Additional
Mathematics
How to Solve
a Problem
Understand
Plan your
Do - Carry out
Check your
the Problem
Strategy
Your Strategy
Answers
•Which Topic /
•Subtopic
?
•What info has been
given?
•What is to be
found?
•Choose suitable
strategy
•Choose the correct
formula
•Carry out the
calculations
•Is the answer
reasonable?
•Graph sketching
• Any other
methods ??
•Creating tables ...
PAPER 1 FORMAT






Objective Test :
No. of Questions :
Total Marks :
Duration :
L.O.D.
:
Additional Materials :






Short Questions
25 questions
80
2 hours
L (15) , M(7-8), H(2-3)
Scientific Calculators,
Mathematical Tables,
Geometrical sets.
PAPER 2 FORMAT

Subjective Questions
No. of Questions : A (6), B (4/5), C (2/4)

Total Marks : 100

Duration

L.O.D


: 2 hours 30 minutes
: L (6) , M(4-5), H(4-5)
Additional Materials : Scientific Calculators,
Mathematical Tables,
Geometrical sets.
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

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
Common Mistakes…
1. The Quadratic equation 3x2 - 4x + 5 = 0
• y = 3x2 + 4x
dy
dx
y = 6x + 4
3
.
6
x

4
dx

3
x

4
x

c

2
4. sin x = 300 , 1500 x = 300 , 1500

5.
AB
1


PQ 3

AB

PQ

1
3
Kesilapan Biasa Calon …
• f ' (x) wrongly interpreted as f – 1(x)
and / or conversely
•
x2 = 4

•
x2 > 4
 2x > ± 2
•
x=
x = 2

•
x

2
4
Common errors…
PA : PB
then
=
2 : 3
2PA = 3 PB
Actually, …
PA : PB =
2 : 3
PA
2

PB
3
3 PA
=
2 PB
More mistakes ……
3
()

()
2
()
()
2

2
2

2



3
 
2
2
2
32 PA2
=
9 PA2 =
2
2
22 PB2
4 PB2
Common mistakes …
loga x + loga y = 0,
then
xy =
0
It should be… xy =
0
a
= 1
Common mistakes …
loga (x – 3) = loga x – loga 3
2x x 2y = 1
x + y = 1
2x x 2y = 20
x
+
y
0
2
= 2
x + y = 0
Common mistakes …
loga x + loga y = 0,
then
So,
loga xy =
0
xy = 0
It should be… xy =
0
a
= 1
Common mistakes …
sin (x + 300) = ½ ,
then sin x + sin
0
30
= ½
…………………gone !
Do NOT use
Sin(A+B) = sin A cos B + cos A sin B !
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 !
?
sin (x + 300) = ½ ,
then x
0
+30
=
0
30
,
0
150
So, x = 00 , 1200 , 3600
,
0
390
Relationship between Functions and Quadratic
Functions
Domain
Codomain
X
Y
y
f(x) = x2
4
1
O
1
2
1
1
2
4
x
Image
Objec
t
(1, 1) , (2, 4). …. form ordered pairs and can be
plotted to obtain a curve.
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
2, 4
(a) 1
(a)
1
1
SPM 2003 Paper 1, Question 2
g: xx
5 1
Answer
(a)
2
5
or 0.4
( x  1)
5
B1 :

25x2 + 2
B1 :
2
h
:x
x

2
x
3
2
or g(x) = 3
2
(5x+1)2 – 2(5x+1) + 3
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.)
B2 :
B1 :
2

(

7
)
(

7
)

4
(
3
)(

2
)
2
(
3
)
3x2 – 7x – 2 = 0
3
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)
B2 :
(p + 3) (p – 5) > 0
B1 :
(1 – p)2 – 4(1)(4) > 0
3
SPM 2003 Paper 1, Question 5
Given that log 2 T - log4 V = 3, express T in
terms of V.
(4 marks)
Answer
4
T= 8V½
log
V
2
log
T

3 B1
2
log
24
log
V
2
log
T

3
2
2
1
2
log
T

log
3 B2
2
2V 
log
2
T
1
2
V
3
B3
T
V
1
2
 23
T  8V
1
2
SPM 2003 Paper 1, Question 6
Solve the equation 42x – 1 = 7x
(4 marks)
Answer
4
x = 1.677
(2x – 1) log 4
= x log 7
2x log 4 – log 4
= x log 7
2x log 4 – x log 7 = log 4
B1
B2
x (2 log 4 – log 7 ) = log 4
l
o
g
4
0
.
6
0
2
1
x

o
r
B3
2
l
o
g
4

l
o
g
7
0
.
3
5
9
1
SPM 2007 (???)
Solve the equation 42x – 1 = 8x
(3 marks)
Answer
22(2x – 1) = 23x
2(2x – 1) = 3x
4x – 1 =
x =
3x
1
4x – 2 = 3x
x =
2
No !!!
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
2
(a) 7
(k + 3) – (k – 3) = (2k + 2) – (k + 3)
B1
6 = k–1
(b) 252
1
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

8

y
x

o
ry


8

4
x
4
Eliminating x or y
K1

8

y

8

y




2


yo

2
r
x

x

(

8

4
x
)

2


4
4


2
Solving the quadratic equation :
K1
N1
x = -2, -3 or y = 0 , 4
y = 0 , 4 or x = -2, -3
N1
SPM 2003 Paper 2, Question 2
The function f(x) = x2 - 4kx + 5k2 + 1 has 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)
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
Equating
q
( q* = r2 + 2k)
(k – 1)2 = r2
r= k–1
(b) Equating (his) - (x – p) = 0
K1
N1
N1
K1

8

y
dy
b




0
atau
x


 


dx
2
a
4


2
k=0,4
r = -1, 3
K1
K1
N1
N1
Eliminating r or k by
any valid method
F4
1. Functions
1. Given f : x
x2 - 2 .
Find the values of x which map onto itself.
f (x) = x
x2 - 2
= x
x2 – x – 2 = 0
(x+1)(x-2) = 0
x = -1 , x = 2
2. Given f : x
x - 3 ,
g:x
f(x) = x – 3, g(x) = 3x
gf (1) = g [ f(1) ]
= g [-2]
= -6
3x
, find gf(1).
T4
F4BAB 1
Functions :
Inverse Functions
4. Given f (x) = 3 – 2x, find f -1.
Method 2
Method 1
Let f
-1(x)
=y
Then x = f (y)
x = 3 – 2y
3 x
y
2
3x

1
f x
2
Let
Then
f (x) = y
3 – 2x = y
3 – y = 2x
3 y
x
2
3y
f 1y
2
3x
f 1x
2
T4
F4BAB 1
Functions :
Applying the Idea of Inverse functions
3x1
f :x ,
4
5. Given
Method 1 (Find f-1 )
Let f -1(x) = y
Then x = f(y)
x=
y=
3y 1
4
4x 1
3
f-1(a) =
find the value of a if f -1(a) = 11
Method 2 ( No need f-1 )
Let
Then
f -1(a) = 11
a = f (11)
= 8
4a  1
3
a= 8
= 11
T4
F4BAB 1
Functions : Given composite function and one function,
find the other function.
6. Given
fx
:
2

x
a
n
d
g
fx
:
2
x

2
, 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
T4
F4BAB 1
**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
3
2
4
F4
2. Quadratic equations:
SPM 2004, K1, Q4
Form the quadratic equation which has
the roots – 3 and ½ .
x =
– 3 ,
x =
(x+3) (2x – 1) = 0
2x2 + 5x – 3 = 0
½
F4
2. Quadratic Equations
ax2 + bx + c = 0
 b
 c
x
x 
0

 a
 a
2
x2 – ( S.O.R) x + (P.O.R.) = 0
b
S.O.R =  a
c
P.O.R. = a
F4
2. The Quadratic Equation : Types of roots
ax2 + bx + c = 0
The quadratic equation
1. Two distinct roots
2. Two equal roots
3. No real roots if
if
if
b2 - 4ac
has
>0
b2 - 4ac = 0
b2 - 4ac
<0
**The straight line y = mx -1 is a tangent to the curve
y = x2 + 2 …….
???
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
-2
6
–2≤ x ≤ 6
x
Back to BASIC
F4
Solve
x2 > 4
x2 – 4 > 0
(x + 2)(x – 2) > 0
x> ±2
???
R.H.S
must be O !
–2
2
x < -2 or x > 2
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
b b 4
ac
2
a
2
F4
5. INDICES
Back to basic… …
Solve ..
2(x
–
1)
3
9
x 1
(–
3x)
3
1
.
x
27
 1
.
= 1
2x – 2 – 3x = 1
– x = 3
Betul
ke
x= –3
???
5. INDICES
F4
Solve
9
x 1
1
.
x
27
32(x – 1) . 3 (– 3x) = 1
32x – 2 +(– 3x) = 30
–x–2 =0
x = –2
 1
F4
5. INDICES
Solve
or…
9
x 1
1
.
x
27
9x-1 = 27x
2(x
–
1)
3x
3
= 3
32x – 2 = 33x
2x – 2 = 3x
x = –2
 1
5. INDICES
F4
Solve
2x + 3 = 2x+2
2x + 3 = 2x . 22
2x + 3 = 4 (2x )
3 =
3(2x )
1 = (2x )
x
=
0
Can U take
log on both
sides ???
WHY?
In the form
u + 3 = 4u
5. INDICES
x+2
x
Solve the equation 3
,
= 32 + 3
give your answer correct to 2 decimal places.
[ 4 marks]
F4
9 (3x)
= 32 + (3x)
8 (3x) = 32
3x = 4
=
x
x
lg 4
lg 3
= 1.26
(Mid-Yr 07)
5. INDICES
F4
Solve
22x
4x
.
5x
= 0.05
.
5x
=
1
20
20x
=
1
20
x
=
–1
ambm = (ab)m
You can
also take
log on both
sides.
5. INDICES & LOGARITHMS
F4
(Mid-Yr 07)
Solve the equation
log 2 ( x - 2) = 2 + 2 log 4 (4 - x)
[ 4 marks]
l
o
g
(
4)

x
2
l
o
g
(
x

2
)

2

2
.
2
l
o
g
4
2
l
o
g
(
x

2
) 2l

o
g
(
4)

x
2
2
log 2 ( x - 2) = log 2 4(4 - x)
x – 2 = 4 (4 – x)
x
= 3.6
5. INDICES & LOGARITHMS
F4
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
F4
Back to basic… …
Solve the equation
log3 4x – log3 (2x – 1) = 1
x
4
lo
g
1
3

x
1
2

4x
3
2x1
4x = 3(2x – 1)
= 6x – 3
2x = 3
3
x =
2
SPM 2005, P1, Q8
F4
5 Indices and Logaritms : Change of base
Given that log3 p = m and log4 p = n. Find logp 36 in
terms of m and n.
logp 36 = logp 9 + logp 4
= 2log p 3 + logp 4
log
3
1
3

2
(
)
log
3p log
4p
logaa = 1
2
1


m
n
K1
K1
K1
N1
Coordinate Geometry
Some extra vitamins 4u …
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)
Coordinate Geometry
Note to candidates:
Solutions
to this
question by scale
drawing will not be
accepted.
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.
Coordinate Geometry
Note to candidates:
 Sketch
a simple diagram to
help you using the required
formula correctly.
6. Coordinate Geometry
6.2.2 Division of a Line Segment
PQ : QR = m : n
Q divides the line segment PR in the ratio
n
m
P(x1, y1)
n
Q(x, y)
●
R(x2, y2)
m
Q(x, y)
P(x1, y1)
 nx1  mx 2 ny1  my 2 
,
Q(x, y) = 

mn 
 mn
R(x2, y2)
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
2
●
P(x, y)
N(6, 2)
1
(
3
)

2
(
6
)1
(
7
)

2
(
2
)


P(x, y) = 
,


1
2

1
 2
=
15 11
 , 
 3 3
=
 11 
 5,

3


M(3, 7)
P(x, y) =
 nx1  mx 2 ny1  my 2 
,


m

n
m

n


6. Coordinate Geometry
Perpendicular lines :
R
m1.m2 = –1
P
Q
S
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]
y
mCB = 2
A(0, 4)
●
●B
mAB = – ½
y = 2x – 1
Equation of AB is
Diagram 5
x
O
●C
y= –½x+4
At B, 2x – 1 = – ½ x + 4
x = 2, y = 3
So, B is the point (2, 3).
6. Coordinate Geometry
Given points P(8,0) and Q(0,-6). Find the equation of the
perpendicular bisector of PQ.
mPQ= 3
4
4
mAB= 
y
K1
O
3
Midpoint of PQ = (4, -3)
4
(y

3
)
(
x

4
)
3
The equation :
or
Q
K1
4x + 3y -7 = 0
4
7
y  x 
3
3
P
N1
x
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)
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)
Let
A(-2,3), B(4,8) and
P = (x, y)
m:n=1:2
B(4, 8)
PA 1
=
PB 2
2 PA = PB
4 PA = PB
2
A(-2, 3)
2
1
2
4 éë( x + 2) 2 + ( y - 3) 2 ùû = ( x - 4) 2 + ( y - 8) 2
3x2 + 3y2 + 24x – 8y – 28 = 0
●
P(x, y)
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
2
2
2
(
x

2
)

(y

3
)

5
●
P(x, y)
2
2
x

y

4
x

6
y

1
20 is the equation of locus of P.
6. Coordinate Geometry
Find 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).
Constraint / Condition :
B(4, 9)
●
PA = PB
PA2
=
PB2
(x+2)2 + (y – 3)2 = (x – 4)2 + (y – 9)2
x + y – 7 = 0 is the equation of
A(-2, 3)
●
● P(x,
y)
Locus of P
locus of P.
Note : This locus is actually the perpendicular bisector of AB
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(-3, 4)
●
y
Diagram 3
C
●
O
x
●
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 yaxis.
[6
marks]
(SPM 2006, P2, Q9) : ANSWERS
y
A(-3, 4)
●
3
C
●
9(a)
0
6

3
01
1

0

2
4

0

0

6

0K1
0

2
4
02
2
= 9
9(b)
N1
(
3
)3
(6
) 2
(4
)3
(
2
)
2
,


32 
 32
2 2
1
 , 
N1
5 5
Diagram 3
2
x
O
●
Use formula
To find area
K1
B(6, -2)
Use formula correctly
 nx1  mx 2 ny1  my 2 
,


m

n
m

n


(SPM 2006, P2, Q9) : ANSWERS
A(-3, 4)
●
y
●P(x, y)
C
9(c) (i)
AP =
2
●
K1
x
O
[ x - (-3)]2 + ( y - 4) 2
1
●
B(6, -2)
Use distance formula
AP = 2PB
K1
AP2 = 4 PB2
(x+3)2
+ (y – 4
)2
= 4 [(x –
6)2
x2 + y2 – 18x + 8y + 45 = 0
+ (y +
2)2
Use AP = 2PB
N1
√
(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 intercept the y-axis.
N1
√
(his locus
& b2 – 4ac)
F4
6. Coordinate Geometry : the equation of locus
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
2
2
2
2
2
(
x

1
)

(
y

2
)

(
x

2
)

(
y

1
)
K1
4[ (x+1)2 + (y+2)2 ] = (x -2 )2 + (y -1)2
J1
3x2 + 3y2 + 12x + 18y + 15 = 0
x2 + y2 + 4x + 6y + 5 = 0
N1
F4
Statistics
Marks
6-10
11-15
16-20
21-25
26-30
31-35
36-40
Total
f
12
20
27
16
13
10
2
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
To estimate median from
Histogram
F5
Graph For Question 6(b)
Number of people
80
70
60
50
40
30
20
10
33.5
0.5
40.5
20.5
Modal age = 33.5
60.5
80.5
100.5
Age
F4 CHAPTER 8
8. CIRCULAR MEASURE
 ‘Radian’
 S =
 ‘Degrees’
rθ
θ
 A =
(θ must be in
RADIANS)
½ r2 θ
Always refer to diagram when answering this question.
F4
8. CIRCULAR MEASURE
Diagram 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.
Area of sector OABC
A
B
= ½ x 42 x 0.8
= 6.4 cm 2
Area of triangle OAC
= ½ x 42 x sin 0.8
= 5.7388
Area of shaded region
cm2
= 6.4 – 5.7388
= 0.6612 cm2
0.8c
O
C
K1
K1
In radians !!!!
K1
N1
F4
DIFFERENTIATION :
Given that
3x1
y
4x5
, find
dy
dx
d
dx
d
y(
4
x

5
)
(
3
)(

31
x

)
(
4
)

2
d
x
(
45
x

)
11

2
(4x5)
æuö
ç ÷
èvø
F4
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
9 Differentiation : The second derivative
Given that
(
)
g ( x) = x + 1
2
5
, find the value of g ” (1) .
g’ (x) = 10x (x2 + 1)4
g’’ (x) = 40x (x2 + 1) 3 . 2x
Ya ke
??
F4- 9
Given that
(
)
g ( x) = x + 1
g’ (x) = 10x
2
5
(x2 +
, find the value of g ” (-1) .
1)4
d
 uv 
dx
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
F4
Differentiation
:
Small increments
dy
Given that y = 2x3 – x2 + 4, find the value of dx at the
point (2, 16). Hence, find the small increment in x
which causes y to increase from 16 to 16.05.
dy
dx
= 6x2 – 2x
= 20 , x = 2
K1
y dy

x dx
0.05
 20
x
x  0.0025
K1
N1
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 ” !!
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 …
100x = 96. 96 96 …..
(2) – (1)
99x = 96
x=
96
=
99
32
33
(1)
(2)
Back to basic… …
F5
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
F5
Linear Law
1. Table for data X and Y
Y
1-2
2. Correct axes and scale used
1
3. Plot all points correctly
1
4. Line of best fit
X
5. Use of Y-intercept to determine value
of constant
6. Use of gradient to determine
another constant
1
2-4
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 !
X
4. Plot ……… against ……….
Vertical Axis
Horizontal
Axis
Linear law
F5
Y
4.5
x
3.5
x
3.0
x
2.5
x
2.5
1.5
x
1.0
x
0.5
Read this value !!!!!
0
2
4
6
8
10
12
x
F5
INTEGRATION
1
.
(3x1)
4
dx =
2
. (2
3
x) dx
4
1
3
.
dx
4
(3
x
1
)
2
4
.
dx
4
(3
x
1
)
=
=
=
(3x1)5
c
15
(23x)5
c
15
1
c
3
9
(3
x
1
)
2
c
3
9(3x1)
F5
INTEGRATION
SPM 2003, P2, Q3(a) 3 marks
dy
Given that dx
= 2x + 2 and y = 6 when x = – 1,
find y in terms of x.
Answer:
dy
= 2x + 2
dx
y
x = -1, y = 6:
Hence
=
(2x2)dx
=
x2 + 2x + c
6
c
= 1 +2 + c
=
3
y = x2 + 2x + 3
F5
INTEGRATION
SPM 2004, K2, S3(a) 3 marks
The gradient function of a curve which passes through
A(1, -12) is 3x2 – 6 .
Answer:
dy
dx
Find the equation of the curve.
= 3x2 – 6
2
(3
x
y =  6)dx
x = 1, y = – 12 :
= x3 – 6x + c
– 12 = 1 – 6 + c
c = –7
Hence
y = x3 – 6 x – 7
Gradient
Function
F5
Vectors :
B
Unit Vectors
Given 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
A
42  32
l AB l =
= 5
Unit vector in the direction of AB =
1
(4i 3j)
5
K1
N1
K1
F5
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=kb
a= b
(m-4) i + 2 j = k (-2i + mj)
m- 4 = -2k
1
mk = 2
2
m=2
K1
K1
N1
F5
5
TRIGONOMETRIC FUNCTIONS
Prove that tan2 x – sin2 x = tan2 x sin2 x
tan2
x–
sin2 x
=
sin2 x

2
kos x
sin 2x
K1
2
2
2
sin
x

kos
x
sin
x

2
kos
x
2
2
sin
x
(1

kos
x
)

2
kos
x
K1
2
2
tan
xsin
x
N1
F5
5 TRIGONOMETRIC FUNCTIONS
Solve the equation
2 cos 2x + 3 sin x - 2 = 0
K1
2( 1 - 2sin2 x) + 3 sin x - 2 = 0
-4 sin2 x + 3 sin x = 0
K1
sin x ( -4 sin x + 3 ) = 0
sin x = 0
x = 00, 1800, 3600
,
sin x =
N1
3
4
x = 48.590, 131.410
N1
F5
5 TRIGONOMETRIC FUNCTIONS (Graphs)
(Usually Paper 2, Question 4 or 5) - WAJIB !
1. Sketch given graph :
(4 marks)
0x 2
(2003) y = 2 cos 3
x
,
2
(2004) y = cos 2x for 00 x 1800
(2005) y = cos 2x ,
0x 2
(2006) y = – 2 cos x , 0x 2
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
=
=
96
3, 6, 8, 9
4 . 4. 3. 2
F5
Find the number of ways the word BESTARI can 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
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
5
X
X
X
X
X
X
4
X
X
X
X
X
X
3
X
X
X
X
X
X
2
X
X
X
X
X
X
1
X
X
X
X
X
X
1
2
3
4
5
6
Constraint : x + y > 4
Draw the line x + y = 4
We need : x + y > 4
6
P( x + y > 4) = 1 –
36
Dice A, x
5
=
6
F5
PROBABILITY DISTRIBUTIONS
The Binomial Distribution
n
r
n

r
P
(
X

r
)

C
(
p
)
(
q
)
r
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
Variance
=
np
=
npq
F5
PROBABILITY DISTRIBUTIONS
The NORMAL Distribution
Candidates must be able to …
f(z)
 determine the Z-score
Z =
x

use the SNDT to find the values
(probabilities)
00
0.5
z
T5
f(z)
f(z)
=
-1.5 0
1
z
1
f(z)
–
–
0
1
z
0
1.5
z
F4
•
Index Numbers
Index Number =
H
I  1 100
H0
wI

I
w
_
•
Composite Index =
•
Problems of index numbers involving
two or more basic years.
Solution of Triangles
•
•
•
•
•
The Sine Rule
The Cosine Rule
Area of Triangles
Problems in 3-Dimensions.
Ambiguity cases (More than ONE
answer)
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...
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
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 ....)
Linear Programming
Maklumat
1. x is at least 10
2. x is not more than 80
3. x is not more than y
4. The value of y is at least twice the value of x
5. The maximum value of x is 100
6. The minimum value of y is 35
7. The maximum value of x+ 2y is 60
8. The minimum value of 3x – 2y is 18
9. The sum of x and y is not less than 50
10. The sum of x and y must exceed 40
11. x must exceed y by at least 10
Ketaksamaan
x ≥ 10
x ≤ 80
x ≤ y
y ≥ 2x
x ≤ 100
y ≥ 35
x + 2y ≤ 60
3x - 2y ≥ 18
x + y ≥ 50
x + y > 40
x ≥ y + 10
12. The ratio of the quantity of Q (y) to the quantity of P (x)
should not exceed 2 : 1
y ≤ 2x
13. The number of units of model B (y) exceeds twice the
number of units of model A (x) by 10 or more.
y - 2x >10
Selamat maju jaya !