Focus KSSM SPM 2021 TG4.5 - Maths DLP

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CC038231
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4∙5
MATHEMATICS KSSM SPM





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CC038231 Textbook
ISBN: 978-967-2930-41-9 NEW SPM ASSESSMENT
KSSM
Mathematics Form 5
Ng Seng How • Ooi Soo Huat FORMAT 2021
PELANGI Samantha Neo • Yong Kuan Yeoh




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Format: 190mm X 260mm TP Focus SPM Maths BI pgi



















MATHEMATICS SPM





Form
4∙5



KSSM

Ng Seng How
Ooi Soo Huat
Samantha Neo
Yong Kuan Yeoh














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TP_Focus_SPM_1PP.indd 4 1/25/21 5:48 PM

CONTENTS









Mathematical Formulae iv
Chapter Linear Inequalities in Two
Form 4 6 Variables 85
6.1 Linear Inequalities in Two Variables 86
Chapter Quadratic Functions and 6.2 System of Linear Inequalities in Two
1 Equations in One Variable 1 Variables 91
SPM Practice 6 97
1.1 Quadratic Functions and Equations 2
SPM Practice 1 13 Chapter
7 Graphs of Motion 101
Chapter 7.1 Distance-Time Graphs 102
2 Number Bases 16 7.2 Speed-Time Graphs 106

SPM Practice 7 113
2.1 Number Bases 17
SPM Practice 2 24 Chapter
8 Measures of Dispersion for 118
Ungrouped Data
Chapter
119
3 Logical Reasoning 26 8.1 Dispersion 121
Measures of Dispersion
8.2
SPM Practice 8 135
3.1 Statements 27
3.2 Argument 36 Chapter
SPM Practice 3 45 9 Probability of Combined
Events 139

Chapter 9.1 Combined Events 140
Dependent Events and
9.2
4 Operations on Sets 48 Independent Events 141
9.3 Mutually Exclusive Events and
4.1 Intersection of Sets 49 Non-Mutually Exclusive Events 147
4.2 Union of Sets 53 9.4 Application of Probability of
4.3 Combined Operation on Sets 59 Combined Events 154
SPM Practice 4 64 SPM Practice 9 157

Chapter
Chapter 10 Consumer Mathematics:
5 Network in Graph Theory 69 Financial Management 161

10.1 Financial Planning and
5.1 Network 70 Management 162
SPM Practice 5 81 SPM Practice 10 172




ii






00 Content Focus SPM Math 2021.indd 2 17/02/2021 5:06 PM

Form 5 Chapter
6 Ratios and Graphs of 260
Trigonometric Functions
Chapter
1 Variation 175 6.1 The Value of Sine, Cosine and Tangent
for Angle q, 0° < q < 360° 261
1.1 Direct Variation 176 6.2 The Graphs of Sine, Cosine and
1.2 Inverse Variation 181 Tangent Functions 270
1.3 Combined Variation 185 SPM Practice 6 277
SPM Practice 1 188
Chapter Measures of Dispersion for

Chapter 7 Grouped Data 283
2 Matrices 191
7.1 Dispersion 284
2.1 Matrices 192 7.2 Measures of Dispersion 294
2.2 Basic Operation on Matrices 194 SPM Practice 7 301
SPM Practice 2 206
Chapter
Chapter 8 Mathematical Modeling 305
3 Consumer Mathematics: 209 8.1 Mathematical Modeling 306
Insurance
SPM Practice 8 316
3.1 Risk and Insurance Coverage 210
SPM Practice 3 219
SPM Model Paper 318

Chapter Answers 335
4 Consumer Mathematics: 222
Taxation

4.1 Taxation 223
SPM Practice 4 232

Chapter
5 Congruency, Enlargement and
Combined Transformations
234
5.1 Congruency 235
5.2 Enlargement 239
5.3 Combined Transformation 247
5.4 Tessellation 254
SPM Practice 5 257







iii






00 Content Focus SPM Math 2021.indd 3 17/02/2021 5:06 PM

Chapter Learning Area: Relationship and Algebra
Form 4

6 Linear Inequalities in Two

Variables






You need more
exercise. My weight is 80 kg and my
height is 1.7 m. Where is my
category in body mass index?








Body Mass Index (BMI)
2.00

1.90

1.80 Underweight
Height (m) 1.70 Normal Overweight


1.60 Obese Clinically obese

1.50

1.40
40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125 130
Weight (kg)




• Boundary line – Garis sempadan
• Half-plane – Separuh satah
• Linear inequality – Ketaksamaan linear
• Linear inequalities in two variables – Ketaksamaan linear dalam dua pemboleh ubah
• Region – Rantau Concept
• System of linear inequalities – Sistem ketaksamaan linear map
• Variable – Pemboleh ubah

Linear inequalities in two variables are often used to find the solution for a system of linear
inequalities that involve some constraints. For example, the manager of a company makes inventory
so that the waste of capital in stock storage would not occur, engineers distribute raw materials


in the best way so that the cost is minimized and so on.







85






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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

6.1 Linear Inequalities in Two (c) Let x = the number of rattan basket ordered
and y = the number of rattan chair ordered.
Variables The number of days needed to produce a rattan
basket is 3 days while to produce a rattan chair
A Representing situations in the form is 5 days. Total number of days to complete the
of linear inequalities order of rattan basket and rattan chairs is less

Linear inequality in two variables is an unequal than 45 days.
relation between two variables where the highest Therefore, 3x + 5y  45.
power of the variables is one. For example, x + y  1,
2x – y  5. Try Question 1 in Try This! 6.1

1 B Verifying the conjecture about the

Represent the following situations in the form of points in the region and the solution Form 4
of certain linear inequalities
linear inequalities.
Form 4
(a) Encik Zaki has two children who are studying 1. The straight line drawn on a Cartesian plane
in Indonesia and Australia respectively. The call divides the plane into two regions (half-plane).
rate to Indonesia is RM0.28 per minute while The straight line is called boundary line.
the call rate to Australia is RM0.66 per minute.
Encik Zaki wants to make a phone call to both of y y
his children and he has prepaid of RM30 in his Region
above
handphone. Region y = mx + c Region y = mx + c
above
(b) In a written quiz, the correct answer for a x below x
multiple choice question will be given 2 marks O Region O
while the correct answer for a structural question below
will be given 5 marks. Participants who earn a
total score of more than 50 are eligible to the y y
second round.
(c) Pak Abu needs 3 days to produce a rattan basket Region above Region Region
and 5 days to produce a rattan chair. He receives y = h on the on the
an order to produce x rattan baskets and y rattan O x left O right x
chairs that need to be completed in less than 45 Region below x = k
days.
Solution
(a) Let e = the duration of call, in minute made to 2. Observe the following diagram. The straight
line y = x – 1 is a straight line that divides the
Indonesia and f = the duration of call, in minute Cartesian plane into two regions..
made to Australia. The call rate to Indonesia is (a) All points that lie on the straight line, for
RM0.28 per minute while to Australia is RM0.66 example, E(–3, –4), satisfy the equation
per minute. Total prepaid is RM30. y = x – 1.
Therefore, 0.28e + 0.66f  30. (b) All points that lie in the region above the
(b) Let x = the number of correct answer of multiple straight line, for example, A(–4, 0), satisfy
choice questions and y = the number of correct the equation y  x – 1.
answer of structural questions. The marks for the (c) All points that lie in the region below the
correct answer for a multiple choice question is 2 straight line, for example, C(5, –2), satisfy
while for a structural question is 5. Participants the equation y  x – 1.
with the total marks of more than 50 is eligible
to the second round.
Therefore, 2x + 5y  50.




86






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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

Point A (–4, 0) Point B (2, 3) (c) For point (–6, –7)
y-coordinate y-coordinate y 2x + 5
p p
0  –4 – 1 3  2 – 1 –7 2(–6) + 5
q q –7 = –7
x-coordinate x-coordinate
y ∴ y = 2x + 5
4 Therefore, point (–6, –7) lies on the straight line
Half-plane B
y = x – 1 y = 2x + 5.
2
A x 3
–6 –4 –2 0 2 4 6 Determine whether each of the following point is the
–2 C Form 4
D solution for the linear inequalities given.
E Half-plane (a) (4, 5); y  –x + 2
Form 4
–4
(b) (–1, 2); y  3x – 4
(c) (–3, –7); 2x + y  –5
Point C (5, –2) Point D (0, –3) (d) (–3, 3); y  x + 6
y-coordinate y-coordinate
p p Solution
–2  5 – 1 –3  0 – 1
q q (a) y  -x + 2
x-coordinate x-coordinate
Left side Right side
3. Points that lie on the straight line, for example 5 –4 + 2
E(–3, –4) are the solutions for the straight line 5 -2
y = x – 1. 

2 Point satisfies
Determine whether each of the following point lies the inequality.
on the straight line, in the region above or below the Therefore, point (4, 5) is the solution of linear
straight line y = 2x + 5. inequality y  –x + 2.
(a) (–4, 2) (b) (1, 3)
(c) (–6, –7) (b)
y  3x – 4
Solution Left side Right side
(a) For point (–4, 2) 2 3(-1) – 4
y 2x + 5 2 -7
2 2(– 4) + 5 
2  -3
Point does not satisfy
∴ y  2x + 5 the inequality.
Therefore, point (–4, 2) lies in the region above Therefore, point (–1, 2) is not the solution of
the straight line y = 2x + 5. linear inequality y  3x – 4.
(b) For point (1, 3)
(c) 2x + y –5
y 2x + 5 
3 2(1) + 5 Left side Right side
3  7 2(-3) + (–7) –5
-13  –5
∴ y  2x + 5
Therefore, point (1, 3) lies in the region below the Therefore, point (–3, –7) is not the solution of
straight line y = 2x + 5. linear inequality 2x + y  –5.

87






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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables
(d) y  x + 6 (c) the points that lie in the region below

Left side Right side the straight line satisfy the equation
y  mx + c.
3 -3 + 6
3 = 3 y
y mx + c
Therefore, point (–3, 3) is the solution of linear y mx + c
inequality y  x + 6.
O x
y mx + c
REMEMBER!
2. For the graph of straight line y = h,
The symbol  means less than or equal to. Form 4
(a) the points that lie on the straight line satisfy
the equation y = h.
Form 4
(b) the points that lie in the region above the
Try Questions 2 – 4 in Try This! 6.1
straight line satisfy the equation y  h.
(c) the points that lie in the region below the
Example of HOTS straight line satisfy the equation y  h.
HOTS Question
In a Cartesian plane, there are two points, P and y
Q, and a straight line R. P(3, –5) lies on the straight
line, y = 4 – 3x and point Q(1, 2) lies in the region y h
above the straight line. Both points P and Q are not y h
the solutions of inequality R. Determine the linear
inequality R. O x
y h
Solution:
Point P lies on the straight line y = 4 – 3x and point
Q lies in the region above the straight line y = 4 – 3x. 3. For the graph of straight line x = k,
Points P and Q are not the solution of linear inequality
R. Therefore, linear inequality R is y  4 – 3x. (a) the points that lie on the straight line satisfy
the equation x = k.
Try this HOTS Question (b) the points that lie in the region on the right
On a Certesian plane, there is a straigh line E, of the straight line satisfy the equation
y = 2x + 5 and three points K, L and M. Given that x  k.
point K lies in the region below the straight line E, (c) the points that lie in the region on the left
point L lies on the straight line E and point M lies in of the straight line satisfy the equation
the region above the straight line E. Determine the x  k.
linear inequality E if only point K is the solution.
Answer: y  2x + 5 y

x k
x
O
C Determining the region that satisfies x k
a linear inequality
x k
1. For the graph of straight line y = mx + c,
(a) the points that lie on the straight line satisfy 4. If the boundary of the shaded region is a dashed
line, then the points on the boundary line are
the equation y = mx + c. not included in the region that satisfies the
(b) the points that lie in the region above the inequality given.
straight line satisfy the equation y  mx + c.



88






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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables
5. If the boundary of the shaded region is a solid (b) y  x + 3
line, then the points on the boundary line are
included in the region that satisfies the inequality y
given.
y x + 3
Type of x
Inequality straight Points on the O
symbol straight line
line

Dashed Not included
 or  line in the solution Solution
region.
(a) y Form 4

Solid line Included in the x 2
Form 4
 or 
solution region.
x
O
6. The following diagrams shows the region that
satisfies the given inequality. (b) y

y y x + 3
The regions do
y mx + c y mx + c not include the O x
points lie on the
x straight line
O y = mx + c.
y mx + c
5


y State the inequality that defines each of the following
shaded regions.
The regions
y mx + c y mx + c include the (a) y
points lie on the
x straight line 1
O y = mx + c. y = x – 1
—
3
y mx + c x
0
4
(b) y
Shade the region that satisfies the given linear
inequality.
x y = 2
(a) x  2
x
0
y
x 2 Solution
1
x (a) Line y = x – 1 is a solid line and the shaded
O 3
region lies above the line. Therefore, the
1
inequality is y  x – 1.
3


89






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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables
(b) Line x + y = 2 is a dashed line and the shaded Try This! 6.1
region lies below the line. Therefore, the
inequality is x + y  2. 1. Represent the following situations in the form of
linear inequalities.
Try Questions 5 – 7 in Try This! 6.1 (a) A patient is at risk of having a heart problem if
the blood pressure in the ankle, k mm Hg, is
6 less than 90% of the blood pressure in the arm,
Draw and shade the region that satisfy the following p mm Hg.
linear inequalities. (b) Aisyah’s father and Haida’s father gives pocket
(a) y  3x + 1 (b) y  –2x + 3 money several times each month respectively.
1 Each time, Aisyah will get RM5 while Haida
(c) y  x + 2 gets RM8. Let’s say Aisyah’s father gave her
2 pocket money for a times while Haida’s father
Solution gave her pocket money for h times last month. Form 4
(a) y = 3x + 1 • Convert the given The total amount of cash they received last
month was more than RM50.
Form 4
x –1 1 linear inequality to (c) Ananda has a 2.4 m of wood to make a
linear equation form
y –2 4 to draw the straight rectangular photo frame for the project of the
line. Living Skills subject.
y
6 2. Determine whether each of the following point lies
• Check the
4 inequality symbol on the straight line, in the region above or in the
y 3x + 1 and draw the region below the straight line y = –2x + 7.
2 straight line. (a) (1, 5)
x • Shade the region (b) (2, 4)
–4 –2 O 2 4 that satisfies the (c) (0, –3)
–2 inequality.
3. Determine whether each of the following points
satisfies y = 4 – 5x, y  4 – 5x or y  4 – 5x.
(b) y = –2x + 3 (a) (–1, 7)
x –1 2 (b) (2, – 6)
y 5 –1 (c) (3, 0)
4. Determine whether each of the following point is the
y
solution for the linear inequalities given.
4 (a) (3, – 5); y  –3x + 8
y = –2x 3 (b) (– 6, 10); y  x + 5
2
(c) (–1, –3); x – 2y  4
x (d) (2, 2); 4 + x  3y
–2 0 2 4
–2
5. Shade the region that satisfies the given inequalities.
(a) y  3x + 2
1
(c) y = x + 2
2 y
x –2 2 y = 3x 2

y 1 3 0 x
y

4
1
2 y = x + 2 (b) y  –1
2 y
x
–2 0 2 4
–2 x
O
y –1
Try Question 8 in Try This! 6.1

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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables
1 (b)
(c) y  x + 1 y
2
y
x
0
1
—
y = x 1 y = 3x – 1
2
0 x

(d) y  2x + 7 (c) y
y 1
—
y = – x 1
y = 2x 7 2
Form 4
0 x
x
0
Form 4

8. Draw and shade the region that satisfy the following
(e) y  –x + 1 linear inequalities.
(a) y  –x – 1
y
(b) y  2x
1
(c) y  – x – 1
y = –x 1 2
0 x (d) 1 x – y  2
3


6. For each of the following graph, determine whether
the shaded region satisfies the inequality given. Systems of Linear Inequalities
(a) y  –2x – 5 6.2 in Two Variables
y

x
0 A Representing situations in the form
y = –2x – 5 of system of linear inequalities
1. System of linear inequalities in two variables
is a set of two or more linear inequalities with
(b) x – y  3 similar variables. For example, x + 2y  3 and

y y  x – 4 is a system of linear inequalities in two
variables that consists of two linear inequalities.
x
0
2. The table below shows the linear inequalities
x – y = 3 used to represent certain situations.
Linear
Example of situation
inequality
7. State the inequality that defines each of the following y is greater than x y  x
shaded regions.
(a) y y is less than x y  x

3 y is not less than x y  x
y = – x
—
2
0 x y is not more than x y  x
y is at least k times x y  kx



91






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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

Linear (ii) Kim does not want to spend more than
Example of situation RM20 000 for the laptop inventory every
inequality
time.
y is at most k times x y  kx
Given the cost of laptop A is RM2 000 and laptop
Minimum of y is k y  k B is RM2 300.
Maximum of y is k y  k Solution
Sum of x and y is greater x + y  k (a) Let x represents the number of notebook and y
than k represents the number of drawing book.
Difference between y
and x is less than k y – x  k First inequality:
3x + 7y  30 Spend not more than the
Sum of x and y is at most x + y  k voucher value of RM30.
k Form 4
Second inequality:
y is more than x by at y – x  k x  y Number of notebook purchased is less
Form 4
least k than drawing book.


7 (b) Let x represents the number of bottles for
Represent each of the following situation in the form pineapple jam and y represents the number of
bottles for strawberry jam.
of system of linear inequalities.
(a) Alia has a RM30 voucher that can be used at 3x + 11 y  11 First inequality:
Total operation time for cooking
a stationery store. A notebook and a drawing 4 process.
book cost RM3 and RM7 respectively. Alia plans
5
to spend not more than the voucher value. The 5 x + y  8 Second inequality:
Total operation time for
number of notebooks to be purchased is less 4 6 cooling process.
than the number of drawing books.
(b) A factory produces pineapple and strawberry (c) Let x represents the number of laptop A and y
jam. The table shows the process and time used represents the number of laptop B.
to produce the jam.
x  2y First inequality:
Pineapple Strawberry The number of laptop A is at most
Process 2 times the number of laptop B.
jam jam
2 000x + 2 300y  20 000
3
Cooking (hour) 3 2 4 Second inequality: Inventory of laptop A
and laptop B is not more than RM20 000.
1 5
Cooling (hour) 1 4 6
Try Question 1 in Try This! 6.2
The cooking and cooling section operate for 11
hours and 8 hours daily respectively. In a day, the
factory is able to produce x bottles of pineapple B Verifying the conjecture about the
jam and y bottles of strawberry jam. points in the region and the solution
(c) Kim sells two types of laptop in his shop. He of linear inequalities system
needs to make inventory for the laptops every
month. The following is his consideration when 1. Points that satisfy all linear inequalities lie
preparing the inventory. in the common region of the inequalities.
(i) Inventory of laptop A is at most two times Therefore, the points are known as the solutions
the inventory of laptop B. of the system of linear inequalities.




92






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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables
1
8 (b) y  x, y  2x − 1 and y  −x + 6
2
Determine whether each of the following point y y 2x – 1
satisfies the inequalities y  –x + 3 and y  2x – 5.
(a) (–1, 2) y x
1
(b) (4, –3) 2
x
O
Solution y –x + 6
(a) Solution
True / True / (a)
Point y  –x + 3 y  2x – 5 • Mark the region that satisfies y  −x + 2 with
False False dots.
(–1, 2) 2  –(–1) + 3 True 2  2(–1) – 5 True • Mark the region that satisfies x  −1 with lines.
• Mark the region that satisfies y  0 with triangles. Form 4
Point (–1, 2) satisfies both inequalities y  –x + 3 • Shade the common region that marked with the
Form 4
and y  2x – 5. three markings.
(b) y
True / True /
Point y  –x + 3 y  2x – 5
False False
(4, –3) –3  –4 + 3 True –3  2(4) – 5 False x
O = 0 is x-axis.
Point (4, –3) does not satisfy both inequalities x –1 y –x + 2
y  –x + 3 and y  2x – 5.
Alternative method
y  −x + 2 ⇒ The region below the line
Try Question 2 in Try This! 6.2
y = −x + 2.
x  −1 ⇒ The region on the right of the line
C Determining and shading the region x = −1.
that satisfies a linear inequality y  0 ⇒ The region above the line y = 0.
system (b)
1
1. The region that satisfies a system of linear • Mark the region that satisfies y  x with A.
2
inequalities can be determined by the following • Mark the region that satisfies y  2x − 1 with B.
• Mark the region that satisfies y  –x + 6 with C.
steps: • Shade the common region that marked with the
(a) Mark the region of each linear inequality three letters.
with different marking.
(b) Identify the common region (intersection y y 2x – 1
of regions) of the markings. A
(c) Shade the common region completely. B A
1
C C A y x
2
9 A B B C B x
Shade the region that satisfies the given system of O y –x + 6
linear inequalities.
(a) y  −x + 2, x  −1 and y  0 Alternative method
1 1
y y  x ⇒ The region above the line y = x.
2 2
y  2x − 1 ⇒ The region below the line
y = 2x − 1.
y  −x + 6 ⇒ The region below the line
x y = −x + 6.
O
x –1 y –x + 2 Try Questions 3 – 5 in Try This! 6.2


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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

D Solving problems involving systems The three linear inequalities that satisfy the
1
of linear inequalities in two variables shaded region are y  – x − 1, y  x + 2 and
2
x  2.
10
State the three inequalities that define the shaded 11
region in the diagram below. Write the three linear inequalities that satisfy the
(a) y given shaded region.
y 3
y

x 4
O
y 2x – 3 2 Form 4
x
–2 O 2 4
(b) y
Form 4
y x + 2
Solution
1 The shaded region is bounded by three straight lines.
y – x – 1
2
y
x
O
4
3
x 2 2 2
x
Solution –2 O 1 2 4
(a) The three straight lines involved are y = 3, x = 0
and y = 2x − 3. Equation of straight line 1
The shaded region is below the line y = 3 and is Straight line on the x-axis, thus y = 0
drawn with dashed line. The shaded region is above the line y = 0 and is drawn
Thus, y  3. with solid line. Thus, y  0.
The shade region is on the right of y-axis. Thus, Equation of straight line 2
x  0.
The shaded region is above the line y = 2x − 3 Gradient, m = 0 – 4 = 2
–2 – 0
and is drawn with solid line. y-intercept, c = 4
Thus, y  2x − 3. Thus, y = 2x + 4.
The three linear inequalities that satisfy the
shaded region are y  3, x  0 and y  2x − 3. The shaded region is below the line y = 2x + 4 and is
drawn with dashed line.
1 Thus, y  2x + 4.
(b) The three straight lines involved are y = – x − 1,
2
y = x + 2 and x = 2. Equation of straight line 3
1
The shaded region is above the line y = − x − 1 Gradient, m = 0 – 4 = −1
2
and is drawn with dashed line. 4 – 0
1
Thus, y  – x − 1. y-intercept, c = 4
2 Thus, y = −x + 4
The shaded region is below the line y = x + 2 and The shaded region is below the line y = −x + 4 and is
is drawn with solid line. drawn with solid line. Thus, y  −x + 4.
Thus, y  x + 2.
The shade region is on the left of line x = 2 The three linear inequalities that satisfy the shaded
and is drawn with solid line. Thus, x  2. region are y  0, y  2x + 4 and y  −x + 4.




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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

12 SPM Highlights

An art centre offers two classes, handicraft and On the graph below, shade the region that satisfies
1
drawing. The number of pupils in the handicraft all the three inequalities y  –x + 7, y  x – 2 and
2
class is x and the number of pupils in the drawing x  2. y
class is y. The maximum number of pupils for both
classes is 30 pupils. The number of pupils in handicraft 6
class is at most twice the number of pupils in drawing 4 y = –x 7
class.
2
(a) Write two linear inequalities other than x  0 and x
y  0 that represent the situation above. O 2 4 1 6
—x – 2
–2 y = Form 4
2
(b) Draw and shade the region that satisfies above
system of linear inequalities. Solution
Form 4
y
(c) From the graph, x = 2
(i) if the number of pupils in handicraft class 6
is 15, find the number of pupils in drawing 4 y = –x 7
class. 2
(ii) can this art centre receive 25 pupils in
handicraft class? Give your justification. 0 2 4 6 x
1
—x – 2
–2 y =
2
Solution
(a) x + y  30 ⇒ y  −x + 30
1
x  2y ⇒ y  x
2 Try This! 6.2
(b) y(drawing) 1. Represent each of the following situations in the
form of system of linear inequalities.
30 x + y 30 (a) Sekolah Setia holds a charity concert in a hall.
20 Two types of concert ticket, VIP and Special,
2y x will be sold among parents. The number of VIP
10 ticket sold is x and the number of Special ticket
sold is y. The hall can provide not more than
x
O 10 20 30 (handicraft) 1 000 seats. The number of Special tickets
must exceed twice the number of VIP tickets.
(b) Lily uses flour and sugar to bake cakes. Given
(c) (i) If the number of pupils in handicraft class the mass of sugar, s kg, being used is less than
is 15, the maximum number of pupils in half the mass of flour, t kg. The price of sugar
is RM2 per kilogram while the price of flour is
drawing class = 15. RM2.50 per kilogram. The total cost of making
a cake is less than RM50.
(ii) No. This is because the value x = 25 is
outside the shaded region. (c) Hassan likes to eat burgers and chicken
nuggets. However, Hassan has been advised
by doctor that his fat and salt intake for
Try Questions 6 – 11 in Try This! 6.2 each meal should not exceed 18 g and 1.6 g
respectively. The following table shows the
fat and salt contain in a burger and chicken
nugget.





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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

Food Fat (g) Salt (g) 6. State three inequalities that define the shaded region
in the diagram below.
Burger 8 1.2
Chicken nugget 3 0.2 y

Let the number of burgers taken is x and the
number of nuggets taken is y. y = x – 2
x
0
2. Determine whether each of the following points is x 2y = 4
the solution of the system of linear inequalities.
(a) (3, –2); y  –2x + 1, x – 2y  4 7. State four inequalities that define the shaded region
1
(b) (–1, 7) ; y  x + 9, y  x in the diagram below.
3
(c) (7, – 4) ; y  4x , y  –2x + 10 y Form 4
y = 2x 2
1
Form 4
—x
3. The diagram below shows three straight lines. y =
2
y x
y = 2x 3 0 y = –x 5

y = –2x 3
0 x
1 8. State three inequalities that define the shaded region
y = – x – 1 in the diagram below.
—
2
y
Shade the region that satisfies the system of linear
1
inequalities y  – x – 1, y  2x + 3 and y  –2x + 3.
2 y = –2x 4
4. On the diagram below, shade the region that x 2y = 4
satisfies the system of linear inequalities y  2x, 0 x
y  –x + 2 and y  0.

y
y = 2x 9. On the diagram below, shade the region that
satisfies the system of linear inequalities y  –x + 4,
y  2x + 4 and y  –1.
x y
0
y = –x 2 y = 2x 4
4
2
5. On the diagram below, shade the region that y = –x 4
satisfies the system of linear inequalities y  1 x, –2 –1 0 2 4 6 x
3
2y  x – 2, y  –x + 4 and y  0. y = –1
y
10. Mr Goh wants to plant two types of flower trees, C
1
y = and D in his house area. The table below shows the
—x
3 prices of a tree C and of a tree D.
2y = x – 2
x
0 Tree Price per tree (RM)
y = –x 4 C 4

D 6



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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables
Given the number of tree C planted should not 11. A patient needs a minimum of 100 units nutrient A
more than the number of tree D by 5 and the cost and less than 120 units nutrient B per day. The table
of planting trees must less than RM120. Let the below shows the content of both nutrients in two
number of tree C planted is x and the number of tree types of food.
D planted is y.
(a) Write two linear inequalities other than x  0 Food Nutrient content in every 1 gram
and y  0 which represent the above situation. Nutrient A Nutrient B
(b) Draw and shade the region that satisfies the P 0.5 unit 0.2 unit
above system of linear inequalities. Q 0.2 unit 0.3 unit
(c) From the graph,
(i) determine the maximum number of tree Given that the intake of food P is not more than
C that can be planted if Mr Goh planted 5 twice the intake of food Q. Let x be the number of
tree D. intake of food P and y be the number of intake of
(ii) can Mr Goh plant 20 tree D? Justify your food Q.
answer. (a) Write a system which consists of three linear Form 4
inequalities that represent the above situation.
(b) Using a scale of 2 cm to 100 grams on both
Form 4
axes, construct and shade the region that
satisfies the system of linear inequalities.
(c) From the graph, if the patient takes 270 grams
of food P, find the maximum number of grams
of food Q that he can take on that day.





SPM Practice 6
SPM Practice


PAPER 1 4. Which of the points is the solution of the linear
inequality x – 3y  4?
1. Nutritionists recommend that fat calories, y kcal, A (1, 3) C (5, –1)
which consumed daily, must be at most 30% of B (2, –1) D (–3, –7)
the total calories, x kcal, taken daily. 5. Choose the inequality that represents the graph
below.
Choose linear inequality that represent the situation y
above.
A y  0.3x 4
B y  30x 2
C y  0.3x
D y  30x –4 –2 0 2 4 x
–2
2. Which of the points lies below the line y = 4x + 3?
A (2, 15) –4
B (–3, 0)
C (–1, –2) A y  1 – 2x C y  1 – 2x
D (1, 7) B y  1 – 2x D y  1 – 2x
6. Which of the following is the correct statement for
3. Point P(2, –3) satisfies inequality Q. Which of the the region of inequality y  2x + 7?
following is inequality Q? A Line y = 2x + 7 is a dashed line and the region
A y  7 + x above the line is shaded.
B y  4x – 6 B Line y = 2x + 7 is a solid line and the region
C y  1 – 3x below the line is shaded.
D y  5 – 2x C Line y = 2x + 7 is a dashed line and the region
below the line is shaded.
D Line y = 2x + 7 is a solid line and the region
above the line is shaded.


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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables
7. A carpenter makes two types of cupboards, P and PAPER 2
2
Q. The production of cupboard P requires 8 m plank
2
while cupboard Q requires 12 m plank. He only 1. Write a linear inequality for the situation below.
has a plank of 200 m . The number of cupboard P An electrical store wants to deliver x refrigerators
2
produced should be twice the number of cupboard and y washing machines, using a truck with a
Q. Let the number of cupboard P is x and the load of 1 000 kg to the customers. The mass of
number of cupboard Q is y, form two inequalities a refrigerator and a washing machine are 110 kg
based on the above situation. and 90 kg respectively.
A 8x + 12y  200, x  2y
B 8x + 12y  200, x  2y
C 8x + 12y  200, x  2y 2. Which of the following points is the solution for each
D 8x + 12y  200, x  2y of the following inequalities.
2
(a) y  – x – 1; point (6, – 4) or (–3, 0)
8. The following points satisfy the system of linear 3 Form 4
inequalities 2y – x  5 and y  3x except (b) y  –4x + 3; point (3, 1) or (–2, 4)
A (–2, –1)
Form 4
B (5, –3) 3. State the inequality that defines each of the following
C (1, 3) shaded region.
D (–4, –2) (a) y

9. The graph below shows a system of linear y = 5 – 2x
inequalities.
y x
0
4
y = –x – 1
2 (b) y
1
x —
–4 –2 0 2 4 y = x
2
1 –2 x
—
y = x 0
2
–4

Choose inequalities that define the shaded region. 4. Shade the region that satisfies
1 (a) y  2x + 7
A y  x, y  –x – 1
2
1 y
B y  x, y  –x – 1
2 y = 2x 7
1
C y  x, y  –x – 1
2
1
D y  x, y  –x – 1
2 0 x

10. Inequality P and inequality Q in a system of linear (b) x – 2y  6
inequalities do not have a solution. Given inequality
P is y  3 – 2x, choose the possible inequality Q. y
A y  –2x – 1 HOTS x – 2y = 6
B y + 2x  –1 Analysing x
0
C y + 2x  5
D y  –2x – 5









98






06 Focus SPM Maths F4.indd 98 17/02/2021 5:24 PM

Mathematics SPM Chapter 6 Linear Inequalities in Two Variables
5. Draw and shade the region that satisfies the 9. On the diagram below, shade the region that
following linear inequality. satisfies the system of linear inequalities x + y  4,
(a) y  x + 1 y  3x + 2 and y  x.
(b) y  –2x + 2 y y = 3x 2

6. Write two linear inequalities based on the situations 4 y = x
below. 2
Siti is controlling her diet. The intake of protein and x y = 4
fat for breakfast should not exceed 15 g and 12 g –2 0 2 4 x
respectively. The table below shows the protein and –2
fat content for Siti’s favourite delicacies.

Food Protein Fat Form 4
10. On the diagram below, shade the region that
Kuih koci (one piece) 3 3 satisfies the system of linear inequalities y  –2x – 5,
Form 4
y – x  –3 and y  1.
Bingka ubi kayu (one piece) 1 4 y
y – x = –3
2
Let the number of kuih koci can be taken is x and y = 1
the number of bingka ubi kayu can be taken is y. –2 0 2 4 x
–2
7. Determine whether each of the following point is the –4
solution of the system of linear inequalities given. y = –2x – 5
(a) (–2, 5); y  –x + 4, y  3x – 2
(b) (3, 1); y  –x + 3, 2x + 2y  5
11. Draw and shade the region that satisfies the
following linear inequalities.
8. Shade the region that satisfies the following system (a) y  2x + 1, y  –x – 1, y  –3x + 5
of linear inequalities on the graph given.
(b) x + 2y  6, 2x – y  1, 2y  x – 4
1 1
(a) y  – x + 4, y  –2x + 4 and y  x
2 3
12. A factory produces two types of furnitures, cupboard
and table. The production of each furniture involves
y two processes which is installing and painting. The
1
—
y = – x 4 table below shows the time spent on the installing
2 1
—x
y = and painting processes.
3
Installing Painting
x
0 y = –2x 4 (minute) (minute)
Cupboard 20 5
1
(b) y  –2x + 4, y  x + 2 and y  x – 1 Table 10 12
2
y The factory produces x units of cupboard and y units
of table daily. Given that the maximum amount of
1
—
y = x 2 time to install both furniture is 720 minutes. The total
2
y = x – 1 time to paint both furniture is over 300 minutes. The
number of tables does not exceed thrice the number
of cupboards.
x
0
y = –2x 4 (a) Write a system which consists of three linear
inequalities, other than x  0 and y  0, that
represents the situation above.


99






06 Focus SPM Maths F4.indd 99 17/02/2021 5:24 PM

Mathematics SPM Chapter 6 Linear Inequalities in Two Variables
(b) Using a scale of 2 cm to 10 units on both axes, (a) Write a system which consists of three linear
construct and shade the region that satisfies inequalities, other than x  0 and y  0 that
the system of linear inequalities. represent the situations above.
(c) From the graph, can the factory produces 20 (b) Using a scale of 2 cm to 10 m on both axes,
units of tables and 20 units of cupboards in a construct and shade the region that satisfies
day? Explain your answer. the system of linear inequalities.
(c) Ahmad fences the land with a width of 32 m.
13. Ahmad wanted to fence a rectangular land with (i) Find the length of the land so that the area
width of x m and length of y m. The installation of is maximum.
this fence should follow the conditions below. (ii) Given that the price of the fence is RM5
HOTS per metre, calculate the maximum cost of
Applying the installation.
(i) The total length of the fence shall be less than
140 m. Form 4
(ii) The width of the fence shall not exceed twice
the length of the fence.
Form 4
(iii) The length of the fence shall not exceed the
wide of the fence by 20 m.






















































100






06 Focus SPM Maths F4.indd 100 17/02/2021 5:24 PM

Mathematics SPM Answers

ANSWERS





(e) (f)
Form 4 f(x) f(x)
0 x
Chapter Quadratic Functions and Equations in –4 4
1 One Variable

x
Try This! 1.1 –2 0
1. (a) Yes, the expression only has one variable, x, and the 14. x = 7
highest power of x is 2.
(b) Yes, the expression only has one variable, u, and the 15. 2 hours
highest power of u is 2. 16. The product of two consecutive odd numbers is 323. x = 17
(c) No, because the highest power of variable x is 1.
(d) No, because the highest power of variable x is 3. 17. RM24
(e) Yes, the expression only has one variable, x, and the 18. (a) No
highest power of x is 2. (b) Yes
(f) No, because the highest power of variable x is 1. (c) No
2. (a) Parabolic shape opens up with a minimum point.
(b) Parabolic shape opens up with a minimum point. SPM Practice 1
(c) Parabolic shape opens down with a maximum point.
(d) Parabolic shape opens down with a maximum point. PAPER 1
3. (a) The shape of the graph is and the width decreased. 1. C 2. A 3. D 4. B 5. A
(b) The shape of the graph is and is on the right of the
y-axis. 6. C 7. B 8. D 9. C 10. A
(c) The shape of the graph is and moves vertically PAPER 2
downwards.
4. The graph moves 12 units vertically upwards. 1. x = 2, –9
5. f(x) = 2x + 16x 3
2
2. x = , –2
2
6. f(x) = 12x + 31x + 9 5
7. (a) f(x) = 10x – 33x – 28 (b) 10x – 33x – 86 = 0 3. 30
2
2
8. (a) f(x) = 7x + 26x – 6 (b) 7x + 26x – 126 = 0 4. 0.2 m
2
2
9. (a) No (b) Yes (c) Yes (d) No 5. P(–1, 0) and Q(4, 0)
10. (a) Yes (b) No (c) Yes (d) No 6.
5 3 f(x)
11. (a) 0, 8 (b) 0, – (c) 0,
2 2
2 2 3 3 x
(d) –6, 6 (e) – , (f) – , –8 0 2
5 5 2 2
7
12. (a) –3, –5 (b) 3, –8 (c) 4, –
2
5 2 2 9 3
(d) , – (e) 6, (f) – ,
2 3 3 2 2 –8
13. (a) f(x) (b) f(x)
10 12
7
7. f(x) = – x – 28x + 112
2
4
8. x = 4
0 2 5 x 9. 304 m 2
x
–3 0 4 10. Bus A is 90 km and bus B is 120 km away from the bus station.
(c) f(x) (d) f(x) 11. 6 seconds
3 12. x = 60, h = 45
13. They have to sell 32 slices of cheese cake with the price of
x RM5 per slice or sell 40 slices of cheese cake with the price
– – 1 0 2 x
3 –1 0 3 of RM4 per slice.
–2 – 4
14. The measurement of the photo is 15 cm × 12 cm
335



12 ANS Focus SPM Math F4_ENG.indd 335 17/02/2021 5:31 PM

Mathematics SPM Answers

Chapter SPM Practice 2
2 Number Bases
PAPER 1
1. B 2. B 3. B 4. D 5. C
Try This! 2.1
6. C 7. B 8. A 9. D 10. A
1. (a) 2 3 11. D 12. C 13. D 14. B 15. D
(b) 4 1
(c) 7 2 16. D 17. B 18. C 19. C 20. A
(d) 6 0
PAPER 2
2. (a) 500
(b) 16 1. (a) 54
(c) 7 (b) 5
(d) 12 (c) 15
(d) 144
3. (a) 55
(b) 95 2. (a) 212 3
(c) 34 (b) 21120 3
(d) 89 3. (a) 210 7
4. (a) 10001 2 (b) 65 7
(b) 100111110 2 4. (a) 293
(c) 1001100 2 (b) 51
5. (a) 201 5 (c) 63
(b) 2113 5 (d) 44
(c) 100 5 5. 210033 6
6. (a) 32 6 6. (a) 2
(b) 133 6 (b) 222 3
(c) 5515 6 7. (a) P = 333, Q = 135
7. (a) 45 8 (b) R = 23 5 , 24 5 , 30 5 , 31 5
(b) 65 8 8. (a) X = 14 5 , 20 5 , 21 5
(c) 160 8
(b) 2
8. (a) 235 8 9. (a) 110001 2
(b) 734 8 (b) 54 9
(c) 1101110 2
(d) 11100111 2 10. (a) 7
(b) 1
9. (a) 120 4
(b) 422 5 11. (a) 5
(c) 1424 6 (b) 3
(d) 1000101 2 12. (a) 3
(e) 26 8 (b) 1
(f) 261 9 13. (a) 242 5
10. (a) 11020 3 (b) 334 5
(b) 4020 5 14. (a) 45 7
(c) 200 6 (b) 51 7
(d) 27 8
11. (a) 212 6
(b) 88 9
12. Y = 121 5 , 122 5 , 123 5 , 124 5













336






12 ANS Focus SPM Math F4_ENG.indd 336 17/02/2021 5:31 PM

Format: 190mm X 260mm Extent= confirm 384 pgs (19mm) (70gsm paper) 3 imp_CRC


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