Spotlight A+ SPM Additional Mathematics Form 4 & 5

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Extra Features of This Book







CHAPTER
1 Functions

Important Learning Standards
sMART SCOPE • Explain function using graphical representations and notations. Page 3 CONCEPT MAP
1.1 Functions • Determine domain and range of a function. 5
• Determine the image of a function when the object is given and 6 The entire content of the
vice versa.
Contains Learning
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• Describe the outcome of composition of two functions. 9 chapter is summarised in
Standards (LS) that need to • Determine the composite functions. 9
1.2 Composite • Determine the image of composite functions when the object is 10 the form of a concept map.
given and vice versa.
Functions
be achieved in each chapter. • Determine a related function when the composite function and 11
another function are given.
• Solve problems involving composite functions. 11
• Describe inverse of a function. 13 Form
4 Additional Mathematics Chapter 7 Coordinate Geometry
1.3 Inverse Functions • Make and verify conjectures related to properties of inverse 13
functions.
Form • Determine the inverse functions. 15
5 Additional Mathematics Chapter 3 Integration
3.1 Integration as the Inverse of Example 1
Differentiation (a) If f(x) = 9x 2 + 5x and f(x) = 18x + 5, find Coordinate Geometry
∫ (18x + 5) dx.
Explaining the relation between • Function / Fungsi • Range / Julat
• Relation / Hubungan and dy = g(x), find ∫ g(x) dx.
differentiation and integration (b) Given y = 4x(1 – x) 3 dx • Domain / Domain
5
• Arrow diagram / Gambar rajah anak panah • Codomain / Kodomain Divisor of a
(c) Given dy ( 3 – x ) = h(x), find ∫ h(x) dx.
CHAP. 1. Given y = x 3 + 7, thus dy = 3x 2 . • Absolute value function / Fungsi nilai mutlak • Object / Objek Line Segment Area of Polygon The Straight Line Equation of Locus
dx
dx
3 Conversely, if given dy = 3x 2 , thus y = x 3 + 7 will • Function notation / Tatatanda fungsi • Image / Imej
Solution:
• Absolute value function graph
• Composite function / Fungsi gubahan
Graf fungsi nilai mutlak
be obtained. dx (a) Given f(x) = 9x 2 + 5x • Inverse function / Fungsi songsangan
f (x) = 18x + 5
2. The reverse process of this differentiation is • Vertical line test / Ujian garis mencancang • Horizontal line test / Ujian garis mengufuk m n Parallel line Perpendicular line
Hence, ∫ 18x + 5 dx = f(x)

called integration. A(x 1 , y 1 ) P(x, y) B(x 2 , y 2 )
∫ 18x + 5 dx = 9x 2 + 5x
3. If d [f(x) = f (x), thus the integral f (x) with Thus, ∫ (18x + 5) dx is 9x 2 + 5x. 1 nx 1 + mx 2 , ny 1 + my 2  Gradient of two Product of gradient
dx
respect to x is ∫ f (x) dx = f(x). (b) Given y = 4x(1 – x) 3 P(x, y) =  m + n m + n lines are the same, for two lines is –1,
m 1 m 2 = –1
m 1 = m 2
dy = g(x)
dx
Differentiation dx d [f(x)] = f (x) Hence, ∫ g(x) dx = y
∫ g(x) dx = 4x(1 – x) 3 CHAP 7 Triangle Quadrilateral Form 5 n-sides of polygon
Integration ∫ f (x) dx = f(x) Thus, ∫ g(x) dx is 4x(1 – x) 3 . Chapter 1 Circular Measure Additional Mathematics
5
(c) Given d ( 3 – x ) = h(x) 1.1 Radian y Example 1 y CHAP. Area = 1 x 1 x 2 … x n x 1 
dx
A(x 1 , y 1 )
B(x 2 , y 2 )
Hence, ∫ h(x) dx = y Convert the angle in the unit of radian to the B(x 2 , y 2 ) 1 2  y 1 y 2 … y n y 1
Comparison between ∫ h(x) dx = 5 Relating angle measurement in radian and degree. [Use π = 3.142]
C(x 3 , y 3 )
differentiation and integration 3 – x degree (a) 1.15 radian (b) 5π radian
bit.ly/2K4y3b0 Thus, ∫ h(x) dx is 5 . 1. In circular measures, the angle can be measured Solution: x 6 D(x 4 , y 4 ) C(x 3 , y 3 )
3 – x
(a)
Try question 1 to 5 in Formative Zone 3.1 in 2 units, which are A(x 1 , y 1 ) π rad = 180° x
(a) degree (°) and minute (ʹ).
1.15 rad = 1.15 × 180°

3.1 (b) unit of radian (in or not in the terms of π). Area = 1 x 1 x 2 y 2 x 3 y 3 x 1  π y 1 180° Area = 1 x 1 x 2 x 3 x 4 x 1 
2  y 1 y 2 y 3 y 4 y 1
2 y 1
2. Radian involves the angle that related with the = 1.15 × 3.142
1. If f(x) = 7 – 3x 4 and f(x) = –12x 3 , find ∫ –12x 3 dx. C1 3. Given d (x 2 – 5x + c) = 2x – 5, find ∫ (2x – 5) dx. radius and circumference of a circle. = 65.89°
Form
C1 dx 3. The diagram below shows the angle in degree (b) π rad = 180°
2. Given dy = 2x – 2 and y = ( x + 1 ) , find Chapter 4 Indices, Surds and Logarithms Additional Mathematics 4 5π rad = 5π × 180°
and minutes and radian while the radius have
2
dx 4. Given d ( 4 + 5x) = g(x), find ∫ g(x) dx. C2 the same length. 6 6 π
x
x 3
dx x 2
x 3)
∫ ( 2x – 2 dx. C1 Calculator 5. Given y = f(x) and dy = 3h(x), find ∫ h(x) dx. C2 Example 25 A constant from a fixed point: A constant ratio from two fixed points:
= 5 × 180°
(x – x 2 ) 2 + (y – y 2 ) 2 = m 2
1. To find the value of antilog 3.26 using a scientific 6 (x – x 1 ) 2 + (y – y 1 ) 2 = r 2 (x – x 1 ) 2 + (y – y 1 ) 2
dx
= 150°
calculator, press SHIFT log 3.26 = . Prove the following. O 57° 17' O 1 rad Alternative Method n 2
(a) log a xy = log a x + log a y
SPOTLIGHT PORTAL Deriving and determining indefinite integral Example 24 (c) ∫ ax n dx = ax n + 1 . (b) y Substitute π =180° into the expression,
2. The screen will be display 1 819.700859
3.2
Indefinite Integral
n + 1 + c such as n ≠ 1 log a x = log a x − log a y
Always equidistant from two fixed points:
5π = 5(180°) = 150°
and c are constants.
6
6
for algebraic functions
Solution:
Find the value of x in each of following equation. Let log In degree and minute In radian Try question 1 in Formative Zone 1.1 (x – x 1 ) 2 + (y – y 1 ) 2 = (x – x 2 ) 2 + (y – y 2 ) 2
2. The following is the steps to find the integral of a a x = m
57° 17ʹ = 1 rad
(b) log 5 x = −2
1. Formula of integration: (a) log 9 27 = x function, ax n with respect to x: log a y = n 4. 1 radian is a measurement of an angle subtended
…❶
x = a m
Step 1: Maintain the value of constant, a.
(c) log 2 (x 2 + 3x) = 2
(d) log x 1 = −6
Scan the QR code to (a) ∫ a dx = ax + c where a and c are (a) log 9 27 = x Step 2: Add 1 to the power or index of x first. (a) y = a n (m + n) log a a = log a xy A CHAP 4 126 Convert Example 2
64
…❷
about the centre of a cirlce such as the arc length
constants.
Step 3: Divide the term with the new index. ❶ × ❷:
a m × a n = xy
is the same as the length of radius of circle.
Solution:
a m + n = xy
(a) 30° into radian unit, in term of π.
(b) ∫ x n dx = x n + 1
n + 1 + c such as n ≠ 1 and c are
Step 4: Add the constant c with the integrals.
(b) log 5 x = −2
(b) 200° into radian unit.
x = 5 −2
m + n = log a xy
browse website or 270 constants. 27 = 9 x For example, ∫ x n dx = x n + 1 n + 1 + c 3.1.1 3.2.1 3.2.2 ❶ ÷ ❷: log a x + log a y = log a xy O r 1 radian B r [Use π = 3.142]
3 3 = 3 2x
= 1
Solution:
2x = 3
a m
5 2
a n = x
(a) 180° = π rad
(b)
r
x = 3
y
video related to the 2 = 1 25 5. Hence, angle subtended about the centre of a 30° = 30° × π 180°
a m − n = x
= π rad
6
y
log a a m − n = log a x
(c) log 2 (x 2 + 3x) = 2
64
y
x 2 + 3x = 2 2 (d) log x 1 = −6 circle, ˙AOB is 1 radian if the arc length of AB is Alternative Method ALTERNATIVE METHOD
subtopics learned. (x + 4)(x − 1) = 0 64 1 = x −6 (m − n) log a a = log a x y 30° = 180° = π rad
equal to the radius of the circle.
Substitute 180° = π into the expression,
x 2 + 3x − 4 = 0
AB = OA = OB = r
6
6
64 −1 = x −6

x = −4 or x = 1
m − n = log a x
(2 6 ) −1 = x −6 6. The relationship between the measurement of (b) 180° = π rad
y
2 −6 = x −6 angle in radian with degree is 200° = 200° × π
x = 2 log a x − log a y = log a x 2π rad = 360° 180°
y
Try question 5 in Formative Zone 4.3 π rad = 180° 200° = 200° × 3.142
180°
7. The conversion of angle measurement in the degree = 3.49 rad Provide alternative
Example 26 Try question 2 and 3 in Formative Zone 1.1
to the radian and vice versa are as follow:
Proving laws of logarithms Prove that log a x n = n log a x.
Calculator
1. The following is the laws of logarithms. Solution: × 180° π Recheck the answer in Example 2(b) by using solutions to certain
• Product rule
CALCULATOR • Quotient rule Let m = log a x (a m ) n = x n Radian Degree calculator,
log a xy = log a x + log a y


a m = x
2. The screen will display 3.490658504 questions.
1. Press 2 0 0 × SHIFT EXP ÷ 1 8 0 =
× π

180°
log a x = log a x – log a y
log a a mn = log a x n
(mn) log a a = log a x n
y (log a x)(n log a a) = log a x n 4 Form
• Power rule n log a x = log a x n Additional Mathematics Chapter 5 Progressions 217

log a x p = p log a x 1.1.1 Example 30
Explains how to use 2. From the laws of logarithms, the following can Example 27 Find the sum of the first 8 terms of the For Example 31, the sum from 5th term to 9th term
be derived:
following geometric progression.
• log a 1 = 0 Given that log 3 2 = 0.631 and log 3 5 = 1.465, is S 9 – S 4 . S 9
find the value for each of the following. 2 , …
(a) 14, 2, 7
• log a a = 1
a scientific calculator • log a a r = r (a) log 3 10 (b) 1.6, 3.2, 6.4, … T 1 + T 2 + T 3 + T 4 + T 5 + T 6 + T 7 + T 8 + T 9
S 4
(b) log 3 2.5
• log a 1 = –log a b (c) log 3 45 Solution: S 9 – S 4
b
in mathematics • log 1 b = –log a b (d) log 3 6 5 (a) a = 14, r = 2 = 1 7 |r|  1
a
14
4.3.1 4.3.2 Sum of the first 8 terms, S 8 Example 32
69
1 8
calculations. = 14  1 –  7   The term of a geometric progression is given
by T n = 2 1 + n . Find
(a) the first term and the common ratio,
Form CHAP = 16.33 1 – 1 7 (b) the sum of the first 8 terms,
of the geometric progression.
5 Additional Mathematics Chapter 6 Trigonometric Functions 5
Solution:
(b) a = 1.6, r = 3.2 = 2 |r|  1 (a) First term, T 1 = 2 1 + 1
Example 19 Example 20 1.6 = 4
Given f(x) = 4 cos 2x for 0 < x < 2π. Sketch the graph of the following trigonometric 2nd term, T 2 = 2 1 + 2

Sum of the first 8 terms, S 8
(a) State the period of the graph function y = f(x). functions in the given range. = 8
Hence, state the number of cycle of the graph (a) y = sin x + 1 for 0 < x < 2π = 1.6(2 8 – 1) Common ratio, r = 8 = 2
2 – 1
in the given range. (b) y = –2 cos x for 0 < x < 2π 4
(b) State the amplitude of the graph. (c) y = | tan x | for 0 < x < 2π = 408 Thus, the first term and the common ratio
(c) Write the coordinates of the maximum and the (d) y = | cos 2x | + 1 for 0 < x < 2π Try question 21 in Formative Zone 5.2 are 4 and 2 respectively.
minimum points.
(d) Sketch the graph of y = f(x). Solution: (b) Sum of the first 8 terms, S 8
(e) Using the same axes, sketch the graph of (a) y = sin x + 1 for 0 < x < 2π = 4(2 8 – 1) BRILLIANT TIPS
2 – 1
function y = –|4 cos 2x| for 0 < x < 2π. 1 Sketch the basic graph, y = sin x. Example 31 = 1 020
EXAMPLE Compare f(x) = 4 cos 2x with the basic cosine 2 The graph moves 1 unit upward, such that Try question 25 to 27 in Formative Zone 5.2
Solution:
It is given that –9, 27, –81, … is a geometric
0
progression. Find the sum from 5th term to 9th
translation ( ) .
function, f(x) = a cos bx + c.
1
term of the geometric progression.
(a) Period 2π = π or 180°. Number of cycle, b = 2.
2
Example 33
(b) Amplitude, a = 4 y Solution: The sum of a geometric progression is given by Useful tips to help
(c) Maximum point: (0, 4), (π, 4) and (2π, 4).
CHAP. Minimum point: ( π , – 4) and ( 3π , – 4) 2 y = sin x + 1 a = −9, r = − 27 = –3 S n = 3(2 n ) − 3. Find
9
2
2
(a) the sum of the first 5 terms,
Examples with 6 (d) To sketch graph function y = 4 cos 2x: 1 O π 3π S 4 = −9(−3) 4 − 1) (b) the 7th term of the geometric progression.
students solve
Number of class = 2 × 2 × 2 = 8

x
−3 − 1
Solution:
2π
Size of class interval = 2π = π 4 –1 π – 2 y = sin x –– 2 = 180 (a) S 5 = 3(2 5 ) − 3
8 = 93
complete solutions x 0 π 2 π 3π 2 2π –2 S 9 = −9((−3) 9 − 1) (b) S 6 = 3(2 6 ) − 3 problems in the related
−3 − 1
= 189
y 4 – 4 4 – 4 4 (b) y = –2 cos x for 0 < x < 2π = − 44 289 S 7 = 3(2 7 ) − 3
Thus, the graph function of y = 4 cos 2x: 1 Sketch graph of y = cos x. The sum from 5th term to 9th term = 381
to enhance students' y 2 Reflect the graph at 1 on x-axis to make 7th term, T 7 = S 7 − S 6 subtopics.
= S 9 − S 4
= − 44 289 − 180

381 − 189
the graph of y = – cos x.
=

= 192
4 y = 4 cos 2x = − 44 469 Try question 28 to 30 in Formative Zone 5.2
Try question 22 to 24 in Formative Zone 5.2
understanding of the 2 O π 3π 2π x 2 y
–2 π – 2 –– 2 1 y = – cos x 94 5.2.3
chapters learned. –4 O π 3π –– 2π x
(e) Steps in sketching the graph of y = –|4 cos 2x| –1 π – 2 y = cos x 2
1 y = |4 cos 2x| is a reflection of graph on –2
negative side of x-axis.
2 y = –|4 cos 2x| is a reflection of graph at 1 3 The value of a is –2. The maximum value is
on x-axis. (π, 2) and the minimum value is (0, –2) and

y (2π, –2).
4 y

2 2 y = –2 cos x
x 1
O π 3π 2π
–– y = – cos x
–2 π – 2 2 x
O π 3π 2π
–4 π – 2 –– 2
y = –|4 cos 2x| –1
–2
Try question 3 in Formative Zone 6.3
354 6.3.1
iv
Prelims Spotlight Add Math Tg4&5.indd 4 23/04/2021 2:10 PM

TAGGING 'Try question ... in
Formative Zone ...”
Chapter 2 Quadratic Functions Additional Mathematics Form 4
Example 6 Solving quadratic inequalities
Given the roots of quadratic equation 1. A quadratic inequality is a function where
The tagging is located at the 2x 2 + (p − 2)x + (q + 1) = 0 are −1 and − 3 , find its degree is 2 and uses the inequality symbols CHAP
which are less than, greater than, less than or
2
the values of p and q.
Solution: equals and greater than or equal. Symbol 2
Inequality
end of the example guides Sum of roots = −1 +  − 3 2  Less than 
2 
−  p − 2 = − 5 2 Greater than  <
Less than or equal to
the students to answer the p − 2 = 5 Greater than or equal to  Chapter 10 Index Numbers Additional Mathematics Form 4
2 2
p − 2 = 5 2. Solving a quadratic inequality means to find the
corresponding questions in Product of roots = −1  − 3 2  3. There are three methods to determine the 1. Find the index number or price index of the 7. The table below shows the price (RM) and the
10.1
range of the values of x that satisfies the inequality.
p = 7
range of values of x that satisfy the quadratic
Formative Zone. q + 1 = 3 2 inequality which are: following quantity or price as 2014 is the base price indices of four items, P, Q, R and S used
(a) Graph sketching
year. C1
in making a dress.
2
(b) Number line
q + 1 = 3 (c) Tabulation (a) Year Price of a tile (RM) Price (RM) Price index for
q = 2 4. The range of values of x can be obtained by 2017 44 Item Year Year the year 2019
Thus, the values of p = 7 and q = 2. considering two types of quadratic inequalities. 2014 55 2016 2019 year 2016
based on the
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Try question 9 to 11 in Formative Zone 2.1 (a) (x − a)(x − b)  0
(b) (x − a)(x − b)  0 (b) Number of visitor P 3.40 x 125
Year
Example 7 A Graph sketching method 2019 (thousand) Q 2.50 3.20 y
24.7
1. In graph sketching method, we have to consider R z 2.30 115
Given that one of the roots of quadratic 2014 19 FORMATIVE ZONE
equation 2x 2 − kx + 54 = 0 is three times the two forms of quadratic equations which are S 2.85 2.80 98.25
other root, find the values of k. y = (x − a)(x − b) and y = −(x − a)(x − b). Find the values of x, y and z. C3
Solution: 2. For quadratic equation y = (x − a)(x − b), the 2. The price of a school bag in the years 2019 and
2017 were RM63.90 and RM45 respectively.
Let the roots be α and 3α. graph is as follows: Find the price index of the school bag for the 8. The price indices of a tube of facial cleanser
Sum of roots = α + 3α y > 0 year 2019 based on the year 2017. C1 in the years 2017 and 2014 based on the year
C3 Questions to test
−k x 2012 are 135 and 120 respectively. Calculate
−  2  = 4α a b 3. The table below shows the price of two type the price index in the year 2017 based on the
y < 0 of food, vegetables and chicken in the year year 2014.
k = 4α (a) The values of a and b are the roots of the 2015 and 2018.
2
quadratic equation.
shown in the table below. students' understanding
α = −k 8 …❶ (b) Thus, Type of food Price (RM) per kg 9. In a boutique, the total number of customers
who came to the boutique are recorded as
Product of roots = α(3α) (i) x  a or x  b when (x − a)(x − b)  0 2015 2018 C3
(ii) a  x  b when (x − a)(x − b)  0
54 = 3α2 3. For quadratic equation y = −(x − a)(x − b), the Vegetable 3.20 3.52 Year 2012 2014 2016
2
3α 2 = 27 …❷ graph is as follows: Form 5 Chicken 6.00 7.20 Number of 3 000 at the end of each
Substitute ❶ into ❷. Chapter 5 Probability Distribution Additional Mathematics customers 3 930
y > 0
k 2 x Calculate the price index of vegetable and (hundred)
3  8  = 27
subtopic.

SPM Simulation HOTS Questions y < 0 a EXAMINER’S b chicken in the year 2018 based on the year (a) Find the index number of the number of
2015. C1
k 2
COMMENT
customers that came to the boutique in CHAP
the year 2014 based on the year 2012.
Paper 1 64 = 9 (a) The values of a and b are the roots of the 4. The number of accident in the year 2017 is (b) Given the index number of the number of 10
k 2 = 576
k = 24 or k = −24 2. The diagram below shows the graph of a
1. A bank auditor claims that credit card’s quadratic equation. 1 840 cases while 1 564 cases in the year
(b) Thus,

2018. Calculate the index number of accident
balances are normally distributed with a mean binomial distribution of X. occur in the year 2018 based on the year 2017. customers in the year 2016 based on the
Try question 9 to 11 in Formative Zone 2.1
of RM2 870 and a variance of RM810 000. (i) a  x  b when (x − a)(x − b)  0 Describe the answer. C2 year 2014 is the same as the index number
What is the probability a randomly selected P (X = x) (ii) x  a or x  b when (x − a)(x − b)  0 in the year 2014 based on the year 2012.
credit card holder has a card balance less than 5. The number of workers in the year 2018 is Estimate how many customers came at the
C4
RM2 500? 2.1.2 2.1.3 0.432 25 2 106 compared to 1 950 in the year 2015. After boutique on 2016?
Examiner’s comment: three years, the index number of workers in the 10. During a dry season, water level in a lake is
year 2020 based on 2018 is 132.6. In what year,
Let X is the credit card’s balance 0.288 the increment of workers is higher? C2 25 m on February 2019 and 27.3 m on March
m = 2 870, s 2 = 810 000 0.216 2019. C3
s = 900 6. The index number of annual income of James (a) Find the water level of the lake on March
So, X ~ N(2 870, 900). 0.064 CHAP. decreases by 6.2% from the year 2017 to the as February is a base month.
Given X is less than RM2 500. x 5 year 2018. If his annual income in the year 2017 (b) If the water level of the lake on April (March
Standardised variable X to Z, 0 0 1 2 3 was RM48 750, what is James’s annual income is a base month) were twice than (a), find
the water level of the lake in April 2019.
in the year 2018? C2
P(X , 2 500) = P ( Z , X – m ) s Find C4
= P ( Z , 2 500 – 2 870 ) (a) the value of probability of ‘success’, 201
(b) P(1 < X , 3).
900
= P(Z , –0.411) Examiner’s comment:
= P(Z . 0.411) (a) Let p is probability of ‘success’
= 0.3405 q is probability of ‘failure’
From the graph, P(X = 3) = 0.216
3 C 3 p 3 q 0 = 0.216
p 3 = 0.216
From standard normal distribution table, p = 0.6
f(z) Thus, the value of probability of ‘success’
is 0.6.
SPM SIMULATION HOTS QUES 0TIONS = 0.288 + 0.432 SPM MODEL PAPER
(b) P(1 < X , 3) = P(X = 1) + P(X = 2)
= 0.72
z
–0.411
Paper 1
3. In normally distributed, the mean and standard
z 1 1 Paper 2 deviation of length of fish is 11 inches and Time: 2 hours
Substract
4 inches respectively. C5 Section A
Provide a complete solutions 0.3409 4 (a) What is the percentage of the length of fish Instruction: Answer all questions
(64 marks)
0.4
are longer than 14 inches?
(b) If 200 fish are randomly selected, how many
function of y = f(x).
Calculator fishes has the length less than 9 inches? 1. (a) Diagram 1 shows a part of graph for a (ii)
with the examiner's comments Examiner’s comment: y y = f(x)
Check the answer by using scientific
calculator.
Let X is the length of fish
1. Press MODE MODE and choose 1 Given m = 11, s = 4, so X ~ N(11, 4)
(a) Given the length of fish is longer than
which is SD.
for the SPM Simulation HOTS 14 inches, X . 14. 0 2 x
2. Press SHIFT 3 and choose 3
Change variable X to Z
represents P(z . a). P(X . 14) = P( Z . X – m s ) –3
questions. 3. Insert 0.411 and the screen will display = P( Z . 14 – 11 ) Diagram 1
4
0.34054 .
= P(Z . 0.75)
SPM MODEL PAPER (ii) has an inverse function or not. [2 marks] domain 0 < x < 4 and its inverse function
= 0.2266 State whether the function of f 2. (a) Diagram 2 shows a graph of function f for
(i) is a discrete or continuous,
335 (b) A function f is derived by f : x ˜ a , x ≠ 0 f –1 . y A (4, 12)
x
such as a is a constant. Given f –1 (2) = 2 , find
(i) the value of a,
(ii) f 17 (8). [3 marks] f
Answer: f –1 B
(a) (i)
Form 0 x
5
Additional Mathematics Chapter 7 Linear Programming –4
(ii) Based on the graph, determine SPM MODEL PAPER
Diagram 2

(b) (i) (i) the domain of f –1 ,
(ii) the coordinate of point B on the graph
of f –1 that corresponding with the point
A on the graph of f.
[2 marks]
Paper 2 SPM format questions according
(i)
1. On the given grid below, show the region that 4. By using the given grid below, Answer:
C2 satisfies all the following inequalities. C3 (a) show that the region is bounded by all the
x > 3, y > 1 and x + y < 5 following inequalities. to the latest SPM 2021
x > 2, y > x and x + y < 6
y y (ii)
Summative Zone 7 6 5 7 6 assessment format cover all the
4 5 418
3 4
3 chapters in Forms 4 and 5.
2
CHAP. 1 2
7
Questions of various levels of 3 4 5 6 7 x 1
2
1
–1 0
–1 –1 0 1 2 3 4 5 6 7 x
–1
thinking skill are provided to (b) The point P with coordinates (x, y) lies inside
2. A green grocer sells bananas and apples. In one
day, he sells
C3
I up to 80 bananas, the region R. x and y are integers. Write down
II up to 90 apples, the coordinates of all the points of R whose
evaluate the understanding 5. The cost of a book is 50 cents and a pen is
III not more than 110 fruits.
coordinates are both integers.
If x be the number of bananas sold and y be
the number of apples sold, show the region that

RM1.30. A student wants to buy x books and
satisfies these inequalities and label the region C4 y pens based on the following conditions: ANSWERS Complete answers
http://bit.ly/2ORHlke
as R.
of each chapter. 3. The diagram below shows the R region that I II The total number of books and pens bought
At least three pens must be bought.
C2
satisfied all three linear inequalities.
must not more than 12.
FORM 4
y III The amount of money spent is at most 9. f(x) 5. – 4 5
RM10. 13 3
10 Write down the three inequalities other than 6. 10
Chapter 1 Functions
8 x > 0 and y > 0 that satisfy all the above 9 7. g(x) = 3x – 2
conditions. 1.1
6 R 3 8. f(x) = x – 2
6. By using the same graph, draw all the inequalities: (a) The relation is a function x 9. (a) 15
1.
4 C3 because each object has –6 – 3 0 5 (b) 13 5
y > x – 1, x > 2 and 2x + y > 8 only one image. 2
2 10. h = –3k
Shade the region that bounded by the (b) The relation is not a Range of f is 0  f(x)  13.
function because there is
x inequalities. Hence, state the minimum value of one object does not have 10. (a) q = 1, p =1 11. p = 12, q = 5 2
0 1 2 3 4 5 y in the region. any image. (b) 3 x x
2. (a) h(x) = |10 – x| or (c) –10 12. (a) 2x + 1 (b) 3x + 1
Define all the three linear inequalities. h(x) = |x – 10| 11. (a) 3 (c) 2x + 1 x (d) 3x + 1 x
(b) h(x) = x 2 – 1 (b) 5 25
3. (a) A function. (c) 2 13. k = 1 , h = 11
(b) A function. 2 2
(c) Not a function.
3 or 1.7321
391 4. (a) {a, b, c, d} 12. (a) (b) 2 or 1.4142 14. (a) RM13 400
(b) RM449 675
FORM 4 ANSWER 5. (a) 7 13. (a) 1 (b) 2 (ii) 3 1. (a) –4 1.3 (b) –2 ANSWER
(b) {–2, 0, 2, 4}
2
(c) a, b, c, d
(d) –2, 0, 2
14. 3
(b) {–3, –2, 2, 7}
15. (a) 3
(d) 2
(c) 4
(b) (i) –12
6. (a) 0, 2, 6, 8
(b) 7, 1, 10
(b) – 5 and 5
3. (a) –7
(b) –2
(c) {7, 1, 10}
(d) 7 16. (a) (i) 5 4 12 (ii) 11 2. (a) –2 (b) 4
4. (a) Has an inverse function
(e) 8 17. (a) (i) 15 because each element in set
P matched with only one
7. (a) Domain = {–2, 0, 2, 4} 4 (b) Does not have an inverse Complete answers are
Codomain = {4, 6} (ii) 3 element in set Q.
Range = {4, 6} 4
(b) Domain of f is –1  x  3. function because from
Codomain of f is 1  f(x)  3. (b) – 31 and 33 horizontal line test, the
Range of f is 1  f(x)  3. 4 4 line cuts the graph on two
points.
8. (a) f(x) 18. –3 and 9 5. (a) Has an inverse function g. provided. Scan the
10 19. (a) 4 4 (b) Does not has an inverse
5 (b) 0  g(x)  10 6. function g. QR Code provided to get
4 y
1.2
x 11
–2 0 4 3 1. (a) fg(x) = 3x – 1 y = x
3 3 g(x)
(b) gf(x) = x – 5
(b) f(x) 2. (a) f 2 (x) = 36x – 7 6 the steps to the solution.
5 (b) g 2 (x) = 16 + 9x 3
(c) gf(x) = 18x + 1 2
3 (d) fg(x) = 18x + 23 –2 0 2 3 6 11 x
3. (a) 111 (b) 731 –2 g –1 (x)
x 4. (a) –4 (b) – 17 Domain of g –1 (x) is 2  x  11.
0 5
1 5 2 Range of g –1 (x) is –2  g –1 (x)  3.
2
434
v
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CONTENTS





FORM 4

Revision Revision
Date Date
Chapter 1 Functions 1 Chapter 6 Linear Law 106
©PAN ASIA PUBLICATIONS
1.1 Functions 3 6.1 Linear and Non-Linear
1.2 Composite Functions 9 Relations 108
1.3 Inverse Functions 13 6.2 Linear Law and
Summative Zone 18 Non-Linear Relations 114
6.3 Application of Linear Law 116
Summative Zone 121
Chapter 2 Quadratic Functions 20

2.1 Quadratic Equations and Chapter 7 Coordinate Geometry 125
Inequalities 22
2.2 Types of Roots of Quadratic 7.1 Divisor of a Line Segment 127
Equations 29 7.2 Parallel Lines and
2.3 Quadratic Functions 31 Perpendicular Lines 130
Summative Zone 42 7.3 Areas of Polygons 135
7.4 Equations of Loci 140
Summative Zone 146
Chapter 3 Systems of Equations 45
3.1 Systems of Linear Equations
in Three Variables 47 Chapter 8 Vectors 151
3.2 Simultaneous Equations 8.1 Vectors 153
involving One Linear 8.2 Addition and Subtraction
Equation and One of Vectors 159
Non-Linear Equation 50 8.3 Vectors in a Cartesian Plane 164
Summative Zone 56 Summative Zone 172



Chapter 4 Indices, Surds and Chapter 9 Solution of Triangles 177
Logarithms 58
9.1 Sine Rule 179
4.1 Laws of Indices 60 9.2 Cosine Rule 183
4.2 Laws of Surds 62 9.3 Area of a Triangle 186
4.3 Laws of Logarithms 68 9.4 Application of Sine Rule,
4.4 Applications of Indices, Cosine Rule and Area of
Surds and Logarithms 75 a Triangle 189
Summative Zone 77 Summative Zone 193


Chapter 5 Progressions 79 Chapter 10 Index Numbers 197
5.1 Arithmetic Progressions 81 10.1 Index Numbers 198
5.2 Geometric Progressions 91 10.2 Composite Index 202
Summative Zone 103 Summative Zone 210


vi




Prelims Spotlight Add Math Tg4&5.indd 6 23/04/2021 2:10 PM

FORM 5

Revision Revision
Date Date
Chapter 1 Circular Measure 215 Chapter 6 Trigonometric Functions 341

1.1 Radian 217 6.1 Positive Angles and
1.2 Arc Length of a Circle 218 Negative Angles 343
1.3 Area of Sector of a Circle 223 6.2 Trigonometric Ratios of
©PAN ASIA PUBLICATIONS
1.4 Application of Circular any Angle 345
Measures 228 6.3 Graphs of Sine, Cosine and
Summative Zone 233 Tangent Functions 351
6.4 Basic Identities 357
6.5 Addition Formulae and
Chapter 2 Differentiation 237 Double Angle Formulae 358
6.6 Application of
2.1 Limit and its Relation to
Differentiation 239 Trigonometric Functions 364
2.2 The First Derivative 242 Summative Zone 371
2.3 The Second Derivative 248
2.4 Application of Chapter 7 Linear Programming 375
Differentiation 249
Summative Zone 266 7.1 Linear Programming
Model 377
7.2 Application of
Chapter 3 Integration 268 Linear Programming 384
Summative Zone 391
3.1 Integration as the
Inverse of Differentiation 270
3.2 Indefinite Integral 270 Chapter 8 Kinematics of Linear
3.3 Definite Integral 273 Motion 394
3.4 Application of Integration 284
Summative Zone 291 8.1 Displacement, Velocity and
Acceleration as a
Function of Time 396
Chapter 4 Permutation and 8.2 Differentiation in Kinematics
Combination 295 of Linear Motion 402
8.3 Integration in Kinematics
4.1 Permutation 297 of Linear Motion 407
4.2 Combination 302 8.4 Applications of Kinematics
Summative Zone 309 of Linear Motion 409
Summative Zone 417

Chapter 5 Probability Distribution 311
SPM Model Paper 418
5.1 Random Variable 313
5.2 Binomial Distribution 317
5.3 Normal Distribution 324
Summative Zone 338 Answer 434





vii




Prelims Spotlight Add Math Tg4&5.indd 7 23/04/2021 2:10 PM

Formulae






Area of a quadrilateral: Chapter 3
FORM 4
• 1 [(x y + x y + x y + x y ) Area under the curve:
2
2 3
1 2
3 4
4 1
– (x y + x y + x y + x y )
∫
∫
b
b
Chapter 2 2 1 3 2 4 3 1 4 • y dx or x dy
a
a
2
–b ± ! b – 4ac Chapter 8
( ©PAN ASIA PUBLICATIONS
• x = Generated volume:
2a • |r| = ! x + y 2 b b
2
~
∫
∫
2
2
r
^
Chapter 4 • r = ~ • πy dx or πx dy
a
a
~
~
• a × a = a m + n |r|
m
n
Chapter 9 Chapter 4
• a ÷ a = a m – n a b c
m
n
• (a ) = a m × n • sin A = sin B = sin C • P = n!
m n
n
r (n – r)!
2
2
• ! a × ! b = ! ab • a = b + c – 2bc cos A n n!
2
2
2
2
r
!
• ! a ÷ ! b = a b • b = a + c – 2ac cos B • C = (n – r)!r!
• c = a + b – 2ab cos C
2
2
2
• log mn = log m + log n Area of a triangle: Identical object:
a a a • P = n!
m
• log = log m – log n 1 1 a!b!c!…
a n a a • ab sin C = ac sin B
2
2
n
• log m = n log m
a a 1
= bc sin A Chapter 5
log b 2
• log b = c
n
r n – r
a log a Heron’s formula: • P(X = r) = C p q , p + q = 1
c r
• ! s(s – a)(s – b)(s – c) • Mean, m = np
Chapter 5 a + b + c
s =
Arithmetic progression: 2 • s = ! npq
• T = a + (n – 1)d Chapter 10
n X – m
n
• S = [2a + (n – 1)d] Q • Z =
n 2 • I = 1 × 100 s
n
• S = [a + l] Q 0
n 2 ∑I w
• I = i i Chapter 6
Geometric progression: ∑w i
2
2
• T = ar n – 1 • sin A + cos A = 1
n n • 1 + tan A = sec A
2
2
• S = a(1 – r ) , r ≠ 1, (|r| , 1) FORM 5
n 1 – r • 1 + kot A = cosec A
2
2
a(r – 1)
n
• S = , r ≠ 1, (|r| . 1) Chapter 1 • sin (A  B) = sin A cos B 
n r – 1 cos A sin B
• S = a • Arc length, s = rq
∞ 1 – r 1 • cos (A  B) = cos A cos B 
• Area of sector, A = r q sin A sin B
2
2
Chapter 7 tan A  tan B
Chapter 2 • tan (A  B) =
Divisor of a line segment: dy 1  tan A tan B
nx + mx ny + my • y = uv, = u dv + v du • sin 2A = 2 sin A cos A
• 1 2 , 1 2 ) dx dx dx
m + n m + n du dv • cos 2A = cos A – sin A
2
2
2
Area of a triangle: • y = , dy = v dx – u dx = 2 cos A – 1
u
2
= 1 – 2 sin A
1
• [(x y + x y + x y ) v dx v 2 2 tan A
2
2 3
1 2
3 1
– (x y + x y + x y ) dy dy du • tan 2A =
2
2 1 3 2 1 3 • = × 1 – tan A
dx du dx
viii
Prelims Spotlight Add Math Tg4&5.indd 8 23/04/2021 2:10 PM

CHAPTER
1 Functions













©PAN ASIA PUBLICATIONS
Important Learning Standards Page

• Explain function using graphical representations and notations. 3

1.1 Functions • Determine domain and range of a function. 5

• Determine the image of a function when the object is given and 6
vice versa.



• Describe the outcome of composition of two functions. 9

• Determine the composite functions. 9

1.2 Composite • Determine the image of composite functions when the object is 10
Functions given and vice versa.

• Determine a related function when the composite function and 11
another function are given.

• Solve problems involving composite functions. 11



• Describe inverse of a function. 13

1.3 Inverse Functions • Make and verify conjectures related to properties of inverse 13
functions.

• Determine the inverse functions. 15








• Function / Fungsi • Range / Julat
• Relation / Hubungan • Domain / Domain
• Arrow diagram / Gambar rajah anak panah • Codomain / Kodomain
• Absolute value function / Fungsi nilai mutlak • Object / Objek
• Function notation / Tatatanda fungsi • Image / Imej
• Absolute value function graph • Composite function / Fungsi gubahan
Graf fungsi nilai mutlak • Inverse function / Fungsi songsangan
• Vertical line test / Ujian garis mencancang • Horizontal line test / Ujian garis mengufuk


1

Form
4 Additional Mathematics Chapter 1 Functions



CHAP
1



Functions




Functions Composite Functions Inverse Functions
©PAN ASIA PUBLICATIONS


A relation where every fg(x) = f [g(x)] Properties of inverse function
object in the domain has (a) A function f that maps set
one and only image in fg A to set B has an inverse
the codomain. function f if f is a one-to-
–1
g f one function.
(b) fg(x) = x where x in the
x g(x) f[g(x)]
Function notation domain of g and gf(x) = x
f : x → y where x in the domain of f.
f(x) = y (c) If two functions, f and g are
inverse to each other, then
gf(x) = g[f(x)] (i) domain of f = range of
Vertical line test gf g and domain of g =
y range of f,
(ii) graph g is a reflection
Vertical line test
f g of graph f.
(d) For any real numbers, a and
f
x f(x) g[f(x)]
b, if the point (a, b) is on
the graph f, then the point
x
0 (b, a) is on the graph g.
Graph is a function
Horizontal line test
Absolute value function y
y f
Horizontal line test
f(x) = |x|
x
x 0
0
x if x  0
|x| =  –x if x  0 Graph is a function


If f : x → y, then f : y → x
–1
Discrete function Continuous function
f(x) f(x)




x x
0 0


2

Form
4
Chapter 1 Functions Additional Mathematics

1.1 Functions 7. If the line intersects the graph at only one CHAP
point, then the graph is a function.
Explaining function by using graphical y 1
representation and notation Vertical line test
f
1. Function is a special relation where for every
object in domain, there is one and only image
in codomain. x
0
2. For example, every element in set X has only
one image in set Y.
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8. If the line intersects the graph at more than
a • • e one points, then the graph is not a function.
b • • f y
c • • g Vertical line test
d • • h
f
Set X Set Y
3. A function f, which maps x to (mx + n) can be 0 x
written as

f : x → mx + n or f(x) = mx + n
9. An absolute value function is a function
where x is the object and f(x) or (mx + n) is the that contains an algebraic expression within
image if x is under the function f. absolute value symbols.
4. Function is a relation of one-to-one or a relation 10. The absolute value function of x is written as
of many-to-one. f(x) = |x| and is defined by
(a) One-to-one relation
x if x  0
f(x) f(x) =  0 if x = 0
–x if x  0
f
f
y x • • y
2 1 1 11. The graph of a linear absolute value function
y
1 x • • y has a V-shaped.
2 2
y
x
0 x x
1 2
(b) Many-to-one relation f(x) = |x|

f(x) x
y = f(x) 0
f
x • 12. The discrete function is a function where the
y 1 • y points on the graph is real, separated and not
x •
2
connected by a straight line or curve.
f(x)
x
x 0 x
1 2
5. If given a graph of a relation, we can determine
whether the graph is a function or not by using
a vertical line test.
6. The vertical line test can be tested by drawing a x
line vertically on the graph. 0

1.1.1 3

Form
4 Additional Mathematics Chapter 1 Functions


13. The continuous function is a function where
CHAP the points on the graph are connected by a Example 2
1 straight line of curve. By using function notation, express f in terms of

f(x) x for each of the following relations.
(a) f
–4 1
2 7
5 10
7 12
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(b) f
x x y
0
–2 •
–1 • • 1
Example 1 1 • • 4
2 •
Determine whether the following relation is a
function or not. Justify your answer.
(a) Solution:
• 10 (a) f : x → x + 5 or f(x) = x + 5
3 • 2 2
• 16 (b) f : x → x or f(x) = x
5 •
• 60 Try question 2 in Formative Zone 1.1
7 •
• 62

(b) Example 3
–1 • By using vertical line test, determine whether
• 1
2 • each of the following graphs is a function or not.
• 4
4 • (a) (b)
• 36 y y
6 •

(c)
• 3 2 x
p • 0
• 6 x
q • 0 2
• 9
r • Solution:
• 12
(a)
(d) y
• 10
10 • Vertical line test
• x
20 •
• y 2
30 •
• z x
0 2
Solution: The graph is a function.
(a) The relation is a function because each (b)
object has only one image. y Vertical line test
(b) The relation is not a function because there
is one object does not have any image.
(c) The relation is not a function because there x
is one object that has two images. 0
(d) The relation is a function because each
object has only one image. The graph is not a function.
Try question 1 in Formative Zone 1.1 Try question 3 in Formative Zone 1.1


4 1.1.1

Form
4 Additional Mathematics Chapter 2 Quadratic Functions


Example 26 (iii) When a  0,
f(x)
Express the quadratic function f(x) = 2(x − 4) − 8
2
in the form of f(x) = a(x − p)(x − q) where a, p 0 x
and q are constants, and q < p. Hence, state the
CHAP values of a, p and q. a = –0.5
2 Solution:

f(x) = 2(x − 4) − 8
2
= 2(x − 8x + 16) − 8 Convert to a = –2 a = –1
2
©PAN ASIA PUBLICATIONS
= 2x − 16x + 32 − 8 general form first • the shape of the graph is ∩,
2
2
= 2x − 16x + 24
• the width of the graph is increasing
= 2(x − 8x + 12) Then, convert to when the value of a is increasing and
2
= 2(x − 2)(x − 6) intercept form vice versa.
Thus, a = 2, p = 6 and q = 2. (b) Changing the values of h
(i) The horizontal movement of the
Al te rnative Me thod graph and the position of axis of
Alternative Method
f(x) = 2(x − 4) − 8 symmetry will be affected.
2
= 2[(x − 4) – 2 ] (ii) When the value of h increases, the
2
2
2
2
Use a − b = (a + b)(a − b) graph will move to the right.
a = (x − 4) and b = −2 (iii) When the value of h decreases, the
f(x) = 2(x − 4 − 2)(x − 4 + 2)
= 2(x − 6)(x − 2) graph will move to the left.
f(x)
Try question 11 in Formative Zone 2.3
h = –2 h = –1 h = 1 h = 2 h = 3
Analysing and making generalisation about
the changes of a, h and k in quadratic
2
functions f (x) = a(x – h) + k towards the
shape and position of the graphs 0 x
1. The change of values of a, h and k in quadratic (c) Changing the values of k
functions f(x) = a(x – h) + k will affect the (i) The vertical movement of the graph
2
shape and position of the graph. and the maximum or minimum
(a) Changing the value of a values of the graph will be affected.
(i) The shape and the width of the graph (ii) When the value of k increases, the
will be affected. graph will move upwards.
(ii) When a  0, (iii) When the value of k decreases, the
graph will move downwards.
f(x)
a = 1 f(x)
a = 2
a = 0.5 k = 2

k = 1

k = –1
x
0 k = –2
• the shape of the graph is ∪, k = –3
• the width of the graph is decreasing x
when the value of a is increasing and 0
vice versa.


36 2.3.3 2.3.4

Form
4
Chapter 2 Quadratic Functions Additional Mathematics

Example 27 (c) The graph with the same shape moves
vertically 3 units upwards and the maximum
The diagram below shows the quadratic function value becomes 6. However, the axis of
f(x) = a(x − h) + k. symmetry remains unchanged.
2
f(x)
f(x) CHAP
(2, 6) 2
(2, 3)
(2, 3)
x
0 –1 x
0 –1
2
f(x) = –(x – 2) + 6
©PAN ASIA PUBLICATIONS
2
f(x) = –(x – 2) + 3
Find the values of h, k and a. Hence, make
generalisation on the effect of change in each
of the following values towards the shape and Try question 3 in Formative Zone 2.3
position of the graph.
(a) The value of a changes to −5.
(b) The value of h changes to −1. Sketching graphs of quadratic functions
(c) The value of k changes to 6.
Solution: 1. The graph of a quadratic function
f(x) = ax + bx + c can be sketched based on the
2
From the graph, we know that h = 2 and following steps:
k = 3. By substituting the coordinates (0, −1),
h = 2 and k = 3 into the quadratic equation Determine the shape of the graph is
f(x) = a(x − h) + k, we get ∪ or ∩ by determining the value of a.
2
−1 = a(0 − 2) + 3
2
−1 = 4a + 3
4a = −4
a = −1 Determine the position of the graph
Thus, a = −1, h = 2 and k = 3. relative to the x-axis by determining the
(a) The shape of the graph and the minimum value of discriminant, b2 – 4ac.
point remain unchanged. However, the
width of the graph decreases.
f(x)
Determine the vertex point
(maximum or minimum point).
x
0
2
f(x) = –(x – 2) + 3
f(x) = –5(x – 2) + 3
2
Determine the intersection point between
the graph and x-axis by solving the
(b) The graph with the same shape moves equation f(x) = 0.
horizontally 3 units to the left and the
equation of axis of symmetry becomes
x = −1. However, the maximum value
remains unchanged. Determine y-intercept by finding the
value f(0).
f(x)
(–1, 3) (2, 3)
x
0 Plot all the points obtained on a Cartesion
–1
f(x) = –(x – 2) + 3 plane. Hence, sketch a smooth parabolic,
2
where symmetry with horizontal line that
. passess through the vertex of graph
2
f(x) = –(x + 1) + 3
2.3.4 2.3.5 37

Form
4 Additional Mathematics Chapter 4 Indices, Surds and Logarithms


4.1 Laws of Indices 1
2
(c) × 32 = 3 × 3
−4
Simplifying algebraic expressions 81 = 3 −4 + 2

involving indices using the laws of indices = 3 −2
1
1. A number can be written in index notation or = 3 2
index form as a where a is a base and n is an 1
n
index or power. = 9
2. An algebraic expression can be simplified using 1 1
the laws of indices. (d) 2 × 32 ÷ 16 = 2 × (2 ) ÷ 2
5
5
5
5 5
4
= ©PAN ASIA PUBLICATIONS
1
4
5
3. The following indices should be known first = 2 × 2 ÷ 2
5 + 1 − 4
before we simplified the algebraic expressions = 2 2
= 2
CHAP using the laws of indices. = 4
4 • a = 1 Try question 1 in Formative Zone 4.1
–1
a (Negative index)
• a = 1 (Zero index) Calculator
0
1
n
• a = a (Fractional index) Based on Example 1(a), the algebraic expression
n
5
2 × 16 can be evaluated using calculator as follows:
−2
m
• a = ( a) m (Fractional index)
n
n
1. Press 2 ^ 5 × 16 ^ (–) 2 =
4. The following shows the laws of indices: 2. The answer will be displayed as 0.125 .
b
• a × a = a n + m 3. Press SHIFT a  to convert the answer to
c
n
m
• a ÷ a = a n – m fraction 1 8 .
m
n
• (a ) = a nm
n m
• (ab) = a b Example 2
n n
n
a n
•   = a b n n Simplify the following algebraic expressions.
b
3
(a) 3 × 27 3 − n (b) 2 n + 1 ÷ 4 × 16 4
2
2
(2p) × p −2 (3x y )
3
3 4 2
Example 1 (c) pq 3 (d) 12x y
5
By using the laws of indices, evaluate the
following algebraic expressions. Solution:
2
2
3 3 − n
(a) 2 × 16 (b) (5 ÷ 5 ) (a) 3 × 27 3 − n = 3 × (3 )
−3
−4 2
−2
5
2
1 1 = 3 × 3 9 − 3n
(c) × 32 (d) 2 × 32 ÷ 16 = 3 2 + 9 − 3n
5
5
81
= 3 11 − 3n
Solution: 3 3
(b) 2 n + 1 ÷ 4 × 16 = 2 n + 1 ÷ (2 ) × (2 )
4
2
2 2
4 4
4 −2
(a) 2 × 16 = 2 × (2 ) = 2 n + 1 − 4 + 3
−2
5
5
= 2 × 2 = 2 n
5
−8
= 2 5 + (−8) a × a = a m+n (2p) × p −2 2 p × p −2
m
n
3
3
3
−3
= 2 (c) pq 3 = pq 3
1
= 8p × p −2
3
2 3 =
1 pq 3
8 = 8p 3−2
pq 3
(b) (5 ÷ 5 ) = (5 ) ÷ (5 )
−3
−3 2
−4 2
−4 2
= 5 ÷ 5 −8 = 8p 3
−6
= 5 −6 − (−8) a ÷ a = a m−n pq
m
n
= 5 2 = 8
= 25 q 3
60 4.1.1

Form
4
Chapter 4 Indices, Surds and Logarithms Additional Mathematics


3 2
4 2
(3x y ) 3 (x ) (y ) (a b ) = a b np Solution:
3 4 2
2
mp
n p
m
(d) =
5
12x y 12x y (a) 7 x + 2 − 7 x + 1 = 6
5
9x y 7 ·7 − 7 ·7 = 6
1
x
x
2
6 8
=
12x y 7 (7 − 7 ) = 6
2
x
1
5
x
3 7 (49 − 7) = 6
y
= x 6 − 5 8 − 1 x
4 7 (42) = 6
3 x 6
= xy 7 7 =
4 42
1 Express the equation
©PAN ASIA PUBLICATIONS
x
7 = of both sided in the
7 same base
Try question 2 in Formative Zone 4.1
7 = 7 −1
x
x = −1 CHAP
4
Example 3
x
–1
Simplify 2 + 2 n + 2 − 2 n + 1 in the form of k(2 ) • If 7 = 7 , then x = –1.
n
n
7
7
where k is a constant. Hence, determine the • If x = 4 , then x = 4.
value of k.
Solution: (b) 81·(3 x + 2 ) = 1
2 + 2 n + 2 − 2 n + 1 = 2 + 2 ·(2 − 2 ) 3 ·(3 x + 2 ) = 1
n
1
2
4
n
n
n
0
1
= 2 (1 + 2 − 2 ) 3 4 + x + 2 = 1 3 = 1
2
= 2 (1 + 4 − 2) 3 6 + x = 3°
n
= 2 (3) 6 + x = 0
n
= 3(2 ) x = −6
n
Thus, k = 3.
Al te rnative Me thod
Alternative Method
Let u = 2 , then
n
n
2 + 2 n + 2 − 2 n + 1 = 2 + 2 ·2 − 2 ·2 1 Change 1, 3 and 81 to a common base, that
n
2
n
n
= u + 4u − 2u is 3. Hence, equate the indices to solve for x.
= 3u
= 3(2 )
n
Try question 5 in Formative Zone 4.1
Try question 3 and 4 in Formative Zone 4.1
Example 5
Show that 5 n + 3 − 5 − 5 n + 2 is divisible by 11
n
Solving problems involving indices for all positive integers of n.
1. When solving an equation involving indices,
we should consider the following statement: Solution:
n
n
3
n
If a = a , then m = n or if a = b , then 5 n + 3 − 5 − 5 n + 2 = 5 ·5 − 5 − 5 ·5 2
n
n
m
m
m
a = b when a > 0 and a ≠ 1. = 5 (5 − 1 − 5 )
n
2
3
= 5 (125 − 1 − 25)
n
= 5 (99)
n
Example 4
Since 5n(99) is divisible by 11. Thus,
Solve the following equations. 5 n + 3 − 5 − 5 n + 2 is divisible by 11.
n
(a) 7x + 2 – 7x + 1 = 6
(b) 81·(3x + 2) = 1 Try question 6 in Formative Zone 4.1
4.1.1 4.1.2 61

Form
4 Additional Mathematics Chapter 5 Progressions



5.2


1. Determine whether the following sequence is 9. The 5th and 8th term of a geometric
a geometric progression or not. C1 1 3
progression are 337 and 42 respectively.
1
(a) 18, 27, 40 , … Find C2 2 16
2
(a) the first term and the common ratio,
1 13
(b) 125, 31 , 7 , … (b) the sum of the 7th and 8th term,
4 16
of the geometric progression.
©PAN ASIA PUBLICATIONS
3
(c) 16, 2 , 9 , …
5 25 10. It is given that 3 and 243 are the 2nd term
3 and 6th term of a geometric progression
(d) –36, 12 , –4.41, … respectively. Find C2
5
(a) the first term and the common ratio,
2. Given that the sequence below is a geometric (b) the sum of the 2nd and 3rd term,
progression. Determine the value of common of the geometric progression.
ratio. C1
CHAP 11. Find the number of terms of the following
5 (a) –144, 72, –36, … geometric progression. C2
(b) –12, –30, –75, …
1
1 (a) 2, 3, 4 , …, 15 3
(c) 8, 1, , …
8 2 16
(d) 15, 3, 0.6, … (b) 120, 60, 30, …, 15
16
3. Given that 3h, 10h – 6 and 25h + 12 are the (c) 12k, 36k, …, 8 748k
1
3 terms in a geometric progression, find the (d) 3 , 2 , …, 57 681
possible values of h. C2 2 4 1 024
4. Find the possible values of k if 2k + 6, 8k – 8 12. It is given that 24, 18, 27 , … is a geometric
and 12k + 4 are the 3 terms of a geometric 2
progression. C2 progression. Determine which term in the
geometric progression is 3 417 . C2
2 048
5. It is given that 4p – 2, 6p + 3 and 12p + 18 are
the 3 terms of a geometric progression. Find 13. Which term in the geometric progression 4,
the value of p. C2 12, 36, … is 26 244? C2

6. Given that 32, 8 and 2 are the first 3 terms of 14. Determine the term in the geometric
a geometric progression. Find C2 progression 28, 7, 1 , … which is equal to
3
(a) the first term and the common ratio, 4
(b) the 8th term. 7 . C2
262 144
3 22
7. It is given that 2p, –21 and –38 are the 15. It is given that k, 18, 54 and h are the four
5 25 terms in a geometric progression. Find the
first 3 terms of a geometric progression. Find value of k and h. C2
C2
(a) the common ratio of the geometric
progression, 16. Given that 10, p, q, 80 are the first four terms
(b) the value of p, in a geometric progression, find the value of p
(c) the 15th term of the geometric and q. C2
progression.
17. Given that u, 15, v, w are the first four terms in
a geometric progression, express v in terms of
8. It is given that 448, –896 and 1 792 are the 6th,
7th and 8th term of a geometric progression. u and w. C3
Find C2
(a) the first term and the common ratio, 18. Given that 4, 10, 25, … is a geometric
(b) the 11th term, progression, what is the first term which its
of the geometric progression. value more than 1 000? C3

98

Form
4
Chapter 5 Progressions Additional Mathematics


1 3 30. It is given that the sum of the first n terms of a
19. It is given that 6, 7 , 9 , … is a geometric
2 8 geometric progression is S = 3 [3 − 1]. C2
2n
progression. Find the minimum value of n if n 8
the value T > 300. C3 (a) Find the sum of the first 5 terms.
n
(b) Express the n term of the geometric
20. The 2nd and 5th terms of a geometric progression, in term of n.
17
progression are 10 and 19 . Find the first
32 31. Given that the sum of the first n terms of the
term which is more than 100. C3 geometric progression 2, −4, 8, … is 21 846,
find the value of n. C2
©PAN ASIA PUBLICATIONS
21. Find the sum of first terms of the following
geometric progression. C1 32. It is given the sequence 3, 1, 1 , … is a
(a) 1, 2, 4, …, to n = 13 3
(b) 5, 15, …, to n = 14 geometric progression. Find the value of
(c) 12, –24, 48, …, to n = 12 n when the sum of the first n terms of the
3
2
(d) 3, 3 , 3 , …, to n = 8 29 524
geometric progression is 4 . C2
59 049
7
22. It is given that , –7, 14, … is a geometric
2 33. Determine the value of n when the sum of the CHAP
progression. Find the sum from 3rd term to 8th first n of the geometric progression 125, –25, 5
term of the geometric progression. C2 104
5, … is 104 . C2
625
23. Find the sum from 4th term to 9th term of the
geometric progression 4, –6, 9, … C2 34. It is given that the 2nd term and the 5th
term of a geometric progression are –27 and
24. It is given that 9, –18, 36, … is a geometric 1 respectively. Find C3
progression. Find the sum from 6th term to
18th term of the geometric progression. C2 (a) the first term and the common ratio,
(b) the sum of the first 13 terms,
25. It is given that the nth term of a geometric of the geometric progression.
5
progression is T = – (–2) . Find C2
n
n 35. Given that the 4th term and the 10th term
2
1
(a) the first term and the common ratio, of a geometric progression are –16 and –
(b) the sum of the first 6 terms, respectively. Find C3 4
of the geometric progression.
(a) the first term and the common ratio,
26. Given that the nth term of a geometric (b) the sum from the 5th term to the 12th
progression is T = 3 3 – n , find the sum of the term,
n
first 10 terms. C2 of the geometric progression.
27. It is given that the nth term of a geometric 36. In a geometric progression, the 3rd term and
625 the sum of 4th and 5th term are 60 and 360
progression is T = . Find the sum from
n n respectively. Find C3
5
the 5th term to the 12th term of the geometric (a) the first term and the common ratio,
progression. C2
(b) the sum of the first 11 terms,
of the geometric progression.
28. The sum of the first nth terms of a geometric
2 2n + 1 – 2
progression is given by S = . Find 37. Find the sum to infinity of the following
n
3 C2 geometry progression. C2
(a) the sum of the first 7 terms,

1
(b) the 8th term, (a) 120, 80, 53 , …
of the geometric progression. 3
2 7
(b) 8, 2 , 1 , …
29. Given that the sum of the first nth terms of a 3 9
geometric progression is given by S = 4(2 n – 1 ). (c) 1 , 1 1 , 1 1 2 , …
   
n
Find C2 5 5 2 5 2
(a) the sum from the 3rd to 7th term, (d) 3x, 1, 1 , …
(b) the nth term of the geometric progression 3x
in terms of n. (in terms of x)
99

Form
4 Additional Mathematics Chapter 6 Linear Law


SPM Simulation HOTS Questions


Paper 1
1. The variable x and y are related by equation y = mx − nx Divide both
2
y = mx – nx where m and n are constant. x side with x .
2
2
x y – y n
y x 2 x 3 = m −
x
Compare to Y = mX + c.
x k + 2
O y 1
©PAN ASIA PUBLICATIONS
Hence Y = , X = , m = –n and c = m.
x 3 x
From the graph, c = k + 2
x m = k + 2
–3k O
k = m − 2 …❷
Diagram (a) To express m in terms of n, eliminate
variable k.
y n
y x – 2 As ❶ = ❷, m − 2 =
3
3 m − 6 = n
x k + 2
n
O m is the m = + 6
subject 3
CHAP 2. The variable x and y are related by an equation
6 O
2
–3k x y = 10 2x + 1 . The diagram below shows a
straight line graph obtained by plotting
Diagram (b) log y against x .
2
10
Diagram (a) and diagram (b) show the straight log y
line graphs obtained by plotting the relations 10
from the equation. Express m in terms of n. (p, 9)
C4
Examiner's Comment:
q
In diagram (a) x 2
O
x is the X-intercept.
y Find the values of p and q. C3
Then, reduce = mx − nx to get x as the
2
x Examiner's Comment:
variable of X. Convert the equation y = 10 2x + 1 into the
2

y = mx − nx Divide both linear form, Y = mX + c.
2
2
x side with x. y = 10 2x + 1 Taking logarithm
y = mx − n log y = log 10 2x + 1 to the base of 10
2
to both side.
x 2 10 10
2
Compare to Y = mX + c. log y = (2x + 1) log 10
10
10
2
y log y = 2x + 1
10
Hence Y = , X = x, m = m and c = −n.
x 2 Compare to Y = mX + c.
From the graph, c = −3k Hence Y = log y, X = x , m = 2 and c = 1.
2
− n = −3k 10
n From the graph, q = y-intercept
k = …❶
3 = 1
In diagram (b) Gradient, m = 2
9 – 1
y is the y-intercept. = 2
x 3 p – 0 p = 8
2
y y
Then, reduce = mx − nx to get as p = 4
2
x x 3
variable of y. Thus, p = 4 and q = 1.
118

Form
4
Chapter 6 Linear Law Additional Mathematics


3. The diagram below shows a part of graph y Paper 2
–x
against x for equation y = pq , where p and q 4. The table below shows the value of variables,
are constant. C4 x and y, obtain from an experiments. The
variables x and y are related by equation
y hy – k
x = where h and k are constants. C4
x
x 0.5 1.0 1.5 2.0 2.5
y = pq –x
y 14 24 32 48 69
©PAN ASIA PUBLICATIONS
(a) Plot a graph of y against x using a scale
2
of 2 cm to 10 units on y-axis and 2 cm to
1 unit on x -axis. Hence, draw the line of
2
x best fit.
O
(b) From the graph from (b), find the values of
(i) h and k,
(a) Find the possible equation of the line of (ii) x when y = 43.
best fit for the non-linear graph.
(b) Based on your answer in (a), find the value Examiner's Comment:
of p and q, if y-intercept is 2 and gradient (a) Build a table of x .
2
is –3.
x 0.5 1.0 1.5 2.0 2.5 CHAP
Examiner's Comment: x 2 0.25 1.0 2.25 4.0 6.25 6

(a) Convert the equation y = pq into the y 14 21 32 48 69
–x

linear form, Y = mX + c.
Graph y against x 2
y = pq –x Taking logarithm to the y
base of 10 to both side.
log y = log p + log q –x 70
10 10 10
log y = log p – x log q 60
10
10
10
log y = –log q(x) + log p
10 10 10 50
Thus, the possible equation of line of 40
best fit is
log y = –log q(x) + log p. 30
10 10 10
20
(b) Compare to Y = mX + c.
10
log y = –log q(x) + log p
10 10 10 x 2
Hence, Y = log y, X = x, m = – log q 0 1 2 3 4 5 6
10
10
and c = log p.
10
When y-intercept = 2 (b) Convert the equation of x = hy – k into
log p = 2 x
10 linear form, Y = mX + c.
p = 10 2 hy – k
= 100 x = x
2
When gradient, m = –3 x = hy – k
2
– log q = –3 hy = x + k
10
log q = 3
10 1 2 k
q = 10 3 y =   x + h
h
= 1 000
Compare to Y = mX + c.
Thus, the values of p and q are 100 and
k
1
1 000 respectively. Hence, Y = y, X = x , m = and c = .
2
h h
119

Form
4 Additional Mathematics Chapter 8 Vectors









Paper 1

1. The diagram shows a quadrilateral ABCD. 5. p = 3a + 4b
→
→
→
C2 It is given AB = ha, CD = ka, CA = hb and C2 ∼ ∼ ∼
q = 2a – b
r = ha + (h – k)b where h and k are constants.
→ ∼ ∼ ∼ ∼ ∼ ∼
©PAN ASIA PUBLICATIONS
DB = 2a + (k + 8)b. ∼ ∼ ∼
∼
∼
A
C Use the information above to find the values of
h and k when r = 4p – 2q.
∼
∼
∼
6. Three points lie on a Cartesian plane which are
C2 the origin, O, K(–2, 6) and L(5, 7). Find
→ x
D (a) KL in terms of   .
B y
→
Find the values of h and k. (b) the unit vector in the direction of KL.
7. The points P, Q and R are collinear. It is given
2. It is given that p = 9i – 6j and q = i – 4j, C2 → ∼ → ∼
PQ = 3i – 2j and QR = (1 – m)i + 4j, determine
∼
∼
∼
C2 1 ∼ ∼ ∼ ∼ the value of m. ∼
p – 2q .
determine  3 ∼ ∼ 
→ →
8. The diagram shows OP = a and OQ = b.
→ → ∼ ∼
3. The diagram shows vector BA, vector BC and C3
→
C2 point D lie on a square grid. It is given BA = u 4
→
CHAP and BC = v. ∼ 3 Q
8 ∼ 2
P 1
C –4 –3 –2 –1 0 1 2
A
On the same square grid, draw the line that
→
B represents the vector OR = 3a + b.
∼
∼
→
→
D 9. The diagram shows two vectors OB and CD
C3 that are parallel to each other.
Express in terms of u and v for each of the y
∼
∼
following vectors. 6
→
(a) BD 5 D
→
(b) DC 4
3 C
→ → 2 B
4. It is given that OA = –3i + 5j and OB = 2i – 7j,
∼
∼
C2 find ∼ ∼ 1
→ x
(a) AB in terms of i + j, –1 O 1 2 3 4 5 6 7 8
∼
∼
→
(b) the unit vector in the direction of AB. It is given that OB = (k + 2)i + 3j and
→
→ ∼ ∼
CD = (2k – 1)i + 4j. Find the value of k.
∼
∼
172

Form
4
Chapter 8 Vectors Additional Mathematics

10. The diagram shows a triangle ABO. 13. The diagram shows a Cartesian plane where O
C3 C4 is Arif’s house, P is a shop and Q is a school.
B y
Q

C


O A
D
P
©PAN ASIA PUBLICATIONS
→ →
It is given that 2BC = 3CO, D is the midpoint
→ → →
of OA, AB = 3u and AO = 4v. Express DC in
∼
∼
terms of u and v.
∼
∼
x
–2 O 2 4 6 8 10
11. The diagram shows a parallelogram ABCD.
C3 Point Q is the intersection of the diagonals.
The shortest distance between Arif’s house and
the shop is 10 km while the shortest distance
B C between the shop and the school is 8 km.
Express the vector of Arif’s house to the school
in terms of a + b.
Q ∼ ∼
14. The points A, B and C are collinear. It is given
→
→
A D C3 that AB = 3x + 2y and BC = (1 – m)x + 4y
∼
∼
such that m is a constant, find ∼ ∼
→ →
It is given that DA = 2x + 3y and AB = 6x – y, (a) the value of m,
∼
∼
→ ∼ ∼
find CQ. (b) the ratio of AB : BC. CHAP
8
→ 15. The diagram shows a regular octagon.
12. The diagram shows OB = 8i – hj and C3
∼
→
C3 BC = 3i – 2j. ∼ F E
∼
∼
p
∼
y G D
q
∼
H C
B
r
∼
A B
C
(a) State the vectors that are equal.
→
(b) Express FA in terms of p, q and r.
∼
x ∼ ∼
O
16. A toy truck was moved at a constant velocity
C3 from point O to point P. It is given that
→
→ OP = (36i – 24j) m and the time taken is
∼
It is given |OB| = 10 units, find 6 seconds, find ∼
(a) the value of h, (a) the velocity of the toy truck,
(b) coordinates of C. (b) the speed of the toy truck.
173

Form
4 Additional Mathematics Chapter 8 Vectors



→
→
→
SPM Clone
SPM Clone
17. The diagram shows a regular decagon with 20. The diagram shows vectors AC , AB and AD
center O. drawn on a square grid with sides of 1 unit.
C3 C3
G F
A
H E
D
I O D
C
B
G C
A B
→ → → → → →
(a) Express BC + CH + HJ + FE as a single (a) It is given that AB = p and AC = q,
∼
∼
→
vector. express AD in terms of mp + nq, where m
→ → ∼ ∼
(b) It is given OE = x, OD = y and the length and n are constants.
∼
∼
of each side of the decagon is 2 units. Find (b) On the diagram above, mark and label
→ → → →
the unit vector in the direction of ED , in point E such that DE + AC = 3AB.
terms of x and y.
∼
∼
21. A school bus depart from Kampung Merbau
→
→
→
SPM Clone
18. The diagram shows vectors QP , QS and QR C4 with a position vector (40i + 30j) km to
∼
∼
SMK Saujana Utama with a position vector
C3 drawn on a square grid with sides of 1 unit. (25i + 16j) km. The school bus travels with
∼
∼
constant velocity and take 1 hour back and
forth to take and drop off students. Find the
Q velocity of the vector.
R
–6
7
∼    
CHAP ©PAN ASIA PUBLICATIONS
22. The equation u =
+ t
represents the
S
8 C4 5 8
P path of a moving particle where t is in seconds
and the distance is in meter. Find
→
(a) Find |– PQ |. (a) the initial position of the particle,
→ → (b) the velocity of the particle,
(b) It is given that QP = x and QS = y,
∼
express in terms of x and y, ∼ (c) the speed of the particle.
→ ∼ ∼ →
(i) PS (ii) QR .
23. In an archery competition, an arrow hits
C4 3 points on archery’s board. The points have
SPM Clone
19. It is given that K(2, m), L(n, 7), p = 7i + 4j, position vectors that marked E, F and G are
∼
∼
∼
→
∼
∼
C3 q = –i – 8j and KL = p – hq where m, n and h ∼ i – j, 5i – 3j and 11i – 6j. Show that E, F and
∼
∼
∼
∼
∼
∼
∼
∼
are constants, express n in terms of m. G are collinear.
174

Form
4
Chapter 9 Solution of Triangles Additional Mathematics


9.2


1. The diagram below shows a triangle ABC. 5. The diagram below shows a quadrilateral ABCD.
D
C
8.1 cm 3.8 cm
x
A C
B 72°
9.6 cm 6.3 cm
©PAN ASIA PUBLICATIONS
4.2 cm 5.7 cm
29°
B
Given that AB = 9.6 cm, BC = 6.3 cm,
A CD = 3.8 cm, AD = 8.1 cm and ∠BCD = 72°.
Find C3
Given that AB = 4.2 cm, AC = 5.7 cm and (a) the length of BD,
∠BAC = 29°, find the value of x. C2
(b) ∠ADB.
6. The diagram below shows a triangle KMN.
2. The diagram below shows a triangle PQR.
Given that KLM is a straight line.
11.2 cm R 2.7 cm
Q L M
14.1 cm
124°
8.1 cm y K 9.8 cm 11.3 cm


x
P
N
Given that PQ = 8.1 cm, QR = 11.2 cm and If KL = 14.1 cm, LM = 2.7 cm, MN = 11.3 cm
∠PQR = 124°, find the value of y. C2 and LN = 9.8 cm, find the value of x. C3
7. In a shooting practice, Syafiq was required to
3. The diagram below shows a triangle ABC. shoot on two boards practice such as board A
is 50 meters and board B is 75 meters from a
B shooting point respectively. If the angle formed
5.7 cm between boards A and B at the shooting point CHAP
is 32°, find the distance between board A and 9
A 6.3 cm
board B. C4
7.4 cm θ
Board
C
B
Given that AB = 5.7 cm, BC = 6.3 cm and 75 m
AC = 7.4 cm, find the value of θ. C2 Board
A
50 m 32°
4. The diagram below shows a triangle KLM.
Shooting
point
K
15.1 cm 8. Zaki drove a boat 25 km to the Nibong's
jetty at the bearing of 118° from Sekinchan's
11.2 cm lighthouse. Then, he drove the boat 38 km to
θ L
Redang Island at the bearing of 232° from the
6.7 cm
M Nibong's jetty. Calculate the distance between
Sekinchan's lighthouse and Redang Island.
Given that KL = 15.1 cm, LM = 6.7 cm and C5
KM = 11.2 cm, find the value of θ. C2

185

Form
4
Chapter 9 Solution of Triangles Additional Mathematics


5. The diagram below shows the position of 6. The diagram below shows the position of two
rambutan tree (R), durian tree (D), mango tree ports, A and B and a ship sailing in an ocean.
(M) and a hut in an orchard.


17 m
Durian Mango
Rambutan

13 m 9 m
1 ©PAN ASIA PUBLICATIONS
67°
32°
Port 25 km Port
Hut A B

Given that the distance of durian tree and The distance from port A to port B is 25 km.
mango tree from the hut are the same. Find the Find the distance between the ship and
distance between durian tree and rambutan port B. C3
tree. C4






SPM Simulation HOTS Questions


Paper 2
1. The diagram below shows a quadrilateral (b) Length of BD can be find by using
ABCD. cosine rule.
2
BD = CD + BC − 2(BC)(CD) cos ∠BCD
2
2
= 6 + 7 − 2(6)(7) cos 72.25°
2
2
B
6 cm = 59.39
C
53° BD = 59.39
= 7.71 cm
7 cm CHAP
(c) Using sine rule, 9
D sin ∠BAD sin 53°
10 cm =
A
7.71 10
The area of ΔBCD is 20 cm and ∠BCD is an sin ∠BAD = sin 53° × 7.71
2
10
acute angled. Calculate C4 = 0.6157
(a) ∠BCD, ∠BAD = 38°
(b) length BD,
(c) ∠ADB, and Thus,
2
(d) the area, in cm , of quadrilateral ABCD. ∠ADB = 180° − 53° − 38°
= 89°
Examiner's Comment:
1
(a) Given ΔBCD = 20 cm2 (d) Area ΔABD = (10)(7.71) sin 89°
2
(6)(7) sin ∠BCD = 20 = 38.54 cm 2
2 2
sin ∠BCD = 20 × Area of equilateral ABCD
42 = Area ΔABD + Area ΔBCD
= 0.9524 = 38.54 + 20
∠BCD = 72.25° = 58.54 cm 2
191

Form
4
Chapter 10 Index Numbers Additional Mathematics

Example 11 If the composite index of the four items in the
year 2019 based on the year 2015 was 136,
The table below shows the price indices for the calculate the value of p.
four ingredients, A, B, C and D used in making a
cake in 2017 based on 2014. Solution:

Ingredient A B C D ∑I w = 115(2) + 150(3) + 135p + 140(1)
i
i
= 820 + 135p
Price index in 2017 80 140 150 100 ∑w = 2 + 3 + p + 1
based on 2014 i
= 6 + p
Bar chart below represents the relative amount ∑Iw i –
©PAN ASIA PUBLICATIONS
i
of the ingredient A, B, C and D, used in making ∑w = I
a cake. i
820 + 135p = 136
Weightage
6 + p
820 + 135p = 136(6 + p)
140
820 + 135p = 816 + 136p
120
p = 4
80
Try question 5 in Formative Zone 10.2
60
Ingredients
0 A B C D
Calculate the composite index for the cost of Solving problems involving index numbers
making a cake in the year 2017 based on the
year 2014. and composite index
Example 13
Solution:
∑I w = 80(140) + 140(80) + 150(120) + 100(60) The table below shows the price indices of
i
i
= 46 400 four raw materials, P, Q, R and S which used to
∑w = 140 + 80 + 120 + 60 produce a tin of drink A in a factory.
i
= 400 Raw Price index in 2019
– ∑Iw material based on 2017
I = i i
∑w
i P 120
46 400
=
400 Q 145
= 116 R
Thus, the composite index for the cost of making
a cake in the year 2017 based on the year 2014 S 130
was 116. Pie chart represents the proportion of the raw CHAP
Try question 4 in Formative Zone 10.2 material used in the drink. 10

P Q
Example 12
85°
140°
The table below shows the price indices and 20°
their corresponding weightage of four items, E, 115° R
F, G and H in 2019 using 2015 as the basis year.
S
Price index in 2019
Item Weightage (a) If the composite index for the cost of
based on 2015
making a tin of drink A in the year 2019
E 115 2 based on the year 2017 is 132, what is the
F 150 3 price index of raw material R?
(b) If the price of raw material R in the year
G 135 p
2017 was RM15, calculate the price of raw
H 140 1 material R in the year 2019.

10.2.1 10.2.2 203

Form
4
Chapter 10 Index Numbers Additional Mathematics


(d) Solution:
2015 2017 2019 (a) For salt,
1.00
123.15 116 x = × 100
0.80
116 = 125
I2019 = × 123.15
2015 100 For sugar,
= 142.854 101.5 = y × 100
≈ 142.85 2.00
The corresponding cost of all the items in y = 101.5 × 2.00
©PAN ASIA PUBLICATIONS
the year 2019 based on the year 2015 is 100
RM142.85. = 2.03
Try question 7 in Formative Zone 10.2 (b) ∑I w = 128.5(60) + 125(25) + 101.5(15)
i i
+ 114(20)
= 14 637.5
∑w = 60 + 25 + 15 + 20
i
= 120
The index number of the T based on T :
2 0 – ∑Iw
I T 2 T T T I = i i
I T 2 = T 1 × I T 1 0 1 2 ∑w i
100
T 0 T 0 14 637.5
I T 1 I T 2 =
T 0 T 1
120
= 121.9792
Example 15 ≈ 121.98
The table below shows the prices and the price (c) Given the composite index for the cost of
indices of four items used in the production of a making the crackers increased by 30% from
packet of cracker. the year 2013 to the year 2019.
130
Price (RM) Price index in
per kg the year 2019
Item
based on the 2013 2016 2019
2016 2019
year 2016
Flour 2.00 2.57 Increase 28.5% 121.98
(i) Price index in the year 2016 based on
Salt 0.80 1.00 x the year 2013:
Sugar 2.00 y Increase 1.5% I2019
I2019 = 2016 × I2016
Oil 5.04 3.5 Increase 14% 2013 100 2013
121.98 CHAP
Given the usage of item flour, salt, sugar and oil 130 = 100 × I2016 10
2013
are 60 g, 25 g, 15 g and 20 g respectively. I2016 = 106.5748
(a) State the values of x and y. 2013
(b) Calculate the composite index for the cost ≈ 106.57
of making the crackers in the year 2019 (ii) The cost of making a crackers in the
based on 2016. year 2016:
(c) It is given that the composite index for the P
2019
cost of making the crackers increased by 54 × 100 = 106.57
30% from the year 2013 to the year 2019. P = 57.5478
(i) Calculate the composite index for the 2016
cost of making a packet of cracker in the ≈ 58 cent per packet
year 2016 based on the year 2013. Number of packet of cracker in 2019
(ii) The cost of making a crackers is 54 = RM120 ÷ RM0.58
cent per packet in the year 2013. Find = 206.90
the maximum number of the packet ≈ 207 packets
of cracker that can produced using an Try question 8 in Formative Zone 10.2
allocation of RM120 in the year 2016.

1 8.2.12 205
0
.
2
.

Form
4 Additional Mathematics Chapter 10 Index Numbers



For chicken, (ii) P 2016 × 100 = 132.5
y = 12.00 × 100 550
8.00 550
= 150 P 2016 = 132.5 × 100
For prawn, = 728.75
z × 100 = 140 Thus, the sale of food at Pak Abu’s
4.00 shop in the year 2016 is RM728.75.
z = 140 × 4.00
100 (c) The production cost is expected to
©PAN ASIA PUBLICATIONS
= 5.60
decrease by k% from the year 2016 to
the year 2019.
(b) (i) For crab, 106
The percentage of sale for the crab
= 360° − 90° − 120° − 80°
2014 2016 2019
= 70°
The ratio of the weightages
132.5 100 – k
= 90° : 70° : 80° : 120°
= 9 : 7 : 8 : 12 132.5 × 100 − k = 106
– 120(9) + 110(7) + 140(8) + 150(12) 100
I =
36 100 − k = 106 × 100
4 770 132.5
=
36 100 − k = 80
= 132.5 k = 20
Thus, the composite index of the The sale of food is expected to decrease
sale in the year 2016 based on the by 20% from the year 2016 to the year
year 2014 is RM132.5. 2019.











Paper 2
CHAP
10 1. The table below shows the unit price of four (b) The price index of component L in the year
C4 components, K, L, M and N which is needed to 2017 based on the year 2014 is 125. The
produce a type of laptop. unit price of component L in the year 2017
was RM30 more than its unit price in the
Unit price (RM) in the year
Component year 2014. Calculate the values of q and r.
2014 2017
(c) The composite index for the cost to
K 40 p produce the laptop in the year 2017 based

L q r on the year 2014 is 139. Calculate
(i) the price of the laptop in the year
M 40 70
2017 if the corresponding price in the
N 50 60 year 2014 was RM1 700,
(a) Given the price index of component K in (ii) the value of m if the ratio of
the year 2017 based on the year 2014 is components used to produce the
130. Find the value of p. laptop is 2 : 1 : 3 : m.

210

Form
4
Chapter 10 Index Numbers Additional Mathematics

2. The table below shows the Sahara’s monthly (ii) The cost of making the health
C4 expenditure and weightage for the year 2013 supplement is RM35.00 per bottle in
and 2018. the year 2011. Find the maximum
number of bottle of health supplement
Expenditure that can be produced using an
Price
Item (RM) index Weightage allocation of RM42 000 in the year
2013 2018 2019.
Transport 500 750 z 6 SPM Clone
4. The table below shows the price indices and
Food 500 800 160 9 C5 change in price indices of the expenditure of
©PAN ASIA PUBLICATIONS
House x vegetables plantation.
rental 360 130 3 Price index Change in
price index
Others y 405 135 2 Expenditure in 2017 based from 2017 to
on 2013
(a) Find the values of x, y and z. 2019
(b) Find the composite index for the year 2018 Seeds 120 15% increase
based on the year 2013.
(c) Hence, find the average monthly Insecticide 110 5% decrease
expenditure for four item which are Fertiliser 135 10% increase
transport, food, house rental and others
in year 2013, if the average monthly Equipment 125 Unchange
expenditure in year 2018 is RM651. maintenance
SPM Clone Bar chart below represents the mass of each
3. The table below shows the price and the price
indices of four item A, B, C and D used in the expenditure used in vegetable plantation.
C5
production of a type of health supplement.
Price (RM) per Mass (kg)
Item bottle for the year Price Weightage
index
2014 2019 35
30
A 40.00 60.00 150 30 25

B 18.00 21.60 120 50
15
C 80.00 90.00 112.5 12
D k 37.50 h 8 Expenditure
(a) The price of item D is increased by 25% Seeds Insecticide Fertiliser Equipment CHAP
from the year 2014 to the year 2019. maintenance 10
(i) State the value of h,
(ii) Find the value of k.
(b) Calculate the composite index for the cost (a) The price of the seeds in 2017 is RM60.
of making the health supplement for the Find the corresponding price in 2013.
year 2019 based on the year 2014.
(b) Find the price index of all the four
(c) It is given that the composite index for expenditure in 2019 based on 2013.
making the health supplement increased by Hence,
54.2% from the year 2011 to the year 2019. (i) calculate the composite index for the
(i) Calculate the composite index for the year 2019 based on the year 2013,
cost of making the health supplement (ii) find the cost price (per kg) in 2013
in the year 2014 based on the year if the corresponding cost in 2019 is
2011. RM6.00 per kg.


211

Tg 4
CHAPTER
1 Circular Measure








SMART


Important Learning Standards Page
©PAN ASIA PUBLICATIONS
1.1 Radian • Relate of angle measurement in radian and degree. 217


• Determine
(i) arc length, 218
(ii) radius, and
(iii) angle subtended at the centre of a circle.
1.2 Arc Length of a • Determine perimeter of segment of a circle. 220
Circle
• Solve problems involving arc length. 220


• Determine
(i) area of sector,
(ii) radius, and 222
(iii) angle subtended at the centre of a circle.
1.3 Area of Sector of a • Determine the area of segment of a circle. 223
Circle
• Solve the problems involving areas of sectors. 224

1.4 Application of • Solve problems involving circular measure. 226
Circular Measures





Words

• Angle / Sudut
• Degree / Darjah
• Arc length / Panjang lengkok
• Radius / Jejari
• Area of sector / Luas sektor
• Circle / Bulatan
• Centre of circle / Pusat bulatan
• Circular measure / Sukatan membulat
• Chord line / Garis perentas
• Circumference of circle / Lilitan bulatan
• Radian / Radian
• Diameter / Diameter
• Angle subtended at the centre of circle / Sudut yang tercangkum di pusat bulatan







215
215



C01 Spotlight Add Math F5.indd 215 16/04/2021 12:50 PM

Form
5 Additional Mathematics Chapter 1 Circular Measure


CHAP.
1 Concept





Circular Measure


©PAN ASIA PUBLICATIONS
Conversion of angle measurement

× 180°
π
Radian Degree

π
×
180°









Arc length of a circle Area of sector of a circle






B Q
s r
r
θ θ A s
O
O r A
r
P
1 2
s = qr, q in radian A = 2 qr , q in radian

Sine rule and cosine rule Heron’s formula can
can be used if involving a be used if involving a
triangle triangle







Application









216




C01 Spotlight Add Math F5.indd 216 16/04/2021 12:50 PM

Form
5
Chapter 1 Circular Measure Additional Mathematics
1.1 Radian Example 1 CHAP.

Relating angle measurement in radian and Convert the angle in the unit of radian to the 1
degree degree. [Use π = 3.142]
(a) 1.15 radian (b) 5π radian
6
1. In circular measures, the angle can be measured Solution:
in 2 units, which are (a) π rad = 180°
(a) degree (°) and minute (ʹ). 180°
(b) unit of radian (in or not in the terms of π). 1.15 rad = 1.15 × π
©PAN ASIA PUBLICATIONS
180°
2. Radian involves the angle that related with the = 1.15 × 3.142
radius and circumference of a circle. = 65.89°
3. The diagram below shows the angle in degree (b) π rad = 180°
and minutes and radian while the radius have 5π rad = 5π × 180°
the same length. 6 6 π
5
= × 180°
6
= 150°
57° 17' 1 rad
O O Alternative Method
Substitute π =180° into the expression,
5π 5(180°)
6 = 6 = 150°
In degree and minute In radian
57° 17ʹ = 1 rad Try question 1 in Formative Zone 1.1

4. 1 radian is a measurement of an angle subtended Example 2
about the centre of a cirlce such as the arc length
is the same as the length of radius of circle. Convert
(a) 30° into radian unit, in term of π.
A (b) 200° into radian unit.
r r [Use π = 3.142]
Solution:
1 radian B
O r (a) 180° = π rad
30° = 30° × π
180°
= π rad
5. Hence, angle subtended about the centre of a 6
circle, ˙AOB is 1 radian if the arc length of AB is Alternative Method
equal to the radius of the circle. Substitute 180° = π into the expression,
180° π
AB = OA = OB = r 30° = = rad
6 6
6. The relationship between the measurement of (b) 180° = π rad
angle in radian with degree is 200° = 200° × π
180°
2π rad = 360° 200° = 200° × 3.142
π rad = 180° 180°
= 3.49 rad
7. The conversion of angle measurement in the degree Try question 2 and 3 in Formative Zone 1.1
to the radian and vice versa are as follow:
180°
× Calculator
π
Recheck the answer in Example 2(b) by using
Radian Degree calculator,
× π 1. Press 2 0 0 × SHIFT EXP ÷ 1 8 0 =
180° 2. The screen will display 3.490658504



1.1.1 217




C01 Spotlight Add Math F5.indd 217 16/04/2021 12:50 PM

Form
5 Additional Mathematics Chapter 2 Differentiation


y 6. In determining the turning point is whether
dy Maximum minimum point or maximum point, two method
— = 0
dx point is used.
dy dy (a) Sketch of tangent method,
CHAP. — > 0 — < 0 (b) Second order derivative.
dx
dx
2
y = f(x) BRILLIANT Tips
• Sketching of tangent method is used to
x

0 x – x x x + x determine the nature of stationary points.

©PAN ASIA PUBLICATIONS
1 1 1
• Second order derivative is used to determine
Value for x x – dx x x + dx the nature of turning points.
dy
Sign for (+) 0 (–)
dx A Sketching of tangent method
Sketch of Example 33
the tangent Given the curve y = 5x + 2x – 3x.
3
2
(a) Find the coordinate of turning points for the
B Minimum point curve.
A stationary point is minimum when the gradient of (b) Hence, determine whether the turning point is
the curve is changes from negative to zero and then maximum point or minimum point.
to positive. Solution:
2
3
y (a) y = 5x + 2x – 3x
dy

2
y = f(x) = 15x + 4x – 3
dx = (5x + 3)(3x – 1)

dy
Turning point, = 0
dy dy dx
— < 0 — > 0
dx dx (5x + 3)(3x – 1) = 0
3
dy
Minimum — = 0 x = – and x = 1
point dx 5 3
( ) ( )
( )
x
3
0 x – x x x + x When x = – , y = 5 – 3 3 + 2 – 3 2 – 3 – 3


2 2 2 5 5 5 5
Value for x x – dx x x + dx = 36
25
dy 1 1
1 2
1 3
Sign for (–) 0 (+) When x = , y = 5 ( ) + 2 ( ) ( )
– 3
dx 3 3 3 3
= – 16
Sketch of 27
(
3 36
the tangent Thus, the turning point is – , ) and
( 1 , – 16 ) . 5 25
C Point of inflection 3 27
3 36
(a) Point of inflection is a stationary point that does (b) At point – , )
(
not change in sign. 5 25
(b) This point is not included in turning points. – 4 – 3 – 2
Value for x 5 5 5

Positive dy 17 0 – 11
gradient Value for dx 5 5
Positive dy

gradient Zero gradient Sign for + 0 –
dx
Sketch of
Zero gradient
Negative the tangent
gradient
Negative Sketch of
gradient
the graph
256 2.4.4
C02 Spotlight Add Math F5.indd 256 23/04/2021 10:53 AM

Form
5
Chapter 3 Integration Additional Mathematics
Explaining relation between the limit of the y
sum of areas of rectangles and the area b
∫
under a curve y = f(x) A = y dx
a

1. The area under curve y = f(x) can be determined A
by integration. O a b x
y
y = f(x) CHAP.
2. For the value of the area bounded by the curve
and the x-axis, 3
©PAN ASIA PUBLICATIONS
A (a) if the region above the x-axis, then the
integral value is positive.
x
O a b
y y
2. Dividing the area under curve between x = a and
x = b into a few rectangles, dL.
y
y = f(x)
x
A A
y x x
2 O O
A y y A = y x

A 3 3
y A 1 2
1 (b) if the region below the x-axis, then the
integral value is negative.
O
x
For each of the following rectangles, y y
b – a
(a) the width is dx = , where n is the
n x x
number of rectangles. O A O A
(b) the height of the rectangle can be obtained
from the function of the curve, y .
i
Area of each rectangle, dA = Height × Width
i
≈ y × dx (c) if the region bounded by below the x-axis
i
≈ y dx and also above the x-axis, then the area of
i the region should be determined separately.
Hence, the total area of the n triangles,
≈ dA + dA + dA + … + dA n y
2
1
3
n
≈ ∑ dA
i = 1 i A
n x
≈ ∑ y dx O a b c
i = 1 i B
3. As the width of the strips of rectangle is become
thinner, the width of each rectangle, dx becomes
narrower and approaching to zero dx ˜ 0. Area of shaded region
Hence, = Area of A + Area of B
∫
∫
c
b
Area under the curve = lim ∑ ydx = y dx + y dx 
dx ˜ 0 a b
∫
b
= y dx
a
Determining the area of a region BRILLIANT Tips
A Area of a region between the curve and the The negative sign only indicates that the region is
x-axis below of x-axis. Therefore, negative signs can be
eleminated by using modulus, |a|.
1. The area under the curve bounded by x-axis = a
and x-axis = b given by:
3.3.2 275




C03 Spotlight Add Math F5.indd 275 23/04/2021 10:57 AM

Form
5 Additional Mathematics Chapter 3 Integration


Determining the area between the curve and Example 14
a straight line
Given the curve y = 16 is intersects the line y = 2x
2
1. at point P. x
The area of shaded region
y y 16
y = f(x) y = ––
x 2 y = 2x
y = g(x) A
CHAP. P
3
x
B O 5

x Find the area of the shaded region.
a O b
Solution:
Area under the Area under 16 y
straight line the curve y = x 2 …1
y y y = f(x) y = 16
––
y = f(x) y = 2x …2 x 2 y = 2x
y = g(x) A y = g(x) A 1 = 2: 16 2 = 2x
1 x P
– 2x = 16 4
3
B B x = 8
3
2 x x
x x = 2 O
a O b a O b When x = 2, y = 2(2) 2 5
= 4
Thus, P is (2, 4).
The shaded area = 1 – 2 Area of the shaded region
∫
∫
b
b
= g(x) dx – f(x) dx = Area of triangle + Area under the curve
a
a
5
∫
b
= [g(x) – f(x) dx] = ( 1 × 2 × 4 + 16 2 dx
) ∫ 2 x
2
a
[ ]
= 4 + – 16 5 2
x
[ ]
2. ©PAN ASIA PUBLICATIONS
16
5
= 4 + –
The area of shaded region x 2
( )]
[
y = 4 + – 16 – – 16
y = f(x) 5 2
24
= 4 +
5
y = g(x) = 44 unit 2
5
x Try question 6 in Formative Zone 3.3
O a b
Area under Area under the
the curve straight line Example 15
y y
Given the curve y = x(4 – x) intersects the straight
y = f(x) y = f(x) line y = x at point A.
1 –
y
y = g(x) 2 y = g(x) y = x
4
A
x x
O a b O a b
2
The shaded area = 1 – 2 y = x(4 – x)
∫
∫
b
b
= f(x) dx – g(x) dx O 2 4
a a
∫
b
= [f(x) – g(x) dx] Find
a (a) the coordinate of point A.
(b) the area of the shaded region.
278 3.3.3
C03 Spotlight Add Math F5.indd 278 23/04/2021 10:57 AM

Form
5 Additional Mathematics Chapter 3 Integration


2. The generated volume of a solid is formed from a (c) Volume of each cylinder, dV i
revolved of x-axis is as follow: = Base area of cylinder × Height of cylinder
(a) Rotate an area of shaded region completely = πx × dy
2
i
through 360° about the x-axis until its = πx dy
2
i
generate a solid, approximately a cylinder. (d) Total volume of n cylinders
= V + V + V + …V
y y 1 2 3 n
n
y = f(x) ≈ ∑ dV
CHAP. i = 1 i
3 x D E O x n
©PAN ASIA PUBLICATIONS
2
≈ ∑ πx dy
i
i = 1
(b) Divide the solid into n vertical cylinders
with a thickness of dx. y
Radius of
Radius of
cylinder
cylinder
y
x
x (e) When the number of cylinders is sufficiently
large, that is n ˜ ∞, then dy ˜ 0. Hence, the
generated volume of the solid:
n b
lim ∑ πx dy = πx dy
(c) Volume of each cylinder, dV i dx ˜ 0 i = 1 i 2 ∫ a 2
= Base area of cylinder × Height of cylinder

= πy × dx
2
i
= πy dx Determining the generated volume of a
2
i
(d) Total volume of n cylinders region revolved at the x-axis or y-axis
= V + V + V + …V
1 2 3 n
n A Generated volume, V through x-axis
≈ ∑ dV
i = 1 i y
n y = f(x)
≈ ∑ πy dx
2
i = 1 i x
(e) When the number of cylinders is sufficiently a b O
large, that is n ˜ ∞, then dx ˜ 0. Hence, the
generated volume of the solid:
The generated volume of a region bounded by the
n
∫
b
2
2
lim ∑ πy dx = πy dx curve y = f(x) is revolved through 360° about the
i
dx ˜ 0 i = 1 a x-axis is given by:
3. The generated volume of a solid is formed from a V = π y dx
∫
b
2
revolved of y-axis is as follow: a
(a) Rotate an area of shaded region completely
through 360° about the y-axis until its Example 17
generate a solid, approximately a cylinder.
Find the generated volume, in terms of π, when
y y the shaded region in each diagram is revolved
through 360° about the x-axis.
D
(a)
y
y = f(x) 2
E y = —
x
x x
O O

(b) Divide the solid into n horizontal cylinders O 1 3 x
with a thickness of dy.
280 3.3.4 3.3.5
C03 Spotlight Add Math F5.indd 280 23/04/2021 10:57 AM

Form
5 Additional Mathematics Chapter 5 Probability Distribution


3. The diagram below shows the normal Determining and interpreting standard
distribution graph and standard normal score, Z
distribution graph.
1. Any continuous random variable X from normal
distribution is standardised by:
Area is Z = X – m
the same s

and produces the standard normal distribution ,
Z ~ N(0, 1), where
©PAN ASIA PUBLICATIONS
a – μ
b – μ
a μ b X z = –––– 0 z = –––– z
X is a continuous random variable,
a σ b σ
Normal distribution Standard normal   m is mean and
graph distribution graph s is standard deviation.
CHAP. 4. A continuous random variable, X can be 2. This changing process is known as distribution
5 standardised by changing it to another standardised.
continuous random variable, Z as follow:
Example 21
X ~ N(m, s )
2
where m = mean and Find z-score if the value is X = 1 and X ~ N(12, 3).
s = standard deviation Solution:
X – m
Standardised Z = s
X – m 1 – 12
Z = s = 3
= –3.367
Z ~ N(0, 1) Try question 1 and 2 in Formative Zone 5.3
where 0 = mean and
1 = standard deviation
Example 22
5. The diagram below shows the percentage of data Given the mass of an apple and orange is normally
distribution graph for each standard normal distributed. Mean and standard deviation of apple
distribution. are 100 g and 15 g respectively while mean and
2
standard deviation of orange are 140 g and 25 g
2
respectively.
(a) Find the z-score if the mass of an apple 110 g
is chosen randomly.
68.3% (b) If the standard score of an orange is –0.16,
of the data
determine the mass of orange.
95.5%
Solution:
of the data
(a) Z = 110 – 100
99.7% 15
of the data
= 0.667
z-score of apple is positive, which is 0.667.
–3SD –2SD –1SD Mean +1SD +2SD +3SD
This shows that the apple has mass more than
From the graph above, the average apple.
(a) 68.3% of the data lies within the standard (b) X – m = –1.6
deviation ±1 from the mean. s
(b) 95.5% of the data lies within the standard X – 140 = –1.6
25
deviation ±2 from the mean. X – 140 = –40
(c) 99.7% of the data lies within the standard X = 100
deviation ±3 from the mean.
Try question 3 in Formative Zone 5.3




326 5.3.2 5.3.3




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Form
5
Chapter 5 Probability Distribution Additional Mathematics
Determining the probability of an event for normal distribution
1. Besides using a scientific calculator or calculation, the probability of z-score for standard normal distribution
can also be determined by the standard normal distribution table.
2. The standard normal distribution table uses the concept that the probability of a normal distribution is given
by the area under the graph with the total area under the graph being 1 unit .
2
3. Since the graph is symmetrical, P(Z > 0) = 0.5, the numeric table only gives the value of the area to the right
starting with 0.5 which is for P(Z . 0) = 0.
4. The table below shows the standard normal distribution table.
©PAN ASIA PUBLICATIONS
Upper Tail Probability Q(z) of the Normal Distribution N(0, 1)
1 2 3 4 5 6 7 8 9
z 0 1 2 3 4 5 6 7 8 9
Substract
0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 4 8 12 16 20 24 28 32 36
0.1 0.4602 0.4562 0.4522 0.4483 0.443 0.4404 0.4364 0.4325 0.4286 0.4247 4 8 12 16 20 24 28 32 36 CHAP.
0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 4 8 12 15 19 23 27 31 35 5
0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.557 0.3520 0.3483 4 7 11 15 19 22 26 30 34
0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 4 7 11 15 18 22 25 29 32
0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 3 7 10 14 17 20 24 27 31
0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.02546 0.2514 0.2483 0.2451 3 7 10 13 16 19 23 26 29
0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148 3 6 9 12 15 18 21 24 27
0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 3 5 8 11 14 16 19 22 25
0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1634 0.1611 3 5 8 10 13 15 18 20 23
1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379 2 5 7 9 12 14 16 19 21
1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 2 4 6 8 10 12 14 16 18
1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 2 4 6 7 9 11 13 15 17
1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 2 3 5 6 8 10 11 13 14
1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681 1 3 4 6 7 8 10 11 13
1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 1 2 4 5 6 7 8 10 11
1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 1 2 3 4 5 6 7 8 9
1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 1 2 3 4 4 5 6 7 8
1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 1 1 2 3 4 4 5 6 6
1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 1 1 2 2 3 4 4 5 5
2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 0 1 1 2 2 3 3 4 4
2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143 0 1 1 2 2 2 3 3 4
2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110 0 1 1 1 2 2 2 3 3
2.3 0.0107 0.0104 0.0102 0 1 1 1 1 2 2 2 2
0.0099 0.00964 0.0093 0.00991 3 5 8 10 13 15 18 20 23
0 9 4
0.0088 0.0088 0.0084 2 5 7 9 12 14 16 16 21
9 66 2
2.4 0.0082 0.0079 0.0077 0.0075 0.0073 2 4 6 8 11 13 15 17 19
0 8 6 5 4
0.0017 0.00695 0.006 0.0065 0.0063 2 4 6 7 9 11 13 5 17
4 76 7 9
2.5 0.0062 0.0060 0.0058 0.0057 0.0055 0.0053 0.00523 0.0050 0.0049 0.0048 2 3 5 5 8 9 11 12 14
1 4 7 0 4 9 8 4 0
2.6 0.0046 0.0045 0.0040 0.0042 0.0041 0.0040 0.00391 0.0037 0.0036 0.0035 1 2 3 4 6 7 9 9 10
6 3 7 5 2 9 8 7
2.7 0.0034 0.0033 0.0032 0.0031 0.0013 0.0029 0.00289 0.0028 0.0272 0.0026 1 2 3 4 5 6 7 8 9
7 6 6 7 07 8 0 4
2.8 0.0025 0.0024 0.0024 0.0023 0.0022 0.0021 0.00212 0.0020 0.0019 0.0019 1 1 2 3 4 4 5 6 6
6 8 0 3 6 9 5 9 3
2.9 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.00154 0.0014 0.0014 0.0013 0 1 1 2 2 3 3 4 4
7 1 5 9 4 9 9 4 9
3.0 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.00111 0.0010 0.0010 0.0010 0 1 1 2 2 2 3 3 4
5 1 6 2 8 4 7 4 0


5.3.4 327




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Form
5 Additional Mathematics Chapter 5 Probability Distribution


5. If Z is a standard normal distribution with mean 8. If the probability is between z-score, which is a
is 0 and standard deviation is 1, thus the value and b:
of z gives the probability of standard normal f(z)
distribution.
f(z)


z
O a b
z
O a
P(a , Z , b) = P(Z . a) – P(Z . b)
©PAN ASIA PUBLICATIONS

Probability by using Standard Example 23
Normal Distribution Table
CHAP. Given Z is a continuous random variable with a
5 bit.ly/2X8SFlz standard normal distribution. Find
(a) P(Z . 0.254)
6. If given the probability is more than z-score and (b) P(Z > 1.056)
(a) the value of a is positive, P(Z . a).
(c) P(Z , –1.386)
f(z)
(d) P(Z . –2.337)
(e) P(Z , 2.337)
(f) P(0 , Z , 1.242)
z
O a (g) P(–2.46 , Z , 1.281)
(b) the value of a is negative, P(Z . –a). (h) P(–2.149 , Z , –0.214)
(i) P(1.331 , Z , 2.147)
f(z)
(j) P(| Z | . 1.471)
Solution:
(a)
z
–a O
f(z)
P(Z . –a) = 1 – P(Z , –a)
= 1 – P(Z . a)

7. If given the probability is less than z-score and z
(a) the value of a is negative, P(Z , –a) 0 0.254
f(z) f(z)
From the standard normal distribution table,
=
z 5 4
z z
–a O O a Substract

P(Z , –a) = P(Z . a) 0.2 0.4013 15
(b) the value of a is positive, P(Z , –a)
f(z)
P(Z . 0.25) = 0.4013

P(Z . 0.254) = 0.4013 – 0.0015
= 0.3998
z
O a
Thus, P(Z . 0.254) = 0.3998.
P(Z , a) = 1 – P(Z . a)

328 5.3.4




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Tg 4
CHAPTER
6 Trigonometric Functions








SMART


Important Learning Standards Page
©PAN ASIA PUBLICATIONS
6.1 Positive Angles and • Represent positive angles and negative angles in a Cartesian Plane. 343
Negative Angles


• Relate secant, cosecant and cotangent with sine, cosine and tangent
6.2 Trigonometric of any angle in a Cartesian plane. 345
Ratios of any Angle
• Determine the values of trigonometric ratios for any angle. 346

• Draw and sketch graphs of trigonometric functions:
(i) y = a sin bx + c
(ii) y = a cos bx + c 351
6.3 Graphs of Sine, (iii) y = a tan bx + c
Cosine and where a, b and c are constants and b . 0.
Tangent Functions
• Solve trigonometric equations using graphical method. 355

• Derive basic identities:
(i) sin A + cos A = 1 357
2
2
2
(ii) 1 + tan A = sec A
2
6.4 Basic Identities (iii) 1 + cot A = cosec A
2
2
• Prove trigonometric identities using basic identities. 358
• Prove trigonometric identities using addition formulae for 358
6.5 Addition Formulae sin (A ± B), cos (A ± B) and tan (A ± B).
and Double Angle • Derive double angle formulae for sin 2A, cos 2A and tan 2A. 360
Formulae
• Prove trigonometric identities using double-angle formulae. 362


6.6 Application of • Solve trigonometric equations. 364
Trigonometric
Functions • Solve problems involving trigonometric functions. 366




Words

• Degree / Darjah • Complementary angle / Sudut pelengkap
• Radian / Radian • Basic identities / Identiti asas
• Trigonometric ratio / Nisbah trigonometri • Addition formulae / Rumus sudut majmuk
• Quadrant / Sukuan • Double angle formulae / Rumus sudut berganda
• Reference angle / Sudut rujukan • Half angle formula / Rumus sudut separuh





341
341



C06 Spotlight Add Math F5.indd 341 16/04/2021 5:39 PM

Form
5 Additional Mathematics Chapter 6 Trigonometric Functions




Concept




Trigonometric Function



Positive Angle Negative Angle Basic Identities
©PAN ASIA PUBLICATIONS
2
2
y y • sin A + cos A = 1
Positive • 1 + tan A = sec A
2
2
angle
• 1 + cot A = cosec A
2
2
θ
x x
θ
Negative Addition Formulae
angle
• sin (A  B) = sin A cos B  cos A sin B
(
)
(
π
CHAP. x° = x × 180 ) rad, q rad = q × 180 ° • cos (A  B) = cos A cos B  sin A sin B
π
tan A  tan B
6 • tan (A  B) = 1  tan A tan B
Trigonometric Ratios Graph of Trigonometric
1
• sin A = Functions Double Angle Formulae
cosec A • Sine, y = sin x • sin 2A = 2 sin A cos B
1
• cosec A = y • cos 2A = cos A – sin A
2
2
sin A 2
1 = 2 cos A – 1
• cot A = 1 = 1 – 2 sin A
2
tan A
x • tan 2A = 2 tan A
2
sin cos –2π π –π π 0 π π π 2π 1 – tan A
–3 –
2 – – 2 – 2 3 – 2
–1
tan 1 cot • Cosine, y = cos x Half Angle Formulae
y
!
q
• sin = ± 1 – cos q
1 2 2
sec cosec
!
q
x • cos = ± 1 + cos q
–2π –3 – π –π – – π 0 π – π 3 – π 2π 2 2
2 2 2 2 q sin q 1 – cos q
–1 • tan = =
Complementary Angles • Tangent, y = tan x 2 1 + cos q sin q
• sin q = cos (90° – q) y
• cos q = sin (90° – q)
• tan q = cot (90° – q) 1 • a = amplitude
• cot q = tan (90° – q) • b = number of cycle in the range
• sec q = cosec (90° – q) –2π –3 – π –π – – π 0 π – π 3 – π 2π x 0 < x < 2π for sine and
• cosec q = sec (90° – q) 2 –1 2 2 cosine graph whereas
2
0 < x < π for tangent graph.
0
• c = translation ( ) from basic
c
Method to Find Trigonometric Ratios graph.
• Use calculator • Period: (i) 2π for sine and
• Use unit circle b
• Use trigonometric ratio of cosine graph.
corresponding reference angle (ii) π for tangent graph.
• Use right-angled triangles b
342
C06 Spotlight Add Math F5.indd 342 16/04/2021 5:39 PM

Form
5
Chapter 6 Trigonometric Functions Additional Mathematics
A Using calculator The following is the trigonometric ratios by using
For the value of cosecant, secant and cotangent, the x-coordinate and y-coordinate in unit circle
value can be calculated by using the reciprocal of the sin q = coordinate-y
following ratios. cos q = coordinate-x
y-coordinate
cosec q = 1 , sec q = 1 , cot q = 1 tan q = x-coordinate
sin q cos q sin q


Example 8 Example 10
©PAN ASIA PUBLICATIONS
Find the value of y
(a) sin 235° (b) sec 505° (–0.643, 0.766) (0.643, 0.766)
Solution: 130°
(a) sin 235° = –0.8192 50°
x
Calculator
1. Press sin 2 3 5 =
2. The screen will display –0.819152044 Based on the unit circle above, state the following CHAP.
value. 6
(b) sec 505° = 1 (a) sin 50° (b) cos 130°
cos 505° (c) cosec 130° (d) cot 50°
= –1.221
Solution:
Try question 3 in Formative Zone 6.2 (a) sin 50° = y-coordinate
= 0.766
(b) cos 130° = x-coordinate
Example 9 = –0.643
1
Find the value of (c) cosec 130° = sin 130°
(a) tan 2.4 rad (b) cosec 2 π rad 1
5 =
Solution: 0.766
= 1.305
(a) tan 2.4 rad = –0.9160 x-coordinate
(d) cot 50° =
y-coordinate
Calculator 0.643
Make sure MODE of calculator is in radian. = 0.766
1. Press tan 2 . 4 = 0.839
Try question 4 in Formative Zone 6.2
2. Press = and the screen will display
–0.916014289
C Using trigonometric ratio of corresponding

(b) cosec 2 π rad = 1 reference angle
5 sin 2 π rad 1. The value of trigonometric ratio of any angle
5 can also be determined using the value of
= 1.0515
the trigonometric ratio of the corresponding
Try question 3 in Formative Zone 6.2 reference angle of the angle.
2. The reference angle, a, is the acute angle formed
B Using unit circle by rotating the ray OP and the x-axis on the
Cartesian plane.
y
y
P (x, y) OP OP
2 1
1
θ y α x
x
O x
OP OP
3 4

6.2.2 347




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Form
5 Additional Mathematics Chapter 6 Trigonometric Functions


Example 19 Example 20
Given f(x) = 4 cos 2x for 0 < x < 2π. Sketch the graph of the following trigonometric
(a) State the period of the graph function y = f(x). functions in the given range.
Hence, state the number of cycle of the graph (a) y = sin x + 1 for 0 < x < 2π
in the given range.
(b) State the amplitude of the graph. (b) y = –2 cos x for 0 < x < 2π
(c) Write the coordinates of the maximum and the (c) y = | tan x | for 0 < x < 2π
minimum points. (d) y = | cos 2x | + 1 for 0 < x < 2π
(d) Sketch the graph of y = f(x). Solution:
(e) Using the same axes, sketch the graph of (a) y = sin x + 1 for 0 < x < 2π
©PAN ASIA PUBLICATIONS
function y = –|4 cos 2x| for 0 < x < 2π.
Solution: 1 Sketch the basic graph, y = sin x.
Compare f(x) = 4 cos 2x with the basic cosine 2 The graph moves 1 unit upward, such that
0
function, f(x) = a cos bx + c. translation ( ) .
(a) Period 2π = π or 180°. Number of cycle, b = 2. 1
2
(b) Amplitude, a = 4 y
(c) Maximum point: (0, 4), (π, 4) and (2π, 4).
CHAP. Minimum point: ( π , –4 and ( 3π , –4 ) 2 y = sin x + 1
)


6 2 2 1
(d) To sketch graph function y = 4 cos 2x:
Number of class = 2 × 2 × 2 = 8 x
Size of class interval = 2π = π O π – 2 π 3π 2π
––
2
8 4 –1 y = sin x

x 0 π π 3π 2π –2
2 2
y 4 – 4 4 – 4 4 (b) y = –2 cos x for 0 < x < 2π
Thus, the graph function of y = 4 cos 2x: 1 Sketch graph of y = cos x.
y 2 Reflect the graph at 1 on x-axis to make
the graph of y = – cos x.
4 y = 4 cos 2x
2 y
x
O π π 3π 2π 2
– ––
–2 2 2 y = – cos x
1
–4 x
O π π 3π 2π
– 2 ––
2
(e) Steps in sketching the graph of y = –|4 cos 2x| –1 y = cos x
1 y = |4 cos 2x| is a reflection of graph on –2
negative side of x-axis.
2 y = –|4 cos 2x| is a reflection of graph at 1 3 The value of a is –2. The maximum value is
on x-axis.
(π, 2) and the minimum value is (0, –2) and
y (2π, –2).
y
4

2 2 y = –2 cos x
x 1
O π π 3π 2π
– –– y = – cos x
–2 2 2 x
O π π 3π 2π
– ––
–4 –1 2 2
y = –|4 cos 2x|
–2
Try question 3 in Formative Zone 6.3
354 6.3.1




C06 Spotlight Add Math F5.indd 354 16/04/2021 5:39 PM

Form
5
Chapter 7 Linear Programming Additional Mathematics

7.1

1. Shade the region which represented by the (c) In a week, Jason makes x cupboard P and y
following linear inequality. C1 cupboard Q. He has a capital of RM2 000.
1
2
(a) y < x + 3 (b) y > x – 1 The cost of making a cupboard P is RM200
2 5 and a cupboard Q is RM100.
y y
4. Represent each of the following linear
6 3 inequalities graphically. C3
©PAN ASIA PUBLICATIONS

1 4 2 (a) x > 4
y = –x + 3 2 y = –x – 1 1
2
2 5 y , 8
x x (b) x , 7
–6 –4 –2 0 2 –3 –2 –1 0 1
–2 –1 y > 1
(c) x > –2
y > 4
4
8
(c) y , x – 5 (d) y < – x – 3 (d) x + y > 2
3 9
y , 6
y y
(e) x + y < 4
2 x x + y . 1
–8 –6 –4 –2 0 (f) x > y
x –1
–2 0 2 4 6 x . 1
4
–2 y = – –x – 3 –2
8 (g) y < 2x CHAP.
–4 y = –x – 5 9 –3
3
–6 –4 y > x + 2 7
(h) y > 2x
y < 3x
2. Determine the linear inequality which satisfies
each of the following regions. C2 5. Shade the region R which satisfied the
(a) (b) following inequalities. C2
x > 0
y y y > x
y < 2x + 1
4 2 6
y = –x + 2 7 x + y < 8.
2 3 y = –x + 5 4
3
x 2
–4 –2 0 2 4 x = 0
–2 x
–3 –2 –1 0 1
–4 –2 y = 2x + 1
8 y = x
(c) (d)
6
y y
4
4 4
3 8 2 2 x + y = 8
2 y = – –x + 4 x
3
–2 0 2 4 6
1 –2 5 y = 0
x –4 y = –x – 4 0 2 4 6 8
–1 0 1 2 3 2

3. Write a mathematical model for each of the 6. Draw y > 2x + 1 and y , 2x – 3 on the
following situations. C1 same graph. Is there any solution for these
(a) The number of participants for course A inequalities? C2
is at most three times than the number of
participants for course B. 7. Is the point (2, 1) a solution of the inequalities
(b) The number of male workers exceed the x + y . 1 and 2x + y , 8? C2
number of female workers by at most 40.

383




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Form
5 Additional Mathematics Chapter 8 Kinematics of Linear Motion


2. The diagram below shows the movement of (c) Displacement is maximum when v = 0, which is
an object thrown by Ahmad and its value of t = 4 s.
displacement and velocity with respect to the time. When t = 4, s = 4 + 8(4) – (4) 2
s = 4 + 32 – 16
t = 2 s s = 20 m
s = 20 m  Maximum Thus, the maximum displacement of the
v = 0 displacement
particle is 20 m.
(d) Velocity decreases, v , 0
t = 1 s t = 3 s
s = 15 m   s = 15 m 8 – 2t , 0
v = +10 m/s v = –10 m/s 2t . 8
©PAN ASIA PUBLICATIONS
t . 4
Initial velocity
Thus, the range of value t is t . 4 s.
t = 0 (e) When the particle moves to the right, v . 0
s = 0  t = 4 s 8 – 2t . 0
v = +20 m/s  s = 0 2t , 8
v = –20 m/s
t , 4
Thus, the time interval of the particle moves to
the right is t , 4 s.
Try question 4 in Formative Zone 8.2
• Particles stop, so velocity, v = 0
• For maximum or minimum displacement,
ds = v = 0 Example 12
dt A particle moves along a straight line with its
• Particles return or through a fixed point O, displacement, s meters, from a fixed point O given
s = 0 by s = t – 12t + 36t + 10, where t is the time, in
2
3
CHAP. • Particles change their direction of movement, seconds. [Assume motion to the right as positive]
8 v = 0 Find –1
(a) the initial velocity, in m s , of the particle,
(b) the values of t, in seconds, when the particle
Example 11 stops instantaneously,
(c) the range of value t when the velocity is
A particle moves along a straight line with its negative.
displacement, s meters, from a fixed point O given
by s = 4 + 8t – t , where t is the time, in seconds. Solution:
2
[Assume motion to the right as positive] Given s = t – 12t + 36t + 10.
2
3
Find Velocity function, v = ds = 3t – 24t + 36
2
(a) the initial velocity, in m s , of the particle, dt
–1
2
(b) the value of t, in seconds, when the particle (a) Initial velocity, t = 0 v = 3(0) – 24(0) + 36
stops instantaneously, = 36
(c) the maximum displacement of the particle, Thus, the initial velocity of the particle is 36 m s .
–1
(d) the range of values t values when the velocity (b) Particle stops, v = 0
decreases, 3t – 24t + 36 = 0
2
(e) the time interval when the particle moves to (t – 6)(t – 2) = 0
the right. t = 6 or t = 2
Thus, the particle stops at t = 6 s and t = 2 s.
Solution: (c) Velocity is negative, v , 0
Given s = 4 + 8t – t . 3t – 24t + 36 , 0
2
2
Velocity function, v = ds = 8 – 2t (t – 6)(t – 2) , 0
dt Let (t – 6)(t – 2) = 0
(a) Initial velocity, t = 0 t = 6 or 2
When t = 0, v = 8 – 2(0)
= 8
–1
Thus, the intial velocity of the particle is 8 m s .
(b) Particle stops, v = 0 t
8 – 2t = 0 2 6
2t = 8 Thus, the range of value t is 2 , t , 6 s.
t = 4
Thus, the particle stops at t = 4 s. Try question 5 in Formative Zone 8.2
404 8.2.2




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Form
5 Additional Mathematics Chapter 8 Kinematics of Linear Motion


Example 20 When t = 5, s = 2(5) 3 – 6(5) + 10(5)
2
3
A particle moves along a straight line through 2
a fixed point O with velocity 10 m s . The s = –16 m
–1
3
acceleration, a m s , is given by a = 4t – 12 where Number line,
–2
t is the time, in seconds, after passing through t = 0
a fixed point O. [Assume motion to the right as
positive] Find t = 1
(a) the time, in seconds, when the acceleration is 2 0 2
zero, –16 – 3 4 – 3
(b) the minimum velocity, in m s , of the particle, t = 5
–1
©PAN ASIA PUBLICATIONS
(c) the time, in seconds, when the particle stop 2 2 2
momentarily, Total distance travelled = 4 + 4 + 16 3
3
3
(d) the total distance, in m, travelled by the particle = 26
in the first 5 seconds.
Solution: Thus, in the first 5th seconds, total distance
travelled of particle is 26 m.
Given the acceleration function, a = 4t – 12
∫
∫
Velocity function, v = a dt = 4t – 12 dt Alternative Method
= 2t – 12t + c (A) Using velocity-time graph
2
Distance can be determine by finding the
When t = 0, v = 10 area under the curve.
10 = 2(0) – 12(0) + c v = 2t – 12t + 10
2
2
c = 10
So, v = 2t – 12t + 10. t 0 1 5
2
∫
Displacement function, s = v dt v 10 0 0
Velocity-time graph for 0 < t < 5:
∫
2
= 2t – 12t + 10 dt
CHAP. v
8 = 2 3 2
t – 6t + 10t + c
3
When t = 0, s = 0 10
2(0) 3
0 = 3 – 6(0) + 10(0) + c v = 2t – 12t + 10
2
2
c = 0 A
So, s = t – 6t + 10t. t
2 3
2
3 0 1 5
(a) Acceleration is zero, a = 0 B
4t – 12 = 0
4t = 12 –8
t = 3
Thus, the acceleration of the particle is zero at 1
∫
2
t = 3 seconds. Area A = 2t – 12t + 10 dt
0
(b) Minimum velocity, dv = 0, a = 0 2 1
3
2
dt = [ t – 6t + 10t ]
4t – 12 = 0 3 0
2
]
3
2
t = 3 = [ 3 (1) – 6(1) + 10(1) – 0
2
When t = 3, v = 2(3) – 12(3) + 10 2 2
= –8 = 4 m
3
Thus, the minimum velocity of the particle is ∫ 5 2
–8 m s . Area B = 2t – 12t + 10 dt
–1
1
(c) Particle stops, v = 0 2 3 2 5
2t – 12t + 10 = 0 = [ 3 t – 6t + 10t ] 1
2
(t – 1)(t – 5) = 0 = 2 (5) – 6(5) + 10(5) – 4 2
3
2
t = 1 and t = 5 [ 3 ] 3
1
Thus, the particle stops at t = 1 s and t = 5 s. = –21 
(d) When t = 0, s = 2(0) 3 – 6(0) + 10(0) 3
2
1
3 = 21 m 2
3
s = 0 m Total distance travelled = Area A + Area B
2(1) 3
When t = 1, s = – 6(1) + 10(1) 2 1
2
3 = 4 + 21 3
3
2
s = 4 m = 26 m
3
410 8.4.1
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Form
5
Chapter 8 Kinematics of Linear Motion Additional Mathematics

(B) Using displacement-time graph When t = 5, v = 6(5) – 18
= 12

s = 2t 3 – 6t + 10t Thus, the acceleration when the particle stops
2
3
ds = 0, v = 0 is –12 m s and 12 m s .
–2
–2
dt (b) The particle changes direction from point A to
t = 1 and t = 5 (Turning point) B, v = 0
2
When t = 1, s = 4 From (a), it is known the particle changes the
3 direction at t = 1 and t = 5. So, when t = 1,
When t = 5, s = –16 2 the particle at point A whereas when t = 5, the
3
When s = 0, particle at point B.
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2t 3 – 6t + 10t = 0 Velocity-time graph:
2
3
t( 2t 2 – 6t + 10) = 0 v
3 15
t = 0, t = 2.21, t = 6.79
Displacement-time graph: v = 3t – 18t + 15
2
s
t
2 0 1 5
4 –
1 3 2
t Point A Point B
0 1 2.21 5 6.79 –12
Distance AB = Area under the curve
3 5
∫
= (3t – 18t + 15) dt
2
2 1
–16 –
3 5
[
= t – 9t + 15t ]
3
2
1
Total distance travelled = 1 + 2 + 3 = [(5 – 9(5) + 15(5)] – CHAP.
2
3
2
2
2
3
2
= 4 + 4 + –16  [(1 – 9(1) + 15(1)] 8
3 3 3
= 26 m = –25 – 7
=  –32 
Try question 3 and 4 in Formative Zone 8.4 = 32 m
∫
(c) Displacement function, s = v dt
Example 21
∫
= (3t – 18t + 15) dt
2
A particle moves along a straight line through a
fixed point P. The velocity of the particle, v m s , = 3t 3 – 18t 2 + 15t + c
–1
is given by v = 3t – 18t + 15 where t is the time, in 3 2
2
2
3
seconds, after passing through a fixed point P. The = t – 9t + 15t + c
particle begins to change direction of movement Initial displacement, v = 0 and t = 0
2
at points A and B. [Assume motion to the right as 0 = (0) – 9(0) + 15(0) + c
3
positive]. Find c = 0
2
3
–2
(a) the acceleration, in m s , when the particle So, s = t – 9t + 15t.
stops, Velocity decreases, a , 0
(b) the distance between point A and B, 6t – 18 , 0
(c) the distance when a particle moves with 6t , 18
decreasing velocity. t , 3
Solution: v
(a) Given velocity function, v = 3t – 18t + 15
2
Particle stops, v = 0 15
3t – 18t + 15 = 0
2
(t – 1)(t – 5) = 0 t
t = 1 and t = 5 O 1 3 5
2
Acceleration function, a = dv v = 3t – 18t – 15
dt 3 2
= d (3t – 18t + 15) t = 1, s = (1) – 9(1) + 15(1) = 7
2
3
2
dt t = 3, s = (3) – 9(3) + 15(3) = –9
= 6t – 18 The distance travelled = 7 + 7 + 9 = 23 m.
When t = 1, a = 6(1) – 18 Try question 5 and 6 in Formative Zone 8.4
= –12
8.4.1 411
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SPM MODEL PAPER




Paper 1

Time: 2 hours
Section A
(64 marks)
Instruction: Answer all questions
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1. (a) Diagram 1 shows a part of graph for a (ii)
function of y = f(x).

y
y = f(x)



x
0 2
–3



Diagram 1
SPM MODEL PAPER (b) A function f is derived by f : x ˜ , x ≠ 0 2. (a) Diagram 2 shows a graph of function f for
State whether the function of f

(i) is a discrete or continuous,
(ii) has an inverse function or not.
domain 0 < x < 4 and its inverse function
[2 marks]
f .
–1
a
x
y

such as a is a constant. Given f (2) = 2, find
(i) the value of a,
17
(ii) f (8). –1 [3 marks] A (4, 12)
f
Answer: –1
(a) (i) f B
x
0
(ii) –4
Diagram 2
Based on the graph, determine
–1
(b) (i) (i) the domain of f ,
(ii) the coordinate of point B on the graph
–1
of f that corresponding with the point
A on the graph of f. [2 marks]
Answer:
(i)




(ii)






418




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SPM Model Paper Additional Mathematics
(b) On the answer space below, sketch the graph (b) The height, h m, of a hit golf ball from a
of y = |2x – 5| for 0 < x < 6. Hence, find the surface of field after t seconds is given by the
range of value x such as y < 3. [3 marks] function h(t) = 20t – 5t . Find
2
(i) the time taken, t when the golf ball
y touches the surface of the field again.
(ii) the maximum height, h reached by the
golf ball.
Hence, in the answer space below,
sketch the graph of h against t for the
movement of the golf ball. [3 marks]
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Answer:
(b) (i)


x
0












3. (a) If a and b are the roots for quadratic (ii)
equations x + px + q = 0, such as p and q
2
are constant, express a  b + b  a in terms of SPM MODEL PAPER
2
2
p and q. [2 marks]
Answer:
(a)









h (meter)



















t (seconds)
0



419




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SPM Model Paper Additional Mathematics
Section B
(16 marks)
Instruction: Answer any two questions only

13. (a) Given that y = (2x + 5)! 4x – 5 .
dy ax
(i) Show that = such as a is a constant and state the value of a.
dx ! 4x – 5
∫
(ii) Hence, find the value of x dx. [4 marks]
! 4x – 5
©PAN ASIA PUBLICATIONS
4
(b) Diagram 8 shows a part of curve y = , line x = 1, x = 2 and y = 4.
x 2
y
5
x = 1
4 y = 4

3 Q
2
x = 2
1 4
P y = — 2
x
x
0 1 2 3 4 5
Diagram 8
4
2
(i) Find the area of region P, in unit , bounded by the curve y = , x-axis and line x = 1 and x = 2.
x 2
2
(ii) Find the area, in unit , of region Q that bounded by the curve of y = 4 and lines x = 2 and y = 4.
x 2 SPM MODEL PAPER
[4 marks]
Answer:
(a) (i) (ii)














(b) (i) (ii)



















425




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Additional Mathematics SPM Model Paper
Paper 2

Time: 2 hours and 30 minutes
Section A
(50 marks)
Instruction: Answer all the questions
1. A married couple has three sons. At one time, the eldest child is 4 years older than the total age of his younger
1
brothers. Two years ago, the age of the youngest child was of the age difference between his older brothers.
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Next seven years later, age of three of them is 51 years old. 4
Determine the age of each son of the couple. [6 marks]


2. Diagram 2 shows the mapping x to y which drerived by the function of f : x ˜ 9x – a and mapping of y to z
is derived by the function of g : y ˜ b , y ≠ 12.
12 – y

x y z


3
3
4
SPM MODEL PAPER (a) Find the values of a and of b. Diagram 2





(b) Express the function that maps the element x to the element z in a similar form.
[6 marks]
(c) Determine the element x that does not change when mapped to the z.
2
2
3. (a) The roots of equation 3x – 2kx + k + 4 = 0 are a and b. If a + b  = 16 , find the possible value of k.
2
9 [3 marks]
2
(b) Diagram 3 shows a part of graph function f(x) = (x – 3) – 2 with minimum point A(h, 2k) and intersect
the f(x)-axis at 7. This graph is moved by 3 units to the right and 4 units upwards with new minimum
point, B as shown in the Diagram 3.
f(x)






7

f(x) = (x – 3) – 2
2
B
0
A (h, 2k)
Diagram 3
Determine
(i) the value of h and of k,
(ii) the coordinate of minimum point, B,
(iii) the new equation of f(x) in the vertex form. [4 marks]

428




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Additional Mathematics SPM Model Paper
7. (a) Solve the equation of 6 tan q + 13 sec q = 2 for 0° < q < 360°. [3 marks]
2
1
(b) Diagram 7 shows the graph of y = a sin x + c, such as a and b are positive integers and c is an integer
b
for 0 < x < 4π.
y
1
y = a sin— x + c
1 b
x
0 π 2π 3π 4π
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–2




–5

Diagram 7
Given the graph that passed through point (0, –2) has maximum point at (π, 1) and minimum point
at (3π, –5).
(i) Find the values of a, b and c.
1
(ii) Sketch the graph of y = | a sin x + c| for 0 < x < 4π. [5 marks]
b Section B

SPM MODEL PAPER 8. Table 8 shows the values for two variables, x and y, obtained from an experiment. The variable x and y are
(30 marks)
Instruction: Answer any three questions only


n
related by the equation y = ax , such as a and n are constants.
y
x

1.2 1.5


1.6 1.8

2.5 2.6


4.0 3.6

6.3 5.0


8.0 5.8

Table 8
(a) Plot log y against log x, by using the scale of 2 cm for 0.1 unit on both axes. Hence, sketch a best line
10
10
of fit. [5 marks]
(b) Based on the graph at 8(a), find the value of
(i) a,
(ii) n. [5 marks]



430




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