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Title: BUDAKCEMERLANG
Description: TAKE NOTES DO WELL

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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

n1

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)
62 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: F
Title: BUDAKCEMERLANG
Description: TAKE NOTES DO WELL