Comprehensive Notes on Add Maths for SPM students

2. 2
PAPER 1 (3472 / 1)
No Items Comments
1 No. of questions 25 questions (Answer ALL)
2 Total Marks
……80 marks………….
3 Time
…….2 hours
4 Level of difficulty Low : Moderate : High
6 : 3 : 1
PAPER 2 (3472 / 2)
No Items Comments
1 Three sections Section A Section B Section C
2 No of questions 6 5 4
(need to answer 12
questions )
Answer
…ALL…..
Choose
…four….
Choose
…two…..
3 Total marks (100) 40 marks 40 marks 20 marks
4 Time
………2 ½ hours……………..
5 Level of difficulty
Low :
4
Moderate :
3
High
3

3. FORMAT&COMPONENT
3
* Extra Time = ………………………………….
ALGEBRA
1. Functions
2. Quadratic Equations
3. Quadratic Functions
4. Indices & Logarithms
5. Simultaneous Equations
6. Progressions
7. Linear Law
Geometry
1. Coordinate Geometry
2. Vector
Calculus
1. Differentiation
2. Integration
Triginometry
1. Circular Measures
2. Trigonometric Functions
STATISTICS
1. Statistics
2. Probability
3. Permutation & Combination
4. Distribution Probability
Social Science
1. Index Numbers
2. Linear Programming
Science & Technology
1. Solutions to Triangle
2. Motion in a Straight Line.
11 mmaarrkk  mmiinnuutteess1.5
Check answers

4. FORM 4 TOPICS
4
Learn with your heart
and you’ll see the wonders …
P(X)

5. FORM 4 TOPICS
5
a b
X  1
TOPIC 1 : FUNCTIONS
 f(a) = b  object = …………………….. image = ………………..
 Given f (x) and gf(x). Find g(x) .  Thus, g(x) =

1
53
)(



x
x
xf ,
TOPIC 2: QUADRATIC EQUATIONS [ ax2
+ bx + c = 0 ]
Types of roots
Sum and Product of Roots [ ax2
+ bx + c = 0 ]
Sum of roots, ( + ) =
a
b
Product of roots ( ) =
a
c
x2
– S x + P = 0
TOPIC 3: QUADRATIC FUNCTIONS
General form CTS form
f (x) = ax2
+ bx + c = a( x + p)2
+ q
Similarity Same value of a Same value of a
Difference
c = y-intercept. q = max/min value
of f(x)
Specialty
Able to find:
- shape
- y intercept
Able to find
- turning point
( - p , q)
- two distinct real roots
- intersects at two points
- two equal roots
- touches / tangent
- no root
- does not intersect
- f(x) is always positive
1.
gf f -1
b2
– 4ac
> 0
=0
< 0
0 Real roots

6. FORM 4 TOPICS
6
2. Sketch Graphs: y = ax2
+ bx + c
(a) From the graph
3. Inequalities : Solving 2 inequalities  use Graph
(x – a) (x – b) > 0 (x – c) (x – d) < 0
1. Let the right hand side = 0
- factorise
2. Find the roots of the equation
3. Sketch the graph
3. Determine the region 
positive or negative a b
………………………. …………….………
TOPIC 4: SIMULTANEOUS EQUATIONS
 Use ……………………………….. method
 To find the ………………………………… points between a straight line and a curve.
i value of a : Positive
ii value of b2
-4ac: < 0
iii type of roots : No roots
iii y-intercept : C
iv
Equation of axis of
symmetry :
X =
a2
b

c d
x < a , x > b, c < x < d
Substitution
intersection

7. FORM 4 TOPICS
7
TOPIC 5: INDICES & LOGARITHM
Use Index rule : (i) 5x+1
. 125x
=
25
1
……………………….
INDEX
Use log : (ii) 8  3x
= 7 ............................ IND
Use substitution : (iii) 3
n+1
+ 3
n
= 12
or can be factorise 3n
. 3 + 3n
= 12
a(3) + a = 12
4a = 12
a = 3
3n
= 3  n = 1
(iv) 3
2n
+ 5. 3
n
= 6
a2
+ 5a – 6 = 0
(a – 1) (a + 6) = 0
a = 1 , a = –6
3n
= 1 3n
 –6
 n = 0
LOGARITHM : Use Rules of logarithms to simplify or to solve logarithmic equations
log2 (x + 9) = 3 + 2 log2 x
log2
(x + 9) – log2
x 2
= 3
log2
(x + 9) = 3
x 2
x + 9 = 23
x 2
x + 9 = 8x 2
8x2
– x – 9 = 0
(8x -9) (x + 1) = 0
x –1, x = 
Steps of solutions
1. separate the index
2. substitute
Change the base to
the same number
Insert log on both
sides
5x +1
. 5 3x
= 5–2
x + 1 + 3x = – 2
x = –

8. FORM 4 TOPICS
8

TOPIC 6: COORDINATE GEOMETRY
 Distance   Ratio theorem 
 Mid point   Equation of straight line y = mx + c
 Area (positive)
Arrange
anticlockwise
: general form ax + by + c = 0
 Gradient : gradient form y = mx + c
- parallel m1
= m2
: intercept form 1
b
y
a
x

- perpendicular m1
m2
= –1
Equation of locus : …use distance formula………………….
Rhombus : ……its diagonal are perpendicular to each other ……….
Parallelogram, square, rectangle, rhombus. its diagonal share the same mid point
TOPIC 7: STATISTCS
EFFECT ON CHANGES TO DATA
The change in
values when
each data is
Mean Mode Median Range Interquartile
range  Variance
Added with k + k +k +k unchange unchange unchange unchange
Multiplied by m  m  m  m  m  m  m  m 2
TOPIC 8: CIRCULAR MEASURES
 …… radian = ………
 For s = r and A = ½ r2
, the value of  is in ……radian…….

 Shaded angle =  – 
Area of segment = ½ r2
(  - sin )

 rad
180 

9. FORM 4 TOPICS
9
TOPIC 9: DIFFERENTIATION
gradient of normal
gradient of tangent
equation of normal
Tangent
equation of tangent
Rate of change
Applications
Small Changes approximate value
y ORIGINAL + y
minimum
Turning points
maximum
TOPIC 10: SOLUTION OF TRIANGLES
 Sine Rule
- Ambiguous Case  two possible angles  acute and obtuse angle
 Cosine Rule
 Area =  ab sin C
TOPIC 11: INDEX NUMBERS
 I A, B  I B, C = I A, C
 Given that the price index of an item is 120. If the price index increases at the same
rate in the next year, what will be the new price index of the item?
………………………………
100
120
120  ……………………………………………
……………………………………………………………………………………….
mT =
dx
dy
y – y1 = m(x – x1)
mN mT = –1
dy = dy  dx
dt dx dt
 y = dy   x
dx
dx
dy
= 0
02
2

dx
yd
02
2

dx
yd

10. 10
:

11. FORM 5 TOPICS
11
T1 T2 T3 T4 T5 T6 T7
TOPIC 1: PROGRESSIONS
Janjang Aritmetik Janjang Geometri
Examples : 20, 15, 10, …..., …. 4, 3, 2.25, ……., ….
Uniqueness : d = T2
– T1
r =
1
2
T
T
Others :
2
)( lan
Sn

 S =
r
a
1
Given Sn find Tn Example: Given Sn = n( 3 + 2n), find T8.
Thus, T8 = S8
– S7
Find the sum from the
3rd
term until the 7th
term.
S 3 to 7 = S7
– S2
TOPIC 2: LINEAR LAW
Convert to linear form
Y = m X + c
ay = x +
x
b
xy =
a
1
x2
+
a
b
xh
x
p
y  xy = h x + p
T + 1 = a2
+ k

1T
= a  + k
y = ax
b
log y = b log x + log a
y = k p
x
log y = log p x + log k
+ + + + + +

12. 12
TOPIC 3: INTEGRATIONS
 To find THE EQUATION OF A CURVE given dy/dx
dx
dy
= ……gradient function ……………………
Equation of CURVE,  dxfunctiongradienty }{  the integrated function must have c
 AREA under a curve: Show how you would find the area of the shaded region.
x = y2
– y
1. Find the intersection
point.
when x = 2,  y = 4
2. Find the area
Area under curve + area 
=

2
0
2
x dx +  (1)(4)
intersection, x = 1
thus, y = 4(1) – 12
= 3
Shaded Area:
Area under curve – area 
=
 
4
1
2)4( xx dx –  (3)(3)
1. Expand y
y = x3
– 3x2
+ 2x
Shaded Area:
Area above =

1
0
dxy
Area below =

1
2
dxy
Total area
1. Formula

1
0
dyxA
 VOLUME : Show the strategy to find the generated volume.
revolved about x-axis revolved about x-axis revolved about y-axis revolved about y-axis
 dxyV 2
 dxyV 2
 dyxV 2
 dyxV 2
where y2
= (x2
)2
where y2
: (4x – x2
)2
where x2
= y + 1 where x2
= (y2
– 1)2

2
0
22
dx)x(
+ Volume cone
 
4
1
22)4( dxxx
– volume cone
I=
 
2
1
dy)1y(


1
1
22
dy)1y(
y = x2
str. line: y = –4x + 12
y = 4x - x2
str. line: y = –x + 4 y = x2
– 1 x = y2
– 1
1
–1412 3
4 –
3–
y = x2
2 3 1 4
0 1 2
1
0
str. line: y = –4x + 12
y = 4x - x2
str.line: y = –x + 4 y = x (x–1) (x–2)
21

13. FORM 5 TOPICS
13
TOPIC 4: VECTORS
 If vectors a and b are parallel, thus, ………a = k b …………………….………..
 If OA = a and OB = b , thus, AB = … OB - OA……= b – a ………………..….
 If T is the mid point of AB thus, OT = ……AB………………………………………...
 Given m = 2i + 3j and n = i – 4j find,
i) m + n = ……2i + 3j + ( i – 4j ) = 3i – j …………….
ii) m + n = …… 10)1(3 22  …………………………………………..
iii) unit vector in the direction of m + n = …… )ji3(
10
1
 ………………………
 If A(1, 3) and B(–2, 5) find AB : …OA = 







3
1
, OB = 







5
2
 AB = 







5
2
– 







3
1
= 







2
3
can also be written in the form of i and j.
 Given CD = 2h x + 5y and CD = 8x – 2hky , find the value of h and of k.
2h x + 5y = 8x – 2hky ……………………… (comparison method)
compare coef. of x , compare coef of y.
TOPIC 5 : TRIGONOMETRIC FUNCTIONS
 Solving equation : SIMPLE
Solve: 2 cos 2x = 3 for 0  x  360
1. Separate coefficient of trigo
cos 2x =
2
3
positive values
 1st
and 4th
quadrant
2. Determine the quadrant
3. Find the reference angle 2x = 30
4. Find new range (if necessary). 0  x  720
5. Find all the angles
2x = 30, 330, 390, 690
x = 15, 165, 195, 345

14. 14
 Solving Equation : Using Identity WHEN?
sin 2x cos x = sin x
cos 2x cos x = 0 ………Angles are not the same………
sin2
x + 3cos x = 3
2 sec2
x + tan2
x = 5 ………have different functions ……….
 Proofing: Use Identity
Remember :


cot
.1
tan,
Acos
Asin
Atan  , sec 2x =
x2cos
1
, cosec A =
 Use of Trigo Ratios: Examples:
From the question given, If sin A =  , A is not acute,
1. Determine the quadrant involved. ……second……………….
2. Determine the values of the other trig. fxn cos and tan = negative
in the quadrant. cos A =
5
3 , tan A = – 
find sin 2A
3. Do you need to use identity? sin 2A = 2 sin A cos A
4. Substitute values = 2 ( ) (
5
3 )
=
25
24

 Sketch Graphs
y = a sin b x + c a =
a cos b x + c
a tan b x + c b =
c =
Basic Graphs
max / minimum point
number of basic shape between 0 and 2
increase / decrease translation of the
2 2 2
1-
–1-
1-
–1-
Asin
1
y = sin x y = cos x y = tan x

15. FORM 5 TOPICS
15
TOPIC 6: PERMUTATIONS & COMBINATIONS
Permutations = …order of arrangement is important Combinations =…order not important….
Three committee members of a society are to be
chosen from 6 students for the position of
president, vice president and secretary. Find
the number of ways the committee can be
chosen.
Permutations: 6P3
Three committee members of a society are to be
chosen from 6 students. Find the number of
ways the committee can be chosen.
Combination: 6C3
with condition:
Find the number of different ways the letters
H O N E S T can be arranged if it must
begin with a vowel.
2 5 4 3 2 1
condition
vowels = 2 choices
Find the number of ways 11 main players of a
football team can be chosen from 15 local
players and 3 imported players on the
condition that not more than 2 imported
players are allowed.
condition  2 Import.
Case : (2 Import, 9 local) or 3
C2. 15
C9
(1 import 10 locals) or + 3
C1. 15
C10
( 0 import 11 locals) + 3
C0.15
C11
TOPIC 7: PROBABILITY
 P(A) =
)S(n
)A(n
- Probability event A or B occurs = P(A) + P(B)
- Probability event C and D occurs = P(A) . P(B)
 Considering several cases:
Probability getting the same colors = Example: (Red and Red) or (Blue and blue)
Probability of at least one win in two matches = (win and lose) or (lose and win) or (win and win)
Or using compliment event = 1 – (lose and lose)

16. 16
TOPIC 8: PROBABILITY DISTRIBUTION
BINOMIAL DISTRIBUTION
- The table shows the binomial probability distribution of
an event with n = 4 .
0 1 2 3 4 X=r
Graph of Binomial Prob Distribution
total = 1
- formula: P(X = r) = n
C r p r
q n – r
- mean,  = np standard deviation = npq variance = npq
NORMAL DISTRIBUTION
- Formula :



X
Z
- Type 1 : Given value of X  find the value of Z  find the probability
[use formula] [use calculator]
- Type 2 : Given the probability Find the value of Z  Find its value of X .
[use log book] [use formula]
TOPIC 9: MOTION IN STRAIGHT LINES
Displacement, s Velocity, v Acceleration, a
s =
 dtv
dt
ds
v 
dt
dv
a  = 2
2
dt
sd
Maximum velocity - 0
dt
dv
 a = 0
return to O s = 0 - -
stops momentarily v = 0
max. acceleration 0
dt
da

X = r r = 0 r = 1 r = 2 r = 3 r = 4
P(X) 0.2 0.15 0.3 0.25 0.1
_
_
_
_
_
_
0.3
0.25
0.2
0.15
0.1
0.05
P(X)

17. FORM 5 TOPICS
17
TOPIC 10: LINEAR PROGRAMMING
Given:
(i) y > x – 2 (ii) x + y  5 (iii) 4x  y
(a) Draw and shade the region, R, that satisfy the three inequalities on the graph paper provided
using 2 cm to 2 units on both axes.
(b) Hence, find, in the region R, the maximum value of 2x + y where x and y are integers.
–3 –2 –1 0 1 2 3 4 5 6
1
2
3
4
5
6
–1
–2
–3
–4
2 possible maximum points (x, y intergers)
(1, 4) and ( 3, 2) . Point (3, 1) cannot be
taken because it is not in R (it’s on dotted line)
2x + y = 2(1) + 4 = 6
= 2(3) + 2 = 8  the max value
y = x – 2
y + x = 5
4x = y
R