Grab Me SPM Add Mathematics Form 4,5

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GRAB






C
M
Y Me
CM
SPM
MY ©PAN ASIA PUBLICATIONS
CY
ADDITIONAL MATHEMATICS
CMY
K
Azizah binti Kamar
(Textbook Author & Guru Cemerlang)





199101016590 (226902-X)

ii ii
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CONTENTS



Form 4 Form 5
CHAPTER 1 Functions ..............................................1 CHAPTER 1 Circular Measure ..............................119
CHAPTER 2 Quadratic Functions ...........................17 CHAPTER 2 Differentiation ...................................129
CHAPTER 3 Systems of Equations .........................31 CHAPTER 3 Integration ........................................147
CHAPTER 4 Indices, Surds and Logarithms ...........37 CHAPTER 4 Permutation and Combination ..........162
CHAPTER 5 Progressions ......................................52
CHAPTER 5 Probability Distribution ......................170
CHAPTER 6 Linear Law ..........................................64
CHAPTER 6 Trigonometric Functions ...................187
CHAPTER 7 Coordinate Geometry .........................72
CHAPTER 7 Linear Programming .........................205
CHAPTER 8 Vectors ...............................................86
CHAPTER 8 Kinematics of Linear Motion .............212
CHAPTER 9 Solution of Triangles ...........................99
CHAPTER 10 Index Number ...................................113







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Algebra
CHAPTER 1 FUNCTIONS




1.1 Functions
• Function is a one-to-one relation or a many-to-one relation such that one object has only one
image.
Functions representation

Arrow diagram Graph Ordered pair
• Functions may be discrete functions or continuous functions.
• Vertical line test is used to determine whether a relation is a function or not.
• Function notation: Alphabets such as f, g or h can be used to
represent a function that maps set A to set B. REMEMBER
For example, f : A ˜ B
f : x ˜ f(x) • Set A is a domain whereas
Let f(x) = 3x + 1 set B is a codomain.
Therefore, f(4) = 3(4) + 1 • x is an object whereas f(x) is
an image.
= 13 • Set of all images is known
• Absolute value function can be written as f : x ˜ |ax + b|. The as range.
shape of the graph is .
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Example 1 Example 2
Determine whether the following relations is a Determine whether the following graphs represents
function or not. Give your justification. a function or not. Give your justification.
(a) a • • 1 (a) y (b) y
➤ ➤
b • ➤ • 2
c • • 3
0 x
(b) {(1, 2), (2, 3), (2, 4), (3, 5)} 0 x
(c) y
3 Solution y
2 ©PAN ASIA PUBLICATIONS
(a) No. The vertical line cuts
1 the graph at more than one
0 x point. 0 x
1 2 3
Solution y
(a) A function. Every object has only one image,
that is, {(a, 2), (b, 2), (c, 1)}. (b) Yes. The vertical line cuts
(b) Not a function. Object 2 has two images. the graph at one point only.
(c) A function. Every object has only one image. 0 x






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Example 4 Solution

Given the function f(x) = |4x – 3|. (a) (i) f(–2) = |4(–2) – 3|
(a) Find = |–11|
(i) the image of –2, = 11
(ii) the objects of 5, (ii) If f(x) = 5, then |4x – 3| = 5
(iii) the value of x when f(x) = 0. 4x – 3 = 5 or 4x – 3 = –5
(b) (i) Sketch the graph of f(x) for the domain 4 x = 8 4x = –2
–2  x  2. x = 2 x = – 1
(ii) Hence, state the corresponding range. 2
(iii) f(x) = 0
4x – 3 = 0
TIPS x = 3
4
f(x) ©PAN ASIA PUBLICATIONS
y
(b) (i)
11
Range  f(x) = |4x – 3|
5
3
0  x –2 0 3 2 x
Domain –
4
(ii) The range of f is 0  f(x)  11.






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Algebra
CHAPTER 5 PROGRESSIONS




5.1 Arithmetic Progressions

First term, The nth term,
T = a T = a + (n – 1)d
1 n
Arithmetic
progressions
Example: Common difference,
2, 4, 6, 8, … d = T – T
n n - 1

• The sum of the first n terms, S is given by the • Given S ,
n
n
formula: (a) a = T = S 1
1
n (b) T = S – S n – 1
n
n
S = [2a + (n – 1)d]
n 2 (c) The sum from the nth term to the (n + k)th
n term, where k is a positive integer
= [a + l], l = T n = S n + k – S n – 1
2

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Example 17 Example 18
The sum of the first three terms of a geometric Given a geometric progression with T = 1 and r = 2.
1
progression with a common ratio – 2 is 63. Find Calculate the minimum value of n if the sum of the
the first term. 3 first n terms is more than 2 000.
Solution Solution
S = 63 S  2 000
n
3
a(1 – (– ) ) 1(2 – 1)  2 000
2
n
3
3 = 63 2 – 1
1 – (– ) 2  2 001
2
n
3 n log 2  log 2 001
35 a = 63( ) 10 10
5
log 2 001
10
27 ©PAN ASIA PUBLICATIONS
3
n 
log 2
27
5
a = 63( )( ) n  10.97 10
3
35
a = 81 Therefore, n = 11.
Another Method
Example 19
T + T + T = 63
3
1
2
a + ( – 2a ) ( )( – 2a ) = 63 The nth term of a geometric progression is
2
+ –
3 3 3 T = 4(4 1 – n ). Calculate
n
9a – 6a + 4a = 63 (a) the common ratio,
9
a = 81 (b) the sum to infinity.
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Example 8 6.3 Application of Linear Law
The diagram below shows part of the graph
log (1 – y) against log x. Given the variables Example 9
10
10
x and y are related by the equation y = 1 – ax .
b
log (1 – y) The table below shows the values of two variables
10
u and v obtained from an experiment. Given that
(3, 2) the variables u and v are related by the equation
uv = pu + qv, where p and q are constants.
u 20 25 30 35 40 45
log x
0 (1, 0) 10 v 20.0 16.7 15.0 14.0 13.3 12.8
Find the constants a and b.
(a) Based on the above table, construct a table for
Solution the values of .
v
2 – 0 u
From the graph, m = = 1. Thus, Y = X + c.
3 – 1 v
At point (1, 0), 0 = 1 + c Ú c = –1 (b) Plot v against v u using a scale of 2 cm to
Hence, log (1 – y) = log x – 1 …  0.2 unit on the u -axis and 2 cm to 2 units on
10
10
Given y = 1 – ax , thus 1 – y = ax b the v-axis. Hence, draw the line of best fit.
b
log (1 – y) = log a + b log x … 
10
10
10
Compare  with : log a = –1 (c) Using the graph in (b), find the values of p and
10 q.
a = 1 , b = 1
10
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Solution (c) uv = pu + qv
v
v = p + q
(a) v 1.00 0.67 0.50 u
u p = v-intercept and q = gradient
v 20.0 16.7 15.0 Therefore, p = c = 10 and q = 20 – 10 = 10
1.0 – 0
v 0.40 0.33 0.28 Example 10
u
v 14.0 13.3 12.8 The table below shows the values of two variables
x and y obtained from an experiment.
(b) v v x 1 2 3 4 5 6
Graph of v against – u

20.0 y 36.0 16.7 7.29 3.28 1.48 0.67
Given that x and y are related by the equation
18.0 ©PAN ASIA PUBLICATIONS
y = ab x – 1 , where p and q are constants.
16.0 (a) Convert y = ab x – 1 to the linear form.
(b) Using a scale of 2 cm to 1 unit on the X-axis and
14.0
2 cm to 0.2 unit on the Y-axis, plot Y against X.
12.0 Hence, draw the line of best fit.
(c) From the graph in (b), find
10.0 (i) the values of a and b,
v
8.0 – u (ii) the value of y when x = 5.5.
0 0.2 0.4 0.6 0.8 1.0 1.2 (d) Find the value of y when x = 7 and the value
of x when y = 0.7.




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Geometry
CHAPTER 8 VECTORS




8.1 Vectors
Scalar quantity Vector quantity
A quantity that has magnitude. A quantity that has magnitude and direction.
Example: Distance, speed, mass and height Example: Velocity and weight
→
• The vector notation is AB or a with magnitude • If a = b, then |a| = |b| and the directions for both
→
~
~
~
~
~
vectors are the same.
|AB| or |a|. ©PAN ASIA PUBLICATIONS
~ → → → →
→ B • AB = lPQ Í AB is parallel to PQ where l is a
AB or a constant.
~ ➤ → →
• Points A, B and C are collinear if AB = lBC,
A → → → →
→ → AB = bAC, AC = mBC and there is a common
• Vector BA is a negative vector of vector AB, point.
→
→
→
→
that is, – AB = BA = –a and |– AB| = |AB| but both • Given a and b are two non-zero vectors and are
~
~
~
are moving in the opposite direction. not parallel.
• Zero vectors, 0 has zero magnitude and no If ha = kb, then h = k = 0.
~
direction. ~ ~
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• Addition and subtraction of non-parallel vectors.
Triangle law Parallelogram law
C C D C
a + b b
~ ~ b ~ ~ –a + b b a + b
~ ~
~ ~ ~
A B A B
a a A B
~ ~ a
→ → → → → → ~
AC = AB + BC BC = BA + AC → → →
→
→
= AC – AB AC = AB + AD
Polygon law
d
E ~ D
e c
~ ~ → → → → → →
F C AF = AB + BC + CD + DE + EF
f b
~ ~
A a B
~
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Geometry
CHAPTER 1 CIRCULAR MEASURE




1.1 Radian • Conversion of degrees to radians and vice versa:
× π
• One radian is a measure of the angle subtended 180°
at the centre of a circle by an arc of equal length Degrees Radians
to the radius of the circle. × 180°
π
B Example:
r r (a) 100° = 100° × π = 1.745 rad
180°
1 rad 180°
O r A p rad = 180° (b) 0.7 rad = 0.7 × π = 40.11°
Example 1
y
The diagram on the right
180° shows a circle with centre
Hence, 1 rad = ≈ 57.29° β
π O which is divided into eight α
π equal parts. x
and 1° = ≈ 0.01746 rad O
180° θ
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State the magnitude of angles (a) a, (b) b and (c) q 1.2 Arc Length of a Circle
(i) in degrees,
(ii) in radians, in terms of π, • The formula for the arc length of a circle is
(iii) in radians. Give the correct answer to four given by:
significant figures.
[Use π = 3.142] Arc length, s = rq
where q is the angle in radians.
Solution
• Note the diagram below.
(a) (i) a = 45°
(ii) a = 45° × π = π rad A
180° 4
3.142 r
(iii) a = 4 = 0.7855 rad
(b) (i) b = 90° ©PAN ASIA PUBLICATIONS C
θ
O
π π
(ii) b = 90° × 180° = 2 rad
3.142
(iii) b = = 1.571 rad
2
(c) (i) q = 135° B
π
(ii) q = 135° × 180° = 3 π rad Perimeter of the segment ACB
4
3 = Arc length ACB + Length of chord AB
(iii) q = 4 (3.142) = 2.357 rad




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Given the arc length AB is 6 cm. Find
REMEMBER (a) the angle q, in radians,
(b) the angle q, in degrees.
Minor
Major sector Minor arc
sector Solution
180°
(a) s = rq (b) q = 0.75 ×
π
8q = 6 q = 42.97°
O
q = 0.75 rad

Major arc Chord Example 3
Segment
The diagram below shows a sector AOB with
Example 2 centre O and a radius of 5 cm.
A O
The diagram below shows a sector AOB with θ 5 cm
centre O and a radius of 8 cm.
A B
8 cm
C
6 cm θ O
Find the perimeter of the shaded segment ACB for
each of the following values of q.
B (a) 2.09 rad (b) 144°
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2.2 The First Derivative
• Differentiation is a process of determining the gradient function of a function y = f(x).
• The gradient function is also known as the first derivative of a function or the derivative function
or the differential coefficient of y with respect to x.
• Techniques of differentiation for an algebraic function:

n
General formula If y = ax where a and n are constants, then dy = nax n – 1
dx
Algebraic sum If y = f(x) ± g(x), then dy = fʹ(x) ± gʹ(x)
and difference dx
(a) If y = f(u) where u = g(x), then dy = dy × du
Chain rule dx du dx
(b) If y = [f(x)] where n is a constant, then dy = n[f(x)] n – 1 fʹ(x)
n
dx
Product rule If y = uv where u = f(x) and v = g(x), then dy = u dv + v du = uvʹ + vuʹ
dx dx dx
v du – u dv
Quotient rule u dy dx dx vuʹ – uvʹ
If y = where u = f(x) and v = g(x), then = 2 =
v dx v v 2






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D. Rates of change for related quantities E. Small changes and approximations of
certain quantities
• If y = f(x) and x = g(t), then dy = dy × dx
(Chain rule). dt dx dt • For a function y = f(x),
dy ≈ dy
t dx dx
TIPS dy
Steps to solve problems involving the rates of dy ≈ dx × dx (dx = x new – x initial )
change for related quantities:
1. Interpret the information in the form of where dy is a small change in y and dx is a small
mathematical symbols. change in x.
2. Determine the appropriate chain rule. (a) When dy  0 or dx  0, there is a small
increase in y or x.
(b) When dy  0 or dx  0, there is a small
• Interpretation of rates of change for related decrease in y or x.
quantities:
• The approximation value of y is given by the
(a) When dy  0, y increases when t formula:
dt
increases. y new = y initial + dy
(b) When dy  0, y decreases when t • If y = f(x) and x changes by dx, then
dt
increases. (a) the percentage change in x = dx × 100%
x
(b) the percentage change in y = dy × 100%
y
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Example 11 TIPS

The diagram below shows a part of the graph dy d y
2
y = f(x). Nature dx Sketch of the graph dx 2
y 0
D(6, 18)
Maximum + –  0
y = f(x) point
B(–2, 4) y = f(x)
2 x
–5 –2 0 6 y = f(x)
C(2, –4)
Minimum = 0  0
point – +
1
A(–5, –16 ) –
2 0
(a) Find the points such that the gradient of the y = f(x) y = f(x)
tangent to the curve is Point of + –
(i) zero, inflection 0 0 = 0
(ii) positive, + –
(iii) negative.
(b) Determine the points such that Solution
2
2
(i) d y  0 (ii) d y  0
dx 2 dx 2 (a) (i) B(–2, 4) and C(2, –4)
1
(c) Hence, state the methods for finding the (ii) A(–5, –16 ) and D(6, 18)
maximum point or the minimum point. (iii) O(0, 0) 2



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Trigonometry
CHAPTER 6 TRIGONOMETRIC FUNCTIONS




6.1 Positive Angles and Negative Angles Example 1
• In trigonometry, Convert each of the following angles into radians.
(a) Angles measured in the anticlockwise Hence, determine the quadrant for each of the
direction from the positive x-axis is angles and represent them in the same Cartesian
known as positive angles. plane.
(b) Angles measured in the clockwise (a) 123° (b) 251° (b) –35°
direction from the positive x-axis is
known as negative angle. Solution
p
• The position of angles in the Cartesian plane: (a) 123° × 180° = 2.147 rad
y 123° lies in quadrant II.
90°
p
Quadrant II Quadrant I (b) 251° × 180° = 4.381 rad
0° 251° lies in quadrant III.
x
180° O 360° (c) –35° × p = –0.6109 rad
180°
Quadrant III Quadrant IV –35° lies in quadrant IV.
270°
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y • From the relation, it is found that:
1 1
A (a) cosec q = sin q
1
(b) sec q = cos q
123° 1
–1 1 x (c) cot q = tan q
251° O –35° • Signs for trigonometric ratios in the quadrants:
y
C Quadrant II Quadrant I
sin (+) All (+)
B –1 cosec (+)
x
O
6.2 Trigonometric Ratios of Any Quadrant III Quadrant IV
Angle
tan (+) cos (+)
• Relation between six trigonometric ratios: cot (+) sec (+)
y r • Complementary angle formulae:
sin q = r cosec q = y (a) sin q = cos (90° – q)
x r r (b) sec q = cosec (90° – q)
cos q = r y sec q = x (c) cos q = sin (90° – q)
θ (d) cosec q = sec (90° – q)
y x x (e) tan q = cot (90° – q)
tan q = x cot q = y (f) cot q = tan (90° – q)






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Example 8 (b) (i) m cos px – 1 = –4
Draw a graph of y = –4 on the graph.
The diagram below shows the graph of There is only 1 point of intersection,
y = m cos px – 1 for 0  x  p. therefore the number of solution is 1.
y (ii) m cos px = –2
y = m cos px – 1 m cos px – 1 = –2 – 1
2 m cos px – 1 = –3
1 Draw a graph of y = –3 on the graph.
x
0 1 1 3 � There are 2 points of intersection, therefore
–1 –� –� –�
–2 4 2 4 the number of solutions is 2.
–3
–4 Example 9
(a) State the values of m and p. The diagram below shows the graph of
(b) Find the number of solutions for each of the y = m sin nx for 0  x  2p.
following. y
(i) y = –4 (ii) m cos px = –2
Solution 3 y = m sin nx
(a) When x = 0 and y = 2, m cos 0 – 1 = 2
m = 3 0 � 2� x
When x = p and y = 2, 3 cos pp – 1 = 2
cos pp = 1 –3
pp = 2p
p = 2 State the values of m dan n.

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