Modul A+1 MM Tingkatan 4

©PAN ASIA PUBLICATIONS

©PAN ASIA PUBLICATIONS













DWIBAHASA
Tingkatan
Matematik 4



Mathematics







Chai Mun











Pan Asia Publications Sdn. Bhd. Bonus Guru
199101016590 (226902-X) • PDF Manual Guru
No. 2-16, Jalan SU 8, • Rancangan
Taman Perindustrian Subang Utama, Seksyen 22, Pengajaran
40300 Shah Alam, Selangor Darul Ehsan, Malaysia. Tahunan
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Cetakan Pertama 2022
MODUL A+1 MATEMATIK Tingkatan 4
ISBN 978-967-466-625-5
Dicetak oleh World Line Marketing Sdn. Bhd. (1115599-K)

BAB 1 Fungsi dan Persamaan Kuadratik BAB 7 Graf Gerakan
BAB
BAB
dalam Satu Pemboleh Ubah Graphs of Motion
Quadratic Functions and Equations in 7.1 Graf Jarak-Masa/Distance-Time Graphs .................107
One Variable 7.2 Graf Laju-Masa/Speed-Time Graphs.......................111
1.1 Fungsi dan Persamaan Kuadratik Soalan Berformat SPM ...................................................120
©PAN ASIA PUBLICATIONS
Quadratic Functions and Equations ...........................1
Soalan Berformat SPM .....................................................13
BAB
BAB 8 Sukatan Serakan Data Tak Terkumpul
Measures of Dispersion for Ungrouped
BAB 2 Asas Nombor Data
BAB
Number Bases 8.1 Serakan/Dispersion .................................................124
2.1 Asas Nombor/Number Bases ....................................15 8.2 Sukatan Serakan/Measures of Dispersion ...............127
Soalan Berformat SPM .....................................................30 Soalan Berformat SPM ...................................................144



BAB 3 Penaakulan Logik Kebarangkalian Peristiwa Bergabung
BAB
BAB
Logical Reasoning BAB 9
3.1 Pernyataan/Statements ...............................................32 Probability of Combined Events
3.2 Hujah/Argument ........................................................42 9.1 Peristiwa Bergabung/Combined Events ..................147
9.2 Peristiwa Bersandar dan Peristiwa Tak Bersandar
Soalan Berformat SPM .....................................................53 Dependent Events and Independent Events.............150
9.3 Peristiwa Saling Eksklusif dan Peristiwa Tidak
BAB
BAB 4 Operasi Set Saling Eksklusif
Mutually Exclusive Events and Non-Mutually
Operations on Sets
4.1 Persilangan Set/Intersection of Sets ..........................58 Exclusive Events ......................................................155
4.2 Kesatuan Set/Union of Sets .......................................63 9.4 Aplikasi Kebarangkalian Peristiwa Bergabung
4.3 Gabungan Operasi Set Application of Probability of Combined Events ......160
Combined Operations on Sets ...................................69 Soalan Berformat SPM ...................................................164
Soalan Berformat SPM .....................................................73

BAB 10 Matematik Pengguna: Pengurusan
BAB
BAB
BAB 5 Rangkaian dalam Teori Graf Kewangan
Network in Graph Theory Consumer Mathematics: Financial
5.1 Rangkaian/Network ...................................................77 Management
Soalan Berformat SPM .....................................................87 10.1 Perancangan dan Pengurusan Kewangan
Financial Planning and Management .....................168
Soalan Berformat SPM ...................................................173
BAB
BAB 6 Ketaksamaan Linear dalam Dua
Pemboleh Ubah
Linear Inequalities in Two Variables
6.1 Ketaksamaan Linear dalam Dua Pemboleh Ubah Jawapan ...................................................................... MG–1
Linear Inequalities in Two Variables .........................91
6.2 Sistem Ketaksamaan Linear dalam Dua Pemboleh Lembaran Pentaksiran Bilik Darjah (PBD) ............ MG–9
Ubah
Systems of Linear Inequalities in Two Variables .......99
Soalan Berformat SPM ...................................................104
















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00_Modul A+ MM Tg4_Kand.indd 2 01/10/2021 3:16 PM

BAB Fungsi dan Persamaan Kuadratik dalam Satu
1 Pemboleh Ubah


Quadratic Functions and Equations in One Variable




1.1 Fungsi dan Persamaan Kuadratik Buku Teks
Quadratic Functions and Equations m.s. 2–29


Nota
©PAN ASIA PUBLICATIONS
1. Ungkapan kuadratik dalam satu pemboleh ubah ialah suatu ungkapan yang berbentuk ax + bx + c
2
dengan a, b, c sebagai pemalar, a ≠ 0 dan x ialah pemboleh ubah.
A quadratic expression in one variable is an expression of the form ax + bx + c with a, b, c as constants, a ≠ 0 and
2
x is a variable.
Contoh/Example,
2
2
2
2x + 3x + 5, x – 7x + 4, 4x – 9, 8x – x 2
2. Ciri-ciri ungkapan kuadratik dalam satu pemboleh ubah.
Characteristics of quadratic expressions in one variable.
• Ungkapan mempunyai hanya satu pemboleh ubah.
The expression has only one variable.
• Kuasa pemboleh ubah ialah suatu nombor bulat.
The power of the variable is a whole number.
• Kuasa tertinggi bagi pemboleh ubah ialah 2.
The highest power of the variable is 2.

Tip SPM

Pemboleh ubah x dalam ungkapan kuadratik juga boleh diwakili oleh huruf-huruf abjab yang lain.
The variable x in quadratic expressions can also be represented by other alphabet letters.

3. Fungsi kuadratik ialah suatu hubungan banyak-kepada-satu.
A quadratic function is a many-to-one relation.

Satu garis lurus mengufuk memotong Fungsi ialah hubungan satu-kepada-
graf fungsi pada satu titik.
A horizontal line cuts the graph of the satu.
function at one point. Ujian garis Function is a one-to-one relation.
mengufuk
Horizontal line
Satu garis lurus mengufuk memotong test Fungsi ialah hubungan banyak-
graf fungsi pada dua titik. kepada-satu.
A horizontal line cuts the graph of the Function is a many-to-one relation.
function at two points.


4. Ciri-ciri fungsi kuadratik.
Characteristics of quadratic functions.
• Graf berbentuk melengkung atau .
Graph is curved shape or .
• Satu titik maksimum atau satu titik minimum.
One maximum point or one minimum point.
• Paksi simetri graf adalah selari dengan paksi-y.
Axis of symmetry of the graph is parallel to the y-axis.











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01_Modul A+ MM Tg4.indd 1 12/10/2021 3:31 PM

BAB 1 Nilai-nilai a yang positif Titik minimum
Positive values of a
Minimum point
Bentuk graf
Shape of graph
y = ax + bx + c
2
Titik maksimum
Nilai-nilai a yang negatif Maximum point
Negative values of a

5. Punca suatu persamaan kuadratik ialah nilai bagi pemboleh ubah yang memuaskan persamaan kuadratik
©PAN ASIA PUBLICATIONS
itu.
A root of a quadratic equation is the value of the variable that satisfies the quadratic equation.
Contoh/Example,
Apabila/When x = 1,
3x – 4x + 1
2
2
= 3(1) – 4(1) + 1
= 3 – 4 + 1
= 0
∴ x = 1 ialah satu punca bagi persamaan kuadratik 3x – 4x + 1 = 0.
2
∴ x = 1 is a root of the quadratic equation 3x – 4x + 1 = 0.
2
6. Punca-punca bagi persamaan kuadratik boleh ditentukan dengan kaedah pemfaktoran.
The roots of quadratic equations can be determined by method of factorisation.
Contoh/Example,
2x + 5x – 12 = 0 2x –3 –3x
2
(2x – 3)(x + 4) = 0 x +4 +8x
2x – 3 = 0 atau/or x + 4 = 0 2x –12 +5x
2
3
x = — atau/or x = –4
2
7.
Graf fungsi kuadratik
Graphs of quadratic functions



y = ax + c y = ax + bx y = a(px + m) 2 y = a(px + m)(qx + n)
2
2
(i) a . 0 (i) a . 0 (i) a . 0 (i) a . 0
y y y y


x
c O b x x m O n
– —
– —


O x – — O m p q
– —
a
p

(ii) a , 0 (ii) a , 0 (ii) a , 0 (ii) a , 0
y y y y
c x
x x O m m O n x
– —
O O b p – — – —
– — p q


a











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01_Modul A+ MM Tg4.indd 2 12/10/2021 3:31 PM

PAUTAN INTERAKTIF
1. Tandakan ✓ bagi ungkapan kuadratik atau ✗ bagi bukan ungkapan kuadratik dalam satu pemboleh ubah.
Mark ✓ for quadratic expression or ✗ for not quadratic expression in one variable. TP 1
Contoh/Example
BAB 1
v
1
5t – 7t – 2 ✓ (a) 4q – — + — ✗ PAUTAN INTERAKTIF
2
2 v
(b) –3w + w ✗ (c) x + y ✗ Persamaan kuadratik dalam
3
2
2
2
satu pemboleh ubah
(d) 8n + 15 ✓ (e) 2m + mn – 4 ✗ https://bit.ly/3d1ZKeM
2
2
2. Padankan setiap yang berikut.
Match each of the following. TP 1
©PAN ASIA PUBLICATIONS
Contoh/Example
4 – a 2
(i) Hubungan satu-kepada-satu
One-to-one relation
(a) 6k + 3 2
(b) 8r + 10r
2
(ii) Hubungan banyak-kepada-satu
Many-to-one relation
(c) 15 – 2v


3. Tandakan ✓ untuk menunjukkan ciri-ciri fungsi kuadratik f (x) = ax + bx + c.
2
Mark ✓ to show the characteristics of quadratic function f(x) = ax + bx + c. TP 2
2
(a) Bentuk graf/Shape of graph
(i) a . 0 (ii) a , 0

✓ ✓


(b) Paksi simetri graf/Axis of symmetry of graph
Selari dengan paksi-x/Parallel to the x-axis Selari dengan paksi-y/Parallel to the y-axis ✓

4. (a) Tentukan nilai-nilai a atau c yang sepadan dengan graf I dan II berikut.
Determine the values of a or c that correspond to the following graphs I and II. TP 2
(i) 2
Graf/Graph I y = ax a . 1
y I

y = x 2
II
Graf/Graph II O x 0 , a , 1



(ii) 2
Graf/Graph I y = x + c c . 0
y
I
y = x 2

II
Graf/Graph II O x c , 0














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01_Modul A+ MM Tg4.indd 3 12/10/2021 3:31 PM

(b) Tandakan ✓ untuk menunjukkan kesan perubahan nilai a dan nilai c ke atas graf fungsi kuadratik di (a).
Mark ✓ to show the effect of the change of values of a and c on the graph of the quadratic functions in (a). TP 2
(i) Apabila nilai a bertambah, lebar graf y = ax
2
BAB 1 When the value of a increases, the width of the graph y = ax 2
2
increases narrower than the graph y = x
bertambah lebih sempit daripada graf y = x ✓
2
bertambah lebih luas daripada graf y = x
2
increases wider than the graph y = x 2
(ii) Apabila nilai c (. 0) bertambah, graf y = x + c
2
When the value of c (. 0) increases, the graph y = x + c
2
berada c unit ke atas graf y = x ✓
2
lies c units above the graph y = x 2
2©PAN ASIA PUBLICATIONS
berada c unit ke bawah graf y = x
2
lies c units below the graph y = x
2
(iii) Apabila nilai c (, 0) berkurang, graf y = x + c
2
When the value of c (, 0) decreases, the graph y = x + c
2
berada c unit ke bawah graf y = x
2
lies c units below the graph y = x 2 ✓
berada c unit ke atas graf y = x
2
lies c units above from the graph y = x 2
5. (a) Rajah menunjukkan graf fungsi y = x + bx. Tentukan nilai-nilai b yang sepadan dengan graf I dan II.
2
Diagram shows the graph of the function y = x + bx. Determine the values of b that correspond to graphs I and II. TP 2
2
Graf/Graph I b , 0
II y y = x 2 I


x
O
Graf/Graph II b . 0


(b) Tandakan ✓ untuk menunjukkan paksi simetri bagi
Mark ✓ to show the axis of symmetry for
(i) graf/graph I (ii) graf/graph II
b b
x = — x = —
2
2
b
b
x = – — ✓ x = – — ✓
2 2
6. Bentukkan satu ungkapan kuadratik bagi luas dalam setiap rajah yang berikut.
Form a quadratic expression for the area in each of the following diagrams. TP 3

Contoh/Example (a) (b) (x + 5) cm
x cm

(x + 3) cm (18 – x) cm x cm
(4x + 8) cm
Luas/Area Luas/Area
Luas/Area = (18 – x) × x = x + 2x(x + 5)
2
1
= — × (4x + 8) × (x + 3) = (18x – x ) cm 2 = x + 2x + 10x
2
2
2
2 2 2
1
= —(4x + 12x + 8x + 24) = (3x + 10x) cm
2
1
= —(4x + 20x + 24)
2
2
= (2x + 10x + 12) cm 2
2




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01_Modul A+ MM Tg4.indd 4 12/10/2021 3:31 PM

7. Bentukkan persamaan kuadratik daripada maklumat yang diberikan.
Form a quadratic equation from the given information. TP 3
Contoh/Example (a) (b)
A (3x + 2) cm B A (x + 7) cm B BAB 1
(x + 5) cm (x + 2) cm
D C
(2x – 3) cm C D x cm C
Luas segi tiga ABC = 85 cm 2 Luas trapezium ABCD
A (x + 6) cm B 2 2
Luas segi empat tepat ABCD Area of triangle ABC = 85 cm = 162 cm 2
Area of trapezium ABCD = 162 cm
1
= 50 cm 2 — × (x + 5) × (3x + 2) = 85
2
1
Area of rectangle ABCD = 50 cm 2 3x + 2x + 15x + 10 = 170 — × 3(x + 7) + x4 × (x + 2) = 162
2
2
(x + 6)(2x – 3) = 50 3x + 17x – 160 = 0 (2x + 7)(x + 2) = 324
2
4 ©PAN ASIA PUBLICATIONS
2x – 3x + 12x – 18 = 50 2x + 4x + 7x + 14 = 324
2
2
2x + 9x – 68 = 0 2x + 11x – 310 = 0
2
2
8. (a) Lengkapkan rajah berikut.
Complete the following diagram. TP 3
Contoh/Example = 1 – 1 – 2
2
= 1 – 1 – 2
= –2
x = 1
2
Nilai bagi/Value of = (–1) – (–1) – 2
(i) x = –1 x – x – 2 = 1 + 1 – 2
2
= 0
(ii) x = 2
2
= 2 – 2 – 2
= 4 – 2 – 2
= 0

(b) Tandakan ✓ bagi nilai-nilai x yang merupakan punca persamaan kuadratik x – x – 2 = 0.
2
Mark ✓ for the values of x that are the roots of the quadratic equation x – x – 2 = 0.
2
x = 1 x = –1 ✓ x = 2 ✓
9. Cari punca-punca bagi persamaan kuadratik yang berikut.
Find the roots of the following quadratic equations. TP 3

Contoh/Example (a) x(2x + 3) = 0 (b) (x – 1)(x – 4) = 0
x = 0 atau/or 2x + 3 = 0 x – 1 = 0 atau/or x – 4 = 0
(x – 7)(3x + 1) = 0 3 x = 1 atau/or x = 4
x – 7 = 0 atau/or 3x + 1 = 0 x = 0 atau/or x = – —
2
1
x = 7 atau/or x = – —
3

10. Selesaikan./Solve. TP 3
Contoh/Example (a) (4f – 1)(3f + 2) = 0
4f – 1 = 0 atau/or 3f + 2 = 0
(4k + 5)(2k – 3) = 0 4f = 1 atau/or 3f = –2
1
4k + 5 = 0 atau/or 2k – 3 = 0 f = — atau/or f = – —
2
4k = –5 atau/or 2k = 3 4 3
3
5
k = – — atau/or k = —
2
(b) (5 – 3p)(1 + 5p) = 0 (c) (3 + 2m)(2 + 7m) = 0
5 – 3p = 0 atau/or 1 + 5p = 0 3 + 2m = 0 atau/or 2 + 7m = 0
3p = 5 atau/or 5p = –1 2m = –3 atau/or 7m = –2
1
5
2
3
p = — atau/or p = – — m = – — atau/or m = – —
3 5 2 7


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01_Modul A+ MM Tg4.indd 5 12/10/2021 3:31 PM

11. Selesaikan dengan kaedah pemfaktoran.
Solve by factorisation method. TP 3
BAB 1 Contoh/Example (a) x – 4x + 3 = 0
2
(x – 1)(x – 3) = 0
2x – 5x – 12 = 0 x – 1 = 0 atau/or x – 3 = 0
2
(2x + 3)(x – 4) = 0 x = 1 atau/or x = 3
2x + 3 = 0 atau/or x – 4 = 0
3
x = – — atau/or x = 4
2


(b) 2k + 7k + 3 = 0 (c) 3p – 2p = 0
2
2
(2k + 1)(k + 3) = 0 p(3p – 2) = 0
©PAN ASIA PUBLICATIONS
2k + 1 = 0 atau/or k + 3 = 0 p = 0 atau/or 3p – 2 = 0
1
2
k = – — atau/or k = –3 p = 0 atau/or p = —
2 3



12. Cari punca-punca bagi persamaan kuadratik yang berikut.
Find the roots of the following quadratic equations. TP 3
Contoh/Example (a) 2x(2x + 1) = 2x + 1
4x + 2x = 2x + 1
2
5x(x – 3) = 2(x – 7) 4x – 1 = 0
2
5x – 15x = 2x – 14 (2x + 1)(2x – 1) = 0
2
5x – 17x + 14 = 0 2x + 1 = 0 atau/or 2x – 1 = 0
2
(5x – 7)(x – 2) = 0 x = – — atau/or x = —
1
1
5x – 7 = 0 atau/or x – 2 = 0 2 2
7
x = — atau/or x = 2
5
(b) (7 – 2f) = 9 (c) (r + 3)(r – 3) = 7
2
49 – 28f + 4f = 9 r – 9 = 7
2
2
4f – 28f + 40 = 0 r – 16 = 0
2
2
f – 7f + 10 = 0 (r + 4)(r – 4) = 0
2
(f – 2)(f – 5) = 0 r + 4 = 0 atau/or r – 4 = 0
f – 2 = 0 atau/or f – 5 = 0 r = –4 atau/or r = 4
f = 2 atau/or f = 5
13. Lakar graf bagi setiap fungsi kuadratik berikut.
Sketch the graph for each of the following quadratic functions. TP 3
Contoh/Example (a) y = –3x 2
y
y = 2x 2 y
x
O


x
O

1
(b) y = 4x + 3 (c) y = – —x – 2
2
2
y 2 y
x
O
–2
3 x
O




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01_Modul A+ MM Tg4.indd 6 12/10/2021 3:31 PM

14. Lakar graf bagi fungsi kuadratik berikut.
Sketch the graph of the following quadratic functions. TP 3
2
Contoh/Example (a) y = 2x – 8 BAB 1
y = x – 9 y
2
y Tip SPM
Nilai c ialah pintasan-y bagi graf x
x –2 O 2
2
–3 O 3 y = x + c.
The value of c is the y-intercept of –8
2
the graph y = x + c.
–9
(b) y = 16 – x 2 (c) y = —(25 – x )
1
©PAN ASIA PUBLICATIONS
2
y 2 y
16
25
—–
2
x x
–4 O 4 –5 O 5


15. Lakar graf bagi setiap fungsi berikut.
Sketch the graph for each of the following fucntions. TP 3

Contoh/Example (a) y = 2x – x 2

y = (x – 1)(x + 3) y
y

x x
–3 O 1 O 2

–3

2
(b) y = 2x + 15x + 25 (c) y = 24 – x – 3x 2
y y

24


x x
8
–5 5 O –3 O —
– —
2 3
16. Lakar graf fungsi berikut.
Sketch the graph of the following functions. TP 3

Contoh/Example (a) y = 2(x + 2) 2
y = (x – 3) 2 y
y
8

9 x
–2 O
x
O 3

(b) y = –x – 8x – 16 (c) y = –4x + 28x – 49
2
2
y y
x O x
7
–4 O —
2
–49
–16


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01_Modul A+ MM Tg4.indd 7 12/10/2021 3:31 PM

17. Rajah yang diberi menunjukkan sebuah segi empat tepat PQRS. Diberi MS = NP = x cm dan luas S M R
segi tiga MNQ ialah 38 cm . Hitung nilai-nilai x yang mungkin. 9 cm
2
The diagram given shows a rectangle PQRS. Given MS = NP = x cm and the area of triangle MNQ is

BAB 1 38 cm . Calculate the possible values of x. TP 4 N
2
Luas segi empat tepat PQRS/ Area of rectangle PQRS = 12 × 9 = 108 cm 2 P 12 cm Q
1
Luas ∆PQN/ Area of ∆PQN = —x(12) = 6x cm 2
2
)
(
1
1
9
Luas ∆MNS/ Area of ∆MNS = —x(9 – x) = —x – — x cm 2
2
2 2 2
(
)
1
9
Luas ∆QRM/ Area of ∆QRM = —(9)(12 – x) = 54 – —x cm 2
2 2
(
)
) (
1
9
9
108 – 6x – —x – —x – 54 – —x = 38
2
2
2
2
x + x – 42 = 0 ©PAN ASIA PUBLICATIONS
1
54 – 6x + —x = 38
2
2
1
—x – 6x + 16 = 0
2
2
x – 12x + 32 = 0
2
(x – 4)(x – 8) = 0
x = 4 atau/or x = 8
18. Sebidang lot banglo yang berbentuk segi empat tepat mempunyai perimeter 120 m dan luas 884 m .
2
Cari panjang dan lebarnya. y m
A rectangular bungalow lot has perimeter 120 m and area 884 m . Find its length and width. TP 5
2
2(x + y) = 120 KBAT Menganalisis x m
x + y = 60
y = 60 – x
xy = 884
x(60 – x) = 884
60x – x = 884
2
x – 60x + 884 = 0
2
(x – 34)(x – 26) = 0
x = 34 atau/or x = 26
Jika / If x = 34, y = 60 – 34 = 26
Jika / If x = 26, y = 60 – 26 = 34
∴ Panjang ialah 34 m dan lebar ialah 26 m.
∴ Length is 34 m and width is 26 m.
19. Tentukan nilai positif x yang ditunjukkan dalam rajah berikut.
Determine the positive value of x as shown in the following diagrams. TP 5
Contoh/Example (a)
x cm (x + 7) cm
(2x + 3) cm
40 cm
(2x + 3) = x + (x + 7) 2
2
2
4x + 12x + 9 = x + x + 14x + 49
2
2
2
2
(x – 1)(x + 2) cm 2x – 2x – 40 = 0
2
(x – 1)(x + 2) = 40 x – x – 20 = 0
2
x + 2x – x – 2 = 40 (x – 5)(x + 4) = 0
2 x – 5 = 0 atau/or x + 4 = 0
(x – 6)(x + 7) = 0 x = 5 atau/or x = –4
x – 6 = 0 atau/or x + 7 = 0 ∴ Nilai positif x ialah 5.
x = 6 atau/or x = –7 ∴ The positive value of x is 5.
∴ Nilai positif x ialah 6.
∴ The positive value of x is 6.


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01_Modul A+ MM Tg4.indd 8 12/10/2021 3:31 PM

20. Rajah menunjukkan graf bagi suatu fungsi kuadratik y = ax + bx + c.
2
Diagram shows the graph of a quadratic function y = ax + bx + c. TP 6
2
y BAB 1


x
–1 O 5
(3, –8)


(a) Tentukan nilai-nilai bagi a, b dan c.
Determine the values of a, b and c.
y = a(x + 1)(x – 5)
x = 3, y = –8,
–8 = a(3 + 1)(3 – 5)
–8 = a(4)(–2)
a = 1
y = (x + 1)(x – 5)
= x – 4x – 5
2
b = –4, c = –5

(b) Seterusnya, lakar graf bagi fungsi kuadratik y = –ax – bx – c.
2
Hence, sketch the graph of the quadratic function y = –ax – bx – c.
2


y


x
–1 O 5

©PAN ASIA PUBLICATIONS




21. Graf fungsi kuadratik f(x) = ax + b mempunyai pintasan-y, 3 dan memotong paksi-x di 1 dan k.
2
The graph of the quadratic function f(x) = ax + b has the y-intercept, 3 and cuts the x-axis at 1 and k. TP 6
2
(a) Lakar graf f(x) = ax + b.
2
Sketch the graph f(x) = ax + b.
2
y
3

x
k O 1




(b) Cari nilai-nilai bagi a, b dan k. Seterusnya, lakar graf y = f(x) – 4.
Find the values of a, b and k. Hence, sketch the graph y = f(x) – 4.
b = 3
f (x) = ax + 3 y
2
x = 1, f (x) = 0,
0 = a(1) + 3 x
2
O
a = –3 –1
k = –1
2
y = –3x + 3 – 4 y = –3x – 1
2
= –3x – 1
2




9




01_Modul A+ MM Tg4.indd 9 12/10/2021 3:31 PM

Uji Kendiri 1.1

BAB 1 1. Rajah menunjukkan graf bagi fungsi kuadratik y = 2x + bx.
2
Diagram shows the graph of the quadratic function y = 2x + bx.

2
y
(a) Tentukan nilai b dan cari koordinat titik M.
Determine the value of b and find the coordinates of point M. y = 2x + bx
2
b
2
Paksi simetri / Axis of symmetry: x = – —– y = 2x – bx
2a
9
b
– —— = – — 9 O 9 x
—
– —
2(2) 4 2 2
b = 9
2
y = 2x + 9x
©PAN ASIA PUBLICATIONS
9
9
9
2
1
1
2
Apabila / When x = – —, y = 2 – — + 9 – — 2
4 4 4
81
81
= —– – —– M
8 4
81
= – —–
8
81
9
1
M – —, – —– 2
4
8
(b) Pada rajah yang sama, lakar graf y = 2x – bx.
2
On the same diagram, sketch the graph y = 2x – bx.
2
2. Rajah menunjukkan sebuah kon tegak dengan jejari r cm dan jumlah luas permukaan 156π cm .
2
Diagram shows a right cone with radius r cm and total surface area 156π cm . 10 cm
2
(a) Bentuk satu persamaan kuadratik dalam bentuk ar + br + c = 0.
2
Form a quadratic equation in the form ar + br + c = 0.
2
πr + 2πr(10) = 156π
2
r + 20r = 156
2
r + 20r – 156 = 0
2
(b) Seterusnya, tentukan nilai r.
Hence, determine the value of r.
(r – 6)(r + 26) = 0 r –6 –6r
r = 6 atau / or r = –26 r +26 +26r
r . 0, ∴ r = 6 r 2 –156 +20r
3. (a) Lengkapkan jadual yang berikut.
Complete the following table.
1
x –3 –2 — 2
2
2x + 5x – 3 0 –5 0 15
2
(b) Daripada jadual di (a), nyatakan punca-punca bagi persamaan 2x + 5x = 3.
2
From the table in (a), state the roots of the equation 2x + 5x = 3.
2
1
x = –3, —
2








10




01_Modul A+ MM Tg4.indd 10 26/10/2021 9:22 AM

4. (a) Tukarkan persamaan (x – 5) = 3x – 17 kepada bentuk ax + bx + c = 0.
2
2
Change the equation (x – 5) = 3x – 17 to the form of ax + bx + c = 0.
2
2
(x – 5) = 3x – 17 BAB 1
2
x – 10x + 25 = 3x – 17
2
x – 13x + 42 = 0
2
2
(b) Seterusnya, cari nilai-nilai x yang memuaskan persamaan (x – 5) = 3x – 17.
Hence, find the values of x that satisfy the equation (x – 5) = 3x – 17.
2
(x – 6)(x – 7) = 0
x = 6 atau / or x = 7
5. Selesaikan setiap persamaan yang berikut.
Solve each of the following equations.
(a) (p + 8) – 10 = 0 w – 12
2
2
2
(b) ———– = w + 5
(p + 8) – 10 = 0 4 w – 12
2
2
2
p + 16p + 64 – 100 = 0 ———– = w + 5
2
4
p + 16p – 36 = 0 w – 12 = 4w + 20
2
2
(p – 2)(p + 18) = 0 w – 4w – 32 = 0
2
p = 2 atau / or p = –18 (w – 8)(w + 4) = 0
w = 8 atau / or w = –4
6. Selesaikan setiap persamaan yang berikut.
Solve each of the following equations.
14
(a) (w + 5) + w = 13 (b) 2x + 3 = —–
2
2
(w + 5) + w = 13 x 14
2
2
w + 10w + 25 + w = 13 2x + 3 = —–
2
2
x
2w + 10w + 12 = 0 2x + 3x – 14 = 0
2
2
w + 5w + 6 = 0 (2x + 7)(x – 2) = 0
2
7
(w + 2)(w + 3) = 0 x = – — atau / or x = 2
2
©PAN ASIA PUBLICATIONS
w = –2 atau / or w = –3
7. Pada satu gambar rajah, lakar graf y = a(x – 1) bagi a = 1, a = 3, a = –2 dan a = –4.
2
On one diagram, sketch the graphs y = a(x – 1) for a = 1, a = 3, a = –2 and a = –4.
2
y
a = 3
4
a = 1
2
x
O
–2
a = –2
–4
a = –4
8. Lakar graf fungsi kuadratik yang berikut.
Sketch the following graphs of quadratic functions.
(a) y = 3x + 2 (b) y = 4 – x 2 (c) y = x + 2x (d) y = x – x – 2
2
2
2
y y y y
4
x x
–2 O –1 O 2
2
x
O x –2
–2 O 2




11




01_Modul A+ MM Tg4.indd 11 12/10/2021 3:31 PM

9.
49
5
1
Titik maksimum/Maximum point —, —– 2
2
8
BAB 1 Melalui titik (0, 3) dan (6, 0)
Passes through points (0, 3) and (6, 0)
Suatu graf fungsi kuadratik f (x) = ax + bx + c mempunyai maklumat yang diberi.
2
The graph of a quadratic function f(x) = ax + bx + c has the given information.
2
(a) Lakar graf itu.
Sketch the graph.
y
5 49
—, —–
2 8
3
©PAN ASIA PUBLICATIONS


x
O 6


(b) Tentukan nilai-nilai bagi a, b dan c.
Determine the values of a, b and c.
c = 3
5
b
– —– = —
2a 2 10a + 4b = 5 ...............
b = –5a ........................... 10a – 20a = 5
y = ax + bx + 3 –10a = 5
2
1
5
49
x = —, y = —–, a = – —
2 8 2
5
49
5
5
2
1 2
1 2
—– = a — + b — + 3 b = —
8 2 2 2
5
25
25
—– = —–a + —b
8 4 2
1
5
5
— = —a + —b
8 4 2
10. Sebuah padang rumput yang berbentuk segi empat tepat mempunyai perimeter 60 m dan luas 216 m . Jika panjang
2
padang rumput itu diwakili oleh x m,
A rectangular grass field has a perimeter of 60 m and an area of 216 m . If the length of the grass field is represented by x m,
2
(a) bentukkan satu persamaan kuadratik dalam sebutan x.
form a quadratic equation in terms of x.
2(x + y) = 60
x + y = 30 ................... y m
xy = 216
x(30 – x) = 216 x m
30x – x = 216
2
x – 30x + 216 = 0
2
(b) cari lebar padang rumput itu.
find the width of the grass field.
(x – 12)(x – 18) = 0
x = 12 atau / or x = 18
Apabila / When x = 12, y = 18
Apabila / When x = 18, y = 12
x . y, ∴ y = 12
Lebar / Width = 12 m







12




01_Modul A+ MM Tg4.indd 12 12/10/2021 3:31 PM

11. Rajah menunjukkan sebidang tanah berbentuk segi empat tepat dengan panjang 32 m 22 m
dan lebar 15 m. Bahagian berlorek dengan luas 342 m ditanami cili.
2
Diagram shows a rectangular piece of land with length 32 m and width 15 m. The shaded
region with an area of 342 m is planted with chillies. KBAT Mengaplikasi 15 m BAB 1
2
x m
(a) Bentukkan satu persamaan kuadratik dalam sebutan x. x m x m
Form a quadratic equation in terms of x.
1
1
32 × 15 – —x – — × (x + 10) × 15 = 342
2
2 2
1
150
15
480 – —x – —–x – —–– = 342
2
2 2 2
960 – x – 15x – 150 = 684
2
x + 15x – 126 = 0
2
©PAN ASIA PUBLICATIONS
(b) Cari nilai x.
Find the value of x.
(x – 6)(x + 21) = 0
x = 6 atau/or x = –24
x . 0, ∴ x = 6
12.
x cm 6 cm
6 cm
x cm (x + 3) cm 8 cm
Dalam rajah, hasil tambah luas segi empat sama, segi empat tepat dan segi tiga ialah 933 cm .
2
In the diagram, the sum of areas of the square, rectangle and triangle is 933 cm . KBAT Menganalisis
2
(a) Persamaan kuadratik yang dibentuk adalah (b) Cari jumlah perimeter bagi segi empat sama, segi

ax + bx + c = 0. Nyatakan nilai-nilai bagi a, b dan c. empat tepat dan segi tiga itu.
2
The quadratic equation formed is ax + bx + c = 0. State Find the total perimeter of the square, rectangle and
2
the values of a, b and c. triangle.
1
2
x + 6(x + 3) + — × 8 × 6 = 933 (x – 27)(x + 33) = 0
2 x = 27 atau / or x = –33
x + 6x + 18 + 24 = 933 x . 0, ∴ x = 27
2
x + 6x – 891 = 0
2
a = 1, b = 6, c = –891 Jumlah perimeter / Total perimeter
= 4x + 2(x + 9) + 24
= 6 (27) + 42
= 204 cm








SPM
Soalan Berformat SPM


Kertas 1 Paper 1


1. Persamaan kuadratik x – kx – 18 = 0 mempunyai satu 3. Rajah di bawah menunjukkan sebuah segi empat tepat
2
punca –3. Cari nilai k. dengan luas 27 cm .
2
The quadratic equation x – kx – 18 = 0 has a root –3. The diagram below shows a rectangle with an area of 27 cm .
2
2
Find the value of k.
A –9 B –3 C 3 D 9 x cm
2
2. Selesaikan persamaan kuadratik 30m – 79m + 44 = 0. (4x – 3) cm
Solve the quadratic equation 30m – 79m + 44 = 0.
2
6 4 11 4 Cari nilai x.
A m = 11 atau/or 5 C m = 6 atau/or 5 Find the value of x.
6 5 11 5 A 3 B 4 C 5 D 6
B m = 11 atau/or 4 D m = 6 atau/or 4

13




01_Modul A+ MM Tg4.indd 13 12/10/2021 3:31 PM

4. Antara graf berikut, yang manakah menunjukkan fungsi 5. Rajah berikut menunjukkan graf bagi fungsi
y = –x + 3? y = x + 4x – 12.
2
2
Which of the following graphs shows the function y = –x + 3? The following diagram shows the graph of the function
2
BAB 1 A y C y y = x + 4x – 12. y
2

3
x x x
O –3 O –6 O 2
L
B y D y
L ialah titik minimum. Tentukan koordinat bagi titik L.
3 L is a minimum point. Determine the coordinates of point L.
©PAN ASIA PUBLICATIONS
x A (–4, –12)
x O 3
O B (–3, –15)
C (–1, –15)
D (–2, –16)




Kertas 2 Paper 2


Bahagian A/Section A 5. (a) Selesaikan persamaan kuadratik berikut.
Solve the following quadratic equation.
1. Selesaikan persamaan kuadratik berikut. 3(x – 1)(x – 3) = (2x – 5) 2
Solve the following quadratic equation.
x + 3x – 12 = 2(x + 4) KBAT Mengaplikasi
2
[4 markah] [5 markah]
[4 marks] [5 marks]
2. Selesaikan persamaan kuadratik yang berikut. (b) Lakar graf bagi setiap fungsi berikut.
Solve the following quadratic equation. Sketch the graph for each of the following functions.
2
8x(x + 3) = 13 + 19x (i) y = x + 3x
2
[4 markah] (ii) y = –x + 4
[4 marks] [4 markah]
[4 marks]
3. Dengan menggunakan pemfaktoran, selesaikan Bahagian C/Section C
persamaan kuadratik berikut.
By using factorisation, solve the following quadratic equation. 6. (a) Selesaikan persamaan yang berikut.
2x(2x – 3) = 14 – 5x KLON Solve the following equations.
2
[4 markah] SPM M (i) (x + 3) = 2(x + 7)
SP
[4 marks] [4 markah]
Bahagian B/Section B 3x 1 [4 marks]
(ii) ———– = ———
4. (a) Selesaikan persamaan kuadratik berikut. x – 14 x – 4 [5 markah]
Solve the following quadratic equation. [5 marks]
6x + 7x = 3(8 – x) (b) Rajah menunjukkan graf bagi fungsi
2
[4 markah] y = ax – 5x + a.
2
[4 marks] Diagram shows the graph of the function
(b) Rajah menunjukkan satu laluan berbentuk segi y = ax – 5x + a.
2
KLON empat tepat yang dibina atas sebuah kolam. Laluan y
SPM M itu dibina dengan 5 keping batu pemijak membulat
SP
yang kongruen.
Diagram shows a rectangular path built on a pond. The 2
path is constructed by 5 congruent circular stepping
stones. x
O k

x m
Tentukan nilai-nilai integer bagi a dan k.
(y + 6) m [6 markah]
Diberi luas laluan itu ialah 45 m , cari diameter, Determine the integer values of a and k. [6 marks]
2
dalam m, bagi sekeping batu pemijak itu.
[5 markah]
Given the area of the path is 45 m , find the diameter, in
2
m, for one piece of the stepping stone. [5 marks]
14




01_Modul A+ MM Tg4.indd 14 12/10/2021 3:31 PM

BAB 1 (c) y (c) y
BAB
1.1 O x O x
7
1. (a) ✗ (b) ✗ (c) ✗ –2 —
2
(d) ✓ (e) ✗ –49
2. (a) i (b) ii (c) i
3. (a) (i) (ii)
©PAN ASIA PUBLICATIONS
(b) Selari dengan paksi–y 14. (a) y 17. x = 4 atau/or 8
Parallel to the y–axis 18. Panjang/Length = 34 m
4. (a) (i) Graf/Graph II: 0 , a , 1 x Lebar/Width = 26 m
(ii) Graf/Graph I: c . 0 –2 O 2 19. (a) x = 5
Graf/Graph II: c , 0 –8 20. (a) a = 1, b = –4, c = –5
(b) (i) bertambah lebih sempit (b) y (b) y
daripada graf y = x 2
increases narrower than the 16
graph y = x 2 O x
(ii) berada c unit ke atas graf x –1 5
y = x 2 –4 O 4
lies c units above the graph
y = x 2 21. (a) y
(iii) berada c unit ke bawah graf (c) y
y = x 2 25 3
2
lies c units below the graph —– x
y = x 2 k O 1
5. (a) Graf/Graph I: b , 0 –5 O 5 x
Graf/Graph II: b . 0 (b) a = –3, b = 3, k = –1
b
b
(b) (i) x = – — (ii) x = – — 15. (a) y y
2 2
6. (a) (18x – x ) cm 2 O x
2
(b) (3x + 10x) cm 2 –1
2
2
7. (a) 3x + 17x – 160 = 0 O 2 x y = –3x – 1
2
(b) 2x + 11x – 310 = 0
2
8. (a) (i) 0 (ii) 0 Uji Kendiri 1.1
(b) x = –1, x = 2 (b) y
(
81
9
3
9. (a) 0, – — (b) 1, 4 1. (a) b = 9, M – —, – —–)
4
8
2 (b)
1 2 5 1 y 2
10. (a) —, – — (b) —, – — y = 2x + bx
4 3 3 5 x y = 2x – bx
2
2
3
– —
(c) – —, – — –5 5 O
2 7 2 9 O x
1
—
11. (a) 1, 3 (b) – —, –3 (c) y – — 9
2
2
2
2
(c) 0, — 24
3
1
1
12. (a) – —, — (b) 2, 5 M
2 2
(c) –4, 4 –3 O 8 x 2. (a) r + 20r – 156 = 0
2
3
13. (a) y — (b) r = 6
16. (a) y 3. (a)
1
x x –3 –2 — 2
O 2
2x + 5x – 3 0 –5 0 15
2
8
1
(b) x = –3, —
x 2
–2 O 4. (a) x – 13x + 42 = 0
2
(b) y (b) y (b) x = 6 atau/or x = 7
5. (a) p = 2 atau/or p = –18
x (b) w = 8 atau/or w = –4
O
3 x –4 6. (a) w = –2 atau/or w = –3
O
–16 7
(b) x = – — atau/or x = 2
2
175
11_Modul A+ MM Tg4_JwpM.indd 175 12/10/2021 5:03 PM

7. y (b) (i) y 3. (a) 1111010 , 1111011 , 1111100 ,
2
2
2
a = 3 1111101 , 1111110
4 a = 1 (b) 330 2 2
2 4. (a) 32 6 (b) 201
x 7
O O x 5. (a) 10000 (b) x = 764
–2 a = –2 –3 6. (a) 3584 5 (b) 160
–4 8
a = –4 7. (a) 32
(b) Bilangan digit/Number of digits = 11
8. (a) y (ii) y (c) w = 103330
(d) y = 5 524
4
8. (a) 1000111 (b) 241 5
2
2 x 9. (a) 432 (b) 101110 3
8
O –2 O 2 x 10. (a) 16336 (b) x = 3
11. (a) (i) a = 4, b = 2, c = 5
©PAN ASIA PUBLICATIONS
(ii) a = 6, b = 5, c = 1
(b) y 6. (a) (i) x = –5 atau/or x = 1 (b) (i) 10142 (ii) 1 714
7
(ii) x = — atau/or x = 2 5 8
4 3 12. (a) 150 (b) 1100 2
8
(b) a = 2, k = 2 13. (a) 1435 (b) 5030 6
6
14. (a) 10301 (b) 2787 9
4
x BAB 2 15. (a) x = 11201 (b) y = 1447
BAB
–2 O 2
16. (a) (i) 2031 (ii) 422
2.1 5 5
(c) y (b) a = 4130
1. (a) 70 (b) 130 (c) 1011 3
9
5
2. (a) 49 (b) 54 (c) 729 Soalan Berformat SPM
3. (a) 2 , 8 (b) 2 , 16 (c) 2 , 64 Kertas 1/Paper 1
3
4
6
x
–2 O 4. (a) 56 (b) 3072 1. C 2. B 3. B 4. B 5. C
5. (a) 210 = 36 (b) 300 = 75 10 6. C 7. B 8. D 9. C 10. C
10
5
4
(c) 215 = 110 11. B 12. C 13. B 14. A 15. B
7 10 16. B 17. B
(d) y 6. (a) 23 (b) 976 10
10
7. (a) 202 (b) 1210 4 Kertas 2/Paper 2
7
8. (a) 1001101 (b) 1011011 2 1. (a) 243 (b) 2023 5
2
x (c) 10000111 2 2. (a) (i) y = 11021
–1 O 2 3
9. (a) 564 (b) 3010 8 (ii) y = 1110011 2
8
–2 10. (a) 223 (b) 1324 5 (b) (i) m = 403 (ii) 100000011 2
5
11. (a) 10011 (b) 100101 2 3. (a) p = 4, 141 = 314 6
2
9
(c) 110000 (b) 202
9. (a) y 2 3
5 49 12. (a) 113 (b) 404 (c) 1460 8 (c) (i) w = 5120 (ii) 1483 9
—, —–
8
2 8
8
13. (a) 230 (b) 331 (c) 1021 (iii) 1187
3 5 5 5 9
14. (a) 11100 (b) 313 (c) 140 9
3
6
15. (a) 32 (b) 527 8 BAB 3
BAB
8
16. (a) 111001000 (b) 11100010001
2 2 3.1
x 17. (a) 110011 (b) 10202 (c) 1223 4
2
3
O 6 (d) 3120 (e) 2044 (f) 4323 1. (a) Bukan/Not (b) Bukan/Not
4
5
(g) 2320 (h) 11510 (i) 4041 5 7 (c) Ya/Yes
6
6
5
1
(b) a = – —, b = —, c = 3 (j) 10120 (k) 6310 (l) 10301 8 2. (a) Palsu/False (b) Benar/True
8
7
2 2 (m) 3080 (n) 15231 (c) Palsu/False
9
10. (a) x – 30x + 216 = 0 18. (a) 11110 (b) 111 9 (c) 33 3. (a) Benar/True (b) Palsu/False
2
4
3
2
(b) 12 m (d) 3102 (e) 434 (f) 2324 (c) Palsu/False
4
5
11. (a) x + 15x – 126 = 0 (g) 3345 (h) 223 (i) 1266 5 4.
2
6
6
(b) x = 6 (j) 2536 (k) 2377 (l) 2644 7 (a) Palsu 240 minit tidak sama Benar
7
12. (a) a = 1, b = 6, c = –891 (m) 5654 (n) 2255 8 9 8 False dengan 6 jam. True
9
(b) 204 cm 19. (a) 1011 (b) 221 (c) 1332 240 minutes is not equal
2
3
(d) 3114 (e) 3053 (f) 4135 4 to 6 hours.
6
5
Soalan Berformat SPM (g) 5134 (h) 2301 (i) 5542 7 9 (b) Benar 300° bukan sudut Palsu
7
8
Kertas 1/Paper 1 20. x = 3423, y = 111101000, z = 1265 True refleks. False
1. C 2. C 3. A 4. B 5. D 21. (a) 260 10 (b) a = 8, b = 9 300° is not a reflex angle.
22. (a) (i) 464 (ii) 5032 5. (a) 1 221 bukan nombor yang
Kertas 2/Paper 2 (b) t = 4235 7 7 terbahagi dengan 7.
1. x = 4 atau/or x = –5 1 221 is not a number divisible by 7.
13
2. x = – —– atau/or x = 1 Uji Kendiri 2.1 (b) {1, 9, 11} tidak mempunyai 6
8 1. (a) 1011 (b) 21 subset.
7
3. x = – — atau/or x = 2 (c) 14 2 (d) 12 5 {1, 9, 11} does not have 6 subsets.
4 7 9 3
4
4. (a) x = — atau/or x = –3 2. (a) 7 3 7 2 7 1 7 0 (c) — adalah tidak kurang daripada
7
3 5
5
3
(b) Diameter/Diameter = 3 m (b) (i) 3 (ii) 7 —–./ — is not less than —–.
14
5. (a) x = 4 (iii) 245 (iv) 686 7 14
176
11_Modul A+ MM Tg4_JwpM.indd 176 12/10/2021 5:03 PM

2
BAB 1 y = 2x – 5x + 2
BAB
x = k, y = 0
0 = 2k – 5k + 2
2
Soalan Berformat SPM (2k – 1)(k – 2) = 0
1
Kertas 2/Paper 2 k = — atau/or k = 2
1. x + 3x – 12 = 2(x + 4) 2
2
x + 3x – 12 = 2x + 8 k ialah integer./k is an integer.
2
x + x – 20 = 0 ∴ k = 2
2
(x – 4)(x + 5) = 0
BAB
x = 4 atau/or x = –5 BAB 2
2. 8x(x + 3) = 13 + 19x
©PAN ASIA PUBLICATIONS
8x + 24x = 13 + 19x Soalan Berformat SPM
2
8x + 5x – 13 = 0
2
(8x + 13)(x – 1) = 0 Kertas 2/Paper 2
13
x = – —– atau/or x = 1 1. (a) Nilai bagi digit 3/Value of digit 3 Baki
2
8 = 3 × 9 5 263 Remainder
3. 2x(2x – 3) = 14 – 5x = 243 5 52 ……3
4x – 6x = 14 – 5x (b) 524 = 5 × 7 + 2 × 7 + 4 × 7 0 5 10 ……2
2
2
1
4x – x – 14 = 0 7 = 245 + 14 + 4
2
……0
(4x + 7)(x – 2) = 0 = 263 5 2 0 …… 2
7
x = – — atau/or x = 2 = 2023 10
4 5 Baki
4. (a) 6x + 7x = 3(8 – x) 2. (a) (i) y = 11021 2 115 Remainder
2
6x + 7x = 24 – 3x 4 3 3 1 2 57 ……1
2
2
6x + 10x – 24 = 0 (ii) y = 3 + 3 + 2 × 3 + 1 2 28 ……1
3x + 5x – 12 = 0 = 81 + 27 + 6 + 1 2 14 ……0
2
= 115
(3x – 4)(x + 3) = 0 = 1110011 2 7 ……0
10
4
x = — atau/or x = –3 2 2 3 ……1
3
(b) y + 6 = 5x … 2 1 ……1
(y + 6)x = 45 … 0 …… 1
5x = 45 (b) (i) 10003 = 1 × 4 + 0 × 4 + 0 × 4 + 0 × 4 + 3 × 4 0
2
4
2
3
1
x = 9 4 = 256 + 0 + 0 + 0 + 3
2
Baki
x = ±3 = 259 8 259 Remainder
x . 0, ∴ x = 3 = 403 10
Diameter/Diameter = 3 m ∴ m = 403 8 8 32 ……3
5. (a) 3(x – 1)(x – 3) = (2x – 5) 2 8 4 ……0
3(x – 4x + 3) = 4x – 20x + 25 (ii) 10003 = 4 0 3 0 …… 4
2
2
3x – 12x + 9 = 4x – 20x + 25 4 8
2
2
x – 8x + 16 = 0 = 100 000 011
2
(x – 4) = 0 2
2
x = 4 3. (a) 1p1 = 1110110 2
9
4
6
5
2
0
1
(b) (i) (ii) y 1 × 9 + p × 9 + 1 × 9 = 1 × 2 + 1 × 2 + 1 × 2 + 0 × 2 3
y
+ 1 × 2 + 1 × 2 + 0 × 2 0
2
1
81 + 9p + 1 = 64 + 32 + 16 + 0 + 4 + 2 + 0
4
9p + 82 = 118
Baki
x 9p = 36 6 118 Remainder
–3 O x p = 4
–2 O 2 6 19 ……4
141 = 118 6 3 ……1
9 = 314 10
6. (a) (i) (x + 3) = 2(x + 7) 6 0 …… 3
2
x + 6x + 9 = 2x + 14 (b) 2110 – 1201 = 202 1 4 0 3
2
x + 4x – 5 = 0 3 3 3 2110 3
2
(x + 5)(x – 1) = 0 1 1 – 1201 3
x = –5 atau/or x = 1 (c) (i) 4203 6 202 3
3x
1
(ii) ——–– = ——– + 513 6
x – 14 x – 4
3x(x – 4) = x – 14 5120 6 ∴w = 5120
3x – 12x = x – 14
2
1
2
3
3x – 13x + 14 = 0 (ii) 5120 = 5 × 6 + 1 × 6 + 2 × 6 + 0 × 6 0
2
6
Baki
(3x – 7)(x – 2) = 0 = 1080 + 36 + 12 + 0 9 1128 Remainder
7
x = — atau/or x = 2 = 1128 10
3 = 1483 9 9 125 ……3
(b) x = 0, y = 2 9 13 ……8
2 = 0 – 0 + a 9 1 ……4
∴ a = 2
0 …… 1
MG-1

Lembaran Pentaksiran Bilik Darjah (PBD)





BAB 1 Fungsi dan Persamaan Kuadratik dalam Satu Pemboleh Ubah

Tahap Penguasaan Tafsiran
1 Mempamerkan pengetahuan asas tentang ungkapan, fungsi dan persamaan kuadratik dalam satu
pemboleh ubah.
2 Mempamerkan kefahaman tentang ungkapan, fungsi dan persamaan kuadratik dalam satu pemboleh
ubah.
3 Mengaplikasikan kefahaman tentang fungsi dan persamaan kuadratik dalam satu pemboleh ubah
©PAN ASIA PUBLICATIONS
untuk melaksanakan tugasan mudah.
4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dan persamaan kuadratik
dalam satu pemboleh ubah dalam konteks penyelesaian masalah rutin yang mudah.
5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dan persamaan kuadratik
dalam satu pemboleh ubah dalam konteks penyelesaian masalah rutin yang kompleks.


1.
k – 8 x + 3x – 6 4p + 9
3
2
2
7
w + 2 –y + 10y 2r – — + 3
2
2
r
Berdasarkan ungkapan-ungkapan di atas, lengkapkan rajah berikut.
Based on the expressions above, complete the following diagram. TP 1
Ungkapan / Expression



(a) Ungkapan kuadratik (b) Bukan ungkapan kuadratik
Quadratic expression Not quadratic expression

3
x + 3x – 6 k – 8
2
4p + 9  + 2
w
2
7
–y +10y 2r – + 3
2
2
r
2. Tiga integer yang berturut-turut ialah n, n + 1 dan n + 2. Tulis satu ungkapan kuadratik, dalam bentuk an + bn + c, bagi
2
The three consecutive integers are n, n + 1 and n + 2. Write a quadratic expression, in the form an + bn + c, for TP 2
2
(a) kuasa dua bagi hasil tambah tiga integer itu. (b) hasil tambah kuasa dua bagi setiap integer itu.
the square for the sum of the three integers. the sum of squares for each of the integers.
(n + n + 1 + n + 2) 2 n + (n + 1) + (n + 2) 2
2
2
= (3n + 3) 2 = n + n + 2n + 1 + n + 4n + 4
2
2
2
= 9n + 18n + 9 = 3n + 6n + 5
2
2















MG-9

3. Selesaikan persamaan kuadratik berikut.
Solve the following quadratic equations. TP 3
(a) (x + 2)(x + 1) = 3(3x – 1) (b) x + 1 = (3x – 7) 2
x + 3x + 2 = 9x – 3 x + 1 = 9x – 42x + 49
2
2
x – 6x + 5 = 0 9x – 43x + 48 = 0
2
2
(x – 1)(x – 5) = 0 (9x – 16)(x – 3) = 0
x – 1 = 0 atau/or x – 5 = 0 9x – 16 = 0 atau/or x – 3 = 0
x = 1 atau/or x = 5 16
x = atau/or x = 3
9
4. (a) Selesaikan.
Solve. TP 3
(i) 12t – 31t – 15 = 0 (ii) 5w + 27w + 8 = 0
2
2
(12t + 5)(t – 3) = 0 (5w + 7)(w + 4) = 0
12t + 5 = 0 atau/or t – 3 = 0 5w + 7 = 0 atau/or w + 4 = 0
t = – 5 atau/or t = 3 7
12 w = – atau/or w = –4
5


(b) Dalam rajah yang diberi, PQRS ialah sebuah segi empat tepat. Sebuah bulatan dan dua Q R
buah semi bulatan yang kongruen terterap dalam segi empat tepat itu. Jejari setiap semi
bulatan itu ialah 1 cm. 2x cm
In the diagram given, PQRS is a rectangle. A circle and two congruent semicircles are inscribed in
the rectangle. The radius of each semicircle is 1 cm. P S
Ungkapkan
Express TP 5 (ii) luas kawasan berlorek, dalam bentuk ax + bx + c.
Lembaran PBD ©PAN ASIA PUBLICATIONS
(i) panjang PS, dalam sebutan x,
2
the length of PS, in terms of x,
the area of the shaded region, in the form ax + bx + c.

2
Luas kawasan berlorek

PS = 1 + 2x + 1
Area of the shaded region

= (2 + 2x) cm

= (2 + 2x)(2x) – p(1) – px
2
2
2
= [(4 – p)x + 4x – p] cm
= 4x + 4x – p – px 2 2
2
5. Rajah yang diberi menunjukkan suatu pepejal yang terdiri daripada gabungan sebuah kuboid
dan sebuah piramid tegak. Tinggi pepejal itu ialah (x + 8) cm.
Diagram given shows a solid consisting of the combination of a cuboid and a right pyramid. The height 3 cm
of the solid is (x + 8) cm.
12 cm
(4x – 5) cm
(a) Bentukkan satu ungkapan kuadratik, dalam bentuk (b) Jika isi padu pepejal itu ialah 476 cm , hitung nilai x.

3
ax + bx + c, bagi isi padu pepejal itu. If the volume of the solid is 476 cm , calculate the value of x.
3
2
Form a quadratic equation, in the form ax + bx + c, for the TP 4
2
volume of the solid. TP 2
16x + 204x – 280 = 476
2
Isi padu pepejal 16x + 204x – 756 = 0
2
Volume of solid 4x + 51x – 189 = 0
2
1 (4x + 63)(x – 3) = 0
= (4x – 5)(12)(3) + (4x – 5)(12)(x + 5)
3
= 144x – 180 + 4(4x + 20x – 5x – 25) x = – 63 atau/or x = 3
2
4
= 144x – 180 + 16x + 60x – 100 5
2
= (16x + 204x – 280) cm 3 x  , x = 3
2
4
MG-10

Book extend: 208pp
Spine: 9.32mm







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AKSES DIGITAL





Siri MODUL A+1 ini menyediakan modul pembelajaran
komprehensif yang dihasilkan berdasarkan Dokumen
Standard Kurikulum dan Pentaksiran (DSKP). Modul ini
telah dirancang dan ditulis oleh guru-guru yang ■ Aktiviti PAK-21
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■ Lembaran Pentaksiran
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Tahap Penguasaan yang perlu dikuasai oleh murid
untuk mengoptimumkan kefahaman mereka.






Dapatkan



DWIBAHASA

Judul-judul dalam 4
siri Modul A+1 sekarang! Matematik

Mata Pelajaran / Tingkatan 4 5 Tingkatan
Mathematics

Sejarah Matematik / Mathematics
Matematik

Matematik Tambahan Chai Mun
Kimia

Fizik
Pentaksiran
Pentaksiran
Perniagaan
Baharu
Ekonomi Tingkatan 4 Baharu
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