1. NOTES AND FORMULAE
SPM MATHEMATICS
FORM 1 – 3 NOTES Cone
1. SOLID GEOMETRY
(a) Area and perimeter V= 1
 r2h
Triangle 3
A= 1
2
 base  height Sphere
1
= 2
bh
V= 4
3
r3
Trapezium
Pyramid
1
A= 2
(sum of two
parallel sides)  height V= 1
3
 base area 
= 1
2
(a + b)  h height
Prism
Circle
V = Area of cross section
Area = r2  length
Circumference = 2r
2. CIRCLE THEOREM
Sector
Area of sector =   Angle at the centre
360
= 2 × angle at the
r2 circumference
Length of arc = x = 2y
  2r
360
Cylinder Angles in the same
segment are equal
Curve surface area x=y
= 2rh
Sphere
Curve surface area = Angle in a
4r2 semicircle
(b) Solid and Volume ACB = 90o
Cube:
Sum of opposite
V = x  x  x = x3 angles of a cyclic
quadrilateral = 180o
a + b = 180o
Cuboid:
The exterior angle
V=lbh
of a cyclic
= lbh
quadrilateral is
equal to the interior
opposite angle.
Cylinder
b=a
V =  r2h
Angle between a
tangent and a radius
= 90o
OPQ = 90o
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2. The angle between a
tangent and a chord 2x2 – 6x + x – 3 = 2x2 – 5x − 3
is equal to the angle
in the alternate (b) (x + 3)2 = x2 + 2 × 3 × x + 32
segment. = x2 + 6x + 9
x=y (c) (x – y)(x + y) = x2 + xy – xy – y2 = x2 – y2
If PT and PS are 6. LAW OF INDICES
tangents to a circle, (a) xm  x n = xm + n
PT = PS
TPO = SPO (b) xm  xn = xm – n
TOP = SOP
(c) (xm)n = x m  n
3. POLYGON
(a) The sum of the interior angles of a n sided polygon
= (n – 2)  180o
1
(d) x-n =
xn
(b) Sum of exterior angles of a polygon = 360o
1
(c) Each exterior angle of a regular n sided polygon = (e) xn  n
x
0
360
n m
(d) Regular pentagon (f) x n  (n x ) m
(g) x0 = 1
7. ALGEBRAIC FRACTION
Each exterior angle = 72o Express 1  10  k as a fraction in its simplest
2
Each interior angle = 108o 2k 6k
form.
(e) Regular hexagon Solution:
1 10  k 1 3k  (10  k )
 
2k 6k 2 6k 2
= 3k  10  k  4k  10  2( k  5)  k  5
6k 2 6k 2 6k 2 3k 2
Each exterior angle = 60o
8. LINEAR EQUATION
Each interior angle = 120o
1
Given that (3n + 2) = n – 2, calculate the value
(f) Regular octagon 5
of n.
Solution:
1
(3n + 2) = n – 2
5
Each exterior angle = 45o 1
Each interior angle = 135o 5 × (3n + 2) = 5(n – 2)
5
4. FACTORISATION 3n + 2 = 5n – 10
2 + 10 = 5n – 3n 2n = 12 n=6
(a) xy + xz = x(y + z)
9. SIMULTANEOUS LINEAR EQUATIONS
(b) x2 – y2 = (x – y)(x + y) (a) Substitution Method:
y = 2x – 5 --------(1)
(c) xy + xz + ay + az 2x + y = 7 --------(2)
= x (y + z) + a (y + z) Substitute (1) into (2)
= (y + z)(x + a) 2x + 2x – 5 = 7 4x = 12 x=3
Substitute x = 3 into (1), y=6–5=1
(d) x2 + 4x + 3 (b) Elimination Method:
= (x + 3)(x + 1) Solve:
3x + 2y = 5 ----------(1)
5. EXPANSION OF ALGERBRAIC x – 2y = 7 ----------(2)
EXPRESSIONS (1) + (2), 4x = 12, x=3
(a) Substitute into (1) 9 + 2y = 5
2y = 5 – 9 = −4
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3. y = −2 3. A bar chart uses horizontal or vertical bars to
represent a set of data. The length or the height of
10. ALGEBRAIC FORMULAE each bar represents the frequency of each data.
Given that k – (m + 2) = 3m, express m in terms of
k.
Solution:
k – (m + 2) = 3m k – m – 2 = 3m
k – 2 = 3m + m = 4m
m=
k 2
4
11. LINEAR INEQUALITIES
1. Solve the linear inequality 3x – 2 > 10.
Solution: 4. A pie chart uses the sectors of a circle to represent
3x – 2 > 10 3x > 10 + 2 the frequency/quantitiy of data.
3x > 12 x>4
2. List all integer values of x which satisfy the
linear inequality 1 ≤ x + 2 < 4
Solution:
1≤x+2<4
Subtract 2, 1−2≤x+2–2<4–2
−1 ≤ x < 2
 x = −1, 0, 1
A pie chart showing the favourite drinks of a group
3. Solve the simultaneous linear inequalities
of students.
1
4p – 3 ≤ p and p + 2  p
2 FORM FOUR NOTES
Solution: 1. SIGNIFICANT FIGURES AND STANDARD
4p – 3 ≤ p 4p – p ≤ 3 3p ≤ 3 FORM
p≤1 Significant Figures
1. Zero in between numbers are significant.
1
p+2 p × 2, 2p + 4  p Example: 3045 (4 significant figures)
2 2. Zero between whole numbers are not
2p – p  −4 p  −4 significant figures.
 The solution is −4 ≤ p ≤ 1. Example: 4560 (3 significant figures)
3. Zero in front of decimal numbers are not
12. STATISTICS significant.
sum of data Example: 0.00324 ( 3 significant figures)
Mean = 4. Zero behind decimal numbers are significant.
number of data Example: 2.140 (4 significant figures)
Mean = sum of(frequency  data) , when the data Standard Form
sum of frequency Standard form are numbers written in the form A ×
has frequency. 10n, where 1 ≤ A < 10 and n are integers.
Mode is the data with the highest frequency Example: 340 000 = 3.4 × 105
Median is the middle data which is arranged in 0.000 56 = 5.6 × 10-4
ascending/descending order. 2. QUADRATIC EXPRESSION AND
1. 3, 3, 4, 6, 8 QUADRATIC EQUATIONS
1. Solve quadratic equations by factorization.
33 4 68
 4.8
Example: Solve 5k  8  2k
Mean = 2
5
Mode = 3 3
Median = 4 5k2 – 8 = 6k 5k2 – 6k – 8 = 0
2. 4, 5, 6, 8, 9, 10, there is no middle number, (5k + 4)(k – 2) = 0
the median is the mean of the two middle k= 4 ,2
numbers. 5
68 2. Solve qudratic equation by formula:
Median = =7 Example: Solve 3x2 – 2x – 2 = 0
2
x = b  b  4ac = 2  4  4(3)( 2)
2
2. A pictograph uses symbols to represent a set of
data. Each symbol is used to represent certain 2a 6
frequency of the data. = 2  28 x = 1.215, −0.5486
January 6
February 3. SET
(a) Symbol
March
 - intersection  - union
Represents 50 books  - subset  - universal set
 - empty set - is a member of
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4. n(A) –number of element in set A.
A – Complement of set A. Type III
Premise 1: If A, then B
(b) Venn Diagram Premise 2: Not B is true.
Conclusion: Not A is true.
5. THE STRAIGHT LINE
(a) Gradient
AB
AB Gradient of AB =
y 2  y1
m =
x 2  x1
(b) Equation of a straight line
A
Example:
Gradient Form:
n(A) = 7 + 6 = 13 y = mx + c
n(B) = 6 + 10 = 16
n(A  B) = 6 m = gradient
c = y-intercept
n(A  B) = 7 + 6 + 10 = 23
n(A  B‟) = 7
n(A‟  B) = 10
n(A  B) = 7 + 10 + 2 = 19
n(A  B) = 2
4. MATHEMATICAL REASONING
(a) Statement
A mathematical sentence which is either true or Intercept Form:
false but not both.
x y
 1
(b) Implication a b
If a, then b
a – antecedent a = x−intercept
b – consequent b = y−intercept
„p if and only if q‟ can be written in two y -int ercept
implications: Gradient of straight line m = 
If p, then q x-intercept
If q, then p
= b
a
(c) Argument
6. STATISTICS
Three types of argument:
(a) Class, Modal Class, Class Interval Size, Midpoint,
Type I
Cumulative frequency, Ogive
Premise 1: All A are B
Example :
Premise 2 : C is A
The table below shows the time taken by 80
Conclusion: C is B
students to type a document.
Type II
Time (min) Frequency
Premise 1: If A, then B
10-14 1
Premise 2: A is true
15-19 7
Conclusion: B is true.
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5. 20-24 12
25-29 21
30-34 19
35-39 12
40-44 6
45-49 2
For the class 10 – 14 :
Lower limit = 10 min
Upper limit = 14 min
Lower boundary = 9.5 min
Upper boundary = 14.5 min
Class interval size = Upper boundary – lower 7. TRIGONOMETRY
boundary = 14.5 – 9.5 = 5 min sin o = Opposite  AB
hypotenuse AC
Modal class = 25 – 29 min
Midpoint of modal class = 25  29 = 27 cos o = adjacent  BC
2 hypotenuse AC
To draw an ogive, a table of upper boundary and
cumulative frequency has to be constructed.
tan o = opposite  AB
Time Upper Cumulative
Frequency adjacent BC
(min) boundary frequency
5-9 0 9.5 0
10-14 1 14.5 1
15-19 7 19.5 8
20-24 12 24.5 20
25-29 21 29.5 42
30-34 19 34.5 60
35-39 12 39.5 72
40-44 6 44.5 78
45-49 2 49.5 80
Acronym: “Add Sugar To Coffee”
Trigonometric Graphs
1. y = sin x
From the ogive : 2. y = cos x
Median = 29.5 min
First quartile = 24. 5 min
Third quartile = 34 min
Interquartile range = 34 – 24. 5 = 9.5 min.
(b) Histogram, Frequency Polygon
Example:
The table shows the marks obtained by a group of 3. y = tan x
students in a test.
Marks Frequency
1 – 10 2
11 – 20 8
21 – 30 16
31 – 40 20
41 – 50 4
8. ANGLE OF ELEVATION AND DEPRESSION
(a) Angle of Elevation
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6. (a) Convert number in base 10 to a number in base 2, 5
or 8.
Method: Repeated division.
Example:
2 34
2 17 0
The angle of elevation is the angle betweeen the 2 8 1
horizontal line drawn from the eye of an observer 2 4 0
and the line joining the eye of the observer to an 2 2 0
object which is higher than the observer. 2 1 0
The angle of elevation of B from A is BAC 0 1
(b) Angle of Depression 3410 = 1000102
8 34
8 4 2
0 4
3410 = 428
The angle of depression is the angle between the
horizontal line from the eye of the observer an the (b) Convert number in base 2, 5, 8 to number in base
line joining the eye of the observer to an object 10.
which is lower than the observer. Method: By using place value
The angle of depression of B from A is BAC. Example: (a) 110112 =
24 23 22 211
9. LINES AND PLANES 1 1 0 1 12
(a) Angle Between a Line and a Plane = 24 + 23 + 21 + 1
= 2710
(b) 2145 =
52 5 1 1
2 1 45
= 2  52 + 1  51 + 4  1
= 5910
(c) Convert between numbers in base 2, 5 and 8.
Method: Number in base m  Number in base
10  Number in base n.
In the diagram,
Example: Convert 1100112 to number in base 5.
(a) BC is the normal line to the plane PQRS.
(b) AB is the orthogonal projection of the line
25 24 23 22 21 1
AC to the plane PQRS.
1 1 0 0 1 12
(c) The angle between the line AC and the plane
= 25 + 24 + 2 + 1
PQRS is BAC
= 5110
5 51
(b) Angle Between Two Planes 5 10 1
5 2 0
0 2
Therefore, 1100112= 2015
(d) Convert number in base two to number in base
eight and vice versa.
Using a conversion table
In the diagram, Base 2 Base 8
(a) The plane PQRS and the plane TURS 000 0
intersects at the line RS. 001 1
(b) MN and KN are any two lines drawn on each 010 2
plane which are perpendicular to RS and 011 3
intersect at the point N. 100 4
The angle between the plane PQRS and the plane 101 5
TURS is MNK. 110 6
111 7
FORM 5 NOTES
Example :
10. NUMBER BASES
10 0112 = 238
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7. (b) Reflection
458 = 100 1012 Description: Reflection in the line __________
11. GRAPHS OF FUNCTIONS Example: Reflection in the line y = x.
(a) Linear Graph
y = mx + c
(b) Quadratic Graph (c) Rotation
y = ax2 + bx + c Description: Direction ______rotation of
angle______about the centre _______.
Example: A clockwise rotation of 90o about the
centre (5, 4).
(c) Cubic Graph
y = ax3+ c
(d) Enlargement
Description: Enlargement of scale factor ______,
(d) Reciprocal Graph with the centre ______.
a
y
x
12. TRANSFORMATION Example : Enlargement of scale factor 2 with the
centre at the origin.
(a) Translastion
Area of image
h
Description: Translastion   k2
k  Area of object
 
 4 
Example : Translastion 
k = scale factor
  3

  (e) Combined Transformtions
Transformation V followed by transformation W is
written as WV.
13. MATRICES
a  c   a  c 
(a)   
 b   d  b  d 

     
(b)  a   ka 
k   
 b   kb 
   
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8.  a b  e f   ae  bg af  bh 
(c) 
 c d  g
  
  h   ce  dg cf  dh 
  
 a b  , then
If M = 
(d)
 

c d 
1  d  b
M-1 =  
ad  bc   c a 
  Gradient = Rate of change of speed
vu
=
(e) If ax + by = h
cx + dy = k
t
= acceleration
 a b  x   h 

 c d  y    k 
   
     Distance = Area below speed-time graph
 x 1  d  b  h 
 
 y  ad  bc   c a  k 
   16. PROBABILITY
     (a) Definition of Probability
(f)
 a c  has no inverse if ad – bc = 0
Matrix 
Probability that event A happen,
 n( A)
b d  P( A) 
n( S )
14. VARIATIONS S = sample space
(a) Direct Variation
If y varies directly as x, (b) Complementary Event
Writtn in mathematical form: y  x, P(A) = 1 – P(A)
Written in equation form: y = kx , k is a constant.
(c) Probability of Combined Events
(b) Inverse Variation
(i) P(A or B) = P(A  B)
If y varies inversely as x,
1 P(A and B) = P(A  B)
Written in mathematical form: y (ii)
x
17. BEARING
k
Written in equation form: y , k is a constant. Bearing
x Bearing of point B from A is the angle measured
clockwise from the north direction at A to the line
(c) Joint Variation joining B to A. Bearing is written in 3 digits.
If y varies directly as x and inversely as z,
x
Written in mathematical form: y ,
z
kx
Written in equation form: y , k is a
z
constant.
Example : Bearing B from A is 060o
15. GRADIENT AND AREA UNDER A GRAPH
(a) Distance-Time Graph 18. THE EARTH AS A SPHERE
(a) Nautical Miles
1 nautical mile is the length of the arc on a great
circle which subtends an angle of 1 at the centre
of the earth.
(b) Distance Between Two Points on a Great Circle.
Distance =   60 nautical miles
 = angle between the parallels of latitude
distance measured along a meridian of longitude.
Gradient = = speed
time
Total distance
Average speed =
Total time
(b) Speed-Time Graph
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9. (ii) the elevation of the combined solid on the
vertical plane parallel to GPS as viewed from
D.
 = angle between the meridians of longitude
measured along the equator.
Solution:
(a)
(c) Distance Between Two Points on The Parallel of
Latitude.
Distance =   60  cos o
 = angle of the parallel of latitude.
Plan
(b) (i)
(d) Shortest Distance
The shortest distance between two points on the
surface of the earth is the distance between the two
points measured along a great circle.
C-elevation
(e) Knot (ii)
1 knot = 1 nautical mile per hour.
19. PLAN AND ELEVATION
(a) The diagram shows a solid right prism with
rectangular base FGPN on a horizontal table. The
surface EFGHJK is the uniform cross section. The
rectangular surface EKLM is a slanting plane.
Rectangle JHQR is a horizontal plane. The edges
EF, KJ and HG are vertical.
Draw to full scale, the plan of the solid. D-elevation
(b) A solid in the form of a cuboid is joined to the solid
in (a) at the plane PQRLMN to form a combined
solid as shown in the diagram. The square base
FGSW is a horizontal plane.
Draw to full scale
(i) the elevation of the combined solid on the
vertical plane parallel to FG as viewed from C,
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