Form 5 Additional Maths Note

FORM FIVE
ADDITIONAL MATHEMATIC NOTE

CHAPTER 1: PROGRESSION Example: Given the 4th term and the 6th term of

Arithmetic Progression 2
Tn = a + (n – 1) d
a G.P. is 24 and 10 respectively. Find the 8th
Sn = n [2a + (n – 1)d]
2 3

term given that all the terms are positive.

n ar3 = 24 ------------------(1)
= 2 [a + Tn]
ar5 = 10 2 = 32 --------(2)
S1 = T1 = a 33
T2 = S2 – S1
ar5  32  1
(2) (1) ar3 3 24

Example : The 15th term of an A.P. is 86 and r2 = 4 r =  2
the sum of the first 15 terms is 555. Find 93
(a) the first term and the common
2
difference
(b) the sum of the first 20 terms Since all the terms are positive, r =
(c) the sum from the 12th to the 20th term.
3

(a) T15 = 86  2 3
a ×  3  = 24
n
Sn = 2 [a + Tn] 27

555 = 15 [a – 86] a = 24 × = 81
2
8
74 = a  86
a = 12  2 7 20
a + 14d = 86 T8 = 81  3  =4
12 + 14d = 86
14d = 98 27
d = 7
Example: Find the least number of terms of the

2

G.P 18, 6, 2, , ... such that the last term is less

3

than 0.0003. Find the last term.

20 a = 18, r = 6  1
(b) S20 = [2(12) + 19(7)] 18 3

2 Tn < 0.0003

= 10[24 – 133]  1 n1
18 ×  3  < 0.0003
= 1090

(c) Sum from 12th to 20th term = S20 – S11  1 n1 0.0003
 3  < 18
= 1090  11 [24 + 10(7)]
2 (n – 1) log 1 < log 0.0003
3 18
= 1090 –(253)
= 837

Geometric Progression log 0.0003

Tn= arn-1 n–1> 18
log 1
a(1 rn ) a(rn 1) 3
Sn = 1 r = r 1
1
For 1 < r < 1, sum to infinity
[Remember to change the sign as log is
S a
1 r 3

negative]
n – 1 > 10.01

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n > 11.01 Example: The diagram shows the line of best fit
 n = 12.
y
 1 11
T12 = 18×  3  = 0.0001016 by plotting against x.

Example: Express each recurring decimals x
below as a single fraction in its lowest term.
(a) 0.7777.... Find the relation between y and x.
(b) 0.151515... Solution:

(a) 0.7777...= 0.7 + 0.07 + 0.007 + .. m = 5 1  1 , passing through (6, 5)
62 2
a = 0.7
5 = 1 (6) + c c = 2
0.07 2

r = = 0.1 The equation is y  1 x  2 , or
x2
0.7
y = 1 x2 + 2x.
S a r  0.7  0.7  7 2
1 1 0.1 0.9 9
Example: The table below shows values of two
(b) 0.151515... = 0.15 + 0.0015 + 0.000015.... variables x and y, obtained from an experiment.
It is known that y is related to x by the equation
a = 0.15, r = 0.0015 = 0.01 y = abx.
x123456
0.15 y 4.5 6.75 10.1 15.2 22.8 34.1
(a) Explain how a straight line can be obtained
S  0.15  0.15  15  5
1 0.01 0.99 99 33 from the equation above.
(b) Plot the graph of log y against x by using a
CHAPTER 2: LINEAR LAW
Characteristic of The Line of Best Fit scale of 2 cm to 1 unit on the x-axis and 2
1. Passes through as many points as possible. cm to 0.2 unit on the y-axis.
2. All the other points are as near to the line of (c) From your graph, find the value of a and b.
Solution:
best fit as possible. (a) y = abx
3. The points which are above and below the log y = log b(x) + log a
By plotting log y against x, a straight line is
line of best fit are equal in number. obtained.
(b)
To Convert from Non Linear to Linear Form x123456
log 0.65 0.83 1.00 1.18 1.36 1.53
To convert to the form Y = mX + c y

Example:

Non Linear mc

Linear log b log a
y = abx log y = log

y = ax2 b(x) + log a a b
+ bx a b
y  ax  b
y = ax + x

xy = ax2 + b

b

x

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x 4  2x x4 2x
x6 x6 x6
 (b) dx =  dx

= x2  2x5dx

= x1  2x4  c
1 4

= 1 1 +c
x 2x4

(c) c = log a = 0.48 The Rule of Integration

a=3 bb

log b = 1.00  0.65 = 0.175  1. kf (x) dx  k f (x) dx
aa
31
g bb
b = 1.5
  2. f (x)  g(x) dx  f (x) dx  g(x) dx
CHAPTER 3: INTEGRATION
a aa
 xndx  xn1  c
n 1 bcc

The area between the graph and the x-axis   3. f (x) dx  f (x) dx  f (x) dx
aba
A = y dx
ba
The area between the graph and the y-axis
 4. f (x) dx   f (x) dx
A = x dy ab

The volume generated when a shaded region is Example:
rotated through 360o about the x-axis
3
V =  y2dx
Given f (x) dx 9 , find the value of
Volume generated when a shaded region is 1
rotated through 360o about the y-axis
3
V =  x2dy
(a) 4 f (x) dx
1
3

(b) [5  f (x)]dx
1

Example: Find 33

(a) 3x2  2x  3dx  (a) 4 f (x) dx = 4 × f (x) dx
 x4  2x 11
= 4 × 9 = 36
(b) x6 dx 3 33

(a) 3x2  2x  3dx   (b) [5  f (x)]dx = 5dx  f (x) dx
1 11
= 3x3  2x2  3x  c
32 = 5x3 + 9
1
= x3 + x2 + 3x + c
= [15 – 5] + 9 = 19

Area Below a Graph
1. The area below a graph and bounded by

the line x = a, x = b and the x-axis is

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Area under a curve = Area of triangle +

5 1 5x2  x3 5
5x  x2dx =  
×3×6+  2 3 3
3
2

= 9 + [125  125 ]   45  9] = 1 unit2
2 3  2
16

3

b Volume of Revolution

A = y dx
a

2. Area between the graph and the line y = c,
y = d and the y-axis is

d Volume generated when a shaded region is
revolved through 360o about the x-axis is
A = x dy
c b

Example: V =  y2dx
Given A is the point of intersection between the a
curve y = 5x – x2 and the line y = 2x, find the
area of the shaded region in the diagram below. Volume generated when a shaded region is
rotated through 360o about the y-axis is

y = 2x d
y = 5x – x2
2x = 5x – x2 V =  x2dy
c
x2  3x = 0
x(x − 3) = 0 CHAPTER 4: VECTORS
Addition of Two Vectors
x = 0 or 3.  A(3, 6) (a) Triangle Law

5x – x2 = 0 , x = 0 or x = 5 AB  BC  AC
x(5 – x) = 0
(b) Parallelogram Law

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AB  AD  AC OP OA AP = x  1 AB
3
Parallel Vectors
= x  1 (y  x) = 1 (2x  y)
AB is parallel to PQ if AB  k , where k is a 33
PQ
(b) (i) OE = k OP = k × 1 (2x  y)
constant. 3

If AB  k , since B is a common point, A, B = 2k x  k y
BC 33

and C are collinear. (ii) OE OA  AE = OA + h AQ
Vector on Cartesian Plane
= x  h(x  1 y)
OA xi  yj 2
OA  x2  y2 = magnitude of vector
= (1 – h) x + h y
OA 2
Unit vector in the direction of
(c) Compare the coefficient of x and y
xi  yj
OA  1 – h = 2k ---------------(1)
3
x2  y2
and h  k , h = 2k -----(2)
Example : 23 3

Given OA = x and OB  y . P is a point on AB Substitute in (1)

such that AP : PB = 1 : 2 and Q is the midpoint 1  2k = 2k
of OB. The line OP intersects AQ at the point E. 33

Given OE = k OP and AE  hAQ , where h 4k k= 3

and k are constants, 1=
34
(a) find OQ and OP in terms of x and/or
y. 2k 2 3 1

(b) Express OE in terms of h= = × = .
(i) k, x and y ,
(ii) h, x and y . 3 342

(c) Hence, find the value of h and k. Example :

(a) OQ = 1 OB  1 y Given OP   3  and OQ   1  .
22  4   5 
   
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(a) Find OP

(b) Find the unit vector in the direction of

OP .
(c) Given OP = m OA  n OQ and A is the

point (2, 7). Find the value of m and n.

(a) OP = 32  (4)2  25 5

(b) Unit vector in the direction of OP =
3i  4 j

5

(c) OP = m OA  n OQ

 3   m  2    1 
 4   7  n 5 
    


5

2m – n = 3 -------(1) The Addition Formulae
7m – 5n = 4 -----(2)
(1) × 5 Sin (A  B) = sin A cos B  cos A sin B
10m – 5n = 15 ----(3) Cos (A  B) = cos A cos B sin A sin B

7m – 5n = 4 ----(4) Tan (A  B) = tan A  tan B
17m = 19 1 tan Atan B

m =  19 The Double Angle Formulae
17
Sin 2A = 2 sin A cos A
substitute in (1), Cos 2A = 2 cos2A − 1

38  n = 3 = 1 – 2 sin2 A
17
n = 38  3 =  13 2 tan A
Tan 2A = 1  tan2 A
17 17
CHAPTER 6: PERMUTATION AND
CHAPTER 5: TRIGONOMETRIC COMBINATION
FUNCTION 1. Arrangement of n different objects without
Angles In The Four Quadrants
repetition.
The Three Trigonometric Functions
nPn = n!
Secant  = sec  = 1
cos 2. Arrangement of r objects from n objects

Cosecant  = cosec  = 1 n Pr  n!
sin (n  r)!
1
Example : Given the word ‘TABLES’. Find
Cotangent  = cot  = tan
(a) the number of ways to arrange all the letters
The Relation Between Trigonometric
Functions in the word.
Sin  = cos (90o − )
Cos  = sin (90o − ) (b) The number of ways of arranging the 6
Tan  = cot (90o − )
letters such that the first letter is a vowel.
Tan  = sin
cos (c) The number of ways of arranging 4 letters

Graphs of Trigonometric Functions out of the 6 letters such that the last letter is
y = asin bx ‘S’.
Amplitude = a
Number of periods = b (a) Number of arrangement = 6P6 = 6! = 720

[email protected] (b) Number of ways of arranging 1 vowel out

of 2 = 2P1

Number or ways of arranging the remaining

5 letters = 5P5 .
Total arrangement = 2P1 × 5P5 = 240.

OR:

254321

Total number of ways = 2 × 5! = 240

(c) If the last letter is ‘S’, the number of ways
of arranging 3 letters out of the remaining 5

letters = 5P3 = 60.

OR:

5431

Number of ways = 5 × 4 × 3 × 1 = 60

6

2. Combination of r objects from n objects (a) P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)
= 6C4 (0.25)4 (0.75)2  6C5 (0.25)5(0.75)
is nCr  n!
r!(n  r)! + (0.25)6

Example: The PTA committee of a school = 0.03760
consists of 8 members. The members are elected
from 7 parents, 6 teachers and the principal of (b) Mean,  = np = 2800 × 0.03760
the school. Find the number of different
committees that can be formed if = 105.3
(a) the principal is one of the member of the
Standard deviation = npq =
committee.
(b) the committee consists of the principal, 3 2800 0.3760 0.6240 = 25.63

teachers and 4 parents. 2. Normal Distribution
(c) the committee consists of at least 2
X 
teachers.
Z = where
(a) number of committees = 1C1  13C7 = 1716

(elect 1 principal, and 7 committee
members from 13 = 7 parents and 6 Z = standard normal score
teachers) X = normal score

(b) number of committees =  = mean
 = standard deviation
1C1  6C3  7C4  700
P( Z < a) = 1 – P(Z >a)
(c) number of committees = total number of P(Z < a) = P(Z > a)
committees – number of committees with P(Z > a) = 1 – P(Z > a)
no teacher – number of committees with 1 P(a < Z < b) = P(Z > a) – P(Z > b)
P(a < Z < b) = 1 – P(Z > a) – P(Z > b)
teacher.
Example: The volume of packet drink produced
= 14C8  6C0  8C8  6C1  8C7 = 2954. by a factory is normal distributed with a mean of
500 ml and standard deviation of 8 ml.
CHAPTER 8: PROBABILITY Determine the probability that a packet drink
DISTRIBUTION chosen at random has a volume of
(a) more than 510 ml
1. Binomial Distribution (b) between 490 ml and 510 ml.
The probability of getting r success in n
trials where p = probability of success and (a) P(X > 510) = P(Z > 510  500 )
q = probability of failure
8
P(X = r) = nCr pr qnr
= P(Z > 1.25) = 0.1056
2. Mean ,  = np
(b) P(490 < X < 510)

490  500 510  500
= P( < Z < )
88
3. Standard deviation = npq
= P(1.25 < Z < 1.25)
Example: = 1 – P(Z > 1.25) – P(Z > 1.25)
In a survey of a district, it is found that one in = 1 – 2(0.1056)
every four families possesses computer.
(a) If 6 families are chosen at random, find the = 0.7888

probability that at least 4 families possess CHAPTER 9: MOTION ALONG A
computers. STRAIGHT LINE
(b) If there are 2800 families in the district, 1. Displacement (S)
calculate the mean and standard deviation
for the number of families which possess Positive displacement  particle at the
computer. right of O

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Negative displacement  particle at the (a) If the particle does not stop in the time
left of O. interval
Return to O again  s = 0
Maximum/minimum displacement b

ds Distance = v dt
a
=0
(b) If the particle stops in the time t = c
dt seconds where c is in the interval a 
b
Distance travelled in the nth second
= Sn – Sn-1 cb
Example: Distance travelled in the third
second = S3 – S2  Distance = v dt  v dt

Example: A particle moves along a straight line ac
and its displacement, s meter, from a fixed point
O, t seconds after leaving O is given by s = 2t – Example: A particle moves along a straight line
t2. Find passing through a fixed point O. Its velocity, v m
(a) the displacement of the particle after 5 s-1, t seconds after passing through O is given by
v = 2t + 3. Find the displacement at the time of 4
seconds, second.
(b) the time at which it returns to O again.
(c) the distance travelled in the fourth second.  s = v dt = 2t  3dt = t2 + 3t + c

(a) s = 2t – t2 If the particle passes through O when t = 0, , s
t = 5, = 0 when t = 0.  c = 0
s = 10 – 25 = 15 m  s = t2 + 3t
When t = 4 s,
(b) Return to O again  s = 0 s = 16 + 12 = 28 m
2t – t2 = 0
t(2 – t) = 0 Example:
t = 0 or t = 2 second
 the particle returns to O again when A particle moves along a straight line passing
t = 2 s. through a fixed point O. Its velocity, v m s-1, t

(c) Distance travelled in the 4th second = S4 – seconds after passing through O is given by v =
S3 = (8 – 16) – (6 – 9) =  8 + 3 =  5 m. t2 + t – 6. Find
 the distance travelled in the 4th second is
5 m. (a) the initial velocity of the particle,

2. Velocity (v) (b) the time when the particle is momentarily at
Velocity is the rate of change of
displacement with respect to time. rest.

ds (a) initial velocity  t = 0
v =  6 m s -1
v=
(b) momentarily at rest  v = 0
dt t2 + t – 6 = 0
(t + 3)(t – 2) = 0
Initial velocity  the value of v when
t=0 t = 3 or t = 2 s
Instantaneously at rest/change direction of
movement  v = 0 Since the time cannot be negative,  t = 2
Moves towards the right  v > 0
Moves towards the left  v < 0 s.
Maximum/minimum velocity
3. Acceleration (a)
 dv = 0
dt Acceleration is the rate of change of

s = v dt velocity with respect to time.

Distance travelled in the time interval a= dv d 2s
t = a until t = b =
dt 2
[email protected] dt

Initial acceleration  a when t = 0

Deceleration  a < 0

Positive acceleration  velocity increasing

Negative acceleration  velocity

decreasing

8

Zero acceleration  uniform velocity. The table above shows the time taken by a tailor
Maximum/minimum velocity  a = 0 to prepare a shirt and a shorts of a school
uniform. In a week, the tailor sells x shirts and y
v = a dt shorts. Given that in a week, the number of shirts
and shorts sold must be at least 10. The time for
Example: A particle moves along a straight line preparation is at most 800 minutes. The ratio of
passing through a fixed point O. Its velocity, v m the number of shorts to the shirts must be at least
s-1, t seconds after leaving O is given by v = t2 – 1 : 2.
6t – 7. Find (a) Write three inequalities other than x ≥ 0
(a) the initial acceleration of the particle
(b) the time when the velocity of the particle is and y ≥ 0 which satisfy the constraints
above.
maximum. (b) By using a scale of 2 cm to 10 units on the
x- and y-axes, draw the graph of all the
(a) a = dv = 2t – 6 inequalities above. Hence, shade the region
dt R which satisfies the constraints above.
(c) The tailor makes a profit of RM5 and RM3
t = 0, a = 6 m s-2 in selling a shirt and a shorts respectively.
Find the maximum profit made by the tailor
(b) Maximum velocity, a = 0 in a week.
2t – 6 = 0
t = 3 s. (a) the inequalities are:
(i) x + y ≥ 10
Example : A particle moves along a straight line (ii) 10x + 20y  800
x + 2y  80
passing through a fixed point O with a velocity
of 5 m s-1. Its acceleration, a ms-2 a the time t (iii) y  1
second after leaving O is given by a = 2t – 4. x2
y≥ 1x
Find the maximum velocity of the particle. 2

a = 2t – 4 (b) To draw x + y = 10, x = 0, y = 10 and y = 0,
x = 10
v = 2t  4dt = t2 – 4t + c To draw x + 2y = 80
x = 0, 2y = 80, y = 40
t = 0, v = 5,  c = 5 y = 0, x = 80
v = t2 – 4t + 5
To draw y = 1 x
For maximum velocity, a = 0 2
2t – 4 = 0, t = 2 s.
vmax = 4 – 8 + 5 = 1 m s-1 x = 0, y = 0
x = 40, y = 20.
CHAPTER 10: LINEAR PROGRAMMING
Steps (c) Profit , k = 5x + 3y
1. Form the linear inequalities. Let k = 150
2. Construct the region which satisfies the 5x + 3y =150
x = 0, 3y = 150, y = 50
constraints. y = 0, 5x = 150, x = 30
3. Form the optimum equation

ax + by = k
4. Find the point in the region which gives the

maximum or minimum value.
5. Substitute the value of x and y to obtain the

optimum value of k.

Example: Time of preparation
School uniform (minutes)
10
Shirt 20
Shorts

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From the graph, maximum profit is achieved
when x = 40 and y = 20.
 maximum profit = 5 × 40 + 3 × 20 = RM 260.

ALL THE BEST FOR YOUR SPM EXAM.

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