2. How to Solve
a Problem
Understand Plan your Do - Carry out Check your
the Problem Strategy Your Strategy Answers
•Which Topic / •Carry out the •Is the answer
•Subtopic ? •Choose suitable calculations reasonable?
•What info has been strategy •Graph sketching
given? •Choose the correct •Any other
•What is to be formula •Creating tables ... methods ??
found?
3. PAPER 1 FORMAT
Objective Test : •Short Questions
No. of Questions : •25 questions
Total Marks : •80
Duration : •2 hours
L.O.D. : •L (15) , M(7-8), H(2-3)
Additional Materials : Scientific Calculators,
•
Mathematical
Tables, Geometrical sets.
4. PAPER 2 FORMAT
•Subjective Questions
•
No. of Questions : A (6), B (4/5), C (2/4)
•
Total Marks : 100
•
Duration : 2 hours 30 minutes
•
L.O.D : L (6) , M(4-5), H(4-5)
•
Additional Materials : Scientific Calculators,
• Mathematical Tables,
• Geometrical sets.
5. Key towards achieving 1A …
•Read question carefully
•Follow instructions
•Start with your favourite question
•Show your working clearly
•Choose the correct formula to be used
+(Gunakannya dengan betul !!!)
•Final answer must be in the simplest form
•The end answer should be correct to 4 S.F.
• (or follow the instruction given in the question)
•
3.142
6. Kunci Mencapai kecemerlangan
•Proper / Correct ways of writing mathematical
notations
•Check answers!
•Proper allocation of time (for each question)
•
•Paper 1 : 3 - 7 minutes for each question
•Paper 2 :
• Sec. A: 8 - 10 minutes for each question
• Sec. B: 15 minutes for each question
• Sec. C: 15 minutes for each question
7. Common Mistakes…
1. The Quadratic equation 3x2 - 4x = 0
• y = 3x2 + 4x
y = 6x + 4
4. sin x = 300 , x = 300 , 1500
5.
8. Kesilapan Biasa Calon …
• f ' (x) wrongly interpreted as f – 1(x)
and / or conversely
• x2= 4 x = 2
•
• x2> 4 x > ±2
•
12. Common mistakes …
logax + loga y = 0,
then xy = 0
It should be… xy = a0= 1
13. Common mistakes …
loga(x – 3) = loga x – loga 3
2x x 2y = 1 2x x 2y = 20
x + y = 1 2 x + y = 20
x + y = 0
14. Common mistakes …
logax + loga y = 0,
then loga xy = 0
So, xy = 0
It should be… xy = a0= 1
15. Common mistakes …
sin (x + 30 0) = ½,
then sin x + sin 300= ½
…………………gone !
Do NOT use
Sin(A+B) = sin A cos B + cos A
sin B !
16. Correct way… …
sin (x + 300) = ½ ,
then x +300= 300 , 1500
So, x = 00 , 1200
?
If 00 is an answer, then 3600 is also an
answer !
17. sin (x + 300) = ½ ,
then x +300= 300 , 1500 , 3900
So, x = 00 , 1200 , 3600
18. Relationship between Functions and Quadratic
Functions
Domain Codomain
y
X Y
f(x) = x
2
4
1 1
1 2 4
x
Image
O 1 2
Objec
t
(1, 1) , (2, 4). …. form ordered pairs and can be
plotted to obtain a curve.
19. SPM 2003 Paper 1, Question 1
P = { 1, 2, 3}
Q = {2, 4, 6, 8, 10}
• The relationship between P and Q is defined by the set
of ordered pairs { (1, 2), (1, 4), (2, 6), (2, 8)}.
• State
• the image of 1,
• The object of 2. [2 marks]
Answer
(a) 2, 4 1
(a) 1 1
20. SPM 2003 Paper 1, Question 2
Answer
(a) or 0.4 2
B1: or g(x) = 3
25x2 + 2 2
B1: (5x+1)2 – 2(5x+1) + 3
21. SPM 2003 Paper 1, Question 3 (SPM 2005,Q5)
Solve the quadratic equation 2x(x – 4) = (1- x)(x+2).
Write your answer correct to four significant figures.
(3 marks)
Answer
2.591, - 0.2573 (both + 4 s.f.) 3
B2:
B1: 3x2– 7x – 2 = 0
22. SPM 2003 Paper 1, Question 4
The quadratic equation x (x+1) = px – 4 has
two distinct roots. Find the range of values of p.
(3 marks)
Answer
p < -3, p > 5 (kedua-duanya) 3
B2: (p + 3) (p – 5) > 0
B1: (1 – p)2– 4(1)(4) > 0
23. SPM 2003 Paper 1, Question 5
Given that log 2T - log4 V = 3, express T in terms
of V. (4 marks)
Answer
T= 8V½ 4
B1
B2
B3
26. SPM 2003 Paper 1, Question 7
The first three terms of an A.P.are k-3, k+3, 2k+2.
Find (a) the value of k,
(b) the sum of the first 9 terms of the progression.
(3 marks)
Answer
(a) 7 2
(k + 3) – (k – 3) = (2k + 2) – (k + 3) B1
6 = k–1
(b) 252 1
27. SPM 2003 Paper 2, Question 1
Solve the simultaneous equation 4x + y = - 8
and x2+ x – y = 2 (5 marks)
Answer
Make x or y the subject P1
Eliminating x or y
K1
Solving the quadratic equation : K1
x = -2, -3 or y = 0 , 4 N1
y = 0 , 4 or x = -2, -3 N1
28. SPM 2003 Paper 2, Question 2
The function f(x) = x2 - 4kx + 5k2 + 1has a minimum
value of r2 + 2k , with r and k as constants.
• By the method of completing the square, show that
r=k–1 (4 marks)
• Hence, or otherwise, find the value of k and the
value of r if the graph of the function is symmetrical
about the line x = r2 -1. (4 marks)
29. SPM 2003 Paper 2, Question 2 ***
Answer
2(a) Writing f(x) in the form (x – p)2 + q
(x – 2k)2 – 4k2 + 5k2 + 1
K1
Equating q ( q* = r2 + 2k) K1
(k – 1)2 = r2 N1
r= k–1 N1
(b) Equating (his) - (x – p) = 0 K1
Eliminating r or k by
K1 any valid method
k=0,4 N1
r = -1, 3 N1
30. F4
1. Functions
1. f:x x2 - 2 . Find the values of x which map onto
Given itself.
f (x) = x
x2 - 2 = x
x 2– x – 2 = 0
(x+1)(x-2) = 0
x = -1 , x = 2
2. f:x x - 3 , g:x 3x , find gf(1).
Given
f(x) = x – 3, g(x) = 3x
gf (1) = g [ f(1) ]
= g [-2]
= -6
31. T4 BAB 1
F4
Functions : Inverse Functions
4. Given f (x) = 3 – 2x, find f -1.
Method 1 Method 2
Let f (x) = y
Let f -1(x) =y
Then 3 – 2x = y
Then x = f (y)
3 – y = 2x
x = 3 – 2y
32. T4 BAB 1
F4
Functions : Applying the Idea of Inverse functions
5. Given , find the value of a if f -1(a) =
11
Method 1 (Find f-1 )
Method 2 ( No need f-1 )
Let f -1(x) = y
Then x = f(y) Let f -1(a) = 11
x=
Then a = f (11)
y=
= 8
f-1(a) = = 11
a= 8
33. T4 BAB 1
F4
Functions : Given composite function and one function,
find the other function.
6. Given find fg.
Remember : you need to find g first !
f(x) =2 - x , gf(x) = 2x-2
Let f(x) = u
Then u = 2 – x or x=2-u
g(u) = 2(2-u) – 2
= 2-2u
g(x) = 2-2x
fg(x) = f(2-2x)
= 2 - (2-2x)
= 2x
34. T4 BAB 1
F4
**Functions :To skecth the graphs of y = |f(x)|
7. Skecth the graph of y = |3-2x|+1 for domain 0 ≤ x ≤ 4
and state the corresponding range.
Tips : Sketch y = |3-2x| first !!!
y
6
5
4
3
Range : 1≤ y ≤ 6
2
1
x
0 4
35. F4
2. Quadratic equations:
SPM 2004, K1, Q4
Form the quadratic equation which
has the roots – 3 and ½ .
x = – 3 , x = ½
(x+3) (2x – 1) = 0
2x2 + 5x – 3 = 0
37. F4
2. The Quadratic Equation : Types of roots
The quadratic ax2 + bx + c= 0 has
equation
1. Two distinct roots if b2 - >0
4ac
2. Two equal roots if b2 - =0
4ac
3. No real roots if b2 - <0
4ac
**The straight line y = mx -1 is a tangent to the curve
y = x2+ 2 ……. ???
38. F4
3 Quadratic Functions : Quadratic Inequalities
SPM 2004, K1, S5
Find the range of values of x for which
x(x – 4) ≤ 12
x (x – 4) ≤ 12
x2 – 4x – 12 ≤ 0
(x + 2)(x – 6) ≤ 0
x
-2 6
–2≤ x ≤ 6
39. F4
Back to
Solve BASIC
x 2 > 4 x> ±2
x2 – 4 > 0 ???
(x + 2)(x – 2) > 0 R.H.S
must be O !
–2 2
x < -2 or x > 2
40. F4
4. Simultaneous Equations
• Solve the simultaneous equations
x + y =1
x2 + 3y2 = 7
Factorisation
• Solve the simultaneous equations, give your answer
correct to three decimal places.
x +y=1
x2 +3y2 = 8
*** P = Q = R
41. F4
5. INDICES
Back to basic… …
Solve ..
32(x – 1) . 3 (– 3x) = 1
2x – 2 – 3x = 1
– x = 3
x= –3 Betul
ke ???
44. F4
5. INDICES
Solve
2x + 3 = 2x+2
Can U take
2x + 3 = 2x . 22
logon both
2x + 3 = 4 (2x ) sides ???
3 = 3(2x ) WHY?
1 = (2x ) In the form
x = 0 u + 3 = 4u
45. F4
5. INDICES
Solve the equation ,
give your answer correct to 2 decimal places.
[ 4 marks]
9 (3x) = 32 + (3x)
8 (3x) = 32
3x = 4
x = 1.26
(Mid-Yr
46. F4
5. INDICES
Solve
22x . 5x = 0.05 ambm = (ab)m
4x . 5x =
You can
also take
20x = log on both
sides.
x = –1
47. F4 5. INDICES &
LOGARITHMS
(Mid-Yr 07)
Solve the equation
[4
marks]
x–2 = 4 (4 – x)
x = 3.6
48. F4 5. INDICES &
LOGARITHMS
Back to basic… …
Solve the the equation
log3(x – 4) + log3 (x + 4) = 2
log3(x-4)(x+4) = 2
x2 – 16 = 9
x = 5
49. F4
Back to basic… … SPM 2005, P1, Q8
Solve the equation
log34x – log3(2x – 1) = 1
4x = 3(2x – 1)
= 6x – 3
2x = 3
x =
50. F4
5 Indices and Logaritms : Change of base
Given that log3 p = m and log4p = n. Find
logp36 in terms of m and n.
logp 36 = logp 9 + logp K1
4 = 2log p 3 + logp 4 K1
K1
N1
logaa 1
=
52. Coordinate Geometry
Distance between two points
Division of line segments : midpoints
+ the ratio theorem
Areas of polygons
Equation of straight lines
Parallel and perpendicular lines
Loci (involving distance between two
points)
54. Coordinate Geometry
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.
55. Coordinate Geometry
Note to candidates:
Sketch a simple diagram
•
to help you using the
required formula
correctly.
56. 6. Coordinate Geometry
6.2.2 Division of a Line Segment
Q divides the line segment PR in the ratio PQ : QR = m : n
R(x2, y2)
n
m n ●
P(x1, y1) m Q(x, y)
Q(x, y) R(x2, y2)
P(x1, y1)
Q(x, y) =
57. 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.
1 N(6, 2)
●
P(x, y) P(x, y) =
2
M(3, 7)
=
=
P(x, y) =
59. 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 CBis y = 2x – 1 .
Find the coordinates of B. [3 marks]
y mCB= 2
y = 2x – 1 mAB = – ½
A(0, 4)
●
Equation of AB is y= –½x+4
●B Diagram 5
At B, 2x – 1 = – ½ x + 4
O x x = 2, y = 3
●C So, Bis the point (2, 3).
60. 6. Coordinate Geometry
Given points P(8,0) and Q(0,-6). Find the equation of
the perpendicular bisector of PQ.
y
mPQ=
mAB= K1
O P x
Midpoint of PQ (4, -3)
= Q
K1
The equation 4x + 3y -7 =
: or 0 N1
61. 6 Coordinate Geometry
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)
62. 6. Coordinate Geometry
Find 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)
B(4, 8)
A(-2, 3) 2
1
●
P(x, y)
3x2 + 3y2 + 24x – 8y– 28 = 0
63. 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)
A(-2,3)
Let P = (x, y) A(-2, 3)
●
5
●
P(x, y)
is the equation of locus of P.
64. 6. Coordinate Geometry
Find the equation of the locus of point Pwhich moves such that
it is always equidistant from points A(-2, 3) and B(4, 9).
Constraint / Condition : B(4, 9)
●
PA = PB
A(-2, 3)
PA2 = PB2 ●
(x+2)2 + (y – 3)2 = (x – 4)2 + (y – 9)2 ● P(x, y)
x + y – 7 = 0 is the equation of Locus of P
locus of P.
Note : This locus is actually the perpendicular bisector of AB
65. Solutions to this question by scale drawing will not be accepted.
(SPM 2006, P2, Q9)
Diagram 3 shows the triangle AOB where Ois the origin.
Point C lies on the straight line AB.
A(-3, 4) y
●
Diagram 3
C
●
x
O
●
B(6, -2)
(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]
• 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]
66. (SPM 2006, P2, Q9): ANSWERS
A(-3, 4) y
●
3
Diagram 3
C
● 2
x
O
9(a) K1 ●
B(6, -2)
N1 Use formula
= 9
To find area
9(b)
K1
Use formula correctly
N1
67. (SPM 2006, P2, Q9): ANSWERS
A(-3, 4) y
● 2
● P(x, y)
9(c) (i) C
● 1
O x
AP = ●
B(6, -2)
K1
Use distance
formula
AP = 2PB
K1
AP2 = 4 PB2
Use AP = 2PB
(x+3)2 + (y – 4 )2 = 4 [(x – 6)2+ (y + 2)2
x2 + y2– 18x + 8y + 45 = 0 N1
√
68. (SPM 2006, P2, Q9): ANSWERS
9(c) (ii) x = 0, y2+ 8y + 45 = 0 K1 Subst. x = 0 into his locus
b2 – 4ac = 82 – 4(1)(45) < 0 K1
Use b2 – 4ac = 0
or AOM
So, the locus does not interceptthe y-axis. N1
√ (his locus
& b2 – 4ac)
69. F4
6. Coordinate Geometry : the equation of locus
Given that A(-1,-2) and B(2,1) are fixed points . Point P movessuch
that the ratio of AP to PB is 1 : 2. Find the equation of locus for P.
2AP = PB
K1
4[ (x+1)2 + (y+2)2 ] = (x -2 )2 + (y -1)2 J1
3x2 + 3y2 + 12x + 18y + 15 = N1
0
x2 + y2 + 4x + 6y + 5 =
0
70. F4
Statistics
From a given set of data,
Marks f (e.g. The frequency distribution
of marks of a group of students)
6-10 12
11-15 20 Students should be able to find ….
16-20 27 • the mean, mode & median
21-25 16 • Q1, Q3 and IQR
• the variance & S.Deviations
26-30 13
•Construct a CFT and draw an ogive
31-35 10 •Use the ogive to solve related
problems
36-40 2
Total 100
71. F5
To estimate median from Graph For Question 6(b)
Histogram
Number of people
80
70
60
50
40
30
20
10
33.5
0.5 20.5 40.5 60.5 80.5 100.5 Age
Modal age = 33.5
72. F4 CHAPTER 8
8. CIRCULAR MEASURE
‘Radian’ ‘Degrees’
S = rθ (θ must be in
θ
RADIANS)
A = ½ r2 θ
Always refer to diagram when answering this question.
73. F4
8. CIRCULAR MEASURE
Diagram shows a sector of a circle A
OABC with centre O and radius 4
B 0.8c O
cm.Given that AOC = 0.8 radians,
find the area of the shaded region.
C
K1
Area of sector = ½ x 42 x 0.8
OABC = 6.4 cm 2
K1
Area of triangle = ½ x 42 x sin 0.8
OAC In radians !!!!
= 5.7388 cm2
K1
Area of shaded = 6.4 –
region 5.7388 N1
= 0.6612 cm2
75. F4
9 Differentiation : The second derivative
Given that f(x) = x3 + x2 – 4x +
5, find the value of f ” (1)
f’ (x) = 3x2 + 2x –
f” (x) = 6x +
2
f” (1) = 8
76. F4
9 Differentiation : The second derivative
Given that , find the value of g
” (1) .
g’ (x) = 10x (x2 + 1)4
g’’ (x) = 40x (x2 + 1) 3 . 2x
Ya
ke ??
77. F4- 9
Given that , find the value of g
” (-1) .
g’ (x) = 10x (x2 +
1)4
g’’ (x) = 10x . 4(x2 + 1) 3.2x +(x2+1)4.
10
g’’ (-1) = 10(-1) . 4[(-1)2 + 1] 3
+[(-1)2+1)4. 10
= 800 Mid-year, Paper 2
78. F4
Differentiation : Small increments
Given that y = 2x3 – x2 + 4, find the value of at
the point (2, 16). Hence, find the small
increment in xwhich causes y to increase from 16
to 16.05.
= 6x2 – 2x
= 20 , x= K1
2
K1
N1
79. F5
Progressions : A.P &
G.P
A.P. : a, a+d, a+2d, a+3d ,…
…..
Most important is “d”
G.P. : a, ar, ar2, ar3,……..
Most important is “ r
” !!
80. F5
Progressions : G.P - Recurring Decimals
SPM 2004, P1, Q12
Express the recurring decimal
0.969696 … as a fraction in the
simplest form.
x = 0. 96 96 96 … (1)
100x = 96. 96 96 ….. (2)
(2) – (1) 99x = 96
x= =
81. F5 Back to basic… …
Progressions
Given that Sn = 5n – n2 , find the sum from
the 5th to the 10th terms of the progression.
Usual Answer :
S10 – S5 = ……. ???
Correct Answer :
S10 – S4
Ans :-
54
82. F5
Linear Law
1. Table for data X and Y
1-2
Y 2. Correct axes and scale used 1
3. Plot all points correctly 1
4. Line of best fit 1
5. Use of Y-intercept to determine
X value of constant
2-4
6. Use of gradient to determine
another constant
83. F5
Linear Law
Bear in mind that …......
1. Scale must be uniform
Y 2. Scale of both axes may defer :
FOLLOW given instructions !
3. Horizontal axis should start from
0 !
4. Plot ……… against ……….
X
Vertical Axis Horizontal
Axis
84. Linear law
F5
Y
4.5
x
3.5
x
3.0
x
2.5
2.5 x
1.5
x
1.0
x
0.5
Read this value !!!!!
0 2 4 6 8 10 12 x
86. F5
INTEGRATION
SPM 2003, P2, Q3(a) 3 marks
Given that = 2x + 2 and y= 6 when x= – 1, find
y in terms of x.
Answer: = 2x + 2
y =
= x2+ 2x + c
x = -1, y = 6: 6 = 1 +2 + c
c = 3
Hence y = x2 + 2x + 3
87. F5
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 = Gradient
= x3– 6x + c Function
x = 1, y = – 12 : – 12 = 1 – 6 + c
c = –7
Hence y = x3 – 6 x – 7
88. F5
Vectors : Unit Vectors
Given that OA = 2i + j and OB = 6i + 4j, find
B the unit vectorin the direction of AB
AB = OB - OA
= ( 6i + 4j ) – ( 2i + j )
A = 4i + 3j K1
l AB l =
= 5
Unit vector in the direction of AB = K1 N1
89. F5
Parallel vectors
Given that a and bare parallel vectors, with
a = (m-4)i +2 j and b= -2i + mj. Find the the value of
m.
a=kb
a= b
(m-4) i + 2 j = k (-2i + mj) K1
m- 4 = -2k 1
mk = 2 2 K1
m=2 N1
90. F5
5 TRIGONOMETRIC FUNCTIONS
Prove that tan2 x – sin2 x = tan2 x sin2 x
tan2 x – sin2 x = sin 2x K1
K1
N1
91. F5
5 TRIGONOMETRIC FUNCTIONS
Solve the equation 2 cos 2x + 3 sin x - 2
=0
2( 1 - 2sin2 x) + 3 sin x - 2 = 0 K1
-4 sin2 x + 3 sin x = 0
sin x( -4 sin x + 3 ) = 0 K1
sin x= 0 sin x =
,
x = 00, 1800, 3600 N1 x = 48.590, 131.410 N1
92. F5
5 TRIGONOMETRIC FUNCTIONS (Graphs)
(Usually Paper 2, Question 4 or 5) - WAJIB
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 ,
93. F5 PERMUTATIONS AND COMBINATIONS
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
94. F5
Find the number of ways the word BESTARIcan
be arranged so that the vowels and consonants
alternate with each other
[ 3 marks ]
Vowels : E, A, I
Consonants : B, S, T, R
Arrangements : C V C V C V C
No. of ways = 4! 3 !
= 144
95. F5
Two unbiased dice are tossed.
Find the probability that the sum of the
two numbers obtained is more than 4.
Dice B, y
n(S) = 6 x 6 = 36
6 X X X X X X Constraint : x + y > 4
5 X X X X X X
Draw the line x + y = 4
4 X X X X X X
3 X X X X X X We need : x + y > 4
2 X X X X X X P( x + y > 4) = 1 –
1 X X X X X X
Dice A, x =
1 2 3 4 5 6
96. F5
PROBABILITY DISTRIBUTIONS
The Binomial Distribution
r = 0, 1, 2, 3, …..n p+q=1
n = Total number of trials
r = No. of ‘successes’
p = Probability of ‘success’
q = probability of ‘failure’ Mean = np
Variance = npq
97. F5
PROBABILITY DISTRIBUTIONS
The NORMAL Distribution
Candidates must be able to … f(z)
determine the Z-score
Z =
z
00 0.5
use the SNDT to find the values
(probabilities)
98. T5
f(z) f(z) f(z)
= 1 – –
z z z
-1.5 0 1 0 1 0 1.5
99. F4
Index Numbers
•
•
Index Number =
•
•
Composite Index =
•
•
Problems of index numbers involving
two or more basic years.
100. Solution of Triangles
The Sine Rule
•
The Cosine Rule
•
Area of Triangles
•
Problems in 3-Dimensions.
•
Ambiguity cases (More than ONE
•
answer)
101. 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...
102. 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
103. Linear Programming
To 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 ....)
104. Linear Programming
Maklumat Ketaksamaan
1. x is at least 10 x ≥ 10
2. x is not more than 80 x ≤ 80
3. x is not more than y x ≤ y
4. The value of y is at least twice the value of x y ≥ 2x
5. The maximum value of x is 100 x ≤ 100
6. The minimum value of y is 35 y ≥ 35
7. The maximum value of x+ 2y is 60 x + 2y ≤ 60
8. The minimum value of 3x – 2y is 18 3x - 2y ≥ 18
9. The sum of x and y is not less than 50 x + y ≥ 50
10. The sum of x and y must exceed 40 x + y > 40
11. x must exceed y by at least 10 x ≥ y + 10
12. The ratio of the quantity of Q (y) to the quantity of P (x) y ≤ 2x
should not exceed 2 : 1
13. The number of units of model B (y) exceeds twice the y - 2x >10
number of units of model A (x) by 10 or more.
