Whole Book Answers f4

1 The diagram shows a rectangle. 2 The diagram shows the curve y − x2 = 2 and the
Rajah menunjukkan sebuah segi empat tepat. straight line y = –2x + 10.

(x – 3) cm Rajah menunjukkan lengkung y − x2 = 2 dan garis lurus
y = –2x + 10.
(y + 1) cm
y – x2 = 2

Given the perimeter is 14 cm and the area is 12 cm2. P
Find the values of x and y. Q y = –2x + 10

Diberi perimeter ialah 14 cm dan luas ialah 12 cm2. Cari Find the coordinates of the points of intersection
nilai x dan y. P and Q.

Perimeter = 14 cm Cari koordinat bagi titik persilangan P dan Q.
2(x − 3) + 2(y + 1) = 14 (÷2)

x−3+y+1=7 Given y − x2 = 2
x + y = 9 ——— 1 or
y = x2 + 2 ——— 1
Area = 12 cm2 and
(x − 3)(y + 1) = 12 y = –2x + 10 —— 2
xy + x − 3y – 3 = 12 x2 + 2 = –2x + 10
xy + x − 3y = 15 ——— 2

From 1 , x2 + 2x − 8 = 0
y=9−x (x + 4)(x − 2) = 0
Substitute y into 2 , x + 4 = 0 or x − 2 = 0
x(9 − x) + x − 3(9 − x) = 15
x = –4 x = 2

9x − x2 + x − 27 + 3x = 15 When x = –4,
–x2 + 13x − 42 = 0 y = –2(–4) + 10 = 18
x2 − 13x + 42 = 0 When x = 2,
(x − 6)(x − 7) = 0 y = –2(2) + 10 = 6
x − 6 = 0 or x − 7 = 0

x = 6 x = 7 Hence,

When x = 6, y = 9 − (6) = 3 P = (–4, 18) and Q = (2, 6)

When x = 7, y = 9 − (7) = 2

3 The diagram shows the circle x2 + y2 = 5 and the Substitute 1 into 2 ,
straight line 2x + y = 4.
x2 + (4 − 2x)2 = 5
Rajah menunjukkan lengkung x2 + y2 = 5 dan garis lurus
2x + y = 4. x2 + 16 − 16x + 4x2 − 5 = 0

2x + y = 4 5x2 − 16x + 11 = 0

P (5x − 11)(x − 1) = 0

Q 5x − 11 = 0 or x − 1 = 0

x2 + y2 = 5 x = 151 x = 1

Find the coordinates of the points of intersection When x = 151,
P and Q.
y = 4 − 2151 = – 25
Cari koordinat bagi titik persilangan P dan Q.
When x = 1,
Given 2x + y = 4 y = 4 − 2(1) = 2
or
y = 4 − 2x ——— 1 Hence,
and
x2 + y2 = 5 ——— 2  P = (1, 2) and Q = 11 , – 52
5

51

4 The diagram below shows a triangle MNT. It is 5 The diagram below shows a triangle EFG with a
given that MN + NT = 32 cm. rectangle PQRS inside it. It is given that
EF = EG = 17 cm, RS = y cm and FG = 16 cm. M is
Rajah di bawah menunjukkan sebuah segi tiga MNT. Diberi the midpoint of FG.
bahawa MN + NT = 32 cm.
Rajah di bawah menunjukkan sebuah segi tiga EFG dengan
M segi empat tepat PQRS di dalamnya. Diberi bahawa EF = EG
= 17 cm, RS = y cm dan FG = 16 cm. M ialah titik tengah
y cm P bagi FG.
5 cm
E

N 9 cm Q x cm T

Find the values of x and y. QR
Cari nilai bagi x dan y.

Given MN + NT = 32 F G
y + (9 + x) = 32 P MS
x + y = 23 ——— 1

MN PQ 2x cm
NT QT
= If the area of rectangle PQRS is 45 cm2, find the

y = 5 values of x and y.
9+x x
Jika luas segi empat tepat PQRS ialah 45 cm2, cari nilai bagi

xy = 5(9 + x) x dan y.

xy = 45 + 5x FM = MG = 16 = 8 cm Substitute y into 2 ,
xy − 5x = 45 ——— 2 2
( )2x
From 1 , y = 23 − x EM = 172 – 82 = 15 cm 120 − 15x = 45
8
( )x
Substitute into 2 , EM = RS 120 − 15x = 45
x(23 − x) − 5x = 45 MG SG 4
 23x − x2 − 5x = 45
x2 − 18x + 45 = 0 15 = y x(120 − 15x) = 180
(x − 15)(x − 3) = 0 8 8−x
x − 15 = 0 or x − 3 = 0 120x − 15x2 = 180
x = 15 x = 3
When x = 15, 15(8 − x) = 8y 15x2 − 120x + 180 = 0
y = 23 − (15) = 8
When x = 5, 120 − 15x = 8y x2 − 8x + 12 = 0
y = 23 − (5) = 18 15x + 8y = 120 ——— 1
∴ x = 15, y = 8 or x = 5, y = 18 (x − 2)(x − 6) = 0

Area of rectangle PQRS: x − 2 = 0 or x − 6 = 0

(2x)(y) = 45 When x = 2,
2xy = 45 ——— 2
y = 120 − 15(2) = 11.25
8
From 1 ,
When x = 6,

y= 120 − 15x y = 120 − 15(6) = 3.75
8 8

6 A lorry travels at an average speed of u km h–1 30v + 20u = 0.9uv (× 10)
for the first 90 km and v km h–1 for the next 60 km 300v + 200u = 9uv
in a journey. The total time taken for the journey 200u + 300v = 9uv (shown)
is 2.7 hours. (b) u − v = 10
Solve
Sebuah lori bergerak pada laju purata u km j–1 bagi 90 km 200u + 300v = 9uv ——— 1
pertama dan v km j–1 bagi 60 km berikutnya dalam satu u − v = 10
perjalanan. Jumlah masa perjalanan yang diambil ialah 2.7 jam. u = 10 + v ——— 2

(a) Show that 200u + 300v = 9uv. Substitute 2 into 1 ,
Tunjukkan bahawa 200u + 300v = 9uv.
200(10 + v) + 300v = 9v(10 + v)
(b) It is given that the average speed of the first
part of the journey is more than the average 2 000 + 200v + 300v = 90v + 9v2
speed of the second part of the journey by
10 km h–1, find the values of u and v. 9v2 − 410v − 2 000 = 0
Diberi bahawa purata laju bagi bahagian pertama
perjalanan melebihi bahagian kedua sebanyak (9v + 40)(v − 50) = 0
10 km j–1, cari nilai bagi u dan v.
9v + 40 = 0 or v − 50 = 0

v = – 40 v = 50
9

(a) Time taken = 2.7 hours Since v > 0, then v = 50

90 + 60 = 2.7 (÷ 3) When v = 50,
u v
u = 10 + v

30 + 20 = 0.9 = 10 + 50
u v
= 60

52

Review 3

Paper 2 Questions

 1 Solve the following simultaneous equations:

Selesaikan persamaan serentak berikut:

5x + 10y + 15z = 32

10x + 15y + 20z = 46

20x + 35y + 30z = 82 [5 marks/markah]

 2 Halim, Daud and Firdaus bought some items for the preparation of Hari Raya. Halim bought 3 packs

of cooking chocolate, 2 packs of dates and a box of candy for RM56.00. Daud bought 4 packs of cooking

chocolate, 3 packs of dates and a box of candy for RM77.00. Firdaus bought 6 packs of cooking chocolate, a

pack of date and 4 boxes of candy for RM83.00. Find the unit price of each type of the item purchased.

Halim, Daud dan Firdaus membeli beberapa barang untuk persiapan Hari Raya. Halim membeli 3 pek coklat masakan, 2 pek kurma

dan sekotak gula-gula dengan harga RM56.00. Daud membeli 4 pek coklat masakan, 3 pek kurma dan sekotak gula-gula dengan

harga RM77.00. Firdaus membeli 6 pek coklat masakan, 1 pek kurma dan 4 kotak gula-gula dengan harga RM83.00. Cari harga

seunit bagi setiap jenis barang yang dibeli. [5 marks/markah]

 3 Yati, Siti and Haili work in a shoes store. Their total sales in the last month was RM1 850. Yati’s sales was

RM150 more than Siti’s sales. The total sales of Siti and Haili was RM950 more than Yati’s sales. Calculate

the value of sales of each saleswoman, in RM.

Yati, Siti dan Haili bekerja di sebuah kedai kasut. Jumlah jualan mereka pada bulan lepas ialah RM1 850. Jualan Yati ialah RM150

lebih daripada jualan Siti. Jumlah jualan Siti dan Haili ialah RM950 lebih daripada jualan Yati. Hitung nilai jualan setiap jurujual,

dalam RM. [5 marks/markah]

 4 Solve the simultaneous equations y − 2x + 4 = 0 and 2x2 + 3y2 − 2xy = 16.

SPM Give your answer correct to three decimal places. 2xy = 16.
Selesaikan persamaan serentak y − 2x + 4 = 0 dan 2x2 + 3y2 −
CLONE
`11
P2Q1 Beri jawapan betul kepada tiga tempat perpuluhan.
[5 marks/markah]

  5 Solve the following simultaneous equations:

SPM Selesaikan persamaan serentak berikut: 3x + y = 1, 4x2 + y2 + 3xy − 7 = 0
CL `O12NE
[5 marks/markah]
P2Q1

  6 Solve the following simultaneous equations:

SPM Selesaikan persamaan serentak berikut:
CLONE
− 2y = 1, x2 + 3xy + 8y2 = 9
x`17
P2Q1
[5 marks/markah]

  7 Solve the following simultaneous equations:

Selesaikan persamaan serentak berikut:

y – 2x + 1 = 0, x2 − 2y2 − y + 4 = 0

Give your answers correct to three decimal places. [5 marks/markah]
Beri jawapan anda betul kepada tiga tempat perpuluhan.

H OTS Zone

  1 Amir planted vegetables on a piece of land. The land is in right-angled triangular shape. Given the longest

SPM side of the land is y metre. The other two sides of the land are 2x metre and (2x + 6) metre respectively. He
CLONE
`16 fenced the land with 72 metres of barbed wire. Find the length, in metre, of each side of the land. HO TS Applying

P2Q3 Amir menanam sayur-sayuran di atas sebidang tanah. Tanah itu berbentuk segi tiga bersudut tegak. Diberi sisi yang paling

panjang tanah ini ialah y meter. Dua sisi yang lain tanah itu ialah masing-masing 2x meter dan (2x + 6) meter. Dia memagarkan

tanah itu dengan dawai berduri sepanjang 72 meter. Cari panjang, dalam meter, bagi setiap sisi tanah.

53

Chapter Learning Area: Algebra

4 Indices, Surds and Logarithms

Indeks, Surd dan Logaritma

4.1 Laws of Indices / Hukum Indeks

Smart Tip

The relationship between index equation and logarithm equation:

Hubungan antara persamaan indeks dan persamaan logaritma:
ax = N ⇔ loga N = x, a > 0, a ≠1

Exercise 1 Express each of the following in the form of man, where a is a prime number.
Ungkapkan setiap yang berikut dalam bentuk man, dengan keadaan a ialah nombor perdana.

TP 1 Mempamerkan pengetahuan asas tentang indeks, surd dan logaritma.

Example 1 1 2(3n) + 4(3n) 2 2n + 2n + 2

2n + 3 − 2n + 1 + 2n = 2(3n) + 4(3n) = 1(2n) + (2n)(22)
= 6(3n) = 1(2n) + 4(2n)
Solution = 5(2n)
2n + 3 – 2n + 1 + 2n 3 3n + 3 – 3n + 2 – 3n + 1
= (2n)(23) − (2n)(21) + (2n) 4 5n + 1 – 5n – 5n – 1
= 8(2n) − 2(2n) + 1(2n)
= 7(2n)

Alternative Method = (3n)(33) − (3n)(32) − (3n)(31) = (5n)(51) − 1(5n) − (5n)(5–1)
= 27(3n) − 9(3n) − 3(3n)
8(2n) – 2(2n) + (2n) = 15(3n) = 5(5n) − 1(5n) − 1  (5n)
5
Let u = 2n, then
= 3 54 (5n) = 159 (5n)
Biarkan u = 2n, maka

8u – 2u + u = 7u

= 7(2n)

Exercise 2 Solve each of the following.
Selesaikan setiap yang berikut.

TP 3 Mengaplikasikan kefahaman tentang indeks, surd dan logaritma untuk melaksanakan tugasan mudah.

Example 2 1 Prove that 3n + 1 + 3n + 2 is a 2 Show that 2n + 2 + 2n + 4 is a
multiple of 4. multiple of 5.
Prove that 3n + 1 + 3n + 3 − 2(3n) Buktikan bahawa 3n + 1 + 3n + 2 ialah Tunjukkan bahawa 2n + 2 + 2n + 4 ialah
is a multiple of 7.
Buktikan bahawa 3n + 1 + 3n + 3 − 2(3n) gandaan bagi 4. gandaan bagi 5.
ialah gandaan bagi 7.
3n + 1 + 3n + 2 2n + 2 + 2n + 4
Solution = 3n(31) + 3n(32) = (2n)(22) + (2n)(24)
3n + 1 + 3n + 3 − 2(3n) = 3(3n) + 9(3n) = 4(2n) + 16(2n)
= 3n(3) + 3n(33) − 2(3n) = 12(3n) = 20(2n)
= 3(3n) + 27(3n) − 2(3n) 12 4 4 = 3 20 4 5 = 4
= 28(3n) 12 is divisible by 4. 20 is divisible by 5.
28 4 7 = 4
28 is divisible by 7. ∴ 3n + 1 + 3n + 2 is a multiple of 4. ∴ 2n + 2 + 2n + 4 is a multiple of 5.
28 boleh dibahagi tepat dengan 7.

∴ 3n + 1 + 3n + 3 − 2(3n) is a
multiple of 7.

∴ 3n + 1 + 3n + 3 – 2(3n) ialah gandaan
bagi 7.

54

3 Show that 6(3n) − 3n + 3n + 2 is 4 Prove that 5n + 2 – 2(5n) + 5n + 1 is 5 Show that 7n + 2 − 2(7n + 1) + 7n is
divisible by 7. divisible by 4. divisible by 9.

Tunjukkan bahawa 6(3n) − 3n + 3n + 2 Buktikan bahawa 5n + 2 – 2(5n) + 5n + 1 Tunjukkan bahawa 7n + 2 – 2(7n + 1) + 7n

boleh dibahagi tepat dengan 7. boleh dibahagi tepat dengan 4. boleh dibahagi tepat dengan 9.

6(3n) − 3n + 3n + 2 5n + 2 − 2(5n) + 5n + 1 7n + 2 − 2(7n + 1) + 7n
= 6(3n) − 1(3n) + 3n(32) = 5n(52) − 2(5n) + 5n(51) = 7n(72) − 2[7n(71)] + 1(7n)
= 6(3n) − 1(3n) + 9(3n) = 25(5n) − 2(5n) + 5(5n) = 49(7n) − 14(7n) + 1(7n)
= 14(3n) = 28(5n) = 36(7n)
14 4 7 = 2 28 4 4 = 7 36 4 9 = 4
∴ 6(3n) − 3n + 3n + 2 is divisible ∴ 5n + 2 − 2(5n) + 5n +1 is divisible ∴ 7n + 2 − 2(7n + 1) + 7n is divisible
by 7. by 4.
by 9.

6 Determine the number of digits of the answer of 2321 × 5324. HOTS Applying
Tentukan bilangan digit bagi jawapan 2321 × 5324.

Multiplication Number of digits
21 × 54 = 2 × 625 = 1 250 4
22 × 55 = 4 × 3 125 = 12 500 5
23 × 56 = 8 × 15 625 = 125 000 6
................. ...
................. ...
2321 × 5324 324

∴ The number of digits of the answer of 2321 × 5324 is 324.

Exercise 3 Solve each of the following.
Selesaikan setiap yang berikut.

TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang indeks, surd dan logaritma dalam konteks penyelesaian masalah rutin yang mudah.

Example 3 1 In an experiment, it is found that the temperature
of a metal increases from 23°C to H°C based
Mr Syaiful deposited RM24 000 in a bank in on the equation H = 23(1.15)k when the metal is
the early 2019 with an interest rate of 8% per heated for k seconds. Calculate the temperature,
annum. After w years, the total amount of in °C, of the metal at the 8th second.
savings of Mr Syaiful is 24 000(1.08)w. Calculate
the total amount of his savings in the early 2025. Dalam satu eksperimen, didapati bahawa suhu sebuah
Encik Syaiful menyimpan sebanyak RM24 000 di sebuah logam meningkat daripada 23°C kepada H°C mengikut
bank pada awal tahun 2019 dengan kadar faedah 8% persamaan H = 23(1.15)k apabila logam itu dipanaskan
setahun. Selepas w tahun, jumlah simpanan Encik Syaiful selama k saat. Hitung suhu, dalam °C, logam itu pada saat
ialah 24 000(1.08)w. Hitung jumlah simpanannya pada kelapan.
awal tahun 2025.

Solution k=8
Number of years from 2019 until 2025 = 6 years H = 23(1.15)8 °C
Bilangan tahun dari 2019 hingga 2025 = 6 tahun = 70.36°C
w=6
Total amount of savings in the early 2025
Jumlah simpanan pada awal tahun 2025
= 24 000(1.08)6 = RM38 084.98

55

2 Mrs Saras deposited RM15 000 in a bank on 3 Determine whether the expression 2n + 3 + 2n + 4 +
1st December 2018 with an interest rate of 4% 2n + 5 is divisible by 7 for all positive integers n.
per annum. After n years, the total amount of
her savings is 15 000(1.04)n. Calculate the total Tentukan sama ada ungkapan 2n + 3 + 2n + 4 + 2n + 5 boleh
amount of her savings on 1st December 2023. dibahagi dengan 7 bagi semua integer positif n.

Puan Saras menyimpan sebanyak RM15 000 pada
1 Disember 2018 di sebuah bank dengan kadar faedah 2n + 3 + 2n + 4 + 2n + 5 = 2n(23) + 2n(24) + 2n(25)
4% setahun. Selepas n tahun, jumlah simpanannya ialah = 2n(8) + 2n(16) + 2n(32)
15 000(1.04)n. Hitung jumlah simpanannya pada = 2n(8 + 16 + 32)
1 Disember 2023. = 56(2n)
56 is multiple of 7. Hence, 56(2n) is divisible by 7
for all positive integers n.
Number of years from 2018 until 2023 = 5 years
Total amount of savings in 1st December 2023
= RM15 000(1.04)5
= RM18 249.79

4 A monkey has a disease. It has 91.5 body cells affected on day 1. The number of body cells affected doubles
every following day. The disease becomes severe when 324 body cells are affected. On which day does the
disease become severe?

Seekor monyet mengidap sejenis penyakit. Monyet itu mempunyai 91.5 sel-sel badan yang terjejas pada hari pertama. Bilangan sel
badan yang terjejas menjadi dua kali ganda pada setiap hari berikutnya. Penyakit itu menjadi teruk apabila 324 sel badan terjejas.

Pada hari keberapakah penyakit tu menjadi teruk?

Day Number of body cells affected Number of body cells affected
(Base 9) (Base 3)
1 91.5 33
2
3 93 36
4
96 312

912 324

The disease becomes severe on the 4th day.

4.2 Laws of Surds / Hukum Surd

Exercise 4 Solve each of the following.
Selesaikan setiap yang berikut.

TP 3 Mengaplikasikan kefahaman tentang indeks, surd dan logaritma untuk melaksanakan tugasan mudah.

Example 4 1 Determine the value of (AB2B ÷ AB5B)2. Hence, make
a generalisation for the value of AB2B ÷ AB5B in the
Determine the value of (AB2B × AB3B)2. Hence, make
a generalisation for the value of AB2B × AB3B in the form of sAnuBi2lrBadi÷.(AAB2B5BB÷daAlaB5mB)2b.eSnettuekrussunryda. , buat generalisasi
bfToaegrnimtunkiolaafni sAnuBi2lrBadi×.(AAB2B3BB×daAlaB3mB)2b.eSnettuekrussunryda. , buat generalisasi Tentukan
bagi nilai
1 2
(AB2B ÷ AB5B)2 = AB2B 2
AB5B
Solution 1 21 2 AB2B AB2B
(AB2B × AB3B)2 = (AB2B × AB3B)(AB2B × AB3B) AB5B AB5B
= AB2B × AB2B × AB3B × AB3B =

= (AB2B)2 × (AB3B)2 Smart Tip = (AB2B)2
(AB5B)2
=2×3

=6 A1 ABaB × ABbB = ABaBbBBaB = 2
2 ABaB ÷ ABbB = 5
(AB2B × AB3B)2 = 6 b

Multiply both sides with square root. (AB2B ÷ AB5B)2 = 2
Darabkan kedua-dua belah dengan punca kuasa dua. 5
Multiply both sides with square root.
Then/Maka, AB2B × AB3B = AB2B×BB3B
= AB6B Then, AB2B ÷ AB5B = A 2
5

56

2 Determine the value of (AB3B × AB5B)2. Hence, make 3 Determine the value of (AB2B × AB5B)2. Hence, make
a generalisation for the value of AB3B × AB5B in the a generalisation for the value of AB2B × AB5B in the

form of sAnuBi3lrBadi×.(AAB3B5BB×daAlaB5mB)2b.eSnettuekrussunryda. , buat generalisasi fTboaegrnimtunkiolaafni sAnuBi2lrBadi×.(AAB2B5BB×daAlaB5mB)2b.eSnettuekrussunryda. , buat generalisasi
Tentukan
bagi nilai (AB2B × AB5B)2 = (AB2B × AB5B)(AB2B × AB5B)
= AB2B × AB2B × AB5B × AB5B

= (AB2B)2 × (AB5B)2
(AB3B × AB5B)2 = (AB3B × AB5B)(AB3B × AB5B)
= AB3B × AB3B × AB5B × AB5B =2×5
= 10
= (AB3B)2 × (AB5B)2
(AB2B × AB5B)2 = 10
=3×5

= 15

(AB3B × AB5B)2 = 15

Multiply both sides with square root. Multiply both sides with square root.

Then, AB3B × AB5B = AB3B×BB5B Then, AB2B × AB5B = AB2B×BB5B
= AB1B5 = AB1B0

4 Determine the value of (ABmB × ABnB)2. Hence, make 5 Determine the value of (AB3B ÷ AB2B)2. Hence, make
a generalisation for the value of ABmB × ABnB in the a generalisation for the value of AB3B ÷ AB2B in the

form of sAnuBimlrBadi ×.(AABmBBnB×daAlaBnmB)2b.eSnettuekrussunryda. , buat generalisasi bfToaegrnimtunkiolaafni sAnuBi3lrBadi÷.(AAB3B2BB÷daAlaB2mB)2b.eSnettuekrussunryda. , buat generalisasi
Tentukan
bagi nilai

1 2(AB3B ÷ AB2B)2 = AB3B 2
AB2B
(ABmB × ABnB)2 = (ABmB × ABnB)(ABmB × ABnB) 1 21 2 AB3B AB3B
= ABmB × ABnB × ABmB × ABnB = AB2B AB2B

= (ABmB)2 × (ABnB)2

= mn = (AB3B)2 = 3
(AB2B)2 2
(ABmB × ABnB)2 = mn

Multiply both sides with square root. (AB3B ÷ AB2B)2 = 3
2
Then, ABmB × ABnB = ABmBnBB

Multiply both sides with square root.

Then, AB3B ÷ AB2B = A 3
2

Exercise 5 Simplify each of the following in the form aABBb .
Permudahkan setiap yang berikut dalam bentuk aABBb .

TP 2 Mempamerkan kefahaman tentang indeks, surd dan logaritma.

Example 5 1 AB5B2 2 AB1B4B7B
(a) AB4B5 = AB4B×BB1B3B = AB4B9B×BB3B
(b) 2AB5B0 – 3AB2B = AB4B × AB1B3 = AB4B9 × AB3B
= 2 × AB1B3 = 7 × AB3B
Solution = 2AB1B3 = 7AB3B

(a) AB4B5 = AB9B×BB5B 3 AB4B8 4 AB7B2
= AB9B × AB5B
2 3
= 3 × AB5B
= 3AB5B = AB1B6B×BB3B = AB9B×BB4BB×B2B
(b) 2AB5B0 – 3AB2B
2 3
= 2AB2B5B×BB2B – 3AB2B
= 2(AB2B5 × AB2B) – 3AB2B = AB1B6 × AB3B = AB9B × AB4B × AB2B
= 2(5 × AB2B) – 3AB2B
= 10AB2B – 3AB2B 2 3
= 7AB2B
= 4 × AB3B = 3 × 2 × AB2B

2 3

= 2 × AB3B = 2 × AB2B
= 2AB3B = 2AB2B

57

5 3AB2B + 2AB8B 6 4AB2B7 + 2AB3B 7 9AB2B – 2AB1B8
= 3AB2B + 2AB4B×BB2B = 4AB9B×BB3B + 2AB3B = 9AB2B – 2AB9B×BB2B
= 3AB2B + 2(AB4B × AB2B) = 4AB9B × AB3B + 2AB3B = 9AB2B – 2(AB9B × AB2B)
= 3AB2B + 2(2 × AB2B) = 4(3) × AB3B + 2AB3B = 9AB2B – 2(3 × AB2B)
= 3AB2B + 4 × AB2B) = 12AB3B + 2AB3B = 9AB2B – 6AB2B
= 3AB2B + 4AB2B) = 14AB3B = 3AB2B
= 7AB2B

Exercise 6 Simplify each of the following.
Permudahkan setiap yang berikut.

TP 3 Mengaplikasikan kefahaman tentang indeks, surd dan logaritma untuk melaksanakan tugasan mudah.

Example 6

(a) 10 (b) AB5B 3 AB2B Smart Tip
–
AB5B Multiply the numerator
and denominators
Solution AB5B AB5B AB2B by conjugate surd to
AB5B AB5B AB2B eliminate surds in the
(a) A1B05B = 10 × (b) AB5B 3 AB2B = AB5B 3 AB2B × + denominator.
– – + Darabkan pengangka dan
AB5B penyebut dengan surd konjugat
= 10AB5B 3(AB5B + AB2B) untuk menghapuskan surd
= 2AB55B = (AB5B)2 – (AB2B)2 daripada penyebut.

= 3(AB5B + AB2B)
5–2
= 3(AB5B + AB2B) = AB5B + AB2B
3

1 12 2 21 3 AB7B 4 AB3B
–
AB3B AB7B

= 12 × AB3B = 21 × AB7B = AB7B 4 AB3B × AB7B + AB3B
AB3B AB7B – AB7B + AB3B
AB3B AB7B
= 12AB3B = 21AB7B 4(AB7B + AB3B) 4(AB7B + AB3B)
3 7 = (AB7B)2 – (AB3B)2 =
7–3
= 4AB3B = 3AB7B
4(AB7B + AB3B) AB7B AB3B
= = +
4

4 4 5 21 6 2AB2B0
3 – AB5B
AB7B + AB3B 3 – AB2B

= AB7B 4 AB3B × AB7B – AB3B = 21 × 3 + AB2B = 2AB4B×BB5B × 3 + AB5B
+ AB7B – AB3B 3 + AB2B 3 – AB5B 3 + AB5B
3 – AB2B
4(AB7B – AB3B) 21(3 + AB2B) AB4B × AB5B)(3 AB5B)
= (AB7B)2 – (AB3B)2 = 32 – (AB2B)2 = (2 × 32 – (AB5B)2 +

= 4(AB7B – AB3B) = 21(3 + AB2B) = (2 × 2 × AB5B)(3 + AB5B)
7–3 9–2 9–5
= 4(AB7B – AB3B) = 21(3 + AB2B) = 4AB5B(3 + AB5B)
= AB7B – 4AB3B = 3(3 + 7AB2B) = 9 + 3AB2B
= AB5B(3 4 AB5B)
+
= 3AB5B + (AB5B)2
= 3AB5B + 5

58

Exercise 7 Solve each of the following problems.
Selesaikan setiap masalah yang berikut.

TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang indeks, surd dan logaritma dalam konteks penyelesaian masalah rutin yang mudah.

Example 7 1 The diagram below shows a rectangle.
Rajah di bawah menunjukkan sebuah segi empat tepat.
The diagram below shows a trapezium ABCD.
Rajah di bawah menunjukkan sebuah trapezium ABCD. (3 + AB5B) cm

A (AB4B5 + 4 ) cm B (3 – AB5B) cm

D AB2B0 cm C Find the area of the rectangle as a surd in the
simplest form.
Find the area of the trapezium as a surd in the
simplest form. Cari luas segi empat tepat dalam surd bentuk termudah.
Cari luas trapezium dalam surd bentuk termudah.

Solution
(3 + AB5B)(3 – AB5B) = 32 – (AB5B)2
Area/Luas = 1 (AB + DC)(AD)
2 =9–5
= 4 cm2
Hence, the area of the rectangle = 4 cm2

= 1 [(AB4B5 + 4) + (AB2B0 )](AB2B0 )
2
1
= 2 (AB2B0 )[(AB4B5 + 4) + (AB2B0 )]

= A21B(52B(A5B5ABB)5(B3+A4B5)B + 4 + 2AB5B)
=
= 5(5) + 4AB5B
= 25 + 4AB5B
Hence, the area of the trapezium = (25 + 4AB5B) cm2
Maka, luas trapezium = (25 + 4AB5B) cm2

2 The diagram below shows a right-angled triangle 3 The diagram below shows a trapezium EFGH.
ABC. Rajah di bawah menunjukkan sebuah trapezium EFGH.

Rajah di bawah menunjukkan sebuah segi tiga bersudut E AB1B2 cm F
tegak ABC.

A

AB8B cm H (4AB3B + 2) cm G

B AB1B0 cm C Determine the area of the trapezium in the form
of surd.
Find the length of AC in the form aABbB, where a Tentukan luas trapezium dalam bentuk surd.

and b are integers. Area = 1 (EF + HG)(EH)
2
Cari panjang AC dalam bentuk aABbB, dengan keadaan a dan 1
2 [AB1B2 (4AB3B +2)](AB1B2 )
b ialah integer.

= +

AC2 = AB2 + BC2 = 1 (AB1B2 )(AB1B2 + 4AB3B +2)
2
= (AB8B)2 + (AB1B0 )2 A12B(32B(A6B3ABB)3(B2+A2B3)B
= + 4AB3B +2)
= 8 + 10 =
= 18 = 6(3) + 2AB3B
= 18 + 2AB3B
AC = AB1B8 Hence, the area of the trapezium = (18 + 2AB3B) cm2
= AB(3B×BB3B×B2B)
= 3AB2B
Hence, AC = 3AB2B cm

59

4 The diagram shows a right-angled triangle ABC. AB4B8 cm A
Rajah menunjukkan sebuah segi tiga bersudut tegak ABC. C
60°
Find the perimeter of the triangle as a surd in the simplest form. B
Cari perimeter segi tiga dalam surd bentuk termudah.

AB = AB4B8 cos 60° = BC
= AB1B6 B×B3B
= AB1B6 × AB3B 4AB3B
= 4AB3B
BC = 4AB3Bcos 60°

AC = 4AB3B1 1 2
2
1 2s in60° =4AB3B = 2AB3B
AC = 4AB3B AB3B
Perimeter = AB + BC + AC
2
= 2(AB3B)2 = 4AB3B + 2AB3B + 6
= 6AB3B + 6
= 2(3) Hence, the perimeter of the triangle = (6AB3B + 6) cm

= 6

4.3 Laws of Logarithms / Hukum Logaritma

Smart Tip

Logarithms/Logaritma:
ax = N if and only if loga N = x, a > 0, a ≠ 1/ax = N jika dan hanya jika loga N = x, a > 0, a ≠ 1

Laws of Logarithms/Hukum Logaritma:

1 loga x + loga y = loga (xy) 4 aloga x = x

1 2 2 loga x − loga y = loga xy 5 IJifklaolgog1010aa==bb,, then a = antilog b
maka a = antilog b

3 loga xn = nloga x 6 loga b = logc b
logc a

Exercise 8 Express each of the following logarithms in the related equation of index. Hence, determine
the value of the logarithm.
Ungkapkan setiap logaritma berikut dalam persamaan indeks yang berkaitan. Seterusnya, tentukan nilai
logaritma tersebut.

TP 2 Mempamerkan kefahaman tentang indeks, surd dan logaritma.

Example 8 1 log2 16 2 log3 9

log2 8 24 = 16 32 = 9
log2 16 = 4 log3 9 = 2
Solution
23 = 8
log2 8 = 3

3 log3 81 4 log10 1 000 5 log25 5

34 = 81 103 = 1 000 1
log3 81 = 4 log10 1 000 = 3
25 2 = 5

log25 5 = 1
2

60

Exercise 9 Prove each of the following.
Buktikan setiap yang berikut.

TP 3 Mengaplikasikan kefahaman tentang indeks, surd dan logaritma untuk melaksanakan tugasan mudah.

Example 9 1 loga a = 1 2 loga 1 = 0

loga PQ = loga P + loga Q a1 = a a0 = 1
loga a = 1 loga 1 = 0
Solution
Let/Katakan
P = ax and/dan Q = ay
Then/Maka,
loga P = x and/dan loga Q = y
loga PQ = loga (ax)(ay)

= loga (ax + y)
=x+y

= loga P + loga Q

3 loga (ax) = x 4 loga M = loga M – loga N 5 Given loga M3 = loga (M × M × M),
N prove loga M3 = 3loga M.
ax = ax
loga (ax) = x Let M = ax and N = ay Diberi loga M3 = loga (M × M × M),
buktikan loga M3 = 3loga M.
Then, loga M = x and loga N = y loga M3 = loga (M × M × M)
ax = loga M + loga M + loga M
loga M = loga ay = 3loga M
N
= loga (ax – y)
=x–y

= loga M – loga N

Exercise 10 Write each of the following logarithmic expressions in the simplest form.
Tuliskan setiap ungkapan logaritma berikut dalam bentuk termudah.

TP 3 Mengaplikasikan kefahaman tentang indeks, surd dan logaritma untuk melaksanakan tugasan mudah.

Example 10 1 2 + log2 b 2 log4 p − log4 q + log4 3

3 − log2 p + log2 16 = 2(log2 2) + log2 b = log4 p × 3
= log2 22 + log2 b q
Solution = log2 22 (b)
= log2 4b 3p
3 − log2 p + log2 16 = log4 q
= 3(1) − log2 p + log2 16
= 3(log2 2) − log2 p + log2 16
= log2 23 − log2 p + log2 16

= log2 23 × 16
p

= log2 128
p

3 1 + log3 m − log3 12 4 logx h + 2 logx k − logx (k − 1) 5 logx ab − logx (ab – a)

= log3 3 + log3 m − log3 12 = logx h + logx k2 – logx (k − 1) = logx ab
ab −
= log3 3×m = logx h × k2 a
12 (k − 1)
= logx ab
m = logx hk2 a(b − 1)
= log3 4 k−1
b
= logx b − 1

61

Exercise 11 Evaluate each of the following.
Nilaikan setiap yang berikut.

TP 3 Mengaplikasikan kefahaman tentang indeks, surd dan logaritma untuk melaksanakan tugasan mudah.

Example 11 1 7log7 0.25 2 4log4 8–2
= 0.25
(a) 2log2 5 = 8–2
(b) log2 4 − log2 2 + log2 32
= 1
82
Solution
(a) 2log2 5 = 5 aloga x = x = 1
64

(b) log2 4 − log2 2 + log2 32 1
5
4 × 32 3 5 3 log5 3  log9 32
2
= log2 = 5log5 33 4 9
= 5log5 27
= log2 64 = 27 1
= log2 26
= 9log9 32 5
= 6 log2 2
1
= 6(1)
= 32 5

=6 = 5 32
=2

5 log3 9 + log3 27 – log3 81 6 log2 8 + log2 16 − log2 4 7 log2 12 + log2 18 − 3 log2 3

= log3 9 × 27 = log2 8 × 16 = log2 12 + log2 18 − log2 33
81 4
12 × 18
= log3 243 = log2 32 = log2 27
81
= log2 25 = log2 8
= log2 23
= log3 3 = 5 log2 2 = 3 log2 2
=1 = 3(1) = 3
= 5(1)

=5

Smart Tip

The logarithm of a number can be determined by changing the base of the logarithm to a suitable base.
Logaritma suatu nombor boleh ditentukan dengan menukar asas logaritma kepada asas yang sesuai.

(a) loga b = logc b 1
logc a            (b) loga b = logb a

Exercise 12 Convert the following to the required base.
Tukarkan setiap yang berikut kepada asas yang diberi.

TP 3 Mengaplikasikan kefahaman tentang indeks, surd dan logaritma untuk melaksanakan tugasan mudah.

Example 12 1 Convert logH 2 to the base 2. 2 Convert log9 (M + 6) to the
Tukarkan logH 2 kepada asas 2. base 3.
Convert log4 T to the base 2.
Tukarkan log4 T kepada asas 2. logH 2 = log2 2 Tukarkan log9 (M + 6) kepada asas 3.
log2 H
log9 (M + 6) = log3 (M + 6)
log3 9
Solution = 1
log2
log4 T = log2 T H = log3 (M + 6)
log24 log3 32

= log2 T = log3 (M + 6)
log2 22 2

= log2 T
2

62

3 Convert loga (8m) to the base m. 4 Convert log16 (2p) to the base 2. 5 Convert log2 (20k) to the base 10.
Tukarkan loga (8m) kepada asas m. Tukarkan log16 (2p) kepada asas 2. Tukarkan log2 (20k) kepada asas 10.

loga (8m) = logm (8m) log16 (2p) = log2 (2p) log2 (20k)
logm a log2 16
= log10 20k
logm 8 + logm m log2 2 + log2 p log10 2
= logma = log2 24
log10 (10 × 2 × k)
logm 8 + 1 log2 2 + log2 p = log10 2
logm a 4
= = log10 10 + log10 2 + log10 k
log10 2
1 + log2 p =
4
= 1 + log10 2 + log10 k
log10 2
=

Exercise 13 Evaluate each of the following. Give the answers correct to 4 decimal places.
Nilaikan setiap yang berikut. Beri jawapan betul kepada 4 tempat perpuluhan.

TP 3 Mengaplikasikan kefahaman tentang indeks, surd dan logaritma untuk melaksanakan tugasan mudah.

Example 13 1 log2 7 2 log7 2

log2 0.842 log10 7 log10 2
log10 2 log10 7
Solution = =

log2 0.842 logc b 0.8451 0.3010
logc a = 0.3010 = 0.8451
= log10 0.842 loga b =
log10 2
= 2.8076 = 0.3562

= –0.07469 = –0.2481
0.3010

3 log3 5.4 4 log5 24.5 5 log3 0.81

= log10 5.4 = log10 24.5 = log10 0.81
log10 3 log10 5 log10 3

0.7324 1.3892 –0.0915
= 0.4771 = 0.6990 = 0.4771

= 1.5351 = 1.9874 = –0.1918

Exercise 14 Solve the following equations
Selesaikan persamaan berikut.

TP 3 Mengaplikasikan kefahaman tentang indeks, surd dan logaritma untuk melaksanakan tugasan mudah.

Example 14 (b) log3 (4x + 1) = 2 + log3 (x − 1) 1 log3 (2x − 5) = 2

(a) log2 (3x − 4) = 3 Solution 2x − 5 = 32
Solution log3 (4x + 1) = 2 + log3 (x − 1) 2 x − 5 = 9
log3 (4x + 1) – log3 (x − 1) = 2 2x = 14
3x − 4 = 23 x = 7
3x – 4 = 8 ( )4x + 1
3x = 12 or
x=4 log3 x − 1 = 2
log3(2x − 5) = log3 9
Common Error 4x + 1 = 32 2x − 5 = 9
log2 (3x – 4) = 3 x−1 2x = 14
log2 3x – log2 4 = 3 4x + 1 = 9(x − 1) x = 7
log2 3x – 2 = 3 4x + 1 = 9x − 9
log2 3x = 5 1 + 9 = 9x − 4x
10 = 5x
x = 2

63

2 log2 (7x + 2) = 4 3 log2 3x − log2 (x − 3) = 2 4 log4 (11x + 3) − log4 (x − 7) = 2

7x + 2 = 24 ( )3x ( )11x + 3
7x + 2 = 16
7x = 14 log2 x − 3 = 2 log4 x − 7 = 2
3x 11x + 3
14
x = 7 x − 3 = 22 x − 7 = 42
=2 3x = 4(x − 3) 11x + 3 = 16(x − 7)
or 3x = 4x − 12 11x + 3 = 16x − 112
log2(7x + 2) = log2 16 12 = 4x − 3x 5x = 115
7x + 2 = 16 12 = x x = 23
7x = 14
x = 2

5 log3 (x + 7) = 2 + log3 (x − 1) 6 log2 (x − 2) = 1 − log2 (x – 3) 7 log2 (√ x − 1) = log2 x – 2
log3 (x + 7) − log3 (x − 1) = 2 log2 (x − 2) + log2 (x − 3) = 1
log2 (x − 2)(x − 3) = 1 log2 x – log2 (√ x − 1) = 2
( )x + 7 (x − 2)(x − 3) = 21 ( )log2
x2 − 5x + 6 = 2 x =2
log3 x − 1 = 2 x2 − 5x + 4 = 0 √ x − 1
(x − 1)(x − 4) = 0 x = 22
x+7 x − 1 = 0 or x − 4 = 0 √ x − 1
x − 1 = 32 x = 1 x = 4 x = 4(√ x − 1)
x + 7 = 9(x − 1)
x + 7 = 9x − 9 If x = 1 is substituted into x = 4√ x − 4
9 + 7 = 9x − x log2 (x – 2), the logarithm is
undefined (due to negative x + 4 = √ x
16 = 8x number). 4
2 = x ∴x=4 ( )x + 4 2
4
=x

( )x2 + 8x + 16 =x
16
x2 + 8x + 16 = 16x

x2 − 8x + 16 = 0

(x − 4)(x − 4) = 0

(x − 4) = 0

∴x=4

5.4 4.4 Applications of Indices, Surds and Logarithms / Aplikasi Indeks, Surd dan Logaritma

Exercise 15 Solve the following problems.
Selesaikan masalah berikut.

TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang indeks, surd dan logaritma dalam konteks penyelesaian masalah rutin yang mudah.

Example 15 1 Kamil bought a plot of land for RM80 000. After

Bala bought a car for RM120 000. After the car is the land is purchased, the value appreciates
purchased, the value depreciates 4% every year.
Find the minimum number of years when the 3.5% every year. Find the minimum number of
value of the car is less than RM40 000.
Bala membeli sebuah kereta dengan harga RM120 000. years when the value of the land is more than
Selepas kereta itu dibeli, nilainya susut sebanyak 4% setiap
tahun. Cari bilangan tahun minimum apabila nilai kereta RM150 000.
itu kurang daripada RM40 000.
Kamil membeli sebidang tanah dengan harga RM80 000.
Solution
Let n = number of years after the car is purchased Selepas tanah itu dibeli, nilainya meningkat sebanyak 3.5%
Katakan n = bilangan tahun selepas kereta dibeli
The value of the car depreciates 4% every year, setiap tahun. Cari bilangan tahun minimum apabila nilai
so the percentage left is 96%.
Nilai harga kereta susut 4% setiap tahun, maka peratus tanah itu melebihi RM150 000.
yang tinggal ialah 96%.
RM120 000 × (0.96)n < RM40 000

(0.96)n < 0.3333 Let n = number of years after the land is
log10 (0.96)n < log10 0.3333
purchased
nlog10 (0.96) < log10 0.3333
n(–0.01773) < –0.4772 The value of the land appreciates 3.5% every

0.01773n > 0.4772 year, so the percentage of the value of land
n > 26.91
n = 27 becomes 103.5%.
1 2 RM80 000 ×
103.5 n
100
> RM150 000

RM80 000 × (1.035)n > RM150 000

(1.035)n > 1.875

log10 (1.035)n > log10 1.875
nlog10 (1.035) > log10 1.875
n(0.01494) > 0.2730

n > 18.273

n = 19

64

2 Johan bought a motorcycle for RM6 000. 3 In the diagram below, ABCD and EFGH are
After the motorcycle is purchased, the value
depreciates 8% every year. Find the minimum rectangles. Given EF = AB3B cm and EH = AB2B cm.
number of years when the value of the
motorcycle is less than RM450. Dalam rajah dEiFb=awAaBh3B, AcmBCdDandEanHE=FAGBH2B ialah segi empat
tepat. Diberi cm.
Johan membeli sebuah motosikal dengan harga RM6 000.
Selepas motosikal itu dibeli, nilainya susut 8% setiap tahun. A AB2B7 cm B
Cari bilangan tahun minimum apabila nilai motosikal itu
kurang daripada RM450. EF

AB8B cm
Let n = number of years after the motorcycle is
purchased HG
The value of the motorcycle depreciates 8% every
year, so the percentage left is 92%. DC
RM6 000 × (0.92)n < RM450
(0.92)n < 0.075 Given the area of the shaded region is kAB6B cm2,
log10 (0.92)n < log10 0.075
nlog10 (0.92) < log10 0.075 find the value of k. berlorek ialah kAB6B cm2, cari
n(–0.03621) < –1.1249 Diberi luas bagi kawasan
n(0.03621) > 1.1249
n > 31.07 nilai k.
n = 32
Area of the shaded region

= (AB2B7 × AB8B) – (AB3B × AB2B)
= (3AB3B × 2AB2B) – (AB3B × AB2B)
= 6AB6B – AB6B
= 5AB6B

∴k=5

Review 4

Paper 1 Questions

 1 Given 2a = 7b = 14m, state m in terms of a and b.  6 Given 8(2x − 2) = 16x, find the value of x.
Diberi 2a = 7b = 14m, nyatakan m dalam sebutan a dan b. Diberi 8(2x − 2) = 16x, cari nilai bagi x.

[3 marks/markah] [3 marks/markah]

 2 Solve 32x − 4(3x + 1) + 27 = 0.  7 Solve the equation/Selesaikan persamaan:
Selesaikan 32x − 4(3x + 1) + 27 = 0.
SPM logx 128 − log x 2x = 3
[3 marks/markah] CLONE [3 marks/markah]

`17
P1Q6

 3 Determine the value of (AB7B ÷ AB5B)2. Hence, make  8 Given log2 3 = m and log2 5 = n. Express log8 90 in
a generalisation for the value of AB7B ÷ AB5B in the terms of m and n.
SPM
bTfoaegrnimtunkiolaafni sAnuBi7lrBadi÷.(AAB7B5BB÷daAlaB5mB)2b.eSnettuekrussunryda. , buat generalisasi CLONE Diberi log2 3 = m dan log2 5 = n. Ungkapkan log8 90 dalam
`10 sebutan m dan n.
[3 marks/markah] P1Q8

[3 marks/markah]

 4 Determine the value of (ABmB ÷ ABnB)2. Hence, make  9 Given 2k − 4 × 8k = 64, find the value of k.
a generalisation for the value of ABmB ÷ ABnB in the Diberi 2k − 4 × 8k = 64, cari nilai bagi k.

fTboaegrnimtunkiolaafni sAnuBimlrBadi ÷.(AABmBBnB÷daAlaBnmB)2b.eSnettuekrussunryda. , buat generalisasi [3 marks/markah]

[3 marks/markah] 10 Solve the equation/Selesaikan persamaan:

 5 Given loga 4 = p and loga 3 = q, express 25x = 8 + 25x − 1
[3 marks/markah]

loga 27a in terms of p and q. 11 Given logα 5 = m and logα 3 = p, express log3 625α3
16 = p dan loga 3 = q, ungkapkan in terms of m and p.
27a SPM
Diberi loga 4 loga 16 dalam CLONE Diberi logα 5 = m dan logα 3 = p, ungkapkan log3 625α3
dalam sebutan m dan p.
sebutan p dan q. `15
P1Q6

[3 marks/markah] [3 marks/markah]

65

12 Given loga 5 = p , express in terms of p: 17 Solve the equation:
Selesaikan persamaan:
SPM Diberi loga 5 = p, ungkapkan dalam sebutan p:
CLONE 1 + log5 x = log5 (x + 7)
`16 [3 marks/markah]
P1Q14 (a) loga 25
(b) log5 125a4
[4 marks/markah]
18 Solve log2 √ x − log4 3 = 3 .
2
13 Given 3n + 3n + 3n = 3x, express n in terms of x. 3
Selesaikan log2 √ x – log4 3 = 2 .
SPM Diberi 3n + 3n + 3n = 3x, ungkapkan n dalam sebutan x. [3 marks/markah]
CLONE
`18 [2 marks/markah]
P1Q16
1
14 Given 9x + 4 = 1, express y in terms of x. 19 Given 9u × 27u – 2 = 81 , find the value of u.
27y − 2 y dalam sebutan x.
SPM 1
CLONE 9x + 4 Diberi 9u × 27u – 2 = 81 , cari nilai bagi u.
`17 Diberi 27y − 2 = 1, ungkapkan
P1Q5 [3 marks/markah]

[3 marks/markah] 20 Given logc 2 = a and logc 5 = b, express log5 32c3 in
terms of a and b.
15 Given log5 x − log25 y = 1, express y in terms of x. Diberi logc 2 = a and logc 5 = b, ungkapkan log5 32c3 in
Diberi log5 x − log25 y = 1, ungkapkan y dalam sebutan x. terms of a and b.
[3 marks/markah]
[3 marks/markah]

16 Solve the equation: 21 Given 33x = g, 3y = h and 33x3y = 8 + 27x. Express g
Selesaikan persamaan: in terms of h.
Diberi 33x = g, 3y = h dan 33x3y = 8 + 27x. Ungkapkan g
16(23x − 5) = 1 dalam sebutan h.
2x
[3 marks/markah]
[3 marks/markah]

Paper 2 Questions

  1 It is given that u = 2x and w = 2y.

SPM Diberi bahawa u = 2x dan w = 2y.
CLONE
`14 (a) Express 8x + y in terms of u and w.

P2Q4 4y

Ungkapkan 8x + y dalam sebutan u dan w. [3 marks/markah]
4y [5 marks/markah]

(b) Express log2 16u7 in terms of x and y.
w3

Ungkapkan log2 16u7 dalam sebutan x dan y.
w3

H OTS Zone

 1 Hawa deposited RM30 000 in a bank. After n years, her savings will become 30 000(1.08)n. Find the number
of years when her savings will exceed RM75 000 for the first time. HOTS Applying

Hawa menyimpan RM30 000 di sebuah bank. Selepas n tahun, wang simpanannya akan menjadi 30 000(1.08)n. Cari bilangan

tahun apabila wang simpanannya akan melebihi RM75 000 buat kali pertama.

 2 (a) Given a2 + 9b2 = 7ab, show that lg(a − 3b) = 1 (lg a + lg b). HO TS Applying
2

Diberi a2 + 9b2 = 7ab, tunjukkan bahawa lg (a − 3b) = 1 (lg a + lg b).
2

(b) Determine the value of 1 + 1 . HO TS Evaluating
logmmn lognmn
1 1
Tentukan nilai bagi logmmn + lognmn .

66

Chapter Learning Area: Algebra

5 Progressions

Janjang

5.1 Arithmetic Progressions / Janjang Aritmetik

Smart Tip

An arithmetic progression is a number sequence such that the difference between each term (after the first term) and
its preceding term is a constant. The constant is called common difference, d.
Janjang aritmetik ialah suatu jujukan nombor dengan keadaan beza antara setiap sebutan (selepas sebutan pertama) dengan sebutan sebelumnya
ialah suatu pemalar. Pemalar ini disebut sebagai beza sepunya, d.

d = Tn + 1 – Tn , n = 1, 2, 3, 4, …

Exercise 1 Determine whether each of the following sequences is an arithmetic progression or not.
Give your reason.
Tentukan sama ada setiap jujukan berikut ialah janjang aritmetik atau bukan. Berikan alasan anda.

TP 1 Mempamerkan pengetahuan asas tentang janjang.

Example 1 1 17, 23, 29, 35, … 2 −7, –3, 1, 5, …

5, 9, 13, 17, … T2 – T1 = 23 – 17 = 6 T2 – T1 = –3 – (–7) = 4
T3 – T2 = 29 – 23 = 6 T3 – T2 = 1 – (–3) = 4
Solution T4 – T3 = 35 – 29 = 6 T4 – T3 = 5 – 1 = 4
T2 – T1 = 9 – 5 = 4
T3 – T2 = 13 – 9 = 4 17, 23, 29, 35, … is an –7, –3, 1, 5, … is an arithmetic
T4 – T3 = 17 – 13 = 4 arithmetic progression with progression with common
common difference = 6. difference = 4.
5, 9, 13, 17, … is an
arithmetic progression with
common difference = 4.
5, 9, 13, 17, … ialah janjang
aritmetik dengan beza sepunya = 4.

3 12, 8, 5, 2, …. 4 1 , 152, 1 , 172, … 5 – 1 , – 158, – 4 , – 1118, …
3 2 9 9
T2 – T1 = 8 – 12 = –4
T3 – T2 = 5 – 8 = –3 T2 – T1 = 5 – 1 = 1 1 2T2– T1 = – 5 – –1 =–1
T4 – T3 = 2 – 5 = –3 12 3 12 18 9 6
1 2T3
12, 8, 5, 2, … is not an T3 – T2 = 1 –5 =1 – T2 = – 4 – –5 =–1
arithmetic progression 2 12 12 9 18 6
because there is no common 1 2T4
difference. T4 – T3 = 7 – 1 = 1 – T3 = – 11 – –4 =–1
12 2 12 18 9 6

1, 5 , 1, 7 , … is an – 1, – 5 , – 4, – 11, … is an
3 12 2 12 9 18 9 18
arithmetic progression with
common difference = 1 . arithmetic progression with
common difference = – 1.
12
6

67

Smart Tip

The nth term, Tn, of an arithmetic progression is written as:
Sebutan ke-n, Tn, bagi suatu janjang aritmetik ditulis sebagai:

Tn = a + (n – 1)d
where/dengan keadaan
a = the first term/sebutan pertama
d = the common difference/beza sepunya
n = number of terms/bilangan sebutan

Exercise 2 Solve each of the following.
Selesaikan setiap yang berikut.

TP 2 Mempamerkan kefahaman tentang janjang aritmetik dan janjang geometri.

Example 2 1 Given the arithmetic progression
Diberi janjang aritmetik
Given the arithmetic progression
Diberi janjang aritmetik 4, 13, 22, 31, …
(a) Determine the nth term.
5, 11, 17, 23, … Tentukan sebutan ke-n.
(a) Determine the nth term. (b) Determine the 26th term.
Tentukan sebutan ke-n.
(b) Determine the 18th term. Tentukan sebutan ke-26.
Tentukan sebutan ke-18.

Solution (a) a = 4
(a) a = 5 d = 13 – 4 = 9
d = 11 – 5 = 6 Tn = a + (n – 1)d
Tn = a + (n – 1)d Tn = 4 + (n – 1)9
Tn = 5 + (n – 1)6 = 4 + 9n – 9
= 5 + 6n – 6 = 9n – 5
= 6n – 1 (b) T26 = 9(26) – 5
(b) T18 = 6(18) – 1 = 107 = 229

2 Given the arithmetic progression 3 Given the arithmetic progression
Diberi janjang aritmetik Diberi janjang aritmetik

42, 39, 36, 33, … 5 + x, 8 + 3x, 11 + 5x, …
(a) Determine the nth term. (a) Determine the nth term in terms of n and x.
Tentukan sebutan ke-n.
(b) Determine the 9th term. Tentukan sebutan ke-n dalam sebutan n dan x.
Tentukan sebutan ke-9. (b) Determine the 13th term in terms of x.

Tentukan sebutan ke-13 dalam sebutan x.
(a) a = 42
d = 39 – 42 = –3
Tn = a + (n – 1)d (a) a = 5 + x
Tn = 42 + (n – 1)(–3) d = (8 + 3x) – (5 + x)
= 42 – 3n + 3 = 8 + 3x – 5 – x
= 45 – 3n = 3 + 2x
(b) T9 = 45 – 3(9) Tn = a + (n – 1)d
= 18 = (5 + x) + (n – 1)(3 + 2x)
(b) T13 = (5 + x) + (13 – 1)(3 + 2x)
= (5 + x) + 12(3 + 2x)
= 5 + x + 36 + 24x
= 25x + 41

68

Exercise 3 Determine the number of terms for each of the following arithmetic progressions.
Tentukan bilangan sebutan bagi setiap janjang aritmetik berikut.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 3 1 9, 16, 23, …, 114

7, 11, 15, …, 191
a = 9, d = 16 – 9 = 7
Solution a + (n – 1)d = Tn
a = 7, d = 11 – 7 = 4 9 + (n – 1)(7) = 114
a + (n – 1)d = Tn
7 + (n – 1)(4) = 191 7(n – 1) = 105
4(n – 1) = 184 n – 1 = 15
n – 1 = 46
n = 16
n = 47

2 31, 25, 19, …, –77 3 1 , 5 , 1 1 , 1 1 ,…., 8 1
2 6 6 2 6

a = 31, d = 25 – 31 = –6
a + (n – 1)d = Tn a = 1, d = 5 – 1 = 1
31 + (n – 1)(–6) = –77 2 62 3
–6(n – 1) = –108
a + (n – 1)d = Tn
n – 1 = 18
n = 19 1 + (n – 1)1 1 2 = 81
2 3 6

1(n – 1) = 23
33

n – 1 = 23

n = 24

Exercise 4 Solve each of the following.
Selesaikan setiap yang berikut.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 4 1 The nth term of an arithmetic 2 Given the arithmetic
progression is given by progression 2, 3.5, 5, 6.5, …,
The nth term of an arithmetic Tn = 5n – 14. Find determine the smallest value
progression is given by of n such that the nth term is
Tn = 3n + 4. Find Sebutan ke-n bagi suatu janjang greater than 130.
Sebutan ke-n bagi suatu janjang aritmetik diberi oleh Tn = 5n – 14.
aritmetik diberi oleh Tn = 3n + 4. Cari Diberi janjang aritmetik 2, 3.5, 5,
Cari (a) the first term, 6.5, …, tentukan nilai terkecil n
(a) the first term, sebutan pertama, dengan keadaan sebutan ke-n adalah
sebutan pertama, (b) the common difference. lebih besar daripada 130.
(b) the common difference. beza sepunya.
beza sepunya. a = 2, d = 3.5 – 2 = 1.5
(a) T1 = 5(1) – 14 Tn > 130
Solution = –9
(a) T1 = 3(1) + 4 (b) T2 = 5(2) – 14 a + (n – 1)d > 130
= –4 2 + (n – 1)1.5 > 130
= 7 d = T2 – T1 1.5(n – 1) > 128
(b) T2 = 3(2) + 4 = –4 – (–9)
= 5 n – 1 > 85.33
= 10 n > 86.33
d = T2 – T1
n = 87
= 10 – 7 Check: T87 = 2 + (87 – 1)(1.5)
=3
= 131

69

3 Given the arithmetic 4 The third term and the 5 The 5th term and 13th term of
progression 213, 209, 205, 201, ninth term of an arithmetic an arithmetic progression are
…, determine the smallest progression are 10 and 28 66 and 34 respectively. Find
value of n such that the nth respectively. Determine the the first term and the common
term is smaller than 50. first term and the common difference.
difference.
Diberi janjang aritmetik 213, 209, Sebutan ke-5 dan sebutan ke-13 bagi
205, 201, …, tentukan nilai terkecil Sebutan ketiga dan sebutan suatu janjang aritmetik masing-
n dengan keadaan sebutan ke-n kesembilan bagi suatu janjang masing ialah 66 dan 34. Cari sebutan
adalah lebih kecil daripada 50. aritmetik masing-masing ialah 10 pertama dan beza sepunya.
dan 28. Tentukan sebutan pertama
a = 213, d = 209 – 213 = –4 dan beza sepunya. Tn = a + (n – 1)d
Tn < 50 Given T5 = 66
Tn = a + (n – 1)d a + (5 – 1)d = 66
a + (n – 1)d < 50 Given T3 = 10 a + 4d = 66 ——— 1
213 + (n – 1)(–4) < 50 a + (3 – 1)d = 10 Given T13 = 34
a + 2d = 10 ——— 1 a + (13 – 1)d = 34
–4(n – 1) < –163 Given T9 = 28 a + 12d = 34 ——— 2
4(n – 1) > 163 a + (9 – 1)d = 28 2 − 1:
n – 1 > 40.75 8 d = –32
a + 8d = 28 ——— 2 d = –4
n > 41.75 2 − 1: Substitute d = –4 into 1 ,
n = 42 6 d = 18 a + 4(–4) = 66
d = 3 a = 82
Check: T42 = 213 + (42 – 1)(–4) Substitute d = 3 into 1 , ∴ a = 82, d = -4
= 49 a + 2(3) = 10

a = 4
∴ a = 4, d = 3

Smart Tip

The sum of the first n terms, Sn, of an arithmetic progression is given by,
Hasil tambah n sebutan pertama, Sn, bagi suatu janjang aritmetik diberi oleh,

Sn = n [2a + (n – 1)d]
2
where/dengan keadaan

a = the first term/sebutan pertama

d = the common difference/beza sepunya

Exercise 5 Solve each of the following.
Selesaikan setiap yang berikut.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 5

Given the arithmetic progression 4, 7, 10, 13, …
Diberi janjang aritmetik 4, 7, 10, 13, ….

(a) Determine Sn.
Tentukan Sn.
(b) Hence, find the sum of the first 20 terms of the progression.
Seterusnya, cari hasil tambah 20 sebutan pertama janjang itu.

Solution (b) S20 = 220[3(20) + 5]

(a) a = 4, d = 7 – 4 = 3 = 10(65)
Sn = n2 [2a + (n – 1)d] = 650
Sn = n2 [2(4) + (n – 1)3]

= n2 [8 + 3n – 3]
= n2 [3n + 5]

70

1 Given the arithmetic progression 2, 9, 16, 23, …. 2 Given the arithmetic progression 2, 7, 12, 17, …
Diberi janjang aritmetik 2, 9, 16, 23, …. Diberi janjang aritmetik 2, 7, 12, 17, …

(a) Determine Sn. (a) Determine Sn.
Tentukan Sn. Tentukan Sn.

(b) If the last term of the progression is 93, find (b) Hence, find the sum of the first 14 terms of
the sum of all the terms. the progression.
Jika sebutan terakhir bagi janjang itu ialah 93, cari Seterusnya, cari hasil tambah 14 sebutan pertama
hasil tambah semua sebutan. janjang itu.

(a) a = 2, d = 9 – 2 = 7 (a) a = 2, d = 7 – 2 = 5

Sn = n [2a + (n – 1)d] Sn = n [2a + (n – 1)d]
2 2

Sn = n [2(2) + (n – 1)7] Sn = n [2(2) + (n – 1)5]
2 2

= n[4 + 7n – 7] = n[4 + 5n – 5]
2 2

= n(7n – 3] = n(5n – 1]
2 2

(b) Tn = 93 (b) S14 = 14(5(14) – 1)
a + (n – 1)d = 93 2

2 + (n – 1)7 = 93 = 483

7(n – 1) = 91

n – 1 = 13

n = 14

S14 = 14(7(14) – 3)
2

= 665

3 Given the arithmetic progression 43, 39, 35, 31, …. 4 Given the arithmetic progression 70, 58, 46, 34, ….

Diberi janjang aritmetik 43, 39, 35, 31, …. Diberi janjang aritmetik 70, 58, 46, 34, ….

(a) Determine Sn. (a) Determine Sn.
Tentukan Sn. Tentukan Sn.

(b) Given the last term of the progression is –17, (b) Determine the sum of all the positive terms.

find the sum of all the terms. Tentukan hasil tambah bagi semua sebutan positif.

Diberi sebutan terakhir janjang itu ialah –17, cari hasil

tambah semua sebutan. (a) a = 70, d = 58 – 70 = –12

(a) a = 43, d = 39 – 43 = –4 Sn = n [2a + (n – 1)d]
2
n [2a
Sn = 2 + (n – 1)d] Sn = n [2(70) + (n – 1)(–12)]
2
n [2(43)
Sn = 2 + (n – 1)(–4)] = n[140 – 12n + 12]
2
= n[86 – 4n + 4]
2 = n(152 – 12n]
2
= n[90 – 4n]
2 (b) Tn > 0
a + (n – 1)d > 0
(b) Tn = –17
a + (n – 1)d = –17 70 + (n –1)(–12) > 0

43 + (n – 1)(–4) = –17 –12(n – 1) > –70

–4(n – 1) = –60 n – 1 < 5.833

n – 1 = 15 n < 6.833

n = 16 n = 6

S16 = 16(90 – 4(16)) S6 = 6(152 – 12(6))
2 2

= 208 = 240

71

5 Given the arithmetic progression 4, 6.5, 9, 11.5, …. 6 Given the arithmetic progression –90, –83, –76,

Diberi janjang aritmetik 4, 6.5, 9, 11.5, …. –69, …

(a) Determine Sn. Diberi janjang aritmetik –90, –83, –76, –69, …
Tentukan Sn. (a) Determine Sn.
Tentukan Sn.
(b) If the last term of the progression is 74, find (b) Hence, find the sum of all the negative terms.
the sum of all the terms.

Jika sebutan terakhir janjang itu ialah 74, cari hasil Seterusnya, cari hasil tambah bagi semua sebutan

tambah semua sebutan. negatif.

(a) a = 4, d = 6.5 – 4 = 2.5 (a) a = –90, d = –83 – (–90) = 7

Sn = n [2a + (n – 1)d] Sn = n [2a + (n – 1)d]
2 2

Sn = n [2(4) + (n – 1)(2.5)] Sn = n [2(–90) + (n – 1)7]
2 2

= n[8 + 2.5n – 2.5] = n[–180 + 7n – 7]
2 2

= n(5.5 + 2.5n] = n(7n – 187]
2 2

(b) Tn = 74 (b) Tn < 0

a + (n – 1)d = 74 a + (n – 1)d < 0

4 + (n – 1)(2.5) = 74 –90 + (n – 1)(7) < 0

2.5(n – 1) = 70 7(n –1) < 90

n – 1 = 28 n –1 < 12.86

n = 29 n < 13.86

S29 = 29[5.5 + 2.5(29)] n = 13
2
S13 = 13(7(13) – 187)
= 1 131 2

= –624

Smart Tip

1 The sum of the first n terms, Sn, of an arithmetic progression is given by,
H asil tambah n sebutan pertama, Sn, bagi suatu janjang aritmetik diberi oleh,
n
Sn = 2 [a + l]

where/dengan keadaan

a = the first term/sebutan pertama, l = the last term/sebutan terakhir.

2 This formula is used when the last term is given.

Rumus ini digunakan apabila sebutan terakhir diberikan.

Exercise 6 Determine the number of terms of each of the following arithemetic progressions. Hence,
find the sum of the progression.
Tentukan bilangan sebutan dalam setiap janjang aritmetik berikut. Seterusnya, cari hasil tambah janjang itu.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 6 1 5, 9, 13, ..., 61

9, 14, 19, … , 84

a = 5, d = 9 – 5 = 4

Solution Tn = a + (n – 1)d
61 = 5 + (n – 1)4
a = 9, d = 14 – 9 = 5
4 (n – 1) = 56
Tn = a + (n – 1)d
84 = 9 + (n – 1)5 n – 1 = 14

5(n – 1) = 75 n = 15

n – 1 = 15 Sn = n [a + l]
2
n = 16
S15 = 125[5 + 61]
Sn = n [a + l]
2
= 495
S16 = 126[9 + 84]

= 744

72

2 11, 17, 23, …, 125 3 45, 42, 39, ..., –9



a = 11, d = 17 – 11 = 6 a = 45, d = 42 – 45 = –3

Tn = a + (n – 1)d Tn = a + (n – 1)d
125 = 11 + (n – 1)6 –9 = 45 + (n – 1)(–3)

6 (n – 1) = 114 3(n – 1) = 54

n – 1 = 19 n – 1 = 18

n = 20 n = 19

Sn = n [a + l] Sn = n [a + l]
2 2

S20 = 20[11 + 125] S19 = 19[45 + (–9)]
2 2

= 1 360 = 342

Exercise 7 Solve each of the following.
Selesaikan setiap yang berikut.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 7 1 Given the first three terms of an arithmetic

Given the first three terms of an arithmetic progression are 4, 8.5, 13, … Find the sum from
progression are 3, 7, 11, … Find the sum from
the 6th term to the 18th term. the 7th term to the 17th term.
Diberi tiga sebutan pertama suatu janjang aritmetik ialah
3, 7, 11, … Cari hasil tambah dari sebutan ke-6 hingga Diberi tiga sebutan pertama suatu janjang aritmetik ialah
sebutan ke-18.
4, 8.5, 13, … Cari hasil tambah dari sebutan ke-7 hingga

sebutan ke-17.



a = 4, d = 8.5 – 4 = 4.5

Solution The sum from T7 to T17
= S17 – S6
a = 3, d = 7 – 3 = 4
127[2(4) 6
The sum from T6 to T18 = + (17 – 1)(4.5)] – 2 [2(4) + (6 – 1)(4.5)]
Hasil tambah dari T6 hingga T18
= S18 – S5 = 680 – 91.5

128[2(3) 5 = 588.5
2
= + (18 – 1)4] – [2(3) + (5 – 1)4]

= 666 – 55

= 611

2 Given the first three terms of an arithmetic progression are 72, 67, 62, … Find the sum from the 10th term

to the 20th term.

Diberi tiga sebutan pertama suatu janjang aritmetik ialah 72, 67, 62, … Cari hasil tambah dari sebutan ke-10 hingga sebutan

ke-20.



a = 72, d = 67 – 72 = –5

The sum from T10 to T20
= S20 – S9

= 220[2(72) + (20 – 1)(–5)] – 9 [2(72) + (9 – 1)(–5)]
2

= 490 – 468

= 22

73

Exercise 8 Solve each of the following.
Selesaikan setiap yang berikut.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 8

Given the first three terms of an arithmetic progression are 5, 8, 11, … Find the number of the first terms
with the sum 258.
Diberi tiga sebutan pertama suatu janjang aritmetik ialah 5, 8, 11, … Cari bilangan sebutan pertama dengan hasil tambah 258.

Solution

a = 5, d = 8 – 5 = 3

Sn = n2 [2a + (n – 1)d] ( 3n + 43)(n – 12) = 0

258 = n[2(5) + (n – 1)3] 3n + 43 = 0, n – 12 = 0

n [10 2 n = – 43 n = 12
258 3
+ 3n – 3] =
2 ∴ Number of terms/Bilangan sebutan = 12
n[3n + 7] = 258
2

3n2 + 7n – 516 = 0

1 The sum of the first n terms of an arithmetic 2 Given the first three terms of an arithmetic
progression is given by Sn = 3n2 + n. Find progression are 4, 9, 14, … Find the number of
the first terms with the sum 216.
Hasil tambah n sebutan pertama bagi suatu janjang Diberi tiga sebutan pertama suatu janjang aritmetik ialah
aritmetik diberi oleh Sn = 3n2 + n. Cari 4, 9, 14, … Cari bilangan sebutan pertama dengan hasil
(a) the first term, tambah 216.
sebutan pertama,
(b) the common difference, a = 4, d = 9 – 4 = 5
beza sepunya,
(c) the nth term. Sn = n [2a + (n – 1)d]
sebutan ke-n. 2
216 = n[2(4) + (n – 1)5]
(a) T1 = S1 = 3(1)2 + (1) = 4 2
(b) T1 + T2 = S2 = 3(2)2 + (2) 432 = n(5n + 3)
(4) + T2 = 14
5n2 + 3n – 432 = 0
T2 = 10
d = 10 – 4 = 6 ( 5n + 48)(n – 9) = 0
(c) Tn = a + (n – 1)d
= 4 + (n – 1)6 5n + 48 = 0, n – 9 = 0
= 4 + 6n – 6
= 6n – 2 n = – 48 n = 9
5

Number of terms = 9

3 Given the first three terms of an arithmetic 4 The sum of the first n terms of an arithmetic
progression are 2, 8, 14, … Find the number of progression is given by Sn = 2n2 – 5n. Find
the first terms with the sum more than 300.
Hasil tambah n sebutan pertama bagi suatu janjang
Diberi tiga sebutan pertama suatu janjang aritmetik ialah aritmetik diberi oleh Sn = 2n2 – 5n. Cari
2, 8, 14, … Cari bilangan sebutan pertama dengan hasil (a) the first term,
tambah melebihi 300. sebutan pertama,
(b) the common difference,
a = 2, d = 8 – 2 = 6 beza sepunya,
(c) the nth term.
n [2a + (n – 1)d] = Sn sebutan ke-n.
2
n[2(2) + (n – 1)6] > 300 (a) T1 = S1 = 2(1)2 – 5(1)
2 = –3
n(4 + 6n – 6) > 600
(b) T1 + T2 = S2 = 2(2)2 – 5(2)
6n2 – 2n – 600 > 0 (–3) + T2 = –2
3n2 – n – 300 > 0 ——— 1
T2 = 1
Let 3n2 – n – 300 = 0 d = 1 – (–3)

–(–1) ± (–1)2 − 4(3)(–300) = 4
n = 2(3) (c) Tn = a + (n – 1)d

n < –9.83 or n > 10.16 = (–3) + (n – 1)4
From 1 , n = 11 = –3 + 4n – 4
= 4n – 7

74

Exercise 9 By using the formula Tn = Sn – Sn – 1, determine Tn for each of the following.
Dengan menggunakan rumus Tn = Sn – Sn – 1, tentukan Tn bagi setiap yang berikut.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 9 1 The sum of the first n terms of an arithmetic
progression is given by Sn = n2 + 3n. Find the
The sum of the first n terms of an arithmetic sixth term.
progression is given by Sn = 2n2 + 5n. Find the
fifth term. Hasil tambah n sebutan pertama bagi suatu janjang
Hasil tambah n sebutan pertama bagi suatu janjang aritmetik diberi oleh Sn = n2 + 3n. Cari sebutan keenam.
aritmetik diberi oleh Sn = 2n2 + 5n. Cari sebutan kelima.

Solution Tn = Sn – Sn – 1
Tn = Sn – Sn – 1 T6 = S6 – S5
T5 = S5 – S4 = [(6)2 + 3(6)] – [(5)2 + 3(5)]
= [2(5)2 + 5(5)] – [2(4)2 + 5(4)] = 54 – 40
= 75 – 52 = 14

= 23

2 The sum of the first n terms of an arithmetic 3 The sum of the first n terms of an arithmetic
progression is given by Sn = 6n2 – 3n. Find the progression is given by Sn = 72n – 2n2. Find the
eighth term. ninth term.

Hasil tambah n sebutan pertama bagi suatu janjang Hasil tambah n sebutan pertama bagi suatu janjang
aritmetik diberi oleh Sn = 6n2 – 3n. Cari sebutan kelapan. aritmetik diberi oleh Sn = 72n – 2n2. Cari sebutan
kesembilan.
Tn = Sn – Sn – 1
T8 = S8 – S7 Tn = Sn – Sn – 1
= [6(8)2 – 3(8)] – [6(7)2 – 3(7)] T9 = S9 – S8
= 360 – 273 = [72(9) – 2(9)2] – [72(8) – 2(8)2]
= 87 = 486 – 448
= 38

Exercise 10 Solve each of the following.
Selesaikan setiap yang berikut.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang janjang aritmetik dan janjang geometri dalam konteks penyelesaian masalah rutin yang mudah.

Example 10

In an arithmetic progression, the fourth term is 18 and the sum of the first eight terms is 160. Find
Dalam suatu janjang aritmetik, sebutan keempat ialah 18 dan hasil tambah bagi lapan sebutan pertama ialah 160. Cari

(a) the first term,
sebutan pertama,

(b) the common difference.
beza sepunya.

Solution Substitute into 2 /Ganti dalam 2 ,
Ga +iv3edn/=D1i8ber—i T—4 =—181 2(18 – 3d) + 7d = 40 ——— 2

Given/Diberi S8 = 160 36 – 6d + 7d = 40
8[2a + (8 – 1)d] = 160 d=4
2
Substitute d = 4 into 1 /Gantikan d = 4 ke dalam 1
2a + 7d = 40 ——— 2 a = 18 – 3(4)
From 1 /Dari 1 ,
=6
a = 18 – 3d ∴ a = 6, d = 4

75

1 In an arithmetic progression, the fifth term is 16 2 In an arithmetic progression, the sixth term is 13
and the sum of the first ten terms is 175. Find and the sum of the first seven terms is 21. Find

Dalam suatu janjang aritmetik, sebutan kelima ialah 16 dan Dalam suatu janjang aritmetik, sebutan keenam ialah 13
hasil tambah bagi sepuluh sebutan pertama ialah 175. Cari dan hasil tambah bagi tujuh sebutan pertama ialah 21. Cari

(a) the first term, (a) the first term,
sebutan pertama, sebutan pertama,

(b) the common difference. (b) the common difference.
beza sepunya. beza sepunya.

Given T5 = 16 1 Given T6 = 13 1
a + 4d = 16 ——— a + 5d = 13 ———

Given S10 = 175 Given S7 = 21
10[2a + (10 – 1)d] = 175 7[2a + (7 – 1)d] = 21
2 2
2a + 6d = 6
2a + 9d = 35 ——— 2 a + 3d = 3 ——— 2

From 1 , From 1 ,

a =16 – 4d a = 13 – 5d

Substitute into 2 Substitute into 2 ,
2(16 – 4d) + 9d = 35 ——— 2
(13 – 5d) + 3d = 3
32 – 8d + 9d = 35
–2d = –10
d = 3
d = 5
Substitute d = 3 into 1 ,
Substitute d = 5 into 1 ,
a = 16 – 4(3)
a = 13 – 5(5)
=4
= –12
∴ a = 4, d = 3
∴ a = –12, d = 5

3 In an arithmetic progression, the second term is 4 A farmer has 5 600 chickens and 4 880 ducks.
35 and the sum of the first ten terms is 140. Find The farmer sells 240 chickens and 180 ducks to
the wholesaler every day. After a few days, the
Dalam suatu janjang aritmetik, sebutan kedua ialah 35 dan number of chickens and the number of ducks left
hasil tambah bagi sepuluh sebutan pertama ialah 140. Cari are the same. On what day did the number of
(a) the first term,
sebutan pertama, chickens and the number of ducks left are the same?
(b) the common difference. Seorang penternak mempunyai 5 600 ekor ayam dan 4 880
beza sepunya.
ekor itik. Penternak itu menjual 240 ekor ayam dan 180 ekor
Ga +ivden=T325=—35—— 2 itik kepada pemborong pada setiap hari. Selepas beberapa
Given S10 = 140 hari, bilangan ayam dan bilangan itik yang tinggal adalah
10[2a + (10 – 1)d] = 140 sama banyak. Pada hari ke berapakah bilangan ayam dan
2 2a + 9d = 28 ——— 2 bilangan itik yang tinggal adalah sama banyak?
From 1 ,
a = 35 – d The number of chickens left: 5 600, 5 360, 5 120, …
Substitute into 2 ,  a = 5 600, d = –240
2(35 – d) + 9d = 28 Tn = a + (n – 1)d
= 5 600 + (n – 1)(–240)
70 – 2d + 9d = 28 = 5 600 – 240n + 240
7d = –42 = 5 840 – 240n
d = –6 The number of ducks left: 4 880, 4 700, 4 520, ….
 a = 4 880, d = –180
Substitute d = –6 into 1 , Tn = a + (n – 1)d
a = 35 – (–6) = 4 880 + (n – 1)(–180)
= 41 = 4 880 – 180n + 180
∴ a = 41, d = -6 = 5 060 – 180n
5 840 – 240n = 5 060 – 180n

60n = 780
n = 13

On the 13th day, the number of chickens and the
number of ducks left are the same.

76

5 The diagram below shows two objects moving toward each other on a straight line, FG. Object P moves

with a distance of 3 cm in the first second, 6 cm in the next second, 9 cm in the third second and so on.

Object Q moves 2 cm in the first second, 4 cm in the next second, 6 cm in the third second and so on. Both

objects start simultaneously and meet at the nth second.

Rajah di bawah menunjukkan dua objek bergerak ke arah satu sama lain pada satu garis, FG. Objek P bergerak sejauh 3 cm pada

saat pertama, 6 cm pada saat kedua, 9 cm pada saat ketiga dan seterusnya. Objek Q pula bergerak sejauh 2 cm pada saat pertama,

4 cm pada saat kedua, 6 cm pada saat ketiga dan seterusnya. Kedua-dua objek bermula serentak dan bertemu pada saat ke-n.

Given FG = 105 cm, find/Diberi FG = 105 cm, cari P Q

(a) the value of n,/nilai bagi n,

(b) the distance travelled by object P before both objects meet. F 105 cm G
jarak yang dilalui oleh objek P sebelum kedua-dua objek bertemu.

(a) The movement of object P: 3 cm, 6 cm, SP + SQ = 105

9 cm, …. (Arithmetic progression) 3 (n2 + n) + (n2 + n) = 105
2
a = 3, d = 3
n
Sn = 2 [2a + (n – 1)d] 3(n2 + n) + 2(n2 + n) = 210

SP = n [2(3) + (n – 1)(3)] 3n2 + 3n + 2n2 + 2n = 210
2
= n(6 + 3n – 3) 5n2 + 5n – 210 = 0

2 n2 + n – 42 = 0

= n(3n + 3) (n + 7) (n – 6) = 0

2 n = –7, n = 6

= 3 (n2 + n) ∴ Both object meet at the 6th second.

2 (b) When n = 6,

The movement of object Q: 2 cm, 4 cm, SP = 3 [(6)2 + 6]
2
6 cm, …. (Arithmetic progression)

a = 2, d = 2 = 63 cm
n(2a + (n – 1)d)
Sn = 2

SQ = n[2(2) + (n – 1)(2)]
2
= n(4 + 2n – 2)
2
= n(2n + 2) = n2 + n
2

5.2 Geometric Progressions / Janjang Geometri

Smart Tip

A geometric progression is a number sequence such that the ratio between each term (after the first term) and its

preceding term is a constant. The constant is called common ratio, r.

Janjang geometri ialah suatu jujukan nombor dengan keadaan nisbah antara setiap sebutan (selepas sebutan pertama) dengan sebutan

sebelumnya ialah suatu pemalar. Pemalar ini disebut sebagai nisbah sepunya, r.

r = Tn + 1 , n = 1, 2, 3, …
Tn

Exercise 11 Determine whether each of the following sequences is a geometric progression or not.
Give your reason.
Tentukan sama ada setiap jujukan berikut ialah janjang geometri atau bukan. Berikan alasan anda.

TP 1 Mempamerkan pengetahuan asas tentang janjang.

Example 11 1 3.5, 7, 14, 28, … 2 288, 144, 48, 16, …

4, 12, 36, 108, … T2 = 7 = 2 T2 = 144 = 1
T1 3.5 T1 288 2
Solution

T2 = 12 = 3 T4 = 108 = 3 T3 = 14 = 2 T3 = 48 = 1
T1 4 T3 36 T2 7 T2 144 3

T3 = 36 = 3 T4 = 28 = 2 T4 = 16 = 1
T2 12 T3 14 T3 48 3

4, 12, 36, 108, … is a

geometric progression with 3.5, 7, 14, 28, … is a geometric 288, 144, 48, 16, … is not a
progression with common geometric progression because
common ratio = 3. ratio = 2. there is no common ratio.

4, 12, 36, 108, … ialah janjang

geometri dengan nisbah sepunya = 3.

77

3 24, 12, 4, 2, … 4 48, 24, 12, 6, … 5 12, –3, 3 , – 136, …
4
T2 12 1 T2 24 1
T1 = 24 = 2 T1 = 48 = 2 T2 = –3 = – 1
T1 12 4
T3 T3
T2 = 4 = 1 T2 = 12 = 1 T3 1 23 1
12 3 24 2 T2 4 4
= –3 = –
T4 2 1 T4 6 1
T3 = 4 = 2 T3 = 12 = 2 – 3
16
T4 = = – 1
T3 3 4
24, 12, 4, 2, … is not a geometric 48, 24, 12, 6, … is a geometric
progression because there is 4
no common ratio. progression with common

ratio = 1 . 12, –3, 3 , – 136, … is a
2 4

geometric progression with

common ratio = – 1 .
4

Smart Tip

The nth term, Tn, of a geometric progression is written as:
Sebutan ke-n, Tn, bagi suatu janjang geometri ditulis sebagai:

Tn = arn – 1
where/dengan keadaan
a = the first term/sebutan pertama
r = the common ratio/nisbah sepunya

Exercise 12 Solve each of the following.
Selesaikan setiap yang berikut.

TP 2 Mempamerkan kefahaman tentang janjang aritmetik dan janjang geometri.

Example 12 1 The first three terms of a geometric progression
are 9, 18, 36, …. Determine
The first three terms of a geometric progression Tiga sebutan pertama bagi suatu janjang geometri ialah 9,
are 4, 12, 36, … Determine 18, 36, …. Tentukan
Tiga sebutan pertama bagi suatu janjang geometri ialah 4, (a) the nth term,
12, 36, … Tentukan
(a) the nth term, sebutan ke-n,
sebutan ke-n,
(b) the seventh term.
(b) the sixth term. sebutan ketujuh.
sebutan keenam.
(a) a = 9

Solution r= T2 = 18 = 2
T1 9
(a) a = 4
Tn = arn – 1
r= T2 = 12 = 3 = 9(2)n – 1
T1 4
(b) T7 = 9(2)7 – 1
Tn = arn – 1 = 9(2)6
= 4(3)n – 1
= 576
(b) T6 = 4(3)6 – 1
= 4(3)5

= 972

78

2 The first three terms of a geometric progression 3 The first three terms of a geometric progression

are 4, 6, 9, …… Determine are 12, 3, 3 , … Determine
4
Tiga sebutan pertama bagi suatu janjang geometri ialah 4,

6, 9, …… Tentukan Tiga sebutan pertama bagi suatu janjang geometri ialah 12,

(a) the nth term, (b) the fifth term. 3, 3 , … Tentukan
4
sebutan ke-n, sebutan kelima.
(a) the nth term, (b) the sixth term.
1 2(b) 3 5–1
(a) a = 4 T5 = 4 2 sebutan ke-n, sebutan keenam.

r = T2 = 6 = 3 41 3 24 (a) a = 12 1 2(b) 1 6–1
T1 4 2 2 T6 = 12 4
= T2 3 1
T1 12 4
Tn = arn – 1 = 81 r= = = = 121 1 25
4 4
1 2 =4 3 n–1
2
Tn = arn – 1 = 3
1 2 1 n–1 256
= 12 4

Exercise 13 Determine the number of terms of each of the following geometric progressions.
Tentukan bilangan sebutan bagi setiap janjang geometri berikut.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 13 1 5, 15, 45 ……, 3 645 2 7, 21, 63, ……, 15 309

6, 12, 24 ……, 6 144 a=5 a=7

Solution r = T2 = 15 = 3 r= T2 = 21 = 3
T1 5 T1 7
a=6
Tn = arn – 1 Tn = arn – 1
r = T2 = 12 = 2 Tn = 5(3)n – 1 Tn = 7(3)n – 1
T1 6
Given Tn = 3 645 Given Tn = 15 309
Tn = arn – 1 5(3)n – 1 = 3 645 7(3)n – 1 = 15 309
Tn = 6(2)n – 1
Given/Diberi Tn = 6 144 (3)n – 1 = 729 (3)n – 1 = 2 187

6(2)n – 1 = 6 144 log10 (3)n – 1 = log10 729 log10 (3)n – 1 = log10 2 187
( n – 1)log10 3 = log10 729 (n – 1)log10 3 = log10 2 187
(2)n – 1 = 1 024 ( n – 1)0.4771 = 2.8627 (n – 1)(0.4771) = 3.3398

log10 (2)n – 1 = log10 1 024 n – 1 = 6 n – 1 = 7
(n – 1)log10 (2) = log10 1 024
(n – 1)(0.3010) = 3.010 n = 7 n = 8

n – 1 = 10

n = 11

3 136, 3 , 3, ……, 49 152 4 3 125, 625, 125, ……, 0.008 5 6 144, 3 072, 1 536, ……, 0.75
4

3 a = 3 125 a = 6 144
16
a = r= T2 = 625 = 0.2 r= T2 = 3 072 = 0.5
T1 3 125 T1 6 144
3

r = T2 = 4 =4 Tn = arn – 1 Tn = arn – 1
T1 3 Tn = 3 125(0.2)n – 1
Tn = 6 144(0.5)n – 1
16 Given Tn = 0.008 Given Tn = 0.75
Tn = arn – 1
3 3 125(0.2)n – 1 = 0.008 6 144(0.5)n – 1 = 0.75
16
Tn = (4)n – 1 (0.2)n – 1 = 0.00000256 (0.5)n – 1 = 0.00012207

Given Tn = 49 152 log10 (0.2)n – 1 = log10 0.00000256 log10 (0.5)n – 1 = log10 0.00012207
3 (n – 1)log10 0.2 = log10 0.00000256 (n – 1)log10 0.5 =log100.00012207
16 (4)n – 1 = 49 152 (n – 1)(–0.6990) = –5.5918 (n – 1)(–0.3010) = –3.9134

(4)n – 1 = 262 144 n – 1 = 8 n – 1 = 13

log10 (4)n – 1 = log10 262 144 n = 9 n = 14
(n – 1)log10 4 = log10 262 144
( n – 1)0.6021 = 5.4185

n – 1 = 9

n = 10

79

Smart Tip

The sum of the first n terms, Sn, of a geometric progression is given by,
Hasil tambah n sebutan pertama, Sn, bagi suatu janjang geometri diberi oleh,

Sn = a(rn – 1) , r > 1
r–1

Exercise 14 Express Sn in terms of n for each of the following geometric progressions. Hence, solve the
problems.
Ungkapkan Sn dalam sebutan n bagi setiap janjang geometri berikut. Seterusnya, selesaikan masalah.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 14

Given the first three terms of a geometric progression are 6, 24, 96, …
Diberi tiga sebutan pertama suatu janjang geometri ialah 6, 24, 96, …

(a) Determine the sum of the first n terms.
Tentukan hasil tambah n sebutan pertama.

(b) If the last term of the progression is 24 576, find the sum of all the terms.
Jika sebutan terakhir janjang itu ialah 24 576, cari hasil tambah semua sebutan.

Solution

(a) a = 6, r = 24 = 4 (b) Tn = 24 576 S7 = 2(47 – 1)
6 arn – 1 = 24 576 = 32 766

Sn = a(rn – 1) 6(4n – 1) = 24 576
r–1 4n – 1 = 4 096
(n – 1)log10 4 = log10 4 096
Sn = 6(4n – 1)
4–1 n – 1 = 6
n = 7
= 2(4n – 1)

1 Given the first three terms of 2 Given the first three terms of 3 Given the first three terms of
a geometric progression are a geometric progression are
2, 10, 50, … 16, 24, 36, … a geometric progression are
Diberi tiga sebutan pertama suatu
Diberi tiga sebutan pertama suatu janjang geometri ialah 16, 24, 36, … 7, 14, 28, …
janjang geometri ialah 2, 10, 50, …
(a) Determine the sum of the Diberi tiga sebutan pertama suatu
(a) Determine the sum of the first n terms.
first n terms. Tentukan hasil tambah n janjang geometri ialah 7, 14, 28, …
Tentukan hasil tambah n sebutan pertama.
sebutan pertama. (a) Determine the sum of the
(b) Find the sum of the
(b) Find the sum of the first seven terms of the first n terms.
first nine terms of the progression.
progression. Cari hasil tambah bagi tujuh Tentukan hasil tambah n
Cari hasil tambah bagi sembilan sebutan pertama janjang itu.
sebutan pertama janjang itu. sebutan pertama.

(b) If the last term of the

progression is 7 168, find

the sum of all the terms.

Jika sebutan terakhir janjang itu

ialah 7 168, cari hasil tambah

10 24 semua sebutan.
2 16
(a) a = 2, r = =5 (a) a = 16, r = = 1.5 (a) a = 7, r = 14 = 2
7
a(rn – 1) a(rn – 1) a(rn – 1)
Sn = r–1 Sn = r–1 Sn = r–1

Sn = 2(5n – 1) Sn = 16(1.5n – 1) Sn = 7(2n – 1)
5–1 1.5 – 1 2–1

= 1 (5n – 1) = 32(1.5n – 1) = 7(2n – 1)
2
(b) S7 = 32(1.57 – 1) (b) Tn = 7 168
(b) S9 = 1 (59 – 1) = 514.75 arn – 1 = 7 168
2
7(2n – 1) = 7 168

= 976 562 2n – 1 = 1 024

(n – 1)log10 2 = log10 1 024
n – 1 = 10

n = 11

S11 = 7(211 – 1)
= 14 329

80

4 Given the first three terms of a geometric 5 Given the first three terms of a geometric

progression are 8, 12, 18, … progression are 3, 6, 12, …

Diberi tiga sebutan pertama suatu janjang geometri ialah Diberi tiga sebutan pertama suatu janjang geometri ialah

8, 12, 18, … 3, 6, 12, …

(a) Determine the sum of the first n terms. (a) Determine the sum of the first n terms.

Tentukan hasil tambah n sebutan pertama. Tentukan hasil tambah n sebutan pertama.

(b) Find the sum from the fourth term to the (b) Find the sum from the fifth term to the ninth

seventh term. term.

Cari hasil tambah dari sebutan keempat hingga sebutan Cari hasil tambah dari sebutan kelima hingga sebutan

ketujuh. kesembilan.

(a) a = 8, r = 12 = 1.5 (a) a = 3, r = 6 =2
8 3

Sn = a(rn – 1) Sn = a(rn – 1)
r–1 r–1

Sn = 8(1.5n – 1) Sn = 3(2n – 1)
1.5 – 1 2–1

= 16(1.5n – 1) = 3(2n – 1)

(b) Sum from T4 to T7 (b) Sum from T5 to T9
= S7 – S3 = S9 – S4
= 16(1.57 – 1) – 16(1.53 – 1) = 3(29 – 1) – 3(24 – 1)

= 257.375 – 38 = 1 533 – 45

= 219.38 = 1 488

Smart Tip

The sum of the first n terms, Sn, of a geometric progression is given by,
Hasil tambah n sebutan pertama, Sn, bagi suatu janjang geometri diberi oleh,

Sn = a(1 – rn) , r < 1
1 – r

Exercise 15 Express Sn in terms of n for each of the following geometric progressions. Hence, solve the
problems.
Ungkapkan Sn dalam sebutan n bagi setiap janjang geometri berikut. Seterusnya, selesaikan masalah.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 15

Given the first three terms of a geometric progression are 384, 96, 24, …
Diberi tiga sebutan pertama suatu janjang geometri ialah 384, 96, 24, …

(a) Determine the sum of the first n terms.
Tentukan hasil tambah n sebutan pertama.

(b) If the last term of the progression is 0.375, find the sum of all the terms.
Jika sebutan terakhir janjang itu ialah 0.375, cari hasil tambah semua sebutan.

Solution

(a) a = 384, r = 96 = 0.25 (b) Tn = 0.375
384 arn – 1 = 0.375

Sn = a(1 – rn) 384(0.25n – 1) = 0.375
1–r
0.25n – 1 = 0.0009766
384(1 – 0.25n)
Sn = 1– 0.25 (n – 1)log10 0.25 = log10 0.0009766
n – 1 = 5
384(1 – 0.25n)
= 0.75 n = 6

= 512(1 – 0.25n) S6 = 512(1 – 0.256)
= 511.88

81

1 Given the first three terms of a 2 Given the first three terms of 3 Given the first three terms of a
geometric progression are a geometric progression are geometric progression are 224,
1 250, 250, 50, … 972, 324, 108, … 112, 56, …
Diberi tiga sebutan pertama suatu Diberi tiga sebutan pertama suatu Diberi tiga sebutan pertama suatu
janjang geometri ialah 1 250, 250, janjang geometri ialah 972, 324, janjang geometri ialah 224, 122,
50, … 108, … 56, …

(a) Determine the sum of the (a) Determine the sum of the (a) Determine the sum of the
first n terms. first n terms. first n terms.
Tentukan hasil tambah n Tentukan hasil tambah n Tentukan hasil tambah n
sebutan pertama. sebutan pertama. sebutan pertama.

(b) Find the sum of the (b) Find the sum of the (b) If the last term of the
first eight terms of the first six terms of the progression is 1.75, find
progression. progression. the sum of all the terms.
Cari hasil tambah bagi lapan Cari hasil tambah bagi enam Jika sebutan terakhir janjang
sebutan pertama janjang itu. sebutan pertama janjang itu. itu ialah 1.75, cari hasil tambah
semua sebutan.
250 324 1
(a) a = 1 250, r = 1 250 = 0.2 (a) a = 972, r = 972 = 3 112
224
a(1 – rn) a(1 – rn) (a) a = 224, r = = 0.5
1–r 1 – r
Sn = Sn = a(1 – rn)
1 –r
1 250(1 – 0.2n) 97231 1 1 2n4 Sn =
1 – 0.2 3
Sn = Sn = – 224(1 – 0.5n)
1 – 0.5
= 1 562.5(1 – 0.2n) 1 – 1 Sn =
3
(b) S8 = 1 562.5(1 – 0.28) 45831 1 1 2n4 = 448(1 – 0.5n)
= 1 562.50 3
= 1 – (b) Tn = 1.75
arn – 1 = 1.75
45831 1 1 264
(b) S6 = 1 – 3 224(0.5n – 1) = 1.75

= 1 456 0.5n – 1 = 0.0078125

(n – 1)log10 0.5 = log10 0.0078125
n – 1 = 7

n = 8

S8 = 448(1 – 0.58)
= 446.25

4 Given the first three terms of a geometric 5 Given the first three terms of a geometric

progression are 192, 96, 48, … progression are 108, 72, 48, …

Diberi tiga sebutan pertama suatu janjang geometri ialah Diberi tiga sebutan pertama suatu janjang geometri ialah

192, 96, 48, … 108, 72, 48, …

(a) Determine the sum of the first n terms. (a) Determine the sum of the first n terms.

Tentukan hasil tambah n sebutan pertama. Tentukan hasil tambah n sebutan pertama.

(b) Find the sum from the fifth term to the eighth (b) Find the sum from the sixth term to the ninth

term. term.

Cari hasil tambah dari sebutan kelima hingga sebutan Cari hasil tambah dari sebutan keenam hingga sebutan

kelapan. kesembilan.

(a) a = 192, r = 96 = 1 (a) a = 108, r = 72 = 2
192 2 108 3

Sn = a(1 – rn) Sn = a(1 – rn)
1–r 1 –r
19231 1 2n4 10831 1 2n4
Sn = – 1 Sn = – 2
2 3

1 – 1 1 – 2
2 3
38431 1 1 2n4 32431 1 2 2n4
= – 2 = – 3

(b) Sum from T5 to T8 (b) Sum from T6 to T9
= S8 – S4 = S9 – S5
38431 1 284 38431 1 244 32431 1 294 32431 1 254
= – 1 – – 1 = – 2 – – 2
2 2 3 3

= 382.5 – 360 = 315.57 – 281.33

= 22.5 = 34.24

82

Smart Tip

When n approaches infinity, n → ∞, then the sum to infinity is
Apabila n menghampiri ketakterhinggaan, n → ∞, maka hasil tambah hingga ketakterhinggaan ialah

S∞ = 1 a r, |r| < 1
–

Exercise 16 Find the sum to infinity, S∞ for each of the following geometric progressions.
Cari hasil tambah hingga ketakterhinggaan,S∞ bagi setiap janjang geometri berikut.

TP 2 Mempamerkan kefahaman tentang janjang aritmetik dan janjang geometri.

Example 16 1 8, 2, 0.5, 0.125, …

(a) 24, 12, 6, 3, … a = 8, r = 2 = 1
(b) 100, –40, 16, –6.4, … 8 4

Solution S∞ = 1 8 1
– 4
(a) a = 24, r = 12 = 0.5 (b) a = 100, r = –40 = – 2
24 100 5 2
a = 10 3
a S∞ = 1–r
S∞ = 1 – r

= 24 = 100
– 0.5
1 1 – 1– 2 2
5
= 48 3
7
= 71

2 36, 12, 4, … 3 12, 9, 247, 8161, … 4 80, –40, 20, –10, …

a = 36, r = 12 = 1 a = 12, r = 9 = 3 a = 80, r = –40 = – 1
36 3 12 4 80 2

S∞ = 36 80
12 1 2S∞ =
1 – 1 S∞ = 1
3 1 – 3 1 – – 2
4
= 54 = 53 1
= 48 3

Exercise 17 Solve the following problems.
Selesaikan masalah yang berikut.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 17

The first term and the sum to infinity of a 9 = 6
geometric progression are 6 and 9 respectively. 1–r
Find the common ratio.
Sebutan pertama dan hasil tambah hingga ketakterhinggaan 1 – r = 6
suatu janjang geometri masing-masing ialah 6 dan 9. Cari 9
nisbah sepunya.
1 – r = 2
3

Solution 3 – 3r = 2

a = 6, S∞ = 9 –3r = –1

S∞ = a r = 1
1–r 3

83

1 The first term and the sum 2 The common ratio and the 3 The common ratio and the
to infinity of a geometric
progression are 14 and sum to infinity of a geometric sum to infinity of a geometric
16 respectively. Find the
common ratio. progression are 2 and 24 progression are – 3 and 15
3 5
Sebutan pertama dan hasil tambah respectively. Find the first term. respectively. Find the first term.
hingga ketakterhinggaan suatu
janjang geometri masing-masing Nisbah sepunya dan hasil tambah Nisbah sepunya dan hasil tambah
ialah 14 dan 16. Cari nisbah
sepunya. hingga ketakterhinggaan suatu hingga ketakterhinggaan suatu

janjang geometri masing-masing janjang geometri masing-masing

ialah 2 dan 24. Cari sebutan pertama. ialah – 3 dan 15. Cari sebutan pertama.
3 5
a = 14, S∞ = 16
r = 2 , S∞ = 24 r = – 3 , S∞ = 15
S∞ = a 3 5
1–r
a = S∞ a = S∞
14 1 – r 1 – r
16 = 1–r
a a
14 – 2 = 24 3 2 = 24
16 3 1– 5
1 – r = 1 1 –

1 – r = 7 a = 24 a = 24
8 1 8

8 – 8r = 7 3 1 5 8
3 5
–8r = –1 a = × 24 a = × 15

r = 1 = 8 = 24
8

Exercise 18 Express the following recurring decimals as a fraction in its simplest form.
Ungkapkan nombor perpuluhan jadi semula berikut sebagai pecahan dalam bentuk termudah.

TP 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah.

Example 18 1 0.5555…

(a) 0.6666 … 0.5555…
(b) 0.694˙ 5˙
= 0.5 + 0.05 + 0.005 + 0.0005 + …
Solution
(a) 0.6666 … a = 0.5, r = 0.1
= 0.6 + 0.06 + 0.006 + …
(b) 0.694˙ 5˙ 0.5555… = 1 0.5
– 0.1
= 0.69454545 …
5
= 0.24 + (0.45 + 0.0045 + = 9

Geometric progression 0.000045 + …)
Janjang geometri 0.4˙ 5˙ is a geometric progression
0.4˙ 5˙ ialah suatu janjang geometri

a = 0.6, r = 0.1 a = 0.45, r = 0.01

0.6666… = a 0.69454545… = 0.24 + 1 a r
1– –
r
0.45
= 0.6 = 0.24 + 1 – 0.01
– 0.1
1 24 0.45
100 0.99
= 2 = +
3
191
Smart Tip = 275

For recurring decimal numbers

Bagi nombor perpuluhan jadi semula
Example/Contoh:
0.3˙ = 0.333333…
0.3˙ 7˙ = 0.373737…
0.43˙ 7˙ = 0.4373737…

84

2 0.15151515… 3 0.1˙ 2˙ 4 0.763˙ 6˙

0.15151515… 0.1˙ 2˙ 0.763˙ 6˙

= 0.15 + 0.0015 + 0.000015 + … = 0.12121212… = 0.7636363636…

a = 0.15, r = 0.01 = 0.12 + 0.0012 + 0.000012 + ... = 0.4 + (0.36 + 0.0036 +

0.15151515… = 0.15 a = 0.12, r = 0.01 0.000036 + ...)
– 0.01 0.3˙ 6˙ is a geometric progression
1 0.12121212… = 0.12
– 0.01
= 5 1 a = 0.36, r = 0.01
33
= 4 0.7636363636…
33
= 0.4 + 0.36
1 – 0.01

= 4 + 0.36
10 0.99

= 42
55

5 0.27272727… 6 1.74˙ 7 2.784˙ 2˙

0.27272727… 1.74˙ 2.784˙ 2˙
= 1 + 0.74˙
= 0.27 + 0.0027 + 0.000027 + … = 1 + 0.3 + 0.4444… = 2 + 0.36 + (0.42 + 0.0042 +

a = 0.27, r = 0.01 0.000042 + …)

0.272727… = 1 0.27 = 1 + 3 + (0.4 + 0.04 + 0.004 + …) = 2 + 36 + 1 0.42
– 0.01 10 100 – 0.01

= 3 = 1 + 3 + 0.4 = 2 + 36 + 0.42
11 10 – 0.1 100 0.99
1 a = 0.4
r = 0.1 9 14
= 1 + 3 + 4 = 2 + 25 + 33
10 9
2 297
= 19507 = 835

Exercise 19 Solve the following problems.
Selesaikan masalah yang berikut.

TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang janjang aritmetik dan janjang geometri dalam konteks penyelesaian masalah rutin yang mudah.

Example 19

Azrul deposited RM600 in a bank in March. From Solution
March to October, the total amount of money
deposited increases by 20% per month. Calculate a = 600, r = 100 + 20 = 1.2
the total amount of money (exclude interest) in 100
Azrul’s savings account until October.
Azrul menyimpan sebanyak RM600 di sebuah bank pada Sn = a(rn – 1)
bulan Mac. Dari bulan Mac hingga Oktober, jumlah wang r–1
yang disimpan bertambah sebanyak 20% setiap bulan. Hitung
jumlah wang (tidak termasuk faedah) dalam akaun simpanan S8 = 600(1.28 – 1)
Azrul sehingga bulan Oktober. 1.2 – 1

= 3 000(1.28 – 1)

= 9 899.45

∴ Total amount of money = RM9 899.45

Jumlah wang = RM9 899.45

85

1 The number of bacterial colonies will double 2 A solar company installs solar panels in a
every 15 hours. If the number of bacterial residential area. In the first year, the company
colonies in a petri dish is initially 90, what is the installed 10 000 m2 solar panels. In each
number of bacterial colonies on day 5? subsequent year, solar panel installation
increases by 15% from the previous year.
Bilangan koloni bakteria akan menjadi dua kali ganda setiap Calculate the total area of solar panel installed by
15 jam. Jika bilangan koloni bakteria di dalam piring petri the company after 7 years.
pada awalnya ialah 90, berapakah bilangan koloni bakteria
pada hari ke-5? Sebuah syarikat solar memasang panel solar di sebuah
kawasan perumahan. Pada tahun pertama, syarikat itu
a = 90, r = 2 memasang panel solar seluas 10 000 m2. Pada setiap tahun
berikutnya, keluasan pemasangan panel solar bertambah
1 day = 24 hours sebanyak 15% daripada tahun sebelumnya. Hitung jumlah
luas kawasan panel solar yang dipasang oleh syarikat itu
n = 5 × 24 =8 selepas 7 tahun.
15

Tn = arn – 1 a = 10 000, r = 1.15, n = 7
T8 = (90)(2)8 – 1
= 11 520 a(rn – 1)
r–1
∴ Number of bacterial colonies = 11 520 Sn =

S7 = 10 000(1.157 – 1)
1.15 – 1

= 110 668

∴ Total area = 110 668 m2

3 Fauzi bought a car for RM80 000. The value of 4 The total collection of a movie on the first day
the car depreciates 14% from the value of the showing was RM150 000. The collection of the
previous year on every year. Calculate the value movie decreased by 20% on the following day.
of the car after 12 years. Calculate the total collection of the movie after
seven days showing.
Fauzi membeli sebuah kereta dengan harga RM80 000.
Nilai kereta itu menyusut sebanyak 14% daripada harga Nilai kutipan sebuah filem pada hari pertama tayangan
tahun sebelumnya pada setiap tahun. Hitung nilai kereta ialah RM150 000. Nilai kutipan filem tersebut menurun
itu selepas 12 tahun. sebanyak 20% pada setiap hari berikutnya. Hitung jumlah
nilai kutipan filem itu selepas 7 hari tayangan.

a = 80 000

r= 100 – 14 = 0.86 a = 150 000
100
r = 100 – 20 = 0.8
Tn = arn – 1 100
T12 = (80 000)(0.86)12 – 1
= 15 225.55 Sn = a(1 – rn)
1–r

∴ The value of the car = RM15 225.55 S7 = 150 000(1 – 0.87)
1 – 0.8

= 592 713.60

∴ Total collection of the movie = RM592 713.60

5 A ball is dropped from a height of 50 cm. After Total distance travelled

hitting the ground, the ball rebounds to 3 of the = 50 + 23501 3 2 + 501 3 22 +501 3 23 + …4
5 5 5 5
height it dropped from and so on for each of the
= 50 + 23 1 a r 4; where a = 501 3 2, r = 3
subsequent bounce. What is the total distance – 5 5

travelled by the ball until it stops? 3 4= 50 + 2 501 3 2
5
Sebiji bola dilepaskan dari ketinggian 50 cm. Selepas

mengena tanah, bola itu melantun setinggi 3 daripada 1 – 3
5 5

ketinggian bola itu dilepaskan dan seterusnya bagi setiap = 50 + 150

lantunan yang berikutnya. Berapakah jumlah jarak yang = 200

dilalui oleh bola itu sehingga ia berhenti?

∴ Total distance travelled = 200 cm

86

Review 5

Paper 1 Questions

 1 In an arithmetic progression, the sum of the first Setiap bahagian dibentuk menjadi sebuah segi empat sama.
Rajah 1 menunjukkan tiga buah segi empat sama pertama
SPM four terms is 30 and the sixth term is –24. Find yang dibentuk oleh lelaki itu.
CLONE
`15 the first term and the common difference of the
P1Q9 progression.

Dalam suatu janjang aritmetik, hasil tambah empat sebutan 9.5 cm

pertama ialah 30 dan sebutan keenam ialah –24. Cari sebutan 6.5 cm

pertama dan beza sepunya janjang itu. 3.5 cm 9.5 cm
3.5 cm
[3 marks/markah] 6.5 cm

  2 A restaurant which sells drinks gives two choices

SPM to the customers. The customers can choose Diagram 1/ Rajah 1
CLONE
`16 either condensed milk or evaporated milk in their How many squares can be formed from the wire?
P1Q21 drinks. On a particular day, the restaurant has Berapa buah segi empat samakah yang boleh dibentuk
daripada dawai itu?
43 cans of condensed milk and 36 cans of [3 marks/markah]

evaporated milk. The restaurant uses 4 cans of

condensed milk and 3 cans of evaporated milk

every day. On what day are the number of cans of   6 It is given that the sum of the first n terms of a

both milk left the same? SPM geometric pogression is Sn = 3 (5n – 1). Find
CLONE 2
Sebuah restoran yang menjual minuman memberi dua
`14
P1Q8 Diberi bahawa hasil tambah n sebutan pertama bagi suatu
pilihan kepada pelanggan. Pelanggan boleh memilih sama
3
ada susu pekat atau susu cair dalam minuman mereka. Pada janjang geometri ialah Sn = 2 (5n – 1). Cari

suatu hari tertentu, restoran itu ada 43 tin susu pekat dan (a) the first term,

36 tin susu cair. Restoran itu menggunakan 4 tin susu pekat

dan 3 tin susu cair pada setiap hari. Pada hari keberapakah sebutan pertama,

bilangan tin kedua-dua susu yang tinggal adalah sama (b) the common ratio.

banyak? nisbah sepunya.

[3 marks/markah] [3 marks/markah]

  3 It is given that x + 4, x – 2 and x – 5 are three   7 It is given that 6, 18, h, j and k are the first five

SPM consecutive terms of a geometric progression. SPM terms of a geometric progression. Find the value
CLONE CLONE of k.
`16 Find
P1Q22 Diberi bahawa x + 4, x – 2 dan x – 5 ialah tiga sebutan `15
P1Q8 Diberi bahawa 6, 18, h, j dan k ialah lima sebutan pertama
berturut-turut bagi suatu janjang geometri. Cari
bagi suatu janjang geometri. Cari nilai k.
(a) the value of x,
[2 marks/markah]
nilai bagi x,

(b) the first term if 3 is the tenth term of the   8 It is given that the nth term of a geometric
x
progression. SPM progression is Tn = 5 (rn – 1), r ≠ h. Find
CLONE Diberi bahawa sebutan 2
sebutan pertama jika 3 ialah sebutan kesepuluh bagi
x `17 ke-n bagi suatu janjang geometri
P1Q7

janjang itu. ialah Tn = 5 (rn – 1), r ≠ h. Cari
2
[4 marks/markah]
(a) the value of h,

  4 It is given that the sum of the first n terms of an nilai h,

SPM arithmetic progression is Sn = n2 (17 – 5n). Find the (b) the first term of the progression.
CLONE nth term.
sebutan pertama bagi janjang itu.
`17
P1Q8 [2 marks/markah]

Diberi bahawa hasil tambah bagi n sebutan pertama bagi

suatu janjang aritmetik ialah Sn = n (17 – 5n). Cari sebutan   9 It is given that that u, 6 and w are the first three
2
ke-n. SPM terms of a geometric progression. Express in
CLONE
[3 marks/markah] `18 terms of w
P1Q14 Diberi bahawa u, 6 dan w ialah tiga sebutan pertama bagi

  5 A man has a wire with the length of 18 m. The suatu janjang geometri. Ungkapkan dalam sebutan w

SPM wire is divided into several parts. Each part of the (a) the first term and the common ratio,
CLONE
`18 wire is formed into a square. Diagram 1 shows sebutan pertama dan nisbah sepunya,
P1Q15 the first three squares formed by the man.
(b) the sum to infinity of the progression.

Seorang lelaki mempunyai seutas dawai dengan panjang hasil tambah hingga ketakterhinggaan janjang itu.

18 m. Dawai itu dibahagikan kepada beberapa bahagian. [4 marks/markah]

87

Paper 2 Questions

 1 At a certain day, a farmer has 4 000 ducks to supply to a wholesaler. The farmer sells 250 ducks every day.

SPM The farmer feeds the ducks before selling them to the wholesaler. If the cost to feed a duck is RM0.50 per
CLONE day, calculate the total cost until the number of ducks left is 1 000.

`15
P2Q4 Pada hari tertentu, seorang penternak mempunyai 4 000 ekor itik untuk dibekalkan kepada pemborong. Penternak itu menjual

250 ekor itik pada setiap hari. Penternak itu memberi makanan kepada itik sebelum dijual kepada pemborong. Jika kos makanan

untuk seekor itik ialah RM0.50 sehari, hitung jumlah kos makanan sehingga bilangan itik yang tinggal ialah 1 000 ekor.

[6 marks/markah]

 2 The sum of the first n terms of an arithmetic progression is given by Sn = 5n(n – 27) . Find
Hasil tambah n sebutan 2
SPM pertama bagi suatu janjang aritmetik diberi oleh Sn = 5n(n – 27)
CLONE 2 . Cari

`18 (a) the sum of the first 12 terms,
P2Q1

hasil tambah bagi 12 sebutan pertama,

(b) the first term and the common difference, [1 mark/markah]
sebutan pertama dan beza sepunya, [3 marks/markah]
[2 marks/markah]
(c) the value of w, if the wth term is the first positive term of the progression.
nilai bagi w, jika sebutan ke-w ialah sebutan positif pertama bagi janjang itu.

 3 A wire is cut into n parts. The length of each part increases in a geometric progression pattern. It is given
that the length of the sixth part is nine times the length of the fourth part of the wire.
Seutas dawai dipotong kepada n bahagian. Panjang setiap bahagian bertambah dalam bentuk janjang geometri. Diberi bahawa
panjang bagi bahagian keenam ialah sembilan kali panjang bagi bahagian keempat dawai itu.

(a) Calculate the common ratio.
Hitung nisbah sepunya.

[2 marks/markah]
(b) If the total length of the wire is 5 465 mm and the length of the first part of the wire is 5 mm, calculate

Jika jumlah panjang dawai itu ialah 5 465 mm dan panjang bahagian pertama dawai itu ialah 5 mm, hitung

(i) the value of n,
nilai n,

(ii) the length, in mm, of the last part of the wire.
panjang, dalam mm, bahagian terakhir dawai itu.

[4 marks/markah]

H OTS Zone

 1 Halim is a fresh graduate. He is offered a job from two companies. Company P offers him an annual salary of

SPM RM31 800 with a 4% yearly increment from the basic salary. Company Q offers him a starting salary of RM27 600
CLONE per annum with an 8% yearly increment from the basic salary. Halim wants to choose the company which

`14
P1Q10 offers a higher income and he wants to save 20% of the income to further his study after working for 10 years.

Which company should he choose? How much is the savings for him to further his study? HOTS Applying

Halim ialah seorang graduan yang baru sahaja menamatkan pengajian. Beliau ditawarkan kerja oleh dua buah syarikat. Syarikat P

menawarkan beliau gaji tahunan sebanyak RM31 800 dengan kenaikan tahunan 4% daripada gaji pokok. Syarikat Q menawarkan

beliau gaji permulaan sebanyak RM27 600 setahun dengan kenaikan tahunan 8% daripada gaji pokok. Halim ingin memilih syarikat

yang menawarkan pendapatan yang lebih tinggi dan beliau ingin menyimpan 20% daripada gajinya untuk melanjutkan pelajaran

selepas bekerja selama 10 tahun. Syarikat yang manakah patut beliau pilih? Berapakah jumlah wang simpanan untuk beliau

melanjutkan pelajaran?

 2 Benedict took 3.5 minutes to complete the first km of a 20 km run. He could not sustain his stamina. Thus,

SPM for each subsequent km, he took 1 more time compared to the time he took for the previous km. The
CLONE 12

`16
P1Q23 participants who exceeded 150 minutes were not qualified for the medal. Did Benedict qualify? Show the

calculation to support your answer. HOTS Applying

Benedict mengambil masa selama 3.5 minit untuk menghabiskan km yang pertama bagi satu acara larian 20 km. Dia tidak dapat

mengekalkan staminanya. Maka, pada setiap km yang berikutnya, dia mengambil 1 lebih masa berbanding dengan masa yang
12

diambil untuk km sebelumnya. Peserta-peserta yang menamatkan larian melebihi 150 minit adalah tidak layak untuk menerima

pingat. Adakah Benedict layak? Tunjukkan pengiraan untuk menyokong jawapan anda.

88

Chapter Learning Area: Algebra

6 Linear Law

Hukum Linear

6.1 Linear and Non-Linear Relations / Hubungan Linear dan Tak Linear

Smart Tip

A graph of linear relation is a graph that forms a straight line while a graph of non-linear relation is a graph that does
not form a straight line.
Graf hubungan linear ialah graf yang membentuk satu garis lurus manakala graf hubungan tak linear ialah graf yang tidak membentuk garis lurus.

Exercise 1 Draw a graph of y against x for each of the following table of values. Hence, determine
whether the graph is a linear relation or a non-linear relation.
Lukis graf y melawan x bagi setiap jadual nilai berikut. Seterusnya, tentukan sama ada graf itu ialah
hubungan linear atau hubungan tak linear.

Example 1 1 x –2 –1 0 1 2
y52125
x024
y123 y

Solution 5
4
y 3
4

22

x 1 12 x
0 24 –2 –1 0
Graph of linear relation/Graf hubungan linear

Graph of non-linear relation

2x01234 3 x –2 –1 0 1 2
y43210 y03430

y y
4
3 4 12 x
2 3
1 2
1
x
0 1234 –2 –1 0

Graph of linear relation Graph of non-linear relation

89

Smart Tip

1 A line of best fit can be drawn such that the straight line passes through as many points as possible on the graph.
Garis lurus penyuaian terbaik boleh dilukis dengan keadaan garis lurus itu melalui seberapa banyak titik yang mungkin pada graf.

2 The number of points which do not lie on the line of best fit should be balanced on both sides of the straight line.
Bilangan titik yang tidak terletak pada garis lurus penyuaian terbaik perlu seimbang di kedua-dua belah garis lurus itu.

Exercise 2 Plot a graph of y against x, using a suitable scale on the x-axis and the y-axis. Hence, draw
the the line of best fit.
Plot graf y melawan x, dengan menggunakan skala yang sesuai pada paksi-x dan paksi-y. Seterusnya, lukis
garis lurus penyuaian terbaik.

TP 2 Mempamerkan kefahaman tentang garis lurus penyuaian terbaik.

Example 2 1 x 1 2 2.5 3 3.6
y 15 28 32 40 46.2
x1 2 2.4 3 3.5 4
y 33 24 19 12 8 4 y
50
Solution

y

50 40

40

30

30

20

20

10 10

x x
0 1234 0 1234

2 x 0.5 2 2.5 3 3.5 4 3 x 1 2 3 4 5

y 32.5 19.5 13.5 11.5 4.5 1.5 y 4 13.5 21.5 31 40.5

y
y 40

40 30

30 20

20 10

10 x
x 0 12345
–10
0 1234

90

Exercise 3 Form an equation of the line of best fit for each of the following.
Bentukkan persamaan bagi garis lurus penyuaian terbaik bagi setiap yang berikut.

TP 2 Mempamerkan kefahaman tentang garis lurus penyuaian terbaik.

1y 2y

Q(7, 6) E(1, 6)

F(5, 4)

0 P(4, 0) x x
0

m= 6–0 = 6 =2 m= 4–6 = – 2 = – 1
7–4 3 5–1 4 2

Equation of the straight line, y = mx + c Equation of the straight line, y = mx + c

At point P(4, 0) At point E(1, 6)

0 = (2)(4) + c 6 = – 1 2(1) + c
2
c = –8
13
∴ y = 2x – 8 c = 2

∴ y = – 1 x + 13
2 2

Exercise 4 Solve the following problems.
Selesaikan masalah yang berikut.

TP 2 Mempamerkan kefahaman tentang garis lurus penyuaian terbaik.

Example 3

The table below shows the values of two variables, x and y, obtained from an experiment.
Jadual di bawah menunjukkan nilai-nilai dua pemboleh ubah, x dan y, yang diperoleh daripada suatu eksperimen.

x 10 20 25 40 50 60
y 12 17 19.5 26 31 35

(a) Plot a graph of y against x, using a scale of 2 cm to 10 units on the x-axis and 2 cm to 5 units on the
y-axis. Hence, draw the line of best fit.
Plot graf y melawan x, dengan menggunakan skala 2 cm kepada 10 unit pada paksi-x dan 2 cm kepada 5 unit pada paksi-y.
Seterusnya, lukis garis lurus penyuaian terbaik.

(b) From the graph, determine the y-intercept and the gradient of the line of best fit.
Daripada graf, tentukan pintasan-y dan kecerunan garis lurus penyuaian terbaik.

(c) Determine the equation of the line of best fit.
Tentukan persamaan garis lurus penyuaian terbaik.

Solution (b) y-intercept/Pintasan-y = 7.5
(a) y

35 Gradient/Kecerunan = 35 – 12 = 0.46
60 – 10

30

(c) The equation of the line of best fit is y = 0.46x + 7.5.

25 Persamaan garis lurus penyuaian terbaik ialah y = 0.46x + 7.5

20

15

10

5

0 x
10 20 30 40 50 60

91

1 The table below shows the values of two variables, g and h, obtained from an experiment.
Jadual di bawah menunjukkan nilai-nilai dua pemboleh ubah, g dan h, yang diperoleh daripada suatu eksperimen.

g 10 20 25 40 50 60
h 131 139 143 156 164 173

(a) Plot a graph of h against g, using a scale of 2 cm to 10 units on the g-axis and 2 cm to 20 units on the
h-axis. Hence, draw the line of best fit.
Plot graf h melawan g, dengan menggunakan skala 2 cm kepada 10 unit pada paksi-g dan 2 cm kepada 20 unit pada
paksi-h. Seterusnya, lukis garis lurus penyuaian terbaik.

(b) From the graph, determine the h-intercept and the gradient of the line of best fit.
Daripada graf, tentukan pintasan-h dan kecerunan garis lurus penyuaian terbaik.

(c) Determine the equation of the line of best fit.
Tentukan persamaan garis lurus penyuaian terbaik.

(a)
h

180

160

140

120

100

80

60

40

20

g
0 10 20 30 40 50 60 70

(b) m = 173 – 131 = 0.84
60 – 10

c = 122
(c) The equation of the line of best fit
h = mg + c
h = 0.84g + 122

92

2 The table below shows the values of two variables, p and q, obtained from an experiment.
Jadual di bawah menunjukkan nilai-nilai dua pemboleh ubah, p dan q, yang diperoleh daripada suatu eksperimen.

p 1.0 2.0 2.5 4.0 5.0 6.0

q 40.0 36.0 33.5 26.5 22.0 17.0

(a) Plot a graph of q against p, using a scale of 2 cm to 1 unit on the p-axis and 2 cm to 5 units on the
q-axis. Hence, draw the line of best fit.
Plot graf q melawan p, dengan menggunakan skala 2 cm kepada 1 unit pada paksi-p dan 2 cm kepada 5 unit pada paksi-q.
Seterusnya, lukis garis lurus penyuaian terbaik.

(b) From the graph, determine the q-intercept and the gradient of the line of best fit.
Daripada graf, tentukan pintasan-q dan kecerunan garis lurus penyuaian terbaik.

(c) Determine the equation of the line of best fit.
Tentukan persamaan garis lurus penyuaian terbaik.

(a)

q

50

45

40

35

30

25

20

15

10

5

0 123456 p

(b) m = 17.0 – 40.0 = –4.6
6.0 – 1.0

c = 44.5
(c) The equation of the line of best fit
q = mp + c
q = –4.6p + 44.5

93

3 The table below shows the values of two variables, x and log10y, obtained from an experiment.
Jadual di bawah menunjukkan nilai-nilai dua pemboleh ubah, x dan log10y, yang diperoleh daripada suatu eksperimen.

x 0.10 0.15 0.35 0.50 0.68
log10 y 0.66 0.74 1.06 1.3 1.58

(a) Plot a graph of log10 y against x, using a scale of 2 cm to 0.1 unit on the x-axis and 2 cm to 0.2 unit on
the log10 y-axis. Hence, draw the line of best fit.
Plot graf log10 y melawan x, dengan menggunakan skala 2 cm kepada 0.1 unit pada paksi-x dan 2 cm kepada 0.2 unit pada
paksi-log10 y. Seterusnya, lukis garis lurus penyuaian terbaik.

(b) From the graph, determine the log10 y-intercept and the gradient of the line of best fit.
Daripada graf, tentukan pintasan-log10 y dan kecerunan garis lurus penyuaian terbaik.

(c) Determine the equation of the line of best fit.
Tentukan persamaan garis lurus penyuaian terbaik.

(a)

log10 y

1.8

1.6
1.4

1.2
1.0
0.8
0.6

0.4

0.2
x

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

(b) m = 1.58 – 0.66 = 1.586
0.68 – 0.1

c = 0.5
(c) The equation of the line of best fit
log10 y = mx + c
log10 y = 1.586x + 0.5

94

Exercise 5 Interpret the information from each of the following lines of best fit.
Tafsirkan maklumat daripada setiap garis lurus penyuaian terbaik berikut.

TP 2 Mempamerkan kefahaman tentang garis lurus penyuaian terbaik.

Example 4 1 Speed (m s–1)

Speed (m s–1) Laju (m s–1)
Laju (m s–1)
P(0, 8)

G(5, 6) Q(6, 3)

F(0, 2) O Time (second)
Masa (saat)

O Time (second) Acceleration = Gradient
Masa (saat)

Solution = 3–8
6–0
Acceleration = Gradient/Pecutan = Kecerunan

= 6 – 2 = 0.8 m s–2 = – 5 m s–2
5 – 0 6
Initial speed/Laju awal = 2 m s–1
Initial speed = 8 m s–1

2 Distance (m) 3 Distance (m)

Jarak (m) Jarak (m)

B(6, 10) K(0, 9)

A(0, 1) L(12, 1)

O Time (second) O Time (second)
Masa (saat) Masa (saat)

Speed = Gradient Speed = Gradient

= 10 – 1 = 1–9
6–0 12 – 0

= 1.5 m s–1 = – 2 m s–1
3
Initial distance = 1 m
Initial distance = 9 m

6.2 Linear Law and Non-Linear Relations / Hukum Linear dan Hubungan Tak Linear

Exercise 6 Convert the following non-linear equations to the linear form, Y = mX + c. Hence, state X,
m and c when Y is given.
Tukarkan persamaan tak linear berikut kepada bentuk linear, Y = mX + c. Seterusnya, nyatakan X, m dan c
apabila Y diberi.

TP 3 Mengaplikasikan kefahaman tentang hukum linear untuk melaksanakan tugasan mudah.

Example 5 1 y = h + 5px
x
y = hx3 + kx
Given/Diberi Y = y Given/Diberi Y = xy
x
1 2Solution  y = h + 5px (× x)
1 x
y = hx3 + kx × x
xy = h + 5px2
y = hx2 + k
x = y X = xy = 5px2 + h
Y x
, x2, m= h, c = k Y = xy, X = x2, m = 5p, c = h

Smart Tip

The equation of the line of best fit is represented by Y = mX + c where,
Persamaan bagi garis lurus penyuaian terbaik diwakili oleh Y = mX + c dengan keadaan,

Y = y-axis/paksi-y
X = x-axis/paksi-x
m = gradient/kecerunan
c = y-intercept/pintasan-y

95

2 y = 10 + x + k 3 hx = py + 6xy 4 y – m = mp
2x x2
Given/Diberi Y = 1
Given/Diberi Y = y – x y Given/Diberi Y = x2y

y = 10 + x + k hx = py + 6xy (× 1 ) y – m = mp (× x2 )
2x xy x2

y – x = k + 10 h = p + 6 (× 1 ) x2y – m x2 = mp
2x y x h
x2y = m x2 + mp
k 1 1 2 1 = p + 6
y – x = 2 x + 10 y hx h Y = x2y, X = x2, m = m , c = mp

1 k 1 = p 1 1 2 + 6
x 2 y h x h
Y = y – x, X = , m = , c = 10

Y = 1 , X = 1 , m = p , c = 6
y x h h

Exercise 7 Convert the following non-linear equations to the linear form, Y = mX + c. Hence, state X,
m and c when Y is given.
Tukarkan persamaan tak linear berikut kepada bentuk linear, Y = mX + c. Seterusnya, nyatakan X, m dan c
apabila Y diberi.

TP 3 Mengaplikasikan kefahaman tentang hukum linear untuk melaksanakan tugasan mudah.

Example 6 1 xy = c 2 y = 50nx
Given/Diberi Y = log10 y Given/Diberi Y = log10 y
y = pxn
Given/Diberi Y = log10 y xy = c y = 50nx
log10 (xy) = log10 c log10 y = log10 (50nx)
Solution log10 x + log10 y = log10 c log10 y = log10 50 + log10 nx
y = pxn log10 y = –log10 x + log10 c log10 y = log10 50 + xlog10 n
log10 y = log10 (pxn) log10 y = (log10 n)x + log10 50
log10 y = log10 p + log10xn Y = log10 y, X = log10 x, m = –1,
log10 y = log10 p + nlog10 x c = log10 c Y = log10 y, X = x, m = log10 n,
log10 y = nlog10 x + log10 p c = log10 50

Y = log10 y, X = log10 x, m = n,
c = log10 p

3 y = 2p 4 y3 = hkx
qx Given/Diberi Y = log10 y

Given/Diberi Y = log10 y y3 = hkx

y = 2p log10 y3 = log10 (hkx)
qx 3 log10 y = log10 h + log10 kx
3log10 y = log10 h + xlog10 k
1 2 log10 y = log10 2p
qx log10 k log10 h
3 3
log10 y = log10 (2p) – log10 qx 1 2 log10 y = x+
log10 y = log10 (2p) – xlog10 q
log10 y = –xlog10 q + log10 (2p) Y = log10 y, X = x, m = log10 k, c = log10 h
log10 y = –(log10 q)x + log10 (2p) 3 3

Y = log10 y, X = x, m = –(log10 q), c = log10 2p

96

Exercise 8 Solve the following problems.
Selesaikan masalah berikut.

TP 3 Mengaplikasikan kefahaman tentang hukum linear untuk melaksanakan tugasan mudah.

Example 7

The diagram below shows a line of best fit obtained Solution
y
by plotting x against x2. m= 6–5 = 1
4–2 2
Rajah di bawah menunjukkan garis lurus penyuaian terbaik
y Let the coordinates of Y-intercept = (0, c)
yang diperoleh dengan memplot x melawan x2.
Anggap koordinat pintasan-Y = (0, c)

y Then/Maka, 6 – c = 1
x 4 – 0 2

(4, 6) 2(6 – c) = 4
6 – c = 2
(2, 5) c = 4

Y = mX + c

y = 1 1 2x2 + 4
x 2
x2
O 1
y = 2 x3 + 4x
Express y in terms of x.
Ungkapkan y dalam sebutan x.

1 The diagram below shows a line of best fit 2 The variables x and y are related by the equation
1 1
obtained by plotting y against x . y = px + qx2, where p and q are constants. The

diagram below shows a straight line graph
y
Rajah di bawah menunjukkan garis lurus penyuaian terbaik obtained by plotting x against x.

yang diperoleh dengan memplot 1 melawan 1 . Pemboleh ubah x dan y dihubungkan oleh persamaan
y x

1 y = px + qx2, dengan keadaan p dan q ialah pemalar. Rajah
y
di bawah menunjukkan graf garis lurus yang diperoleh
y
dengan memplot x melawan x.

y

F(6, 6) x

(2, 10)

2E 1 (6, 4)
O x

Express y in terms of x. x
Ungkapkan y dalam sebutan x. O

m = 6 – 2 = 2 Calculate the values of p and q.
6 – 0 3 Hitung nilai p dan q.

c=2 y = px + qx2 (÷ x)

Y = mX + c xy = p + qx

1 = 1 2 2 1 + 2 yx = qx +p
y 3 x Y = m=
y
1 = 2 + 2 x , q, X = x, c = p
y 3x
q = 4 – 10 = – 3
1 2 + 6x 6–2 2
y = 3x
Let the coordinates of Y-intercept = (0, p)

y = 2 3x p – 10 = – 3
+ 6x 0–2 2

21 p – 10 2 = –3
–2

p – 10 = 3

p = 13

97

3 In the diagram below, the first diagram shows a 4 The variables x and y are related by the equation
curve y = –4x2 + 6. The second diagram shows a y = px5, where p is a constant. The diagram below
straight line graph obtained when y = –4x2 + 6 is shows a straight line graph obtained by plotting
expressed in the linear form, Y = 6X + c. Express log10 y against log10 x.
X and Y in terms of x and/or y.
Pemboleh ubah x dan y dihubungkan oleh persamaan
Dalam rajah di bawah, rajah pertama menunjukkan y = px5, dengan keadaan p ialah pemalar. Rajah di bawah
lengkung y = –4x2+ 6. Rajah kedua menunjukkan graf garis menunjukkan graf garis lurus yang diperoleh dengan
lurus yang diperoleh apabila y = –4x2 + 6 diungkapkan memplot log10 y melawan log10 x.
dalam bentuk linear, Y = 6X + c. Ungkapkan X dan Y
dalam sebutan x dan/atau y. log10 y

Y (3, k)

y

y = –4x2 + 6

O X

O x –4 (0, 2) log10 x
O

y = –4x2 + 6 (÷ x2) Find the value of
Cari nilai
xy2 = 6
x2 (a) log10 p (b) k

–4 + y = px5 (a) log10 p = 2
l og10 y = log10 (px5)
1 2 xy2 = 6 1 –4 log10 y = log10 p + log10 x5 (b) k–2 = 5
x2 l og10 y = 5log10 x + log10 p 3–0
Y = log10 y, m = 5, X = log10 x,
Y = y , X = 1 c = log10 p k – 2 = 15
x2 x2
k = 17

5 The diagram below shows a straight line graph 6 The diagram below shows a straight line graph
obtained by plotting log10 y against x. obtained by plotting log2 y against log2 x.

Rajah di bawah menunjukkan graf garis lurus yang Rajah di bawah menunjukkan graf garis lurus yang
diperoleh dengan memplot log10 y melawan x. diperoleh dengan memplot log2 y melawan log2 x.
log2 y
log10 y
(6, 2)

(4, 6) log2 x

O2

O (6, 0) x

Express y in terms of x. Express y in terms of x.
Ungkapkan y dalam sebutan x. Ungkapkan y dalam sebutan x.

m= 2–0 = 1
6–2 2
Y = log10y, X = x 1
Let the coordinates log2 y = 2 log2 x + (–1)
0–6 of Y-intercept = (0, c)
m= 6–4 = –3 1
log2 y = log2 x2 – 1
62 – c = 1
Let the coordinates of Y-intercept = (0, c) – 0 2 1
log2 y – log2 x2 = –1
0c – 6 = –3 2 – c = 3
– 4 1 2 log2y
c = –1 = –1
c – 6 = 12 1
x 2
c = 18
y = 2–1

1
x2 1
log10y = –3x + 18 y = x2(2–1)
y = 10–3x + 18
y = 1 x 1
2 2

y = x
2

98

6.3 Application of Linear Law / Aplikasi Hukum Linear

Exercise 9 Solve the following problems.
Selesaikan masalah berikut.

TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum linear dalam konteks penyelesaian masalah rutin yang mudah.
TP 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum linear dalam konteks penyelesaian masalah rutin yang kompleks.

Example 8

It is given that the object distance, u, the image distance, v, and the focal length, f, of a lens are related by the

equation 1 + 1 = 1 . The table below shows the values of variables, u and v, obtained from an experiment.
u v f f, bagi suatu kanta dihubungkan oleh persamaan 1 + 1 = 1 . Jadual

Diberi bahawa jarak objek, u, jarak imej, v, dan jarak fokus, uv f

di bawah menunjukkan nilai-nilai pemboleh ubah, u dan v, yang diperoleh daripada suatu eksperimen.

u 10 16 20 25 36
v 10.87 7.81 7.14 6.67 6.17

(a) Construct a table for the values of 1 and 1 .
Bina satu jadual bagi nilai-nilai 1 dan 1 v
u
.
u uovnnitbpoatdhaakxedeusa,-pdluoatpaakgsria, pplhotkoafn1vgraagf a1inmsetlau1w.aHne1n.cSee,tderruaswnytah, elulkiinsegaorfisbleusrtufsit.
(b) Using a scale of 2 cm to 0.02 unit vu
Menggunakan skala
2 cm kepada 0.02

penyuaian terbaik.

(c) Use the graph in (b) to find the value of

Gunakan graf di (b) untuk mencari nilai

(i) 1 when/apabila 1 = 0, (ii) f.
f u

Solution

(a) 1 0.1 0.0625 0.05 0.04 0.0278
u 0.092 0.128 0.14 0.15 0.162

1
v

(b) 1 (c) 1 + 1 = 1
u v f
v
0.20 1 = – 1 + 1
0.189 v u f
0.18
(i) When/Apabila 1 = 0,
0.16 u

0.14 1 = 1 = 0.189
f v
0.12
(ii) f = 1
0.10 0.189

= 5.29

0.08

0.06

0.04

0.02

1
0 0.02 0.04 0.06 0.08 0.10 0.12 u

99

1 The period, T, of a simple pendulum and its length, L are related by the equation T = 2pA L , where g is
g

the acceleration due to gravity. The table below shows the values of variables, L and T, obtained from an

experiment. L , dengan keadaan
g
Tempoh ayunan, T, bagi sebuah bandul ringkas dan panjangnya, L, dihubungkan oleh persamaan T = 2pA

g ialah pecutan disebabkan oleh graviti. Jadual di bawah menunjukkan nilai-nilai pemboleh ubah, L dan T, yang diperoleh

daripada suatu eksperimen.

L (cm) 20 30 50 70 80

T (s) 0.898 1.1 1.42 1.68 1.80

(a) Construct a table for the values of T2. Give your answer correct to two decimal places.
Bina satu jadual bagi nilai-nilai T2. Beri jawapan anda betul kepada dua tempat perpuluhan.

(b) Using a scale of 2 cm to 10 units on the L-axis and 2 cm to 0.5 unit on the T2-axis, plot a graph of T2
against L. Hence, draw the line of best fit.
Menggunakan skala 2 cm kepada 10 unit pada paksi-L dan 2 cm kepada 0.5 unit pada paksi-T2, plotkan graf T2 melawan L.
Seterusnya, lukis garis lurus penyuaian terbaik.

(c) Using the graph in (b), find the value of
Menggunakan graf di (b), cari nilai

(i) g,
(ii) T when/apabila L = 200 cm.

(a) L (cm) 20 30 50 70 80
T2 (s2) 0.81 1.21 2.02 2.82 3.24

(b)

T2
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5

L
0 10 20 30 40 50 60 70 80

(c) (i) T = 2pA L 0.0405 = 4p2 (ii) T = 2pA L
g g 975
4p2
T2 = 4p2 L g = 0.0405 When L = 200,
g
4p2 200
Gradient = g g = 975 cm/s2 T = 2pA 975

3.8240 – 0.81 = 4p2 = 2.846 seconds
– 20 g

100