kebarangkalian
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Kunci Mencapai Kecemerlangan
Baca soalan dengan teliti Ikut arahan Mula dengan soalan yang mudah/pilihan Tunjukkan semua jalan kerja dengan jelas Pilih formula yang betul+(Gunakannya
dengan betul !!!) Berikan jawapan dalam bentuk termudah Jawapan akhir betul kepada 4 angka bererti. (atau ikut arahan dalam soalan)
3.142
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Kunci Mencapai Kecemerlangan
Tulis simbol matematik dengan betul Semak Jawapan! Peruntukkan masa yang sesuai untuk setiap soalan
Kertas 1 : 3 - 7 minit untuk setiap soalanKertas 2 : Bhg. A : 8 - 10 minit untuk setiap soalan Bhg. B : 15 minit untuk setiap soalan Bhg. C : 15 minit untuk setiap soalan
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Kesilapan Biasa Calon…
cxxdxx 4346.3 2
4. sin x = 300 ,
1500
2. y = 3x2 + 4x
y = 6x + 4
1. The Quadratic equation 3x2 - 4x
+ 5= 0
dxdy
x = 300 , 1500
31
PQ
AB5. 1
3
AB
PQ
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Kesilapan Biasa Calon …• f ' (x) diinterpretasi sebagai f – 1(x) atau
sebaliknya
• x2 = 4 x = 2
• x2 = 4 x = ± 2
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Kesilapan Biasa Calon…
PA : PB = 2 : 3
maka 2PA = 3 PB
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Sebenarnya, …
PA : PB = 2 : 3
3 PA = 2 PB
32
PBPA
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Kesilapan calon……
32 PA2 = 22 PB2
9 PA2 = 4 PB2
2222 )()(2)()(3
2222 23
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Kesilapan biasa calon…loga x + loga y = 0, loga xy = 0 maka xy = 0
Sepatutnya… xy = a0 = 1
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Kesilapan biasa…
loga (x – 3) = loga x – loga 3 2x x 2y = 1 x + y = 1
2x x 2y = 20
2x + y = 20
x + y = 0
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Fungsi
F4
f : x x - 3 , g : x 3x , cari gf(1).1. Diberi
f(x) = x – 3, g(x) = 3xgf (1) = g [ f(1) ] = g [-2] = -6
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Jika f -1(x) = y
Maka x = f (y)
x = 3 – 2y
Cara 1
2
31 xxf
23 xy
T4 BAB 1
Fungsi Songsang
2. Diberi f (x) = 3 – 2x, find f -1.
Cara 2
Jika x = 3 – 2f -1(x)
Maka 2f -1(x) = 3 – x
231 xxf
F4
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T4 BAB 1
3. Diberi f:x 2 – x dan gf:x 2x – 2. Cari fg.
Ingat : Cari g dahulu !
f(x) =2 - x , gf(x) = 2x-2Jika f(x) = u
Maka u = 2 – x atau x = 2 - u g(u) = 2(2-u) – 2
= 2-2u g(x) = 2-2x
fg(x) = f(2-2x) = 2 - (2-2x)
= 2x
F4
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Persamaan KuadratikBina persamaan kuadratik yang mempunyai punca-punca – 3 dan ½ .
x = – 3 , x = ½ (x+3) (2x – 1) = 0 2x2 + 5x – 3 = 0
F4
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Persamaan Kuadratik
02
acx
abx
x2 – ( HTP) x + (HDP) = 0
ax2 + bx + c = 0
ac
HTP = HTP = ab
F4
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Persamaan Kuadratik ax2 + bx + c = 0 mempunyai
1. Dua punca berbeza jika
2. Dua punca yang sama jika
3. Tiada punca jika
b2 -
4ac
b2 -
4ac b2
- 4ac
> 0
< 0
= 0
Persamaan Kuadratik: Jenis-jenis Punca
F4
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Fungsi Kuadratik : Ketaksamaan KuadratikCari julat nilai x jika x(x – 4) ≤ 12
x (x – 4) ≤ 12 x2 – 4x – 12 ≤ 0 (x + 2)(x – 6) ≤ 0
– 2 ≤ x ≤ 6
6 x -2
F4
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Selesaikan:
x2 > 4
Back to BASIC
x> ±2???x2 – 4 > 0
R.H.S must be O !
(x + 2)(x – 2) > 0
x < -2 or x > 2
– 2 2
F4
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1. Selesaikan persamaan serentak dan berikan jawapan anda kepada 3 tempat perpuluhan.
x + y =1
x2 + 3y2 = 8a
acbb2
42
Persamaan Serentak
F4
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32(x – 1) . 3 (– 3x) = 1 2x – 2 – 3x = 1
– x = 3 x = – 3
Selesaikan .. 1
271.9 1
xx
Betul ke ???
5. INDEKS & LOGF4
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32(x – 1) . 3 (– 3x) = 1 32x – 2 +(– 3x) = 30
– x – 2 = 0 x = – 2
Selesaikan 1
271.9 1
xx
F4 5. INDEKS & LOG
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atau… 9x-1 = 27x
32(x – 1) = 3 3x 32x – 2 = 33x
2x – 2 = 3x x = – 2
Selesaikan 1
271.9 1
xx
F4 5. INDEKS & LOG
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Selesaikan
2x + 3 = 2x+2
2x + 3 = 2x . 22
x = 0
2x + 3 = 4 (2x )
Dalam bentuku + 3 = 4u
3 = 3(2x )
1 = (2x )
F4 5. INDEKS & LOG
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Selesaikan persamaan , beri jawapan anda betul kepada 2 tempat perpuluhan. [ 4 markah]
F4
9 (3x) = 32 + (3x)
8 (3x) = 32 3x = 4 x = 1.26
lg 4lg 3
x
23 32 3x x 5. INDEKS & LOG
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Selesaikan:22x . 5x = 0.05
4x . 5x = 201
20x = 201
x = – 1
ambm = (ab)m
F4 5. INDEKS & LOG
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pp 43
3
log1)
log3log(2
Diberi bahawa log3 p = m dan log4 p = n. Cari logp 36 dalam sebutan m dan n.
= 2log p 3 + logp 4
logp 36 = logp 9 + logp 4
nm12
Menukar asas logaritma
logaa =1
F4 5. INDEKS & LOG
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Note to candidates:
Solutions to this question by scale drawing will not be accepted.
Coordinate Geometry
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Note to candidates:
A diagram is usually given (starting from SPM 2004). You SHOULD make full use of the given diagram while answering the question.
Coordinate Geometry
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Note to candidates:
Sketch a simple diagram to help you using the required formula correctly.
Coordinate Geometry
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6. Coordinate Geometry6.2.2 Division of a Line Segment
Q divides the line segment PR in the ratio PQ : QR = m : n
nmP(x1, y1) R(x2, y2)Q(x, y)
●
n
m
R(x2, y2)
P(x1, y1)
Q(x, y)
nmmyny
nmmxnx 2121 ,Q(x, y) =
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6. Coordinate Geometry (Ratio Theorem)
The point P divides the line segment joining the point M(3,7) and N(6,2) in the ratio 2 : 1. Find the coordinates of point P.
nmmyny
nmmxnx 2121 ,P(x, y) =
●1
2
N(6, 2)
M(3, 7)
P(x, y)
12)2(2)7(1,
12)6(2)3(1
311,
315
115,3
=
=
P(x, y) =
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6. Coordinate Geometry
m1.m2 = –1 P
Q
R
S
Perpendicular lines :
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6. Coordinate Geometry(SPM 2006, P1, Q12)
Diagram 5 shows the straight line AB which is perpendicular to the straight
line CB at the point B.
The equation of CB is y = 2x – 1 .Find the coordinates of B. [3 marks]
mCB = 2
mAB = – ½
Equation of AB is y = – ½ x + 4
At B, 2x – 1 = – ½ x + 4
x = 2, y = 3
So, B is the point (2, 3).
x
y
O
A(0, 4)
C
Diagram 5B
●
●
●y = 2x – 1
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6. Coordinate Geometry
Given points P(8,0) and Q(0,-6). Find the equation of the perpendicular bisector of PQ.
34
43
)4(34)3( xy
mPQ=
mAB=
Midpoint of PQ =
(4, -3)
The equation :
4x + 3y -7 = 0
K1
K1
N137
34
xyor
P
Q
x
y
O
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TASK : To find the equation of the locus of the moving point P such that its distances from the points A and B are in the ratio m : n
(Note : Sketch a diagram to help you using the distance formula correctly)
6 Coordinate Geometry
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6. Coordinate GeometryFind the equation of the locus of the moving point P such that its distances from the points A(-2,3) and B(4, 8) are in the ratio 1 : 2.(Note : Sketch a diagram to help you using the distance formula correctly)
A(-2,3), B(4,8) and m : n = 1 : 2 Let P = (x, y)
●
2
1
B(4, 8)
A(-2, 3)
P(x, y)2 2
2 2 2 2
12
2
4
4 ( 2) ( 3) ( 4) ( 8)
PAPBPA PB
PA PB
x y x y
3x2 + 3y2 + 24x – 8y – 28 = 0
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6. Coordinate Geometry
Find the equation of the locus of the moving point P such that its distance from the point A(-2,3) is always 5 units. (≈ SPM 2005)
●
5
A(-2, 3)
P(x, y)
● A(-2,3) Let P = (x, y)
is the equation of locus of P.2 2 4 6 12 0x y x y
2 2 2( 2) ( 3) 5x y
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6. Coordinate GeometryFind the equation of the locus of point P which moves such that it is always equidistant from points A(-2, 3) and B(4, 9).
A(-2, 3)●
B(4, 9)●
Locus of P
● P(x, y)
Constraint / Condition :
PA = PB
PA2 = PB2
(x+2)2 + (y – 3)2 = (x – 4)2 + (y – 9)2
x + y – 7 = 0 is the equation of
locus of P.
Note : This locus is actually the perpendicular bisector of AB
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Solutions to this question by scale drawing will not be accepted.(SPM 2006, P2, Q9) Diagram 3 shows the triangle AOB where O is the origin. Point C lies on the straight line AB.
(a) Calculate the area, in units2, of triangle AOB. [2 marks](b) Given that AC : CB = 3 : 2, find the coordinates of C. [2 marks](c) A point P moves such that its distance from point A is always twice its distance from point B.
(i) Find the equation of locus of P, (ii) Hence, determine whether or not this locus intercepts the y-axis.
[6 marks]
x
y
O
A(-3, 4)
Diagram 3C
●
●
●
B(6, -2)
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(SPM 2006, P2, Q9) : ANSWERS
9(a)
= 9
0 6 3 01 1 0 24 0 0 6 00 2 4 02 2
x
y
O
A(-3, 4)
Diagram 3C
●
●
●
B(6, -2)
3
2
9(b) 2( 3) 3(6) 2(4) 3( 2),3 2 3 2
12 2,5 5
K1
N1
nmmyny
nmmxnx 2121 ,
Use formula correctly
N1K1
Use formulaTo find area
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(SPM 2006, P2, Q9) : ANSWERS
√
AP = 2PB
AP2 = 4 PB2
(x+3)2 + (y – 4 )2 = 4 [(x – 6)2 + (y + 2)2
x2 + y2 – 18x + 8y + 45 = 0
N1
K1 Use distance formula
K1Use AP = 2PB
x
y
O
A(-3, 4)
C
●
●
●
B(6, -2)
2
1
P(x, y)●
AP = 2 2[ ( 3)] ( 4)x y
9(c) (i)
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(SPM 2006, P2, Q9) : ANSWERS
9(c) (ii) x = 0, y2 + 8y + 45 = 0
b2 – 4ac = 82 – 4(1)(45) < 0
So, the locus does not intercept the y-axis.
Use b2 – 4ac = 0or AOM
K1
K1 Subst. x = 0 into his locus
N1√ (his locus & b2 – 4ac)
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Given that A(-1,-2) and B(2,1) are fixed points . Point P moves such that the ratio of AP to PB is 1 : 2. Find the equation of locus for P.
2 AP = PB
x2 + y2 + 4x + 6y + 5 = 0
K1
J14[ (x+1)2 + (y+2)2 ] = (x -2 )2 + (y -1)2
6. Coordinate Geometry : the equation of locus
N13x2 + 3y2 + 12x + 18y + 15 = 0
2222 )1()2()2()1(2 yxyx
F4
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Marks f6-10 12
11-15 2016-20 2721-25 1626-30 1331-35 1036-40 2Total 100
From a given set of data,(e.g. The frequency distribution of marks of a group of students)
Students should be able to find ….
• the mean, mode & median• Q1, Q3 and IQR• the variance & S.Deviations
• Construct a CFT and draw an ogive• Use the ogive to solve related problems
Statistics
F4
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To estimate median from Histogram
F5
10
20
30
40
50
60
70
80
0.5 20.5 40.5 60.5 80.5 100.5Modal age = 33.5
Age
Number of people
33.5
Graph For Question 6(b)
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F4 CHAPTER 8
8. CIRCULAR MEASURE
‘Radian’ ‘Degrees’
S = rθ (θ must be in RADIANS) A = ½ r2 θ
Always refer to diagram when answering this question.
θ
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F4
8. CIRCULAR MEASUREDiagram shows a sector of a circle OABC with centre O and radius 4 cm. Given that AOC = 0.8 radians, find the area of the shaded region.
C
A
B O0.8c
Area of sector OABC
= ½ x 42 x 0.8= 6.4 cm 2
= ½ x 42 x sin 0.8= 5.7388 cm2
Area of triangle OAC
Area of shaded region
= 6.4 – 5.7388 = 0.6612 cm2
K1
N1
K1
K1
In radians !!!!
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DIFFERENTIATION :
2
( (3) (3 1)(4)(4 5
5)
4 )dy xxdx x
2)54(11
x
dxdy
5413
xxyGiven that , find
d udx v
F4
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9 Differentiation : The second derivative
Given that f(x) = x3 + x2 – 4x +
5 , find the value of f ” (1)
f’ (x) = 3x2 + 2x – 4 f” (x) = 6x + 2
f” (1) = 8
F4
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9 Differentiation : The second derivative
Given that , find the value of g ” (1) .
g’ (x) = 10x (x2 + 1)4
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52( ) 1g x x
g’’ (x) = 40x (x2 + 1) 3 . 2x
Ya ke ??
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g’ (x) = 10x (x2 + 1)4
F4- 9
Given that , find the value of g ” (-1) .
52( ) 1g x x
d uvdx
g’’ (x) = 10x . 4(x2 + 1) 3.2x +(x2+1)4. 10g’’ (-1) = 10(-1) . 4[(-1)2 + 1] 3 +[(-1)2+1)4. 10
= 800 Mid-year, Paper 2
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Given that y = 2x3 – x2 + 4, find the value of at the point (2, 16). Hence, find the small increment in x which causes y to increase from 16 to 16.05.
dxdy
Differentiation : Small increments
K1
K1
N1
= 6x2 – 2x
= 20 , x = 2
F4
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Progressions : A.P & G.P
A.P. : a, a+d, a+2d, a+3d , …….. Most important is “d”
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G.P. : a, ar, ar2, ar3, ……..
Most important is “ r ” !!
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Progressions : G.P - Recurring Decimals
SPM 2004, P1, Q12Express the recurring decimal 0.969696 … as a fraction in the simplest form.
9699
x = 0. 96 96 96 … (1) 100x = 96. 96 96 ….. (2)(2) – (1) 99x = 96 x = = 32
33
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Usual Answer : S10 – S5 = ……. ???
Correct Answer : S10 – S4
ProgressionsGiven that Sn = 5n – n2 , find the sum from the 5th to the 10th terms of the progression.
Back to basic… …
Ans :-54
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Y
X
1. Table for data X and Y
2. Correct axes and scale used
3. Plot all points correctly4. Line of best fit5. Use of Y-intercept to determine value of constant6. Use of gradient to determine another constant
1
1-2
1
1
2-4
Linear LawF5
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Y
X
Bear in mind that …......
1. Scale must be uniform 2. Scale of both axes may defer : FOLLOW given instructions !
3. Horizontal axis should start from 0 !4. Plot ……… against ……….
Linear Law
Vertical Axis Horizontal Axis
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0 2 4 6 8 10 12 x
0.5
1.0
1.5
2.5
2.5
3.0
3.5
4.5Y
x
x
x
x
x
x
Linear lawF5
Read this value !!!!!
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dxx 4)13(.1
dxx 4)32(.2
dx
x 4)13(1.3
dxx 4)13(
2.4
=
=
=
=
cx
15)32( 5
cx
15
)13( 5
cx
3)13(9
1
cx
3)13(9
2
INTEGRATION F5
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INTEGRATION SPM 2003, P2, Q3(a) 3 marks
Given that = 2x + 2 and y = 6 when x = – 1, find y in terms of x.
dydx
Answer: = 2x + 2
y =
= x2 + 2x + c
x = -1, y = 6: 6 = 1 + 2 + c c = 3
Hence y = x2 + 2x + 3
dydx
(2 2)x dx
F5
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INTEGRATION SPM 2004, K2, S3(a) 3 marks
The gradient function of a curve which passes through
A(1, -12) is 3x2 – 6 . Find the equation of the curve.
Answer: = 3x2 – 6
y = = x3 – 6x + cx = 1, y = – 12 : – 12 = 1 – 6 + c c = – 7
Hence y = x3 – 6 x – 7
dydx
2(3 6)x dx Gradient Function
F5
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22 34
A
BGiven that OA = 2i + j and OB = 6i + 4j, find the unit vector in the direction of AB
AB = OB - OA = ( 6i + 4j ) – ( 2i + j )
= 4i + 3j
l AB l =
= 5
Unit vector in the direction of AB = )34(51 ji
Vectors : Unit Vectors
K1 N1
K1
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Parallel vectors
Given that a and b are parallel vectors, with a = (m-4)i +2 j and b= -2i + mj. Find the the value of m.
a = k b
(m-4) i + 2 j = k (-2i + mj)
m- 4 = -2k
mk = 2
1
2
a = b
m = 2
K1
N1
K1
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Prove that tan2 x – sin2 x = tan2 x sin2 x
xkosx
2
2sin sin 2x
xkosxxkosx
2
222 sinsin
xx 22 sintan
xkosxkosx
2
22 )1(sin
tan2 x – sin2 x = K1
N1
K1
5 TRIGONOMETRIC FUNCTIONSF5
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Solve the equation 2 cos 2x + 3 sin x - 2 = 0
sin x ( -4 sin x + 3 ) = 0
sin x = 0 ,
2( 1 - 2sin2 x) + 3 sin x - 2 = 0
-4 sin2 x + 3 sin x = 0
sin x =43
x = 00, 1800, 3600 x = 48.590, 131.410
K1
N1
K1
N1
5 TRIGONOMETRIC FUNCTIONSF5
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5 TRIGONOMETRIC FUNCTIONS (Graphs)
(Usually Paper 2, Question 4 or 5) - WAJIB !
F5
1. Sketch given graph : (4 marks) (2003) y = 2 cos x , (2004) y = cos 2x for (2005) y = cos 2x , (2006) y = – 2 cos x ,
0 2x
0 00 180x
0 2x 32
0 2x
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Find the number of four digit numbers exceeding 3000 which can be formed from the numbers 2, 3, 6, 8, 9 if each number is allowed to be used once only.
No. of ways = 4 . 4. 3. 2 = 96
3, 6, 8, 9
F5 PERMUTATIONS AND COMBINATIONS
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Vowels : E, A, IConsonants : B, S, T, R Arrangements : C V C V C V C
No. of ways = 4! 3 ! = 144
Find the number of ways the word BESTARI can be arranged so that the vowels and consonants alternate with each other
[ 3 marks ]
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Two unbiased dice are tossed. Find the probability that the sum of the two numbers obtained is more than 4.Dice B, y
4
1
5
6
2
3
Dice A, x2 3 4 51 6
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
n(S) = 6 x 6 = 36
Constraint : x + y > 4
Draw the line x + y = 4
We need : x + y > 4
P( x + y > 4) = 1 – 366
= 65
F5
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The Binomial Distributionrnr
rn qpCrXP )()()(
r = 0, 1, 2, 3, …..n
n = Total number of trials
q = probability of ‘failure’
p = Probability of ‘success’
r = No. of ‘successes’
p + q = 1
F5
PROBABILITY DISTRIBUTIONS
Mean = npVariance = npq
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The NORMAL Distribution
F5
PROBABILITY DISTRIBUTIONS
Candidates must be able to … determine the Z-score
Z =
use the SNDT to find the values (probabilities)
x
z
f(z)
0 0.50
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z
f(z)
0 1.5z
z
f(z)
0-1.5 1
= – 1
f(z)
0 1
–
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Index Numbers
• Index Number =
• Composite Index =
• Problems of index numbers involving two or more basic years.
1000
1 HHI
wwI
I_
F4
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Solution of Triangles• The Sine Rule• The Cosine Rule• Area of Triangles• Problems in 3-Dimensions.• Ambiguity cases (More than
ONE answer)
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Motion in a Straight Line
Initial displacement, velocity, acceleration... Particle returns to starting point O... Particle has maximum / minimum velocity.. Particle achieves maximum displacement... Particle returns to O / changes direction... Particle moves with constant velocity...
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Motion in a Straight Line
Question involving motion of TWO particles. ... When both of them collide / meet ??? … how do we khow both particles are of the same
direction at time t ??? The distance travelled in the nth second. The range of time at which the particle returns …. The range of time when the particle moves with
negative displacement Speed which is increasing Negative velocity Deceleration / retardation
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Linear ProgrammingTo answer this question, CANDIDATES must be able to ..... form inequalities from given mathematical information
draw the related straight lines using suitable scales on both axes
recognise and shade the region representing the inequalities
solve maximising or minimising problems from the objective function (minimum cost, maximum profit ....)
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Linear Programming
y ≤ 2x12. The ratio of the quantity of Q (y) to the quantity of P (x) should not exceed 2 : 1
x ≥ y + 1011. x must exceed y by at least 10
y - 2x >1013. The number of units of model B (y) exceeds twice the number of units of model A (x) by 10 or more.
x + y > 4010. The sum of x and y must exceed 40x + y ≥ 509. The sum of x and y is not less than 50
3x - 2y ≥ 188. The minimum value of 3x – 2y is 18 x + 2y ≤ 607. The maximum value of x+ 2y is 60
y ≥ 356. The minimum value of y is 35 x ≤ 1005. The maximum value of x is 100
y ≥ 2x4. The value of y is at least twice the value of xx ≤ y3. x is not more than y
x ≤ 802. x is not more than 80 x ≥ 101. x is at least 10
KetaksamaanMaklumat
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THANK Q & Selamat maju jaya !