Modul A+1 Matematik Tambahan Tingkatan 5

Book extend: 184pp
Spine: 8.36mm







PAN ASIA

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 ■ Jawapan Lengkap Soalan
©PAN ASIA PUBLICATIONS
Berformat SPM
berpengalaman dalam membantu proses PdPc dengan ■ Lembaran Pentaksiran
lebih efektif. Latihan yang disediakan mencakupi Bilik Darjah (PBD)
Tahap Penguasaan yang perlu dikuasai oleh murid
untuk mengoptimumkan kefahaman mereka.






Dapatkan



DWIBAHASA

Judul-judul dalam 5
siri Modul A+1 sekarang! Matematik

Mata Pelajaran / Tingkatan 4 5 Tambahan Tingkatan


Sejarah Additional Mathematics Matematik Tambahan
Matematik Additional Mathematics

Matematik Tambahan
Kimia
Penulis Buku Teks
Fizik
Azizah binti Kamar (Guru Cemerlang)
Pentaksiran
Perniagaan Tingkatan 5 Dr. M. K. Wong Pentaksiran
Baharu
Ekonomi Nurbaiti binti Ahmad Zaki Baharu
SPM
SPM
English

Pelengkap kepada SPM
buku teks

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Penerbit asal buku teks

©PAN ASIA PUBLICATIONS




5







DWIBAHASA
Matematik Tingkatan




Tambahan




Additional Mathematics





Penulis Buku Teks
Azizah binti Kamar (Guru Cemerlang)
Dr. M. K. Wong
Nurbaiti binti Ahmad Zaki


Pan Asia Publications Sdn. Bhd. Bonus Guru
199101016590 (226902-X) • PDF Manual Guru
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Cetakan Pertama 2022
MODUL A+1 MATEMATIK TAMBAHAN Tingkatan 5
ISBN 978-967-466-631-6
Dicetak oleh Metrobay Industry Sdn. Bhd. (430102-W)

BAB 1 Sukatan Membulat BAB 5 Taburan Kebarangkalian
Circular Measure Probability Distribution
1.1 Radian 5.1 Pemboleh Ubah Rawak
Radian .......................................................................... 2 Random Variable ......................................................... 72
1.2 Panjang Lengkok Suatu Bulatan 5.2 Taburan Binomial
Binomial Distribution .................................................. 75
©PAN ASIA PUBLICATIONS
Arc Length of a Circle .................................................... 3 5.3 Taburan Normal
1.3 Luas Sektor Suatu Bulatan Normal Distribution .................................................... 82
Area of Sector of a Circle ............................................... 8 Soalan Berformat SPM .................................................... 88
1.4 Aplikasi Sukatan Membulat
Application of Circular Measures.................................. 12
Soalan Berformat SPM .................................................... 13 BAB 6 Fungsi Trigonometri
Trigonometric Functions
6.1 Sudut Positif dan Sudut Negatif
BAB 2 Pembezaan Positive Angles and Negative Angles ............................. 95
Differentiation 6.2 Nisbah Trigonometri bagi Sebarang Sudut
2.1 Had dan Hubungannya dengan Pembezaan Trigonometric Ratios of Any Angle ................................ 97
Limit and Its Relation to Differentiation ......................... 19 6.3 Graf Fungsi Sinus, Kosinus dan Tangen
2.2 Pembezaan Peringkat Pertama Graphs of Sine, Cosine and Tangent Functions ............. 100
The First Derivative ..................................................... 21 6.4 Identiti Asas
Basic Identities .......................................................... 103

2.3 Pembezaan Peringkat Kedua 6.5 Rumus Sudut Majmuk dan Rumus Sudut Berganda
The Second Derivative ................................................. 26 Addition Formulae and Double Angle Formulae .......... 105
2.4 Aplikasi Pembezaan 6.6 Aplikasi Fungsi Trigonometri
Application of Differentiation ....................................... 28 Application of Trigonometric Functions ....................... 108
Soalan Berformat SPM .................................................... 37 Soalan Berformat SPM ...................................................110



BAB 3 Pengamiran BAB 7 Pengaturcaraan Linear
Integration Linear Programming
3.1 Pengamiran sebagai Songsangan Pembezaan 7.1 Model Pengaturcaraan Linear
Integration as the Inverse of Differentiation ................... 42 Linear Programming Model ....................................... 113
3.2 Kamiran Tak Tentu 7.2 Aplikasi Pengaturcaraan Linear
Indefinite Integral ........................................................ 43 Application of Linear Programming ............................ 119
3.3 Kamiran Tentu Soalan Berformat SPM .................................................. 123
Definite Integral .......................................................... 45
3.4 Aplikasi Pengamiran 8 Kinematik Gerakan Linear
Application of Integration ............................................ 56 BAB Kinematics of Linear Motion
Soalan Berformat SPM .................................................... 57 8.1 Sesaran, Halaju dan Pecutan sebagai Fungsi Masa
Displacement, Velocity and Acceleration as a Function
of Time ..................................................................... 126
BAB 4 Pilih Atur dan Gabungan 8.2 Pembezaan dalam Kinematik Gerakan Linear
Permutation and Combination Differentiation in Kinematics of Linear Motion ............ 129

4.1 Pilih Atur 8.3 Pengamiran dalam Kinematik Gerakan Linear
Permutation ................................................................ 62 Integration in Kinematics of Linear Motion ................. 133
4.2 Gabungan 8.4 Aplikasi Kinematik Gerakan Linear
Combination ............................................................... 67 Application of Kinematics of Linear Motion ................. 136
Soalan Berformat SPM .................................................... 69 Soalan Berformat SPM .................................................. 138

Kertas Model SPM ......................................................... 141

Jawapan ..................................................................... MG–1
Lembaran PBD........................................................ MG–14


ii




00 Kand Modul A+ MateTam Tg5.indd 2 04/10/2021 2:11 PM

BAB
1 Sukatan Membulat

Circular Measure




Nota



1.1 Radian
1. Satu radian ditakrifkan sebagai sudut yang dicangkum di pusat bulatan oleh satu lengkok A
©PAN ASIA PUBLICATIONS
yang sama panjang dengan jejari bulatan, j seperti yang ditunjukkan dalam rajah. j
One radian is defined as an angle subtended at the centre of the circle by an arc that has a j
length equal to the radius of the circle, j as shown in the diagram. O 1 rad

2. 360° = 2p rad B
× π
180°

q° (dalam darjah)/(in degrees) q (dalam radian)/(in radians)

180°
×
π

1.2 Panjang Lengkok Suatu Bulatan/Arc Length of a Circle
1. Rumus berikut boleh digunakan:
The following formula can be used: j

q ° = q rad = Panjang lengkok/Length of arc, s θ s
360° 2π rad 2πj


Jika diringkaskan, ia menjadi s = j q (q dalam radian/in radians)
When simplified, it becomes

2. Petua kosinus digunakan untuk mencari panjang perentas AB.
The cosine rule is used to find the length of the chord AB.
AB = j + j – 2j kos q
2
2
2
2
AB = j + j – 2j cos q O j
2
2
2
2
θ B
dengan keadaan j ialah jejari dan q ialah sudut dalam darjah.
where j is the radius and q is the angle in degrees. A
1.3 Luas Sektor Suatu Bulatan/Area of Sector of a Circle
1. Hubungan di bawah boleh digunakan bergantung kepada maklumat yang diberi.
The relationship below can be used depending on the information given.
q ° = q rad = Panjang lengkok/Length of arc, s = Luas sektor/Area of sector, A
360° 2π rad 2πj πj 2 O
θ B
j
2. Jika sudut ialah dalam radian, maka Luas sektor/Area of a sector = j q
1 2
If the angle is in radians, then 2 A

Luas tembereng = j q – j sinq , dengan keadaan q ialah dalam radian bagi
1 2
1 2
2 2
rumus j q dan q ialah dalam darjah bagi
1 2
2 O
rumus j sin q.
1 2
2 j θ
A B
1 2
1 2
Area of a segment = j q – j sin q , where q is in radians in the formula j q and
1 2
2 2 2
1 2
q is in degrees in the formula j sin q.
2
1

01 Modul Series MateTam Tg5.indd 1 04/10/2021 2:15 PM

1.1 Radian Buku Teks
Radian m.s. 2-4

1. Tukarkan setiap sudut dalam darjah yang berikut kepada radian.
Convert each of the following angles in degrees into radians. TP 1
BAB 1 Contoh/Example (a) 75° (b) 230°


π
π
150° q = 75° × 180° q = 230° × 180°
q = 150° × π = 1.31 rad = 4.01 rad
180°
= 2.62 rad
©PAN ASIA PUBLICATIONS



(c) 315° 15′ (d) 41° (e) 175° 30′
q = 315.25° × π q = 41° × π q = 175° 30′ × π
180° 180° 180°
= 5.50 rad = 0.72 rad = 3.06 rad







(f) 85° 10′ (g) 178° 20′ (h) 228° 48′

q = 85.167° × π q = 178.333° × π q = 228.8° × π
180° 180° 180°
= 1.49 rad = 3.11 rad = 3.99 rad







2. Tukarkan setiap sudut dalam radian yang berikut kepada darjah.
Convert each of the following angles in radians into degrees. TP 1

Contoh/Example (a) 0.5 rad (b) 1.3 rad
q° = 0.5 × 180° q° = 1.3 × 180°
2.1 rad π π
= 28° 39′ = 74° 29′
q ° = 2.1 × 180°
π
= 120° 19′



(c) 1.8π rad (d) 2 π rad (e) 1 π rad
q° = 1.8π × 180° 5 3
1
2
π q° = π × 180° q° = π × 180°
= 324° 5 π 3 π
= 72° = 60°





(f) π rad (g) 2.5 rad (h) 4 π rad
2 q° = 2.5 × 180° 5
4
π
q° = × 180° π q° = π × 180°
2 π = 143° 14′ 5 π
= 90° = 144°





2




01 Modul Series MateTam Tg5.indd 2 04/10/2021 2:15 PM

Uji Kendiri 1.1

1. Tukarkan 451° 15′ kepada radian. Berikan jawapan 2. Tukarkan 4π rad kepada darjah.
betul kepada tiga tempat perpuluhan. 3
Convert 451° 15′ into radians. Give answer correct to three Convert 4π rad into degrees.
3
decimal places. 4π 180°
q = 451° 15′ × π q° = 3 × π BAB 1
180°
= 7.876 rad = 240°

3. Rajah menunjukkan sebuah bulatan berpusat O dengan 4. Rajah menunjukkan sebuah sektor AOC berpusat O.
keadaan sudut minor AOB ialah 108° 42′. AOB ialah segi tiga bersudut tegak di B.
©PAN ASIA PUBLICATIONS
The diagram shows a circle with centre O where the minor The diagram shows a sector AOC with centre O. AOB is a
angle AOB is 108° 42′. right-angled triangle at B. KBAT Mengaplikasi
A
θ
O 5 4
B
108° 42’ θ
O C
B
A
Cari nilai q, dalam radian.
Cari nilai q, dalam radian. Find the value of q, in radians.
Find the value of q, in radians.
360° – 108° 42′ = 251° 18′ sin q = 4
5
π
q = 251° 18′ × 180° q = 53.13°
π
= 4.386 rad   q = 53.13° × 180°
= 0.927 rad







1.2 Panjang Lengkok Suatu Bulatan Buku Teks
Arc Length of a Circle m.s. 5-11


1. Hitung panjang lengkok bagi setiap yang berikut.
Calculate the arc length for each of the following. TP 3
Contoh/Example (a)
s

B B
A
14 cm s = jq 1.8 rad s = 7 × 1.8
s 7 cm = 12.6 cm
1.1 rad = 14 × 1.1
O O
= 15.4 cm
A




(b) (c)
s s


B
A A
2.5 cm
O 0.4 rad
O 1.5 cm B
240°
s = 1.5 × (π – 0.4)
p = 4.11 cm
s = 2.5 × (360° – 240°) ×
180°
= 5.24 cm

3




01 Modul Series MateTam Tg5.indd 3 04/10/2021 2:15 PM

2. Hitung jejari, j bagi setiap yang berikut.
Calculate the radius, j for each of the following. TP 2

Contoh/Example (a)
24.5 cm
10.2 cm Q
BAB 1 P Q 2.5 rad

j 1.5 rad
j O
O P
j = 10.2 j = 24.5
1.5 2.5
= 6.8 cm = 9.8 cm
©PAN ASIA PUBLICATIONS

(b) (c)
15.7 cm
Q
j
0.7 rad
30.5 cm O P
O Q
j
P 3.1 rad
15.7
j = 2π – 3.1
j = 30.5
2π – 0.7 = 4.93 cm
= 5.46 cm

3. Hitung sudut, dalam radian, bagi setiap yang berikut.
Calculate the angle, in radians, for each of the following. TP 2

Contoh/Example

Gunakan/Use s = jq
14 cm Tip SPM
16 = 14q
θ 16 cm
  q = 16 Simbol q dibaca sebagai “téta”.
14 The symbol q is read as “théta”.
= 8 rad
7


(a) (b)
P 30.2 cm
P
12.3 cm 12 cm
θ
θ
O O
7.5 cm Q
Q 30.2
q = 12.3 q = 12
7.5 = 2.52 rad
= 1.64 rad


(c) Q (d)
θ 2π – q = 35 Q
11
11 cm
O q = 2π – 35 5 cm
P 11 θ
= 3.1 rad O P
π
35 cm q = rad
3





4




01 Modul Series MateTam Tg5.indd 4 04/10/2021 2:15 PM

4. Hitung perimeter bagi setiap kawasan berlorek.
Calculate the perimeter for each of the shaded region. TP 3

Contoh/Example (a) P
6.5 cm
D O 0.45 rad
A BAB 1
5 cm Q
O 1.6 rad
Perimeter = 6.5 + 6.5 + 6.5(0.45)
B = 13 + 2.93
4 cm = 15.93 cm
C
©PAN ASIA PUBLICATIONS
Panjang lengkok AB/Length of arc AB = 5(1.6) = 8 cm
Panjang lengkok CD/Length of arc CD = 9(1.6) = 14.4 cm
Perimeter kawasan berlorek/Perimeter of the shaded region
= AB + BC + CD + AD
= 8 + 4 + 14.4 + 4
= 30.4 cm


(b) S (c) P
4 cm S
P T
6.2 cm
O
1.8 rad 0.5 rad
Q O Q 3 cm R

R 4
OP =
Panjang lengkok SR = 12.4(1.8) 0.5
= 22.32 cm = 8 cm
Panjang lengkok PQ = 6.2(1.8) Panjang lengkok SR = 11(0.5)
= 11.16 cm = 5.5 cm
Perimeter = 11.16 + 6.2 + 22.32 + 6.2 Perimeter = 3 + 5.5 + 3 + 4 + 4 + 8 + 8
= 45.88 cm = 35.5 cm






5. Hitung jejari, j bagi setiap yang berikut.
Calculate the radius, j for each of the following. TP 4

Contoh/Example (a) Q


1
— rad
T 5 O
V 13 cm
j 8 cm
U
W 1.2 rad

Perimeter rantau berlorek = 20 cm O j P
Perimeter of the shaded region = 20 cm
VT = WU = 5 cm Perimeter rantau berlorek = 22.4 cm
1
1
5 + j + ( j + 5) + 5 = 20 Perimeter of the shaded region = 22.4 cm

5 5
2 j = 9 1.2j + (13 – j) + 8 = 22.4
5 1.2j – j + 21 = 22.4
j = 45 0.2j = 1.4
2 j = 7 cm
= 22.5 cm



5




01 Modul Series MateTam Tg5.indd 5 04/10/2021 2:15 PM

(b) S (c)
Q
P

O 1.5 rad θ
1.5 rad
BAB 1 j R O j P

Q Perimeter rantau berlorek = 16π + 24
Perimeter of the shaded region = 16π + 24
R OP = OQ = QP
OR = 2OP π
Perimeter rantau berlorek = 21 cm º q = 60° = 3
©PAN ASIA PUBLICATIONS
π
π
2
Perimeter of the shaded region = 21 cm j + πj + j + j + j = 16π + 24
2j + 2j + 1.5(2j) + j + j + 1.5j = 21 3 3 3 4
10.5j = 21 2j + πj = 4(4π + 6)
3
j = 2 cm 6j + 4πj = 12(4π + 6)
j(6 + 4π) = 12(4π + 6)
j = 12 cm


6. Tentukan perimeter bagi tembereng berlorek berdasarkan maklumat yang diberi.
Determine the perimeter of the shaded segment based on the given informations. TP 4

Contoh/Example (a) (b)
P K
A 11.2 cm
8.2 cm O j 8.5 cm

O 1.2 rad
2.3 rad
1.5 rad
B O
L
Q
2.3 rad = 2.3 × 180° j(1.5) = 8.5
π
1.2 rad = 1.2 × 180° j = 5.67 cm
π = 131.78° 1.5 rad = 1.5 × 180°
= 68.75° Panjang perentas PQ π
Panjang perentas AB = ! 11.2 + 11.2 – 2(11.2) kos 131.78° = 85.94°
2
2
2
Length of chord AB = ! 418.034 Panjang perentas KL
2
2
2
= ! 8.2 + 8.2 – 2(8.2) kos/cos 68.75° = 20.45 cm = ! 5.67 + 5.67 – 2(5.67) kos 85.94°
2
2
2
= ! 85.74 Panjang lengkok PQ = ! 59.745
= 9.26 cm = 11.2 × 2.3 = 7.73 cm
Panjang lengkok AB = 25.76 cm Perimeter
Length of arc AB Perimeter = 7.73 + 8.5
= 8.2 × 1.2 = 25.76 + 20.45 = 16.23 cm
= 9.84 cm = 46.21 cm
Perimeter
= 9.84 + 9.26
= 19.1 cm


Tip SPM

Gunakan petua kosinus.
Use the cosine rule.
a = b + c – 2bc kos q
2
2
2
a = b + c – 2bc cos q
2
2
2


6




01 Modul Series MateTam Tg5.indd 6 04/10/2021 2:15 PM

7. Selesaikan masalah yang melibatkan panjang lengkok berikut.
Solve the following problems which involve the length of arc. TP 5
(a) Dane ingin menghasilkan beberapa buah topi seperti
Contoh/Example
yang ditunjukkan dalam rajah untuk digunakan
Aminah sedang menjahit tampalan pada sehelai semasa jamuan hari jadi.
selimut. Dia menggunakan corak seperti yang Dane wants to make some decorative hats as shown in the
diagram to be used in a birthday party.
ditunjukkan dalam rajah berikut. Ketiga-tiga kawasan BAB 1
berlorek adalah serupa dengan diameter bulatan itu O
ialah 12 cm. Jika dia perlu menjahit di sekeliling
kawasan berlorek, hitung panjang benang, dalam m,
yang diperlukan untuk menjahit 15 keping tampalan 25 cm
yang sama pada selimut tersebut.
©PAN ASIA PUBLICATIONS
Aminah is sewing some patchwork on a blanket. She uses
the pattern as shown in the following diagram. The three 9 cm P
shaded regions are identical with the diameter of the circle
is 12 cm. If she needs to sew around each shaded region, Dia merancang untuk menggunakan sekeping kadbod
find the length of the thread, in m, needed to sew 15 pieces yang berbentuk segi empat tepat dengan ukuran
of same patchwork on the blanket. 25 cm × 40 cm untuk menghasilkan setiap topi.
Tentukan sama ada kadbod itu mencukupi untuk
P menghasilkan topi tersebut. Tunjukkan langkah kerja
anda.
He plans to use a piece of rectangular cardboard with a
dimension of 25 cm × 40 cm to make each hat. Determine
Q
whether the cardboard is sufficient to make the hat. Show
your working.
R
S O 25 cm P
Jejari/Radius = 6 cm
∠POQ = 120° P 25 cm 39.6°
PQ = QR = PR
120°
6 cm O Q
Panjang PQ Q
Length of PQ R
R
= ! 6 + 6 – 2(6) kos/cos 120° Panjang lilitan topi = 2π(9)
2
2
2
= 10.39 cm Maka, panjang lengkok PQ = 18π cm
25 × ˙QOP × p = 18π
Panjang lengkok PQ 180°
Length of arc PQ ∠ QOP = 18 × 180°
25
= 6 × 120° × π = 129.6°
180°
= 12.57 cm ∠QOR = 129.6° – 90°
= 39.6°
Jumlah panjang benang bagi setiap bulatan Maka, OS = 25 sin 39.6° = 15.94 cm
Total length of thread for each circle Panjang minimum kadbod
= 3(10.39 + 12.57) = (25 + 15.94) cm
= 68.88 cm = 40.94 cm
≈ 41 cm
15 tampalan/patches = 15 × 68.88
= 1 033.2 cm ÷ 100 Saiz minimum kadbod
= 10.332 m = (25 × 41) cm
Oleh kerana saiz minimum kadbod tersebut ialah
(25 × 41) cm, maka saiz sebenar kadbod (25 × 40) cm
itu tidak mencukupi untuk menghasilkan topi tersebut.


















7




01 Modul Series MateTam Tg5.indd 7 04/10/2021 2:15 PM

Uji Kendiri 1.2

1. Rajah menunjukkan sebuah sektor AOB berpusat O dan 2. Rajah menunjukkan sebuah semibulatan berpusat O
berjejari 8 cm. dan sebuah sektor APB dengan pusat P yang berjejari
The diagram shows a sector AOB with centre O and a radius 10 cm.
The diagram shows a semicircle with centre O and a sector APB
BAB 1 of 8 cm. KBAT Mengaplikasi with centre P and a radius of 10 cm. KBAT Mengaplikasi
A

12 cm
8 cm
B
θ A O
O 1.4 rad 10 cm
B
P
Diberi panjang lengkok AB ialah 12 cm, cari
Given the length of arc AB is 12 cm, find Cari
Find
(a) nilai q, dalam radian,
the value of q, in radians, (a) jejari semibulatan itu,
(b) panjang perentas AB. the radius of the semicircle,
the length of the chord AB. (b) perimeter kawasan berlorek itu.
the perimeter of the shaded region.
(a) 8q = 12 (a) 1.4 rad = 1.4 × 180°
q = 12 π
8 = 80.21°
= 1.5 rad
(b) 1.5 rad = 1.5 × 180° sin ( 80.21° ) = OB
2
10
π
= 85.94° OB = 6.44 cm
Panjang perentas AB (b) Perimeter = 6.44π + 10(1.4)
= ! 8 + 8 – 2(8) kos 85.94° = 34.23 cm
2
2
2
= ! 118.937
= 10.91 cm
©PAN ASIA PUBLICATIONS

1.3 Luas Sektor Suatu Bulatan Buku Teks
Area of Sector of a Circle m.s. 12-19

1. Hitung luas kawasan berlorek yang berikut berdasarkan syarat-syarat yang diberi.
Calculate the area of the following shaded regions based on the given conditions. TP 3

Contoh/Example (a)
Luas
Gunakan/Use = 1 (3.5) (2.2)
2
1.4 = Luas/Area 3.5 cm 2
2π πj 2 O = 13.48 cm 2
10 cm 2.2 rad
O B Luas/Area
1.4 rad 1.4
= (10) 2
2
= 70 cm 2
A
(b) (c) 4.5 cm O
7.2 cm θ q = π
3θ 4
0.4 rad 3q = 3π
4
Luas Luas
( )
= 1 (7.2) (0.4) 1 2 3π
2
2 = 2 (4.5) 4
= 10.37 cm 2
= 23.86 cm 2


8




01 Modul Series MateTam Tg5.indd 8 04/10/2021 2:15 PM

2. Hitung jejari, j dan/atau sudut q, dalam radian, bagi setiap yang berikut.
Calculate the radius, j and/or the angle q, in radians, for each of the following. TP 3

Contoh/Example (a)
Luas kawasan berlorek
Luas kawasan berlorek Area of the shaded region
7 cm
Area of the shaded region = 20 cm 2
θ = 24 cm 2 1.3 rad BAB 1
j

1 (1.3)j = 20
2
Luas/Area = j q 2
1 2
2 j = 20 × 2
2
1
©PAN ASIA PUBLICATIONS
24 = (7) q 1.3
2
2 j = 5.55 cm
  q = 48
49
= 0.98 rad


(b) (c)
Luas kawasan berlorek 10 cm Luas kawasan berlorek
Area of the shaded region Area of the shaded region
O = 100 cm 2 = 30 cm 2
θ j
10 cm θ

1
j q = 30 .................a
100 = (10) (2π – q) 1 2
2
2
2 jq = 10 .................b
2 = 2π – q 1
q = 2π – 2 a ÷ b 2 j = 3
= 4.28 rad j = 6 cm
ºq = 10
6
= 5 rad
3





3. Hitung luas bagi setiap kawasan berlorek yang berikut.
Calculate the area for each of the following shaded regions. TP 4

Contoh/Example (a) P
2.3 cm
O
5 cm S 1.4 rad Q
1.6 rad D R
3 cm
A C O

B

1
Luas/Area OBC = (8) (1.6)
2
2 Luas kawasan berlorek
1
Luas/Area OAD = (5) (1.6) 1 1
2
2 = (4.6) (2π – 1.4) – (2.3) (2π – 1.4)
2
2
2
2
Luas kawasan berlorek = 38.75 cm 2
Area of the shaded region
1
1
= (8) (1.6) – (5) (1.6)
2
2
2 2
= 31.2 cm 2
9
01 Modul Series MateTam Tg5.indd 9 04/10/2021 2:15 PM

(b) (c)
B B

C
7 cm C 8.2 cm O A
BAB 1 ∠BOA = rad
π
3
1
O θ 6 cm A Luas semibulatan = (8.2) π
2
tan q = 7 2
6 = 105.62 cm 2
q = 49.4°
( )
1
1
Luas ∆OAB = (6)(7) Luas sektor AOB = (8.2) 2 π
©PAN ASIA PUBLICATIONS
2 2 3
= 21 cm 2 = 35.21 cm 2
(
1
Luas sektor = (6) 49.4° × π )
2
2 180° Luas kawasan berlorek
= 15.52 cm 2 = 105.62 – 35.21
Luas kawasan berlorek = 70.41 cm 2
= 21 – 15.52
= 5.48 cm 2
4. Cari luas tembereng bagi setiap yang berikut.
Find the area of segment for each of the following. TP 5
Contoh/Example
Tip SPM
Luas tembereng
Area of the segment Luas segi tiga/Area of triangle = 1 AB × h
= Luas sektor AOB – Luas segi tiga AOB 2
O 4 cm Area of segment AOB – Area of triangle AOB sin q = AC
B 2 j
1 2
1 2
1.4 rad h = j q – j sin q° º AC = j sin q , AB = 2j sin q
2
2
C 2 2
Jika q = 1.4 rad dan j = 4 cm, luas tembereng q q h
/
If q = 1.4 rad and j = 4 cm, the area of the kos cos =
A 2 2 j
segment h = j kos q j cos q
/
(
1
1
Tip SPM = (4) (1.4) – (4) sin 1.4 × 180° ) 2 2
2
2
2 2 π Luas segi tiga/Area of triangle
= 3.32 cm 2 q q q q
1
q hendaklah dalam darjah = 1 ( 2j sin kos )/ ( 2j sin cos )
2

2
untuk trigonometri sinus. 2 2 2 2 2 2
q must be in degrees for = 1 2
j sin q
trigonometric sine. 2
(a) (b)
O Q
1.1 rad
8 cm —
π
3
O 9 cm P
P Q π 2π
∠POQ = π – = rad
3 3
1.1 rad = 1.1 × 180° = 63.02° Luas tembereng
π
( )
1
1
1
2
Luas sektor = (8) (1.1) = 35.2 cm 2 = (9) 2 2π – (9) sin ( 2π × 180° )
2
2 2 3 2 3 π
1
2
Luas segi tiga = (8) sin 63.02° = 49.75 cm 2
2
= 28.52 cm 2
Luas tembereng
= 35.2 – 28.52
= 6.68 cm 2
10
01 Modul Series MateTam Tg5.indd 10 04/10/2021 2:15 PM

(c) (d)
P
B
6.5 cm
O
O
C
5 cm
30° BAB 1
A
D
Q
π
∠POQ = 180° – 60° = 120° ∠BOC = ∠AOB = rad
3
3
120° × π = 2π rad Luas tembereng
180° 3
( )
1 ©PAN ASIA PUBLICATIONS
1
2
Luas tembereng = 3 [ 1 (5) 2 π – (5) sin 1 π × 180° 2]
2
π
3
2
3
( )
1
1
= (6.5) 2 2π – (6.5) sin 120° 2
2
2 3 2 = 6.79 cm
= 25.95 cm 2
Uji Kendiri 1.3
1. Rajah menunjukkan sebuah kipas tangan berbentuk 2. Rajah menunjukkan dua sektor bulatan, OACB dan
sektor OABCD dan bahagian berlabel ABCD diperbuat CBOA, masing-masing dengan pusat O dan C.
daripada kain. The diagram shows two sectors, OACB and CBOA, with centres
The diagram shows a hand fan in a form of a sector OABCD O and C respectively. KBAT Mengaplikasi
and the area labelled ABCD is made of cloth. B
KBAT Mengaplikasi
C

B M
13 cm O
A
14 cm Kain A
Cloth
O θ Diberi bahawa OC adalah berserenjang dengan AB dan
D melalui titik tengah M. Jika OB = 8 cm,
C
Given that OC is perpendicular to AB and passes through the
midpoint M. If OB = 8 cm,
)
2
Diberi perimeter kain ialah ( 41 p + 26 cm, cari (a) tunjukkan bahawa sudut AOB ialah π,
3
4
2
)
3
Given the perimeter of the cloth is ( 41 p + 26 cm, find show that the angle AOB is π,
4
(a) nilai q, dalam radian, (b) cari luas kawasan berlorek itu.
the value of q, in radians, find the area of the shaded region.
(b) luas kain itu, dalam sebutan π. (a) OB = 8 cm, OM = 4 cm
the area of the cloth, in terms of π.
4
kos ∠MOB =
8
∠MOB = 60°
(a) 14q + 27q = 41 p + 26 – 26 p 2
4 ∠AOB = 120° × 180° = π rad
3
( )
41q = 41 π 1 2 2
4 (b) Luas sektor AOB = (8) 3 π
2
1
q = π rad = 67.02 cm 2
4 1
2
(b) Luas kain Luas ∆OAB = (8) sin 120°
2
( )
( )
1
= (27) 2 1 π – (14) 2 1 π = 27.71 cm 2
2 4 2 4 Luas kawasan berlorek = 67.02 – 27.71
1
1
= 91 π – 24 π = 39.31 cm 2
8 2 1 2 π 1
5
2
3
2
= 66 π cm 2 Luas sektor ACB = 2 [ 2 (8) ( ) – (8) sin 60° ]
8
= 11.60 cm 2
Luas kawasan berlorek = 39.31 + 11.60
= 50.91 cm 2
11
01 Modul Series MateTam Tg5.indd 11 04/10/2021 2:15 PM

1.4 Aplikasi Sukatan Membulat Buku Teks
Application of Circular Measures m.s. 20-22

1. Selesaikan setiap masalah yang berikut.
Solve each of the following problems. TP 6
BAB 1 Contoh/Example (a) Rajah menunjukkan sebuah sektor dengan pusat O

dan berjejari 9.3 cm. Diberi bahawa Q terletak pada
Rajah menunjukkan sekeping jubin berbentuk segi lilitan bulatan.
empat sama ABCD. Diberi bahawa panjang sisi AB The diagram shows a sector with centre O and a radius of
ialah 14 cm. Keempat-empat kawasan berlorek itu 9.3 cm. Given that Q lies on the circumference.
adalah serupa. OPQ dan BPQ adalah sektor masing-
masing dengan pusat O dan B.
©PAN ASIA PUBLICATIONS
The diagram shows a square tile ABCD. Given that the length Q R
of side AB is 14 cm. The 4 shaded regions are identical. OPQ O
and BPQ are sectors with centres O and B respectively. 140°
48°
A P B
P
Cari
S O Q Find
(i) perimeter kawasan berlorek,
the perimeter of the shaded region,
(ii) luas kawasan berlorek.
D C
R the area of the shaded region.
Jika terdapat 150 keping jubin yang sama pada suatu
lantai, cari luas yang dilitupi oleh kawasan berlorek itu. R
If there are 150 same tiles on the floor, find the area covered Q 22°
by the shaded regions. 48° O
140°
P
48°
H
P
(i) ∠POQ = 180° – 2(48°)
= 84°
O Q
Luas sektor OPQ Panjang perentas PQ
Area of the sector OPQ = ! 9.3 + 9.3 – 2(9.3) kos 84°
2
2
2
1
= × π(7) 2 = 12.45 cm
4
Luas ∆OPQ Panjang lengkok PQ
p
Area of ∆OPQ = 9.3 × 84° × 180°
1
= × 7 × 7 = 13.63 cm
2
1
= (7) 2 ∠QOR = 360° – 84° – 140°
2 = 136°
Luas tembereng PHQ Panjang perentas QR
Area of the segment PHQ = ! 9.3 + 9.3 – 2(9.3) kos 136°
2
2
2
1
1
= × π(7) – (7) 2 = 17.25 cm
2
4 2
Maka, luas kawasan berlorek setiap jubin Panjang lengkok QR
p
Hence, the area of the shaded region of each tile = 9.3 × 136° × 180°
1
= 8 [ 1 × π(7) – (7) 2 ] = 22.07 cm
2
4
2
= 8(13.98) Maka, perimeter kawasan berlorek
= 111.88 cm 2 = 12.45 + 17.25 + 13.63 + 22.07
= 65.4 cm
Jumlah luas kawasan berlorek
Total area of the shaded regions (ii) Luas kawasan berlorek
]
1
= 111.88 × 150 = [ 1 × 9.3 × 84° × p – (9.3) sin 84°
2
2
= 1.6782 × 10 cm 2 2 180° 2
4
p
1
+ [ 1 × 9.3 × 136° × 180° – (9.3) sin 136° ]
2
2
2
2
= 93 cm 2
12
01 Modul Series MateTam Tg5.indd 12 04/10/2021 2:15 PM

Uji Kendiri 1.4

1. Rajah menunjukkan sebuah bulatan berjejari 6.5 cm 2. Rajah menunjukkan sepotong tembikai dengan keratan
dan berpusat O. Diberi bahawa panjang AB adalah rentas seragamnya berbentuk sektor berpusat O.
sama dengan panjang AC. The diagram shows a uniform cross-section of a piece of
The diagram shows a circle with a radius of 6.5 cm and watermelon in the shape of a sector with centre O.
centre O. Given that the length of AB is the same as AC. KBAT Menilai BAB 1
KBAT Menilai
10 cm
B
O
A θ
C
O D
50°
A B
C
Diberi bahawa panjang lengkok AB dan CD masing-
masing ialah 20 cm dan 16 cm dengan bahagian ABCD
Cari/Find tidak boleh dimakan. Jika AC = 2 cm dan panjang
(a) perimeter kawasan berlorek,
the perimeter of the shaded region, potongan tembikai itu ialah 10 cm, cari
(b) luas kawasan berlorek. Given that the arc length AB and CD is 20 cm and 16 cm
respectively with the portion ABCD is not edible. If AC = 2 cm
the area of the shaded region. and the length of a piece of the watermelon is 10 cm, find
(a) ∠BAO = 25° (a) nilai q, dalam radian,
Panjang AB = 2(6.5 kos 25°) the value of q, in radians,
= 11.78 cm (b) isi padu bahagian yang boleh dimakan.
Panjang lengkok AB (a) the volume of the area that can be eaten.
OC(q) = 16 ..............a
p
= 6.5 × 130° × 180° (OC + 2)q = 20
= 14.75 cm OC(q) + 2q = 20 ..............b
Perimeter kawasan berlorek b – a: 2q = 20 – 16
q = 2 rad
= 2(11.78) + 2(14.75) 16
= 53.06 cm (b) OC = 2 = 8 cm
(b) Luas kawasan berlorek Isi padu = (8) (2) × 10
1
2
]
[
p
1
2
1
– (6.5) sin 130°
©PAN ASIA PUBLICATIONS
× 6.5 × 130° ×
= 2
2
2
180°
2
2
= 640 cm


3
= 63.5 cm 2
SPM
Soalan Berformat SPM
Kertas 1 Paper 1
Bahagian A/Section A
1. Rajah menunjukkan sebuah bulatan berpusat O dengan jejari 5 cm. Diberi OQTC ialah C T
sebuah segi empat tepat dengan luas 40 cm , cari B
2
The diagram shows a circle with centre O and a radius of 5 cm. Given OQTC is a rectangle with an θ
2
area of 40 cm , find KBAT Menilai O A Q
(a) nilai q, dalam radian,
the value of q, in radians, [2 markah/marks]
(b) luas sektor OAB,
the area of the sector OAB, [2 markah/marks]
(c) perimeter kawasan berlorek.
the perimeter of the shaded region. [2 markah/marks]
)
(a) OC × CT = 40 (c) Panjang lengkok AB = 5 ( π – 1.01 = 2.80 cm
5 × CT = 40 2
º CT = 8 cm Panjang BT = ! 8 + 5 – 5 = 4.43 cm
2
2
tan q = 8 Perimeter = 2.80 + 4.43 + 3 + 5 = 15.23 cm
5
q = 58° × π = 1.01 rad
180°
(
)
1
(b) Luas sektor OAB = (5) 2 π – 1.01 = 7.01 cm 2
2 2
13
01 Modul Series MateTam Tg5.indd 13 04/10/2021 2:15 PM

2. Rajah menunjukkan sebuah sektor AOB berpusat O. 3. Rajah menunjukkan dua sektor, POQ dan SOR,
The diagram shows a sector AOB with centre O. berpusat O.
KBAT Menilai The diagram shows two sectors, POQ and SOR, with centre O.
KBAT Menganalisis
S
A
BAB 1 130° 7.5 cm B O θ P
O
Cari
Find Q
(a) ∠AOB, dalam radian, R
∠AOB, in radians, [2 markah/marks] Diberi bahawa OP = 6 cm, OP : OS = 2 : 3 dan luas
©PAN ASIA PUBLICATIONS
(b) perimeter, dalam cm, sektor AOB. kawasan berlorek ialah 33.75 cm , cari
2
the perimeter, in cm, of the sector AOB. Given that OP = 6 cm, OP : OS = 2 : 3 and the area of the
[3 markah/marks] shaded region is 33.75 cm , find
2
(a) panjang OS,
(a) 130° × π = 2.27 rad the length of OS, [2 markah/marks]
180° (b) nilai q, dalam radian.
(b) Perimeter = 7.5 + 7.5 + 7.5(2.27) the value of q, in radians. [3 markah/marks]
= 32.03 cm
2
(a) OP = = 6 = OS = 9 cm
OS 3 OS
1
(b) 1 (9) q – (6) q = 33.75
2
2
2 2
22.5q = 33.75
q = 1.5 rad
Bahagian B/Section B
4. Rajah menunjukkan dua bulatan sepusat berpusat O. 5. Rajah menunjukkan sebuah segi tiga sama sisi OPQ dan
KLON Sudut yang dicangkum pada pusat O oleh lengkok titik S yang terletak di dalam segi tiga itu dengan keadaan
SPM M major PQ ialah 8p rad dan perimeter bagi seluruh rajah OS = SP = SQ.
SP
ialah 48 cm. The diagram shows an equilateral triangle OPQ and the
The diagram shows two concentric circles with centre O. point S lies inside the triangle such that OS = SP = SQ.
The angle subtended at the centre O by the major arc PQ is KBAT Menganalisis
8p rad and the perimeter for the whole diagram is 48 cm. O
KBAT Menganalisis
P
12 cm
A S
P Q
O
B Q
X

Diberi OP = 3 OA, OA = j cm dan ∠AOB = 4p rad, Diberi bahawa S ialah pusat bagi sektor SPXQ dan
2 panjang lengkok PXQ ialah 12 cm. Cari perimeter
ungkapkan j dalam sebutan p. kawasan berlorek, dalam sebutan π dan ! 3.
3
Given OP = OA, OA = j cm and ∠AOB = 4p rad, express Given that S is the centre of the sector SPXQ and the length of
2
j in terms of p. [8 markah/marks] arc PXQ is 12 cm. Find the perimeter of the shaded region, in
3
OP = OA terms of π and ! 3 . [8 markah/marks]
2 kos ∠OQS = 6
3
= j SQ
2 kos 30° = 6
3
Panjang lengkok major PQ = j(8p) SQ
2
6
6
12
= 12jp SQ = kos 30° = ! 3 = ! 3 = 4! 3 cm
Panjang lengkok AB = j(4p) 2
= 4jp Perimeter kawasan berlorek O
Jumlah perimeter: ( π )
3
12jp + 4jp + 2( j – j) = 48 = 4! 3 + 4! 3 + (4! 3) 120° × 180°
2 S
16jp + j = 48 = 8! 3 + 8! 3 π 30° 120° 6
j(16p + 1) = 48 3 P 12 Q
(
j = 48 = 8! 3 1 + π ) cm
16p + 1 3
14



01 Modul Series MateTam Tg5.indd 14 04/10/2021 2:15 PM

Kertas 2 Paper 2

Bahagian A/Section A


1. Rajah menunjukkan sektor POR dengan pusat O. Diberi bahawa panjang lengkok PQ ialah S
KLON 3.4 cm. Hitung P
SPM M The diagram shows a sector POR with centre O. It is given that the length of arc PQ is 3.4 cm. BAB 1
SP
Q
Calculate KBAT Menganalisis
(a) ∠POQ, dalam radian, 8.6 cm
∠POQ, in radians, [3 markah/marks]
(b) luas, dalam cm , kawasan berlorek. 50°
2
the area, in cm , of the shaded region. [3 markah/marks] O R
2
©PAN ASIA PUBLICATIONS
2. Rajah menunjukkan sebuah kebun yang berbentuk sektor OAB dengan pusat O dan A
berjejari 12 m. Diberi OPQR ialah kawasan tanaman sayur yang berbentuk trapezium
dengan keadaan OR = 10 m, PQ = 8 m dan QR adalah berserenjang dengan OA. Diberi R
bahawa ∠AOB = 1.2 rad, cari Q
The diagram shows a garden in the shape of a sector OAB with centre O and a radius of 12 m.
Given OPQR is a trapezoidal vegetable growing area such that OR = 10 m, PQ = 8 m and QR is
perpendicular to OA. Given that ∠AOB = 1.2 rad, find KBAT Mengaplikasi 1.2 rad
(a) panjang pagar yang digunakan untuk memagar kawasan tanaman sayur, O P B
the length of the fence used to fence around the vegetable growing area, [3 markah/marks]
(b) luas kawasan yang tidak ditanam dengan sayur.
the area not planted with vegetables. [3 markah/marks]


Bahagian B/Section B


3. Rajah menunjukkan sebuah bulatan berpusat O. Diberi D ialah titik tengah bagi AB dan C
luas kawasan berlorek ialah 200π cm , cari A
2
3 D
The diagram shows a circle with centre O. Given D is the midpoint of AB and the area of the shaded 60° B
region is 200π cm , find KBAT Mengaplikasi O
2
3 θ
(a) nilai q, dalam radian,
the value of q, in radians, [2 markah/marks]
(b) jejari bulatan itu,
the radius of the circle, [2 markah/marks]
(c) perimeter sektor berlorek,
the perimeter of the shaded sector, [3 markah/marks]
(d) luas tembereng ADBC.
the area of the segment ADBC. [3 markah/marks]



4. Rajah menunjukkan sebuah semibulatan berpusat O dengan jejari 8 cm. TSQ ialah sebuah
sektor dengan pusat T dan berjejari 12 cm. Diberi bahawa OR berserenjang dengan POQ. S
Hitung
The diagram shows a semicircle with centre O and a radius of 8 cm. TSQ is a sector with centre T R
and a radius of 12 cm. Given that OR is perpendicular to POQ. Calculate KBAT Menganalisis
(a) nilai q, dalam radian,
the value of q, in radians, [2 markah/marks] θ
(b) perimeter kawasan berlorek, P T O Q
the perimeter of the shaded region, [4 markah/marks]
(c) luas kawasan berlorek.
the area of the shaded region. [4 markah/marks]










15




01 Modul Series MateTam Tg5.indd 15 04/10/2021 2:15 PM

5. Rajah menunjukkan sebuah sektor AOB berpusat O dengan jejari 15 cm. Titik C terletak A
pada OB dengan keadaan OC = CA = 9 cm. Hitung
The diagram shows a sector AOB with centre O and a radius of 15 cm. Point C lies on OB such
that OC = CA = 9 cm. Calculate KBAT Menganalisis
(a) nilai q, dalam radian, θ
the value of q, in radians, [2 markah/marks] O C B
BAB 1 (b) perimeter kawasan berlorek, [4 markah/marks]
the perimeter of the shaded region,
(c) luas kawasan berlorek.
the area of the shaded region. [4 markah/marks]


6. Rajah menunjukkan sebuah semibulatan dengan pusat O dan sebuah sukuan bulatan
©PAN ASIA PUBLICATIONS
RPQ berpusat P. Hitung
The diagram shows a semicircle with centre O and a quadrant RPQ with centre P. Calculate
KBAT Menilai R
(a) nilai q, dalam radian, θ 6 cm 3 cm
the value of q, in radians, [2 markah/marks] A O P Q
(b) perimeter kawasan berlorek,
the perimeter of the shaded region, [4 markah/marks]
(c) luas kawasan berlorek.
the area of the shaded region. [4 markah/marks]


7. Rajah menunjukkan dua sektor, AOB dan APQ, masing-masing berpusat O dan P.
Diberi bahawa OB selari dengan PQ. Jika luas sektor APQ sama dengan luas kawasan A
berlorek, cari
The diagram shows two sectors, AOB and APQ, with centre O and P respectively. Given that OB P
is parallel to PQ. If the area of the sector APQ is the same as the area of the shaded region, find j 2 Q
KBAT Menganalisis θ
(a) nisbah j kepada j , O j 1 B
1
2
the ratio of j to j , [5 markah/marks]
1
2
(b) nisbah perimeter sektor APQ kepada AOB.
the ratio of the perimeter of the sector APQ to AOB. [5 markah/marks]
8. Rajah menunjukkan suatu corak berbentuk bulatan yang dilukis oleh Donny bagi projek
KLON lukisannya. Diberi bahawa lapan kawasan yang berlorek itu adalah serupa. Panjang C B
SPM M lengkok ACB dan ADB adalah sama dan perimeter bagi kawasan berlorek itu ialah A
SP
28π cm. Cari D
The diagram shows the circular design drew by Donny for his art project. Given that the eight
shaded regions are congruent. The lengths of arc ACB and ADB are the same and the perimeter
of the shaded region is 28π cm. Find KBAT Mengaplikasi
(a) jejari bulatan itu,
the radius of the circle, [5 markah/marks]
(b) luas kawasan berlorek itu.
the area of the shaded region. [5 markah/marks]



























16




01 Modul Series MateTam Tg5.indd 16 04/10/2021 2:15 PM

2.3 Pembezaan Peringkat Kedua Buku Teks
The Second Derivative m.s. 49-50

dy d y
2
1. Cari dan bagi setiap yang berikut.
dx dx 2
dy d y
2
Find and for each of the following. TP 2
dx dx 2
Contoh/Example Tip SPM
y = 4x + 2x – 5 d y d dy
3
2
x • = ( )
y = 4x + 2x – 5x –1 dx 2 dx dx
3
©PAN ASIA PUBLICATIONS
d
BAB 2 dy = 12x + 2 + 5x d y 2 = 24x – 10x –3 • f(x) = (f(x))
2
–2
2
dx
dx
dx
2
5
dy 2
≠
= 12x + 2 + = 24x – 10 • d y 2 ( )
2
x 2 x 3 dx dx
(a) y = (1 – 2x ) (b) y = 2x – 5
2 4
3
dy 2 3 x
dx = 4(1 – 2x ) (– 4x) y = 2x – 5x –1
3
= –16x(1 – 2x ) dy 2 5
2 3
d y 2 2 2 3 dx = 6x + x 2
2
2
dx 2 = –16x(3)(1 – 2x ) (– 4x) + (1 – 2x ) (–16) = 6x + 5x –2
= (1 – 2x ) (192x + 32x – 16) = 6x + 5
2 2
2
2
2
= (1 – 2x ) (224x – 16) x 2
2
2 2
d y = 12x – 10
2
dx 2 x 3
2. Selesaikan masalah berikut.
Solve the following problems. TP 3
Contoh/Example (a) Diberi bahawa f(x) = (5x – 2) , cari f '(1) dan f ʺ(0).
5
Given that f(x) = (5x – 2) , find f '(1) and f ʺ(0).
5
Diberi g(x) = 3x + 2x – 9x – 7, cari gʹ(4) dan gʺ(–1).
2
3
Given g(x) = 3x + 2x – 9x – 7, find gʹ(4) and gʺ(–1). f(x) = (5x – 2) 5
3
2
f ʹ(x) = 5(5x – 2) (5)
4
g(x) = 3x + 2x – 9x – 7 = 25(5x – 2) 4
3
2
gʹ(x) = 9x + 4x – 9 f ʹ(1) = 25(5(1) – 2) 4
2
gʹ(4) = 9(4) + 4(4) – 9 = 2 025
2
= 151 f ʺ(x) = 25(4)(5x – 2) (5)
3
gʺ(x) = 18x + 4 = 500(5x – 2) 3
gʺ(–1) = 18(–1) + 4 f ʺ(0) = 500(5(0) – 2) 3
= –14 = – 4 000










26




02 Modul Series MateTam Tg5.indd 26 04/10/2021 2:42 PM

4. Nilai yang mungkin bagi X ditunjukkan dalam jadual berikut berserta dengan kebarangkalian yang sepadan. Cari nilai p
dan seterusnya lukis graf taburan kebarangkaliannya.
The possible values of X are shown in the following table together with the corresponding probabilities. Find the value of p and hence
draw the probability distribution graph. TP 3
(a) X = r
Contoh/Example 3 4 5 6 7
P(X = r) 1 p 0.25 p 2 p 0.1
X = r 1 3 5 7 9 2 3
1 1 1 2
P(X = r) p 2p p 2 p 2 p 2 p + 0.25 + p + p + 0.1 = 1
3
1
1
p + 2p + p + p + p = 1 13 p = 0.65
6
2 2
5p = 1 p = 0.3
©PAN ASIA PUBLICATIONS
p = 1
5 X = r 3 4 5 6 7
X = r 1 3 5 7 9 P(X = r) 0.15 0.25 0.3 0.2 0.1
P(X = r) 0.2 0.4 0.2 0.1 0.1

P(X = r)
P(X = r)
0.4 x
0.4
0.3 0.3 x
x
0.2 x
0.2 x x x
0.1 x
PAUTAN INTERAKTIF r
0.1 x x 0 3 4 5 6 7

r
0 1 3 5 7 9

PAUTAN INTERAKTIF
Graf taburan kebarangkalian
The graph of the probability
BAB 5 distribution








Uji Kendiri 5.1


1. Tulis dalam tatatanda set bagi pemboleh ubah rawak 2. Berikut merupakan sebahagian daripada sistem merit
X yang mewakili jisim sebiji durian dari kebun Ronny dan demerit di sebuah sekolah:
yang jisimnya di antara 0.8 kg dengan 3 kg. The following is a part of the merit and demerit system in a
Write down in a set notation of the random variable X that school:
represents the mass of a durian from Ronny’s farm which has • Tolak 1 markah jika lewat datang ke sekolah.
a mass between 0.8 kg and 3 kg. Minus 1 mark if late to school.
• Tambah 2 markah bagi setiap kerja rumah yang
X = {x : 0.8 , x , 3} selesai.
Add 2 marks for each finished homework.
Tulis dalam tatatanda set bagi pemboleh ubah rawak X
yang mewakili sistem itu apabila Isah mempunyai kerja
rumah bagi 5 subjek pada suatu hari.
Write down in a set notation of the random variable X that
represents the system when Isah has a homework for 5
subjects on a certain day.
X = {–1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}







74




05 Modul A+ MateTam Tg5.indd 74 05/10/2021 12:29 PM

π
BAB 1 RQ = (8) = 4π
2
Perimeter kawasan berlorek
Kertas 2/Paper 2
Perimeter of the shaded region
Bahagian A/Section A = 12 – ! 80 + 13.32 + 4π
1. (a) 3.4 = 8.6q = 28.94 cm
q = 3.4 (c) Luas sektor STQ/Area of sector STQ
8.6 1
2
= 0.4 rad = 2 (12) (1.11)
180° ©PAN ASIA PUBLICATIONS
∴∠POQ = 0.4 rad = 79.92 cm 2
1
(b) tan 50° = SR Luas/Area ∆ = (4)(8) = 16 cm 2
OR 2
SR = 8.6 tan 50° 1 2 2
= 10.25 cm Luas/Area ROQ = π(8) = 16π cm
4
Luas kawasan berlorek Luas kawasan berlorek
Area of the shaded region Area of the shaded region
1
1
= 1 (8.6)(10.25) – (8.6) sin 50° + (8.6) (0.4) = 79.92 – 16 – 16π
2
2
2 2 2 = 13.65 cm 2
= 30.54 cm 2 7.5
2. (a) 1.2 rad = 68.75° 5. (a) kos/cos q = 9
tan 68.75° = QR q = 33.56°
2 q = 0.586 rad
QR = 2 tan 68.75° (b) Perimeter kawasan berlorek
QR = 5.14 m Perimeter of the shaded region
kos/cos 68.75° = 2 = 9 + 6 + 15(0.586)
OP = 23.79 cm
OP = 2
kos/cos 68.75° (c) Luas kawasan berlorek
OP = 5.52 m Area of the shaded region
1
Perimeter pagar = 1 (15) (0.586) – (15)(9) sin 33.56°
2
Perimeter of the fence 2 2
= 5.14 + 5.52 + 10 + 8 = 28.61 cm 2
= 28.66 m 3
(b) Luas kawasan yang tidak ditanam dengan sayur 6. (a) sin ∠ROP = 6
Area of the area not planted with vegetables ∠ROP = 30°
1
= 1 (12) (1.2) – (10 + 8)(5.14) 30° × π = π
2
2 2 180° 6
= 40.14 m 2 ∠AOR = q
q = π – π
Bahagian B/Section B 6
q = 2.62 rad
3. (a) q = 360° – 2 × 60° 2 2
q = 240° (b) OP = ! 6 – 3
π
q = 240° × 180° = 3! 3 cm
q = 4π rad Perimeter kawasan berlorek
Perimeter of the shaded region
3
π
π
( )
+ 3
j
(b) 1 2 4π = 200π = 6 ( ) ( ) + 3 – (6 – 3! 3)
6
2
2 3 3
2
j = 100 = 10.05 cm
j = 10 cm (c) Luas sektor ROP/Area of sector ROP
( )
(c) Perimeter sektor berlorek = 1 (6) 2 π = 3π cm 2
Perimeter of the shaded sector 2 6
4π
1
= 10 + 10 + 10 ( ) Luas/Area ∆ROP = (3! 3)(3)
3
2
= 61.89 cm 9! 3
(d) ∠AOB = 120° = 2 cm 2
120° × π = 2π Luas sektor RPQ/Area of sector RPQ
3
( )
Luas tembereng ADBC/Area of the segment ADBC = 1 (3) 2 π
( )
1
= 1 (10) 2 2π – (10) sin 120° 2 2
2
2 3 2 = 9 π cm 2
4
= 61.42 cm 2 Luas kawasan berlorek
8
4. (a) tan q = Area of the shaded region

4
9! 3
q = 63.43° = 9 π – (3π – )
q = 1.11 rad 4 2
= 5.44 cm 2
2
2
(b) TR = ! 8 + 4 = ! 80
SQ = 12(1.11) = 13.32
MG-1
Jwpn_G Modul A+ MateTam Tg5.indd 1 27/10/2021 11:11 AM

Lembaran Pentaksiran Bilik Darjah (PBD)



BAB 1 Sukatan Membulat

Tahap Penguasaan Tafsiran
5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sukatan membulat dalam
konteks penyelesaian masalah rutin yang kompleks.
6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sukatan membulat dalam
konteks penyelesaian masalah bukan rutin secara kreatif.
©PAN ASIA PUBLICATIONS
1. Rajah menunjukkan sebuah segi tiga bersudut tegak PQR. Diberi A ialah titik tengah PQ dengan keadaan PQ = 50 cm,
∠PRQ = π rad, PBQ ialah satu lengkok daripada sebuah sektor berjejari r cm dan AB = 10 cm.
6
The diagram shows a right-angled triangle PQR. Given that A is the midpoint of PQ where PQ = 50 cm, ∠PRQ = π rad, PBQ is an
6
arc from a sector with radius r cm and AB = 10 cm. TP 5
P



A B


Q R

Cari
Find
(a) nilai r,
the value of r,
(b) panjang lengkok PBQ,
the arc length PBQ,
(c) luas kawasan berlorek.
the area of the shaded region.
(a) OA = r – 10 P
OP = OA + AP 2
2
2
r = (r – 10) + 25 2 r cm
2
2
r = r – 20r + 100 + 625 θ 10 cm
2
2
20r = 725 O A B
r = 36.25 cm 25 cm
30°
25 Q R
(b) tan ∠POA = 26.25
∠POA = 43.6° = 0.7610 rad
∠POQ = 1.52 rad
Panjang lengkok PBQ/Arc length PBQ
= 36.25(1.52)
= 55.1 cm

(c) Luas tembereng PBQ/Area of the segment PBQ
1 1
= 2 (36.25) (1.52) – 2 (36.25) sin 87.2°
2
2
= 342.44 cm 2
Luas ∆PQR/Area of ∆PQR
1
= (50)(100) sin 60° = 2 165.06 cm 2
2
Luas kawasan berlorek/Area of the shaded region
= 2 165.06 – 342.44
= 1 822.62 cm 2




MG-14




Lemb PBD Modul A+ Maths Tg5.indd 14 27/10/2021 11:13 AM

BAB 4 Pilih Atur dan Gabungan

Tahap Penguasaan Tafsiran
3 Mengaplikasikan kefahaman tentang pilih atur dan gabungan untuk melaksanakan tugasan mudah.
4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pilih atur dan gabungan dalam
konteks penyelesaian masalah rutin yang mudah.
5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pilih atur dan gabungan dalam
konteks penyelesaian masalah rutin yang kompleks.
6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pilih atur dan gabungan dalam
konteks penyelesaian masalah bukan rutin secara kreatif.

©PAN ASIA PUBLICATIONS
1. (a) Diberi C = 10, nyatakan nilai
n
1
n
Given C = 10, state the value of TP 3
1
(i) n
(ii) C 9
n
(b) Jika C = C , ungkapkan n dalam sebutan r dan s.
n
n
s
r
If C = C , express n in terms of r and s. TP 5
n
n
r
s
(a) (i) 10
(ii) Apabila/When n = 10, C = 10
10
9
(b) C = C = C
n
n
n
r n – r s
n – r = s
n = s + r
2. Cari bilangan cara untuk membina kata laluan yang terdiri daripada 6 abjad jika menggunakan semua abjad dan
Find the number of ways to form a password consisting of 6 alphabets chosen from all the alphabets if TP 3
(a) ulangan abjad tidak dibenarkan,
repetition of alphabets is not allowed,
(b) ulangan abjad dibenarkan,
repetition of alphabets is allowed, Lembaran PBD
(c) huruf-huruf vokal digunakan dan huruf A digunakan dua kali.
the vowel letters are used and the letter A is used twice.
(a) 26 P atau/or 26 × 25 × 24 × 23 × 22 × 21
6
= 165 765 600
(b) 26 atau/or 26 × 26 × 26 × 26 × 26 × 26
6
= 308 915 776
(c) 6! = 360
2!
3. Seorang jurulatih ingin memilih 6 orang pemain yang terdiri daripada 3 orang lelaki dan 3 orang perempuan untuk
membentuk satu pasukan badminton. 6 orang pemain itu dipilih daripada sekumpulan 4 orang lelaki dan 5 orang
perempuan.
A coach wants to choose 6 players consisting of 3 boys and 3 girls to form a badminton team. These 6 players are chosen from a group
of 4 boys and 5 girls. TP 5
(a) Cari
Find
(i) bilangan cara pasukan itu dapat dibentuk,
the number of ways the team can be formed,
(ii) bilangan cara menyusun ahli pasukan itu dalam satu baris untuk satu sesi bergambar jika lelaki dan perempuan
duduk berselang-seli.
the number of ways the team members can be arranged in a row for a group photograph if the boys and the girls sit
alternately.
(b) Semasa taklimat, ahli pasukan duduk dalam bentuk bulatan. Cari bilangan pilih atur yang mungkin.
During briefing, the team members sit in circular form. Find the possible number of permutations.
(a) (i) 4 C × C = 40
5
3
3
(ii) 3! × 3! × 2 = 72
(b) (6 – 1)! = 120



MG-21




Lemb PBD Modul A+ Maths Tg5.indd 21 27/10/2021 11:13 AM

Book extend: 184pp
Spine: 8.36mm







PAN ASIA

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 ■ Jawapan Lengkap Soalan
©PAN ASIA PUBLICATIONS
Berformat SPM
berpengalaman dalam membantu proses PdPc dengan ■ Lembaran Pentaksiran
lebih efektif. Latihan yang disediakan mencakupi Bilik Darjah (PBD)
Tahap Penguasaan yang perlu dikuasai oleh murid
untuk mengoptimumkan kefahaman mereka.






Dapatkan



DWIBAHASA

Judul-judul dalam 5
siri Modul A+1 sekarang! Matematik

Mata Pelajaran / Tingkatan 4 5 Tambahan Tingkatan


Sejarah Additional Mathematics Matematik Tambahan
Matematik Additional Mathematics

Matematik Tambahan
Kimia
Penulis Buku Teks
Fizik
Azizah binti Kamar (Guru Cemerlang)
Pentaksiran
Perniagaan Tingkatan 5 Dr. M. K. Wong Pentaksiran
Baharu
Ekonomi Nurbaiti binti Ahmad Zaki Baharu
SPM
SPM
English

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buku teks

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