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FORM FIVE
ADDITIONAL MATHEMATIC NOTE
CHAPTER 1: PROGRESSION
Arithmetic Progression
Tn = a + (n – 1) d
n
[2a + (n – 1)d]
2
n
= [a + Tn]
2
Sn =
S1 = T1 = a
T2 = S2 – S1
Example : The 15th term of an A
...
is 86 and
the sum of the first 15 terms is 555
...
(a)
n
[a + Tn]
2
15
555 =
[a – 86]
2
Sn =
74 = a 86
a = 12
a + 14d = 86
12 + 14d = 86
14d = 98
d = 7
20
[2(12) + 19(7)]
2
= 10[24 – 133]
= 1090
(c)
Sum from 12th to 20th term = S20 – S11
= 1090
11
[24 + 10(7)]
2
= 1090 –(253)
= 837
Geometric Progression
Tn= arn-1
Sn =
a(1 r n ) a(r n 1)
=
1 r
r 1
For 1 < r < 1, sum to infinity
S
a
1 r
zefry@sas
...
my
a G
...
is 24 and 10
2
respectively
...
ar3 = 24 ------------------(1)
2 32
=
--------(2)
3
3
ar 5 32 1
(2) (1)
ar 3 3 24
4
2
r2 =
r=
9
3
ar5 = 10
Since all the terms are positive, r =
2
3
3
2
= 24
3
27
a = 24 ×
= 81
8
7
20
2
T8 = 81 = 4
27
3
a×
T15 = 86
(b) S20 =
Example: Given the 4th term and the 6th term of
Example: Find the least number of terms of the
G
...
such that the last term is less
3
than 0
...
Find the last term
...
0003
1
3
n1
18 ×
1
3
< 0
...
0003
18
1
0
...
0003
log
18
n–1>
1
log
3
<
[Remember to change the sign as log
1
is
3
negative]
n – 1 > 10
...
01
n = 12
...
0001016
3
Example: Express each recurring decimals
below as a single fraction in its lowest term
...
7777
...
151515
...
7777
...
7 + 0
...
007 +
...
7
y
against x
...
Solution:
5 1 1
, passing through (6, 5)
62 2
1
5 = (6) + c c = 2
2
y 1
The equation is
x 2 , or
x 2
1
y = x2 + 2x
...
07
= 0
...
7
a
0
...
7 7
S
1 r 1 0
...
9 9
r=
(b) 0
...
= 0
...
0015 + 0
...
0
...
01
0
...
15
0
...
01 0
...
15, r =
CHAPTER 2: LINEAR LAW
Characteristic of The Line of Best Fit
1
...
2
...
3
...
To Convert from Non Linear to Linear Form
To convert to the form Y = mX + c
Example:
Non
Linear
m
c
Linear
y = abx
log y = log
log b
log a
b(x) + log a
y = ax2
a
b
y
ax b
+ bx
Example: The table below shows values of two
variables x and y, obtained from an experiment
...
x
1
2
3
4
5
6
y
4
...
75 10
...
2 22
...
1
(a) Explain how a straight line can be obtained
from the equation above
...
2 unit on the y-axis
...
Solution:
(a) y = abx
log y = log b(x) + log a
By plotting log y against x, a straight line is
obtained
...
65 0
...
00 1
...
36 1
...
edu
...
b
kf ( x) dx k f ( x) dx
a
(c)
c = log a = 0
...
00 0
...
175
3 1
b = 1
...
3
...
A=
a
a
f ( x) dx f ( x) dx
b
Example:
x dy
Given
f ( x) dx 9 , find the value of
1
3
y dx
(a)
1
3
2
x dy
4 f ( x) dx
(b)
[5 f ( x)]dx
1
3
2
(a)
3
4 f ( x) dx = 4 ×
1
Example: Find
(a)
(b)
(a)
3x
2
3x 2 2 x 3 dx
3x3 2 x 2
=
3x c
3
2
f ( x) dx
1
= 4 × 9 = 36
2 x 3 dx
x4 2 x
dx
x6
a
c
b
a
Volume generated when a shaded region is
rotated through 360o about the y-axis
V=
c
3
The volume generated when a shaded region is
rotated through 360o about the x-axis
V=
f ( x) dx f ( x) dx f ( x) dx
y dx
The area between the graph and the y-axis
A=
b
f ( x) dx g ( x) dx
a
a
b
n
The area between the graph and the x-axis
b
f ( x) g ( x) dx
a
b
CHAPTER 3: INTEGRATION
x n 1
x dx
c
n 1
a
g
3
(b)
3
[5 f ( x)]dx = 5dx
1
1
3
f ( x) dx
1
= 5x 1 + 9
3
= [15 – 5] + 9 = 19
Area Below a Graph
1
...
edu
...
Area between the graph and the line y = c,
y = d and the y-axis is
Volume generated when a shaded region is
revolved through 360o about the x-axis is
b
d
A=
x dy
c
V=
y dx
2
a
Volume generated when a shaded region is
rotated through 360o about the y-axis is
Example:
Given A is the point of intersection between the
curve y = 5x – x2 and the line y = 2x, find the
area of the shaded region in the diagram below
...
A(3, 6)
5x – x2 = 0
x(5 – x) = 0
, x = 0 or x = 5
CHAPTER 4: VECTORS
Addition of Two Vectors
(a) Triangle Law
AB BC AC
(b) Parallelogram Law
zefry@sas
...
my
4
AB AD AC
Parallel Vectors
AB is parallel to PQ if
AB
k , where k is a
PQ
constant
...
Vector on Cartesian Plane
1
OP OA AP = x AB
3
1
1
= x ( y x) = (2 x y )
3
3
1
(b) (i) OE = k OP = k × (2 x y )
3
2k
k
=
x y
3
3
(ii) OE OA AE = OA + h AQ
1
= x h( x y )
2
h
= (1 – h) x + y
2
(c) Compare the coefficient of x and y
2k
---------------(1)
3
h k
2k
and , h =
-----(2)
2 3
3
1–h=
OA xi yj
OA x 2 y 2 = magnitude of vector
Substitute in (1)
OA
Unit vector in the direction of
OA
2k
2k
=
3
3
4k
3
1=
k=
4
3
2k
2 3 1
h=
=
×
=
...
4
5
Given OP
Find OP
Given OA = x and OB y
...
The line OP intersects AQ at the point E
...
(b) Express OE in terms of
(i) k, x and y ,
OP
...
(a)
1
1
OQ = OB y
2
2
zefry@sas
...
my
OP =
32 (4)2 25 5
(b) Unit vector in the direction of OP =
(ii) h, x and y
...
Find the value of m and n